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Hypna velox är en fjärilsart som beskrevs av Butler 1866. Hypna velox ingår i släktet Hypna och familjen praktfjärilar. Inga underarter finns listade i Catalogue of Life.
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velox | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 6,727 |
\section{Introduction}
Bayes' theorem plays an important role today in various fields such as AI, neuroscience, cognitive science, statistical physics and bioinformatics. It underlies most modern approaches to uncertain reasoning in AI systems \cite{russell:09}. In neuroscience, it is often successfully used as a metaphor for functions of the cerebral cortex, which is the outer portion of the brain in charge of higher-order functions such as perception, memory, emotion and thought \cite{lee:03,knill:04,george:05,colombo:12,funamizu:16}. These successes of Bayesian methods give rise to the Bayesian brain hypothesis that the brain is a Bayesian machine \cite{friston:12,sanborn:16}.
\par
Logic concerns entailment (i.e. a consequence relation) whereas learning concerns prediction. They are both practices of the human brain. The Bayesian brain hypothesis thus leads to another hypothesis that there is a common Bayesian interpretation of entailment and prediction, which are traditionally studied in different disciplines. The interpretation is important for the following reasons. First, it gives a more unified view to critically assess the existing formalisms of entailment and prediction. Second, it has a potential to give a better explanation of how the human brain performs them. Third, it backs up the Bayesian brain hypothesis emerging from the field of neuroscience. In spite of the values, few research has focused on the unified interpretation in terms of Bayesian perspectives (see Section \ref{sec:discussion}).
\par
In this paper, we give a formal account of the process of how the truth value of a sentence is probabilistically generated from the probability distribution over states of the world. Our model based on this idea, often called a generative model, begins by assuming a probability distribution over states of the world, e.g. valuation functions in propositional logic. The probability of each state of the world represents how much it is natural, normal or typical. We then formalise the causal relation between each state of the world and each sentence. Let $w$ and $\alpha$ denote a state of the world and a sentence, respectively. The probability that $\alpha$ is true, denoted by $p(\alpha)$, will be shown to have
\begin{eqnarray*}
p(\alpha)=\sum_{w}p(\alpha,w)=\sum_{w}p(\alpha|w)p(w).
\end{eqnarray*}
The equation states that the probability of the truth value of $\alpha$ is the weighted average of the products of likelihood $p(\alpha|w)$ and prior $p(w)$ over all states of the world. Given a set $\Delta$ of sentences, we will show to have
\begin{eqnarray*}
p(\alpha|\Delta)=\sum_{w}p(\alpha|w)p(w|\Delta).
\end{eqnarray*}
This equation is known as a form of Bayesian learning \cite{russell:09}. It states that the probability of the truth value of $\alpha$ is the weighted average of the products of likelihood $p(\alpha|w)$ and posterior $p(w|\Delta)$ over all states of the world.
\par
We define Bayesian entailment using a conditional probability with a fixed probability threshold. Several important logical and machine learning properties are derived from the simple idea. The Bayesian entailment is shown to be identical to the classical consequence relation in reasoning with consistent knowledge. In addition, it is a paraconsistent consequence relation in reasoning with inconsistent knowledge, and it is a nonmonotonic consequence relation in deterministic situations. We moreover show that the Bayesian entailment outperforms several representative classification algorithms in predictive accuracy and complexity on the Kaggle Titanic dataset.
\par
This paper contributes to the field of commonsense reasoning by providing a simple inference principle that is correct in terms of classical logic, paraconsistent logic, nonmonotonic logic and machine learning. It gives a more general answer to the questions such as how to logically infer from inconsistent knowledge, how to rationally handle defeasibility of everyday reasoning, and how to probabilistically infer from noisy data without a conditional dependence assumption, which are all studied and explained individually.
\par
This paper is organised as follows. Section 2 gives a simple generative model for a Bayesian consequence relation. Section 3 shows logical and machine learning correctness of the generative model. Section 4 concludes with discussion of related work.
\section{Method}
We assume a syntax-independent logical language, denoted by $L$. It is logical in the sense that it is defined using only usual logical connectives such as $\lnot$, $\land$, $\lor$, $\rightarrow$, $\leftarrow$ and $\leftrightarrow$. It is syntax independent in the sense that it specifies no further syntax such as propositional or first-order language.
\par
An interpretation is an assignment of truth values to well-formed formulas. It is given by a valuation function in propositional logic, and is given by a structure and variable assignment in first-order logic. In this paper, we call them a possible world to make our discussion general. We assume a probability distribution over possible worlds to quantify the uncertainty of each possible world. Let $W$ denote a random variable for possible worlds, $w_{i}$ the $i$-th possible world, and $\phi_{i}$ the probability of the occurrence of $w_{i}$, i.e., $p(W=w_{i})=\phi_{i}$. Then, the probability distribution over possible worlds can be modelled as a categorical distribution with parameter $(\phi_{1},\phi_{2},...,\phi_{N})$ where $\sum_{i=1}^{N}\phi_{i}=1$ and $\phi_{i}\in[0,1]$, for all $i$. That is, we have
\begin{eqnarray*}
p(W)=(\phi_{1},\phi_{2},...,\phi_{N}).
\end{eqnarray*}
We assume that its prior distribution is statistically estimated from data. For all natural numbers $i$ and $j$, $\phi_{i}>\phi_{j}$ intuitively means that the interpretation specified by possible world $w_{i}$ is more natural, typical or normal than that of $w_{j}$, according to given data.
\par
In formal logic, truth values of formulas depend on possible worlds. The interpretation uniquely given in each possible world indeed assigns a certain truth value to every formula. In this paper, we consider the presence of noise in interpretation. We assume that every formula is a random variable whose realisations are 0 and 1, meaning false and true, respectively. Variable $\mu\in[0,1]$ denotes the probability that a formula is interpreted as being true (resp. false) in a possible world when it is actually true (resp. false) in the same possible world. $1-\mu$ is thus the probability that a formula is interpreted as being true (resp. false) in a possible world when it is actually false (resp. true) in the same possible world. For any possible worlds $w$ and formulas $\alpha$, we thus define the conditional probability of each truth value of $\alpha$ given $w$, as follows.
\begin{eqnarray*}
p(\alpha=1|W=w)=
\begin{cases}
\mu & \text{if } w\in\llbracket\alpha=1\rrbracket\\
1-\mu & \text{otherwise }
\end{cases}
\\%
p(\alpha=0|W=w)=
\begin{cases}
\mu & \text{if } w\in\llbracket\alpha=0\rrbracket\\
1-\mu & \text{otherwise }
\end{cases}
\end{eqnarray*}
Here, $\llbracket\alpha=1\rrbracket$ denotes the set of all possible worlds in which $\alpha$ is true, and $\llbracket\alpha=0\rrbracket$ the set of all possible worlds in which $\alpha$ is false. The above expressions can be simply written as a Bernoulli distribution with parameter $\mu$ where $0\leq \mu\leq 1$. That is, we have
\begin{eqnarray*}
p(\alpha|W=w)=\mu^{\llbracket\alpha\rrbracket_{w}}(1-\mu)^{1-\llbracket\alpha\rrbracket_{w}}.
\end{eqnarray*}
Here, $\llbracket\alpha\rrbracket$ is either $\llbracket\alpha=0\rrbracket$ or $\llbracket\alpha=1\rrbracket$, and $\llbracket\alpha\rrbracket_{w}$ denotes a function of $w$ and $\alpha$ that returns 1 if $w\in\llbracket\alpha\rrbracket$ and 0 otherwise.
\par
In formal logic, the truth values of formulas are independently determined from each possible world. In probabilistic terms, the truth values of any two formulas $\alpha_{1}$ and $\alpha_{2}$ are conditionally independent given a possible world $w$, i.e., $p(\alpha_{1},\alpha_{2}|w)=p(\alpha_{1}|w)p(\alpha_{2}|w)$\footnote{Note that this equation holds not only for atomic formulas but also for compound formulas. The independence, i.e., $p(\alpha_{1},\alpha_{2})=p(\alpha_{1})p(\alpha_{2})$, however, holds only for atomic formulas.}. Let $\Delta=\{\alpha_{1},\alpha_{2},...,\alpha_{N}\}$ be the set of $N$ formulas. We thus have
\begin{eqnarray*}
p(\Delta|W=w)=\prod_{n=1}^{N}p(\alpha_{n}|W=w).
\end{eqnarray*}
\par
So far, we defined prior distribution $p(W)$ as a categorical distribution with parameter $(\phi_{1},\phi_{2},...,\phi_{N})$ and model likelihood $p(\Delta|W)$ as Bernoulli distributions with parameter $\mu$. Given all of the parameters, they give the full joint distribution over all of the random variables. We call $\{p(\Delta|W)$, $p(W)\}$ the probabilistic-logical model, or simply the logical model. When the parameters of the logical model need to be specified, we write the logical model as $\{p(\Delta|W,\mu)$, $p(W|\phi_{1},\phi_{2},...,\phi_{N})\}$.
\par
Now, let $Pow(L)$ denote the powerset of logical language $L$. On the logical model, we define a consequence relation called a Bayesian entailment.
\begin{definition}[Bayesian entailment]\label{def:BE}
Let $\theta\in[0,1]$. $\mathrel{\scalebox{1}[1.5]{$\shortmid$}\mkern-3.1mu\raisebox{0.1ex}{$\approx$}}_{\theta}\subseteq Pow(L)\times L$ is a Bayesian entailment with probability threshold $\theta$ if $\Delta\mathrel{\scalebox{1}[1.5]{$\shortmid$}\mkern-3.1mu\raisebox{0.1ex}{$\approx$}}_{\theta}\alpha$ holds if and only if $p(\alpha|\Delta)\geq \theta$ holds.
\end{definition}
It is obvious from the definition that $\mathrel{\scalebox{1}[1.5]{$\shortmid$}\mkern-3.1mu\raisebox{0.1ex}{$\approx$}}_{\theta_{1}}\subseteq\mathrel{\scalebox{1}[1.5]{$\shortmid$}\mkern-3.1mu\raisebox{0.1ex}{$\approx$}}_{\theta_{2}}$ holds, for all $\theta_{1}\in[0,1]$ and $\theta_{2}\in[0,\theta_{1}]$.
\par
The Bayesian entailment is actually Bayesian in the sense that it involves the following form of Bayesian learning where the probability of consequence $\alpha$ is weighted averages over the posterior distribution of all possible worlds in which premise $\Delta$ is true.
\begin{eqnarray*}
p(\alpha |\Delta)=\sum_{w}p(\alpha|w,\Delta)p(w|\Delta)=\sum_{w}p(\alpha|w)p(w|\Delta)
\end{eqnarray*}
Therefore, the Bayesian entailment is an application of Bayesian prediction on the logical model.
\par
On the logical model, we also define a consequence relation called a maximum a posteriori (MAP) entailment.
\begin{definition}[Maximum a posteriori entailment]
$\mathrel{\scalebox{1}[1.5]{$\shortmid$}\mkern-3.1mu\raisebox{0.1ex}{$\approx$}}_{MAP}$ $\subseteq$ $Pow(L) $ $\times$ $L$ is a maximum a posteriori entailment if $\Delta\mathrel{\scalebox{1}[1.5]{$\shortmid$}\mkern-3.1mu\raisebox{0.1ex}{$\approx$}}_{MAP}\alpha$ holds if and only if there is $w_{MAP}\in\argmax_{w}p(w|\Delta)$ such that $w_{MAP}\in\llbracket\alpha\rrbracket$.
\end{definition}
Here, $w_{MAP}\in\argmax_{w}p(w|\Delta)$ is said to be a maximum a posteriori estimate. It is intuitively the most likely possible world given $\Delta$. The maximum a posteriori entailment can be seen as an approximation of the Bayesian entailment. They are equivalent under the assumption that posterior distribution $p(W|\Delta)$ has a sharp peak, meaning that a possible world is very normal, natural or typical. Under the assumption, we have $p(W|\Delta)\simeq 1$ if $W=w_{MAP}$ and $0$ otherwise, where $\simeq$ denotes an approximation. We thus have
\begin{eqnarray*}
p(\alpha|\Delta)&=&\sum_{w}p(\alpha|w)p(w|\Delta)\\
&\simeq& p(\alpha|w_{MAP})=
\begin{cases}
\mu & (w_{MAP}\in\llbracket\alpha\rrbracket)\\
1-\mu & (w_{MAP}\notin\llbracket\alpha\rrbracket)
\end{cases}
\end{eqnarray*}
Note that both the Bayesian entailment and the maximum a posteriori entailment are general in the sense that the parameters, i.e., $\mu$ and $(\phi_{1},\phi_{2},...,\phi_{N})$, of the logical model are all unspecified.
\par
The probability of the truth value of each formula is not primitive in the logical model. We thus guarantee that it satisfies the Kolmogorov axioms.
\begin{proposition}\label{kolmogorov}
Let $\alpha,\beta\in L$.
\begin{enumerate}
\item $0\leq p(\alpha=i)$ holds, for all $i\in\{0,1\}$.
\item $\sum_{i\in\{0,1\}}p(\alpha=i)=1$ holds.
\item $p(\alpha\lor\beta=i)=p(\alpha=i)+p(\beta=i)-p(\alpha\land\beta=i)$ holds, for all $i\in\{0,1\}$.
\end{enumerate}
\end{proposition}
\begin{proof}
See Appendix.
\end{proof}
\par
The next proposition shows that the logical model is sound in terms of logical negation.
\begin{proposition}\label{negation}
For all $\alpha\in L$, $p(\alpha=0)=p(\neg\alpha=1)$ holds.
\end{proposition}
\begin{proof}
See Appendix.
\end{proof}
In what follows, we thus replace $\alpha=0$ by $\lnot\alpha=1$ and then abbreviate $\lnot\alpha=1$ to $\lnot\alpha$.
Now, let's see an example in propositional logic.
\begin{table}[t
\caption{Possible-world distribution and truth-value likelihoods.}
\label{ex.Int}
\begin{center}
\begin{tabular}{c|c|cc|cc}
& $p(W)$ & $rain$ & $wet$ & $p(rain|W)$ & $p(wet|W)$\\\hline
$w_{1}$ & $0.4$ & $0$ & $0$ & $1-\mu$ & $1-\mu$\\
$w_{2}$ & $0.2$ & $0$ & $1$ & $1-\mu$ & $\mu$\\
$w_{3}$ & $0.1$ & $1$ & $0$ & $\mu$ & $1-\mu$\\
$w_{4}$ & $0.3$ & $1$ & $1$ & $\mu$ & $\mu$
\end{tabular}
\end{center}
\end{table}
\begin{example}\label{ex:BE
Let $rain$ and $wet$ be two propositional symbols meaning ``it is raining'' and ``the grass is wet'', respectively. The second column of Table \ref{ex.Int} shows the probability distribution over all valuation functions. The fifth and sixth columns show the likelihoods of the atomic propositions being true given a valuation function. Given $\mu=1$, predictive probability $p(rain|wet)$ is calculated as follows.
\begin{eqnarray*}
p(rain|wet)&=&\frac{\sum_{w}p(w)p(rain|w)p(wet|w)}{\sum_{w}p(w)p(wet|w)}\\
&=&\frac{\mu^{2}\phi_{4}+\mu(1-\mu)(\phi_{2}+\phi_{3})+(1-\mu)^{2}\phi_{1}}{\mu(\phi_{2}+\phi_{4})+(1-\mu)(\phi_{1}+\phi_{3})}\\
&=&\frac{0.3\mu^{2}+(0.2+0.1)\mu(1-\mu)+0.4(1-\mu)^{2}}{(0.2+0.3)\mu+(0.4+0.1)(1-\mu)}\\
&=&\frac{0.4\mu^{2}-0.5\mu+0.4}{0.5}=0.6
\end{eqnarray*}
Therefore, $\{wet\}\mathrel{\scalebox{1}[1.5]{$\shortmid$}\mkern-3.1mu\raisebox{0.1ex}{$\approx$}}_{\theta} rain$ thus holds, for all $\theta\leq 0.6$. Figure \ref{dependency2} shows the Bayesian network visualising the dependency of the random variables and parameters used in this calculation.
\end{example}
\begin{figure}[t]
\begin{center}
\includegraphics[scale=0.35]{figs/BN6.png}
\caption{Bayesian network visualising the dependency of the elements of the probabilistic model.}
\label{dependency2}
\end{center}
\end{figure}
\section{Correctness}
This section discusses logical and machine learning correctness of the logical model. The logical model is specialised in several ways to show that the Bayesian entailments defined on the specialised models perform key logical and machine learning tasks.
\subsection{Classicality}
Recall that a set $\Delta$ of formulas entails a formula $\alpha$ in classical logic, denoted by $\Delta\models\alpha$, if and only if $\alpha$ is true in every possible world in which $\Delta$ is true. In this paper, we call the Bayesian entailment defined on the logical model $\{p(\Delta|W,\mu=1),p(W|\phi_{1}=1/N,\phi_{2}=1/N,...,\phi_{N}=1/N)\}$ the Bayesian classical entailment. The model can be seen as an ideal specialisation of the logical model in the absence of data and noise. Each formula is interpreted without noise effect, i.e., $\mu=1$, in possible worlds that are equally likely, i.e., $(\phi_{1}=1/N,\phi_{2}=1/N,...,\phi_{N}=1/N)$. The following two theorems state that the Bayesian classical entailment $\mathrel{\scalebox{1}[1.5]{$\shortmid$}\mkern-3.1mu\raisebox{0.1ex}{$\approx$}}_{1}$ is a proper fragment of the classical entailment, i.e., $\mathrel{\scalebox{1}[1.5]{$\shortmid$}\mkern-3.1mu\raisebox{0.1ex}{$\approx$}}_{1}\subseteq\models$.
\begin{theorem}\label{thrm:1}
Let $\alpha\in L$, $\Delta\subseteq L$ and $\mathrel{\scalebox{1}[1.5]{$\shortmid$}\mkern-3.1mu\raisebox{0.1ex}{$\approx$}}_{1}$ be the Bayesian classical entailment. If there is a model of $\Delta$ then $\Delta\mathrel{\scalebox{1}[1.5]{$\shortmid$}\mkern-3.1mu\raisebox{0.1ex}{$\approx$}}_{1}\alpha$ if and only if $\Delta\models\alpha$.
\end{theorem}
\begin{proof}
Let $|\Delta|$ denote the cardinality of $\Delta$. Dividing possible worlds into the models of $\Delta$ and the others, we have
\begin{eqnarray*}
p(\alpha|\Delta)&=&\frac{\displaystyle{\sum_{w}p(\alpha|w)p(\Delta|w)p(w)}}{\displaystyle{\sum_{w}p(\Delta|w)p(w)}}\\
&=&\frac{\displaystyle{\sum_{w\in\llbracket\Delta\rrbracket}p(w)p(\alpha|w)\mu^{|\Delta|}+\sum_{w\notin\llbracket\Delta\rrbracket}p(w)p(\alpha|w)p(\Delta|w)}}{\displaystyle{\sum_{w\in\llbracket\Delta\rrbracket}p(w)\mu^{|\Delta|}+\sum_{w\notin\llbracket\Delta\rrbracket}p(w)p(\Delta|w)}}.
\end{eqnarray*}
$p(\Delta|w)=\prod_{\beta\in\Delta}p(\beta|w)=\prod_{\beta\in\Delta}\mu^{\llbracket\beta\rrbracket_{w}}(1-\mu)^{1-{\llbracket\beta\rrbracket_{w}}}$. For all $w\notin\llbracket\Delta\rrbracket$, there is $\beta\in\Delta$ such that $\llbracket\beta\rrbracket_{w}=0$. Thus, $p(\Delta|w)=0$ when $\mu=1$, for all $w\notin\llbracket\Delta\rrbracket$. We thus have
\begin{eqnarray*}
p(\alpha|\Delta)&=&\frac{\sum_{w\in\llbracket\Delta\rrbracket}p(w)\mu^{\llbracket\alpha\rrbracket_{w}}(1-\mu)^{1-\llbracket\alpha\rrbracket_{w}}\mu^{|\Delta|}}{\sum_{w\in\llbracket\Delta\rrbracket}p(w)\mu^{|\Delta|}}\\
&=&\frac{\sum_{w\in\llbracket\Delta\rrbracket}p(w)\llbracket\alpha\rrbracket_{w}}{\sum_{w\in\llbracket\Delta\rrbracket}p(w)}=\frac{\sum_{w\in\llbracket\alpha,\Delta\rrbracket}p(w)}{\sum_{w\in\llbracket\Delta\rrbracket}p(w)}.
\end{eqnarray*}
Now, $p(\alpha|\Delta)=\frac{\sum_{w\in\llbracket\alpha,\Delta\rrbracket}p(w)}{\sum_{w\in\llbracket\Delta\rrbracket}p(w)}=1$ if and only if $\llbracket\alpha\rrbracket\supseteq\llbracket\Delta\rrbracket$, i.e., $\Delta\models\alpha$.
\end{proof}
\begin{theorem}\label{thrm:2}
Let $\alpha\in L$, $\Delta\subseteq L$ and $\mathrel{\scalebox{1}[1.5]{$\shortmid$}\mkern-3.1mu\raisebox{0.1ex}{$\approx$}}_{1}$ be the Bayesian classical entailment. If there is no model of $\Delta$ then $\Delta\mathrel{\scalebox{1}[1.5]{$\shortmid$}\mkern-3.1mu\raisebox{0.1ex}{$\approx$}}_{1}\alpha$ implies $\Delta\models\alpha$, but not vice versa.
\end{theorem}
\begin{proof}
($\Rightarrow$) If $\llbracket\Delta\rrbracket=\emptyset$ then $\Delta\models\alpha$, for all $\alpha$, in classical logic. ($\Leftarrow$) Definition \ref{def:BE} implies that $\Delta\mathrel{\scalebox{1}[1.5]{$\shortmid$}\mkern-3.1mu\raisebox{0.1ex}{$\approx$}}_{\theta}\alpha$ if $p(\alpha|\Delta)\geq\theta$ holds, and $\Delta\not\mathrel{\scalebox{1}[1.5]{$\shortmid$}\mkern-3.1mu\raisebox{0.1ex}{$\approx$}}_{\theta}\alpha$ if $p(\alpha|\Delta)<\theta$ holds or $p(\alpha|\Delta)$ is undefined. Given $\Delta=\{\beta,\lnot\beta\}$, the following derivation exemplifies that the predictive probability of a formula $\alpha$ is undefined due to division by zero.
\begin{eqnarray*}
p(\alpha|\beta,\lnot\beta)&=&\frac{\sum_{w}p(w)p(\alpha|w)p(\beta|w)p(\lnot\beta|w)}{\sum_{w}p(w)p(\beta|w)p(\lnot\beta|w)}\\
&=&\frac{\mu(1-\mu)\sum_{w}p(w)p(\alpha|w)}{\mu(1-\mu)\sum_{w}p(w)}~~~(\text{undefined if }\mu=1)
\end{eqnarray*}
\end{proof}
In classical logic, everything can be entailed from a contradiction. However, Theorem \ref{thrm:2} implies that nothing can be entailed from a contradiction using the Bayesian classical entailment. In the next section, we study a logical model that allows us to derive something useful from a contradiction.
\subsection{Paraconsistency}
In classical logic, the presence of contradictions in a knowledge base and the fact that the knowledge base entails everything are inseparable. In practice, this fact calls for truth maintenance of the knowledge base, which makes it difficult to scale up the knowledge base toward a useful AI application beyond toy problems.
\par
In this section, we consider the logical model with specific parameters such that $\mu$ approaches 1 and $(\phi_{1},\phi_{2},...,\phi_{N})$ is a uniform distribution, i.e, $\mu\rightarrow 1$ and $\phi_{n}=1/N$, for all $n$. Then, the specific logical model is written as $\{\lim_{\mu\rightarrow 1}$ $p(\Delta|W,\mu)$, $p(W|\phi_{1}=1/N$, $\phi_{2}=1/N$ ,..., $\phi_{N}=1/N)\}$. We call the Bayesian entailment defined on the logical model the Bayesian paraconsistent entailment. Similar to the classical model, the model is an ideal specialisation of the logical model in the absence of data, where formulas are interpreted without noise effect in every possible world that is equally likely.
\par
The following two theorems state that the Bayesian paraconsistent entailment $\mathrel{\scalebox{1}[1.5]{$\shortmid$}\mkern-3.1mu\raisebox{0.1ex}{$\approx$}}_{1}$ is also a proper fragment of the classical entailment, i.e., $\mathrel{\scalebox{1}[1.5]{$\shortmid$}\mkern-3.1mu\raisebox{0.1ex}{$\approx$}}_{1}\subseteq\models$.
\begin{theorem}\label{thrm:3}
Let $\alpha\in L$, $\Delta\subseteq L$ and $\mathrel{\scalebox{1}[1.5]{$\shortmid$}\mkern-3.1mu\raisebox{0.1ex}{$\approx$}}_{1}$ be the Bayesian paraconsistent entailment. If there is a model of $\Delta$ then $\Delta\mathrel{\scalebox{1}[1.5]{$\shortmid$}\mkern-3.1mu\raisebox{0.1ex}{$\approx$}}_{1}\alpha$ if and only if $\Delta\models\alpha$.
\end{theorem}
\begin{proof}
The proof of Theorem \ref{thrm:1} still holds under the presence of the limit operation.
\end{proof}
\begin{theorem}\label{thrm:4}
Let $\alpha\in L$, $\Delta\subseteq L$ and $\mathrel{\scalebox{1}[1.5]{$\shortmid$}\mkern-3.1mu\raisebox{0.1ex}{$\approx$}}_{1}$ be the Bayesian paraconsistent entailment. If there is no model of $\Delta$ then $\Delta\mathrel{\scalebox{1}[1.5]{$\shortmid$}\mkern-3.1mu\raisebox{0.1ex}{$\approx$}}_{1}\alpha$ implies $\Delta\models\alpha$, but not vice versa.
\end{theorem}
\begin{proof}
($\Rightarrow$) The proof of Theorem \ref{thrm:2} still holds. ($\Leftarrow$) Suppose $p(\alpha)<1$. The following derivation exemplifies $p(\alpha|\beta\land\lnot\beta)<1$.
\begin{flalign*}
p(\alpha|\beta\land\lnot\beta)&=\frac{\sum_{w}p(w)\lim_{\mu\rightarrow 1}p(\alpha|w)\lim_{\mu\rightarrow 1}p(\beta\land\lnot\beta|w)}{\sum_{w}p(w)\lim_{\mu\rightarrow 1}p(\beta\land\lnot\beta|w)}\\
&=\lim_{\mu\rightarrow 1}\frac{(1-\mu)\sum_{w}p(w)p(\alpha|w)}{(1-\mu)\sum_{w}p(w)}=\lim_{\mu\rightarrow 1}\frac{\sum_{w}p(w)p(\alpha|w)}{\sum_{w}p(w)}\\
&=\sum_{w}p(w)\lim_{\mu\rightarrow 1}p(\alpha|w)=p(\alpha)
\end{flalign*}
\end{proof}
\par
The Bayesian paraconsistent entailment handles reasoning with inconsistent knowledge in a proper way described below. Let $\alpha\in L$ and $\Delta\subseteq L$. For simplicity, we use symbol $\#_{w}$ to denote the number of formulas in $\Delta$ that are true in $w$, i.e. $\#_{w}=\sum_{\beta\in\Delta}\llbracket\beta\rrbracket_{w}$, and symbol $(\!(\Delta)\!)$ to denote the set of possible worlds in which the maximum number of formulas in $\Delta$ are true, i.e., $(\!(\Delta)\!)=\argmax_{w}(\#_{w})$. $(\!(\Delta)\!)$ is thus the set of models of $\Delta$, i.e., $(\!(\Delta)\!)=\llbracket\Delta\rrbracket$, if and only if there is a model of $\Delta$. Now, by case analysis of the possible worlds in $(\!(\Delta)\!)$ and the others, we have
\begin{flalign*}
p(\alpha|\Delta)&=\lim_{\mu\rightarrow 1}\frac{\sum_{w}p(\alpha|w)p(w)p(\Delta|w)}{\sum_{w}p(w)p(\Delta|w)}\\
&=\lim_{\mu\rightarrow 1}\frac{\displaystyle{\sum_{\hat{w}\in(\!(\Delta)\!)}p(\alpha|\hat{w})p(\hat{w})p(\Delta|\hat{w})+\sum_{w\notin(\!(\Delta)\!)}p(\alpha|w)p(w)p(\Delta|w)}}{\displaystyle{\sum_{\hat{w}\in(\!(\Delta)\!)}p(\hat{w})p(\Delta|\hat{w})+\sum_{w\notin(\!(\Delta)\!)}p(w)p(\Delta|w)}}.
\end{flalign*}
$p(\Delta|w)=\prod_{\beta\in\Delta}p(\beta|w)=\mu^{\#_{w}}(1-\mu)^{|\Delta|-\#_{w}}$ holds, for all $w$. Since $\#_{\hat{w}}$ has the same value for all $\hat{w}\in(\!(\Delta)\!)$, we can simplify the fraction by dividing the denominator and numerator by $(1-\mu)^{|\Delta|-\#_{\hat{w}}}$. The fraction inside of the limit operator is now given by
\begin{eqnarray*}
\frac{\displaystyle{\sum_{\hat{w}\in(\!(\Delta)\!)}p(\alpha|\hat{w})p(\hat{w})\mu^{\#_{\hat{w}}}+\sum_{w\notin(\!(\Delta)\!)}p(\alpha|w)p(w)\mu^{\#_{w}}(1-\mu)^{\#_{\hat{w}}-\#_{w}}}}{\displaystyle{\sum_{\hat{w}\in(\!(\Delta)\!)}p(\hat{w})\mu^{\#_{\hat{w}}}+\sum_{w\notin(\!(\Delta)\!)}p(w)\mu^{\#_{w}}(1-\mu)^{\#_{\hat{w}}-\#_{w}}}}.
\end{eqnarray*}
Applying the limit operation to the second terms of the denominator and numerator, we have
\begin{eqnarray*}
p(\alpha|\Delta)&=&\lim_{\mu\rightarrow 1}\frac{\sum_{\hat{w}\in(\!(\Delta)\!)}p(\alpha|\hat{w})p(\hat{w})\mu^{\#_{\hat{w}}}}{\sum_{\hat{w}\in(\!(\Delta)\!)}p(\hat{w})\mu^{\#_{\hat{w}}}}\\
&=&\lim_{\mu\rightarrow 1}\frac{\sum_{\hat{w}\in(\!(\Delta)\!)}\mu^{\llbracket\alpha\rrbracket_{\hat{w}}}(1-\mu)^{1-\llbracket\alpha\rrbracket_{\hat{w}}}p(\hat{w})\mu^{\#_{\hat{w}}}}{\sum_{\hat{w}\in(\!(\Delta)\!)}p(\hat{w})\mu^{\#_{\hat{w}}}}=\frac{\sum_{\hat{w}\in(\!(\Delta)\!)}\llbracket\alpha\rrbracket_{\hat{w}}p(\hat{w})}{\sum_{\hat{w}\in(\!(\Delta)\!)}p(\hat{w})}
\end{eqnarray*}
From the above derivation, $\Delta\mathrel{\scalebox{1}[1.5]{$\shortmid$}\mkern-3.1mu\raisebox{0.1ex}{$\approx$}}_{1}\alpha$ holds if and only if $\llbracket\alpha\rrbracket\supseteq (\!(\Delta)\!)$. For the sake of intuition, let us say that $\Delta$ is almost true in a possible world $w$ if $w\in(\!(\Delta)\!)\setminus\llbracket\Delta\rrbracket$. Then, $\Delta\mathrel{\scalebox{1}[1.5]{$\shortmid$}\mkern-3.1mu\raisebox{0.1ex}{$\approx$}}_{1}\alpha$ states that
\begin{itemize}
\item if $\Delta$ has a model then $\alpha$ is true in every possible world in which $\Delta$ is true, i.e., $\Delta\models\alpha$, and
\item if $\Delta$ has no model then $\alpha$ is true in every possible world in which $\Delta$ is almost true.
\end{itemize}
\begin{example}
Let $a$ and $b$ be two propositional variables. Suppose the uniform prior distribution over possible worlds of the two variables.
\begin{itemize}
\item $p(a|a,b,\lnot b)=1$ holds. Therefore, $a,b,\lnot b\mathrel{\scalebox{1}[1.5]{$\shortmid$}\mkern-3.1mu\raisebox{0.1ex}{$\approx$}}_{\theta}a$ holds if and only if $\theta=1$.
\item $p(a|a\land b,\lnot b)=\frac{2}{3}$ holds. Therefore, $a\land b,\lnot b\mathrel{\scalebox{1}[1.5]{$\shortmid$}\mkern-3.1mu\raisebox{0.1ex}{$\approx$}}_{\theta}a$ holds if and only if $\theta\leq \frac{2}{3}$.
\item $p(a|a\land b\land\lnot b)=\frac{1}{2}$ holds. Therefore, $a\land b\land\lnot b\mathrel{\scalebox{1}[1.5]{$\shortmid$}\mkern-3.1mu\raisebox{0.1ex}{$\approx$}}_{\theta}a$ holds if and only if $\theta\leq \frac{1}{2}$.
\end{itemize}
\end{example}
\par
Let us examine abstract inferential properties of the Bayesian paraconsistent entailment. Mathematically, let $\alpha,\beta\in L$, $\Delta\subseteq L$ and $\vdash$ be a consequence relation over logical language $L$, i.e., $\vdash\subseteq Pow(L)\times L$. We call tuple $(L,\vdash)$ a logic. A logic is said to be non-contradictory, non-trivial, and explosive if it satisfies the following respective principles.
\begin{itemize}
\item Non-contradiction: $\exists\Delta\forall\alpha(\Delta\not\vdash\alpha~or~\Delta\not\vdash\lnot\alpha)$
\item Non-triviality: $\exists\Delta\exists\alpha(\Delta\not\vdash\alpha)$
\item Explosion: $\forall\Delta\forall\alpha\forall\beta(\Delta,\alpha,\lnot\alpha\vdash\beta)$
\end{itemize}
A logic is paraconsistent if and only if it is not explosive, and is sometimes called dialectical if it is contradictory \cite{carnielli}. The following theorem states that the Bayesian paraconsistent entailment is paraconsistent, but not dialectical.
\begin{theorem}
Let $\theta\in(0.5,1]$. The Bayesian paraconsistent entailment $\mathrel{\scalebox{1}[1.5]{$\shortmid$}\mkern-3.1mu\raisebox{0.1ex}{$\approx$}}_{\theta}$ satisfies the principles of non-contradiction and non-triviality, but does not satisfy the principle of explosion.
\end{theorem}
\begin{proof}
(1) It is sufficient to show $\nexists\alpha(\mathrel{\scalebox{1}[1.5]{$\shortmid$}\mkern-3.1mu\raisebox{0.1ex}{$\approx$}}_{\theta}\alpha$ and $\mathrel{\scalebox{1}[1.5]{$\shortmid$}\mkern-3.1mu\raisebox{0.1ex}{$\approx$}}_{\theta}\lnot\alpha)$. From definition, we show there is no $\alpha$ such that $p(\alpha)\geq\theta$ and $p(\lnot\alpha)\geq\theta$. We have
\begin{flalign*}
p(\alpha)&=\sum_{w}\lim_{\mu\rightarrow 1}p(\alpha|w)p(w)=\sum_{w}\lim_{\mu\rightarrow 1}\mu^{\llbracket\alpha\rrbracket_{w}}(1-\mu)^{1-\llbracket\alpha\rrbracket_{w}}p(w)\\
&=\lim_{\mu\rightarrow 1}\{\mu\sum_{w\in\llbracket\alpha\rrbracket} p(w)+(1-\mu)\sum_{w\notin\llbracket\alpha\rrbracket} p(w)\}=\sum_{w\in\llbracket\alpha\rrbracket} p(w)\\
p(\lnot\alpha)&=\sum_{w}\lim_{\mu\rightarrow 1}p(\lnot\alpha|w)p(w)=\sum_{w}\lim_{\mu\rightarrow 1}\mu^{1-\llbracket\alpha\rrbracket_{w}}(1-\mu)^{\llbracket\alpha\rrbracket_{w}}p(w)\\%=\sum_{w\notin\llbracket\alpha\rrbracket} p(w)
&=\lim_{\mu\rightarrow 1}\{\mu\sum_{w\notin\llbracket\alpha\rrbracket} p(w)+(1-\mu)\sum_{w\in\llbracket\alpha\rrbracket} p(w)\}=\sum_{w\notin\llbracket\alpha\rrbracket} p(w)
\end{flalign*}
Now, $p(\alpha)+p(\lnot\alpha)=\sum_{w\in\llbracket\alpha\rrbracket} p(w)+\sum_{w\notin\llbracket\alpha\rrbracket} p(w)=\sum_{w}p(w)=1$.
(2) It is sufficient to show $\exists\alpha\not\mathrel{\scalebox{1}[1.5]{$\shortmid$}\mkern-3.1mu\raisebox{0.1ex}{$\approx$}}_{\theta}\alpha$. We show there is $\alpha$ such that $p(\alpha)\leq 0.5$. Using proof by contradiction, we assume $p(\alpha)> 0.5$ holds, for all $\alpha$. This contradicts (1).
(3) It is sufficient to show $\exists\alpha\exists\beta(\alpha,\lnot\alpha\not\mathrel{\scalebox{1}[1.5]{$\shortmid$}\mkern-3.1mu\raisebox{0.1ex}{$\approx$}}_{\theta}\beta)$ holds. $p(\beta|\alpha,\lnot\alpha)=p(\beta)$ is shown as follows.
\begin{eqnarray*}
p(\beta|\alpha,\lnot\alpha)&=&\lim_{\mu\rightarrow 1}\frac{\sum_{w}p(\alpha|w)p(\lnot\alpha|w)p(\beta|w)p(w)}{\sum_{w}p(\alpha|w)p(\lnot\alpha|w)p(w)}\\
&=&\lim_{\mu\rightarrow 1}\frac{\mu(1-\mu)\sum_{w}p(\beta|v)p(w)}{\mu(1-\mu)\sum_{w}p(w)}=p(\beta)
\end{eqnarray*}
The principle of explosion does not hold when $p(\beta)<1$.
\end{proof}
\subsection{Non-monotonicity}
In classical logic, whenever a sentence is a logical consequence of a set of sentences, then the sentence is also a consequence of an arbitrary superset of the set. This property called monotonicity cannot be expected in commonsense reasoning where having new knowledge often invalidates a conclusion. A practical knowledge-based system with this property is possible under the unrealistic assumption that every rule in the knowledge base sufficiently covers possible exceptions.
\par
A preferential entailment \cite{shoham:87} is a general approach to a nonmonotonic consequence relation. It is defined on a preferential structure $({\cal W},\succ)$, where ${\cal W}$ is a set of valuation functions of propositional logic and $\succ$ is an irreflexive and transitive relation on ${\cal W}$. $w_{1}\succ w_{2}$ represents that $w_{1}$ is preferable\footnote{For the sake of simplicity, we do not adopt the common practice in logic that $w_{2}\succ w_{1}$ denotes $w_{1}$ is preferable to $w_{2}$.} to $w_{2}$ in the sense that $w_{1}$ is more normal, typical or natural than $w_{2}$. Given a preferential structure $({\cal W},\succ)$, $\alpha$ is preferentially entailed by $\Delta$, denoted by $\Delta\mathrel{\scalebox{1}[1.5]{$\shortmid$}\mkern-3.1mu\raisebox{0.1ex}{$\sim$}}_{({\cal W},\succ)}\alpha$, if $\alpha$ is true in all $\succ$-maximal\footnote{$\succ$ has to be smooth (or stuttered) \cite{kraus:90} so that a maximal model certainly exists.} models of $\Delta$.
\par
Given a preferential structure $({\cal W},\succ)$, we consider the logical model with specific parameters $\mu\rightarrow 1$ and $(\phi_{1},\phi_{2},...,\phi_{N})$ such that, for all $w_{1}$ and $w_{2}$ in ${\cal W}$, if $w_{1}\succ w_{2}$ then $\phi_{1}\geq\phi_{2}$.\footnote{If we assume a function mapping $w_{n}$ to $\phi_{n}$, for all $n$, then the function satisfying the condition is said to be order-preserving.} We call the maximum a posteriori entailment defined on the logical model the maximum a posteriori entailment with respect to $({\cal W}, \succ)$. The following two theorems show the relationship between the maximum a posteriori entailment and preferential entailment.
\begin{theorem}\label{thrm:MAP1}
Let $({\cal W},\succ)$ be a preferential structure and $\mathrel{\scalebox{1}[1.5]{$\shortmid$}\mkern-3.1mu\raisebox{0.1ex}{$\approx$}}_{MAP}$ be a maximum a posteriori entailment with respect to $({\cal W},\succ)$. If there is a model of $\Delta$ then $\Delta\mathrel{\scalebox{1}[1.5]{$\shortmid$}\mkern-3.1mu\raisebox{0.1ex}{$\sim$}}_{({\cal W},\succ)}\alpha$ implies $\Delta\mathrel{\scalebox{1}[1.5]{$\shortmid$}\mkern-3.1mu\raisebox{0.1ex}{$\approx$}}_{MAP}\alpha$.
\end{theorem}
\begin{proof}
Since $\geq$ is a linear extension of $\succ$ given ${\cal W}$, if $w_{1}\succ w_{2}$ then $\phi_{1}\geq \phi_{2}$, for all $w_{1}, w_{2}\in{\cal W}$. Thus, if $w_{i}$ is $\succ$-maximal then $\phi_{i}$ is maximal or there is another $\succ$-maximal $w_{j}$ such that $\phi_{j}\geq \phi_{i}$. Therefore, there is $w^{*}$ such that $w^{*}$ is a $\succ$-maximal model of $\Delta$ and $w^{*}\in\argmax_{w}p(w|\Delta)$. $\alpha$ is true in $w^{*}$ since $\Delta\mathrel{\scalebox{1}[1.5]{$\shortmid$}\mkern-3.1mu\raisebox{0.1ex}{$\sim$}}_{({\cal W},\succ)}\alpha$.
\end{proof}
\begin{theorem}\label{thrm:MAP2}
Let $({\cal W},\succ)$ be a preferential structure and $\mathrel{\scalebox{1}[1.5]{$\shortmid$}\mkern-3.1mu\raisebox{0.1ex}{$\approx$}}_{MAP}$ be a maximum a posteriori entailment with respect to $({\cal W},\succ)$. If there is no model of $\Delta$ then $\Delta\mathrel{\scalebox{1}[1.5]{$\shortmid$}\mkern-3.1mu\raisebox{0.1ex}{$\approx$}}_{MAP}\alpha$ implies $\Delta\mathrel{\scalebox{1}[1.5]{$\shortmid$}\mkern-3.1mu\raisebox{0.1ex}{$\sim$}}_{({\cal W},\succ)}\alpha$, but not vice versa.
\end{theorem}
\begin{proof}
($\Rightarrow$) From the definition, $\Delta\mathrel{\scalebox{1}[1.5]{$\shortmid$}\mkern-3.1mu\raisebox{0.1ex}{$\sim$}}_{({\cal W},\succ)}\alpha$ holds, for all $\alpha$, when $\Delta$ has no model. ($\Leftarrow$) Let $\alpha,\beta\in L$. Suppose $(\phi_{1},\phi_{2},..,\phi_{N})$ such that $w_{1}\notin\llbracket\alpha\rrbracket$ and $\phi_{n}>\phi_{n+1}$, for all $1\leq n\leq N-1$. Now, $p(W|\beta,\lnot\beta)=p(W)$ is shown as follows.
\begin{eqnarray*}
p(W|\beta,\lnot\beta)&=&\frac{p(\beta|W)p(\lnot\beta|W)p(W)}{\sum_{w}p(\beta|w)p(\lnot\beta|w)p(w)}\\
&=&\frac{\mu(1-\mu)p(W)}{\mu(1-\mu)\sum_{w}p(w)}=p(W)
\end{eqnarray*}
Although $w_{1}=\argmax_{w}p(w|\beta,\lnot\beta)$, $w_{1}\notin\llbracket\alpha\rrbracket$.
\end{proof}
\par
When a preferential structure is assumed to be a total order, the maximum a posteriori entailment with respect to the preferential structure becomes a fragment of the preferential entailment.
\begin{theorem}\label{thrm:MAP3}
Let $({\cal W},\succ)$ be a totally ordered preferential structure and $\mathrel{\scalebox{1}[1.5]{$\shortmid$}\mkern-3.1mu\raisebox{0.1ex}{$\approx$}}_{MAP}$ be a maximum a posteriori entailment with respect to $({\cal W},\succ)$. If there is a model of $\Delta$ then $\Delta\mathrel{\scalebox{1}[1.5]{$\shortmid$}\mkern-3.1mu\raisebox{0.1ex}{$\sim$}}_{({\cal W},\succ)}\alpha$ if and only if $\Delta\mathrel{\scalebox{1}[1.5]{$\shortmid$}\mkern-3.1mu\raisebox{0.1ex}{$\approx$}}_{MAP}\alpha$.
\end{theorem}
\begin{proof}
Same as Theorem \ref{thrm:MAP1}. The only difference is that such model $w^{*}$ exists uniquely.
\end{proof}
\begin{theorem}\label{thrm:MAP4}
Let $({\cal W},\succ)$ be a totally ordered preferential structure and $\mathrel{\scalebox{1}[1.5]{$\shortmid$}\mkern-3.1mu\raisebox{0.1ex}{$\approx$}}_{MAP}$ be a maximum a posteriori entailment with respect to $({\cal W},\succ)$. If there is no model of $\Delta$ then $\Delta\mathrel{\scalebox{1}[1.5]{$\shortmid$}\mkern-3.1mu\raisebox{0.1ex}{$\approx$}}_{MAP}\alpha$ implies $\Delta\mathrel{\scalebox{1}[1.5]{$\shortmid$}\mkern-3.1mu\raisebox{0.1ex}{$\sim$}}_{({\cal W},\succ)}\alpha$, but not vice versa.
\end{theorem}
\begin{proof}
Same as Theorem \ref{thrm:MAP2}.
\end{proof}
\begin{example}
Suppose preferential structure $(\{w_{1}$, $w_{2}$, $w_{3}$, $w_{4}\}$, $\{(w_{1}$, $w_{2})$, $(w_{1}$, $w_{3})$, $(w_{1}$, $w_{4})$, $(w_{3}$, $w_{2})$, $(w_{4}$, $w_{2})\}$ depicted on the left hand side in Figure \ref{fig:tab}. On the right hand side, you can see the probability distribution over valuation functions that preserves the preference order.
\par
Now, $\{a\lor\lnot b\}\mathrel{\scalebox{1}[1.5]{$\shortmid$}\mkern-3.1mu\raisebox{0.1ex}{$\sim$}}_{({\cal W},\succ)}\lnot b$ holds because $\lnot b$ is true in $w_{1}$, which is the $\succ$-maximal model of $\{a\lor\lnot b\}$. Meanwhile, $\{a\lor\lnot b\}\mathrel{\scalebox{1}[1.5]{$\shortmid$}\mkern-3.1mu\raisebox{0.1ex}{$\approx$}}_{MAP}\lnot b$ holds because $w_{1}\in \argmax_{w}p(w|a\lor\lnot b)$ and $w_{1}\in\llbracket\lnot b\rrbracket$.
\par
In contrast, $\{a\}\not\mathrel{\scalebox{1}[1.5]{$\shortmid$}\mkern-3.1mu\raisebox{0.1ex}{$\sim$}}_{({\cal W},\succ)}\lnot b$ holds because $\lnot b$ is false in $w_{4}$, which is a $\succ$-maximal model of $a$. However, $\{a\}\mathrel{\scalebox{1}[1.5]{$\shortmid$}\mkern-3.1mu\raisebox{0.1ex}{$\approx$}}_{MAP}\lnot b$ holds because $w_{3}=\argmax_{w}p(w|a)$ and $w_{3}\in\llbracket\lnot b\rrbracket$.
\end{example}
\begin{figure}[t
\begin{center}
\includegraphics[scale=0.4]{figs/PS6.png}
\caption{The left graph shows the map from the preferential structure to the prior distribution over valuation functions given in the right table.}
\label{fig:tab}
\end{center}
\end{figure}
\subsection{Predictive Accuracy}
In this section, we specialise the logical model so that the Bayesian entailment can deal with classification tasks. Correctness of the specialisation is empirically discussed in terms of machine learning using the Titanic dataset available in Kaggle \cite{titanic}, which is an online community of machine learning practitioners. The dataset is used in a Kaggle competition aimed to predict what sorts of people were likely to survive in the Titanic disaster in 1912. Each of 891 data in the dataset contains nine attributes (i.e. ticket class, sex, age, the number of spouses aboard, the number of children aboard, ticket number, passenger fare, cabin number and port of embarkation) and one goal (i.e. survival). In contrast to Table \ref{ex.Int}, the attributes of the Titanic dataset are not generally Boolean variables. We thus treat each attribute with a certain value as a Boolean variable. For example, for the ticket class attribute (abbreviated to $TC$), we assume three Boolean variables $TC=1$, $TC=2$ and $TC=3$, meaning the 1st, 2nd and 3rd class, respectively. In this way, we replace each value of all categorical data with a distinct integer value for identification purpose.
\par
Mathematically, let $D$ be a set of tuples $(\Delta,\alpha)$ where $\Delta$ is a set of formulas and $\alpha$ is a formula. We call $D$ a dataset, $(\Delta,\alpha)$ data, $\Delta$ attributes, and $\alpha$ a goal. The dataset is randomly split into three disjoint sets: 60\% training set, 20\% cross validation set and 20\% test set, denoted by $D_{\text{train}}$, $D_{\text{cv}}$ and $D_{\text{test}}$, respectively.
\par
We consider the logical model with parameter $(\phi_{1},\phi_{2},...,\phi_{N})$ given by a MLE (maximum likelihood estimate) using the training set and parameter $\mu$ given by a model selection using the cross validation set. Concretely, the MLE is calculated as follows.
\begin{eqnarray*}
(\hat{\phi}_{1},...,\hat{\phi}_{N})&\in&\argmax_{\phi_{1},...,\phi_{N}}p(D_{\text{train}}|\phi_{1},...,\phi_{N})\\
&=&\argmax_{\phi_{1},...,\phi_{N}}\prod_{(\Delta,\alpha)\in D_{\text{train}}}p(\alpha,\Delta|\phi_{1},...,\phi_{N})\\
&=&\argmax_{\phi_{1},...,\phi_{N}}\prod_{(\Delta,\alpha)\in D_{\text{train}}}p(\alpha|\phi_{1},...,\phi_{N})\prod_{\beta\in\Delta}p(\beta|\phi_{1},...,\phi_{N})
\end{eqnarray*}
In practice, we regard each data in the training set as a possible world, and directly use the training set as a uniform distribution of possibly duplicated possible worlds. This technique results in the same Bayesian predictive entailment although it reduces the cost of the MLE calculation.
\par
Given $(\hat{\phi}_{1},\hat{\phi}_{2},...,\hat{\phi}_{N})$, the model selection is calculated as follows.
\begin{eqnarray*}
\hat{\mu}=\argmax_{\mu}\sum_{(\Delta,\alpha)\in D_{\text{cv}}}\llbracket\Delta\mathrel{\scalebox{1}[1.5]{$\shortmid$}\mkern-3.1mu\raisebox{0.1ex}{$\approx$}}_{0.5}\alpha\rrbracket,
\end{eqnarray*}
where $\llbracket\Delta\mathrel{\scalebox{1}[1.5]{$\shortmid$}\mkern-3.1mu\raisebox{0.1ex}{$\approx$}}_{0.5}\alpha\rrbracket=1$ if $\Delta\mathrel{\scalebox{1}[1.5]{$\shortmid$}\mkern-3.1mu\raisebox{0.1ex}{$\approx$}}_{0.5}\alpha$ holds and $\llbracket\Delta\mathrel{\scalebox{1}[1.5]{$\shortmid$}\mkern-3.1mu\raisebox{0.1ex}{$\approx$}}_{0.5}\alpha\rrbracket=0$ otherwise. We call the Bayesian entailment defined on the logical model the Bayesian predictive entailment.
\par
We investigate learning performance of the Bayesian predictive entailment in terms of whether or to what extent $\Delta\mathrel{\scalebox{1}[1.5]{$\shortmid$}\mkern-3.1mu\raisebox{0.1ex}{$\approx$}}_{\theta}\alpha$ holds, for all $(\Delta,\alpha)\in D_{\text{test}}$. Several representative classifiers are compared in Table \ref{tab:comp} in terms of accuracy, AUC (i.e. area under the ROC curve) and the runtime associated with one test datum prediction.
\par
The experimental results were calculated using a MacBook (Retina, 12-inch, 2017) with 1.4 GHz Dual-Core Intel Core i7 processor and 16GB 1867 MHz LPDDR3 memory. We assumed $\theta=0.5$ for the accuracy scores and $\theta\in[0,1]$ for the AUC scores. The best parameter $\mu$ of the Bayesian predictive entailment was selected from $\{0, 0.2, 0.4, 0.6, 0.8, 1\}$. The best number of trees in the forest of the random forest classifier was selected from $\{25, 50, 75, 100, 125, 150\}$. The best additive smoothing parameter of the categorical naive Bayes classifier was selected from $\{0, 0.2, 0.4, 0.6, 0.8, 1\}$. The best number of neighbours of the K-nearest neighbours classifier was selected from $\{5, 10, 15, 20, 25, 30\}$. The best regularisation parameter of the support vector machine classifier was selected from $\{0.001, 0.01, 0.1, 1, 10, 100\}$. All of the remaining parameters were set to be defaults given in scikit-learn 0.23.2.
\par
\begin{table}[t
\caption{Learning performance averaged over one-hundred random splits of training, cross validation and test sets.}
\label{tab:comp}
\begin{center}
\begin{tabular}{c|ccc}
Classifier & Accuracy (std. dev.) & AUC (std. dev.) & Runtime (sec.) \\\hline
Bayesian entailment & 0.785 (0.034) & \textbf{0.857} (0.032) & \textbf{0.004} \\
Random forest & \textbf{0.790} (0.032) & 0.844 (0.029) & 0.092 \\
Naive Bayes & 0.707 (0.037) & 0.826 (0.034) & 0.009 \\
K-nearest neighbours & 0.718 (0.034) & 0.676 (0.043) & 0.005 \\
Support vector machine & 0.696 (0.035) & 0.652 (0.041) & 0.085
\end{tabular}
\end{center}
\end{table}
\section{Discussion and Conclusions}\label{sec:discussion}
There are a number of attempts to combine logic and probability theory, e.g., \cite{adams:98,Fraassen:81b,Fraassen:83,Morgan:83a,Cross:93,Leblanc:79,Leblanc:83b,Pearl:91,Goosens:79,Richardson:06,matthias:13}. They are commonly interested in the notion of probability preservation, rather than truth preservation, where the uncertainty of the conclusion preserves the uncertainty of the premises. They all presuppose and extend the classical entailment. In contrast, this paper gives an alternative entailment without presupposing it.
\par
Besides the preferential entailment, various other semantics for non-monotonic consequence relations have been proposed such as plausibility structure \cite{friedman:96}, possibility structure \cite{dubois:90,Benferhat:03}, ranking structure \cite{goldszmidt:92} and $\varepsilon$-semantics \cite{adams:75,pearl:89}. The common idea of the first three approaches is that $\Delta$ entails $\alpha$ if $\llbracket\Delta\land\alpha\rrbracket\succeq\llbracket\Delta\land\lnot\alpha\rrbracket$ holds given preference relation $\succeq$. However, as discussed in \cite{brewka:97}, it is still unclear how to encode preferences among abnormalities or defaults. A benefit of our approach is that the preferences can be encoded via Bayesian updating, where the distribution over possible worlds is dynamically updated within probabilistic inference in accordance with observations. Meanwhile, the idea of $\varepsilon$-semantics is that $\Delta$ entails $\alpha$ if $p(\alpha|\Delta)$ is close to one, given a probabilistic knowledge base quantifying the strength of the causal relation or dependency between sentences. They are fundamentally different from our work as we probabilistically model the interaction between models and sentences. The same holds true in the approaches \cite{adams:75,pearl:89,Hawthorne:07a,Hawthorne:07b}.
\par
Naive Bayes classifiers and Bayesian network classifiers work well under the assumption that all or some attributes in data are conditionally independent given another attribute. However, it is rare in practice that the assumption holds in real data. In contrast to the classifiers, our logical model does not need the conditional independence assumption. This is because the logical model always evaluates dependency between possible worlds and attributes, but not dependency among attributes.
\par
In this paper, we introduced a generative model of logical entailment. It formalised the process of how the truth value of a formula is probabilistically generated from the probability distribution over possible worlds. We discussed that it resulted in a simple inference principle that was correct in terms of classical logic, paraconsistent logic, nonmonotonic logic and machine learning. It allowed us to have a general answer to the questions such as how to logically infer from inconsistent knowledge, how to rationally handle defeasibility of everyday reasoning, and how to probabilistically infer from noisy data without a conditional dependence assumption.
\section*{Appendix}
\begin{proof}[Proposition 1]
We abbreviate $W=w$ to $w$ for simplicity. Since $\llbracket\alpha=0\rrbracket_{w}=1-\llbracket\alpha=1\rrbracket_{w}$, we have
\begin{eqnarray*}
p(\alpha=0|w)+p(\alpha=1|w)&=&\mu^{\llbracket\alpha=0\rrbracket_{w}}(1-\mu)^{1-\llbracket\alpha=0\rrbracket_{w}}+\mu^{\llbracket\alpha=1\rrbracket_{w}}(1-\mu)^{1-\llbracket\alpha=1\rrbracket_{w}}\\
&=&\mu^{1-\llbracket\alpha=1\rrbracket_{w}}(1-\mu)^{\llbracket\alpha=1\rrbracket_{w}}+\mu^{\llbracket\alpha=1\rrbracket_{w}}(1-\mu)^{1-\llbracket\alpha=1\rrbracket_{w}}.
\end{eqnarray*}
(1) holds because both $p(\alpha|w)$ and $p(w)$ cannot be negative. If $\llbracket\alpha=1\rrbracket_{w}=1$ then $p(\alpha=0|w)+p(\alpha=1|w)=(1-\mu)+\mu=1$. If $\llbracket\alpha=1\rrbracket_{w}=0$ then $p(\alpha=0|w)+p(\alpha=1|w)=\mu+(1-\mu)=1$. Now, (2) is shown as follows.
\begin{eqnarray*}
p(\alpha=0)+p(\alpha=1)&=&\sum_{w}p(\alpha=0|w)p(w)+\sum_{w}p(\alpha=1|w)p(w)\\
&=&\sum_{w}p(w)\{p(\alpha=0|w)+p(\alpha=1|w)\}\\
&=&\sum_{w}p(w)=1
\end{eqnarray*}
(3) is shown as follows. From (2), it is sufficient to show only case $i=1$ because case $i=0$ can be developed as follows.
\begin{eqnarray*}
1-p(\alpha\lor\beta=1)&=&1-\{p(\alpha=1)+p(\beta=1)-p(\alpha\land\beta=1)\}.
\end{eqnarray*}
Now, it is sufficient to show $p(\alpha\lor\beta=1|w)=p(\alpha=1|w)+p(\beta=1|w)-p(\alpha\land\beta=1|w)$ since case $i=1$ can be developed as follows.
\begin{eqnarray*}
\sum_{w}p(\alpha\lor\beta=1|w)p(w)=\sum_{w}\{p(\alpha=1|w)+p(\beta=1|w)-p(\alpha\land\beta=1|w)\}p(w)
\end{eqnarray*}
By case analysis, the right expression is shown to have
\begin{eqnarray}
(1-\mu)+(1-\mu)-(1-\mu)&=&1-\mu\label{1}\\
(1-\mu)+\mu-(1-\mu)&=&\mu \label{2}\\
\mu +(1-\mu)-(1-\mu)&=&\mu\label{3}\\
\mu+\mu-\mu&=&\mu \label{4}
\end{eqnarray}
where (\ref{1}), (\ref{2}), (\ref{3}) and (\ref{4}) are obtained in the cases ($\llbracket\alpha=1\rrbracket_{w}=\llbracket\beta=1\rrbracket_{w}=0$), ($\llbracket\alpha=1\rrbracket_{w}=0$ and $\llbracket\beta=1\rrbracket_{w}=1$), ($\llbracket\alpha=1\rrbracket_{w}=1$ and $w\in\llbracket\beta=1\rrbracket_{w}=0$), and ($\llbracket\alpha=1\rrbracket_{w}=\llbracket\beta=1\rrbracket_{w}=1$), respectively. All of the results are consistent with the left expression, i.e., $p(\alpha\lor\beta=1|w)$.
\end{proof}
\begin{proof}[Proposition 2]
For all $w$, $p(\alpha=1|w)=\mu$ if and only if $p(\lnot\alpha=1|w)=1-\mu$, and $p(\alpha=1|w)=1-\mu$ if and only if $p(\lnot\alpha=1|w)=\mu$. Therefore, $p(\alpha=1|w)=1-p(\lnot\alpha=1|w)$. From (2) of Proposition \ref{kolmogorov}, we have
\begin{eqnarray*}
p(\alpha=1)&=&\sum_{w}p(\alpha=1|w)p(w)\\
&=&\sum_{w}\{1-p(\lnot\alpha=1|w)\}p(w)\\
&=&\sum_{w}p(\lnot\alpha=0|w)p(w)=p(\lnot\alpha=0).
\end{eqnarray*}
\end{proof}
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Shalini R. Urs is an information scientist who pioneered the digital library movement in India, particularly the Electronic Thesis and Dissertations (ETD). She was awarded the Networked Digital Library of Theses and Dissertations–Adobe Leadership award in 2004 for her initiatives. She built a multilingual online digital library called Vidyanidhi in 2000, long before the digital era captured the imagination of many in India. She currently serves on the Board of Directors of Gooru and is an Associate Editor of Information Matters.
Career
Dr. Urs started her career with University of Mysore in 1976 which she served till her retirement in 2016. She founded the first iSchool in India—the International School of Information Management, at the University of Mysore. She was a Fulbright scholar and a visiting professor at the Department of Computer Science, Virginia Tech, USA during 2000 -2001. She was a visiting professor at the Documentation Research and Training Centre, Indian Statistical Institute from 1998 -1999. She was an adjunct faculty at the International Institute of Information technology, Bangalore (IIIT-B) from 2005-2008. She was awarded the Mortenson Distinguished Lecturer 2010 by the University of Illinois Urbana Champaign, USA and was invited to deliver their annual lecture series.
Research
Her research areas include Information Retrieval, Ontology Development, and Social Media and Network Analysis. As a leading digital library expert, she was commissioned to carry out studies and projects by UNESCO, Educational Development Council, Washington, DC, USA and others. Some of her well-known projects/studies are:
UNESCO Guide to Electronic Theses and Dissertations (ETDs), 2002
UNESCO Directory of Open Access Education and Training Opportunities, 2011
UNESCO sponsored Training Program called UN4IM Training Library and Information Professionals in India.
UNESCO Workshop on eBooks, ISiM. September 2004
Digital Library of Learning Educational Development Council, Washington, DC, USA Project the Dot-EDU digital library of audiovisual learning materials for underprivileged school children developed under the project 'Technology Tools for Teaching and Training in India
Unicode Implementation for Indian Scripts funded by Microsoft
She has published more than 200 papers in peer-reviewed journals and prestigious academic conferences and also edited five books published by Springer. Her Google Scholar profile has more than 650 citations. She received the Emerald Research Fund Award 2007/2008 for her research "Networks and Turning Points: An Exploration of the Dynamics of Academic Networks and Social Movements in India".
Vidyanidhi
Vidyanidhi, the digital library project that was conceptualized and developed by Dr. Shalini Urs was funded by Government of India, Department of Scientific and Industrial Research under their National Information System for Science and Technology (NISSAT). After further funding from the Ford Foundation, Vidyanihi became a national initiative. Vidyanidhi project was featured as an innovative case in the book - Digital Libraries: Principles and Practice in a Global Environment. Vidyanidhi was an endeavor to establish a national repository and a collaborative consortium of participating universities and academic institutions for the creation, submission, archiving and accessing of Indian theses and dissertations.
Conferences
Dr. Urs was instrumental in bringing the International Conference on Asian Digital Library (ICADL), one of the biggest digital library conferences, to India and organized it in Bangalore in 2001. She brought the series again in 2013. She also initiated and has organized a well-known Academia-Industry Summit series called the InfoVision-Knowledge Summit in collaboration with industries such as Rediff.com, Microsoft Research, and industry bodies such as CII and BCIC since 2005 and has brought speakers from across the globe including star speakers such as Jimmy Wales (founder of Wikipedia) and Prabhakar Raghavan, Vice President of Engineering at Google and others.
International School of Information Management (ISiM)
The International School of Information Management (ISiM) is the first Information School (iSchool) founded by Shalini Urs with seed funding from the Ford Foundation to explore the feasibility of establishing an interdisciplinary institute focused on training and educating students to develop knowledge and skill sets in the broad-based domain of the science, technology, and management of information at the intersection of 'information' 'technology' and 'society' in line with the motto of iSchools.. ISiM was established in 2005 in partnership with the University of Pittsburgh, International Institute of Information Technology, Bangalore and Informatics India as an industry partner.
MYRA
Shalini Urs founded the Mysore Royal Academy (MYRA) School of Business in 2012 as a boutique business school focused on the Triple Bottom Line. MYRA is committed to excellence in everything they do-faculty, pedagogy, and campus. MYRA created a unique learning platform called the Immersion Learning Model by bringing faculty from across the globe to teach one course in an immersive two-week-long program.
InfoFire podcast series
As an Associate Editor of Information Matters, Shalini Urs conceptualized and initiated a video podcast series called InfoFire: Fireside chat series in January 2022. Each episode is a conversation with an expert on a topical theme. InfoFire brings visionaries and experts across different domains to share their perspectives on the chosen topic.
InfoFire episodes are also published as part of the eJournal Information Matters and distributed by Information Systems & eBusiness network, a division of Social Science Electronic Publishing (SSEP) and Social Science Research Network.
References
Living people
Information scientists
University of Mysore alumni
Year of birth missing (living people) | {
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{"url":"https:\/\/web2.0calc.com\/questions\/a-bag-contains-different-colored-beads","text":"+0\n\n# A bag contains different colored beads.\n\n0\n177\n13\n+83\n\nA bag contains different colored beads. The probability of drawing two black beads from the bag without replacement is 3\/35 , and the probability of drawing one black bead is 3\/10 .\n\nWhat is the probability of drawing a second black bead, given that the first bead is black?\n\nRedBlue \u00a0Dec 21, 2017\n\n#1\n+85937\n+2\n\nSo we have\n\n(3\/35)\u00a0 \u00a0=\u00a0 (3\/10)\u00a0 \u00a0* P(Black Bead on Second Draw)\n\nMultiply both sides by (10\/3)\n\n(10\/3)(3\/35)\u00a0 =\u00a0 P(Black Bead on Second Draw)\u00a0 =\n\n10\/35\n\nWe can confirm this\n\nLet\u00a0 A\u00a0 = P(Black Bead on first Draw)\u00a0 =\u00a0 3\/10\n\nLet\u00a0B\u00a0 = P(Black Bead on the Second Draw)\n\nLet\u00a0 P(A and B)\u00a0 =\u00a0 3\/35\n\nSo\n\nP(B \/ A)\u00a0 =\u00a0 P(A and B)\u00a0 \/ P(A)\u00a0 \u00a0=\u00a0 \u00a0(3\/35)\u00a0 \/ (3\/10)\u00a0 =\u00a0 (3\/35)(10\/3)\u00a0 =\u00a0 10\/35\n\nCPhill \u00a0Dec 21, 2017\nSort:\n\n#1\n+85937\n+2\n\nSo we have\n\n(3\/35)\u00a0 \u00a0=\u00a0 (3\/10)\u00a0 \u00a0* P(Black Bead on Second Draw)\n\nMultiply both sides by (10\/3)\n\n(10\/3)(3\/35)\u00a0 =\u00a0 P(Black Bead on Second Draw)\u00a0 =\n\n10\/35\n\nWe can confirm this\n\nLet\u00a0 A\u00a0 = P(Black Bead on first Draw)\u00a0 =\u00a0 3\/10\n\nLet\u00a0B\u00a0 = P(Black Bead on the Second Draw)\n\nLet\u00a0 P(A and B)\u00a0 =\u00a0 3\/35\n\nSo\n\nP(B \/ A)\u00a0 =\u00a0 P(A and B)\u00a0 \/ P(A)\u00a0 \u00a0=\u00a0 \u00a0(3\/35)\u00a0 \/ (3\/10)\u00a0 =\u00a0 (3\/35)(10\/3)\u00a0 =\u00a0 10\/35\n\nCPhill \u00a0Dec 21, 2017\n#2\n+83\n0\n\nThat's not one of the answer choices.\n\n1\/5\n\n2\/7\n\n9\/350\n\n6\/35\n\nProbably should have included that in the post.\n\nRedBlue \u00a0Dec 21, 2017\n#3\n+85937\n0\n\nSorry.....i'm not great at probability....maybe someone else can help.....\n\nCPhill \u00a0Dec 21, 2017\n#4\n+973\n+3\n\nSir CPhill's method is correct.\n\n$$P(B|A)={\\dfrac {P(A\\cap B)}{P(A)}} \\\\\\\\ \\text { Solve for } P(A\\cap B) =\\dfrac{3}{35} \\text{ and } P(A) =\\dfrac{3}{10}\\\\ P(B|A)= \\Large( \\small \\frac{3\/10}{3\/35} \\Large) = \\small \\dfrac{2}{7}\\\\$$\n\nHis final answer has an arithmetic error.\n\nJust an itsy-bitsy, teeny-weeny, microscopic, nearly infinitesimally small, but calculable f\u00fackup!\n\nI did that once, when I was a toddler chimp, because someone substituted plantains for some of the bananas I was counting.\n\nGingerAle \u00a0Dec 21, 2017\n#5\n+85937\n+3\n\nDuh!!....10\/35\u00a0 =\u00a0 2\/7\n\n[ I always tell Sisyphus to reduce ...unfortunately.....dieting isn't part of his regimen\u00a0]\n\nThanks, GA!!!!\n\nCPhill \u00a0Dec 21, 2017\n#7\n+973\n+3\n\nAs many times as Sisyphus has pushed that bolder up the hill, you\u2019d think both he and the rock would be reduced to near ZERO by now.\n\nGingerAle \u00a0Dec 22, 2017\n#6\n0\n\nGA: It looks like you effed up again! Maybe you should eat boiled plantains instead of bananas! Or maybe crow! Happy chowing!.\n\nGuest\u00a0Dec 21, 2017\n#9\n+973\n0\n\nBl\u00f2\u00f3dy amazing!\n\nGingerAle \u00a0Dec 22, 2017\nedited by GingerAle \u00a0Dec 22, 2017\n#10\n+973\n+2\n\nWell, thank you for the catered meal, Mr. BB! Crow is a delicacy for us chimps, and plantains are a fine complement for such a meal, especially when boiled and reduced in blarney sauce.\u00a0\u00a0 I usually dress for fine dining, so I will don my gay apparel and Troll the ancient Mr. BB\u2019s carol: Troll-la-Troll-la-Troll, la-Troll-Troll, Troll!\n\nLancelot Link had Stu, and Nauseated had Sorasyn, but they left, taking their dumbness elsewhere.\u00a0 I have you, Mr. BB, the stubborn, relentless, intractable Blarney Banker of lore: a pseudo intellectual with a multiplicity of advanced dimwit degrees in arrogant stup\u00ecdity, a professor of misinformation, who teaches with authority and irritation.\n\nOne usually needs a scanning electron microscope to find an error by Sir CPhill, That\u2019s why the errors are \u201citsy-bitsy, teeny-weeny, microscopic, nearly infinitesimally small,\u201d but still calculable. For you, I need no such instrument. A casual stroll through the forum shows a path, clearly marked by stup\u00ecdity, m\u00f2ronic analysis, and toxic misinformation, leading right to your hovel.\n\nYou may not believe this, but I\u2019m glad you\u2019re here.\u00a0 The most successful Trolls have a reliable parade of stubborn d\u00f9mb d\u00f9mbs, or a relentless, indefatigable d\u00f9mb d\u00f9mb who always gives the wrong answer to \u201cWhat is the airspeed velocity of an unladen swallow?\u201d\u00a0 No matter how many times my troll character tosses you over the bridge railing into the river, you always come back for more.\u00a0 With this, you have personified a modern attachment to the ancient legends of Camelot and its Trolls.\n\nWe all owe you a debt of pseudo gratitude for the unwitting humor you cause.\n\nGA\n\nGingerAle \u00a0Dec 22, 2017\nedited by GingerAle \u00a0Dec 22, 2017\n#11\n+92254\n+4\n\nA bag contains different colored beads. The probability of drawing two black beads from the bag without replacement is 3\/35 , and the probability of drawing one black bead is 3\/10 .\n\nWhat is the probability of drawing a second black bead, given that the first bead is black?\n\nThe prob of drawing one black bead is 3\/10 so\n\nThere are 3n black beads and 7n non-black beads where n is a positive integer making 10n altogether:\n\nP(BB with no replacement) =\n\n$$\\frac{3}{10}\\times \\frac{3n-1}{10n-1}=\\frac{3}{35}\\\\ \\frac{3n-1}{10n-1}=\\frac{3}{35}\\times \\frac{10}{3}\\\\ \\frac{3n-1}{10n-1}=\\frac{2}{7}\\\\ 21n-7=20n-2\\\\ n=5$$\n\nSo there is 15 black and 35 non-black and 50 altogether\n\nIf one black has already been selected then there will be 14 black and 49 altogether remaining.\n\nSo the probability of selecting a second black is\u00a0$$\\frac{14}{49}=\\frac{2}{7}$$\n\nI seem to have done it the long way but our answers are all the same. :\/\n\nMelody \u00a0Dec 23, 2017\n#12\n+85937\n+1\n\nI like that method, Melody.......!!!\n\nCPhill \u00a0Dec 23, 2017\n#13\n+92254\n+1\n\nThanks Chris,\n\nI get confused with all 'Ginger's'\u00a0set symbols. I should become more familiar with them I suppose.\n\nMy way is using basic algebra that most of us are more comfortable with.\n\nThen again, your way is just using basic algebra too and it is much easier than what I did.\n\nI like your approach much better than mine. :)\u00a0 I think Ginger's is the same as yours. :)\n\nMelody \u00a0Dec 23, 2017\n\n### 33 Online Users\n\nWe use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners. \u00a0See details","date":"2018-04-26 17:24:21","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\": 0.7629196643829346, \"perplexity\": 9678.781092206826}, \"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-2018-17\/segments\/1524125948426.82\/warc\/CC-MAIN-20180426164149-20180426184149-00362.warc.gz\"}"} | null | null |
Sartana nella valle degli avvoltoi é uma produção cinematográfica italiana de 1970, do subgênero western spaghetti, dirigida por Roberto Mauri.
Trata-se de um dos títulos não-oficiais entre os quais o personagem Sartana esteve presente. O protagonista adota no filme o nome de Lee Calloway. Porém a denominação Sartana se empregou nas mais diversas titulações como forma de atração de lucro.
Sinopse
Sartana, um célebre pistoleiro do oeste, é contratado por dois indivíduos misteriosos para que liberte um criminoso de um presídio. A recompensa será a metade do ouro que o foragido havia roubado. Contudo, Sartana cairá numa armadilha, se salvando graças à intervenção de um mexicano.
Elenco
William Berger - Lee Calloway (Sartana)
Wayde Preston - Anthony Douglas
Aldo Berti - George Douglas
Carlo Giordana - Slim Douglas
Franco De Rosa - Peter Douglas
Luciano Pigozzi (creditado como Alan Collins) - Paco
Jolanda Modio - Juanita
Pamela Tudor - Esther
Josiane Tanzilli - Carmencita
Bibliografia
Fridlund, Bert: The Spaghetti Western. A Thematic Analysis. Jefferson, NC and London: McFarland & Company Inc., 2006.
Fridlund, Bert: 'A First Class Pall-bearer!' The Sartana/Sabata Cycle in Spaghetti Westerns. Film International Vol. 6 No. 3 / 2008.
Joyeux, Francois: Sartana Gianni Garko Anthony Ascott. Vampirella 13. Publicness Paris 1975.
Marco Giusti Dizionario del Western all'italiana, Mondadori ISBN 978-88-04-57277-0.
Filmes da Itália de 1970
Filmes em língua italiana
Filmes de faroeste da Itália
Filmes sobre fuga da prisão
Filmes dirigidos por Roberto Mauri | {
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Maxville is an unincorporated community in Perry County, in the U.S. state of Ohio. Maxville Limestone, a natural limestone formation, was named after the community.
History
A post office was established at Maxville in 1855, and remained in operation until 1904. The community's name is derived from shortening and alteration of the name of William McCormick, a first settler.
Notable person
Edwin D. Ricketts, a U.S. Representative from Ohio, was born near Maxville in 1867.
References
Unincorporated communities in Perry County, Ohio
Unincorporated communities in Ohio
1855 establishments in Ohio
Populated places established in 1855 | {
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The Buddh International Circuit () is an Indian motor racing circuit situated in National Capital Region at Greater Noida, Uttar Pradesh. The track shares its name with Gautama Buddha, as does the district the track is located in.
The track was officially inaugurated on 18 October 2011. The long circuit was designed by German racetrack designer Hermann Tilke.
The circuit was best known as the venue for the annual Formula One Indian Grand Prix, which was first hosted in October 2011. However, the Grand Prix was suspended for 2014 and subsequently cancelled due to a tax dispute with the Government of Uttar Pradesh during the Akhilesh Yadav administration.
History
2000s
In 2007, a tentative agreement to host the Indian Grand Prix was reached between the Indian Olympic Association and Bernie Ecclestone, then chief-executive of the Formula One Group. A site in Greater Noida in the north Indian state of Uttar Pradesh was chosen as the location of the track that would host the race. Delays in the procurement of the land and construction were reported in early 2009 owing in part to the ongoing financial crisis. The design of the track was ultimately revealed in November 2009.
The venue was slated to debut during the 2010 Formula One season, with construction scheduled to be completed in time. However, this date was pushed back and the inaugural race eventually took place the following year.
2010s
The first Indian Grand Prix was held at the Buddh International Circuit on 30 October 2011. It was initially scheduled to take place in December 2011. However, following scheduling conflicts with the Bahrain Grand Prix and its subsequent cancellation, the race was rescheduled to October. The inaugural race was won by Red Bull Racing's Sebastian Vettel, who started from pole position. He also set the fastest lap and therefore race lap record that year, which stood for the following two Grands Prix.
During the 2012 Formula One season and the second Formula One race to take place at the circuit, Sebastian Vettel took pole once again, led every lap and went on to win the race. The fastest lap of the race was set by McLaren's Jenson Button.
The following year, the Indian Grand Prix was the sixteenth race in a nineteen-race season. Sebastian Vettel started on pole and subsequently won the race. Vettel also secured his fourth consecutive Formula One World Drivers' Championship at this race, beating his closest challenger, Ferrari's Fernando Alonso, who finished 11th. Red Bull also secured its fourth consecutive Constructors' Championship. The fastest lap of the race was set by Lotus' Kimi Räikkönen. This was the last Formula One race held at the circuit.
The track was not slated to be part of the 2014 season. One reason given was the rescheduling of the Indian Grand Prix to March for 2014, which made little sense after a race in October 2013. Its appearance in the 2015 season was subsequently ruled out in mid-2014, citing contractual and taxation issues.
The owners were unable to significantly recover their investments through all three seasons the races were held, and were forced to write off losses worth at least $25.1 million. In 2016, the owners reiterated their desire to not sell the circuit to other buyers, despite its high maintenance costs, their own financial distress, and the lack of future scheduled international sporting events. The circuit has continued to host smaller local racing series and championships, including the JK Tyre National Racing Championship which includes open-wheel racing as well as motorcycle racing, and open track days.
Tax dispute
The Buddh International Circuit is located in the state of Uttar Pradesh, subject to its local taxes as well as national customs duties. The first signs of a dispute arose in 2009: in a letter to the promoters JPSK Sports, the Indian Ministry of Youth Affairs and Sports denied JPSK Sports permission to remit $36.5 million in licensing fees to Formula One administration headquarters in London. The reason given, by the then Chief Minister of Uttar Pradesh - Akhilesh Yadav, was the nature of Formula One, considered not to be a sport but rather entertainment, and its perceived lack of impact on the development of sports in the country. Customs fees for imported components including engines and tyres were not waived, and tax exemptions given to other sports were not offered to the organisers. Fees worth $51.3 million, meant to be paid by Jaypee Sports to Formula One World Championship Limited (FOWC), were still pending as of Liberty Media's acquisition of the Formula One Group in 2016. The government's expectation of tax revenue meant that Jaypee was prevented from paying its dues to FOWC.
In a judgement issued in April 2017, the Supreme Court of India ruled that the circuit constituted a 'permanent establishment', and as such FOWC was liable to pay taxes on any income accrued by it in India, estimated at 40% of business income. It considered royalty payments made by Jaypee to FOWC to be business income as well, subject to tax, which contradicted the original agreement between Jaypee and FOWC that stipulated that any fees would be paid free of taxes. Liberty Media was prepared to settle the due amount in July 2017, setting aside $14.8 million.
2020s
For 2022, it was planned that Buddh International Circuit would host events for the Formula Regional Indian Championship and the F4 Indian Championship, however both championships were cancelled in 2022. The circuit will also host the Indian motorcycle Grand Prix from 2023 as part of the MotoGP World Championship.
Design
Formula One racing's governing body, the FIA, had announced the inclusion of the Indian Grand Prix for 30 October 2011. Estimated to cost about 20 billion ($400 million) to build, the circuit has a length of and is spread over an area of and is another creation of Hermann Tilke. The circuit was officially inaugurated on 18 October 2011, just about two weeks before the first race. The seating capacity is initially expected to be 110,000 with provisions to increase it to 200,000 later on.
Circuit
The circuit is part of the Jaypee Greens Sports City, which has increasingly delayed plans to include a 100,000 seating capacity international cricket stadium, 18-hole golf course, 25,000 seat field hockey stadium and a sports academy. The circuit design incorporated feedback from the teams on how the circuit could be altered to improve overtaking. This resulted in some minor changes before 2010: the planned hairpin at turn seven was removed, and the track at turn three was widened to allow drivers to take different lines throughout the corner. More information was released in August 2010, revealing that there were plans to make the circuit one of the most challenging for drivers, with the circuit rising fourteen metres within the first three corners alone and a banked double-apex bend on the far side of the circuit. The track has since been praised by drivers, including Lewis Hamilton who compared it to the classic Circuit de Spa-Francorchamps.
The banked multi-apex turn 10–11–12 sequence is one of the most notable sections of the circuit. It has been likened to the long, fast Turn 8 at Turkey's Istanbul Park circuit as it is a challenging sequence that generates high tyre loadings. Unlike Turkey's Turn 8, it tightens on exit and is a clockwise right-hander. Although it is not one of the main overtaking points, it is the corner that shows F1 cars to their full cornering potential. The circuit's main straight, at , is among the longest in F1, with a key overtaking point at its end. The pitlane is also one of the longest in F1, at more than 600 metres: as in most races with pit stops, time spent in the pitlane is an important factor in determining race strategies.
Before the opening weekend, the expected lap time for a Formula One car around the track was 1 minute, 27.02 seconds, at an average speed of 210.03 km/h (131 mph). At the end of the long straight between corners 3 and 4, Formula One cars were expected to reach a top speed of about . In the inaugural qualifying session, Sebastian Vettel turned in a lap time of 1 minute, 24.178 seconds, beating the predicted lap times from tyre manufacturer Pirelli. Scuderia Toro Rosso driver Jaime Alguersuari posted the top speed through the speed trap, reaching .
The relatively compact circuit was due to host a GT1 World Championship round as the season finale in December 2012, but the event was cancelled.
The circuit was due to host a Superbike World Championship round for four seasons, starting in 2013. However, the 2013 race was cancelled due to operational charges of Buddh International Circuit.
From 2014 to 2022, there were no international championships have raced at the circuit except the Asia Road Racing Championship race in 2016. From 2023, the circuit will host the Indian motorcycle Grand Prix as part of the MotoGP World Championship.
Reception
The reception among drivers was positive, with praise for the high-speed layout and challenging corner combinations that Jenson Button described as difficult to drive in a consistently quick fashion.
The circuit has been often compared by F1 drivers to the classic Circuit de Spa-Francorchamps, a track known for high speeds and the type of corner-to-corner flow that comes from natural terrain. Inaugural winner Sebastian Vettel praised the track saying that "there is a lot of elevation change around the lap which adds to the fun, from as much as 8% downhill and up to 10% uphill; it's like a roller coaster. It really has emerged as one of the most challenging circuits on the calendar for the drivers."
Name and the logo
Originally known as the Jaypee Group Circuit or the Jaypee International Circuit after the circuit's owners, the circuit was officially named the Buddh International Circuit in April 2011. According to Sameer Gaur, the managing director and chief executive officer of Jaypee Sports International Limited, the name 'Buddh International Circuit' has been chosen with reference to the area where the racetrack is situated – Gautam Buddh Nagar district (also known as Greater Noida). Because of its location, naming the circuit 'Buddh International Circuit' was a logical choice for the company.
The logo consists of a stylized 'B' in the shape of a heart, which stands for 'Buddh' and 'Bharat' (the native name of India). The saffron, green and white colours used in the logo are representative of the Indian flag, while the curves in the stylized 'B' in the logo represent the lines of a racetrack.
Awards and recognition
Buddh International Circuit, which hosted India's first Formula One Grand Prix on 30 October 2011, has been awarded the '2011 Motorsport Facility of the Year' award at the Professional Motorsport World Expo 2011. BIC has also been honored with the 'Best Promoter Trophy' for the successful conduct of Formula One races in 2011 & 2012 at the FIA prize-giving gala.
Events
Current
September: Grand Prix motorcycle racing Indian motorcycle Grand Prix
Former
Asia Road Racing Championship (2016)
Formula One Indian Grand Prix (2011–2013)
JK Racing Asia Series (2011–2012)
MRF Challenge Formula 2000 Championship (2012–2013, 2017)
Lap records
The official race lap records at the Buddh International Circuit are listed as:
Gallery
See also
Madras Motor Race Track
Hyderabad Street Circuit
Notes
References
External links
Official Website
BIC News
Sports venues in Noida
Formula One circuits
Grand Prix motorcycle circuits
Motorsport venues in Uttar Pradesh
Indian Grand Prix
Jaypee Group
Racing circuits designed by Hermann Tilke
2011 establishments in Uttar Pradesh
Sports venues completed in 2011 | {
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\section{Introduction}
\label{sec:intro}
\subsection{IDH mutation status}
Gliomas are the most common malignant brain tumors with remarkable heterogeneity leading to distinct clinical outcomes~\cite{li2019intratumoral}. It is established that the mutation status of isocitrate dehydrogenase (IDH) genes is one of the most critical indicators for glioma stratification and prognosis determination. Specifically, IDH mutants generally have longer survival time than wild-types~\cite{louis20162016}. Due to the clinical significance, IDH mutation status is recommended as a routine assessment for glioma patients by the World Health Organization classification of tumors of the Central Nervous System~\cite{louis20162016}. However, current approaches of IDH mutation detection, i.e., immunohistochemistry and gene sequencing, are invasive and expensive~\cite{louis20162016}. Further, these approaches usually require the tumor tissue obtained from resection or biopsy, which could be challenging in some patients.
\subsection{Radiogenomics}
Radiogenomics is an approach to bridge radiomic and genomic features for tumor phenotyping. Mounting numbers of studies have shown that radiomic features could predict IDH mutations based on routinely collected multi-modal MRI, i.e., pre-contrast and post-contrast T1, T2, and T2-weighted fluid-attenuated inversion recovery (FLAIR)~\cite{hyare2019modelling,liang2018multimodal}. In addition to the commonly used texture and intensity features, novel radiomic features are further developed to describe tumor geometric information and successfully characterize various types of cancer, including gliomas \cite{wu2021radiological}.
These results could highlight the usefulness of geometric information in tumor characterization and genotype prediction.
\subsection{Geometric deep learning}
In parallel, deep learning models provide end-to-end training schemes, which have shown better performance in prediction tasks with improved robustness than traditional models. In particular, geometric deep learning approaches, based on graph neural networks (GNN), have shown promising performance in computer vision tasks, e.g., 3D object classification. A typical geometric learning model relies on the 3D object scanner to produce 3D geometric data, e.g., meshes and points cloud, which can be readily reconstructed from MRI for model training.
\subsection{Proposed method}
We hypothesize that geometric deep learning could be employed to extract useful geometric information from brain MRI for predicting glioma genotype, i.e., IDH mutation status. Further, as gliomas frequently demonstrate significant intra-tumoral heterogeneity, a model capturing relevant information from both tumor boundary and internal bulk tumor content could achieve better model performance.
This study proposes a collaborative learning framework consisting of three components: an image learner based on 3D convolutional neural networks (CNN), a geometric learner based on GNN, and a collaborative learning objective function that maximizes the agreement of two networks and optimizes the prediction performance. To our best knowledge, this is the first study that
\begin{itemize}
\item investigates the usefulness of geometric deep learning to predict tumor genotype.
\item collaborates GNN and CNN to classify brain tumors.
\end{itemize}
\section{Methods}
\subsection{Data preparation}
We collect multi-modal MRI of 389 glioma patients from three cohorts available from The Cancer Imaging Archive (TCIA) website \cite{tcgalgg,tcgagbm,ivygap} and another in-house cohort with 117 patients. The imaging modalities include routinely available pre-operative anatomical MRI: pre-contrast T1, post-contrast T1, T2, and FLAIR. We exclude 17 patients due to the missing IDH mutation status or incomplete MRI scans, and finally include 372 patients (IDH mutant: 103; IDH wild-type: 269).
\begin{figure}[h]
\begin{center}
\centerline{\includegraphics[width=\columnwidth]{FIG1_data_prepare.png}}
\caption{Input data for the collaborative framework. Tumor segmentation is performed on the multi-modal MRI to generate the tumor mask (red). The MRI voxels within the tumor mask are extracted as the input image data for the CNN. The binarized tumor masks are converted to surface meshes sampled to points cloud as the input geometric data for the GNN. }
\label{Fig:experiment}
\end{center}
\end{figure}
MR images are processed following a standard pipeline described previously~\cite{bakas2017advancing}, including co-registration, brain extraction, histogram-matching, smoothing, and tumor segmentation, to generate a binary mask of core tumor. Then, the tumor masks, manually corrected by a neurosurgeon and a researcher after initial training, are evaluated using the DICE score. Finally, the tumor mask bounded multi-modal MRI are fed into CNN for model training. In generating the geometric data of the tumor, we convert the tumor mask into mesh data, and the vertices of the mesh are randomly sampled into points cloud as the geometric input data for GNN. We adopted this approach because graphs generated from points cloud reflect hierarchical topological information which may be omitted by mesh-based graph data (Figure~\ref{Fig:experiment}). Points cloud coordinates are normalized to [-1, 1], and 512 points are randomly sampled from the points cloud of each tumor due to the memory limitation of our local computation resources.
\subsection{Collaborative framework}
\begin{figure}[h]
\begin{center}
\centerline{\includegraphics[width=\columnwidth]{FIG2_NeuralNetworks.png}}
\caption{The architecture of the proposed framework. \textbf{(a)} The collaborative framework consists of a CNN, a GNN and a collaborative objective function. The CNN and GNN output latent features for KL divergence (KLDiv) loss and a label prediction for binary cross entropy (BCE) loss. The CNN consists of four 3D-convolutional layers (3DConv) and three fully connected (FC) layers. The GNN consists of four points cloud convolutional layers and three FC layers. \textbf{(b)} The points cloud convolution of the GNN involves three steps: graph generation from points cloud, graph convolution and pooling. }
\label{Fig:neuralnetworks}
\end{center}
\end{figure}
The collaborative framework combines a CNN as the image learner and a GNN as the geometric learner to predict the IDH mutation status using the binary cross entropy loss (Equation~\ref{Eq:BCE}) while maximizing the concordance between the latent features of two networks using the Kullback-Leibler (KL) divergence loss to optimize the prediction performance (Equation~\ref{Eq:KLD}):
\begin{align}
\begin{split}
L_{BCE} &= -[y \cdot \log{x_u} + (1 - y) \cdot \log{(1 - x_u)}]
\\ & -[y \cdot \log{x_v} + (1 - y) \cdot \log{(1 - x_v)}]
\end{split}
\label{Eq:BCE}
\end{align}
where $y$ represents the label of the IDH mutation status; $x_u$ and $x_v$ are the output from the CNN and the GNN, respectively.
\begin{align}
\begin{split}
L_{KLdiv} = z_u (\log{z_u} - z_v) + z_v (\log{z_v} - z_u)
\end{split}
\label{Eq:KLD}
\end{align}
where $z_u$ and $z_v$ are latent features before the last FC layer of the CNN and the GNN, respectively. The dimension of both features is 128. Due the asymmetry between different data modalities, we included both graph to image and image to graph KL divergence here.
For the image learner (Figure~\ref{Fig:neuralnetworks}a), the adopted 3D CNN consists of three-dimensional convolutional layers (3DConv) with four input channels corresponding to the four MRI modalities. The three FC layers generate latent features and make predictions. Each 3DConv layer is followed by 3D batch normalization and max pooling for stabilizing training. Meanwhile, the GNN of the geometric learner utilizes specialized points cloud convolution operations, including graph generation, graph convolution and pooling (Figure~\ref{Fig:neuralnetworks}b). Specifically, the graph generation step firstly converts points cloud into a graph by generating links between points and their neighbors within a predefined radius distance (0.25, 0.5, 0.75, 1 for four points cloud convolution layers, respectively). Secondly, graph convolution operators aggregate node features (euclidean coordinates of points) and edge features (difference between node features) to the center node. We applied the NNConv operator defined in PyG library ~\cite{fey2019fast} as follows:
\begin{align}
\begin{split}
x_i^\prime = w x_i + \sum_{j\in\mathcal{N}(i)} x_j \cdot h(e_{i,j})
\end{split}
\label{Eq:graphconv}
\end{align}
where $x_i$ and $e_{i,j}$ represent node and edge features, respectively; $j\in\mathcal{N}(i)$ represents all nodes $j$ that are connecting to node $i$; $h$ denotes the multi-layer perceptron that encodes edge features to number of node features; $w$ is the trainable weight. Batch normalization is applied to the multi-layer perceptron in the operator. Finally, we applied the fps pooling of the PyG library to sample points that are the most distant from other points (sampling ratio is set to 0.5 for all four layers).
Furthermore, we interpret the prediction of both the CNN and the GNN. We use Grad-CAM~\cite{selvaraju2017grad} to visualize the activation map of the CNN on tumor voxels. Moreover, we use the GNNExplainer~\cite{ying2019gnnexplainer} to identify the points essential for predicting labels in the points cloud. Additionally, importance of the surface tumor voxels indicated by the Grad-CAM are projected to the corresponding points cloud.
\subsection{Experimental details}
For model training, we first split our dataset into training and testing sets at a 4:1 ratio. The training set was further divided into 4:1 for cross-validation. Volume rotation was applied to augment the minority class in both image and geometric data to mitigate the effect of label imbalance. Adam optimiser was used for training. Learning rate was scheduled from 0.001 to 0.0001 in 200 training epochs to stabilize the training process. Early stopping mechanism, weight decay, and dropout layers were used to prevent over-fitting.
We adopted a 3D Densely Connected Convolutional Networks (3D-DenseNet) as the benchmark for model comparison. Specifically, the classic 121-layer version of 3D-DenseNet described in~\cite{liang2018multimodal} was used, with MRI as the only input. In addition, we conducted ablation experiments by removing the CNN or the GNN individually.
\section{Results}
\begin{table}[htbp]
\caption{Performance of predicting IDH status}
\label{tab:performance}
\resizebox{\columnwidth}{!}{%
\begin{tabular}{cccc}
\hline
Models & Accuracy (\%) & Sensitivity (\%) & Specificity (\%) \\ \hline
\multicolumn{4}{c}{Cross-validation} \\ \hline
3D-DenseNet121 & 85.9 & 89.2 & 82.6 \\
CNN only& 80.1 & 78.4 & 81.8 \\
GNN only& 83.2 & 81.1 & 85.2 \\
Collaborative framework & \textbf{90.1} & \textbf{93.2} & \textbf{86.6} \\ \hline
\multicolumn{4}{c}{Test} \\ \hline
3D-DenseNet121 & 85.1 & 86.5 & 83.8 \\
CNN only& 79.7 & 78.7 & 81.1 \\
GNN only& 82.4 & 81.1 & 83.8 \\
Collaborative framework & \textbf{89.2} & \textbf{91.9} & \textbf{86.5} \\ \hline
\end{tabular}%
}
\end{table}
The numerical results show that the proposed collaborative framework outperforms the benchmark models (Table~\ref{tab:performance}). In addition, the ablation experiments show that the CNN and GNN models could be complementary for better model performance.
\begin{figure}[htbp]
\begin{center}
\centerline{\includegraphics[width=\columnwidth]{FIG3_Interpretation.png}}
\caption{Intepretation of the geometric learner and the image learner. \textbf{(a)} The Grad-CAM output of the image learner. \textbf{(b)} The Grad-CAM output mapped to points cloud. \textbf{(c)} The GNNExplainer output for the geometric learner. The arrows indicate the point clusters with points of importance larger than 0.5 (green) or 0.8 (yellow)}
\label{Fig:intepretation}
\end{center}
\end{figure}
The interpretation results show that the image learner and the geometric learner identify common tumor regions (enhanced and irregular tumor boundaries) important for model prediction (Figure~\ref{Fig:intepretation}b\&c), which may indicate that these regions are critical to differentiate IDH mutation status. In addition, both learners identify their unique regions for model prediction. Particularly, the geometric learner could identify those tumor regions with irregular surface (Figure~\ref{Fig:intepretation}c), while the image learner mainly focuses on the tumor content (Figure~\ref{Fig:intepretation}a\&b). Combining these identified regions could help to better understand the imaging hallmarks of IDH mutant and wild-type.
\section{Conclusion and discussion}
In this paper, we propose a method to collaboratively learn from MRI images and geometrics for predicting IDH mutation status in gliomas. Performance comparison demonstrates that the proposed method outperforms the benchmark models. In our future work, we will systematically screen different combinations of deeper CNN and GNN models to enhance the prediction performance further. Extra ablation experiments on the selection of collaborative loss will be investigated in the future. In addition, we could analyze the interaction between geometrics and focal tumor regions to understand tumor growth and invasion.
To conclude, collaborative learning of CNN and GNN promises to boost deep learning model development in radiogenomic studies.
\section{Compliance with Ethical Standards}
This study was performed in line with the principles of the Declaration of Helsinki. This study was approved by the local institutional review board. Informed written consent was obtained from all institutional patients.
\bibliographystyle{IEEEbib}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 7,881 |
function im_upvpl = im_rgb2upvpl(im_rgb )
%IM_RGB2UPVPL Convert RGB to u'v'L colorspace
cf = makecform('srgb2xyz');
cf2 = makecform('xyz2upvpl');
im_upvpl = applycform(applycform(im_rgb, cf), cf2);
end
| {
"redpajama_set_name": "RedPajamaGithub"
} | 96 |
{"url":"https:\/\/www.ias.ac.in\/listing\/bibliography\/boms\/MOHSEN_ASADI_ASADABAD","text":"Articles written in Bulletin of Materials Science\n\n\u2022 Synthesis and optical characterization of copper nanoparticles prepared by laser ablation\n\nThe remarkable size-tunable properties of nanoparticles (NPs) make them a hot research topic with applications in a wide range of fields. Hence, copper (Cu) colloidal NPs were prepared using laser ablation (Nd:YAG, 1064 nm, 7 ns, 10 Hz, 6000 pulses) of a coppermetal plate at different laser fluences (LFs) in the range of 1\u20132.5 J cm$^{\u22122}$ in ethylene glycol (EG), at room temperature. Analysis of NPs was carried using different independent techniques such as ultraviolet\u2013visible (UV\u2013vis) spectroscopy; transmission electron microscopy (TEM) and Fourier transform infrared (FTIR) spectroscopy. TEM analysis showed that the NPs were spherical with a bimodal distribution and an average particle size of 5 and 16nm influence of 1.2 J cm$^{\u22122}$, and 9 and 22 nm at 2 J cm$^{\u22122}$. The UV\u2013vis spectra of colloidal NPs revealed the maximum absorbance at around 584 nm, indicating the formation of Cu NPs, which supported using FTIR spectra. Furthermore, the absorption spectra confirmed the metallic nature of Cu NPs. FTIR spectroscopy was utilized to verify information about the NPs surface state and chemical bonds constructed in the atom groups apparent on their surface.\n\n\u2022 # Bulletin of Materials Science\n\nVolume 45, 2022\nAll articles\nContinuous Article Publishing mode\n\n\u2022 # Dr Shanti Swarup Bhatnagar for Science and Technology\n\nPosted on October 12, 2020\n\nProf. Subi Jacob George \u2014 Jawaharlal Nehru Centre for Advanced Scientific Research, Jakkur, Bengaluru\nChemical Sciences 2020\n\nPhysical Sciences 2020\n\n\u2022 # Editorial Note on Continuous Article Publication\n\nPosted on July 25, 2019","date":"2022-10-04 22:44:54","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\": 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\": 0.4618052542209625, \"perplexity\": 9365.985548329654}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 5, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2022-40\/segments\/1664030337529.69\/warc\/CC-MAIN-20221004215917-20221005005917-00699.warc.gz\"}"} | null | null |
.DEFAULT_GOAL := build
PACKAGES = $(shell go list ./... | grep -v /vendor/)
TESTARGS ?= -race
CURRENTDIR = $(shell pwd)
SOURCEDIR = $(CURRENTDIR)
SOURCES := $(shell find $(SOURCEDIR) -name '*.go')
APP_SOURCES := $(shell find $(SOURCEDIR) -name '*.go' -not -path '$(SOURCEDIR)/vendor/*')
PATH := $(CURRENTDIR)/bin:$(PATH)
COVERAGEDIR = $(CURRENTDIR)/coverage
VERSION = $(shell cat .goxc.json | python -c "import json,sys;obj=json.load(sys.stdin);print obj['PackageVersion'];")
TEMPDIR := $(shell mktemp -d)
LD_FLAGS = -ldflags '-w' -ldflags "-X main.VERSION=$(VERSION)"
TEST_TARGETS = $(PACKAGES)
all: build
deps:
@./install_consul.sh
@mkdir -p $(COVERAGEDIR)
@which gover > /dev/null || \
(go get github.com/modocache/gover)
@which goxc > /dev/null || \
(go get github.com/laher/goxc)
@which goimports > /dev/null || \
(go get golang.org/x/tools/cmd/goimports)
build-deps: deps format
@mkdir -p bin/
build: check build-deps test
CGO_ENABLED=0 go build $(LD_FLAGS) -o bin/marathon-consul
build-linux: build-deps
CGO_ENABLED=0 GOOS=linux go build -a -tags netgo $(LD_FLAGS) -o bin/marathon-consul
docker: build-linux
docker build -t allegro/marathon-consul .
test: deps $(SOURCES) $(TEST_TARGETS)
gover $(COVERAGEDIR) $(COVERAGEDIR)/gover.coverprofile
$(TEST_TARGETS):
go test -coverprofile=coverage/$(shell basename $@).coverprofile $(TESTARGS) $@
check-deps: deps
@which golangci-lint > /dev/null || \
(GO111MODULE=on go get -v github.com/golangci/golangci-lint/cmd/golangci-lint@v1.25.0)
check: check-deps $(SOURCES) test
golangci-lint run --config=golangcilinter.yaml ./...
format:
goimports -w -l $(APP_SOURCES)
FPM-exists:
@fpm -v || \
(echo >&2 "FPM must be installed on the system. See https://github.com/jordansissel/fpm"; false)
deb: FPM-exists build-linux
mkdir -p dist/$(VERSION)/
cd dist/$(VERSION)/ && \
fpm -s dir \
-t deb \
-n marathon-consul \
-v $(VERSION) \
--url="https://github.com/allegro/marathon-consul" \
--vendor=Allegro \
--architecture=amd64 \
--maintainer="Allegro Group <opensource@allegro.pl>" \
--description "Marathon-consul service (performs Marathon Tasks registration as Consul Services for service discovery) Marathon-consul takes information provided by the Marathon event bus and forwards it to Consul agents. It also re-syncs all the information from Marathon to Consul on startup and repeats it with given interval." \
--deb-priority optional \
--workdir $(TEMPDIR) \
--license "Apache License, version 2.0" \
../../bin/marathon-consul=/usr/bin/marathon-consul \
../../debian/marathon-consul.service=/etc/systemd/system/marathon-consul.service \
../../debian/marathon-consul.upstart=/etc/init/marathon-consul.conf \
../../debian/config.json=/etc/marathon-consul.d/config.json
release: deb deps
goxc
version: deps
goxc -wc -pv=$(v)
git add .goxc.json
git commit -m "Release $(v)"
.PHONY: all bump build release deb
| {
"redpajama_set_name": "RedPajamaGithub"
} | 3,465 |
Q: ASP.NET masterpage events I am subscribing to an event in a masterpage from one of the pages in the page_load event i.e. I have the code
this.Master.Event1 += new Event1EventHandler(Master_Event1);
in page_load event.
If I dont unsubscribe from the event can this cause a memory leak? What would be an appropriate way to unsubscribe? Should I do it in the page_unload event? That will handle the user action of exit from the page, but what would be the correct way to handle it if the user closes the browser? Session_end in global.asax?
Thanks
A: Event handler lists are in essence weak references, so that will not cause a memory leak. You're not explicitly removing all your Click and whatever handlers either, are you?
Also, the Unload event has nothing to do with the user closing the browser window. To me, it sounds like you've misunderstood some fundamentals of the web and/or ASP.NET.
A: Unlike Windows Forms or WPF, ASP.NET is stateless (as is the rest of all web server solutions). In short, that means that when a GET request comes in, the web server (IIS) handles the request and generates a html page based on the aspx page that you built (with or w/o a Master Page).
So, when you hook an event handler to an event fired from the Master Page, you can only handle it in the same request, in the content page or maybe any user controls that you load into that content page.
Why would you fire any event in your Master page? If you tell us more about which functionality you're looking for, we might be able to provide an alternative - maybe even better - solution.
Read more about ASP.NET on the following MSDN pages, or simply Google yourself to knowlegde :)
ASP.NET Page Life Cycle Overview
Events in ASP.NET Master and Content Pages
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 7,375 |
\section{Introduction}
\setcounter{equation}{0}
The main purpose of this paper is to study the homogenization and boundary layers
for the Neumann problem
with first-order oscillating boundary data,
\begin{equation}\label{NP-0}
\left\{
\aligned
& \mathcal{L}_\varepsilon (u_\varepsilon) =0 &\quad & \text{ in } \Omega,\\
& \frac{\partial u_\varepsilon}{\partial \nu_\varepsilon}
= T_{ij}\cdot \nabla_x \big\{ g_{ij}(x, x/\varepsilon) \big\} +g_0(x, x/\varepsilon)
& \quad & \text{ on } \partial \Omega,
\endaligned
\right.
\end{equation}
where $T_{ij}=(n_i e_j -n_j e_i)$ is a tangential vector field on $\partial\Omega$,
\begin{equation}\label{operator}
\mathcal{L}_\varepsilon =-\text{\rm div}(A(x/\varepsilon)\nabla)
\end{equation}
a second-order elliptic system in divergence form with oscillating periodic coefficients, and
$\partial u_\varepsilon/\partial\nu_\varepsilon=n\cdot A(x/\varepsilon)\nabla u_\varepsilon$ denotes the conormal derivative associated with
the operator $\mathcal{L}_\varepsilon$.
Throughout this paper, unless otherwise stated,
we assume that
$\Omega$ is a bounded smooth, strictly convex domain in ${\mathbb R}^d$ and that
$g_0(x, y), \, g_{ij}(x, y)$ are smooth in $(x, y)\in {\mathbb R}^d \times {\mathbb R}^d$ and
1-periodic in $y$,
\begin{equation}\label{periodicity-g}
g(x, y+z)=g(x,y) \quad \text{ for any } x, y\in {\mathbb R}^d \text{ and } z\in \mathbb{Z}^d.
\end{equation}
We also assume that the coefficient matrix $A=A(y)=\big( a_{ij}^{\alpha\beta}\big)$,
with $1\le i, j\le d$ and $1\le \alpha, \beta\le m$, is smooth and satisfies the ellipticity condition,
\begin{equation}\label{ellipticity}
\mu |\xi|^2 \le a_{ij}^{\alpha\beta} (y) \xi_i^\alpha \xi_j^\beta \le \frac{1}{\mu} |\xi|^2 \quad
\text{ for any } \xi=(\xi_i^\alpha)\in \mathbb{R}^{m\times d},
\end{equation}
where $\mu>0$, and the periodicity condition,
\begin{equation}\label{periodicity}
A(y+z)=A(y) \quad \text{ for any } y\in {\mathbb R}^d \text{ and } z\in \mathbb{Z}^d.
\end{equation}
Under these conditions we will show that as $\varepsilon\to 0$,
the unique solution of (\ref{NP-0}) with
$\int_{\Omega} u_\varepsilon =0$ converges strongly in $L^2(\Omega)$ to $u_0$, where
$u_0$ is a solution of
\begin{equation}\label{NP-h}
\left\{
\aligned
& \mathcal{L}_0 (u_0) =0 &\quad & \text{ in } \Omega,\\
&\frac{\partial u_0}{\partial \nu_0}
= T_{ij}\cdot \nabla_x \overline{g}_{ij} +\langle g_0 \rangle & \quad & \text{ on } \partial \Omega.
\endaligned
\right.
\end{equation}
The operator $\mathcal{L}_0$ is given by
$\mathcal{L}_0 =-\text{\rm div}(\widehat{A}\nabla)$, with $\widehat{A}$ being the usual homogenized
matrix of $A$, and
\begin{equation}\label{g-bar}
\langle g_0\rangle (x) =-\!\!\!\!\!\!\int_{\mathbb{T}^d} g_0 (x, y)\, dy.
\end{equation}
The formula for the function $\{ \overline{g}_{ij}\} $ on $\partial\Omega$, which is much more involved and
will be given explicitly in Section 6,
shows that its value at $x\in \partial\Omega$ depends only on $A$, $\{ g_{ij}(x, \cdot)\}$,
and the outward normal $n$ to $\partial\Omega$
at $x$.
Moreover, we establish a convergence rate in $L^2$, which is optimal
(up to an arbitrarily small exponent) for $d\ge 3$.
\begin{theorem}\label{main-theorem-1}
Let $\Omega$ be a bounded smooth, strictly convex domain in ${\mathbb R}^d$,
$d\ge 3$.
Assume that $A(y)$, $g_0 (x, y)$ and $g_{ij}(x, y)$ are smooth and
satisfy conditions (\ref{periodicity-g}), (\ref{ellipticity}) and (\ref{periodicity}).
Let $u_\varepsilon$ and $u_0$ be the solutions of (\ref{NP-0}) and (\ref{NP-h}), respectively,
with $\int_{\Omega} u_\varepsilon=\int_{ \Omega} u_0=0$.
Then for any $\sigma \in (0,1/2)$ and $\varepsilon\in (0,1)$,
\begin{equation}\label{rate-0}
\| u_\varepsilon -u_0 \|_{L^2(\Omega)}
\le C_\sigma\, \varepsilon^{\frac12-\sigma},
\end{equation}
where $C_\sigma$ depends only on $d$, $m$, $\sigma$, $A$, $\Omega$, and $g=\{g_0, g_{ij}\}$.
Furthermore, the function $\overline{g}=\{ \overline{g}_{ij}\}$ in (\ref{NP-h}) satisfies
\begin{equation}\label{Sobolev}
\| \overline{g} \|_{ W^{1, q} (\partial\Omega)} \le C_q\, \sup_{y\in \mathbb{T}^d}
\| g(\cdot, y)\|_{C^1(\partial\Omega)} \quad \text{ for any } q< d-1,
\end{equation}
where $C_q$ depends only on $d$, $m$, $\mu$, $q$ and $\|A\|_{C^k(\mathbb{T}^d)}$ for some
$k=k(d)\ge 1$.
\end{theorem}
In recent years there has been considerable interest in the homogenization of
boundary value problems with oscillating boundary data
\cite{Masmoudi-2011, Masmoudi-2012, Prange-2013, ASS-2013, KLS4, ASS-2014, Feldman-2014, Kim-2014, Kim-2015, ASS-2015, Armstrong-Prange-2016} (also see related
earlier work in \cite{Santosa-Vogelius, Santosa-Vogelius-erratum, Moskow-Vogelius-1997, Moskow-Vogelius-2,
Allaire-Amar-1999}.
In the case of Dirichlet problem,
\begin{equation}\label{DP-0}
\mathcal{L}_\varepsilon (u_\varepsilon) =0 \quad \text{ in } \Omega \quad \text{ and } \quad
u_\varepsilon =g(x, x/\varepsilon) \quad \text{ on } \partial\Omega,
\end{equation}
where $g(x, y)$ is assumed to be periodic in $y$,
major progress was made in \cite{Masmoudi-2012} and more recently in \cite{Armstrong-Prange-2016}.
Let $u_\varepsilon$ be the solution of (\ref{DP-0}).
Under the assumption that $\Omega$ is smooth and strictly convex in ${\mathbb R}^d$,
$d\ge 2$, it was proved in \cite{Masmoudi-2012} that
$$
\| u_\varepsilon -u_0\|_{L^2(\Omega)} \le C \, \varepsilon^{\frac{(d-1)}{3d+5}-\sigma}
$$
for any $\sigma \in (0,1)$, where $u_0$ is the solution of the homogenized problem,
\begin{equation}\label{DP-h}
\mathcal{L}_0 (u_0) =0 \quad \text{ in } \Omega \quad \text{ and } \quad
u_0 =\overline{g} \quad \text{ on } \partial\Omega,
\end{equation}
and the homogenized data $\overline{g}$ at $x$ depends on $g(x, \cdot)$, $A$, and $n(x)$.
A sharper rate of convergence in $L^2$ was obtained recently in \cite{Armstrong-Prange-2016} for
the Dirichlet problem (\ref{DP-0}), with $O(\varepsilon^{\frac12-})$ for $d\ge 4$,
$O(\varepsilon^{\frac{1}{3}-})$ for $d=3$, and
$O(\varepsilon^{\frac16-})$ for $d=2$.
As demonstrated in \cite{ASS-2015} in the case of elliptic equations with constant coefficients,
the optimal rate would be $O(\varepsilon^{1/2})$ for $d\ge 3$ (up to a factor of $\ln \varepsilon$ in the case of $d=3$),
and $O(\varepsilon^{1/4})$ for $d=2$.
Thus the convergence rates obtained in \cite{Armstrong-Prange-2016}
for (\ref{DP-0}) is optimal, up to an arbitrarily small exponent, for $d\ge 4$.
With the sharp results established in this paper for the homogenized data (see Theorem \ref{main-theorem-2} below),
the method used in \cite{Armstrong-Prange-2016} also leads to the optimal rate of convergence
for $d=2$ or $3$.
In the case of the Neumann problem with only zero-order oscillating data $g_0(x, x/\varepsilon)$, i.e., $g_{ij}(x, y)=0$,
the homogenization of (\ref{NP-0}) is well understood, mostly due to the fact that the Neumann function
$N_\varepsilon(x, y)$ for $\mathcal{L}_\varepsilon$ in $\Omega$
converges pointwise to $N_0 (x, y)$, the Neumann function for the homogenized
operator $\mathcal{L}_0$ in $\Omega$. In fact, it was proved in \cite{KLS4} that if $\Omega$ is a bounded
$C^{1,1}$ domain in ${\mathbb R}^d$ and $d\ge 3$, then
\begin{equation}\label{N-function-0}
|N_\varepsilon (x, y) -N_0(x, y)|\le \frac{C\, \varepsilon \ln \big[ \varepsilon^{-1}|x-y| +2\big]}{|x-y|^{d-2}}
\end{equation}
for any $x, y\in \Omega$.
This effectively reduces the problem to the case of operators with constant coefficients,
which may be handled by the method of oscillatory integrals \cite{ASS-2013, ASS-2015}.
Thus the real challenge for the Neumann problem starts with the first-order oscillating boundary
data that includes terms
in the form of $\varepsilon^{-1} g(x, x/\varepsilon)$.
Since the Neumann data involves the first-order derivatives of the solution,
the problem (\ref{NP-0}) seems quite natural.
We remark that one of the main motivations for studying Dirichlet problem (\ref{DP-0})
is its applications to the higher-order convergence in the two-scale expansions of solutions to
Dirichlet problems for $\mathcal{L}_\varepsilon$ with non-oscillating boundary data.
As we will show in Section 9,
in the study of the higher-order convergence of solutions to the Neumann problems
for $\mathcal{L}_\varepsilon$ with non-oscillating boundary data,
one is forced to deal with a Neumann problem in the form of (\ref{NP-0}).
We now describe the main steps in the proof of Theorem \ref{main-theorem-1}
as well as the organization of the paper.
Our general approach is inspired by the recent work \cite{Armstrong-Prange-2016}.
The starting point is the following
asymptotic expansion for the Neumann function $N_\varepsilon(x, y)$,
\begin{equation}\label{N-function-a}
\Big| \frac{\partial}{\partial y_k} \big\{ N_\varepsilon (x, y) \big\}
-\frac{\partial}{\partial y_k} \big\{ \Psi^*_{\varepsilon, \ell} (y) \big\}
\frac{\partial}{ \partial {y_\ell}} \big\{ N_0 (x, y)\big\}\Big|
\le \frac{C_\sigma\, \varepsilon^{1-\sigma}}{|x-y|^{d-\sigma}},
\end{equation}
for any $x, y\in \Omega$ and $\sigma \in (0,1)$.
The function $\Psi_{\varepsilon, \ell}^*$ in (\ref{N-function-a})
is the so-called Neumann corrector, defined by
(\ref{N-corrector}), for the adjoint operator $\mathcal{L}_\varepsilon^*$.
The estimate (\ref{N-function-a}) was proved in \cite{KLS4}
for bounded $C^{2, \alpha}$ domains, using boundary Lipschitz estimates
for solutions with Neumann data \cite{KLS1} (also see \cite{AS}, which extends the estimates to
operators without the symmetry condition $A^*=A$).
Using (\ref{N-function-a}), one reduces the problem to the study of the tangential derivatives of
$\Psi_{\varepsilon, \ell}^*$ on the boundary.
This preliminary reduction is carried out in Section 2.
Next, to analyze the Neumann corrector on the boundary,
we approximate it in a neighborhood of a point $x_0\in \partial\Omega$,
by a solution of a Neumann problem in a half-space,
\begin{equation}\label{half-space-NP-0}
\left\{
\aligned
&\mathcal{L}_\varepsilon^* (\phi^*_{\varepsilon, k})=0 &\quad & \text{ in } \mathbb{H}_n^d(a),\\
& \frac{\partial}{\partial \nu_\varepsilon^*} \big(\phi^*_{\varepsilon, k}\big)
=-n_\ell b^*_{\ell k} (x/\varepsilon) &\quad & \text{ on } \partial\mathbb{H}_n^d(a),
\endaligned
\right.
\end{equation}
where $n=n(x_0)$,
$\partial\mathbb{H}_n^d(a)$ is the tangent plane for $\partial\Omega$ at $x_0$, and
$b^*_{\ell k} (y)$ are some 1-periodic functions depending only on $A$.
Under the assumption that $n$ satisfies the Diophantine condition (\ref{D-condition}),
a solution to (\ref{half-space-NP-0})
is constructed in Section 3 by solving a Neumann problem for a degenerated elliptic system
in $\mathbb{T}^d\times \mathbb{R}_+$, in a similar fashion as in the case of the Dirichlet problem \cite{Masmoudi-2011, Masmoudi-2012}.
In Section 4 we prove two refined estimates with bounding constants
$C$ independent of the constant $\kappa$ in the Diophantine condition (\ref{D-condition}).
The first estimate shows that the solution constructed for (\ref{half-space-NP-0})
satisfies
\begin{equation}\label{decay-0}
|\nabla \phi_{\varepsilon, k}^* (x)|\le \frac{C \, \varepsilon}{|x\cdot n +a|},
\end{equation}
where $C$ depends only on $d$, $m$, $\mu$ and some H\"older norm of $A$.
We remark that the estimate (\ref{decay-0}) plays the same role as the maximum principle
in the Dirichlet problem.
The other estimates in Section 4 concern with weighted norm inequalities for $\mathcal{L}_1$
in a half-space $H^d_n(a)$.
In particular, we show that if $u$ is a solution of
$\mathcal{L}_1 (u)=\text{\rm div}(f)$ in $\mathbb{H}_n^d(a)$,
with either Dirichlet condition $u=0$ or the Neumann condition
$\frac{\partial u}{\partial \nu} =-n \cdot f$ on $\partial\mathbb{H}_n^d(a)$, then
\begin{equation}\label{weighted-0}
\int_{\mathbb{H}_n^d (a)} |\nabla u (x) |^2 \big[\delta (x)\big]^\alpha\, dx
\le C_\alpha \int _{\mathbb{H}_n^d (a)} |f (x) |^2 \big[\delta (x)\big]^\alpha\, dx,
\end{equation}
where $-1<\alpha<1$ and
$$
\delta(x) =\text{dist}(x, \partial \mathbb{H}_n^d (a)) =|x\cdot n +a|.
$$
We point out that it is this weighted estimate with $-1<\alpha<0$
that allows us to prove the optimal
result (\ref{Sobolev}) for the homogenized data.
We further remark that the weighted estimate (\ref{weighted-0}), which is obtained
by using a real-variable method and the theory of $A_p$ weights in harmonic analysis,
also leads to the optimal
result for the oscillating Dirichlet problem (\ref{DP-0}).
\begin{theorem}\label{main-theorem-2}
Let $\Omega$ be a bounded smooth, strictly convex domain in ${\mathbb R}^d$, $d\ge 2$.
Assume that $A(y)$ and $g(x, y)$ are smooth and satisfy conditions
(\ref{periodicity-g}), (\ref{ellipticity}) and (\ref{periodicity}).
Let $\overline{g}$ be the homogenized data in (\ref{DP-h}).
Then
\begin{equation}\label{Sobolev-D}
\| \overline{g}\|_{W^{1, q}(\partial\Omega)} \le C_q\, \sup_{y\in \mathbb{T}^d} \| g(\cdot, y)\|_{C^1(\partial\Omega)}
\quad \text{ for any } q<d-1,
\end{equation}
where $C_q$ depends only on $d$, $m$, $\mu$, $q$ and $\|A\|_{C^k(\mathbb{T}^d)}$ for some
$k=k(d)\ge 1$.
Moreover,
\begin{equation}\label{rate-DP-low}
\| u_\varepsilon -u_0\|_{L^2(\Omega)}
\le C_\sigma \left\{
\aligned
& \varepsilon^{\frac12-\sigma} & \quad & \text{ if } d=3,\\
& \varepsilon^{\frac14-\sigma} & \quad & \text{ if } d=2,
\endaligned
\right.
\end{equation}
where $u_\varepsilon$ and $u_0$ are solutions of Dirichlet problems (\ref{DP-0})
and (\ref{DP-h}), respectively.
\end{theorem}
Estimate (\ref{Sobolev-D}) improves a similar
result in \cite{Armstrong-Prange-2016}, where it was proved that
$\nabla \overline{g}\in L^{q, \infty}(\partial\Omega)$ for $q=\frac{2(d-1)}{3}$ and $d\ge 3$,
and that $\overline{g}\in W^{1, s}(\partial\Omega)$ for $s<\frac{2}{3}$ and $d=2$.
In an earlier work \cite{Masmoudi-2012}, it was shown that $\nabla\overline{g}\in L^{q, \infty}(\partial\Omega)$
for $q=\frac{d-1}{2}$. We believe that our results (\ref{Sobolev}) and (\ref{Sobolev-D})
are optimal, since the estimate $\overline{g}\in W^{1, q}(\partial\Omega)$
for some $q>d-1$ would imply that $\overline{g}$ is continuous on $\partial\Omega$.
As indicated before,
the convergence rate (\ref{rate-DP-low}), which is optimal,
improves the main results in \cite{Armstrong-Prange-2016} for $d=2$ and $3$.
One of the main technical lemmas in this paper is given in Section 5.
Motivated by \cite{Armstrong-Prange-2016} for Dirichlet problem, we show that for any $\sigma\in (0,1)$
and $\varepsilon\in (0,1)$,
\begin{equation}\label{main-estimate-0-1}
\aligned
& \| \nabla \Big( \Psi_{\varepsilon, k}^* - x_k -\varepsilon \chi_k^* (x/\varepsilon) -\phi_{\varepsilon, k}^* \Big)\|_{L^\infty(B(x_0, r)\cap\Omega)}\\
&\qquad \qquad
\le C\sqrt{\varepsilon} \big\{ 1+ |\ln \varepsilon|\big\}
+C \varepsilon^{-1-\sigma} r^{2+\sigma},
\endaligned
\end{equation}
where $\varepsilon\le r\le \sqrt{\varepsilon}$.
This is done by using boundary Lipschitz estimates and representations by Neumann functions $N_\varepsilon(x, y)$.
In contrast to the case of Dirichlet problem (\ref{DP-0}),
in order to fully utilize the decay estimates for derivatives of Neumann functions,
the key insight here is to transfer,
through integration by parts on the boundary, the derivatives from the Neumann data
to the Neumann functions.
In Section 6 we use the weighted estimates in Section 4 to prove some weighted estimates
for solutions of the degenerated elliptic systems in $\mathbb{T}^d\times \mathbb{R}_+$.
This, together with a partition of unity for
$\partial\Omega$ that adapted to the constant $\kappa$ in the Diophantine condition (\ref{D-condition}),
leads to the proof of (\ref{Sobolev}) and (\ref{Sobolev-D}) for the homogenized boundary data.
We remark that the partition of unity for $\partial\Omega$
we construct in Section 7 is in the same spirit as that in \cite{Armstrong-Prange-2016}.
Our $L^p$-based approach with a finite $p>d-1$, which is motivated
by \cite{Fefferman-1996, Shen-1998},
seems to make the proof much more transparent.
Moreover, it gives us a pointwise estimate for
$|u_\varepsilon (x) - u_0 (x)|$ in terms of the Hardy-Littlewood maximal function of
$\kappa^{-q}$ with $q<d-1$ on the boundary, when $x$ is not in a boundary layer
$\Omega_\varepsilon$ of measure less than $C \varepsilon^{1-\sigma}$ (see Remark \ref{remark-8.1}).
Our main result on the convergence rate in Theorem \ref{main-theorem-1} is proved
in Section 8, using a similar line of argument as in \cite{Armstrong-Prange-2016}.
We mention that the convergence rates in $L^p$ for $2<p<\infty$, even in
$H^\alpha(\Omega)$ for $0<\alpha<1/2$, follow from (\ref{rate-0})
by interpolation (see Remark \ref{remark-2.10}).
Finally, in Section 9, we use Theorem \ref{main-theorem-1} to establish
a higher-order convergence in the two-scale expansions
of solutions to Neumann problems with non-oscillating data.
Although some related work may be found in \cite{Moskow-Vogelius-2},
to the best of the authors' knowledge,
our $O(\varepsilon^{3/2})$ rate in $L^2(\Omega)$ with a homogenized first-order term
seems to be
the first result on the higher-order convergence for $\mathcal{L}_\varepsilon$ with Neumann conditions.
The summation convention will be used throughout.
We will use $C$ and $c$ to denote constants that depend at most on $d$, $m$, $A$ and $\Omega$
as well as other parameters,
but never on $\varepsilon$ or the constant $\kappa$ in (\ref{D-condition}).
We will always assume that $d\ge 3$ (unless otherwise indicated).
Our method, with some modifications, should also apply to the two-dimensional Neumann problem,
which will be handled in a separate work.
\medskip
\noindent{\bf Acknowledgement.}
The first author would like to thank David G\'erard-Varet for bringing the problem of homogenization
of the Neumann problem (\ref{NP-0}) to his attention.
\section{Preliminaries}
\setcounter{equation}{0}
Under the conditions (\ref{ellipticity})-(\ref{periodicity}) and $A\in C^\sigma(\mathbb{T}^d)$ for some $\sigma \in (0,1)$,
one may construct an $m\times m$ matrix of Neumann functions
$N_\varepsilon (x, y)=(N_\varepsilon^{\alpha\beta} (x, y))$ in a bounded $C^{1, \alpha}$ domain $\Omega$,
such that
\begin{equation}\label{N-def}
\left\{
\aligned
&\mathcal{L}_\varepsilon \big\{ N_\varepsilon (\cdot, y)\big\} = \delta_ y(x)I
& \quad & \text{ in } \Omega,\\
& \frac{\partial}{\partial \nu_\varepsilon} \big\{ N_\varepsilon (\cdot, y) \big\}=- |\partial\Omega|^{-1} I
&\quad & \text{ on } \partial\Omega,\\
& \int_{\partial\Omega} N_\varepsilon (x, y)\, dx=0,
\endaligned
\right.
\end{equation}
where $I=I_{m\times m} $ and
the operator $\mathcal{L}_\varepsilon$ acts on each column of $N_\varepsilon(\cdot, y)$.
Let $u_\varepsilon\in H^1(\Omega; \mathbb{R}^m)$ be a solution to $\mathcal{L}_\varepsilon (u_\varepsilon)=F$ in $\Omega$
with $\frac{\partial u_\varepsilon}{\partial \nu_\varepsilon}= h$ on $\partial\Omega$,
then
\begin{equation}\label{N-representation}
u_\varepsilon (x) --\!\!\!\!\!\!\int_{\partial\Omega} u_\varepsilon
=\int_{\partial\Omega} N_\varepsilon (x, y) F(y)\, dy
+\int_{\partial\Omega} N_\varepsilon(x, y) h(y)\, d\sigma (y)
\end{equation}
for any $x\in \Omega$.
If $d\ge 3$,
the Neumann functions satisfy the following estimates,
\begin{equation}\label{N-estimate}
\aligned
|N_\varepsilon (x, y)| & \le C\, |x-y|^{2-d}\\
|\nabla_x N_\varepsilon (x, y)| +|\nabla_y N_\varepsilon (x, y)| & \le C |x-y|^{1-d},\\
|\nabla_x \nabla_y N_\varepsilon (x, y)| & \le C |x-y|^{-d},
\endaligned
\end{equation}
for any $x, y\in \Omega$.
This was proved in \cite{KLS1}, using boundary Lipschitz estimates with Neumann conditions,
which require the additional assumption $A^*=A$.
This additional assumption for the boundary Lipschitz estimates was removed later in \cite{AS}.
As a result, the estimates in (\ref{N-estimate}) hold if
$A$ satisfies (\ref{ellipticity})-(\ref{periodicity}) and is H\"older continuous.
Note that if $x, y, z\in \Omega$ and $|x-z|\le (1/2) |x-y|$, it follows from (\ref{N-estimate}) that
\begin{equation}\label{N-estimate-1}
\aligned
|N_\varepsilon (x, y)-N_\varepsilon (z, y) | & \le \frac{C|x-z|}{|x-y|^{d-1}},\\
|\nabla_y \big\{ N_\varepsilon (x, y)-N_\varepsilon (z, y)\big\} | & \le \frac{C|x-z|}{|x-y|^{d}}.
\endaligned
\end{equation}
Let $\chi_j (y) =(\chi_j^{\alpha\beta} (y))$ denote the matrix of correctors for $\mathcal{L}_\varepsilon$ in
${\mathbb R}^d$, defined by
\begin{equation}\label{corrector}
\left\{
\aligned
& \mathcal{L}_1\big(\chi_j^\beta + P_j^\beta\big) =0 \quad \text{ in } {\mathbb R}^d,\\
& \chi_j^\beta \text{ is 1-periodic and }
\int_{\mathbb{T}^d} \chi_j^\beta =0,
\endaligned
\right.
\end{equation}
where $\chi_j^\beta =(\chi_j^{1 \beta}, \dots,
\chi_j^{m\beta})$
and $P_j^\beta (y) =y_j (0, \dots, 1, \dots, 0)$ with $1$ in the $\beta^{th}$ position.
The homogenized operator $\mathcal{L}_0$ for $\mathcal{L}_\varepsilon$ is given by
$\mathcal{L}_0=-\text{\rm div} (\widehat{A}\nabla)$, where $\widehat{A}=\big( \widehat{a}_{ij}^{\alpha\beta} \big)$
is the homogenized matrix with
\begin{equation}\label{homo}
\widehat{a}_{ij}^{\alpha\beta}
=-\!\!\!\!\!\!\int_{\mathbb{T}^d} \Big\{ a_{ij}^{\alpha\beta}
+ a_{ik}^{\alpha\gamma} \frac{\partial}{\partial y_k} \big( \chi_j^{\gamma\beta} \big)\Big\}.
\end{equation}
To study the boundary regularity for solutions of Neumann problems,
the matrix of Neumann correctors
$\Psi_{\varepsilon, j}^\beta =\big( \Psi_{\varepsilon, j}^{\alpha\beta}\big)$ for $\mathcal{L}_\varepsilon$ in $\Omega$, defined by
\begin{equation}\label{N-corrector}
\mathcal{L}_\varepsilon \big(\Psi_{\varepsilon, j}^\beta\big)=0 \quad \text{ in } \Omega
\quad \text{ and } \quad
\frac{\partial}{\partial \nu_\varepsilon} \big( \Psi_{\varepsilon, j}^\beta\big)
=\frac{\partial}{\partial \nu_0} \big( P_j^\beta \big) \quad \text{ on } \partial\Omega,
\end{equation}
was introduced in \cite{KLS1}, where $\partial u/\partial\nu_0$ denotes the conormal derivative
associated with $\mathcal{L}_0$. One of the main estimates in \cite{KLS1} is the following Lipschitz estimate for
$\Psi_{\varepsilon, j}^\beta$,
\begin{equation}\label{Lip-Psi}
\|\nabla \Psi_{\varepsilon, j}^\beta\|_{L^\infty(\Omega)} \le C.
\end{equation}
Let $N_0(x, y)$ denote the matrix of Neumann functions for $\mathcal{L}_0$ in $\Omega$.
It was proved in \cite{KLS4} that if $\Omega$ is $C^{1,1}$,
\begin{equation}\label{size-diff}
|N_\varepsilon (x, y) -N_0(x, y)|\le \frac{C\varepsilon \ln \big[ \varepsilon^{-1}|x-y| +2\big]}{|x-y|^{d-1}}
\end{equation}
for any $x, y\in \Omega$, and that if $\Omega$ is $C^{2, \alpha}$ for some $\alpha\in (0, 1)$,
\begin{equation}\label{derivative-diff}
\Big|\frac{\partial}{\partial y_i} \big\{ N^{\gamma \alpha}_\varepsilon (x, y) \big\}
-\frac{\partial}{\partial y_i} \big\{ \Psi_{\varepsilon, j}^{*\alpha\beta} (y) \big\}
\cdot
\frac{\partial}{\partial y_j } \big\{ N_0^{\gamma\beta} (x, y) \big\}\Big|
\le \frac{C_\sigma\, \varepsilon^{1-\sigma}}{|x-y|^{d-\sigma}}
\end{equation}
for any $x, y\in \Omega$ and $\sigma\in (0,1)$.
The functions $(\Psi_{\varepsilon, j}^{*\alpha\beta})$ in (\ref{derivative-diff})
are the Neumann correctors, defined as in (\ref{N-corrector}),
for the adjoint operator $\mathcal{L}_\varepsilon^*$
in $\Omega$.
We remark that these estimates as well as (\ref{Lip-Psi})
were proved in \cite{KLS1} under the additional assumption
$A^*=A$. As in the case of (\ref{N-estimate}), with the results in \cite{AS},
they continue to hold without this assumption.
The estimates (\ref{size-diff}) and (\ref{derivative-diff}) mark the starting point of our investigation of
the Neumann problem (\ref{NP-0}) with oscillating data.
Indeed, let $u_\varepsilon$ be the solution of (\ref{NP-0}) with $\int_{\partial\Omega} u_\varepsilon=0$.
It follows by (\ref{N-representation}) that
\begin{equation}\label{p-1}
u_\varepsilon (x) =\int_{\partial\Omega}
N_\varepsilon (x, y) (T_{ij}(y)\cdot \nabla_y) \big\{ g_{ij}(y, y/\varepsilon)\big\}\, d\sigma (y)
+\int_{\partial\Omega} N_\varepsilon(x, y) g_0(y, y/\varepsilon)\, d\sigma (y),
\end{equation}
where $T_{ij}=n_i e_j -n_j e_i$, $n=(n_1, \cdots, n_d)$ is the outward normal to $\partial\Omega$, and
$e_i =(0, \dots, 1, \dots, 0)$ with $1$ in the $i^{th}$ position.
\begin{lemma}\label{lemma-p1}
Let $\Omega$ be a bounded Lipschitz domain in ${\mathbb R}^d$.
Then, for $u, v\in C^1({\partial\Omega})$,
\begin{equation}\label{parts}
\int_{\partial\Omega} ( (n_i e_j -n_j e_i)\cdot \nabla u ) \, v \, d\sigma
=-\int_{\partial\Omega} u \, ((n_ie_j -n_j e_i)\cdot \nabla v)\, d\sigma.
\end{equation}
\end{lemma}
\begin{proof}
This may be proved by localizing the functions $u$ and $v$ to a small neighborhood of
$x_0\in\partial\Omega$ and reducing the problem to the case of flat boundary $\mathbb{R}^{d-1}$.
One may also prove (\ref{parts}) by using the divergence theorem.
\end{proof}
It follows from (\ref{p-1}) and (\ref{parts}) that
\begin{equation}\label{p-2}
\aligned
u_\varepsilon(x) &
=-\int_{\partial\Omega} \big( T_{ij} (y)\cdot \nabla_y \big)N_\varepsilon (x, y) \cdot
g_{ij}(y, y/\varepsilon)\, d\sigma (y)\\
&\qquad + \int_{\partial\Omega}
N_\varepsilon (x, y) g_0 (y, y/\varepsilon)\, d\sigma (y).
\endaligned
\end{equation}
In view of (\ref{N-estimate}) this implies that
\begin{equation}\label{p-2-0}
\aligned
|u_\varepsilon (x)| &\le C \, \| g\|_\infty \int_{\partial\Omega} \frac{d\sigma (y)}{|x-y|^{d-1}}
+C \, \| g\|_\infty \int_{\partial\Omega} \frac{d\sigma (y)}{|x-y|^{d-2}}\\
&\le C\, \| g\|_\infty \big\{ 1 + |\ln \delta(x) | \big\},
\endaligned
\end{equation}
where $g=\{ g_{ij}, g_0\}$ and $\delta (x)=\text{dist}(x, \partial\Omega)$.
\begin{remark}\label{remark-2.10}
{\rm
It follows from (\ref{p-2-0}) that for any $1<q<\infty$,
$$
\| u_\varepsilon\|_{L^q(\Omega)} \le C _q\, \| g\|_\infty,
$$
where $C_q$ depends on $q$, $A$ and $\Omega$.
By interpolation, this, together with (\ref{rate-0}), implies that
\begin{equation}\label{rate-q}
\| u_\varepsilon -u_0\|_{L^q(\Omega)}
\le C_{q, \sigma} \, \varepsilon^{\frac{1}{q}-\sigma}
\end{equation}
for any $2<q<\infty$ and $\sigma \in (0, \frac{1}{q})$.
Moreover, if $A^*=A$, it follows from \cite{KLS2} that
\begin{equation}\label{square}
\| u_\varepsilon\|_{H^{1/2}(\Omega)} +\left(\int_\Omega |\nabla u_\varepsilon (x)|^2 \, \delta(x)\, dx\right)^{1/2}
\le C \| g\|_{L^2(\partial\Omega)}.
\end{equation}
Thus, by interpolation, we may deduce from (\ref{rate-0}) and (\ref{square}) that
\begin{equation}\label{rate-h}
\| u_\varepsilon -u_0\|_{H^{\alpha} (\Omega)}
\le C_{\alpha, \sigma} \varepsilon^{\frac12 -\alpha -\sigma}
\end{equation}
for any $\alpha \in (0, 1/2)$ and $\sigma\in (0, (1/2)-\alpha)$.
}
\end{remark}
Using (\ref{size-diff}) and (\ref{derivative-diff}), we obtain
\begin{equation}\label{p-3}
\aligned
& u^\gamma_\varepsilon (x)=
-\int_{\partial\Omega}
\big( T_{ij}(y)\cdot \nabla_y \big)\Psi_{\varepsilon, k}^{*\alpha\beta} (y)
\cdot \frac{\partial}{\partial y_k} \big\{ N_0^{\gamma \beta} (x, y)\big\} \cdot
g_{ij}^\alpha (y, y/\varepsilon)\, d\sigma (y)\\
&\qquad\qquad +\int_{\partial\Omega}
N^{\gamma \alpha}_0(x, y) g_0^\alpha(y, y/\varepsilon)\, d\sigma (y)
+ R_\varepsilon^\gamma(x),
\endaligned
\end{equation}
where the remainder $R_\varepsilon$ satisfies
\begin{equation}\label{R-1}
|R_\varepsilon(x)|
\le C\varepsilon^{1-\sigma} \| g\|_\infty \int_{\partial\Omega} \frac{d\sigma (y)}{|x-y|^{d-\sigma}}
+ C\varepsilon \| g\|_\infty
\int_{\partial\Omega} \frac{\ln [\varepsilon^{-1}|x-y| +2]}{|x-y|^{d-1}} d\sigma (y).
\end{equation}
\begin{lemma}\label{lemma-p2}
Let $\Omega$ be a bounded $C^{2, \alpha}$ domain for some $\alpha\in (0,1)$.
Then the function $R_\varepsilon$, given by (\ref{p-3}), satisfies
\begin{equation}\label{p2-0}
\| R_\varepsilon\|_{L^q(\Omega)} \le C \, \varepsilon^{\frac{1}{q}} ( 1+ |\ln \varepsilon|) \| g\|_\infty,
\end{equation}
for any $1<q<\infty$,
where $C$ depends only on $q$, $A$ and $\Omega$.
\end{lemma}
\begin{proof}
Let $x\in \Omega$.
If $\delta (x)=\text{dist}(x, \partial\Omega)\ge \varepsilon$, we may use (\ref{R-1}) to show that
\begin{equation}\label{p2-1}
| R_\varepsilon (x) |\le C_\sigma \left(\frac{\varepsilon}{\delta (x)}\right)^{1-\sigma} \| g\|_\infty
\end{equation}
for any $\sigma \in (0,1)$.
If $\delta(x)\le \varepsilon$, the estimates in (\ref{N-estimate}), as in (\ref{p-2-0}), lead to
\begin{equation}\label{p2-2}
|R_\varepsilon (x)|
\le C\, \| g\|_\infty ( 1+|\ln \delta(x)|).
\end{equation}
It is not hard to verify that (\ref{p2-0}) follows from (\ref{p2-1}) and (\ref{p2-2}).
\end{proof}
As $\varepsilon \to 0$, the second term in the RHS of (\ref{p-3}) converges to
\begin{equation}\label{p-4}
w^\gamma_0(x)
=\int_{\partial\Omega}
N^{\gamma \alpha}_0(x, y) \langle g_0^\alpha\rangle (y)\, d\sigma (y),
\end{equation}
where
\begin{equation}\label{g-average}
\langle g_0 \rangle (y) = -\!\!\!\!\!\!\int_{\mathbb{T}^d} g_0(y, z)\, dz.
\end{equation}
More precisely, the following results on the convergence rate were obtained in \cite{ASS-2015}.
\begin{lemma}\label{lemma-p3}
Let $w_\varepsilon$ denote the second term in the RHS of (\ref{p-3}).
Assume that $\Omega$ is a bounded smooth, uniformly convex domain in ${\mathbb R}^d$.
Then, for any $1\le q <\infty$,
\begin{equation}\label{p3-0}
\| w_\varepsilon -w_0\|_{L^q(\Omega)}
\le C_q
\left\{
\aligned
& \varepsilon^{\frac{1}{q}} & \quad &\text{ if d=3},\\
& \varepsilon^{\frac{3}{2q}} & \quad & \text{ if d=4}, \\
& \varepsilon^{\frac{2}{q}} (1+|\ln \varepsilon|)^{\frac{1}{q}} & \quad & \text{ if } d\ge 5,
\endaligned
\right.
\end{equation}
where $w_0$ is given by (\ref{p-4}).
\end{lemma}
Much of the rest of paper is devoted to the study of the first term in
the RHS of (\ref{p-3}).
To this end
we first replace the function $\Psi_{\varepsilon, k}^{*\alpha\beta}$ by
\begin{equation}\label{psi}
\psi_{\varepsilon, k}^{*\alpha\beta} (x)
=\Psi^{*\alpha\beta}_{\varepsilon, k} (x) -P_k^{\alpha\beta} (x)
-\varepsilon \chi_{k}^{*\alpha\beta} (x/\varepsilon),
\end{equation}
where $(\chi_{k}^{*\alpha\beta} (y))$ denotes the matrix of correctors for $\mathcal{L}_\varepsilon^*$
in ${\mathbb R}^d$. Note that
\begin{equation}\label{psi-star}
\mathcal{L}^*_\varepsilon (\psi_{\varepsilon, k}^{*\beta}) =0 \quad \text{ in } \Omega,
\end{equation}
where $\psi_{\varepsilon, k}^{*\beta}=(\psi_{\varepsilon, k}^{*1\beta}, \dots, \psi_{\varepsilon, k}^{*m\beta})$.
We end this section with some observations on its conormal derivatives.
\begin{lemma}\label{lemma-p5}
Let $\psi_{\varepsilon, k}^{*\alpha\beta}$ be defined by (\ref{psi}).
Then
\begin{equation}
\left(\frac{\partial}{\partial\nu_\varepsilon^*} \big\{ \psi_{\varepsilon, k}^{*\beta} \big\} \right)^\alpha (x)
=-n_i (x) b_{ik}^{*\alpha\beta} (x/\varepsilon) \quad \text{ for } x\in \partial\Omega,
\end{equation}
where
\begin{equation}\label{b}
b_{ik}^{*\alpha\beta} (y) =a_{ik}^{*\alpha\beta} (y)+ a_{ij}^{*\alpha\gamma} (y)\frac{\partial}{\partial y_j}
\big( \chi_{k}^{*\gamma \beta}\big) -\widehat{a}_{ik}^{*\alpha\beta}
\end{equation}
and $\widehat{A^*} =(\widehat{a}_{ij}^{*\alpha\beta}) = (\widehat{A})^*$ is the homogenized
matrix of $A^*$.
\end{lemma}
\begin{proof}
See e.g. \cite[p.1023]{KLS2}.
\end{proof}
Note that by the definitions of correctors $\chi_k^{*\alpha\beta}$ and the homogenized matrix $\widehat{A^*}$,
\begin{equation}\label{p101}
\frac{\partial}{\partial y_i} \big\{ b_{ik}^{*\alpha\beta}\big\} =0 \quad
\text{ and } \quad \int_{\mathbb{T}^d} b_{ik}^{*\alpha\beta} =0.
\end{equation}
This implies that there are 1-periodic functions $f_{\ell i k}^{\alpha\beta}$ with mean value zero
such that
\begin{equation}\label{p102}
b_{ik}^{*\alpha\beta} =\frac{\partial}{\partial y_\ell} \big\{ f_{\ell i k}^{\alpha\beta}\big\}
\quad \text{ and } \quad
f_{\ell i k}^{\alpha\beta} =- f_{i \ell k}^{\alpha\beta}.
\end{equation}
(see e.g. \cite{KLS2}).
As a result, we obtain
\begin{equation}\label{p103}
\aligned
n_i (x) b_{ik}^{* \alpha\beta} (x/\varepsilon)
=\frac12 ( n_i e_j -n_j e_i) \cdot \nabla_x \big\{ \varepsilon f_{j i k}^{\alpha\beta} (x/\varepsilon)\big\}.
\endaligned
\end{equation}
This shows that $\varepsilon^{-1}\psi^{*\beta}_{\varepsilon, k} $ is a solutions of the Neumann problem (\ref{NP-0})
with $g_{ij} (x, y)=(1/2)f^\beta_{j i k }(y)$ and $g_0=0$.
\section{Neumann problems in a half-space}
\setcounter{equation}{0}
For $n\in \mathbb{S}^{d-1}$ and $a\in \mathbb{R}$, let
\begin{equation}\label{half}
\mathbb{H}^d_n (a)
=\big\{ x\in {\mathbb R}^d: x\cdot n <-a \big\}
\end{equation}
denote a half-space with outward unit normal $n$.
Consider the Neumann problem
\begin{equation}\label{half-NP}
\left\{
\aligned
\text{\rm div} (A\nabla u) & = 0 & \quad & \text{ in } \mathbb{H}_n^d(a),\\
n \cdot A\nabla u &=T\cdot \nabla g & \quad & \text{ on } \partial \mathbb{H}_n^d(a),
\endaligned
\right.
\end{equation}
where $T\in \mathbb{R}^d$, $|T|\le 1$ and $T\cdot n=0$.
We will assume that $g\in C^\infty(\mathbb{T}^d)$ with mean value zero and
$n$ satisfies the Diophantine condition
\begin{equation}\label{D-condition}
| ( I-n \otimes n) \xi|\ge \kappa |\xi|^{-2} \quad \text{ for any } \xi\in \mathbb{Z}^d\setminus \{ 0\},
\end{equation}
where $\kappa>0$ and $n\otimes n=(n_i n_j)_{d\times d}$.
We emphasize again that all constants $C$ will be independent of $\kappa$.
Let $M$ be a $d\times d$ orthogonal matrix such that $Me_d=-n$.
Note that the last column of $M$ is $-n$. Let $N$ denote the $d\times (d-1)$
matrix of the first $d-1$ columns of $M$. Since $MM^T=I$, we see that
\begin{equation}\label{2.0}
N N^T + n \otimes n =I,
\end{equation}
where $M^T$ denotes the transpose of $M$.
Suppose that a solution of (\ref{half-NP}) is given by
\begin{equation}\label{V}
u(x)= V (x- (x\cdot n) n, -x\cdot n),
\end{equation}
where $V=V(\theta, t)$ is a function of $(\theta, t)\in \mathbb{T}^d\times [a, \infty)$.
Note that
\begin{equation}\label{2.1}
\nabla_x u = \Big( I - n \otimes n, -n \Big) \left(\begin{array}{c} \nabla_\theta V\\ \partial_t V \end{array}\right)
=M \left( \begin{array} {c}N^T \nabla_\theta \\ \partial_t \end{array} \right)V,
\end{equation}
where we have used (\ref{2.0}).
It follows from (\ref{half-NP}) and (\ref{2.1}) that $V$ is a solution of
\begin{equation}\label{half-NP-V}
\left\{
\aligned
\left( \begin{array} {c}N^T \nabla_\theta \\ \partial_t \end{array}\right)
\cdot B \left( \begin{array} {c}N^T \nabla_\theta \\ \partial_t \end{array} \right)V
& =0 &\quad & \text{ in } \mathbb{T}^d \times (a, \infty),\\
-e_{d+1} \cdot B \left( \begin{array} {c}N^T \nabla_\theta \\ \partial_t \end{array} \right)V
& =T\cdot \nabla_\theta \widetilde{g} & \quad & \text{ on } \mathbb{T}^d \times \{ a \},
\endaligned
\right.
\end{equation}
where
\begin{equation}\label{B}
B=B(\theta, t)=M^T A(\theta-tn) M,
\end{equation}
$\widetilde{g}(\theta,t) =g(\theta -tn)$, and we have used the assumption that $T\cdot n=0$ to obtain
$T\cdot \nabla_x g =T\cdot \nabla_\theta \widetilde{g}$.
Observe that if $V^0$ is a solution of (\ref{half-NP-V}) with $a=0$ and
$$
V^a (\theta, t)= V^0(\theta-an, t-a) \quad \text{ for } a\in \mathbb{R},
$$
then $V^a$ is a solution of (\ref{half-NP-V}).
This follows from the fact that
$$
B(\theta-an, t-a)=B(\theta, t) \quad \text{ and } \quad \widetilde{g}(\theta-an, t-a)=\widetilde{g}(\theta, t).
$$
As a result, it suffices to study the boundary value problem (\ref{half-NP-V}) for $a=0$.
To this end, we shall consider the Neumann problem
\begin{equation}\label{half-NP-V-G}
\left\{
\aligned
-\left( \begin{array} {c}N^T \nabla_\theta \\ \partial_t \end{array}\right)
\cdot B \left( \begin{array} {c}N^T \nabla_\theta \\ \partial_t \end{array} \right)V -\lambda \Delta_\theta V
& = \left( \begin{array} {c}N^T \nabla_\theta \\ \partial_t \end{array} \right)G
&\quad & \text{ in } \mathbb{T}^d \times \mathbb{R}_+,\\
-e_{d+1} \cdot B \left( \begin{array} {c}N^T \nabla_\theta \\ \partial_t \end{array} \right)V
& =T\cdot \nabla_\theta g +e_{d+1} \cdot G
& \quad & \text{ on } \mathbb{T}^d \times \{ 0 \},
\endaligned
\right.
\end{equation}
where $\lambda>0$
and the term $-\lambda \Delta_\theta V $ is added to regularize the system.
Let
\begin{equation}\label{H}
\mathcal{H}=\left\{ f\in H^1_{\text{\rm loc}} (\mathbb{T}^d\times \mathbb{R}_+):
\int_0^\infty \int_{\mathbb{T}^d} \left( | \nabla_\theta f|^2 +|\partial_t f|^2 \right)<\infty \right\}.
\end{equation}
We call $V\in \mathcal{H}$ a weak solution of (\ref{half-NP-V-G})
with $g\in H^1(\mathbb{T}^d)$ and $G \in L^2(\mathbb{T}^d \times \mathbb{R}_+)$, if
\begin{equation}\label{weak}
\aligned
&\int_0^\infty \int_{\mathbb{T}^d} \left\{
B\left(\begin{array}{c} N^T\nabla_\theta \\ \partial_t \end{array}\right) V
\cdot
\left(\begin{array}{c} N^T\nabla_\theta \\ \partial_t \end{array}\right)W
+\lambda
\left(\begin{array}{c} \nabla_\theta \\ 0 \end{array}\right) V
\cdot
\left(\begin{array}{c} \nabla_\theta \\ 0 \end{array}\right)W\right\} \, d\theta dt \\
& =-\int_{\mathbb{T}^d} (T\cdot \nabla_\theta g) \cdot W(\theta, 0)\, d\theta
-\int_0^\infty \int_{\mathbb{T}^d} G \cdot \left( \begin{array} {c}N^T \nabla_\theta \\ \partial_t \end{array}\right) W\, d\theta dt
\endaligned
\end{equation}
for any $W\in \mathcal{H}$.
\begin{prop}\label{prop-2.1}
Let $g\in H^1(\mathbb{T}^d)$ and $G\in L^2(\mathbb{T}^d\times \mathbb{R}_+)$.
Then the boundary value problem (\ref{half-NP-V-G})
has a solution, unique up to a constant, in $\mathcal{H}$.
Moreover, the solution $V$ satisfies
\begin{equation}\label{2.1-0}
\int_0^\infty \int_{\mathbb{T}^d} \left( |N^T \nabla_\theta V|^2 +|\partial_t V|^2 \right)
\le C \Big\{ \|g\|^2_{H^1(\mathbb{T}^d)} +\| G\|^2_{L^2(\mathbb{T}^d\times \mathbb{R}_+)} \Big\},
\end{equation}
and
\begin{equation}\label{2.1-000}
\lambda \int_0^\infty \int_{\mathbb{T}^d}
|\nabla_{\theta} V |^2 \le C \Big\{ \|g\|^2_{H^1(\mathbb{T}^d)} +\| G\|^2_{L^2(\mathbb{T}^d\times \mathbb{R}_+)} \Big\},
\end{equation}
where $C$ depends only on $d$, $m$ and $\mu$.
\end{prop}
\begin{proof}
This follows readily from the Lax-Milgram theorem.
One only needs to observe that
\begin{equation}\label{2.1-00}
\Big|\int_{\mathbb{T}^d} (T\cdot \nabla_\theta g) \cdot W(\theta, 0)\, d\theta \Big|
\le C \| g\|_{H^1(\mathbb{T}^d)}
\left(\int_0^1 \int_{\mathbb{T}^d} \left( |N^T \nabla_\theta W|^2 +|\partial_t W|^2 \right)\right)^{1/2}
\end{equation}
for any $W\in \mathcal{H}$.
Indeed, write
\begin{equation}\label{2.1-1}
\aligned
& \int_{\mathbb{T}^d} (T\cdot \nabla_\theta g) \cdot W(\theta, 0)\, d\theta= \\
& \int_0^1\int_{\mathbb{T}^d} (T\cdot \nabla_\theta g) \cdot ( W(\theta, 0) -W(\theta, t) )\, d\theta dt
+\int_0^1\int_{\mathbb{T}^d} (T\cdot \nabla_\theta g) \cdot W(\theta, t)\, d\theta dt.
\endaligned
\end{equation}
It is easy to see that the first term in the RHS of (\ref{2.1-1}) is bounded by
$$
C\| \nabla_\theta g \|_{L^2(\mathbb{T}^d)} \left(\int_0^1 \|\partial_t W\|_{L^2(\mathbb{T}^d)}^2\, dt\right)^{1/2}.
$$
To handle the second term in the RHS of (\ref{2.1-1}), we use
$$
\int_0^1\int_{\mathbb{T}^d} (T\cdot \nabla_\theta g) \cdot W(\theta, t)\, d\theta dt
=-\int_0^1 g \cdot (T\cdot \nabla_\theta W) (\theta, t)\, d\theta dt
$$
and
$$
T\cdot \nabla_{\theta} W =NN^T T \cdot \nabla_{\theta} W = T\cdot NN^T \nabla _\theta W
$$
to bound it by
$$
C \| g\|_{L^2(\mathbb{T}^d)} \left(\int_0^1 \| N^T\nabla_\theta W\|_{L^2(\mathbb{T}^d)}^2 \, dt \right)^{1/2}.
$$
The estimate (\ref{2.1-00}) now follows.
\end{proof}
\begin{prop}\label{prop-2.2}
Let $g\in H^k(\mathbb{T}^d)$ and $G\in L^2(\mathbb{R}_+, H^{k-1}(\mathbb{T}^d))$
for some $k\ge 1$. Then the solution of (\ref{half-NP-V-G}), given by Proposition
\ref{prop-2.1}, satsfies
\begin{equation}\label{2.2-0}
\aligned
& \int_0^\infty \left( \| N^T \nabla_\theta V \|^2_{H^{k-1} (\mathbb{T}^d)}
+ \| \partial_t V\|_{H^{k-1}(\mathbb{T}^d)}^2 +\lambda \| V\|^2_{H^k(\mathbb{T}^d)} \right)\, dt\\
&
\qquad\qquad
\le C_k \left\{
\|g\|^2_{H^k(\mathbb{T}^d)}
+ \int_0^\infty \| G\|^2_{H^{k-1}(\mathbb{T}^d)} \right\}\, dt,
\endaligned
\end{equation}
where $C_k$ depends on $d$, $m$, $k$, $\mu$ and $\| A\|_{C^{k-1}(\mathbb{T}^d)}$.
\end{prop}
\begin{proof}
The proof is standard.
The case $k=1$ is given in Proposition \ref{prop-2.1}.
To prove the estimate for $k=2$, one applies the estimate for $k=1$ to the quotient of difference
$ \left\{ V(\theta+ s e_j , t) -V (\theta, t)\right\} s^{-1}$ and lets $s\to 0$.
The general case follows similarly by an induction argument on $k$.
\end{proof}
\begin{prop}\label{prop-2.3}
Let $g\in H^{k+\ell-1}(\mathbb{T}^d)$ for some $k, \ell\ge 1$.
Suppose that
$$
\partial_t^\alpha G \in L^2(\mathbb{R}_+, H^{k+\ell-2-\alpha}(\mathbb{T}^d))
\quad \text{ for } \ 0\le \alpha \le \ell-1.
$$
Then the solution of (\ref{half-NP-V-G}), given by Proposition
\ref{prop-2.1}, satsfies
\begin{equation}\label{2.3-0}
\int_0^\infty \| \partial_t^\ell V \|^2_{H^{k-1} (\mathbb{T}^d)} \, dt
\le C \left\{
\|g\|^2_{H^{k+\ell-1} (\mathbb{T}^d)}
+ \sum_{0\le \alpha\le \ell-1}
\int_0^\infty \| \partial_t^\alpha G\|^2_{H^{k+\ell -2-\alpha}(\mathbb{T}^d)} \right\}\, dt,
\end{equation}
where $C$ depends on $d$, $m$, $k$, $\ell$, $\mu$ and $\| A\|_{C^{k+\ell-2}(\mathbb{T}^d)}$.
\end{prop}
\begin{proof}
The case $\ell=1$ is contained in Proposition \ref{prop-2.2}.
To see the case $\ell=2$, we observe that the second-order equation in (\ref{half-NP-V-G})
allows us to obtain
\begin{equation}\label{2.3-1}
\aligned
\partial_t^2 V
&= \text{\rm a linear combination of }\\
& \qquad
\nabla_\theta (N^T\nabla_\theta)V, N^T\nabla_\theta V,
\partial_t \nabla_\theta V, \partial_tV, \nabla_\theta G, \lambda\Delta_\theta V,
\partial_t G
\endaligned
\end{equation}
with smooth coefficients. It follows that
$$
\aligned
\|\partial_t^2 V\|_{H^{k-1}(\mathbb{T}^d)}
\le C & \Big\{ \| N^T \nabla_\theta V \|_{H^k(\mathbb{T}^d)}
+\|\partial_t V \|_{H^k(\mathbb{T}^d)}
+ \| G \|_{H^{k}(\mathbb{T}^d)}\\
&\qquad\qquad
+\| \partial_t G \|_{H^{k-1}(\mathbb{T}^d)}
+\lambda \| V\|_{H^{k+1}(\mathbb{T}^d)}
\Big\}.
\endaligned
$$
This, together with the estimate (\ref{2.2-0}), gives (\ref{2.3-0}) for $\ell=2$.
The general case follows by differentiating (\ref{2.3-1}) in $t$ and using an induction argument on $\ell$.
\end{proof}
\begin{prop}\label{prop-2.4}
Suppose that $n$ satisfies the Diophantine condition (\ref{D-condition}) with constant $\kappa>0$.
Let $V$ be the solution of (\ref{half-NP-V-G}), given by Proposition \ref{prop-2.1}.
Let
$$
\widetilde{V}(\theta, t) = V(\theta, t) --\!\!\!\!\!\!\int_{\mathbb{T}^d} V(\cdot, t).
$$
Then
\begin{equation}\label{2.4-0}
\int_0^\infty \kappa^2 \| \widetilde{V} \|_{H^k(\mathbb{T}^d)}^2 dt \le C \Big\{
\| g\|_{H^{k+3}(\mathbb{T}^d)}^2 +\int_0^\infty \| G\|_{H^{k+2}(\mathbb{T}^d)}^2 \, dt \Big\},
\end{equation}
where $C$ depends on $d$ and $k$ .
\end{prop}
\begin{proof}
It follows from (\ref{D-condition}) and (\ref{2.0}) that
$|N^T \xi|\ge \kappa |\xi|^{-2}$ for any $\xi\in \mathbb{Z}^d\setminus \{ 0\}$.
This implies that
$$
\| N^T \nabla_\theta V \|_{H^{k+2}(\mathbb{T}^d)} \ge C \kappa \| \widetilde{V}\|_{H^k(\mathbb{T}^d)},
$$
which, together with (\ref{2.2-0}), gives the estimate (\ref{2.4-0}).
\end{proof}
\begin{remark}\label{remark-2.1}
{\rm
Suppose that $g\in C^\infty(\mathbb{T}^d)$,
$G\in C^\infty(\mathbb{T}^d\times \mathbb{R}_+)$ and $\partial_t^k \partial_\theta^\alpha G
\in L^2(\mathbb{T}^d\times \mathbb{R}_+)$ for any $k$ and $\alpha$.
For $\lambda>0$, let
$V_\lambda$ be the solution of (\ref{half-NP-V-G}), given by Proposition \ref{prop-2.1}.
By subtracting a constant we may assume that $\int_{\mathbb{T}^d} V_\lambda (\theta, 0)d \theta=0$ and
thus
$$
V_\lambda(\theta, t) =\widetilde{V_\lambda} (\theta, t) +\int^t_0 \int_{\mathbb{T}^d} \partial_s V_\lambda(\theta, s) d\theta ds.
$$
It follows from Propositions \ref{prop-2.3} and \ref{prop-2.4} that the $L^2(\mathbb{T}^d \times (0, L))$ norm
of
$\partial_t^k \partial^\alpha_\theta {V_\lambda}$
is uniformly bounded in $\lambda$, for any $k$, $\alpha$ and $L\ge 1$.
Hence, by Sobolev imbedding, the $C^k(\mathbb{T}^d\times (0, L))$ norm
of ${V_\lambda}$ is uniformly bounded in $\lambda$, for any $k\ge 0$ and $L\ge 1$.
By a simple limiting argument this allows us to show that the Neumann problem (\ref{half-NP-V-G})
with $\lambda=0$ has a solution $V$, unique up to a constant,
in $C^\infty(\mathbb{T}^d\times [0, \infty))$.
Furthermore, by passing to the limit, estimates (\ref{2.1-0}), (\ref{2.2-0}),
(\ref{2.3-0}) and (\ref{2.4-0}) continue to hold for this solution.
}
\end{remark}
\begin{prop}\label{prop-2.5}
Suppose that $n$ satisfies the Diophantine condition (\ref{D-condition}) with constant $\kappa>0$.
Let $V$ be the solution of (\ref{half-NP-V-G})
with $\lambda=0$, $g\in C^\infty(\mathbb{T}^d)$ and $G=0$, given by Remark \ref{remark-2.1}. Then
there exists a constant $V_\infty$ such that for any $\ell\ge 1$,
\begin{equation}\label{2.5-0}
| \partial_\theta^\alpha (V-V _\infty) (\theta, t) |\le \frac{C_{\alpha, \ell}}{ \kappa (1+\kappa t)^\ell},
\end{equation}
for any $\alpha=(\alpha_1, \dots, \alpha_d)$.
Moreover, we have
\begin{equation}\label{2.5-1}
|N^T \nabla_\theta (\partial_\theta^\alpha V) (\theta, t)|
+|\partial_t^k \partial_\theta^\alpha V(\theta, t)| \le \frac{C_{\alpha, \ell, k}}{(1+\kappa t)^\ell},
\end{equation}
where $k\ge 1$.
\end{prop}
\begin{proof}
It follows from Propositions \ref{prop-2.2} and \ref{prop-2.3} by Sobolev imbedding that
$$
|N^T \nabla_\theta (\partial_\theta^\alpha V) (\theta, t)|
+|\partial_t^k \partial_\theta^\alpha V(\theta, t)| \le C_{\alpha, k}
$$
for any $\alpha=(\alpha_1, \dots, \alpha_d)$ and $k\ge 1$.
Next we note that the decay estimate in (\ref{2.5-1})
follows by the exact argument as in the case of Dirichlet boundary conditions, given in \cite{Masmoudi-2012}.
Indeed, let
$$
F(s)= \int_{s}^\infty \int_{\mathbb{T}^d} \Big( |N^T \nabla_\theta (\partial_\theta^\alpha V)|^2
+ |\partial_t^k \partial_\theta^\alpha V|^2\Big)\, d\theta dt.
$$
An inspection of the proof of Proposition 2.6 in \cite{Masmoudi-2012} shows that
$$
F(s)\le \frac{C_\ell}{(\kappa s)^\ell} \quad \text{ for any } \ell\ge 1
$$
(the proof does not use the boundary condition at $t=0$).
By Sobolev imbedding this gives
$$
|N^T \nabla_\theta (\partial_\theta^\alpha V) (\theta, t)|
+|\partial_t^k \partial_\theta^\alpha V(\theta, t)| \le \frac{C_{\alpha, \ell, k}}{(\kappa t)^\ell}.
$$
Finally, we note that a solution of (\ref{half-NP-V-G}) is also a solution of the same system with Dirichlet condition
$V(\theta, 0)\in C^\infty(\mathbb{T}^d)$. It follows from \cite{Masmoudi-2011, Masmoudi-2012} (also see \cite{Prange-2013})
that $V(\theta,t)$ has a limit $V_\infty$, as $t\to \infty$.
Moreover,
\begin{equation}\label{estimate-V}
\aligned
|\partial_\theta^\alpha (V-V_\infty) (\theta, t)|
& \le \int_t^\infty |\partial_t \partial_\theta^\alpha V(\theta, s)|\, ds
\le C \int_t^\infty \frac{ds}{(1+\kappa s)^{\ell+2}}\\
&\le \frac{C}{(1+\kappa t)^\ell} \int_t^\infty \frac{ds}{(1+\kappa s)^2}\\
&\le \frac{C}{\kappa (1+\kappa t)^\ell},
\endaligned
\end{equation}
where we have used (\ref{2.5-1}) for the second inequality.
\end{proof}
We now state and prove the main result of this section.
\begin{theorem}\label{theorem-2.1}
Let $n\in \mathbb{S}^{d-1}$ and $a\in \mathbb{R}$, where $d\ge 2$.
Let $T\in {\mathbb R}^d$ such that $|T| \le 1$ and $T\cdot n=0$.
Suppose that $n\in \mathbb{S}^{d-1}$ satisfies the Diophantine condition (\ref{D-condition})
with constant $\kappa>0$.
Then for any $g\in C^\infty(\mathbb{T}^d)$, the Neumann problem (\ref{half-NP}) has
a smooth solution $u$ satisfying
\begin{equation}\label{2.6-0}
\aligned
| u(x)| & \le \frac{C}{\kappa ( 1+ \kappa |x\cdot n +a|)^\ell},\\
|\partial_x^\alpha u(x)| & \le \frac{C}{( 1+ \kappa |x\cdot n +a|)^\ell},
\endaligned
\end{equation}
for any $|\alpha|\ge 1$ and $\ell\ge 1$. The constant
$C$ depends at most on $d$, $m$, $\mu$, $\alpha$, $\ell$ as well as the $C^k(\mathbb{T}^d)$ norms of $A$ and $g$
for some $k=k(d, \alpha, \ell)$.
\end{theorem}
\begin{proof}
Let $V$ be the solution of (\ref{half-NP-V-G})
with $\lambda=0$, $g\in C^\infty(\mathbb{T}^d)$ and $G=0$, given by Remark \ref{remark-2.1}.
Let
$$
u(x)=V(x-(x\cdot n +a)n, -(x\cdot n +a)) -V_\infty.
$$
Then $u$ is a solution of the Neumann problem (\ref{half-NP}).
The first inequality in (\ref{2.6-0}) follows directly from (\ref{2.5-0}).
To see the second inequality, one uses (\ref{2.1}) and (\ref{2.5-1}).
\end{proof}
\section{Some refined estimates in a half-space}
\setcounter{equation}{0}
Throughout this section we fix $n\in \mathbb{S}^{d-1}$ and $a\in \mathbb{R}$.
We assume that $n\in \mathbb{S}^{d-1}$ satisfies the Diophantine condition (\ref{D-condition})
with constant $\kappa>0$.
However, we will be only interested in estimates that are independent of $\kappa$.
Our first result plays the same role as the maximum principle in the case of Dirichlet
problem.
\begin{theorem}\label{theorem-r}
Let $T\in {\mathbb R}^d$ such that $|T| \le 1$ and $T\cdot n=0$.
Then for any $g\in C^\infty(\mathbb{T}^d)$, the solution $u$ of Neumann problem (\ref{half-NP}), given by
Theorem \ref{theorem-2.1}, satisfies
\begin{equation}\label{r-1-0}
| \nabla u(x)|\le \frac{C\, \| g\|_\infty }{|x\cdot n +a|},
\end{equation}
for any $x\in \mathbb{H}_n^d(a)$,
where $C$ depends only on $d$, $m$ and $\mu$ as well as some H\"older norm of $A$.
\end{theorem}
\begin{proof}
By translation we may assume that $a=0$.
We choose a bounded smooth domain $D$ such that
$$
\aligned
& B(0, 1)\cap \mathbb{H}_n^d(0)\subset D \subset B(0, 2) \cap \mathbb{H}_n^d (0),\\
&\overline{B(0,1)} \cap\partial \mathbb{H}_n^d(0) = \partial D\cap \partial\mathbb{H}_n^d(0).
\endaligned
$$
Let $v_\varepsilon (x)=\varepsilon u (x/\varepsilon)$.
Since $\mathcal{L}_\varepsilon (v_\varepsilon)=0$ in $D$,
$$
v_\varepsilon (x) -v_\varepsilon (z)
=\int_{\partial D} \big\{ N_\varepsilon (x, y) -N_\varepsilon (z, y) \big\} \frac{\partial v_\varepsilon}{\partial \nu_\varepsilon} (y)\, d\sigma (y)
$$
for any $x, z\in D$,
where $N_\varepsilon(x, y)$ denotes the matrix of Neumann functions for $\mathcal{L}_\varepsilon$ in $D$.
By a change of variables it follows that
$$
u(x)-u(z)=
\varepsilon^{d-2}\int_{\partial D_{1/\varepsilon}}
\big\{ N_\varepsilon (\varepsilon x, \varepsilon y)-N_\varepsilon (\varepsilon z, \varepsilon y) \big\} n(\varepsilon y) \cdot A(y)\nabla u(y)\, d\sigma (y),
$$
where $D_{1/\varepsilon} =\big\{ \varepsilon^{-1} y: y\in D\big\}$.
Fix $x, z\in \mathbb{H}_n^d(0)$ such that $|x-z|< (1/2)|x\cdot n|=(1/2)\text{dist}(x, \partial \mathbb{H}_n^d(0))$.
Choose $\eta_\varepsilon \in C_0^1(B(0, \varepsilon^{-1}))$ such that $0\le \eta_\varepsilon \le 1$,
$\eta_\varepsilon=1$ on $B(0, \varepsilon^{-1}-1)$ and $|\nabla \eta_\varepsilon|\le 1$, where $\varepsilon< 1/10$.
Let $u(x)-u(z) =I_1 +I_2$,
where
$$
\aligned
I_1 &=\varepsilon^{d-2}\int_{\partial D_{1/\varepsilon}} \eta_\varepsilon (y)
\big\{ N_\varepsilon (\varepsilon x, \varepsilon y)-N_\varepsilon (\varepsilon z, \varepsilon y) \big\} n (\varepsilon y) \cdot A(y)\nabla u(y)\, d\sigma (y)\\
&=\varepsilon^{d-2}\int_{\partial D_{1/\varepsilon}} \eta_\varepsilon (y)
\big\{ N_\varepsilon (\varepsilon x, \varepsilon y)-N_\varepsilon (\varepsilon z, \varepsilon y) \big\} T\cdot \nabla g (y)\, d\sigma (y)\\
&=-\varepsilon^{d-2}\int_{\partial B(0, \varepsilon^{-1})\cap \partial \mathbb{H}_n^d (0)}
T\cdot \nabla_y \Big\{ \eta_\varepsilon (y) (N_\varepsilon (\varepsilon x, \varepsilon y)-N_\varepsilon (\varepsilon z, \varepsilon y)) \Big\} g(y) \, d\sigma (y),
\endaligned
$$
where we have used the Neumann condition for $u$ as well as an integration by parts on
the boundary. We now apply the estimates in (\ref{N-estimate-1}).
This gives
$$
\aligned
|I_1| &\le C |x-z| \| g\|_\infty\int_{\partial \mathbb{H}_n^d(0)} \frac{d\sigma (y)}{|x-y|^d}
+ C |x-z|\| g\|_\infty \int_{\frac{1}{\varepsilon}-1 \le |y|\le \frac{1}{\varepsilon}}
\frac{ d\sigma (y)}{|x-y|^{d-1}}\\
& \le C_0 \|g\|_\infty + C \varepsilon\| g\|_\infty |x-z|,
\endaligned
$$
if $\varepsilon$, which may depend on $|x|$, is sufficiently small.
We point out that the constant $C_0$ in the estimate above
depends only on $d$, $m$, $\mu$ and some H\"older norm of $A$.
Next, to handle $I_2$, we use the estimate
$$
|\nabla u(y)|\le \frac{C}{(1+\kappa |y\cdot n|)^2}
$$
from (\ref{2.6-0}). This, together with (\ref{N-estimate-1}), leads to
$$
\aligned
| I_2| &=\varepsilon^{d-2}\Big| \int_{\partial D_{1/\varepsilon}} (1-\eta_\varepsilon (y))
\big\{ N_\varepsilon (\varepsilon x, \varepsilon y)-N_\varepsilon (\varepsilon z, \varepsilon y) \big\} n(\varepsilon y) \cdot A(y)\nabla u(y)\, d\sigma (y)\Big|\\
& \le C |x-z| \int_{\partial D_{1/\varepsilon}\cap \mathbb{H}_n^d(0)}
\frac{d\sigma (y)}{|x-y|^{d-1} (1+\kappa |y\cdot n|)^2}\\
&\le C_{x, z} \int_{\partial D_{1/\varepsilon}\cap \mathbb{H}_n^d(0)}
\frac{d\sigma (y)}{|y|^{d-1} (1+\kappa |y\cdot n|)^2},
\endaligned
$$
which shows that $I_2\to 0$, as $\varepsilon\to 0$.
As a result, we have proved that for any $x, z\in \mathbb{H}_n^d(0)$ with
$|x-z|\le \text{dist}(x, \partial\mathbb{H}_n^d(0))$,
$$
|u(x)-u(z)|
=\lim_{\varepsilon\to 0} | I_1 +I_2|
\le C_0 \| g\|_\infty.
$$
Since $\mathcal{L}_1 (u)=0$ in $\mathbb{H}_n^d(0)$,
by the interior Lipschitz estimates \cite{AL-1987} for $\mathcal{L}_1$,
we obtain
$$
|\nabla u(x)|\le \frac{C_0 \| g\|_\infty}{|x\cdot n|},
$$
which completes the proof.
\end{proof}
Let $\Omega=\mathbb{H}_n^d(a)$ and $\mathcal{L} =-\text{\rm div} (A(x)\nabla )$ .
In the rest of this section we consider Dirichlet problem,
\begin{equation}\label{DP-w}
\left\{
\aligned
\mathcal{L} (u) &=\text{div} (f) +h &\quad &\text{ in } \Omega,\\
u & =0 &\quad & \text{ on }\partial\Omega,
\endaligned
\right.
\end{equation}
and the Neumann problem,
\begin{equation}\label{NP-w}
\left\{
\aligned
\mathcal{L} (u) &=\text{div} (f) &\quad &\text{ in } \Omega,\\
\frac{\partial u}{\partial \nu} & =-n\cdot f &\quad & \text{ on }\partial\Omega,
\endaligned
\right.
\end{equation}
where $A$ is assumed to satisfy the ellipticity condition (\ref{ellipticity}) and
$A\in C^\sigma (\mathbb{T}^d)$ for some $\sigma\in (0,1)$.
We shall be interested in the weighted $L^2$ estimate,
\begin{equation}\label{w-estimate-1}
\int_\Omega |\nabla u(x)|^2 \big[\delta(x)\big]^\alpha\, dx
\le C \int_\Omega | f(x) |^2 \big[\delta(x)\big]^\alpha\, dx
+C \int_\Omega |h(x)|^2 \big[\delta(x)\big]^{\alpha+2}\, dx,
\end{equation}
where $-1<\alpha<0$ and
\begin{equation}\label{weight}
\delta(x)=\text{dist} (x, \partial\Omega)=|a + (x\cdot n)|.
\end{equation}
We start with some observations on the weight $\omega (x) =\big[\delta(x)\big]^\alpha$.
\begin{lemma}\label{lemma-weight}
Let $\omega (x) =\big[\delta(x)\big]^\alpha$, where $-1<\alpha<0$ and $\delta(x)$ is defined by (\ref{weight}).
Then $\omega(x)$ is an $A_1$ weight, i.e., for any ball $B\subset {\mathbb R}^d$,
\begin{equation}\label{A-1}
-\!\!\!\!\!\!\int_B \omega \le C \inf_{B} \omega ,
\end{equation}
where $C$ depends only on $d$ and $\alpha$. Moreover, $w$ satisfies the reverse H\"older's inequality,
\begin{equation}\label{reverse-H}
\left(-\!\!\!\!\!\!\int_B \omega ^p\, dx \right)^{1/p}
\le C -\!\!\!\!\!\!\int_B \omega\, dx,
\end{equation}
where $1<p< \infty$ and $\alpha p>-1$.
\end{lemma}
\begin{proof}
This is more or less well known and may be verified directly by reducing
the problem to the case $\Omega={\mathbb R}^d_+$ and $\delta (x)=|x_d|$.
\end{proof}
It follows from (\ref{reverse-H}) by H\"older's inequality that if $E\subset B$, then
\begin{equation}\label{measure}
\frac{\omega(E)}{\omega(B)}
\le C \left( \frac{|E|}{|B|}\right)^{1-\frac{1}{p}},
\end{equation}
where $\omega(E)=\int_E \omega\, dx$.
Since (\ref{measure}) implies that $\omega$ satisfies the doubling condition,
$\omega(2B)\le C \omega(B)$,
it is easy to see that (\ref{measure})
also holds if one replaces ball $B$ by cube $Q$.
In fact, if $\omega$ is an $A_p$ weight in ${\mathbb R}^d$,
i.e.,
\begin{equation}\label{A-p}
-\!\!\!\!\!\!\int_B \omega \cdot \left(-\!\!\!\!\!\!\int_B \omega^{-\frac{1}{p-1}}\right)^{p-1}
\le C,
\end{equation}
then there exist some $\sigma>0$ and $C>0$
such that
\begin{equation}\label{A-infty}
\frac{\omega(E)}{\omega(Q)}
\le C \left( \frac{|E|}{|Q|}\right)^{\sigma} \quad \text{ for any } E\subset Q.
\end{equation}
Functions satisfying (\ref{A-infty}) are called $A_\infty$ weights.
In the following we will also need the well known fact that if $\omega$ is an $A_p$ weight for some $1<p<\infty$, then
\begin{equation}\label{weighted-max}
\int_{{\mathbb R}^d} |\mathcal{M}(f)|^p\, \omega\, dx \le C \int_{{\mathbb R}^d} | f|^p\, \omega \, dx,
\end{equation}
where $\mathcal{M}(f)$ denotes the Hardy-Littlewood maximal function of $f$.
This is the defining property of the $A_p$ weights.
Note that if $\omega$ is $A_1$, then $\omega$ is $A_p$ for any $p>1$.
We refer the reader to \cite{Stein-1993} for the theory of weights in harmonic analysis.
The next lemma gives the Lipschitz estimates for $\mathcal{L}$ in $\Omega$ with
either Dirichet or Neumann boundary conditions.
\begin{lemma}\label{lemma-w-1}
Let $x_0\in \overline{\Omega}$ and $r>0$.
Let $u\in H^1(B(x_0, 2r)\cap\Omega)$ be a weak solution of
$\mathcal{L} (u)=0$ in $B(x_0, 2r)\cap\Omega$
with either $u=0$ or
$\frac{\partial u}{\partial \nu}=0$ on $B(x_0, 2r)\cap \partial\Omega$.
Then
\begin{equation}\label{w-1-0}
\| \nabla u\|_{L^\infty(B(x_0, r)\cap \Omega)}
\le C -\!\!\!\!\!\!\int_{B(x_0, 2r)\cap\Omega} |\nabla u|,
\end{equation}
where $C$ depends only on $d$, $m$, $\mu$ and $\| A\|_{C^\sigma(\mathbb{T}^d)}$.
\end{lemma}
\begin{proof}
By translation and dilation we may reduce (\ref{w-1-0}) to the uniform Lipschitz estimate
for operator $\mathcal{L}_\varepsilon$ with $\varepsilon=r^{-1}$
in $B(0, 2)\cap\Omega$.
In the case with Dirichlet
condition $u=0$ on $B(0, 2)\cap\partial\Omega$, the Lipschitz estimate
was proved in \cite{AL-1987}.
The Lipschitz estimate for solutions with Neumann condition $\frac{\partial u}{\partial \nu}=0$
on $B(0, 2)\cap\partial\Omega$ was obtained in \cite{KLS1, AS}.
The fact that
$\Omega$ has a flat boundary is crucial here.
Otherwise, the constant $C$ will depend on $r$, if $r$ is large.
This is true even for harmonic functions in smooth domains.
\end{proof}
\begin{theorem}\label{theorem-w-2}
Let $\omega$ be an $A_1$ weight in ${\mathbb R}^d$.
Let $u\in H^1_{\text{\rm loc}}(\Omega)$ be a weak solution of Dirichlet problem(\ref{DP-w}) with $h=0$.
Assume that
\begin{equation}\label{w-2-00}
\omega(B(x_0, R)\cap \Omega)
-\!\!\!\!\!\!\int_{B(x_0, R)\cap\Omega} |\nabla u|^2 \to 0 \quad \text{ as } R\to \infty,
\end{equation}
for some $x_0\in \partial\Omega$.
Then
\begin{equation}\label{w-2-0}
\int_\Omega |\nabla u|^2 \, \omega \, dx
\le C \int_\Omega |f|^2 \, \omega \, dx,
\end{equation}
where $C$ depends only on $d$, $m$, $\mu$, $\|A\|_{C^\sigma(\mathbb{T}^d)}$, and the constant in (\ref{A-1}).
\end{theorem}
\begin{proof}
This is essentially proved in \cite{Shen-2005} by a real-variable method, originated in \cite{CP-1998}.
We provide a proof here for the reader's convenience.
By translation we may assume that $a=0$.
We will also assume that $n=-e_d$ and thus $\Omega={\mathbb R}^d_+$ for simplicity of exposition.
We point out that the periodicity of the coefficient matrix
is not used particularly in the proof, only the estimates in Lemma \ref{lemma-w-1}.
Fix $1<p<2$.
Let $\rho\in (0,1)$ be a small constant to be determined and $A=\rho^{-\sigma/2}$, where $\sigma$ is given in
(\ref{A-infty}). Let
$$
\Omega_{R}= (-R, R)\times \cdots \times (-R, R)\times (0, 2R)\subset {\mathbb R}^d_+.
$$
W fix $R>1$ and consider the set
\begin{equation}\label{E-lambda}
E(\lambda )= \big\{ x\in \Omega_R: \mathcal{M}_{{R}} (|\nabla u|^p)(x) > \lambda \big\},
\end{equation}
where $\mathcal{M}_R(F)$ is a localized Hardy-Littlewood maximal function of $F$, defined by
$$
\mathcal{M}_{{R} }(F) (x) =\sup _{x\in Q\subset \Omega_{2R}} -\!\!\!\!\!\!\int_Q |F|.
$$
Let
\begin{equation}\label{lambda-0}
\lambda_0 =\frac{C_0}{|\Omega_{2R}|}\int _{\Omega_{2R}}
|\nabla u|^p,
\end{equation}
where $C_0$ is a large constant depending on $d$.
For each $\lambda>\lambda_0$,
we perform a Calder\'on-Zygmund decomposition to $E(A\lambda)\subset \Omega_R$.
This produces a sequence of disjoint dyadic subcubes
$\{ Q_k\}$ of $\Omega_R$ such that
\begin{equation}\label{w-10}
\aligned
& |E(A\lambda)\setminus \cup_k Q_k|=0,\\
& |E(A\lambda)\cap Q_k|>\rho |Q_k|,\ \ \
|E(A\lambda) \cap \overline{Q}_k|\le \rho |\overline{Q}_k|,
\endaligned
\end{equation}
where $\overline{Q}_k$ denotes the dyadic parent of $Q_k$, i.e., $Q_k$ is obtained by
bisecting $\overline{Q}_k$ once.
We claim that it is possible to choose $\rho, \gamma\in (0,1)$ so that
\begin{equation}\label{w-11}
\text{ if } \
\big\{ x\in \overline{Q}_k: \mathcal{M}_{R} (|f|^p)(x)\le \gamma \lambda \big\} \neq \emptyset,
\ \text{ then } \ \overline{Q}_k\subset E(\lambda).
\end{equation}
The claim is proved by contraction.
Suppose that there exists some $x_0$ such that
$x_0\in \overline{Q}_k\setminus E(\lambda)$.
Then, if a cube $Q$ contains $\overline{Q}_k$ and $Q\subset \Omega_{2R}$,
\begin{equation}\label{w-12}
-\!\!\!\!\!\!\int_Q |f|^p \le \gamma \lambda \quad \text{ and } \quad -\!\!\!\!\!\!\int_{Q} |\nabla u|^p \le \lambda.
\end{equation}
This implies that for any $x\in Q_k$,
\begin{equation}\label{w-13}
\mathcal{M}_{R} (|\nabla u|^p) (x)
\le \max ( \mathcal{M}_{2\overline{Q}_k}( |\nabla u|^p) (x), 5^d \lambda),
\end{equation}
where
$$
\mathcal{M}_{2\overline{Q}_k} (|F|)(x) =\sup_{x\in Q \subset 2\overline{Q}_k\cap\Omega}-\!\!\!\!\!\!\int_Q |F|.
$$
We now write $u=v+w$, where $v$ is a function such that
\begin{equation}\label{w-14}
\mathcal{L} (v)=\text{\rm div} (f)\quad \text{ in } \Omega\cap 5\overline{Q}_k
\quad \text{ and } \quad v=0 \quad \text{ on } \partial\Omega\cap 5\overline{Q}_k ,
\end{equation}
\begin{equation}\label{w-15}
\int_{\Omega\cap 4\overline{Q}_k} |\nabla v|^p \le C \int_{\Omega \cap 5 \overline{Q}_k} |f|^p,
\end{equation}
and $C$ depends only on $d$, $m$, $p$, $\mu$ and $\|A\|_{C^\sigma(\mathbb{T}^d)}$.
The existence of such $v$ follows from the boundary $W^{1, p}$ estimates for $\mathcal{L}_\varepsilon$ \cite{AL-1987, AL-1991}.
Note that if $A>5^d$,
\begin{equation}\label{w-16}
\aligned
|Q_k\cap E(A\lambda)|
&\le |\big\{ x\in Q_k: \mathcal{M}_{2\overline{Q}_k} (|\nabla u|^p)(x)> A\lambda \big\}|\\
&\le |\big\{ x\in Q_k: \mathcal{M}_{2\overline{Q}_k} (|\nabla v|^p)(x)> (1/4)A\lambda \big\}|\\
&\qquad
+ |\big\{ x\in Q_k: \mathcal{M}_{2\overline{Q}_k} (|\nabla w|^p)(x)> (1/4)A\lambda \big\}|.\\
\endaligned
\end{equation}
For the first term in the RHS of (\ref{w-16}), we use the fact that the operator $\mathcal{M}$
is bounded from $L^1$ to weak-$L^1$. This, together with (\ref{w-15}) and (\ref{w-12}), shows that
the term is dominated by
$$
\frac{C}{A\lambda}\int_{\Omega\cap 5 \overline{Q}_k} |f|^p
\le C\gamma A^{-1} |Q_k|,
$$
where $C$ depends only on $d$, $m$, $\mu$ and $\|A\|_{C^\sigma(\mathbb{T}^d)}$.
Since $\mathcal{L} (w)=0$ in $\Omega\cap 5 \overline{Q}_k$ and $w=0$ on $\partial\Omega\cap 5\overline{Q}_k$,
in view of Lemma \ref{lemma-w-1},
we obtain
$$
\aligned
\|\nabla w\|_{L^\infty(\Omega\cap2\overline{Q}_k)}
&\le C -\!\!\!\!\!\!\int_{\Omega\cap 4\overline{Q}_k} |\nabla w| \\
& \le C \left(-\!\!\!\!\!\!\int_{\Omega\cap 4\overline{Q}_k} |\nabla u|^p\right)^{1/p}
+C \left(-\!\!\!\!\!\!\int_{\Omega\cap 4\overline{Q}_k} |\nabla v|^p\right)^{1/p} \\
& \le C\lambda^{1/p} +
C \left(-\!\!\!\!\!\!\int_{\Omega\cap 5\overline{Q}_k} | f|^p\right)^{1/p}\\
& \le C \lambda^{1/p},
\endaligned
$$
where we have also used estimates (\ref{w-12}) and (\ref{w-15}).
It follows that the second term in the RHS of (\ref{w-16}) is zero, if
$A$ is large. As a result, we have proved that
$$
|Q_k\cap E(A\lambda)|\le C \gamma A^{-1} |Q_k|
=C\gamma \rho^\sigma |Q_k|,
$$
if $\rho\in (0,1)$ is sufficiently small.
By choosing $\gamma\in (0,1)$ so small that $C\gamma \rho^\sigma<\rho$,
we obtain $|Q_k\cap E(A\lambda)|< \rho |Q_k|$,
which is in contradiction with (\ref{w-10}).
This proves the claim (\ref{w-11}).
We should point out that the choices of $\rho$ and $\gamma$ are uniform for all $\lambda>\lambda_0$.
To proceed, we use (\ref{A-infty}) and (\ref{w-10}) to obtain
\begin{equation}\label{w-17}
\omega(E(A\lambda)\cap Q_k) \le C \rho^\sigma \omega(Q_k).
\end{equation}
This, together with (\ref{w-11}), leads to
\begin{equation}\label{w-18}
\aligned
\omega(E(A\lambda))
&\le \omega \big( E(A\lambda)\cap \big\{x\in \Omega_R: \mathcal{M}_R (|f|^p)(x)\le \gamma \lambda\big\} \big)\\
&\qquad\qquad
+\omega\big\{ x\in \Omega_R: \mathcal{M}_R (|f|^p)(x)>\gamma \lambda \big\}\\
& \le \sum_{k} \omega \big\{x\in E(A\lambda)\
\cap Q_k: \ \mathcal{M}_R (|f|^p)(x)\le \gamma \lambda\big\} \\
&\qquad\qquad
+ \omega\big\{ x\in \Omega_R: \mathcal{M}_R (|f|^p)(x)>\gamma \lambda \big\}\\
& \le C\rho^\sigma\sum_{k} \omega (Q_k)
+ \omega\big\{ x\in \Omega_R: \mathcal{M}_R (|f|^p)(x)>\gamma \lambda \big\},
\endaligned
\end{equation}
for any $\lambda >\lambda_0$,
where the last sum is taken only over those $Q_k$'s such that
$\{ x\in Q_k: \mathcal{M}_R (|f|^p)(x)\le \gamma \lambda \} \neq \emptyset$.
By the claim (\ref{w-11}) this gives
\begin{equation}\label{w-19}
\omega(E(A\lambda)) \le C \rho^\sigma \omega (E(\lambda))
+ \omega\big\{ x\in \Omega_R: \mathcal{M}_R (|f|^p)(x)>\gamma \lambda \big\}
\end{equation}
for any $\lambda >\lambda_0$.
Finally, we multiply both sides of (\ref{w-19}) by $\lambda^t$ with $t=\frac{2}{p}-1\in (0,1)$,
and integrate the resulting inequality in $\lambda$ over the interval $(\lambda_0, \Lambda)$ to obtain
$$
A^{-1-t}\int_{A\lambda_0}^{A\Lambda} \lambda^t
\omega(E(\lambda))\, d\lambda
\le C\rho^\sigma \int_{\lambda_0}^\Lambda \lambda^t \omega(E(\lambda))\, d\lambda
+ C_\gamma \int_{\Omega_R} \Big\{ \mathcal{M}_{2R} (|f|^p) \Big\}^{\frac{2}{p}}
\omega\, dx.
$$
Since $CA^{1+t}\rho^\sigma =C\rho^{-\frac{\sigma}{2} (1+t)}\rho^{\sigma}<(1/2)$ if $\rho>0$ is small, this gives
$$
\int_0^\Lambda \lambda^t \omega(E(\lambda))\, d\lambda
\le C \int_0^{A\lambda_0} \lambda^t \omega(E(\lambda))\, d\lambda
+ C \int_{\Omega_{2R}} |f|^2\omega\, dx,
$$
where we have used the weighted norm inequality (\ref{weighted-max}) as well
as the fact that $2/p>1$.
By letting $\Lambda\to \infty$ we obtain
\begin{equation}\label{w-20}
\aligned
\int_{\Omega_R} |\nabla u|^2 \omega \, dx
& \le C \lambda_0^{\frac{2}{p}} \omega(\Omega_R) + C \int_{\Omega_{2R}} |f|^2\omega\, dx\\
& = \frac{C\, \omega(\Omega_{R})}{|\Omega_{2R}|}
\int_{\Omega_{2R}} |\nabla u|^2\, dx
+ C \int_{\Omega_{2R}} |f|^2\omega\, dx,
\endaligned
\end{equation}
where we have used the fact $|F|\le \mathcal{M}(F)$.
We complete the proof by letting $R\to \infty$ in (\ref{w-20}) and
using the assumption (\ref{w-2-00}).
\end{proof}
The next theorem treats the Neumann problem (\ref{NP-w}).
\begin{theorem}\label{theorem-w-3}
Let $\omega$ be an $A_1$ weight in ${\mathbb R}^d$.
Let $u\in H^1_{\text{\rm loc}}(\Omega)$ be a weak solution of the Neumann problem(\ref{NP-w}).
Assume that $u$ satisfies the condition (\ref{w-2-00}).
Then the estimate (\ref{w-2-0}) holds with constant $C$
depending only on $d$, $m$, $\mu$, $\|A\|_{C^\sigma(\mathbb{T}^d)}$, and the constant in (\ref{A-1}).
\end{theorem}
\begin{proof}
The proof is similar to that of Theorem \ref{theorem-w-2}.
The only difference is that in the place of (\ref{w-14}), we need to find a function $v$ such that,
\begin{equation}\label{w-3-1}
\mathcal{L}(v) =\text{\rm div}(f\varphi) \quad \text{ in } \Omega\cap 5\overline{Q}_k
\quad \text{ and } \quad \frac{\partial v}{\partial \nu} =n\cdot (\varphi f) \quad \text{ on } \partial\Omega\cap
5\overline{Q}_k,
\end{equation}
where $\varphi\in C_0^\infty(5\overline{Q}_k)$,
$0\le \varphi\le 1$, and $\varphi=1$ on $4\overline{Q}_k$.
The existence of functions satisfying (\ref{w-3-1}) and
the estimate (\ref{w-15}) follows from the boundary $W^{1, p}$ estimates for $\mathcal{L}_\varepsilon$
with Neumann conditions \cite{KLS1, AS}.
We omit the details and refer the reader to \cite{Geng-2012} for
related $W^{1, p}$ estimates for Neumann problems.
\end{proof}
Finally, we go back to the case where $\omega(x)=\big[\delta(x)\big]^\alpha$.
\begin{theorem}\label{theorem-w-4}
Let $-1<\alpha<0$, $\Omega=\mathbb{H}^d_n(a)$ and $\delta(x)$ be given by (\ref{weight}).
Let $u\in H^1_{\text{\rm loc}}(\Omega)$ be a weak solution of (\ref{DP-w}) in $\Omega$.
Assume that
\begin{equation}\label{w-4-0}
R^\alpha \int_{B(x_0, R)\cap\Omega} |\nabla u|^2 \to 0 \quad \text{ as } R \to \infty,
\end{equation}
for some $x_0\in \partial\Omega$.
Then estimate (\ref{w-estimate-1}) holds with a constant $C$ depending only on $d$, $m$, $\mu$,
$\alpha$ and $\|A\|_{C^\sigma(\mathbb{T}^d)}$.
\end{theorem}
\begin{proof}
By translation we may assume that $a=0$.
Recall that $\omega(x)=\big[\delta(x)\big]^\alpha$ is an $A_1$ weight for $-1<\alpha<0$.
Also observe that in this case the assumption (\ref{w-2-00}) is reduced to (\ref{w-4-0}).
We may assume that
$$
\int_\Omega |h(x)|^2 \big[\delta(x)\big]^{\alpha+2}\, dx
=\int_{\partial\mathbb{H}_n^d(0)}
\left\{ \int_0^\infty | h(x^\prime-tn)|^2 t^{\alpha +2} dt \right\} d\sigma (x^\prime)
<\infty.
$$
For otherwise there is nothing to prove. It follows that for a.e. $x^\prime \in \partial\mathbb{H}_n^d(0)$,
$$
\int_s^\infty |h(x^\prime-tn)|dt
\le \left(\int_s^\infty |h(x^\prime-tn)|^2 t^{\alpha+2}dt\right)^{1/2}
\left(\int_s^\infty t^{-\alpha-2} dt\right)^{1/2}<\infty,
$$
if $s>0$. This allows us to write
$$
h=\text{\rm div}(H),
$$
where $H=(H_1, \dots, H_d)$ and
$$
H_i (x)= n_i\int_0^\infty h(x-tn)\, dt.
$$
As a result, we obtain $\mathcal{L}(u) =\text{\rm div} ( f+H)$ in $\Omega$.
It follows by Theorem \ref{theorem-w-2} that
\begin{equation}\label{w-4-10}
\int_\Omega |\nabla u|^2 \omega\, dx
\le C \int_\Omega |f|^2\omega\, dx +C \int_\Omega |H|^2\omega\, dx.
\end{equation}
Finally, we observe that for $x^\prime\in \partial\mathbb{H}_n^d(0)$,
$$
\aligned
\int_0^\infty |H(x^\prime-tn)|^2 t^\alpha\, dt
&\le \int_0^\infty \Big| \int_0^\infty |h(x^\prime-tn -sn)|\, ds \Big|^2 t^\alpha\, dt \\
&\le \int_0^\infty \Big| \int_t^\infty |h(x^\prime-sn)|\, ds \Big|^2 t^\alpha\, dt\\
& \le \frac{4}{(\alpha+1)^2} \int_0^\infty |h(x^\prime-tn)|^2 t^{\alpha +2}\, dt,
\endaligned
$$
where $\alpha>-1$ and a Hardy's inequality was used for the last step.
By integrating above inequalities in $x^\prime$ over $\partial\mathbb{H}_n^d(0)$,
we obtain
$$
\int_\Omega |H(x) |^2 \big[\delta(x)\big]^\alpha \, dx
\le C \int_\Omega |h (x)|^2 \big[\delta(x)\big]^{\alpha+2} \, dx.
$$
This, together with (\ref{w-4-10}), gives the weighted estimate (\ref{w-estimate-1}).
\end{proof}
The next theorem establishes (\ref{w-estimate-1}) for the Neumann problem (\ref{NP-w}).
\begin{theorem}\label{theorem-w-5}
Let $-1<\alpha<0$, $\Omega=\mathbb{H}^d_n(a)$ and $\delta(x)$ be given by (\ref{weight}).
Let $u\in H^1_{\text{\rm loc}}(\Omega)$ be a weak solution of Neumann problem (\ref{NP-w}) in $\Omega$.
Suppose that $u$ satisfies the condition (\ref{w-4-0})
for some $x_0\in \partial\Omega$.
Then
\begin{equation}\label{w-5-0}
\int_{\Omega} |\nabla u(x)|^2 \, \big[\delta(x)\big]^\alpha\, dx
\le C \int_{\Omega} |f(x)|^2 \, \big[\delta(x)\big]^\alpha\, dx,
\end{equation}
where $C$ depending only on $d$, $m$, $\mu$,
$\alpha$ and $\|A\|_{C^\sigma(\mathbb{T}^d)}$.
\end{theorem}
\begin{proof}
This follows directly from Theorem \ref{theorem-w-3}.
\end{proof}
\begin{remark}\label{remark-w-100}
{\rm
Let $\Omega=\mathbb{H}^d_n(a)$.
Let $u\in H^1_{\text{\rm loc}}(\Omega)$ be a solution of $-\text{\rm div}(A(x)\nabla u)=\text{\rm div}(f)$
in $\Omega$, with either Dirichlet condition $u=0$ or Neumann condition
$\frac{\partial u}{\partial\nu}=-n\cdot f$ on $\partial\Omega$.
Let
\begin{equation}\label{Omega-R}
\Omega_R = \big\{ x\in \Omega: |x-(a+x\cdot n)n|\le R \text{ and } |a+x\cdot n|\le 2R \big\}.
\end{equation}
It follows from (\ref{w-20}) that for $R\ge 1$
\begin{equation}\label{w-101}
\int_{\Omega_R} |\nabla u|^2 \, \big[\delta(x)\big]^\alpha \, dx
\le C R^\alpha \int_{\Omega_{2R}}|\nabla u|^2\, dx
+C \int_{\Omega_{2R}} |f|^2\, \big[\delta(x)\big]^\alpha \, dx,
\end{equation}
where $-1<\alpha<0$ and
$C$ depends only on $d$, $m$, $\mu$, $\alpha$, and some H\"older norm of $A$.
This will be used in Section 6.
}
\end{remark}
\begin{remark}\label{remark-4.100}
{\rm
The weighted estimates in Theorems \ref{theorem-w-4} and \ref{theorem-w-5}
also hold for the range $0<\alpha<1$, which is not used in the paper.
This may be proved by a duality argument.
}
\end{remark}
\section{Approximation of Neumann correctors}
\setcounter{equation}{0}
Throughout this section we assume that $\Omega$ is a bounded smooth, strictly convex domain
in ${\mathbb R}^d$, $d\ge3$, and that $A$ is smooth and satisfies (\ref{ellipticity})-(\ref{periodicity}).
For $g\in C^\infty(\mathbb{T}^d; \mathbb{R}^m)$, consider the Neumann problem
\begin{equation}\label{NP-M}
\left\{
\aligned
& \mathcal{L}_\varepsilon (u_\varepsilon) =0 & \quad &\text{ in } \Omega,\\
&\frac{\partial u_\varepsilon}{\partial\nu_\varepsilon} = T(x) \cdot \nabla g (x/\varepsilon) &\quad & \text{ on } \partial\Omega,
\endaligned
\right.
\end{equation}
where $T(x)=T_{ij}(x)=n_i (x) e_j -n_j(x) e_i$ for some $1\le i, j\le d$ is a tangential vector field
on $\partial\Omega$,
and $n(x)=(n_1(x), \dots, n_d(x))$ denotes the outward normal to $\partial\Omega$
at $x\in \partial\Omega$.
Fix $x_0 \in \partial\Omega$.
Assume that $n=n(x_0)$ satisfies the Diophantine condition (\ref{D-condition}) with constant
$\kappa>0$ (all constants $C$ will be independent of $\kappa$).
To approximate $u_\varepsilon$ in a neighborhood of $x_0$, we solve the
Neumann problem in a half-space
\begin{equation}\label{NP-A-1}
\left\{
\aligned
& \mathcal{L}_\varepsilon (v_\varepsilon) =0 & \quad &\text{ in } \mathbb{H}_{n}^d (a),\\
& \frac{\partial v_\varepsilon}{\partial\nu_\varepsilon} = T(x_0) \cdot \nabla g (x/\varepsilon) &\quad & \text{ on } \partial \mathbb{H}_{n}^d (a),
\endaligned
\right.
\end{equation}
where $a=-x_0\cdot n$ and $\partial \mathbb{H}_{n}^d (a)$ is the tangent plane of $\partial\Omega$
at $x_0$.
Note that if $v_\varepsilon (x)=\varepsilon w(x/\varepsilon)$, then $w$ is a solution of
\begin{equation}\label{NP-m}
\left\{
\aligned
&\mathcal{L}_1 (w) =0 & \quad &\text{ in } \mathbb{H}_{n}^d (a\varepsilon^{-1}),\\
&\frac{\partial w}{\partial\nu_1} = T(x_0) \cdot \nabla g(x) &\quad & \text{ on } \partial \mathbb{H}_{n}^d (a\varepsilon^{-1}).
\endaligned
\right.
\end{equation}
It then follows by Theorem \ref{theorem-2.1} that (\ref{NP-A-1}) has a bounded smooth solution $v_\varepsilon$ satisfying
\begin{equation}\label{4.10}
\|\partial_x^\alpha v_\varepsilon\|_\infty \le C_\alpha\, \varepsilon^{1-|\alpha|}
\quad \text{ for any } |\alpha|\ge 1.
\end{equation}
In particular, $\|\nabla v_\varepsilon\|_\infty\le C$.
In view of Theorem \ref{theorem-r}, we also obtain the estimate
\begin{equation}\label{refined-main-1}
|\nabla v_\varepsilon (x)|\le \frac{C\, \varepsilon}{|x\cdot n +a|}.
\end{equation}
The goal of this section is prove the following.
\begin{theorem}\label{theorem-4.1}
Let $u_\varepsilon$ be a solution of (\ref{NP-M}) and $v_\varepsilon$ a solution of (\ref{NP-A-1}),
constructed above.
Let $\varepsilon \le r\le \sqrt{\varepsilon}$.
Then, for any $\sigma\in (0,1)$,
\begin{equation}\label{4.1-0}
\| \nabla (u_\varepsilon -v_\varepsilon)\|_{L^\infty(B(x_0, r)\cap\Omega)}
\le C \sqrt{\varepsilon} \big\{ 1+|\ln \varepsilon |\big\}
+C \varepsilon^{-1 -\sigma } r^{2+\sigma},
\end{equation}
where $C$ depends on $d$, $m$, $\mu$, $\sigma$, $\Omega$, $\|A\|_{C^k(\mathbb{T}^d)}$ and
$\| g\|_{C^k(\mathbb{T}^d)}$ for some $k=k(d)>1$.
\end{theorem}
Let $N_\varepsilon (x, y)$ denote the matrix of Neumann functions for the operator $\mathcal{L}_\varepsilon$ in $\Omega$.
Since $\Omega\subset \mathbb{H}_{n}^d(a)$ and $\mathcal{L}_\varepsilon (u_\varepsilon -v_\varepsilon)=0$ in $\Omega$,
we obtain the representation,
\begin{equation}
\aligned
& u_\varepsilon (x)-v_\varepsilon (x) -\big\{ u_\varepsilon (z) -v_\varepsilon (z)\big\}\\
&=\int_{\partial\Omega}
\Big\{ N_\varepsilon (x, y)- N_\varepsilon (z, y) \Big\}
\left\{ \frac{\partial u_\varepsilon}{\partial\nu_\varepsilon}
-\frac{\partial v_\varepsilon}{\partial\nu_\varepsilon} \right\} \, d\sigma (y)
\endaligned
\end{equation}
for any $x, z\in \Omega$.
Fix a cut-off function $\eta=\eta_\varepsilon \in C_0^\infty(B(x_0, 5\sqrt{\varepsilon}))$ such that $0\le \eta\le 1$,
$\eta=1$ in $B(x_0, 4 \sqrt{\varepsilon})$ and $|\nabla \eta|\le C \varepsilon^{-1/2}$.
Let
\begin{equation}\label{I}
I(x, z)=\int_{\partial\Omega}
\eta (y) \Big\{ N_\varepsilon (x, y)- N_\varepsilon (z, y) \Big\}
\left\{ \frac{\partial u_\varepsilon}{\partial\nu_\varepsilon}
-\frac{\partial v_\varepsilon}{\partial\nu_\varepsilon} \right\} \, d\sigma (y),
\end{equation}
\begin{equation}\label{J}
J(x, z)=\int_{\partial\Omega}
(1-\eta (y)) \Big\{ N_\varepsilon (x, y)- N_\varepsilon (z, y) \Big\}
\left\{ \frac{\partial u_\varepsilon}{\partial\nu_\varepsilon}
-\frac{\partial v_\varepsilon}{\partial\nu_\varepsilon} \right\} \, d\sigma (y).
\end{equation}
We begin with the estimate of $I(x, z)$ in (\ref{I}).
Here it is essential to take advantage of the fact that the Neumann data of $v_\varepsilon$
agrees with the Neumann data of $u_\varepsilon$ at $x_0$.
Furthermore, to fully utilize the decay estimates for the derivatives of Neumann functions,
we need to transfer the derivative from $\partial (u_\varepsilon -v_\varepsilon)/\partial \nu_\varepsilon$
to the Neumann functions.
For $x\in \mathbb{H}_{n}^d (a)$, we use
\begin{equation}\label{hat}
\widehat{x}=x- ((x-x_0)\cdot n)) n \in \partial \mathbb{H}_n^d(a)
\end{equation}
to denote its projection onto the tangent plane of $\partial\Omega$ at $x_0$.
Observe that if $x\in \partial\Omega$, then $|x-\widehat{x}|\le C|x-x_0|^2$.
\begin{lemma}\label{lemma-4.2}
Suppose that $x\in\partial\Omega$ and $|x-x_0|\le c_0$. Then
\begin{equation}\label{4.2-0}
\aligned
& n(x)\cdot A^\varepsilon(x)\nabla v_\varepsilon (x)
-n(x_0) \cdot A^\varepsilon (\widehat{x})\nabla v_\varepsilon (\widehat{x})=\\
&
T_{i\ell} (x)\cdot
\nabla_x
\left\{ \int_0^1 a^\varepsilon_{ij}(x) \frac{\partial v_\varepsilon}{\partial x_j}
\big( sx +(1-s)\widehat{x}\big) \, ds \, ((x-x_0)\cdot n) n_\ell \right\}
+ R(x),
\endaligned
\end{equation}
where $T_{i\ell}(x)=n_i (x) e_\ell -n_\ell (x) e_i$, $n=n(x_0)=(n_1, \dots, n_d)$, and
\begin{equation}\label{4.2-1}
|R(x)|\le C \big\{ |x-x_0| +\varepsilon^{-1} |x-x_0|^3 \big\}.
\end{equation}
\end{lemma}
\begin{proof}
In view of (\ref{4.10}) we have $\| \nabla v_\varepsilon\|_\infty \le C$ and
$\|\nabla^2 v_\varepsilon\|_\infty \le C \varepsilon^{-1}$.
It follows that
$$
\aligned
& n(x)\cdot A^\varepsilon(x)\nabla v_\varepsilon (x) -n(x_0 )\cdot A^\varepsilon(\widehat{x})\nabla v_\varepsilon (\widehat{x})\\
&=n(x_0)\cdot A^\varepsilon(x)\nabla v_\varepsilon (x) -n(x_0)\cdot A^\varepsilon(\widehat{x})\nabla v_\varepsilon (\widehat{x}) +
O(|x-x_0|)\\
&=n_i(x_0) \int_0^1 \frac{\partial}{\partial s}
\left\{ a_{ij}^\varepsilon \frac{\partial v_\varepsilon}{\partial x_j}
\big(sx +(1-s)\widehat{x}\big) \right\} ds + O(|x-x_0|)\\
& =n_i(x_0) \int_0^1 \frac{\partial}{\partial x_\ell}
\left( a_{ij}^\varepsilon \frac{\partial v_\varepsilon}{\partial x_j} \right)
\big( sx +(1-s)\widehat{x}\big) ((x-x_0)\cdot n) n_\ell (x_0)\, ds
+ O(|x-x_0|)\\
&= \int_0^1 \left(n_i(x_0)\frac{\partial}{\partial x_\ell}
-n_\ell (x_0) \frac{\partial}{\partial x_i} \right)
\left( a_{ij}^\varepsilon \frac{\partial v_\varepsilon}{\partial x_j} \right)
\big( sx +(1-s)\widehat{x}\big) ((x-x_0)\cdot n) n_\ell (x_0)\, ds \\
& \qquad\qquad
+ O(|x-x_0|),
\endaligned
$$
where we have used the equation $\mathcal{L}_\varepsilon (v_\varepsilon)=0$ in the last step.
Using the observation that
$$
\aligned
& \left(n_i(x_0)\frac{\partial}{\partial x_\ell}
-n_\ell (x_0) \frac{\partial}{\partial x_i} \right)
\Big( F(sx +(1-s)\widehat{x})\Big)\\
&\qquad
=\left(n_i(x_0)\frac{\partial}{\partial x_\ell}
-n_\ell (x_0) \frac{\partial}{\partial x_i} \right)
F(sx +(1-s)\widehat{x}),
\endaligned
$$
we then obtain
$$
\aligned
& n(x)\cdot A^\varepsilon(x)\nabla v_\varepsilon (x) -n(x_0 )\cdot A^\varepsilon(\widehat{x})\nabla v_\varepsilon (\widehat{x})\\
&= \left(n_i(x_0)\frac{\partial}{\partial x_\ell}
-n_\ell (x_0) \frac{\partial}{\partial x_i} \right)\left( \int_0^1
\left( a_{ij}^\varepsilon \frac{\partial v_\varepsilon}{\partial x_j} \right)
\big( sx +(1-s)\widehat{x}\big) ((x-x_0)\cdot n) n_\ell (x_0)\, ds \right)\\
& \qquad\qquad
+ O(|x-x_0|)\\
&= \left(n_i(x)\frac{\partial}{\partial x_\ell}
-n_\ell (x) \frac{\partial}{\partial x_i} \right)\left( \int_0^1
\left( a_{ij}^\varepsilon \frac{\partial v_\varepsilon}{\partial x_j} \right)
\big( sx +(1-s)\widehat{x}\big) ((x-x_0)\cdot n) n_\ell (x_0)\, ds \right)\\
& \qquad\qquad
+ O(|x-x_0|) +O(\varepsilon^{-1} |x-x_0|^3),
\endaligned
$$
where we have used the fact that $|(x-x_0)\cdot n| \le C |x-x_0|^2$
as well as the estimate $|\nabla (A^\varepsilon \nabla v_\varepsilon)|\le C \varepsilon^{-1}$
for the last step.
\end{proof}
Lemma \ref{lemma-4.2} allows us to carry out an integration by parts on the boundary
for $I(x, z)$.
\begin{lemma}\label{lemma-4.3}
Let $I(x, z)$ be given by (\ref{I}). Suppose that $x, z\in B(x_0, 3r)\cap \Omega$
for some ${\varepsilon}\le r\le \sqrt{\varepsilon}$ and $|x-z|\le (1/2)\delta (x)$, where $\delta (x)=dist(x, \partial\Omega)$.
Then
\begin{equation}\label{4.3-0}
|I(x, z)|\le C\, r\sqrt{\varepsilon}.
\end{equation}
\end{lemma}
\begin{proof}
Let $y\in \partial\Omega$ and $|y-x_0|\le 5 \sqrt{\varepsilon}$.
Using the Neumann conditions for $u_\varepsilon$, $v_\varepsilon$ and Lemma \ref{lemma-4.2},
we see that
$$
\aligned
& \frac{\partial u_\varepsilon}{\partial \nu_\varepsilon} (y) -\frac{\partial v_\varepsilon}{\partial \nu_\varepsilon} (y)\\
&= T(y)\cdot \nabla g (y/\varepsilon)-T(x_0)\cdot \nabla g (\widehat{y}/\varepsilon)
+n(x_0)\cdot A^\varepsilon (\widehat{y}) \nabla v_\varepsilon (\widehat{y})
- n(y)\cdot A^\varepsilon (y) \nabla v_\varepsilon (y)\\
&=T(y)\cdot \nabla_y \Big\{ \varepsilon g(y/\varepsilon) -\varepsilon g(\widehat{y}/\varepsilon) \Big\}
-T_{i\ell} (y) \cdot \nabla_y \big\{ f_{i \ell} (y) \big\}
+ O(|y-x_0|),
\endaligned
$$
where
$$
f_{i\ell} (y)
=\int_0^1 a^\varepsilon_{ij}(y) \frac{\partial v_\varepsilon}{\partial y_j}
\big( sy +(1-s)\widehat{y}\big) \, ds \, ((y-x_0)\cdot n) n_\ell
$$
is given by Lemma \ref{lemma-4.2}. We have also used the observation,
$$
T(x_0)\cdot \nabla_y \big\{ \varepsilon g(\widehat{y}/\varepsilon)\big\}
=T(x_0) \cdot \nabla g (\widehat{y}/\varepsilon),
$$
in the computation above.
This, together with (\ref{parts}), gives
$$
I(x, z)=I_1 +I_2 +I_3,
$$
where
$$
\aligned
&I_1=-\int_{\partial\Omega}
T(y) \cdot\nabla_y \Big\{ \eta(y) \Big(N_\varepsilon x, y)-N_\varepsilon (z,y)\Big) \Big\}
\Big\{ \varepsilon g(y/\varepsilon) - \varepsilon g(\widehat{y}/\varepsilon) \Big\}\, d\sigma (y),\\
& I_2 = \int_{\partial\Omega}
T_{i\ell} (y) \cdot\nabla_y \Big\{ \eta(y) \Big(N_\varepsilon x, y)-N_\varepsilon (z,y)\Big) \Big\}
f_{i\ell} (y)\, d\sigma (y),\\
& |I_3|
\le C \int_{\partial\Omega}
|\eta(y)| |N_\varepsilon (x, y)-N_\varepsilon (z, y)| |y-x_0|\, d\sigma (y).
\endaligned
$$
Since $|x-z|\le (1/2)\delta (x)\le (1/2) |y-x|$ for any $y\in \partial\Omega$,
by (\ref{N-estimate-1}), we obtain
\begin{equation}\label{4.3-10}
\aligned
|I_1| &\le C \int_{\partial\Omega}
|\nabla_y \big\{ \eta(y) \big(N_\varepsilon (x, y)-N_\varepsilon (z,y)\big) \big\}| |y-x_0|^2\,d\sigma (y)\\
&\le C \sqrt{\varepsilon} \delta(x) \int_{4\sqrt{\varepsilon}\le |y-x_0|\le 5\sqrt{\varepsilon}} \frac{d\sigma(y)}{|y-x|^{d-1}}
+ C \, \delta (x) \int_{|y-x_0|\le C \sqrt{\varepsilon}}
\frac{ |y-x_0|^2}{|y-x|^d} \, d\sigma (y),
\endaligned
\end{equation}
where we have used the fact $|y-\widehat{y}|\le C |y-x_0|^2$ and $|\nabla\eta(y)|\le C \varepsilon^{-1/2}$.
For the first term in the RHS of (\ref{4.3-10}), we note that
if $|y-x_0|\ge 4\sqrt{\varepsilon}$, then
$$
|y-x|\ge |y-x_0|-|x-x_0|\ge 4\sqrt{\varepsilon} -3r\ge \sqrt{\varepsilon}.
$$
For the second term, we use $|y-x_0|\le |y-x| +r$.
This leads to
$$
\aligned
|I_1| & \le C \sqrt{\varepsilon} \, \delta(x) +C \delta (x) \int_{|y-x_0|\le C\sqrt{\varepsilon}} \frac{d\sigma (y)}{|y-x|^{d-2}}
+Cr^2 \delta (x)\int_{\partial\Omega} \frac{d\sigma (y)}{|y-x|^{d}}\\
&\le C \sqrt{\varepsilon} \, \delta(x) + C r^2\\
&\le C r\sqrt{\varepsilon}.
\endaligned
$$
Since $|f_{i\ell}|\le C |y-x_0|^2$, the estimate of $I_2$ is the exactly same as that of $I_1$.
Finally, to handle $I_3$, we use (\ref{N-estimate-1}) as well as $|y-x_0|\le |y-x| +r$
again to obtain
$$
\aligned
|I_3|
&\le C \int_{|y-x_0|\le C \sqrt{\varepsilon}}
\frac{|x-z|}{|y-x|^{d-1}}
\big\{ |y-x| + r \big\} \, d\sigma (y)\\
&\le C\, r \sqrt{\varepsilon}.
\endaligned
$$
This completes the proof.
\end{proof}
To estimate $J(x,z)$ in (\ref{J}), we split it as $J(x,z)=J_1 -J_2$,
where
\begin{equation}\label{J-1-2}
\aligned
J_1 (x, z) & =\int_{\partial\Omega}
(1-\eta (y)) \Big\{ N_\varepsilon (x, y)- N_\varepsilon (z, y) \Big\}
\frac{\partial u_\varepsilon}{\partial\nu_\varepsilon} \, d\sigma (y),\\
J_2 (x, z) & =\int_{\partial\Omega}
(1-\eta (y)) \Big\{ N_\varepsilon (x, y)- N_\varepsilon (z, y) \Big\}
\frac{\partial v_\varepsilon}{\partial\nu_\varepsilon} \, d\sigma (y).
\endaligned
\end{equation}
\begin{lemma}\label{lemma-4.4}
Let $J_1 (x, z)$ be given by (\ref{J-1-2}). Suppose that $x, z\in B(x_0,3 r)\cap \Omega$
for some ${\varepsilon}\le r\le \sqrt{\varepsilon}$ and $|x-z|\le (1/2)\delta (x)$, where $\delta (x)=dist(x, \partial\Omega)$.
Then
\begin{equation}\label{4.4-0}
|J_1 (x, z)|\le C\, r\sqrt{\varepsilon}.
\end{equation}
\end{lemma}
\begin{proof}
Using the Neumann condition for $u_\varepsilon$ and (\ref{parts}), we see that
$$
J_1(x, z)=-\int_{\partial\Omega}
T\cdot \nabla_y \Big\{ (1-\eta(y)) \big( N_\varepsilon (x, y)-N_\varepsilon (z, y) \big) \Big\} \varepsilon g(y/\varepsilon)\, d\sigma (y).
$$
It follows that
$$
\aligned
|J_1(x, z)| & \le
C \varepsilon \| g\|_\infty \int_{\partial\Omega}
|\nabla_y \Big\{ (1-\eta(y)) \big( N_\varepsilon (x, y)-N_\varepsilon (z, y) \big) \Big\}|\, d\sigma (y)\\
&\le C \sqrt{\varepsilon} |x-z|
+ C\varepsilon |x-z| \int_{ |y-x_0|\ge 5\sqrt{\varepsilon}}
\frac{d\sigma (y)}{|x-y|^d}\\
& \le C\, r\sqrt{\varepsilon},
\endaligned
$$
where we have used (\ref{N-estimate-1}) for the second inequality.
\end{proof}
It remains to estimate $J_2 (x, z)$.
\begin{lemma}\label{lemma-4.6}
Let $J_2 (x, z)$ be given by (\ref{J-1-2}). Suppose that $x, z\in B(x_0,3 r)\cap \Omega$
for some ${\varepsilon}\le r\le \sqrt{\varepsilon}$ and $|x-z|\le (1/2)\delta (x)$, where $\delta (x)=dist(x, \partial\Omega)$.
Then
\begin{equation}\label{4.6-0}
|J_2 (x, z)|\le C\, r\sqrt{\varepsilon}\big\{ 1+|\ln \varepsilon|\big\}.
\end{equation}
\end{lemma}
\begin{proof}
It follows by the divergence theorem that
$$
\aligned
J_2(x,z)
&=-\int_{\Omega} (1-\eta (y))\nabla_y
\big\{ N_\varepsilon (x, y) -N_\varepsilon (z, y) \big\} \cdot A (y/\varepsilon) \nabla v_\varepsilon (y)\, dy\\
& \qquad -\int_{\Omega} \big\{ N_\varepsilon (x, y)-N_\varepsilon (z, y) \big\} \nabla_y (1-\eta (y)) \cdot A (y/\varepsilon)
\nabla v_\varepsilon (y)\, dy\\
&= \int_{\Omega}
A^{*} (y/\varepsilon) \nabla_y \big\{ N_\varepsilon (x, y) -N_\varepsilon (z, y) \big\} \cdot \nabla_y (1-\eta (y)) \, (v_\varepsilon (y)-E)\, dy\\
& \qquad -\int_{\Omega} \big\{ N_\varepsilon (x, y)-N_\varepsilon (z, y) \big\} \nabla_y (1-\eta (y)) \cdot A (y/\varepsilon)
\nabla v_\varepsilon (y)\, dy,
\endaligned
$$
where $E\in \mathbb{R}^m$ is a constant to be chosen. Here
we have used $\mathcal{L}_\varepsilon (v_\varepsilon)=0$ in $\Omega$ for the first equality and
$$
\aligned
\mathcal{L}_\varepsilon^* \big\{ N_\varepsilon (x, \cdot) -N_\varepsilon (z, \cdot) \big\} &=0 \quad \text{ in } \Omega \setminus B(x_0, 3\sqrt{\varepsilon}),\\
\frac{\partial}{\partial \nu_\varepsilon^*} \big\{ N_\varepsilon (x, \cdot) -N_\varepsilon (z, \cdot) \big\} &=0 \quad \text{ on } \partial\Omega
\endaligned
$$
for the second.
As before, we apply the estimates in (\ref{N-estimate-1}) to obtain
\begin{equation}\label{4.6-10}
\aligned
|J_2 (x, z)|
&\le \frac{C |x-z|}{(\sqrt{\varepsilon})^{d+1}} \int_{B(x_0, 5\sqrt{\varepsilon})\cap \mathbb{H}_n^d(a)}
|v_\varepsilon -E|
+ \frac{C |x-z|}{(\sqrt{\varepsilon})^{d}} \int_{B(x_0, 5\sqrt{\varepsilon})\cap \mathbb{H}_n^d(a)}
| \nabla v_\varepsilon|\\
& \le \frac{C r}{(\sqrt{\varepsilon})^{d}} \int_{B(x_0, 5\sqrt{\varepsilon})\cap \mathbb{H}_n^d(a)}
| \nabla v_\varepsilon|,
\endaligned
\end{equation}
where we have chosen $E$ to be the average of $v_\varepsilon$ over $B(x_0, 5\sqrt{\varepsilon})\cap \mathbb{H}_n^d(a)$
and used a Poincar\'e type inequality for the last step.
Finally, to estimate the integral in the RHS of (\ref{4.6-10}), we split the region
$B(x_0, 5\sqrt{\varepsilon})\cap \mathbb{H}_n^d (a)$ into two parts.
If $|x\cdot n+a|\le \varepsilon$, we use the estimate $\|\nabla v_\varepsilon\|_\infty \le C$.
If $|x\cdot n+a|\ge \varepsilon$, we apply the refined estimate (\ref{refined-main-1}).
This yields that
$$
|J_2 (x, z)|\le C r \sqrt{\varepsilon} \big\{ 1+|\ln \varepsilon|\big\},
$$
which completes the proof.
\end{proof}
We are now ready to give the proof of Theorem \ref{theorem-4.1}.
\begin{proof}[\bf Proof of Theorem \ref{theorem-4.1}]
Let $\varepsilon\le r\le \sqrt{\varepsilon}$.
In view of Lemmas \ref{lemma-4.3}, \ref{lemma-4.4} and \ref{lemma-4.6},
we have proved that if $x, z\in \Omega\cap B(x_0, 3r)$
and $|x-z|<(1/2)\delta(x)$, where $\delta(x)= \text{dist}(x, \partial\Omega)$, then
\begin{equation}\label{4.7-1}
|u_\varepsilon (x)-v_\varepsilon(x) - \{ u_\varepsilon (z)-v_\varepsilon (z)\}|
\le C r \sqrt{\varepsilon} \big\{ 1+|\ln \varepsilon |\big\}.
\end{equation}
Since $\mathcal{L}_\varepsilon (u_\varepsilon -v_\varepsilon)=0$ in $\Omega$,
by the interior Lipschitz estimate for $\mathcal{L}_\varepsilon$ \cite{AL-1987},
it follows that for any $x\in B(x_0, 2r)$,
\begin{equation}\label{4.7-2}
|\nabla u_\varepsilon (x) -\nabla v_\varepsilon (x)|\le
C r\sqrt{\varepsilon} \big\{ 1+|\ln \varepsilon |\big\} \big[\delta(x)\big]^{-1}.
\end{equation}
Thus, if $0<p<1$,
\begin{equation}\label{4.7-3}
\left(-\!\!\!\!\!\!\int_{B(x_0, 2r)\cap\Omega} |\nabla (u_\varepsilon -v_\varepsilon)|^p \right)^{1/p}
\le C \sqrt{\varepsilon} \big\{ 1+|\ln \varepsilon |\big\}.
\end{equation}
Next, we estimate the $C^\sigma(B(x_0, 2r)\cap\partial \Omega)$ norm of
$$
F(y)=\frac{\partial u_\varepsilon}{\partial \nu_\varepsilon} -\frac{\partial v_\varepsilon}{\partial \nu_\varepsilon}.
$$
As in the proof of Lemma \ref{lemma-4.3}, we write
$$
\aligned
F(y) &=\Big\{ T(y)\cdot \nabla g (y/\varepsilon)-T(x_0)\cdot \nabla g (\widehat{y}/\varepsilon)\Big\}\\
& \qquad+\Big\{ n(x_0)\cdot A (\widehat{y}/\varepsilon) \nabla w (\widehat{y}/\varepsilon)
- n(y)\cdot A (y/\varepsilon) \nabla w (y/\varepsilon)\Big\}\\
&=F_1 (y) + F_2(y),
\endaligned
$$
where we have used the fact $v_\varepsilon(x)=\varepsilon w(x/\varepsilon)$ and $w$ is a solution of (\ref{NP-m}).
Using $|y-\widehat{y}|\le C |y-x_0|^2$ and $\| \nabla w\|_\infty +\|\nabla^2 w\|_\infty
\le C$,
it is easy to see that if $y\in B(x_0, 2r)\cap\partial\Omega$,
\begin{equation}\label{4.7-10}
|F_1(y)| +|F_2(y)|
\le C |y-x_0| + C \varepsilon^{-1} |y-x_0|^2\le C\, \varepsilon^{-1} r^2,
\end{equation}
where we also used the assumption $\varepsilon\le r$ for the last step.
By extending $n(y)$ smoothly to a neighborhood of $\partial\Omega$,
we may assume that $F(y)$ is defined in $B(x_0, c_0)\cap \mathbb{H}_n^d$.
A computation shows that
\begin{equation}\label{4.7-11}
|\nabla_y F(y)|\le C \big\{ 1 +\varepsilon^{-1} |y-x_0| +\varepsilon^{-2} |y -x_0|^2 \big\}
\le C \varepsilon^{-2} r^2,
\end{equation}
where we have used the estimate $\|\nabla^3 w\|_\infty\le C$.
By interpolation it follows from (\ref{4.7-10}) and (\ref{4.7-11}) that
\begin{equation}\label{4.7-12}
\| F\|_{C^{0, \sigma}(B(x_0, 2r)\cap\partial\Omega)}
\le C\, (\varepsilon^{-1} r^2)^{1-\sigma} (\varepsilon^{-2} r^2)^{\sigma}
=C \, \varepsilon^{-1-\sigma} r^2
\end{equation}
for any $\sigma \in (0, 1)$.
Finally, since $\mathcal{L_\varepsilon}(u_\varepsilon -v_\varepsilon)=0$
in $\Omega \cap B(x_0, 2r)$,
we apply the boundary Lipschitz estimate for solutions with Neumann data \cite{KLS1, AS}
to obtain
$$
\aligned
& \| \nabla (u_\varepsilon -v_\varepsilon)\|_{L^\infty(\Omega\cap B(x_0, r))}\\
&\le C \left(-\!\!\!\!\!\!\int_{B(x_0, 2r)\cap\Omega} |\nabla (u_\varepsilon -v_\varepsilon)|^p \right)^{1/p}
+ C \|F \|_{L^\infty(B(x_0, 2r)\cap\partial\Omega)}
+ C r^\sigma \| F\|_{C^{0, \sigma} (B(x_0, 2r)\cap \partial\Omega)}\\
&\le C \sqrt{\varepsilon} \big\{ 1+|\ln \varepsilon |\big\}
+C \varepsilon^{-1} r^2
+C \varepsilon^{-1 -\sigma } r^{2+\sigma}\\
&\le C \sqrt{\varepsilon} \big\{ 1+|\ln \varepsilon |\big\}
+C \varepsilon^{-1 -\sigma } r^{2+\sigma}.\\
\endaligned
$$
This completes the proof.
\end{proof}
Recall that the function $\psi_{\varepsilon, k}^{*\beta}
=(\psi_{\varepsilon, k}^{* 1\beta} (y), \dots, \psi_{\varepsilon, k}^{* m \beta}(y)$ in (\ref{psi}) is a solution of the Neumann
problem
\begin{equation}\label{NP-E}
\left\{
\aligned
&\mathcal{L}_\varepsilon^* (\psi_{\varepsilon, k}^{*\beta})=0 &\quad & \text{ in }\Omega,\\
& \frac{\partial}{\partial \nu_\varepsilon^*}
\big\{ \psi_{\varepsilon, k}^{*\beta} \big\} = - \frac12 (n_i(x) e_\ell -n_\ell (x) e_i)\cdot
\nabla f_{\ell i k}^\beta (x/\varepsilon) &\quad &\text{ on }\partial\Omega,
\endaligned
\right.
\end{equation}
where the 1-periodic functions $f_{\ell i k}^\beta\in C^\infty(\mathbb{T}^d; \mathbb{R}^m)$ are given by (\ref{p102}).
For each $x_0\in \partial\Omega$ fixed and satisfying (\ref{D-condition}),
in view of Theorem \ref{theorem-4.1},
we may approximate this function in a small neighborhood of $x_0$
by a solution of
\begin{equation}\label{NP-A}
\left\{
\aligned
&\mathcal{L}_\varepsilon^* (\phi_{\varepsilon, k }^{*\beta, x_0 })=0 &\quad & \text{ in } \mathbb{H}_n^d(a),\\
& \frac{\partial}{\partial \nu_\varepsilon^*}
\big\{ \phi_{\varepsilon, k}^{*\beta, x_0} \big\} = - \frac12 (n_i e_\ell -n_\ell e_i)\cdot
\nabla f_{\ell i k}^\beta (x/\varepsilon) &\quad &\text{ on }\partial\mathbb{H}_n^d(a),
\endaligned
\right.
\end{equation}
where $n=n(x_0)$ and $\mathbb{H}_n^d(a)$ is the tangent plane of $\partial\Omega$ at $x_0$.
Recall that by a change of variables, a solution of (\ref{NP-A}) is given by
\begin{equation}\label{E-1}
\phi_{\varepsilon, k}^{\alpha\beta, x_0}
(x)=\varepsilon V_k^{*\alpha\beta, n}\left(\frac{x-(x\cdot n+a)n}{\varepsilon}, -\frac{x\cdot n +a}{\varepsilon} \right),
\end{equation}
where
$V^*=V_{k}^{*\beta, n} (\theta, t)=(V_k^{* 1\beta, n} (\theta, t), \dots, V_k^{*m\beta, n}(\theta, t))$ is the smooth solution of
\begin{equation}\label{E-2}
\left\{
\aligned
& \left( \begin{array} {c}N^T \nabla_\theta \\ \partial_t \end{array}\right)
\cdot M^TA^*(\theta -tn)M \left( \begin{array} {c}N^T \nabla_\theta \\ \partial_t \end{array} \right)V^*
=0 &\quad & \text{in } \mathbb{T}^d \times \mathbb{R}_+,\\
&-e_{d+1} \cdot M^T A^*(\theta)M \left( \begin{array} {c}N^T \nabla_\theta \\ \partial_t \end{array} \right)V^*
=-\frac12 (n_i e_\ell -n_\ell e_i)\cdot \nabla_\theta f_{lik}^\beta & \quad & \text{on } \mathbb{T}^d \times \{ 0\},
\endaligned
\right.
\end{equation}
given by Remark \ref{remark-2.1}. As a result, we may deduce the following from Theorem \ref{theorem-4.1}
\begin{theorem}\label{main-lemma}
Let $\varepsilon\le r\le \sqrt{\varepsilon}$ and $\sigma\in (0,1/2)$.
Then for any $x\in B(x_0, r)\cap\Omega$,
\begin{equation}\label{main-estimate-4}
\aligned
& \Big| \nabla \left(\Psi_{\varepsilon, k}^{*\alpha\beta} (x) - P_k^{\alpha\beta} (x)
-\varepsilon \chi_k^{*\alpha\beta} \left(\frac{x}{\varepsilon}\right)
-\varepsilon V_k^{*\alpha\beta, n}\left(\frac{x-(x\cdot n+a)n}{\varepsilon}, -\frac{x\cdot n +a}{\varepsilon} \right)\right) \Big|\\
& \le C\sqrt{\varepsilon} \big\{ 1+|\ln \varepsilon|\big\}
+C \varepsilon^{-1-\sigma} r^{2+\sigma},
\endaligned
\end{equation}
where $C$ depends only on $d$, $m$, $\sigma$, $\mu$, $\Omega$ and
$\|A\|_{C^k(\mathbb{T}^d)}$ for some $k=k(d)>1$.
\end{theorem}
\section{Estimates of the homogenized data}
\setcounter{equation}{0}
Observe that by (\ref{E-1}),
\begin{equation}\label{E-3}
\aligned
T_{ij}(x_0) \cdot \nabla_x \phi_{\varepsilon, k}^{*\alpha\beta, x_0} (x)
&=T(x_0) \cdot (I-n\otimes n, -n) \left( \begin{array}{c} \nabla_\theta \\ \partial_t\end{array}\right)
V_k^{*\alpha\beta, n} \left(\frac{x}{\varepsilon}, 0\right)\\
& =T_{ij}(x_0) \cdot \nabla_\theta V_k^{*\alpha\beta, n} \left(\frac{x}{\varepsilon}, 0\right),
\endaligned
\end{equation}
where $x\in \partial\mathbb{H}_n^d(a)$ and
we have used the fact $T_{ij}(x_0)\cdot n(x_0)=0$.
For $x\in \partial\Omega$, define
\begin{equation}\label{mean}
\aligned
\widetilde{g}_{k}^\beta (x)
&= T_{ij}(x)\cdot\left\langle \nabla_\theta (V_k^{*\alpha\beta,n} +\chi_k^{*\alpha\beta})
(\cdot, 0) g_{ij}^\alpha(x, \cdot)\right\rangle
+(T_{ij}\cdot \nabla_x) x_k \langle g_{ij}^\beta (x, \cdot) \rangle\\
& = T_{ij}(x)\cdot \int_{\mathbb{T}^d} (I-n\otimes n)\nabla_\theta (V_k^{*\alpha\beta, n}+\chi_k^{*\alpha\beta})
(\theta, 0) g_{ij}^\alpha(x, \theta)\, d\theta\\
&\qquad\qquad
+(n_i \delta_{jk}-n_j \delta_{ik}) \int_{\mathbb{T}^d} g_{ij}^\beta (x, \theta) d\theta\\
&=T_{ij}(x)\cdot
\int_{\mathbb{T}^d}
\Big( e_k \delta^{\alpha\beta}
+\nabla_\theta \chi_k^{*\alpha\beta} (\theta)
+\nabla_\theta V_k^{*\alpha\beta} (\theta, 0) \Big) g_{ij}^\alpha (x, \theta)\, d\theta,
\endaligned
\end{equation}
where $n=n(x)$.
Using the estimate $\|(I-n\otimes n)\nabla_\theta V^*\|\le C$ in Proposition \ref{prop-2.5}, we obtain
$\|\widetilde{g}\|_\infty \le C \| g\|_\infty$.
Let
\begin{equation}\label{v}
v^\gamma_\varepsilon (x)
=-\int_{\partial\Omega}
\big( T_{ij}(y)\cdot \nabla_y \big)\Psi_{\varepsilon, k}^{*\alpha\beta} (y)
\cdot \frac{\partial}{\partial y_k} \big\{ N_0^{\gamma \beta} (x, y)\big\} \cdot
g_{ij}^\alpha (y, y/\varepsilon)\, d\sigma (y)
\end{equation}
be the first term in the RHS of (\ref{p-3}).
In the next section we will show that as $\varepsilon \to 0$,
\begin{equation}\label{p-200}
v_\varepsilon^\gamma (x)
\to v_0^\gamma (x)=-\int_{\partial\Omega}
\frac{\partial}{\partial y_k} \big\{ N_0 ^{\gamma \beta} (x, y)\big\}
\widetilde{g}_{k}^\beta (y)\, d\sigma (y).
\end{equation}
Fix $1\le \gamma\le m$ and $1\le k\le d$.
Using
\begin{equation}\label{p-201}
\aligned
& n_in_j\widehat{a}_{ij}^{*\alpha\beta} \frac{\partial}{\partial y_k} \big\{ N_0^{\gamma\beta} (x, y) \big\}\\
&=n_i\widehat{a}_{ij}^{*\alpha\beta}
\Big( (n_j e_k-n_k e_j)\cdot \nabla_y \Big) N_0^{\gamma \beta} (x, y)
+n_k \left(\frac{\partial}{\partial \nu_0^* (y)} \big\{ N^\gamma_0(x, y) \big\}\right)^\alpha,
\endaligned
\end{equation}
where $N_0^\gamma (x, y)=(N_0^{\gamma 1} (x, y), \cdots, N_0^{\gamma m}(x,y))$,
we may write
\begin{equation}\label{p-202}
\frac{\partial}{\partial y_k} \big\{ N_0^{\gamma\beta} (x, y)\big\}
=h^{*\beta\alpha} n_i \widehat{a}^{*\alpha t}_{ij}
\Big( (n_j e_k-n_k e_j)\cdot \nabla_y \Big) N_0^{ \gamma t} (x, y)
- h^{*\beta \gamma } n_k |\partial\Omega|^{-1},
\end{equation}
where $h^*(y)=(h^{*\alpha\beta}(y))$ is the inverse of the $m\times m$ matrix
$(\widehat{a}_{ij}^{*\alpha\beta} n_i n_j)$ and we have used the fact that
the conormal derivative of the matrix of Neumann functions is $-|\partial\Omega|^{-1} I_{m\times m}$.
It follows that
\begin{equation}\label{v-0}
\aligned
v_0^\gamma (x)
= &-\int_{\partial\Omega}
\big[ (n_j e_k-n_k e_j)\cdot \nabla_y\big] N_0^{\gamma t} (x, y)
\cdot h^{*\beta \alpha} n_i \widehat{a}^{*\alpha t}_{ij} \widetilde{g}_k^\beta (y)\, d\sigma (y)\\
&\qquad\qquad
+-\!\!\!\!\!\!\int_{\partial\Omega} h^{* \beta\gamma} n_k (y) \widetilde{g}^\beta_k (y)\, d\sigma (y)\\
=&\int_{\partial\Omega} N_0^{\gamma t} (x,y)
\Big[ (n_j e_k-n_k e_j)\cdot \nabla_y\Big]
\Big( n_i \widehat{a}^{t\alpha}_{ij} h^{\alpha\beta} \widetilde{g}^\beta_k (y)\Big) \, d\sigma (y)
+ \text{constant},
\endaligned
\end{equation}
where $h=(h^*)^*=(h^{\alpha\beta})$ is the inverse of the matrix $(\widehat{a}_{ij}^{\alpha\beta}n_in_j)$.
This shows that $v_0$ is a solution of the following Neumann problem,
\begin{equation}\label{NP-v}
\left\{
\aligned
&\mathcal{L}_0 (v_0)=0 &\quad & \text{ in } \Omega,\\
&\left(\frac{\partial v_0}{\partial \nu_0}\right)^\gamma
=\big( T_{jk}\cdot \nabla\big)
\Big( n_i \widehat{a}_{ij}^{\gamma \alpha} h^{\alpha\beta} \widetilde{g}^\beta_k (y)\Big)
&\quad & \text{ on } \partial\Omega.
\endaligned
\right.
\end{equation}
Thus the homogenized data $\overline{g}_{jk}^\gamma$ in (\ref{NP-h}) is
given by
\begin{equation}\label{h-data-ij}
\overline{g}_{jk}^\gamma
=n_i\widehat{a}_{ij}^{\gamma\alpha} h^{\alpha\beta} \widetilde{g}_k^\beta,
\end{equation}
where $\widetilde{g}_k^\beta$ is given by (\ref{mean})
and $(h^{\alpha\beta})$ is the inverse of the matrix $(\widehat{a}_{ij}^{\alpha\beta}n_in_j)$.
The rest of this section is devoted to the proof of the following.
\begin{theorem}\label{theorem-E}
Let $x, y\in \partial\Omega$ and $|x-y|\le c_0$.
Suppose that $n(x)$ and $n(y)$ satisfy the Diophantine condition (\ref{D-condition})
with constants $\kappa (x)$ and $\kappa (y)$, respectively.
Let $\overline{g}= (\overline{g}_k^\beta)$ be defined by (\ref{h-data-ij}).
Then, for any $\sigma\in (0,1)$,
\begin{equation}\label{E-1-0}
|\overline{g}(x)-\overline{g}(y)|\le \frac{C_\sigma |x-y|}{ \kappa^{1+\sigma}}
\left(\frac{|x-y|}{\kappa} + 1\right) \sup_{z\in \mathbb{T}^d} \| g(\cdot, z)\|_{C^1(\partial\Omega)},
\end{equation}
where $\kappa=\max(\kappa(x), \kappa(y))$ and
$C_\sigma$ depends only on $d$, $m$, $\sigma$, $\mu$,
and $\|A\|_{C^k(\mathbb{T}^d)}$ for some $k=k(d)>1$.
\end{theorem}
Assume that $n, \widetilde{n}\in \mathbb{S}^{n-1}$ satisfy the condition (\ref{D-condition}).
Choose two orthogonal matrices $M_n$, $M_{\widetilde{n}}$ such that $M_n (e_d)=-n$,
$M_{\widetilde{n}} e_d=-\widetilde{n}$ and $|M_n -M_{\widetilde{n}}|\le C |n-\widetilde{n}|$.
Let $N_n$ and $N_{\widetilde{n}}$ denote the $d\times (d-1)$ matrices of the first $d-1$ columns of
$M_n$ and $M_{\widetilde{n}}$, respectively.
Let $V_n^* (\theta, t) $ and $V_{\widetilde{n}}^*(\theta, t)$ be the corresponding solutions of (\ref{E-2}).
We will show that
\begin{equation}\label{E-1-1}
\int_{\mathbb{T}^d} |N_n^T \nabla_\theta \big(V_n^*(\theta, 0) -V_{\widetilde{n}}^*( \theta, 0)\big)|\, d\theta
\le \frac{C_\sigma | n-\widetilde{n}|}{ \kappa^{1+\sigma}}
\left(\frac{|n-\widetilde{n}|}{\kappa} + 1\right),
\end{equation}
where $\kappa>0$ is the constant in the Diophantine condition (\ref{D-condition}) for $\widetilde{n}$.
Using $N_nN_n^T=I -n\otimes n$,
$N_{\widetilde{n}} N_{\widetilde{n}}^T=I -\widetilde{n}\otimes \widetilde{n}$, and the estimate
$|\nabla_\theta V^*_{\widetilde{n}} |\le C\kappa^{-1}$ from Proposition \ref{prop-2.5},
it is not hard to see that (\ref{E-1-0}) follows from (\ref{E-1-1}).
Furthermore, let
\begin{equation}\label{W}
W(\theta, t)= V_n^*(\theta, t)-V_{\widetilde{n}}^*(\theta, t).
\end{equation}
By Sobolev imbedding it suffices to show that
\begin{equation}\label{E-1-1-0}
\int_0^1 \int_{\mathbb{T}^d}
\Big\{ |N_n^T \nabla_\theta W|^2
+| \nabla_\theta \partial_t W|^2 \Big\} d\theta dt
\le C_\sigma \left\{ \frac{|n-\widetilde{n}|^2}{\kappa^{2+\sigma}}
+\frac{|n-\widetilde{n}|^4}{\kappa^{4+\sigma}}\right\},
\end{equation}
for any $\sigma \in (0,1)$.
Let
$$
B^*_n(\theta, t)=M_n^T A^*(\theta-tn) M_n\quad \text{ and } \quad
B^*_{\widetilde{n}}(\theta, t)=M_{\widetilde{n}}^T A^*(\theta-t\widetilde{n}) M_{\widetilde{n}}.
$$
To prove (\ref{E-1-1-0}), as in the case of Dirichlet condition \cite{Masmoudi-2012, Armstrong-Prange-2016}, we first note that
$W$ is a solution of the Neumann problem,
\begin{equation}\label{E-1-2}
\left\{
\aligned
& -\left( \begin{array} {c}N_{n}^T \nabla_\theta \\ \partial_t \end{array}\right)
\cdot B^*_n \left( \begin{array} {c}N_{n}^T \nabla_\theta \\ \partial_t \end{array} \right) W
= \left( \begin{array} {c}N_{n}^T \nabla_\theta \\ \partial_t \end{array} \right) G
+H
&\quad & \text{ in } \mathbb{T}^d \times \mathbb{R}_+,\\
&-e_{d+1} \cdot B_n^*\left( \begin{array} {c}N_{n}^T \nabla_\theta \\ \partial_t \end{array} \right) W
=h + e_{d+1} \cdot G
& \quad & \text{ on } \mathbb{T}^d \times \{ 0\},
\endaligned
\right.
\end{equation}
where $G=G_1 +G_2 +G_3$ , $H$, and $h$ are given by
\begin{equation}
\aligned
G_1 &= \big(B_n -B_{\widetilde{n}} \big)
\left( \begin{array} {c}(N_{n}-N_{\widetilde{n}})^T \nabla_\theta \\ 0 \end{array} \right)V_{\widetilde{n}},\\
G_2&=
\big(B_n-B_{\widetilde{n}} \big)
\left( \begin{array} {c}N_{\widetilde{n}}^T \nabla_\theta \\ \partial_t \end{array} \right)V_{\widetilde{n}},\\
G_3&=B_{\widetilde{n}}
\left( \begin{array} {c}(N_{n}-N_{\widetilde{n}})^T \nabla_\theta \\ 0 \end{array} \right)V_{\widetilde{n}},\\
H&=\left( \begin{array} {c}(N_{n}-N_{\widetilde{n}})^T \nabla_\theta \\ 0 \end{array} \right)
B_{\widetilde{n}}
\left( \begin{array} {c}N_{\widetilde{n}}^T \nabla_\theta \\ \partial_t \end{array} \right)V_{\widetilde{n}},\\
h & =-\frac12 \big[ (n_i -\widetilde{n}_i)e_\ell -(n_\ell -\widetilde{n}_\ell)e_i\big] \cdot \nabla_\theta f_{i\ell}.
\endaligned
\end{equation}
Note that $|\partial_t^k \partial_\theta^\alpha (B_n -B_{\widetilde{n}} )|\le
C (1+t)|n-\widetilde{n}|$. This, together with Proposition \ref{prop-2.5}, gives
\begin{equation}\label{G-1}
|\partial_t^k \partial_\theta^\alpha G_1(\theta, t)|
\le \frac{C( t +1) |n-\widetilde{n}|^2}{\kappa (1+\kappa t)^\ell},
\end{equation}
\begin{equation}\label{G-2}
|\partial_t^k \partial_\theta^\alpha G_2(\theta, t)|
\le \frac{C(t+1) |n-\widetilde{n}|}{(1+\kappa t)^\ell},
\end{equation}
\begin{equation}\label{G-3}
|\partial_t^k \partial_\theta^\alpha G_3(\theta, t)|
\le \frac{C |n-\widetilde{n}|}{ (1+\kappa t)^\ell},
\end{equation}
\begin{equation}\label{H-1}
|\partial_t^k \partial_\theta^\alpha H (\theta, t)|
\le \frac{C |n-\widetilde{n}|}{ (1+\kappa t)^\ell},
\end{equation}
for any $\alpha$, $k$, $\ell$,
where $C$ depends on $d$, $m$, $\alpha$, $k$, $\ell$ and $A$.
To deal with the growth factor $t+1$ in (\ref{G-1}), (\ref{G-2}) as well as the term $H$,
we rely on the following weighted estimates.
\begin{lemma}\label{lemma-E1}
Suppose that $n\in \mathbb{S}^{n-1}$ satisfies the Diophantine condition (\ref{D-condition}).
Let $U$ be a smooth solution of
\begin{equation}\label{E-1-100}
\left\{
\aligned
& -\left( \begin{array} {c}N_{n}^T \nabla_\theta \\ \partial_t \end{array}\right)
\cdot B^*_n \left( \begin{array} {c}N_{n}^T \nabla_\theta \\ \partial_t \end{array} \right) U
= \left( \begin{array} {c}N_{n}^T \nabla_\theta \\ \partial_t \end{array} \right) F
&\quad & \text{ in } \mathbb{T}^d \times \mathbb{R}_+,\\
&-e_{d+1} \cdot B_n^*\left( \begin{array} {c}N_{n}^T \nabla_\theta \\ \partial_t \end{array} \right) U
=e_{d+1} \cdot F
& \quad & \text{ on } \mathbb{T}^d \times \{ 0\}.
\endaligned
\right.
\end{equation}
Assume that
\begin{equation}\label{E-1-A}
(1+t)\| \nabla_{\theta, t} U(\cdot, t)\|_{L^\infty(\mathbb{T}^d)}
+(1+t)\| F(\cdot, t)\|_{L^\infty(\mathbb{T}^d)} <\infty.
\end{equation}
Then, for any $-1<\alpha<0$,
\begin{equation}\label{E-1-101}
\int_0^\infty \int_{\mathbb{T}^d}
\Big\{ |N^T_n \nabla_\theta U|^2 +|\partial_t U|^2 \Big\}\, t^\alpha \, d\theta dt
\le C_\alpha \int_0^\infty \int_{\mathbb{T}^d} |F|^2\, t^\alpha\, d\theta dt,
\end{equation}
where $C_\alpha$ depends only on $d$, $m$, $\mu$, $\alpha$ as well as some H\"older norm of $A$.
\end{lemma}
\begin{proof}
We will reduce the weighted estimate (\ref{E-1-101})
to the analogous estimates we proved in Section 4 in a half-space.
Let
$$
u(x)=U(x-(x\cdot n)n , -x\cdot n) \quad \text{ and } \quad f(x)=F(x-(x\cdot n)n, -x\cdot n).
$$
Then $u$ is a solution of the Neumann problem,
$$
-\text{\rm div}(A(x)\nabla u)=\text{div}(f) \quad \text{ in } \mathbb{H}_n^d(0)
\quad \text{ and } \quad n\cdot A(x)\nabla u=-n\cdot f \quad \text{ on } \partial \mathbb{H}_n^d(0).
$$
It follows from the estimate (\ref{w-101}) that
\begin{equation}\label{E-1-102}
\aligned
& \frac{1}{R^{d-1}}\int_{\Omega_R} |N_n^T \nabla_\theta U (x-(x\cdot n) n, -x\cdot n) |^2 \, |x\cdot n|^\alpha \, dx\\
&\qquad +\frac{1}{R^{d-1}}\int_{\Omega_R} |\partial_t U (x-(x\cdot n) n, -x\cdot n) |^2 \, |x\cdot n|^\alpha \, dx\\
& \le C R^{\alpha+1-d} \int_{\Omega_{2R}}|\nabla u|^2\, dx
+\frac{C}{R^{d-1}} \int_{\Omega_{2R} }|F (x-(x\cdot n) n, -x\cdot n) |^2 \, |x\cdot n|^\alpha \, dx,
\endaligned
\end{equation}
where
$$
\Omega_R=\big\{ x\in \mathbb{H}_n^d (0): |x-(x\cdot n)n|\le R \text{ and } |x\cdot n|\le 2R \big\}.
$$
Next, we compute the limit for each term in (\ref{E-1-102}), as $R\to \infty$.
In view of (\ref{E-1-A}) it is clear that the first term in the RHS of (\ref{E-1-102})
goes to zero.
For the second term in the RHS of (\ref{E-1-102}), we write it as
\begin{equation}\label{E-1-1020}
C\int_0^{2R} t^\alpha \left\{ \frac{1}{R^{d-1}}\int_{\substack{ x\cdot n=t\\ |x+tn| <R}}
|F(x+tn, t)|^2 \, d\sigma (x) \right\} dt.
\end{equation}
Since $F(\theta, t)$ is 1-periodic in $\theta$ and $n$ satisfies the Diophantine condition,
\begin{equation}\label{E-1-103}
\frac{1}{R^{d-1}}\int_{\substack{ x\cdot n=t\\ |x+tn| <R}}
|F(x+tn, t)|^2 \, d\sigma (x)
\to C_d \int_{\mathbb{T}^d} |F(\theta, t)|^2\, d\theta
\end{equation}
for each $t>0$, as $R\to \infty$.
With the assumption (\ref{E-1-A}) at our disposal, we apply the Dominated Convergence Theorem
to deduce that the last integral in (\ref{E-1-102}) converges to
$$
C_d \int_0^\infty \int_{\mathbb{T}^d} |F(\theta, t)|^2\, t^\alpha\, d\theta dt.
$$
A similar argument also shows that the LHS of (\ref{E-1-102}) converges to
$$
C_d \int_0^\infty \int_{\mathbb{T}^d}
\Big\{ |N_n^T \nabla_\theta U|^2 +|\partial_t U|^2 \Big\}\, t^\alpha\, d\theta dt.
$$
As a result, we have proved the estimate (\ref{E-1-101}).
\end{proof}
\begin{remark}\label{D-remark}
{\rm
The same argument as in the proof of Lemma \ref{lemma-E1} also gives a weighted estimate for
Dirichlet problem.
More precisely,
suppose that $n\in \mathbb{S}^{n-1}$ satisfies the Diophantine condition (\ref{D-condition}).
Let $U$ be a smooth solution of
\begin{equation}\label{E-1-100-D}
\left\{
\aligned
-\left( \begin{array} {c}N_{n}^T \nabla_\theta \\ \partial_t \end{array}\right)
\cdot B^*_n \left( \begin{array} {c}N_{n}^T \nabla_\theta \\ \partial_t \end{array} \right) U
& = \left( \begin{array} {c}N_{n}^T \nabla_\theta \\ \partial_t \end{array} \right) F +H
&\quad & \text{ in } \mathbb{T}^d \times \mathbb{R}_+,\\
U
& =0
& \quad & \text{ on } \mathbb{T}^d \times \{ 0\}.
\endaligned
\right.
\end{equation}
Assume that
$$
(1+t)\| \nabla_{\theta, t} U(\cdot, t)\|_{L^\infty(td)}
+ (1+t)\| F(\cdot, t)\|_{L^\infty(\mathbb{T}^d)}
+ (1+t)^2 \| H(\cdot, t)\|_{L^\infty(td)} <\infty.
$$
Then, for any $-1<\alpha<0$,
\begin{equation}\label{E-1-1010}
\int_0^\infty \int_{\mathbb{T}^d}
\Big\{ |N^T_n \nabla_\theta U|^2 +|\partial_t U|^2 \Big\}\, t^\alpha \, d\theta dt
\le C_\alpha \int_0^\infty \int_{\mathbb{T}^d} \Big\{ |F|^2 +|H|^2 t^2 \Big\}\, t^\alpha\, d\theta dt,
\end{equation}
where $C_\alpha$ depends only on $d$, $m$, $\mu$, $\alpha$ as well as some H\"older norm of $A$.
}
\end{remark}
We are now in a position to give the proof of Theorem \ref{theorem-E}.
\begin{proof}[\bf Proof of Theorem \ref{theorem-E}]
Recall that it suffices to prove (\ref{E-1-1-0}) with $W$ given by (\ref{W}).
To do this we split $W$ as
$W=W_1 +W_2 +W_3$,
where $W_1$ is the solution of (\ref{E-1-2}) with $G=0$ and $H=0$,
$W_2$ the solution with $H=0$ and $h=0$, and
$W_3$ the solution with $G=0$ and $h=0$.
To estimate $W_1$, we use the energy estimate in Proposition \ref{prop-2.1}.
This gives
\begin{equation}\label{E-10-1}
\int_0^1 \int_{\mathbb{T}^d} \Big\{ |N_n^T \nabla_\theta W_1|^2 +|\partial_t W_1|^2 \Big\} d\theta dt\\
\le C\, \| h\|^2_{H^1(\mathbb{T}^d)}
\le C |n-\widetilde{n}|^2.
\end{equation}
To handle the function $W_2$, we use the weighted estimate in Lemma \ref{lemma-E1}
with $\alpha =\sigma -1$.
This, together with estimates (\ref{G-1})-(\ref{H-1}), leads to
\begin{equation}\label{E-10-2}
\aligned
& \int_0^1 \int_{\mathbb{T}^d} \Big\{ |N_n^T \nabla_\theta W_1|^2 +|\partial_t W_1|^2 \Big\} d\theta dt\\
&\le C \int_0^\infty\int_{\mathbb{T}^d} |G|^2 t^{\sigma-1}\, d\theta dt\\
&\le C \int_0^\infty
\left\{ \frac{(t+1)^2 |n-\widetilde{n}|^4}{\kappa^2 (1+\kappa t)^4}
+\frac{(t+1)^2 |n-\widetilde{n}|^2}{(1+\kappa t)^4}
+\frac{|n-\widetilde{n}|^2}{(1+\kappa t)^4} \right\} t^{\sigma-1}\, dt\\
&\le C\left\{ \frac{ |n-\widetilde{n}|^4}{\kappa^{4+\sigma}}
+\frac{|n-\widetilde{n}|^2}{\kappa^{2+\sigma}} \right\}.
\endaligned
\end{equation}
Finally, we note that by writing $H(\theta, t)=\partial_t \widetilde{H}(\theta, t)$, where
$$
\widetilde{H}(\theta, t)=- \int_t^\infty H(\theta, s)ds,
$$
we may reduce the estimate of $W_3$ to the previous two cases.
Indeed, we split $W_3$ as $W_{31} +W_{32}$, where
$W_{31}$ is a solution of (\ref{E-1-2}) with $G=0$, $H=0$, $h=e_d \cdot \widetilde{H}(\theta, 0)$,
and $W_{32}$ a solution of (\ref{E-1-2}) with $G=(0, \widetilde{H})$, $H=0$, and $h=0$.
Observe that by (\ref{H-1}),
$$
\| \widetilde{H}(\cdot, 0)\|_{H^1(\mathbb{T}^d)}^2 \le \frac{C |n-\widetilde{n}|^2}{\kappa^2}
$$
and
$$
|\partial_t^k \partial_\theta^\alpha \widetilde{H}(\theta, t)|\le \frac{C |n -\widetilde{n}|}{\kappa (1+\kappa t)^\ell},
$$
for any $\alpha$, $k$ and $\ell$.
As in the cases of $W_1$ and $W_2$, by Proposition \ref{prop-2.1} and Lemma \ref{lemma-E1}, we obtain
$$
\int_0^1 \int_{\mathbb{T}^d} \Big\{ |N_n^T \nabla_\theta W_2|^2 +|\partial_t W_2|^2 \Big\} \, d\theta dt
\le \frac{C |n-\widetilde{n}|^2}{\kappa^{2+\sigma}}.
$$
Consequently, we have proved that
\begin{equation}\label{E-10-10}
\int_0^1 \int_{\mathbb{T}^d} \Big\{ |N_n^T \nabla_\theta W|^2 +|\partial_t W|^2 \Big\}\, d\theta dt
\le C\left\{ \frac{ |n-\widetilde{n}|^4}{\kappa^{4+\sigma}}
+\frac{|n-\widetilde{n}|^2}{\kappa^{2+\sigma}} \right\}.
\end{equation}
Finally, we note that by differentiating the system (\ref{E-1-1}),
the function $\nabla_\theta W$ is a smooth solution to a boundary value problem of same type
as $W$.
As a result, we may replace $W$ in (\ref{E-10-10}) by $\nabla_\theta W$.
This gives the desired estimate for $\nabla_\theta \partial_t W$
and completes the proof of (\ref{E-1-1-0}).
\end{proof}
\section{A partition of unity}
\setcounter{equation}{0}
For $x\in \partial\Omega$, let
\begin{equation}\label{k-x}
\aligned
\kappa (x)=\sup
\big\{ \kappa\in [0,1]: \text{the Diophantine condition (\ref{D-condition}) holds}\\
\text{ for $n(x)$ with constant $\kappa$} \big\}.
\endaligned
\end{equation}
It is known that if $\Omega$ is a bounded smooth, strictly convex domain in ${\mathbb R}^d$,
then $\kappa(x)>0$ for a.e. $x\in \partial\Omega$. Moreover,
$1/\kappa(x)$ belongs to the space $ L^{d-1, \infty}(\partial\Omega)$ \cite{Masmoudi-2012}.
This means that there exists $C>0$ such that
\begin{equation}\label{weak-L}
|\big\{ x\in \partial\Omega: \big[ \kappa (x)\big]^{-1} >\lambda \big\}| \le C\, \lambda^{1-d}
\quad \text{ for any } \lambda>0.
\end{equation}
\begin{prop}\label{prop-7.0}
Let $0<q<d-1$. Then for any $x\in \partial\Omega$ and $0<r<\text{\rm diam}(\Omega)$,
\begin{equation}\label{kappa-7}
\left(-\!\!\!\!\!\!\int_{B(x, r)\cap\partial\Omega} \big( \kappa (y)\big)^{-q}\, d\sigma (y) \right)^{1/q}
\le \frac{C}{r},
\end{equation}
where $C$ depends only on $d$, $q$ and $\Omega$.
\end{prop}
\begin{proof}
Note that
$$
\aligned
\int_{B(x, r)\cap\partial\Omega}
\big( \kappa (y)\big)^{-q}\, d\sigma (y)
&=q \int_0^\infty \lambda^{q-1} |\big\{ x\in B(x, r)\cap\partial\Omega: \big[ \kappa (x)\big]^{-1} >\lambda \big\}|\, d\lambda\\
& \le C \int_0^\Lambda \lambda^{q-1} \cdot r^{d-1}\, d\lambda
+C \int_\Lambda^\infty
\lambda^{q-d}\, d\sigma\\
&\le C r^{d-1}\cdot \Lambda^q + C \Lambda^{q-d +1},
\endaligned
$$
where we have used (\ref{weak-L}) for the first inequality.
The proof is finished by optimizing $\Lambda$ with $\Lambda=r^{-1}$.
\end{proof}
In this section we construct a partition of unity for $\partial\Omega$, which is adapted to the function
$\kappa(x)$. We mention that a similar partition of unity, which plays an important role in the analysis of
the oscillating Dirichlet problem,
was given in \cite{Armstrong-Prange-2016}. Here we provide a more direct $L^p$-based approach.
We first describe such construction in the flat space.
\begin{lemma}\label{lemma-7.1}
Let $Q_0$ be a cube in $\mathbb{R}^{d-1}$
and $F\in L^p(12Q_0)$ for some $p>d-1$. Let
$\tau>0$ be a number such that
\begin{equation}\label{7.1-0}
\left(\int_{12Q_0}|F|^p\right)^{1/p}> \frac{\tau}{\big[\ell(Q_0)\big]^{1-\frac{d-1}{p}}},
\end{equation}
where $\ell(Q_0)$ denotes the side length of $Q_0$.
Then there exists a finite sequence $\{ Q_j\}$ of dyadic sub-cubes of $Q_0$ such that
the interiors of $Q_j$'s are mutually disjoint,
\begin{equation}\label{7.1-1}
Q_0=\cup Q_j,
\end{equation}
\begin{equation}\label{7.1-2}
\left(\int_{12Q_j}|F|^p\right)^{1/p} \le \frac{\tau}{\big[\ell(Q_j)\big]^{1-\frac{d-1}{p}}},
\end{equation}
\begin{equation}\label{7.1-3}
\left(\int_{12Q^+_j}|F|^p\right)^{1/p} > \frac{\tau}{\big[\ell(Q^+_j)\big]^{1-\frac{d-1}{p}}},
\end{equation}
where $Q_j^+$ denotes the dyadic parent of $Q_j$, i.e., $Q_j$ is obtained by bisecting $Q_j^+$ once. Moreover,
if $4Q_j\cap 4Q_k \neq \emptyset$, then
\begin{equation}\label{7.1-4}
(1/2) \ell (Q_k)\le \ell (Q_j) \le 2 \ell (Q_k).
\end{equation}
\end{lemma}
\begin{proof}
The lemma is proved by using a stoping time argument (Calder\'on-Zygmund decomposition).
We begin by bisecting the sides of $Q_0$ and obtaining $2^{d-1}$ dyadic sub-cubes $\{ Q^\prime\}$.
If a cube $Q^\prime$ satisfies
\begin{equation}\label{7.1-5}
\left(\int_{Q^\prime} |F|^p\right)^{1/p}
\le \frac{\tau}{\big[\ell(Q^\prime)\big]^{1-\frac{d-1}{p}}}
\end{equation}
we stop and collect the cube. Otherwise, we repeat the same procedure on $Q^\prime$.
Since the RHS of (\ref{7.1-5}) goes to $\infty$ as $\ell(Q^\prime)\to 0$, the procedure is stopped
in a finite time. As a result, we obtain a finite number of sub-cubes with mutually disjoint interiors
satisfying (\ref{7.1-1})-(\ref{7.1-3}).
We point out this decomposition was performed in the whole space in \cite{Shen-1998}
in the study of negative eigenvalues for the Pauli operator.
The inequalities in (\ref{7.1-4}) were proved in \cite{Shen-1998}
by adapting an argument found in \cite{Fefferman-1996}.
The same argument works equally well in the setting of a finite cube $Q_0$.
We omit the details.
\end{proof}
Fix $x_0\in \partial\Omega$.
Let $c_0>0$ be sufficiently small so that
$B(x_0, 10c_0\sqrt{d})\cap\partial\Omega$ is given by the graph of a smooth function in a coordinate system,
obtained from the standard system through rotation and translation.
Let $\mathbb{H}_n^d(a)$ denote the tangent plane for $\partial\Omega$ at $x_0$, where $n=n(x_0)$ and
$a=x_0\cdot n$.
For $x\in B(x_0, 5c_0\sqrt{d})\cap \partial\Omega$, let
\begin{equation}\label{proj}
P(x; x_0)= x-x_0 -((x-x_0)\cdot n)n
\end{equation}
denote its projection to $\mathbb{H}_n^d(a)$.
The projection $P$ is one-to-one in $B(x_0, 10c_0\sqrt{d})\cap\partial\Omega$.
To construct a partition of unity for $B(x_0, c_0)\cap\partial\Omega$, adapted to the function $\kappa(x)$,
we use the inverse map $P^{-1}$ to lift a partition on the tangent plane, given in Lemma \ref{lemma-7.1},
to $\partial\Omega$.
More precisely,
fix a cube $Q_0$ on the tangent plane $H_n^d(a)$ such that
$$
B(x_0, 5c_0\sqrt{d})\cap \partial\Omega \subset P^{-1}(Q_0) \subset B(x_, 10c_0\sqrt{d})\cap \partial\Omega.
$$
We apply Lemma \ref{lemma-7.1} to $Q_0$ with the bounded function $F(x)=\kappa(P^{-1}(x))$ and
some $p>d-1$.
For each $0<\tau<c_1$,
this generates a finite sequence of sub-cubes $\{Q_j\}$ with the
properties (\ref{7.1-1})-(\ref{7.1-4}).
Let $x_j$ denote the center of $Q_j$ and $r_j$ the side length.
Let $\widetilde{x}_j=P^{-1}(x_j)$ and $\widetilde{Q}_j =P^{-1}(Q_j)$.
We will use the notation $t\widetilde{Q}_j =P^{-1} (t Q_j)$ for $t>0$
and call $\widetilde{Q}_j$ a cube on $\partial\Omega$.
Then
$$
\widetilde{Q}_0=P^{-1}(Q_0)=\cup_{j= 1}^N \widetilde{Q}_j.
$$
For each $\widetilde{Q}_j$ with $j\ge 1$, we choose $\eta_j\in C_0^\infty({\mathbb R}^d)$ such that
$0\le \eta_j \le 1$, $\eta_j=1$ on $\widetilde{Q}_j$,
$\eta_j =0$ on $\partial\Omega \setminus 2\widetilde{Q}_j$,
and $|\nabla\eta_j|\le C r_j^{-1}$. Note that by the property (\ref{7.1-4}),
$1\le \sum_j \eta_j \le C_0$ on $\widetilde{Q}_0$, where $C_0$ is a constant depending only
on $d$ and $\Omega$.
Let
$$
\varphi_j (x) =\frac{\eta_j(x)}{\sum_{k=1}^N \eta_k (x)}.
$$
Then
$$
\sum_{j=1}^N\varphi_j (x)=1\quad \text{ for any } x\in \widetilde{Q}_0.
$$
Observe that $0\le \varphi_j\le 1$, $\varphi_j \ge C_0^{-1}$ on $\widetilde{Q}_j$,
$\varphi_j=0$ on $\partial\Omega\setminus 2\widetilde{Q}_j$, and
$|\nabla \varphi_j|\le Cr_j^{-1}$.
Furthermore, by the property (\ref{7.1-5}),
there are positive constants $c_1$ and $c_2$,
depending only on $d$, $p$ and $\Omega$, such that
\begin{equation}\label{7.2-10}
\left(-\!\!\!\!\!\!\int_{12\widetilde{Q}_j}(\kappa(x))^p d\sigma (x)\right)^{1/p} \le
\frac{c_1\tau}{r_j},
\end{equation}
\begin{equation}\label{7.2-11}
\left(-\!\!\!\!\!\!\int_{36\widetilde{Q}_j} (\kappa (x))^p\, d\sigma(x) \right)^{1/p} \ge
\frac{c_2\tau}{r_j}.
\end{equation}
In particular, since $\kappa\le 1$,
it follows from (\ref{7.2-11}) that
\begin{equation}
r_j \ge c_2\, \tau.
\end{equation}
Also, by (\ref{7.2-11}), there exists some $z_j\in 36\widetilde{Q}_j$
such that
\begin{equation}\label{z-j}
\kappa (z_j) \ge \frac{c_2 \tau}{r_j}.
\end{equation}
\begin{prop}\label{prop-7.2}
There exists a constant $C$, depending only on $d$, $p$ and $\Omega$, such that
\begin{equation}\label{7.2-0}
r_j\le C \sqrt{\tau}.
\end{equation}
\end{prop}
\begin{proof}
By H\"older's inequality,
\begin{equation}\label{7.2-20}
\aligned
1 &\le \left(-\!\!\!\!\!\!\int_{12 \widetilde{Q}_j } \kappa^{-p^\prime}\, d\sigma \right)^{1/p^\prime}
\left(-\!\!\!\!\!\!\int_{12 \widetilde{Q}_j} \kappa^{p}\, d\sigma \right)^{1/p}\\
&\le C\, r_j^{-1}
\left(-\!\!\!\!\!\!\int_{12 \widetilde{Q}_j } \kappa^{p}\, d\sigma \right)^{1/p},
\endaligned
\end{equation}
where we have used Proposition \ref{prop-7.0} for the last step.
Note that the condition $p>d-1$ is equivalent to $p^\prime<\frac{d-1}{d-2}$,
which is less or equal to ${d-1}$ if $d\ge 3$.
In view of (\ref{7.2-10}) and (\ref{7.2-20}) we obtain (\ref{7.2-0}).
\end{proof}
\begin{prop}\label{prop-7.3}
Let $0\le \alpha<d-1$.
Then
\begin{equation}\label{7.3-0}
\sum_j r_j^{\alpha +d-1}
\le C_\alpha\, \tau^\alpha,
\end{equation}
where $C_\alpha$ depends only on $d$, $p$, $\alpha$ and $\Omega$.
\end{prop}
\begin{proof}
It follows from the first inequality in (\ref{7.2-20}) and (\ref{7.2-10}) that
\begin{equation}\label{7.3-1}
r_j \le C \tau \left(-\!\!\!\!\!\!\int_{12\widetilde{Q}_j}
\kappa^{-p^\prime} d\sigma \right)^{1/p^\prime}.
\end{equation}
Let $\mathcal{M}_{\partial\Omega}(f)$ denote the Hardy-Littlewood maximal function of $f$ on $\partial\Omega$,
defined by
\begin{equation}\label{HL}
\mathcal{M}(f)(x)
=\sup\left\{ -\!\!\!\!\!\!\int_{B(x, r)\cap\partial\Omega} |f|\, d\sigma: 0<r<\text{diam}(\Omega) \right\}
\end{equation}
for $x\in \partial\Omega$.
By (\ref{7.3-1}) we obtain
\begin{equation}\label{7.3-2}
r_j^\alpha \le C \tau^\alpha \left(\inf_{x\in 12\widetilde{Q}_j}
\mathcal{M}_{\partial\Omega} ( \kappa^{-p^\prime})(x) \right)^{\alpha/p^\prime}
\end{equation}
Hence,
$$
\aligned
\sum_j r_j^{\alpha +d-1}
&\le C \tau^\alpha
\sum_j \int_{\widetilde{Q}_j}
\left[ \mathcal{M}_{\partial\Omega} ( \kappa^{-p^\prime}) \right]^{\alpha/p^\prime}\\
&\le C \tau^\alpha\int_{\partial\Omega}
\left[ \mathcal{M}_{\partial\Omega} ( \kappa^{-p^\prime}) \right]^{\alpha/p^\prime}.
\endaligned
$$
Finally,
recall that $p^\prime<\frac{d-1}{d-2}$ and $0\le \alpha<d-1$.
Choose $t>1$ so that $p^\prime<t\alpha<d-1$.
Then
$$
\aligned
\int_{\partial\Omega}
\left[ \mathcal{M}_{\partial\Omega} ( \kappa^{-p^\prime}) \right]^{\alpha/p^\prime} d\sigma
&\le C \left(\int_{\partial\Omega}
\left[ \mathcal{M}_{\partial\Omega} ( \kappa^{-p^\prime}) \right]^{\alpha t/p^\prime} d\sigma \right)^{1/t}\\
& \le C\left(\int_{\partial\Omega}
(\kappa^{-1})^{\alpha t} d\sigma \right)^{1/t}
<\infty,
\endaligned
$$
where we have used the fact that the operator $\mathcal{M}_{\partial\Omega}$
is bounded on $L^q(\partial\Omega)$ for $q>1$.
This completes the proof.
\end{proof}
\begin{theorem}\label{h-data-theorem}
Let $ \overline{g}=(\overline{g}_k^\beta)$, where $\overline{g}_k^\beta$ is defined by
(\ref{h-data-ij}).
Then $\overline{g} \in W^{1, q}(\partial\Omega)$ for any $1<q< d-1$ and
\begin{equation}\label{h-data-1}
\| \overline{g}\|_{W^{1, q}(\partial\Omega)}
\le C_q\, \sup_{y\in \mathbb{T}^d} \| g(\cdot, y)\|_{C^1(\partial\Omega)},
\end{equation}
where $C_q$ depends only on $d$, $m$, $\mu$, $q$, and
$\|A\|_{C^k(\mathbb{T}^d)}$ for some $k=k(d)\ge 1$.
\end{theorem}
\begin{proof}
With Theorem \ref{theorem-E} at our disposal, the line of argument is similar to that in \cite{Armstrong-Prange-2016}.
Without loss of generality we assume that
$\sup_{y\in \mathbb{T}^d}\| g(\cdot, y)\|_{C^1(\partial\Omega)}=1$ and that for any $y\in {\mathbb R}^d$,
supp$(g(\cdot, y))\subset B(x_0, c_0)\cap \partial\Omega$
for some $x_0\in \partial\Omega$ and $c_0>0$. Fix $\tau\in (0,1/2)$.
Let
$$
\overline{g}_\tau (x)=\sum_j \overline{g}(z_j) \varphi_j (x),
$$
where $\{ \varphi_j\}$ is a finite sequence of smooth functions constructed in the partition of unity for
$B(x_0, c_0)\cap\partial\Omega$, associated with $\kappa (x)$ at the level $\tau$,
and $z_j$ is given by (\ref{z-j}).
Note that for $x\in \widetilde{Q}_\ell$,
$$
\nabla \overline{g}_\tau (x)=\sum_j \overline{g}(z_j) \nabla \varphi_j (x)
=\sum_j \big( \overline{g} (z_j) -\overline{g}(y_\ell)\big) \nabla \varphi_j (x),
$$
where $y_\ell $ is a point in $\widetilde{Q}_\ell$ that satisfying the condition (\ref{D-condition}).
In view of Theorem \ref{theorem-E} it follows that if $x\in \widetilde{Q}_\ell$,
$$
\aligned
|\nabla \overline{g}_\tau (x)|
& \le \sum_j |\overline{g} (z_j) -\overline{g}(y_\ell)|\, | \nabla \varphi_j (x)|\\
& \le C \left(\frac{r_\ell}{\tau}\right)^{1+\sigma} \left\{ \frac{r_\ell^2}{\tau} +1\right\}\\
&\le C \left(\frac{r_\ell}{\tau}\right)^{1+\sigma},
\endaligned
$$
where we have also used the estimates $|\nabla \varphi_j|\le Cr_j^{-1}$,
$\kappa(z_j)\ge c\, \tau r_j^{-1}$, and $r_\ell \le C \sqrt{\tau}$.
Hence, if we choose $\sigma\in (0,1)$ so small that $(1+\sigma)q<d-1$,
\begin{equation}
\int_{B(x_0, c_0)\cap\partial\Omega} |\nabla \overline{g}_\tau|^q \, d\sigma
\le C \sum_\ell \left(\frac{r_\ell}{\tau}\right)^{(1+\sigma)q} r_\ell ^{d-1}
\le C,
\end{equation}
where we have used (\ref{7.3-0}) for the last step.
This shows that the set
$\{ \overline{g}_\tau: 0<\tau<1\} $ is uniformly bounded in $W^{1, q}(B(x_0, c_0)\cap\partial\Omega)$
for any $q<d-1$.
Finally, we observe that a similar argument also gives
$$
\aligned
\int_{B(x_0, c_0)\partial\Omega}
|\overline{g}_\tau (x) -\overline{g}(x)|^q\, d\sigma
& \le
\sum_\ell \left(\frac{r_\ell^{2+\sigma}}{\tau^{1+\sigma}} \right)^q r_\ell^{d-1}\\
&\le C\tau^{\frac{q}{2}} \sum_\ell \left(\frac{r_\ell}{\tau}\right)^{(1+\sigma)q} r_\ell^{d-1}\\
&\le C\tau^{\frac{q}{2}}.
\endaligned
$$
Hence, $\| \overline{g}_\tau -\overline{g}\|_{L^q(B(x_0, c_0)\cap\partial\Omega)} \to 0$ as $\tau\to 0$,
for any $1<q<d-1$.
As a result, we have proved that $\overline{g}\in W^{1, q}(\partial\Omega)$ for any $1<q<d-1$.
\end{proof}
\begin{proof}[\bf Proof of Theorem \ref{main-theorem-2}]
With the weighted norm inequality (\ref{E-1-1010}) for Dirichlet problem (\ref{E-1-100-D}) at our disposal,
the estimate in Theorem \ref{theorem-E} continues to hold
for the homogenized Dirichlet boundary data for $d\ge 2$.
The rest of argument for (\ref{Sobolev-D}) and (\ref{rate-DP-low})
is exactly the same as in \cite{Armstrong-Prange-2016}.
We omit the details.
\end{proof}
\section{Proof of Theorem \ref{main-theorem-1}}
\setcounter{equation}{0}
With the estimates in Sections 5, 6 and 7,
the line of argument is similar to that in \cite{Armstrong-Prange-2016}
for the oscillating Dirichlet problem.
Recall that
$$
v^\gamma_\varepsilon (x)= -\int_{\partial\Omega}
\frac{\partial}{\partial y_k} \big\{ N_0^{\gamma \beta} (x, y)\big\}\cdot
\big(T_{ij}(y)\cdot \nabla_y \big)\Psi_{\varepsilon, k}^{*\alpha\beta} (y)
\cdot
g_{ij}^\alpha (y, y/\varepsilon)\, d\sigma (y)
$$
and
\begin{equation}
v_0^\gamma (x)=-\int_{\partial\Omega}
\frac{\partial}{\partial y_k} \big\{ N_0 ^{\gamma \beta} (x, y)\big\}
\widetilde{g}_{k}^\beta (y)\, d\sigma (y),
\end{equation}
where the function $\widetilde{g}_k^\beta$ is given by (\ref{mean}).
We will show that for any $\sigma \in (0,1)$,
\begin{equation}\label{final-goal}
\int_\Omega |v_\varepsilon -v_0|^2\, dx \le C_\sigma \, \varepsilon^{1-\sigma}.
\end{equation}
This would imply that if $u_\varepsilon$ and $u_\varepsilon$ are solutions of (\ref{NP-0})
and (\ref{NP-h}) respectively, then there exists some constant $E$ such that
$$
\| u_\varepsilon -u_0-E\|_{L^2(\Omega)} \le C_\sigma \, \varepsilon^{\frac12 -\sigma}.
$$
It then follows that
$$
\| u_\varepsilon - u_0 --\!\!\!\!\!\!\int_\Omega (u_\varepsilon-u_0)\|_{L^2(\Omega)}
\le C_\sigma \, \varepsilon^{\frac12 -\sigma},
$$
which gives (\ref{rate-0}) in the case $\int_\Omega u_\varepsilon =\int_\Omega u_0=0$.
Recall that the estimate (\ref{Sobolev}) is already proved in Section 7 (see Theorem \ref{h-data-theorem}).
To prove (\ref{final-goal})
we first note that by using a partition of unity for $\partial\Omega$,
without loss of generality, we may assume that there exists some $x_0\in \partial\Omega$
such that for any $y\in \mathbb{T}^d$,
supp$(g(\cdot, y))\subset B(x_0, c_0)$, where $c_0>0$ is sufficiently small so that
$B(x_0, 10c_0\sqrt{d})\cap\partial\Omega$ is given by the graph of a smooth function in a coordinate system,
obtained from the standard system by rotation and translation.
We construct another partition of unity for $B(x_0, 5c_0\sqrt{d})\cap \partial\Omega$, as described in Section 7,
with
\begin{equation}\label{tau}
\tau = \varepsilon^{1-\sigma},
\end{equation}
adapted to the function $\kappa(x)$.
Thus there exist a finite sequence $\{ \varphi_j\}$ of $C_0^\infty$ functions in ${\mathbb R}^d$
and a finite sequence $\{\widetilde{Q}_j \}$ of "cubes" on $\partial\Omega$,
such that $\sum_j \varphi_j =1$ on $B(x_0, 5c_0\sqrt{d})\cap\partial\Omega$.
Next, observe that by the estimate $|\nabla_y N_0(x, y)|+|\nabla_y N_\varepsilon (x, y)| \le C |x-y|^{1-d}$,
$$
|v_\varepsilon(x)| +|v_0(x)| \le C \big\{ 1+ |\ln \delta(x) |\big\},
$$
where $\delta(x)=\text{dist}(x, \partial\Omega)$.
This implies that
\begin{equation}\label{D-out}
\aligned
\sum_j \int_{B(\widetilde{x}_j, C r_j)\cap \Omega}
|v_\varepsilon -v_0|^2\, dx
&\le C \sum_j \int_{B(\widetilde{x}_j, C r_j)\cap\Omega}
\big( 1+ |\ln \delta(x) |\big)^2\, dx\\
& \le C \sum_j r_j^d (1+|\ln r_j|)^2\\
&\le C \varepsilon^{1-\sigma} (1+|\ln \varepsilon|)^2,
\endaligned
\end{equation}
where we have used Propositions \ref{prop-7.2} and \ref{prop-7.3}
(see Section 7 for the definitions of $\widetilde{x}_j$ and $r_j$).
Also note that
\begin{equation}\label{volume}
| \cup_j B(\widetilde{x}_j, Cr_j)|
\le C \sum_j r_j^{d}
\le C \tau,
\end{equation}
where we have used Proposition \ref{prop-7.3}.
To estimate the $L^2$ norm of $v_\varepsilon -v_0$ on the set
\begin{equation}\label{D-8}
D= D_\varepsilon=\Omega \setminus \cup_j B(\widetilde{x}_j , C r_j),
\end{equation}
we introduce a function
\begin{equation}\label{Theta}
\Theta_t (x)
=\sum_j \frac{ r_j^{d-1+t}}{|x-\widetilde{x}_j |^{d-1}},
\end{equation}
where $0\le t<d-1$.
\begin{lemma}\label{lemma-8.1}
Let $\Theta_t (x)$ be defined by (\ref{Theta}).
Then, if $ q\ge 1$ and $0\le q t<d-1$,
\begin{equation}\label{8.1-00}
\int_D \left( \Theta_t (x)\right)^q \, dx \le C \, \tau^{q t}.
\end{equation}
\end{lemma}
\begin{proof}
Observe that if
$x\notin B(\widetilde{x}_j , C r_j)$, then
$$
\frac{r_j^{d-1}}{|x-\widetilde{x}_j|^{d-1}}
\le C \int_{\widetilde{Q}_j} \frac{d\sigma (y)}{|x-y|^{d-1}}.
$$
Hence, for $x\in D$,
$$
\aligned
\Theta_t (x)
&\le C \int_{\partial\Omega} \frac{ f_t (y)}{|x-y|^{d-1}} \, d\sigma (y)\\
&\le C \left(\int_{\partial\Omega} \frac{|f_t(y)|^q}{|x-y|^{d-1}}\, d\sigma (y) \right)^{1/q}
( 1+|\ln \delta(x)|)^{1/q^\prime},
\endaligned
$$
where $f_t(y)=\sum_j r_j^t \varphi_j (y)$, $\delta (x)=\text{dist}(x, \partial\Omega)$, and
we have used H\"older's inequality for the last step.
It follows that
$$
\aligned
\int_D |\Theta_t (x)|^q \, dx
&\le C \int_{\partial\Omega} |f_t (y)|^q \, d\sigma\\
& \le C \sum_j r_j^{qt} r_j^{d-1}\\
&\le C \tau^{q t},
\endaligned
$$
where have used Proposition \ref{prop-7.3}.
\end{proof}
As in the case of Dirichlet problem in \cite{Armstrong-Prange-2016}, we split $v_\varepsilon -v_0$ into several
parts,
\begin{equation}
\aligned
& -\big(v^\gamma_\varepsilon (x) -v^\gamma_0(x)\big) \\
& =
\sum_j \int_{\partial\Omega}
\partial_{y_k} N_0^{\gamma \beta} (x, y)
\left\{ \big(T_{i\ell}(y)\cdot \nabla_y \big)\Psi_{\varepsilon, k}^{*\alpha\beta} (y)
\cdot g_{i\ell}^\alpha (y, y/\varepsilon)
-\widetilde{g}^\beta_k (y) \right\} \varphi_j (y) \, d\sigma (y)\\
& =
I_1+I_2+I_3+I_4+I_5,
\endaligned
\end{equation}
where $I_1, I_2, \dots, I_5$ are defined below and handled separately.
We will show that for $k=1,2, \dots , 5$,
\begin{equation}\label{estimate-I}
\int_D | I_k(x)|^2\, dx \le C_\sigma\, \varepsilon^{1-4\sigma},
\end{equation}
which, together with (\ref{D-out}), gives (\ref{final-goal}), as $\sigma\in (0,1/4)$ is arbitrary.
\bigskip
\noindent{\bf Estimate of $I_1$}, where
\begin{equation}\label{I-1}
\aligned
&I_1=\sum_j \int_{\partial\Omega}
\partial_{y_k} N_0^{\gamma \beta} (x, y)
\big(T_{i\ell}(y)\cdot \nabla_y \big)\big( \Psi_{\varepsilon, k}^{*\alpha\beta}
-\Phi_{\varepsilon, k}^{* \alpha\beta, z_j} \big) \cdot g_{i\ell}^\alpha (y, y/\varepsilon)
\varphi_j (y) \, d\sigma (y),\\
&\Phi_{\varepsilon, k}^{*\alpha\beta, z_j} (y)
=y_k \delta^{\alpha\beta}+\varepsilon \chi_{k}^{*\alpha\beta}(y/\varepsilon) +\phi_{\varepsilon, k}^{*\alpha\beta, z_j}(y),
\endaligned
\end{equation}
and $z_j$ is given in (\ref{z-j}).
Here we use Theorem \ref{main-lemma} to obtain that for any $\rho\in (0,1/2)$,
\begin{equation}\label{I-10}
\Big|\nabla \Big(\Psi_{\varepsilon, k}^{*\alpha\beta} - \Phi_{\varepsilon, k}^{*\alpha\beta, z_j}\Big)\Big|
\le C \sqrt{\varepsilon} \big\{ 1+ |\ln \varepsilon|\big\}
+C \varepsilon^{-1-\rho} r_j^{2+\rho},
\end{equation}
for $y\in 2\widetilde{Q}_j$.
It follows from (\ref{I-10}) that for $x\in D$,
\begin{equation}\label{I-11}
|I_1 (x)|
\le C \sqrt{\varepsilon} (1+|\ln \varepsilon|)\sum_j \frac{r_j^{d-1}}{|x-\widetilde{x}_j|^{d-1}}
+ C\varepsilon^{-1-\rho}
\sum_j \frac{r_j^{2+\rho +d-1}}{|x-\widetilde{x}_j|^{d-1}}.
\end{equation}
We now use Lemma \ref{lemma-8.1} to estimate the $L^2$ norm of $I_1$ on $D$.
The first term in the RHS of (\ref{I-11})
is harmless.
For the second term we use the fact $r_j \le C \sqrt{\tau}$ to bound it by
$$
C \varepsilon^{-1-\rho} \tau^{\frac12 +\rho} \sum_j \frac{r_j^{1-\rho +d-1}}{|x-\widetilde{x}_j|^{d-1}}.
$$
Since $2(1-\rho)<d-1$ for $d\ge 3$, we obtain
$$
\aligned
\int_D |I_1(x)|^2\, dx & \le C \varepsilon (1+|\ln \varepsilon|)^2 +C \varepsilon^{-2-2\rho} \tau^{1+2\rho}
\tau^{2(1-\rho)}\\
& \le C\, \varepsilon^{1-4\sigma},
\endaligned
$$
if $\rho$ is sufficiently small.
\bigskip
\noindent{\bf Estimate of $I_2$}, where
\begin{equation}\label{I-2}
\aligned
& I_2 = \sum_j \int_{\partial\Omega}
\partial_{y_k} N_0^{\gamma \beta} (x, y)
\big(T_{ij}(y)\cdot \nabla_y \big)\big( \Phi_{\varepsilon, k}^{* \alpha\beta, z_j} \big) \cdot g_{ij}^\alpha (y, y/\varepsilon)
\varphi_j (y) \, d\sigma (y)\\
&
- \sum_j \int_{\partial\mathbb{H}_j^d}
\partial_{y_k} N_0^{\gamma \beta} (x, P^{-1}_j (y))
\big(T_{i\ell}(P_j^{-1}(y))\cdot \nabla_y \big)\big( \Phi_{\varepsilon, k}^{* \alpha\beta, z_j} (y)\big) \cdot g_{i\ell}^\alpha (y, y/\varepsilon)
\varphi_j (y) \, d\sigma (y),
\endaligned
\end{equation}
where $\partial\mathbb{H}^d_j$ denotes the tangent plane for $\partial\Omega$ at $z_j$ and
$P_j^{-1}$ is the inverse of the projection map from $B(z_j, C r_j)\cap\partial\Omega$ to $\partial \mathbb{H}^d_j$.
Here we rely on the estimates
\begin{equation}\label{I-21}
\aligned
|\nabla_y^2 N_0 (x, y)| & \le C |x-y|^{-d},\\
|\nabla^2 \Phi_{\varepsilon, k}^{*\alpha\beta, z_j} | &\le C \varepsilon^{-1},
\endaligned
\end{equation}
as well as the observation that $|y- P^{-1}_j (y)|\le C r_j^2$ for $y\in B(\widetilde{x}_j, Cr_j)\cap\partial\Omega$.
It is not hard to see that for $x\in D$,
\begin{equation}\label{I-22}
|I_2 (x)|
\le C\, \varepsilon^{-1} \sum_j \frac{r_j^{2+d-1}}{|x- \widetilde{x}_j|^{d-1}}
\le C \varepsilon^{-1} \tau^{\frac{1+\rho}{2}} \sum_j \frac{r_j^{1-\rho+d-1}}{|x- \widetilde{x}_j|^{d-1}},
\end{equation}
which, by Lemma \ref{lemma-8.1},
leads to (\ref{estimate-I}) for $k=2$.
\bigskip
\noindent{\bf Estimate of $I_3$}, where
\begin{equation}\label{I-3}
\aligned
& I_3 = \sum_j \int_{\partial\mathbb{H}_j^d}
\partial_{y_k} N_0^{\gamma \beta} (x, P^{-1}_j (y))
\big(T_{i\ell} (y)\cdot \nabla_y \big)\big( \Phi_{\varepsilon, k}^{* \alpha\beta, z_j} \big) \cdot g_{i\ell}^\alpha (y, y/\varepsilon)
\varphi_j (y) \, d\sigma (y)\\
& - \sum_j \int_{\partial\mathbb{H}_j^d}
\partial_{y_k} N_0^{\gamma \beta} (x, P^{-1}_j (y))
\big(T_{i\ell} (z_j )\cdot \nabla_y \big)\big( \Phi_{\varepsilon, k}^{* \alpha\beta, z_j} \big) \cdot g_{i\ell}^\alpha (z_j, y/\varepsilon)
\varphi_j (y) \, d\sigma (y).
\endaligned
\end{equation}
It is easy to see that for $x\in D$,
$$
|I_3(x)|\le C \sum_j \frac{r_j^{1+d-1}}{|x-\widetilde{x}_j|^{d-1}}
\le C\, \tau^{\frac{\rho}{2}} \sum_j \frac{r_j^{1-\rho +d-1}}{|x-\widetilde{x}_j|^{d-1}},
$$
which may be handled by Lemma \ref{lemma-8.1}.
\bigskip
\noindent{\bf Estimate of $I_4$}, where
\begin{equation}\label{I-4}
\aligned
I_4 = & \sum_j \int_{\partial\mathbb{H}_j^d}
\partial_{y_k} N_0^{\gamma \beta} (x, P^{-1}_j (y))
\big(T_{i\ell} (z_j)\cdot \nabla_y \big)\big( \Phi_{\varepsilon, k}^{* \alpha\beta, z_j} \big) \cdot g_{i\ell}^\alpha (z_j, y/\varepsilon)
\varphi_j (y) \, d\sigma (y)\\
& -\sum_j \int_{\partial\mathbb{H}_j^d}
\partial_{y_k} N_0^{\gamma \beta} (x, P^{-1}_j (y))
\widetilde{g}_k^\beta (z_j)
\varphi_j (y) \, d\sigma (y).
\endaligned
\end{equation}
The estimate of $I_4$ uses the fact that for each $j$, the function
$$
\big(T_{i\ell} (z_j)\cdot \nabla_y \big)\big( \Phi_{\varepsilon, k}^{* \alpha\beta, z_j} \big) \cdot g_{i\ell}^\alpha (z_j, y/\varepsilon)
$$
is of form $U(y/\varepsilon)$, where $U(x)$ is a smooth 1-periodic function whose mean value is
given by $\widetilde{g}_k^\beta (z_j)$.
Furthermore, the normal to the hyperplane $\partial \mathbb{H}_j^d$ is $n(z_j)$, which satisfies the
Diophantine condition (\ref{D-condition}) with constant
$$
\kappa (z_j)\ge \frac{c\, \tau}{r_j}.
$$
It then follows from Proposition 2.1 in \cite{Armstrong-Prange-2016} that if $x\in D$,
$$
|I_4(x)|\le C_N (\tau^{-1} \varepsilon)^{-N} \sum_j \frac{r_j^{d-1}}{|x-\widetilde{x}_j|^{d-1}},
$$
for any $N\ge 1$.
Since $\tau =\varepsilon^{1-\sigma}$, this implies that
$$
\int_D |I_ 4(x)|^2\, dx \le C_N \varepsilon^{ N\sigma}.
$$
\bigskip
\noindent{\bf Estimate of $I_5$}, where
\begin{equation}\label{I-5}
\aligned
I_5= & \sum_j \int_{\partial\mathbb{H}_j^d}
\partial_{y_k} N_0^{\gamma \beta} (x, P^{-1}_j (y))
\overline{g}_k^\beta (z_j)
\varphi_j (y) \, d\sigma (y)\\
&-
\sum_j \int_{\partial\Omega}
\partial_{y_k} N_0^{\gamma \beta} (x, y)
\overline{g}_k^\beta (y)
\varphi_j (y) \, d\sigma (y).
\endaligned
\end{equation}
Finally, to estimate $I_5$, we use the regularity estimate for $\overline{g}$ in
Theorem \ref{theorem-E} to obtain
$$
\aligned
|\overline{g}(y) -\overline{g}(z_j)|
&\le C \left\{ \frac{r_j^2}{\big[\kappa(z_j)\big]^{2+\rho}} +\frac{r_j}{\big[\kappa(z_j)\big]^{1+\rho} }\right\}\\
&\le C \left\{ \frac{r_j^{4+\rho}}{\tau^{2+\rho}}
+\frac{r_j^{2+\rho}}{\tau^{1+\rho}} \right\}\\
&\le \frac{C r_j^{2+\rho}}{\tau^{1+\rho}},
\endaligned
$$
for any $y\in B(\widetilde{x}_j, Cr_j)\cap\partial\Omega$, where $\rho\in (0, 1/2)$
and we also used the fact $r_j\le C\sqrt{\tau}$.
It follows that for any $x\in D$,
$$
\aligned
|I_5(x)|
&\le C \sum_j \frac{r_j^d}{|x-\widetilde{x}_j|^{d-1}}
+C\tau^{-1-\rho} \sum_j \frac{r_j^{2+\rho +d-1}}{|x-\widetilde{x}_j|^{d-1}}\\
& \le C \sum_j \frac{r_j^d}{|x-\widetilde{x}_j|^{d-1}}
+C
\tau^{-\frac12}
\sum_j \frac{r_j^{1-\rho +d-1}}{|x-\widetilde{x}_j|^{d-1}}.
\endaligned
$$
As before, by applying Lemma \ref{lemma-8.1} and choosing $\rho>0$ sufficiently small,
we obtain the desired estimate for $I_5$.
This completes the proof of (\ref{final-goal}) and thus of Theorem \ref{main-theorem-1}.
\begin{remark}\label{remark-8.1}
{\rm
Let $\Theta_t (x)$ be defined by (\ref{Theta}).
It follows from the proof of Proposition \ref{prop-7.3} that for $x\in D$,
$$
\Theta_t (x)\le
C \tau^t \int_{\partial\Omega}
\frac{\big[ \mathcal{M}_{\partial\Omega} (\kappa^{-q} )\big]^{t/q} (y)}
{|x-y|^{d-1}} d\sigma (y),
$$
where $q=p^\prime<\frac{d-1}{d-2}$ and $t\ge 0$.
Let $u_\varepsilon$ and $u_0$ be solutions of (\ref{NP-0}) and (\ref{NP-h}), respectively.
An inspection of our proof of Theorem \ref{main-theorem-1} shows that for any $\sigma\in (0,1/2)$,
there exists a neighborhood $\Omega_\varepsilon$ of $\partial\Omega$ in $\Omega$
such that
\begin{equation}\label{error-8.0}
|\Omega_\varepsilon|\le C\, \varepsilon^{1-\sigma},
\end{equation}
and for $x\in \Omega \setminus \Omega_\varepsilon$,
\begin{equation}\label{error-8.1}
|u_\varepsilon (x) -u_0 (x)-E |\le
C \varepsilon^{\frac12 -4\sigma}
\int_{\partial\Omega}
\frac{\big[ \mathcal{M}_{\partial\Omega} (\kappa^{-q} )\big]^{\frac{1-\rho}{q}} (y)}
{|x-y|^{d-1}} d\sigma (y),
\end{equation}
where $1<q<d-1$, $\rho=\rho(\sigma)>0$ is small, and $E$ is a constant.
The boundary layer $\Omega_\varepsilon$, which is given locally by the union of $B(\widetilde{x}_j, Cr_j)\cap\Omega$,
depends only on the function $\kappa$ and $\Omega$.
Furthermore, if $F(x)$ denotes the integral in (\ref{error-8.1}),
then
$$
|F(x)|\le C(1+|\ln \delta(x)|)^{1/s^\prime} \left(\int_{\partial\Omega}
\frac{\big[ \mathcal{M}_{\partial\Omega} (\kappa^{-q} )\big]^{\frac{s(1-\rho)}{q}} (y)}
{|x-y|^{d-1}} d\sigma (y)\right)^{1/s},
$$
where $q<s(1-\rho)<d-1$. Since $\kappa^{-1}\in L^s(\partial\Omega)$ for any $1<s<d-1$,
$\mathcal{M}_{\partial\Omega} (\kappa^{-q})\in L^{s/q}(\partial\Omega)$
for any $q<s<d-1$.
It follows that $F\in L^s(\Omega)$ for any $q<s\le d-1$.
This, together with (\ref{error-8.1}),
\begin{equation}\label{error-8.3}
\| u_\varepsilon -u_0-E\|_{L^s(\Omega\setminus \Omega_\varepsilon)}
\le C\, \varepsilon^{\frac12-4\sigma} \quad \text{ for any } 1<s\le d-1.
\end{equation}
Finally, assume that $\int_\omega u_\varepsilon =\int_\Omega u_0=0$.
Since $\| u_\varepsilon -u_0\|_{L^2(\Omega)}\le C_\sigma\, \varepsilon^{\frac12-\sigma}$ by Theorem \ref{main-theorem-1},
it follows from (\ref{error-8.3}) that
$|E|\le C\, \varepsilon^{\frac12-4\sigma}$.
As a result, estimates (\ref{error-8.1}) and (\ref{error-8.3}) hold with $E=0$.
}
\end{remark}
\section{Higher-order convergence}
\setcounter{equation}{0}
In this section we use Theorem \ref{main-theorem-1} to establish a higher-order rate of convergence
in the two-scale expansion for the Neumann problem,
\begin{equation}\label{N-High}
\left\{
\aligned
\mathcal{L}_\varepsilon (u_\varepsilon) &=F &\quad &\text{ in } \Omega,\\
\frac{\partial u_\varepsilon}{\partial \nu_\varepsilon} &= g &\quad &\text{ on } \partial\Omega,
\endaligned
\right.
\end{equation}
where $F$ and $g$ are smooth functions.
Our goal is to prove the following.
\begin{theorem}\label{theorem-9.1}
Suppose that $A$ and $\Omega$ satisfy the same conditions as in Theorem \ref{main-theorem-1}.
Let $u_\varepsilon$ be the solution of (\ref{N-High}) with $\int_{\Omega} u_\varepsilon=0$, and
$u_0$ the solution of the homogenized problem. Then there exists a function $v^{bl}$, independent of $\varepsilon$,
such that
\begin{equation}\label{9.1}
\| u_\varepsilon - u_0 -\varepsilon \chi_k (x/\varepsilon)\frac{\partial u_0}{\partial x_k}
-\varepsilon v^{bl} \|_{L^2(\Omega)}
\le C_\sigma \varepsilon^{\frac32-\sigma} \| u_0\|_{W^{3, \infty}(\Omega)},
\end{equation}
for any $\sigma\in (0,1/2)$,
where $C_\sigma$ depends only on $d$, $m$, $\sigma$, $A$ and $\Omega$.
Moreover, the function $v^{bl}$ is a solution to the Neumann problem
\begin{equation}\label{N-bl}
\mathcal{L}_0(v^{bl}) = F_* \quad \text{ in } \Omega \quad \text{ and } \quad
\frac{\partial v^{bl}}{\partial \nu_0} =g_*\quad \text{ on }\partial\Omega,
\end{equation}
where $F_*= -\overline{c}_{ki\ell}\frac{\partial^3 u_0}{\partial x_k \partial x_i \partial x_\ell}$
for some constants $\overline{c}_{ki\ell}$, and
$g_*$ satisfies
\begin{equation}\label{9.2}
\| g_*\|_{L^q(\partial\Omega)} \le C_q \| u_0\|_{W^{2, \infty}(\Omega)},
\end{equation}
for any $1<q<d-1$.
\end{theorem}
For simplicity of exposition we will drop the superscripts in this section.
Let
\begin{equation}\label{b-ij}
b_{ij} =a_{ij} +a_{ik}\frac{\partial \chi_j}{\partial y_k} -\widehat{a}_{ij},
\end{equation}
where
$(\chi_k$) are the (first-order) correctors for $\mathcal{L}_\varepsilon$ in ${\mathbb R}^d$.
Since
$$
\int_{\mathbb{T}^d} b_{ij}=0 \quad \text{ and } \quad \frac{\partial }{\partial y_i} \big( b_{ij}\big)=0,
$$
there exist 1-periodic functions $(\phi_{kij})$ such that
\begin{equation}\label{9.3}
b_{ij}=\frac{\partial }{\partial y_k} \big( \phi_{kij} \big) \quad
\text{ and } \quad \phi_{kij}=-\phi_{ikj}.
\end{equation}
The second-order correctors $(\chi_{k\ell})$ with $1\le k, \ell \le d$ are defined by
\begin{equation}\label{second}
\left\{
\aligned
& -\frac{\partial}{\partial y_i}
\left\{ a_{ij}\frac{\partial \chi_{k\ell}}{\partial y_j} \right\}
=b_{k\ell} +\frac{\partial}{\partial y_i}
\big( a_{i\ell} \chi_k\big) \quad \text{ in } {\mathbb R}^d,\\
& \chi_{k\ell} \text{ is 1-periodic and } \int_{\mathbb{T}^d} \chi_{k\ell}=0.
\endaligned
\right.
\end{equation}
Let
\begin{equation}\label{w-9}
w_\varepsilon =u_\varepsilon -u_0 -\varepsilon \chi^\varepsilon_k\frac{\partial u_0}{\partial x_k}
-\varepsilon^2 \chi^\varepsilon_{k\ell} \frac{\partial^2 u_0}{\partial x_k \partial x_\ell},
\end{equation}
where we have used the notation $f^\varepsilon(x)= f(x/\varepsilon)$.
A direct computation shows that
\begin{equation}\label{L-w}
\aligned
\mathcal{L}_\varepsilon (w_\varepsilon)
= &-\varepsilon \Big( \phi_{kij}^\varepsilon \delta_{j\ell}
-a_{ij}^\varepsilon \chi_k^\varepsilon \delta_{j\ell}
-a_{ij}^\varepsilon \big(\frac{\partial \chi_{k\ell}}{\partial x_j}\big)^\varepsilon \Big)
\frac{\partial^3 u_0}{\partial x_i\partial x_k \partial x_\ell}\\
&\qquad
+\varepsilon^2 \frac{\partial}{\partial x_i}
\left\{ a_{ij}^\varepsilon \chi_{k\ell}^\varepsilon \frac{\partial^3 u_0}{\partial x_j \partial x_k \partial x_\ell} \right\}.
\endaligned
\end{equation}
Let
\begin{equation}\label{c-k}
\left\{
\aligned
& c_{ki\ell}=
\phi_{kij} \delta_{j\ell}
-a_{ij} \chi_k \delta_{j\ell}
-a_{ij} \frac{\partial \chi_{k\ell}}{\partial x_j}, \\
& \overline{c}_{ki\ell}
=-\!\!\!\!\!\!\int_{\mathbb{T}^d} c_{ki\ell}.
\endaligned
\right.
\end{equation}
Note that by the definition of $\chi_{k\ell}$,
$$
\frac{\partial}{\partial x_i} \big( c_{ki\ell} \big)=0.
$$
It follows that there exist 1-periodic functions $f_{mki\ell}$ with
$1\le m, k, i, \ell \le d$ such that
\begin{equation}\label{f-m}
c_{ki\ell}-\overline{c}_{ki\ell} =\frac{\partial}{\partial y_m} \big( f_{m k i \ell}\big)
\quad \text{ and } \quad
f_{mki\ell}=-f_{ikm\ell}.
\end{equation}
This allows us to rewrite (\ref{L-w}) as
\begin{equation}\label{L-w-1}
\aligned
\mathcal{L}_\varepsilon (w_\varepsilon) & =-\varepsilon \overline{c}_{ki\ell}\frac{\partial^3 u_0}
{\partial x_i\partial x_k \partial x_\ell}
-\varepsilon \left(\frac{\partial f_{mki\ell}}{\partial x_m}\right)^\varepsilon
\frac{\partial^3 u_0}
{\partial x_i\partial x_k \partial x_\ell}\\
&\qquad
+\varepsilon^2 \frac{\partial}{\partial x_i}
\left\{ a_{ij}^\varepsilon \chi_{k\ell}^\varepsilon \frac{\partial^3 u_0}{\partial x_j \partial x_k \partial x_\ell} \right\}.
\endaligned
\end{equation}
Next we compute the conormal derivative of $w_\varepsilon$.
Again, a direct computation gives
\begin{equation}\label{c-normal}
\aligned
\frac{\partial w_\varepsilon}{\partial \nu_\varepsilon}
&=-n_i b_{ij}^\varepsilon \frac{\partial u_0}{\partial x_j}
-\varepsilon n_i a_{ij}^\varepsilon \chi_k^\varepsilon \frac{\partial^2 u_0}{\partial x_j\partial x_k}
-\varepsilon n_i a_{ij}^\varepsilon \left(\frac{\partial \chi_{k\ell}}{\partial x_j}\right)^\varepsilon \frac{\partial^2 u_0}{\partial x_k\partial x_\ell}\\
& \qquad\qquad
-\varepsilon^2 n_i a_{ij}^\varepsilon \chi_{k\ell}^\varepsilon \frac{\partial^3 u_0}{\partial x_j\partial x_k\partial x_\ell}.
\endaligned
\end{equation}
Using (\ref{9.3}) and (\ref{c-k}),
we further obtain
\begin{equation}\label{c-normal-1}
\frac{\partial w_\varepsilon}{\partial \nu_\varepsilon}
=-\varepsilon n_i \frac{\partial}{\partial x_k} \left( \phi_{kij}^\varepsilon \frac{\partial u_0}{\partial x_j} \right)
+\varepsilon n_i c^\varepsilon_{ki\ell} \frac{\partial^2 u_0}{\partial x_k\partial x_\ell}
-\varepsilon^2 n_i a_{ij}^\varepsilon \chi_{k\ell}^\varepsilon \frac{\partial^3 u_0}{\partial x_j\partial x_k\partial x_\ell}
\end{equation}
In view of (\ref{L-w-1}) and (\ref{c-normal-1}),
we split $w_\varepsilon--\!\!\!\!\!\!\int_{\Omega} w_\varepsilon$ as $w_\varepsilon^{(1)} +w_\varepsilon^{(2)} +w_\varepsilon^{(3)} +w_\varepsilon^{(4)}$,
where
\begin{equation}\label{w-9-1}
\left\{
\aligned
& \mathcal{L}_\varepsilon (w^{(1)}_\varepsilon) =0
&\quad & \text{ in } \Omega,\\
&\frac{\partial}{\partial \nu_\varepsilon}
\big(w_\varepsilon^{(1)}\big)
=-\varepsilon n_i \frac{\partial}{\partial x_k}
\left(\phi_{kij}^\varepsilon \frac{\partial u_0}{\partial x_j}\right)
& \quad & \text{ on } \partial\Omega,
\endaligned
\right.
\end{equation}
\begin{equation}\label{w-9-2}
\left\{
\aligned
& \mathcal{L}_\varepsilon (w^{(2)}_\varepsilon) =-\varepsilon \overline{c}_{ki\ell} \frac{\partial^3 u_0}{\partial x_i \partial x_k \partial x_\ell }
&\quad & \text{ in } \Omega,\\
&\frac{\partial}{\partial \nu_\varepsilon}
\big(w_\varepsilon^{(2)}\big)
=\varepsilon n_i \overline{c}_{ki\ell} \frac{\partial^2 u_0}
{\partial x_k\partial x_\ell} & \quad & \text{ on } \partial\Omega,
\endaligned
\right.
\end{equation}
\begin{equation}\label{w-9-3}
\left\{
\aligned
& \mathcal{L}_\varepsilon (w^{(3)}_\varepsilon) =
-\varepsilon \left(\frac{\partial f_{mki\ell}}{\partial x_m}\right)^\varepsilon
\frac{\partial^3 u_0}
{\partial x_i\partial x_k \partial x_\ell} &\quad & \text{ in }\Omega,\\
& \frac{\partial}{\partial \nu_\varepsilon}
\big(w_\varepsilon^{(3)}\big)
=\varepsilon n_i \big( c^\varepsilon_{ki\ell} -\overline{c}_{ki\ell} \big)
\frac{\partial^2 u_0}{\partial x_k\partial x_\ell}& \quad & \text{ on } \partial\Omega,
\endaligned
\right.
\end{equation}
and
\begin{equation}\label{w-9-4}
\left\{
\aligned
& \mathcal{L}_\varepsilon (w^{(4)}_\varepsilon) =
\varepsilon^2 \frac{\partial}{\partial x_i}
\left\{ a_{ij}^\varepsilon \chi_{k\ell}^\varepsilon \frac{\partial^3 u_0}{\partial x_j \partial x_k \partial x_\ell} \right\}
&\quad & \text{ in }\Omega,\\
& \frac{\partial}{\partial \nu_\varepsilon}
\big(w_\varepsilon^{(4)}\big)
=-\varepsilon^2 n_i a_{ij}^\varepsilon \chi_{k\ell}^\varepsilon \frac{\partial^3 u_0}{\partial x_j\partial x_k\partial x_\ell}
&\quad &\text{ on }\partial\Omega.
\endaligned
\right.
\end{equation}
We further require that
\begin{equation}\label{mean-w}
\int_{\Omega} w_\varepsilon^{(1)}=
\int_{\Omega} w_\varepsilon^{(2)} =\int_{\Omega} w_\varepsilon^{(3)}
=\int_{\Omega} w_\varepsilon^{(4)}=0.
\end{equation}
To proceed, we first note that by Poincar\'e inequality, (\ref{mean-w}) and energy estimates,
\begin{equation}\label{w-4-estimate}
\|w_\varepsilon^{(4)}\|_{L^2(\Omega)}
\le C\, \|\nabla w^{(4)}_\varepsilon\|_{L^2(\Omega)}
\le C\, \varepsilon^2 \| \nabla^3 u_0\|_{L^2(\Omega)}.
\end{equation}
The solution $w_\varepsilon^{(3)}$ may be handled in a similar manner.
To see this we use the skew-symmetry of $f_{mki\ell}$ in $m$ and $i$ to write
the RHS of the equation as
$$
-\varepsilon^2\frac{\partial}{\partial x_m}
\left( f_{m k i \ell}^\varepsilon \frac{\partial^3 u_0}{\partial x_i \partial x_k \partial x_\ell}\right),
$$
while the Neumann data for $w_\varepsilon^{(3)}$ may be written as
$$
\frac {\varepsilon^2}{2} \left(n_i \frac{\partial }{\partial x_m}
-n_m \frac{\partial}{\partial x_i} \right)
\left(f^\varepsilon_{m k i \ell} \frac{\partial^2 u_0}{\partial x_k \partial x_\ell}\right)
-\varepsilon^2 n_i f_{mki\ell}^\varepsilon \frac{\partial^3 u_0}{\partial x_k \partial x_\ell \partial x_m}.
$$
As a result, we obtain
\begin{equation}\label{w-3-estimate}
\aligned
\| w_\varepsilon^{(3)}\|_{L^2(\Omega)}
&\le C \|\nabla w_\varepsilon^{(3)} \|_{L^2(\Omega)}
\le C \, \varepsilon^2 \left\{ \|\nabla^3 u_0\|_{L^2(\Omega)}
+\| f^\varepsilon \nabla^2 u_0\|_{H^{\frac12}(\partial\Omega)}\right\}\\
&\le C \varepsilon^{\frac32} \| u_0\|_{W^{3, \infty}(\Omega)}.
\endaligned
\end{equation}
Next, we observe that $w_\varepsilon^{(2)}$ may be dealt with by the classical homogenization results for
$\mathcal{L}_\varepsilon$. Indeed, let $v_0^{(2)}$ be the solution of
\begin{equation}\label{v-0-1}
\left\{
\aligned
& \mathcal{L}_0 (v^{(2)}_0)=-\overline{c}_{ki\ell} \frac{\partial^3 u_0}{\partial x_i \partial x_k \partial x_\ell}
&\quad & \text{ in } \Omega,\\
&
\frac{\partial v_0^{(2)}}{\partial \nu_0}
=n_i \overline{c}_{k i \ell} \frac{\partial^2 u_0}{\partial x_k \partial x_\ell}
&\quad & \text{ on } \partial\Omega,
\endaligned
\right.
\end{equation}
with $\int_\Omega v_0^{(2)}=0$. It is well known that
\begin{equation}\label{v-1-estimate}
\| w_\varepsilon^{(2)} -\varepsilon v_0^{(2)}\|_{L^2(\Omega)} \le C\, \varepsilon^2 \| u_0\|_{W^{3, \infty}(\Omega)}.
\end{equation}
It remains to estimate the solution $w_\varepsilon^{(1)}$, which will be handled by using
Theorem \ref{main-theorem-1}.
Observe that by the skew-symmetry of $\phi_{kij}$ in $k$ and $i$,
the Neumann data of $ w_\varepsilon^{(1)}$ may be written as
\begin{equation}\label{data-9}
-\frac{\varepsilon}{2} \Big(T_{ik}\cdot \nabla\Big)
\left(\phi_{kij}^\varepsilon \frac{\partial u_0}{\partial x_j}\right),
\end{equation}
where $T_{ik}=n_ie_k -n_k e_i$.
This allows us to apply Theorem \ref{main-theorem-1} to deduce that
\begin{equation}\label{w-1-estimate}
\| w_\varepsilon -\varepsilon v_0^{(1)}\|_{L^2(\Omega)}
\le C_\sigma \, \varepsilon^{\frac32 -\sigma} \| u_0\|_{W^{2, \infty}(\Omega)},
\end{equation}
for any $\sigma\in (0, 1/2)$,
where $v_0^{(1)}$ is a solution of the Neumann problem
\begin{equation}\label{N-v-1}
\left\{
\aligned
&\mathcal{L}_0 (v_0^{(1)}) =0 &\quad & \text{ in } \Omega,\\
&\frac{\partial}{\partial \nu_0} \big( v_0^{(1)}\big)
=(T_{ij}\cdot \nabla \big)(\overline{g}_{ij}) & \quad &\text{ in } \partial\Omega,
\endaligned
\right.
\end{equation}
and $\overline{g}_{ij}\in W^{1, q}(\partial\Omega)$ for any $1<q<d-1$.
We remark that the explicit dependence on the $W^{2, \infty}(\Omega)$ norm of
$u_0$ in the RHS of (\ref{w-1-estimate})
follows from the proof of Theorem \ref{main-theorem-1}.
The key observation is that the fast variable $y$ in the Neumann data (\ref{data-9})
is separated from the slow variable $x$.
Let $v^{bl}= v_0^{(1)} + v_0^{(2)}$.
In view of (\ref{w-4-estimate}), (\ref{w-3-estimate}), (\ref{v-1-estimate}) and
(\ref{w-1-estimate}), we have proved that
\begin{equation}\label{9-100}
\| w_\varepsilon --\!\!\!\!\!\!\int_{\Omega} w_\varepsilon -\varepsilon v^{bl}\|_{L^2(\Omega)} \le C _\sigma \, \varepsilon^{\frac32-\sigma}
\| u_0\|_{W^{3, \infty}(\Omega)}.
\end{equation}
Finally, we note that since $\int_\Omega u_\varepsilon =\int_\Omega u_0=0$,
$$
\aligned
\Big|-\!\!\!\!\!\!\int_{\Omega} w_\varepsilon\Big|
&\le C \, \varepsilon\, \Big|\int_{\Omega} \chi_k(x/\varepsilon) \frac{\partial u_0}{\partial x_k}\, dx \Big|
+C \varepsilon^2 \| \nabla^2 u_0\|_\infty\\
&\le C\, \varepsilon^{2} \|u_0\|_{W^{2, \infty}(\Omega)},
\endaligned
$$
where the last step follows from the fact that $\chi_k$ is periodic with mean value zero.
This, together with (\ref{9-100}), yields the estimate (\ref{9.1})
and thus completes the proof of Theorem \ref{theorem-9.1}.
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 1,190 |
\section*{}
\vspace{-1cm}
\footnotetext{\textit{$^{\textsf{a}}$Department of Chemistry, University of Oxford, Inorganic Chemistry Laboratory, South Parks Road, Oxford OX1 3QR, U.K.; Tel: +44 1865 272137; E-mail: andrew.goodwin@chem.ox.ac.uk\\
$^{\textsf{b}}$School of Physical Sciences, University of Kent, Canterbury CT2 7NH, U.K.\\
$^{\textsf{c}}$School of Chemistry, University of Birmingham, Edgbaston, Birmingham B15 2TT, U.K.; Email: h.yeung@bham.ac.uk}}
\footnotetext{\dag~Electronic Supplementary Information (ESI) available. See DOI: 10.1039/b000000x/}
\section{Introduction}
It is a remarkable recent discovery that some dabconium hybrid perovskites are ferroelectric (FE) with spontaneous polarisations comparable to that of BaTiO$_3$.\cite{Ye_2018,Zhang_2017,Merz_1953} Remarkable, because polarisation is a measure of displaced charge density, and both the magnitude of localised charges and ABX$_3$ density are substantially reduced in hybrid perovskites relative to their longer-established inorganic cousins.\cite{Li_2017,Kieslich_2017} The possibility of combining the electrical performance of conventional ceramics with the mechanical flexibility and low-temperature processibility of hybrids is exciting because it offers many potential advantages in the development of \emph{e.g.}\ wearable electronics, flexible devices, and bionics.\cite{You_2017,Ye_2018,Li_2018,Guo_2019}
The basic phenomenology of this family is well exemplified by [MDABCO]RbI$_3$ (MDABCO$^{2+}$ = {\it N}-methyl-{\it N}$^\prime$-diazabicyclo-[2.2.2]octonium) [Fig.~\ref{fig1}(a)].\cite{Zhang_2017} At high temperatures this system adopts the aristotypic (cubic) ABX$_3$ perovskite structure, with the high crystal symmetry reflecting orientational disorder of polar MDABCO A-site cations. On cooling below $\sim$400\,K an orientational ordering transition occurs such that each MDABCO cation now aligns along a common $\langle111\rangle$ direction, giving a structure characterised by the polar space-group $R3$. This phase is FE as its polarisation can be reversed under an applied field. The high-profile ``metal-free'' congener [MDABCO](NH$_4$)I$_3$ behaves analogously, but with NH$_4^+$ orientations amplifying the saturation polarisation in the FE phase.\cite{Ye_2018,Ehrenreich_2019}
\begin{figure}
\centering
\includegraphics{fig1.png}
\caption{(a) The FE phase transition in [MDABCO]RbI$_3$ involves orientational ordering of MDABCO ions (space filling representation) on the A-site of the ABX$_3$ perovskite lattice. The low-temperature form has rhombohedral $R3$ symmetry, and the disordered high-temperature (paraelectric) form has $Pm\bar3m$ symmetry.\cite{symnote} (b) Candidate microscopic driving forces for FE order: (left--right) the tendency for each iodide anion (purple spheres) to hydrogen-bond with a single N-H proton (orange sphere), dipolar interactions, and strain coupling.}
\label{fig1}
\end{figure}
Since relatively few amongst the diverse and extensive family of hybrid perovskites are polar,\cite{Li_2017,Gebhardt_2019,Ning_2019} it is natural to question what makes this particular family so special. Ultimately, of course, the goal is to develop design principles that allow the targeted synthesis and optimisation of hybrid ferroelectrics. On the simplest level, the use of polar A-site cations is considered important: indeed this was a key conclusion of Ref.~\citenum{Zhang_2017} on finding that replacing the $C_{3v}$-symmetric [MDABCO]$^{2+}$ cation by [DABCO]$^{2+}$ (= {\it N,N}$^\prime$-diazabicyclo-[2.2.2]octonium, $D_{3h}$ point symmetry), the corresponding perovskites do not have polar ground states. This principle has now matured into the so-called ``quasispherical theory'', whereby the local polarisation arising from asymmetric substitution of spherical cations is clearly linked empirically to FE response in perovskites and beyond:\cite{Zhang_2019,Yang_2019,Morita_2019,Wei_2020} \emph{e.g.}\ the DABCO cation is non-polar and roughly spherical, while MDABCO is polar and ovoid. But if A-site polarisation is a necessary condition it is not a sufficient one: many hybrid perovskites with polar A-site cations---from the famous photovoltaic [CH$_3$NH$_3$]PbI$_3$ to the antiferroelectric (AFE) multiferroics [NH$_2$(CH$_3$)$_2$]M(HCOO)$_3$---are not themselves polar.\cite{Weller_2015,Weller_2015b,Jain_2008}
So the question of what couples dipole orientations in dabconium perovskites to generate bulk polarisation is an important and open problem.\cite{Shahrokhi_2020} The elongated shape of MDABCO itself is understood to favour alignment along $\langle111\rangle$, with this orientation also allowing hydrogen bonding between the R$_3$N--H proton and three nearby halide ions held in a \emph{fac} arrangement around a common B-site cation.\cite{Zhang_2017} The electronic implications of this hydrogen bonding---in essence a requirement that each halide interact strongly with only a single MDABCO cation---and the steric implications of coupling between distortions of the [BX$_3$]$^{2-}$ lattice have been proposed in general terms as coupling mechanisms [Fig.~\ref{fig1}(b)].\cite{Zhang_2017,Ye_2018} The logical difficulty one faces is that similar considerations are at play in most hybrid perovskites, including the many non-polar examples.
Here we seek to understand whether the different microscopic ingredients of hydrogen-bonding rules, distortion strain coupling, and the always-present through-space dipolar interactions---either on their own or in combination---can account for the observed FE transition observed experimentally in dabconium perovskites. We use a coarse-graining approach supplemented by straightforward calculations and Monte Carlo (MC) simulations to identify ground states for various combinations of these ingredients. Our key result is that the pairing of strain coupling and dipolar interactions stabilises the $R3$ polar ground state when the two interactions are not too dissimilar in strength. Using density functional theory (DFT) calculations on a range of suitably-chosen [MDABCO]RbI$_3$ polymorphs, we demonstrate that our simple theoretical model provides a sensible coarse graining of the energetics of this particular system, and hence provides a first-order explanation for its FE behaviour. Finally, we show that this crucial dipolar--strain combination drives collective polarisation only when there is a driving force for alignment of the A-site dipole along $\langle111\rangle$; the corresponding ground states for $\langle100\rangle$ and $\langle110\rangle$ are non-polar. In this way we rationalise the absence of polarisation in many hybrid perovskites, and arrive at a detailed set of design rules for generating FE examples beyond the dabconium family alone.
\section{Results}
\begin{figure}[b]
\centering
\includegraphics{fig2.png}
\caption{Representative ground states for each of the interactions depicted in Fig.~\ref{fig1}(b). Arrows denote the orientations of MDABCO $C_{3v}$ axes (\emph{i.e.}\ N--H bond vectors) and are coloured accordingly. (a) Hydrogen-bonding interactions give a correlated-disordered ground state with $Pm\bar3m$ symmetry. (b) Dipolar interactions drive an ordered state with $I23$ symmetry. (c) Strain coupling gives a partially-ordered (nematic) state, the average structure of which has $R\bar3$ symmetry. None of the three states is polar.}
\label{fig2}
\end{figure}
Our starting point is to consider in turn the implications of each of these three interaction types by themselves. Taking first the case of hydrogen-bonding, we recall that each MDABCO$^{2+}$ ion forms its strongest hydrogen bonds with a triplet of halide ions that share a triangular face of the AX$_{12}$ cuboctahedron (see SI for further discussion).\cite{Zhang_2017,Ye_2018} Hydrogen bonding reduces the effective charge on each of these three anions, such that they are now less available for additional bonding to the A-site cations of neighbouring perovskite cages. Hence the orientation of one MDABCO constrains those of its neighbours, generating a `hydrogen-bonding rule' that couples MDABCO orientations and at face value may imply long-range FE order. We identify the implications for crystal symmetry by carrying out MC simulations driven by the coarse-grained energy
\begin{equation}\label{hamil1}
E_{\rm HB}=H\sum_{j\in\{\rm X\}}(n_j-1)^2.
\end{equation}
Here the sum is over all X-site anions in the MC configuration (a supercell of the aristotypic ABX$_3$ cell), $n_j$ is the number of MDABCO cations strongly hydrogen-bonded to the $j^{\rm th}$ anion, and $H>0$ is the energy penalty for double-bonding. The ground state of Eq.~\eqref{hamil1} is the set of configurations for which each anion is hydrogen-bonded to a single MDABCO cation, and the form of this equation can be rationalised as the leading term in the expansion of the crystal energy around this minimum. Our MC simulations show the ground state to be disordered---but strongly correlated---such that its average crystal symmetry remains $Pm\bar3m$ [Fig.~\ref{fig2}(a)]. There is a strong conceptual parallel to the disordered `chain' structure of paraelectric BaTiO$_3$, where Ti--O covalency drives a strongly-correlated nonpolar state.\cite{Comes_1970,Senn_2016}
In a similar manner, MC simulations also allow us to test the symmetry-breaking implications of dipole--dipole interactions. The corresponding configurational energy is now
\begin{equation}\label{hamil2}
E_{\rm dip}=D\sum_{i\ne j}\frac{\mathbf S_i\cdot\mathbf S_j-3(\mathbf S_i\cdot\hat{\mathbf r}_{ij})(\mathbf S_j\cdot\hat{\mathbf r}_{ij})}{(r_{ij}/a)^3},
\end{equation}
where $D$ is the dipolar interaction strength and the sum is over all distinct A-site cations $i,j$ with MDABCO orientations $\mathbf S_i,\mathbf S_j\in\langle111\rangle$. The vector $\hat{\mathbf r}_{ij}$ is the normalised vector between the two cations, $r_{ij}$ their absolute separation, and $a$ the unit-cell length. Our MC implementation\cite{Paddison_2015} takes into account the need for Ewald summation as dipolar interactions are long-range. We find the ground state of Eq.~\eqref{hamil2} to be the fully-ordered but non-polar arrangement with $I23$ crystal symmetry shown in Fig.~\ref{fig2}(b). This particular arrangement, the generalised form of which is known from theory,\cite{Belobrov_1983} is related to that found in the low-temperature phases of dipolar solids such as CO$_{({\rm s})}$.\cite{Lipscomb_1974}
The third and final interaction type is that of strain coupling. Formally, strain is a rank-two tensor, but because the ovoid shape of the MDABCO cation strains the corresponding perovskite cell along the same $\langle111\rangle$ axis along which the MDABCO is oriented, we can simplify strain coupling in terms of the biquadratic interaction\cite{Coates_2019}
\begin{equation}\label{hamil3}
E_{\rm strain}=-J\sum_{i,j}(\mathbf S_i\cdot\mathbf S_j)^2.
\end{equation}
This sum is over neighbouring A-sites $i,j$, with orientations $\mathbf S_i,\mathbf S_j\in\langle111\rangle$ as above. The magnitude of $J$ determines the strength of strain coupling, and if $J>0$ (as it likely is for the dabconium perovskites; see further discussion below) this coupling is ferroelastic. The ground state of Eq.~\eqref{hamil3} is obvious by inspection: it is the set of configurations with all A-site cations oriented parallel or antiparallel to the same $\langle111\rangle$ axis. This state is disordered with $R\bar3$ symmetry [Fig.~\ref{fig2}(c)] and so again is non-polar. In the context of magnetic and liquid-crystal statistical mechanics, this is a `nematic' phase with quadrupolar (\emph{cf} strain)---but not dipolar---order.
Consequently, we can conclude that none of the three microscopic interactions proposed in the literature can by themselves account for inversion-symmetry breaking in the dabconium perovskites. Hence, we consider the different combinations of the interactions in turn, in order to examine whether applied together they explain the polar $R3$ ground state observed experimentally.
We next show that hydrogen-bonding rules, applied in conjunction with dipole--dipole interactions or strain coupling, also fail to produce a macroscopic polarisation. To do this, we compare
three very simple (ordered) models of MDABCO orientations that all strictly obey the hydrogen-bonding rules discussed above (\emph{i.e.} $E_{\rm{HB}}=0$) [Fig.~\ref{fig3}]. Model 1 is the $R3$ structure itself; Model 2 is an AFE variant of this structure with $R\bar3$ symmetry where the MDABCO orientation alternates in a checkerboard fashion from cell to cell (R-type AFE order); and Model 3 is a ``ferri''-electric $2\times1\times1$ supercell of the $R3$ structure with monoclinic $Pc$ symmetry in which neighbouring planes of A-site cations point alternately along $[111]$ and $[11\bar1]$ axes.
\begin{figure}
\centering
\includegraphics{fig3.png}
\caption{Three model systems used to consider the importance of hydrogen-bonding rules in stabilising the polar $R3$ phase. (a) {\bf Model 1:} The FE $R3$ phase itself. (b) {\bf Model 2:} An (R-point) AFE variant with $R\bar3$ symmetry; note that the dipole orientation alternates from cell to cell. (c) {\bf Model 3:} A ferrielectric arrangement in which successive planes switch one component of their polarisation. In each case the underlying dipole arrangements are shown on the left, and the corresponding DFT-relaxed structures are shown on the right. All models satisfy the hydrogen-bonding rule that each iodide anion is strongly bonded to one and only one MDABCO cation.}
\label{fig3}
\end{figure}
Since Models 1 and 2 involve alignment of A-site cations parallel and antiparallel to a single common axis, the corresponding $E_{\rm strain}$ terms are identical, irrespective of the value of $J$. Hence both are ground states of any interaction model based on hydrogen bonding and strain coupling together---and there is nothing to favour the observed $R3$ state over the $R\bar3$ AFE alternative. By a similar argument, the dipolar interaction energy of Model 3 ($E_{\rm dip}=-2.204D$; determined by numerical evaluation of Eq.~\eqref{hamil2} using Ewald summation) is always more favourable than that of Model 1 ($E_{\rm dip}=-2.094D$) irrespective of the value of $D$. Thus, interaction models based on hydrogen bonding and dipole--dipole interactions are also unable to stabilise the $R3$ state.
By contrast, the combination of dipole--dipole interactions and strain-coupling \emph{does} give a polar ground state with $R3$ symmetry. We obtain this result using MC simulations driven by the combined interaction energy
\begin{equation}\label{dualhamil}
E_{\rm MC}=E_{\rm dip}+E_{\rm strain},
\end{equation}
setting $J\simeq D$. Fig.~\ref{fig4}(a) illustrates our finding that this model exhibits a phase transition at $T_{\rm c}\sim1.5D(,J)$ by tracking the average relative polarisation $P=|\langle\mathbf S\rangle|$ as a function of reduced temperature $T^\prime=T/D$; the low-temperature phase is precisely the $R3$ state. This polar phase is the ground state for all finite $D$ until $D\gtrsim4.6J$, beyond which the $I23$ phase is the more stable. Moreover, since the $R3$ state obeys our hydrogen-bonding rules, it is stable regardless of the importance or otherwise of an additional $E_{\rm HB}$ term. We summarise in Fig.~\ref{fig4}(b) this key result of our analysis. The emergence of polarisation \emph{via} the interplay of two competing interactions---neither of which can by itself break inversion symmetry---has strong conceptual parallels to the phenomenology of hybrid-improper ferroelectrics.\cite{Bousquet_2008,Benedek_2011,Wolpert_2019}
\begin{figure}
\centering
\includegraphics{fig4.png}
\caption{Emergence of spontaneous polarisation in the ground state of Eq.~\eqref{dualhamil} (a) Temperature dependence of the bulk polarisation as determined using Monte Carlo simulation. The three traces correspond to the simulations in which local orientations are confined to $\langle111\rangle$ (black), $\langle110\rangle$ (blue), and $\langle100\rangle$ axes. The corresponding ground states are represented schematically; only that of the $\langle111\rangle$ system is polar. Error bars are smaller than the symbols. (b) Summary of the symmetry-breaking implications of the three microscopic orientation-coupling mechanisms discussed in the text. For perovskites with polar B-site cations aligned preferentially along $\langle111\rangle$ axes, the combination of strain-coupling and dipole--dipole interactions (black region) is sufficient to stabilise the polar $R3$ ground state observed experimentally in FE dabconium perovskites.}
\label{fig4}
\end{figure}
But how relevant is this coarse-grained model to a physical system such as [MDABCO]RbI$_3$? To answer this question we use DFT calculations to compare the energies of a range of [MDABCO]RbI$_3$ polymorphs with different MDABCO orientational ordering patterns. We consider five such configurations in total: Models 1--3 as described above, and the two simplest additional AFE orderings (formally, corresponding to X-point and M-point AFE order) [Fig.~\ref{fig5}(a)]. On the one hand, the coarse-grained energy of each model can be determined analytically in terms of the two parameters $J$ and $D$ (see SI). And, on the other hand, we can calculate the (real) energies using DFT. Equating the two we find good agreement for the values $D=833$\,K and $J=912$\,K; \emph{i.e.} $D\simeq J$ as above [see comparison of coarse-grained and DFT energies in Fig.~\ref{fig5}(b)]. Hence---to a first approximation---the energetics of this system are captured by the coarse graining implied by Eq.~\eqref{dualhamil}, which finally explains the driving force for FE order.
The variation in DFT energies amongst Models 2, 4, and 5---all of which have the same relative coarse-grained energy $E=+2.094D$---likely reflects the importance of deviations away from $\langle111\rangle $ in A-site orientations. The DFT-relaxed structure of model 4 in particular shows relatively large deviations of this type, which reduce the dipolar energy cost of its AFE ordering pattern and contribute to its stability relative to the other AFE models. Importantly, if we re-run our MC simulations allowing the vectors $\mathbf S_i$ to adopt any orientation---yet retaining some bias towards the $\langle111\rangle$ axes\cite{sianote}---the same FE transition is observed, albeit at lower critical temperature.\cite{tempnote} So strict adherence to $\langle111\rangle$-type orientational anisotropy is not required for inversion-symmetry breaking.
\begin{figure}
\centering
\includegraphics{fig5.png}
\caption{(a) The additional AFE models of [MDABCO]RbI$_3$ used to determine the relative importance of dipolar and strain interactions as described in the text. {\bf Model 4:} An X-point variant of the $R3$ FE ground state with triclinic $P\bar1$ space-group symmetry. {\bf Model 5:} The corresponding M-point AFE structure, which also has $P\bar1$ symmetry, albeit with a different unit cell. In both models, MDABCO orientations are parallel or antiparallel to the $[111]$ axis of the cubic aristotype. The structures shown are those obtained following DFT relaxation; we use the same representation as in Fig.~\ref{fig3}. (b) Comparison between \emph{ab initio} and coarse-grained energies for each of the models 1--5 of [MDABCO]RbI$_3$, given relative to the global energy minimum (Model 1). The reasonable match obtained suggests that the energetics of this system are well captured by the simple coarse-grained model, with similar energy scales for both dipolar and strain interactions.}
\label{fig5}
\end{figure}
But some anisotropy of this type \emph{is} important. As a final calculation, we carried out a set of MC simulations using Eq.~\eqref{dualhamil} but with A-site orientations confined to either $\langle100\rangle$ or $\langle110\rangle$ orientations. While both systems undergo phase transitions on cooling, the ground states of the two models are AFE rather than FE [Fig.~\ref{fig4}(a)].\cite{Luttinger_1946,Schildknecht_2019} So the fact that MDABCO orients along the body diagonal of the perovskite cage---a function of its shape and hydrogen-bonding characteristics---is a crucial ingredient in its own right.
\section{Discussion and concluding remarks}
We might now claim to understand the origin of the FE phase in systems such as [MDABCO]RbI$_3$. The dipole moment of MDABCO$^{2+}$ is certainly important, but equally important is that this moment is aligned along a $\langle111\rangle$ body-diagonal of the perovskite cage. The displacement of MDABCO$^{2+}$ ions along this same axis amplifies the corresponding effective dipole moment.\cite{Wang_2019} The through-space interactions between neighbouring dipoles are always present, and turn out to be a crucial ingredient for collective polarisation. The final necessary component is strain coupling---the tendency for the strains in neighbouring cells to coalign. While hydrogen-bonding or covalency effects may help stabilise the FE state, they cannot explain its occurrence.
From a design perspective, there are two aspects one might wish to control---(i) A-site dipoles with a tendency to orient preferentially along $\langle111\rangle$, and (ii) ferroelastic strain coupling. The former is likely the easier (or at least more obvious): use ovoid A-site cations with three-fold rotational symmetry, small enough to stabilise the perovskite structure\cite{Mitzi_2001,Kieslich_2014,Burger_2018} but large enough to strain the perovskite cage along $\langle111\rangle$. The balance between ferroelastic and antiferroelastic strain coupling is less obviously navigated,\cite{Evans_2016} although we expect that for strains along $\langle111\rangle$ the ferroelastic case may be the more common. Antiferroelastic strain ($J<0$ in Eq.~\eqref{hamil3}) stabilises the same $I23$ ground state observed for dipolar interactions alone. An important distinction between this structure and the polar $R3$ structure is the extent of conformational degree of freedom of the BX$_6$ octahedra. In the ferroelastic $R3$ case, symmetry allows rotations of the BX$_6$ octahedra around the three-fold axis and distortions of their internal X--B--X angles---both of which help to accommodate the MDABCO$^{2+}$ ion and optimise hydrogen-bonding (as observed experimentally).\cite{Zhang_2017} By contrast, the B-site point symmetry of the antiferroelastic $I23$ structure forbids rotations and bond-angle distortions, allowing only modulation of the lengths of B--X bonds, which are much more expensive to distort.
With these design principles in mind, it is straightforward to rationalise why many well-known hybrid perovskites with polar A-site cations are nevertheless themselves apolar: the A-site dipoles are not aligned along $\langle111\rangle$ and/or the strain coupling is antiferroelastic. But the rules we develop are by no means exhaustive, since (obviously) other mechanisms can and do also drive FE order (see Refs.~\citenum{Pan_2017,Ferrandin_2019} and the ODABCO[NH$_4$]Cl$_3$ example in Ref.~\citenum{Ye_2018} for relevant counterexamples). So the key remaining challenge is to develop a more general understanding of which combinations of local symmetry breaking and various short- and long-range interactions can break inversion symmetry in order/disorder FE materials.
\section*{Conflicts of interest}
There are no conflicts to declare.
\section*{Acknowledgements}
A.L.G. gratefully acknowledge the E.R.C. for funding (Advanced Grant 788144). H.H.-M.Y. thanks the University of Birmingham for Startup funds.
\renewcommand\refname{Notes and references}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 4,977 |
<?php
namespace Symfony\Component\DomCrawler\Field;
/**
* TextareaFormField represents a textarea form field (an HTML textarea tag).
*
* @author Fabien Potencier <fabien@symfony.com>
*
* @api
*/
class TextareaFormField extends FormField
{
/**
* Initializes the form field.
*
* @throws \LogicException When node type is incorrect
*/
protected function initialize()
{
if ('textarea' != $this->node->nodeName) {
throw new \LogicException(sprintf('A TextareaFormField can only be created from a textarea tag (%s given).', $this->node->nodeName));
}
$this->value = null;
foreach ($this->node->childNodes as $node) {
$this->value .= $this->document->saveXML($node);
}
}
}
| {
"redpajama_set_name": "RedPajamaGithub"
} | 3,325 |
\section{Introduction}
\label{sec:intro}
The discovery of large numbers of interstellar and circumstellar species
regularly refreshes our understanding of the physical conditions of the sources of
astrochemical interest \citep{herb06}. Astronomical observations along with laboratory investigations
of various meteoritic samples have discovered the presence of numerous organic molecules of
biological interest \citep{cron93}. It is also believed that the production of such molecules in star-
and planet-forming regions of interstellar clouds, which tend to be partially saturated species
containing the elements nitrogen and/or oxygen in addition to carbon and hydrogen, should be connected in some manner with the production
of terrestrial bio-molecules. Other types of organic molecules are also present in the ISM.
There is strong evidence for species astronomers refer to as ``carbon chains" in cold and dense
interstellar clouds. These carbon chains are unsaturated and linear species, which can be
simple hydrocarbons or species with
other heavy atoms such as cyanopolyynes, which contain a terminal cyano (CN) group. Various
infrared (UIR) emission bands in the $3-15$ $\mu m$ range have been observed
in different astrophysical sources \citep{alla85,tiel87}. Laboratory investigations along
with theoretical calculations led to the hypothesis that the carriers of these bands are aromatic in nature, consisting most probably
of free molecular polycyclic aromatic hydrocarbons (PAHs), possibly with other heavy atoms
such as nitrogen
\citep{sala14,nobl15}.
Other suggestions include surface functional groups on small grains,
quenched carbonaceous composites,
amorphous carbon, hydrogenated amorphous carbon and condensed phase PAHs
\citep{bren92, jage09}.
Recently, two fullerenes,
C$_{60}$ and C$_{70}$, have been discovered in infrared emission in post-stellar objects \citep{cami10} while the cation C$_{60}^{+}$ has been confirmed in near-infrared absorption in a diffuse cloud \citep{walk15}.
In order to understand the synthesis of PAHs, either in interstellar or circumstellar regions,
it is essential to understand
the formation of the six-member aromatic species, benzene
($\mathrm{C_6H_6}$). So far there are only two experimentally studied pathways that might result in the synthesis
of interstellar or circumstellar benzene. The first is the addition of three
acetylene ($\mathrm{C_2H_2}$) molecules \citep{zhou10} and the second is the
recombination of two propargyl (C$_{3}$H$_{3}$) radicals \citep{wils03}. The formation of these radicals could occur in a number of ways.
\cite{shar14} carried out an experiment to study the
thermal decomposition of PA, and found the products to include
$\mathrm{OH}$ and $\mathrm{C_3H_3}$, suggesting that PA could be a precursor to benzene formation.
In addition, \cite{siva15} found that benzene is the major product from PA
irradiation, and suggested that the dissociation of PA plays a key role in the
synthesize of benzene in interstellar icy mantles.
Since PA might play a crucial role in the formation of PAH molecules, it is of interest
to explore various aspects of its interstellar chemistry and spectroscopy
in detail. Although PA has not been detected unambiguously in the ISM,
propenal (CH$_{2}$CHCHO), one of it isomers,
has been detected \citep{holl04} towards the star-forming region Sgr B2(N).
\cite{requ08} estimated the abundance of CH$_{2}$CHCHO to be around $0.3-2.3 \times 10^{-9}$ with respect to the
$\rm {H_2}$ molecule in the galactic center.
Moreover, PA has a well-known rotational spectrum.
Depending upon the internal motion of the OH group, PA could possess two
stable conformers, named $gauche$ and $trans$.
However, microwave studies of PA show that the molecule exists only
as the $gauche$ isomer, in which the hydroxyl H atom lies $\sim 60^{\circ}$
out of the $\mathrm{H-C\equiv C-C-O}$ plane \citep{hiro68}. \cite{pear05} summarized other rotational studies of PA, extended the experimental
work of \cite{hiro68} through 600 GHz and obtained rotational and distortional constants for the gauche form of PA and its -OD singly deuterated
isotopomer. According to their studies, the $gauche$ state is split by inversion into two states, separated by 652.4 GHz for normal PA and 213.5 GHz for the -OD isotopomer. Other spectroscopic work on PA and related species has also been undertaken. \cite{nyqu71} recorded and assigned Infrared and
Raman spectra for PA, and its deuterated isotopomers, while
\cite{deve13} carried out experiments to
determine the structure of the Ar..PA complex and its two deuterated isotopologues.
They found PA to have a $gauche$ structure, with Ar located in between the -OH and -C$\equiv$C-H groups.
In another study, \cite{deve14} carried out experiments for the pure rotational spectra of the PA
dimer and its three deuterium isotopologues.
\begin{table*}
\vbox{
\vskip -3cm
\addtolength{\tabcolsep}{-5pt}
\centering{
\scriptsize
\caption{Formation and destruction pathways of PA and its related species.}
\renewcommand{\arraystretch}{0}
\begin{tabular}{|p{0.8in}|p{2.6in}|p{0.7in} p{0.35in} p{0.3in} p{1.1in}|p{1.2in}|}
\hline
{\bf Reaction number}&{\bf Reaction } &{\leavevmode\color{black}$\alpha$}&{\leavevmode\color{black}$\beta$}&{\leavevmode\color{black}$\gamma$}&{\bf Gas phase}& {\bf Ice phase} \\
{\bf (type)}&&&&&{\bf rate coefficient }& {\bf rate coefficient }\\
{}&&&&&{\bf at $T=10$ K ($100$ K)}& {\bf at $T=10$ K ($100$K)}\\
&&&&&&\\
\multicolumn{3}{p{2in}}{\bf Formation pathways}\\
&&&&&&\\
R1 (NR)&$\mathrm{O(^{3}P)+C_3H_3\rightarrow HC_{2}CHO+H}$($-252.30^a$,$-231.11^b$) &{\leavevmode\color{black}$2.3 \times 10^{-10}$} &{\leavevmode\color{black}0.0}&{\leavevmode\color{black}0.0}&$2.3 \times 10^{-10}$($2.3 \times 10^{-10}$)$^x$&$2.83\times 10^{-7}$($2.51 \times 10^{4}$)$^y$\\
R2 (NR)&$\mathrm{C_2H+H_2CO\rightarrow HC_{2}CHO+H}$($-109.17^a$,$-106.307^b$) &{\leavevmode\color{black}$1.00 \times 10^{-10}$} &{\leavevmode\color{black}0.0}&{\leavevmode\leavevmode\color{black}0.0}& $1.00 \times 10^{-10}$($1.00 \times 10^{-10}$)$^x$&$2.84\times 10^{-17}$($3.03\times 10^{3}$)$^y$\\
R3 (NR)&$\mathrm{HC_2CHO+H\rightarrow HC_2CH_2O}$ ($-86.69^a$,$-86.23^b$)&{\leavevmode\color{black}-}&{\leavevmode\color{black}-}&{\leavevmode\color{black}-}& $0$($1.96 \times 10^{-15}$)$^x$&$6.24 \times 10^{-9}$ ($8.94\times 10^{-3}$)$^y$\\
R4 (RR)&$\mathrm{HC_2CH_2O+H\rightarrow HC_{2}CH_2OH}$ ($-420.35^a$,$-421.67^b$) &{\leavevmode\color{black}$1.00 \times 10^{-10}$} &{\leavevmode\color{black}0.0}&{\leavevmode\color{black}0.0}&$1.00 \times 10^{-10}$($1.00 \times 10^{-10}$)$^x$&$1.77 \times 10^{-1}$ ($2.54\times 10^{5}$)$^y$\\
R5 (RR)&$\mathrm{C_3H_3+OH\rightarrow HC_2CH_2OH}$ ($-301.35^a$,$-298.97^b$)&{\leavevmode\color{black}$1.00 \times 10^{-10}$}&{\leavevmode\color{black}0.0}&{\leavevmode\color{black}0.0}& $1.00\times 10^{-10}$($1.00 \times 10^{-10}$)$^x$&$7.86 \times 10^{-29}$($2.19\times 10^{2}$)$^y$ \\
{ R6 (DR)}&$\mathrm{ {HC_2CH_2OH_2}^+ + e^- \rightarrow HC_2CH_2OH+H}$ &{\leavevmode\color{black}$2.67 \times 10^{-8}$}&{\leavevmode\color{black}$-0.59$}&{\leavevmode\color{black}0.0}& { $2.00 \times 10^{-7}$($5.16 \times 10^{-8}$)$^x$} &-\\
&&&&&&\\
\multicolumn{3}{p{2in}}{\bf Destruction pathways}\\
&&&&&&\\
{ R7 (NR)}&$\mathrm{ {HC_2CH_2OH} + OH \rightarrow HOC_2HHCHOH}$ &{\leavevmode\color{black}$9.20 \times 10^{-12}$}&{\leavevmode\color{black}0.0}&{\leavevmode\color{black}0.0}& $9.20 \times 10^{-12}$($9.20 \times 10^{-12}$)$^x$&$3.58 \times 10^{-38}$($3.64\times 10^{1}$)$^y$ \\
{ R8 (NR)}&$\mathrm{ {HC_2CH_2OH} + OD \rightarrow HOC_2HHCHOD}$ &{\leavevmode\color{black}$9.20 \times 10^{-12}$}&{\leavevmode\color{black}0.0}&{\leavevmode\color{black}0.0}& $9.20 \times 10^{-12}$($9.20 \times 10^{-12}$)$^x$&$3.48 \times 10^{-38}$($3.54\times 10^{1}$)$^y$ \\
R9 (IN)&$\mathrm{HC_2CH_2OH + C^{+} \rightarrow HC_2CH_2O^{+} + CH}$ &{\leavevmode\color{black}$2.03 \times 10^{-09}$}&{\leavevmode\color{black}-0.5}&{\leavevmode\color{black}0.0}& $1.22\times 10^{-08}$($ 4.63 \times 10^{-09}$)$^x$&-\\
R10 (IN)&$\mathrm{HC_2CH_2OH + C^{+} \rightarrow C_3H_3^{+} + HCO}$ &{\leavevmode\color{black}$2.03 \times 10^{-09}$}&{\leavevmode\color{black}-0.5}&{\leavevmode\color{black}0.0}& $1.22\times 10^{-08}$($ 4.61\times 10^{-09}$)$^x$&-\\
R11 (IN)&$\mathrm{HC_2CH_2OH + H_3O^{+}\rightarrow {HC_2CH_2OH_2}^{+} + H_2O}$ &{\leavevmode\color{black}$1.70 \times 10^{-09}$}&{\leavevmode\color{black}-0.5}&{\leavevmode\color{black}0.0}& $1.02\times 10^{-08}$($ 3.85\times 10^{-09}$)$^x$&- \\
R12 (IN)&$\mathrm{HC_2CH_2OH + HCO^{+}\rightarrow {HC_2CH_2OH_2}^{+} + CO }$ &{\leavevmode\color{black}$1.46 \times 10^{-09}$}&{\leavevmode\color{black}-0.5}&{\leavevmode\color{black}0.0}& $8.79\times 10^{-09}$($ 3.32 \times 10^{-09}$)$^x$&- \\
R13 (IN)&$\mathrm{HC_2CH_2OH + H_3^{+}\rightarrow C_3H_3^{+} + H_2O + H_2}$ &{\leavevmode\color{black}$3.78 \times 10^{-09}$}&{\leavevmode\color{black}-0.5}&{\leavevmode\color{black}0.0}& $2.27\times 10^{-08}$($ 8.56\times 10^{-09}$)$^x$&- \\
R14 (IN)&$\mathrm{HC_2CH_2OH + H_3^{+}\rightarrow {HC_2CH_2OH_2}^{+} + H_2}$ &{\leavevmode\color{black}$3.78 \times 10^{-09}$}&{\leavevmode\color{black}-0.5}&{\leavevmode\color{black}0.0}& $2.27\times 10^{-08}$($ 8.56\times 10^{-09}$)$^x$&- \\
R15 (IN)&$\mathrm{HC_2CH_2OH + He^{+}\rightarrow C_3H_3^{+} + He + OH}$ &{\leavevmode\color{black}$3.31 \times 10^{-09}$}&{\leavevmode\color{black}-0.5}&{\leavevmode\color{black}0.0}& $ 1.99\times 10^{-08}$($7.51\times 10^{-09}$)$^x$&- \\
R16 (IN)&$\mathrm{HC_2CH_2OH + He^{+}\rightarrow C_3H_3 + He + OH^+}$ &{\leavevmode\color{black}$3.31 \times 10^{-09}$}&{\leavevmode\color{black}-0.5}&{\leavevmode\color{black}0.0}& $ 1.99\times 10^{-08}$($ 7.51 \times 10^{-09}$)$^x$&- \\
R17 (IN)&$\mathrm{HC_2CH_2OH + CH_3^{+}\rightarrow HC_2CH_2O^{+} + CH_4}$ &{\leavevmode\color{black}$1.86 \times 10^{-09}$}&{\leavevmode\color{black}-0.5}&{\leavevmode\color{black}0.0}& $1.12\times 10^{-08}$($ 4.21\times 10^{-09}$)$^x$&- \\
R18 (IN)&$\mathrm{HC_2CH_2OH + H^{+}\rightarrow C_3H_3^{+} + H_2O}$ &{\leavevmode\color{black}$6.43 \times 10^{-09}$}&{\leavevmode\color{black}-0.5}&{\leavevmode\color{black}0.0}& $3.86\times 10^{-08}$($ 1.46 \times 10^{-08}$)$^x$&- \\
R19 (IN)&$\mathrm{HC_2CH_2OH + H^{+}\rightarrow HC_2CH_2O^{+} + H_2}$ &{\leavevmode\color{black}$6.43 \times 10^{-09}$}&{\leavevmode\color{black}-0.5}&{\leavevmode\color{black}0.0}& $3.86\times 10^{-08}$($ 1.46\times 10^{-08}$)$^x$&- \\
R20 (IN)&$\mathrm{HC_2CH_2OH + H^{+}\rightarrow HC_3O^{+} + H_2+H_2}$ &{\leavevmode\color{black}$6.43 \times 10^{-09}$}&{\leavevmode\color{black}-0.5}&{\leavevmode\color{black}0.0}& $3.86\times 10^{-08}$($ 1.46 \times 10^{-08}$)$^x$&- \\
R21 (IN)&$\mathrm{HC_2CH_2OH + H^{+}\rightarrow HC_2CH_2OH^{+} + H}$ &{\leavevmode\color{black}$6.43 \times 10^{-09}$}&{\leavevmode\color{black}-0.5}&{\leavevmode\color{black}0.0}& $3.86\times 10^{-08}$($ 1.46\times 10^{-08}$)$^x$&- \\
R22 (IN)&$\mathrm{HC_2CH_2OH + H_2D^{+}\rightarrow HC_2CH_2OHD^+ + H_2}$ &{\leavevmode\color{black}$3.21 \times 10^{-09}$}&{\leavevmode\color{black}-0.5}&{\leavevmode\color{black}0.0}& $1.93\times 10^{-08}$($ 7.29\times 10^{-09}$)$^x$&- \\
R23 (IN)&$\mathrm{HC_2CHO + C^{+} \rightarrow HC_3O^{+} + CH}$ &{\leavevmode\color{black}$3.80 \times 10^{-09}$}&{\leavevmode\color{black}-0.5}&{\leavevmode\color{black}0.0}& $2.19\times 10^{-08}$($7.64 \times 10^{-09}$)$^x$&-\\
R24 (IN)&$\mathrm{HC_2CHO + C^{+} \rightarrow C_3H^{+} + HCO}$ &{\leavevmode\color{black}$3.80 \times 10^{-09}$}&{\leavevmode\color{black}-0.5}&{\leavevmode\color{black}0.0}& $2.19\times 10^{-08}$($ 7.64 \times 10^{-09}$)$^x$&-\\
R25 (IN)&$\mathrm{HC_2CHO + H_3O^{+}\rightarrow {HC_2CH_2O}^{+} + H_2O}$ &{\leavevmode\color{black}$3.17 \times 10^{-09}$}&{\leavevmode\color{black}-0.5}&{\leavevmode\color{black}0.0}& $1.83\times 10^{-08}$($ 6.38 \times 10^{-09}$)$^x$&- \\
R26 (IN)&$\mathrm{HC_2CHO + HCO^{+}\rightarrow {HC_2CH_2O}^{+} + CO }$ &{\leavevmode\color{black}$2.74 \times 10^{-09}$}&{\leavevmode\color{black}-0.5}&{\leavevmode\color{black}0.0}& $1.58\times 10^{-08}$($ 5.51\times 10^{-09}$)$^x$&- \\
R27 (IN)&$\mathrm{HC_2CHO + H_3^{+}\rightarrow C_3H^{+} + H_2O + H_2}$ &{\leavevmode\color{black}$7.03 \times 10^{-09}$}&{\leavevmode\color{black}-0.5}&{\leavevmode\color{black}0.0}& $4.04\times 10^{-08}$($ 1.41 \times 10^{-08}$)$^x$&- \\
R28 (IN)&$\mathrm{HC_2CHO + H_3^{+}\rightarrow {HC_2CH_2O}^{+} + H_2}$ &{\leavevmode\color{black}$7.03 \times 10^{-09}$}&{\leavevmode\color{black}-0.5}&{\leavevmode\color{black}0.0}& $4.04\times 10^{-08}$($ 1.41\times 10^{-08}$)$^x$&- \\
R29 (IN)&$\mathrm{HC_2CHO + He^{+}\rightarrow C_3H^{+} + He + OH}$ &{\leavevmode\color{black}$6.17 \times 10^{-09}$}&{\leavevmode\color{black}-0.5}&{\leavevmode\color{black}0.0}& $ 3.55\times 10^{-08}$($ 1.24\times 10^{-08}$)$^x$&- \\
R30 (IN)&$\mathrm{HC_2CHO + He^{+}\rightarrow C_3H + He + OH^+}$ &{\leavevmode\color{black}$6.17 \times 10^{-09}$}&{\leavevmode\color{black}-0.5}&{\leavevmode\color{black}0.0}& $ 3.55\times 10^{-08}$($ 1.24 \times 10^{-08}$)$^x$&- \\
R31 (IN)&$\mathrm{HC_2CHO + CH_3^{+}\rightarrow HC_2CH_2O^{+} + CH_2}$ &{\leavevmode\color{black}$3.47 \times 10^{-09}$}&{\leavevmode\color{black}-0.5}&{\leavevmode\color{black}0.0}& $2.00\times 10^{-08}$ ( $ 6.98 \times 10^{-09}$)$^x$&- \\
R32 (IN)&$\mathrm{HC_2CHO + H^{+}\rightarrow C_3H^{+} + H_2O}$ &{\leavevmode\color{black}$1.20 \times 10^{-08}$}&{\leavevmode\color{black}-0.5}&{\leavevmode\color{black}0.0}& $6.89\times 10^{-08}$ ( $ 2.41 \times 10^{-08}$)$^x$&- \\
R33 (IN)&$\mathrm{HC_2CHO + H^{+}\rightarrow C_3H_2 + OH^+}$ &{\leavevmode\color{black}$1.20 \times 10^{-08}$}&{\leavevmode\color{black}-0.5}&{\leavevmode\color{black}0.0}& $6.89\times 10^{-08}$ ( $ 2.41 \times 10^{-08}$)$^x$&- \\
R34 (IN)&$\mathrm{HC_2CHO + H^{+}\rightarrow HC_3O^{+} + H_2}$ &{\leavevmode\color{black}$1.20 \times 10^{-08}$}&{\leavevmode\color{black}-0.5}&{\leavevmode\color{black}0.0}& $6.89\times 10^{-08}$ ( $ 2.41 \times 10^{-08}$)$^x$&- \\
R35 (IN)&$\mathrm{HC_2CHO + H^{+}\rightarrow C_3O^{+} + H_2}$ &{\leavevmode\color{black}$1.20 \times 10^{-08}$}&{\leavevmode\color{black}-0.5}&{\leavevmode\color{black}0.0}& $6.89\times 10^{-08}$ ($ 2.41\times 10^{-08}$)$^x$&- \\
R36 (IN)&$\mathrm{HC_2CHO + H_2D^{+}\rightarrow HC_2CH_2O^+ + HD}$ &{\leavevmode\color{black}$5.98 \times 10^{-09}$}&{\leavevmode\color{black}-0.5}&{\leavevmode\color{black}0.0}& $3.44\times 10^{-08}$ ($ 1.20\times 10^{-08}$)$^x$&- \\
R37 (IN)&$\mathrm{HC_2CH_2O + C^{+} \rightarrow HC_3O^{+} + CH_2}$ &{\leavevmode\color{black}$1.41 \times 10^{-09}$}&{\leavevmode\color{black}-0.5}&{\leavevmode\color{black}0.0}& $8.86\times 10^{-09}$ ($ 3.57 \times 10^{-09}$)$^x$&-\\
R38 (IN)&$\mathrm{HC_2CH_2O + C^{+} \rightarrow C_3H_2^{+} + HCO}$ &{\leavevmode\color{black}$1.41 \times 10^{-09}$}&{\leavevmode\color{black}-0.5}&{\leavevmode\color{black}0.0}& $8.86\times 10^{-09}$ ($ 3.57 \times 10^{-09}$)$^x$&-\\
R39 (IN)&$\mathrm{HC_2CH_2O + H_3O^{+}\rightarrow {HC_2CH_2O}^{+} + H_2O}$ &{\leavevmode\color{black}$1.18 \times 10^{-09}$}&{\leavevmode\color{black}-0.5}&{\leavevmode\color{black}0.0}& $7.40\times 10^{-09}$ ($ 2.92 \times 10^{-09}$)$^x$&- \\
R40 (IN)&$\mathrm{HC_2CH_2O + HCO^{+}\rightarrow {HC_2CH_2O}^{+} + HCO }$ &{\leavevmode\color{black}$1.02 \times 10^{-09}$}&{\leavevmode\color{black}-0.5}&{\leavevmode\color{black}0.0}& $6.38\times 10^{-09}$ ($ 2.58 \times 10^{-09}$)$^x$&- \\
R41 (IN)&$\mathrm{HC_2CH_2O + H_3^{+}\rightarrow C_3H_2^{+} + H_2O + H_2}$ &{\leavevmode\color{black}$2.62 \times 10^{-09}$}&{\leavevmode\color{black}-0.5}&{\leavevmode\color{black}0.0}& $1.64\times 10^{-08}$ ($ 6.62 \times 10^{-09}$)$^x$&- \\
R42 (IN)&$\mathrm{HC_2CH_2O + H_3^{+}\rightarrow {HC_2CH_2O}^{+} + H_2+H}$ &{\leavevmode\color{black}$2.62 \times 10^{-09}$}&{\leavevmode\color{black}-0.5}&{\leavevmode\color{black}0.0}& $1.64\times 10^{-08}$ ($ 6.62 \times 10^{-09}$)$^x$&- \\
R43 (IN)&$\mathrm{HC_2CH_2O + He^{+}\rightarrow C_3H^{+} + He + H_2O}$ &{\leavevmode\color{black}$2.30 \times 10^{-09}$}&{\leavevmode\color{black}-0.5}&{\leavevmode\color{black}0.0}& $ 1.44\times 10^{-08}$ ($ 5.80 \times 10^{-09}$)$^x$&- \\
R44 (IN)&$\mathrm{HC_2CH_2O + He^{+}\rightarrow C_4H + He + OH^+}$ &{\leavevmode\color{black}$2.30 \times 10^{-09}$}&{\leavevmode\color{black}-0.5}&{\leavevmode\color{black}0.0}& $ 1.44\times 10^{-08}$ ($ 5.80 \times 10^{-09}$)$^x$&- \\
R45 (IN)&$\mathrm{HC_2CH_2O + CH_3^{+}\rightarrow HC_2CH_2O^{+} + CH_3}$ &{\leavevmode\color{black}$1.29 \times 10^{-09}$}&{\leavevmode\color{black}-0.5}&{\leavevmode\color{black}0.0}& $8.09\times 10^{-09}$ ($ 3.26 \times 10^{-09}$)$^x$&- \\
R46 (IN)&$\mathrm{HC_2CH_2O + H^{+}\rightarrow C_3H_2^{+} + H_2O}$ &{\leavevmode\color{black}$4.46 \times 10^{-09}$}&{\leavevmode\color{black}-0.5}&{\leavevmode\color{black}0.0}& $2.80\times 10^{-08}$ ($ 1.13 \times 10^{-08}$)$^x$&- \\
R47 (IN)&$\mathrm{HC_2CH_2O + H^{+}\rightarrow C_3H_2 +OH^+ + H}$ &{\leavevmode\color{black}$4.46 \times 10^{-09}$}&{\leavevmode\color{black}-0.5}&{\leavevmode\color{black}0.0}& $2.80\times 10^{-08}$ ($ 1.13 \times 10^{-08}$)$^x$&- \\
R48 (IN)&$\mathrm{HC_2CH_2O + H^{+}\rightarrow HC_3O^{+} + H_2+H}$ &{\leavevmode\color{black}$4.46 \times 10^{-09}$}&{\leavevmode\color{black}-0.5}&{\leavevmode\color{black}0.0}& $2.80\times 10^{-08}$ ($ 1.13 \times 10^{-08}$)$^x$&- \\
R49 (IN)&$\mathrm{HC_2CH_2O + H^{+}\rightarrow HC_3O^{+} + H_2}$ &{\leavevmode\color{black}$4.46 \times 10^{-09}$}&{\leavevmode\color{black}-0.5}&{\leavevmode\color{black}0.0}& $2.80\times 10^{-08}$ ($1.13\times 10^{-08}$)$^x$&- \\
R50 (IN)&$\mathrm{HC_2CH_2O + H_2D^{+}\rightarrow HC_2CH_2O^+ + HD + H}$ &{\leavevmode\color{black}$2.23 \times 10^{-09}$}&{\leavevmode\color{black}-0.5}&{\leavevmode\color{black}0.0}& $1.40\times 10^{-08}$ ($ 5.63\times 10^{-09}$)$^x$&- \\
R51 (PH)&$\mathrm{HC_2CH_2OH + PHOTON \rightarrow C_3H_3 + OH}$ &{\leavevmode\color{black}$6.0\times 10^{-10}$}&{\leavevmode\color{black}0.0}&{\leavevmode\color{black}1.8}& $9.14 \times 10^{-18}$($9.14 \times 10^{-18}$)$^y$&$9.14\times 10^{-18}$ ($9.14\times 10^{-18}$)$^y$\\
R52 (PH)&$\mathrm{ HC_2CH_2OH + PHOTON \rightarrow HC_2CH_2O + H}$ &{\leavevmode\color{black}$6.0\times 10^{-10}$}&{\leavevmode\color{black}0.0}&{\leavevmode\color{black}1.8}& $9.14 \times 10^{-18}$($9.14 \times 10^{-18}$)$^y$&$9.14\times 10^{-18}$ ($9.14\times 10^{-18}$)$^y$\\
R53 (PH)&$\mathrm{HC_2CH_2O + PHOTON \rightarrow C_3H_2 + OH}$ &{\leavevmode\color{black}$6.0\times 10^{-10}$}&{\leavevmode\color{black}0.0}&{\leavevmode\color{black}1.8}&$9.14 \times 10^{-18}$($9.14 \times 10^{-18}$)$^y$&$9.14\times 10^{-18}$ ($9.14\times 10^{-18}$)$^y$\\
R54 (PH)&$\mathrm{HC_2CH_2O + PHOTON \rightarrow H_2CO + C_2H }$ &{\leavevmode\color{black}$6.0\times 10^{-10}$}&{\leavevmode\color{black}0.0}&{\leavevmode\color{black}1.8}& $9.14 \times 10^{-18}$($9.14 \times 10^{-18}$)$^y$&$9.14\times 10^{-18}$ ($9.14\times 10^{-18}$)$^y$\\
R55 (PH)&$\mathrm{ HC_2CH_2O + PHOTON \rightarrow HC_2CHO + H }$ &{\leavevmode\color{black}$6.0\times 10^{-10}$ }&{\leavevmode\color{black}0.0}&{\leavevmode\color{black}1.8}& $9.14 \times 10^{-18}$($9.14 \times 10^{-18}$)$^y$&$9.14\times 10^{-18}$ ($9.14\times 10^{-18}$)$^y$\\
R56 (PH)&$\mathrm{HC_2CHO + PHOTON \rightarrow C_2H + HCO }$ &{\leavevmode\color{black} $6.0\times 10^{-10}$}&{\leavevmode\color{black}0.0}&{\leavevmode\color{black}1.8}& $9.14 \times 10^{-18}$($9.14 \times 10^{-18}$)$^y$&$9.14\times 10^{-18}$ ($9.14\times 10^{-18}$)$^y$\\
R57 (PH)&$\mathrm{HOC_2HHCHOH + PHOTON \rightarrow HC_2CH_2OH + OH}$ &{\leavevmode\color{black}$6.0\times 10^{-10}$}&{\leavevmode\color{black}0.0}&{\leavevmode\color{black}1.8}& $9.14 \times 10^{-18}$($9.14 \times 10^{-18}$)$^y$&$9.14\times 10^{-18}$ ($9.14\times 10^{-18}$)$^y$\\
R58 (PH)&$\mathrm{HOC_2HHCHOD + PHOTON \rightarrow HC_2CH_2OH + OD}$ &{\leavevmode\color{black}$6.0\times 10^{-10}$}&{\leavevmode\color{black}0.0}&{\leavevmode\color{black}1.8}& $9.14 \times 10^{-18}$($9.14 \times 10^{-18}$)$^y$&$9.14\times 10^{-18}$ ($9.14\times 10^{-18}$)$^y$\\
R59 (CR)&$\mathrm{ HC_2CH_2OH + CRPHOT \rightarrow C_3H_3 + OH}$ &{\leavevmode\color{black}$1.3\times 10^{-17}$}&{\leavevmode\color{black}0.0}&{\leavevmode\color{black}752}& $2.44 \times 10^{-14}$($2.44 \times 10^{-14}$)$^y$&$2.44 \times 10^{-14}$ ($2.44\times 10^{-14}$)$^y$\\
R60 (CR)&$\mathrm{ HC_2CH_2OH + CRPHOT \rightarrow HC_2CH_2O + H}$ &{\leavevmode\color{black}$1.3\times 10^{-17}$}&{\leavevmode\color{black}0.0}&{\leavevmode\color{black}752}& $2.44 \times 10^{-14}$($2.44 \times 10^{-14}$)$^y$&$2.44 \times 10^{-14}$ ($2.44\times 10^{-14}$)$^y$\\
R61 (CR)&$\mathrm{ HC_2CH_2O + CRPHOT \rightarrow C_3H_2 + OH}$ &{\leavevmode\color{black}$1.3\times 10^{-17}$}&{\leavevmode\color{black}0.0}&{\leavevmode\color{black}752}& $2.44 \times 10^{-14}$($2.44 \times 10^{-14}$)$^y$&$2.44 \times 10^{-14}$ ($2.44\times 10^{-14}$)$^y$\\
R62 (CR)&$\mathrm{ HC_2CH_2O + CRPHOT \rightarrow H_2CO + C_2H }$ &{\leavevmode\color{black}$1.3\times 10^{-17}$}&{\leavevmode\color{black}0.0}&{\leavevmode\color{black}752}& $2.44 \times 10^{-14}$($2.44 \times 10^{-14}$)$^y$&$2.44\times 10^{-14}$ ($2.44\times 10^{-14}$)$^y$\\
R63 (CR)&$\mathrm{ HC_2CH_2O + CRPHOT \rightarrow HC_2CHO + H }$ &{\leavevmode\color{black}$1.3\times 10^{-17}$}&{\leavevmode\color{black}0.0}&{\leavevmode\color{black}752}& $2.44 \times 10^{-14}$($2.44 \times 10^{-14}$)$^y$&$2.44\times 10^{-14}$ ($2.44\times 10^{-14}$)$^y$\\
R64 (CR)&$\mathrm{ HC_2CHO + CRPHOT \rightarrow C_2H + HCO}$ &{\leavevmode\color{black}$1.3\times 10^{-17}$}&{\leavevmode\color{black}0.0}&{\leavevmode\color{black}752}& $2.44 \times 10^{-14}$($2.44 \times 10^{-14}$)$^y$&$2.44 \times 10^{-14}$ ($2.44\times 10^{-14}$)$^y$\\
R65 (CR)&$\mathrm{ HOC_2HHCHOH + CRPHOT \rightarrow HC_2CH_2OH + OH}$ &{\leavevmode\color{black}$1.3\times 10^{-17}$}&{\leavevmode\color{black}0.0}&{\leavevmode\color{black}752}& $2.44 \times 10^{-14}$($2.44 \times 10^{-14}$)$^y$&$2.44\times 10^{-14}$ ($2.44\times 10^{-14}$)$^y$\\
R66 (CR)&$\mathrm{ HOC_2HHCHOD + CRPHOT \rightarrow HC_2CH_2OH + OD}$ &{\leavevmode\color{black}$1.3\times 10^{-17}$}&{\leavevmode\color{black}0.0}&{\leavevmode\color{black}752}& $2.44 \times 10^{-14}$($2.44 \times 10^{-14}$)$^y$&$2.44\times 10^{-14}$ ($2.44\times 10^{-14}$)$^y$\\
R67 (DR)&$\mathrm{ HC_2CH_2O^+ + e^- \rightarrow CO+C_2H_2+H}$&{\leavevmode\color{black}$2.0\times 10^{-7}$}&{\leavevmode\color{black}-0.5}&{\leavevmode\color{black}0.0}&{$1.10 \times 10^{-6}$($3.46 \times 10^{-7}$) $^x$}&-\\
R68(DR)&$\mathrm{ HC_2CH_2O^+ + e^- \rightarrow HCO+C_2H+H}$ &{\leavevmode\color{black}$2.0\times 10^{-7}$}&{\leavevmode\color{black}-0.5}&{\leavevmode\color{black}0.0}&{$1.10 \times 10^{-6}$}($3.46 \times 10^{-7}$)$^x$&- \\
R69(DR)&$\mathrm{ HC_2CH_2O^+ + e^- \rightarrow H_2CO+C_2H}$ &{\leavevmode\color{black}$2.0\times 10^{-7}$}&{\leavevmode\color{black}-0.5}&{\leavevmode\color{black}0.0}&{$1.10 \times 10^{-6}$}($3.46 \times 10^{-7}$)$^x$&- \\
R70(DR)&$\mathrm{ {HC_2CH_2OH_2}^+ + e^- \rightarrow C_3H_3+H_2O}$ &{\leavevmode\color{black}$8.01\times 10^{-8}$}&{\leavevmode\color{black}-0.59}&{\leavevmode\color{black}0.0}&{$5.95 \times 10^{-7}$($1.52 \times 10^{-7}$)$^x$}&- \\
R71(DR)&$\mathrm{ {HC_2CH_2OH_2}^+ + e^- \rightarrow C_3H_3+OH+H}$ &{\leavevmode\color{black}$4.54\times 10^{-7}$}&{\leavevmode\color{black}-0.59}&{\leavevmode\color{black}0.0}& {$3.34 \times 10^{-6}$($8.6 \times 10^{-7}$)$^x$}&- \\
R72(DR)&$\mathrm{ {HC_2CH_2OH_2}^+ + e^- \rightarrow C_3H_2+H_2O+H}$ &{\leavevmode\color{black}$1.87\times 10^{-7}$}&{\leavevmode\color{black}-0.59}& {\leavevmode\color{black}0.0}&{$1.41 \times 10^{-6}$($3.63 \times 10^{-7}$)$^x$}&-\\
R73(DR)&$\mathrm{ {HC_2CH_2OH_2}^+ + e^- \rightarrow H_2CO+C_2H_2+H}$ &{\leavevmode\color{black}$8.9 \times 10^{-8}$} &{\leavevmode\color{black} -0.59}&{\leavevmode\color{black}0.0}& {$6.6 \times 10^{-7}$($1.7 \times 10^{-7}$)$^x$}&-\\
R74(DR)&$\mathrm{ {HC_2CH_2OH}^+ + e^- \rightarrow C_3H_3+OH}$ &{\leavevmode\color{black}$3.00\times 10^{-7}$}&{\leavevmode\color{black}-0.5} & {\leavevmode\color{black}0.0}&{$1.64 \times 10^{-6}$($5.20 \times 10^{-7}$)$^x$}&- \\
R75(DR)&$\mathrm{ {HC_2CH_2OH}^+ + e^- \rightarrow HC_2CH_2O+H}$ &{\leavevmode\color{black}$3.00\times 10^{-7}$}&{\leavevmode\color{black}-0.5}&{\leavevmode\color{black}0.0}&{$1.64 \times 10^{-6}$($5.20 \times 10^{-7}$)$^x$}&- \\
\hline
\multicolumn{7}{c}{ Notes: For the two-body reactions (R1-R2, R4-R8, R67-R75), rate coefficients are tabulated as $\alpha \ (\frac{T}{300})^{\beta} \ exp(-\frac{\gamma}{T})$. For the photo-dissociation reactions with}\\
\multicolumn{7}{c}{ external interstellar photons (R51-R58), rate coefficients are tabulated as $\alpha \ exp(-\gamma A_V)$. For photo-dissociation by cosmic ray induced photons (R59-R66),} \\
\multicolumn{7}{c}{\color{black} rate coefficients are tabulated as $\alpha \ \frac{\gamma'}{1-\omega}$. For the rate coefficients of ion-polar neutral reactions (R9-R50), we use the relation discussed in \cite{su82}.} \\
\multicolumn{7}{c}{\color{black} Tabulated rate coefficients for these ion-neutral reactions in terms of $\alpha \ (\frac{T}{300})^{\beta} \ exp(-\frac{\gamma}{T})$ are valid for the low temperature regime only.}\\
\multicolumn{3}{c}{$^a$ gas phase enthalpy in kJ/mol}\\
\multicolumn{3}{c}{$^b$ ice phase enthalpy in kJ/mol}\\
\multicolumn{3}{c}{$^x$ rate coefficient in cm$^3$s$^{-1}$}\\
\multicolumn{3}{c}{$^y$ rate coefficient in s$^{-1}$}\\
\end{tabular}}}
\end{table*}
In this paper, we report the use of our interstellar chemical model to explore various pathways for the
formation and destruction of PA (gauche form), and to estimate the possibility of detecting
this molecule in a dense molecular cloud.
Since there are some existing laboratory results for the spectrum of the -OD deuterated form of
PA and since some observational evidence for deuterium fractionation of large complex species exists
(see for instance $\mathrm{DCOOCH_3/HCOOCH_3}$, \cite{demy10}), we also consider the -OD isotopomer of PA.
Various vibrational transitions of PA are computed and compared with the existing experimental results.
The remainder of this paper is organized as follows. In Section 2, we discuss various reaction pathways and
their rate coefficients for the formation and destruction of PA and OD-PA. In {\color{black} Section 3.1,
modeling details are presented while in Section 3.2 we discuss modeling results..} {\color{black} LTE radiative transfer results are presented in
{\color{black} Section 4.1}},
while computed vibrational spectra for PA and OD-PA are
discussed in {\color{black} Section 4.2}. Finally, in {\color{black} Section 5}, we draw our conclusions.
\section{Chemical network}
\label{sec:chem}
\subsection{Formation pathways}
In Table 1, all formation and destruction pathways of PA utilized
are presented with rate coefficients, if applicable, in both the gas and dust phases. The rate coefficients are shown for two temperatures ($T=10$ K and $100$ K)
to represent the temperature dependency (if any). The determination of the rate coefficients actually used is discussed in the next few subsections.
Most rate coefficients for the case of deuterated PA are not very different, and are not tabulated. Reaction numbers R1-R6 of Table 1 represent various possible
pathways for the formation of PA ($\mathrm{HC_2CH_2OH}$). Reaction numbers R1-R5 are found to be
exothermic in both phases and are included in our network. The reaction
exothermicities or endothermicities for
all these reactions have been calculated by using the Gaussian 09 \citep{fris09} program with a
B3LYP functional \citep{beck88,lee88} and basis set 6-311g++(d,p).
Note that reaction exothermicities or endothermicities do not differ significantly between the
gaseous and ice mantle phases. We calculated the endothermicity/exothermicity ($\Delta H$) of a reaction
by taking the difference between the total optimized enthalpy including zero point corrections
of the products
and reactants.
If $\Delta H$ is positive, we label the reaction endothermic and if $\Delta H$ is negative, we label the reaction
exothermic.
Another formation reaction, R6, is considered only in the gas phase.
The individual formation reactions are discussed in the following paragraphs.
Reaction R1 ($\mathrm{O + C_3H_3}$) in the gas phase was
studied by \cite{kwon06}, who carried out an experiment as
well as ab initio statistical calculations.
They found that the reaction is barrier-less and can produce
propynal (HC$_{2}$CHO) and H. The conversion of propynal into PA then can occur
via two association reactions (R3, R4) with atomic hydrogen. In the gas phase,
the process occurs via radiative association, in which emission of a photon stabilizes the
intermediate reaction complex. \cite{lee06} predicted that the barrier-less
addition of O($^3$P) to propargyl radical ($\mathrm{C_3H_3}$) on the lowest doublet potential energy
surface could produce several energy-rich intermediates, which undergo subsequent
isomerization and decomposition steps to generate various exothermic reaction products.
Their statistical calculation also suggests that the primary reaction channel leads to
the formation of propynal.
Reaction R2, in which the radical CCH and formaldehyde produce propynal + H, was studied by \cite{dong05} and by \cite{petr95}. \cite{dong05} calculated a very small barrier of 2.1 kcal/mol at the highest level of theory, while \cite{petr95} assumed the channel to be barrier-less based on similar reactions. The propynal product can then also be hydrogenated to PA via R3 and R4.
As an alternative to two successive H-atom association reactions involving
atomic hydrogen, we checked the reaction of $\mathrm{H_2}$ with propynal ($\mathrm{HC_2CHO}$)
to form PA but found it to be highly endothermic.
We tried a few other pathways for the formation of PA
via single step reactions, sometimes involving a radical.
In this effort, we considered the reaction between $\mathrm{C_2H_4}$ and CO, the
reaction between propynal and $\mathrm{H_2O}$, and the reaction between
the propargyl radical and OH (reaction number R5). The reactions between $\mathrm{C_2H_4}$ and CO and
between propynal and $\mathrm{H_2O}$ are found to be highly endothermic
in nature whereas reaction R5 is highly exothermic and is likely barrier-less,
since the reactants are both radicals.
Since both reactants in R5 ($\mathrm{C_3H_3}$ and OH) are reasonably abundant in the ISM,
we think that this reaction can contribute towards the formation of interstellar PA,
although, in the gas phase, it must proceed via a radiative route, so that it must be looked at closely.
On ice mantles, however, radical-radical association reactions are normally quite efficient.
As can be seen in Table 1, R4 and R5 are the sole radical-radical reactions leading directly to the formation of PA,
both in the gas and on the ice. A gas phase dissociative recombination (DR) reaction (R6), in which protonated PA and
an electron recombine to form smaller neutral products, is also included. This reaction may contribute significantly to the gas phase formation
of PA unless the two-body product channel shown is unimportant if ion-neutral processes can produce protonated PA efficiently. Similar pathways to
all those considered for the synthesis of PA are assumed to be responsible for the production of OD-PA.
\subsection{Destruction pathways}
As shown in Table 1, the destruction of gaseous PA occurs via ion-neutral (IN) and photo-dissociative (PH \& CR) pathways, as well as via two radical-neutral
(NR) reactions. The destruction of gas phase PA also occurs via adsorption onto ice, but the reverse process of desorption also occurs. These processes
are in our network, but not listed in Table 1. The rate coefficient for the gas phase NR reaction between the hydroxyl radical (OH) and PA was measured by
\cite{upad01}, who used laser photolysis combined with the laser induced fluorescence technique at room temperature. According to their study, this
reaction (R7) produces an adduct, $\mathrm{HOC_2HHCHOH}$.
Since the abundances of OH and OD are comparable in a dense cloud \citep{mill89}, a similar destruction mechanism
with OD (R8) is also considered here. In addition to the gas-phase, reactions R7 and R8 are included in the ice phase as well. For the ion-neutral
(IN) destruction of gas phase PA, we include reactions R9-R22 by following the similar gas phase ion-neutral (IN) destruction pathways available
for methanol in \cite{wood07}. Similar IN destruction reactions (R23-R50) for $\rm{HC_2CHO}$ and $\rm{HC_2CH_2O}$ are also considered.
Photo-dissociation reactions (direct or cosmic ray induced) (R51-R66) are also responsible for the destruction of PA and its associated species in both phases.
In Table 1, DR reactions R6 and R67-R75 involve the ions ($\mathrm{HC_2CH_2O^+}$, $\mathrm{HC_2CH_2OH_2}$$^+$ and $\mathrm{HC_2CH_2OH^+}$); these ions are
produced by various formation/destruction pathways of PA and its related species. For the destruction of OD-PA and other associated species similar
destruction pathways to all those considered for PA are included.
\subsection{Rate coefficients}
\subsubsection{Gas phase rate coefficients}
\cite{slag91} experimentally obtained a temperature independent rate coefficient of
$\sim 2.31 \times 10^{-10}$ cm$^3$~s$^{-1}$
for reaction R1. According to their study, this reaction
proceeds through a fast and irreversible association-fragmentation process.
We utilize the rate coefficient obtained by \cite{slag91}.
\cite{petr95} estimated a rate coefficient of $1.0 \times 10^{-10}$
cm$^3$~s$^{-1}$ for reaction R2 with the assumption that it occurs without a barrier in the gas phase. Although this assumption contradicts the calculation of a small barrier by \cite{dong05}, we assume the reaction to be barrier-less, and use the estimated rate coefficient of \cite{petr95}.
As can be seen in Table 1, reactions R3, R4, and R5 are highly exothermic in nature.
Based on the high exothermicity of these reactions, one might assume that these
reactions can process without barriers.
The Hydrogen addition reaction of $\rm{HC_2CHO}$ (reaction R3) may occur in two ways. First,
$\rm{H}$ addition may occur with the $\rm{O}$ atom of the $\rm{CHO}$ group and produce
${\rm HC_2CHOH}$ and secondly H addition may occur with the $\rm{C}$ atom of the $\rm{CHO}$ group
and produce ${\rm HC_2CH_2O}$. Our quantum chemical calculation found that the hydrogen addition
to carbon is more favourable than the hydrogen addition to oxygen in the $\rm{CHO}$ group.
Using the DFT/6-31+G(d,p) method, we found that the gas-phase activation barrier ($\Delta E\ddag$) and Gibbs energy
of activation ($\Delta {G}\ddag$) for the second possibility
of reaction R3 (${\rm H+HC_2CHO\rightarrow HC_2CH_2O}$) is $3.74$ kcal/mol and $9.63$ kcal/mol respectively.
For the computation of the gas phase rate coefficient for reaction R3, we use transition state theory, which leads
to the Eyring equation \citep{eyri35}:
\begin{equation}
k= (k_BT/hc) \exp(-\Delta {G}\ddag/RT).
\end{equation}
{\leavevmode\color{black}
The rate coefficient calculated by the above equation thus has a strong temperature dependence.
In Table 1, we have included the rate coefficient for two temperatures; $10$ K and $100$ K.
At $10$ K, rate coefficient is $\sim 0$ and at $100$ K it becomes $1.96 \times 10^{-15}$ cm$^3$ s$^{-1}$.
Thus in the low temperature regime, R3 does not contribute at all to the gas phase formation of PA,
while it could play a role for the formation of PA in the high temperature regime.
}
In the case of reactions R4 and R5, we were unable to locate any suitable transition state
and assume that these two reactions are barrier-less, as is customary for radical-radical reactions. The rate coefficients of these
two reactions are assumed to be $10^{-10}$ $\rm{cm^3 \ s^{-1}}$.
The formation of OD-PA is treated with similar pathways and rate coefficients.
For the formation of PA by the DR mechanism, which involves the destruction of
$\mathrm{HC_2CH_2OH_2}^+$ (R6, R67-R75),
we follow the destruction of
$\mathrm{CH_3OH_2}^+$ from \cite{wood07} for the rate coefficients and product channels. Rate coefficients between two species are standardly parameterized as a function of temperature by the equation
\begin{equation}
k = {\alpha}(T/300)^{\beta} \exp^{-\gamma/T}.
\end{equation}
In this particular case, the values of the parameters for the formation of PA are
$\alpha=2.67 \times 10^{-8}$ cm$^{3}$ s$^{-1}$,
$\beta=-0.59$ and $\gamma=0$. Similar
product channels and rate coefficients are used for the formation of OD-PA by a DR reaction.
Now let us consider the destruction of gaseous PA. \cite{upad01} studied the rate coefficient of the NR reaction between PA and the OH radical (R7). According to their study,
it produces an adduct with a rate coefficient of $(9.2\pm1.4) \times 10^{-12}$
$\rm {cm^3 s^{-1}}$, which we use with no temperature dependence. Here also, we assume a similar rate coefficient
for reaction R8, which involves OD. Similar destruction reactions and rate coefficients are adopted for
OD-PA as well.
Ion Neutral (IN) reactions are the dominant means for the destruction of
neutral interstellar species. If the neutral species is non-polar, we use the
Langevin collision rate coefficient \citep{herb06,wake10}.
If the neutral species is polar, we employ the trajectory scaling relation found in
\cite{su82} and \cite{woon09}.
From our quantum chemical calculations, we find that the polarizability $\alpha_d=5.62 \times 10^{-24}$ cm$^3$
and dipole moment $\mu_D$=1.6548 Debye for PA.
For the destruction reactions of the other two associated species (${\rm HC_2CHO,\ HC_2CH_2O}$),
in reaction numbers R23-R36,
we use $\alpha_d=5.26 \times 10^{-24}$ cm$^3$
and $\mu_D$=3.08 Debye, and for R37-R50 we use
$\alpha_d=5.93 \times 10^{-24}$ cm$^3$
and $\mu_D$=1.1480 Debye. For deuterated PA, the ion-neutral destruction reactions have similar rate
coefficients to those of normal PA, the only differences being due to the
reduced mass, which are rather small. Deuterated reactions and their rate coefficients
are not tabulated here.
For the photo-dissociation reactions of PA and its associated species by external interstellar photons and cosmic ray-induced photons, we use analogous products and
the same first-order rate coefficients (s$^{-1}$) as the case of $\mathrm{CH_3OH}$ \citep{wood07}.
For the case of external photons,
we use the following relation for the rate coefficients:
\begin{equation}
k = \alpha \exp(-\gamma A_V )
\end{equation}
where $\alpha$ represents the rate coefficient (s$^{-1}$) in the unshielded interstellar ultraviolet radiation field, $A_V$ is
the visual extinction, for which we use a value of 10, and $\gamma$ is
used to take into account the increased
extinction of dust in the UV. Here, following \cite{wood07}, we use $\alpha = 6.0 \times 10^{-10}$ s$^{-1}$,
and $\gamma = 1.8$ in our model.
Incorporating all the parameters into the above equation, we obtain a photo-dissociation rate coefficient of
about $9.14 \times 10^{-18}$ s$^{-1}$.
For cosmic-ray-induced photo-reactions, we use the following relation, which was originally developed by \citep{gred89}:
\begin{equation}
k=\alpha \gamma'/(1- \omega)
\end{equation}
where $\alpha$ is the cosmic-ray ionization rate (s$^{-1}$), $\gamma'$ is the number of photo-dissociative events that take place per cosmic-ray ionization
and $\omega$ is the dust grain albedo in the far ultraviolet. Here, we use $\omega$ =0.6,
$\alpha=1.3 \times 10^{-17}$ s$^{-1}$, and $\gamma'=752.0$ by following the cosmic-ray-induced
photo-reactions of $\rm{CH_3OH}$ in \cite{wood07}.
By including these parameters into the above equation, we obtain a rate coefficient of $2.44 \times 10^{-14}$ s$^{-1}$, which is
roughly $4$ orders
higher than the rate of external photo-dissociative reactions. Cosmic ray induced photo-reactions can play an important role in
interstellar chemistry. The choices of these parameters are highly reaction specific and a wrong estimation
might lead to misleading results. In the UMIST 2006 database, $\gamma'$ ranges as high
as $5290$ and as low as $25.0$. Though the higher values of $\gamma'$ are more reliable for the larger molecules because of the increasing overlap between the cross sections for photo-dissociation and the cosmic-ray-induced emission spectrum \citep{gred89},
we use {\leavevmode\color{black} $\gamma'=752.0$} {\color{black} (in Table 1, we presented it under the column head marked $\gamma$)}
for our calculations of the cosmic-ray-induced photo-dissociative reactions.
The same photo-dissociation rate coefficients are adopted for the destruction of
OD-PA and its associated species.
Dissociative recombination (DR) reactions and rate coefficients for some of the associated ions of PA are shown in reactions R67-R75.
Pathways and rate coefficients of these reactions are adopted by
following the similar DR pathways available for $\mathrm{CH_3O}^+$, $\mathrm{CH_3OH_2}^+$, and
$\rm{CH_3OH^+}$ in \cite{wood07}. Since for all these reactions, $\beta \ne 0$, the reactions are temperature dependent.
The same DR rate coefficients are used for the associated ions of OD-PA.
{\leavevmode\color{black}
For two or three-body gas-phase reactions in Table 1, the rate coefficients are represented in terms of the three
rate coefficients, $\alpha$, $\beta$ and $\gamma$.
Most of the gas phase rate coefficients adopted here are either estimated or taken
from similar kind of reactions. For reactions R1-R2, R4-R5 and R7-R8,
we assign $\beta$ and $\gamma$ to be zero.
For the dissociative recombination reactions (R6 and R68-R75), these three coefficients are
estimated based on similar reactions listed in \cite{wood07}.
For reaction R3,
we calculate the gas phase rate coefficient by using transition state theory, which leads to
the Eyring equation \citep{eyri35}. Thus for reaction R3, these three parameters are not shown.
For destruction by photo-reactions, we supply these three coefficients following the
similar type of reactions available in \cite{wood07}.
For the destruction of polar neutral species by ions, we use the two relations discussed in
\cite{su82}. These relations cannot be represented over the whole temperature range in terms of one set of three coefficients.
However, we can tabulate $\alpha$, $\beta$ and $\gamma$ for reactions R9-R50
in the low temperature regime.}
\begin{figure}
\hskip -1cm
\centering{
\vbox{
\includegraphics[height=6cm,width=9cm]{physical.pdf}}}
\caption{ {\leavevmode\color{black} Adopted physical conditions for the warm-up phase in which the density increases as the temperature increases.
.}}
\end{figure}
\subsubsection{Ice phase rate coefficients}
Chemical enrichment of interstellar grain mantles depends on the binding energies ($E_d$)
and barriers against diffusion ($E_b$) of the adsorbed species. The binding energies of these
species are often available from past studies \citep{alle77,tiel87,hase93,hase92}. But
these binding energies mostly pertain to silicates. Binding energies of the
most important surface species on ice, which mostly control the
chemical composition of interstellar grain mantles, are available from some
recent studies \citep{cupp07,garr13}. We use these latter energies in our model. For the rest of
the species for which binding energies are still unavailable,
we use the same values as in past studies or estimate new values. For barriers against diffusion, which are poorly known, we use values
equal to $0.35 E_d$ \citep{garr13}.
\begin{figure}
\hskip -1cm
\centering{
\vbox{
\includegraphics[height=9cm,width=9cm]{isothermal.pdf}}}
\caption{ \color{black} Chemical evolution of (a) PA and OD-PA and (b) their related species during
the isothermal phase at $n_H = 10^4$
cm$^{-3}$ with a constant visual extinction of $10$.}
\end{figure}
\begin{figure}
\hskip -1cm
\centering{
\vbox{
\includegraphics[height=9cm,width=9cm]{afac.pdf}}}
\caption{ Chemical evolution of PA during the constant density isothermal phase at $n_H = 10^4$
cm$^{-3}$ with a constant visual extinction of $10$ and
different values of the non-thermal reactive desorption parameter $a$.}
\end{figure}
For the formation of PA in the ice phase, we include surface reactions R1-R5.
The rate coefficients ($R_{diff}$) of these reactions were calculated by the method
described in \cite{hase92}, which is based on thermal diffusion.
They defined a probability factor $\kappa$ to differentiate between exothermic reactions
without activation energy barriers and reactions with activation energy barriers ($E_{\rm a}$) in such a way that
the effective rate coefficient becomes $R_{ij}=\kappa \times R_{diff}$.
The factor $\kappa$ is unity in the absence of a barrier.
For reactions with activation energy barriers, $\kappa$ is defined as the quantum mechanical
probability for tunneling through a rectangular barrier of thickness a($=1$ $\AA$), which is calculated from the equation
\begin{equation}
\kappa= \exp [-2(a/ \hbar)(2\mu E_a)^{1/2}].
\end{equation}
{\leavevmode\color{black}
Binding energies to the surface for some
complex organics with a hydroxyl group are normally considered to be higher due to the phenomenon of hydrogen bonding \citep{coll04,latt11,garr13}.
Since PA has an -OH group, creating hydrogen bonds with a
water substrate, this molecule is therefore expected to have a higher binding energy,
close to that of water. Here, we assume the binding energy of both PA and OD-PA to be the same as methanol ($5530$ K).
Since $\mathrm{HC_2CHO}$forms by the reaction between $\rm{C_3H_3}$ and $\rm{O}$,
we add the binding energies of $\mathrm{C_3H_3}$ ($2220$ K) and O ($800$ K) to determine
a binding energy value $3020$ K. In the case of the binding energy of
$\mathrm{HC2CH2O}$,
we add the binding energies of HC2CHO ($3020$ K) and H ($450$ K) to obtain
$3470$ K.}
For other binding energies for species in reaction R1, R2, R3, R4 and R5,
we use $E_d(\mathrm{C_2H})= 1460$ K,
$E_d(\mathrm{H_2CO})=2050$ K and $E_d(\mathrm{OH})=2850$ K.
Our transition state theory calculation found that reaction R3 (hydrogen addition to the carbon atom of the CHO group)
contains an activation barrier
of about $3.59$ kcal/mol ($1807$ K) in the ice phase.
\begin{figure}
\hskip -1cm
\centering{
\vbox{
\includegraphics[height=10cm,width=9cm]{warm-up.pdf}}}
\caption{ {\color{black} Chemical evolution of (a) PA and OD-PA (b) $\mathrm{C_3H_3}$, $\mathrm{HC_2CHO}$
and $\mathrm{HC_2CH_2O}$ {\color{black} (c) ${\rm H_2O}$, ${\rm CH_3OH}$ and ${\rm CH_4}$} for the warm-up phase, during which both density and temperature increase. }}
\end{figure}
Destruction of ice phase PA occurs by various means; thermal desorption,
non-thermal desorption from interstellar dust grains via exothermic surface reactions,
cosmic ray induced desorption, photo-dissociation, and reaction with OH or OD radicals.
Intact photo-desorption is not included.
Thermal desorption plays a crucial role only at high temperatures, depending upon the adsorbed species.
For thermal desorption, we use the relation prescribed by \cite{hase92}.
At low temperatures, non-thermal desorption by reaction exothermicity, photons, and cosmic rays serve as
important means for transferring the surface molecules to the gas phase. For
non-thermal reactive desorption \citep{garr06}, we consider that all
surface reactions which result in a single product break the surface-molecule bond with some
fraction $a$. Here, we use a fiducial value for the fraction of $a=0.03$. For the
cosmic ray induced desorption rates, we follow the expression developed by \cite{hase93}.
Photo-dissociation by direct (R51-R58) as well as cosmic ray induced photons (R59-R66) are
considered for the destruction of
ice phase PA and its two associated species, and the rate coefficients were assumed similar to
their gas phase rate coefficients.
The dissociated products remain on the grain surface and are thus able to react
and desorb depending on their binding energies.
For the destruction of ice phase PA by reactions with OH and OD (R7 and R8), we consider the destruction pathways
proposed by \cite{upad01} for the gas-phase, but we calculate the rate coefficients of these ice phase reaction pathways
by using diffusive reactions with barriers against diffusion, which have already been discussed.
Selected ice phase rate coefficients used for PA are shown in Table 1 for $T=10$ K and $100$ K. Desorption and accretion rate coefficients are not included.
\begin{table*}
\centering{
\scriptsize
\caption{Initial elemental abundances with respect to total hydrogen nuclei.}
\begin{tabular}{|c|c|}
\hline
Species&Abundance\\
\hline\hline
$\mathrm{H_2}$ & $5.00 \times 10^{-01}$\\
$\mathrm{He}$ & $1.40 \times 10^{-01}$\\
$\mathrm{N}$ & $2.14 \times 10^{-05}$\\
$\mathrm{O}$ & $1.76 \times 10^{-04}$\\
$\mathrm{e^-}$ & $7.31 \times 10^{-05}$\\
$\mathrm{C^+}$ & $7.30 \times 10^{-05}$\\
$\mathrm{S^+}$ & $8.00 \times 10^{-08}$\\
$\mathrm{Si^+}$& $8.00 \times 10^{-09}$\\
$\mathrm{Fe^+}$& $3.00 \times 10^{-09}$\\
$\mathrm{Na^+}$& $2.00 \times 10^{-09}$\\
$\mathrm{Mg^+}$& $7.00 \times 10^{-09}$\\
$\mathrm{HD}$ & $ 1.6 \times 10^{-05}$\\
\hline
\end{tabular}}
\end{table*}
\section{Chemical modeling}
\label{sec:chem}
\subsection{\leavevmode\color{black}Model}
We use our large gas-grain chemical network \citep{das15a,das15b} for the purpose of chemical modeling.
We assume that the gas and grains are coupled through accretion and thermal and non-thermal
desorption processes. A visual extinction of $10$ and a cosmic ray ionization rate of
$1.3\times 10^{-17}$ $\rm s^{-1}$ are used.
{\leavevmode\color{black}Including the deuterated reactions and deuterated species,}
our present gas phase chemical network
consists of $6384$ reactions involving $641$ gas phase
species and the surface chemical network consists of $358$ reactions involving $291$ surface species.
The gas phase chemical network is mainly adopted from the UMIST 2006 database \citep{wood07}, but
contains some deuterated reactions as well. To minimize the huge
computational time and avoid difficulty in handling a large chemical network, we have considered
only some dominant deuterated reaction pathways.
{\leavevmode\color{black}
Deuteration reactions important for the increase of deuteration in the cold gas phase are included
in our network.}
Our selection was based on some earlier
studies by \cite{rodg96,robe00,robe03}.
{\leavevmode\color{black} All the deuterated reactions, which were added/updated and identified
as the dominant pathways for the formation of essential deuterated species in the tables (Table 13-18)
of \cite{albe13} are also included here.}
For the grain surface reaction network,
we primarily follow \cite{hase92} and \cite{cupp07}, while for the ice phase deuterium fractionation reactions,
we follow \cite{case02} and \cite{caza10}. In addition to this, we consider some reactions involving PA and coupled species, which are
described in Section 2.
Initial elemental abundances with respect to total H nuclei in all forms are shown in
Table 2. These ``low metal" abundances are often used for dense clouds, where
hydrogen is mostly in the form of molecular hydrogen and ionization of the medium
is mainly governed by cosmic rays. We adopted these values from \cite{leun84}.
It is assumed that initially all deuterium is locked up in the form of HD.
The initial abundance of HD is assumed to be $1.6 \times 10^{-5}$ \citep{robe00}
with respect to the total number of hydrogen nuclei.
{\leavevmode\color{black} In order to consider suitable physical conditions for a star forming region,
we adopt a model with an initial phase of constant density ($\mathrm{n_H}$=$1.0\times 10^{4} \rm cm^{-3}$)
and temperature ($T=10$ K) followed by a warm-up phase (temperature increases from $10$ K to
$200$ K).
The warm-up phase is assumed to be accompanied by an increase of density from
$n_H = 10^4$ cm$^{-3}$ to $n_H = 10^7$ cm$^{-3}$, relevant when the material approaches
the inner protostellar regions and appropriate for hot-core conditions.
Both phases have a visual extinction of $10$.
The initial phase is assumed to be sustained for $10^6$ years and
the subsequent phase lasts for another $10^5$ years {\leavevmode\color{black}(a typical lifetime for a young embedded stage
\citep{evan09})}.
In Fig. 1, the adopted time-dependent physical conditions for the second stage of evolution are shown.
For the rise in density and temperature, we assume constant slopes $m_{den}$ and $m_{temp}$
respectively for the density and temperature as determined from the following equations:
\begin{equation}
\rm m_{den}= \frac{\rho_{max}-\rho_{min}}{Time_f-Time_i}=99.9~{\rm cm^{-3} yr^{-1}},
\end{equation}
\begin{equation}
\rm m_{temp}= \frac{T_{max}-T_{min}}{Time_f-Time_i}=1.9 \times 10^{-3}~{\rm K yr^{-1}}.
\end{equation}}
\subsection{\leavevmode\color{black}Modeling Results}
Figure 2a,b shows the chemical evolution of {\color{black} PA, OD-PA and their related species} in the gaseous
(solid line) and ice phase (dashed line) during the cold isothermal phase.
The ice phase production
of PA is dominated by the successive H association reactions (R3 and R4).
Similarly, for the case of OD-PA, the association reactions are found to be dominant.
{\color{black} The peak abundances of PA and OD-PA and their related species are listed in Table 3, for both phases of our calculation.
For example, in the isothermal cold phase, the peak fractional abundances of PA and OD-PA in the
ice p are \color{black} $5.03 \times 10^{-10}$
and \color{black} $1.33 \times 10^{-11}$ respectively and for the gas phase, these values
are \color{black} $1.07 \times 10^{-12}$ and $1.32 \times 10^{-13}$ Note that the ice and gas phase peaks do not occur at the same time}.
Non-thermal desorption processes significantly contribute to the maintenance of the
gas-phase abundances
of PA and OD-PA with our fiducial value $a$ of $0.03$ for reactive desorption. Fig. 3 show the gas
phase abundance of PA versus time with $a=0$, $a=0.03$ and $a=0.3$ {\color{black} in the isothermal phase}.
For the case of Fig. 3, throughout some but not all of the time of the evolutionary stage,
the PA abundance for the $a=0.03$ case is slightly
higher than in the case without reactive desorption ($a=0$).
For the highest `a' value ($a=0.3$), with the most rapid reactive desorption, the gas phase peak
abundance is significantly larger than that with $a = 0.03$ and also with $a=0$.
{\color{black} In the isothermal phase, the peak abundance of gas phase PA for these three values of
$a$ is found to be $8.69 \times 10^{-13}$, $1.07 \times 10^{-12}$ and $1.12 \times 10^{-10}$
for $a=0.00, \ 0.03$ and $0.30$ respectively.}
{\color{black} At the initial stages of evolution, the models with $a=0$ and $a=0.03$ give
very similar gas phase abundances
This means that the gas phase
production of PA in our fiducial model
is actually dominated by gas phase chemistry (by reaction R5) as long as the $a$ parameter is not
too high.}
{\color{black} Figure 4a,b,c refers to the warm-up portion of our simulation.
{\color{black} Fig. 4a represents the evolution of PA and OD-PA whereas Fig. 4b represents the
evolution of PA related species such as ${\rm C_3H_3}$, $\rm{HC_2CHO}$ and
$\rm{HC_2CH_2O}$ and Fig. 4c represents the evolution of the major icy species water,
methane and methanol for the gas and ice phases.}
{\color{black} The temporal evolution of gas phase and ice species shown in Fig. 4a,b,c
appears to differ slightly with similar models found in \cite{garr08}.
This difference is solely due to the adopted physical condition mentioned in section 3.1
in comparison to that in \cite{garr08}.}
{\color{black} From Fig. 4c it is clear that among the main icy species, methane disappears
first from the grain mantle due to its low adsorption energy ($1300$ K)
followed by methanol (adsorption energy $5530$ K) and water (adsorption energy $5700$ K).}
As can be seen, the gas phase production of PA and OD-PA is found to be favourable around
the high temperature region. This happens
due to the increase in the rate coefficient of reaction R3 with the temperature.
For the ice phase reactions, H or D addition reactions become irrelevant beyond $40$ K,
beyond which the $\rm{C_3H_3}$ and OH or OD radicals become mobile enough to react with
each other on grain surfaces. {\color{black} At the same time, the destruction of PA
and OD-PA by reactions with OH and OD increases due to the increase in the
destruction rates with increasing temperature.}
Above $100$ K, $\rm{C_3H_3}$ and OH or OD radicals roughly disappear and the only source of ice phase PA or OD-PA is accretion
from the gas phase. Since the density increases with time, accretion of the gas phase
species becomes more favourable.
Above $140$ K,
the gas phase abundances of PA and OD-PA are roughly invariant.
In Table 3, we list
the peak abundances of PA, OD-PA and their related species for the warm-up phase along with their
corresponding times and temperatures.
In this phase, the peak abundances of ice phase PA and OD-PA are found to be decreased in comparison
with the isothermal phase. The peak fractional abundances
of ice phase PA and OD-PA are found to be $7.09 \times 10^{-11}$ at
$32.2$ K, and {\color{black} $7.23 \times 10^{-12}$ at $47.3$ K}, respectively, whereas,
for the gas phase, the corresponding peak abundances are $1.20 \times 10^{-9}$
and $2.14 \times 10^{-12}$ respectively.
Although reality is more complex
due to the consideration of variable density and temperature in the second phase,
the basic explanation for the higher abundances of
PA and OD-PA in the gas phase is that instead of chemical desorption, thermal desorption
becomes much larger at higher temperatures, and indeed the ice abundances become
infinitesimal at temperatures above $100$ K.
In the first phase of our simulation (Fig. 2), which occurs while the
temperature is $10$ K, only non-thermal slow desorption occurs,
and indeed the ice abundances are higher.
For the gas-phase at higher temperatures, at which there is very little ice, fractionation occurs
via gas-phase formation and destruction reactions, which lead to low fractionation at temperatures
above $50$ K due to endothermic reactions between selected deuterated ions with H$_2$ which
return deuterated species to normal ones. After all, the high temperature limit of OD-PA/PA is
$10^{-4}$, which is close to the D/H elemental value of $10^{-5}$.
At temperatures under $50$ K, the abundance
ratio of gaseous OD-PA/PA approaches unity, a very high degree of fractionation, while the ice ratio
is $10^{-2}$.
These high ratios probably stem from large values of the atomic abundance ratio D/H in the gas
and on the grains at low temperatures.
}
\begin{table*}
\scriptsize{
{\leavevmode\color{black}\caption{Peak abundances of PA, OD-PA and their related species with respect to H$_2$.}
\begin{tabular}{|p{0.40in}|p{0.50in}p{0.50in}|p{0.50in}p{0.50in}|p{0.50in}|p{0.30in}|p{0.50in}|p{0.50in}|p{0.30in}|p{0.50in}|}
\hline
{\bf Species}&\multicolumn{4}{|c|}{\bf Isothermal phase}&\multicolumn{6}{|c|}{\bf Warm-up phase}\\
\cline{2-11}
&\multicolumn{2}{|c|}{\bf gas phase}&\multicolumn{2}{|c|}{\bf ice phase}&\multicolumn{3}{|c|}{\bf gas phase}&\multicolumn{3}{|c|}{\bf ice phase}\\
\cline{2-11}
&{\bf Time (year)}& {\bf Abundance}& {\bf Time (year)}& {\bf Abundance}& {\bf Time (year)}& {\bf T (K)} & {\bf Abundance}& {\bf Time (year)}& {\bf T (K)} & {\bf Abundance}\\
\hline
{\bf PA}&$3.29 \times 10^5$&$1.07 \times 10^{-12}$&$7.40 \times 10^5$&$5.03 \times 10^{-10}$&$1.10 \times 10^6$&200.0&$1.20 \times 10^{-09}$&$1.01 \times 10^{6}$&32.2&$7.09 \times 10^{-11}$\\
{\bf OD-PA}&$2.89 \times 10^5$&$1.32 \times 10^{-13}$&$5.66 \times 10^5$&$1.33 \times 10^{-11}$&$1.10 \times 10^0$&200.0&$2.14 \times 10^{-12}$&${\color{black}1.02 \times 10^{6}}$&${\color{black}47.3}$&${\color{black}7.23 \times 10^{-12}}$\\
{$\rm{\bf C_3H_3}$}&$9.36 \times 10^5$&$2.52 \times 10^{-10}$&$1.85 \times 10^5$&$4.32 \times 10^{-13}$&$1.034 \times 10^6$&75.0&$1.92 \times 10^{-9}$&$1.022 \times 10^{6}$&50.9&$3.25 \times 10^{-14}$\\
{$\rm{\bf HC_2CHO}$}&$8.22 \times 10^4$&$6.71 \times 10^{-10}$&$2.17 \times 10^5$&$2.60 \times 10^{-10}$&$1.028 \times 10^0$&63.5&$1.66 \times 10^{-9}$&$1.01 \times 10^{6}$&29.2&$2.05 \times 10^{-9}$\\
{$\rm{\bf HC_2CH_2O}$}&$5.14 \times 10^5$&$5.26 \times 10^{-15}$&$1.85 \times 10^5$&$9.43 \times 10^{-18}$&$1.054 \times 10^6$&114.1&$2.52 \times 10^{-13}$&$1.01 \times 10^{6}$&29.2&$7.23 \times 10^{-17}$\\
\hline
\end{tabular}}}
\end{table*}
In the warm-up case, the predicted fractional abundance of PA under hot-core conditions lies at approximately
$10^{-9}$.
Our results argue strongly for the possibility of detection, especially at temperatures above
{\color{black} 150 K}. Interestingly,
this prediction implies that PA does not occupy the same regions as does its isomer propenal.
(\cite{requ08} measured its abundance to $\sim 2.30 \times 10^{-9}$ with respect to H$_2$), but the species
is seen in absorption against Sgr B2(N),
and so is more likely to be present in colder foreground gas \citep{holl04}. For OD-PA,
the predicted hot-core abundance is {\color{black}$\approx 10^{-12}$ at
temperatures above $150$.} For the cold core
case, even the peak fractional abundance of gas phase PA with the fiducial value of $a$ is sufficiently low to rule out
detection unless the highest value of the parameter
$a$ for reactive desorption is used. We conclude that PA and possibly OD-PA can most
likely be detected in warm-up regions at temperatures above 100 K.
\section{\color{black} Astronomical Spectroscopy}
\label{sec:astro}
\subsection{\color{black} Detectability of PA and OD-PA in the millimeter-wave regime}
{\leavevmode\color{black}
To estimate the possibility of detecting PA and OD-PA with present
astronomical facilities, we use the CASSIS radiative transfer model [http://cassis.irap.omp.eu] at LTE with the JPL molecular database [http://spec.jpl.nasa.gov].
Propenal ($\mathrm{CH_2CHCHO}$) had already been successfully detected towards the high mass star forming region
Sagittarius B2(N) with the $100$-meter Green Bank
Telescope operating in the frequency range $18-26$ GHz by \cite{holl04}.
They had detected the $2_{11}-1_{10}$ line of $\mathrm{CH_2CHCHO}$ with $14$ mK intensity
($\approx$ 7$\sigma$ detection). From our chemical model, we found that the peak abundances
for PA and OD-PA are respectively $1.2\times10^{-9}$ and $
2.14\times10^{-12}$ with respect to $\rm{H_2}$ and for propenal, we use an abundance of
$2.3\times10^{-9}$ with respect to $\mathrm{H_2}$ \citep{requ08}.
As the input parameters, we use the following parameters, which
represent a typical high mass star forming region analogous to Sgr B2 (N): \\\\
\noindent Column density of $\mathrm{H_2}$= $10^{24}$cm$^{-2}$, \\
\noindent Excitation temperature ($T_{\rm ex}$)= $100$ K, \\
\noindent FWHM= $5$ km/s, \\
\noindent $\mathrm{V_{LSR}}$= $74$ km/s, \\
\noindent Source size to take into account the beam dilution= $3^{''}$
\begin{table*}
\centering{
\scriptsize
{\leavevmode\color{black}\caption{Calculated line parameters for mm-wave transitions of Propenal, PA and OD-PA with ALMA}
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline
{\bf Species} & {\bf ALMA}& {\bf Frequency}&${\bf J_{k{_a}^{'}K_c^{'}}-J_{k{_a}^{''}K_c^{''}}}$&${\bf E_{up}}$&${\bf A_{ij}}$&{\bf Intensity}\\
&{\bf Band}&{\bf (GHz)}& & {\bf (K)}&&{\bf (mK)}\\
\hline
{\bf Propenal}&Band 1&44.4975&$5_{05}-4_{04}$&6.4&$4.33\times10^{-6}$&287\\
\hline
{\bf Propenal}&Band 2{\color{black}/Band 3}&89.05935&$10_{46}-9_{45}$&56.44&$3.06\times10^{5}$&2320\\
&&&$10_{47}-9_{46}$&&&\\
\hline
{\bf Propenal}&Band 3&115.80341&$13_{410}-12_{49}$&71.83&$7.34\times10^{-5}$&3920\\
&&&$13_{49}-12_{48}$&&&\\
\hline
{\bf PA}&Band 1&44.54410&$5_{05}-4_{04}$&6.42&$5.03\times10^{-7}$&8.18\\
\hline
{\bf PA}&Band 2{\color{black}/Band 3}&89.39184&$10_{65}-9_{64}$&72.01&$2.74\times10^{-6}$&102\\
&&&$10_{64}-9_{63}$&&&\\
&&89.39212&$10_{56}-9_{55}$&57.23&$3.21\times10^{-6}$&102\\
&&&$10_{55}-9_{54}$&&&\\
\hline
{\bf PA}&Band 3&116.23546&$13_{59}-12_{58}$&72.68&$8.10\times10^{-6}$&160\\
&&&$13_{58}-12_{57}$&&&\\
&&116.23586&$13_{86}-12_{85}$&125.04&$5.97\times10^{-6}$&160\\
&&&$13_{85}-12_{84}$&&&\\
\hline
{\bf OD-PA}&Band 1&44.88953&$5_{14}-4_{13}$&7.68&4.93E-7&0.013\\
\hline
{\bf OD-PA}&Band 2{\color{black}/Band 3}&87.65146&$10_{55}-9_{54}$&53.43&$3.43\times10^{-6}$&0.112\\
\hline
{\bf OD-PA}&Band 3&113.96386&$13_{59}-12_{58}$&68.58&$8.59\times10^{-6}$&0.206\\
&&113.96429&$13_{58}-12_{57}$&&&\\
\hline
\multicolumn{3}{c}{Band 1= 31-45 GHz}\\
\multicolumn{3}{c}{Band 2= 67-90 GHz}\\
\multicolumn{3}{c}{Band 3= 84-{\color{black}116} GHz}\\
\end{tabular}}}
\end{table*}
From our radiative transfer model, the intensity for $2_{11}-1_{10}$ line ($18.28065$ GHz) of PA
is found to be $0.011$ mK, which is far below than the
detection limit of GBT (3$\sigma$$\approx$6mK).
Because of the higher dipole moment components of propenal ($\mu_a=3.052$ D,
$\mu_b = 0.630$ D and $\mu_c = 0$ D) compared with PA ($\mu_a=1.037$ D, $\mu_b = 0.147$ D
and $\mu_c = 0.75$ D), intensity of PA is much lower compared with propenal.
This prompted us to suggest the use of high-spatial and high-spectral resolution observations from the
Atacama Large Millimeter/Sub-millimeter Array (ALMA). In particular, we can use lower frequency
bands of ALMA -- Bands 1, 2 and 3 ($31-116$ GHz) --
since the observed rotational spectra for PA and OD-PA are expected to be cleaner in these frequency
ranges given that low energy transitions of
light molecules fall at much higher frequencies. This is why we have listed only the
intense transitions for PA along with OD PA in Table 4
that fall in ALMA Bands 1, 2 and 3.
{\color{black} Currently ALMA Band 3 is in operation while Bands 1 and 2 are not yet available.
The strongest transitions pointed out in Table 4 for PA (at $116.23546$ GHz and $116.23586$ GHz)
are just beyond ALMA Band 3 and are not available in the ALMA time estimator. Thus, we
have focused on the next strongest transitions of Band 3 at
$89.39184$ GHz and $89.39212$ GHz, which also have an overlap with Band 2.
Using Band 3, we can detect these transitions of PA with approximately
$7.5$ hours of integration time by assuming a spectral resolution of $0.5$ km/s,
a sensitivity of $17$ mK, a signal-to-noise ratio of $6$, an angular resolution $3$ arcseconds
and $40$ antennas of $12$ meter array in the ALMA time estimator.
But it is not possible to detect any lines for OD-PA using ALMA Band 3 with any reasonable integration time.}
\subsection{\color{black} Vibrational spectra}
Here we use the MP2 method with an aug-cc-pVTZ basis set for the computation of the vibrational
transition frequencies and intensities of various normal modes of vibration of PA and OD-PA.
The prefix aug stands for diffuse functions and cc-pVTZ refers to Peterson and Dunning's correlation consistent basis set \citep{pete02} for
performing our calculations. The influence of the solvent in vibrational spectroscopy is taken into
account using the Polarizable
Continuum Model (PCM) with the integral equation formalism variant (IEFPCM) as a
default Self-consistent Reaction Field (SCRF) method \citep{toma05, mier81}. The IEFPCM model is considered to be a
convenient one, because the second energy derivative is available for
this model and its analytic form is also available. Our calculation was performed in the presence of a
solvent (water molecule) by placing the PA or OD-PA solute in the cavity within the
reaction field of the solvent.
Since the dielectric constant of ice ($85.5$) is slightly higher than that of water ($78.5$),
the vibrational frequencies reported here are not exactly those that pertain it ice but are close to these values.
In Table 5, we present the vibrational frequencies and intensities of PA and OD-PA in both the gas
and ice phases.
We show the band assignments and compare
our results with the existing low resolution experimental results \citep{siva15, nyqu71}.
Vibrational detections of molecules in the
interstellar medium may increase in importance with the impending launch of the the James Webb Space Telescope (JWST) although the spectroscopic
resolution may not be sufficient for rotational substructure in the gas.
We find that the most intense mode of
PA in the gas phase appears at $1070.33 \ \mathrm{cm^{-1}}$, which corresponds to CO stretching with an integral absorbance coefficient of 1.70$\times 10^{-17}$
cm $\mathrm{molecule^{-1}}$.
This peak is shifted downward in the ice water phase by nearly $17$ cm$^{-1}$ and appears at $1053.89 \ \mathrm{cm^{-1}}$ with an
integral absorbance coefficient 2.46 $\times 10^{-17}$ cm $\mathrm{molecule^{-1}}$. Our frequencies can be compared with the experimental ice phase value of $1030.7~\mathrm{cm^{-1}}$ \citep{siva15} and the vapor phase value of $1051~\mathrm{cm^{-1}}$ \citep{nyqu71}. Other strong modes
of vibrations are the OH torsion, CCC bending, $\mathrm{CH_2}$ wagging, CH stretching, and OH stretching.
In the case of OD-PA, the CCO stretching mode is the most intense mode and appears at $1068.32$ cm$^{-1}$
in the gas phase and $1050.98$ cm$^{-1}$ in the ice phase with integral absorbance coefficients of
$1.61 \times 10^{-17}$ and $2.49 \times 10^{-17}$ cm molecule$^{-1}$ respectively.
Table 5 clearly shows the differences between spectroscopic
parameters computed for propargyl alcohol between the two phases. These differences can be explained due to electrostatic effects, which are often
much less important for species placed in a solvent such as water with high dielectric constant than they are in the gas phase but these changes will be
significant if both the solvent and
the molecule are polar.
The two sets of results for PA are in good agreement as regards frequencies
with the existing experimental results, as shown above for the single case of the strong CO stretching.
Differences between the results could be attributed to multiple reasons, one of which concerns the
Gaussian 09 program, with which we are unable to consider the mixing of different modes under
harmonic oscillator approximations. Another reason is that in our quantum
chemical simulation of ice spectra, we considered a single propargyl alcohol molecule
inside a spherical cavity and immersed it in a continuous medium with a dielectric constant, while in the experiment
propargyl alcohol was deposited atop the ice, which could lead to the formation of clusters.
\begin{table*}
\scriptsize{
\centering
\vbox{
\caption{Vibrational frequencies and intensities of PA and OD-PA in the gas phase and in water
ice at the MP2/aug-cc-pVTZ level of theory}
\begin{tabular}{|c|c|c|c|c|c|c|c|c|}
\hline
\hline
{\bf Species}&{\bf Peak }&{\bf Integral absorbance}&{\bf Peak }&{\bf Integral absorbance}&{\bf Band}&{\bf Experimental}\\
{}&{\bf positions }&\bf coefficient)&{\bf positions }& {\bf coefficient}&{\bf assignments}&{\bf values}\\
{}&{\bf (Gas phase)}&{($\mathrm{cm/molecule}$)}&{\bf (H$_2$O ice)}&{\bf( $\rm {cm/molecule}$})&&{\bf ($\mathrm{cm^{-1}}$)}\\
&\bf ($\mathrm{cm^{-1}}$)& & {\bf( $\mathrm{cm^{-1}}$})& &&\\
&&&& && \\
\hline
& 186.09 &1.48$\times 10^{-18}$ & 192.71 &1.09$\times 10^{-18}$ & CCC out of plane bending&$240^a$\\
& 257.59 &1.02$\times 10^{-17}$ &258.48 &1.59$\times 10^{-17}$ & OH torsion &\\
& 319.40 &9.89 $\times 10^{-18}$ & 330.49 &1.64$\times 10^{-17}$ & CCC bending&$305^a$\\
& 554.56 &1.91$\times 10^{-18}$ & 555.37& 2.97$\times 10^{-18}$ & CCO bending &$558.6^b$ $550^a$\\
& 640.33 &7.43 $\times 10^{-18}$ & 647.20 &1.08$\times 10^{-17}$ & CH bending & $639^a$\\
& 665.86 & 6.09$\times 10^{-18}$ & 672.76 &8.99$\times 10^{-18}$ & CH out of plane bending &$654^b$ $657^a$\\
& 922.65 & 4.37$\times 10^{-18}$ & 922.89 & 5.97$\times 10^{-18}$ &C-C stretching&$916.8^b$ $918^a$\\
& 1002.52 & 2.89$\times 10^{-18}$ & 987.69 & 6.28$\times 10^{-18}$ & CH$_2$ rocking&$987.2^b$ $980^a$\\
& 1070.33& 1.70$\times 10^{-17}$ & 1053.89 & 2.46$\times 10^{-17}$ & C-O stretching&$1030.7^b$ $1051^a$\\
{\bf PA}& 1225.86 & 2.61$\times 10^{-18}$& 1224.44 &3.85$\times 10^{-18}$& CH$_2$ twisting &$1228^a$\\
& 1355.10 & 4.50$\times 10^{-19}$ & 1371.30 & 9.84$\times 10^{-19}$ & CH/OH bending&---\\
& 1417.76& 8.47$\times 10^{-18}$& 1409.79 & 1.03$\times 10^{-17}$ & CH$_2$ wagging &$1362^b$\\
& 1517.33 & 2.70 $\times 10^{-19}$& 1505.02 & 5.81$\times 10^{-19}$ & CH$_2$ scissoring&$1480^a$\\
& 2131.06 &2.92$\times 10^{-19}$ & 2125.63 & 4.23$\times 10^{-19}$ & CC stretching& $2185^a$\\
& 3066.90 & 4.64$\times 10^{-18}$ & 3084.99 &4.92$\times 10^{-18}$ & CH$_2$ symmetric stretching&$2922^b$ $2930^a$\\
& 3145.17 & 8.51$\times 10^{-19}$& 3154.35 & 1.05$\times 10^{-18}$ & CH$_2$ antisymmetric stretching&$2940^a$\\
& 3483.14 & 9.39$\times 10^{-18}$ & 3364.13 &1.45$\times 10^{-17}$ & C-H stretching&$3282^b$ $3322^a$\\
& 3831.50 & 6.97 $\times 10^{-18}$& 3813.81 &1.32$\times 10^{-17}$ & O-H stretching&--- \\
\hline
&164.70 &3.97$\times 10^{-18}$ & 173.05 &5.27$\times 10^{-18}$ & OD torsion&\\
& 221.58 &5.59$\times 10^{-18}$ &221.28 &9.19$\times 10^{-18}$ & CCC out of plane bending &\\
& 307.17 &3.08 $\times 10^{-18}$ & 313.58 &5.01$\times 10^{-18}$ & CCC bending&\\
& 543.25 &2.13$\times 10^{-18}$ & 547.08& 3.24$\times 10^{-18}$ & CCO bending &\\
& 640.32 &7.35 $\times 10^{-18}$ & 647.20 &1.08$\times 10^{-17}$ & CH bending &\\
& 665.39& 6.12$\times 10^{-18}$ & 672.72 &9.03$\times 10^{-18}$ & CH out of plane bending &\\
& 885.79 &5.89$\times 10^{-18}$ & 852.99 & 8.03$\times 10^{-18}$ &OH bending&\\
& 961.44 &2.40$\times 10^{-18}$ & 944.47 & 3.55$\times 10^{-18}$ & CC stretching &\\
& 1068.32&1.61$\times 10^{-17}$ & 1050.98 & 2.49$\times 10^{-17}$ & CCO stretching&\\
{\bf OD-PA}& 1086.19 & 5.05$\times 10^{-19}$& 1086.55 &1.06$\times 10^{-18}$& CH$_2$ rocking&\\
& 1303.40 & 2.16$\times 10^{-20}$ & 1313.89 & 8.66$\times 10^{-20}$ & CH$_2$ twisting&\\
& 1396.85& 4.39$\times 10^{-18}$& 1396.93 & 6.64$\times 10^{-18}$ & CH$_2$ wagging & \\
& 1517.29 & 2.63 $\times 10^{-19}$&1504.99 & 5.71$\times 10^{-19}$ & CH$_2$ scissoring&\\
& 2131.04 &2.94$\times 10^{-19}$ & 2125.63 & 4.25$\times 10^{-19}$ & CC stretching&\\
& 2789.51 & 4.50$\times 10^{-18}$ & 2776.81 &8.25$\times 10^{-18}$ & OD stretching&\\
& 3066.98 & 4.54$\times 10^{-18}$& 3085.06& 4.83$\times 10^{-18}$ & CH$_2$ symmetric stretching &\\
& 3145.44 & 7.29$\times 10^{-19}$ & 3154.61 &8.76$\times 10^{-19}$ &CH$_2$ antisymmetric stretching &\\
& 3483.14& 9.40 $\times 10^{-18}$& 3464.13 &1.46$\times 10^{-17}$ & CH stretching& \\
\hline
\multicolumn{4}{c}{$^a$ \cite{nyqu71}} (vapor phase), $^b$ \cite{siva15} (ice phase experiment)\\
\end{tabular}}}
\end{table*}
\section{Conclusions}
\label{sec:concl}
Earlier results suggest that PA could be treated as a precursor of benzene formation \citep{wils03}.
The interstellar identification of propenal, which is an isomer of PA, encouraged us to check
the possibility of detecting PA and OD-PA in the ISM. Our quantum chemical calculations, combined with existing rotational transition frequencies for PA and OD-PA, allow us to achieve the following major results for this paper:
\noindent {$\bullet$ A complete reaction network has been prepared for the formation and destruction of
PA. Various formation routes are discussed and identified based on calculated exothermicities and endothermicities.
}
\noindent {$\bullet$ The predicted abundances of PA and OD-PA in the gas and ice phases yield some
information for both cold and warm interstellar sources on the possibility of observing this molecule in the ISM, especially in hot-core regions such as Sgr B2(N). }
\noindent{$\bullet$ \color{black} A simple radiative transfer model has been employed to discuss the
possibility of detecting PA in the ISM through millimeter-wave observations. The most likely environments for detection of PA are hot-core regions with a temperature above 100 K.
}
\noindent {$\bullet$ The frequencies and intensities of vibrational transitions of PA and OD-PA
both in the gas and in a mimetic for an ice environment are
calculated and compared with existing low resolution experimental frequencies. The calculated intensities should be particularly useful.}
\section{Acknowledgments}
PG, AD \& SKC are grateful to DST (Grant No. SB/S2/HEP-021/2013) and ISRO Respond
(Grant No. ISRO/RES/2/402/16-17) for financial support.
LM thanks the ERC for a starting grant (3DICE, grant agreement 336474). E. H. wishes to
acknowledge the support of the National Science Foundation for his astrochemistry program.
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 335 |
Johan Daniel Drissel, född 16 februari 1751 i Bred, död 1823, var en svensk jurist och universitetslärare.
Drissel fick professors fullmakt 1799 och blev 1805 professor i juridik. Han var 1813 rektor vid Uppsala universitet.
Källor
Svenska professorer i juridik
Rektorer vid Uppsala universitet
Födda 1751
Avlidna 1823
Män
Personer från Breds socken | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 5,910 |
La Serra del Sastret és una serra del terme municipal de Sant Esteve de la Sarga, del Pallars Jussà. A ponent enllaça amb el Serrat de Castellnou, i està situada a llevant, una mica decantada cap al sud, del poble de Castellnou de Montsec. La seva continuïtat pel costat de llevant és la serra del Coscó, ja dins del terme de Castell de Mur, també del Pallars Jussà.
Referències
Vegeu també
Masia del Castell.
Sastret | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 6,443 |
HWI – codice aeroportuale IATA dell'aeroporto civile di Hawk Inlet (Stati Uniti) | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 8,730 |
De Polymoog is een polyfone synthesizer die werd geproduceerd door Moog van 1975 tot 1980. De klankopwekking is gebaseerd op een frequentiedeler, vergelijkbaar met elektronische orgels uit die tijd.
Dit instrument bevat voorgeprogrammeerde geluiden zoals die van een piano of een viool. Deze geluiden zijn aanpasbaar, maar kunnen niet worden opgeslagen in een geheugen.
Externe link
Polymoog op vintagesynth.com
Zie ook
Minimoog
Moog-synthesizer | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 5,502 |
Q: libgdx dragging & placing actors I'm stuck on making widgets grab-able / movable & I was hoping someone here on stack could point me in the direction of achieving this. I Was originally thinking using movement from com.badlogic.gdx.scene.scene2d.ui.Window & was curious if there was an easier way. I've looked into all the methods inherited from other classes on GDXwiki & still have no luck.
A: I think that what you are looking for is Box2D's QueryAABB method.
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 7,846 |
\section{Introduction}
Dust, neutral gas and ionized gas are the major components
of the interstellar medium (ISM) in galaxies. Observations of
their properties and inter-relationships can give important
clues to the physics governing star formation.
Relationships between components in the ISM are to be expected. Observations have shown that in the Galaxy dust and neutral gas are well mixed. In dense clouds of molecular gas mixed with cold dust most of the stars are formed. They subsequently heat the dust and gas in their surroundings and ionize the atomic gas. As the major coolants of the ISM are continuum emission and line emission at various frequencies, a close comparison of these emissions could shed light on spatial and physical connections between the emitting components. Present-day IR and radio telescopes have produced sensitive high resolution maps of several nearby galaxies, which are ideal laboratories to study the interplay between the ISM and star formation \citep[e.g. ][]{Kennicutt_07,Bigiel_08,Verley_09}.
The spiral galaxy nearest to us, the Andromeda Nebula
(NGC224), is a highly inclined Sb galaxy of low surface brightness.
Table 1 lists the positional data on M~31. Its proximity
and large extent on the sky ($> 5^{\circ} \times 1^{\circ}$) enable detailed studies
of the ISM over a large radial range.
Surveys of M~31 at high angular resolution ($< 1\arcmin$) are available at
many wavelengths. In the HI line the galaxy was mapped by \cite{Brinks}
at $24\arcsec \times 36\arcsec$ resolution, the northeastern half by
\cite{Braun_90} at 10$\arcsec$ resolution and, most recently,
the entire galaxy with high sensitivity by \cite{Braun_09} at
a resolution of 15$\arcsec$. \cite{Nieten} made a survey in the
$^{12}$CO(1-0) line at a resolution of 23$\arcsec$. \cite{Devereux_etal_94b} observed M~31 in the H$\alpha$ line to obtain the distribution of the ionized gas. The dust emission from M~31 was recently observed by the multiband imaging photometer Spitzer \citep[MIPS, ][]{Rieke} with high sensitivity at 24\,$\mu$m, 70\,$\mu$m, and 160\,$\mu$m at resolutions~$\leq 40\arcsec$.
The relationships between gas and dust as well as between gas and
star formation in M~31 have been studied in the past at resolutions
of several arcminutes. \cite{Walterbos_87} derived a nearly
constant dust temperature across M~31 using the IRAS 100\,$\mu$m and
60\,$\mu$m maps. They also found a strong increase in the atomic gas-
to-dust surface density ratio with increasing radius. This increase
was confirmed by \cite{Walterbos_88} who used optical extinction as dust
tracer, and by \cite{Nieten} using the ISO map at 175\,$\mu$m
\citep{Haas}. Interestingly, the latter authors did not find
a radial increase in the molecular gas-to-dust ratio.
The dependence of star formation on HI surface density in M~31 has been
studied by a number of authors \citep{Emerson_74,Berkhuijsen_77,Tenjes,
Unwin,Nakai_82,Nakai_84} using the number density of HII regions or of OB
stars as star formation tracers. They obtained power-law indices between
0.5 and 2, possibly depending on the region in M~31, the star formation
tracer and the angular resolution. \cite{Braun_09} plotted the star formation
density derived from the brightnesses at 8$\mu$m, 24$\mu$m and UV against
the surface densities of molecular gas, HI and total gas, but did not fit
power laws to their data.
The high-resolution data available for M~31 show the morphologies of the emission from dust and gas components in detail. We apply a 2-D wavelet analysis technique \citep{Frick_etal_01} to the MIPS IR data \citep{Gordon_06} and the gas (HI, H$_2$, and H$\alpha$) maps to study the scale distribution of emission power and to separate the diffuse emission components from compact sources. We then compare the wavelet-decomposed maps at various spatial scales. We also use pixel-to-pixel (Pearson) correlations to derive quantitative relations not only between different ISM components but also between them and the present-day star formation rate.
Following \cite{Walterbos_87} and \cite{Haas}, we derive the dust temperature
assuming a $\lambda^{-2}$ emissivity law for the MIPS bands at which the emission
from the big grains and hence the LTE condition is relevant, and present a map of
the dust color temperature. We also obtain the distribution of the optical
depth and analyze the gas-to-dust surface-density ratio at a resolution
of 45$\arcsec$ (170~pc\,$\times$\,660~pc along the major and minor axis,
respectively, in the galaxy plane), 9 times higher than
before \citep{Walterbos_87}. We use the optical depth map to
de-redden the H$\alpha$ emission observed by \cite{Devereux_etal_94b}
yielding the distribution of the absorption-free emission from the ionized gas,
and use this as an indicator of massive star formation. We compare it with the
distributions of neutral gas to obtain the dependence of the star
formation rate on gas surface density.
The paper is organized as follows: The relevant data sets are described
in Sect. 2. In Sect. 3 we derive maps of the dust color temperature and
optical depth, and correct the H$\alpha$ emission for absorption
by dust. Radial profiles of the dust and gas emission and of the various
gas-to-dust ratios are obtained in Sect. 4. Sect. 5 is devoted to
wavelet decompositions and wavelet spectra of the dust and gas distributions,
and their cross correlations. Complementary, we discuss in
Sect. 6 classical correlations between gas and dust. In Sect. 7
the dependence of the star formation rate on the gas surface density is presented.
Finally, in Sect. 8 we summarize our results.
\begin{table}
\begin{center}
\caption{Positional data adopted for M~31.}
\begin{tabular}{ l l }
\hline
\hline
Position of nucleus & RA\,=\,$00^{h}42^{m}45.97^{s}$ \\
\,\,\,(J2000) & DEC\,=\,$41^{\circ}16\arcmin11.64\arcsec$\\
Position angle of major axis & 37$^{\circ}$ \\
Inclination$^{1}$ & 75$^{\circ}$ (0$^{\circ}$=face on)\\
Distance$^{2}$ & 780$\pm$40\,kpc$^3$\\
\hline
\noalign {\medskip}
\multicolumn{2}{l}{$^{1}$ \cite{Berkhuijsen_77} and \cite{Braun_91}}\\
\multicolumn{2}{l}{$^{2}$ \cite{Stanek} }\\
\multicolumn{2}{l}{$^{3}$ 1$\arcmin$=\,227$\pm$12\,pc along major axis}\\
\end{tabular}
\end{center}
\end{table}
\section{Data}
Table 2 summarizes the data used in this work.
M~31 was mapped in IR (at 24\,$\mu$m, 70\,$\mu$m, and 160\,$\mu$m) by MIPS in August 2004 covering a region of about 1$^{\circ} \times 3^{\circ}$ \citep{Gordon_06}. The
basic data reduction and mosaicing was performed using the MIPS instrument team Data
Analysis Tool versions 2.90 \citep{Gordon_05}.
M~31 was observed in the $^{12}$CO(1-0) line with the IRAM telescope
by \cite{Nieten} at a resolution of 23$\arcsec$. They derived the distribution of the molecular gas using a constant conversion factor of X$_{\rm CO} =\,1.9\,\times \, 10^{20}$\,mol.\,K$^{-1}$\,km$^{-1}$\,s. The galaxy was observed in the 21-cm HI
line with the Westerbork interferometer by \cite{Brinks}
at a resolution of 24$\arcsec \times 36\arcsec$. The HI survey has been corrected
for missing spacings. The H$\alpha$ observations of \cite{Devereux_etal_94b} were carried out on the Case Western Burell-Schmidt telescope
at the Kitt Peak National Observatory, providing a 2$^{\circ} \times 2^{\circ}$ field
of view.
Although the resolution of 40$\arcsec$ of the 160\,$\mu$m image is the lowest
of the data listed in Table 2, we smoothed all maps to a Gaussian
beam with a half-power width of 45$\arcsec$ for a comparison with radio
continuum data at 20~cm \citep{Hoernes_etal_98} in a forthcoming study
(Tabatabaei et al. in prep.). As the point spread function (PSF) of
the MIPS data is not Gaussian, we convolved the MIPS images using
custom kernels created with Fast Fourier transforms to account for
the detailed structure of the PSFs. Details of the kernel creation
can be found in \cite{Gordon_07}.
\begin{table*}
\begin{center}
\caption{M~31 data used in this study. }
\begin{tabular}{ l l l l}
\hline
Wavelength & Resolution & Telescope & Ref. \\
\hline
160\,$\mu$m & $40\arcsec$ & Spitzer& \cite{Gordon_06}\\
70\,$\mu$m & $18\arcsec$ & Spitzer&\cite{Gordon_06}\\
24\,$\mu$m & $6\arcsec$ & Spitzer & \cite{Gordon_06} \\
2.6\,mm\,$^{12}$CO(1-0) & $23\arcsec$ &IRAM 30-m & \cite{Nieten} \\
21\,cm\,HI & 24$\arcsec \, \times \,30\arcsec$ & WSRT &\cite{Brinks}\\
6570\AA{}\,H$\alpha$ & $2\arcsec$ (pixel size) & KPNO& \cite{Devereux_etal_94b}\\
\hline
\end{tabular}
\end{center}
\end{table*}
After convolution, the maps were transformed to the same grid of 15$\arcsec$ width
with the reference coordinates and position angle of the major axis
given in Table 1. Finally, they were cut to a common extent
of 110$\arcmin \times 38.\arcmin5$, for which most data sets are complete.
The field is not centred on the nucleus of M~31, but extends to 56.$\arcmin25$
along the northern major axis (corresponding to a radius of $R$\,=\,12.8\,kpc) and to
53.$\arcmin75$ along the southern major axis ($R$\,=\,12.2\,kpc). The H$_2$ map of Nieten et
al. (2006) extends to 48.$\arcmin5$ along the southern major axis ($R$\,=\,11.0\,kpc).
With an extent of 19.$\arcmin25$ along the minor axis in both directions, the field
covers radii of $R <$\,16.9\,kpc in the plane of M~31. Hence, radial profiles derived by
averageing the data in circular rings in the plane of the galaxy (equivalent to
elliptical rings in the plane of the sky) are incomplete at $R>$\,12\,kpc because
of missing data near the major axis.
\section{Dust temperature and opacity}
\cite{Walterbos_87} extensively studied the distributions
of the dust temperature and opacity in M~31 using the IRAS data at
60\,$\mu$m and 100\,$\mu$m smoothed to a resolution of 4.$\arcmin3\times 6.\arcmin9$.
Assuming a $\lambda^{-2}$ emissivity law, they found a remarkably constant dust temperature (21-22\,K) across the disk between 2\,kpc and 15\,kpc radius. Using this temperature, they obtained the opacity distribution at 100\,$\mu$m. Below we apply a similar method to the
70\,$\mu$m and 160\,$\mu$m MIPS maps to derive the distributions of dust
temperature and optical depth at the H$\alpha$ wavelength at higher
resolution and sensitivity.
\subsection{Dust temperature}
We derived the color temperature of the dust, $T_{\rm d}$, between 70\,$\mu$m
and 160\,$\mu$m assuming a $\lambda^{-2}$ emissivity law that should be
appropriate for interstellar grains emitting at these wavelengths
\citep{andriesse,Draine}. The resulting dust temperature
map (Fig.~1) and a histogram of the temperatures (Fig.~\ref{fig:thistring}a)
show that $T_{\rm d}$ varies between 15.6\,$\pm$\,0.8\,K and 30.8\,$\pm$\,0.3\,K. The mean value of 18.7\,$\pm$\,1.4\,K\,(standard deviation) is lower than that obtained by \cite{Walterbos_87} and close to the ISO measurements (16\,$\pm$\,2\,K) of \cite{Haas}.
Figure~1 shows that dust of {\bf $\sim$\,18\,K} exists all over M~31.
Warmer dust with $T_{\rm d}>20$\,K dominates in star forming regions and in an
extended area around the center of the galaxy, while cooler dust
dominates in interarm regions.
Figure~\ref{fig:thistring}b shows the dust temperature averaged in rings of 0.2~kpc width in the plane of M~31 against radius R\footnote{Because the spiral structure is different in the northern and southern half (northeast and southwest of the minor axis, i.e. left and right of the minor axis, respectively), we present all radial profiles for each half separately.}. On both sides of the center the dust temperature falls very fast from about 30\,K near
the nucleus to 19~K at $R\simeq$\,4.5\,kpc. To the outer parts of the galaxy,
it then stays within a small range of about 17\,K-19\,K in the north and
16\,K-19\,K in the south. This indicates different radiation characteristics
between the inner 4~kpc and beyond. In {\bf the ring of bright emission, the so called `10~kpc ring',} the temperature
is clearly enhanced, especially in the northern half.
Thus, in contrast to the finding of \cite{Walterbos_87}, $T_{\rm d}$ is not constant in the range $R$\,=\,2-15\,kpc but varies between 22.5\,$\pm$\,0.5\,K and 17.2\,$\pm$\,0.7\,K.
It is interesting that the mean dust temperature obtained between
70\,$\mu$m and 160\,$\mu$m is about 3\,K lower in M~31 than in M~33 \citep{Tabatabaei_3_07}. The emission from cold dust in M~31 is stronger than in
M~33, which can also be inferred from the total emission spectra based
on IRAS and ISO observations \citep[see ][]{Haas,Hippelein}.
\begin{figure*}
\begin{center}
\resizebox{\hsize}{!}{\includegraphics*{m31-Tdust.color2.eps}}
\caption{Dust temperature in M~31 obtained from the ratio I$_{70\mu m}$/I$_{160\mu m}$ based on the Spitzer MIPS data. Only pixels with intensity above 3$\times$ noise level were used. The angular resolution of 45$\arcsec$ is shown in the lower right-hand corner of the map. The cross indicates the location of the center. The bar at the top gives the dust temperature in Kelvin.}
\end{center}
\label{fig:tempere}
\end{figure*}
\subsection{Dust opacity distribution}
\label{sec:tau}
The total dust optical depth along the line of sight $\tau_{160}$
was obtained from the dust intensity at 160\,$\mu$m and the derived temperature.
Following \cite{Tabatabaei_3_07}, $\tau_{160}$ was converted into the dust
optical depth at the wavelength of the H$\alpha$ line, $\tau_{{\rm H}\alpha}$, by
multiplying it by the ratio of the dust extinction coefficient per unit mass at the
corresponding wavelengths, $\tau_{{\rm H}{\alpha}} \simeq 2200\, \tau_{160}$ \citep[see e.g.
Figure 12.8 of][]{krugel}. Figure~3 shows the distribution of $\tau_{{\rm H}\alpha}$
across the disk of M~31 at an angular resolution of 45$\arcsec$. Regions with
considerable dust opacity ($\tau_{{\rm H}\alpha}>\,0.6$) follow the spiral arms,
even the inner arms which are either weak or not detected in H$\alpha$ emission.
The high opacity clumps ($\tau_{{\rm H}\alpha}>\,1.5$), however, only occur in the
arms at $\simeq$\,5\,kpc and in the `10\,kpc ring'. On average, the optical depth
is largest in the '10\,kpc ring'. Hence, in this ring, dust has the highest density like the atomic gas \citep{Brinks}.
Fig.~\ref{fig:tauhistring}a shows the histogram of $\tau_{{\rm H}\alpha}$, indicating
a most probable value of 0.5 and a mean value of 0.7\,$\pm$\,0.4 across the galaxy. This
agrees with the value of $\tau_{{\rm H}\alpha}$ = 0.5\,$\pm$\,0.4 that follows from the mean
total extinction obtained by Barmby et al (2000) towards 314 globular clusters.
The variation of the mean dust optical depth $\tau_{{\rm H}\alpha}$ with galactocentric
radius is shown in Fig.~\ref{fig:tauhistring}b.
In the north, $\tau_{{\rm H}\alpha}$ peaks not only in the `10\,kpc ring' (with two
maxima at $R=$\,9.9 and 10.9\,kpc) but also near 5\,kpc (with two maxima at $R=$\,4.3
and 5.9\,kpc). Beyond 11\,kpc $\tau_{{\rm H}\alpha}$ drops with an exponential scale
length of 2.48$\pm$0.07\,kpc in the north and 5.06$\pm$0.22\,kpc in the south (Sect.~4,
eq.~1).
In Fig.~\ref{fig:extinction} we compare the radial variation of $\tau_{{\rm H}\alpha}$ for the
total area with earlier determinations. The various estimates agree well given the
large uncertainties. \cite{Xu_96} derived $\tau_{\rm V}$ from high-resolution IRAS data
using a dust heating/cooling model and a sandwich configuration of dust and stars.
Although they left out the discrete sources, their values may be too high because
they did not include inter-arm regions in their study. \cite{Montalto}
calculated the extinction $A_{\rm FUV}$ from the total infrared TIR-to-FUV intensity ratio
and a sandwich model for stars and dust. They note that at $R <$\,8\,kpc the geometry
of M~31 may differ from the sandwich model due to the stars in the bulge, making
the inner points less reliable. This would also affect the results of \cite{Xu_96} at $R <$\,8\,kpc. The TIR-to-FUV ratio is applicable if the dust is mainly heated
by young stars, but in M~31 about 70\% of the cold dust is heated by the ISRF
\citep{Xu_96}. Therefore, $A_{\rm FUV}$ is overestimated, as was also found for M33
\citep{Verley_09}. The curve of \cite{Tempel} closely agrees with our
data. They derived $\tau_{\rm V}$ from MIPS data and a star-dust model. Their smooth curve
underestimates $\tau_{\rm V}$ in the brightest regions by about 0.1 and may overestimate
$\tau_{\rm V}$ in regions of low brightness.
\begin{figure*}
\begin{center}
\resizebox{\hsize}{!}{\includegraphics*{m31-Tdust_hist.ps}
\includegraphics*{m31.Tdust.ring.eastwest.ps}}
\caption[]{{\it a}) Histogram of the dust temperature shown in Fig.~1. {\it b}) Distribution of the dust temperature in rings of 0.2\,kpc in the galactic plane in the northern and southern halves of M~31.}
\label{fig:thistring}
\end{center}
\end{figure*}
The opacity map in Figure~3 can be used to correct the H$\alpha$ emission for the extinction by dust. In general, extinction depends on the relative distribution of emitting regions and dust along the line of sight and changes with the geometry \citep[e.g. well mixed diffuse medium
or shell-like in HII regions ][]{Witt}. In this study, individual HII regions are rarely resolved and the geometry is close to a mixed diffuse medium.
Furthermore, there is no information about the relative position of emittors and absorbers along the line of sight.
For the Milky Way, \cite{Dickinson} found indications of a non-uniform mixing by comparing the z-distribution of atomic gas and dust. They adopted one third of the total dust optical depth as the effective extinction as a first-order approach.
This is also in agreement with \cite{Kruegel_09} taking scattering into account. Moreover, \cite{Magnier} found that on average the extinction comes from dust associated with only one-third of average N(HI) in their study of OB associations along the eastern spiral arm regions of M~31. Therefore, we use an effective optical depth $\tau_{\rm eff} = 0.33\,\times\, \tau_{{\rm H}\alpha}$ in this paper. The attenuation factor for the H$\alpha$ intensity then is $e^{-\tau_{\rm eff}}$ and we derive the intrinsic H$\alpha$ intensity $I_0$ from the observed H$\alpha$ intensity $I=I_0\,e^{-\tau_{\rm eff}}$. Integration of the H$\alpha$ map out to a radius of 16\,kpc yields a ratio of corrected-to-observed total H$\alpha$ flux density of 1.29, thus about 30\% of the total H$\alpha$ emission is obscured by dust within M~31. The corrected H$\alpha$ map is shown in Fig.~\ref{fig:wave}a.
\begin{figure*}
\begin{center}
\resizebox{\hsize}{!}{\includegraphics*{m31-tau_CO.color3.eps}}
\caption[]{ Distribution of the dust optical depth at H$\alpha$ wavelength in
M~31. The bar at the top shows $\tau_{{\rm H}_{\alpha}}$. The angular
resolution of 45$\arcsec$ is shown in the lower right-hand corner of the map. The cross indicates the location of the center. Overlayed are contours of molecular gas column density N(H$_2$) with levels of 250 and 800\,$\times 10^{18}$\,mol.\,cm$^{-2}$. Note that maxima in $\tau_{{\rm H}_{\alpha}}$ not always coincide with maxima in N(H$_2$). }
\end{center}
\label{fig:tau}
\end{figure*}
\begin{figure*}
\resizebox{\hsize}{!}{\includegraphics*{m31-tauha_hist.ps}
\includegraphics*{m31.tauHa.ring.eastwest.ps}}
\caption{{\it a}) Histogram of the dust optical depth shown in Fig.~3, {\it b}) radial distribution of the mean optical depth at the H$\alpha$ wavelength in rings of width of 0.2\,kpc in the galactic plane in the north and south of M~31. The errors are smaller than the size of the symbols.}
\label{fig:tauhistring}
\end{figure*}
Near the center ($R\,<\,1$\,kpc), $\tau_{\rm eff}$ varies between 0.03 and 0.13,
corresponding to an extinction of $A({\rm H}\alpha)=1.086\,\times\,\tau_{\rm eff}\simeq$\,
0.03-0.14 mag. At larger radii, the mean extinction increases, particularly in dense
clouds and star forming regions, reaching a maximum of $A({\rm H}\alpha)\simeq$\,1.2\,mag
at the densest dust cloud in the south-east of the `10\,kpc ring' (RA\,=\,00$^h$
41$^m$ 05.10$^s$ and DEC\,=\,+40$^{\circ}$ 38$\arcmin$ 17.73$\arcsec$). The range of
extinction values agrees with that derived from the optical study of dust lanes by
\cite{Walterbos_88} and the photometric study of \cite{Williams}.
\section{Radial distributions of dust and gas emission}
\subsection{Radial profiles}
\label{sec:profile}
In this section, we present the mean surface brightness along the line of sight
of dust and gas components as a function of galactocentric radius $R$. The surface
brightnesses are averaged in {\bf 200\,pc-wide} circular rings about the nucleus in the plane
of M~31. This is equivalent to averaging in elliptical rings of 53$\arcsec$ width in the plane of the sky.
For simplicity we used a constant inclination angle of 75$^{\circ}$ at all radii,
appropriate for the emission at $R > 30\arcmin$ (6.8\,kpc), although in H$\alpha$ and HI
the inner regions are seen more face-on \citep{Ciardullo,Braun_91,Chemin}. However, using i\,=\,75$^{\circ}$ for $R < 30\arcmin$ instead of i\,=\,68$^{\circ}$ \citep[the area-weighted mean of the inclinations for the interval $R\,=\,1.9 - 6.8$\,kpc given by ][]{Chemin} does not change our results. The smaller inclination shifts the radial positions of the inner arms about 0.5\,kpc inwards but the general
shape of the profiles remains the same, and as all profiles change in a similar
way their inter-comparison is not affected. Furthermore, the results of the classical
correlations for $R < 30\arcmin$ presented in Sect.~6 are the same within the errors
for i\,=\,68$^{\circ}$ and i\,=\,75$^{\circ}$.
\begin{figure}
\begin{center}
\resizebox{7.8cm}{!}{\includegraphics[angle=-90]{m31.R-tau.ps}}
\caption{Radial variation of the (total) mean optical depth
in H$\alpha$ along the line of sight for the full area (north+south).
{\it Plusses}: our data averaged in {\bf 0.2\,kpc-wide} rings in the plane
of M~31 using i\,=\,75$^{\circ}$; {\it stars}: same but with i\,=\,77.5$^{\circ}$ for
comparison with other work; {\it triangles}: \cite{Xu_96},
averages in {\bf 2\,kpc-wide} rings with i\,=\,77$^{\circ}$ (scaled to D\,=\,780\,kpc);
{\it circles}: \cite{Montalto}, averages in {\bf 2\,kpc-wide} rings
with i\,=\,77.6$^{\circ}$; {\it solid line}: \cite{Tempel}, semi-major
axis cut through model with i\,=\,77.5$^{\circ}$. The errors in our data
and in the curve of \cite{Tempel} are about 10\% of the
mean values. }
\label{fig:extinction}
\end{center}
\end{figure}
Figure~\ref{fig:surfir} shows the mean IR intensities and the gas surface densities versus the galactocentric radius $R$ for the northern and southern halves
of M~31. The radial profiles of the IR emission at 24\,$\mu$m and 70\,$\mu$m are similar. The 160\,$\mu$m emission, representing the colder dust emission, however, shows a generally flatter radial distribution than the 24\,$\mu$m and 70\,$\mu$m emission. In particular, the fast decrease of the 24\,$\mu$m and 70\,$\mu$m profiles from the center to $R\,\simeq$\,2\,kpc does not occur at 160\,$\mu$m. This is in agreement with \cite{Haas} who concluded from their ISO 175\,$\mu$m map and IRAS data that the dust near the center is relatively warm.
The fast central decrease of warmer dust emission may be attributed to a decrease in the UV radiation field outside the nucleus, as a similar trend is seen in the GALEX UV profiles presented by \cite{Thilker_05}.
At all three IR wavelengths the arms are visible, even the weak inner arms. The bright arms forming the `10\,kpc ring', are pronounced in the north and followed by an exponential decrease toward larger radii.
\begin{figure*}
\begin{center}
\resizebox{\hsize}{!}{\includegraphics*{surfbright_IR.east.ps}
\includegraphics*{surfbright_IR.west.ps}}
\resizebox{\hsize}{!}{\includegraphics*{surfbright_gas.east.ps}
\includegraphics*{surfbright_gas.west.ps}}
\caption[]{{\it Top}: Radial profiles of the Spitzer IR emission from the northern ({\it left}) and the southern ({\it right}) halves of M~31. {\it Bottom}: Radial profiles of the surface densities of the atomic, molecular and total neutral gas together with that of the ionized gas (de-reddened H$\alpha$) for the northern ({\it left}) and southern ({\it right}) halves of M~31. The units are $10^{18}$\,atoms\,cm$^{-2}$ for HI and HI+2H$_2$, $10^{18}$\,molecules\,cm$^{-2}$ for H$_2$, and 10$^{10}$\,erg\,s$^{-1}$\,cm$^{-2}$\,sr$^{-1}$ for H$\alpha$ profiles. The profiles show intensities along the line of sight averaged in circular rings of 0.2\,kpc width in the plane of M~31 against the galactocentric radius. The errors are smaller than 5\% for all profiles, only for H$_2$ they increase from 10\% to 25\% at $R>12.3$\,kpc and in the inner arms for $R<4.5$\,kpc. }
\label{fig:surfir}
\end{center}
\end{figure*}
Although the general trend of the warm dust surface brightness
(at 24\,$\mu$m and 70\,$\mu$m) resembles more that of H$\alpha$ than those of
the neutral gas profiles (Fig.~\ref{fig:surfir}, lower panels), small variations
(e.g. in the inner 5\,kpc and for $R$\,=\,11-12\,kpc in the south)
follow those in the total gas distribution due to variations in the molecular gas. Beyond about 5\,kpc, the radial profile of the cold dust (160\,$\mu$m) is similar to that
of the molecular gas, but with smoother variations. The minimum between 5~kpc and
10~kpc radius at 24\,$\mu$m and 70\,$\mu$m is less deep at 160\,$\mu$m and is missing
in the HI profile.
We obtained radial scale lengths between the maximum in the `10\,kpc
ring' and R=14.9~kpc for the northern ($l_{\rm N}$) and southern ($l_{\rm S}$) halves
of M~31 separately as well as for the total area ($l$). We fit an
exponential function of the form
\begin{equation}
I(R)= I_{0}exp(-R/l),
\end{equation}
where $I_0$ is
the intensity at $R$\,=\,10.9\,kpc for the total area and in the north, and
$R$\,=\,8.9\,kpc in the south. The resulting scale lengths are listed in
Table~\ref{table:slength}.
In each half of M~31, the scale lengths of the warm dust emission
are smaller than that of the cold dust. This confirms that the
warm dust is mainly heated by the UV photons from the star forming
regions in the `10\,kpc ring' and the cold dust mainly by the interstellar
radiation field (ISRF) from old stars
\citep{Xu_96}.
\begin{table}
\begin{center}
\label{table:scalelength}
\caption{Exponential scale lengths of dust and gas emissions from M~31. The scale lengths were calculated from $R$=\,10.9\,kpc to $R$=\,14.9\,kpc for the whole galaxy ($l$) and the northern half ($l_{\rm N}$) and from $R$=\,8.9\,kpc to $R$=\,14.9\,kpc for the southern half ($l_{\rm S}$). The scale length of $\tau_{\rm H\alpha}$ is also shown for comparison. }
\begin{tabular}{ l l l l}
\hline
& $l_{\rm N}$\,(kpc) & $l_{\rm S}$\,(kpc) & $l$\,(kpc)\\
\hline
\hline
IR 160$\mu$m & 1.86\,$\pm$\,0.06 & 3.87\,$\pm$\,0.18 & 2.29\,$\pm$\,0.11 \\
IR 70$\mu$m & 1.38\,$\pm$\,0.08 & 3.07\,$\pm$\,0.17 & 1.66\,$\pm$\,0.09\\
IR 24$\mu$m & 1.38\,$\pm$\,0.07 & 3.44\,$\pm$\,0.21 & 1.57\,$\pm$\,0.06 \\
H$_2$ & 1.14\,$\pm$\,0.07 & 2.40\,$\pm$\,0.22 & 1.27\,$\pm$\,0.11 \\
HI & 4.54\,$\pm$\,0.24 & 15.90\,$\pm$\,1.70 & 5.90\,$\pm$\,0.20 \\
HI+2H$_2$ & 3.80\,$\pm$\,0.12 & 9.00\,$\pm$\,0.69 & 4.73\,$\pm$\,0.21 \\
H$\alpha$ & 1.08\,$\pm$\,0.03 & 2.92\,$\pm$\,0.10 & 1.39\,$\pm$\,0.04 \\
\hline
$\tau_{\rm H\alpha}$ & 2.48$\pm$0.07 & 5.06$\pm$0.22 & 3.17\,$\pm$\,0.15\\
\hline
\end{tabular}
\label{table:slength}
\end{center}
\end{table}
The scale lengths of the 24\,$\mu$m and 70\,$\mu$m emission are nearly the same and the 24\,$\mu$m-to-70\,$\mu$m intensity ratio (Fig.~\ref{fig:ratio}) hardly varies between $R\simeq10$\,kpc and $R\simeq15$\,kpc. This indicates a similar distribution of their origins. Assuming that the main source of the 24\,$\mu$m emission is very small dust grains and of the 70\,$\mu$m and 160\,$\mu$m emission is big grains, as argued by \cite{Walterbos_87}, a constant intensity ratio of the 24\,$\mu$m-to-70\,$\mu$m and 24\,$\mu$m-to-160\,$\mu$m emission suggests that the very small and big grains are well mixed in the interstellar medium. Other possible origins of the 24\,$\mu$m emission are stars with dust shells like evolved AGB stars or Carbon stars. Using the IRAS data, \cite[][]{Soifer} attributed the 25\,$\mu$m emission from the bulge (central 8$\arcmin$) of M~31 to circumstellar dust emission from late-type stars. In the disk of M~31, \cite{Walterbos_87} found no direct evidence for a contribution from stars with dust shells \citep[contrary to that in the Milky Way, ][]{Cox_86}.
The higher resolution and sensitivity of the MIPS IR intensity ratios (Fig.\ref{fig:ratio}), however, provide more information. Although at $R\,>\,3$\,kpc the variations in the IR intensity ratios are not large, their radial behavior is not the same. For instance, the 24-to-70\,$\mu$m intensity ratio peaks between 5\,kpc and 10\,kpc radius, whereas the 70-to-160\,$\mu$m intensity ratio peaks in the `10\,kpc ring'. The latter can be explained by the higher temperature of the dust heated by OB associations in the {\bf `10\,kpc ring'}. The fact that the 24-to-70\,$\mu$m intensity ratio is not enhanced in the {\bf `10\,kpc ring'} (and in the central region) shows the invalidity of this ratio for temperature determination due to the important contribution from the very small grains. On the other hand, the enhancement of the 24-to-70\,$\mu$m in regions where there is no strong radiation field (between the arms) reveals possibly different origins of the 24\,$\mu$m and 70\,$\mu$m emission.
The stellar origin, e.g. photosphere of cool stars or dust shell of the evolved stars, may provide the enhancement of the 24-to-70\,$\mu$m intensity ratio in the inter-arm region. In M~33, \cite{Verley_09} attributed a similar enhancement of the diffuse 24\,$\mu$m emission to dusty circumstellar shells of unresolved, evolved AGB stars. For M~31, this needs to be quantified through a more detailed study and modelling of the spectral energy distribution, which is beyond the scope of this paper.
\begin{figure}
\begin{center}
\resizebox{7.5cm}{!}{\includegraphics*{IRtoIR.ps}}
\caption[]{Ratio of the MIPS IR intensities against galactocentric radius in M~31.}
\label{fig:ratio}
\end{center}
\end{figure}
\begin{figure}
\resizebox{7cm}{!}{\includegraphics*{m31.HI-to-dust.color.ps}}
\resizebox{7cm}{!}{\includegraphics*{m31.H2-to-dust.color.ps}}
\resizebox{7cm}{!}{\includegraphics*{m31.gas-to-dust.color.ps}}
\caption[]{Radial profiles of the gas-to-dust ratios in M~31, the northern half and the southern half. {\it Top}: N(HI)/$\tau_{{\rm H}\alpha}$, {\it middle}: N(2H$_2$)/$\tau_{{\rm H}\alpha}$ , {\it bottom}: N(HI+2H$_2$)/$\tau_{{\rm H}\alpha}$. In the middle panel, the northern profile is shifted by 300 units for clarity. The errors are smaller than 5\% everywhere only for N(2H$_2$)/$\tau_{{\rm H}\alpha}$ they increase from 10\% to 25\% at $R>12.3$\,kpc and in the inner arms for $R<4.5$\,kpc. }
\label{fig:gasdustratio}
\end{figure}
\subsection{Gas-to-dust ratio}
The gas-to-dust mass ratio and its variation across the galaxy can provide information about the metallicity distribution \citep[e.g. ][]{Viallefond_82} and hence about the evolutionary history of the galaxy.
The relative amount of dust and gas is expected to be correlated with the abundance of the heavy elements \citep{Draine_07}.
A number of authors has studied the gas-to-dust ratio in M~31 by
comparing HI column densities and optical or UV extinction \citep{Walterbos_88,
Bajaja_77,Xu_96,Nedialkov,Savcheva}. All authors found an increase of the atomic
gas-to-dust ratio with radius. \cite{Walterbos_87} derived the HI gas-to-dust ratio
using dust optical depth from IRAS 60\,$\mu$m and 100\,$\mu$m data. They found a
radial gradient that is 4-5 times larger than the abundance gradient
of \cite{Blair}. After adding the molecular and atomic gas
column densities, \cite{Nieten} obtained a strong radial
increase in the total gas-to-175\,$\mu$m intensity resulting from the
increase in the atomic gas-to-175\,$\mu$m intensity. As the dust
optical depth is a better measure for the dust column density than
the temperature-dependent dust emission, we re-investigated the
gas-to-dust ratio in M~31 taking advantage of the high resolution
of the Spitzer MIPS data.
We calculated the radial profiles of the three gas-to-dust ratios
from the mean column densities of N(HI), N(2H$_2$), N(HI+2H$_2$) and $\tau_{{\rm H}\alpha}$ in circular rings of 0.2\,kpc width in the plane of the galaxy. Figure \ref{fig:gasdustratio} (upper panel) shows that the atomic gas-to-dust ratio increases exponentially with
radius by more than a factor of 10 from about $0.6\times10^{21}$ at\,cm$^{-2}$ at
the center to about $6.5\times10^{21}$ at\,cm$^{-2}$ at $R$=\,15\,kpc. The increase is
surprisingly smooth and, at least up to $R$=\,13\,kpc, nearly the same for
the northern and southern half, indicating little variation between
arm and inter-arm regions and within the arms. In contrast, the
molecular gas-to-dust ratio (Fig.~\ref{fig:gasdustratio}, middle panel) does not increase systematically with radius but shows clear enhancements of a factor
2-3 in the spiral arms and the `10\,kpc ring'. The minima in the inter-arm
regions are due to a stronger decrease in N(2H$_2$) than in $\tau_{{\rm H}\alpha}$.
Figure~3 shows that along the arms N(2H$_2$)/$\tau_{{\rm H}\alpha}$ also varies significantly because maxima in H$_2$ emission and $\tau_{{\rm H}\alpha}$ are often not coincident.
The variations in N(2H$_2$)/$\tau_{{\rm H}\alpha}$ are visible in the profile of the
total gas-to-dust ratio (Fig.~\ref{fig:gasdustratio}, bottom panel) as weak enhancements
at the positions of the arms near $R\,=\,6$\,kpc and $R\,=\,8-12$\,kpc. As the atomic
gas is the dominant gas phase in M~31, dust mixed with HI gas largely
determines the optical depth. Inspection of the distribution of the
total gas-to-dust ratio across M~31 (not shown) reveals small-scale
variations along the arms of typically a factor of 2.
\begin{table*}
\begin{center}
\label{table4}
\caption{Exponential scale lengths $L$ and radial gradients of dust-to-gas ratios and the abundance [O/H] combined data from \cite{Blair} and \cite{Dennefeld}, where Ratio($R$)\,=\,C.\,exp$(-R/L)$. N(HI) and N(gas) are in 10$^{21}$\,at\,cm$^{-2}$. Errors are standard deviations. }
\begin{tabular}{ l l l l l }
\hline
Ratio & $R$ & \,\,\,\,\,\,\,\,\,\, \,C & $L$ & Gradient \\
& (kpc) & & (kpc) &(dex/kpc) \\
\hline
\hline
$\tau_{{\mathrm H}\alpha}$/N(HI) & 0-16 & 1.95\,$\pm$\,0.08 & 6.2\,$\pm$\,0.1 & 0.070\,$\pm$\,0.001 \\
& 5-15 & 1.97\,$\pm$\,0.08 & 6.1\,$\pm$\,0.2 & 0.071\,$\pm$\,0.001 \\
& & &\\
$\tau_{{\rm H}\alpha}$/N(gas) & 0-16 & 1.35\,$\pm$\,0.05 &7.3\,$\pm$\,0.2 & 0.059\,$\pm$\,0.002 \\
& 5-15 & 1.27\,$\pm$\,0.06 & 7.4\,$\pm$\,0.2 & 0.059\,$\pm$\,0.002 \\
& & &\\
$[{\rm O/H}]\times 10^{-4}$ & 5-15 & 2.95\,$\pm$\,0.35 & 9.7\,$\pm$\,2.6 & 0.045\,$\pm$\,0.012 \\
\hline
\end{tabular}
\end{center}
\end{table*}
We conclude that the radial increase in the total gas-to-dust ratio of
more than a factor 10 between the center and $R=\,15$\,kpc is entirely due to
that of the atomic gas-to-dust ratio, whereas the molecular gas-to-dust
ratio is only increased in the arms. This confirms the conclusion of
\cite{Nieten} based on the same gas data and the 175\,$\mu$m intensity.
At which radius in M~31 would the gas-to-dust ratio observed in the solar
neighborhood occur? \cite{Bohlin} and \cite{Diplas_94}
derived N(HI)/E(B-V)=$4.8 \times 10^{21}$ at\,cm$^{-2}$\,mag$^{-1}$ and $(4.9 \pm 0.3) \times 10^{21}$ at\,cm$^{-2}$\,mag$^{-1}$, respectively, using the extinction towards large samples
of stars to determine the color excess E(B-V). Since E(B-V)\,=\,A$_{\rm V}$/R$_{\rm V}$,
where the visual extinction A$_{\rm V}$\,=\,1.234\,$\tau_{{\rm H}\alpha}$\,mag
\citep[e.g. ][]{krugel} and the total/selective extinction R$_{\rm V}$=\,2.8\,$\pm$\,0.3 in M~31 \citep{Walterbos_88}, we have E(B-V)=\,0.44\,$\tau_{{\rm H}\alpha}$ mag$^{-1}$. Hence, a value of N(HI)/E(B-V)=\,$4.9 \times 10^{21}$ at\,cm$^{-2}$\,mag$^{-1}$ corresponds to N(HI)/$\tau_{{\rm H}\alpha}$\,=\,$2.2 \times 10^{21}$ at\,cm$^{-2}$, which occurs in M~31 near $R\,=\,$8.5\,kpc (Fig.~\ref{fig:gasdustratio}, top panel), just in the bright emission ring. The total gas-to-dust ratio near the sun of $5.8 \times 10^{21}$ at\,cm$^{-2}$\,mag$^{-1}$ \citep{Bohlin} corresponding to N(gas)/$\tau_{{\rm H}\alpha}$=\,$2.6 \times 10^{21}$ at\,cm$^{-2}$ occurs at nearly the same radius (Fig.~\ref{fig:gasdustratio}, bottom panel). Thus the gas-to-dust ratio near the sun is similar to that in the {\bf `10\,kpc ring'} in M~31, in agreement with earlier studies \citep{Genderen,Walterbos_87}.
In contrast to Fig.~\ref{fig:gasdustratio}, we present in Fig.~9
the radial profiles of the dust-to-gas ratios, here for the total area in M~31. The two lower curves closely follow exponentials with scale lengths of 6.1$\pm$\,0.2\,kpc and
7.4$\pm$\,0.2\,kpc for $\tau_{{\rm H}\alpha}$/N(HI) and $\tau_{{\rm H}\alpha}$/N(gas), respectively, between $R$=\,5\,kpc and $R$=\,15\,kpc (see Table 4). For nearly the same radial range ($R$=\,3-15\,kpc), \cite{Walterbos_87} derived a scale length
of $\tau_{100\mu{\rm m}}$/N(HI)$\simeq$\,4\,kpc from data near the major axis.
\cite{Walterbos_88} obtained a scale length of A$_{\rm B}$/N(HI)$\simeq$\,9\,kpc
for the inner and outer dust lanes, and the $\tau_{\rm V}$/N(HI) ratio of \cite{Xu_96}
for diffuse spiral arm regions also indicates a scale length of about 4\,kpc (all scale
lengths were scaled to D=\,780\,kpc). Since our scale lengths
are not restricted to specific areas, our results are more representative for the
mean dust-to-gas ratios in the disk of M~31.
As dust consists of heavy elements and both dust and heavy elements are found
in star formation regions, the radial variations in the dust-to-gas ratio and the
metal abundance are expected to be similar (e.g. \citep{Hirashita_99,Hirashita_02}. This has
indeed been observed in several nearby galaxies \citep{Issa}. In M~31 the
variation in the metallicity with radius is not well established. Measurements of the
element abundance strongly depend on the empirical method and calibration applied
\citep{Trundle}. \cite{Pagel} showed that the ([OII] +[OIII])/H$\beta$ ratio (the so
called 'R$_{23}$') is a good probe of oxygen abundance and radial trends of this ratio
have been studied in many nearby galaxies \citep[e.g. ][]{Pagel_81,Evans,Henry,Garnett}. \cite{Blair} and \cite{Dennefeld} derived R$_{23}$ for HII regions in M~31. We combined their results and derived a scale length of log[O/H] of 9.7$\pm$\,2.6\,kpc corresponding to a gradient of 0.045$\,\pm \,0.012$\,dex/kpc. Comparing four different calibrations, \cite{Trundle} derived gradients of 0.027 - 0.013\,dex/kpc using the 11 HII regions of \cite{Blair}, with Pagel's calibration giving 0.017\,$\pm$0.001\,dex/kpc. Since our value of the [O/H] gradient is based on 19 HII regions, we expect it to be more reliable than that of \cite{Trundle}.
Table~4 shows that the radial gradient in $\tau_{{\rm H}\alpha}$/N(gas) best matches the
metallicity gradient. In view of the large uncertainties, the gradients in the dust-to-gas
surface-density ratio and the oxygen abundance in M~31 may indeed be comparable. A much larger
sample of abundance measurements of HII regions is needed to verify this similarity. Our
result agrees with the approximately linear trend between gradients in dust-to-gas ratios
and [O/H] in nearby galaxies noted by \cite{Issa}.
\begin{figure}
\begin{center}
\resizebox{7.5cm}{!}{\includegraphics[angle=-90]{m31.R-dust-to-gas0-16kpc.ps}}
\caption[]{Dust-to-gas ratios as function of galactocentric radius
for M~31, calculated from the radial
profiles of $\tau_{{\rm H}\alpha}$, N(HI), N(2H$_2$) and N(HI+2H$_2$)=~N(gas). }
\end{center}
\label{fig:dusttogas}
\end{figure}
\section{Wavelet analysis of dust and gas emission}
To investigate the physical properties of different phases of the interstellar medium as a function of the size of emitting regions, wavelet transformation is an ideal tool.
We use the {\it Pet Hat} wavelet \citep[see ][]{Frick_etal_01,Tabatabaei_1_07} to decompose the emissions of IR, HI, H$2$, HI+2H$2$, and de-reddened H$\alpha$ into 10 spatial scales starting at 0.4\,kpc (about twice the resolution).
The central 2\,kpc was subtracted from all images before the wavelet transformation to prevent a strong influence of the nucleus on the results. As an example, we show the extinction-corrected H$\alpha$ map and the H$\alpha$ emission for 3 different scales in Fig.~\ref{fig:wave}. On the scale of 0.4\,kpc, the distribution of HII complexes and large HII regions is borne out. The scale of 1.6\,kpc (the typical width of spiral arms) shows connected HII complexes along the arms, and on the scale of 4\,kpc we see the extended emission from the '10\,kpc ring'.
\begin{figure*}
\begin{center}
\resizebox{13cm}{!}{\includegraphics*{Hacorr_a.eps}}
\resizebox{13cm}{!}{\includegraphics*{Hacorr1_b.ps}}
\resizebox{13cm}{!}{\includegraphics*{Hacorr4_c.ps}}
\resizebox{13cm}{!}{\includegraphics*{Hacorr6_d.ps}}
\caption[]{Distribution of the de-reddened H$\alpha$ emission ({\it a}) and the wavelet decomposition for scales 0.4, 1.6, 4.0\,kpc ({\it b} to {\it d}). The central 2\,kpc was subtracted from the H$\alpha$ map before the decomposition. The cross in the H$\alpha$ map indicates the location of the center. }
\label{fig:wave}
\end{center}
\end{figure*}
\subsection{Wavelet spectra}
The wavelet spectrum, {\it M(a)}, represents the distribution of the
emitting power as function of the scale $a$. The wavelet spectrum
will smoothly increase towards larger scales if most of the
emission is coming from diffuse structures forming the largest
scales, here up to 25\,kpc. On the other hand, the spectrum will decrease
with increasing scale if compact structures are the dominant source
of emission. The spectra of the IR and gas emission are shown in
Fig.~\ref{fig:wave1}.
All IR and gas spectra are intermediate between the two cases
described above. Only the spectra of the HI gas and the 160\,$\mu$m
emission generally increase with scale indicating the importance
of diffuse HI and cold dust emission. In addition, the HI spectrum
exhibits a dominant scale at $a\,\simeq\,4$\,kpc corresponding to the width of
the '10\,kpc ring', where strong diffuse emission occurs in interarm
regions. The large width of the HI `ring' is also visible in the
radial profiles in Fig.~\ref{fig:surfir}. The dominant scale of the emission from
warm dust, molecular gas and H$\alpha$ is near 1\,kpc, where
complexes of giant molecular clouds and star forming regions show up. The IR spectra at 24\,$\mu$m and 70\,$\mu$m on scales $a<$\,6\,kpc look most similar indicating that the star forming regions are the main heating sources at both wavelengths. On the other hand, the effect of the ISRF heating the cold dust is well indicated in the 160\,$\mu$m spectrum where a general increase towards larger scales is found. All spectra, apart from that of HI, show a minimum near $a$\,=\,6\,kpc corresponding to the large, weak interarm region inside the `10\,kpc ring'. The spectrum of H$\alpha$ is most similar to that of 70\,$\mu$m, which may explain why the H$\alpha$ emission correlates better with 70\,$\mu$m emission than with that at 24\,$\mu$m (see Sect. 6.2 and Table 6).
The spectrum of the H$\alpha$ emission is
flat on small scales up to 1.6\,kpc, the width of the spiral arms
in the H$\alpha$ map. This is understandable as the emission from
very compact HII regions is unresolved at our resolution and not many large
HII complexes exist especially in the south (see the decomposed map in
Fig.~\ref{fig:wave}b of $a$=\,0.4\,kpc).
\subsection{Wavelet cross-correlations}
We derive the cross-correlation coefficients, $r_w(a)$,
for different scales following \cite{Tabatabaei_1_07}. The correlation coefficients are plotted in terms of scale in Fig.~\ref{fig:gas-IR}. They show that IR emission correlates
with the emission from different gas phases on most scales.
In all cases,
emission from structures on scales larger than 10\,kpc are best
correlated. This corresponds to scales of the diameter of the
'10\,kpc ring' and the over-all structure of the galaxy. On medium
scales, the weakest correlation occurs between HI and dust emission
on $a=$\,6\,kpc. This scale includes areas of significant diffuse HI emission
where the dust emission is weak interior to the `10\,kpc ring' (compare also
Fig.~\ref{fig:wave1}).
On the smallest scale of 0.4\,kpc, the cold dust emission is best correlated with that of the total neutral gas, while the warm dust emission at 70\,$\mu$m is best correlated with the ionized gas emission.
Note that on this scale, the 24\,$\mu$m and 70\,$\mu$m (warm dust) emissions hardly correlate with HI ($r_{w(a)}<0.5$) because only a small fraction of the HI emission occurs on this scale (see Fig.~\ref{fig:wave1}). Furthermore, the coefficients of the 70\,$\mu$m--H$\alpha$ correlation are higher than those of the 70\,$\mu$m--neutral gas correlation on scales $a\,<\,6.3$\,kpc.
\begin{figure*}
\begin{center}
\resizebox{\hsize}{!}{\includegraphics*{spectIR.centsub.ps}
\includegraphics*{spectgas.centsub.ps}}
\caption[]{Wavelet spectra of MIPS IR ({\it left}) and gas ({\it right}) emission in M~31, shown in arbitrary units. The data points correspond to the scales 0.4, 0.6, 1.0, 1.6, 2.5, 4.0, 6.3, 10.0, 15.9, 25.1\,kpc. }
\label{fig:wave1}
\end{center}
\end{figure*}
\begin{figure*}
\begin{center}
\resizebox{\hsize}{!}{\includegraphics*{f_HI-IRcentsub.ps}
\includegraphics*{f_CO-IRcentsub.ps}}
\resizebox{\hsize}{!}{\includegraphics*{f_gas-IRcentsub.ps} \includegraphics*{f_Hacorrect-IRcentsub.ps}}
\caption[]{Wavelet cross correlations of atomic gas ({\it top-left}), molecular gas ({\it top-right}), and total neutral gas ({\it bottom-left}) with IR emission in M~31. The IR correlation with the ionized gas ({\it bottom-right}) is also shown. The data points correspond to the scales 0.4, 0.6, 1.0, 1.6, 2.5, 4.0, 6.3, 10.0, 15.9, and 25.1\,kpc. }
\label{fig:gas-IR}
\end{center}
\end{figure*}
\section{Classical correlations between dust and gas}
The wavelet cross-correlations for different scales in
Fig.~\ref{fig:gas-IR} show on which scales the distributions of the
various types of emission are significantly correlated. However, because the scale maps are normalized and information about absolute intensities is lost, they cannot be used to find quantitative relations between components of the ISM. Hence, to obtain numerical equations relating two distributions, we need classical correlations. Classical cross-correlations contain all scales that exist in
a distribution. For example, the high-intensity points of the H$\alpha$-70$\mu$m correlation in Fig.~\ref{fig:ha70} represent high-emission peaks on small scales in the spiral arms (compare
Fig.~\ref{fig:wave}b), whereas low-intensity points represent weak emission
around and between the arms on larger scales (compare Fig.~\ref{fig:wave}d).
The correlation coefficient of 79\% is a mean of all scales,
consistent with Fig.~\ref{fig:gas-IR}.
We made
pixel-to-pixel correlations between the distributions of
$\tau_{{\rm H}_{\alpha}}$ and H$_2$, HI, total gas as well as between
de-reddened H$\alpha$ and 24\,$\mu$m, 70\,$\mu$m, and 160\,$\mu$m. We
restricted the comparisons to radii where all data sets are complete,
$R<50\arcmin$ (or 11.4\,kpc), and
to intensities above 2\,$\times$\, rms noise. To reduce the influence
of the gradient in the gas-to-dust ratio (see Sect. 4.2),
we calculated correlations for two radial ranges: $0\arcmin<R<30\arcmin$
and $30\arcmin<R<50\arcmin$. We obtained sets of independent data points,
i.e. a beam area overlap of $<5\%$, by choosing pixels spaced
by more than 1.67$\times$ the beamwidth. Since the correlated variables
are not directly depending on each other, we fitted a power law
to the bisector in each case \citep{Isobe}.
We also calculated the correlation coefficient, $r_c$, to show how well two
components are correlated, and the student-t test to indicate the statistical
significance of the fit. For a number of independent points of n\,$>100$, the fit
is significant at the 3$\sigma$ level if $t>3$. Errors in intercept $a_c$
and slope $b$ of the bisector are standard deviations (1\,$\sigma$).
We first discuss the correlations between the neutral gas and dust extinction, scaled
from $\tau_{{\rm H}_{\alpha}}$. Then we
investigate the relationships between the emissions from dust
and ionized gas. The results are
given in Tables~\ref{table:dustgas1} and \ref{table:dustgas2}, and examples of correlation plots are shown in Figs.~\ref{fig:class1} to \ref{fig:sfr1}.
\begin{figure}
\begin{center}
\resizebox{8cm}{!}{\includegraphics[angle=-90]{m31loggas-ext+bisectorsbw.ps}}
\caption[]{Classical cross-correlations between gas column densities and dust extinction A$_{\rm H\alpha}$ in the radial ranges $0\arcmin<R<30\arcmin$ and $30\arcmin<R<50\arcmin$ (or 6.8~kpc~$<R<$~11.4~kpc). Only independent data points (separated by 1.67$\times$ beamwidth) with values above 2$\times$ rms noise were used. {\it Top}. As an example, the scatter plot between N(gas) and A$_{\rm H\alpha}$. {\it Bottom}. Power-law fits to the various correlations given in Table 5. Thick lines: $30\arcmin<R<50\arcmin$; thin lines: $0\arcmin<R<30\arcmin$. The shift between these intervals of the fits to the N(HI)-A$_{\rm H\alpha}$ and N(gas)-A$_{\rm H\alpha}$ correlations is due to the radial increase in the atomic gas-to-dust ratio.}
\label{fig:class1}
\end{center}
\end{figure}
\subsection{Correlation between neutral gas and dust extinction}
In search of a general relationship between neutral gas and dust
extinction, a number of authors employed scatter plots between
gas column densities and extinction, optical depth or FIR surface
brightness \citep[e.g. ][]{Savage_78,Boulanger,Walterbos_88,Xu_96,Neininger_98,Nieten}. They obtained nearly linear relationships between these quantities. As the studies
on M~31 have various shortcomings (lower limits for extinction, H$_2$
data not included and/or low angular resolution), we calculated
classical correlations between the distribution of dust extinction A$_{\rm H\alpha}$= 1.086\,$\tau_{\rm eff}$ and those of N(HI), N(2H$_2$) and N(gas) at our resolution of 45$\arcsec$.
As the correlations are restricted to gas column densities above 2$\times$
the rms noise, values of $\tau_{{\rm H}_{\alpha}} <$ 0.04 are not included (see
upper panel of Fig.~\ref{fig:class1}). The bisector fits given in Table 5 are
plotted in the bottom panel of Fig.~\ref{fig:class1}.
The relationships between A$_{\rm H\alpha}$ and N(2H$_2$) for the two radial
ranges are the same within errors, so the two areas can be combined.
With a correlation coefficient of $r_c \simeq$\,0.6, the correlation is not
very good, indicating that only a small part of the extinction is caused by dust in molecular clouds. This is not surprising
in view of the low molecular gas fraction in M~31 (see lower panels of
Fig.~\ref{fig:surfir}) and the small area filling factor of the molecular gas compared to that of the atomic gas.
\begin{table*}
\begin{center}
\caption{Power-law relations and correlation coefficients $r_c$ between dust extinction and gas components. Ordinary least-squares fits of bisector log(Y)=$a_c+b$\,log(X) through n pairs of (logX, logY), where n is the number of independent points (Isobe et al. 1990); $t$ is the student-t test.}
\begin{tabular}{ l l l l l l l l}
\hline
X & Y & $R$ & \,\,\,\,\,\, $a_c$ &\,\,\,\,\,\, $b$ & n & \,\,\,\,\,\, $r_c$ & $t$\\
($10^{18}$ at~cm$^{-2}$)& (mag) & ($\arcmin$) &&&&& \\
\hline
\hline
N(2H$_2$) & A$_{\rm H\alpha}$ & 0-30 & -1.88$\pm$0.06 & 0.53$\pm$0.03 & 207& 0.60$\pm$0.06 & 11\\
& & 30-50 & -1.83$\pm$0.04 & 0.52$\pm$0.02 & 610 & 0.64$\pm$0.03 & 20 \\
& & 0-50 & -1.85$\pm$0.03 & 0.52$\pm$0.01 & 817 & 0.63$\pm$0.03 & 23\\
& & & & & & &\\
N(HI) & A$_{\rm H\alpha}$ & 0-30 & -3.29$\pm$0.11 & 0.94$\pm$0.04 & 354 & 0.69$\pm$0.04& 18\\
& & 30-50 & -3.26$\pm$0.07 & 0.84$\pm$0.02 & 768 & 0.72$\pm$0.03 & 29\\
&&&&&&&\\
N(gas) & A$_{\rm H\alpha}$ & 0-30 & -2.97$\pm$0.08 & 0.79$\pm$0.03 & 350 & 0.80$\pm$0.03 & 25\\
& & 30-50 & -3.05$\pm$0.06 & 0.76$\pm$0.02 & 766 & 0.77$\pm$0.02 & 33\\
\hline
\end{tabular}
\label{table:dustgas1}
\end{center}
\end{table*}
The correlations between A$_{\rm H\alpha}$ and N(HI) are indeed better
($r_c \simeq\,0.7$) than those between A$_{\rm H\alpha}$ and N(2H$_2$), but the
relationships for the two radial intervals are not the same. Although
both are nearly linear (power-law exponent $b \simeq$\,0.9), their power laws are shifted (see Fig.~\ref{fig:class1}) in the sense that the values of A$_{\rm H\alpha}$ in $R=30\arcmin- 50\arcmin$ are about a factor of 2 lower than those inside $R=30\arcmin$.
This difference is caused by the radial decrease of $\tau_{{\rm H}_{\alpha}}$/N(HI)
discussed in Sect. 4.2. The variation of this ratio within each of the
radial intervals contributes to the spread in the scatter plots and
reduces the correlation coefficients.
The correlations between total gas N(gas) and A$_{\rm H\alpha}$ are best
($r_c\simeq\,0.8$), as A$_{\rm H\alpha}$ represents dust mixed with both HI and H$_2$. They are close to linear ($b \simeq$\,0.8) and differ by nearly
a factor 2 in A$_{\rm H\alpha}$. The scatter plots for the two intervals are shown in the
upper panel of Fig.~\ref{fig:class1}. In linear plots both power-law fits are going
through zero, suggesting that the dust causing the extinction and
neutral gas are mixed down to very low densities.
Interestingly, extinction (or dust opacity) is proportional to the square root of N(2H$_2$), while it is about linearly related to the atomic gas density. This is due to the quadratic dependence of N(2H$_2$) on N(HI) in M~31 observed by \citep[][]{Nieten}. This dependence is expected if in cool, dense, and dusty HI clouds the formation and destruction rates of H$_2$ are balanced \citep{Reach}.
\subsection{Correlation between ionized gas and dust}
Because the emission from ionized gas is a good tracer of the
present-day star formation rate and massive stars both heat the
dust and ionize the gas, a correlation between the emissions from
warm dust and ionized gas is expected. Relationships between the emission at 24\,$\mu$m
and Pa$\alpha$ or H$\alpha$ emission from HII regions in nearby galaxies as well as
relationships between global luminosities of galaxies have been reported
\citep[see ][ and references therein]{Kennicutt_09}.
For M~31, the correlation between the emission from dust and ionized gas was first
tested by \cite{Hoernes_etal_98}, who found a good, nearly linear
correlation between warm dust emission and free-free radio emission
for the radial range $30\arcmin <R<90\arcmin$, using HIRAS data and multi-wavelength
radio data. Here we correlate the extinction-corrected H$\alpha$
emission presented in Fig.~\ref{fig:wave}a with dust emission in the MIPS maps.
The wavelet correlations in Fig.~\ref{fig:gas-IR} (bottom-right) show
that in M~31 H$\alpha$ emission is best correlated with dust emission at
70\,$\mu$m. This suggests that of the MIPS bands, the 70\,$\mu$m emission could best be used as the tracer of present-day star formation, making a numerical
relation between the emissions at 70\,$\mu$m and H$\alpha$ of interest.
Table~\ref{table:dustgas2} gives the bisector fits for the two radial ranges, which are
very similar. Therefore, we present this correlation in Fig.~\ref{fig:ha70} for
the entire radial range of $0\arcmin <R<50\arcmin$. The power-law fit for this radial
interval is
\begin{eqnarray}
\nonumber
{\rm log}(I_{70})=(-0.98 \pm 0.02)+(1.09 \pm 0.02)\,{\rm log}(I_{{\rm H}\alpha}), \nonumber
\end{eqnarray}
where $I_{70}$ is in MJy/sr and $I_{{\rm H}\alpha}$ in 10$^{-7}$\,erg~s$^{-1}$~cm$^{-2}$~sr$^{-1}$.
The correlation is quite good ($r_c \simeq 0.8$) and nearly linear. In a linear plot the power-law
fit goes through zero suggesting that the correlation is also valid
for the lowest intensities. The good correlation indicates that the
heating sources that power dust emission at 70\,$\mu$m and ionize the gas
must indeed be largely the same.
Naturally, H$\alpha$ emission is less correlated with the emission from cold dust at 160\,$\mu$m than with emission from warm dust seen at the shorter wavelengths. This is
especially so at $R<30\arcmin$ where the radial profiles differ most (see Fig.~\ref{fig:surfir}). Moreover, the relation between the emission from cold dust and H$\alpha$ is non-linear (see the bisector slope $b$ in Table~\ref{table:dustgas2}).
Table~\ref{table:dustgas2} shows that the correlations with 24\,$\mu$m are slightly
worse than those with 70\,$\mu$m. In contrast, in M~33 the 24\,$\mu$m--H$\alpha$ correlation is better than the 70\,$\mu$m--H$\alpha$ correlation \citep{Tabatabaei_1_07}. This may suggest that in early-type galaxies like M~31 the contribution from evolved AGB stars to the 24\,$\mu$m
emission is larger than in late-type galaxies like M~33.
A significant stellar contribution to the 24\,$\mu$m emission from M~31 is also
indicated by the enhancement of the 24\,$\mu$m-to-70\,$\mu$m intensity ratio in inter-arm regions
where the radiation field is weak (Fig.~\ref{fig:ratio}).
Across M~31, the 24\,$\mu$m emission is linearly proportional to the extinction-corrected
H$\alpha$ emission ($b = 0.98 \pm 0.02$). A linear relationship was also found between the
luminosities at 24\,$\mu$m and extinction-corrected Pa$\alpha$ of HII regions in M~51 \citep{Calzetti_05} and between the luminosities at 24\,$\mu$m and extinction-corrected H$\alpha$
of HII regions in M~81 \citep{Perez}. Comparing the 24\,$\mu$m luminosities
and corrected H$\alpha$ luminosities of HII regions in 6 nearby galaxies (including
M~51 and M~81), \cite{Relano_07} obtained a somewhat steeper power law with index
$1.21 \pm 0.01$, in agreement with the index for global luminosities of galaxies \citep[see
also ][]{Calzetti_07}. Thus, while the L$_{24}$ - L$_{\rm H\alpha}$ relationship
is linear within a single galaxy, the relationships for HII regions in a sample of
galaxies and for global luminosities are non-linear. According to \cite{Kennicutt_09}, the steepening is due to variations between galaxies in the contribution from
evolved, non-ionizing stars to the heating of the dust that emits at 24\,$\mu$m.
\begin{figure}
\begin{center}
\resizebox{8cm}{!}{\includegraphics[angle=-90]{m31loghacorrnew-mu70r0-50.ps}}
\caption[]{Scatter plot between the surface brightnesses of ionized gas
and dust emission at 70\,$\mu$m for the radial range 0$\arcmin$-50$\arcmin$. Only independent data points (separated by 1.67$\times$ beamwidth) with values above 2$\times$ rms noise are included. The line shows the power-law fit given in Table 6,
which has an exponent close to 1. In a linear frame, this fit
goes through the zero point of the plot. }
\label{fig:ha70}
\end{center}
\end{figure}
\begin{table*}
\begin{center}
\caption{Power-law relations and correlation coefficients $r_c$ between the emission from dust and ionized gas. Ordinary least-squares fits of bisector log(Y)=$a_c+b$\,log(X) through n pairs of (logX, logY), where n is the number of independent points (Isobe et al. 1990); $t$ is the student-t test.}
\begin{tabular}{ l l l l l l l l}
\hline
X & Y & $R$ & \,\,\,\,\,\, $a_c$ & \,\,\,\,\,\,$b$ & n & \,\,\,\,\,\,$r_c$ & $t$\\
($10^{-7}$ erg\,s$^{-1}$\,cm$^{-2}$\,sr$^{-1}$) & (MJy/sr) & ($\arcmin$)& & & & &\\
\hline
\hline
I$_{{\rm H}\alpha}$ &I$_{24\mu{\rm m}}$ & 0-30 & -1.76$\pm$0.05 & 0.94$\pm$0.03 & 417 & 0.79$\pm$0.03 & 27\\
& & 30-50 & -1.75$\pm$0.05 & 1.00$\pm$0.03 & 677 & 0.75$\pm$0.03 & 30\\
& & 0-50 & -1.75$\pm$0.05 &0.98$\pm$0.02 & 1094 & 0.76$\pm$0.02 & 38\\
& & & & & & &\\
I$_{{\rm H}\alpha}$ & I$_{70\mu{\rm m}}$ & 0-30 & -1.04$\pm$0.03 & 1.10$\pm$0.03 & 412 & 0.83$\pm$0.03 & 30\\
& & 30-50 &-0.96$\pm$0.03 &1.09$\pm$0.03 & 677 & 0.78$\pm$0.02 & 32 \\
& & 0-50 &-0.98$\pm$0.02 &1.09$\pm$0.02 & 1089 & 0.79$\pm$0.02 & 43 \\
& & & & & & &\\
I$_{{\rm H}\alpha}$& I$_{160\mu{\rm m}}$ & 0-30 & 0.26$\pm$0.05 & 0.76$\pm$0.03 & 417 & 0.54$\pm$0.04 & 13\\
& & 30-50 &0.47$\pm$0.04 &0.72$\pm$0.02 & 678 & 0.73$\pm$0.03 & 28 \\
\hline
\end{tabular}
\label{table:dustgas2}
\end{center}
\end{table*}
\section{Star formation rate and efficiency}
Over the last 40 years many authors have studied the relationship
between the rate of star formation and gas density in M~31 by
comparing the number surface density of massive young stars or
of HII regions with that of HI \citep{Berkhuijsen_77,Tenjes,Unwin,Nakai_84}.
They found power-law exponents
near 2 as was also obtained for the solar neighborhood by \cite{Schmidt}, who first proposed this relationship with HI volume density. \cite{Kennicutt_98} showed that a similar relationship is expected between SFR and gas column densities. The early studies
suffered from the effects of dust absorption and could not consider
molecular gas \citep[apart from ][]{Tenjes}. As the necessary
data are now available, we again address this issue.
We compared the distribution of the H$\alpha$ emission corrected
for dust attenuation (see Fig.~\ref{fig:wave}a) with those of HI, H$_2$ and total
gas. The corrected H$\alpha$ emission is a good measure for the
present-day star formation rate (SFR) that we first estimate for
the total area observed using the relation of \citep{Kennicutt}:
\begin{equation}
{\rm SFR}({\rm M}_{\odot} {\rm yr}^{-1})=\, \frac{L_{\rm H_{\alpha}} ({\rm erg\,s^{-1}})}{1.26 \times 10^{41} } \ ,
\end{equation}
where $L_{\rm H_{\alpha}}$ is the ${\rm H_{\alpha}}$ luminosity. In an area of $110.0^{\arcmin} \times 38.5^{\arcmin}$ ($R<17$\,kpc), the luminosity of the de-reddened ${\rm H_{\alpha}}$ emission is $L_{\rm H_{\alpha}}\,=\,4.75 \times\,10^{40}$\,erg\,s$^{-1}$ or $L_{\rm H_{\alpha}}\,=\,1.7 \times 10^7\,L_{\odot}$ for the distance
to M~31 of 780\,kpc (see Table~1), giving SFR\,=\,0.38\,${\rm M}_{\odot} {\rm yr}^{-1}$.
However, this value is rather uncertain for two reasons. First, the contribution from
the inner disk ($R <\,25\arcmin$, nearly 6\,kpc) is overestimated because in this area the number
of ionizing stars is low and the gas must be mainly heated by other sources (see Sect. 7.1).
Second, our H$\alpha$ map is limited to about 55$\arcmin$ (12.5\,kpc) along the major axis, so
some of the emission between $R$\,=\,12.5\,kpc and $R$\,=\,17\,kpc is missing. Subtracting the
luminosity from the area $R <$\,6\,kpc gives a lower limit to the SFR of 0.27\,M$_{\odot}$\,yr$^{-1}$ for the radial range 6\,$< R <$\,17\,kpc. Earlier estimates of the recent SFR for a larger
part of the disk indicated 0.35 - 1\,M$_{\odot}$\,yr$^{-1}$ \citep{Walterbos_94,Williams,Barmby}. Recently, \cite{Kang} derived a SFR of 0.43\,M$_{\odot}$\,yr$^{-1}$
(for metallicity 2.5\,$\times$\,Solar) from UV observations of young star forming regions
($<$10\,Myr) within 120$\arcmin$ from the center ($R <$\,27\,kpc). Their Fig.~13 suggests that about 20\% of this SFR is coming from $R >$ 17\,kpc and a negligible amount from $R <$\,6\,kpc.
So for the range 6\,kpc\,$< R <$\,17\,kpc they find a SFR of about 0.34 M$_{\odot}$\,yr$^{-1}$,
which is consistent with our lower limit of 0.27\,M$_{\odot}$\,yr$^{-1}$.
A SFR of 0.3\,M$_{\odot}$\,yr$^{-1}$ yields a mean face-on surface density of
$\Sigma_{\rm SFR}$\,=\,0.4 M$_{\odot}$\,Gyr$^{-1}$\,pc$^{-2}$ between $R$\,=\,6\,kpc and $R$\,=\,17\,kpc. This is about 6 times lower than the value of $\Sigma_{\rm SFR}$=\,2.3\,M$_{\odot}$\,Gyr$^{-1}$\,pc$^{-2}$ that \cite{Verley_09} obtained for the disk of M~33 ($R <$\,7kpc), also using de-reddened H$\alpha$ data.
We can also calculate the star formation efficiency between $R$=\,6\,kpc and $R$=\,17\,kpc in M31. The total molecular gas mass in the entire area of $R<17$\,kpc in M~31 is
M(H$_2$)\,=\,3.6\,$\times\,10^8$\,M$_{\odot}$ \citep{Nieten} and that in the area
6\,kpc\,$<\,R\,<$17\,kpc is M(H$_2$)\,=\,2.9\,$\times\,10^8$\,M$_{\odot}$. Hence the
star formation efficiency SFE\,=\,SFR/M(H$_2$) between 6\,kpc and 17\,kpc radius is
SFE\,=\,0.9\,Gyr$^{-1}$. It is equivalent to a molecular depletion
time scale of 1.1\,Gyr. {\bf Hence, the disk of M~31 is about three times less
efficient in forming young massive stars than the northern part of the disk of M~33 \citep{Gardan}. }
\subsection{Star formation rate in the `10~kpc ring'}
The radial distributions of H$\alpha$ emission in the
bottom panels of Fig.~\ref{fig:surfir} show a steep decrease from the center
to $R \simeq$\,2\,kpc, followed by a shallower decrease to a minimum
near $R$\,=\,6\,kpc. In the radial profile of $\Sigma_{\rm SFR}$, shown in the
upper panel of Fig.~\ref{fig:sfr2} for the total area, the inner arms at
$R$\,=\,2.5\,kpc and $R$\,=\,5.5\,kpc are only visible as little wiggles
superimposed onto a high background. Clearly, the starforming
regions in these arms hardly contribute to the ionization of the
gas at $R <$\,6\,kpc. \cite{Devereux_etal_94b} noted that at these
radii the H$\alpha$ emission is filamentary and unlike that in
starforming regions, and since not many young, massive, ionizing
stars are found interior to the {\bf `10\,kpc ring'} \citep{Berkhuijsen_89,Tenjes,Kang} the gas must be ionized by other sources. Naturally, the same holds for the
heating of the warm dust, the emission of which also strongly increases
towards the center. In an extensive discussion, \cite{Devereux_etal_94b} concluded that a collision with another galaxy in the past
may explain the ionization of the gas and the heating of the dust
as well as several other peculiarities (e.g. the double nucleus)
in the inner disk of M~31 \citep[see also ][]{Block}. We note that the UV emission may also
be influenced by this event, because it shows a similarly steep
increase towards the center as the H$\alpha$ emission. Furthermore,
all radial profiles that increase towards the center are anti-
correlated with the radial profiles of HI and total gas (see Fig.~\ref{fig:surfir},
bottom panels). This leads to apparent deviations in the Kennicutt-Schmidt
law in the inner disk if $\Sigma_{\rm SFR}$ is calculated from the usual
star formation tracers \citep[see ][]{Boissier,Yin}.
If massive stars are not responsible for the ionization of the
gas and the heating of the dust, we can neither use the H$\alpha$
emission nor the infrared emission as tracers of present-day
star formation at $R <$\,6\,kpc in M~31, as was also pointed out by
\cite{Devereux_etal_94b}. Therefore, we investigated the relationship
between SFR and neutral gas only for the interval $30\arcmin <R< 50\arcmin$
($R$=\,6.8-11.4\,kpc) containing the '10\,kpc ring'.
The correlation plots in Fig.~\ref{fig:sfr1} and the results in Table~\ref{table:sfr}
show that $\Sigma_{\rm SFR}$ is not well correlated with the surface densities of
either H$_2$, HI or total gas ($r_c \sim$ 0.45-0.59). In spite of this, the fitted
bisectors are statistically significant ($t>3$). Interestingly, we find a linear
relationship between $\Sigma_{\rm SFR}$ and $\Sigma_{\rm 2H_2}$
(exponent $b=0.96\pm$0.03), which closely agrees with the average
relationship for 7 nearby galaxies, much brighter than M~31 (see
Fig.~\ref{fig:sfr1}a), analyzed by \cite{Bigiel_08}. While in these galaxies
molecular hydrogen is the dominant gas phase, most of the neutral
gas in M~31 is atomic (compare Fig.~\ref{fig:sfr1}a,b). Hence, the surface density
of SFR is linearly related to that of molecular gas, irrespective of
the fraction of molecular gas or the absolute value of the total
gas surface density in a galaxy. \cite{Bigiel_08} arrived at the
same conclusion after comparing the galaxies in their sample.
The correlation between $\Sigma_{\rm SFR}$ and total gas surface density is slightly better than that between $\Sigma_{\rm SFR}$ and molecular gas surface density. The bisector fit in Table~7 yields the Kennicutt-Schmidt-law
\begin{equation}
\Sigma_{\rm SFR} = (0.076 \pm 0.005)\,\Sigma_{\rm GAS}^{1.30 \pm 0.05},
\end{equation}
where $\Sigma_{\rm GAS}$ and $\Sigma_{\rm SFR}$ are in M$_{\odot}$\,pc$^{-2}$ and M$_{\odot}$\,Gyr$^{-1}$\,pc$^{-2}$, respectively.
The exponent of $1.30 \pm 0.05$ is well in the range of
1.1-2.7 derived by \cite{Bigiel_08}. As a galaxy of low surface brightness,
the SFRs in M~31 are correspondingly low. Our $\Sigma_{\rm SFR}$--$\Sigma_{\rm GAS}$
relationship nicely fits on the low-brightness extension of the compilation
of available galaxy data in Fig.~15 of \cite{Bigiel_08}, formed by the
outer parts of their 7 galaxies and the global values for 20 galaxies of low
surface brightness.
\begin{figure*}
\begin{center}
\resizebox{\hsize}{!}{\includegraphics[angle=-90]{m31.logSFR-h2h1gasscr30+.ps}}
\caption[]{Scatter plots between the surface density of the star formation rate and neutral gas surface densities for the radial interval $30\arcmin <R<50\arcmin$ (6.8~kpc~$<R<$~11.4~kpc). All surface densities are face-on values. Only independent data points (separated by 1.67$\times$ beamwidth) above 2$\times$ rms noise were used. {\bf a}. SFR versus molecular gas; {\bf b}. SFR versus atomic gas; {\bf c}. SFR versus total gas. Full lines indicate the power-law fits given in Table 6. The relationship in a. is linear and nearly the same as the average relationship for 7 bright galaxies (dashed line) derived by \cite{Bigiel_08}. Note that the cut off in $\Sigma_{\rm HI}$ and $\Sigma_{\rm GAS}$ is near 10\,M$_{\odot}$~pc$^{-2}$. }
\label{fig:sfr1}
\end{center}
\end{figure*}
Very recently, \cite{Braun_09} also studied the dependence of SFR
on gas density in M~31 using the new Westerbork HI survey and the
CO survey of \cite{Nieten}. They estimated the SFR from the
surface brightnesses at IRAC 8\,$\mu$m, MIPS 24\,$\mu$m and GALEX FUV following
the procedure of \cite{Thilker_05}. Our Fig.~\ref{fig:sfr1}a is comparable
to the radial range 8-16\,kpc in their Fig.~20D that shows the same range
in $\Sigma_{\rm SFR}$ as we find. Note that the molecular
gas densities of \cite{Braun_09} are a factor of 1.6 larger (+0.21 dex)
and have a larger dynamic range than our values due to differences
in scaling of the CO data, inclination, angular resolution and radial
range. Scaling our relationship to the assumptions of \cite{Braun_09}
gives ${\rm log}(\Sigma_{\rm SFR})=\, -0.44\, +\, 0.96\,{\rm log}(\Sigma_{\rm 2H_2})$, which is in good agreement with their Fig.~20D.
The dependencies of SFR surface density on total gas surface density in
Fig.~\ref{fig:sfr1}c and in Fig.~20E of \cite{Braun_09} have the same pear-like
shape characterized by a broadening towards lower $\Sigma_{\rm SFR}$ and a rather sharp
cut-off near $\Sigma_{\rm GAS}$\,=\,10\,M$_{\odot}$\,pc$^{-2}$. The cut-off comes from the
$\Sigma_{\rm SFR}$-$\Sigma_{\rm HI}$ relation (see Fig.~15b) and occurs at the same value as in the bright galaxies analyzed by \cite{Bigiel_08}, who interpreted the lack of higher surface mass densities as a saturation effect. \cite{Braun_09} show that in M~31 this truncation indeed
vanishes after correcting the HI data for opacity, which could lead to somewhat
steeper slopes in Figs.~15b and c.
\begin{table*}
\begin{center}
\caption{Kennicutt-Schmidt law in M~31 for $30\arcmin<R<50\arcmin$. Ordinary-least squares fits of the bisector log($\Sigma_{\rm SFR}$)=$a_c+b$\,log($\Sigma$), where $\Sigma_{\rm SFR}$ is the face-on surface density of the star formation rate in M$_{\odot}$\,Gyr$^{-1}$\,pc$^{-2}$ and $\Sigma$ the face-on value of the gas surface density in M$_{\odot}$\,pc$^{-2}$; n is the number of independent points; $r_c$ is the correlation coefficient and $t$ the student-t test.}
\begin{tabular}{ l l l l l l l}
\hline
$\Sigma$(M$_{\odot}$~pc$^{-2}$) & \,\,\,\,\,\,\,\, $a_c$ &\,\,\,\,\,\, $b$ & n & \,\,\,\,\,\, $r_c$& $t$\\
\hline
\hline
$\Sigma_{\rm 2H_{2}}$ & -0.15$\pm$0.01 & 0.96$\pm$0.03 & 537 & 0.45$\pm$0.04& 12 \\
$\Sigma_{\rm HI}$ & -1.08$\pm$0.03 & 1.40$\pm$0.05 & 670 & 0.55$\pm$0.03& 17 \\
$\Sigma_{\rm GAS}$ & -1.12$\pm$0.03 & 1.30$\pm$0.05 & 668 & 0.59$\pm$0.03& 19 \\
\hline
\end{tabular}
\label{table:sfr}
\end{center}
\end{table*}
\subsection{Radial variations in the Kennicutt-Schmidt law}
In Fig.~\ref{fig:KS}a, we plot the mean values in {\bf 0.5\,kpc-wide} rings
in the plane of M~31 of $\Sigma_{\rm SFR}$ against those of $\Sigma_{\rm GAS}$ {\bf from $R$\,=\,6\,kpc to $R$\,=\,16\,kpc}. The points form a big loop with a horizontal
branch for $R$\,=\,6 - 8.5\,kpc and a maximum $\Sigma_{\rm SFR}$ in the ring $R$\,=\,10.5 - 11.0\,kpc (see also Fig.~\ref{fig:sfr2}). This behavior was already noted by \cite{Berkhuijsen_77}
and \cite{Tenjes}, who used the number density of HII regions as
tracer of SFR and HI gas, and was recently confirmed by \cite{Boissier}
from GALEX UV data and total gas. Both \cite{Berkhuijsen_77} and \cite{Tenjes} showed that the differences between the slopes inside and outside
the maximum of the starforming ring is greatly reduced when the increase
in the scale height of the gas with increasing radius is taken into account. We calculated
the scale height, h, from the scale height of the HI gas given by Eq.~13 of
\cite{Braun_91}, scaled to D\,=\,780\,kpc, assumed half this value for that of the
H$_2$ gas, and a constant scale height for the ionizing stars. Fig.~\ref{fig:KS}b shows
$\Sigma_{\rm SFR}$ as a function of gas volume density $n_{\rm GAS}$\,= N(HI)/2h + N(2H$_2$)/h.
The points have moved towards each other, but the horizontal branch remained
and the behavior on the starforming ring has become more complicated.
The variations in slope in Fig.~\ref{fig:KS} are clear evidence for radial
variations in the index of the Kennicutt-Schmidt law. Such variations are
not specific to M~31 as they are also seen in some of the galaxies
analyzed by \cite{Bigiel_08}. In order to quantify the variations,
we determined the bisectors in scatter plots for three circular rings:
$R$\,=\,7 - 9\,kpc, $R$\,=\,9 - 11\,kpc and $R$\,=\,11 - 13\,kpc, covering the horizontal
branch, the increasing part inside the maximum and the decreasing part
outside the maximum, respectively. The results are given in Table~8.
The index for the star fromation law for surface densities is unity
for the 7-9\,kpc ring and about 1.6 for the other two rings. Hence, the slope of $b$\,=\,1.30\,$\pm$\,0.05 obtained for the {\bf `10\,kpc ring'} ($R$\,=\,6.8 - 11.4\,kpc) in Sect.~7.1 represents the mean value of the first two rings considered here.
The scatter plots between $\Sigma_{\rm SFR}$ and gas volume density yield bisector
slopes that are about 0.2 smaller than those for surface density. The
correlation coefficients are all close to $r_c = 0.59 \pm 0.03$ for the
{\bf `10\,kpc ring'} (see Table~7), indicating that even in 2\,kpc-wide rings
the intrinsic scatter is considerable. This implies that on scales
of a few hundred parsec significant variations in the index of the
Kennicutt-Schmidt law and in the star formation efficiency occur.
\begin{figure*}
\begin{center}
\resizebox{13cm}{!}{\includegraphics[angle=-90]{newfig16.ps}}
\caption[]{ Mean face-on values of $\Sigma_{\rm SFR}$, averaged in 0.5\,kpc-wide
circular rings in the plane of M~31, plotted against the corresponding
mean values of {\it (a)} gas surface density $\Sigma_{\rm GAS}$, and {\it (b)} gas volume
density $n_{\rm GAS}$. The upper left point is for the ring $R$\,=\,6.0-6.5\,kpc, the
maximum in $\Sigma_{\rm SFR}$ is in ring 10.5-11.0\,kpc and the minimum in ring
$R$\,=\,14.5-15.0\,kpc. Typical errors are 0.01\,M$_{\odot}$\,Gyr$^{-1}$\,pc$^{-2}$ in $\Sigma_{\rm SFR}$, 0.02\,M$_{\odot}$\,pc$^{-2}$ in $\Sigma_{\rm GAS}$, and 3\,$\times$\,10$^{-5}$\,M$_{\odot}$\,pc$^{-3}$ in $n_{\rm GAS}$. Points for $R>$\,12kpc suffer from
missing data points near the major axis, the number of which increases
with radius. }
\label{fig:KS}
\end{center}
\end{figure*}
\begin{table*}
\begin{center}
\caption{Kennicutt-Schmidt law in three radial intervals in M~31.
Ordinary least-squares fits of the bisector log($\Sigma_{\rm SFR}$)=$a_c\,+\,b$\,log(X),
where $\Sigma_{\rm SFR}$ is the face-on surface density of the star formation rate
in M$_{\odot}$\,Gyr$^{-1}$\,pc$^{-2}$, X\,=\,$\Sigma_{\rm GAS}$ is the face-on gas surface density in M$_{\odot}$\,pc$^{-2}$ and X\,=\,$n_{\rm GAS}$ is the gas volume density in at cm$^{-3}$.
n is the number of independent points; $r_c$ is the correlation coefficient
and t the student-t test. }
\begin{tabular}{ l l l l l l l }
\hline
\,\,\,$R$(kpc) & \,\,\ X &\,\,\,\,\,\, $a_c$ & \,\,\,\,\,\, $b$ & n & \,\,\,\,\,\, $r_c$& $t$\\
\hline
\hline
7-9 & $\Sigma_{\rm GAS}$ & -0.90$\pm$0.04 &1.03$\pm$0.06 & 216 & 0.54$\pm$0.06 & 9 \\
9-11 & & -1.43$\pm$0.06 & 1.67$\pm$0.08 & 356 & 0.63$\pm$0.04 & 15 \\
11-13 & & -1.46$\pm$0.06 & 1.55$\pm$0.08 & 297 & 0.62$\pm$0.05& 13 \\
&&&& & &\\
7-9 & $n_{\rm GAS}$ & 0.18$\pm$0.04 & 0.88$\pm$ 0.06 & 218 & 0.51$\pm$0.06 & 9\\
9-11 & & 0.45$\pm$0.04 & 1.50$\pm$0.07 & 356 & 0.62$\pm$0.04 & 15\\
11-13 & & 0.35$\pm$0.04 & 1.35$\pm$0.07 & 297 & 0.62$\pm$0.05 & 14\\
\hline
\end{tabular}
\label{table:KS}
\end{center}
\end{table*}
\subsection{Radial variations of SFR and SFE}
In Fig.~\ref{fig:sfr2} (upper panel) we present the radial profile of the SFR
surface density between 6\,kpc and 17\,kpc, averaged in 0.5\,kpc-wide
circular rings in the plane of M~31. The face-on values vary between about
0.1 and 1\,M$_{\odot}$\, Gyr$^{-1}$\, pc$^{-2}$. \cite{Boissier} and \cite{Braun_09}
obtained similar values for $\Sigma_{\rm SFR}$ in this radial range from GALEX UV data.
They are about 10 times smaller than the~surface densities of SFR between R\,=~1.5~kpc and R\,=~7\,kpc in the northern part of M~33 observed by \citep{Gardan}.
In the lower panel of Fig.~\ref{fig:sfr2} we show the radial profiles of the
surface density of the molecular gas and of the star formation
efficiencies SFE\,=\,$\Sigma_{\rm SFR}$/$\Sigma_{\rm 2H_2}$ and $\Sigma_{\rm SFR}$/$\Sigma_{\rm GAS}$. Although the maximum $\Sigma_{\rm SFR}$ occurs on a relative maximum in the molecular gas density (in ring 10.5-11.0\,kpc), $\Sigma_{\rm SFR}$ is only about 70\% of its maximum value where the molecular gas density is highest (in ring 9.0-9.5\,kpc).
Consequently, SFE varies significantly with radius. Between $R=6$\,kpc and
$R=15$\,kpc SFE fluctuates around a value of 0.9\,Gyr$^{-1}$, with a minimum
of 0.46\,$\pm$\,0.01\,Gyr$^{-1}$ near $R$=\,9\,kpc. Thus SFE is smallest where $\Sigma_{\rm 2H_2}$
is highest! Up to $R$=12\,kpc the efficiency $\Sigma_{\rm SFR}$/$\Sigma_{\rm GAS}$ shows the
same trend as SFE. The increase in SFE between 12~kpc and 15~kpc radius of
a factor 1.5 results from the difference in radial scale lengths of $\Sigma_{\rm SFR}$
(or H$\alpha$ emission) and the molecular gas density (see Fig.~\ref{fig:surfir} and Table~3).
Interestingly, in M~33 \citep{Gardan} found a radial increase
in SFE of a factor 2 between 2~kpc and 6~kpc radius with similar
fluctuations around the mean as we observe in M~31, but the mean value
in M~31 is about three times lower than in M~33. Furthermore, \cite{Leroy} found significant variations in the efficiency $\Sigma_{\rm SFR}/\Sigma_{\rm GAS}$ on a linear scale of 800\,pc in the sample of 12 spiral galaxies analyzed by them.
That large, small-scale variations in SFE exist in galaxies is also
clear from the large spread in the scatter plots of $\Sigma_{\rm SFR}$--$\Sigma_{\rm 2H_2}$
visible in Fig.~\ref{fig:sfr1}a and in several figures of \cite{Bigiel_08}. The
same value of $\Sigma_{\rm SFR}$ can occur in a range of $\Sigma_{\rm 2H_2}$ spanning more
than a factor of 10.
We may conclude that neither the present-day star formation rate
$\Sigma_{\rm SFR}$ nor the star formation efficiency SFE is well correlated
with the molecular gas surface density. Hence, other factors than molecular
gas density must play an important role in the star formation process.
\cite{Bigiel_08} argue that local environmental circumstances
largely determine the SFE in spiral galaxies. These factors are
extensively discussed by e.g. \cite{Leroy}.
\begin{figure}
\begin{center}
\resizebox{8cm}{!}{\includegraphics[angle=-90]{newfig17.ps}}
\caption[]{Radial variation of the face-on surface density of the star formation rate
$\Sigma_{\rm SFR}$ and star formation efficiency SFE in M~31, averaged in 0.5\,kpc-wide
rings in the plane of the galaxy. Beyond $R=$\,12\,kpc the data are not complete,
because the observed area is limited along the major axis (see Fig.~10a). {\it Top}:
Radial profile of $\Sigma_{\rm SFR}$. {\it Bottom}: {\it full line} - SFE\,=\,$\Sigma_{\rm SFR}$/$\Sigma_{\rm 2H_2}$; {\it long dashed line} - $\Sigma_{\rm SFR}$/$\Sigma_{\rm GAS}$;
{\it dots} - radial profile of the molecular gas surface density $\Sigma_{\rm 2H_2}$ seen face-on.
Note that $\Sigma_{\rm SFR}$ and $\Sigma_{\rm 2H_2}$ do not peak at the same radius.
Statistical errors in $\Sigma_{\rm SFR}$ and $\Sigma_{\rm 2H_2}$ are smaller than the symbols and those in SFE and $\Sigma_{\rm SFR}$/$\Sigma_{\rm GAS}$ are smaller than the thickness of the lines. Only beyond $R$\,=\,12\,kpc the error in SFE slowly increases to $0.3\,{\rm Gyr}^{-1}$ at $R$\,=\,15\,kpc. }
\label{fig:sfr2}
\end{center}
\end{figure}
\section{Summary}
In this paper, we studied the emission from dust,
neutral gas and ionized gas in the disk of M~31, and the relationships
between these components on various linear scales.
We compared the Spitzer MIPS maps at 24\,$\mu$m, 70\,$\mu$m and 160\,$\mu$m \citep{Gordon_06} to the distributions of atomic gas seen in the HI line \citep{Brinks},
molecular gas as traced by the $^{12}$CO(1-0) line \citep{Nieten} and
ionized gas observed in H$\alpha$ \citep{Devereux_etal_94b}. All data were smoothed to an angular resolution of 45$\arcsec$ corresponding to 170\,pc\,$\times$\,660\,pc in the plane of the galaxy.
For each of the dust and gas maps, we calculated the mean intensity
distribution as a function of radius (Fig.~\ref{fig:surfir}), separately for the
northern and the southern half of M~31. Using wavelet analysis, we decomposed the dust and gas distributions in spatial scales and calculated cross-correlations as a function of scale. We also used classical correlations to derive quantitative relations between the various dust and gas components.
Using the MIPS 70\,$\mu$m and 160\,$\mu$m maps, we derived the distributions of the dust
temperature and optical depth. The dust optical depth at the H$\alpha$ wavelength was used
to a) investigate the dust-to-gas ratio, b) derive scaling relations between extinction and
neutral gas emission, and c) de-redden the H$\alpha$ emission in order to estimate the
recent star formation rate. We also presented the Kennicutt-Schmidt law indices obtained for the bright emission ring {\bf near $R$=\,10\,kpc} in M~31.
We summarize the main results and conclusions as follows.\\
\\
1. Dust temperature and opacity:\\
$\bullet$ The dust temperature steeply drops from about 30\,K in the center to about
19\,K near $R=$\,4.5\,kpc, and stays between about 17~K and 20~K beyond this
radius (Fig.~2).
The mean dust temperature in the area studied is about 18.5~K. This is
3~K less than the temperature obtained by \cite{Walterbos_87}
between the IRAS maps at 60\,$\mu$m and 100\,$\mu$m that both trace warmer dust
than the MIPS maps at 70\,$\mu$m and 160\,$\mu$m used here.\\
$\bullet$ The dust optical depth at H$\alpha$ along the line of sight varies in
a range between about 0.2 near the center and about 1 in the `10~kpc ring' (Fig.~4) with a mean value of 0.7$\pm$0.4 (the error is standard deviation) and a most probable value of $\simeq$\,0.5, indicating that M~31 is mostly optically thin to the H$\alpha$ emission. The total flux density of the H$\alpha$ emission increases by 30\% after correction for extinction.\\
\\
2. Radial distributions:\\
$\bullet$
The radial scale lengths between the maximum in the `10~kpc ring' and
$R=$~15~kpc of the warm dust are smaller than that of the cold dust, as is
expected if the warm dust is mainly heated by UV photons from star
forming regions and cold dust by the ISRF. With the largest
scale length, atomic gas has the largest radial extent of the dust and gas
components considered here.\\
$\bullet$ The radial gradient of the total gas-to-dust ratio is consistent
with that of the oxygen abundance in M~31. The gas-to-dust
ratios observed in the solar neighborhood \citep{Bohlin} occur
near $R=$~8.5~kpc in the disk of M~31 where N(gas)/$\tau_{{\rm H}\alpha}$=
$2.6 \times 10^{21}$ at\,cm$^{-2}$.\\
\\
3. Properties as a function of scale:\\
$\bullet$
Spatial scales larger than about 8~kpc contain most of the emitted power from
the cold dust and the atomic gas, whereas the emissions from warm dust,
molecular gas and ionized gas are dominated by scales near 1~kpc,
typical for complexes of star forming regions and molecular clouds in spiral arms (Fig.~11). \\
%
$\bullet$ Dust emission is correlated ($r_w \ge 0.6$) with both neutral and ionized gas on scales~$>1$\,kpc. \\
$\bullet$ On scales\,$<1$\,kpc, ionized gas is best correlated with
warm dust and neutral gas (both HI and H$_2$) with cold dust.
On the smallest scale of 0.4\,kpc, an HI--warm dust correlation
hardly exists ($r_w \simeq 0.4$) because not much HI occurs on the
scale of star forming regions (see Fig.~11).\\
\\
4. Relationships between gas and dust:\\
$\bullet$
H$\alpha$ emission is slightly better correlated with the
emission at 70\,$\mu$m than at 24\,$\mu$m (Fig.~13, Table 6), especially
on scales~$<$\,2 kpc (Fig.~12). As in M~33 the 24\,$\mu$m--H$\alpha$ correlation
is best, this suggests that in early-type galaxies like M~31 the contribution from
evolved AGB stars to the 24\,$\mu$m emission is larger than in late-type galaxies like M~33.\\
$\bullet$ Dust extinction A$_{H\alpha}$ is not well correlated with N(2H$_2$)
indicating that dust mixed with molecular clouds does not
contribute much to the total extinction. Although the
correlation with N(HI) is better, A$_{H\alpha}$ is best correlated
with N(HI+2H$_2$).\\
$\bullet$ Dust opacity is proportional to the square root
of N(2H$_2$) but about linearly related to N(HI), as was also
found by \cite{Nieten} at 90$\arcsec$ resolution. This is an
indirect indication of a balance between the formation and
destruction rates of H$_2$ in cool, dusty HI clouds.\\
$\bullet$ In the central 2 kpc both the dust opacity and the HI
column density are very low and the dust temperature is high.
This combination may explain the lack of H$_2$ in this region.\\
\\
5. SFR and SFE:\\
$\bullet$
The SFR in M~31 is low. The total SFR in the observed field between $R=$~6~kpc and $R=$~17~kpc
is $0.27\,{\rm M}_{\odot} {\rm yr}^{-1}$ and the star formation efficiency is 0.9\,Gyr$^{-1}$,
yielding a molecular depletion time scale of 1.1~Gyr. This is about three
times longer than observed in the northern part of M~33 (Gardan et al. 2007).
The radial distribution of $\Sigma_{\rm SFR}$ in 0.5\,kpc-wide rings in the plane of
the galaxy (Fig.~17) varies between about 0.1 and 1\,M$_{\odot}$\, Gyr$^{-1}$\, pc$^{-2}$, values that are about 10 times smaller than in the northern part of M~33 \citep{Gardan}. Between
$R=$~6~kpc and $R=$~15~kpc, SFE varies between about 0.5\,Gyr$^{-1}$ and 1.5\,Gyr$^{-1}$,
whereas the efficiency with respect to the total gas surface density
slowly decreases from about 0.18\,Gyr$^{-1}$ to about 0.03\,Gyr$^{-1}$. \\
$\bullet$ SFR is not well correlated with neutral gas and worst of all with molecular gas in the radial range $30\arcmin-50\arcmin$ containing the `10\,kpc ring' (Fig.~15, Table 7). In spite of this, the power-law fits are statistically significant. We find a linear
relationship between the surface densities of SFR and molecular gas
(power-law exponent 0.96~$\pm$~0.03), and a power law with index 1.30~$\pm$~0.05 between the surface densities of SFR and total gas. These results agree with the average relationship
for 7 nearby galaxies much brighter than M~31 \citep{Bigiel_08}. While
in these galaxies molecular hydrogen is the dominant gas phase, most of
the neutral gas in M~31 is atomic. Thus, the surface density of SFR depends
linearly on that of molecular gas irrespective of the fraction of molecular
gas or the absolute value of the total gas surface density in a galaxy.
Some important implications of this study are:
\begin{itemize}
\item[-] Precaution is required in using the total IR luminosity (TIR) as an indicator
of recent SFR or to derive dust opacity for an early-type galaxy like M~31,
because the cold dust is mainly heated by the ISRF and the warm dust emission
at 24\,$\mu$m is partly due to evolved stars (especially in the bulge of the galaxy).
\item[-] Neither the present-day SFR nor SFE is well correlated
with the surface density of molecular gas or total gas. Therefore, other factors than
gas density must play an important role in the process of star
formation in M~31.
\end{itemize}
\begin{acknowledgements}
We are grateful to E. Kr\"ugel for valuable and stimulating comments. We thank K.M. Menten and R. Beck for comments and careful reading of the manuscript. The Spitzer MIPS data were kindly provided by Karl D. Gordon. E.Tempel kindly sent us a table of extinction values that we used
for Fig.~5. We thank an anonymous referee for extensive comments leading to improvements
in the manuscript. FT was supported through a stipend from the Max Planck Institute for Radio Astronomy (MPIfR).
\end{acknowledgements}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 6,435 |
\section{Introduction}
The production of charm quarks is expected to be well described by
perturbative Quantum Chromodynamics (pQCD) due to the presence of a
hard scale provided by the charm mass. The evolution of an ``off-shell''
charm quark via gluon radiation until it is ``on-shell'' can be calculated
in pQCD in fixed order of the strong coupling or by summing all orders in
the leading-log approxi\-mation. The transition of an on-shell charm quark
into a charmed hadron is, however, not calculable within the framework of pQCD and is thus usually described by phenomenological models. One of the
major characteristics of this transition process is the longitudinal
momentum fraction transferred from the quark to the hadron, the
distribution of which is parametrised by a fragmentation function.
\par
Several phenomenological models are available to describe the
transition of a quark into hadrons, for example the independent
fragmentation~\cite{feynman-field}, the string~\cite{string}, and the
cluster model~\cite{cluster_model}.
The fragmentation function is unambiguously defined
in a given context of a phenomenological model together with a pQCD
calculation. Universality is then only expected to hold within this context.
\par
The fragmentation function is not a directly measurable quantity as the momentum of the heavy quark is experimentally not accessible. Also the
momentum distribution of the heavy hadron can only be measured within
a restricted phase space. The momentum spectrum is further affected by
the fact that some heavy hadrons are not produced directly, but are the
result of decays of still heavier excited states, whose contribution is
not well known.
\par
The production of charmed hadrons has been measured
and parameters of fragmentation functions have been determined in $e^+e^-$ annihilation experiments~\cite{epem-articles}.
The H1 and ZEUS collab\-orations have published total cross sections for
the production of various charmed hadrons in deep-inelastic $ep$ scattering (DIS)~\cite{h1-fragfrac} and in
photoproduction~\cite{zeus-fragfrac}.
These data show that the probabilities of charm quarks to fragment
into various final state hadrons are consistent, within experimental
uncertainties, for $e^+e^-$ and $ep$ collisions.
\par
In this paper the transition of a charm quark into a $D^{*\pm}$ meson
in DIS is further investigated. The normalised differential cross
sections are measured as a function of two observables with different sensitivity to gluon emissions. The momentum of the charm quark is
approximated either by the momentum of the jet, which includes the
$D^{*\pm}$ meson, or by the sum of the momenta of particles belonging
to a suitably defined hemisphere containing the $D^{*\pm}$ meson.
The measurements are performed for two different event samples. The DIS phase space and the kinematic requirements on the $D^{* \pm}$ meson are
the same for both samples. In the first sample, referred to as the
``$D^{* \pm}$ jet sample", the presence of a jet containing the
$D^{* \pm}$ meson and exceeding a minimal transverse momentum is required
as a hard scale. In the second sample no such jet is allowed to be present. This
sample, referred to as the ``no $D^{* \pm}$ jet sample", allows the investigation
of charm fragmentation in a region close to the kinematic
threshold of charm production.
\par
The normalised differential cross sections are used to fit
parameters of different fragmentation functions:
for QCD models as implemented in the
Monte Carlo (MC) programs RAPGAP~\cite{RAPGAP} and CASCADE~\cite{CASCADE},
which use the Lund string model for fragmentation as implemented in
PYTHIA~\cite{PYTHIA62}, and
for a next-to-leading order (NLO) QCD calculation as implemented in HVQDIS~\cite{hvqdis} with the addition of independent fragmentation of
charm quarks to $D^{* \pm}$ mesons.
\par
The paper is organised as follows. Section~\ref{Section:H1Detector} gives
a brief description of the H1 detector. It is followed by the details of
the event selection, the $D^{*\pm}$ meson signal extraction and the jet
selection in section~\ref{Section:DataSelection}. The experimental
fragmentation observables are defined in section~\ref{Section:Observables}.
The QCD models and calculations used for data corrections and for the
extraction of fragmentation functions are described in
section~\ref{Section:QCDModels}. The data correction procedure and the
determination of systematic uncertainties is explained in
section~\ref{Section:CorrectionsSystematics}. In
section~\ref{Section:Results} the results of the measurements and of
the fits of the fragmentation parameters are given.
\section{H1 Detector}
\label{Section:H1Detector}
The data were collected with the H1 detector at HERA in the years 1999
and 2000. During this period HERA collided positrons of energy
$E_{e}=27.5$~GeV with protons of energy $E_{p}=920$~GeV, corresponding to
a centre-of-mass energy of $\sqrt{s}=319$~GeV. The data sample used for
this analysis corresponds to an integrated luminosity of $47$~pb$^{-1}$.
\par
A right handed Cartesian coordinate system is used with the origin at the nominal primary $ep$ interaction vertex. The direction of the proton beam
defines the positive $z$-axis (forward direction). Transverse momenta are
measured in the $x$-$y$ plane. Polar ($\theta$) and azimuthal ($\phi$)
angles are measured with respect to this reference system. The pseudorapidity is defined as $\eta = - \ln (\tan\frac{\theta}{2})$.
\par
A detailed description of the H1 detector can be found
in~\cite{H1detector}. Here only the components relevant for this
analysis are described. The scattered positron is identified and measured
in the SpaCal~\cite{h1spacal}, a lead-scintillating fibre calorimeter
situated in the backward region of the H1 detector, covering the
pseudorapidity range $-4.0 < \eta < -1.4$. The SpaCal also provides
information to trigger on the scattered positron in the kinematic region
of this analysis. Hits in the backward drift chamber (BDC) are used to
improve the identification of the scattered positron and the measurement
of its angle~\cite{h1bdc}. Charged particles emerging from the
interaction region are measured by the Central Silicon Track detector
(CST)~\cite{h1cst} and the Central Tracking Detector (CTD), which
covers a range $-1.74 < \eta < 1.74$. The CTD comprises two large
cylindrical Central Jet drift Chambers (CJCs) and two $z$-chambers
situated concentrically around the beam-line, operated within a solenoidal
magnetic field of $1.16$~T. The CTD also provides triggering information
based on track segments measured in the $r$-$\phi$-plane of the CJCs and
on the $z$-position of the event vertex obtained from the double layers
of two Multi-Wire Proportional Chambers (MWPCs). The tracking detectors
are surrounded by a finely segmented Liquid Argon calorimeter
(LAr)~\cite{h1cal}. It consists of an electromagnetic section with lead
absorbers and a hadronic section with steel absorbers and covers the
range $-1.5 < \eta < 3.4$.
\par
The luminosity determination is based on the measurement of the
Bethe-Heitler process $ep \rightarrow ep\gamma$, where the photon is
detected in a calorimeter close to the beam pipe at $z = -103$~m.
\section{Data Selection and Analysis}
\label{Section:DataSelection}
The events selected in this analysis are required to contain a scattered
positron in the SpaCal and at least one $D^{*\pm}$ meson candidate. The
scattered positron must have an energy above $8$~GeV. The virtuality of
the photon $Q^{2}$ and the inelasticity $y$ are determined from the
measured energy $E_e^\prime$ and the polar angle $\theta_e^{\prime}$ of
the scattered positron via the relations:
\begin{equation}
Q^2=4E_eE^\prime_e\cos^2\left(
\frac{\theta_e^{\prime}}{2}\right) \; \; \; {\rm and} \; \; \;
y=1-\frac{ E^\prime_e}{ E_e} \sin^{2} \left(\frac{\theta_e^{\prime}}
{2}\right) \; .
\end{equation}
In addition, the energy $W$ of the $\gamma^{*}p$ rest-frame is determined
using:
\begin{equation}
W^2=ys - Q^2 \; ,
\end{equation}
where $s=4E_eE_p$ is the centre-of-mass energy squared of the $ep$ system.
The photon virtuality is required to be in the range
$2 < Q^{2} < 100$~GeV$^{2}$. This kinematic range is determined by the
geometric acceptance of the SpaCal. The inelasticity is required to lie
in the region $0.05 < y < 0.7$. The difference between the total energy
$E$ and the longitudinal component $P_{z}$ of the total
momentum, as calculated from the scattered positron and the hadronic final
state, is restricted to $40 < E-P_{z} < 75$~GeV. This requirement
suppresses photoproduction background, where a hadron is misidentified
as the scattered positron. It also reduces the contribution of DIS events
with hard initial state photon radiation, where the positron or photon
escapes in the negative $z$-direction. This leads to values of $E-P_{z}$
significantly lower than the expectation $2E_e = 55$~GeV.
\par
The $D^{*\pm}$ mesons are reconstructed from tracks using the decay channel
$D^{*+} \rightarrow D^{0} \pi_{\rm s}^{+} \rightarrow (K^{-}\pi^{+})\pi_{\rm s}^{+}$ and its charge conjugate, where $\pi_{\rm s}$ denotes the
low momentum pion from the $D^{*\pm}$ meson decay. Requirements on the
transverse momentum and pseudorapidity of the $D^{*\pm}$ meson candidate
and its decay products, as well as on particle identification using
d$E/$d$x$, are similar to those used in previous H1
analyses~\cite{h1-dstar}. A summary of the most important requirements
is given in table~\ref{table:dstar-cuts}.
\renewcommand{\arraystretch}{1.15}
\begin{table}[htdp]
\begin{center}
\begin{tabular}{|c|l|}
\hline
$D^{0}$ & $P_{\rm T}(K) > 0.25$~GeV \\
& $P_{\rm T}(\pi) > 0.25$~GeV \\
& $P_{\rm T}(K) + P_{\rm T}(\pi) > 2$~GeV \\
& $| M(K\pi) - M(D^{0}) | < 0.07$~GeV\\
\hline
$D^{* \pm}$ & $P_{\rm T}(\pi_{\rm s}) > 0.12$~GeV \\
& $| \eta(D^{*\pm}) | < 1.5$ \\
& $1.5 < P_{\rm T}(D^{*\pm}) < 15$~GeV \\
\hline
\end{tabular}
\caption{ Kinematic requirements for the selection of $D^{*\pm}$ meson
candidates. }
\label{table:dstar-cuts}
\end{center}
\end{table}
To select $D^{*\pm}$ meson candidates the invariant mass difference
method~\cite{feldman-deltam} is used. The distribution of
$\Delta M_{D^{*\pm}} = M(K\pi\pi_{\rm s}) - M(K\pi)$ is shown in
figure~\ref{fig:deltam} for the full data sample, together with the wrong
charge $K^{\pm}\pi^{\pm}\pi_{\rm s}^{\mp}$ combinations, using
$K^{\pm}\pi^{\pm}$ pairs in the accepted $D^{0}$ mass range.
Detailed studies show that the wrong
charge $\Delta M_{D^{*\pm}}$ distribution provides a good description of
the right charge $K^{\mp}\pi^{\pm}\pi_{\rm s}^{\pm}$ combinatorial
background.
\par
The signal is extracted using a simultaneous fit to the
$\Delta M_{D^{*\pm}}$ distribution of the right and wrong charge
combinations. The signal is fitted using a modified Gaussian
function~\cite{modgauss}
\begin{equation}
G_{\rm mod} \propto N_{D^{*\pm}}\, \exp \left[ -0.5\, x^{1+1/(1+0.5\, x)}
\right] \; ,
\end{equation}
where $x=|\Delta M_{D^{*\pm}} - M_0|/\sigma$. The signal position $M_0$
and width $\sigma$ as well as the number of $D^{*\pm}$ mesons
$N_{D^{*\pm}}$ are free parameters of the fit. The background is
parametrised as a power function of the form
$N (a+1)\: (\Delta M_{D^{*\pm}}-m_{\pi})^{a}/(M_\textrm{max}-m_{\pi})^{a+1}$,
with the fit boundaries given by the charged pion mass $m_{\pi}$ and $M_\textrm{max}=0.17$~GeV. The two free parameters $a$ and $N$ determine the shape and
normalisation of the background, respectively. The total event sample is
fitted using six free parameters: three for the modified Gaussian, two for
the normalisation of the right and wrong charge $\Delta M_{D^{*\pm}}$
background distributions and one for the background shape, common for the right and
wrong charge combinatorial background. In total $2865 \pm 89$~(stat.)
$D^{*\pm}$ mesons are obtained. For the differential distributions, the
number of $D^{*\pm}$ mesons in each measurement bin is extracted using the
same procedure, except that the position of the signal peak and its width
are fixed to the values determined from the fit to the total sample.
\par
The hadronic final state is reconstructed in each event using an energy
flow algorithm. The algorithm combines charged particle tracks and calorimetric energy
clusters, taking into account their respective resolution and geometric
overlap, into so called hadronic objects while avoiding double counting of
energy~\cite{hadroo2}. The hadronic objects corresponding to the three
decay tracks forming the $D^{*\pm}$ meson are removed from the event and
replaced by one hadronic object having the four-momentum vector of the
reconstructed $D^{*\pm}$ meson candidate. The energy of the $D^{*\pm}$ meson is calculated using $M(D^{*\pm}) = 2.010$~GeV~\cite{pdg}.
\par
Jets are found in the $\gamma^{*}p$ rest-frame using the inclusive
$k_{\rm T}$ cluster algorithm~\cite{incl-kt-algo} with the distance
parameter $R=1$ in the $\eta$-$\phi$ plane. In order to combine hadronic
objects into jets, the E-recombination scheme is applied using the
four-momenta of the objects. The jet containing the $D^{*\pm}$ meson
candidate is referred to as the $D^{*\pm}$ jet and is required to have a jet transverse energy $E_{\rm T}^* > 3$~GeV in the $\gamma^{*}p$
rest-frame\footnote{Kinematic variables with the superscript $^*$ refer
to the rest-frame of the virtual photon ($\gamma^{*}$) and proton. In
this frame the photon direction is taken as the direction of the $z$-axis.
The four-vector of the virtual photon used in the boost calculation is
determined from the measurement of the scattered positron.}.
According to MC simulations, the $D^{*\pm}$ jet is found to be well
correlated with the original direction of the charm or anti-charm quark. The distance
in azimuth-pseudorapidity,
$\Delta r = \sqrt{\Delta\eta^2 + \Delta\phi^2}$, between the charm quark
jet, found using final state partons (``parton level"), and the
$D^{*\pm}$ jet, found using final state hadrons (``hadron level"), is
below $0.3$ for $90$\% of all events. The correlation between the
$D^{*\pm}$ jet at hadron level and the $D^{*\pm}$ jet found using charged
particle tracks and calorimetric clusters (``detector level") is even
better, since most of the energy of these jets is reconstructed from
tracks, which are well measured in the tracking system. The number of
$D^{*\pm}$ mesons is $1508 \pm 68$~(stat.) in the $D^{*\pm}$ jet sample
and $1363 \pm 54$~(stat.) in the no $D^{*\pm}$ jet sample.
\section{Definition of Experimental Observables}
\label{Section:Observables}
A standard method to study fragmentation is to measure the differential production
cross section of a heavy hadron (H) as a function of a scaled momentum or
energy. In $e^+e^-$ experiments a customary experimental definition of
the scaled energy is
${\rm z}_{e^+e^-} = E_{\rm H}/E_{\rm beam}$, where $E_{\rm beam}$ is the
energy of the beams in the centre-of-mass system. In leading order (LO),
i.e. without gluon emissions, the beam energy is equal to the energy of
the charm or anti-charm quark, which are produced in a colour singlet
state. The differential cross section of heavy hadron production as a
function of ${\rm z}_{e^+e^-}$ is directly related to the fragmentation
function.
\par
In the case of $ep$ interactions the situation is more complex. In DIS
the dominant process for $D^{*\pm}$ meson production at HERA is
photon-gluon fusion $\gamma^{*}g \rightarrow c\bar{c}$~\cite{h1-dstar}.
In this case the $c\bar{c}$ pair is produced in a colour octet state.
The energy of the charm quark pair depends on the energy of the incoming
photon and gluon. Hadrons produced by initial state gluon emissions and
by fragmentation of the proton remnant are also present in the final
state.
\par
In this analysis charm fragmentation is studied by measuring the
differential cross sections of $D^{*\pm}$ meson production as a function
of two different observables ${\rm z}_{\rm hem}$ and ${\rm z}_{\rm jet}$
defined below, which are sensitive to the fraction of momentum inherited
by the $D^{*\pm}$ meson from the initial charm quark \cite{zuzana}.
\subsection*{The hemisphere method:
z ${\mathbf =}$ z$_{\rm {\bold{hem}}}$ }
In the LO photon-gluon fusion process, which dominates charm
production at HERA, the charm and anti-charm quarks are moving in the
direction of the virtual photon in the $\gamma^*p$ rest-frame of reference. This is due to the fact that the photon is more energetic than the gluon,
which typically carries only a small fraction of the proton's momentum.
Assuming no further gluon radiation in the initial and final state, the
charm and anti-charm quarks are balanced in transverse momentum
(figure~\ref{fig:hem}, left). This observation suggests to divide the
event into hemispheres, one containing the fragmentation products of the
charm quark, the other one those of the anti-charm quark. In order to suppress contributions from initial state radiation and the proton
remnant, particles pointing in the proton direction of the $\gamma^{*}p$
rest-frame ($\eta^*<0$) are discarded. The projections of the momenta of
the remaining particles onto a plane perpendicular to the $\gamma^*p$-axis
are determined. Using the projected momenta, the thrust-axis in this plane,
i.e. the axis maximising the sum of the momenta projections onto it, is
found. A plane perpendicular to the thrust-axis allows the division of the
projected event into two hemispheres, one of them containing the $D^{*\pm}$
meson and usually other particles (figure~\ref{fig:hem}, right). The
particles belonging to the same hemisphere as the $D^{*\pm}$ meson are
attributed to the fragmentation of the charm or anti-charm quark. The
fragmentation observable is defined as:
\begin{equation}
{\rm z}_{\rm hem} = \frac{(E^*+P_{\rm L}^*)_{D^{*\pm}}}{\sum_{\rm hem}
(E^*+P^*)} \; ,
\label{eq:zhem}
\end{equation}
where in the denominator the energy $E^*$ and the momentum $P^*$ of all
particles of the $D^{*\pm}$ meson hemisphere are summed. The longitudinal
momentum $P^*_{{\rm L}\,D^{*\pm}}$ is defined with respect to the direction
of the three-momentum of the hemisphere, defined as the vectorial sum of
the three-momenta of all particles belonging to the hemisphere. The
variable ${\rm z}_{\rm hem}$ is invariant with respect to boosts along the
direction of the sum of the momenta of all particles in the hemisphere.
Neglecting the mass of the $D^{*\pm}$ meson and of the hemisphere, this definition of ${\rm z}_{\rm hem}$ simplifies to the ratio of their
energies.
\subsection*{ The jet method: z ${\mathbf =}$ z$_{\rm {\bold{jet}}}$ }
In the case of the jet method the energy and direction of the charm quark
are approximated by the energy and direction of the reconstructed jet,
which contains the $D^{*\pm}$ meson. The fragmentation observable is
defined in analogy to ${\rm z}_{\rm hem}$ as:
\begin{equation}
{\rm z}_{\rm jet} = \frac{(E^*+P_{\rm L}^*)_{D^{*\pm}}}{(E^*+P^*)_{\rm
jet}} \; ,
\label{eq:zjet}
\end{equation}
where the longitudinal momentum $P^*_{{\rm L}\,D^{*\pm}}$ is defined with
respect to the direction of the three-momentum of the jet. The jet finding
and the determination of ${\rm z}_{\rm jet}$ are performed in the
$\gamma^{*}p$ rest-frame.
\par
Both fragmentation observables are defined in such a way that they would
lead to similar distributions, assuming independent fragmentation and no
gluon radiation. The measured distributions, however, are expected to
differ, as they have different sensitivities to gluon radiation and charm
quarks, which are colour connected to the partons of the proton remnant.
The hemisphere method
typically includes more energy around the charm quark direction than the
jet method. The parameters of fragmentation functions should however be
the same, if extracted for a QCD model, which provides a very good
description of the underlying physics over the full phase space of this
analysis. A comparison of both methods thus may provide a consistency
check and a test of the perturbative and non-perturbative physics as
encoded in the models.
\par
The measurement is restricted to the regions
$0.2 < {\rm z}_{\rm hem} \leq 1.0$ and
$0.3 < {\rm z}_{\rm jet} \leq 1.0$, as at lower ${\rm z}$ it is not
possible to separate the $D^{*\pm}$ meson signal from the large
combinatorial background. In order to minimise the sensitivity of the
analysis to the total $D^{*\pm}$ meson cross section, and to reduce
systematic errors, normalised differential cross sections are measured as
a function of the fragmentation observables ${\rm z}_{\rm hem}$ and
${\rm z}_{\rm jet}$. The normalisations are chosen such that their
integrals over the respective ${\rm z}$-regions yield unity.
\section{QCD Models and Calculations}
\label{Section:QCDModels}
The MC programs RAPGAP and CASCADE are used to generate events containing
charm and beauty quarks, which are passed through a detailed simulation of
the detector response, based on the GEANT simulation
program~\cite{Brun:1987ma}. They are reconstructed using the same
software as used for the data. These event samples are used to
determine the acceptance and efficiency of the detector and to estimate
the systematic errors associated with the measurements. In addition, these
models are fitted to the data in order to determine the parameters of the
fragmentation functions.
\par
The Monte Carlo program RAPGAP~\cite{RAPGAP}, based on collinear
factorisation and DGLAP~\cite{dglap} evolution, is used to generate the
direct process of photon-gluon fusion to a heavy $c\bar{c}$~pair, where
the photon acts as a point-like object. In addition, RAPGAP allows the
simulation of charm production via resolved processes, where the photon
fluctuates into partons, one of which interacts with a parton in the
proton, and the remaining partons produce the photon remnant. The program
uses LO matrix elements with massive (massless) charm quarks for the
direct (resolved) processes. Parton showers based on DGLAP evolution are used to model higher order QCD effects.
\par
The CASCADE program~\cite{CASCADE} is based on the
$k_{\rm T}$-factorisation approach. Here, the calculation of the
photon-gluon fusion matrix element takes into account the charm quark mass
and the virtuality and the transverse momentum of the incoming gluon.
Gluon radiation off the incoming gluon as well as parton showers off the
charm or anti-charm quark are implemented including angular ordering
constraints. The gluon density of the proton is evolved according to the
CCFM equations~\cite{ccfm}. The $k_{\rm T}$-unintegrated gluon density function A0~\cite{CASCADE-gluon}, extracted from inclusive DIS data, is
used.
\par
In both RAPGAP and CASCADE the hadronisation of partons is performed
using the Lund string model as implemented in PYTHIA~\cite{PYTHIA62}.
In the Lund model, the heavy hadron is produced in the process of string
breaking. The fraction of the string longitudinal momentum $z$ carried
by the hadron is generated according to different choices of adjustable
fragmentation functions $D^{\rm H}_{ \rm Q}(z)$. Within this analysis
three widely used parametrisations are employed, of which two depend on
a single free parameter, and one depends on two free parameters. The
parametrisation suggested by Peterson et al.~\cite{peterson-ff} has the
functional form:
\begin{equation}
D^{\rm H}_{ \rm Q}(z) \varpropto \frac{1}{{z}[1-(1/{z})-\varepsilon /
(1-{z})]^2} \; ,
\label{eq:peterson}
\end{equation}
and the one by Kartvelishvili et al.~\cite{kartvelishvili-ff} is given
by:
\begin{equation}
D^{\rm H}_{ \rm Q}(z) \varpropto {z}^{\alpha} (1-{z}) \; .
\label{eq:kartvelishvili}
\end{equation}
The free parameters $\varepsilon$ and $\alpha$ determine the ``hardness"
of the fragmentation function and are specific to the flavour of the heavy
quark, i.e. charm in the case of $D^{*\pm}$ meson production. The
parametrisation inspired by Bowler and Morris~\cite{bowler-ff} (referred
to as the Bowler parametrisation) has the functional form:
\begin{equation}
D^{\rm H}_{ \rm Q}(z) \varpropto \frac{1}{z^{1+r_{\rm Q}bm_{\rm Q}^2}}
(1-z)^a \exp{(- \frac{bM_{\rm T}^2}{z})} \; .
\label{eq:bowler}
\end{equation}
The shape of the fragmentation function is determined by two free
parameters $a$ and $b$, $m_{\rm Q}$ is the mass of the heavy quark,
$M_{\rm T}= \sqrt{M^{2}_{\rm H}+P_{\rm T}^{2}}$ the transverse mass of
the heavy hadron, and $r_{\rm Q}=1$ as default in PYTHIA.
\par
For data corrections the parameter setting tuned by the ALEPH
collaboration\cite{ALEPH-steering} together with the Peterson fragmentation
function is used for the fragmentation of
partons in PYTHIA. It includes higher excited charm states, of which some
also decay to $D^{*\pm}$ mesons and contribute significantly to the
$D^{*\pm}$ meson yield. When extracting parameters of the fragmentation
functions also the default parameter setting of PYTHIA is used as an
alternative. In this case no higher excited charm states are produced.
Both parameter settings are indicated in table~\ref{table:steering}.
\par
The parameters for the Kartvelishvili and Peterson fragmentation functions
are also extracted for the HVQDIS program~\cite{hvqdis}. HVQDIS is based on
the NLO, i.e. $\mathcal{O}(\alpha_{\rm s}^2)$, calculation in the fixed
flavour number scheme, with three light active flavours as well as gluons
in the proton. The proton parton density functions (PDFs) of the light
quarks and the gluon are evolved according to the DGLAP equations.
Massive charm quarks are assumed to be produced only
perturbatively via photon-gluon fusion and higher order processes. The
final state charm quarks are fragmented independently into $D^{*\pm}$ mesons in the $\gamma^* p$ rest-frame. Kartvelishvili and Peterson
parametrisations are used to generate the charm quark's momentum fraction
transferred to the $D^{*\pm}$ meson. The energy of the charm quark is calculated using the on-mass-shell condition. In addition, the $D^{*\pm}$
meson can be given a transverse momentum $P_{\rm T}$ with respect to the
charm quark, according to the function $P_{\rm T} \exp(-\beta P_{\rm T})$.
The value used for the parameter $\beta$ corresponds to an average
$P_{\rm T}(D^{*\pm})$ of $350$~MeV.
\par
The Monte Carlo programs RAPGAP and HERWIG~\cite{HERWIG} are used to
estimate the size of the hadronisation corrections to the data for
comparison with HVQDIS predictions. While the perturbative QCD model of
HERWIG is similar to the one of RAPGAP, the HERWIG program employs the
cluster hadronisation model, which is quite different from the Lund string
model used by PYTHIA.
\par
The basic parameter choices for the QCD models and the NLO calculation
are summarised in table~\ref{table:models}.
\section{Data Corrections and Systematic Errors}
\label{Section:CorrectionsSystematics}
In this analysis, the differential cross section for the production of
$D^{*\pm}$ mesons, which result from the fragmentation of charm quarks
either directly or via decays from higher excited charm states, is
measured. The small contribution of $D^{*\pm}$ mesons originating from
B-hadron decays is estimated with RAPGAP and is subtracted from the data. It is on the level of $1$ to $2$\%. The data are corrected for detector
and QED radiative effects. The transverse momentum and pseudorapidity
distributions of the $D^{*\pm}$ mesons of the Monte Carlo event samples,
which are used to correct the data samples for detector effects, are
reweighted to the corresponding distributions of the data in order to
achieve an improved description. The $\eta$ and $P_{\rm T}$ dependent
reweighting
factors differ from unity by typically $10-30$\%. After this
reweighting, both RAPGAP and CASCADE provide a good description of
the data as shown in figure~\ref{fig:controlplots}.
The description of the no $D^{*\pm}$ event sample by the reweighted MC models,
as shown in figure~\ref{fig:controlplots-noDsJetsample}, is worse.
The measurement bins are defined
in such a way that the purity in each bin, defined as the fraction of
events reconstructed in a ${\rm z}_{\rm hem}$ or ${\rm z}_{\rm jet}$ bin
that originate from that bin on hadron level, is between $40$ and $70$\%.
\par
The correction for the detector effects is done using regularised
deconvolution, taking into account migrations between measurement
bins~\cite{deconvolution}. The detector response matrix is generated using
RAPGAP, and the value of the regularisation parameter is determined through
decomposition of the data into eigenvectors of the detector response
matrix. As a check, the detector response matrix was also generated using
CASCADE and found to be consistent with the one from RAPGAP. Statistical
errors are calculated by error propagation using the covariance matrix.
The data are then corrected for migrations into the visible phase space
using RAPGAP and CASCADE. The effects of QED radiation are corrected for
using the HERACLES~\cite{heracles} program, which is interfaced with
RAPGAP. Correction factors are calculated from the ratio between cross
sections obtained from the model including and not including QED
radiation. The corrections are applied bin-by-bin in ${\rm z}_{\rm hem}$
and ${\rm z}_{\rm jet}$. In the case of ${\rm z}_{\rm jet}$ the
corrections are about $2$\%. In the case of ${\rm z}_{\rm hem}$,
where photons radiated into the $D^{*\pm}$ meson hemisphere can be mistaken
as fragmentation products of the charm quark, the corrections reach
$10$\% for the lowest value of ${\rm z}_{\rm hem}$.
\par
In contrast to the QCD models discussed so far, the HVQDIS program
provides only a partonic final state with the exception of the additional
$D^{*\pm}$ meson from charm fragmentation. In order to compare the NLO
predictions to the measurements, the data are corrected to the parton
level by means of hadronisation corrections, which are estimated using
the MC generators RAPGAP and HERWIG. While the quantity
$(E^*+P_{\rm L}^*)_{D^{*\pm}}$ in equation~\ref{eq:zhem} and~\ref{eq:zjet}
is calculated using the momentum of the $D^{*\pm}$ meson, the jet finding
and the calculation of the jet and hemisphere quantities, the denominators
in equations~\ref{eq:zhem} and~\ref{eq:zjet}, are performed using the
partonic final state. All partons after parton
showering are considered, and the same jet and hemisphere finding
algorithms are applied at parton and hadron level. For each $\rm z$-bin
the hadronisation correction factor is calculated as the ratio of parton
to hadron level cross section. The arithmetic mean of the hadronisation
correction factors of both models is used to multiply the data cross
section. In case of $\rm z_{hem}$ the hadronisation corrections differ
from unity by typically $\pm 40$\%. For $\rm z_{jet}$ they differ from
unity by typically $\pm 20$\%, except for the highest $\rm z$-bin, where
they are about $50$\%. The hadronisation corrections as determined by HERWIG and RAPGAP are similar for most of the measurement bins, with
the exception of the lowest bin in $\rm z_{jet}$, where they differ by
about $60 $\%.
\par
The following systematic uncertainties on the normalised differential
cross sections are considered:
\begin{itemize}
\item
The energy uncertainty of the scattered positron varies linearly from
$\pm 3$\% for an energy of $8$~GeV to $\pm 1$\% for $27$~GeV.
\item
The polar angle of the scattered positron has an estimated uncertainty
of $\pm 1$~mrad.
\item
The uncertainty of the energy scale of the hadronic objects is made up
of $\pm 0.5$\% due to tracks and $\pm 4$\% ($\pm 7$\%) due to LAr
(Spacal) clusters.
\item
The effect of the uncertainty of the tracking efficiency on reconstructing
the $D^{*\pm}$ meson is determined by changing the nominal efficiency
in the simulation as a function of track $\eta$ and $P_{\rm T}$. In the
central region of the accepted $\eta$--$P_{\rm T}$ phase space the
estimated uncertainty of the nominal efficiency is $\pm 2$\%, in the
regions of large $|\eta|$ but not small $P_{\rm T}$ it is $\pm 3$\%,
and for large $|\eta|$ and small $P_{\rm T}$ it is $\pm 4$\%.
\item
The value of d$E/$d$x$ of the $D^{*\pm}$ meson decay products has an
estimated uncertainty of $\pm 8$\%, which is of similar size as the
experimental resolution in d$E/$d$x$.
\item
The uncertainty of the $D^{*\pm}$ meson signal extraction is estimated
using different $D^{*\pm}$ counting techniques and by using different
fit functions for the background parametrisation.
The largest uncertainty comes from the background description, which
determines the systematic error on the signal extraction.
\item
The uncertainty of beauty production by the RAPGAP MC is assumed to be
$\pm 100$\%. The resulting small uncertainty of the normalised $D^{*\pm}$
meson cross sections is taken to be symmetrical.
\item
The effect of using different MC models for the small correction for
migrations into the visible phase space is studied using RAPGAP and
CASCADE.
The factors used to correct the data are determined as the average of the
correction factors obtained from the two models.
Half the difference is taken
as systematic uncertainty.
\item
For parton level corrected distributions half the difference between
the hadronisation correction factors of RAPGAP and HERWIG is taken as
the uncertainty due to the different fragmentation models.
\end{itemize}
\par
Other systematic effects, which are investigated and found to be negligible,
are: the effect of reflections, i.e. wrongly or incompletely reconstructed
$D^{*\pm}$ meson decays, on the shape of the fragmentation observables,
the effect on acceptance and reconstruction efficiency from including
diffractive events, the effect of using different MC models for the
deconvolution of the data and the uncertainty of the QED radiative
effects.
\par
Each source of systematic error is varied in the Monte Carlo simulation
within its uncertainty. In each measurement bin, the corresponding
deviation of the normalised cross sections from the central value is taken
as the systematic error. Among the systematic errors the uncertainties due to
the scattered positron energy scale, the hadronic energy scale, and the
beauty fraction are correlated amongst the bins in ${\rm z}$. In the
extraction of the parameters of the fragmentation functions, the
statistical and systematic errors with their correlations are taken into
account. The average effect of various systematic errors on the
${\rm z}_{\rm hem}$ and ${\rm z}_{\rm jet}$ distributions is summarised
in table~\ref{table:systematics}. Since the distributions of
${\rm z}_{\rm hem}$ and ${\rm z}_{\rm jet}$ are normalised, the effect of
many systematic uncertainties is reduced and the statistical error
dominates the uncertainty of the measurement.
\section{Results}
\label{Section:Results}
\subsection{Normalised differential cross sections and comparison with
predictions}
\label{Subsection:results_default}
The differential cross sections of $D^{*\pm}$ meson production as a
function of the fragmentation observables ${\rm z}_{\rm hem}$ and
${\rm z}_{\rm jet}$ are shown in figure~\ref{fig:results-defaults} for
the $D^{*\pm}$ jet sample. They refer to the visible phase space given
by $2 < Q^2 < 100$~GeV$^2$, $0.05 < y < 0.7$,
$1.5 < P_{\rm T}(D^{*\pm}) < 15$~GeV and $ |\eta(D^{*\pm}) | < 1.5$. In
addition, a $D^{*\pm}$ jet with $E_{\rm T}^* > 3$~GeV in the $\gamma^{*}p$
rest-frame is required in order to have the same hard scale in the event for both distributions ${\rm z}_{\rm hem}$ and ${\rm z}_{\rm jet}$. The
measurements and the corresponding predictions are normalised such that
their integrals over the respective ${\rm z}$-regions yield unity.
The normalised cross sections and
their errors are given in table~\ref{table:results-hem-dsjet} for the
hemisphere observable and in table~\ref{table:results-jet} for the jet observable.
\par
Figure~\ref{fig:results-defaults} also includes predictions of RAPGAP with
three commonly used fragmentation parameter settings for PYTHIA
(described in table~\ref{table:steering}), obtained from $e^{+}e^{-}$ annihilation. The
settings and the corresponding values of $\chi^2$/n.d.f., as calculated
from the data and the model predictions, are summarised in
table~\ref{table:results-parameters-default}. In general, there is
reasonable agreement between data and the QCD model with all settings
for both the jet and the hemisphere observables.
The large difference between
the two distributions observed in the highest ${\rm z}_{\rm jet}$ bin is
mainly due to a significant fraction of $D^{*\pm}$ jets consisting of a
$D^{*\pm}$ meson only, for which ${\rm z}_{\rm jet}$ equals unity.
CASCADE provides a
similar description of the data as RAPGAP.
\subsection{Extraction of parameters for the Kartvelishvili and
Peterson fragmentation functions}
\label{subsection:fit}
The normalised $D^{*\pm}$ meson differential cross sections as a
function of ${\rm z}_{\rm hem}$ and ${\rm z}_{\rm jet}$ are used to
extract optimal parameters for the Peterson and Kartvelishvili
fragmentation functions described in section~\ref{Section:QCDModels}.
Both parametrisations have a single free parameter.
\par
The parameter extraction is done by comparing different model
configurations with the data. A configuration is defined by one of the QCD
calculations (RAPGAP, CASCADE or HVQDIS), by one of the fragmentation
functions (Peterson or Kartvelishvili) and by a possible value for the
corresponding fragmentation parameter, $\varepsilon$ or $\alpha$. For
RAPGAP and CASCADE the configuration also depends on the PYTHIA parameter settings used (ALEPH and default, see table~\ref{table:steering}).
In order to be able to compare all configurations to the data, a reweighting
procedure is applied. For each of the QCD calculations large event samples
with $D^{*\pm}$ mesons are generated using the Peterson fragmentation
function. For these events the $z$-value of the fragmentation function,
used by the model to generate the fraction of charm quark or string
momentum transferred to the $D^{*\pm}$ meson, is stored such that each
event can be reweighted to another fragmentation function or any other
parameter value. For each configuration the predicted and measured distributions of the fragmentation observables are used to determine a
$\chi^2$ as a function of the fragmentation parameter. In the calculation
of the $\chi^2$ the full covariance matrix is used, taking into account
correlated and uncorrelated statistical and systematic errors.
The fragmentation parameter is determined at the minimum of the $\chi^2$.
The shape of the $\chi^2$ distribution is used to determine the
$\pm 1\sigma$ error (using $\chi^2_{\rm min} + 1$) of the extracted parameter. As an example, in figure~\ref{fig:results-rapgap-kart} the
data are compared to the prediction of RAPGAP with the ALEPH setting for
PYTHIA as given in table~\ref{table:steering} but using the Kartvelishvili
parametrisation. The two lines indicate the $\pm 1 \sigma$ total
uncertainty around the best fit value of $\alpha$. The description of
the data by CASCADE is similar.
\par
The parameters $\alpha$ and $\varepsilon$, which are extracted using
RAPGAP and CASCADE, with and without higher excited charmed hadrons,
are summarised in table~\ref{table:results-parameters-hem-jet} together
with their corresponding values of $\chi^2$/n.d.f.. With the fitted
parameters the model predictions using either the Peterson or the
Kartvelishvili parametrisations describe the data reasonably well, with
the Kartvelishvili parametrisation being in all cases slightly preferable,
as indicated by the values of $\chi^2$/n.d.f.. When using the same PYTHIA
parameter setting, the fragmentation parameters $\alpha$ and
$\varepsilon$, extracted from the ${\rm z}_{\rm hem}$ and
${\rm z}_{\rm jet}$ observables, are in good agreement. Both RAPGAP and
CASCADE lead to statistically compatible parameters $\varepsilon$ and
$\alpha$. A priori, agreement in the fragmentation function parameters
for RAPGAP and CASCADE is not required, since the models differ in terms
of simulated processes (direct and resolved in case of RAPGAP compared to
direct only for CASCADE) and in their implementation of perturbative QCD.
\par
The fragmentation parameters $\alpha$ and $\varepsilon$ depend
significantly on the PYTHIA parameter settings used, i.e. whether
$D^{*\pm}$ mesons are assumed to be produced only via direct fragmentation
of charm quarks or additionally originate from decays of higher excited
charm states. In the latter case the $D^{*\pm}$ mesons carry a smaller
fraction of the original charm or anti-charm quark momentum in comparison
with the directly produced ones. Both the default PYTHIA setting and the
setting containing higher excited charm states describe the data equally
well. The values of the Peterson parameter $\varepsilon$ extracted for the
PYTHIA setting containing higher charm states, see table~\ref{table:results-parameters-hem-jet}, are in agreement with the
value $\varepsilon = 0.04$ tuned by ALEPH\cite{ALEPH-steering}.
This result is consistent with
the hypothesis of fragmentation universality in $ep$ and $e^+e^-$
processes.
\par
The NLO calculation as implemented in HVQDIS with the Kartvelishvili
fragmentation function leads to a good fit of the data, corrected for
hadronisation effects, as shown in figure~\ref{fig:results-hvqdis-kart}.
On the other hand HVQDIS provides a rather poor description of
${\rm z}_{\rm hem}$ and ${\rm z}_{\rm jet}$, if the Peterson fragmentation
function is used (the $\chi^2$ values of the fit are shown in
table~\ref{table:results-parameters-hem-jet}). Simulating a
$P_{\rm T}$ of the $D^{*\pm}$ meson with respect to the charm quark
direction, as explained in section~\ref{Section:QCDModels}, has only a little
effect on the extracted value of $\alpha$.
\par
The distributions of ${\rm z}_{\rm hem}$ and ${\rm z}_{\rm jet}$ are also
measured in two bins of $Q^2$ and $W$. In the case of $Q^2$, the bins are
defined as $2 < Q^{2} < 10$~GeV$^2$ and $10 < Q^{2} < 100$~GeV$^2$. The
accessible range in $W$, determined by the cuts on $Q^2$ and $y$,
corresponds to $70<W<270$~GeV, and the bins are defined as $70 < W < 170$~GeV
and $170 \le W < 270$~GeV. Correction factors and systematic uncertainties for
these samples are determined in the same way as for the full $D^{*\pm}$
jet sample. The data are compared to the QCD models with the PYTHIA
parameter setting including higher excited charm states and using the Kartvelishvili fragmentation function. For the low and high $Q^{2}$
bins the measured distributions are found to be almost the same and well
described by the QCD models. The distributions of ${\rm z}_{\rm jet}$
for the low and high $W$ regions are also similar. A difference is
observed for the ${\rm z}_{\rm hem}$ distribution, which is softer at
high $W$ as shown in figure~\ref{fig:w-dependence}. RAPGAP and CASCADE
show the same behaviour as a function of $W$ as observed in data. This
behaviour can be understood as being partly due to enhanced gluon
radiation at high $W$ and partly due to the kinematic effect of the
requirement $P_{\rm T}(D^{*\pm}) > 1.5$~GeV. For events at low $W$,
where the charm quark tends to have smaller energy than at high $W$, a
$D^{*\pm}$ meson needs to carry a large fraction of the original quark
momentum in order to pass the $P_{\rm T}$ requirement.
\par
The hemisphere observable allows an investigation of charm fragmentation
close to the kinematic threshold, at the limit of applicability of the
concept of fragmentation functions, by selecting events without a
$D^{*\pm}$ jet with $E_{\rm T}^* > 3$~GeV. As estimated by MC, the mean
centre-of-mass energy squared of the $\gamma^*g$ system $\hat{s}$
for this sample is about $36$~GeV$^2$, to be compared with
about $100$~GeV$^2$ for the $D^{*\pm}$ jet sample. This event sample
(no $D^{*\pm}$ jet sample) has no overlap with the $D^{*\pm}$ jet sample investigated so far. The normalised cross section as a function of
${\rm z}_{\rm hem}$ for the no $D^{*\pm}$ jet sample is shown in
figure~\ref{fig:results-rapgap-kart-twosamples} and listed in table~\ref{table:results-hem-nodsjet}. Predictions of RAPGAP with the three
commonly used fragmentation parameter settings for PYTHIA (see
table~\ref{table:steering}), which provide a reasonable description of the
$D^{*\pm}$ jet sample (see table~\ref{table:results-parameters-default}),
fail for this sample. Also the prediction using the fragmentation
parameters obtained from the $D^{*\pm}$ jet sample is not able to describe
these data.
\par
The fragmentation parameters for RAPGAP, CASCADE and the NLO calculation
are extracted from the no $D^{*\pm}$ jet sample using the same procedure
as for the $D^{*\pm}$ jet sample. The fit results are summarised in
table~\ref{table:results-parameters-nodsjet}.
The predictions of RAPGAP, showing the $\pm 1\sigma$ total uncertainty
around the fitted value of $\alpha$ are also presented in
figure~\ref{fig:results-rapgap-kart-twosamples}. The fragmentation parameters obtained for RAPGAP and CASCADE are statistically
compatible. The fragmentation parameters fitted to the no $D^{*\pm}$ jet sample are found to be significantly different from those for
the $D^{*\pm}$ jet sample. They indicate that the fragmentation function
for an optimal description of the sample without a $D^{*\pm}$ jet needs
to be significantly harder than for the $D^{*\pm}$ jet sample. The NLO
calculation as implemented in HVQDIS fails to describe the no $D^{*\pm}$
jet sample as shown in figure~\ref{fig:results-hvqdis-kart-nodssample}.
\par
Several parameters of the QCD models, for example those influencing
parton showers, have been varied, in trying to describe both samples
using the same value for the fragmentation function parameter. However,
it was not possible to find MC parameters leading to a consistent
fragmentation function for the two samples.
Furthermore, the effect of diffractive production of $D^{*\pm}$ mesons
was not able to explain the difference between the fragmentation
parameters observed for the two samples.
These investigations indicate that QCD models, together with simple parametrisations of the fragmentation functions, are not able to
describe charm fragmentation consistently in the full phase space down
to the kinematic threshold.
\section{Conclusions}
The fragmentation of charm quarks into $D^{*\pm}$ mesons in DIS is studied
using the H1 detector at the HERA collider. The normalised $D^{*\pm}$
meson differential cross sections are measured as a function
of two observables sensitive to fragmentation, the hemisphere observable ${\rm z}_{\rm hem}$
and the jet observable ${\rm z}_{\rm jet}$, in the visible
DIS phase space defined by $2 < Q^{2} < 100$~GeV$^{2}$ and $0.05 < y < 0.7$,
and the $D^{*\pm}$ meson phase space defined by $1.5 < P_{\rm T}(D^{*\pm}) <
15$~GeV and
$| \eta(D^{*\pm}) | < 1.5$. An additional jet with $E_{\rm T}^* > 3$~GeV,
containing the $D^{*\pm}$ meson, is required in the $\gamma^{*}p$
rest-frame in order to provide a hard scale for the events.
\par
The data are compared with predictions of RAPGAP with three widely used PYTHIA parameter settings and the Peterson and the Bowler parametrisations
for the fragmentation of heavy flavours obtained from $e^{+}e^{-}$
annihilation. They provide a reasonable description of the $ep$ data
presented.
\par
The normalised differential cross sections are used to fit the parameters
of the Kartvelishvili and Peterson fragmentation functions within the
framework of the QCD models RAPGAP and CASCADE. The fragmentation
parameters extracted using the ${\rm z}_{\rm hem}$ and ${\rm z}_{\rm jet}$
observables are in good agreement with each other. Both QCD models lead
to statistically compatible parameters. The value of the Peterson
parameter $\varepsilon$ extracted for the PYTHIA parameter setting,
which includes not only $D^{*\pm}$ mesons from direct fragmentation of
charm quarks but also from the decays of higher excited charm states,
is in agreement with the value of $\varepsilon = 0.04$ tuned by
ALEPH.
This result is consistent with the hypothesis of fragmentation
universality between $ep$ and $e^+e^-$ collisions.
\par
The QCD models, with the fragmentation parameters fitted to the data, also
provide a good description of the $Q^2$ and $W$ dependence of the
fragmentation observables.
\par
The data, corrected to the parton level, are also compared to the NLO
calculation as implemented in HVQDIS, with the addition of independent fragmentation of charm quarks to $D^{*\pm}$ mesons. A good fit to the data is obtained when using the fragmentation function by Kartvelishvili
et al., while using the one of Peterson et al. results in a poor fit.
\par
Finally, the hemisphere method is used to study the fragmentation of
charm produced close to the kinematic threshold, by selecting a data
sample fulfilling the nominal requirements on the DIS and $D^{*\pm}$
meson phase space, but without a $D^{*\pm}$ jet having
$E_{\rm T}^* > 3$~GeV in the event. The fragmentation parameters
extracted for the QCD models, using this sample of events, are
significantly different from those fitted to the $D^{*\pm}$ jet sample.
Furthermore, the fit for the NLO calculation using the no $D^{*\pm}$
jet sample fails. Both observations can be interpreted as an inadequacy
of the QCD models and the NLO calculation to provide a consistent
description of the full phase space down to the kinematic threshold.
\section*{Acknowledgements}
We are grateful to the HERA machine group whose outstanding efforts
have made this experiment possible. We thank the engineers and
technicians for their work in constructing and maintaining the H1
detector, our funding agencies for financial support, the DESY
technical staff for continual assistance and the DESY directorate for
support and for the hospitality which they extend to the non DESY
members of the collaboration.
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 1,509 |
\section{Introduction}
The physics of black holes has been at the center stage of
theoretical physics for over thirty years, as black holes
raise puzzles that directly challenge many cherished fundamental
physical properties such as as unitarity and locality.
Defining questions in this area have been
\begin{itemize}
\item Why does a black hole have entropy proportional to its horizon area?
\item Is there information loss because of black holes?
\item How does one resolve spacetime singularities, such as those inside black
holes or in Big Bang cosmologies?
\end{itemize}
The fact that black holes appear to have entropy is puzzling on
many counts. Typically in a quantum system the correspondence
principle relates the quantum states to the classical phase space
and the entropy to the volume of this phase space in Planck units.
Black holes however are uniquely fixed in terms of the conserved
charges they carry (black holes have ``no hair'' in classical general
relativity) so the
classical phase space is zero dimensional\footnote{Strictly speaking
such uniqueness theorems have only been proven for very special
systems and in general black holes do carry hair. However the
volume of the corresponding classical phase space is still far too
small to account for black hole entropy upon quantization.}.
One could argue that perhaps the states counted by the Bekenstein-Hawking
formula are purely quantum mechanical with no classical limit,
but even in this case one is faced with the puzzle that
the entropy is proportional to the area of the horizon, rather than
the volume enclosed in it, as one might have anticipated based on the fact that the entropy is an
extensive quantity. This has been taken as a hint of a fundamental
property of quantum gravitational theories, namely that they
are {\it holographic} \cite{'t Hooft:1993gx}: any $(d{+}1)$-dimensional gravitational theory should have a
description in terms of $d$-dimensional quantum field theory without
gravity with one degree of freedom per Planck area.
We will discuss holography and its realization in the AdS/CFT
correspondence extensively in this report.
Classically black holes are completely black, but
semi-classically they thermally radiate \cite{Hawking:1974sw}. This fact considerably
strengthens the case for taking seriously the analogy between the black hole
laws and thermodynamics and searching for an underlying statistical
explanation of this thermodynamic behavior. More importantly, the thermal nature of the radiation
has led to one of the biggest recent conundrums in theoretical
physics: the information loss paradox \cite{Hawking:1976ra}.
Matter in a pure state may be thrown into black hole
but only thermal radiation comes out. So it would appear as if a pure state
has evolved into a mixed state,
thus violating unitarity. This issue has been debated vigorously over the
years and has been taken as an indication that well accepted physical
principles, such as unitarity or locality, may have to be abandoned,
see \cite{ILreviews} for reviews. As we will review later,
recent progress based on the AdS/CFT correspondence implies a
unitary evolution and the holographic nature of the correspondence
also implies that spacetime locality is only approximate.
Black holes have a curvature singularity hidden behind their
horizon. Near these singularities Einstein gravity breaks down
and a fundamental question is how (and when) the quantum theory of
gravity resolves these singularities and what effect the
resolution has. When considering black holes
with macroscopic horizons, one might anticipate that semi-classical
computations, such as those implying Hawking radiation,
would be applicable. This conclusion has also been challenged in the
literature, see \cite{tHooft,Giddings:2006sj,Mazur:2001fv,Corley:1996ar} for
a (small) sample of works in this direction.
Here recent developments also offers a new perspective: non-singular
spacetimes would generically differ from the black hole
background up to the horizon scale, rather than only in the
neighborhood of the singularity.
Over the last 15 years great progress has been achieved in
string theory in addressing black hole issues, as we now briefly
review. Firstly, for
a class of supersymmetric black holes the black hole entropy
was understood using D-branes. The basic idea is simple:
supersymmetric states (generically) exist for all values
of the parameters of the underlying theory.
Changing the gravitational strength one can interpolate
between the description of the system as a black hole
and the description in terms of bound states
of D-branes. Thus one can compute the degeneracy in the
D-brane description and thanks to supersymmetry extrapolate this
result to the black hole regime. Starting from \cite{Strominger:1996sh}
such computations were done for a class
of black holes and exact agreement was found with the Bekenstein-Hawking
entropy formula. In more recent times, the agreement was extended to subleading
orders where on the gravitational side one takes into account the effect
of higher derivative terms and on the D-brane side one computes subleading
terms in the large charge limit of the degeneracy formulas, see \cite{Sen:2007qy} for a
review. These developments show that the gravitational entropy indeed
has a statistical origin, but they do not directly address any of the
other black hole issues since the computations involve in an essential way
an extrapolation from weak to strong coupling.
Further progress was achieved with the advent of the AdS/CFT correspondence.
The asymptotically flat black holes under consideration have a
near-horizon region that contains an AdS factor, so AdS/CFT
is applicable. The black hole microstates
can then be understood as certain supersymmetric states of the dual CFT.
Since the AdS/CFT duality is (conjectured to be) an exact equivalence
and the boundary theory is unitary, the dynamics of the black hole
microstates is unitary. Although this in principle shows that
there is no information loss it still does not
explain what is the gravitational nature of the black hole microstates,
nor does it show where Hawking's original argument goes wrong.
The AdS/CFT correspondence however implies more: for every stable
state of the CFT, there should exist a corresponding regular
asymptotically AdS geometry that encodes in its asymptotics the vevs
of gauge invariant operators in that state. Thus, for every CFT
state that one counts to account for the black hole entropy there
should exist a corresponding asymptotically AdS geometry. These solutions
will generically be stringy in the interior although some will be well
described in the (super)gravity limit.
Now since these solutions have the same behavior near the AdS
boundary as the near-horizon limit of the original asymptotically
flat black hole, one can attach the asymptotically flat
region as in the original black hole. We thus find that associated with the original black hole
there is an exponential number of solutions that look like the black
hole up to the horizon scale but differ from it in the interior; the
interior region is replaced by the asymptotically AdS solutions just discussed and
there is one such solution per microstate. The
ones that have a good supergravity description everywhere
should not have horizons, since for those the semiclassical
arguments that relate horizons to entropy should be applicable and
each of these solutions should correspond to a pure state. This is the
fuzzball proposal for black holes, formulated in works of Mathur and
collaborators in \cite{Lunin:2001jy,Lunin:2002qf,Mathur:2002ie,Lunin:2002bj}.
The purpose of this report is to review and
make a critical appraisal of the fuzzball proposal. Other reviews on
this subject include
\cite{Mathur:2005zp,Mathur:2005ai,Mathur:2008wi,Bena:2007kg}.
In the next section we introduce the fuzzball proposal, discuss
its relation with the AdS/CFT correspondence, explaining more fully the argument in the preceding
paragraph, and sketch why this proposal would resolve the black hole puzzles.
In section 3 we explain how holography works. In particular, we discuss
how the vevs of gauge invariant operators are encoded in supergravity
solutions. Then in section 4 we discuss the best understood and
simplest example, namely
the two charge D1-D5 system where one can explicitly find all fuzzball
geometries (visible in supergravity)
and test the general arguments.
To make further progress with the
fuzzball proposal, one would like to understand black holes which have
macroscopic horizons, such as the 3-charge D1-D5-P system.
Section 5 contains a discussion of what is
known about the 3-charge system, in particular the candidate fuzzball
geometries which have been constructed. Much of the work in the
current literature has been focused on constructing explicit examples
of fuzzball geometries, with a view to reproducing the entropy of the
black hole. Many open questions remain as to how the fuzzball
proposal would address black hole puzzles, and we discuss these issues
in section 6. Our emphasis throughout is on connecting the different
developments and emphasizing open problems and directions for further research
rather than reviewing exhaustively the literature.
\section{Generalities}
\subsection{What is the fuzzball proposal?}
{ Consider a black hole solution with associated gravitational entropy $S$.
According to the fuzzball proposal associated with this black hole there are
$\exp S$ {\it horizon-free non-singular} solutions that asymptotically look
like the black hole but generically differ from the black hole up to the
horizon scale. These solutions, the fuzzballs, are considered to be the
black hole microstates while the original black hole represents the
average description of the system.}
To complete this definition we should specify
which class of theories we consider and what precisely we
mean by ``solutions''. We will work within the framework of
string theory; black holes are then solutions of the
corresponding low energy effective action. Lower dimensional
black holes, such as for example the 4d Schwarzchild black hole,
are viewed as resulting from a corresponding 10d solution upon
compactification. Black hole solutions
typically involve only a very small number of the low energy fields,
e.g. only the metric for the Schwarzchild black hole, or the metric plus
a gauge field for the Reissner-Nordstr\"{o}m solution.
The corresponding fuzzball solutions would however in general
be solutions of the full theory, not just its low
energy approximation. Only a subset of fuzzballs would solve the
low energy field equations
and in general these solutions would involve all low energy fields, not
just the fields participating in the black hole solution.
In fact, as it will become clear later on, it is crucial that
the fuzzball solutions involve many other fields. This explains in part
why people have not stumbled upon such an exponential number
of regular solutions that resemble black holes.
Fuzzballs that involve string scale physics
would in general only have a sigma model description or, if
string field theory were adequately developed, they
would be non-singular solutions of the string field equations.
Some of these string solutions, however, may have an
extrapolation to low energies, i.e. there would exist a
corresponding supergravity solution but it would contain
small regions of high curvature. Furthermore, there would also
be cases where the differences between fuzzball supergravity solutions
are comparable to the corrections coming from
the leading higher derivative corrections to the string theory effective
action. In such cases these solutions will not be reliably distinguishable
within supergravity.
To properly define the dynamics of a system on a non-compact
spacetime we have to specify boundary conditions for
all fields. We will loosely refer to the boundary conditions
as non-normalizable modes and we will refer to the leading part of the asymptotics
which is affected by dynamics as normalizable modes.
All fuzzball solutions would share the same non-normalizable
modes with the original black hole spacetime but they would differ in
the normalizable modes. The precise notion of
normalizable and non-normalizable modes
depends on the asymptotics under consideration, and will be clear
in the specific examples of interest later.
\subsection{Fuzzballs and black hole puzzles}
The fuzzball proposal has the potential of resolving all black hole
related puzzles, although more work is required in order to demonstrate this
with sufficient precision.
Firstly, the black hole entropy becomes of a standard statistical origin:
there is a corresponding solution for every black hole microstate.
In other words, the entropy is related to the volume of the classical
phase space. The fact that the entropy grows like an area will be
seen to be a direct consequence of holography, at least in the examples
where AdS/CFT is applicable.
Since the geometries have no horizons, there is no
information loss either. Matter coming from infinity
will escape back to infinity at late times. A typical
fuzzball geometry is expected to look like the black hole
asymptotically, but it would differ from it up to
the horizon scale (although this has not been demonstrated
to date for solutions with macroscopic non-extremal horizons).
One might anticipate that this difference in the ``inner horizon
region'' is responsible for obtaining a different answer than
in Hawking's original computation.
Boundary conditions and regularity in the interior are
expected to fix the fuzzball solutions and so the resolution of the
black hole singularity is already built into this proposal.
Notice also that the ``size'' of the fuzzball is determined
dynamically from these requirements. We will discuss these issues
in more detail in section \ref{open},
after reviewing the current results and literature on the fuzzball
proposal.
\subsection{AdS/CFT and the fuzzball proposal}
The black holes whose entropy we best understand microscopically
have an near-horizon region that contains an AdS factor.
For these black holes one can use the AdS/CFT correspondence and
for this reason the general discussion in the previous section
can be made much more precise.
Let us consider for example the 3-charge D1-D5-P system which was the
first to be understood quantitatively. The near-horizon region is
$AdS_3 \times S^3 \times X_4$, with $X_4$ either $T^4$ or K3 (more properly,
the near-horizon region is $BTZ \times S^3 \times X_4$ \cite{skenderis_sfetsos}) .
The entropy of this system was originally computed in \cite{Strominger:1996sh}
by finding the degeneracy of the D1-D5-P bound states at weak coupling and then
extrapolating the result to the black hole phase. It was later realized that
this computation is part of the AdS/CFT duality: what one counts is
the degeneracy of
certain supersymmetric states of the dual CFT.
Using gravity/gauge theory duality however one can say more. Given a
state in (a deformation of) the CFT, the duality implies that there
is a corresponding asymptotically AdS spacetime with non-trivial
matter fields capturing the parameters of deformation and the vevs
of gauge invariant operators in the given state. The detailed
correspondence will be discussed in the next section, but for the
current argument one only needs the existence of such a
correspondence. Now consider one of the supersymmetric states
counted in accounting for the entropy of, say, the Strominger-Vafa
black hole. Associated with this state there should exist a regular
asymptotically AdS solution. We thus arrive at the conclusion that
there should exist $\exp S$ regular solutions which asymptotically
look like the near-horizon region of the original black hole. These
solutions share the same non-normalizable modes as the near-horizon
limit of the original black hole solution, since we are considering
states and not deformations of the CFT, but differ in their
normalizable modes, which capture the non-trivial vevs in the state
under consideration. One can now attach back\footnote{As emphasized
the fuzzball solutions and near-horizon region of the black hole
have the same leading behavior near the AdS boundary (since they
have the same non-normalizable modes) so the gluing is the same in
both cases.} the asymptotically flat region to arrive at the
conclusion that there should exist $\exp S$ regular solutions that
look like the original black hole up to the horizon scale but differ
in the interior; the interior has been replaced by the asymptotically
AdS solution corresponding to each state. These are the fuzzball
solutions. The place where each solution starts to
differ from the black hole is controlled by the vev of the lowest
dimension operator in this state and solutions corresponding to
different states are distinguished by the vevs of higher
dimension operators \cite{Skenderis:2006ah, Kanitscheider:2006zf,
Kanitscheider:2007wq}.
What this argument emphatically does {\it not} imply is that the
solutions would be supergravity solutions and indeed the majority of
fuzzball solutions will not be, although some will be well
described by supergravity.
For states where operators dual to supergravity fields acquire large
vevs the solution will differ appreciably from the black hole
solution already at the supergravity level (but still look like the
black hole at the asymptotically flat infinity). For such solutions
with everywhere small curvatures, one would anticipate that standard
treatments that associate entropy with horizons would be valid.
Since each of these solutions is meant to correspond to a pure
state, the corresponding geometry should therefore be horizonless.
Much of the current fuzzball literature has focused on finding such
supergravity solutions.
On the other hand there would be many cases/states where none
of the operators dual to supergravity fields acquire a vev, or the vev is of string
scale: the corresponding solutions will then agree with the original solution up to
the string scale. One would not expect to
find fuzzball solutions representing these states in supergravity, and
indeed we will see this behavior exemplified in the 2-charge system in section
4.
There will also be cases where a large fraction of the microstates
of the original black hole have large vevs of operators dual to
supergravity fields (chiral primaries) but these vevs differ from
each other very little. Such states should not be distinguishable in
supergravity. One might find supergravity fuzzball solutions
corresponding to these states, which cannot be reliably
distinguished, as the differences between them are of the same order
as the corrections due to leading higher derivative terms. Then the
relevant fuzzball solutions have an extrapolation to the
supergravity regime, but one cannot really trust the distinctions
between similar solutions. Again we will see this behavior occurring
in the 2-charge system, and on general grounds this must persist to
other black holes with macroscopic horizons. There is also the
possibility that states sharing the same vevs of chiral primaries
and differing in the vevs of string states are best described in the
low energy regime by solutions with ``hair'' capturing the vevs of the
chiral primaries, which have either singularities or a horizon, with
area smaller than the horizon area of the original black hole. The latter
indicate that these states can not be distinguished in supergravity.
For instance, one could relate the black ring solutions to corresponding
black holes with the same charges in this way.
Clearly the low energy approximation will not suffice to describe such
cases, and to make progress one will need to work with backgrounds of the full string
theory. In particular, to even define the properties of a generic fuzzball,
one will need to address the question of the
definition of entropy outside the geometric regime: when does a string
theory background have entropy and how is this defined given the
worldsheet theory?
To date, one has relied on the
geometric definition of entropy, with the entropy being associated
with horizons. In
the supergravity limit, one uses the Bekenstein-Hawking entropy of the
horizon, with the generalization by Wald \cite{Wald2} being used when
working perturbatively with higher derivative corrections. This
geometric definition breaks down when dealing with string scale
solutions, be they fuzzballs or black holes, and needs to be
replaced.
Note that this issue can also not be circumvented in the case of
so-called small black holes which do not have horizons in
supergravity. In recent work (reviewed in \cite{Sen:2007qy}) such black holes have been treated
by evaluating the leading corrections to supergravity on (singular)
supergravity solutions, assuming that the corrected solution has a
horizon, and then computing the Wald entropy. This approach is a
priori unjustified, as the neglected higher corrections are
not small on a string scale solution, and it works unreasonably
well at reproducing the entropy of the dual CFT microstates.
Such small black holes should properly be described by a background of
the full string theory, in which the entropy would need to be defined
from the worldsheet theory as above.
Given the state of current technology, most work on the fuzzball
proposal has so far been in the context of supergravity solutions, as
will be reviewed in sections 4 and 5. The limitations of
the supergravity approximation will however be a recurrent theme
throughout, and we will return to discuss the need to
go beyond supergravity in the final sections. Next however we will
review the evidence for the fuzzball proposal obtained from finding
and analyzing fuzzball supergravity solutions.
One should note here that the fuzzball proposal has been developed
most in the context of asymptotically flat black holes in four and five
dimensions, for which the near horizon region contains $AdS_2$ or
$AdS_3$ factors. It is fuzzball solutions for these black holes which
will be discussed in the following sections. Clearly the proposal is
more generally applicable, and one could hope to make comparable
progress at finding fuzzball solutions
with other supersymmetric black hole systems. In particular,
asymptotically $AdS_5$ black holes are a natural system to explore, as
these fall into the best understood AdS/CFT duality, that with ${\cal
N} = 4$ SYM in four dimensions.
Whilst one could carry out a detailed parallel fuzzball discussion for such black
holes, we will not explore this case here for several reasons. First
of all, it is natural to understand first black holes which are
asymptotically flat, and thus closer to astrophysical black
holes. Secondly, it turns out that the $AdS_5$ case is technically
harder than the cases we discuss. The LLM bubbling solutions
\cite{Lin:2004nb} describe all $1/2$ BPS states of the
system (visible in supergravity) but the corresponding ``black hole''
does not have a macroscopic horizon in supergravity \cite{Myers:2001aq}, so is not a good
test case for the fuzzball proposal.
In the asymptotically $AdS_5$ case it seems that one needs to break
the supersymmetry to $1/16$ to
obtain a black hole with macroscopic horizon area in
supergravity, see for example \cite{Gutowski:2004yv}.
With so little supersymmetry, even the counting of
the black hole microstates is rather more subtle, as the degeneracy
depends on the coupling, and the black hole entropy has not yet been
reproduced from field theory. Moreover, finding explicit fuzzball
solutions with
so little supersymmetry is likely to be hard. Thus, whilst the
fuzzball proposal should be applicable for this system, and indeed many
other interesting black hole systems, we will focus on
the better understood case of asymptotically flat black holes.
One should note that in using AdS/CFT arguments to support the
fuzzball proposal there may be certain additional subtleties that we
have not yet mentioned. Firstly, one should be slightly more careful
in relating geometries to states: the fuzzball solutions for the D1-D5 system
should be in correspondence with states in
the Higgs branch of a $(1+1)$-dimensional theory. Due to the strong infrared fluctuations
in 1+1 dimensions one should consider wavefunctions
that spread over the whole of Higgs branch
rather than continuous moduli spaces of the quantum states.
So more properly one should view the fuzzball solutions as dual to
wavefunctions on the Higgs branch. These wavefunctions, however, may be localized
around specific regions in the large $N$ limit and indeed this issue
does not seem to play a key role in any of the subsequent discussions.
Secondly, in AdS/CFT one can have multiple saddle points of the bulk
action, with the same boundary conditions, the most well known example
being thermal AdS and the Schwarzschild black hole. In this case the
onshell (renormalized) action determines the thermodynamically
preferred solution, and there is a Hawking-Page transition between the
two phases at a critical temperature. One
might wonder whether such multiple saddle points could complicate the
relationship between a given CFT microstate and a corresponding
asymptotically AdS string background. In the current context however where we
specify the state, this information determines the vevs of chiral primaries, so the boundary conditions
include both the source and vev part of the solution. As we will review
in the next section this data determines a point in the phase space of the
gravitational theory and thus there should be a unique regular
solution (if such a solution exists), see also \cite{math} and
references therein, for the corresponding discussion in
the mathematics literature on hyperbolic manifolds.
Nonetheless one should bear in mind both caveats as an
issue to be addressed in future when sharpening the definition of the fuzzball
proposal in the string regime.
\section{Holographic methods}
In this section we will summarize the status of holographic methods,
with the emphasis being on summarizing results rather on derivations.
The aim is to provide a handbook of holographic formulae and
associated prescriptions that can be readily used without having
to delve into their derivation.
The basic principles of holography were laid out in the original
papers \cite{Maldacena:1997re,Gubser:1998bc,Witten:1998qj}. In particular, the
duality maps the spectrum of string theory on asymptotically $AdS
\times X$, where $X$ is a compact space, to the spectrum of gauge
invariant operators of the dual QFT, and the string theory partition
function, which is a function of boundary conditions posed on the
conformal boundary of the spacetime, to the generating functional of
correlation functions of the dual QFT, with the fields parameterizing
the boundary conditions mapped to sources of the dual operators.
The duality relation in full generality is still very difficult to probe,
so different approximations have been developed over the years, e.g.
the low energy limit, the limit of long operators, the plane wave limit etc.
In this report we focus on the low energy limit, where string theory is
well approximated by supergravity. This limit typically corresponds
to a strong coupling limit of the boundary theory. We will further
consider the leading saddle point approximation of the bulk
path integral, where the (logarithm of the) bulk partition function
becomes equal to the on-shell supergravity action. In other words,
we suppress supergravity loops. This typically corresponds to the
large N limit in the dual theory. Within these approximations
the gravity/gauge theory duality equates the supergravity on-shell
action to the generating functional of connected QFT correlators
at strong coupling and large $N$.
Let us discuss first the case where the bulk solution is exactly
$AdS_{d+1} \times X_q$, e.g. $AdS_5 \times S^5$ or $AdS_3 \times S^3
\times X_4$. In such cases the dual theory is a $d$-dimensional
conformal field theory ($CFT_d$) and there is a one to one
correspondence between the supergravity KK spectrum and primary
operators of the dual CFT. One can use the duality to compute
correlation functions of primary operators at strong coupling and
large $N$. Conformal field theories have vanishing 1-point
functions, so the first non-trivial computation is that of a 2-point
function and the latter can be computed holographically by solving
the linearized fluctuation equation around $AdS_p \times X_q$ with
prescribed boundary conditions at the conformal boundary of $AdS_p$.
Higher $n$-point functions can be obtained by solving the
$(n{-}1)$-th order fluctuation equations.
Solutions that are asymptotically $AdS_{d+1} \times X_q$ describe
either deformations of the $CFT_d$ or the CFT in a non-trivial
state. In such cases the most elementary question is what is
the deformation parameter and/or the state. The state of the CFT
is uniquely specified if one knows the expectation values of
all gauge invariant operators in that state. Within the supergravity
approximation we only have access to primary operators: one can
reliably describe only deformations of the original CFT
by primary operators and one can only compute the vevs of primary operators.
The latter gives partial information about the state, but this information
should be enough to specify the state within the approximations used
(strong coupling, large $N$).
The parameters of deformation and the vevs of primary operators can be
extracted by means of {\it algebraic manipulations only} from the asymptotic
expansion of the supergravity solution. The solutions we discuss
in this report describe states rather than deformations, so we will not
discuss any further the case of deformations. In the next two subsections
we will discuss the issues involved in obtaining the vevs (1-point functions)
of primary operators and describe how these are resolved leading to
general formulae for the 1-point functions. Afterwards we
focus on the case of interest, namely asymptotically $AdS_3 \times S^3$
solutions and present explicit formulae for the vevs of operators
up to dimension 2. Higher point functions can be obtained
by solving the fluctuation equations around the original
solution \cite{Bianchi:2001de,Bianchi:2001kw,Skenderis:2002wp,
Papadimitriou:2004rz} but this will not be reviewed here.
\subsection{Holographic renormalization}
The first issue one needs to address when attempting to carry out
holographic computations is that the on-shell action diverges,
essentially due to the infinite volume of spacetime. This issue is
dealt with by the formalism of holographic renormalization
\cite{Henningson:1998gx,Balasubramanian:1999re,deHaro:2000xn,
Skenderis:2000in,Bianchi:2001de,Bianchi:2001kw,Papadimitriou:2004ap,
Papadimitriou:2004rz}; for a review see \cite{Skenderis:2002wp}, and
amounts to adding local boundary covariant counterterms to cancel
the infinities. Actually the local boundary counterterms are
required, irrespectively of the issue of finiteness, by the more
fundamental requirement of the appropriate
variational problem being well posed \cite{Papadimitriou:2005ii}. As is well known
the conformal boundary of asymptotically $AdS$ spacetimes have a
well-defined conformal class of metrics rather than an induced
metric. This means that the appropriate variational problem involves
keeping fixed a conformal class and not an induced metric as in the
usual Dirichlet problem for gravity in a spacetime with a boundary.
The new variational problem requires the addition of further
boundary terms, on top of the Gibbons-Hawking term, which turn out
to be precisely the boundary counterterms, see
\cite{Papadimitriou:2005ii} for the details and a discussion of the
subtleties related to conformal anomalies.
The subject of holographic renormalization is extensively discussed
in the literature, so we will only highlight a few points and
introduce the notation to be used in later sections. The points that
should be emphasized are:
\begin{itemize}
\item To obtain renormalized correlators, the main object of interest
is the radial canonical momentum, rather than the on-shell action.
\item The source and renormalized 1-point function are a conjugate
pair and can be considered as coordinates in the phase space of the
gravitational theory.
\end{itemize}
Let us briefly discuss these points. Firstly, the asymptotic form of
bulk fields, specialized to the $D=3$ case of interest, is
\begin{eqnarray}
ds_{3}^{2} &=& \frac{dz^2}{z^2} + \frac{1}{z^2} \left(g_{(0)uv} + z^2
\left(g_{(2)uv} + {\rm{log}}(z^2) h_{(2) uv} + ({\rm{log}}(z^2))^2
\tilde{h}_{(2) uv}\right) + \cdots\right) dx^{u} dx^v \nonumber} \def\bd{\begin{document}} \def\ed{\end{document} \\
\Psi^1 &=& z ({\rm{log}}(z^2) \Psi^1_{(0)}(x) + \tilde{\Psi}^1_{(0)}(x) +
\cdots ); \label{as_exp} \\
\Psi^{k} &=& z^{2-k} \Psi^{k}_{(0)}(x) + \cdots + z^{k}
\Psi^{k}_{(2k-2)}(x) + \cdots, \hspace{0.5cm} k \neq 1. \nonumber} \def\bd{\begin{document}} \def\ed{\end{document}
\end{eqnarray}
In these expressions $(g_{(0)uv}, \Psi^1_{(0)}(x), \Psi^{k}_{(0)}(x))$
are sources for the stress energy tensor and scalar operators of dimension
one and $k$ respectively; as usual one must treat separately the operators of
dimension $\Delta = d/2$, where $d$ is the dimension of the boundary. Note that
the 2-dimensional boundary coordinates are labeled by $(u,v)$. We will
also use the notation $[\psi]_n$ to denote the coefficient of the $z^n$
term in the asymptotic expansion of the field $\psi$.
Correlation functions can be computed using the basic holographic
dictionary that relates the on-shell gravitational action to the
generating functional of correlators. The first variation can be
done in all generality \cite{deHaro:2000xn} yielding a relation
between the 1-point function (in the presence of sources, so higher
point functions can be obtained by further functional differentiation
w.r.t. sources) and non-linear combinations of the asymptotic
coefficients in (\ref{as_exp}). The underlying structure of the
correlators is best exhibited in the radial Hamiltonian formalism,
which is a Hamiltonian formulation with the radius playing the role
of time. The Hamilton-Jacobi theory, introduced in this context in
\cite{de Boer:1999xf}, then relates the variation of the on-shell
action w.r.t. boundary conditions, thus the holographic 1-point
functions, to radial canonical momenta. It follows that that one can
bypass the on-shell action and directly compute renormalized
correlators using radial canonical momenta $\pi$
\cite{Papadimitriou:2004ap,Papadimitriou:2004rz}.
A fundamental property of asymptotically (locally) AdS spacetimes is
that dilations are part of their asymptotic symmetries. This implies
that all covariant quantities can be decomposed into a sum of terms
each of which has definite scaling. For example, the radial
canonical momentum $\pi^k$ of scalar $\Psi^k$ has the expansion
\begin{equation}\label{momentum_exp}
\pi^k=\pi_{(2-k)}^k+ \pi_{(1-k)}^k + \cdots
+ \pi_{(k)}^k+\tilde{\pi}_{(k)}^k\ \log z^2 +\cdots, \end{equation}
where the
various terms have the dilation weight indicated by the
subscript\footnote{Note that that the coefficient $\pi_{(k)}^k$ has
an anomalous scaling transformation, $\d \pi_{(k)}^k = -k
\pi_{(k)}^k -2 \tilde{\pi}_{(k)}^k$, due to conformal anomalies,
see \cite{Papadimitriou:2004ap}.}, i.e. $\pi_{{n}}^k$ has scaling
weight $n$. These coefficients are in one to one correspondence with the
asymptotic coefficients in (\ref{as_exp}) with the exact relation
being in general non-linear. The advantage of working with dilation
eigenvalues rather than with asymptotic coefficients is that the
former are manifestly covariant while the latter in general are not:
the asymptotic expansion (\ref{as_exp}) singles out one coordinate
so it is not covariant. Holographic 1-point functions can be
expressed most compactly in terms of eigenfunctions of the dilation
operator, and this explains the non-linearities found in explicit
computations of 1-point functions. In particular, if one considers
the case of a single scalar $\Psi^k$ in an asymptotically (locally)
AdS spacetime, the 1-point function is simply given by
\begin{equation}
\< O^k \> = \pi_{(k)}^k
\end{equation}
Thus, we indeed see that source and the
renormalized 1-point are conjugate variables and one may consider
them as coordinates in the classical phase space of the theory.
\subsection{Kaluza-Klein holography}
The discussion in the previous section involved
a $(d{+}1)$ dimensional supergravity theory which admits an $AdS_{d+1}$ solution.
However the string theory backgrounds of interest
which contain an $AdS$ factor typically
also involve a compact space, for example, $AdS_5 \times S^5$,
$AdS_3 \times S^3 \times X_4$ etc. On general grounds, one expects
that there is an effective $(d+1)$ dimensional description. Thus provided one can
obtain such a description one can use holographic renormalization
to obtain the 1-point functions of gauge invariant operators. The
method of Kaluza-Klein holography provides an explicit algorithm
for constructing the corresponding $(d+1)$ dimensional action
and extracting the vevs.
Note that generically the spheres appearing in these solutions have
a radius which is of the same order as the $AdS$ radius, so
the higher KK modes are not suppressed relative to the zero modes
and one cannot ignore them. In some cases it is
nevertheless possible to only keep a subset of modes
because the equations of motion admit a solution with all modes,
except the ones kept, set equal to zero, i.e.
there exists a ``consistent truncation''.
The existence of such a truncation signifies the existence of a
subset of operators of the dual theory that are closed under OPEs.
The resulting theory is a $(d+1)$-dimensional gauged supergravity and such
gauged supergravity theories have been the starting point for
many investigations in AdS/CFT.
However, starting from
a lower dimensional gauged supergravity is unsatisfactory on many grounds.
Firstly, gauged supergravity captures the physics of only a very small
subset of operators, typically that of the stress tensor supermultiplet.
An infinite number of other operators, namely the ones dual to KK
modes, are in principle accessible within the low energy limit,
but one excludes them a priori.
Secondly, the higher dimensional solutions are more fundamental
as reduction of regular solutions may result in singular solutions.
Thirdly, even in the case where a consistent truncation to a lower
dimensional supergravity is possible, it is often very difficult or
indeed unknown how to express known interesting higher dimensional
solutions in a non-linear reduction ansatz that produces
a corresponding solution of the gauged supergravity. Finally, a key conceptual
question for holography is how the compact part of the
geometry is encoded in QFT data and answering this question requires
keeping all modes.
For these reasons we will keep all KK modes in the reduction
(so there is also never an issue of consistency). Naively, this
results in an intractable problem of an infinite number of fields
all coupled together. Recall however that the 1-point functions are
extracted from the asymptotic expansion near the AdS boundary,
and the fall off of the fields near the AdS boundary is fixed by their
mass. This implies that to compute the
holographic 1-point function of any given operator only a finite number
of fields and a finite number of interactions are relevant. The method of
Kaluza-Klein holography systematically constructs the lower dimensional
action in a way that only the fields and interactions needed are kept
at each step.
The steps involved in this construction are the following. The starting
point is a $D$ dimensional action that admits an $AdS_{d+1} \times X^q$
solution. We further assume that the harmonic analysis on the compact
$X^q$ is known, i.e. the set of spherical harmonics is known. This clearly
is the case for $X^q= S^q$. In the first step we consider fluctuations
around this solution and expand the fluctuations in the harmonics of the
compact space. Let $\phi$ denote collectively all fields,
$\phi_o$ be the $AdS_{d+1} \times X^q$ solution and
$\delta \phi$ the fluctuation, then schematically,
\begin{eqnarray}
\phi(x,y)&=& \phi_o(x,y) + \d \phi(x,y) \nonumber \\
\d \phi(x,y) &=& \sum_I \psi^I(x) Y^I(y) \label{har_exp}
\end{eqnarray}
where $x$ is a coordinate in the $(d+1)$ non-compact directions, $y$
is a coordinate in the compact directions and $Y^I$ denotes collectively
all spherical harmonics (scalar, vector, tensor and their covariant
derivatives). Precise formulae for the case of interest will be
presented in the next subsection.
The decomposition (\ref{har_exp}) is not unique because there are
coordinate transformations,
\begin{equation} \label{trns}
X^M{}' = X^M - \xi^M(x,y)
\end{equation}
where $X^M=\{x,y\}$ that transform the fluctuations $\psi^I$ to each over
or the background solution $\phi_o$. In the supergravity literature on
KK reduction one often imposes a gauge condition, most notably the
de Donder gauge condition, to eliminate this issue. We instead
construct gauge invariant combination that have the property
that in the de Donder gauge they coincide with gauge fixed variables.
This is done by working out
perturbatively in the number of fluctuations how each $\psi^I$
transformations under (\ref{trns}) and then one constructs combinations
$\hat{\psi}^I$ that transform as tensors, i.e. scalar
combinations are invariant under (\ref{trns}), $(d{+}1)$-dimensional
vector combinations $A_\mu(x)$, transform
as vectors, the metric $g_{\m {\sst{(n)}}}$ transforms as metric etc.
Details of this procedure can be found in \cite{Skenderis:2006uy}.
The aim is now to derive the equations that the gauge invariant modes
$\hat{\psi}^I$ satisfy by substituting (\ref{har_exp}) in the $D$ dimensional
equations and working perturbatively in the number of fields. To linear
order one obtains the spectrum, to quadratic order the cubic interactions
etc. Note however that not all cubic (or higher) interactions are relevant
for the computation of the 1-point function of any given operator. Only the
ones that could modify the asymptotic coefficients that determine the
vev need to be retained.
Expanding perturbatively in fluctuations one finds
\begin{equation}
{\cal L}_{\cal I} \hat{\psi}^{\cal I} = {\cal L}_{ {\cal I J K} }
\hat{\psi}^{\cal J} \hat{\psi}^{\cal K}
+ {\cal L}_{ {\cal I J K L}}
\hat{\psi}^{\cal J} \hat{\psi}^{\cal K} \hat{\psi}^{\cal L} + \cdots,
\end{equation}
where $ {\cal L}_{{\cal I}_1 \cdots {\cal I}_n}$ is an appropriate differential
operator. $ {\cal L}_{{\cal I}_1 \cdots {\cal I}_n}$ involves higher
derivative terms and the set of field equations cannot generically be
integrated into an action. However, one can always define $(d+1)$-dimensional
fields $\Psi^{\cal I}$ by a non-linear Kaluza-Klein reduction map of
the fields $\psi^{\cal I}$:
\begin{equation} \label{KKmap}
\Psi^{\cal I} = \hat{\psi}^{\cal I} + {\cal K}^{I}_{ {\cal J K } }
\hat{\psi}^{\cal J} \hat{\psi}^{\cal K} + \cdots,
\end{equation}
where ${\cal K}^I_{\cal J K}$ contains appropriate derivatives. The
reduction map is such that the fields $\Psi^{\cal I}$
do satisfy field equations which can be integrated into an action.
Given this $(d+1)$-dimensional action, it is then straightforward to obtain the
one point functions of operators in terms of the asymptotics of the
fields $\Psi^{\cal I}$, using the well-developed techniques of
holographic renormalization.
This results in the following general formula for an operator
of dimension $k$
\begin{equation} \label{Ovev}
\< O_k^I \> = \pi_{(k)}^I
+ \sum_{JK} a^I_{JK} \pi^J_{(k_1)} \pi^K_{(k-k_1)} + \cdots
\end{equation}
where $a^I_{JK}$ are numerical constants and the ellipses indicate
higher powers of canonical momenta.
The non-linear terms are related to extremal correlators, i.e.
these terms are possible when the theory contains operators
with dimension $k_i$, such that $\sum k_i =k$, and the
numerical constants $a^I_{JK}$ are related to the extremal
3-point functions at the conformal point. Note that these
formulae for the vevs apply to any solution of the same
action. Given any solution one can evaluate them to find the
QFT data encoded by this solution.
To summarize, the radial canonical momenta are related
(in general non-linearly) to the coefficients in the asymptotic expansion
of the $(d+1)$-dimensional fields $\Psi$. These fields in turn are related
to the $D$-dimensional coefficients $\psi$ via the non-linear Kaluza-Klein
map (\ref{KKmap}) which relates $\Psi$ to the gauge invariant version
$\hat{\psi}$ of $\psi$, which itself is a non-linearly related to $\psi$'s.
One can now combine all these maps to produce a final formula
for the vevs which is of the schematic form
\begin{equation} \label{Ovevf}
\< O_k^I (\vec{x})\> = [\psi^I(\vec{x}) ]_k
+ \sum_{JK} b^I_{JK} [\psi^J(\vec{x})]_{k_1} [\psi^K(\vec{x})]_{(k-k_1)}
+ \cdots
\end{equation}
where $b^I_{JK}$ are numerical coefficients,
$\vec{x}$ are $d$-dimensional (boundary) coordinates, $z$ is the
Fefferman-Graham radial coordinates and the coefficients $[\psi^I]_k$
are asymptotic coefficients in
\begin{equation} \label{dphi}
\d \phi(z,\vec{x},y) = \sum_{I,m} [\psi^I(\vec{x})]_m z^m Y^I(y)
\end{equation}
Thus, if we are interested in extracting the vevs of gauge invariant
operators from a given solution that is asymptotically
$AdS_{d+1} \times X^q$, the procedure is to write it as the deviation
from $AdS_{d+1} \times X^q$ in the form (\ref{dphi}), extract the
coefficients $ [\psi^I(\vec{x})]_m$ and insert those in (\ref{Ovevf}).
This procedure was carried out for asymptotically $AdS_5 \times S^5$
solutions describing the Coulomb branch of ${\cal N}=4$ SYM in
\cite{Skenderis:2006uy,Skenderis:2006di} and for the LLM solutions
in \cite{Skenderis:2007yb}, resulting in strong tests of gravity/gauge
theory duality away from the conformal point
and for asymptotically $AdS_3 \times S^3$ solutions relevant for the
fuzzball program in \cite{Kanitscheider:2006zf,Kanitscheider:2007wq}.
In the next section we present the detailed results for the case
of interest.
\subsection{Asymptotically $AdS_3 \times S^3$ solutions}
In what follows we will be interested in black hole and fuzzball
solutions whose decoupling regions are asymptotic to
$AdS_3 \times S^3 \times X_4$, where $X_4$ is $T^4$ or $K3$. Therefore
one can use AdS/CFT methods to extract holographic data from these
geometries and, in particular, the asymptotics of the six-dimensional
solutions near the $AdS_3 \times S^3$ boundary encode the vevs of
chiral primary operators in the dual field theory.
In the solutions of interest only zero modes of the compact space
$X_4$ are excited, so it
is convenient to first compactify the solution over $X_4$. Solutions of
type IIB supergravity compactified on $K3$
give rise to solutions of
$d=6$, $N=4b$ supergravity coupled to 21 tensor multiplets,
constructed by Romans in \cite{Romans:1986er}. Corresponding fuzzball solutions
of type IIB on $T^4$ can also be expressed as solutions of $d=6$,
$N=4b$ coupled to 5 tensor multiplets. These theories admit an
$AdS_3 \times S^3$ solution, so the program of Kaluza-Klein holography
can be applied to obtain an effective three dimensional description
of all relevant KK modes.
The bosonic field content of the $d=6$ $N =4b$ theory with $n_t$
tensor multiplets is the
graviton $g_{MN}$, 5 self-dual and $n_t$ anti-self dual tensor fields
and an $O(5,n_t)$ matrix of scalars ${\cal M}$ which can be written in terms
of a vielbein ${\cal M}^{-1} = V^T V$.
Following the notation of \cite{Sez98}
the bosonic field equations may be written as
\begin{eqnarray} \label{sugraIIBK3}
R_{MN} &=& 2 P^{nr}_M P^{nr}_N + H^n_{MPQ} {H^n_N}^{PQ}
+ H^r_{MPQ} {H^r_N}^{PQ}, \nonumber \\
\na^M P_M^{nr} &=& Q^{M nm} P_M^{mr} + Q^{M rs} P_M^{ns}
+ \frac{\sqrt{2}}{3} H^{n MNP} H^r_{MNP},
\end{eqnarray}
along with Hodge duality conditions on the 3-forms
\begin{equation}
\label{sugraIIBK3_hd}
\ast_6 H^n_3 = H^n_3, \qquad \ast_6 H^r_3 = -H^r_3,
\end{equation}
In these equations $(m,n)$ are $SO(5)$ vector indices running from
1 to 5 whilst $(r,s)$ are
$SO(n_t)$ vector indices running from 6 to $(5+n_t)$.
The 3-form field strengths are given by
\begin{equation}
\label{G3forms}
H^{n} = G^{A} V_{A}^n; \hspace{0.5cm}
H^{r} = G^{A} V_{A}^r,
\end{equation}
where $A \equiv \{n,r\} = 1, \cdots, (5+n_t)$; $G^{A} = db^A$ are closed and the vielbein on the
coset space $SO(5,n_t)/(SO(5) \times SO(n_t))$ satisfies
\begin{equation}
V^T \eta V = \eta, \qquad V = \left(\begin{array}{c} {V^n}_A \\
{V^r}_A \end{array}\right), \qquad \eta = \left(\begin{array}{cc}
I_5 & 0 \\ 0 & -I_{21} \end{array}\right).
\end{equation}
The associated connection is
\begin{equation}
\label{IIBK3conn}
d V V^{-1} = \left ( \begin{array} {c c} Q^{mn} & \sqrt{2} P^{ms} \\
\sqrt{2} P^{rn} & Q^{rs} \end{array} \right ),
\end{equation}
where $Q^{mn}$ and $Q^{rs}$ are antisymmetric and the off-diagonal
block matrices $P^{ms}$ and $P^{rn}$ are transposed to each other.
$Q^{mn}$ and $Q^{rs}$ are composite $SO(5)$ and $SO(n_t)$ connections
which are solvable in terms of $5 n_t$ physical scalars $\phi^{m r}$
via the Cartan-Maurer equation.
The six-dimensional field equations (\ref{sugraIIBK3})
admit an $AdS_3 \times S^3$ solution, such that
\begin{eqnarray}
ds_6^2 &=& \sqrt{Q_1 Q_5} \left ( \frac{1}{z^2} (-dt^2 + dy^2 + dz^2)
+ d\Omega_3^2 \right ); \label{background} \\
G^{5} &=& H^{5} \equiv G^{o5} = \sqrt{Q_1 Q_5} (\frac{dz}{z^3} \wedge dt \wedge
dy + d\Omega_3), \nonumber} \def\bd{\begin{document}} \def\ed{\end{document}
\end{eqnarray}
with the vielbein being diagonal and all other three forms (both
self-dual and anti-self dual) vanishing. In what follows it is
convenient to absorb the curvature radius $\sqrt{Q_1 Q_5}$ into an
overall prefactor in the action, and work with the unit radius $AdS_3
\times S^3$. Perturbations of six-dimensional supergravity fields
relative to an $AdS_3 \times S^3$ background may be expressed as
\begin{eqnarray}
g_{MN} &=& g^{o}_{MN} + h_{MN}; \hspace{0.5cm}
G^{A} = G^{oA} + g^A; \\
V^{n}_{A} &=& \d^{n}_{A} + \f^{nr} \d_{A}^r + {\textstyle{1\over2}} \f^{nr} \f^{mr}
\d_{A}^m; \nonumber} \def\bd{\begin{document}} \def\ed{\end{document} \\
V^{r}_A &=& \d^{r}_{A} + \f^{nr} \d_{A}^n + {\textstyle{1\over2}} \f^{nr} \f^{ns}
\d_{A}^s. \nonumber} \def\bd{\begin{document}} \def\ed{\end{document}
\end{eqnarray}
These fluctuations can then be expanded in a basic of spherical harmonics as
follows:
\begin{eqnarray}
h_{\m {\sst{(n)}}} &=& \sum h_{\m{\sst{(n)}}}^I (x) Y^{I} (y), \label{flc1} \\
h_{\m a} &=& \sum (h_{\m}^{I_v} (x) Y_a^{I_v} (y) +
h_{(s)\m}^{I} (x) D_a Y^I (y) ), \nonumber} \def\bd{\begin{document}} \def\ed{\end{document} \\
h_{(a b)} &=& \sum (\rho^{I_t} (x) Y_{(ab)}^{I_t} (y) +
\rho_{(v)}^{I_v} (x) D_a Y_b^{I_v} (y) +
\rho_{(s)}^{I} (x) D_{(a} D_{b)} Y^{I} (y) ), \nonumber} \def\bd{\begin{document}} \def\ed{\end{document} \\
h^{a}_{a} &=& \sum \pi^{I} (x) Y^{I} (y), \nonumber} \def\bd{\begin{document}} \def\ed{\end{document} \\
g^{A}_{\m{\sst{(n)}} \r} &=& \sum 3 D_{[\m} b_{{\sst{(n)}} \r]}^{(A)I} (x) Y^{I} (y), \nonumber} \def\bd{\begin{document}} \def\ed{\end{document}
\\
g^{A}_{\m {\sst{(n)}} a} &=& \sum ( b_{\m {\sst{(n)}}}^{(A)I} (x) D_{a} Y^{I} (y) + 2
D_{[\m} Z_{{\sst{(n)}}]}^{(A) I_{v}} (x) Y_{a}^{I_v} (y)); \nonumber} \def\bd{\begin{document}} \def\ed{\end{document} \\
g^{A}_{\m a b} &=& \sum (D_{\m} U^{(A)I}(x) \epsilon_{abc} D^{c} Y^I(y) +
2 Z_{\m}^{(A) I_v} D_{[b} Y^{I_v}_{a]}); \nonumber} \def\bd{\begin{document}} \def\ed{\end{document} \\
g^{A}_{a b c} &=& \sum (- \epsilon_{abc} \L^{I} U^{(A)I}(x) Y^{I} (y)) ; \nonumber} \def\bd{\begin{document}} \def\ed{\end{document} \\
\phi^{mr} &=& \sum \phi^{(mr) I} (x) Y^{I} (y), \nonumber} \def\bd{\begin{document}} \def\ed{\end{document}
\end{eqnarray}
Here $(\mu, \nu)$ are AdS indices and $(a,b)$ are $S^3$ indices,
with $x$ denoting AdS coordinates and $y$ denoting sphere coordinates.
$\L^I$ is defined in (\ref{sph_Y}). The
subscript $(ab)$ denotes symmetrization of indices $a$ and $b$ with
the trace removed. Relevant properties of the spherical harmonics are
reviewed in appendix \ref{sphere}. We will often use a notation
where we replace the index $I$ by the degree of the harmonic $k$
or by a pair of indices
$(k,I)$ where $k$ is the degree of the harmonic and $I$ now parameterizes
their degeneracy, and similarly for $I_v, I_t$.
Imposing the de Donder gauge condition $D^{A} h_{aM} = 0$
on the metric fluctuations removes the fields with subscripts $(s,v)$.
In deriving the spectrum and computing correlation functions, this is
therefore a convenient choice. The de Donder gauge choice is however not
always a convenient choice for the asymptotic expansion of solutions;
indeed the natural coordinate choice in our applications takes us
outside de Donder gauge. As discussed in \cite{Skenderis:2006uy}
this issue is straightforwardly dealt with by
working with gauge invariant combinations of the fluctuations.
The linearized spectrum of the fluctuations was derived in
\cite{Sez98}, with the cubic interactions obtained in
\cite{Arutyunov:2000by}.
Let us briefly review the linearized spectrum
derived in \cite{Sez98}, focusing on fields dual to chiral
primaries. Consider first the scalars. It is
useful to introduce the following combinations
which diagonalize the linearized equations of motion:
\begin{eqnarray}
s^{(r) k}_{I} &=& \frac{1}{4(k+1)} ({\phi}^{(5r) k}_{I} +2 (k+2)
{U}^{(r)k}_{I}), \label{diageqm} \\
\s^k_{I} &=& \frac{1}{12 (k+1)} (6 (k+2) \hat{U}^{(5)k}_{I} -
\hat{\pi}_I^{k}), \nonumber} \def\bd{\begin{document}} \def\ed{\end{document}
\end{eqnarray}
The fields $s^{(r)k}$ and $\s^k$ correspond to scalar chiral
primaries, with the masses of the scalar fields being
\begin{equation} \label{masses}
m_{s^{(r)k}}^2 = m_{\s^k}^2 = k (k-2), \hspace{0.5cm}
\end{equation}
The index $r$ spans $6 \cdots 5 + n_t$ with $n_t = 5, 21$ respectively
for $T^4$ and $K3$.
Note also that $k \ge 1$ for $s^{(r)k}$; $k \ge 2$ for $\s^k$. The hats
$(\hat{U}^{(5)k}_{I}, \hat{\pi}_I^{k})$ denote the following. As
discussed in \cite{Skenderis:2006uy}, the equations of motion for
the gauge invariant fields are precisely the same as those in de
Donder gauge, provided one replaces all fields with the corresponding
gauge invariant field. The hat thus denotes the appropriate gauge
invariant field, which reduces to the de Donder gauge field
when one sets to zero all fields with subscripts $(s,v)$. For our
purposes we will need these gauge invariant quantities only to
leading order in the fluctuations, with the appropriate
combinations being
\begin{eqnarray}
\hat{\pi_2}^{I} &=& \pi_2^{I} + \Lambda^{2} \rho_{2(s)}^{I}; \label{gaug}
\\
\hat{U}^{(5)I}_2 &=& U^{(5)I}_2 - {\textstyle{1\over2}} \rho^{I}_{2(s)}; \nonumber} \def\bd{\begin{document}} \def\ed{\end{document} \\
\hat{h}^{0}_{\m{\sst{(n)}}} &=& h^{0}_{\m{\sst{(n)}}} - \sum_{\a,\pm} h_{\m}^{1 \pm \a}
h_{{\sst{(n)}}}^{1 \pm \a}. \nonumber} \def\bd{\begin{document}} \def\ed{\end{document}
\end{eqnarray}
Next consider the vector fields. It is useful to introduce
the following combinations which diagonalize the equations of motion:
\begin{equation}
h^{\pm}_{\m I_v} = {\textstyle{1\over2}} (C_{\m I_v}^{\pm} - A^{\pm}_{\m
I_v}), \hspace{0.5cm}
Z_{\m I_v}^{(5) \pm} = \pm {\textstyle{1\over 4}} (C_{\m I_v}^{\pm} + A_{\m I_v}^{\pm}).
\end{equation}
For general $k$ the equations of motion are Proca-Chern-Simons
equations which couple $(A_{\m}^{\pm}, C^{\pm}_{\m})$ via a first
order constraint \cite{Sez98}. The three dynamical fields at each degree $k$
have masses $(k-1, k+1, k+3)$, corresponding to dual operators of
dimensions $(k,k+2,k+4)$ respectively; the operators of dimension $k$
are vector chiral primaries. The lowest dimension operators
are the R symmetry currents, which couple to the $k=1$ $A^{\pm \a}_{\m }$
bulk fields. The latter satisfy the Chern-Simons equation
\begin{equation} \label{cs11}
F_{\m {\sst{(n)}}}(A^{\pm \a}) = 0,
\end{equation}
where $F_{\m{\sst{(n)}}}(A^{\pm \a})$ is the curvature of the connection and
the index $\a = 1,2,3$ is an $SU(2)$ adjoint index. We will here only
discuss the vevs of these vector chiral primaries.
Finally there is a tower of KK gravitons with $m^2 = k (k+2)$ but
only the massless graviton, dual to the stress energy tensor, will
play a role here. Note that it is the
combination $\hat{H}_{\m {\sst{(n)}}} = \hat{h}^{0}_{\m {\sst{(n)}}} + \pi^{0} g^{o}_{\m {\sst{(n)}}}$ which
satisfies the Einstein equation; moreover one needs the appropriate
gauge covariant combination $\hat{h}^0_{\m{\sst{(n)}}}$ given in (\ref{gaug}).
\subsection{Vevs of gauge invariant operators}
Let us now give the expressions for the vevs of gauge invariant operators
up to dimension two. These are expressed in terms of coefficients in
the asymptotic expansions of the fields near the conformal boundary at $z=0$.
Let us denote by $({\cal O}_{S^{(r) k}_I}, {\cal O}_{\Sigma^{k}_I})$
the chiral primary operators dual to the fields $(s^{(r) k}_I,
\s^{k}_I)$ respectively. The vevs of the scalar operators with dimension two or less
can then be expressed in terms of the coefficients in the
asymptotic expansion as
\begin{eqnarray}
\left < {\cal O}_{S^{(r)1}_i} \right > &=& \frac{2 N}{\pi} \sqrt{2}
[s^{(r)1}_i]_1; \hspace{0.5cm}
\left < {\cal O}_{S^{(r)2}_I} \right > = \frac{2 N}{\pi} \sqrt{6}
[s^{(r)2}_I]_2; \label{sc-1} \\
\left < {\cal O}_{\Sigma^{2}_I} \right > &=& \frac{N}{\pi} \left (2
\sqrt{2} [\s^2_I]_2 - \frac{1}{3} \sqrt{2} a_{Iij}
\sum_{r} [s^{(r) 1}_{i}]_1 [s^{(r) 1}_{j}]_1 \right ). \nonumber} \def\bd{\begin{document}} \def\ed{\end{document}
\end{eqnarray}
Here $[\psi]_n$ denotes the coefficient of the $z^n$ term in the
asymptotic expansion of the field $\psi$. The coefficient $a_{Iij}$
refers to the triple overlap between spherical harmonics, defined in
(\ref{ap-ov0}). Note that dimension one scalar spherical harmonics
have degeneracy four, and are thus labeled by $i = 1, \cdots 4$.
In deriving these vevs one needs to follow the steps outlined in the
previous section, in particular making use of the cubic coupling and
Kaluza-Klein reduction formulae obtained in \cite{Arutyunov:2000by}.
Now consider the stress energy tensor and the R symmetry currents. The
three dimensional metric, which is given by $g^{o}_{\m{\sst{(n)}}} + \hat{H}_{\m{\sst{(n)}}}$,
and the Chern-Simons gauge fields admit the
following asymptotic expansions:
\begin{eqnarray}
ds_{3}^{2} &=& \frac{dz^2}{z^2} + \frac{1}{z^2} \left(g_{(0)\bar{\mu}\bar{\nu}} + z^2
\left(g_{(2) \bar{\mu}\bar{\nu}} + {\rm{log}}(z^2) h_{(2)
\bar{\mu}\bar{\nu}} + ({\rm{log}}(z^2))^2
\tilde{h}_{(2) \bar{\mu}\bar{\nu} }\right) + \cdots\right) dx^{\bar{\mu}} dx^{\bar{\nu}}; \nonumber} \def\bd{\begin{document}} \def\ed{\end{document} \\
A^{\pm \a} &=& {\cal A}^{\pm \a} + z^2 A_{(2)}^{\pm \a} + \cdots
\end{eqnarray}
The vevs of the R symmetry currents $J^{\pm \a}_u$ are then given in terms
of terms in the asymptotic expansion of $ A^{\pm \a}_{\m}$ as
\begin{equation}
\left < J^{\pm
\a}_{\bar{\mu}} \right > = \frac{N}{4 \pi} \left (g_{(0) \bar{\mu} \bar{\nu}} \pm \epsilon_{\bar{\mu}
\bar{\nu}} \right ) {\cal A}^{\pm \a \bar{\nu}}, \label{j1}
\end{equation}
where $\epsilon_{\bar{\mu} \bar{\nu}}$ is the constant antisymmetric tensor
such that $\epsilon_{01} = 1$.
The vev of the stress energy tensor $T_{\bar{\mu}\bar{\nu}}$ is given by
\begin{eqnarray}
\< T_{\bar{\mu}\bar{\nu}} \> &=& \frac{N}{2 \pi} \left (
g_{(2)\bar{\mu}\bar{\nu}} + {\textstyle{1\over2}} R g_{(0) \bar{\mu}\bar{\nu}}
+ 8 \sum_r [\tilde{s}^{(r)1}_{i}]_1^2 g_{(0) \bar{\mu}\bar{\nu}} + {\textstyle{1\over 4}} ({\cal A}^{+ \a}_{(\bar{\mu}}
{\cal A}^{+ \a}_{\bar{\nu})} + {\cal A}^{- \a}_{(\bar{\mu}}
{\cal A}^{- \a}_{\bar{\nu})}) \right) \label{t1}
\end{eqnarray}
where parentheses denote the symmetrized traceless combination of indices.
\bigskip
This summarizes the expressions for the vevs of chiral primaries with
dimension two or less which were derived in
\cite{Kanitscheider:2006zf}. Note that these operators correspond to
supergravity fields which are at the bottom of each Kaluza-Klein
tower. The supergravity solution of course
also captures the vevs of operators dual to the other fields in each
tower. Expressions for these vevs involving all contributing non-linear
terms were not explicitly derived in
\cite{Kanitscheider:2006zf}: in general the vev of a dimension $p$ operator will include
contributions from terms involving up to $p$ supergravity
fields. Computing these in turn requires the field equations (along
with gauge invariant combinations, KK reduction maps etc) up to
$p$th order in the fluctuations.
Note that using only the linear term in a vev generically gives
qualitatively the wrong answer. For example, when supersymmetry is
unbroken only operators which are the bottom components of
supermultiplets can acquire vevs. Evaluating only the linearized
expressions for the vevs of operators which are not bottom
components of a supermultiplet in a supersymmetric background
however usually gives a non-zero answer, which cannot be correct. In
\cite{Alday:2005xj} linearized expressions for holographic vevs of
operators were evaluated in the decoupling $AdS_3 \times S^3 \times
X_4$ limit of the 2-charge supersymmetric black ring background.
Non-zero vevs for found for the operators ${\cal O}_{\rho^{I_t}}$
dual to the scalars $\rho^{I_t}$ given in (\ref{flc1}). However,
these operators (for dimension $\Delta \ge 3$) sit in the middle
component of the spin 2 supermultiplet whose lowest component is a
vector chiral primary (as can inferred from Table 3 and Figure 1 of
\cite{Sez98}) and cannot acquire a vev in a supersymmetric state.
Thus the linearized expressions indeed give qualitatively wrong
answers in this case.
Given an asymptotically $AdS$ supergravity solution, one can extract
the vevs, and the deformation parameters, by purely algebraic
manipulations. This information characterizes the state of the dual
theory and is the natural information to extract first. Higher point
functions can be extracted by extending the holographic one-point functions to include sources,
and then solving fluctuation equations in the geometry to sufficient order in the sources.
\section{Two charge system}
\subsection{Naive geometry and D-branes}
The 2-charge supersymmetric D1-D5 supergravity solution is
\begin{eqnarray}
ds^2 &=& \frac{1}{\sqrt{h_1 h_5}} (- dt^2 + dz^2)
+ \sqrt{h_1 h_5} dx^m dx^m + \sqrt{\frac{h_1}{h_5}} ds^2(X_4); \\
C &=& (h_1^{-1} - 1) dt \wedge dz + \ast_{4} d h_5, \qquad
e^{-2 \Phi} = \frac{h_5}{h_1}. \nonumber} \def\bd{\begin{document}} \def\ed{\end{document}
\end{eqnarray}
Here $ds^2(X_4)$ denotes the metric on $T^4$ or $K3$ respectively
and the functions $(h_1,h_5)$ are harmonic on $R^4$, which has
coordinates $x^m$. The notation $\ast_{4}$ denotes the Poincar\'{e}
dual in the flat $R^4$ metric. Our conventions for the supergravity field
equations are given in appendix \ref{conv}. The solution describes
D5-branes wrapping $X_4$ and intersecting D1-branes over the string
directions $(t,z)$, with $z$ being a periodic coordinate\footnote{Note
we called $z$ the Fefferman-Graham radial coordinate in the previous
section. It should be clear from the context whether $z$ is a Fefferman-Graham
coordinate or the compact $S^1$ coordinate that the D1 and D5 brane wraps.
I hope this wil not cause any confusion.}
with radius $R_z$.
Consider now single-centered solutions such that $h_i = (1+ Q_i/r^2)$;
the charges $Q_i$ are related to the integral charges
$N_i$ as
\begin{equation}
Q_1 = \frac{(\a')^3 N_1}{V}; \qquad
Q_5 = \a' N_5,
\end{equation}
where we have set the string coupling $g =1$ and
the volume $v$ of $X_4$ is $(2 \pi)^4 V$.
Reducing over $X_4$ and $z$ leads to the five dimensional metric
\begin{equation} \label{5dmetric}
ds^2 = \l^{-2/3} dt^2 +\l^{1/3} (dr^2 + r^2 d \Omega_3^2)
\end{equation}
where
\begin{equation}
\l= h_1 h_5 = (1 + \frac{Q_1}{r^2})(1 + \frac{Q_5}{r^2})
\end{equation}
This is an asymptotically flat spacetime which has a naked singularity
at $r=0$: the Ricci scalar is given by
\begin{equation} \label{curv1}
R = - \frac{1}{6} \left ( \frac{2 \pa_r^2 \l}{\l^{4/3}} + \frac{6
\pa_r \l}{r \l^{4/3}} - \frac{(\pa_r \l)^2}{\l^{7/3}} \right ),
\end{equation}
and thus for $\l \sim r^{-2p}$ near $r=0$
\begin{equation} \label{curv2}
R = \frac{4}{3} p (p+1) r^{2p/3 - 2}.
\end{equation}
In the 2-charge case, $p = 2$ and the curvature diverges at $r=0$.
As we shall see when we discuss the 3-charge case the
corresponding $5d$ metric is of exactly the same form but in that
case $\l$ contains the product of 3-harmonic functions. Thus for this
case $p=3$ and the curvature is finite at $r=0$: there is an
extremal horizon rather than a naked singularity at $r=0$.
Consider now the decoupling limit:
\begin{equation}
\alpha' \to 0, \qquad \r = \frac{r}{\a'}\ {\rm fixed}, \qquad R_z,
\frac{v}{\a'^2}\ {\rm fixed}
\end{equation}
In this limit the constant in the harmonic functions drops out and one is focusing
on the region near $r=0$. After rescaling the coordinates as
\begin{equation}
\r \to \r \frac{R_z}{\sqrt{Q_1 Q_5}}, \qquad
z \to \frac{z}{R_z}, \qquad t \to \frac{t}{R_z}
\end{equation}
the metric becomes
\begin{equation} \label{D1D5nearhor}
ds^2 = \sqrt{Q_1 Q_5} \left ( \rho^2 (-dt^2 + dz^2) +
\frac{d\rho^2}{\rho^2} + d \Omega_3^2 \right )
+ \sqrt{\frac{Q_1}{Q_5}} ds^2(X_4),
\end{equation}
where $d\Omega_3^2$ is the metric on a unit 3-sphere and $z$ has
period $2 \pi$ (and as usual we have set $\a'=1$ after the limit
was taken). The
six-dimensional metric is thus the product of a locally $AdS_3$
spacetime and a 3-sphere, with $AdS$ and sphere radii both
equal to $(Q_1 Q_5)^{1/4}$. Had the $z$ coordinate been
non-compact the non-compact part would have been $AdS_3$ in Poincar\'{e}
coordinates and the apparent horizon at $\rho = 0$ would have
merely been a coordinate horizon, which can be removed by introducing global
coordinates for $AdS_3$. Since $z$ is compact, however, the three
dimensional non-compact spacetime is instead the zero mass BTZ black
hole. The spacetime (\ref{D1D5nearhor}) does not have a curvature
singularity but there are geodesics that terminate at $r=0$.
Extremal BTZ black holes have a horizon radius proportional to their
mass and a singularity at $r=0$. In the massless limit the position
of the horizon coincides with the singularity, so we end up
with a naked singularity. In the 3-charge case one instead obtains
a massive extremal BTZ black hole with horizon radius proportional to
the new charge. The area of the horizon is proportional to the horizon radius
and thus vanishes for the 2-charge system. Note that reducing the decoupled
region over $X^4$ and $z$ leads to a spacetime with a curvature
singularity, explaining the singularity structure
of (\ref{5dmetric}), but this is due to the fact that we reduce over a
circle of vanishing proper length.
Thus the 2-charge black hole does not have a macroscopic horizon
within supergravity. As we will momentarily review, however, the entropy
of the 2-charge system is $S \sim \sqrt{N_1 N_5}$
and one would expect that the D1-D5 solution with higher derivative
corrections taken into account should
have a horizon which accounts for this entropy. In this section we
will see that supergravity fuzzball solutions for this case
can be explicitly constructed, and matched with CFT
microstates. However, the average fuzzball in this system is necessarily
string scale, and thus the system can only provide an indicative toy
example for the proposal as a whole.
\subsection{CFT microstates}
The dual CFT describing the low
energy physics of the D1-D5 system is believed to be a deformation of
the ${\cal N}=(4,4)$ supersymmetric sigma model with target space
$S^N(X_4)$, where $N= N_1 N_5$ and the compact space $X_4$
is either $T^4$ or $K3$, see \cite{David:2002wn} for a review.
Most of the discussion below will
be for the case of $T^4$, although the results extend simply to $K3$.
The orbifold point is roughly the analogue
of the free field limit of ${\cal N}=4$ SYM in the context of $AdS_5/CFT_4$
duality.
The Hilbert space of the orbifold theory decomposes into twisted
sectors, which are labeled by the conjugacy classes of the symmetric
group $S(N)$, the latter consisting of cyclic groups of various
lengths. The various conjugacy classes that occur and their multiplicity are
subject to the constraint
\begin{equation}
\sum_i n_i m_i = N,
\end{equation}
where $n_i$ is the length of the cycle (or the twist) and $m_i$ is the
multiplicity of the cycle. As emphasized in \cite{Moore} there is a
direct correspondence between the combinatorial description of the
conjugacy classes and a second quantized string theory, the long and
short string picture advocated in \cite{Das:1996ug,Mald}.
The point is that one can view the supersymmetric sigma model on
$S^N(X_4)$ as the transversal fluctuations of a string with target
space $R \times S^1 \times S^N(X_4)$, in lightcone gauge, with the
string winding once around $S^1$. However, the partition function of
this single string decomposes into several distinct topological
sectors, corresponding to the number of ways the string can be
disentangled into separate strings that wind one or more times around
$X_4 \times S^1$. The factorization of the conjugacy classes into a
product of irreducible cyclic permutations $n_i$ determines the
decomposition into several strings of winding number $n_i$. Whilst the
string picture is intuitive and useful for qualitative features, one
will need the explicit orbifold CFT
description for computations and it is therefore the latter we will use.
The chiral primaries and R ground states can be precisely described
at the orbifold point. In particular the R ground states are
associated with the cohomology of the internal manifold, $T^4$ or
$K3$, and therefore the corresponding chiral primaries in the NS
sector obtained from spectral flow are also labeled by the cohomology.
For our discussions only the states associated with the even
cohomology will be relevant; let the NS sector chiral primaries
be labeled as ${\cal{O}}_{n}^{(p,q)}$ where $n$ is the twist and
$(p,q)$ labels the associated cohomology class. The degeneracy of the
operators associated with the $(1,1)$ cohomology is $h^{1,1}$ of the
compact manifold.
The complete set of chiral primaries associated with the even cohomology
is then built from products of the form
\begin{equation}
\prod_{l}({\cal{O}}_{n_l}^{p_l,q_l})^{m_l}, \hspace{0.5cm}
\sum_{l} n_l m_l = N, \label{oper-1}
\end{equation}
where symmetrization over the N copies of the CFT is implicit.
Spectral flow maps these chiral primaries in the NS sector to R ground states, where
\begin{eqnarray}
h^R &=& h^{NS} - j_3^{NS} + \frac{c}{24}; \nonumber \\
j_3^R &=& j_3^{NS} - \frac{c}{12}, \label{spe_fl}
\end{eqnarray}
where $c$ is the central charge of the CFT. Each of the operators in (\ref{oper-1})
is mapped by spectral flow to a (ground state) operator of definite R-charge
\begin{eqnarray} \label{Rop}
& & \prod_{l=1} ({\cal {O}}^{(p_l,q_l)}_{n_l})^{m_l} \quad
\rightarrow \qquad
\prod_{l=1} ({\cal {O}}^{R (p_l,q_l)}_{n_l})^{m_l}, \\
j_3^R &=& {\textstyle{1\over2}} \sum_l (p_l - 1) m_l, \qquad
\bar{j}^R_3 ={\textstyle{1\over2}} \sum_l (q_l - 1) m_l. \nonumber} \def\bd{\begin{document}} \def\ed{\end{document}
\end{eqnarray}
Note that Ramond operators which are obtained from spectral flow of primaries
associated with the $(1,1)$ cohomology have zero R charge. Explicit
free field representations of these operators are reviewed in
\cite{David:2002wn} but these will not be needed in what follows.
The degeneracy of the Ramond ground states can be computed by counting
the partitions of $N$ into pairs of integers; for large $N$ this
results in a total degeneracy $d$ such that
\begin{equation}
\log (d) \approx 2 \pi \sqrt{\frac{C}{6} N},
\end{equation}
where $C = 12$ for $T^4$ and $C = 24$ for K3. (In counting the former
one needs to include the odd dimensional cohomology.) Note that the
degeneracies are clearly those of type IIB strings on $T^4$ and
heterotic strings on $T^4$ respectively, with left moving excitation level $N$. It
is also useful to note here that the distribution of ground states is
highly localized around those with zero R charge: the degeneracy of ground states with zero R charge,
$d_0$, is given by \cite{Kanitscheider:2007wq}, see also \cite{Russo:1994ev},
\begin{equation}
d_0 \approx d/N
\end{equation}
for large $N$. Thus as already mentioned the 2-charge black hole
has an entropy of order $\sqrt{N}$. By adding momentum one obtains a
black hole with a regular horizon in supergravity; the corresponding
microstates are left moving excited and right moving ground states,
whose degeneracy for high excitation level can be computed via the
Cardy formula. We will postpone detailed discussions of these states
until section \ref{3cft}.
\bigskip
The AdS/CFT dictionary relates the supergravity spectrum about the
$AdS_3 \times S^3 \times X_4$ background to these chiral primary
operators. In particular in what follows we will make use of the
relationship between (scalar) supergravity fields and
chiral primaries. This map was established recently
\cite{Taylor:2007hs} by matching three
point functions computed from supergravity with those computed in the
orbifold CFT \cite{Jevicki:1998bm,Lunin:2001pw}, assuming non-renormalization of the correlators. Such
non-renormalization is supported by the matching of the correlators computed from the
worldsheet theory with those computed in the orbifold CFT
\cite{Gaberdiel:2007vu,Dabholkar:2007ey,Pakman:2007hn}. Labeling
the orbifold operators by their twist and $(j_3,\bar{j_3})$ charges
$(m, \bar{m})$ respectively, the correspondence
is
\begin{equation}
\left ( \begin{array} {c} {\cal O}^{S}_{(n-1) m \bar{m}} \\
{\cal O}^{\Sigma}_{(n-1) m \bar{m}} \end{array} \right )
\leftrightarrow {\cal M}
\left ( \begin{array} {c} {\cal O}^{0,0}_{n m \bar{m}} \\
{\cal O}^{2,2}_{(n-2) m \bar{m}} \end{array} \right ),
\end{equation}
and
\begin{equation}
{\cal O}^{(r)1,1}_{n m \bar{m}} \leftrightarrow {\cal O}^{S^{(r)}}_{n m
\bar{m}};
\end{equation}
with the matrix ${\cal M}$ being
\begin{equation}
{\cal M} = \frac{1}{\sqrt{2\D}} \left ( \begin{array} {c c }
(\D + 1)^{1/2} & - (\D - 1)^{1/2} \\
(\D - 1)^{1/2} & (\D + 1)^{1/2}
\end{array} \right )
\end{equation}
for $\D \ge 2$. Here ${\cal O}^{\Phi}_{\Delta m \bar{m}}$ denotes the
operator of dimension $\Delta$ and R charges $(m,\bar{m})$ coupling to
the bulk supergravity field $\Phi$ of corresponding mass and $SO(4)$ charges.
The formulae for the holographic vevs allow us to extract the
expectation values of these operators, along with those of the
stress energy tensor and R currents, from a given asymptotically
$AdS_3 \times S^3 \times X_4$ supergravity solution. Given a proposed
correspondence between the supergravity solution and a field theory
state, one can use this data to test the correspondence. More
generally $n$-point functions extracted from the bulk can also
be compared $n$-point functions in a given state, but there is no reason to
anticipate that such correlation functions are non-renormalized.
\subsection{Fuzzball solutions}
The key observation in constructing fuzzball solutions for the 2-charge system
is that D1-D5 is related by dualities to a fundamental string carrying
momentum, and supergravity solutions for the latter have been known
for more than a decade: The relevant solutions are general chiral null models. For the heterotic
string, the general such model takes the form
\begin{eqnarray} \label{het1}
ds^2 &=& H^{-1}(-dudv + (K-2\alpha^\prime
H^{-1}N^{(c)}N^{(c)})dv^2+ 2A_I dx^I dv) + dx_I dx^I \nonumber} \def\bd{\begin{document}} \def\ed{\end{document} \\
{B}^{(2)}_{uv} &=& \frac{1}{2}(H^{-1}-1), \qquad {B}^{(2)}_{vI} = H^{-1} A_{I}, \\
\qquad {\Phi} &=& - \frac{1}{2} \ln H, \qquad
{V}_v^{(c)} = H^{-1} N^{(c)}, \nonumber} \def\bd{\begin{document}} \def\ed{\end{document}
\end{eqnarray}
where $I=1,\cdots,8$ labels the transverse directions and ${V}_m^{(c)}$ are
Abelian gauge fields, with
$((c)=1,\cdots,16)$ labeling the elements of the Cartan of the gauge group.
The equations of motion for the heterotic string are given in appendix
\ref{conv}; here the defining functions satisfy
\begin{equation}
\Box H (x,v) = \Box K (x,v) = \Box A_I (x,v) = (\pa_I A^I (x,v) - \pa_v
H (x,v)) = \Box N^{(c)} = 0.
\end{equation}
For the solution to correspond to a charged heterotic
string with generic wave profile, one takes the following solutions
\begin{eqnarray}
\label{K3harmonics}
H &=& 1+\frac{Q}{|x-F(v)|^6},
\qquad A_I = - \frac{Q\dot{F}_I(v)}{|x-F(v)|^6}, \qquad N^{(c)} =
\frac{q^{(c)}(v)}{|x-F(v)|^6}, \nonumber \\
K &=& \frac{Q^2 \dot{F}(v)^2+2\alpha^\prime q^{(c)}q^{(c)}(v)}{Q|x-F(v)|^6},
\end{eqnarray}
where $F^I(v)$ is an arbitrary null curve in $R^8$; $q^{(c)}(v)$ is an
arbitrary charge wave and $\dot{F}_I(v)$ denotes $\pa_v F_I(v)$. Such
solutions were first discussed in
\cite{Callan:1995hn,Dabholkar:1995nc}, although the above has
a more generic charge wave, lying in $U(1)^{16}$ rather than $U(1)$.
Chiral null model solutions of type IIB are obtained by setting
$N^{(c)} = 0$. More general solutions in which the fundamental strings
carry condensates of fermion bilinears were considered in
\cite{Taylor:2005db}; such solutions are necessary to account for all D1-D5
microstates, but for brevity we will not discuss them in
detail here.
\bigskip
The F1-P solutions described by such chiral null models can be
dualized to give corresponding solutions for the D1-D5 system as
follows \cite{Lunin:2001jy}. Compactify four of the transverse directions on a torus,
such that $x^i$ with $i=1,\cdots,4$ are coordinates
on $R^4$ and $x^{\r}$ with $\r = 5,\cdots,8$ are coordinates on $T^4$.
Then let $v = (t - z)$ and $u = (t+ z)$ with the coordinate $z$ being
periodic with length $L_z \equiv 2 \pi R_z$,
and smear all harmonic functions over both this circle and
over the $T^4$, so that they satisfy
\begin{equation} \label{equ_harmonics}
\Box_{R^4} H (x) = \Box_{R^4} K (x) = \Box_{R^4} A_I (x) = \Box_{R^4}
N^{(c)} = 0, \hspace{0.5cm}
\pa_{i} A^i = 0.
\end{equation}
Thus the harmonic functions appropriate for describing strings with
only bosonic condensates are \cite{Lunin:2001fv}
\begin{eqnarray}
H &=& 1 + \frac{Q}{L_z} \int_0^{L_z} \frac{dv}{|x-F(v)|^2};
\qquad A_I = -\frac{Q}{L_z}\int_0^{L_z} \frac{dv \dot{F}_I(v)}{|x-F(v)|^2}; \\
N_{(c)} &=& -\frac{Q}{L_z} \int_0^{L_z} \frac{dv q_{c}(v)} {|x-F(v)|^2}; \qquad
K = \frac{Q }{L_z} \int_0^{L_z} \frac{dv (\dot{F}_i(v)^2 +
\dot{ F}_{\rho} (v)^2)} {|x-F(v)|^2}. \nonumber} \def\bd{\begin{document}} \def\ed{\end{document}
\end{eqnarray}
Here $|x-F(v)|^2$ denotes $\sum_{i} (x_i - F_i(v))^2$.
Next one follows an appropriate chain of dualities to obtain a D1-D5
solution. For D1-D5 on $T^4$ one starts with type IIB F1-P solutions on $T^4$
whilst for D1-D5 on $K3$ one starts with heterotic F1-P solutions and
uses heterotic/IIA duality. The computational details of this duality
chain are given in \cite{Kanitscheider:2007wq}.
The final result in both cases can be written as
\begin{eqnarray}
\label{equ_D1D5K3pot}
ds^2 &=& \frac{f_1^{1/2}}{\tilde{f_1} f_5^{1/2} }[-(dt-A_i
dx^i)^2+(dz-B_i dx^i)^2] + f_1^{1/2} f_5^{1/2} dx_i dx^i
+ f_1^{1/2} f_{5}^{-1/2} ds^2(X_4), \nonumber} \def\bd{\begin{document}} \def\ed{\end{document} \\
e^{2\Phi} &=& \frac{f_1^2}{f_5 \tilde{f_1} }, \qquad B^{(2)}_{tz} =
\frac{{\cal A}}{f_5 \tilde{f}_1},
\qquad B^{(2)}_{\bar{\mu} i} = \frac{{\cal A} {\cal B}^{\bar{\mu}}_i}{f_5 \tilde{f_1} }, \\
B^{(2)}_{ij} &=& \l_{ij} +
\frac{2 {\cal A} A_{[i} B_{j]}} {f_5 \tilde{f}_1 }, \qquad
B^{(2)}_{\rho\sigma} = f_{5}^{-1} k^{\gamma}
\w_{\rho \sigma}^{\gamma}, \qquad C^{(0)} = - f_{1}^{-1} \cal A, \nonumber \\
C^{(2)}_{tz} &=& 1-\tilde{f}_1^{-1}, \qquad C^{(2)}_{\bar{\mu}i} = -
\tilde{f}_1^{-1} {\cal B}^{\bar{\mu}}_i, \qquad C^{(2)}_{ij} = c_{ij} - 2 \tilde{f}_1^{-1}
A_{[i}B_{j]}, \nonumber \\
C^{(4)}_{tzij} &=& \l_{ij} + \frac{{\cal A}}{f_5 \tilde{f_1}}(c_{ij} + 2 A_{[i}B_{j]}),
\qquad C^{(4)}_{\bar{\mu}ijk} = \frac{3 {\cal A}}{f_5 \tilde{f}_1}
{\cal B}^{\bar{\mu}}_{[i}c_{jk]}, \nonumber \\
C^{(4)}_{tz \rho \sigma}
&=& f_{5}^{-1} k^{\gamma} \w^{\gamma}_{\rho \sigma},
\qquad C^{(4)}_{ij \rho \sigma} = (\l^{\gamma}_{ij} +
f_{5}^{-1} k^{\gamma}c_{ij})\w^{\gamma}_{\rho \sigma}, \qquad
C^{(4)}_{\rho \sigma \tau \pi} = f_{5}^{-1} \cal A \e_{\rho
\sigma \tau \pi}, \nonumber} \def\bd{\begin{document}} \def\ed{\end{document}
\end{eqnarray}
where we introduce a basis of self-dual and anti-self-dual 2-forms
$\w^{\gamma} \equiv (\w^{\a_+}, \w^{\a_-})$ with $\gamma =
1,\cdots,b^2$ on the compact manifold $X_4$. For both $T^4$ and $K3$
the self-dual forms are labeled by
$\a_{+} = 1,2,3$ whilst the anti-self-dual forms are labeled by
$\a_{-} =1,2,3$ for $T^4$ and $\a_{-} = 1,\cdots 19$ for $K3$. The
intersections of these forms are defined by
\begin{equation}
\label{K3dABdef}
d_{\g\d} = \frac{1}{(2 \pi)^4 V} \int_{X_4} \w_2^{\g} \wedge \w_2^{\d}.
\end{equation}
The solutions are expressed
in terms of the following combinations of harmonic functions
$(H,K,A_{i},{\cal A}, {\cal A}^{\a_-})$
\begin{eqnarray}
\label{D1D5K3aux}
f_{5} &=& H; \qquad \tilde{f}_1 = 1 + K - H^{-1} ({\cal A}^2 + {\cal A}^{\a_-} {\cal A}^{\a_-}); \qquad
f_1 = \tilde{f}_1 + H^{-1} {\cal A}^2; \nonumber} \def\bd{\begin{document}} \def\ed{\end{document} \\
k^{\gamma} &=& (0_3, \sqrt{2} {\cal A}^{\a_-});
\qquad dB = -\ast_4 dA; \qquad
dc = - \ast_4 df_5; \\
d\l^{\gamma} &=& \ast_4 dk^{\gamma}; \qquad d\l = \ast_4 d {\cal A}; \qquad
{\cal B}^{\bar{\mu}}_i = (-B_i,A_i), \nonumber} \def\bd{\begin{document}} \def\ed{\end{document}
\end{eqnarray}
where $\bar{\mu} = (t,z)$ and the Hodge dual $\ast_{4}$ is defined over (flat)
$R^4$, with the Hodge dual in the Ricci flat metric on
the compact manifold being denoted by $\epsilon_{\r \s\t \pi}$.
The constant term in $C_{tz}^{(2)}$ is chosen so that the
potential vanishes at asymptotically flat infinity.
We are interested in solutions for which the defining harmonic functions are given by
\begin{eqnarray}
H &=& 1 + \frac{Q_5}{L} \int_0^L \frac{dv}{|x-F(v)|^2};
\qquad A_i = -\frac{Q_5}{L}\int_0^L \frac{dv \dot{F}_i(v)}{|x-F(v)|^2},
\label{harm} \\
{\cal A} &=& -\frac{Q_5}{L}\int_0^L \frac{dv
\dot{\cal F} (v)}{|x-F(v)|^2}; \qquad
{\cal A}^{\a_-} = -\frac{Q_5}{L} \int_0^L \frac{dv \dot{\cal{F}}^{\a_-}(v)}{|x-F(v)|^2}, \nonumber \\
K &=& \frac{Q_5}{L} \int_0^L \frac{dv (\dot{F}(v)^2 +
\dot{\cal F}(v)^2 + \dot{\cal{F}}^{\a -}(v)^2 )}{|x-F(v)|^2}. \nonumber} \def\bd{\begin{document}} \def\ed{\end{document}
\end{eqnarray}
In these expressions $Q_5$ is the 5-brane charge and
$L$ is the length of the defining curve in the D1-D5 system, given
by
\begin{equation}
L = 2 \pi Q_5/R_z,
\end{equation}
where $R_z$ is the radius of the $z$ circle. Note that $Q_5$ has
dimensions of length squared and is related to the integral charge via
\begin{equation} \label{q5}
Q_5 = \a' N_5
\end{equation}
(where $g_s$ has been set to one).
The D1-brane charge $Q_1$ is given by
\begin{equation} \label{d1-charg}
Q_1 = \frac{Q_5}{L} \int_0^L dv (\dot{F}(v)^2 +
\dot{\cal F}(v)^2 + \dot{\cal{F}}^{\a_-}(v)^2 ),
\end{equation}
and the corresponding integral charge is given by
\begin{equation} \label{q1}
Q_1 = \frac{N_1 (\a')^3}{V},
\end{equation}
where $(2 \pi)^4 V$ is the volume of the compact manifold.
\bigskip
Solutions with no internal
excitations, that is $({\cal F}(v), {\cal F}^{\a_-}(v)) = 0$, were
obtained in \cite{Lunin:2001jy}. These Lunin-Mathur solutions involve
only the graviton, dilaton and RR 2-form:
\begin{eqnarray}
ds^2 &=& \frac{1 }{ (f_1 f_5)^{1/2} }[-(dt-A_i
dx^i)^2+(dz-B_i dx^i)^2] + (f_1 f_5)^{1/2} dx_i dx^i
+ f_1^{1/2} f_{5}^{-1/2} ds^2(X_4), \nonumber} \def\bd{\begin{document}} \def\ed{\end{document} \\
e^{2\Phi} &=& \frac{f_1}{f_5}, \label{lun-mat} \\
C^{(2)}_{tz} &=& 1- {f}_1^{-1}, \qquad C^{(2)}_{\bar{\mu}i} = -
{f}_1^{-1} {\cal B}^{\bar{\mu}}_i, \qquad C^{(2)}_{ij} = c_{ij} - 2 {f}_1^{-1}
A_{[i}B_{j]}, \nonumber
\end{eqnarray}
and are defined in terms of harmonic functions sourced on the curve
$F^{i}(v)$ in $R^4$:
\begin{equation}
f_5 = 1 + \frac{1}{L} \int^{L}_{0} \frac{ Q_5 dv}{ |x -
F|^2} ; \qquad
f_1 = 1 + \frac{1}{L} \int^{L}_{0} \frac{ Q_5 (\pa_v F)^2 dv}{ |x -
F|^2} ; \label{lm-func}
\qquad A = \frac{1}{L} \int_{0}^{L} \frac{ Q_5 \pa_v F}{| x- F |^2},
\end{equation}
with $dB = - \ast_4 dA$. Note that corresponding 2-charge
solutions carrying KK monopole charge and momentum which are related
by dualities to these solutions were found in \cite{Srivastava:2006xn}.
The mapping of the parameters from the original F1-P systems to
the D1-D5 systems was discussed in \cite{Lunin:2001jy}.
The fact that the solutions take exactly the same form, regardless of
whether the compact manifold is $T^4$ or $K3$, is unsurprising given
that only zero modes of the compact manifold are excited.
The solutions defined in terms of the harmonic functions (\ref{harm})
describe the complete set of two-charge fuzzballs for the D1-D5 system on
$K3$. In the case of $T^4$, these describe fuzzballs with only
bosonic excitations; the most general solution would include fermionic
excitations and thus more general harmonic functions of the type
discussed in \cite{Taylor:2005db}.
Given a generic fuzzball solution, one would first like to check whether
the geometry
is indeed smooth and horizon-free. There are no horizons, since the
defining functions are positive definite. For the fuzzballs with no internal
excitations the question of singularity was discussed in
\cite{Lunin:2002iz}, the
conclusion being that the solutions are non-singular unless the
defining curve $F^i(v)$ is non-generic and self-intersects. A generic fuzzball
solution with internal excitations is also non-singular provided that the
defining curve $F^i(v)$ does not self-intersect and $\dot{F}_i(v)$
only has isolated zeroes \cite{Kanitscheider:2007wq}.
One can illustrate that the fuzzballs are non-singular as
follows \cite{Lunin:2002iz}. The fuzzballs are defined in terms of harmonic functions, and
potential singularities of the solutions are at the sources of these
harmonic functions. Consider
for simplicity the case of fuzzballs with no internal excitations,
for which the defining harmonic functions are sourced on a closed
curve $F^{i}(v)$ in $R^4$. In the neighborhood of a specific point on the
curve, the harmonic function picks up contributions from a line of
sources, and thus effectively behaves like a three-dimensional
harmonic function. That is, parameterizing the $R^4$ as
\begin{equation}
ds_4^2 = dw^2 + dr^2 + r^2 d \Omega_2^2,
\end{equation}
the direction $w$ is aligned with the curve. The defining
harmonic functions behave as
\begin{equation}
\int_{-\infty}^{\infty} \frac{dw}{w^2 + r^2} = \frac{\pi}{r},
\end{equation}
so that $f_a = Q_a/r$, whilst the defining one form behaves as
\begin{equation}
w = a \frac{dz}{r}.
\end{equation}
Then the metric behaves as
\begin{eqnarray}
ds^2 &=& - \frac{2a}{\sqrt{Q_1 Q_5}} dw dt + \sqrt{Q_1 Q_5}
\frac{dw^2}{r} (1 - \frac{a^2}{Q_1 Q_5}) \\
&& + \frac{r}{\sqrt{Q_1 Q_5}} (dz + a \cos \q d \phi)^2 + \frac{\sqrt{Q_1
Q_5}}{r} (dr^2 + r^2 d \Omega_2^2) + \frac{\sqrt{Q_1}}{\sqrt{Q_5}} ds^2(X_4). \nonumber} \def\bd{\begin{document}} \def\ed{\end{document}
\end{eqnarray}
When $a = \sqrt{Q_1 Q_5}$, which is exactly the condition that follows
from the explicit forms of the full harmonic functions, this metric is
the product of $R^{1,1} \times X_4 \times TN$ with $TN$ Taub-Nut and
thus the metric is non-singular at the curve location. We should
contrast this behavior with that of the naive solution (\ref{D1D5nearhor}).
In the naive geometric the circle $S^1$ parameterized by $z$ is
non-contractible but it has vanishing proper length in the interior
leading to singular behavior. Instead, in the fuzzball geometries
the $z$ circle is contractible and part of a Taub-NUT geometry, so
the vanishing of its radius in the interior
is akin to the behavior of polar coordinates
near the origin. A generalization of this
analysis shows that generic fuzzballs defined by non-intersecting
curves are also non-singular \cite{Lunin:2002iz,Kanitscheider:2007wq}.
Note that if there are no transverse excitations, $F^i(v) = 0$, the solution
collapses to the naive singular solution. When there are only internal
excitations, the corresponding microstates in the CFT are such that no
supergravity operators acquire vevs. Thus fuzzballs corresponding to
such states are not visible in the supergravity approximation at all.
\bigskip
Fuzzballs for which the defining curve is a circle in $R^4$ have been
extensively used in the literature; these solutions are a special case
of the Lunin-Mathur solutions (\ref{lun-mat}).
One chooses $F^i(v)$ to be a (multiwound) circle
\cite{Balasubramanian:2000rt,Maldacena:2000dr,Lunin:2001fv},
\begin{equation} \label{fcirc}
F^1 = \mu_n \cos \frac{2 \pi n v}{L}, \qquad F^2 =
\mu_n \sin \frac{2 \pi n v}{L},
\qquad F^3=F^4=0,
\end{equation}
with the internal excitations being zero.
In this case one can integrate the
expressions for the harmonic functions to obtain the solution in the
form
\begin{eqnarray}
ds^2 &=& f_{1}^{- 1/2} f_{5}^{- 1/2} \left ( - (d{t} - \frac{\mu_n
\sqrt{Q_1 Q_5}}{{r}^2 + \mu_n^2 \cos^2 {\theta}} \sin^2 {\q}
d \phi)^2 + (d{z} -
\frac{\mu_n \sqrt{Q_1 Q_5}}{{r}^2 + \mu_n^2 \cos^2 {\theta}}
\cos^2 {\q} d \psi)^2
\right ) \nonumber} \def\bd{\begin{document}} \def\ed{\end{document} \\
&& + f_{1}^{1/2} f_{5}^{1/2} \left ( ({r}^2 + \mu_n^2 \cos^2 {\q}) (
\frac{d{r}^2}{{r}^2 + \mu_n^2} + d {\q}^2) + {r}^2 \cos^2
{\q} d \psi^2 + ({r}^2 + \mu_n^2) \sin^2 {\q} d \phi^2 \right )
\nonumber} \def\bd{\begin{document}} \def\ed{\end{document} \\
&& +
f_{1}^{1/2} f_{5}^{- 1/2} ds^2(X_4); \label{ez2} \\
e^{2 \Phi} &=& f_1 f_5^{-1}, \nonumber} \def\bd{\begin{document}} \def\ed{\end{document} \\
f_{1,5} &=& 1 + \frac{Q_{1,5}}{{r}^2 + \mu_n^2 \cos^2 {\q}}, \nonumber} \def\bd{\begin{document}} \def\ed{\end{document}
\end{eqnarray}
whilst the RR 2-form field is as in (\ref{equ_D1D5K3pot}) with
\begin{equation}
A = \mu_n \frac{\sqrt{Q_1 Q_5}}{({r}^2 + \mu_n^2 \cos^2 {\q})} \sin^2
{\q} d\phi; \hspace{0.5cm}
B = - \mu_n \frac{\sqrt{Q_1 Q_5}}{({r}^2 + \mu_n^2 \cos^2 {\q})} \cos^2
{\q} d\psi.
\end{equation}
The parameters $(n,\mu_n)$ labeling the curve are
related to the charges via
\begin{equation} \label{ch-ab}
n \mu_n = \frac{L}{2 \pi} \sqrt{\frac{Q_1}{Q_5}} = \frac{\sqrt{Q_1
Q_5}}{R_z} \equiv \mu.
\end{equation}
These solutions have been used for illustrative purposes, since
the defining functions are in analytic form and the geometries
preserve a $U(1)^2$ symmetry of the transverse $R^4$.
Furthermore, the fluctuation equations are separable, due to the
symmetry, and thus two point functions can be obtained analytically.
One needs to bear in mind, however, that these solutions are not
representative of generic fuzzballs. In this case the defining
curves are multi-wound, and therefore self-intersect, violating
the condition for non singularity given above. Indeed except for the
case of $n=1$ the solutions have orbifold singularities at the curve
location.
\bigskip
{}From the asymptotics one
can see that the fuzzball solutions have the same mass and D1-brane,
D5-brane charges as the naive solution; the latter are given in
(\ref{q5}) and (\ref{q1}) whilst the ADM mass is
\begin{equation}
M = \frac{R_z V}{(\a')^4} (Q_1 + Q_5).
\end{equation}
The two charge fuzzball geometries
have the correct charges to describe Ramond ground states in the
CFT. At the same time, to make quantitative progress with the fuzzball
program, one would like to understand the detailed correspondence
between these geometries and microstates. Holographic methods provide
a powerful tool to address this question by extracting field theory
data from a given geometry.
The first step is thus to extract the decoupling asymptotically $AdS$
regions of the fuzzball geometries; this decoupling region is obtained
by dropping the constant from $(\tilde{f}_1,f_5)$. Note that the very existence of
such a region is a test that a given geometry corresponds to a CFT
ground state. If by contrast a geometry with the correct D1 and D5
charges does not have such a throat region, then it should not be
interpreted as a 2-charge black hole microstate; it fails the first
test.
It may be useful to comment here that {\it all} CFT quantities
should be interpreted in the asymptotically $AdS$ region of the
geometry, be they conserved charges or scattering amplitudes. CFT data
reconstructs the inner $AdS$ region of the asymptotically flat
geometry, but there is no known holographic description which
reconstructs the full geometry. Indeed, if there were, an immediate
corollary would be a holographic description of flat spacetime!
Given the asymptotically $AdS_3 \times S^3 \times X_4$ type IIB
solutions, one can compactify on $X_4$ to obtain asymptotically $AdS_3
\times S^3$ solutions of six dimensional ${\cal N} = 4b$
supergravity. (Note that whilst a general $T^4$ compactification would
give a solution of ${\cal N} =8$ supergravity in six dimensions the solutions
of interest here solve the equations of motion of the truncated
theory.) These six-dimensional solutions can be analyzed in detail using
holographic technology.
\subsection{Map between geometries and microstates}
The key question one would like to address is the correspondence
between a given fuzzball solution and a Ramond ground state in the
CFT. Motivated by the duality relation of the fuzzball solutions
to the fundamental string system, a precise proposal was made in
\cite{Skenderis:2006ah,Kanitscheider:2006zf,Kanitscheider:2007wq} for the
correspondence between
geometries and ground states; see also \cite{Alday:2006nd}.
The fuzzball geometries are determined by giving a curve
$F^I \equiv (F^i(v),{\cal F}(v), {\cal F}^{\a_-}(v))$ in an
$N$ dimensional space, where $N=8$ when $X_4 = T^4$ and $N=24$
when $X_4 = K3$. The corresponding state is now determined
via the following steps.
\begin{enumerate}
\item Fourier expand the curve
\begin{equation}
F^I(v) = \sum_{n > 0} \frac{1}{\sqrt{n}} (\a^I_n e^{-i n \s^+} +
(\a^I_n)^{\ast} e^{in \s^+}),
\end{equation}
and consider the coherent state
\begin{equation}
\left | F^I \right ) = \prod_{n,I} \left | \a^I_n \right ),
\end{equation}
where $\left | \a^I_n \right )$ is a standard coherent state, i.e.
it satisfies
\begin{equation}
\hat{a}^I_n \left | \a^I_n \right ) =
\a^I_n \left | \a^I_n \right )
\end{equation}
At this stage $\hat{a}^I_n$ are just auxiliary harmonic oscillators.
\item Expressing the coherent states in terms of Fock states,
$\left | \a \right ) = e^{-|\a|^2/2} \sum_k \frac{\a^k }{k!}
(\hat{a}^\dagger)^k
|0 \rangle$, we identify in $\left | F^I \right )$ the Fock
states that satisfy the constraint
\begin{equation}
\prod (\hat{a}^I_{-n_I})^{m_I} \left | 0 \right >, \hspace{0.5cm}
\sum n_I m_I = N_1 N_5, \quad (n_I>0). \label{constr}
\end{equation}
\item We retain only these terms from $\left | F^I \right )$ and map the
harmonic oscillators to CFT R operators via the dictionary
\begin{eqnarray}
\frac{1}{\sqrt{2}} (\hat{a}^1_n \pm i \hat{a}^2_n) & \leftrightarrow &
\hat{\cal O}^{R (\pm 1 + 1),(\pm 1 + 1)}_{n}; \label{fp-d} \\
\frac{1}{\sqrt{2}} (\hat{a}^3_n \pm i \hat{a}^4_n) & \leftrightarrow &
\hat{\cal O}^{R (\pm 1 + 1),(\mp 1 + 1)}_{n}; \nonumber} \def\bd{\begin{document}} \def\ed{\end{document} \\
\hat{a}^{\tilde{\a}}_{n} & \leftrightarrow &
\hat{\cal O}^{R (1,1)}_{\tilde{\a} n}. \qquad \nonumber} \def\bd{\begin{document}} \def\ed{\end{document}
\end{eqnarray}
where $F^{\tilde{\a}} =({\cal F}(v), {\cal F}^{\a_-}(v))$.
This state is proposed to correspond via AdS/CFT to the fuzzball geometry,
i.e. the vevs of gauge invariant operators in this state should
agree with the one extract from the asymptotics of the solution.
\end{enumerate}
The motivation for this map comes from the duality between the fuzzball
solutions and the supergravity solutions describing a
fundamental string carrying momentum. The FP solutions (\ref{het1})
solve the supergravity equations coupled to a macroscopic classical
string source with the profile given by the curve $F^I$. On general grounds,
one expects that
the classical string source is produced by a coherent state of
string oscillators. For the case at hand, these are BPS solutions
and the auxiliary oscillator $\hat{a}_n^I$ are
identified with the left moving string oscillators. We now want to
apply the duality to map the system to the D1-D5 frame. The FP states
satisfying (\ref{constr}) are precisely the states that map across to
D1-D5 states and this is the reason we retained only this part
of the coherent state. The one to one correspondence between string oscillators
and R operators of the symmetric orbifold CFT in (\ref{fp-d}) goes back to
\cite{Vafa:1994tf}. Note however that while the stringy oscillators
satisfy a Heisenberg algebra, the operator algebra of R ground states
is more complicated.
\subsection{Tests of correspondence}
In this section we will describe the methods used to probe the proposed
correspondence between geometries and black hole microstates.
\subsubsection{Holographic one point functions}
Holographic methodology can be used to systematically test the correspondence.
In particular, one can extract the holographic one point
functions for chiral primaries from the geometry, and compare
them to the vevs in the proposed dual state. Such computations were
carried out in \cite{Skenderis:2006ah,Kanitscheider:2006zf,Kanitscheider:2007wq}.
One first needs to reduce the asymptotically $AdS_3 \times S^3 \times X_4$ type IIB
solutions on $X_4$ to obtain asymptotically $AdS_3
\times S^3$ solutions of six dimensional ${\cal N} = 4b$
supergravity (\ref{sugraIIBK3}), and then one can apply the holographic formulae given in
(\ref{sc-1}), (\ref{j1}), (\ref{t1}) to extract the vevs from the spacetime asymptotics.
Let us briefly sketch the steps involved. The compactification is
straightforward although the explicit reduction
formulae are rather complicated, particularly for $X_4 = K3$; details may be found in
\cite{Kanitscheider:2006zf,Kanitscheider:2007wq}. The resulting
six-dimensional metric and three form fields are given by
\begin{eqnarray}
\label{D1D56d}
ds^2 &=& \frac{1}{\sqrt{f_5 \tilde{f}_1}} [ -(dt-A_idx^i)^2 + (dy-B_i
dx^i)^2)] + \sqrt{f_5 \tilde{f}_1}
dx_idx^i, \\
G_{tyi}^A &=& \pa_i \left ( \frac{m^A}{f_5 \tilde{f}_1 } \right ) , \qquad
G_{\bar{\mu}ij}^A = - 2 \pa_{[i} \left( \frac{m^A}{f_5 \tilde{f}_1 }
{\cal B}^{\bar{\mu}}_{j]}\right), \nonumber \\
G_{ijk}^A &=& \e_{ijkl} \pa^l m^A + 6\pa_{[i} \left(\frac{m^A}{f_5
\tilde{f}_1 } A_j B_{k]}\right), \nonumber} \def\bd{\begin{document}} \def\ed{\end{document}
\end{eqnarray}
with
\begin{eqnarray}
m^{n} &=& \left(0_4,{\textstyle{1\over 4}} (f_5 + F_1) \right),\\
m^{r} &=& \frac{1}{4}\left((f_5 - F_1),
- 2 A_{\a} ,
- \sqrt{2} N^{(c)}, 2 A_5 \right) \nonumber \\
&\equiv & \frac{1}{4}\left((f_5 - F_1), - 2 {{\cal A}}^{\a_-}, 2 {\cal A}
\right). \nonumber} \def\bd{\begin{document}} \def\ed{\end{document}
\end{eqnarray}
Here $n=1 \cdots 5$ and $r = 6 \cdots (n_t+5)$ with
the index $\a=6,7,8$ and $F_1 = (1 + K)$. The corresponding $SO(5,n_t)$ matrix of scalar fields may be found in
\cite{Kanitscheider:2006zf,Kanitscheider:2007wq}.
Before giving the vevs let us remark that in solutions with only
transverse excitations there are only two non-zero tensors
$(G^5,G^6)$. Such solutions involve only an $SO(1,1)$ truncation of
the ${\cal N} = 4b$ fields. Moreover, if one sets $f_1 = f_5$ then
$G^6 = 0$, and the resulting solution solves the equations of minimal
supergravity in six dimensions (i.e. with no additional tensors coupled).
Looking ahead to the next section, we will see that
almost all known fuzzball solutions for three charge black holes
involve only the $SO(1,1)$ truncation of
the ${\cal N} = 4b$ fields; there are no analogues of the fuzzballs
with internal excitations. It would be interesting to explore whether
one can generate more general three charge solutions using the
$SO(5,n_t)$ symmetry of the ${\cal N} = 4b$ theory.
The decoupling asymptotically $AdS_3 \times S^3$ region of the geometry is
obtained by dropping the constant terms in the functions $(f_1,
\tilde{f}_1, f_5,F_1)$. Next one can expand the defining functions for
large radius and extract the appropriate fields (\ref{flc1})
characterizing the perturbation with respect to $AdS_3 \times
S^3$. Substituting these perturbations in the formulae (\ref{sc-1},
\ref{j1},\ref{t1}) gives the holographic vevs.
The vevs of the stress energy tensor and of the R symmetry currents are
\begin{equation}
\left < T_{{\mu}{\nu}} \right > = 0; \qquad
\left < J^{\pm \a} \right > = \pm \frac{N}{2 \pi} a^{\a \pm} (dz \pm
dt), \label{j-char}
\end{equation}
where $a^{\a \pm}$ is a constant defined below in (\ref{a_def}).
The vanishing of the stress energy tensor is as anticipated, since
these solutions should be dual to R vacua; however, the cancellation is very
non-trivial. The vevs of the low lying chiral primaries were also
computed, and expressed in terms of the harmonics of the curves
defining the solutions:
\begin{eqnarray}
f_{kI}^5 &=& \frac{1}{L (k+1)} \int_{0}^L dv (C^I_{i_1\cdots i_k}
F^{i_1} \cdots F^{i_k}) , \label{coeff} \\
f_{kI}^1 &=& \frac{Q_5}{L (k+1)Q_1} \int_{0}^L dv \left (\dot{F}^2 + \dot{\cal{F}}^2 +
(\dot{\cal{F}}^{\a_-})^2 \right ) C^I_{i_1\cdots i_k} F^{i_1} \cdots
F^{i_k} , \nonumber \\
(A_{kI})_i &=& -\frac{1}{L (k+1)} \int_{0}^L dv
\dot{F}_i C^I_{i_1\cdots i_k} F^{i_1} \cdots F^{i_k}, \nonumber \\
({\cal A}_{kI}) &=& -\frac{\sqrt{Q_5}}{\sqrt{Q_1} L(k+1)} \int_{0}^L
dv \dot{\cal{F}} C^I_{i_1\cdots i_k} F^{i_1} \cdots F^{i_k} , \nonumber \\
{\cal A}^{\a_-}_{kI} &=& -\frac{\sqrt{Q_5}}{\sqrt{Q_1} L(k+1)} \int_0^L dv
\dot{\cal{F}}^{\a_-} C^I_{i_1\cdots i_k} F^{i_1} \cdots F^{i_k} .\nonumber} \def\bd{\begin{document}} \def\ed{\end{document}
\end{eqnarray}
Here the $C^I_{i_1\cdots i_k}$ are orthogonal symmetric traceless rank $k$
tensors on $\mathbb{R}^4$ which are in one-to-one correspondence with
the (normalized) spherical harmonics $Y_k^I(\theta_3)$ of degree $k$
on $S^3$. Fixing the center of mass of the whole system implies that
\begin{equation}
(f_{1i}^1 + f_{1i}^5) = 0.
\end{equation}
The leading term in the asymptotic expansion of the transverse gauge
field $A_i$ can be written in terms of degree one vector harmonics as
\begin{equation}
A = \frac{Q_5}{r^2}(A_{1j})_iY_1^j dY_1^i \equiv
\frac{\sqrt{Q_1Q_5}}{r^2}(a^{\a-}Y_1^{\a-} + a^{\a+} Y_1^{\a+}),
\end{equation}
where $(Y_1^{\a-},Y_1^{\a+})$ with $\a=1,2,3$ form a basis for
the $k=1$ vector harmonics and we have defined
\begin{equation} \label{a_def}
a^{\a\pm}
= \frac{\sqrt{Q_5}}{\sqrt{Q_1}} \sum_{i > j} e^\pm_{\a ij} (A_{1j})_i,
\end{equation}
where the spherical harmonic triple overlap $e^\pm_{\a ij}$ is
defined in (\ref{ap-ov3}).
Note that the vevs of the R symmetry currents increase with the
radius of the curve in $R^4$. For example, the vevs for geometries
based on the circular curves (\ref{fcirc}) are
\begin{equation}
\left < J^{\pm 3} \right > = \pm \frac{N}{2 \pi n} (dz \pm
dt), \label{j-char2}
\end{equation}
where the symmetry implies that only the $J^{\pm 3}$ components
acquire vevs. In this case the R symmetry charges are proportional to
the radius of the curve. Typical Ramond states have small R charges,
and necessarily correspond to geometries sourced by curves of small
radii.
The vevs of the scalar operators dual to the fields $(s^{(6)k}_I,\s^k_I)$
are given in terms of the transverse fluctuations:
\begin{eqnarray}
\left < {\cal O}_{S^{(6)1}_i} \right > &=& \frac{N}{4 \pi}
(- 4 \sqrt{2} f^{5}_{1i}); \label{vv2} \\
\left < {\cal O}_{S^{(6)2}_I} \right > &=& \frac{N}{4 \pi} ( \sqrt{6} (f^1_{2I} -
f^5_{2I}) ); \nonumber} \def\bd{\begin{document}} \def\ed{\end{document} \\
\left < {\cal O}_{\Sigma^2_I} \right > &=& \frac{N}{4 \pi} \sqrt{2}
( - (f^1_{2I} +f^5_{2I}) + 8 a^{\a -} a^{\b +} f_{I\a\b} ). \nonumber} \def\bd{\begin{document}} \def\ed{\end{document}
\end{eqnarray}
The internal excitations of the fuzzball solutions are
captured by the vevs of operators dual to the fields
$s^{(r)k}_I$ with $r > 6$:
\begin{eqnarray}
\left < {\cal O}_{S^{(5 + n_t)1}_i} \right > &=& - \frac{N}{\pi}
\sqrt{2} ({\cal A}_{1i}); \hspace{0.5cm}
\left < {\cal O}_{S^{(6+ \a_-)1}_i} \right > =
\frac{N}{\pi} \sqrt{2} {\cal A}^{\a_-}_{1i}; \label{vv-new} \\
\left < {\cal O}_{S^{(5+n_t)2}_I} \right > &=& - \frac{N}{2 \pi}
\sqrt{6} ({\cal A}_{2I}); \hspace{0.5cm}
\left < {\cal O}_{S^{(6+ \a_-)2}_I} \right > =
\frac{N}{2 \pi} \sqrt{6} {\cal A}^{\a_-}_{2I}. \nonumber} \def\bd{\begin{document}} \def\ed{\end{document}
\end{eqnarray}
Here $n_{t} = 5,21$ for $T^4$ and $K3$ respectively, with $\a_{-} =
1,\cdots,b^{2-}$ with $b^{2-} = 3,19$ respectively.
Thus each curve $({\cal F}(v), {\cal F}^{\a_-}(v))$ induces
corresponding vevs of operators associated with the middle cohomology
of $M^4$. Note the sign difference for the vevs of operators which are
related to the distinguished harmonic function ${\cal F}(v)$.
\bigskip
One would like to reproduce these vevs from the field theory, using
the proposed correspondence between geometries and Ramond ground
states. Substantial progress in this direction was obtained in
\cite{Skenderis:2006ah,Kanitscheider:2006zf,Kanitscheider:2007wq}
where the detailed structure of these
vevs was reproduced from the corresponding field theory
calculations. The point is that one point functions in a given state
may be related to three point functions at the conformal
point. Consider a general state such that
$|\Psi\rangle = O_\Psi(0)|0\rangle$. Then the vev of an operator
$O_k$ of dimension $k$ in the this state is given by
\begin{equation}
\langle \Psi| O_k(\l^{-1}) |\Psi \rangle
= \langle 0| (O_\Psi(\infty))^\dagger O_k(\l^{-1}) O_\Psi(0)|0\rangle,
\end{equation}
where $\l$ is a mass scale.
For scalar operators the 3-point function is uniquely determined by
conformal invariance and the above computation yields
\begin{equation} \label{fusion}
\langle \Psi| O_k(\l^{-1}) |\Psi \rangle = \l^k C_{\Psi k \Psi}
\end{equation}
where $C_{\Psi k \Psi}$ is the fusion coefficient. Similarly, the
expectation value of a symmetry current measures the charge
of the state
\begin{equation}
\langle \Psi| j(\l^{-1}) |\Psi \rangle
= \langle 0| (O_\Psi(\infty))^\dagger j(\l^{-1}) O_\Psi(0)|0\rangle
= q \l \langle \Psi|\Psi \rangle
\end{equation}
where $q$ is the charge of the operator $ O_\Psi$ under $j$.
To reproduce the vevs one needs to consider states $|\Psi \rangle$
which are ground states in the Ramond sector, or equivalently chiral
primaries in the NS sector. The stress energy tensor and the R current
expectation values are immediately reproduced from the proposed
correspondence between curves and superpositions of Ramond ground states.
Although the relevant
three point functions of scalar chiral primaries have not been computed in
full generality, the structure of the vevs given above is reproduced
using selection rules for the three point functions, along with the large
$N$ expansion.
Selection rules are responsible for determining
which scalar operators acquire a vev in a given superposition of Ramond
ground states. For example, one can see from (\ref{vv-new}), (\ref{coeff})
that primaries associated with the $(1,1)$ cohomology of $X_4$ acquire a
vev only if both internal and transverse fluctuations of the fuzzball
are non-zero. Moreover, the internal and transverse curves must share
common Fourier modes for the dimension one operators to acquire vevs;
further selection rules on the Fourier modes are needed for dimension
two operators to acquire vevs. This structure can be reproduced in the field
theory, using selection rules for the three point functions. A basic property of such
three point functions is that they are only
non-zero when the total number of operators ${\cal O}^{(1,1)}_{\tilde{\a}}$ with
a given index $\tilde{\a}$ in the correlation function is even. This
implies that the $(1,1)$ chiral primaries only acquire vevs when the fuzzball has internal
excitations. When there are no internal internal fluctuations, there
are no ${\cal O}^{(1,1)}_{\tilde{\a}}$ operators in the superposition defining
the state, so total number of operators ${\cal O}^{(1,1)}_{\tilde{\a}}$ with
a given index $\tilde{\a}$ in the three point function is never even. By
contrast when there are internal excitations there are
${\cal O}^{(1,1)}_{\tilde{\a}}$ operators in the superposition defining
the state, and thus $(1,1)$ chiral primaries generically acquire vevs.
One can also easily see why the operator only acquires a vev if there
are transverse excitations as well. All Ramond ground states
associated with the middle cohomology have
zero R charge, with the corresponding chiral primaries in the NS
sector having the same charge $j_{3}^{NS} = {\textstyle{1\over2}}
N$. Thus a superposition involving only ${\cal O}^{(1,1)}$ operators has a
definite R charge, and a charged operator cannot acquire a
vev. Including transverse excitations means that the superposition
of Ramond ground states contains charged operators, associated
with the other cohomology, does not have definite R
charge and therefore a charged operator can acquire a vev. Note that
this point also explains why superpositions
involving only ${\cal O}^{(1,1)}$ operators
do not have dual fuzzball descriptions within supergravity: no
operator which is dual to a supergravity mode can acquire a vev.
In fact one can go even further in matching field theory one point
functions to the holographic vevs, using the fact that the most
disconnected component of any correlator dominates at large $N$. In
practice this implies that for many calculations the operators involved in the
superposition behave like free harmonic oscillators - the deviation of
the algebra from the free algebra is subleading in $N$ - and thus
even the numerical values of the vevs can be reproduced by harmonic oscillator
algebra! This point was illustrated in detail in \cite{Kanitscheider:2006zf,Kanitscheider:2007wq}
using specific examples, such as elliptic curves.
Given the recently refined dictionary relating supergravity fields to chiral
primaries \cite{Taylor:2007hs}, and the proposed non-renormalization of the chiral ring,
one could hope to push these calculations even further.
Another issue which is nicely demonstrated by the calculations in
\cite{Kanitscheider:2006zf,Kanitscheider:2007wq} is the
indistinguishability of many solutions with supergravity: typically
the vevs extracted from a solution defined by any given curve are
small, comparable with the next order corrections to
supergravity. Thus even fuzzballs which are non-singular in
supergravity generally suffer from indistinguishability.
To summarize: the detailed holographic data extracted from these two charge
geometries supports their interpretation as superpositions of Ramond
ground states. Moreover, the data reinforces the general expectation
that supergravity does not suffice to describe all the fuzzballs. Some
cannot be seen at all in the supergravity approximation, whilst others
cannot be reliably distinguished.
As we shall see later there is as yet no complete understanding of the
space of fuzzball solutions for the 3-charge system: candidate
solutions within supergravity seem to be characterized by a (constrained) set of curves along
with additional parameters and the correspondence with states in the
CFT is unclear. Extracting vevs from such geometries may
thus provide a useful indicator as to the correspondence with black hole
microstates.
\subsubsection{Flight time and mass gap}
Apart from matching the one-point functions, one could also test the
correspondence by matching higher point functions. To make precision
tests is rather hard in general. Firstly to compute even a two-point
function one needs to solve a partial differential equation on the
asymptotically $AdS_3 \times S^3$ background, which for a general
fuzzball solution is not even separable due to lack of
symmetry. Secondly, $n$-point functions in a given geometry are
related to $(n+2)$-point functions at the conformal point, and there
is no reason to expect that these are non-renormalized even for chiral
primaries. Thus the
correlation functions computed in the orbifold CFT will not necessarily match
those in the geometry.
Bearing in mind these caveats, estimates of the two point functions in
symmetric geometries can give indicative information about the
correspondence with CFT states, see
\cite{Lunin:2001jy,Lunin:2002qf,Giusto:2004ip,Mathur:2005zp,Mathur:2005ai}. For example, in the geometric
approximation for the two point function of a massless scalar one
considers null geodesics propagating inwards and then returning back
to the boundary. The (finite) affine length of such a geodesic defines
a characteristic scale or inverse mass gap.
To give an example: one can consider the geometries obtained from a
circular curve in $R^4$ given in (\ref{ez2}). The decoupled region in the
geometry has a metric:
\begin{eqnarray}
ds^2 &=& \sqrt{Q_1 Q_5} \left ( - (r^2 + \mu_n^2) d\hat{t}^2 + r^2 d\hat{z}^2 +
\frac{dr^2}{(r^2 + \mu_n^2)} \right ) \label{nh} \\
&& + \sqrt{Q_1 Q_5}
\left ( d{\q}^2 + \sin^2 {\q} (d \phi + \mu_n d \hat{t})^2 +
\cos^2 {\q} (d \psi - \mu_n d \hat{z} )^2 \right ) + \sqrt{Q_1/Q_5}
ds^2 (X_4), \nonumber} \def\bd{\begin{document}} \def\ed{\end{document}
\end{eqnarray}
where $\mu_n = \mu/n = \sqrt{Q_1 Q_5}/(nR_z)$ and $\hat{z}$ is periodic
with periodicity $2\pi \mu^{-1}$. Null radial geodesics such that $\q
= 0$ and $(\hat{z},\psi)$ are constant satisfy
\begin{equation}
\left ( \frac{dr}{d \hat{t}} \right )^2 = (r^2 + \mu_n^2)^2,
\end{equation}
and thus the total affine length of a geodesic propagating inwards from radial
infinity and back out again is
\begin{equation}
\hat{T} = \int_{0}^{\infty} \frac{dr}{(r^2 + \mu_n^2)} = \frac{\pi}{2
\mu_n}.
\end{equation}
The corresponding energy scale $\mu_n$ characterizes the
mass gap of the spectrum in this sector
of the CFT and since $\mu^{-1}$ is the scale of $\hat{z}$,
the dimensionless energy scale is $1/n$. (This is
the same energy scale that is obtained from the holographic
computation of the R charges, see (\ref{j-char2}).)
This fits with the conjectured
correspondence between these geometries and dual R charge eigenstates,
$(\hat{\cal O}^{R(p,q)}_n)^{N/n}$: excitations in the twist $n$ sector
are of the scale $1/n$. In the effective string picture,
the excitation energy is also
related to the time taken by excitations to
propagate around the circle wound by the string.
In such a simple example, one does not acquire new information from
this computation. In more complicated fuzzball geometries, however,
such as 3-charge bubbling solutions to be discussed later, this
estimation of the mass gap may be a useful indicator of the dual black
hole microstate. At the same time, in geometries which break
rotational symmetry and in which there are many parameters, the flight
time is sensitive to the specific geodesic under consideration and has
to be interpreted with care.
\subsubsection{Geometric quantization} \label{geo-quant}
The fuzzballs are characterized by a curve in an $8$ ($T^4$) or $24$
($K3$) dimensional space. One would like to enumerate them and compare
them with the number of dual microstates, but naively one has an
infinite number of solutions since any classical curve determines a solution.
The map between fuzzball solutions and microstates effectively discretizes the space
of fuzzball solutions. From our earlier discussion it is clear that
the classical curves are due to the states being approximately coherent; the
coefficients in the Fourier expansion of the curve are eigenvalues of quantum
harmonic oscillators.
Alternatively, note that the space of curves is the classical phase space of these
gravitational solutions.
The number of the corresponding quantum states should be obtained by appropriate
quantization. These states should be in correspondence with the states in the
dual CFT. One can thus consider
quantizing this moduli space, and comparing the latter to the
Ramond ground states in the CFT. For fuzzballs with only
transverse excitations, namely (\ref{lun-mat}), such that the
solutions are determined by
a smooth closed non-intersecting curve in four dimensions, the
geometric quantization was carried out in \cite{Rychkov:2005ji}.
It was found that the Fourier coefficients of the curve satisfy chiral boson
commutation relations, with the value of the effective Planck constant
agreeing with that obtained from dualising the fundamental string.
The basic idea of this quantization procedure is as follows. Consider
any classical dynamical system with phase space coordinates satisfying
the standard Poisson brackets
\begin{equation}
\{ q^I, p^J \} = \d^{IJ}.
\end{equation}
Now restrict to a subspace $M$ of the full phase space, parameterized by
some coordinates $x^A$. The induced Poisson brackets can be extracted
from pullback of the symplectic structure onto this subspace. The
symplectic structure on the phase space is given by
\begin{equation}
\Omega = d p^{I} \wedge dq^{I},
\end{equation}
and the pullback $\Omega_M$ of this symplectic structure onto a subspace is given
by
\begin{eqnarray}
\Omega_M &=& \omega_{AB}(x) d x^{A} \wedge dx^B, \\
\omega_{AB} &=& \left ( \frac{\pa p^I}{\pa x^{A}} \frac{\pa q^I}{\pa
x^{B}} - \frac{\pa p^I}{\pa x^{B}} \frac{\pa q^I}{\pa
x^{A}} \right ) . \nonumber} \def\bd{\begin{document}} \def\ed{\end{document}
\end{eqnarray}
Then the induced Poisson brackets are given by the inverse of $\w$,
\begin{equation}
\{ x^A, x^B \} = {\textstyle{1\over2}} \w^{AB}.
\end{equation}
Once one has extracted the Poisson brackets from the symplectic form, one
can quantize in the standard way to find the commutation relations.
In the case of interest, one considers the symplectic form of
supergravity, restricted to the fuzzball solutions of interest, and
extracts the appropriate Poisson brackets. Starting from the
relevant part of the IIB supergravity action (in Einstein frame)
\begin{equation}
S_{IIB} = \frac{1}{2 \k_{10}^2} \int d^{10}x \sqrt{-g} (R - {\textstyle{1\over2}} (\pa
\Phi)^2 - {\textstyle{1\over2}} e^{\Phi} F_3^3),
\end{equation}
the symplectic form is
\begin{equation}
\Omega = \frac{1}{2 \k_{10}^2} \int d\Sigma_{m} J^m
\end{equation}
where the integration is over a Cauchy surface $\Sigma$ and the
symplectic current $J^m$ is given by \cite{Crnkovic,Wald,Rychkov:2005ji}
\begin{eqnarray}
J^{m} &=& - \d \G^{m}_{ab} \wedge \d (\sqrt{-g} g^{ab}) + \d \G^{b}_{ab}
\wedge \d (\sqrt{-g} g^{am}) - \d (\sqrt{-g} e^{-\Phi} F^{mab} )
\wedge \d C_{ab} \nonumber} \def\bd{\begin{document}} \def\ed{\end{document} \\
&& - \delta (\sqrt{-g} \pa^m \Phi) \wedge \d \Phi.
\end{eqnarray}
Restricting to the fuzzball solutions (\ref{lun-mat}) characterized by a curve
$F_i(v)$ in $R^4$, the pullback $\Omega_M$ of the symplectic form can be
explicitly evaluated to give
\begin{equation} \label{op-s}
\Omega_M = \frac{1}{2 \a} \int \d (\pa_v F_i(v)) \wedge \d F_{i}(v) dv
\end{equation}
which implies the Poisson bracket
\begin{equation}
\{ F_i(v), \pa_v F_j(v') \} = \a \d_{ij} \d (v - v').
\end{equation}
The parameter $\a$ is found by explicit computation of the symplectic
form to be given by
\begin{equation}
\a = \pi \frac{g_s^2}{R_z^2 V},
\end{equation}
in agreement with the value obtained by duality from the fundamental
string. Thus as previously stated the defining curve of the fuzzball
solution satisfies the standard Poisson bracket relations of chiral
bosons. One can next promote the Poisson bracket to a quantum
commutation relation, and introduce harmonic oscillators
$(\hat{a}^i_n, (\hat{a}^i_n)^{\dagger})$ for the
curve. To count the number of black hole microstates obtained from
these geometries, one must then restrict to curves satisfying
the relation constraining the total D1-brane charge:
\begin{equation}
Q_1 = \frac{Q_5}{L} \int_{0}^L | \pa_v F^i |^2 dv.
\end{equation}
As expected this reduces to counting the degeneracy of states at level $N_1
N_5$ in a system of four chiral bosons, namely counting the degeneracy
of states satisfying
\begin{equation}
N_1 N_5 = \sum_{n =1}^{\infty} n \langle (\hat{a}^i_{n})^{\dagger} \hat{a}^i_n \rangle,
\end{equation}
which gives an entropy
\begin{equation}
S \approx 2 \pi \sqrt{\frac{2}{3} N_1 N_5}.
\end{equation}
It would be interesting to see if the quantization of general fuzzball
geometries with internal excitations gives an analogous
answer, corresponding to eight or twenty-four bosons for $T^4$ and
$K3$ respectively. Given that fuzzballs with only internal excitations
are not visible in supergravity, it seems possible that the symplectic
form in the general case would have a more complicated structure than
in (\ref{op-s}), but the calculation has not been carried out.
Whilst the geometric quantization reproduces the expected entropy of
the 2-charge microstates, one must clearly view this calculation with
some caution. The 2-charge system does not have a horizon in
supergravity, so the average fuzzball geometry must have a scale
which is sub-string scale and thus higher derivative
corrections to supergravity are non-negligible for these
geometries. It is not clear that quantizing the extrapolation of
fuzzball geometries to the supergravity regime will in general give the
correct counting, even though it does in this case.
Note that in this case the number of microstates accounted for by the
fuzzball solutions was also estimated using (probe) supertubes
\cite{Mateos:2001qs} in the dual D0-F1 system. In \cite{Palmer:2004gu} and \cite{Bak:2004kz},
the quantum states of the supertube were counted
by directly quantizing the linearized Born-Infeld action near a
round tube. This gives an entropy $S = 2\pi \sqrt{2
N_{D0}N_{F1}}$, where $N_{D0}$ is the D0-brane charge and $N_{F1}$
is the fundamental string charge. Backreacting a supertube with given
profile in $R^4$ onto
the supergravity fields, and then dualizing, should give the D1-D5 fuzzball
solutions characterized by the same profile in $R^4$. Thus the entropy
obtained from quantizing the Born-Infeld action indicates the number
of 2-charge fuzzball solutions, with the entropy matching the expected
value. This approach is less direct than that discussed above, and is
hard to generalize to the 3-charge system where there is no such
duality to a supertube description.
\subsection{Open questions}
Even given the detailed analysis already made in this system, there
remain many interesting open questions and lessons for black holes
with macroscopic horizons. Firstly, the recent demonstration that the chiral ring is
non-renormalized, along with the matching between orbifold CFT
operators and supergravity fields \cite{Taylor:2007hs}, would allow the vevs to be computed
systematically at large $N$, and for the proposed correspondence to be
tested to even higher precision using the holographic methods.
Furthermore, there is a detailed map from smooth geometries to black
hole microstates, but not vice versa. Put differently, it is
not understood directly in the D1-D5 system how a curve emerges.
It is clearly important to understand this better,
not least because this might lead to clues as to the defining data of
3-charge D1-D5-P fuzzball solutions. Note that a given R charge
eigenstate does not in general appear to have a smooth supergravity
description, whereas the smooth geometries correspond to large
superpositions of R charge eigenstates with the specific superposition
determined by the classical curve. Thus the natural geometric basis
does not coincide with the natural field theory basis, and the
geometric basis does not sample well states with very small R charge.
Given the matching between geometries and microstates, one could
consider coarse-graining over the fuzzball geometries, to understand
how black hole properties emerge. Of course in this specific system
the resulting black hole does not have a regular horizon in
supergravity, so the coarse-graining can at best provide an indication
of what happens for macroscopic black holes, but nonetheless it may be
informative to explore this issue.
In general the fact that many of the fuzzball geometries have very
small parameters and are not well described using only supergravity
could be interpreted as an indication that they will develop a horizon
when higher derivative corrections are taken into account. For fuzzballs
characterized by a generic curve this however seems unlikely: there is
no singularity in the supersymmetric supergravity solution, so the
higher derivative corrections are bounded.
Results in this section indicate that a more complete understanding of this
system rests on understanding fuzzballs in the stringy regime. This is
not unexpected, as the black hole does not have a macroscopic horizon,
and thus one would clearly like to develop the fuzzball proposal for
black holes with macroscopic horizons. As we shall see, however, even
for large black holes, one is unlikely to evade the need to go beyond
supergravity. Thus this 2-charge system may be useful as the simplest test case
for both sharpening the definition of fuzzballs outside the
supergravity regime and for developing calculational techniques to
handle stringy fuzzballs.
\section{Three and four charge systems}
In the previous section we have discussed fuzzball solutions for the two
charge system, and shown how AdS/CFT methodology can be used to make
detailed tests of the correspondence between such geometries and CFT
states. At the same time, the two charge black hole does not have a
macroscopic horizon, so one would like to make progress with a system
which does have a horizon in supergravity.
As a first step one would like to develop the fuzzball picture for
asymptotically flat, (near) supersymmetric,
black holes in four and five dimensions which have
finite area horizons in supergravity. This class
includes static five-dimensional black holes with three charges, the
first black hole whose entropy was explained microscopically in string
theory by Strominger and Vafa \cite{Strominger:1996sh}, along with the rotating BMPV
generalization \cite{Breckenridge:1996is}.
In four dimensions black holes with a macroscopic horizon have four
charges; for reviews of black hole solutions see \cite{Youm:1997hw}.
There are a variety of reasons for developing first the picture for
supersymmetric black holes. Perhaps most important is that
in constructing candidate fuzzball
solutions for supersymmetric black holes, one is not forced to solve
the (highly non-linear) supergravity equations. One instead first solves the
first order supersymmetry equations and then enforces where
necessary additional restrictions arising from the equations of
motion. Moreover, one can make use of the systematic classifications
of supersymmetric solutions of supergravity that have been derived in
recent years; in fact, almost all candidate three charge solutions
derive from the classification of minimal supergravity in five
dimensions \cite{Gauntlett:2002nw}. Once supersymmetry is broken, one loses these powerful
tools, and, as we will later review, very few fuzzball solutions are
known.
Detailed microscopic explanations of the entropy are also only
currently possible for supersymmetric black holes. Following the
steps of Strominger and Vafa, one relates states in a D-brane system at weak
coupling to a black hole at strong coupling, but the degeneracy of
states in the weakly coupling theory can only be compared to the
entropy of the black hole at strong coupling if supersymmetry implies
non-renormalization. This is the case for (the leading order term in)
the entropy of supersymmetric black holes with large charges.
Without detailed knowledge of the properties of black hole
microstates, it is hard to probe whether candidate fuzzball
geometries do indeed correspond to such microstates. Even when
one has a description in terms of D-branes at weak coupling, one
cannot generically make a precise map between all data in this
theory and that extracted from the fuzzball geometry.
One can however develop a precision map in cases where one can use AdS/CFT
technology. Thus to push the fuzzball proposal further it is natural
to consider black holes which admit an $AdS_3$ factor
in their near horizon regions, and relate the fuzzball geometries to
states in the dual two dimensional conformal field theories. AdS/CFT
technology is rather less developed for five and four dimensional
black holes with $AdS_2$ near horizon regions, and thus we will not
focus on such black holes here, although candidate fuzzball geometries
have been constructed in these cases.
Before moving on to reviewing relevant properties of the black holes,
one should note that whilst supersymmetric solutions can have finite
horizon area they do not have non-zero temperature. Thus one does not
expect to address how temperature and Hawking radiation emerges in
such a system. We will return to this issue in section \ref{open}.
\subsection{Black hole geometry and D-branes}
In this section we will review relevant
properties of 3-charge black hole solutions. For what follows it is
convenient to discuss solutions both of M theory compactified on a
Calabi-Yau manifold, and of type IIB compactified on $X_4$ which is
either $T^4$ or $K3$. Our conventions for the supergravity field
equations are given in appendix \ref{conv}.
\subsubsection{D1-D5 solutions}
Let us begin with the type IIB solutions. Supersymmetric rotating (BMPV)
strings wrapping a circle and carrying D1 charge $Q_1$, D5 charge
$Q_5$ along with momentum charge
$Q_p$ along the circle have a metric of the form
\begin{eqnarray}
ds^2 &=& \frac{1}{\sqrt{h_1 h_5}} \left ( - (dt^2 - dz^2) +
\frac{Q_p}{r^2} (dz - dt)^2 \right ) \nonumber} \def\bd{\begin{document}} \def\ed{\end{document} \\
&&
+ \sqrt{h_1 h_5} \left ( dr^2 +
r^2 (d \q^2 + \cos^2 \q d \psi^2 + \sin^2 \q d \phi^2) \right )
\label{bmpv1} \\
&&
+ \frac{2 \sqrt{Q_1 Q_5 Q_p}}{r^2 \sqrt{h_1 h_5}} (L_1 - L_2) ( dt -dz ) (
\cos^2 \q d \psi - \sin^2 \q d \phi)
+ \sqrt{\frac{h_1}{h_5}} ds^2(X_4), \nonumber} \def\bd{\begin{document}} \def\ed{\end{document}
\end{eqnarray}
with $h_i = 1 + Q_i/r^2$ and where
$ds^2(X_4)$ denotes the metric on $T^4$ or $K3$ respectively. The
corresponding RR 2-form potential
and dilaton are given by
\begin{eqnarray}
e^{-2 \Phi} &=& \frac{h_5}{h_1}; \\
C &=& (h_1^{-1} - 1) dt \wedge dz - Q_5 \cos^2 \q d \psi \wedge d
\phi \nonumber} \def\bd{\begin{document}} \def\ed{\end{document} \\
& & + \frac{\sqrt{Q_1 Q_5 Q_p}}{(r^2 + Q_1)} (L_2 - L_1) (dt -dz) (
\cos^2 \q d \psi - \sin^2 \q d \phi). \nonumber} \def\bd{\begin{document}} \def\ed{\end{document}
\end{eqnarray}
The
angular momentum charge is proportional to $(L_1 - L_2)$ and will be given
below.
This solution follows from the supersymmetric limit of general non-extremal 3-charge black
hole (actually black string) solutions of type IIB with rotation constructed
in \cite{Cvetic:1996xz}. The metric for these solutions is
\begin{eqnarray}
ds^2 &=& \frac{1}{\sqrt{H_1 H_5}} \left ( - f(dt^2 - dz^2) + m (s_p dz
- c_p dt)^2 + m (a_1 \cos^2 \q d \psi + a_2 \sin^2 \q d \phi)^2 \right
) \nonumber} \def\bd{\begin{document}} \def\ed{\end{document} \\
&&
+ \sqrt{H_1 H_5} \left ( \frac{r^2 dr^2}{ (r^2 + a_1^2)(r^2 + a_2^2) -
m r^2} + d \q^2 + \cos^2 \q d \psi^2 + \sin^2 \q d \phi^2 \right )
\label{cvetic-youm} \\
&&
+ \frac{2m}{\sqrt{H_1 H_5}} \left ( \cos^2 \q ( (a_1 c_1 c_5 c_p - a_2
s_1 s_5 s_p) dt + (a_2 s_1 s_5 c_p - a_1 c_1 c_5 s_p) dz) d \psi\right
) \nonumber} \def\bd{\begin{document}} \def\ed{\end{document} \\
&& + \frac{2m}{\sqrt{H_1 H_5}} \left (
\sin^2 \q ( (a_2 c_1 c_5 c_p - a_1
s_1 s_5 s_p) dt + (a_1 s_1 s_5 c_p - a_2 c_1 c_5 s_p) dz) d \phi
\right ) \nonumber} \def\bd{\begin{document}} \def\ed{\end{document} \\
&& + (a_2^2 - a_1^2) \frac{H_1 + H_5 - f}{\sqrt{H_1 H_5}} (\sin^4 \q d
\phi^2 - \cos^4 \q d \psi^2) + \sqrt{\frac{H_1}{H_5}} ds^2(X_4), \nonumber} \def\bd{\begin{document}} \def\ed{\end{document}
\end{eqnarray}
where the solution is parameterized by $(\d_1,\d_5,\d_p)$ and we use
the abbreviations $s_i = \sinh \d_i$, $c_i = \cosh \d_i$.
The functions $(H_1,H_5,f)$ are given by
\begin{equation}
H_i = f + m \sinh^2 \d_i, \qquad
f = r^2 + a_1^2 \sin^2 \q + a_2^2 \cos^2 \q.
\end{equation}
The RR 2-form potential and the dilaton are given by
\begin{eqnarray}
C &=& \frac{m \cos^2 \q}{H_1} \left ( ( a_2 c_1 s_5 c_p - a_1
s_1 c_5 s_p) dt + (a_1 s_1 c_5 c_p - a_2 c_1 s_5 s_p) dz \right )
\wedge d\psi \\
&& + \frac{m \sin^2 \q}{H_1} \left ( ( a_1 c_1 s_5 c_p - a_2
s_1 c_5 s_p) dt + (a_2 s_1 c_5 c_p - a_1 c_1 s_5 s_p) dz \right )
\wedge d\phi \nonumber} \def\bd{\begin{document}} \def\ed{\end{document} \\
&& - \frac{m s_1 c_1}{H_1} dt \wedge dz - \frac{m s_5 c_5}{H_1} (r^2 +
a_2^2 + m s_1^2) \cos^2 \q d\psi \wedge d\phi; \nonumber} \def\bd{\begin{document}} \def\ed{\end{document} \\
e^{-2 \Phi} &=& \frac{H_5}{H_1}. \nonumber} \def\bd{\begin{document}} \def\ed{\end{document}
\end{eqnarray}
Thus the solution is labeled by six parameters $(m,\d_i,a_1,a_2)$
along with the coupling constants and the
two moduli $(R_z,V)$. The former is the radius of the $z$ circle and
the latter is related to the volume $v$ of $X_4$ by $v = (2 \pi)^4 V$. The
solution is asymptotically flat, approaching $R^{4,1} \times S^1
\times X_4$ as $r \rightarrow \infty$; here $S^1$ is the $z$ circle.
The horizons are located at
\begin{equation}
(r^2 + a_1^2) (r^2 + a_2^2) = m r^2,
\end{equation}
and the topology of the horizon is $S^1 \times S^3 \times X_4$ where
again the $S^1$ direction is associated with the $z$ circle. This
solution is hence a black string, and compactifying to five dimensions
over $S^1 \times X_4$ produces a five-dimensional three charge
rotating black hole.
The conserved charges and angular momenta are given by
\begin{eqnarray}
Q_1 &=& m s_1 c_1 = \frac{g \a^{'3} N_1}{V}; \label{charges} \\
Q_5 &=& m s_5 c_5 = g \a' N_5; \nonumber} \def\bd{\begin{document}} \def\ed{\end{document} \\
Q_p &=& m s_p c_p = \frac{g^2 (\a')^4}{V R_z^2} N_p, \nonumber} \def\bd{\begin{document}} \def\ed{\end{document} \\
J_{\psi} &=& - \frac{\pi m}{4 G_5} (a_1 c_1 c_5 c_p - a_2 s_1 s_5
s_p) \nonumber} \def\bd{\begin{document}} \def\ed{\end{document} \\
J_{\phi} &=& - \frac{\pi m}{4 G_5} (a_2 c_1 c_5 c_p - a_1 s_1 s_5
s_p) , \nonumber} \def\bd{\begin{document}} \def\ed{\end{document}
\end{eqnarray}
where $(N_1,N_5,N_p)$ are the integral quantized charges and
\begin{equation}
\frac{\pi}{4 G_5} = \frac{R_z V}{g^2 (\a')^4}.
\end{equation}
The mass is given by
\begin{equation}
M = \frac{\pi m}{4 G_5} (s_1^2 + s_5^2 + s_p^2 + \frac{3}{2}).
\end{equation}
Regular extremal black holes which saturate a BPS bound
are obtained in the limit $m \rightarrow 0$ and $\d_i \rightarrow
\infty$ with $Q_i = m s_i^2 $, $L_1 = a_1/\sqrt{m_1}$ and $L_2 =
a_2/\sqrt{m_2}$ held fixed. The solution in this limit becomes
(\ref{bmpv1}).
The decoupling limit of these solutions is obtained by focusing on the
region $r^2 \ll Q_1, Q_5$; then $H_1$ and $H_5$ may be approximated by
$H_1 \approx Q_1$ and $H_5 \approx Q_5$. One also focuses on near
extremal solutions for which $Q_1, Q_5 \gg m,a_1^2,a_2^2$ so that
$m \sinh \d_1 \sinh \d_5 \approx m \cosh \d_1 \cosh \d_5 \approx
\sqrt{Q_1 Q_5}$. In this limit the metric can be written as
\begin{eqnarray} \label{btzz}
ds^2 &=& - l^2 (\rho^2 - M_3 + \frac{J_3^2}{4 \rho^2}) d\t^2 +
({\rho^2} - M_3 + \frac{J_3^2}{4 \rho^2})^{-1} l^2 d\rho^2 +
l^2 \rho^2 (d\varphi - \frac{J_3}{2 \rho^2} d\t)^2 \nonumber} \def\bd{\begin{document}} \def\ed{\end{document} \\
&& + l^2 d \q^2 + \sin^2 \q (d \phi + \frac{R_z}{l} (a_1 c_p - a_2
s_p) d\varphi + \frac{R_z}{l} (a_2 c_p - a_1
s_p) d\tau)^2 \\
&& + l^2 \cos^2 \q (d\psi + \frac{R_z}{l} (a_2 c_p - a_1
s_p) d\varphi + \frac{R_z}{l} (a_1 c_p - a_2
s_p) d\tau)^2 + \frac{\sqrt{Q_1}}{\sqrt{Q_5}} ds^2(X_4), \nonumber} \def\bd{\begin{document}} \def\ed{\end{document}
\end{eqnarray}
with new coordinates $\varphi = z/l R_z$ and $\tau = t /l R_z$ with $l^2
= \sqrt{Q_1 Q_5}$ and
\begin{equation}
\rho^2 = \frac{R_z^2}{l^2} (r^2 +(m - a_1^2 - a_2^2) s_p^2 + 2 a_1 a_2
s_p c_p).
\end{equation}
The non-trivial six-dimensional metric is a twisted fibration of $S^3$
over the BTZ black hole, with
the mass and angular momentum parameters $(M_3,J_3)$ of the BTZ metric being
\begin{eqnarray}
M_3 &=& \frac{R_z^2}{l^2} ( (m-a_1^2 - a_2^2) \cosh 2 \d_p + 2 a_1
a_2 \sinh 2 \d_p); \\
J_3 &=& \frac{R_z^2}{l^2} ( (m-a_1^2 - a_2^2) \sinh 2 \d_p + 2 a_1
a_2 \cosh 2 \d_p). \nonumber} \def\bd{\begin{document}} \def\ed{\end{document}
\end{eqnarray}
Restricting to the extremal limit in which $M_3 = J_3$ gives
\begin{eqnarray}
ds^2 &=& - l^2 ({\rho} - \frac{M_3}{2 \rho})^2 d\t^2 +
({\rho} - \frac{M_3}{2 \rho})^{-2}
l^2 d\rho^2 +
l^2 \rho^2 (d\varphi - \frac{M_3}{2 \rho^2} d\t)^2 \nonumber} \def\bd{\begin{document}} \def\ed{\end{document} \\
&& + l^2 (d \q^2 + \sin^2 \q (d \phi + \frac{R_z}{l} \sqrt{Q_p} (L_1
- L_2) (d\varphi - d\tau ) )^2 \\
&& + l^2 \cos^2 \q (d\psi + \frac{R_z}{l} \sqrt{Q_p} (L_1 - L_2)
(d\tau - d \varphi))^2 + \frac{\sqrt{Q_1}}{\sqrt{Q_5}} ds^2(X_4), \nonumber} \def\bd{\begin{document}} \def\ed{\end{document}
\end{eqnarray}
which is a twisted fibration of $S^3$ over the extremal BTZ black
hole.
The entropy of black hole solutions is given by
\begin{equation}
S = \frac{2 \pi \rho_+}{4 G_3}
= 2 \pi \left ( \sqrt{ N N_p - {\textstyle{1\over 4}} (J_{\phi} - J_{\psi})^2} + \sqrt{N
\bar{N}_p - {\textstyle{1\over 4}} (J_{\phi} + J_{\psi})^2} \right ). \label{entropy1}
\end{equation}
where the effective three dimensional Newton constant is
\begin{equation}
\frac{1}{4 G_3} = N = N_1 N_5
\end{equation}
and $\rho_+$ is the location of the outer event horizon. The Hawking
temperature of the black hole is given by
\begin{equation} \label{btz-temp}
T_{H} = \frac{(\r_{+}^2 - \rho_{-}^2)}{2 \pi \r_{+}},
\end{equation}
where $\rho_{-}$ is the location of the inner event horizon of the black hole.
\bigskip
Given the asymptotically $AdS_3 \times S^3$ region, one can obtain the
corresponding data in the dual conformal field theory. In particular,
making use of the formulae for the holographic vevs previously derived, one
can extract the non-zero components of the stress energy tensor and R
symmetry currents. Recall that the vevs are
\begin{eqnarray}
\langle T_{ij} \rangle &=& \frac{N}{2 \pi} (g_{(2) ij} + {\textstyle{1\over 4}} {\cal
A}^{+\a}_{(i} {\cal A}^{+\a}_{j)} + {\textstyle{1\over 4}} {\cal
A}^{-\a}_{(i} {\cal A}^{-\a}_{j)} ); \nonumber} \def\bd{\begin{document}} \def\ed{\end{document} \\
\langle J^{i \pm \a} \rangle &=& \frac{N}{8 \pi} (g_{(0) ij} \pm
\epsilon_{ij}){\cal
A}^{\pm\a j} ,
\end{eqnarray}
where $g_{(n) ij}$ and ${\cal A}^{\pm \a}_{ij}$ are terms in the
asymptotic expansion of the three dimensional metric and gauge fields
respectively. The relevant three dimensional metric is the BTZ metric
in the first line of (\ref{btzz}) which in Fefferman-Graham form is
\begin{eqnarray}
ds^2 &=& \frac{l^2}{z^2} (dz^2 + dx^i dx^j (g_{(0) ij} + z^2 g_{(2)ij}
+ \cdots)); \\
&=& \frac{l^2}{z^2} (dz^2 + (-d\t^2 + d\varphi^2) + {\textstyle{1\over2}} M_3 z^2 (d\t^2 +
d\varphi^2) + J_3 z^2 d\t d\varphi + \cdots), \nonumber} \def\bd{\begin{document}} \def\ed{\end{document}
\end{eqnarray}
whilst the three dimensional gauge
fields are respectively
\begin{eqnarray}
{A}^{+ 3} &=& {\cal A}^{+3} + \cdots = \frac{R_z}{l} e^{\d_p} (a_1
-a_2) (d\varphi - d\t); \\
{A}^{- 3} &=& {\cal A}^{-3} + \cdots = \frac{R_z}{l} e^{-\d_p} (a_2
+ a_1) (d\varphi + d\t). \nonumber} \def\bd{\begin{document}} \def\ed{\end{document}
\end{eqnarray}
Putting these values into the holographic formulae gives
\begin{eqnarray} \label{h-vevs2}
\langle T \rangle &=& \frac{1}{2\pi} (N_p (dx^+)^2 + \bar{N}_p
(dx^-)^2); \\
\langle J^{+3} \rangle &=& \frac{\sqrt{N N_p}}{2\pi} (L_1 - L_2) dx^+ \equiv
\frac{1}{2 \pi} j^{+} dx^+; \nonumber} \def\bd{\begin{document}} \def\ed{\end{document} \\
\langle J^{-3} \rangle &=& - \frac{\sqrt{N \bar{N}_p}}{2 \pi} (L_1 + L_2) dx^- \equiv
\frac{1}{2 \pi} j^{-} dx^-, \nonumber} \def\bd{\begin{document}} \def\ed{\end{document}
\end{eqnarray}
where we define
\begin{equation}
N_p = \frac{V R_z^2}{4 g^2 (\a')^4} m e^{2 \d_p}; \qquad
\bar{N}_p = \frac{V R_z^2}{4 g^2 (\a')^4} m e^{-2 \d_p},
\end{equation}
and introduce light cone coordinates $dx^{\pm} = l (d\tau \pm d
\varphi)$, which have periodicity $2\pi$. Define $h = \int \langle
T_{++} \rangle$ and $\bar{h} = \int \langle T_{--} \rangle$, and similarly
$j^{+} = \int \langle J^{+3} \rangle dx^{+}$, $j^{-} = \int \langle
J^{-3} \rangle dx^-$. Then for the product of global $AdS_3$ with
$S^3$, $j^{+} = j^{-} = L_1 = L_2 = 0$ and
$h = \bar{h} = N_p = \bar{N}_p = - N/4$. For an extremal BTZ
black hole, such that $M_3 = J_3$ one obtains
$\bar{N}_p = 0$ and
\begin{equation}
h = N_p; \qquad
j^{+} = \sqrt{N N_p} (L_1 - L_2) \equiv J_{\phi}; \qquad
\bar{h} = j^- = 0,
\end{equation}
where $J_{\phi}$ is the angular momentum with respect to
asymptotically flat infinity given in (\ref{charges}).
\bigskip
One can reduce the black string solutions over the $z$
circle and $X_4$ to produce a three charge rotating
black hole solution of five-dimensional supergravity, as was indeed
done in the original BMPV paper \cite{Breckenridge:1996is}.
For the supersymmetric case this leads to a five dimensional metric
of the form
\begin{equation} \label{5d3metric}
ds^2 = \l^{-2/3} (dt + k) ^2 + \l^{1/3} (dr^2 + r^2 d \Omega_3^2)
\end{equation}
where
\begin{eqnarray}
\l &=& h_1 h_5 h_P = (1 + \frac{Q_1}{r^2})(1 + \frac{Q_5}{r^2}) ( 1 +
\frac{Q_p}{r^2}); \\
k &=& \frac{\sqrt{Q_1 Q_5 Q_p}}{r^2} (L_2 - L_1) (
\cos^2 \q d \psi - \sin^2 \q d \phi). \nonumber} \def\bd{\begin{document}} \def\ed{\end{document}
\end{eqnarray}
Setting $k=0$ gives the 3-charge Strominger-Vafa black hole \cite{Strominger:1996sh},
with the metric being of the same form as the 2-charge metric
(\ref{5dmetric}). In this case there is no curvature singularity at $r=0$,
see (\ref{curv1}), (\ref{curv2}), but instead a regular horizon. The
same is true for the rotating BMPV 3-charge black hole. The near
horizon region of the five-dimensional solutions is $AdS_2 \times S^3$
which is less amenable to detailed holographic analysis\footnote{
Note though that in this case the $AdS_2$ factor originates from a
reduction of the extremal BTZ over the compact boundary direction,
so one would anticipate the dual CFT to be a chiral
CFT \cite{Boonstra:1998yu} and thus that there is
better control over the duality.}
than the
six-dimensional $AdS_3 \times S^3$ near horizon region.
For this reason, we will focus on fuzzball geometries
for the black string solutions.
\subsubsection{M theory solutions}
It is also useful to review briefly certain three charge solutions of M theory
compactified on a Calabi-Yau. For simplicity let us first give
the supersymmetric eleven-dimensional solution for a toroidal
compactification describing orthogonally intersecting M2-branes:
\begin{eqnarray}
ds^2 &=& - \left ( \frac{1}{H_1 H_2 H_3} \right )^{2/3} (dt + k)^2 + (H_1 H_2
H_3)^{1/3} dx^m dx^m \\
&& + \left ( \frac{H_2 H_3}{H_1^2} \right )^{1/3} (dy_1^2 + dy_2^2) +
\left ( \frac{H_1 H_3}{H_2^2} \right )^{1/3} (dy_3^2 + dy_4^2)
+ \left ( \frac{H_1 H_2}{H_3^2} \right )^{1/3} (dy_5^2 + dy_6^2), \nonumber} \def\bd{\begin{document}} \def\ed{\end{document}
\end{eqnarray}
where $x^m$ are coordinates on $R^4$ and the four form is
\begin{equation}
F = F^1 \wedge dy_1 \wedge dy_2 + F^2 \wedge dy_3 \wedge dy_4 +
F^3 \wedge dy_5 \wedge dy_6.
\end{equation}
where the two forms can be written as
\begin{equation}
F^{a} = \frac{1}{2} d (H_{a}^{-1} (dt + k)),
\end{equation}
with $a = 1,2,3$. The solutions are defined by three harmonic
functions $(H_1,H_2,H_3)$ on $R^4$ along with the one form $k$ on $R^4$:
\begin{eqnarray}
H_a &=& \left (1 + \frac{Q_a}{r^2} \right ); \label{data222} \\
k &=& \sqrt{Q_1 Q_2 Q_3} \frac{\w}{r^2} (\cos^2 \q d \psi - \sin^2 \q d\phi),
\nonumber} \def\bd{\begin{document}} \def\ed{\end{document}
\end{eqnarray}
where the metric on $R^4$ is
\begin{equation}
ds^2 = dr^2 + r^2 d \q^2 + \cos^2 \q d\psi^2 + \sin^2 \q d \phi^2
\end{equation}
and $dk$ is self-dual in this metric, namely $dk - \ast_4 dk = 0$.
The charges $Q_a$ are related to the
integral M2-brane charges via $Q_a = l_p^2 V_a/(V_1 V_2 V_3)$ where
$(2\pi)^2 V_a$ is the volume of each $T^2$ and $l_p$ is the Planck
length, related to the eleven-dimensional Newton constant $G_{11}$ via
\begin{equation}
\frac{1}{16 \pi G_{11}} = \frac{1}{(2 \pi)^8 l_p^9}.
\end{equation}
This metric describes an extremal 3-charge
rotating black hole, with the decoupling region geometry being
$AdS_2 \times S^3 \times T^6$. The 3 charges are the $Q_a$ with the
angular momenta being
\begin{equation}
J_{\psi} = - J_{\phi} = \frac{\pi}{4 G_5} \w \sqrt{Q_1 Q_2 Q_3} = \w
\sqrt{N_1 N_2 N_3},
\end{equation}
and the entropy is given by
\begin{equation}
S = 2 \pi \sqrt{N_1 N_2 N_3 (1 - \w^2)}.
\end{equation}
Clearly this is the same formula as the extremal BPS limit of
(\ref{entropy1}). Indeed the type IIB on $T^4$ and M theory on $T^6$ solutions are
related by dualities; first reduce the latter on the M theory circle $y_6$
to obtain a $(D2_{y_1 y_2} \perp D2_{y_3 y_4} \perp F1_{y_5})$ solution of type IIA and
then T dualise on $(y_3,y_4,y_5)$ to obtain a $(D4_{y_1 y_2 y_3 y_4 y_5}
\perp D1_{y_5} \perp P_{y_5})$ solution of type IIB. Here $Dp_{y_1
\cdots y_p}$ denotes the spatial directions wrapped by the
$Dp$-brane.
Analogous three charge rotating five-dimensional black hole
solutions for Calabi-Yau compactifications are well-known; the entropy
of such wrapped M-brane configurations was first discussed in
\cite{Maldacena:1997de}. Moreover by
replacing $R^4$ by a Taub-NUT space and compactifying on the NUT circle one
can obtain an extremal four-charge solution in four
dimensions. However, for us the key feature of all of these solutions
is that they give a near horizon
geometry with an $AdS_2$ factor, rather than $AdS_3$.
Candidate fuzzball solutions have been constructed for
these black holes; indeed as we shall see almost all supersymmetric
fuzzball solutions
are constructed from a set of defining data, which can be used to
build fuzzballs for both type IIB and M theory black holes.
Detailed holographic analysis of type IIA and M theory systems
is however more subtle than in the IIB solutions which have $AdS_3$ decoupling
regions. One can easily build other three charge M theory solutions which
do have $AdS_3$ near horizon regions, for example, by intersecting
$(M2 \perp M5 \perp P)$ along a string.
However the dual conformal
field theory is far less understood in such cases, so again it is more
natural to explore first the D1-D5-P system of type IIB.
\subsection{CFT microstates} \label{3cft}
To find the microstates of the (non-extremal) 3-charge D1-D5-P black
hole, we need to consider states in the D1-D5 CFT
with $h = N_{p}$ and $\bar{h} = \bar{N}_p$. The asymptotic number of
distinct states of this CFT can be obtained immediately by Cardy's formula \cite{Cardy}
\begin{eqnarray}
d &=& \Omega \bar{\Omega}; \\
\Omega &=& \exp (2 \pi \sqrt{c N_p/6}); \qquad
\bar{\Omega} = \exp (2 \pi \sqrt{c \bar{N}_p/6}), \nonumber} \def\bd{\begin{document}} \def\ed{\end{document}
\end{eqnarray}
where $c = 6 N_1 N_5$ is the central charge. Clearly this formula
exactly reproduces the Bekenstein-Hawking entropy of the non-rotating
black hole, which is somewhat surprising, as the states are far from
supersymmetric when $\bar{N}_p \gg 1$.
Cardy's formula gives the entropy of a 3-charge
black hole in a canonical ensemble, in which the R-charges are not
fixed. However, just as for the 2-charge black hole, the majority of
states being counted have zero R charge. Thus the difference
between the density of states with zero R charge $d_0$ and the total
density of states $d$ is subleading for large $N$. The density of
(supersymmetric) states with fixed and large $j^{+}$ and $h =
N_{p}$ was computed by \cite{Breckenridge:1996is}, in the case where
$X_4 = K3$. This gives
\begin{equation}
S = 2 \pi \sqrt{N N_p - (j^{+})^2},
\end{equation}
exactly reproducing the Bekenstein-Hawking entropy (\ref{entropy1}).
Many subsequent works have been devoted to refining the computation of
the black hole entropy; in particular, in recent times, substantial
effort has gone into reproducing subleading terms in the entropy on
both sides of the correspondence, see the review \cite{Sen:2007qy}. In the bulk this involves evaluating
higher derivative corrections on the leading order solution, whilst
in the CFT one needs subleading terms in the asymptotic expansion of
the degeneracy of states. In computing the latter, one can use an appropriate
supersymmetric index, the elliptic genus in the case of $K3$ and a
topological partition function in the case of $T^4$.
Whilst the index is clearly a useful tool in the counting of BPS states,
it is important to derive more properties of
the microstates being counted, in order to deduce the corresponding
properties of
fuzzball solutions. There is surprisingly little literature on the
relevant CFT states; for example, no correlation functions for
non-primary operators have been explicitly computed even in the
orbifold theory.
At the same time, one can already infer information about typical
black hole microstates from the long/short string picture introduced in
\cite{Mald}, which more formally corresponds to the decomposition of
the Hilbert space of the orbifold theory into twisted sectors. The
supersymmetric 3-charge microstates have left moving excitation level
$N_p$ and no right moving excitations. In the sector of twist $n_i$
the momentum is quantized in units of
$1/n_i$, although the total momentum is necessarily integral. This
immediately implies \cite{Mald} that the entropy is dominated by the highest twist
(twist $N$) sector, or equivalently by the longest strings, as the momentum
$N_p$ can be partitioned maximally in this sector. Thus to establish
properties of a typical microstate of a Strominger-Vafa black hole it
may suffice to explore states in this sector.
\subsection{Survey of fuzzball solutions}
The key step in constructing fuzzball solutions for the two charge
D1-D5 system was to dualize known supergravity solutions for a
fundamental string carrying an arbitrary profile wave. The three and
four charge systems cannot however be dualized to any analogous
system, and thus there is no systematic construction of fuzzball
solutions corresponding to black hole microstates.
Instead families and isolated examples of horizon-free, non-singular
solutions with the correct conserved charges have been constructed
using standard techniques for finding supergravity solutions. In
particular, the two principle construction techniques are:
\bigskip
\noindent {\bf 1. Supersymmetric classification techniques:}
In recent years there has been considerable progress in classifying
supersymmetric solutions of supergravity theories. In the context of
the fuzzball proposal, the classification of solutions of
minimal (ungauged) supergravity in five dimensions
\cite{Gauntlett:2002nw}, and its extension to minimal supergravity
coupled to Abelian vector multiplets, has proved particularly useful.
As we will review below, every supersymmetric
solution is built from a set of defining data, a four-dimensional
hyper-K\"{a}hler space along with certain functions and forms on this
space. Specific choices of this defining data reproduce the previously known
supersymmetric rotating black hole and black ring solutions, but more
general choices of defining data lead to horizon-free non-singular
fuzzball solutions. Note that the classifications of six dimensional
supergravity solutions found in
\cite{Gutowski:2003rg,Cariglia:2004kk} have also proved useful.
The five-dimensional fuzzball or, as they are often called,
{\it bubbling}, solutions can be embedded into
eleven-dimensional supergravity compactified on a Calabi-Yau or torus,
in which case they should be related to microstates of M-brane black
holes. By taking the hyper-K\"{a}hler space to be asymptotically
locally flat (Taub-NUT), and reducing on the Taub-NUT circle one also
obtains candidate geometries for four-dimensional four-charge black
holes. Moreover, using a chain of dualities, the same defining data
generates a D1-D5-P solution of type IIB on $T^4$ or $K3$, and thus
gives candidate fuzzball geometries for the microstates of this system.
Below we will discuss in detail which black hole microstates are
likely to be captured by such solutions. Let us remark here, however,
that it is clear a priori that
solutions of minimal supergravity coupled to Abelian vector multiplets
will not be sufficient to
capture all fuzzball solutions for a given black hole. In the two
charge case we saw that a finite fraction of the black hole entropy
was associated with fuzzballs which also had internal excitations,
along the compact part of the geometry. Such fuzzballs with internal
excitations excite all type II supergravity fields, and when compactified along the
compact manifold, give rise to generic solutions of {\it extended} rather than
minimal supergravities in five or six dimensions.
In the three charge system we would also expect to find fuzzball
solutions with internal excitations, as solutions of extended
supergravity theories in five or six dimensions. However, there is to
date no complete classification of solutions of the relevant
extended supergravity theories, so finding fuzzballs with internal
excitations is an open problem. We should emphasize here that such
fuzzballs are likely to have qualitatively different properties to
those without internal excitations (as they did in the two charge
system) and thus one will need to know these properties before
successfully coarse-graining over geometries.
\bigskip
\noindent {\bf 2. Horizon-less limits of known black hole solutions:}
The second construction technique is to begin with known supergravity
solutions describing rotating charged black objects, and then take
careful limits of the parameters in the solutions to obtain
horizon-free non-singular solutions. Note that this technique is applicable to
non-supersymmetric solutions, representing microstates of non-extremal
black holes, and has been used to generate the few known fuzzball
solutions for non-extremal black holes.
We will discuss below which black hole microstates are likely to be
captured by such techniques, but let us already mention the main
limitation of the technique. The most general
non-extremal black hole and black ring
solutions in five dimensions are labeled by a discrete number of
parameters, and admit three Killing vectors (time, plus two rotational
symmetries). The fuzzball geometries obtained by careful limits of the
parameters are thus also highly symmetric, and are labeled by only
a few parameters in addition to their conserved charges. Thus one does
not obtain families of solutions, parameterized by arbitrary functions,
as in the two charge case; one obtains only a discrete (small) number
of fuzzball geometries. Moreover, the high degree of symmetry often
allows one to identify uniquely the dual microstate, and it is found
to be atypical.
\bigskip
There is by now a substantial literature on fuzzball solutions for the
three and four charge cases. The aim of this section will not be to
give a comprehensive discussion of every known solution, but instead
to give an overview of the known solutions, emphasizing the
connections between them. A comprehensive
review of the known three charge fuzzball geometries was
given in the review \cite{Bena:2007kg}.
Solutions for three charge black holes and black rings have been
discussed in
\cite{Mathur:2003hj,Bena:2004wt,Lunin:2004uu,Giusto:2004id,Giusto:2004ip,Bena:2004de,
Giusto:2004kj,Bena:2005ay,Giusto:2006zi,Ford:2006yb}.
In particular bubbling solutions for black holes and black rings, to be described below,
were developed in
\cite{Bena:2005va,Bena:2005zy,Berglund:2005vb,Bena:2006is,Bena:2006kb,Cheng:2006yq,Bena:2007ju,Bena:2007qc,
Gimon:2007mha,Bena:2008wt}.
Four charge solutions in four dimensions have been discussed in
\cite{Saxena:2005uk,Balasubramanian:2006gi}.
Non-supersymmetric solutions were found and their properties explored in
\cite{Jejjala:2005yu,Gimon:2007ps,Giusto:2007tt}.
\subsubsection{Supersymmetric solutions of M theory}
Let us begin with supersymmetric solutions of eleven-dimensional
supergravity compactified on $T^6$, which were obtained in \cite{Bena:2004de}
using the classification
of solutions of minimal ungauged supergravity coupled to Abelian
vector multiplets in five dimensions.
The eleven-dimensional solutions of interest have a metric of the form
\begin{eqnarray}
ds^2 &=& - \left ( \frac{1}{Z_1 Z_2 Z_3} \right )^{2/3} (dt + k)^2 + (Z_1 Z_2
Z_3)^{1/3} h_{mn} dx^{m} dx^{n} \label{m-bub} \\
&& + \left ( \frac{Z_2 Z_3}{Z_1^2} \right )^{1/3} (dy_1^2 + dy_2^2) +
\left ( \frac{Z_1 Z_3}{Z_2^2} \right )^{1/3} (dy_3^2 + dy_4^2)
+ \left ( \frac{Z_1 Z_2}{Z_3^2} \right )^{1/3} (dy_5^2 + dy_6^2), \nonumber} \def\bd{\begin{document}} \def\ed{\end{document}
\end{eqnarray}
with the four form being
\begin{equation}
F = F^1 \wedge dy_1 \wedge dy_2 + F^2 \wedge dy_3 \wedge dy_4 +
F^3 \wedge dy_5 \wedge dy_6.
\end{equation}
Thus the solutions are defined by three functions $(Z_1,Z_2,Z_3)$, one
form $k$ and three two-forms $(F^1,F^2,F^3)$. The two forms can be
written as
\begin{equation}
F^{a} = \theta^{a} - \frac{1}{2} d (Z_{a}^{-1} (dt + k)),
\end{equation}
with $a = 1,2,3$. Then supersymmetric solutions are such that
$h_{mn}$ is hyper-K\"{a}hler, and the forms $\theta^{a}$ are
self-dual and closed on the hyper-K\"{a}hler manifold $H_4$, namely
\begin{equation}
\theta^{a} = \ast_4 \theta^{a}; \hspace{0.5cm}
d \theta^a = 0,
\end{equation}
with the Hodge dual taken in the metric $h_{mn}$. As we will see later
in this section, smooth fuzzball
solutions will require choosing pseudo hyper-K\"{a}hler base spaces,
in which the signature changes in certain regions of the space.
The functions $Z_a$ and the one form $k$ then satisfy
\begin{eqnarray}
\Box Z_a &=& 2 | \epsilon^{abc}| ( \theta_b \cdot \theta_c); \\
dk + \ast_4 dk &=& 2 \sum_{a} \theta^a Z_a, \nonumber} \def\bd{\begin{document}} \def\ed{\end{document}
\end{eqnarray}
where $|\epsilon^{abc} |$ is the absolute value of the totally
antisymmetric tensor; $\Box$ is the Laplacian on $H_4$ and for
2-forms on $H_4$, $(\alpha \cdot \beta) \equiv \frac{1}{2} \a^{mn} \b_{mn}$.
Note that setting $Z_1 = Z_2 = Z_3$ and reducing on the torus gives a
solution of minimal ungauged supergravity in five dimensions; the
effective five-dimensional solution is thus within the framework of that general
classification \cite{Gauntlett:2002nw}.
The Killing spinors of these solutions are given by
\begin{equation}
\epsilon = (Z_1 Z_2 Z_3)^{-1/6} \epsilon_{0},
\end{equation}
where the $32$-dimensional Majorana spinor satisfies the following
projection conditions
\begin{equation} \label{proj}
\G^{0 56} \epsilon_{0} = \G^{0 78 } \epsilon_0 = \G^{0 9 (10)} \epsilon_0 = - \epsilon_0,
\end{equation}
and furthermore the spinor $\epsilon_{0}$ is covariantly constant on the
hyper-K\"{a}hler space. Note that since $\G^{0 1234 56789 (10)} = 1$
the projection conditions also imply that $\G^{1234} \epsilon_0 =
\epsilon_0$. The proof is reviewed in appendix \ref{spinors}.
\bigskip
Reducing on the torus to five dimensions for general $(Z_1,Z_2,Z_3)$
gives solutions of ${\cal N} = 8$ ungauged supergravity in five
dimensions, which involve only three scalars, three gauge fields and the
metric. Thus the solutions involve only a subset of the
${\cal N} = 8$ fields; more general fuzzball solutions involving
excitations along the $T^6$ would need to involve additional ${\cal N}
= 8$ fields and have not yet been constructed in the three charge
case.
Replacing $|\epsilon_{abc}|$ by the intersection form $C_{abc}$ of a Calabi-Yau gives
a solution of ${\cal N} = 2$ ungauged supergravity, corresponding to
eleven-dimensional supergravity compactified on a Calabi-Yau. Further
reducing the five-dimensional solutions along a compact isometry in
the hyper-K\"{a}hler manifold results in a four-dimensional solution,
corresponding to reduction of eleven-dimensional supergravity solutions on
the product of a Calabi-Yau and a circle.
Solutions of this type, built from the defining data of a
4-dimensional hyper-K\"{a}hler space, along with harmonic forms and
functions on this space, account for almost all known bubbling
solutions for M theory and type IIA two, three and four charge black
holes. The main class of exceptions are the two charge
fuzzballs with internal excitations \cite{Kanitscheider:2007wq},
given in (\ref{equ_D1D5K3pot}), which switch on many additional fields
in the compactified theory. More recently in \cite{Bena:2008wt}
coordinate transformations have also been used
to generate from known bubbling solutions new smooth three charge
solutions which involve internal excitations.
In what follows it is useful to work directly
with the eleven-dimensional solution, rather than its dimensional
reduction to five or four dimensions. In
the next section we will see how these solutions are dualized to the type IIB
frame. Moreover, when analyzing regularity
conditions, it is regularity in the uplifted solution which is
relevant, not that in the dimensionally reduced solution. There are
many known examples where the dimensionally reduced solution is
singular, where the uplifted solution is not. Conditions for
the absence of closed timelike curves can differ, and where they do
again it is the conditions in the uplifted solution which are relevant.
\subsubsection{Bubbling solutions of type IIB}
The 11d solutions can be related to solutions of type IIB on $T^4$ by
a simple chain of dualities. First reduce on the $y_6$ circle to
obtain an F1-D2-D2 solution of type IIA, then T-dualize on
$(y_3,y_4,y_5)$ to obtain a D1-D5-P type IIB solution on $T^4$.
The solutions in the type IIB frame can then be written in the form:
\begin{eqnarray}
ds^2 &=& - \frac{1}{ \sqrt{Z_1 Z_2} Z_{3}} (dt + k)^2 + \frac{Z_3}{
\sqrt{Z_1 Z_2}} (dz + {\cal A}_3)^2 + \sqrt{ Z_1 Z_2 } dx_4^2 + \sqrt{ \frac{Z_2}{Z_1}}
dz_4^2, \nonumber} \def\bd{\begin{document}} \def\ed{\end{document} \\
e^{2 \Phi} &=& \frac{Z_2}{Z_{1}}; \qquad
F^{(3)} = (Z_1^{-2} Z_2 Z_3)^{-2/3} \ast_{5} F_1 + F_{2} \wedge (dz +
{\cal A}_3), \label{gen1}
\end{eqnarray}
where $dz_4^2$ is the metric on $T^4$ or $K3$
and the dual $\ast_5$ is defined in the five-dimensional metric
\begin{equation}
ds_5^2 = - (Z_1 Z_2 Z_3)^{-2/3} (dt + k)^2 + (Z_1 Z_2 Z_3)^{1/3}
dx_4^2.
\end{equation}
They are therefore defined in terms of three functions $Z_a$ with $a =
1,2,3$ and four one forms $(\w,{\cal A}_a)$ with ${\cal F}_a = d{\cal
A}_a$, along with the metric on the four-dimensional
hyper-K\"{a}hler space. Introducing
three harmonic self-dual one forms $\eta_a$ on the base space
the connections ${\cal A}_a$ are
\begin{equation}
{\cal A}_a = - Z_a^{-1} (dt + k) + \eta_a.
\end{equation}
The form $k$ satisfies
\begin{equation}
dk + \ast dk = \sum_a Z_a d \eta_a, \label{p2}
\end{equation}
whilst the functions $Z_a$ satisfy
\begin{equation}
\Box Z_a = {\textstyle{1\over2}} | \epsilon_{abc} | (d\eta_b)_{ij}
(d\eta_c)^{ij} , \label{p3}
\end{equation}
where $\Box$ is the Laplacian on the hyper-K\"{a}hler space. Clearly
the defining data is the same as for the M theory and type IIA solutions.
\bigskip
The Killing spinors in this case can be written as
\begin{equation}
\epsilon = (Z_1 Z_2)^{-1/8} (Z_3)^{-1/4} \epsilon_0, \label{spiniib}
\end{equation}
where the spinor $\epsilon_0$ is covariantly constant on the
hyper-K\"{a}hler space and satisfies the projection conditions
\begin{equation}
\epsilon_0 = \G^{01} \epsilon_0 = \G^{016789} \epsilon_0 = - i \epsilon_0^{\ast}.
\end{equation}
These projection conditions along with the Majorana-Weyl condition
imply that the covariantly constant (complex) spinor $\epsilon_0 = \epsilon^1_0
+ i \epsilon^2_0$ is such that $\epsilon^1_0 = - \epsilon^2_0 \equiv \eta_0$ and
\begin{equation}
\G^{01} \eta_0 = \G^{016789} \eta_{0} = \G^{2345} \eta_0 = \eta_0.
\end{equation}
Again we include for convenience the derivation of these spinors in appendix
\ref{spinors}.
\subsubsection{Types of supersymmetric solutions}
We have seen that the supersymmetric solutions of both type IIB and M theory
are specified by a set of data, the metric $h_{mn}$ on the
hyper-K\"{a}hler space, along with three functions $Z_a$ and one-forms
$(k,\eta_a)$, satisfying (\ref{p2}) and (\ref{p3}). We will now review
various black hole and fuzzball solutions characterized by different choices of the forms and
functions. In view of what will follow we will focus on solutions in
the type IIB frame.
\bigskip
\noindent {\bf a. Explicit integration for Gibbons-Hawking base space}
\noindent The most general defining data can be given
in terms of three harmonic functions $h_a$ and three harmonic one
forms $\eta_a$ on the hyper-K\"{a}hler base space. An
additional one form $k_{-}$ on the hyper-K\"ahler space, whose field
strength is anti-self-dual, appears as an integration constant.
That is, the scalar functions $Z_a$ can be formally expressed as
\cite{Bena:2004de,Bena:2005ay}
\begin{equation}
Z_a = h_a + {\textstyle{1\over2}} | \epsilon_{abc} | \Box^{-1} \left ( (d\eta_b)_{ij}
(d\eta_c)^{ij} \right ),
\end{equation}
whilst the three forms ${\cal A}_a$ are given by
\begin{equation}
{\cal A}_a = Z_a^{-1} (dt + k) + \eta_a.
\end{equation}
For the asymptotically flat limit of the metric to be manifest one can
use a (constant) gauge transformation on
${\cal A}_3$ and write it as
\begin{equation}
{\cal A}_3 = (Z_{3}^{-1} - 1) (dt + k) + (k + \eta_3).
\end{equation}
Since the form $k$ satisfies
\begin{equation} \label{wq1}
dk + \ast d k = - \sum_a Z_a d \eta_a.
\end{equation}
the solution implicitly includes an
integration constant. There is always
the freedom to add to $k$ any form $k_{-}$ which satisfies
\begin{equation} \label{int-const}
d k_{-} + \ast_4 d k_{-} = 0.
\end{equation}
Explicit integrated solutions are not known in general. In the case of
hyper-K\"ahler base spaces with a $U(1)$ isometry which is preserved,
the equations can however be integrated \cite{Gauntlett:2004wh}.
Writing the base space in the
Gibbons-Hawking form \cite{Gibbons:1979zt,Gibbons:1987sp}, the metric is
\begin{equation} \label{gh2}
ds_4^2 = V^{-1} (d \psi + A)^2 + V dx^{i} dx^{i}
\end{equation}
where $x^i$ with $i=2,3,4$ are coordinates on $R^3$ and the connection $A$ satisfies
$\ast_3 dA = dV$. Then the solution may be written in terms
of seven harmonic functions $(K_a,L_a,M)$ on $R^3$ as
\begin{eqnarray}
Z_a &=& V^{-1} K_b K_c + L_a; \label{u1-sol} \\
\eta_a &=& V^{-1} K_a (d \psi + A) - \ast_3 d K_a, \nonumber} \def\bd{\begin{document}} \def\ed{\end{document} \\
k &=& k_{\psi} (d \psi + A) + \hat{k}_i dx^i; \nonumber} \def\bd{\begin{document}} \def\ed{\end{document} \\
k_{\psi} &=& V^{-2} K_a K_b K_c + {\textstyle{1\over2}} V^{-1} \sum_a K_a L_a + M; \nonumber} \def\bd{\begin{document}} \def\ed{\end{document}
\\
\ast_3 d \hat{k} &=& V d M - M dV + {\textstyle{1\over2}} \sum_a (K_a d L_a - L_a d
K_a). \nonumber} \def\bd{\begin{document}} \def\ed{\end{document}
\end{eqnarray}
Here $\ast_3$ denotes the Hodge dual on $R^3$. Note that the harmonic
functions $(K_a,M)$ define $(\eta_a,k_-)$ respectively. This explicit integration
is central to most of the fuzzball solutions which have been
constructed, with the strategy being to choose harmonic data on $R^3$
such that regularity conditions are satisfied.
Finding explicit solutions for the case in which the
hyper-K\"ahler base space does not have a $U(1)$ isometry is an open
problem. In the 2-charge system fuzzballs in which the
symmetry of the transverse $R^4$ is completely broken form a less
singular and more representative basis. One might expect the same to
be true in the 3-charge system, and thus one would like to solve
explicitly in the non-symmetric case.
\bigskip
\noindent {\bf b. Black holes and black rings}
\noindent Supersymmetric three charge rotating black hole and black ring
geometries can be realized as specific solutions within this
framework. In fact the defining data for the supersymmetric rotating
black holes was already given in (\ref{data222}): three harmonic
functions sourced at the origin of an $R^4$ base space, along with one anti-self
dual form (the integration constant $k_{-}$).
Let us now consider the supersymmetric black rings found and analyzed in
\cite{Elvang:2004rt,Gauntlett:2004wh,Bena:2004de, Elvang:2004ds,Gauntlett:2004qy}.
The defining functions can be written as
\begin{eqnarray}
Z_a &=& 1 + \frac{Q_a}{\Sigma} - {\textstyle{1\over2}} | \epsilon_{abc} | \frac{q_b q_c R^2 \cos 2 \q}{\Sigma^2}; \\
\Sigma &=& (r^2 + R^2 \cos^2 \q); \nonumber} \def\bd{\begin{document}} \def\ed{\end{document} \\
{\cal A}_{a} &=& Z_a^{-1} (dt + k) + \frac{q_a R^2}{\Sigma} (\sin^2 \q d \varphi -
\cos^2 \q d \phi); \nonumber} \def\bd{\begin{document}} \def\ed{\end{document} \\
k_{\phi} &=& - \frac{r^2 \cos^2 \q}{2 \Sigma^2} \left (\sum_a q_a Q_a - q_1 q_2
q_3 ( 1 + \frac{2 R^2 \cos^2 2 \q}{\Sigma}) \right ); \nonumber} \def\bd{\begin{document}} \def\ed{\end{document} \\
k_{\varphi} &=& - \frac{\sum_a q_a R^2 \sin^2 \q}{\Sigma} + (1 + \frac{R^2}{r^2})
\tan^2 \q k_{\phi}, \nonumber} \def\bd{\begin{document}} \def\ed{\end{document}
\end{eqnarray}
with $a = 1,2,3$ and the base space being $R^4$ with the metric
\begin{equation} \label{met-r4}
dx_4^2 = \Sigma \left (\frac{dr^2}{(r^2 + R^2)} + d \q^2 \right ) + (r^2 + R^2) \sin^2
\q d\varphi^2 + r^2 \cos^2 \q d\phi^2.
\end{equation}
The defining harmonic functions and harmonic forms in this case are
simply
\begin{eqnarray}
h_a &=& \left (1 + \frac{Q_a}{\Sigma} \right ); \\
\eta_a &=& \frac{q_a R^2}{\Sigma} (\sin^2 \q d \varphi - \cos^2 \q d \phi). \nonumber} \def\bd{\begin{document}} \def\ed{\end{document}
\end{eqnarray}
In the form $k$ is included an integration constant (\ref{int-const})
\begin{equation}
k_{-} = - \frac{\sum_a q_a R^2}{2 \Sigma} (\sin^2 \q d \varphi + \cos^2 \q d \phi).
\end{equation}
In contrast to the black hole solution, the harmonic functions in the
black ring are sourced on a circle in $R^4$, located at $r = 0$, $\q = \pi/2$. Since the solution
preserves $U(1)^2$ isometries of the hyper-K\"{a}hler base ($R^4$), it
can also be rewritten as a solution of type (a) on a Gibbons-Hawking
base space. The metric on $R^4$ given in (\ref{met-r4}) can be rewritten in
Gibbons-Hawking form via the coordinate transformations
\begin{equation}
r \cos \theta = 2 \sqrt{\rho} \cos ({\textstyle{1\over2}} \theta_3); \qquad
r^2 \sin^2 \theta + R^2 = 4 \rho; \qquad
\phi = {\textstyle{1\over2}} (\psi + \phi_3); \qquad
\varphi = {\textstyle{1\over2}} (\psi - \phi_3),
\end{equation}
to give the Gibbons-Hawking metric
\begin{equation}
dx_4^2 = V^{-1} (d \psi + \cos \q_3 d \phi_3)^2 + V dx^i dx_i,
\end{equation}
where the $R^3$ metric is written as
\begin{equation}
dx_i dx^i = d \rho^2 + \rho^2 (d \theta_3^2 + \sin^2 \theta_3 d \phi_3^2)
\end{equation}
and the Gibbons-Hawking potential is $V = 1/\rho$. Thus the black ring solutions
written in this coordinate system are such that the harmonic functions
are sourced at $\rho = {\textstyle{1\over 4}} R^2$ and $\q_3 = \pi$. The source circle thus wraps
the fibre, and is located at a point on $R^3$. Letting the Cartesian
coordinates be $x^1 = \rho \cos \theta_3$, $x^2 = \rho \sin \theta_3
\cos \phi_3$ and $x^3 = \rho \sin \theta_3 \sin \phi_3$, then the
source is located at $\vec{x}^i_0 \equiv (- {\textstyle{1\over 4}} R^2,0,0)$. Noting that
\begin{equation}
(r^2 + R^2 \cos^2 \q) = 4 (\rho^2 + {\textstyle{1\over2}} R^2 \rho \cos \theta_3 +
\frac{R^4}{16})^{1/2} = 4 |x - x_0|,
\end{equation}
the defining harmonic functions in $R^3$ are given by
\begin{eqnarray}
h &=& \frac{1}{|x - x_0|}; \qquad
L_a = 1 + {\textstyle{1\over 4}} (Q_a - q_b q_c) h; \\
K_a &=& -{\textstyle{1\over2}} q_a h; \qquad M = {\textstyle{1\over 4}} \sum_a q_a (1 - {\textstyle{1\over 4}} R^2 h). \nonumber} \def\bd{\begin{document}} \def\ed{\end{document}
\end{eqnarray}
Multi-centered supersymmetric black rings, and black saturns, may
then be immediately constructed by choosing instead defining harmonic
functions in $R^3$ which are multi-centered
\cite{Gauntlett:2004wh,Bena:2004de,Gauntlett:2004qy}. Note however that each
point source in $R^3$ corresponds to a circle of sources in $R^4$,
except when the source is located at the origin of $R^3$, in which
case the radius of the wrapped fiber is zero, and the harmonic
function is sourced at the origin of $R^4$. The latter choice of
harmonic function gives as above a supersymmetric black hole.
Let us consider the solutions in the type IIB frame, where they are
understood in the context of the D1-D5-P system.
The single supersymmetric black ring is characterized by seven
parameters, $(Q_a,q_a,R)$ along with the moduli $(R_z,V)$, where
$R_z$ is the radius of the $z$ circle and $v = (2 \pi)^4 V$ is the volume of
$X_4$. The integral charges $N_a$ of the black ring are given by
\cite{Elvang:2004rt,Bena:2004de, Elvang:2004ds}
\begin{equation}
Q_1 = \frac{g (\a')^3}{V} N_1; \qquad
Q_2 = g \a' N_5; \qquad
Q_3 = \frac{g^2 (\a')^4}{V R_z^2} N_p,
\end{equation}
where $(N_1,N_5,N_p)$ are the D1-brane, D5-brane and momentum charges respectively.
The dimensionless (non-conserved) dipole charges $n_a$ are given by
\begin{equation}
q_1 = \frac{g\a'}{R_z} n_1; \qquad
q_2 = \frac{g (\a')^3}{V R_z} n_2; \qquad
q_3 = R_z n_3,
\end{equation}
and the angular momenta at asymptotically flat infinity are
\begin{equation}
J_{\phi} = {\textstyle{1\over2}} \sum_a n_a N_a - {\textstyle{1\over2}} n_1 n_2 n_3; \qquad
J_{\varphi} = J_{\phi} + \frac{R_z V}{(\a')^4 g^2} (q_1 + q_2 + q_3) R^2.
\end{equation}
The entropy of the black ring can be written as
\begin{equation}
S = 2 \pi \sqrt{n_1 n_2 n_3 \d - \g^2}
\end{equation}
where
\begin{eqnarray}
\g &=& {\textstyle{1\over2}} (n_3 N_3 - n_1 N_1 - n_2 N_2 + n_1 n_2 n_3); \\
\d &=& \frac{N_1 N_2}{n_3} - n_1 N_1 - n_2 N_2 + n_1 n_2 n_3 -
\frac{q_3 R^2}{C}, \nonumber} \def\bd{\begin{document}} \def\ed{\end{document} \\
C &=& \frac{(\a')^4}{R_z V}. \nonumber} \def\bd{\begin{document}} \def\ed{\end{document}
\end{eqnarray}
Given that the supersymmetric black ring has the same charges as the
three charge black hole, it should be interpreted as a specific
mixed state in the D1-D5 CFT, with momentum $N_p$, characterized by
the parameters $(n_a,R)$. Explicitly identifying this mixed state is
an open problem. Presumably candidate 3-charge fuzzball geometries
with the same angular momentum as the black ring should correspond to
microstates of both the 3-charge black hole and the 3-charge black
ring. However, identifying which geometries would be relevant for the black ring
is likely to be rather difficult, unless the corresponding CFT mixed
state can be identified.
In the context of the fuzzball proposal, one might wonder whether
there exist horizon-less non-singular limits of the black ring
solutions. These would have D1-D5-P charges and could correspond to
microstates of both the black hole and of the black ring.
The case $\d = 0 = \g$ is known to give a solution without a horizon,
but with an orbifold singularity \cite{Bena:2004tk}. Clearly other restrictions on the
parameters also give solutions with zero entropy, such as for example
$\g = \d = n_1 n_2 n_3$, and it is possible that specific restrictions
could give rise to regular horizon-free geometries. This issue has not
yet been systematically investigated.
\bigskip
\noindent {\bf c. Lunin-Mathur solutions}
\noindent It is useful to show explicit how
the two-charge Lunin-Mathur fuzzball solutions (\ref{lun-mat}) are contained within this
solution set. Let us first restrict the defining data to three harmonic functions
$h_a$ and a single one form $\eta^3 \equiv \eta$ on the
hyper-K\"{a}hler space, taken now to be $R^4$.
In these solutions the remaining three one-forms
${\cal A}_a$ are defined in terms of $(h_a,\eta)$ as
\begin{eqnarray}
{\cal A}_1 &=& h_1^{-1} (dt + k); \qquad {\cal A}_2 = h_{2}^{-1} (dt + k);
\label{res1} \\
{\cal A}_3 &=& (h_3^{-1} - 1) (dt + k) + (k - \eta) \equiv
(h_3^{-1} - 1) (dt + k) + b, \nonumber} \def\bd{\begin{document}} \def\ed{\end{document}
\end{eqnarray}
Substituting these expressions into (\ref{gen1}) the three form can be
written as
\begin{equation} \label{above}
F^{(3)} = d (h_{2}^{-1} (dt + k) \wedge (dz + b)) + \ast_4 dh_1.
\end{equation}
This is the same form as in the Lunin-Mathur solutions,
as one would expect, since the gravitational wave does not couple to
the RR field strengths. However, the form $k$ satisfies the relation
\begin{equation} \label{eq2}
d k + \ast_4 d k = h_3 d \eta,
\end{equation}
and therefore $dk$ {\it cannot} be harmonic except when $h_3$ is
constant. On setting $\eta = 0$ and choosing the harmonic
functions to be of the standard (single-centered) form
\begin{equation}
h_a = \left (1 + \frac{Q_a}{r^2} \right )
\end{equation}
one clearly recovers the three charge static extremal black
string. Choosing $\eta$ to be non-zero, for the same choice of
harmonic functions, reproduces the three charge rotating extremal
BMPV black string.
On setting $h_3 =1$ one recovers the Lunin-Mathur geometries as
follows. Let $d \eta = d A + \ast_4 d A \equiv dA + dB$, where the
definition of $B$ is that $dB = \ast_4 dA$, so that $\eta = A + B$.
Then (\ref{eq2}) is solved by letting $k = A$ which implies that in (\ref{above})
$b = B$. Choosing the explicit forms of the harmonic
functions to be those given in (\ref{lm-func}) one indeed obtains the
solutions given in (\ref{lun-mat}).
It is important to note however that the two
charge fuzzball solutions with internal excitations cannot be obtained within
this framework, as they involve many additional fields in type IIB, and
hence in the compactified theory.
\bigskip
\noindent{\bf d. Bubbling geometries}
\noindent We now turn our attention to the main class of fuzzball
geometries which have been constructed, the so-called bubbling
solutions
\cite{Bena:2005va,Bena:2005zy,Berglund:2005vb,Bena:2006is,Bena:2006kb,Cheng:2006yq,Bena:2007ju,Bena:2007qc,
Gimon:2007mha,Bena:2008wt,Balasubramanian:2006gi}.
Given the detailed discussion of these backgrounds in other works,
for example the review \cite{Bena:2007kg}, we will simply
highlight their main features here.
The basic idea of the bubbling geometries is to use a non-trivial
``ambipolar'' $U(1)$ invariant
hyper-K\"{a}hler base space, whose signature changes in
different regions. That is, the potential for the Gibbons-Hawking
space is
\begin{equation}
V = \sum_{i} \frac{q_i}{|x - x_i|}, \qquad \sum_i {q_i} = 1.
\end{equation}
The constraint on the total charge is necessary for the space to be
asymptotically flat. In regions where $V$ changes sign the signature
of the base space changes from $(+,+,+,+)$ to $(-,-,-,-)$. Denote the
2-dimensional surfaces where $V$ changes sign by $\Sigma_{\a}$; then
$V(\Sigma_{\a}) = 0$. For the
complete metric signature to remain unchanged this means that
\begin{equation} \label{po2}
Z_a V \ge 0
\end{equation}
everywhere, which in turn requires that the $Z_a$ change sign at
$\Sigma_{\a}$ also. Taking the remaining seven harmonic functions
$(L_a,K_a,M)$ defined in (\ref{u1-sol})
to be sourced at the same locations $x_i$, so that
\begin{equation}
L_a = 1 + \sum_{i} \frac{l_a^i}{|x - x_i|}; \qquad
K_a = \sum_{i} \frac{k_a^i}{|x - x_i|}; \qquad
M = m + \sum_{i} \frac{m^i}{|x - x_i|},
\end{equation}
one then chooses the parameters $(l_a^i,k_a^i,m^i)$ so that
$(Z_a,\eta_a,k)$ are finite at the locations of the harmonic function
sources, $x_i$. In particular this implies that
\begin{equation} \label{po3}
l_a^i = - \frac{k_b^i k_c^i}{q_i}; \qquad m^i = \frac{k_1^i k_2^i
k_3^i}{2 (q_i)^2}; \qquad m = - {\textstyle{1\over2}} \sum_{i,a} k_a^i.
\end{equation}
These values are chosen so as to cancel the poles in the
defining functions.
Note that the functions $Z_a$ for large $|x|$ are expanded as
\begin{equation}
Z_a = 1 + \frac{1}{|x|} \left (\sum_i l_a^i + \sum_{i,j} k_b^i k_c^i
\right ) +
\cdots \equiv 1 - \frac{1}{|x|} \sum_{i} \frac{\tilde{k}_b^i
\tilde{k}_c^i}{q_i} + \cdots,
\end{equation}
where
\begin{equation}
\tilde{k}_a^i = k_a^i - q_i \sum_{j} k_a^j; \qquad \sum_{i} \tilde{k}_a^i =
0.
\end{equation}
One can show that the supergravity solutions are invariant under
\begin{equation}
K_a \rightarrow K_a + c_a V,
\end{equation}
for any constant $c_a$, see \cite{Bena:2005va,Berglund:2005vb}, and thus physical quantities such as mass
which are expressed in terms of $Z_a$ must depend on the invariant quantities
$\tilde{k}_a^i$ rather than $k_a^i$.
For the supergravity solutions to be regular requires:
\begin{enumerate}
{\item{Absence of singularities, in particular (a) at the locations of the sources
in the harmonic functions and (b) where the base space signature
changes.}}
{\item{Absence of closed timelike curves and Dirac-Misner strings.}}
\end{enumerate}
Conditions (\ref{po2}) and (\ref{po3}) are
sufficient to ensure that the solution is non-singular, see the
detailed analysis of \cite{Berglund:2005vb}. Note that although the
defining functions in the solution are built
from harmonic functions which have sources at the Gibbons-Hawking
centers there are no delta function sources in the defining
functions \cite{Bena:2005va,Berglund:2005vb}. The solution is therefore regular at
the locations of the sources. One can also show that the solution is
regular where the base space signature changes, namely where $V
\rightarrow 0$. Consider the five-dimensional part of the metric
\begin{equation}
ds_5^2 = - (Z_1 Z_2 Z_3)^{-2/3} (dt + k)^2 + (Z_1 Z_2 Z_3)^{1/3} \left
( V^{-1} (d \psi + A)^2 + V dx^i dx^i \right )
\end{equation}
in a neighborhood of a hypersurface
$\Sigma_a$ where $V(\Sigma_a) = 0$. Using the
explicit forms of the functions one can show that in the neighborhood of a
point on $\Sigma_a$
\begin{equation} \label{sigma-beh}
ds^2_5 \approx - 2 dt d \varphi + dx^i dx^i.
\end{equation}
Here $d{\varphi} = d \psi + A_{\Sigma_a}$ with
$A_{\Sigma_a}$ the Gibbons-Hawking gauge field and
$(t,x^i)$ have been appropriately rescaled.
The metric is regular at $\Sigma_a$, with the
Killing vector $\pa_t$ becoming null on this surface. A detailed
analysis of the regularity of the solutions on these hypersurfaces can
be found in \cite{Berglund:2005vb}.
Removing closed timelike curves restricts the parameters further; for
example, in the eleven-dimensional geometries one has to ensure that
\begin{equation}
Z_1 Z_2 Z_3 V^2 - k_{\psi}^2 V \ge 0
\end{equation}
globally. Recall that the $k_{\psi}$ was given in (\ref{u1-sol}).
Solving the constraints this equation
makes on the parameters $(k_a^i,q_i,\vec{x}_i)$ in general is rather difficult.
However, one can derive constraints which remove Dirac-Misner
strings from $\hat{k}$, defined in (\ref{u1-sol}); these so-called {\it bubble equations} are
\begin{eqnarray}
\sum_{j \neq i} \Pi_{ij}^1 \Pi_{ij}^2 \Pi_{ij}^3 \frac{q_i q_j}{r_{ij}}
&=& - 2 (m q_i + {\textstyle{1\over2}} \sum_{a} k_a^i); \\
\Pi_{ij}^a &=& \left ( \frac{k_a^j}{q_j} - \frac{k_a^i}{q_i} \right );
\qquad r_{ij} = |x_i - x_j|. \nonumber} \def\bd{\begin{document}} \def\ed{\end{document}
\end{eqnarray}
In specific examples one finds that satisfying these equations is sufficient to
guarantee the global absence of CTCs, but this is not always the case
and identifying
systematically the conditions required to remove closed timelike
curves remains an open problem.
The three conserved charges with respect to asymptotically flat infinity can
be expressed in terms of the defining data
$(\tilde{k}_a^i,q_i,\vec{x}_i)$ as
\begin{equation}
Q_a = - 2 {\cal N} | \epsilon_{abc} | \sum_{i=1}^N \frac{\tilde{k}^b_i
\tilde{k}^c_i}{q_i},
\end{equation}
where the prefactor ${\cal N}_a$ depends on whether one is considering
the type IIB or M theory case. In the former case, the $Q_a$ are the conserved
D1, D5 and momentum charges, and the normalization is
\begin{equation}
{\cal N}_{IIB} = \frac{R_z V}{(\a')^4 g^2}.
\end{equation}
In the M theory case, the $Q_a$ are the conserved membrane charges,
and the appropriate normalization is
\begin{equation}
{\cal N}_{M} = \frac{\pi}{4 G_5} = \frac{\pi V_{6}}{4 G_{11}},
\end{equation}
with $G_{d}$ the $d$-dimensional Newton constant and $V_6$ the volume
of the six torus (or of the Calabi-Yau). The solutions also have conserved
angular momenta with respect to asymptotically flat infinity. The
generic solution breaks the $SO(4)$ rotational invariance of the
transverse $R^4$ at infinity to $U(1)$, and therefore there are
non-zero angular momenta $(J_{\psi},\vec{J})$, where $\vec{J}$ defines
a three dimensional vector in the $R^3$ of the Gibbons-Hawking metric
(\ref{gh2}). Then one finds that
\begin{eqnarray}
J_{\psi} &=& \frac{4}{3} {\cal N} | \epsilon_{abc} | \sum_{i=1}^N
\frac{\tilde{k}^a_i \tilde{k}^b_i \tilde{k}^c_i}{q_i^2}; \nonumber} \def\bd{\begin{document}} \def\ed{\end{document} \\
\vec{J} &=& 8 {\cal N} \vec{D} , \qquad
\vec{D} \equiv \sum_{i=1}^{N} \vec{D}_i, \qquad
\vec{D}_i \equiv \sum_{a} \tilde{k}^a_i \vec{x}^i. \nonumber} \def\bd{\begin{document}} \def\ed{\end{document}
\end{eqnarray}
$J_{\psi}$ and $J_{\phi_3}$ correspond to the left and right moving components
$J_{+}$ and $J_{-}$ of the CFT R symmetry currents,
respectively. More generally $\vec{J}$ should correspond to the right
moving $SU(2)$ CFT R current $J^a_{-}$.
Thus all the bubbling solutions should correspond to
R charge eigenstates in the left moving (excited) sector, but need not
be R charge eigenstates in the right moving (ground state) sector. Note
however that the angular momenta given above are with respect to
asymptotically flat infinity and do not automatically coincide precisely
with the vevs of the R symmetry currents, extracted from the decoupling
region in the geometry; one needs to extract the vevs of the R
symmetry currents using the holographic formulae.
The main goal of the literature has been to find data sets
$(\tilde{k}_a^i,q_i,\vec{x}_i)$ such that the supergravity solution is
regular and has the same charges as a black hole or black ring with macroscopic
horizon area. For example, recall that a supersymmetric
BMPV black hole with integral D1-D5-P
charges $(N_1,N_5,N_p)$ and angular momentum $J_{+} = j_{+}$ has entropy
\begin{equation}
S = 2 \pi \sqrt{N_1 N_5 N_p - j_{+}^2},
\end{equation}
so for a black hole with macroscopic horizon area one needs $j_{+}^2 \ll
N_1 N_5 N_p$. The regular solutions first constructed had $j_{+}^2 \sim
N_1 N_5 N_p$, and thus could not correspond to microstates of a
macroscopic BMPV black hole. Later, however, it was observed that
$j_{+}^2 = k N_1 N_5 N_p$ (with the fraction $k$ being smaller than one)
could be achieved in scaling solutions \cite{Bena:2006kb,Bena:2007ju,Bena:2007qc}, namely
solutions in which all the centers are at
\begin{equation}
\vec{x}_i = \mu \vec{y}_i,
\end{equation}
with $\mu \rightarrow 0$ and $\vec{y_i}$ finite. These so-called deep
microstate solutions were explored numerically and analytically in
\cite{Bena:2006kb,Bena:2007ju,Bena:2007qc}.
Recalling the
behavior in the 2-charge system it is perhaps unsurprising that the bubbling
points need to be clustered at the origin. In the 2-charge geometries
the radius of the curve in $R^4$ determining the solution was related
to the angular momenta, with solutions of small angular momenta
deriving from curves of small radii, see the discussion around
(\ref{j-char2}). Most of the Ramond ground states
have small R charge, and thus the typical fuzzball should be
characterized by a curve of small radius. Here we see analogous
behavior in the 3-charge system: the defining harmonic functions
are sourced on curves in the four-dimensional base space, which must
have small radii for the fuzzball to have typical R charges.
It also seems natural that one needs to cluster the centers. In
the 2-charge system it seems unlikely that solutions characterized by
disconnected curves describe bound states of D1 branes and D5
branes. They can have the same charges and angular momenta as typical
black hole microstates, but are most likely related to Coulomb branch
physics. Here in the 3-charge system one sees that a scaling solution
in which the centers cluster is needed to even obtain a solution with
typical charges.
Note that candidate fuzzball geometries for black
rings can be found by taking a single center at the origin with $q_1 =
1$ and clustering the remaining points at some distance $\r$ from the
origin. The scale $\r$ determines the radius of the corresponding
black ring. In section \ref{corresp} we will consider the
correspondence between such geometries and D1-D5-P microstates.
\subsubsection{Limits of black hole solutions and spectral flow}
The second technique used to find candidate fuzzball geometries
involves starting with known non-extremal charged rotating black hole
solutions, and then taking careful limits of the parameters to obtain
solutions with no horizons or singularities. More generally one could
start with any rotating solution of Einstein equations, apply
boosts and dualities to obtain solutions carrying the required charges
and then restrict the parameters to obtain horizon-free non-singular
solutions. Thus as previously mentioned one could also look for fuzzball limits
of black ring solutions. A systematic exploration of the possibilities
has not yet been carried out.
The principal advantage
of this technique is that it does not rely on supersymmetry, and thus
allows one to find fuzzball solutions for non-extremal black
holes. The main drawback, however, is that the known black hole and
black ring solutions are highly symmetric and characterized by only a
small number of
parameters, which are further restricted by demanding regularity and
absence of horizons. Thus the resulting fuzzball solutions typically
have only a few parameters in
addition to the required conserved charges, and correspond to rather
atypical black hole microstates.
Another related solution generating technique is ``spectral flow'': one
begins with a given near horizon geometry and makes a coordinate
transformation, which preserves the $AdS$ asymptotics.
In particular, one can generate certain 3-charge D1-D5-P geometries
from 2-charge D1-D5 geometries in this way
\cite{Lunin:2004uu,Giusto:2004id,Giusto:2004ip,Giusto:2004kj}.
Gluing back the asymptotically flat region then gives a fuzzball
geometry of the 3-charge D1-D5-P black hole. One should note that these
solution generating
techniques are often called spectral flow transformations, although it
is unclear whether all such named transformations indeed correspond to
spectral flow in the CFT.
Before the development of the bubbling solutions using the
classification of supergravity solutions, most candidate fuzzball
solutions for the 3-charge and 4-charge systems were found by
techniques of this kind. Supersymmetric fuzzballs for the D1-D5-P system were found in
\cite{Lunin:2004uu,Giusto:2004id,Giusto:2004ip,Giusto:2004kj}, whilst
supersymmetric fuzzballs for the D1-D5-KK system were found in
\cite{Bena:2005ay} and for the 3-charge supersymmetric black ring in
\cite{Giusto:2006zi}. In \cite{Ford:2006yb} time dependent solutions
carrying 3 charges were found and in \cite{Giusto:2007tt}
non-supersymmetric fuzzball solutions of the D1-D5-KK system were
found and analyzed.
To illustrate these techniques, let us focus on the example of
the non-extremal D1-D5-P solutions found in \cite{Jejjala:2005yu} and
analysed further in \cite{Gimon:2007ps}. General non-extremal 3-charge black
hole solutions with rotation were given in equation (\ref{cvetic-youm}).
Smooth geometries with no horizons can be obtained by demanding that
the singularity where $g_{rr}^{-1} = 0$ be nothing but a
coordinate singularity, analogous to that of polar coordinates at the
origin of $R^2$. There are four conditions on the parameters needed to
ensure regularity (assuming the momentum charge is non-zero):
\begin{eqnarray} \label{con-1}
m &=& a_1^2 + a_2^2 - a_1 a_2 \frac{c_1^2 c_5^2 c_p^2 + s_1^2 s_5^2
s_p^2}{s_1 c_1 s_5 c_5 s_p c_p}; \\
\frac{j + j^{-1}}{s + s^{-1}} &=& (l - n), \qquad
\frac{j - j^{-1}}{s - s^{-1}} = (l + n), \qquad (l,n \in Z); \nonumber} \def\bd{\begin{document}} \def\ed{\end{document} \\
R_z &=& \frac{m s_1 c_1 s_5 c_5 \sqrt{s_1 c_1 s_5 c_5 s_p
c_p}}{\sqrt{a_1 a_2} (c_1^2 c_5^2 c_p^2 - s_1^2 s_5^2 s_p^2)},
\nonumber} \def\bd{\begin{document}} \def\ed{\end{document}
\end{eqnarray}
where $R_z$ is the radius of the $z$ direction and
the parameters $(j,s)$ are given by
\begin{equation}
j \equiv (\frac{a_2}{a_1})^{1/2}, \qquad
s \equiv \left (\frac{s_1 s_5 s_p}{c_1 c_5 c_p} \right )^{1/2} \le 1.
\end{equation}
The conserved charges and angular momenta are given by
\begin{eqnarray}
Q_1 &=& m s_1 c_1 = \frac{g \a^{'3} N_1}{V}; \qquad
Q_5 = m s_5 c_5 = g \a' N_5; \\
Q_p &=& m s_p c_p = \frac{g^2 (\a')^4}{V R_z^2} N_p, \qquad N_p = N_1
N_5 l n; \nonumber} \def\bd{\begin{document}} \def\ed{\end{document} \\
J_{\psi} &=& - \frac{\pi m}{4 G_5} (a_1 c_1 c_5 c_p - a_2 s_1 s_5
s_p) = - N_1 N_5 l \nonumber} \def\bd{\begin{document}} \def\ed{\end{document} \\
J_{\phi} &=& - \frac{\pi m}{4 G_5} (a_2 c_1 c_5 c_p - a_1 s_1 s_5
s_p) = N_1 N_5 n, \nonumber} \def\bd{\begin{document}} \def\ed{\end{document}
\end{eqnarray}
where $(N_1,N_5,N_p)$ are the integral charges and
\begin{equation}
\frac{\pi}{4 G_5} = \frac{R_z V}{g^2 (\a')^4}.
\end{equation}
The first expressions for $(Q_p,J_{\psi},J_{\phi})$ hold generally,
with the restrictions to the solutions of interest being given in
terms of the integers $(l,n)$. The mass is given by
\begin{equation}
M = \frac{\pi m}{4 G_5} (s_1^2 + s_5^2 + s_p^2 + \frac{3}{2}).
\end{equation}
Restricting the parameters to ensure regularity, and focussing on near
extremal configurations, this can be rewritten as
\begin{equation}
M = \frac{\pi}{4 G_5} (Q_1 + Q_5 + Q_p) + \frac{1}{2 R_z} N_1 N_5 (l^2 + n^2
- 1) \equiv M_{BPS} + \Delta M.
\end{equation}
Thus one can see that if one fixes the three charges $(Q_1,Q_5,Q_p)$,
along with the moduli, then the solution is characterized by only one
other (integral) parameter. This parameter controls both the
non-extremality and the angular momenta, so the solution gives
precisely one microstate of a non-extremal D1-D5-P system with fixed
non-extremality.
It is useful to rewrite the six-dimensional metric as a fibration over
a four-dimensional base space, in order to facilitate comparison with
the general forms of supersymmetric solutions. Of course in the
non-supersymmetric case the base space does not have any special
character, but the hyper-K\"{a}hler structure is recovered in the
supersymmetric limit. The six-dimensional part of the metric can thus
be written as
\begin{eqnarray}
ds^2 &=& \frac{1}{\sqrt{H_1 H_5}} \left ( - (f - m) (d \tilde{t} - (f -
m)^{-1} m c_1 c_5 (a_1 \cos^2 \q d \psi + a_2 \sin^2 \q d \phi))^2
\right . \nonumber} \def\bd{\begin{document}} \def\ed{\end{document} \\
&& + \left . f (d \tilde{z} + f^{-1} m s_1 s_5 (a_2 \cos^2 \q d \psi + a_1 \sin^2 \q
d \phi))^2 \right ) \\
&& + \sqrt{H_1 H_5} \left ( \frac{r^2 dr^2}{ (r^2 + a_1^2)(r^2 + a_2^2) -
m r^2} + d \q^2 \right . \nonumber} \def\bd{\begin{document}} \def\ed{\end{document} \\
&& + (f (f-m))^{-1} [f (f-m) + f a_2^2 \sin^2 \q - (f-m) a_1^2 \sin^2 \q
d\phi^2 \nonumber} \def\bd{\begin{document}} \def\ed{\end{document} \\
&& + 2 m a_1 a_2 \sin^2 \q \cos^2 \q d \psi d\phi \nonumber} \def\bd{\begin{document}} \def\ed{\end{document} \\
&& + \left . (f (f-m) + f a_1^2 \cos^2 \q - (f-m) a_2^2 \cos^2 \q)
d\psi^2 ] \right ), \nonumber} \def\bd{\begin{document}} \def\ed{\end{document}
\end{eqnarray}
where $\tilde{t} = t c_p - z s_p$ and $\tilde{z} = z c_p - t s_p$. In the
supersymmetric limit this six-dimensional metric can indeed be rewritten in
terms of an ambipolar hyper-K\"{a}hler base and harmonic functions on
this base \cite{Giusto:2004kj}.
The decoupling region of this geometry is obtained by
replacing $\tilde{H}_1 = Q_1$ and $\tilde{H}_5 = Q_5$ and also approximating
$m s_1 s_5 = m c_1 c_5 = \sqrt{Q_1 Q_5}$. This gives
\begin{eqnarray}
ds^2 &=& \frac{1}{\sqrt{Q_1 Q_5}} \left ( - (f - m) (d \tilde{t} - (f -
m)^{-1} \sqrt{Q_1 Q_5} (a_1 \cos^2 \q d \psi + a_2 \sin^2 \q d \phi))^2
\right . \nonumber} \def\bd{\begin{document}} \def\ed{\end{document} \\
&& + \left . f (d \tilde{z} + f^{-1} \sqrt{Q_1 Q_5} (a_2 \cos^2 \q d \psi + a_1 \sin^2 \q
d \phi))^2 \right ) \\
&&+ \sqrt{Q_1 Q_5} \left ( \frac{r^2 dr^2}{ (r^2 + a_1^2)(r^2 + a_2^2) -
m r^2} + d \q^2 \right . \nonumber} \def\bd{\begin{document}} \def\ed{\end{document} \\
&& + (f (f-m))^{-1} [f (f-m) + f a_2^2 \sin^2 \q - (f-m) a_1^2 \sin^2 \q
d\phi^2 \nonumber} \def\bd{\begin{document}} \def\ed{\end{document} \\
&& + 2 m a_1 a_2 \sin^2 \q \cos^2 \q d \psi d\phi \nonumber} \def\bd{\begin{document}} \def\ed{\end{document} \\
&& + \left . (f (f-m) + f a_1^2 \cos^2 \q - (f-m) a_2^2 \cos^2 \q)
d\psi^2 ] \right ), \nonumber} \def\bd{\begin{document}} \def\ed{\end{document}
\end{eqnarray}
This metric in turn can be rewritten as the twisted fibration of $S^3$
over BTZ as in (\ref{btzz}).
The regularity conditions of (\ref{con-1}) then translate to $J_3 =
0$ and $M_3 = -1$, in which case the BTZ part of the metric becomes
global $AdS_3$. Imposing both the regularity conditions and setting $m
\rightarrow 0$ restricts the fibration further, as we shall see below.
\bigskip
If one focuses on this decoupled region of the geometry it is
easy to understand why the non-extremality is pinned to the angular
momentum, and moreover what the corresponding CFT state is.
Consider the geometry
\begin{eqnarray}
ds^2 &=& \sqrt{Q_1 Q_5} \left ( - (r^2 + \g_1^2 \mu^2) dt^2 + r^2 dy^2 +
\frac{dr^2}{r^2 + \g_1^2 \mu^2} \right ) \\
&& + \sqrt{Q_1 Q_5} \left (d\q^2 + \cos^2 \q (d\psi + j_{\psi})^2 +
\sin^2 \q (d\phi + j_{\phi})^2 \right ) + (Q_1/Q_5)^{1/2} dz_4^2
\nonumber} \def\bd{\begin{document}} \def\ed{\end{document}
\\
F &=& \sqrt{Q_1 Q_5} \left (dt \wedge dy \wedge dr + \cos \q \sin \q d \q
\wedge (d\phi + j_{\phi}) \wedge (d\psi + j_{\psi}) \right ), \nonumber} \def\bd{\begin{document}} \def\ed{\end{document}
\end{eqnarray}
with constant dilaton $e^{-2 \Phi} = Q_5/Q_1$ and $y \sim y + 2 \pi
\mu^{-1}$, $\mu = \sqrt{Q_1 Q_5}/R$. Let
\begin{equation}
j_{\phi} = j_{+} dx^{+} + j_- dx^{-}; \qquad
j_{\psi} = j_{+} dx^{+} - j_{-} dx^{-},
\end{equation}
where $dx^{\pm} = \mu (dt \pm dy)$.
The geometry is regular if $\g_1^2 =1$, and has permissible conical
singularities if $\g_1 = 1/n$ with $n \in Z$. One can immediately
extract the holographic stress energy tensor and R symmetry currents
using the formulae for the holographic vevs given in (\ref{h-vevs2})
\begin{equation}
\< T \> = \frac{N}{2\pi} \left (
( (j_{+})^2 - \g_1^2) (dx^{+})^2 + ( (j_{-})^2 - \g_1^2) (dx^{-})^2 \right
); \qquad
\< J^{\pm 3} \> = \frac{N}{2\pi} j_{\pm} dx^{\pm},
\end{equation}
with $N = N_1 N_5$. The vevs of scalar operators vanish.
For the case of interest, one finds that $\g_1 = 1$ and
\begin{equation}
j_{\pm} = {\textstyle{1\over2}} (l \pm n),
\end{equation}
and thus
\begin{equation}
h = N ({\textstyle{1\over 4}} (l+n)^2 ); \qquad
\bar{h} = N ({\textstyle{1\over 4}} (l-n)^2); \qquad
j_+ = {\textstyle{1\over2}} N (l+n); \qquad
{j}_- = {\textstyle{1\over2}} N (l-n).
\end{equation}
Note that the momentum charge at asymptotically flat infinity
is proportional to $(h - \bar{h})$
whilst the mass depends on $(h + \bar{h})$. The CFT state with
these charges corresponds to the spectral flow of the NS-NS
vacuum. One can see this by recalling that
under a spectral flow in the left sector by $\a_L = (l+n)$
units and in the right sector by $\a_R = (l-n)$ units,
\begin{equation}
h \rightarrow h - \a_L j_+ + \frac{c}{24} \a_L^2; \qquad
j_+ \rightarrow j_+ + \frac{c}{12} \a_L,
\end{equation}
with corresponding expressions for $(\bar{h},{j}_-)$, where $c = 6 N$.
Letting $\a_R = 1$ and $\a_L = 2 p + 1$ gives a BPS state in the RR
sector which is in the right moving vacuum and in a left moving
excited state. When $(\a_R,\a_L)$ are odd one obtains a
non-supersymmetric state in the RR sector which is a microstate of
the non-extremal black hole.
Since these states are related by spectral flow to the identity, it is
rather simple to compute properties of these states, for example,
decay rates of the non-supersymmetric states. (We will defer discussion
of this issue to section \ref{open}). At the same time, they are atypical black hole
microstates, and thus their properties cannot be assumed to be
representative. To be more precise, consider the microstates of the
BPS D1-D5-P (Strominger-Vafa) static black hole. Counting states with fixed
$(N_1,N_5,N_p)$ and either states with
$(j_+ = {j}_- = 0)$ or {\it all} states gives the same leading
contribution to the entropy, namely the famous result
\begin{equation}
S = 2 \pi \sqrt{N_1 N_5 N_p}.
\end{equation}
The point is that most states have $(j_+ = {j}_- = 0)$, so the
difference between the total degeneracy of states $d_t$ with fixed $(N_1,N_5,N_p)$
and arbitrary R-charges, and the degeneracy of states $d_0$ with
fixed $(N_1,N_5,N_p)$ and $(j_+ = {j}_- = 0)$ is subleading.
Fixing $j_+$ and ${j}_-$ corresponds to working in the canonical
ensemble, whilst allowing for all $(j_+,{j}_-)$ corresponds to working
in the grand canonical ensemble. Clearly the BPS state discussed
above, which has $j_+ \gg 1$, does not contribute to the canonical
ensemble for the Strominger-Vafa black hole, and it is a highly
non-representative state in the grand canonical ensemble, which is
peaked around states with zero R-charge. The BPS state above could be
interpreted as a microstate of a rotating BMPV black hole with this value of
$j_+$. However, recalling the BMPV entropy formula, and noting that
$j_{+}^2 = N_1 N_5 N_p$ one sees there are not enough microstates with this value of $j_+$
to give a black hole with macroscopic horizon area.
More generally, even if the state did not have atypical R-charges,
since it is obtained by spectral flow of the identity only
the stress energy tensor and R currents can acquire expectation values. By
contrast, a typical microstate will be characterized by the vevs of
operators within that state, and these vevs will determine the scale
at which the geometry starts to differ from the naive black hole
geometry.
Before moving to the correspondence between geometries and microstates
we should comment that the known non-extremal fuzzball solutions
have superradiant instabilities; see for example
\cite{Cardoso:2004nk,Cardoso:2005gj,Gimon:2007ps,Cardoso:2007ws}. This
is another indication that these solutions are not representative. As
we will discuss in section 6, one expects non-extremal fuzzballs to be
unstable, with their decay rate matching the decay rate of the
corresponding non-BPS microstates in the CFT. The radiation emitted by
a typical fuzzball should be very similar to that of the non-extremal
black hole. However, the known non-extremal fuzzball solutions decay
much faster; there is no contradication, as the decay rate matches
that of the corresponding CFT states, see
\cite{Chowdhury:2007jx,Dias:2007nj}, but these atypical fuzzballs are not
representative.
\subsection{Correspondence between geometries and microstates} \label{corresp}
One of the most important outstanding issues is to derive the
correspondence between the candidate fuzzball geometries and the black
hole microstates. One would first like to match sufficient data to be sure
that the geometries do indeed correspond to black hole microstates,
rather than other states with the same charges. Then one would like to
explore what fraction of the black hole microstates are captured by
the known fuzzball geometries. No systematic analysis of the
correspondence has yet been carried
out, and a number of issues in matching geometries to microstates are rather
puzzling.
The bubbling geometries
\cite{Bena:2005va,Berglund:2005vb,Bena:2006is,Bena:2006kb,Cheng:2006yq,Bena:2007ju,Bena:2007qc,
Gimon:2007mha,Bena:2008wt} are viewed as the most promising candidates
for duals to typical black hole microstates. First of all, they have
the correct charges to correspond to three charge black holes and
in scaling solutions one
can obtain angular momenta comparable to that of a macroscopic BMPV black hole. By
replacing the asymptotically flat base space by an asymptotically
locally flat (Taub-NUT) space such that $V \rightarrow 1$ as $r
\rightarrow \infty$ one can also obtain candidate geometries for four charge
black holes in four dimensions, as were discussed in \cite{Balasubramanian:2006gi},
again with the same charges as typical black hole microstates.
Secondly, the scaling solutions
\cite{Bena:2006kb,Bena:2007ju,Bena:2007qc} have throat regions, such that there
exists a decoupling region with the same asymptotics as in the black
hole geometry. In the context of the type IIB solutions, this implies
that bubbling solutions like the corresponding
black hole or black ring admit an $AdS_3 \times S^3 \times X_4$ decoupling
region. This is a necessary condition for the bubbling solution to
correspond to a black hole microstate: the AdS/CFT dictionary implies
that any microstate is dual to an asymptotically
$AdS_3 \times S^3 \times X_4$ geometry, with the information encoded
in the asymptotics determining the specific microstate.
Solutions of M theory compactified on a Calabi-Yau admit decoupling
$AdS_2 \times S^3$ or $AdS_3 \times S^2$ regions, depending on the
Taub-NUT charge. In the former case there has been
considerable progress on counting the relevant BPS states, via the counting of curves
in the Calabi-Yau, and on understanding the quiver quantum
mechanics of the D0-D2-D4-D6 branes which wrap these curves and
generate the black hole. At the same time, it is hard to use detailed AdS/CFT technology
to analyze these bubbling geometries as one is forced to
work within the rather poorly understood $AdS_2/CFT_1$
correspondence and there is no precise dictionary
developed between the decoupling $AdS_2$ regions and the dual theory.
In the other case, where one does have an $AdS_3$ region in an M theory
solution, one does not have a good description of the corresponding
2-dimensional CFT, given the limiting understanding of M-brane
theories.
Thus, to make full use of AdS/CFT technology to identify specific
bubbling solutions, it is natural to work instead with the type IIB solutions
which admit an $AdS_3 \times S^3 \times X_4$ decoupling
region and correspond to states in the relatively well
understood D1-D5 conformal field theory. A clear advantage of working
in this system is that one already has an identification of the geometries
dual to the microstates of the 2-charge black hole, namely the
Ramond ground states.
\bigskip
Let us turn to the correspondence between bubbling geometries in the
type IIB frame and D1-D5-P microstates in the conformal field
theory. Recall first the correspondence between two charge geometries
and Ramond ground states: the former are characterized by a curve
whose Fourier coefficients determine the corresponding dual
superposition of Ramond ground states. Restricting to ground states
built from the universal cohomology of $X_4$, the curve is a generic
closed curve in $R^4$. In the special case where the curve is a
circle, the corresponding dual is a specific R charge eigenstate
Ramond ground state.
Note that this dictionary determines the dual superposition of Ramond
ground states given the curve defining a smooth supergravity
solution. As discussed in \cite{Kanitscheider:2006zf, Kanitscheider:2007wq}
the geometric dual of a generic R charge eigenstate is not
known; most likely it is not well described in supergravity,
but one can also not exclude that the dual consists of multi-center
concentric circular curves. The latter is consistent with both the symmetry
of the CFT state and the charges, and cannot be excluded by comparing
vevs of operators, since these are just too small for reliable
comparisons \cite{Kanitscheider:2006zf}. A priori, however, it seems rather unnatural that the
geometric dual should be multi-center, as this generically signals
Coulomb branch, rather than Higgs branch, physics.
Moving on to the three charge bubbling geometries, recall that the defining data
for these geometries is the data at the centers of the Gibbons-Hawking
metric, namely $(\tilde{k}^a_i,q_i,\vec{x}_i)$. Here $q_i$ is the nut
charge of the $i$th center; $\vec{x}_i$ is its position in $R^3$, and
$\tilde{k}^a_i$ determines the associated contributions to the
charges. Lifting this data into four-dimensional language, to make
contact with the two charge geometries, the defining data is
associated with a set of circles on the four-dimensional base space,
with $\vec{x}_i$ determining their radii and relative orientation. In
particular, if the $\vec{x}_i$ are parallel, the circles are
concentric and $|\vec{x}_i|$ determines their radii. The latter
solutions preserve a $U(1)^2$ isometry group of the base space, and
should correspond to R charge eigenstates in the conformal field
theory.
The basic question is therefore: how does the data
$(\tilde{k}^a_i,q_i,\vec{x}_i)$ capture the dual D1-D5-P microstate?
Consider first the case where the $\vec{x}_i$ are parallel, so the
corresponding microstate must be an R charge eigenstate. Then as a
first guess one might think that the multi-center data corresponds to
fractionation in the CFT. Recall that the Ramond ground states in
the orbifold CFT language as
\begin{equation}
\prod_{j} {\cal O}^{a_j \bar{a}_j}_{R n_j} | 0 \rangle, \qquad \sum_j
n_j = N,
\end{equation}
where each twist $n_i$ operator ${\cal O}^{a_i \bar{a}_i}_{R n_i}$
is associated with the $(a_i,\bar{a}_i)$ cohomology and
corresponds on spectral flow to an NS chiral primary.
The same operator can occur multiple times, with appropriate symmetrization.
The left moving excited states with total momentum $N_p$ are obtained
by exciting these ground states, with the momentum being distributed
between the different twist sectors. In the
string picture this corresponds to distributing the momentum
between each group of multiwound strings. Note that in such a
distribution the momenta $p_j$ given to each twist sector $n_j$ is non-negative.
It would be rather natural if this fractionation was also present in
the corresponding geometries. One might first try to identify each effective string with a
Gibbons-Hawking center, but this naive identification does not seem to be
correct. Recall that the total charges are given by
\begin{equation}
Q_a = -2 {\cal N} |\epsilon_{abc}| \sum_{i} \frac{\tilde{k}^b_i \tilde{k}^c_i}{q_i} \equiv
\sum_i (Q_a)_i,
\end{equation}
with $\sum_i \tilde{k}^a_i = 0$ and $\sum_i q_i = 1$. In particular,
the contributions to the three charges from centers with positive
$q_i$ cannot all be positive. At most, two out of the three $(Q_a)_i$
are positive, with the other negative, or only one of the three is
non-zero and positive. This does not however agree with the CFT
fractionation, where the total D1-D5-P charges are sums over
only positive contributions.
Actually this disagreement is not really surprising:
one should be rather careful about associating charge contributions to
each center, as there are no sources there. Whilst the solution is
built from harmonic functions sourced at the Gibbons-Hawking centers,
all the defining functions appearing in the solution are
finite at these centers. The distinguished hypersurfaces in the full
solution are clearly those where the Killing vector which is timelike
at infinity becomes null (\ref{sigma-beh}), which happens when the base space signature
changes. Thus it may be physically more natural to parameterize the
solutions in terms of functions which have poles at these
surfaces.
Indeed, this is explicitly demonstrated by the analysis of \cite{Giusto:2004kj} for the
specific 3-charge geometries obtained from limits of Cvetic-Youm
black hole solutions. Whilst one can rewrite these solutions in terms of six harmonic
functions on an ambipolar Gibbons-Hawking base space, the physical
interpretation is more manifest in the original Cvetic-Youm
coordinates, in which the defining functions are harmonic on $R^4$
with poles at the sources for the D-brane charges.
An important open question is thus whether
one can reparameterize the bubbling solutions in such a way
that the connection with fractionation in the CFT is manifest, and
moreover such that one can systematically solve the equations for
absence of closed timelike curves. For recent progress along related
lines see \cite{Bena:2008wt}: here coordinate transformations are used to
relate smooth three charge solutions with particular hyper-K\"{a}hler
bases to solutions in which several of the Gibbons-Hawking centers are
replaced by two charge (Lunin-Mathur) geometries. Such techniques may
help to understand which geometries correspond to
bound states and how the space of smooth three-charge solutions is
parameterized.
One should mention here that it is possible that the parameterization
in terms of the ambipolar Gibbons-Hawking space may still be natural
for the M theory geometries; perhaps the centers are related to the
nodes in the quiver quantum mechanics as has been
suggested in \cite{Bena:2007ju,Bena:2007qc}. Nonetheless, a systematic solution of the regularity
conditions in the M theory system is also still lacking.
Furthermore, without solving such conditions, it would be hard to
carry out geometric quantization for these fuzzball geometries; one
clearly needs to determine these constraints before quantization. Of
course, even if one could carry out the geometric quantization, it would be
hard to make further progress without determining which black hole
microstates the bubbling geometries sample, i.e. without carrying out the
matching.
Another related puzzle is the flight time calculation in the scaling
bubbling geometries: this gives a finite answer for symmetric
geometries, which can be interpreted in terms of the field theory mass
gap, but for non-symmetric geometries the time can become infinite
\cite{Bena:2007ju,Bena:2007qc} although one would still anticipate an
energy gap in the field theory. Thus either such geometries do not
correspond to microstates or the calculation needs to be interpreted
more carefully.
\section{Fuzzballs and black hole physics} \label{open}
The previous sections have focused on the explicit construction of
fuzzball solutions in supergravity for near supersymmetric black holes, and the
matching of these solutions with black hole microstates. Most of the
literature on the fuzzball proposal has concentrated on this
issue, since finding and matching candidate geometries for
black holes with macroscopic horizons has turned out to be difficult
even in the supersymmetric case. Results to date can be regarded as
evidence for the fuzzball proposal, but to make further detailed and
quantitative progress on issues such as coarse-graining one would
anticipate the need to match better the candidate bubbling geometries
with CFT data and to understand fuzzball solutions in the stringy regime.
At the same time, one can already envisage how many key conceptual
questions could be answered, and there are many interesting and
suggestive results in the current literature. Therefore in this section we will
consider how longstanding issues in black hole physics are addressed
by the fuzzball proposal.
\subsection{Quantizing fuzzball geometries}
Suppose one finds fuzzball geometries in supergravity with the correct
charges to correspond to microstates of a given black hole. Typically
the geometries will be characterized by continuous functions, and will
not be countable. Knowledge of the holographic map between these
solutions and the CFT microstates should determine how the continuous
functions are discretized, as in the case of 2-charge geometries.
Alternatively, one could consider quantizing the
geometries, as in the discussion in section (\ref{geo-quant}).
This would allow one to estimate the number of states accounted for by the
geometries, and, if it
reproduced the black hole entropy, this could be interpreted as evidence
that the geometries do describe black hole microstates.
At the same time, it is not clear why quantizing the fuzzball
geometries {\it visible in supergravity} should reproduce the full black hole
entropy in general. In the 2-charge case, quantizing the extrapolation
of the fuzzball solutions to supergravity (for the subset of solutions that this
was done) indeed reproduced the black
hole entropy, even though the average solution was string scale, but
one cannot assume that this agreement will
persist in other, less supersymmetric systems.
If geometric quantization of the fuzzball solutions found in
supergravity does reproduce the black hole entropy, it would suggest that
the extrapolation of the fuzzballs to supergravity is more
representative than one may have anticipated. Moreover, one would then
be able to address the important issue of coarse-graining, to
demonstrate explicitly how black hole properties emerge.
\subsection{Finite temperature} \label{temp}
Much of the discussion of previous sections has been restricted to
the case of supersymmetric (zero temperature) solutions, since one
can then use powerful tools to find supergravity solutions. Clearly
restricting to zero temperature removes an important feature of black
hole physics: extremal black holes do not Hawking radiate. So an
important question is how the fuzzball proposal works for a
non-extremal black hole.
Given the very few known candidate fuzzball geometries for non-extremal
black holes, the issue of Hawking radiation has not been discussed in
generality. However, in cases where AdS/CFT is applicable, one can
understand how the non-extremal fuzzballs will radiate, and moreover
use the specific known solutions as a testing ground.
First let us recall how Hawking radiation of asymptotically $AdS_{d+1}$
black holes is computed in the dual conformal field theory. Consider
radiation of a scalar field $\phi$, which corresponds to the dual
gauge invariant operator ${\cal O}_{\phi}$ of dimension
$\Delta_{\phi}$. Let the emitted radiation
have $d$-dimensional momentum $\vec{p}$.
Now consider a state $|i \rangle$ in the boundary field theory which
is dual to an asymptotically $AdS_{d+1}$ spacetime. The rate of
emission of the scalar field $\phi$ with momentum $\vec{p}$ in the
spacetime is computed from the decay rate of the corresponding field
theory state $| i \rangle$. The differential decay rate $d \Gamma(i)$
is given by
\begin{equation}
d \Gamma(i) = \sum_{f} | M_{fi} |^2 d \Omega
\end{equation}
where $| f \rangle$ are possible final states, $d \Omega$ is an
appropriate phase space factor, and the matrix element $M_{fi}$ is
given by
\begin{equation}
{\cal N} \langle f | {\cal O}_{\phi} (\vec{p}) | i \rangle,
\end{equation}
with ${\cal N}$ a normalization constant (fixed by two point functions
at the conformal point). In practice one computes this by using the
optical theorem to relate the sum of $|M_{fi}|^2$ over final states
into an appropriate discontinuity of the analytically continued
Euclidean two point function.
Let us now focus on the case of interest, a 2d CFT.
To compute the decay rate of the BTZ black hole corresponding to a thermal
state in a (grand) canonical ensemble, one must average over initial
states weighted by the appropriate Boltzmann factors.
The analytic continuation of the
Euclidean thermal two point function for the scalar operator of dimension
$\Delta_{\phi}$ is given by
\begin{equation}
\Pi(x^+ x^{-}) \equiv \langle {\cal
O}^{\dagger}_{\phi} (0,0) {\cal O}_{\phi} (x^+,x^-)
\rangle_{\rm thermal} = {\cal C} \left [\frac{\pi T_+}{\sinh (\pi T_{+} x^+)} \right ] ^{\Delta_{\phi}}
\left [ \frac{\pi T_-}{\sinh (\pi T_{-} x^-)} \right ]^{\Delta_{\phi}},
\end{equation}
where ${\cal C}$ is a normalization factor, $x^{\pm} = t \pm z$ are lightcone coordinates and
$T_{\pm}$ are the left and right moving temperatures
respectively. Expressed in terms of the inverse temperature $\beta = T^{-1}$ conjugate to
the energy $\omega$ and the chemical potential $\mu_z$ conjugate to the
momentum $p_z$,
$2 T^{-1}_{\pm} = \beta \pm \mu_z$. In terms
of the BTZ temperature $T_{H}$ given in (\ref{btz-temp}) and the inner and outer horizon
locations $\r_{\pm}$ these quantities are
\begin{equation}
T_{+}^{-1} = l T_{H}^{-1} \left ( 1 + \frac{\r_{-}}{\r_{+}} \right ); \qquad
T_{-}^{-1} = l T_{H}^{-1} \left (1 - \frac{\r_-}{\r_{+}} \right ).
\end{equation}
Then the emission rate $\G_e$ is proportional to the discontinuity in the two
point function
\begin{equation}
\G_e = {\cal F} \int dt dz e^{i \omega t - i p_z z} \Pi(t + i\epsilon,z)
\end{equation}
where ${\cal F}$ is the flux of the emitted particle.
For example, in the case of $\Delta_{\phi} = 2$ and $p_z \rightarrow
0$ and inserting appropriate normalization factors one obtains
\begin{equation} \label{decay}
\G_e = \frac{\pi^3 Q_1 Q_5 \omega} {(e^{\frac{\omega}{2T_+}} -1)
(e^{\frac{\omega}{2T_-}} -1)}
\end{equation}
in agreement with the corresponding Hawking radiation rate computed in
the bulk. More details of this calculation may be found in
\cite{emission}. Note that detailed balance implies that the emission rate
$\G_e = e^{-\omega/T_H} \G$ where $\G$ is the absorption rate.
To compute instead the decay (or absorption) rate of a particular state $| i \rangle $
in an ensemble one needs to compute the appropriate discontinuity in the
two point function in that
state. For a typical member of a thermal ensemble one would
expect the variance from the thermal spectral density to be small. For
an atypical member of the ensemble the decay rate can however
differ substantially.
Now suppose one has a candidate asymptotically $AdS_{d+1}$ fuzzball
geometry dual to the specific state $| i \rangle$: how should the
instability of the state to decay by the operator ${\cal O}_{\phi}$ be
reflected in the bulk, which has no horizon so does not Hawking
radiate? The answer is that modes of the bulk field
$\phi$ with real momentum $\vec{p}$ should be tachyonic, with
the imaginary part of the frequency reflecting the decay rate.
This is precisely what has been found for the specific known examples of
non-extremal fuzzball solutions, such as that discussed in the
previous section. The instability of the bulk solutions
precisely matches the computed decay rate of the corresponding CFT
microstate \cite {Chowdhury:2007jx,Dias:2007nj}. However, the CFT
microstate under consideration
is a highly atypical member of the thermal ensemble, and thus the
emitted radiation is not even approximately thermal: $\G_e \approx
\w^2$, compared to (\ref{decay}). This rapid decay rate is responsible
for the superradiant behavior in the bulk solution.
\subsection{Path integral approach}
Despite its computational and conceptual difficulties, the Euclidean
path integral approach to gravity rather naturally accounts for black
hole thermodynamics. The Euclidean path integral is
\begin{equation}
{\cal Z} = \int D[g] D [\Phi] e^{- I_{E} (g,\Phi)},
\end{equation}
where $g$ is the metric, $\Phi$ collectively denotes matter fields and
$I_{E}$ is the Euclidean action. Whilst making sense of the
integration over geometries remains an open problem, it is interesting
that in the saddle
point approximation the onshell Euclidean action for a solution of the
Einstein (supergravity) equations behaves as the free energy $F$, i.e.
\begin{equation}
I_{E} = \beta F = \beta (E - \mu_{i} Q_i) - S,
\end{equation}
with $\beta$ the inverse temperature, $\mu_i$ the conjugate
potentials to conserved charges $Q_i$ (angular momentum, electric
charge etc) and $S$ the Bekenstein-Hawking entropy. This
identification is consistent with standard thermodynamic relations,
for example
\begin{equation}
S = - \left ( \frac{\pa F}{\pa T} \right )_{Q_i} = \left (\beta
\frac{\pa I_{E}}{\pa \beta} - I_E \right ).
\end{equation}
Note that extremal black holes should be treated as a limiting case of
non-extremal black holes, in which case their entropy is non-zero. If
one works directly at extremality ($T=0$) the
Euclidean action is linear in $\beta$, and thus the entropy would seem
to vanish.
The fuzzball solutions are however solutions of the
gravitational field equations with the same asymptotics as the black
hole, and are thus also saddle points of the path integral. Fuzzball
solutions which are stationary in the Lorentzian correspond to
topologically trivial Euclidean solutions, since the imaginary time
Killing vector has no fixed point sets. There is a natural
thermodynamic interpretation of the fuzzball solutions: each solution
has the same mass $E$ and conserved charges $Q_i$ as the black hole, and
quantization of the solutions should give a total degeneracy
${\cal D} = e^{S}$. Therefore the Euclidean action $I_{f_a}$ associated with each fuzzball
$f_a$ should be
\begin{equation}
I_{f_a} = \beta F = \beta (E - \mu_{i} Q_i).
\end{equation}
Clearly by construction one finds that
\begin{equation}
\int_{f_a} D [g_a] D [\Phi_a] e^{-I_{f_a}} = e^{-I_{f_a}
+ S} = e^{-I_{BH}},
\end{equation}
where the functional integral is over all fuzzball geometries $f_a$,
with metric $g_a$ and matter fields $\Phi_a$, and
$I_{BH}$ is the action for the black hole of the same mass and
charges.
Evidently one would like to derive these formulae explicitly in a
specific example, by quantizing the set of fuzzball solutions. However
the argument above suggests that in the integration over geometries
one should not include both black holes and fuzzballs simultaneously,
as this is over counting. One should either include all topologically
trivial configurations with the same charges (namely the fuzzballs) or
one should include the topologically non-trivial configuration (namely
the black hole). In cases where AdS/CFT is applicable, this fits
exactly with field theory expectations: one either sums over
pure states or in equilibrium configurations typical states may be
represented by thermal states, but one does not simultaneously
include both pure and mixed states when computing a partition
function.
Hawking's proposed resolution of the information loss paradox
\cite{Hawking:2005kf} also makes use of the path integral over
asymptotically AdS metrics. However both topologically trivial and
non-trivial configurations are included simultaneously in the path
integral. The former are unitary and the latter are not, but it is
argued that contributions to correlation functions from topologically
non-trivial configurations fall off rapidly, and do not contribute at
late times. Thus just as in the fuzzball proposal one effectively only
includes the topologically trivial configurations in the path
integral although of course the justification is rather different.
\subsection{Distinguishing between a black hole and a fuzzball}
Given that the black hole curvature is small at the horizon scale, it
may initially seem surprising that the fuzzball solutions start to
differ from the black hole already at this scale. One might have
argued that deviations from the black hole spacetime which resolve its
singularity should be localized near the curvature singularity.
At the same time, standard results in black
hole physics already indicate that the horizon itself is deeply
connected with quantum gravity effects. Firstly, the existence of a
closed trapped surface is an input into the singularity theorems
relevant to black holes. As we review below, it is hard to evade these
theorems within supergravity when a horizon is present. Secondly,
Hawking radiation is associated with the existence of a horizon, and
thus with the well known trans-Planckian and backreaction issues
indicating the breakdown of semiclassical approximations.
One might worry that the small differences between the fuzzball geometry
and that of the black hole at the horizon scale manifest as
substantial differences in the measurements of an observer at
infinity. In asymptotically AdS examples, however, we have discussed in detail
how a black hole microstate is characterized by vevs and higher point
functions; the deviation of these from the corresponding ensemble
average is small for a typical microstate, and thus distinguishing
between black hole and fuzzball requires a large sequence of
measurements over a long timescale. Essentially the same point was
made by Hawking in \cite{Hawking:2005kf}: an observer at spatial infinity would
need to make measurements over an infinite time to be certain that a black hole
formed and then evaporated.
Inside the outer future event horizon of a non-extremal black hole the Killing
vector $\pa_{t}$ is spacelike and the radius necessarily decreases along
future directed timelike geodesics. In a corresponding fuzzball
solution the geometry at sub-horizon scales may be radically
different, with for example $\pa_t$ remaining timelike, so cannot an
observer at infinity detect this difference? Again this question
needs to be phrased more precisely in terms of measurements accessible
to the observer: scattering of particles and measurement of emitted radiation.
The former corresponds to excitation and subsequent decay by radiation
of the black hole with the latter corresponding to spontaneous radiation.
A typical non-extremal fuzzball should emit radiation
with only small deviations from the thermal spectrum of the black
hole, which would take the observer a long time to measure.
Discussions of the small scale of deviations of correlation functions in a typical
state from that of the thermal state can be found in
\cite{Vijay,Balasubramanian:2007qv}; see also \cite{Balasubramanian:2005mg}.
Note that atypical non-extremal fuzzballs can contain ergoregions, in which a
Killing vector which is timelike at infinity becomes spacelike. Such
ergoregions are responsible both for superradiance processes
(mirroring excitation and decay of the black hole) and for
spontaneous emission (corresponding to Hawking radiation). However
these atypical fuzzballs decay much faster than the corresponding
black hole, as we saw in section (\ref{temp}). One can also understand
Hawking radiation from static black holes in terms of spontaneous emission
from an ergoregion bounded by the horizon. The key
difference between the black hole and the atypical fuzzball in this respect is
that in former case the ergoregion straddles the horizon, whereas for
the fuzzball there is no horizon to cloak it.
\subsection{Gravitational collapse}
Whilst eternal black holes represent a simplified system for explicit
analysis, one would also like to apply the fuzzball proposal to
astrophysical gravitational collapse. In astrophysical collapse it is
believed that given sufficiently dense and generic initial data the formation
of first a closed trapped surface and then a (cloaked) singularity is
inevitable. The fuzzball proposal in this context would be that supergravity
solutions exist which describe the collapse of initial data,
avoiding horizon and singularity formation. The late-time
quasi-stationary solution should differ from the corresponding black hole spacetime
only at sub-horizon scales.
To date only stationary fuzzball solutions have been found, and
it would be difficult to find the time dependent solutions
needed to describe gravitational collapse. However, in cases where one
can use AdS/CFT, the initial data corresponding to a pure state in the
field theory should collapse to form a fuzzball. Recall that the
fuzzball solutions generically involve all supergravity fields (and beyond)
reflecting the fact that in a generic state all primary (and other) operators
acquire a vev. If one specifies initial data only for a subset
of fields then one is restricting to a subspace of the phase space,
and this could be interpreted as tracing over the complementary part of the
phase space, so this initial data should correspond to mixed states.
Such initial data should collapse to form black holes.
It would clearly be interesting to investigate this further, and
understand how this connects to results from
numerical simulations of gravitational collapse. In particular, it is
known that even for spherically symmetric collapse horizon formation
can be avoided for non-generic initial data
\cite{Choptuik:1992jv}. For the non spherically symmetric collapse
relevant to fuzzballs fewer results are known (see the review \cite{Gundlach:2002sx})
but perhaps the conditions on initial data needed
to avoid horizon formation can be related to conditions on the dual
theory being in a pure state.
It is also interesting to note here that many of the results of
numerical relativity support the fuzzball proposal: by adding just one
scalar field to the action one can find numerically solutions of the
Einstein equations which are horizonless, non-singular and asymptote
to the black hole solution. In particular, in a recent paper
\cite{Krueger:2008nq} numerical solutions were found which could
be interpreted as fuzzballs for the Schwarzschild black hole.
\subsection{Singularity theorems}
The fuzzball proposal states that each black hole microstate
corresponds to a non-singular horizon-free geometry. In this section
we will consider how the proposal is reconciled with the general
proofs of singularity theorems.
We have seen in explicit examples that fuzzball solutions for
spherically symmetric black holes break the spherical symmetry, and
involve many more of the supergravity fields than the black hole solution.
This at first sight seems somewhat reminiscent of proposals for
singularity resolution in general relativity suggested in the
1960s. The first singularity
theorems showed that there would be singularities in symmetric
solutions under certain reasonable conditions, but the results depended
on the symmetry being exact. It was thus suggested by a number of
authors (in particular, the Russian school of Lifshitz, Khalatanikov and
co-workers \cite{Lifshitz:1963ps}) that singularities were the result of symmetries and would not
occur in more general solutions. Indeed this idea seemingly fits with the fuzzball
proposal, as the non-singular solutions are less symmetric and involve
more fields.
However, the idea that singularities were non-generic was soon
disfavored, in part because the Russian school found that explicit
families of non-symmetric solutions did contain singularities \cite{Khalatnikov:1969eg}, but
principally because of the Penrose-Hawking
proofs \cite{Hawking:1969sw} of singularity theorems without assumptions of symmetry. So how
is the fuzzball proposal reconciled with the singularity theorems?
To understand this it is useful to recall the three main conditions
required to prove the existence of a singularity, or to be more precise,
geodesic incompleteness of the spacetime:
\begin{enumerate}
\item{An energy condition;}
\item{Condition on global structure;}
\item{Gravity strong enough to trap a region.}
\end{enumerate}
The exact conditions depend on the theorem under consideration, see
\cite{Hawking:1973uf}; for
example, the second condition can be the existence of a non-compact
Cauchy surface or that the chronology condition holds throughout the
spacetime. More importantly for our purposes, the third condition can
be that the spatial cross-section of the universe is closed or that
there is a closed trapped surface in the spacetime. The proofs input
these conditions into the
Raychaudhuri equation to show that timelike or null geodesics converge
within finite affine length, and thus that the spacetime is
geodesically incomplete.
The fuzzball solutions which are non-singular in supergravity evade
the singularity theorems via the third condition: the supergravity
fields do satisfy the weak and/or strong energy condition, and there is an
appropriate non-compact Cauchy surface, but there is no closed trapped
surface or horizon in the spacetime!
Of course, as we have emphasized throughout this report, a generic
fuzzball solution should be described as a solution of the
full string theory equations of motion, not just as a supergravity
background. In such cases the singularity theorems are simply not applicable.
For the same reason, whilst the singularity theorems imply that any black object in
supergravity has a singularity behind its horizon, stringy corrections
are expected to resolve these singularities.
\subsection{Emergence of horizons}
The main objective of the fuzzball proposal is to show that black hole
properties emerge upon coarse-graining over fuzzball geometries. In
particular, one would like to understand the emergence of the
defining property of a macroscopic black hole, its horizon. Given the
incomplete understanding of fuzzball geometries for even
supersymmetric macroscopic black holes, such a coarse-graining has not
yet been carried out.
One can however use rather general arguments to infer the scale of the
coarse-grained geometry. One first needs to estimate how much a typical
fuzzball geometry differs from the naive black hole solution, but this
is a question which is easy to answer, having set up the AdS/CFT
dictionary. Near the conformal boundary, in the far UV of the field
theory, all fuzzball solutions should look like the naive solution,
since in the UV we see only the conformal behavior. The solutions
should start differing at energies comparable to the scale set by the
vev of the most relevant gauge invariant operator. Using AdS/CFT this
translates into a certain radial distance scale.
Suppose one estimates the vev of the lowest dimension operator in a
typical microstate of the black hole, and then translates this value
into a characteristic radial scale $r_c$. Now compute the area of a
spatial surface of constant radius $r_c$: the area of this so-called
stretched horizon (in the naive metric)
should approximately reproduce the black hole horizon area. A version of this
argument was given in \cite{Lunin:2001jy,Lunin:2002qf,
Mathur:2007jn,Mathur:2007sc}, although here we use the AdS/CFT
dictionary and the vevs to determine more systematically the relevant
radial scale.
It is straightforward to carry out this calculation for the 2-charge
black hole. In this case the vev of the most relevant gauge invariant
operator is expressed in terms of the curve characterizing the
fuzzball solution. Thus the radial scale $r_c$ is determined by the
scale of a typical curve. Now recall that the defining curve $F^a(v)$
in $k$-dimensional space, with $a=1,\cdots k$, behaves
as
\begin{equation}
F^a(v) = \mu \sum_{n > 0} (f^a_n \exp(2 \pi i nv/L) + (f^a_n)^{\ast} \exp(-2 \pi i nv/L)),
\end{equation}
with $\mu = \sqrt{Q_1 Q_5}/R_z$, $L =2 \pi Q_5/R_z$ and
\begin{equation}
Q_1 = \frac{Q_5}{L} \int^L_0 (\pa_v F^a)^2 dv \qquad \rightarrow
\qquad 2 \sum_{a, n >0} n (f_n^a (f_n^{a})^{\ast}) = 1.
\end{equation}
A typical curve $F^a(v)$ satisfying this constraint will have a scale
of the order of
\begin{equation} \label{r-c}
r_c \sim \frac{\mu}{\sqrt{N}}.
\end{equation}
Now recall that the naive 2-charge metric (in the string frame) in the
decoupling region is
\begin{equation}
ds^2 = \frac{r^2}{\sqrt{Q_1 Q_5}} (-dt^2 +dz^2) + \sqrt{Q_1 Q_5}
(\frac{dr^2}{r^2 } + d\Omega_2^2) + \frac{\sqrt{Q_1}}{\sqrt{Q_5}} ds^2(X_4),
\end{equation}
with $e^{2 \Phi} = Q_1/Q_5$. Then the area of a spatial surface at
constant $r_c$ in the Einstein frame metric ($g_{ein} = e^{-\Phi/2}
g$) is ${\cal A}$ where
\begin{equation}
{\cal A} = (2 \pi)^6 \pi V R_z \sqrt{Q_1 Q_5} r_c \sim (2\pi)^6 \pi
(\a')^4 \sqrt{N} .
\end{equation}
Putting the area into the Bekenstein-Hawking entropy formula gives
\begin{equation}
S = \frac{2 {\cal A}}{(2 \pi)^6 (\a')^4} \sim 2 \pi \sqrt{N}.
\end{equation}
Hence the area of the stretched horizon reproduces the microscopic
entropy of the 2-charge black hole. Of course one should be a little
cautious of this computation, as the typical scale implied by
(\ref{r-c}) is
\begin{equation}
r_c \sim \frac{(\a')^2}{R_z \sqrt{V}},
\end{equation}
which is clearly substring scale for the compactification radii of
interest, as indeed it must be for the 2-charge black hole which
has no horizon in supergravity.
\bigskip
Whilst these general arguments can be used to infer the characteristic
scale of a typical fuzzball geometry, explicitly demonstrating that a
horizon emerges upon coarse-graining remains an open problem. One
first needs to construct a representative basis of fuzzball geometries
for a black hole with a macroscopic horizon before one can explore
quantization and coarse-graining. The three charge system seems a
promising test case for exploring such issues.
Note that if the fuzzball proposal is correct then coarse-graining
over horizonless backgrounds of trivial topology should result in the
black hole background, whose topology is non-trivial. The black hole
should have a global feature, a horizon, which none of the individual
fuzzballs possess. On the Euclidean section this translates into
periodicity in imaginary time, and non-trivial topology.
It may at first seem surprising that the black hole has a property which is not
shared by any of the fuzzballs. Of course, given that this property is
associated with entropy, this must be the case if the fuzzball
proposal is to be correct. Moreover, the emergence of non-trivial
topology under coarse-graining precisely mirrors the corresponding
field theory behavior: thermal states in the field theory are
characterized by periodicity in imaginary time, which is not a
property of any state in the thermal ensemble.
\subsection{Stringy fuzzballs}
Throughout this report we have emphasized that many of the fuzzballs
for any given black hole will not be well-described by
supergravity. To develop the fuzzball proposal, and indeed to
understand more rigorously small black holes, one will most likely
need to work with backgrounds of the full string theory.
Clearly there are many technical obstacles to overcome in this area:
string theory in backgrounds with Ramond-Ramond fluxes is still rather
poorly understood, even given the progress with the pure spinor
formalism, and solving the worldsheet string theory in curved
backgrounds is generically very hard.
At the same time, not all issues relating to stringy fuzzballs are
likely to be intractable. Suppose one considers a worldsheet theory
in an asymptotically $AdS_3 \times S^3 \times X_4$ fuzzball
background, which is sufficiently strongly curved that the
supergravity approximation does not hold. Just as in the
case of $AdS_3 \times S^3 \times X_4$, the information that one
extracts from the worldsheet theory is the (holographic) correlation
functions.
Throughout this report, we have expressed the correspondence between
fuzzballs and black hole microstates in terms of these correlation
functions: the one-point functions are used to identify the specific
microstate whilst the two-point functions characterize the decay rate
of non-BPS fuzzballs etc. Even when a geometric description is
accessible, the correspondence is not expressed in terms of the local
geometry, but rather in terms of measurements made in the asymptotic
region. Thus the absence of a geometric description does not
necessitate a complete reformulation of the fuzzball proposal:
the information one computes from the worldsheet theory is
exactly the information needed to probe the fuzzball
proposal.
In the supergravity regime, one could compute holographic one-point
functions from a given supergravity solution from algebraic
manipulations even when solving fluctuation equations to extract
higher correlation functions is intractable.
The same should be true for the worldsheet theory: given the
string spectrum in the $AdS_3 \times S^3$ background one would expect
that one can extract holographic one-point functions from the worldsheet theory
in asymptotically $AdS_3 \times S^3$ backgrounds
even if one cannot fully solve the
worldsheet theory. Moreover, one would anticipate being able to
determine whether a given worldsheet theory corresponds to a black
hole (i.e. there is entropy) or to a fuzzball. One should
be able to address in general terms
how one point functions and entropy are encoded in
the worldsheet theory, and this would be interesting in itself.
\section{Conclusions}
The fuzzball proposal is a promising idea which has the potential to
resolve longstanding black hole related puzzles. In the few years since
this proposal was suggested a body of evidence
in support of the proposal has been found. In particular, families of
fuzzball solutions visible within supergravity have been found for
various black hole systems in string theory, and
holographic technology has been used to match their properties with
those of black hole microstates. All computations, be they of
holographic one point functions, scattering or geometric quantization,
support the interpretation of these geometries as black hole
microstates. The aim of this report was to collect in one place
all current evidence, emphasizing interconnections and pointing out
open problems and avenues for further research. We further sketched how
key issues in black hole physics could be addressed by the fuzzball proposal.
A recurrent theme in this report was the use of the AdS/CFT correspondence
to test, motivate and perhaps even explain why (a form) of the fuzzball
proposal should hold, at least for black holes which admit AdS near horizon
regions. In our view, precision holography would be needed if this interesting
idea is to become a physical model that would either be falsified or explain
black holes at a quantitative level and for this reason we placed
particular emphasis on this topic.
Whilst the fuzzball proposal has the potential to address black hole
issues, substantial work and technical progress
is still needed to demonstrate explicitly how the key issues are
resolved. One would like to show in concrete examples
how black hole properties, such as
the horizon and Hawking radiation, emerge upon coarse-graining over
fuzzballs. To address these issues one will need to understand better
the holographic map between solutions and microstates
and how to coarse-grain geometries. Most likely
one will also be forced to work with backgrounds of the full string
theory, rather than just supergravity.
The fuzzball proposal is a concrete idea which could explain black
hole physics and is very natural from a holographic perspective.
Given its potential and the increasing body of supporting
evidence, it merits further development and investigation.
\section*{Acknowledgments}
The authors acknowledge support from NWO, KS via the Vernieuwingsimplus grant
``Quantum gravity and particle physics'' and
MMT via the Vidi grant ``Holography, duality and time dependence in
string theory''. We would like
to thank the 2007 Simons Workshop and the 2007 Paris Summer
Workshop for their hospitality and the stimulating environment they
provided. We have
benefited from interesting discussions with many colleagues, including
S. Mathur, I. Bena, N. Warner, E. Rabinovici, R. Brustein, D. Marolf,
I. Kanitscheider, P. Kraus, F. Larsen, M. Berkooz and G. Horowitz.
| {
"redpajama_set_name": "RedPajamaArXiv"
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Amazing è un singolo della cantante italiana Francesca Michielin, pubblicato il 28 marzo 2014 come unico estratto dalla colonna sonora del film The Amazing Spider-Man 2 - Il potere di Electro.
Il singolo è stato successivamente inserito nella lista tracce del secondo album in studio della Michielin, di20.
Descrizione
Ad inizio marzo, la cantante ha annunciato attraverso Facebook di far parte degli artisti che hanno composto la colonna sonora del film The Amazing Spider-Man 2 - Il potere di Electro, annunciando il titolo del brano da lei realizzato. Amazing ha visto la cantante sia nei ruoli di interprete sia in quello di autrice di musica e testo insieme a Negin Djafari e Fausto Cogliati.
Pochi giorni dopo, sempre attraverso Facebook, la cantante ha ufficializzato il 28 marzo come data per la pubblicazione del singolo sia per il download digitale che per la rotazione radiofonica. Nello stesso annuncio, la Michielin si è detta «felice per aver avuto la possibilità di scrivere un pezzo per questo film» e di essere contenta per essersi messa in gioco con un brano in inglese.
Video musicale
Il videoclip, diretto da Gaetano Morbioli, è stato reso disponibile a partire dall'11 aprile e alterna scene tratte da The Amazing Spider-Man 2 - Il potere di Electro con altre in cui la Michielin canta, ambientate in un interno.
Tracce
Note
Collegamenti esterni | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 4,657 |
Суперкубок Німеччини з футболу 1989 — 3-й офіційний розіграш турніру (6-й розіграш, враховуючи офіційні та неофіційні матчі турніру). Матч відбувся 25 липня 1989 року між чемпіоном Німеччини «Баварією» та володарем кубка Німеччини «Боруссією» (Дортмунд).
Матч
Деталі
Посилання
Матч на rsssf
Матч на transfermarkt
ФРН
1989
1989 у німецькому спорті | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 8,664 |
{"url":"https:\/\/math.libretexts.org\/TextMaps\/Calculus\/Book%3A_Calculus_(OpenStax)\/17%3A_Second-Order_Differential_Equations\/17.0%3A_Prelude_to_Second-Order_Differential_Equations","text":"\n17.0: Prelude to Second-Order Differential Equations\n\n$$\\newcommand{\\vecs}[1]{\\overset { \\scriptstyle \\rightharpoonup} {\\mathbf{#1}} }$$\n\n$$\\newcommand{\\vecd}[1]{\\overset{-\\!-\\!\\rightharpoonup}{\\vphantom{a}\\smash {#1}}}$$\n\nWe have already studied the basics of differential equations, including separable first-order equations. In this chapter, we go a little further and look at second-order equations, which are equations containing second derivatives of the dependent variable. The solution methods we examine are different from those discussed earlier, and the solutions tend to involve trigonometric functions as well as exponential functions. Here we concentrate primarily on second-order equations with constant coefficients.\n\nFigure $$\\PageIndex{1}$$: A motorcycle suspension system is an example of a damped spring-mass system. The spring absorbs bumps and keeps the tire in contact with the road. The shock absorber damps the motion so the motorcycle does not continue to bounce after going over each bump. (credit: nSeika, Flickr)\n\nSuch equations have many practical applications. The operation of certain electrical circuits, known as resistor\u2013inductor\u2013capacitor (RLC\u00a0) circuits, can be described by second-order differential equations with constant coefficients. These circuits are found in all kinds of modern electronic devices\u2014from computers to smartphones to televisions. Such circuits can be used to select a range of frequencies from the entire radio wave spectrum, and are they commonly used for tuning AM\/FM radios. We look at these circuits more closely in\u00a0Applications.\n\nSpring-mass systems, such as motorcycle shock absorbers, are a second common application of second-order differential equations. For motocross riders, the suspension systems on their motorcycles are very important. The off-road courses on which they ride often include jumps, and losing control of the motorcycle when landing could cost them the race. The movement of the shock absorber depends on the amount of damping in the system. In this chapter, we model forced and unforced spring-mass systems with varying amounts of damping.","date":"2018-06-25 11:39:40","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\": 0.4051613211631775, \"perplexity\": 570.5082856797417}, \"config\": {\"markdown_headings\": false, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2018-26\/segments\/1529267867666.97\/warc\/CC-MAIN-20180625111632-20180625131632-00546.warc.gz\"}"} | null | null |
{"url":"https:\/\/math.stackexchange.com\/questions\/3160936\/recurrence-formality-question","text":"# Recurrence formality question\n\nI have tried to solve this recurrence relation using induction. $$T(n) = T(\\lfloor \\log_2 n \\rfloor) +1$$ It is clear that I should get something similar to $$\\log *n$$, but I don't know how to formalize this kind of questions. Thank you for your kind help.\n\nThat $$\\lfloor \\log_2 n \\rfloor$$ suggests you should plug in powers of $$2$$ and see if something interesting happens: (Assume $$T(1)$$ is given) \\begin{align} T(2) &= T(\\lfloor \\log_2 2 \\rfloor) +1 \\\\ &= T(1) +1; \\\\ T(4) &= T(\\lfloor \\log_2 4 \\rfloor) +1 \\\\ &= T(2) +1 \\\\ &= T(1) +2; \\\\ T(8) &= T(\\lfloor \\log_2 8 \\rfloor) +1 \\\\ &= T(3) +1 \\\\ &= T(1) +3; \\\\ T(16) &= T(\\lfloor \\log_2 16 \\rfloor) +1 \\\\ &= T(4) +1 \\\\ &= T(1) +4; \\end{align} and so on. Then, you notice that $$T(2^m) = T(1) +m$$; you may prove it by induction (the inductive step will make use of the recurrence relation.)","date":"2019-05-23 19:35:32","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\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 7, \"wp-katex-eq\": 0, \"align\": 1, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 1.0000100135803223, \"perplexity\": 53.51569081556682}, \"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-2019-22\/segments\/1558232257361.12\/warc\/CC-MAIN-20190523184048-20190523210048-00157.warc.gz\"}"} | null | null |
{"url":"https:\/\/community.wolfram.com\/web\/kmartin\/home?p_p_id=user_WAR_userportlet&p_p_lifecycle=0&p_p_state=normal&p_p_mode=view&p_p_col_id=column-1&p_p_col_count=1&tabs1=Discussions","text":"# User Portlet\n\nKyle Martin\nDiscussions\nRen\u00e9, I'd recommend sending a minimized example of your notebook to support@wolfram.com for them to look over for regressions.\nHussain, Of course I would first check that you filled out the form correctly with a valid payment information (I'm sure you already did this). Otherwise, I think your best chance of getting a reply from someone who can help will be to fill out...\nDid you restart Mathematica after deleting the palette?\nIf you're evaluating Wolfram Language code, why not do so in Mathematica, the Wolfram Engine, or the Wolfram Cloud (or some other WL environment)? I always treated Wolfram|Alpha handling any raw WL code as me getting lucky rather than a feature I...\nThanks for the pointer, I've reported this to the appropriate development team.\nThanks for the details!\nI can confirm that this is a regression in 12.1. We're looking into it, I'll update this thread if I come across a workaround.\nDoes this help? http:\/\/support.wolfram.com\/47123\nIncreasing PlotPoints might help, but it will be difficult to definitively say much more without reproducible code.\nPerhaps this is trivial, but I want to be sure: It is essential that your cursor be _within_ the cell when you hit SHIFT+ENTER. In your screenshot, you have the line below the cell, and if you hit SHIFT+ENTER in this scenario, nothing will happen. ...","date":"2021-10-21 09:25:33","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\": 0.27907776832580566, \"perplexity\": 1225.2941799142995}, \"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\/1634323585382.32\/warc\/CC-MAIN-20211021071407-20211021101407-00508.warc.gz\"}"} | null | null |
{"url":"https:\/\/dbpedia.org\/page\/Lambda","text":"An Entity of Type: radio station, from Named Graph: http:\/\/dbpedia.org, within Data Space: dbpedia.org\n\nLambda (\/\u02c8l\u00e6mbd\u0259\/; uppercase \u039b, lowercase \u03bb; Greek: \u03bb\u03ac\u03bc(\u03b2)\u03b4\u03b1, l\u00e1m(b)da) is the 11th letter of the Greek alphabet, representing the sound \/l\/. In the system of Greek numerals lambda has a value of 30. Lambda is derived from the Phoenician Lamed . Lambda gave rise to the Latin L and the Cyrillic El (\u041b). The ancient grammarians and dramatists give evidence to the pronunciation as [la\u02d0bda\u02d0] (\u03bb\u03ac\u03b2\u03b4\u03b1) in Classical Greek times. In Modern Greek the name of the letter, \u039b\u03ac\u03bc\u03b4\u03b1, is pronounced [\u02c8lam.\u00f0a].\n\nProperty Value\ndbo:abstract\n\u2022 Lambda (\/\u02c8l\u00e6mbd\u0259\/; uppercase \u039b, lowercase \u03bb; Greek: \u03bb\u03ac\u03bc(\u03b2)\u03b4\u03b1, l\u00e1m(b)da) is the 11th letter of the Greek alphabet, representing the sound \/l\/. In the system of Greek numerals lambda has a value of 30. Lambda is derived from the Phoenician Lamed . Lambda gave rise to the Latin L and the Cyrillic El (\u041b). The ancient grammarians and dramatists give evidence to the pronunciation as [la\u02d0bda\u02d0] (\u03bb\u03ac\u03b2\u03b4\u03b1) in Classical Greek times. In Modern Greek the name of the letter, \u039b\u03ac\u03bc\u03b4\u03b1, is pronounced [\u02c8lam.\u00f0a]. In early Greek alphabets, the shape and orientation of lambda varied. Most variants consisted of two straight strokes, one longer than the other, connected at their ends. The angle might be in the upper-left, lower-left (\"Western\" alphabets) or top (\"Eastern\" alphabets). Other variants had a vertical line with a horizontal or sloped stroke running to the right. With the general adoption of the Ionic alphabet, Greek settled on an angle at the top; the Romans put the angle at the lower-left. The HTML 4 character entity references for the Greek capital and small letter lambda are Λ and λ respectively. The Unicode code points for lambda are U+039B and U+03BB. (en)\ndbo:thumbnail\ndbo:wikiPageID\n\u2022 17851 (xsd:integer)\ndbo:wikiPageLength\n\u2022 15542 (xsd:nonNegativeInteger)\ndbo:wikiPageRevisionID\n\u2022 1019291765 (xsd:integer)\ndbp:map\ndbp:map2char\n\u2022 A2 (en)\n\u2022 8.0\ndbp:map3char\n\u2022 B6 (en)\n\u2022 E5 (en)\ndbp:map4char\n\u2022 CB (en)\n\u2022 EB (en)\ndbp:name\n\u2022 Coptic Capital Letter Laula (en)\n\u2022 Coptic Small Letter Laula (en)\n\u2022 Greek Capital Letter Lamda (en)\n\u2022 Greek Letter Small Capital Lamda (en)\n\u2022 Greek Small Letter Lamda (en)\ndbp:namedref\ndbp:ref1char\n\u2022 \\lambda (en)\n\u2022 \\Lambda (en)\ndbp:wikiPageUsesTemplate\ndct:subject\ngold:hypernym\nrdf:type\nrdfs:comment\n\u2022 Lambda (\/\u02c8l\u00e6mbd\u0259\/; uppercase \u039b, lowercase \u03bb; Greek: \u03bb\u03ac\u03bc(\u03b2)\u03b4\u03b1, l\u00e1m(b)da) is the 11th letter of the Greek alphabet, representing the sound \/l\/. In the system of Greek numerals lambda has a value of 30. Lambda is derived from the Phoenician Lamed . Lambda gave rise to the Latin L and the Cyrillic El (\u041b). The ancient grammarians and dramatists give evidence to the pronunciation as [la\u02d0bda\u02d0] (\u03bb\u03ac\u03b2\u03b4\u03b1) in Classical Greek times. In Modern Greek the name of the letter, \u039b\u03ac\u03bc\u03b4\u03b1, is pronounced [\u02c8lam.\u00f0a]. (en)\nrdfs:label\n\u2022 Lambda (en)\nowl:differentFrom\nowl:sameAs\nprov:wasDerivedFrom\nfoaf:depiction\nfoaf:isPrimaryTopicOf\nis dbo:award of\nis dbo:knownFor of\nis dbo:wikiPageDisambiguates of\nis dbo:wikiPageRedirects of","date":"2021-10-16 21:31:41","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\": 0.8811057806015015, \"perplexity\": 13335.965735386339}, \"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-2021-43\/segments\/1634323585025.23\/warc\/CC-MAIN-20211016200444-20211016230444-00455.warc.gz\"}"} | null | null |
Q: Do something after a button is clicked in a ListView I've tried looking at various various various examples of trying to explain this (what I thought would be) simple concept.
I have a ListView that uses a BaseAdapter for displaying rows of information. I have a button called btnEdit (R.id.Fragment_Edit_Member_RoleButton) that I need to be able to do work after the button is clicked.
I already have this....
@Override
public View getView(int position, View convertView, ViewGroup parent) {
Member m = memberItems.get(position);
LayoutInflater mInflater = (LayoutInflater) context
.getSystemService(Activity.LAYOUT_INFLATER_SERVICE);
convertView = mInflater.inflate(R.layout.fragment_forum_members_item, null);
TextView lblMemberText = (TextView) convertView.findViewById(R.id.memberText);
TextView lblMemberRole = (TextView) convertView.findViewById(R.id.memberRole);
lblMemberRole.setText(m.getRole());
lblMemberText.setText(m.getUsername());
Button btnEdit = (Button) convertView.findViewById(R.id.Fragment_Edit_Member_RoleButton);
btnEdit.setTag(position);
btnEdit.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View v) {
int wantedPosition = (Integer) v.getTag();
});
Now I thought that the wantedPosition would work for me, but apparently I'm suppose to do some math based off the actual visual position or something? This is where i'm getting lost.
I tried referencing to: Android: Access child views from a ListView but this page isn't making any sense to me. He is pointing to int firstPosition = listView.getFirstVisiblePosition() - listView.getHeaderViewsCount(); but I don't have the foggiest clue on how to reference my actual listView from within the BaseAdapter? it doesn't even make sense to me to even do this in the first place since the Adapter is being called from some other fragement that we would have no clue which fragment called it.
I really need to be able to pull the memberText and the memberRole out of the selected row that the button is located in. This is my goal.
Help is greatly appreciated
A: I guess i was overthinking it. I just needed to do...
View view = (View) v.getParent();
TextView lblMemberText = (TextView) view.findViewById(R.id.memberText);
TextView lblMemberRole = (TextView) view.findViewById(R.id.memberRole);
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 2,213 |
<?php
namespace Forumdash\Entity;
use Doctrine\ORM\Mapping as ORM;
/**
* Topicsmagicmaman
*
* @ORM\Table(name="topicsmagicmaman")
* @ORM\Entity
*/
class Topicsmagicmaman
{
/**
* @var integer
*
* @ORM\Column(name="Id", type="integer", nullable=false)
* @ORM\Id
* @ORM\GeneratedValue(strategy="IDENTITY")
*/
private $id;
/**
* @var string
*
* @ORM\Column(name="Link", type="string", length=255, nullable=true)
*/
private $link;
/**
* @var string
*
* @ORM\Column(name="Topic", type="text", nullable=true)
*/
private $topic;
/**
* @var \DateTime
*
* @ORM\Column(name="PostingTime", type="datetime", nullable=true)
*/
private $postingtime;
/**
* @var boolean
*
* @ORM\Column(name="PostedStatus", type="boolean", nullable=false)
*/
private $postedstatus;
/**
* @var string
*
* @ORM\Column(name="ProductKeyword", type="string", length=100, nullable=true)
*/
private $productkeyword;
/**
* @var string
*
* @ORM\Column(name="LongURL1", type="string", length=255, nullable=true)
*/
private $longurl1;
/**
* @var string
*
* @ORM\Column(name="LongURL2", type="string", length=255, nullable=true)
*/
private $longurl2;
/**
* @var string
*
* @ORM\Column(name="ShortURL1", type="string", length=100, nullable=true)
*/
private $shorturl1;
/**
* @var string
*
* @ORM\Column(name="ShortURL2", type="string", length=100, nullable=true)
*/
private $shorturl2;
/**
* @var string
*
* @ORM\Column(name="Response", type="text", nullable=true)
*/
private $response;
/**
* @var string
*
* @ORM\Column(name="CustResponse", type="text", nullable=true)
*/
private $custresponse;
/**
* @var integer
*
* @ORM\Column(name="AffprogramId1", type="integer", nullable=false)
*/
private $affprogramid1;
/**
* @var integer
*
* @ORM\Column(name="AffprogramId2", type="integer", nullable=false)
*/
private $affprogramid2;
/**
* Get id
*
* @return integer
*/
public function getId()
{
return $this->id;
}
/**
* Set link
*
* @param string $link
* @return Topicsmagicmaman
*/
public function setLink($link)
{
$this->link = $link;
return $this;
}
/**
* Get link
*
* @return string
*/
public function getLink()
{
return $this->link;
}
/**
* Set topic
*
* @param string $topic
* @return Topicsmagicmaman
*/
public function setTopic($topic)
{
$this->topic = $topic;
return $this;
}
/**
* Get topic
*
* @return string
*/
public function getTopic()
{
return $this->topic;
}
/**
* Set postingtime
*
* @param \DateTime $postingtime
* @return Topicsmagicmaman
*/
public function setPostingtime($postingtime)
{
$this->postingtime = $postingtime;
return $this;
}
/**
* Get postingtime
*
* @return \DateTime
*/
public function getPostingtime()
{
return $this->postingtime;
}
/**
* Set postedstatus
*
* @param boolean $postedstatus
* @return Topicsmagicmaman
*/
public function setPostedstatus($postedstatus)
{
$this->postedstatus = $postedstatus;
return $this;
}
/**
* Get postedstatus
*
* @return boolean
*/
public function getPostedstatus()
{
return $this->postedstatus;
}
/**
* Set productkeyword
*
* @param string $productkeyword
* @return Topicsmagicmaman
*/
public function setProductkeyword($productkeyword)
{
$this->productkeyword = $productkeyword;
return $this;
}
/**
* Get productkeyword
*
* @return string
*/
public function getProductkeyword()
{
return $this->productkeyword;
}
/**
* Set longurl1
*
* @param string $longurl1
* @return Topicsmagicmaman
*/
public function setLongurl1($longurl1)
{
$this->longurl1 = $longurl1;
return $this;
}
/**
* Get longurl1
*
* @return string
*/
public function getLongurl1()
{
return $this->longurl1;
}
/**
* Set longurl2
*
* @param string $longurl2
* @return Topicsmagicmaman
*/
public function setLongurl2($longurl2)
{
$this->longurl2 = $longurl2;
return $this;
}
/**
* Get longurl2
*
* @return string
*/
public function getLongurl2()
{
return $this->longurl2;
}
/**
* Set shorturl1
*
* @param string $shorturl1
* @return Topicsmagicmaman
*/
public function setShorturl1($shorturl1)
{
$this->shorturl1 = $shorturl1;
return $this;
}
/**
* Get shorturl1
*
* @return string
*/
public function getShorturl1()
{
return $this->shorturl1;
}
/**
* Set shorturl2
*
* @param string $shorturl2
* @return Topicsmagicmaman
*/
public function setShorturl2($shorturl2)
{
$this->shorturl2 = $shorturl2;
return $this;
}
/**
* Get shorturl2
*
* @return string
*/
public function getShorturl2()
{
return $this->shorturl2;
}
/**
* Set response
*
* @param string $response
* @return Topicsmagicmaman
*/
public function setResponse($response)
{
$this->response = $response;
return $this;
}
/**
* Get response
*
* @return string
*/
public function getResponse()
{
return $this->response;
}
/**
* Set custresponse
*
* @param string $custresponse
* @return Topicsmagicmaman
*/
public function setCustresponse($custresponse)
{
$this->custresponse = $custresponse;
return $this;
}
/**
* Get custresponse
*
* @return string
*/
public function getCustresponse()
{
return $this->custresponse;
}
/**
* Set affprogramid1
*
* @param integer $affprogramid1
* @return Topicsmagicmaman
*/
public function setAffprogramid1($affprogramid1)
{
$this->affprogramid1 = $affprogramid1;
return $this;
}
/**
* Get affprogramid1
*
* @return integer
*/
public function getAffprogramid1()
{
return $this->affprogramid1;
}
/**
* Set affprogramid2
*
* @param integer $affprogramid2
* @return Topicsmagicmaman
*/
public function setAffprogramid2($affprogramid2)
{
$this->affprogramid2 = $affprogramid2;
return $this;
}
/**
* Get affprogramid2
*
* @return integer
*/
public function getAffprogramid2()
{
return $this->affprogramid2;
}
} | {
"redpajama_set_name": "RedPajamaGithub"
} | 9,889 |
You are at:Home»Dr Vivek Baliga»What Are Artificial Pacemakers?
Pacemaker – have you heard of this device before? Here I will discuss what it is, how it is implanted and what precautions you need to take if you have one.
The human heart beats at a rate of 60 to 100 beats per minute at a regular pace. At times, the heart can slow down to a rate that is lower than the normal. When this happens, patients may experience symptoms such as dizziness, breathlessness or lightheadedness. Some patients may even collapse.
Patients in whom the heart rate is too slow or irregular often require pacemakers. Here we will take a look at what pacemakers are.
A pacemaker (sometimes called an artificial pacemaker) is a small device that measures around 1.5 inches in length (see picture above) that sits in a small pocket underneath the skin near the heart. It is a metallic device to which is attached one or two wireless called leads. These leads are placed in the chambers of the heart and the pacemaker controls the rate and rhythm of the heartbeat.
Pulse generator – This is a small oval metallic device that generates electrical impulses that are transmitted to the heart.
Leads or electrodes – These are flexible insulated wires that are attached to the pulse generator and are inserted into the heart chambers. Up to 3 wires may be attached to the pulse generator. Electricity from the pulse generator is transmitted through these wires into the heart and this in turn controls the rate at which the heartbeats.
Pacemakers are inserted when the heartbeat is either too slow or is irregular. The use of the pacemaker is to maintain a normal and a regular heart rhythm.
Before we discuss why pacemakers are inserted, it may be worthwhile reviewing how electricity is conducted through the heart.
Advanced heart failure accompanied by life-threatening irregular heart rhythms such as ventricular tachycardia (here the pacemaker is often a part of a treatment system called an implantable cardioverter defibrillator or ICD).
Here, a single wire i.e. lead is attached to the pacemaker pulse generator and is inserted into the heart. The wire is usually placed in the right ventricle.
Here, two leads that are attached to the pacemaker pulse generator are inserted into the heart. The first lead is inserted into the right ventricle while the second is often inserted into the right atrium.
This is a specialised pacemaker that is used in the treatment of patients who have advanced heart failure and in whom the normal electrical conduction is disrupted.
The pacemaker leads are inserted into both ventricles and the contraction of both the chambers is synchronized with the help of the pacemaker. For this reason, this treatment is sometimes called cardiac resynchronisation therapy or CRT.
This treatment can help improve the function of the heart and the way the heart contracts.
Pacemaker insertion is often performed under local anaesthesia. Depending on whether the individual is right handed or left-handed, the device is often placed in a pocket on the non-dominant side. What this means is that if an individual is right-handed, the pacemaker will be inserted on the left-hand side and vice versa.
Pacemaker insertion is performed with the individual awake and under local anaesthesia. If required, patients may request a small amount of sedation if they feel nervous.
Method of insertion of a pacemaker. The blue tube is the subclavian vein that connects directly to the heart.
The site where an incision is to be made is cleaned with antiseptic solution and the patient is covered in sterile drapes. local anaesthetic is injected into the skin and subcutaneous tissues.
A small incision of about 1 to 1.5 inches is made just under the collar bone near the shoulder joint. A pocket is created by separating out the tissues and a vein called the subclavian vein is located using a needle that is attached to a syringe filled with saltwater (normal saline).
Once this vein is located, a wire is passed into the vein and along this the required number of pacemaker leads are passed.
The procedure is performed under x-ray guidance so the doctor can visualise the heart and the leads at all times.
Once the cardiologist is satisfied that the leads are in place, the leads are sutured to the nearby tissues and the ends of the leads are attached to the pulse generator box. This box is then placed in a small pocket underneath the skin and tissues which are sutured close.
Pacemaker implantation is performed as a day case procedure. Patients are admitted on the day of the procedure and are discharged the following day.
A chest x-ray may be performed the day after the procedure to confirm the position of the leads and to ensure that the pacemaker has not moved overnight.
Prior to discharge, a pacemaker check is conducted to check the battery, the position of the leads and the amount of electricity being passed through the leads. These pacemaker checks are performed from time to time for a number of years.
For a few days after insertion of the pacemaker, patients are advised not to lift anything heavy and to avoid strenuous exercise. The wound needs to heal adequately and a follow-up appointment may be arranged to ensure this has happened.
A small amount of bleeding or bruising at the site is common. Patients may experience some pain and this can be easily treated with painkillers.
As mentioned above, some individuals may notice a small amount of bleeding and bruising at the site of the pacemaker implant. This often passes after a few days.
Any pain can be easily treated with over-the-counter medication such as Paracetamol (Dolo, Crocin) or Non-Steroidal Anti-Inflammatory Drugs ( Ibuprofen).
Rarely, an infection may occur at the site of the surgical incision. If this happens, it is essential that the patient to see the doctor for a complete assessment.
Minor infections can be treated with antibiotics where as deeper infections that have extended into the pocket may require surgical drainage.
Another rare complication that may occur during pacemaker insertion is accidental puncture of the lung.
As a needle is inserted into this subclavian vein (that lies under the collar bone and close to the lung), it is sometimes possible for the tip of the needle to damage the lining of the lung.
When this happens, a small amount of air may enter the space around the lung resulting in a condition called a pneumothorax. This pneumothorax can be easily detected on a chest x-ray.
If the pneumothorax is small, no treatment is required and the air gets absorbed by itself. However, if the pneumothorax is large, patients may require the insertion of a chest drain to remove the air through a tube.
Finally, another rare complication that may occur is puncture of the heart muscle during placement of the pacemaker leads. This usually requires emergency management.
It is uncommon for any of these complications to be life-threatening.
Sometimes,when going through security checks at the airport, it is possible that the pacemaker triggers the alarm system.
In order to avoid any confusion and apprehension, always carry some form of an identification card that describes the type of device you have implanted under the skin.
This would have been provided to you when being discharged from hospital after pacemaker implantation. When the security personnel are using a handheld metal detector as a part of their search, request them not to hover the device over your pacemaker for too long.
If you are of the habit of placing your cellphone in your shirt pocket, make sure you do not do so after you had a pacemaker implanted.
While it is uncommon for cellphones to interfere with pacemaker signals, it is possible for it to happen and the circumstance is best avoided completely.
Cellphone signals can sometimes interfere with how the pacemaker works and may withhold the generation of electrical signals within the pacemaker. This can in turn reduce the heart rate and may lead to the development of symptoms.
Prevent any problems by keeping your cell phone at least 6 inches away from your pacemaker.
Sometimes, doctors may arrange certain investigations such as an MRI scan or shock treatment which can interfere with the function of pacemakers. These days, MRI safe pacemakers are available but are not implanted for most patients due to the higher cost.
High-voltage equipment such as transformers and generators can sometimes affect the functioning of your pacemaker. It is recommended that you stand at least 2 to 3 feet away from these.
The headphones used for MP3 players contain a magnetic material within them that can interfere with the functioning of pacemakers. It is recommended that they be placed at least 3 cm away the device. Wearing headphones to listen to music will not affect the pacemaker.
A short while after pacemaker implantation, patients can restart their normal exercise and physical activity. However, it is recommended they refrain from participating in contact sports as a direct hit to the chest can dislodge the pacemaker.
Once your pacemaker has been implanted, you are likely to require annual checks to ensure that it is working correctly and that the battery life is good. As the battery approaches its end of life, the hospital personnel will recommend a change of the pulse generator. The leads do not need to be changed. There are rare cases when both the pulse generator and the leads need to be changed.
Below are some questions that patients ask us after they have pacemakers implanted.
The battery life of most pacemakers ranges between 5 to 10 years. This is constantly monitored by a hospital personnel.
Can I participate in sports and similar activities?
If you are an active individual who participate in sporting activities, you may be reassured to know that pacemakers do not interfere with your daily activities. If you wish to participate in athletic activities or aerobic activity such as running, swimming and cycling (after getting the all clear from your doctor), then you can do so.
Once the pacemaker wound has healed and the initial checks have confirmed the pacemaker to be working properly, there should be no problem re-commencing normal sexual activity.
If an individual with a pacemaker is terminally ill or has died, the cells within the heart muscle can no longer conduct the electricity that is generated by the pacemaker. Consequently, a working pacemaker will not generate heart contractions. In such a situation, the it will be switched off or deactivated. Contrary to popular belief, pacemakers do not prolong life if a patient is terminally ill.
There may be rare cases when patients are so ill that having a working pacemaker may be prolonging their suffering. Doctors may discuss turning off the the device with family members and the patient if such a situation arises.
Previous ArticleDo I Have Heart Blockages? Here Is How You Can Find Out. | {
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Polska Partia Przyjaciół Piwa (PPPP) – polska partia polityczna, założona w 1990 przez satyryka Janusza Rewińskiego. Sądownie została zarejestrowana 28 grudnia 1990. Prowadziła działalność do 1993.
Historia
Geneza partii
Pomysł utworzenia ugrupowania powstał w kręgu osób realizujących program telewizyjny z końca lat 80. Skauci piwni. Brali w nim udział m.in. Janusz Rewiński, Krzysztof Piasecki, Bohdan Smoleń i Stanisław Zygmunt. Na fali popularności programu powstało stowarzyszenie Skauci Piwni, które przeobraziło się w partię polityczną, wzorowaną w swoim charakterze na Pomarańczowej Alternatywie i innych ruchach quasi-politycznych z tamtego okresu.
Sukces polityczny PPPP
Pomysłodawcą przekształcenia stowarzyszenia w partię była redakcja nieistniejącego już miesięcznika dla mężczyzn "Pan". Początkowo powstanie partii było traktowane przez jej twórców jako żart, ale umożliwiło dostanie się do Sejmu osobom zajmującym się na poważnie biznesem i polityką. Oficjalnie celem PPPP było promowanie kulturalnego stylu picia piwa w pubach (stylizowanych na angielskie), zamiast wódki, a tym samym zwalczanie alkoholizmu.
Fragment programu wyborczego PPPP:
Władzą naczelną partii były walne zgromadzenie i ława najwyższa. W skład ławy wchodził prezydent partii oraz 10–15 członków. Działały również okręgowe ławy piwne. Pierwszym prezydentem PPPP został Janusz Rewiński, a wiceprezydentem Adam Halber.
Niecodzienna nazwa i rozczarowanie transformacją ustrojową w Polsce sprawiły, że w wyborach parlamentarnych w 1991 partia uzyskała względnie wysokie poparcie, tj. 3,27% głosów (dziesiąty wynik na 111 komitetów wyborczych), co pozwoliło PPPP na wprowadzenie do Sejmu 16 posłów.
Rozpad partii
Partia wkrótce podzieliła się na dwie frakcje – Małe Piwo i Duże Piwo, mimo oświadczeń Rewińskiego, że piwo nie jest ani ciemne, ani jasne, jest po prostu smaczne.
tzw. Małe Piwo skupiło 3 posłów (Adam Halber, Krzysztof Ibisz i Adam Piechowicz), którzy działali w Sejmie przez kilka miesięcy jako Koło Poselskie Spolegliwość, a następnie (po odejściu Adama Halbera do SLD) jako Koło Poselskie Partii Emerytów i Rencistów "Nadzieja".
tzw. Duże Piwo obejmowało 13 posłów (Tomasz Bańkowski, Tomasz Brach, Leszek Bubel, Sławomir Chabiński, Andrzej Czernecki, Jerzy Dziewulski, Zbigniew Eysmont, Tomasz Holc, Marek Kłoczko, Janusz Rewiński, Cezary Urbaniak, Jan Zylber i Andrzej Zakrzewski) i funkcjonowało w Sejmie pod nazwą Polski Program Gospodarczy. Kilka miesięcy po utworzeniu rządu Hanny Suchockiej posłowie PPG razem z posłami KLD utworzyli 6 listopada 1992 wspólny klub poselski pod nazwą Polski Program Liberalny, do którego nie wstąpili Andrzej Czernecki i Leszek Bubel.
Od 1992 prezydentem PPPP był Leszek Bubel, pod wodzą którego w wyborach parlamentarnych w 1993 partia uzyskała jedynie 0,1% głosów, w efekcie czego ugrupowanie zaprzestało jakiejkolwiek działalności (nie zostało ponownie zarejestrowane zgodnie z wymogami nowej ustawy z 1997). Głównym powodem tak słabego rezultatu wyborczego był fakt, że posłowie PPPP kandydowali do Sejmu z innych list wyborczych, głównie KLD. Jako członkowie PPPP z listy Samoobrony kandydowali m.in. ludzie sportu: Kazimierz Górski, Andrzej Supron czy Władysław Komar.
W 2007 działacze związani z Leszkiem Bublem rozpoczęli próby ponownej rejestracji PPPP, jednak bez powodzenia.
Zobacz też
Kompozycja marketingowa
Przypisy
Linki zewnętrzne
Nieistniejące partie i ugrupowania III Rzeczypospolitej
Organizacje piwne | {
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Q: Join data from more tables to single ADF table I am creating ADF application in JDeveloper 12c with Oracle 11g XE database.
I need join more tables to one ADF table. It is possible? I created this model in Toad Data Modeler:
Model is in Czech language, but it doesn't matter. This is a simple film database. Important is the concept of the model.
I want to have an ADF table that will contain data from multiple tables.
Unfortunately I did not come to that.
I beg you for advice.
A: You don't say whether you are using ADF BC to not. Assuming you are, I would start here.
Specifically;
Create new VO based on EO as the Base, up-dateable EO. In the next page of the Wizard, select the other tables as the reference EOs.
You will want to have created Associations that represent FKs to make this work.
This may help too.
Or, create a view in MySql and base the View Object on that.
| {
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class Error(Exception):
"""Base class for exceptions in this module."""
pass
class KeywordArgument(Error):
pass
class AuthError(Error):
pass
class MultipleRecordsFound(Error):
pass
class Exists(Error):
pass
| {
"redpajama_set_name": "RedPajamaGithub"
} | 7,798 |
Karl-Heinz Bauer (* 19. August 1958 in Neuötting) ist ein deutscher Mediziner. Er ist seit dem 1. Oktober 2003 Chefarzt der Klinik für Chirurgie am Knappschaftskrankenhaus Dortmund im Klinikum Westfalen. 2006 wurde Bauer zum Ärztlichen Direktor des Hauses berufen. Er ist seit 1999 Vereinsarzt für den VfL Bochum und unterrichtet an der Technischen Universität Dortmund.
Werdegang
Nach dem Abitur war Bauer zunächst als Medizinisch-technischer Assistent (MTA) (Labor) tätig, bevor er das Medizinstudium an der Ruhr-Universität Bochum aufnahm, das er 1986 abschloss. Mit einer Doktorarbeit über die Kolostomie wurde er 1987 zum Dr. med. promoviert. Ebenfalls zu einem Thema aus dem Bereich des Stoffwechsels habilitierte er sich 2001, indem er die Beeinflussbarkeit der operationsbedingten Immunsuppression durch präoperative enterale Immunnutrition bei elektiven Oberbaucheingriffen analysierte. Er ist Facharzt für Chirurgie und führt die Schwerpunktbezeichnungen Unfallchirurgie und Viszeralchirurgie sowie die Zusatzbezeichnungen Sportmedizin und Spezielle Unfallchirurgie.
In der Chirurgischen Klinik der Universitätsklinik St. Josef-Hospital war Bauer ab 1992 zunächst als Funktionsoberarzt tätig. In seiner Zeit als Oberarzt behandelte er Patienten in der Unfallchirurgischen Klinik – Unfallklinik Dortmund und im Universitätsklinikum St. Josef-Hospital, Bochum, wo er 1998 Leitender Oberarzt wurde. Erstmals Chefarzt wurde Bauer im Jahr 2000 in der Giradetklinik, Essen. Seit 2003 ist er Chefarzt der Chirurgischen Klinik am Knappschaftskrankenhaus Dortmund. Dort wurde er auf Vorschlag seiner chefärztlichen Kollegen vom Krankenhausträger zum Ärztlichen Direktor ernannt. 2016 verlieh ihm die Technische Universität Dortmund die Professorenwürde. Im selben Jahr wurde Bauer zum Präsidenten der Vereinigung Niederrheinisch-westfälischer Chirurgen gewählt.
Wirken
Bauer baute die vier Chirurgischen Abteilungen des Knappschaftskrankenhauses Dortmund auf:
Allgemein- und Viszeralchirurgie
Unfallchirurgie
Sportmedizin und
Thoraxchirurgie
Er entwickelte sie fachlich, baulich, technisch und personell auf das Niveau von Universitätskliniken weiter. Dies fand seinen Niederschlag in der Anerkennung des Knappschaftskrankenhauses Dortmund als Akademisches Lehrkrankenhaus der Ruhruniversität Bochum. Zunächst entstand das Darmzentrum Dortmund, in dem die für die Behandlung von Darmtumoren erforderlichen Fachabteilungen fachübergreifend zusammenwirken.
Schwerpunkte
Intensivmedizin
Ernährung von Patienten mit bösartigen Erkrankungen
Hernien
Minimalinvasive Chirurgie
Mitgliedschaften
Deutsche Gesellschaft für Chirurgie
Vereinigung der Bayerischen Chirurgen
Vereinigung Niederrheinisch-Westfälischer Chirurgen
Deutsche Gesellschaft für Ernährungsmedizin
Deutsche Gesellschaft für Visceralchirurgie
Deutsche Kontinenz Gesellschaft
Berufsverband der Deutschen Chirurgen e.V.
Einzelnachweise
Mediziner (20. Jahrhundert)
Mediziner (21. Jahrhundert)
Chirurg
Geboren 1958
Deutscher
Mann
Absolvent der Ruhr-Universität Bochum
Ehrenprofessor
Person (Dortmund) | {
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{"url":"https:\/\/icml.cc\/Conferences\/2021\/ScheduleMultitrack?event=12847","text":"Timezone: \u00bb\n\nLearning Pareto-Optimal Policies in Low-Rank Cooperative Markov Games\nAbhimanyu Dubey \u00b7 Alex Sandy' Pentland\nWe study cooperative multi-agent reinforcement learning in episodic Markov games with $n$ agents. While the global state and action spaces typically grow exponentially in $n$ in this setting, in this paper, we introduce a novel framework that combines function approximation and a graphical dependence structure that restricts the decision space to $o(dn)$ for each agent, where $d$ is the ambient dimensionality of the problem. We present a multi-agent value iteration algorithm that, under mild assumptions, provably recovers the set of Pareto-optimal policies, with finite-sample guarantees on the incurred regret. Furthermore, we demonstrate that our algorithm is {\\em no-regret} even when there are only $\\cO(1)$ episodes with communication, providing a scalable and provably no-regret algorithm for multi-agent reinforcement learning with function approximation. Our work provides a tractable approach to multi-agent decision-making that is provably efficient and amenable to large-scale collaborative systems.\n\n#### Author Information\n\n##### Abhimanyu Dubey (Massachusetts Institute of Technology)\n\nI am a PhD student in the Human Dynamics group at MIT, advised by Professor Alex Pentland. My research interests are in robust and cooperative machine learning, including problems in multi-agent decision-making and transfer learning. Prior to this, I received a master's degree in Computer Science and bachelor's degree in Electrical Engineering at IIT Delhi, where I was advised by Professor Sumeet Agarwal. I've also spent time as a research intern at Facebook AI, and was a post-baccalaureate fellow at the Department of Economics at Harvard, under Professor Ed Glaeser. My research has been supported by a Snap Research Scholarship (2019) and an Emerging Worlds Fellowship (2017).","date":"2022-09-25 05:46:54","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\": 0.4387727975845337, \"perplexity\": 1089.112069812557}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2022-40\/segments\/1664030334514.38\/warc\/CC-MAIN-20220925035541-20220925065541-00721.warc.gz\"}"} | null | null |
La Drake University è un'università privata situata a Des Moines, Iowa, Stati Uniti d'America. Offre corsi undergraduate e graduate in economia, legge e farmacia. La facoltà di legge della Drake University è tra le 25 più antiche degli Stati Uniti.
Storia
Venne fondata nel 1881 da Francis Marion Drake, un generale dell'Unione durante la guerra civile americana, e George T. Carpenter, un insegnante e pastore protestante. I primi corsi iniziarono nel 1881, con 77 studenti ospitati provvisoriamente nella Student's Home. Nel 1883 venne completato l'Old Main, il primo edificio permanente.
L'Old Main rimane il più importante edificio dell'università; vi sono situati gli uffici amministrativi, la Levitt Hall e lo Sheslow Auditorium. L'Old Main ha ospitato molti dibattiti politici delle elezioni presidenziali americane.
Nel 2013 la Drake University è diventata sede del "The Harkin Institute for Public Policy & Citizen Engagement", intitolato al senatore Tom Harkin.
La Drake University è suddivisa in sette facoltà e scuole:
College of Arts and Sciences
College of Business and Public Administration
John Dee Bright College
School of Education
School of Journalism and Mass Communication
Law School
College of Pharmacy and Health Sciences
Note
Altri progetti
Collegamenti esterni
Drake University | {
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Q: Read and write an ArrayList to a .txt file I can't get this part of my project to work. (fileOut() and fileIn() methods). I would appreciate any help!
I am trying to construct a method to open a text file and to write the details of an ArrayList of type to it.
I am also trying to construct a second method to open the text file containing details of all the bank's accounts and uses the incoming data to create BankAccount objects using a BankAccount Constructor and storing each account in an ArrayList.
Please note that other methods in the class need to use the information stored in the arraylist(s).
Here is the main bit of code to focus on(reading and writing):
public void fileOut()
{
File fileName = new File("BankAccountFiles.txt");
try{
FileWriter fw = new FileWriter(fileName);
Writer output = new BufferedWriter(fw);
int numEntries = bankAccArrayList.size();
for (int i = 0; i < numEntries; i++) {
output.write(bankAccArrayList.get(i).toString() + "\n");
}
output.close();
}
catch(Exception e) {
JOptionPane.showMessageDialog(null,"File cannot be created");
}
}
public void fileIn()
{
ArrayList<BankAccount> aList = new ArrayList<BankAccount>();
String line;
try {
BufferedReader input = new BufferedReader(new FileReader("BankAccountFiles.txt"));
if(!input.ready()) {
throw new IOException();
}
while ((line = input.readLine()) != null) {
aList.add(line);
}
input.close();
} catch (IOException e) {
JOptionPane.showMessageDialog(null,e);
}
}
Here is the code for the whole class:
import javax.swing.JOptionPane;
import java.util.ArrayList;
import java.io.*;
import java.lang.*;
import java.util.*;
import javax.swing.*;
public class MyBankController
{
private ArrayList<BankAccount> bankAccArrayList;
public MyBankController()
{
bankAccArrayList = new ArrayList<BankAccount>();
}
public void createAccount (String accNum, String custName)
{
bankAccArrayList.add(new BankAccount( accNum,custName));
printAccountDetails(bankAccArrayList.get(bankAccArrayList.size()-1));
}
private void printAccountDetails(BankAccount incomingAcc)
{
JOptionPane.showMessageDialog(null,incomingAcc.toString(),"Account Details",JOptionPane.PLAIN_MESSAGE);
}
public void listAllAccounts()
{
String outputStr = "List of all accounts :\n\n";
for(BankAccount account : bankAccArrayList){
outputStr += account + "\n\n";
}
JOptionPane.showMessageDialog(null,outputStr,"List of all Accounts",JOptionPane.PLAIN_MESSAGE);
}
public void listAllActiveAccounts()
{
String outputStr = "List of all active accounts :\n\n";
for(BankAccount account : bankAccArrayList){
if(account.getActive()) {
outputStr += account + "\n\n";
}
}
JOptionPane.showMessageDialog(null,outputStr,"List of all Active Accounts",JOptionPane.PLAIN_MESSAGE);
}
private int getIndex(String bankAccNum)
{
for (int i = 0; i < bankAccArrayList.size(); i++) {
if (bankAccNum.equals(bankAccArrayList.get(i).getAccNumber()))
{
return i;
}
}
JOptionPane.showMessageDialog(null,"Bank Account Number: " + bankAccNum + " is invalid","Error wrong account number",JOptionPane.ERROR_MESSAGE);
return -1;
}
public void makeWithdrawal(String accNumber, double amount)
{
if(getIndex(accNumber) != -1){
bankAccArrayList.get(getIndex(accNumber)).makeWithdraw(amount);
}
}
public void makeLodgement(String accNumber, double amount)
{
if(getIndex(accNumber) != -1){
bankAccArrayList.get(getIndex(accNumber)).makeLodgement(amount);
}
}
public void displayAccount(String accNumber)
{
if(getIndex(accNumber) != -1){
JOptionPane.showMessageDialog(null,(bankAccArrayList.get(getIndex(accNumber)).toString()),"Accounts Details",JOptionPane.PLAIN_MESSAGE);
}
}
public void closeAccount(String accNumber)
{
if(getIndex(accNumber) != -1){
bankAccArrayList.get(getIndex(accNumber)).closeAccount();
}
}
public void removeAccount(String accNumber)
{
int index = getIndex(accNumber);
if(index != -1){
BankAccount myAcc = bankAccArrayList.get(index);
if ((myAcc.getActive() == false) && (myAcc.getAccBalance() == 0)){
bankAccArrayList.remove(index);
}else if((myAcc.getActive() == false) && (myAcc.getAccBalance() > 0)){
int dialogResult = JOptionPane.showConfirmDialog(null," This account has a balance,do you wish to withdraw this balance " +
"so as to remove the account ?"," Balance in inactive account",JOptionPane.PLAIN_MESSAGE);
if(dialogResult == JOptionPane.YES_OPTION){
myAcc.setActive();
myAcc.makeWithdraw(myAcc.getAccBalance());
myAcc.setActive();
bankAccArrayList.remove(index);
JOptionPane.showMessageDialog(null,"Account Removed","Confirmation",JOptionPane.PLAIN_MESSAGE);
}
} else if((myAcc.getActive() == true) && (myAcc.getAccBalance() == 0)){
int dialogResult = JOptionPane.showConfirmDialog(null," This account still has an active status, do you wish to change its status so as to remove account ?"
,"Account Active",JOptionPane.PLAIN_MESSAGE);
if(dialogResult == JOptionPane.YES_OPTION){
myAcc.setActive();
bankAccArrayList.remove(index);
JOptionPane.showMessageDialog(null,"Account Removed","Confirmation",JOptionPane.PLAIN_MESSAGE);
}
} else{
int dialogResult = JOptionPane.showConfirmDialog(null," This account still has an active status and a balance, do you wish to close the account so as remove it"
,"Account Active with Balance",JOptionPane.PLAIN_MESSAGE);
if(dialogResult == JOptionPane.YES_OPTION){
myAcc.closeAccount();
bankAccArrayList.remove(index);
JOptionPane.showMessageDialog(null,"Account Removed","Confirmation",JOptionPane.PLAIN_MESSAGE);
}
}
}
}
public void customerInterface()
{
String accNumber = JOptionPane.showInputDialog(null,"Please enter your account number","Account Login",JOptionPane.PLAIN_MESSAGE);
if(getIndex(accNumber) != -1)
{
String numOption = JOptionPane.showInputDialog(null,"Please select an option below:\n\n" +
" [1] Make a lodgment:\n\n [2] Make a withdrawal:\n\n [3] Display account details:\n\n" +
" [4] Close account:\n\n [5] Exit","MyBank ATM",JOptionPane.PLAIN_MESSAGE);
if (numOption == null) //User presses cancel or 'x'.
{
JOptionPane.showMessageDialog(null,"Goodbye.","MyBank System",JOptionPane.INFORMATION_MESSAGE);
}
else if(Integer.parseInt(numOption) == 1) //Converts numOption from String to Integer.
{
//Brings up lodgement interface
String amount = JOptionPane.showInputDialog("Please enter the amount you would like to lodge.");
makeLodgement(accNumber, Double.parseDouble(amount)); // Lodges amount into account.
customerInterface();
}
else if(Integer.parseInt(numOption) == 2)
{
String amount = JOptionPane.showInputDialog("Please enter the amount you wish to withdraw.");
makeWithdrawal(accNumber, Double.parseDouble(amount)); //Call on makeWithdrawl method.
customerInterface();
}
else if(Integer.parseInt(numOption) == 3)
{
displayAccount(accNumber); //Calls on displayAccount method.
customerInterface();
}
else if(Integer.parseInt(numOption) == 4)
{
closeAccount(accNumber); //Call on close account method.
customerInterface();
}
else if(Integer.parseInt(numOption) == 5)
{
return; //Exits system.
}
else if(Integer.parseInt(numOption) > 5 || Integer.parseInt(numOption) < 1) //If number enter is outside of 1-5.
{
JOptionPane.showMessageDialog(null,"Please enter a number between 1-5 and try again.","Error",JOptionPane.ERROR_MESSAGE);
customerInterface();
}
}
}
public void bankInterface()
{
String numOption = JOptionPane.showInputDialog(null,"Please select an option below:\n\n" +
" [1] Display All Accounts:\n\n [2] Display All Active Accounts:\n\n [3] Open a New Account:\n\n" +
" [4] Close an Existing Account:\n\n [5] Run Start of Day:\n\n [6] Run End of Day:\n\n [7] Exit:","MyBank System",JOptionPane.PLAIN_MESSAGE);
if (numOption == null)
{
JOptionPane.showMessageDialog(null,"Goodbye.","MyBank System",JOptionPane.INFORMATION_MESSAGE); //User presses cancel or 'x'.
}
else if(Integer.parseInt(numOption) == 1) //Converts numOption from String to Integer.
{
//Displays all accounts.
listAllAccounts();
bankInterface();
}
else if(Integer.parseInt(numOption) == 2)
{
//Display all active accounts.
listAllActiveAccounts();
bankInterface();
}
else if(Integer.parseInt(numOption) == 3)
{
//Open a new account
String accNum = JOptionPane.showInputDialog("Please allocate an account number:");
String custName = JOptionPane.showInputDialog("Please enter the customers name:");
createAccount (accNum, custName);
bankInterface();
}
else if(Integer.parseInt(numOption) == 4)
{
//Close an existing account.
String accNumber = JOptionPane.showInputDialog("Please enter the account number you would like to close:");
closeAccount(accNumber);
bankInterface();
}
else if(Integer.parseInt(numOption) == 5)
{
//Run start of day file.
fileIn();
bankInterface();
}
else if(Integer.parseInt(numOption) == 6)
{
//Run end of day file.
fileOut();
bankInterface();
}
else if(Integer.parseInt(numOption) == 7)
{
//Exits system.
return;
}
else if(Integer.parseInt(numOption) > 7 ||Integer.parseInt(numOption) < 1) //If user enters number outside of 1-7.
{
JOptionPane.showMessageDialog(null,"Please enter a number between 1-7 and try again.","Error",JOptionPane.ERROR_MESSAGE);
bankInterface();
}
}
public void fileOut()
{
File fileName = new File("BankAccountFiles.txt");
try{
FileWriter fw = new FileWriter(fileName);
Writer output = new BufferedWriter(fw);
int numEntries = bankAccArrayList.size();
for (int i = 0; i < numEntries; i++) {
output.write(bankAccArrayList.get(i).toString() + "\n");
}
output.close();
}
catch(Exception e) {
JOptionPane.showMessageDialog(null,"File cannot be created");
}
}
public void fileIn()
{
ArrayList<BankAccount> aList = new ArrayList<BankAccount>();
String line;
try {
BufferedReader input = new BufferedReader(new FileReader("BankAccountFiles.txt"));
if(!input.ready()) {
throw new IOException();
}
while ((line = input.readLine()) != null) {
aList.add(line);
}
input.close();
} catch (IOException e) {
JOptionPane.showMessageDialog(null,e);
}
}
}
A: In your fileIn method, you're trying to add a String to an ArrayList of type BankAccount. Instead of aList.add(line), you should use aList.add(new BankAccount(line)).
Assuming your BankAccount constructor takes only a single String parameter, this should work.
Full method:
public void fileIn()
{
List<BankAccount> aList = new ArrayList<BankAccount>();
String line;
try {
BufferedReader input = new BufferedReader(new FileReader("BankAccountFiles.txt"));
if(!input.ready()) {
throw new IOException();
}
while ((line = input.readLine()) != null) {
aList.add(new BankAccount(line));
}
input.close();
} catch (IOException e) {
JOptionPane.showMessageDialog(null,e);
}
aList.forEach(System.out::println); // Java8
}
I've also changed your aList to use the List interface, and added a lambda to print each value in aList at the end.
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 1,512 |
\section*{ABSTRACT}
Array-RQMC has been proposed as a way to effectively apply randomized quasi-Monte Carlo (RQMC) when simulating a Markov chain over a large number of steps to estimate an expected cost or reward. The method can be very effective when the state of the chain has low dimension. For pricing an Asian option under an ordinary geometric Brownian motion model, for example, Array-RQMC reduces the variance by huge factors. In this paper, we show how to apply this method and we study its effectiveness in case the underlying process has stochastic volatility. We show that Array-RQMC can also work very well for these models, even if it requires RQMC points in larger dimension. We examine in particular the variance-gamma, Heston, and Ornstein-Uhlenbeck stochastic volatility models, and we provide numerical results.
\section{INTRODUCTION}
\label{sec:intro}
Quasi-Monte Carlo (QMC) and randomized QMC (RQMC) methods can improve efficiency significantly
when estimating an integral in a moderate number of dimensions, but their use for simulating
Markov chains over a large number of steps has been limited so far.
The array-RQMC method, developed for that purpose, has been show to work well for some
chains having a low-dimensional state.
It simulates an array of $n$ copies of the Markov chain so that each chain follows its exact distribution,
but the copies are not independent, and the empirical distribution of the states at any given step
of the chain is a ``low-discrepancy'' approximation of the exact distribution.
At each step, the $n$ chains (or states) are matched one-to-one to a set of $n$ RQMC points
whose dimension is the dimension of the state plus the number of uniform random numbers required
to advance the chain by one more step. The first coordinates of the points are used to match the
states to the points and the other coordinates provide the random numbers needed to determine the next state.
When the chains have a large-dimensional state, the dimension used for the match can be reduced via
a mapping to a lower-dimensional space. Then the matching is performed by sorting both the points and
the chains. When the dimension of the state exceeds 1, this matching is done via a multivariate sort.
The main idea is to evolve the array of chains in a way that from step to step, the empirical distribution
of the states keeps its low discrepancy.
For further details on the methodology, sorting strategies, convergence analysis, applications,
and empirical results, we refer the reader to
\shortciteN{vLEC04m,vDEM05a,vLEC06a,vLEC07b,vLEC08a,vELH08a,vLEC09d,vELH10a,vDIO10a,vLEC10c,tGER15a,vLEC18b},
and the other references given there.
The aim of this paper is to examine how Array-RQMC can be applied for option pricing under a
stochastic volatility process such as the variance gamma, Heston, and Ornstein-Uhlenbeck models.
We explain and compare various implementation alternatives, and report empirical experiments to
assess the (possible) gain in efficiency and convergence rate.
A second objective is for the WSC community to become better aware of this method,
which can have numerous other applications.
Array-RQMC has already been applied for pricing Asian options when the underlying process
evolves as a geometric Brownian motion (GBM) with fixed volatility \def\citeauthoryear##1##2##3{##2, ##3}\@internalcite{vLEC09d,vLEC18b}.
In that case, the state is two-dimensional (it contains the current value of the GBM and its running average)
and a single random number is needed at each step, so the required RQMC points are three-dimensional.
In their experiments, \shortciteN{vLEC18b} observed an empirical variance of the average payoff
that decreased approximately as $\cO(n^{-2})$ for Array-RQMC, in a range of reasonable values of $n$,
compared with $\cO(n^{-1})$ for independent random points (Monte Carlo).
For $n=2^{20}$ (about one million chains), the variance ratio between Monte Carlo
and Array-RQMC was around 2 to 4 millions.
In view of this spectacular success, one wonders how well the method would
perform when the underlying process is more involved, e.g., when it has stochastic volatility.
This is relevant because stochastic volatility models are more
realistic than the plain GBM model \def\citeauthoryear##1##2##3{##2, ##3}\@internalcite{fMAD90a,fMAD98a}.
Success is not guaranteed
because the dimension of the required RQMC points is larger.
For the Heston model, for example, the RQMC points must be five-dimensional instead of three-dimensional,
because the state has three dimensions and we need two uniform random numbers at each step.
It is unclear a priori if there will be any significant variance reduction for reasonable values of $n$.
The remainder is organized as follows.
In Section~\ref{sec:background}, we state our general Markov chain model and provide background
on the Array-RQMC algorithm, including matching and sorting strategies.
In Section \ref{sec:experimental}, we describe our experimental setting,
and the types of RQMC point sets that we consider.
Then we study the application of Array-RQMC under the variance-gamma model in Section~\ref{sec:vg},
the Heston model in Section~\ref{sec:heston}, and
the Ornstein-Uhlenbeck model in Section~\ref{sec:ou}.
We end with a conclusion.
\iffalse
\color{gray}
The Randomized quasi-Monte Carlo (RQMC) \citeN{sLEC09a} methods turn QMC into variance reduction methods by carefully randomizing the points set. They are hybrid techniques that combine
clever integration rules with some randomness in order to obtain unbiased estimators with a better convergence rate
for the variance than for MC. These methods work
well for estimating the integral of a smooth function in a small or moderate number of
dimensions. That is, when the cost (whose expectation is to be estimated) depends only
on a few random numbers in the simulation, and is a smooth function of these random
numbers. But when the function depends on the sample path of a discrete-event model
that evolves over a large number of time steps, which is often the case in applications,
the dimension of the integrand is very large and the RQMC methods are ineffective when
applied in a standard way.
Some examples of RQMC point sets are randomly shifted lattice rules, scrambled digital nets, digital nets with a random digital shift, and stratified sampling. For the stratified sampling, the unit cube is partitioned into $ n $ cubes of equal volume and one point is generated randomly and uniformly in each box.
One of the interesting applications is pricing complex financial derivatives, such as contracts whose payoff depends on several correlated assets or on the entire sample path of an asset price,
when there are no
known closed-form analytic solution for the price of the options.
The pricing of an Asian option combined with array-RQMC, to estimate the expectation of its payoff, was already studied by
\citeN{vLEC18b} and they have observed a $ O(n^{-2}) $ convergence rate for the variance.
In this paper, we extend this empirical analysis to price this option under a variance gamma (VG) process.
The VG process is a Brownian motion with
a random time change that follows a stationary
gamma process. In the first view, the model appears like having a three dimensional Markov chain state that depends on the value of the gamma but the difference of the game at two sequential time is follows a gamma distribution that can be written as an inversion of an uniform random over the unit cube, Thus we don't require to store the value of the gamma in the state and then we have only a two dimensional state instead of two.
Also we will examine the Heston volatility model which is one of the most popular stochastic volatility option pricing
models. It is motivated by the widespread evidence that volatility is stochastic
and that the distribution of risky asset returns has tail(s) longer than
that of a normal distribution. This model requires the storage of the value of the volatility at each step and thus it needs a three dimensional state. We will apply the array-RQMC method to this model in order to price Asian and European options, and we will compare the results obtained with the RQMC method.
and Finally we will examine the Ornstein volatility model in order to price European option.
\color{black}
\fi
\section{BACKGROUND: MARKOV CHAIN MODEL, RQMC, AND ARRAY-RQMC}
\label{sec:background}
The option pricing models considered in this paper fit the following framework,
which we use to summarize the Array-RQMC algorithm.
We have a discrete-time \emph{Markov chain} $\{X_j,\, j\ge 0\}$ defined by a stochastic recurrence over a measurable
state space $\cX$:
\begin{equation}
\label{eq:markov-chain}
X_{0} = x_{0}, \qquad\text{and}\qquad X_{j} = \varphi_{j} (X_{j-1},\bU_{j}), \quad j = 1,\dots,\tau.
\end{equation}
where $x_0\in\cX$ is a deterministic initial state,
$\bU_{1},\bU_{2},...$ are independent random vectors uniformly distributed over the
$d$-dimensional unit cube $(0,1)^{d}$,
the functions $\varphi_{j} : \cX \times(0,1)^{d}\rightarrow\cX$ are measurable,
and $\tau$ is a fixed positive integer (the time horizon).
The goal is:
\[
\mbox{Estimate} \quad \mu_{\rmy} = \EE[Y], \qquad\mbox{ where \ } Y = g(X_{\tau})
\]
and $g : \cX \to \RR$ is a \emph{cost} (or reward) function.
Here we have a cost only at the last step
but in general there can be a cost function for each step and $Y$ would be the sum of
these costs \def\citeauthoryear##1##2##3{##2, ##3}\@internalcite{vLEC08a}.
\emph{Crude Monte Carlo} estimates $\mu$ by the average
$
\bar Y_n = \frac{1}{n} \sum_{i=0}^{n-1} Y_{i},
$
where $Y_0,\dots,Y_{n-1}$ are $n$ independent realizations of $Y$.
One has $\EE[{\bar Y_n}] = \mu_{\rmy}$ and
$\Var[{\bar Y_n}] = \Var[Y]/n$, assuming that $\EE[Y^2] = \sigma^2_{\rmy} < \infty$.
Note that the simulation of each realization of $Y$ requires a vector
$\bV = (\bU_1,\dots,\bU_\tau)$ of $d\tau$ independent uniform random variables over $(0,1)$,
and crude Monte Carlo produces $n$ independent replicates of this random vector.
\emph{Randomized quasi-Monte Carlo} (RQMC) replaces the $n$ independent realizations
of $\bV$ by $n$ \emph{dependent} realizations, which form an RQMC point set in $d\tau$ dimensions.
That is, each $\bV_i$ has the uniform distribution over $[0,1)^{d\tau}$, and
the point set $P_{n}=\{V_{0},...,V_{n-1}\} $ covers $[0,1)^{d\tau}$ more evenly than typical
independent random points.
With RQMC, ${\bar Y_n}$ remains an unbiased estimator of $\mu$, but its variance can be much smaller,
and can converge faster than $\cO(1/n)$ under certain conditions.
For more details, see \shortciteN{rDIC10a,vLEC00b,vLEC09f,vLEC18a}, for example.
However, when $d\tau$ is large, standard RQMC typically becomes ineffective, in the sense that it
does not bring much variance reduction unless the problem has special structure.
\emph{Array-RQMC} is an alternative approach developed specifically for Markov chains
\def\citeauthoryear##1##2##3{##2, ##3}\@internalcite{vLEC06a,vLEC08a,vLEC18b}.
To explain how it works, let us first suppose for simplicity (we will relax it later) that there is a mapping
$h : \cX \to\RR$, that assigns to each state a \emph{value} (or score) which summarizes in a single
real number the most important information that we should retain from that state
(like the value function in stochastic dynamic programming).
This $h$ is called the \emph{sorting function}.
The algorithm simulates $n$ (dependent) realizations of the chain ``in parallel''.
Let $X_{i,j}$ denote the state of chain $i$ at step $j$, for $i=0,\dots,n-1$ and $j=0,\dots,\tau$.
At step $j$, the $n$ chains are sorted by increasing order of their values of $h(X_{i,j-1})$,
the $n$ points of an RQMC point set in $d+1$ dimensions are sorted by their first coordinate,
and each point is matched to the chain having the same position in this ordering.
Each chain $i$ is then moved forward by one step, from state $X_{i,j-1}$ to state $X_{i,j}$,
using the $d$ other coordinates of its assigned RQMC point.
Then we move on to the next step, the chains are sorted again, and so on.
The sorting function can in fact be more general and have the form $h : \cX \to\RR^c$
for some small integer $c \ge 1$. Then the mapping between the chains and the points must be
realized in a $c$-dimensional space, i.e., via some kind of $c$-dimensional multivariate sort.
The RQMC points then have $c+d$ coordinates, and are sorted with the same
$c$-dimensional multivariate sort based on their first $c$ coordinates, and mapped to the
corresponding chains. The other $d$ coordinates are used to move the chains ahead by one step.
In practice, the first $c$ coordinates of the RQMC points do not have to be randomized at each step;
they are usually fixed and the points are already sorted in the correct order based on these coordinates.
Some multivariate sorts are described and compared by \shortciteN{vELH08a,vLEC09d,vLEC18a}.
For example, in a \emph{multivariate batch sort}, we select positive integers
$n_1, \dots, n_c$ such that $n = n_1 \dots n_c$.
The states are first sorted by their first coordinate in ${n_1}$ packets of size ${n/n_1}$,
then each packet is sorted by the second coordinate into ${n_2}$ packets of size ${n/n_1 n_2}$, and so on.
The RQMC points are sorted in exactly the same way, based on their first $c$ coordinates.
In the \emph{multivariate split sort}, we assume that $n = 2^e$
and we take $n_1 = n_2 = \cdots = n_e = 2$.
That is, we first split the points in 2 packets based on the first coordinate,
then split each packet in two by the second coordinate, and so on.
If $e > c$, after $c$ splits we get back to the first coordinate and continue.
Examples of heuristic sorting functions $h : \cX\to\RR$ are given in \def\citeauthoryear##1##2##3{##2, ##3}\@internalcite{vLEC08a,vLEC18b}.
\citeN{vWAC08a} and \citeN{tGER15a} suggested to first map the $c$-dimensional states to $[0,1]^c$
and then use a space filling curve in $[0,1]^c$ to map them to $[0,1]$, which provides a total order.
\citeN{tGER15a} proposed to map the states to $[0,1]^c$ via a component-wise rescaled logistic transformation,
then order them with a Hilbert space-filling curve. See \shortciteN{vLEC18b} for a more detailed discussion.
Under smoothness conditions, they proved that the resulting unbiased Array-RQMC estimator has $o(1/n)$ variance,
which beats the $\cO(1/n)$ Monte Carlo rate.
Algorithm~\ref{algo:array-rqmc} states the Array-RQMC procedure in our setting.
Indentation delimits the scope of the \textbf{for} loops.
For any choice of sorting function $h$, the average $\hat\mu_{\arqmc,n} = \bar Y_n$
returned by this algorithm is always an \emph{unbiased} estimator of $\mu$.
An unbiased estimator of $\Var[\bar Y_n]$ can be obtained by making $m$ independent realizations
of $\hat\mu_{\arqmc,n}$ and computing their \emph{empirical variance}.
\begin{algorithm}[ht]
\caption{: Array-RQMC Algorithm for Our Setting}
\label{algo:array-rqmc}
\begin{algorithmic}
\State {\textbf{for} $i=0,\dots,n-1$ \textbf{do} ${X_{i,0}} \g x_0$;}
\For {${j} = 1, 2, \dots, \tau$}
\State Sorting: Compute an appropriate permutation $\pi_{j}$ of the $n$ chains, based on
\State $\qquad$ the $h(X_{i,j-1})$, to match the $n$ states with the RQMC points;
\State Randomize afresh
the RQMC points $\{\bU_{0,j},\dots,\bU_{n-1,j}\}$;
\State {\textbf{for} $i=0,\dots,n-1$ \textbf{do} ${X_{i,j}} = \varphi_j(X_{\pi_j(i),j-1}, \bU_{i,j})$;}
\EndFor
\Return the average $\hat\mu_{\arqmc,n} = \bar Y_n = (1/n) \sum_{i=0}^{n-1} g(X_{i,\tau})$
as an estimate of $\mu_{\rmy}$.
\end{algorithmic}
\end{algorithm}
\section{EXPERIMENTAL SETTING}
\label{sec:experimental}
For all the option pricing examples in this paper, we have an asset price that evolves
as a stochastic process $\{S(t),\, t\ge 0\}$ and a payoff that depends on the values of
this process at fixed observation times $0 = t_{0} <t_{1}<t_{2}< ... < t_{c}=T$.
More specifically, for given constants $r$ (the interest rate) and $K$ (the strike price),
we consider an \emph{European option} whose payoff is
\[
Y = Y_{\rme} = g(S(T)) = e^{-rT} \max (S(T)-K, 0)
\]
and a discretely-observed \emph{Asian option} whose payoff is
\[
Y = Y_{\rma} = g(\bar S) = e^{-rT} \max (\bar S-K, 0)
\]
where $\bar S = (1/c) \sum _{j=1}^{c} S(t_{j})$.
In this second case, the running average
$\bar S_j = (1/j)\sum_{\ell=1}^j S(t_{\ell})$ must be kept in the state of the Markov chain.
The information required for the evolution of $S(t)$ depends on the model and is given
for each model in forthcoming sections. It must be maintained in the state.
For the case where $S$ is a plain GBM, the state of the Markov chain
at step $j$ can be taken as $X_j = (S(t_j), \bar S_j)$, a two-dimensional state, as was done in
\shortciteN{vLEC09d} and \shortciteN{vLEC18b}.
In our examples, the states are always multidimensional.
To match them with the RQMC points, we will use a split sort, a batch sort, and a Hilbert-curve sort,
and compare these alternatives. The Hilbert sort requires a transformation of the $\ell$-dimensional states
to the unit hypercube $[0,1]^\ell$. For this, we use a \emph{logistic transformation} defined by
$\psi(x)=(\psi_{1}(x_{1}),...,\psi_{\ell}(x_{\ell})) \in[0,1]^\ell$ for all $x = (x_1,\dots,x_\ell) \in\cX$, where
\begin{equation}
\psi_{j}(x_{j})=
\left [1+\exp \left(-\frac{x_{j}-\underline{x}_{j}}{\bar{x}_{j}-\underline{x}_{j}} \right) \right]^{-1},
\quad j=1,...,\ell,
\end{equation}
with constants $\bar{x}_{j}=\mu_{j}+2 \sigma_{j} $ and $\underline{x}_{j}=\mu_{j}-2 \sigma_{j}$
in which $ \mu_{j} $ and $ \sigma_{j}$ are estimates of the mean and the variance of the distribution
of the $j$th coordinate of the state.
In Section~\ref{sec:vg}, we will also consider just taking a linear combination
of the two coordinates, to map a two-dimensional state to one dimension.
For RQMC, we consider
\vskip-25pt\nul
\begin{verse}
(1) Independent points, which corresponds to crude Monte Carlo (MC);\\
(2) Stratified sampling over the unit hypercube (Stratif);\\
(3) Sobol' points with a random linear matrix scrambling
and a digital random shift (Sobol+LMS);\\
(4) Sobol' points with nested uniform scrambling (Sobol+NUS);\\
(5) A rank-1 lattice rule with a random shift modulo 1 followed by a baker's transformation (Lattice+baker).
\end{verse}
\vskip-12pt
The first two are not really RQMC points, but we use them for comparison.
For stratified sampling, we divide the unit hypercube into $n = k^{\ell+d}$ congruent subcubes
for some integer $k > 1$, and we draw one point randomly in each subcube.
For a given target $n$, we take $k$ as the integer for which $k^{\ell+d}$ is closest to this target $n$.
For the Sobol' points, we took the default direction numbers in SSJ,
which are from \shortciteN{iLEM04a}.
The LMS and NUS randomizations are explained in \citeN{vOWE03a} and \citeN{vLEC09f}.
For the rank-1 lattice rules, we used generating vectors found by Lattice Builder \def\citeauthoryear##1##2##3{##2, ##3}\@internalcite{vLEC16a},
using the $\cP_2$ criterion with order-dependent weights $(0.8)^k$ for projections of order $k$.
For each example, each sorting method, each type of point set, and each selected value of $n$,
we ran simulations to estimate $\Var[\bar Y_n]$.
For the stratified and RQMC points, this variance was estimated by replicating the RQMC scheme $m = 100$ times
independently.
For a fair comparison with the MC variance $\sigma^2_{\rmy} = \Var[Y]$, for these point sets we used the
\emph{variance per run}, defined as $n \Var[\bar Y_n]$.
We define the \emph{variance reduction factor} (VRF) for a given method compared with MC
by $\sigma^2_{\rmy} / (n \Var[\bar Y_n])$.
In each case, we fitted a linear regression model for the variance per run as a function of $n$,
in log-log scale. We denote by $\hat\beta$ the regression slope estimated by this linear model.
In the remaining sections, we explain how the process $\{S(t),\, t\ge 0\}$ is defined in each
case, how it is simulated. We show how we can apply Array-RQMC and we provide numerical results.
All the experiments were done in Java using the SSJ library \cite{sLEC05a,iLEC16j}.
\section{OPTION PRICING UNDER A VARIANCE-GAMMA PROCESS}
\label{sec:vg}
The \emph{variance-gamma} (VG) model was proposed for option pricing by \citeN{fMAD90a} and \citeN{fMAD98a},
and further studied by \shortciteN{fFU98a,fAVR03a,fAVR06a}, for example.
A VG process is essentially a Brownian process for which the time clock runs
at random and time-varying speed driven by a gamma process.
The VG process with parameters $(\theta,\sigma^2,\nu)$
is defined as $Y = \{Y(t) = X(G(t)),\, t\ge 0\}$ where
$X = \{X(t),\, t\ge 0\}$ is a Brownian motion with drift and variance parameters $\theta$ and $\sigma^2$, and
$G = \{G(t),\, t\ge 0\}$ is a gamma process with drift and volatility parameters $1$ and $\nu$,
independent of $X$.
This means that $X(0) = 0$, $G(0)=0$, both $B$ and $G$ have independent increments,
and for all $ t \geq 0 $ and $ \delta > 0 $, we have
$X(t+\delta)-X(t)\sim {\rm Normal}(\delta\theta,\delta\sigma^{2})$,
a normal random variable with mean $\delta\theta$ and variance $\delta\sigma^{2}$,
and $G(t+\delta)-G(t) \sim {\rm Gamma}(\delta/\nu, \nu)$, a gamma random variable
with mean $\delta$ and variance $\delta \nu$.
The gamma process is always non-decreasing, which ensures that the time clock never goes backward.
In the VG model for option pricing, the asset value follows the \emph{geometric variance-gamma} (GVG)
process $S = \{S(t),\, t\ge 0\}$ defined by
\[
S(t) = S(0) \exp\left[ (r+\omega) t + X(G(t)) \right],
\]
where $\omega = \ln (1-\theta\nu-\sigma^2 \nu/2)/\nu$.
To generate realizations of $\bar S$
for this process, we must generate $S(t_1),\dots,S(t_{\tau})$, and there are many ways of doing this.
With Array-RQMC, we want to do it via a Markov chain with a low-dimensional state.
The running average $\bar S_j$ must be part of the state, as well as sufficient information
to generate the future of the path.
A simple procedure for generating the path is to sample sequentially $G(t_1)$,
then $Y(t_1) = X(G(t_1))$ conditional on $G(t_1)$,
then $G(t_2)$ conditional on $G(t_1)$,
then $Y(t_2) = X(G(t_2))$ conditional on $(G(t_1), G(t_2), Y(t_1))$, and so on.
We can then compute any $S(t_j)$ directly from $Y(t_j)$.
It is convenient to view the sampling of $(G(t_j),\, Y(t_j))$ conditional on $(G(t_{j-1}),\, Y(t_{j-1}))$
as one step (step $j$) of the Markov chain.
The state of the chain at step $j-1$ can be taken as $X_{j-1} = (G(t_{j-1}),\, Y(t_{j-1}),\, \bar S_{j-1})$,
so we have a three-dimensional state, and we need two independent uniform random numbers at each step,
one to generate $G(t_j)$ and the other to generate $Y(t_j) = X(G(t_j))$ given $(G(t_{j-1}), G(t_j), Y(t_{j-1}))$,
both by inversion.
Applying Array-RQMC with this setting would require a five-dimensional RQMC point set at each step,
unless we can map the state to a lower-dimensional representation.
However, a key observation here is that the distribution of the increment
$\Delta Y_j = Y(t_j) - Y(t_{j-1})$ depends only on the increment $\Delta_j = G(t_j) - G(t_{j-1})$
and not on $G(t_{j-1})$. This means that there is no need to memorize the latter in the state!
Thus, we can define the state at step $j$ as the two-dimensional vector $X_{j} = (Y(t_{j}),\, \bar S_{j})$,
or equivalently $X_{j} = (S(t_{j}),\, \bar S_{j})$,
and apply Array-RQMC with a four-dimensional RQMC point set if we use a two-dimensional sort for the states,
and a three-dimensional RQMC point set if we map the states to a one-dimensional representation
(using a Hilbert curve or a linear combination of the coordinates, for example).
At step $j$, we generate $\Delta_{j} \sim \dGamma((t_{j}-t_{j-1})/\nu, \nu)$ by inversion using a
uniform random variate $U_{j,1}$, i.e., via $\Delta_{j} = F_j^{-1}(U_{ j,1})$ where $F_j$ is the cdf
of the $\dGamma((t_{j}-t_{j-1})/\nu, \nu)$ distribution, then $\Delta Y_j$ by inversion from the
normal distribution with mean $\theta \Delta_{j}$ and variance $\sigma^2 \Delta_{j}$,
using a uniform random variate $U_{j,2}$.
Algorithm~\ref{algo:sequential} summarizes this procedure.
The symbol $\Phi$ denotes the standard normal cdf.
We have
\[
X_j = (Y(t_{j}),\bar{S}_{j}) = \varphi_{j}(Y(t_{j-1}), \bar{S}_{j-1}, U_{j,1}, U_{j,2})
\]
where $\varphi_j$ is defined by the algorithm.
The payoff function is $g(X_c) = \bar S_c = \bar S$.
\begin{algorithm}
\caption{Computing $X_j = (Y(t_{j}),\bar{S}_{j})$ given $(Y(t_{j-1}), \bar{S}_{j-1})$, for $1\leq j\leq \tau$.}
\label{algo:sequential}
\begin{algorithmic}
\State {Generate \ $U_{j,1}, U_{j,2} \sim \dUnif(0,1)$, independent;}
\State {$\Delta_{j} = F_j^{-1}(U_{j,1}) \sim {\rm Gamma}((t_{j}-t_{j-1})/\nu, \nu)$;}
\State {$Z_{j} = \Phi^{-1}(U_{j,2}) \sim \dNormal(0,1)$;}
\State {$Y(t_{j})\leftarrow Y(t_{j-1}) + \theta\Delta_{j}+ \sigma \sqrt{\Delta_{j}} Z_j$;}
\State {$S(t_j) \leftarrow S(0) \exp[(r+\omega) t_j + Y(t_j)]$;}
\State {$\bar S_j = [(j-1) \bar S_{j-1} + S(t_j)]/j$;}
\end{algorithmic}
\end{algorithm}
With this two-dimensional state representation, if we use a split sort or batch,
we need four-dimensional RQMC points.
With the Hilbert-curve sort, we only need three-dimensional RQMC points.
We also tried a simple linear mapping $h_j : \RR^2\to\RR$ defined by
$h_j(S(t_j), \bar S_j) = b_j \bar S_j + (1-b_j) S(t_j)$
where $b_j = (j-1)/(\tau-1)$.
At each step $j$, this $h_j$ maps the state $X_j$ to a real number $h_j(X_j)$,
and we sort the states by increasing order of their value of $h_j(X_j)$.
It uses a convex linear combination of $S(t_j)$ and $\bar S_j$ whose coefficients depend on $j$.
The rationale for the (heuristic) choice of $b_j$
is that in the late steps (when $j$ is near $\tau$), the current average $\bar{S}_{j}$
is more important (has more predictive power for the final payoff) than the current $S(t_{j})$,
whereas in the early steps, the opposite is true.
We made an experiment with the following model parameters, taken from \citeN{fAVR06a}:
$\theta = -0.1436 $, $ \sigma = 0.12136 $, $\nu = 0.3 $, $r = 0.1 $, $T = 240/365$, $\tau = 10$,
$t_j = 24j/365$ for $j=1,\dots,\tau$, $K = 100$, and $S(0) = 100$.
The time unit is one year, the horizon is 240 days, and there is an observation time every 24 days.
The exact value of the expected payoff for the Asian option is $\mu \approx 8.36 $,
and the MC variance per run is $\sigma_{\rmy}^{2} = \Var[Y_{\rma}] \approx 59.40 $.
\begin{table}[ht ]
\centering
\small
\caption{Regression slopes $\hat\beta$ for $\log_{2} \Var[\mua]$ vs $\log_2(n)$,
and VRF compared with MC for $n=2^{20}$, denoted VRF20, for the Asian option under the VG model }
\label{tab:vg-array}
\begin{tabular}{l|r|c|r}
\hline
Sort & Point sets & $ \hat{\beta} $ & VRF20 \\
\hline
\multirow{4}{2cm}{Split sort } & MC & -1 & 1 \\
& Stratif & -1.17 & 42 \\
& Sobol'+LMS & -1.77 & 91550 \\
& Sobol'+NUS & -1.80 & 106965 \\
& Lattice+baker & -1.83 & 32812 \\
\hline
\multirow{4}{2cm}{Batch sort ($ n_{1}=n_{2} $) } & MC & -1 & 1 \\
& Stratif & -1 & 42 \\
& Sobol'+LMS & -1.71 & 100104 \\
& Sobol'+NUS & -1.54 & 90168 \\
& Lattice+baker & -1.95 & 58737 \\
\hline
\multirow{4}{2cm}{Hilbert sort (with logistic map) } & MC & -1 & 1 \\
& Stratif & -1.43 & 204 \\
& Sobol'+LMS & -1.59 & 68297 \\
& Sobol'+NUS & -1.67 & 79869 \\
& Lattice+baker & -1.55 & 45854 \\
\hline
\multirow{4}{2cm}{Linear map sort } & MC & -1 & 1 \\
& Stratif & -1.35 & 192 \\
& Sobol'+LMS & -1.64 & 115216 \\
& Sobol'+NUS & -1.75 & 166541 \\
& Lattice+baker & -1.72 & 68739 \\
\hline
\end{tabular}
\end{table}
\begin{figure}[!hbt] \footnotesize
\centering
\begin{tikzpicture}
\begin{axis}[
cycle list name=defaultcolorlist,
legend style={at={(1.22,1.85)}, anchor={north east}},
xlabel=$\log_2(n)$,
grid,
width=4cm,
height=4cm,
]
\addplot table[x=log(n),y=log(MISE)] {
log(n) log(MISE)
16.0 -10.151201808434037
17.0 -11.374817074545694
18.0 -11.994381098726798
19.0 -13.208370282151458
20.0 -13.949334962385464
};
\addlegendentry{MC}
\addplot table[x=log(n),y=log(MISE)] {
log(n) log(MISE)
16.0 -14.959378903560621
16.99171005377434 -15.88089020607308
18.094247824228052 -17.14658716084826
19.019550008653873 -18.564637351701542
20.0 -19.513185167970356
};
\addlegendentry{Stratif}
\addplot table[x=log(n),y=log(MISE)] {
log(n) log(MISE)
16.0 -23.633704459633822
17.0 -25.580879149864206
18.0 -27.124986360860927
19.0 -29.348723118567527
20.0 -30.58977405394141
};
\addlegendentry{Sobol+LMS}
\addplot table[x=log(n),y=log(MISE)] {
log(n) log(MISE)
16.0 -23.55879443464881
17.0 -25.452065755021675
18.0 -26.978058126434465
19.0 -28.994972675511697
20.0 -30.81428988989172
};
\addlegendentry{Sobol+NUS}
\addplot table[x=log(n),y=log(MISE)] {
log(n) log(MISE)
16.0 -22.28254034764946
17.0 -23.01761769453436
18.0 -23.731399349454925
19.0 -27.640591609428213
20.0 -29.109451903531756
};
\addlegendentry{Lattice+baker}
\end{axis}
\end{tikzpicture}
\begin{tikzpicture}
\begin{axis}[
cycle list name=defaultcolorlist,
legend style={at={(1.22,1.85)}, anchor={north east}},
xlabel=$\log_2(n)$,
grid,
width=4cm,
height=4cm,
]
\addplot table[x=log(n),y=log(MISE)] {
log(n) log(MISE)
16.0 -10.125167088824263
17.0 -11.391734398723784
18.0 -12.030243862257686
19.0 -13.19124273647122
20.0 -13.957222521780116
};
\addlegendentry{MC}
\addplot table[x=log(n),y=log(MISE)] {
log(n) log(MISE)
16.0 -14.967680095056181
16.99171005377434 -15.872803090894319
18.094247824228052 -17.11791662799947
19.019550008653873 -18.618933749651113
20.0 -19.491670326261605
};
\addlegendentry{Stratif}
\addplot table[x=log(n),y=log(MISE)] {
log(n) log(MISE)
16.0 -23.742038334288367
17.0 -25.18529149145997
18.0 -26.575981540352245
19.0 -28.349253143579467
20.0 -30.71863728507717
};
\addlegendentry{Sobol +LMS}
\addplot table[x=log(n),y=log(MISE)] {
log(n) log(MISE)
16.0 -24.05405256317996
17.0 -25.551335850072785
18.0 -26.532683096004593
19.0 -27.949058481148693
20.0 -30.56782525433717
};
\addlegendentry{Sobol+NUS}
\addplot table[x=log(n),y=log(MISE)] {
log(n) log(MISE)
16.0 -23.354622638739173
17.0 -21.675118811834484
18.0 -24.319492844187447
19.0 -27.946213398912015
20.0 -29.94949071310452
};
\addlegendentry{Lattice+baker}
\end{axis}
\end{tikzpicture}
\begin{tikzpicture}
\begin{axis}[
cycle list name=defaultcolorlist,
legend style={at={(1.22,1.85)}, anchor={north east}},
xlabel=$\log_2(n)$,
grid,
width=4cm,
height=4cm,
]
\addplot table[x=log(n),y=log(MISE)] {
log(n) log(MISE)
16.0 -9.965506807807301
17.0 -11.205344676647975
18.0 -11.967883997584597
19.0 -13.384159678031219
20.0 -14.196724538935994
};
\addlegendentry{MC}
\addplot table[x=log(n),y=log(MISE)] {
log(n) log(MISE)
15.965784284662087 -16.08499538939687
17.017276025914487 -17.56037697635942
18.0 -19.018210419986847
19.019550008653873 -20.572247692334237
20.017276025914487 -21.7649559791986
};
\addlegendentry{Stratif}
\addplot table[x=log(n),y=log(MISE)] {
log(n) log(MISE)
16.0 -23.93571019224644
17.0 -25.832152908036722
18.0 -27.33008775845543
19.0 -29.091177852445593
20.0 -30.578431433321636
};
\addlegendentry{Sobol +LMS}
\addplot table[x=log(n),y=log(MISE)] {
log(n) log(MISE)
16.0 -24.120858868108805
17.0 -25.058596678607625
18.0 -27.841113569201394
19.0 -28.632361952794387
20.0 -30.308379158193635
};
\addlegendentry{Sobol+NUS}
\addplot table[x=log(n),y=log(MISE)] {
log(n) log(MISE)
16.0 -23.42830483146864
17.0 -25.702792296644837
18.0 -27.16182898294017
19.0 -28.83775332640567
20.0 -29.59227100834182
};
\addlegendentry{Lattice+baker}
\end{axis}
\end{tikzpicture}
\begin{tikzpicture}
\begin{axis}[
cycle list name=defaultcolorlist,
legend style={at={(1.22,1.85)}, anchor={north east}},
xlabel=$\log_2(n)$,
grid,
width=4cm,
height=4cm,
]
\addplot table[x=log(n),y=log(MISE)] {
log(n) log(MISE)
16.0 -9.82305725456555
17.0 -11.099014117169833
18.0 -12.044117067884285
19.0 -13.140553493058876
20.0 -14.440549409485193
};
\addlegendentry{MC}
\addplot table[x=log(n),y=log(MISE)] {
log(n) log(MISE)
15.965784284662087 -16.08023037439075
17.017276025914487 -17.658391548343992
18.0 -19.220332169166976
19.019550008653873 -20.058388378525507
20.017276025914487 -21.710452609694315
};
\addlegendentry{Stratif}
\addplot table[x=log(n),y=log(MISE)] {
log(n) log(MISE)
16.0 -24.301030870661574
17.0 -26.19935628004586
18.0 -27.716727137785195
19.0 -29.385458064674975
20.0 -30.921744935733827
};
\addlegendentry{Sobol +LMS}
\addplot table[x=log(n),y=log(MISE)] {
log(n) log(MISE)
16.0 -24.23697989879872
17.0 -26.19235071537741
18.0 -28.169357039010745
19.0 -29.216652851994684
20.0 -31.453280459358083
};
\addlegendentry{Sobol+NUS}
\addplot table[x=log(n),y=log(MISE)] {
log(n) log(MISE)
16.0 -23.236130419924017
17.0 -26.186668822233617
18.0 -27.56543498454785
19.0 -29.55463426017593
20.0 -30.17659115439563
};
\addlegendentry{Lattice+baker}
\end{axis}
\end{tikzpicture}
\caption{Plots of empirical $\log_{2}\Var[\mua]$ vs $\log_2(n)$ for various sorts and point sets,
based on $m=100$ independent replications.
Left to right: split sort, batch sort, Hilbert sort, linear map sort.}
\label{fig:vg-array}
\end{figure}
Table~\ref{tab:vg-array} summarizes the results.
For each selected sorting method and point set, we report the estimated slope $\hat\beta$ for the
linear regression model of $\log_{2}\Var[\mua]$ as a function of $\log_2(n)$
obtained from $m=100$ independent replications with $n=2^{e}$ for $e=16,...,20$,
as well as the variance reduction factors (VRF) observed for $n=2^{20}$ (about one million samples),
denoted VRF20.
For MC, the exact slope (or convergence rate) $\beta$ is known to be $\beta = -1$.
We see from the table that Array-RQMC provides much better convergence rates (at least empirically),
and reduces the variance by very large factors for $n = 2^{20}$.
Interestingly, the largest factors are obtained with the Sobol' points combined with our
heuristic linear map sort, although the other sorts are also doing quite well.
Figure~\ref{fig:vg-array} shows plots of $\log_{2}\Var[\mua]$ vs $\log_2(n)$ for selected sorts.
It gives an idea of how well the linear model fits in each case.
There are other ways of defining the steps of the Markov chain for this example.
For example, one can have one step for each $\dUnif(0,1)$ random number
that is generated. This would double the number of steps, from $c$ to $2c$.
We generate $\Delta_1$ in the first step, $Y(t_1)$ in the second step,
$\Delta_2$ in the third step, $Y(t_2)$ in the fourth step, and so on.
Generating a single uniform per step instead of two reduces by 1 the dimension of
the required RQMC point set. At odd step numbers, when we generate a $\Delta_j$,
the state can still be taken as $(Y(t_{j-1}), \bar{S}_{j-1})$ and we only need three-dimensional
RQMC points, so we save one dimension.
But at even step numbers, we need $\Delta_j$ to generate $Y(t_j)$,
so we need a three-dimensional state $(Y(t_{j-1}), \Delta_j, \bar{S}_{j-1})$
and four-dimensional RQMC points.
We tried this approach and it did not perform better than the one described earlier,
with two uniforms per step. It is also more complicated to implement.
\shortciteN{fAVR03a,fAVR06a} describe other ways of simulating the VG process, for instance
Brownian and gamma bridge sampling (BGBS) and difference of gammas bridge sampling (DGBS).
BGBS generates first $G(t_c)$ then $Y(t_c)$, then conditional on this it generates
$G(t_{c/2})$ then $Y(t_{c/2})$ (assuming that $c$ is even), and so on.
DGBS writes the VG process $Y$ as a difference of two independent gamma processes
and simulate both using the bridge idea just described: first generate the values of the two
gamma processes at $t_{c}$, then at $t_{c/2}$, etc.
When using classical RQMC, these sampling methods brings an important variance reduction
compared with the sequential one we use here for our Markov chain.
With Array-RQMC, however, they are impractical, because the dimension of the state
(the number of values that we need to remember) grows up to about $c$, which is much to high,
and the implementation is much more complicated.
For this VG model, we do not report results on the European option with Array-RQMC,
because the Markov chain would have only one step: We can generate directly
$G(t_c)$ and then $Y(t_c)$. For this, ordinary RQMC works well enough \def\citeauthoryear##1##2##3{##2, ##3}\@internalcite{vLEC18a}.
\section{OPTION PRICING UNDER THE HESTON VOLATILITY MODEL}
\label{sec:heston}
The Heston volatility model is defined by the following two-dimensional stochastic differential equation:
\begin{eqnarray*}
\d S(t) &=& rS(t)\d t+ V(t)^{1/2}S(t)\d B_{1}(t),\\
\d V(t) &=& \lambda(\sigma^{2}-V(t))\d t + \xi V(t)^{1/2}\d B_{2}(t),
\end{eqnarray*}
for $t \geq 0$, where $(B_{1},B_{2})$ is a pair of standard Brownian motions with correlation
$\rho $ between them, $ r $ is the risk-free rate, $ \sigma^{2} $ is the long-term average variance parameter,
$\lambda $ is the rate of return to the mean for the variance,
and $\xi$ is a volatility parameter for the variance.
The processes $S = \{S(t),\, t\ge 0\}$ and $V = \{V(t),\, t\ge 0\}$ represent the asset price
and the volatility, respectively, as a function of time.
We will examine how to estimate the price of European and Asian options with Array-RQMC under this model.
Since we do not know how to generate $(S(t+\delta),\, V(t+\delta))$ exactly from its conditional
distribution given $(S(t),\, V(t))$ in this case, we have to discretize the time.
For this, we use the Euler method with $\tau$ time steps of length $\delta = T/\tau$ to generate a skeleton
of the process at times $w_j = j\delta$ for $j=1,\dots,\tau$, over $[0,T]$.
For the Asian option, we assume for simplicity that the observation times $t_1,\dots,t_c$
used for the payoff are all multiples of $\delta$, so each of them is equal to some $w_j$.
\hpierre{Here we should avoid the notation $t_j$, because $j$ is the step number of the Markov chain,
so we would have two different meanings for the index $j$.}
\begin{table}[hbt]
\centering
\small
\caption{Regression slopes $\hat\beta$ for $\log_{2} \Var[\mua]$ vs $\log_2(n)$,
and VRF compared with MC for $n=2^{20}$, denoted VRF20, for the Asian option under the Heston model. }
\label{tab:heston-array}
\begin{tabular}{l|r|r|r|r|r}
\hline
\multicolumn{2}{c|}{} & \multicolumn{2}{c|}{European}& \multicolumn{2}{c}{Asian} \\
\hline
Sort & Point sets & $ \hat{\beta} $ & VRF20 & $ \hat{\beta} $ & VRF20 \\
\hline
\multirow{5}{*}{Split sort } & MC & -1 & 1 & -1 & 1 \\
& Stratif & -1.26 & 103 & -1.29 & 38 \\
& Sobol'+LMS & -1.59 & 44188 & -1.48 & 6684 \\
& Sobol'+NUS & -1.46& 30616 & -1.46 & 5755 \\
& Lattice+baker & -1.50 & 26772 & -1.55 & 5140 \\
\hline
\multirow{5}{2.5cm}{Batch sort
} & MC & -1 & 1 & -1 & 1 \\
& Stratif & -1.24 & 91 & -1.25 & 33 \\
& Sobol'+LMS & -1.66 & 22873 & -1.23 & 815 \\
& Sobol'+NUS & -1.72 & 30832 & -1.38& 1022 \\
& Lattice+baker & -1.75 & 12562 & -1.22& 762 \\
\hline
\multirow{5}{2.5cm}{Hilbert sort (with logistic map)
} & MC & -1& 1 & -1 & 1 \\
& Stratif & -1.26 & 43 & -1.05 & 29 \\
& Sobol'+LMS & -1.14 & 368 & -0.87 & 39 \\
& Sobol'+NUS & -1.06 & 277 & -1.11& 49 \\
& Lattice+baker & -1.12 & 250 & -0.89 & 42\\
\hline
\end{tabular}
\end{table}
\begin{figure}[htb] \footnotesize
\centering
\small
\begin{tikzpicture}
\begin{axis}[
cycle list name=defaultcolorlist,
legend style={at={(1.12,1.95)}, anchor={north east}},
xlabel=$\log_2(n)$,
ylabel=$\log_2(\Var)$,
grid,
width=5cm,
height=4cm,
]
\addplot table[x=log(n),y=log(MISE)] {
log(n) log(MISE)
16.0 -10.0393965896146
17.0 -11.122034035538492
18.0 -12.203482113338744
19.0 -13.588402242719132
20.0 -14.578709496795367
};
\addlegendentry{MC}
\addplot table[x=log(n),y=log(MISE)] {
log(n) log(MISE)
15.84962500721156 -14.111302800188136
17.297158093186486 -16.028321372198608
17.92481250360578 -16.774827814961732
19.03677461028802 -18.304195272754914
20.0 -19.43172105454755
};
\addlegendentry{Stratif}
\addplot table[x=log(n),y=log(MISE)] {
log(n) log(MISE)
16.0 -21.050533059699816
17.0 -22.395066360734926
18.0 -23.75575975421682
19.0 -25.512215835716276
20.0 -26.881037454834505
};
\addlegendentry{Sobol' +LMS}
\addplot table[x=log(n),y=log(MISE)] {
log(n) log(MISE)
16.0 -20.730649960589506
17.0 -22.67690957689726
18.0 -24.255429521603652
19.0 -25.40529102791617
20.0 -26.665055006634283
};
\addlegendentry{Sobol'+NUS}
\addplot table[x=log(n),y=log(MISE)] {
log(n) log(MISE)
16.0 -20.438241859943144
17.0 -21.49545519899545
18.0 -22.060176245547
19.0 -24.903082129918214
20.0 -26.502180414676182
};
\addlegendentry{Lattice+baker}
\end{axis}
\end{tikzpicture}
\begin{tikzpicture}
\begin{axis}[
cycle list name=defaultcolorlist,
legend style={at={(1.12,1.95)}, anchor={north east}},
xlabel=$\log_2(n)$,
grid,
width=5cm,
height=4cm,
]
\addplot table[x=log(n),y=log(MISE)] {
log(n) log(MISE)
16.0 -10.024845458352644
17.0 -11.159811274387627
18.0 -12.195612913788786
19.0 -13.590093217796193
20.0 -14.467680580335442
};
\addlegendentry{MC}
\addplot table[x=log(n),y=log(MISE)] {
log(n) log(MISE)
15.84962500721156 -14.174715605814937
17.297158093186486 -15.965674741102829
17.92481250360578 -16.629129192425264
19.03677461028802 -18.38704490691805
20.0 -19.230981344472582
};
\addlegendentry{Stratif}
\addplot table[x=log(n),y=log(MISE)] {
log(n) log(MISE)
16.0 -18.710312874636735
17.0 -19.7195535149438
18.0 -21.545707235411527
19.0 -21.710845947842163
20.0 -23.844994987304258
};
\addlegendentry{Sobol'+LMS}
\addplot table[x=log(n),y=log(MISE)] {
log(n) log(MISE)
16.0 -18.471130506776717
17.0 -19.687071984460964
18.0 -21.2895781658613
19.0 -22.077391393023678
20.0 -24.171378533597295
};
\addlegendentry{Sobol'+NUS}
\addplot table[x=log(n),y=log(MISE)] {
log(n) log(MISE)
16.0 -18.65403828216375
17.0 -19.716095401701306
18.0 -21.742953862467
19.0 -21.729154800441403
20.0 -23.7487392043938
};
\addlegendentry{Lattice+baker}
\end{axis}
\end{tikzpicture}
\begin{tikzpicture}
\begin{axis}[
cycle list name=defaultcolorlist,
legend style={at={(1.12,1.95)}, anchor={north east}},
xlabel=$\log_2(n)$,
grid,
width=5cm,
height=4cm,
]
\addplot table[x=log(n),y=log(MISE)] {
log(n) log(MISE)
16.0 -10.386671027956352
17.0 -10.906323234868157
18.0 -12.02420612295912
19.0 -13.030561924022795
20.0 -14.133836473244045
};
\addlegendentry{MC}
\addplot table[x=log(n),y=log(MISE)] {
log(n) log(MISE)
15.965784284662087 -14.624122625895069
17.017276025914487 -15.926972640736638
18.0 -16.85182456550578
19.019550008653873 -17.704148569387396
20.017276025914487 -19.060495580354043
};
\addlegendentry{Stratif}
\addplot table[x=log(n),y=log(MISE)] {
log(n) log(MISE)
16.0 -16.141755010858162
17.0 -16.589960669586148
18.0 -17.93502264226188
19.0 -18.621379892942045
20.0 -19.459860334907876
};
\addlegendentry{Sobol' +LMS}
\addplot table[x=log(n),y=log(MISE)] {
log(n) log(MISE)
16.0 -15.615982736205153
17.0 -16.31525007722051
18.0 -17.572558128094023
19.0 -19.077760800136044
20.0 -19.794678952215136
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log(n) log(MISE)
16.0 -16.068445597149932
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18.0 -17.67312220237025
19.0 -18.652494876004223
20.0 -19.574632883313445
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16.0 -13.936914603457174
16.99171005377434 -15.218176303204794
18.094247824228052 -16.614081351574377
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16.0 -21.395980476569648
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18.0 -24.72212845665191
19.0 -26.248759822966957
20.0 -27.845069680521398
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20.0 -27.122142807184268
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log(n) log(MISE)
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20.0 -27.32581432571954
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width=5cm,
height=4cm,
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log(n) log(MISE)
16.0 -8.53934570240081
17.0 -9.476899800633442
18.0 -10.40309577892365
19.0 -11.505510626641803
20.0 -12.178521386358495
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17.017276025914487 -14.197559620420003
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20.017276025914487 -17.855870032217776
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log(n) log(MISE)
16.0 -16.28738135858736
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18.0 -18.883818810162555
19.0 -19.44211050869576
20.0 -20.93951862124304
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log(n) log(MISE)
16.0 -16.117677581400883
17.0 -17.776532878357298
18.0 -18.24325663944454
19.0 -19.5599986210728
20.0 -20.53025947107574
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log(n) log(MISE)
16.0 -16.172081320286516
17.0 -17.111259607558328
18.0 -18.81875995025613
19.0 -19.908594307500554
20.0 -20.378405160721815
};
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\end{axis}
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\caption{Plots of empirical $\log_{2}\Var[\mua]$ vs $\log_2(n)$ for various sorts and point sets,
based on $m=100$ independent replications, for the Heston model.
Asian option (above) and European option (below),
with split sort (left), batch sort (middle), and Hilbert sort (right).}
\label{fig:heston-array}
\end{figure}
Following \citeN{vGIL08a}, to reduce the bias due to the discretization, we make the change of variable
$W(t)=e^{\lambda t}(V(t)-\sigma^{2})$, with $\d W(t) = e^{\lambda t} \xi V(t)^{1/2} \d B_{2}(t)$,
and apply the Euler method to $(S,W)$ instead of $(S,V)$.
The Euler approximation scheme with step size $\delta$ applied to $W$ gives
\[
\widetilde{W}(j\delta) = \widetilde{W}((j-1)\delta)
+ e^{\lambda (j-1)\delta} \xi (\widetilde{V}((j-1)\delta) \delta)^{1/2} Z_{j,2}.
\]
Rewriting it in terms of $V$ by using the reverse identity $V(t)=\sigma^{2}+e^{-\lambda t} W(t)$,
and after some manipulations, we obtain the following discrete-time stochastic recurrence,
which we will simulate by Array-RQMC:
\begin{eqnarray*}
\widetilde{V}(j\delta) &=& \max\left[0,\, \sigma^{2} + e^{-\lambda\delta}
\left(\widetilde{V}((j-1)\delta)-\sigma^{2} + \xi (\tilde{V}((j-1)\delta) \delta)^{1/2} Z_{j,2}\right)\right],\\
\widetilde{S}(j\delta) &=& (1 + r\delta) \widetilde{S}((j-1)\delta)
+ (\widetilde{V}((j-1)\delta) \delta)^{1/2} \widetilde{S}((j-1)\delta) Z_{j,1},
\end{eqnarray*}
where $(Z_{j,1},Z_{j,2})$ is a pair of standard normals with correlation $\rho$.
We generate this pair from a pair $(U_{j,1}, U_{j,2})$ of independent $\dUnif(0,1)$ variables via
$Z_{j,1} = \Phi^{-1}(U_{j,1})$ and $Z_{j,2} = \rho Z_{j,1} + \sqrt {1-\rho^{2}} \,\Phi^{-1}(U_{j,2})$.
We then approximate each $S(j\delta)$ by $\widetilde S(j\delta)$.
The running average $\bar S_j$ at step $j$ must be the average of the $S(t_k)$ at the observation times
$t_k \le w_j = j\delta$. If we denote $N_j = \sum_{k=1}^c \II[t_k \le j\delta]$, we have
$\bar S_j = (1/N_j) \sum_{k=1}^{N_j} S(t_k)$, which we approximate by
$\overline{\bar S}_j = (1/N_j) \sum_{k=1}^{N_j} \widetilde S(t_k)$.
\iffalse
having in mind that the true payoff could be the continuous average
$(1/T) \int_{0}^T S(t) \d t$, and that we might want to approximate that,
we use the slightly different average
\hpierre{I do not think it is a good idea to use a different average; it makes the paper less
uniform and more complicated for nothing.
Now, $\bar S_j$ must have a different definition as well!
In real life, the number of time steps for Euler would be much larger than the
number of observation times to determine the payoff. }
\[
\bar{S}_d = h \left(\dfrac{\widetilde{S}(0)}{2} + \dfrac{\widetilde{S}(T)}{2}
+ \sum _{j=1}^{d-1} \widetilde{S}(t_j) \right).
\]
\fi
Here, the state of the chain is $X_{j} = (\widetilde S(j\delta), \widetilde V(j\delta))$
when pricing the European option and
$X_{j} = (\widetilde S(j\delta), \widetilde V(j\delta), \overline{\bar{S}}_{j})$
when pricing the Asian option.
And two uniform random numbers, $(U_{j,1}, U_{j,2})$, are required at each step of the chain.
We thus need four-dimensional RQMC point sets for the European option and
five-dimensional RQMC point sets for the Asian option, if we do not map the state to a
lower-dimensional representation.
If we map the state to one dimension, as in the Hilbert curve sort, then we only need three-dimensional
RQMC points for both option types.
We ran experiments with $T=1$ (one year), $ K=100 $, $ S(0)=100 $, $ V(0)=0.04 $, $ r=0.05 $, $ \sigma=0.2 $,
$\lambda=5 $, $ \xi=0.25 $, $ \rho=-0.5 $, and $c = \tau=16$.
\hpierre{Maybe these numbers will change? Perhaps $\tau=256$ and $c=12$ or 24?
This gives $\delta = 1/16$,
so the time discretization for Euler is very coarse, but a smaller $\delta$ gives similar results
in terms of variance reduction by Array-RQMC.
\hpierre{We should report a few experiments with much smaller $h$, for example $h = 1/256$ and $h = 1/1024$,
and compare the estimates (for expected payoff) and variance reductions.
\hamal{it's takes a lot of times may be I will have the results for tomorrow for $h = 1/256$
Table~\ref{tab:heston-array} reports the estimated slopes $\hat\beta$ and VRF20,
as in Table~\ref{tab:vg-array}.
Again, we observe large variance reductions and improved convergence rates from Array-RQMC.
The best results are obtained with the split sort.
Figure~\ref{fig:heston-array} shows plots of $\log_{2}\Var[\mua]$ vs $\log_2(n)$ for selected sorts.
We tried an alternative Markov chain definition in which the chain advances by one step
each time a uniform random number is used, as in the VG example, to reduce the dimension of the RQMC points,
but this gave no improvement.
\hpierre{Well, here I am surprised that it does not help, because the dimension of the RQMC points
is really reduced by 1 at all steps, if done properly!}
\hpierre{We also tried a map that combines the two state variables
$\widetilde S(j\delta)$ and $\overline{\bar{S}}_j$ to a single variable
$Z_j = h_j(\widetilde S(j\delta),\overline{\bar{S}}_j)$ before performing a split sort based on
$(Z_j, \widetilde V(j\delta)))$ at step $j$,
to reduce the dimension of the sort by 1, as in the VG example, but it did not help.}
\hpierre{The VRF of MC with respect to MC should be 1. Why are we getting 1.27 and 1.28? Very suspicious.}
\section{OPTION PRICING UNDER THE ORNSTEIN-UHLENBECK VOLATILITY MODEL}
\label{sec:ou}
\begin{table}[hbt]
\centering
\small
\caption{
Regression slopes $\hat\beta$ for $\log_{2} \Var[\mua]$ vs $\log_2(n)$,
and VRF compared with MC for $n=2^{20}$, denoted VRF20, for the European and Asian
options under the Ornstein-Uhlenbeck model.}
\label{tab:ou-array}
\begin{tabular}{l|c|c|r|c|r}
\hline
\multicolumn{2}{c|}{} & \multicolumn{2}{c|}{European}& \multicolumn{2}{c}{Asian} \\
\hline
Sort & Point sets & $ \hat{\beta} $ & VRF20 & $ \hat{\beta} $ & VRF20 \\
\hline
\multirow{5}{2.5cm}{Batch sort
} & MC & -1 & 1 & -1 & 1 \\
& Stratif & -1.28 & 111 & -1.23 & 29. \\
& Sobol'+LMS& -1.35 & 61516 & -1.22 & 4558 \\
& Sobol'+NUS & -1.31 & 56235 & -1.22& 5789 \\
& Lattice+baker & -1.37 & 61318 & -1.20& 5511 \\
\hline
\multirow{5}{2.5cm}{Hilbert sort (with logistic map)
} & MC & -1 & 1 & -1 & 1 \\
& Stratif & -1.40 & 440 & -1.37 & 250 \\
& Sobol'+LMS & -1.52 & 194895 & -1.40 & 41100 \\
& Sobol'+NUS & -1.68 & 191516 & -1.37 & 39861 \\
& Lattice+baker & -1.59 & 165351 & -1.47 & 37185 \\
\hline
\end{tabular}
\end{table}
The Ornstein-Uhlenbeck volatility model is defined by the following
stochastic differential equations:
\begin{eqnarray*}
\d S(t) &=& rS(t)\d t + e^{V(t)} S(t) \d B_{1}(t),\\
\d V(t) &=& \alpha (b-V(t))\d t + \sigma \d B_{2}(t),
\end{eqnarray*}
for $ t \geq 0 $, where $ (B_{1},B_{2}) $ is a pair of standard Brownian motions with correlation $ \rho $ between them, $ r $ is the risk-free rate, $ b $ is the long-term average volatility,
$ \alpha $ is the rate of return to the average volatility,
and is $ \sigma $ a variance parameter for the volatility process.
The processes $S = \{S(t),\, t\ge 0\}$ and $V = \{V(t),\, t\ge 0\}$ represent the asset price and
the volatility process.
We simulate these processes using Euler's method with $\tau$ time steps of length $\delta$,
as we did for the Heston model, but without a change of variable.
The discrete-time approximation of the stochastic recurrence is
\begin{eqnarray*}
\widetilde{S}(j\delta) &=& \widetilde{S}((j-1)\delta) + r\delta \widetilde{S}((j-1)\delta)
+ \exp\left[{\widetilde{V}((j-1)\delta)}\right] \sqrt {\delta} Z_{j,1},\\
\widetilde{V}(j\delta) &=& \alpha \delta b + (1-\alpha \delta ) \widetilde{V}((j-1)\delta)
+ \sigma \sqrt {\delta} Z_{j,2},
\end{eqnarray*}
where $(Z_{j,1},Z_{j,2})$ is a pair of standard normals with correlation $\rho$.
To generate this pair, we generate independent $\dUnif(0,1)$ variables $(U_{j,1}, U_{j,2})$, and put
$Z_{j,1} = \Phi^{-1}(U_{j,1})$ and $Z_{j,2} = \rho Z_{j,1} + \sqrt {1-\rho^{2}} \,\Phi^{-1}(U_{j,2})$.
For either the European or Asian option, the state of the Markov chain and the dimension of the RQMC points
are the same as for the Heston model.
We ran a numerical experiment with $ T=1 $, $ K=100 $, $ S(0)=100 $, $ V(0)=0.04 $, $ r=0.05 $,
$ b=0.4 $, $ \alpha=5 $, $ \sigma=0.2 $, $ \rho=-0.5$, and $c = \tau =16 $ (so $\delta = 1/16$).
Table~\ref{tab:ou-array} reports the estimated regression slopes $\hat\beta$ and VRF2.
\section*{CONCLUSION}
\label{sec:conclusion}
We have shown how Array-RQMC can be applied for pricing options under stochastic volatility
models, and gave detailed examples with the VG, Heston, and Ornstein-Uhlenbeck models.
With the models, the method requires higher-dimensional RQMC points than with the simpler GBM model
studied previously, and when time has to be discretized to apply Euler's method, the number of steps
of the Markov chain is much larger.
For these reasons, it was not clear a priori if Array-RQMC would be effective.
Our empirical results show that it brings very significant variance reductions compared with
crude Monte Carlo.
\section*{ACKNOWLEDGMENTS}
This work has been supported by a discovery grant from NSERC-Canada, a Canada Research Chair,
and a Grant from the IVADO Fundamental Research Program, to P. L'Ecuyer.
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xmlns:app="http://schemas.android.com/apk/res-auto"
xmlns:tools="http://schemas.android.com/tools"
android:layout_width="match_parent"
android:layout_height="match_parent"
android:paddingBottom="@dimen/activity_vertical_margin"
android:paddingLeft="@dimen/activity_horizontal_margin"
android:paddingRight="@dimen/activity_horizontal_margin"
android:paddingTop="@dimen/activity_vertical_margin"
tools:context="com.example.user.communication.Third"
android:background="@drawable/wall7">
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android:layout_width="100dp"
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android:layout_toEndOf="@+id/imageButton4"
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<ImageButton
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android:layout_width="100dp"
android:layout_height="100dp"
android:scaleType="fitXY"
app:srcCompat="@drawable/chocklate"
android:layout_alignTop="@+id/imageButton6"
android:layout_alignLeft="@+id/imageButton7"
android:layout_alignStart="@+id/imageButton7" />
<ImageButton
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android:layout_width="100dp"
android:layout_height="100dp"
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android:layout_above="@+id/imageButton9"
android:layout_alignParentRight="true"
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android:layout_width="100dp"
android:layout_height="100dp"
android:scaleType="fitXY"
app:srcCompat="@drawable/mangoe"
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android:layout_alignParentRight="true"
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<ImageButton
android:id="@+id/imageButton10"
android:layout_width="100dp"
android:layout_height="100dp"
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app:srcCompat="@drawable/pizza"
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android:layout_alignLeft="@+id/imageButton6"
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<ImageButton
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android:layout_width="100dp"
android:layout_height="100dp"
android:scaleType="fitXY"
app:srcCompat="@drawable/khichuri"
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app:srcCompat="@drawable/friedrice"
android:layout_below="@+id/imageButton9"
android:layout_alignParentRight="true"
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<ImageView
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android:layout_height="wrap_content"
android:id="@+id/imageView8"
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android:layout_height="wrap_content"
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android:layout_below="@+id/imageView8"
android:layout_marginLeft="12dp"
android:layout_marginStart="12dp"
android:layout_marginTop="31dp"
android:text="Foods"
android:textColor="@android:color/background_light"
android:textSize="35sp" />
<ImageButton
android:id="@+id/imageButton6"
android:layout_width="100dp"
android:layout_height="100dp"
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app:srcCompat="@drawable/fruits"
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android:layout_alignLeft="@+id/imageButton4"
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app:srcCompat="@drawable/chicken"
android:layout_marginLeft="16dp"
android:layout_marginStart="16dp"
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android:layout_toRightOf="@+id/imageView8"
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<Button
android:id="@+id/button7"
android:layout_width="wrap_content"
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android:text="Next"
android:layout_alignParentBottom="true"
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| {
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Q: Resizing a button I have a "button" that I wish to use all throughout my site, but depending on where in the site the button is, I want it to display at different sizes.
In my HTML I have tried (but its not working):
<div class="button" width="60" height="100">This is a button</div>
And the matching CSS:
.button {
background-color: #000000;
color: #FFFFFF;
float: right;
padding: 10px;
border-radius: 10px;
-moz-border-radius: 10px;
-webkit-border-radius: 10px;
}
I assumed that if each time I call this class I can just pass in a size and hey-presto!....but not....
By adding the width and height as I call the button class seems to do nothing to the size of it. Does anyone know how I can do this? And if I put the width and height in the CSS then the button will be the same size everywhere.
A: If you want to call a different size for the button inline, you would probably do it like this:
<div class="button" style="width:60px;height:100px;">This is a button</div>
Or, a better way to have different sizes (say there will be 3 standard sizes for the button) would be to have classes just for size.
For example, you would call your button like this:
<div class="button small">This is a button</div>
And in your CSS
.button.small { width: 60px; height: 100px; }
and just create classes for each size you wish to have. That way you still have the perks of using a stylesheet in case say, you want to change the size of all the small buttons at once.
A: Another alternative is that you are allowed to have multiple classes in a tag. Consider:
<div class="button big">This is a big button</div>
<div class="button small">This is a small button</div>
And the CSS:
.button {
/* all your common button styles */
}
.big {
height: 60px;
width: 100px;
}
.small {
height: 40px;
width: 70px;
}
and so on.
A: You should not use "width" and "height" attributes directly, use the style attribute like style="some css here" if you want to use inline styling:
<div class="button" style="width:60px;height:30px;">This is a button</div>
Note, however, that inline styling should generally be avoided since it makes maintenance and style updates a nightmare. Personally, if I had a button styling like yours but also wanted to apply different sizes, I would work with multiple css classes for sizing, like this:
.button {
background-color: #000000;
color: #FFFFFF;
padding: 10px;
border-radius: 10px;
-moz-border-radius: 10px;
-webkit-border-radius: 10px;
margin:10px
}
.small-btn {
width: 50px;
height: 25px;
}
.medium-btn {
width: 70px;
height: 30px;
}
.big-btn {
width: 90px;
height: 40px;
}
<div class="button big-btn">This is a big button</div>
<div class="button medium-btn">This is a medium button</div>
<div class="button small-btn">This is a small button</div>
jsFiddle example
Using this way of defining styles removes all style information from your HTML markup, which in will make it easier down the road if you want to change the size of all small buttons - you'll only have to change them once in the CSS.
A: Use inline styles:
<div class="button" style="width:60px;height:100px;">This is a button</div>
Fiddle
A: try this one out resizeable button
<button type="submit me" style="height: 25px; width: 100px">submit me</button>
| {
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} | 7,701 |
Great interview with Brian Eno on the Guardian today. Lots of interesting stuff, but I just want to pull two quotes. The first is a good reminder about succeeding on your own terms.
Instead of shooting arrows at someone else's target, which I've never been very good at, I make my own target around wherever my arrow happens to have landed. You shoot your arrow and then you paint your bullseye around it, and therefore you have hit the target dead centre.
The second quote is about the death of the recording industry, but it's really about much more than that.
I think records were just a little bubble through time and those who made a living from them for a while were lucky. There is no reason why anyone should have made so much money from selling records except that everything was right for this period of time. I always knew it would run out sooner or later. It couldn't last, and now it's running out. I don't particularly care that it is and like the way things are going. The record age was just a blip. It was a bit like if you had a source of whale blubber in the 1840s and it could be used as fuel. Before gas came along, if you traded in whale blubber, you were the richest man on Earth. Then gas came along and you'd be stuck with your whale blubber. Sorry mate – history's moving along. Recorded music equals whale blubber. Eventually, something else will replace it.
Be sure and check out the quote about Frank Zappa as well. | {
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Roger Valentin Iribarne Contreras (born 2 January 1996) is a Cuban athlete specialising in the high hurdles. He represented his country at the 2017 World Championships reaching the semifinals. In addition, he won the silver medal at the 2015 Pan American Junior Championships.
His personal best in the 110 metres hurdles is 13.39 seconds (-0.1 m/s) set in Havana in 2017.
International competitions
1Disqualified in the final
References
1996 births
Living people
Cuban male hurdlers
World Athletics Championships athletes for Cuba
Central American and Caribbean Games bronze medalists for Cuba
Competitors at the 2018 Central American and Caribbean Games
Athletes (track and field) at the 2019 Pan American Games
Pan American Games competitors for Cuba
Central American and Caribbean Games medalists in athletics
21st-century Cuban people | {
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package com.appcelerator.titanium.desktop.documentation;
import org.eclipse.help.AbstractTocProvider;
import org.eclipse.help.IToc;
import org.eclipse.help.ITocContribution;
import com.appcelerator.titanium.core.documentation.SDKListenerProxy;
import com.appcelerator.titanium.core.mobile.SDKLocator;
import com.appcelerator.titanium.core.mobile.SDKManager;
import com.appcelerator.titanium.desktop.DesktopPlugin;
import com.appcelerator.titanium.desktop.TitaniumDesktopSDKLocator;
/**
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*/
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return new ITocContribution[] { contribution };
}
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* Update TOC when content changes
*/
public void sdkChanged(Class<? extends SDKLocator> clazz)
{
super.contentChanged();
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}
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{"url":"https:\/\/planetmath.org\/montecarlomethods","text":"# Monte Carlo methods\n\nMonte Carlo methods are the systematic use of samples of random numbers in order to estimate parameters of an unknown distribution by statistical simulation. Methods based on this principle of random sampling are indicated in cases where the dimensionality and\/or complexity of a problem make straightforward numerical solutions impossible or impractical. The method is ideally adapted to computers, its applications are varied and many, its main drawbacks are potentially slow convergence (large variances of the results), and often the difficulty of estimating the statistical error (variance) of the result.\n\nMonte Carlo problems can be formulated as integration of a function $f=(\\vec{x})$ over a (multi-dimensional) volume $V$, with the result\n\n $\\int_{V}fdV=V\\overline{f}$\n\nwhere $\\overline{f}$ , the average of $f$, is obtained by exploring randomly the volume $V$.\n\nMost easily one conceives a simple (and inefficient) hit-and-miss Monte Carlo: assume, for example, a three-dimensional volume $V$ to be bounded by surfaces difficult to intersect and describe analytically; on the other hand, given a point $(x,y,z)\\in V$, it is easy to decide whether it is inside or outside the boundary. In this case, a simply bounded volume which fully includes $V$ can be sampled uniformly (the components $x$,$y$,$z$ are generated as random numbers with uniform probability density function), and for each point a weight is computed, which is zero if the point is outside $V$, 1 otherwise. After $N$ random numbers, $n$ ($\\leq N$) will have been found inside $V$, and the ratio $n\/N$ is the fraction of the sampled volume which corresponds to $V$.\n\nAnother method, crude Monte Carlo, may be used for integration: assume now the volume $V$ is bounded by two functions $z(x,y)$ and $z^{\\prime}(x,y)$, both not integrable, but known for any $x$,$y$, over an interval $\\Delta x$ and $\\Delta y$ . Taking random pairs $(x,y)$, evaluating $\\Delta z=|z(x,y)=z^{\\prime}(x,y)|$ at each point, averaging to $\\left<\\Delta z\\right>$ and forming $\\Delta x\\Delta y\\left<\\Delta z\\right>$ , gives an approximation of the volume (in this example, sampling the area with quasirandom numbers or, better, using standard numerical integration methods will lead to more precise results).\n\nOften, the function to be sampled is, in fact, a probability density function , e.g. a matrix element in phase space. In the frequent case that regions of small values of the probability density function dominate, unacceptably many points will have to be generated by crude Monte Carlo, in other words, the convergence of the result to small statistical errors will be slow. Variance reducing techniques will then be indicated, like importance sampling or stratified sampling. For more reading, see [Press95], [Hammersley64], [Kalos86].\n\nReferences\n\n\u2022 Originally from The Data Analysis Briefbook (http:\/\/rkb.home.cern.ch\/rkb\/titleA.htmlhttp:\/\/rkb.home.cern.ch\/rkb\/titleA.html)\n\n\u2022 Press95\n\nW.H. Press, S.A. Teukolsky, W.T. Vetterling, and B.P. Flannery, Numerical Recipes in C, Second edition, Cambridge University Press, 1995. (The same book exists for the Fortran language). There is also an Internet version which you can work from.\n\n\u2022 Hammersley64\n\nJ.M. Hammersley and D.C. Handscomb, Monte Carlo Methods, Methuen, London, 1964.\n\n\u2022 Kalos86\n\nM.H. Kalos and P.A. Whitlock, Monte Carlo Methods, Wiley, New York, 1986.\n\nTitle Monte Carlo methods MonteCarloMethods 2013-03-22 12:07:10 2013-03-22 12:07:10 akrowne (2) akrowne (2) 7 akrowne (2) Topic msc 65C05 msc 11K45 Monte Carlo hit-and-miss Monte Carlo","date":"2018-11-18 18:47:43","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 30, \"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\": 0.9095361828804016, \"perplexity\": 753.9642606689215}, \"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-2018-47\/segments\/1542039744561.78\/warc\/CC-MAIN-20181118180446-20181118202446-00399.warc.gz\"}"} | null | null |
Are your courses open for all students from all countries?
Do you have accommodation for participants?
What recreational facilities do you have in your campus?
Do you have internet facility open for participants?
Do I need to bring my Laptop with me?
What kind of electric sockets do you have in your Center?
What type of uniform do I need to bring with me if I belong to a military organization?
What type of uniform/dress do I need to bring with me if I am civilian?
What could you tell me about weather in Jordan?
What is the dead line for applying a course?
What is the maximum number for participants per irritation per course?
When reaching the maximum number for a certain course, would you accept extra participants?
What is your courses calendar for the year 2015-2016?
Can you hold another course out of your calendar (POTC Courses)?
Am I responsible for my Visa arrangements?
Which Airport you recommend me to be in upon arrival Jordan?
Do you provide transportation from / to Airport?
Do you have a dining facility?
Do you provide a special food diets?
What are the working days in Jordan?
What are the working hours in your Center?
What are the breakfast/lunch/dinner timings?
Do I need to pay for meals, tea breaks? How much in detail?
Are USA dollars accepted in Jordan?
How many Jordanian Dinar/s in one USA Dollar?
What places do you recommend me to visit in Jordan in weekends?
Do you offer e-learning courses?
Can I use your Center as a venue to hold a certain course? | {
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The John Howard Society of Canada > About Us > Our Role
Respect for, and confidence in, the criminal justice system is directly related to social justice policy that is effective, just and humane. To encourage the development of these characteristics, we have a continuing presence and involvement in the field of reform and policy development. The John Howard Society is interested in the relationship between correctional policy, criminal justice policy and other areas of social policy generally. We are aware of the fact that policy initiatives in other areas can have profound effects on the criminal justice system. We are concerned, therefore, that policy-makers consider the implications for criminal justice, and in particular, corrections, when they revise or draft a policy. The Society has a long history of commitment, involvement and knowledge in this field and can keep policy-makers in touch with the practical implications of policy. We can bring to their work our broad-based experience and knowledge in community-based corrections and justice.
The John Howard Society has actively participated in consultative forums of government and made submissions to legislative committees with respect to sentencing, conditional release, crime prevention, restorative justice, and young offenders. The John Howard Society of Canada's advice and expertise is regularly sought out by a variety of government, voluntary and professional organizations.
The Society supports the provincial offices and local branches in their community education efforts through the publication and distribution of information, support for community forums and the facilitation of discussion throughout the Society as a whole on relevant issues.
A significant aspect of the society's education mandate at both the national and provincial levels, is research and public policy development on justice issues.
The national organization has accumulated a library of peer-reviewed best evidence on correctional and criminal justice policy, including primary research, Senate and House of Commons submissions, and links to various related resources. Some of the larger provincial societies are also significantly involved in a broad range of research activity designated to promote and enhance the society's mandate. The Ontario website, for example, has a large database of information on current criminal justice policy issues.
The John Howard Society of Canada is responsible for the administration of the national grant which is received by the Society from the Public Safety Canada. This includes the receipt and distribution of funds to the qualifying provincial societies as well as the collection and compilation of all information and documentation required for the submission of the application for the grant.
The John Howard Society of Canada provides administrative support and advice to provincial offices and local branches as required.
The John Howard Society of Canada successfully negotiated a comprehensive national insurance policy which provides optional coverage to John Howard Societies which meet minimum standard requirements and which comply with National Policies which have been developed and approved by the membership. These include: "National Risk Management Policies", and "Preparation for and Responses to Serious Incidents".
The John Howard Society of Canada holds the registered patent for the name "John Howard Society". | {
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\section{Introduction}
\label{sec:intro}
It is still debatable whether common cause explanations are appropriate to account for EPR correlations. Addressing this issue is of both interesting and relevant from the philosophical point of view. Philosophers wanting to understand causation, for instance, need to face the adequacy of the diverse methods of causal inference available. Reichenbach's Principle of the Common Cause (RPCC) is one such method, and it is precisely in the quantum mechanical context that RPCC faces one of its most significant threats.
This paper aims to provide an analysis that recalls our most robust intuitions about causation (assumed to be well reflected in RPCC), with the ultimate purpose of testing them in the context of quantum mechanics. I will not discuss the specific problems related to the philosophical status of RPCC ---these are diverse and interesting in themselves even if quantum mechanics is not brought into the picture. RPCC will simply be assumed to hold.
The received view takes it that (Reichenbachian) common cause accounts of EPR correlations are to be ruled out.\footnote{See for instance \cite{vanfraassen1982a}, \cite{cartwright1987} or \cite{butterfield1989}. \cite{grasshoff2005} and \cite{butterfield2007} provide updated discussions on the matter.} The standard argument views causes as hidden variables onto which several constraints are set, intended to reflect standard requirements typical of any physical system, including temporal order of causal relations and considerations about locality. As a result, some version of Bell's \emph{factorizability} ---and therefore of a Bell-type inequality--- is derived. The strength of such arguments relies, as it stands, on the plausibility of the conditions imposed on the common causes.
\emph{No-conspiracy} is one such condition present, even if implicitly, in the usual derivations of the Bell inequalities. The motivation behind it is that the postulated common causes be independent of the measurement settings. Violations of such independence are standardly interpreted as to entail certain strange ``conspiratorial behaviour'', unless backwards in time causation is brought into the picture.\footnote{Cf.~\cite{price1994}, \cite{berkovitz2002} and \cite{szabo2000,szabo2008}.}
I shall challenge this standard reading of no-conspiracy-type conditions. I will do this in two steps. I will first look at the very formal structure of such conditions, while remaining neutral as to whether their violation entail any kind of conspiracy. I will then look at the EPR correlations from a purely phenomenological point of view and assess the role of measurement in the EPR experiment. This will suggest a re-interpretation of no-conspiracy-type conditions, under which violations of these can be accommodated without any conspiratorial implications, neither backwards in time causation. This new view will also provide grounds for a common cause model. It will not be hidden-variable model for the EPR correlations as such, though. Instead, the model responds to a purely explanatory motivation, i.e.~it constitutes what I take to be a plausible causal explanation of the EPR correlations.
The structure of the paper is as follows. Sections~\ref{sec:epr} and~\ref{sec:rpcc} provide the necessary background on the EPR experiment and Reichenbach's Principle of the Common Cause respectively. In Section~\ref{sec:structure} the formal structure of the problem is stated and the idea of \emph{no-conspiracy} introduced. Section~\ref{sec:phenomenological} looks at the EPR experiment again but from a purely phenomenological point of view this time, and examines the role of measurement in it. This analysis will eventually lead to the common cause model of EPR, outlined in Section~\ref{sec:themodel}. The paper closes with a brief discussion on the implications of the model as regards locality and states some open questions for further investigation.
\section{The EPR Correlations}
\label{sec:epr}
Einstein, Podolsky and Rosen first introduced the so-called EPR thought experiment in 1935\footnote{\cite{epr1935}.} as an argument to suggest that quantum theory did not provide a complete description of reality. In a later refined version presented by David Bohm, two \emph{entangled} electrons are emitted from a source in opposite directions. The spin component of each of the electrons can be later detected (measured) in three different directions $\theta_i ~(i=1,2,3)$ after having passed through an inhomogeneous magnetic field.
Let us first denote $L_{i}$ and $R_{j}~ (i,j = 1,2,3)$ the different measurement settings in each wing of the experiment. Also, I shall denote by $a$ and $b$ the value of the spin variable of each electron in a given measurement direction which, in the singlet state, can be either `spin-up' ($+$) or `spin-down' ($-$) with probability $\frac{1}{2}$. We can then express the corresponding measurement outcome events on each particle along each of the three different measurement directions as $L^{a}_i$ and $R^{b}_{j}$, with $a,b = +,-$ and $i,j = 1,2,3$. It is assumed that the state of the entangled electron pair is the spin \emph{singlet} state:
\[
\Psi_s = \frac{1}{\sqrt{2}} (\psi^{+}_{L} \otimes \psi^{-}_{R} - \psi^{-}_{L} \otimes \psi^{+}_{R})
\]
Furthermore, it is assumed that measurement events at each wing of the experiment, $L^{a}_{i}$ and $R_{j}^{b}$, are space-like separated events. This is represented in Figure~\ref{fig:epr} by noting that no time-like world-line can reach from $R_{j}^{b}$ to $L^{a}_{i}$ or \emph{vice versa}. Under a conventional albeit controversial interpretation of special relativity, such events can not be causally connected.\footnote{See~\cite{maudlin1994} for a critical discussion.}
\begin{figure}
\begin{center}
\includegraphics[scale=.54]{epr_lightcones.eps}
\caption{Space-time representation of a typical EPR experiment.}\label{fig:epr}
\end{center}
\end{figure}
Quantum mechanics allows us to calculate single and joint conditional probabilities for the different possible outcomes in both wings. When those calculations are performed on the entangled pair in the singlet state, correlations between these outcomes are derived:
\[
p(L^{a}_{i} \land R_{j}^{b}) \neq p(L^{a}_{i}) \cdot p(R_{j}^{b}),
\]
where $a,b = +,-$ and $i,j = 1,2,3$.
These are the EPR correlations, which have been often positively tested in experiment, and for which we would like to know whether they are the result of underlying causal processes. At this point already, it seems rather intuitively appealing to attempt at an answer to this question by recalling the idea of common cause. For the space-like separation of the outcomes at each wing of the experiment seems to rule out, at least in principle, direct causal interactions between these. And this is in fact the typical situation common cause explanations are usually suitable for.
\section{Reichenbach's Principle of the Common Cause}
\label{sec:rpcc}
The principle of the common cause was first introduced by \citet{reichenbach1956}. It states, in short, that there are \emph{no correlations without causal explanation} (either in terms of direct causal interactions or by means of a common cause).
The idea of correlation is defined within the framework of classical Kolmogorovian probability spaces:\footnote{The definition here is of positive correlation. A completely symmetrical definition may be given for negative correlations. Distinguishing between positive and negative correlations will not be important for the argument here. Thus, if not stated otherwise, positive correlations will be assumed throughout the paper.}
\begin{defn}
\label{defn:corr}
Let $(\mathcal{S},p)$ be a classical probability measure space with Boolean algebra $\mathcal{S}$ representing the set of random events and with the probability measure $p$ defined on $\mathcal{S}$. If $A,~ B \in \mathcal{S}$ are such that
\begin{equation}
\label{eqn:def_corr}
Corr_{p}(A,B) = p(A \land B) - p(A) \cdot p(B) > 0,
\end{equation}
then the event types $A$ and $B$ are said to be (positively) correlated.
\end{defn}
The principle can then be formalised, following \citeauthor{reichenbach1956}'s own ideas, as follows:
\begin{defn}[RPCC]
For any two (positively) correlated event types $A$ and $B$ ($Corr_p(A, B) > 0$), if $A$ is not a cause of $B$ and neither $B$ is a cause of $A$, there exists a common cause $C$ of $A$ and $B$ such that the following independent conditions hold:
\begin{align}
\label{eq:rcc1}
p(A \land B \vert C) & = p(A \vert C) \cdot p(B \vert C) \\
\label{eq:rcc2}
p(A \land B \vert \neg C) & = p(A \vert \neg C) \cdot p(B \vert
\neg C) \\
\label{eq:rcc3}
p(A \vert C) & > p(A \vert \neg C) \\
\label{eq:rcc4}
p(B \vert C) & > p(B \vert \neg C)
\end{align}
\noindent where $p(A \vert B) = p(A \land B)/p(B)$ denotes the probability of $A$ conditional on $B$ and it is assumed that none of the probabilities $p(X)~ (X = A, B, C, \neg C)$ is equal to zero.
\end{defn}
The definition above incorporates two quite distinct claims. On the one hand, one may identify a metaphysical existential claim suggesting that `common causes' exist (or are to be provided) for correlations with no direct causal explanation. As such I shall refer to it as the \emph{Postulate of the Common Cause} (PosCC).\footnote{This terminology is borrowed from \cite{sanpedro-suarez2009}.} This claim \emph{per se} says nothing about how common causes are to be characterised.
This is precisely the role of the second claim in the definition, which provides probabilistic relations~\eqref{eq:rcc1}-\eqref{eq:rcc4} for the characterisation of the common causes. Following \citet{suarez2007}, I shall refer to expressions~\eqref{eq:rcc1}-\eqref{eq:rcc4} as \emph{Reichenbach's Criterion for Common Causes} (CritCC).
The two last expressions of CritCC are just statistical relevance relations aimed to represent the causal dependence between $A$ and $C$ on the one hand, and $B$ and $C$ on the other. Whether statistical relevance does indeed reflect causal dependence is of course a debatable issue but I will assume so here, as it is standard in most theories of probabilistic causation. As for the first two probabilistic conditions, they express a new restriction on the postulated common cause $C$. They require, specifically, that if the presence (or the absence) of the common cause is taken into account the correlated events $A$ and $B$ are rendered probabilistically independent. The common cause $C$ is then said to \emph{screen-off} the correlation $Corr (A, B)$.
In contrast to the metaphysical character of PosCC, Reichenbach's criterion (CritCC) is a purely methodological claim. However, properly distinguishing the two claims is crucial for the assessment of the status of RPCC as a whole. We do not need to address here the various problems associated to the status of either PosCC or CritCC ---they each have indeed their own problematic issues. I shall just assume that the PosCC holds, i.e.~that common causes may be provided for any given correlation, and that CritCC is the right characterisation of such common cause events. I shall call such common causes irrespectively \emph{Reichenbachian common causes}, \emph{screening-off common causes} or simply \emph{common causes}.
Despite the controversies, endorsing RPCC may be motivated by at least two reasons. First, note that for Reichenbach the role of the principle as a whole, and of the screening-off condition in particular, is mainly explanatory.\footnote{Cf.~\cite[p.~159]{reichenbach1956}.} The explanatory power of screening-off common causes may thus be taken as a good methodological reason to support the adequacy of CritCC for the inference of causal relations from probabilistic facts, even if it can be patently shown no to hold as a necessary nor as a sufficient condition for common causes.
Second, recent results show that, at least formally, it is always possible to provide a Reichenbachian common cause for any given correlation.\footnote{Cf.~\cite{hofer-szabo1999,hofer-szabo2000a}.} These results build on the intuition that any probability space $\mathcal S$ containing a set of correlations and which does not include (Reichenbachian) common causes of these, may be extended in such a way that the new probability space $\mathcal{S}'$ does include (Reichenbachian) common causes for each of the original correlations. This is formalised in so-called \emph{extendibility} and \emph{common cause completability} theorems.
Thus (Reichenbachian) \emph{common cause completability} constitutes a very powerful tool if we are to provide common cause explanations of generic correlations. Common cause completability however faces its own problems, especially when it comes to the physical interpretation of either the enlarged probability space $\mathcal{S}'$ or the new common causes contained in it. In particular, it seems a fair criticism to the program to claim that common cause completability is a merely a formal device, which is likely to lack physical meaning in many (perhaps too many) cases. I shall not discuss such matters here but just point out that such criticisms may successfully be dealt with.\footnote{I point the reader to \cite{sanpedro-suarez2009} for a recent assessment the significance of common cause completability, possible criticisms to it and possible strategies to avoid these.} I shall assume here that common cause completability is in fact a reliable and powerful tool when aiming at common cause explanations.
\begin{figure}
\begin{center}
\includegraphics[scale=.6]{individual-common_cc.eps}
\caption{\emph{Individual}-common causes versus \emph{common}-common causes.}\label{fig:icc_ccc}
\end{center}
\end{figure}
There is an important further remark as regards \emph{common cause completability}. It has to do with the distinction between what I'll be referring as \emph{individual}-common causes on the one hand, and \emph{common}-common causes on the other. The fundamental difference between these is that \emph{individual}-common causes are defined to screen-off a single (individual) correlation \emph{only}, while \emph{common}-common causes screen-off a set of two or more correlations (Figure~\ref{fig:icc_ccc}). This is indeed an important distinction precisely due to the fact that, while \emph{individual}-common cause completability holds in general for every classical probability space $\mathcal S$, this is not the case for \emph{common}-common causes. In other words, while \emph{common cause completability} guarantees that \emph{individual}-common causes may be provided (at least formally) for any given correlation, this is not true in general for \emph{common}-common causes.\footnote{Cf.~\cite{hofer-szabo2002}.} In what follows and if not stated otherwise, when common cause is written it will mean \emph{individual}-common cause.
\section{Structure of the Problem}
\label{sec:structure}
We now seem to be in a position then to attempt at an answer to the question as to whether an explanation of the EPR correlations can be given in terms of common causes. However we should note first that the description of the EPR experiment in Section~\ref{sec:epr} involves statements about `quantum probabilities', while probabilities in Section~\ref{sec:rpcc} are classical. This very fact renders the whole issue about common cause explanations of EPR correlations hardly sensible. For `quantum probabilities' do not express probabilities of real events --they cannot be interpreted as relative frequencies, for instance-- and do not therefore fit in the formulation of classical Kolmogorovian probabilities, as it should be the case for Reichenbach's Criterion for Common Causes (CritCC) to make sense.\footnote{Cf.~\cite[p.4]{szabo2000}. The fact that I refer here specifically to relative frequencies is motivated by \citet{szabo2000}'s remarks, and is not crucial for the soundness of the foregoing argument. Indeed, I believe that the arguments provided here do not hinge on a particular interpretation of probability.}
We then seem to have to options. We may want to redefine CritCC in terms of `quantum probabilities', or we may alternatively interpret `quantum probabilities' so that they fit the formulation of classical Kolmogorovian probabilities.
I will take the later option here and interpret quantum mechanical probabilities, i.e~trace-like quantities $Tr (\hat{W} \hat{A})$, as \emph{classical conditional} probabilities ---conditional on the measurement operations, that is.\footnote{The difference between `measurement operations' and `measurement setting operations' (or `measurement settings' for short) will not play any major role for the purposes of the discussion here. I shall thus use those two expressions equivalently.} In other words, the `quantum probability' that the particle in the left wing of the experiment, for instance, is measured with spin-up will be given by $p(L_i^+ \vert L_i)$, i.e.~$p(L_i^+) = p(L_i^+ \vert L_i) \cdot p(L_i)$.
Within this framework ``Can a common cause explanation be provided for EPR correlations?'' is indeed a sensible question. And considering \emph{common cause completability}, the answer to it would in principle seem quite straightforwardly a positive one. The issue is not so simple, though. For we cannot overlook the fact that if our postulated common causes are to be physically sensible at all, they will need to fulfil certain requirements ---besides those in their definition, i.e.~equations~\eqref{eq:rcc1}-\eqref{eq:rcc4}. The problem becomes thus more complex since we first need to identify and characterise these extra requirements to which (physically sensible) Reichenbachian common causes must conform.
Conditions of this sort typically include those intended to capture in some sense or another the idea of physical locality ---so as to avoid conflict with special relativity. Bell's \emph{factorizability}\footnote{Cf.~\cite{bell1964}.} is perhaps the most famous and influential of these conditions:
\begin{equation}
\label{eq:factorizabilty}
p(L_i^a \land R_j^b \vert L_i \land R_j \land C_{ij}^{ab}) = p(L_i^a \vert L_i \land C_{ij}^{ab}) \cdot p(R_j^b \vert R_j \land C_{ij}^{ab}).
\end{equation}
Consistently with the notation in Section~\ref{sec:epr}, $C_{ij}^{ab}$ designs here a (Reichenbachian) common cause of the correlation $Corr(L_i^a, R_j^b)$ between two generic outcomes of an EPR-Bohm experiment.
Requiring \emph{factorizability} leads, of course, to the Bell inequalities, which are well known to be violated by experiment. Whether Bell's \emph{factorizability} indeed reflects physical locality is not completely settled. I shall not look at such issues in detail here and just point the reader to \cite{fine1981, fine1986}, \cite{wessels1985}, \cite{maudlin1994} or \cite{butterfield2007} for discussions of such issues. In any case, if we are to provide a Reichenbachian common cause explanation of the correlations while avoiding the charge of Bell's theorem, sensible constraints other than \emph{factorizability} are to be required for the postulated common causes.
The strategy is then to define a set of conditions that altogether does not amount to \emph{factorizability} which, along with the idea of Reichenbachian common cause, provides a satisfactory explanation of the correlations. Ideally we would like to make sure that such restrictions do not entail the idea of \emph{common}-common cause either. For, as I have pointed out it is not likely that \emph{common}-common cause explanations may be provided for any set of correlations in general.
One of the most remarkable attempts to provide common cause model for EPR following this strategy is \citet{szabo2000}. \citeauthor{szabo2000}'s model is explicitly construed by recursively applying the ideas of \emph{extensibility} and \emph{common cause completability} but it turns out that it features unwanted dependencies between certain combinations of the postulated common causes and the measurement setting events. These dependencies are usually interpreted as conspiratorial.
Avoiding such conspiratorial features is indeed the role of so-called \emph{no-conspiracy} conditions which are, in one way or another, standard in the derivations of the Bell inequalities, along with other restrictions ---intended to reflect locality, etc.--- on the common causes. Roughly, \emph{no-conspiracy} conditions require statistical independence between the common causes and the measurement settings:
\begin{align}
p(L_i \land C) & = p(L_i) \cdot p(C),\label{eq:simple-no-cons1} \\
p(R_j \land C) & = p(R_j) \cdot p(C).\label{eq:simple-no-cons2}
\end{align}
The usual interpretation being that the postulated common causes must not have an influence, or be otherwise influenced, by the choices of measurement settings. Existence of such influences ---i.e.~violations of the equations~\eqref{eq:simple-no-cons1} and~\eqref{eq:simple-no-cons2}--- would be conspiratorial in the sense that the taking place of the common cause would somehow `force' or determine to some extent the presumably free independent decisions of the experimenter about measurement.
Equations~\eqref{eq:simple-no-cons1} and~\eqref{eq:simple-no-cons2} may be regarded as `simple' \emph{no-conspiracy} conditions, in that they involve only one common cause event at a time. More complex conditions may involve combinations (both conjunctions and disjunctions) of the different postulated \emph{individual}-common causes, e.g.~$p(L_i \land C_{ij}^{ab} \land C_{i'j'}^{a'b'})$.\footnote{These later complex no-conspiracy conditions are what \citet{szabo2000}'s model fails to satisfy}. We need not discuss the later case here. For I will be arguing that violations of the `simple' case, i.e. of equations~\eqref{eq:simple-no-cons1} and~\eqref{eq:simple-no-cons2}, are consistent with a plausible common cause explanation of EPR correlations. But, of course, violations of the `simple' no-conspiracy conditions already entail violations of the more complex relations.
The significance of \emph{no-conspiracy} assumptions relies heavily on another underlying implicit assumption. This is the assumption that the postulated common causes take place (or happen) \emph{before} the apparatus are set for measurement, and therefore also \emph{before} (always in the rest-frame of the laboratory) measurement is performed. And it is only by assuming this specific time-order framework that one can make sense of the standard interpretation involving claims about world conspiracies and free will (or the lack of it).\footnote{In fact, I think this is why discussions about free will and backwards in time causation are so much entangled. For those defending backwards causation still assume the (timely) priority of common causes in relation to measurement operations. (Only, in those cases, time order and causal order are not assumed to coincide.) See, for instance, \cite{price1994} or \cite{berkovitz2002}. More generally, whether \emph{no-conspiracy}-type assumptions capture any intuition related to free will is a subtle issue which would deserve a whole paper on its own.}
This is a crucial assumption but I think it is not completely warranted. Indeed, I shall challenge it and suggest therefore that violations of~\eqref{eq:simple-no-cons1} and~\eqref{eq:simple-no-cons2} do not necessarily entail that there being world conspiracies. In order to keep neutral as regards the possible conspiratorial implications of the violation of~\eqref{eq:simple-no-cons1} and~\eqref{eq:simple-no-cons2}, I shall refer to them as \emph{measurement independence} conditions.
\section{EPR from a Phenomenological Point of View}
\label{sec:phenomenological}
\subsection{Pure Phenomenological Data Correlations}
\label{ssec:phenomenological_data_correlations}
I pointed out in Section~\ref{sec:rpcc} that Reichenbach's Principle of the Common Cause (RPCC) was originally proposed to account for classical correlations (and common causes). In dealing with quantum correlations I suggested, following Szab\'o, that we should take quantum probabilities, i.e.~trace-like quantities, as classical conditional probabilities (conditional on the corresponding measurement operations, that is) so that the quantum formalism is adapted to the classical talk of RPCC. This led to a reformulation of the problem in which measurement operations were explicitly present in the corresponding probability relations.
I would now like to tackle the issue from a somehow different point of view, however. More specifically, I shall not look at the EPR correlations as `quantum correlations' ---i.e.~as correlations defined between conditional probabilities, as \citeauthor{szabo2000} suggested---, and then postulate the corresponding \emph{individual}-common causes. Instead, I shall look at the EPR correlations from a completely classical perspective first, postulate common causes for them, and then provide these with an appropriate quantum mechanical interpretation. To illustrate this let us consider a simple example:
\label{flashing_ex}Suppose that a good friend of us, who is well familiar with quantum theory, has designed a device which collects the information of an EPR experiment and `translates' it into light flashes. The device (Figure~\ref{fig:panels}) has two identical panels (\textsc{panel~A} and \textsc{panel~B}) with green and red lights in three rows. Each row, in particular, has a pair of lights, one green and the other red. The device is somehow connected to the EPR detectors. Spin measurement outcomes in the left wing of the EPR experiment then make the lights in \textsc{panel~A} flash. Say, for instance, that if the outcome $L_3^+$ has been measured, then the green light in the third row of \textsc{panel~A} flashes; or that if $L_1^-$ is obtained in the EPR experiment, the first row's red light in `\textsc{panel~A}' flashes. Similarly for the right wing outcomes. Say, for instance, that the outcome $R_1^-$ has occurred, then the red light of the first row in \textsc{panel~B} flashes. Moreover, the device is such that the causal connections between the EPR events and their corresponding flashing light events are known to be deterministic and work completely independently for each panel.\footnote{Say for instance that the mechanisms connecting each wing with its corresponding panel are completely reliable and operate independently.} It is clear then that the correlations between the two panels' flashing light events exactly mirror those of the EPR experiment.
\begin{figure}
\begin{center}
\includegraphics[scale=.44]{panels.eps}
\end{center}
\caption{A sequence of red ($r$) and green ($g$) flashing lights.}\label{fig:panels}
\end{figure}
Suppose, on the other hand, that we do not know anything about quantum EPR, nor do we know what is the origin of the correlations observed between the lights. We just observe the correlation. Say however that we do know about Reichenbach's Principle of the Common Cause (RPCC) and are then ready to postulate a common cause for each of the observed correlations (especially if we make use of \emph{common cause completability}). I shall stress that to us the whole issue is entirely classical, i.e.~classical correlations (between lights) are observed and classical Reichenbachian common causes are to be postulated.
Our good friend intends to use the information that we may provide about our postulated common causes and attempt at an explanation of the EPR correlations. Yet, she is aware of an extra issue that might be problematic, namely that unlike the correlations between the flashing lights observed by us, those in the EPR experiment are space-like separated. On reflection, however, she realizes that it should not make a difference. For she knows that each EPR outcome is the one and only cause of its corresponding flashing light event. (Say for instance that she knows the detailed causal structure that relates each of the outcome events in the EPR experiment and the corresponding flashing light event). She has then come to the conclusion that the postulated common causes for the flashing light correlations may very well be the same that the common causes for the corresponding EPR correlations.
Our friend thus sets for a common cause explanation of her EPR correlations in terms of our postulated common causes. Only, she will later need to give an appropriate interpretation of such common causes in order to accommodate the specific features of the EPR experiment, such as the space-like separation of the to wings, etc.
\subsection{Common Causes of EPR}
\label{ssec:commoncauses}
She may then ask what is the structure of the $\mathsf{C_{ij}^{ab}}$, or in what sense these common causes are associated to the quantum mechanical system, i.e.~to the singlet state.
The first thing to note is that our (classically) postulated common causes are Reichenbachian \emph{individual}-common causes. (This is so by construction). This is to say, they are common causes which screen-off one, and only one, of the correlations. Consequently, the postulated common causes have a label which identifies the specific correlation (between flashing lights in our case, or the corresponding EPR outcomes in her case).
Let us denote such a common cause, for instance, $\mathsf{C_{ij}^{ab}}$.\footnote{I am following here basically the same notation to that in the rest of the paper. I am just denoting the common cause of the classical correlations with a \textsf{\small{serif}}~font~$\mathsf{C_{ij}^{ab}}$ instead of the usual \textit{italic}~$C_{ij}^{ab}$ in order to stress the fact that they are indeed different events.} We will then have:
\begin{align}
p(L_i^a \land R_j^b \vert \mathsf{C_{ij}^{ab}}) & = p(L_i^a \vert \mathsf{C_{ij}^{ab}}) \cdot p(R_j^b \vert \mathsf{C_{ij}^{ab}}), \label{eq:model_RPCC_01}\\
p(L_i^a \land R_j^b \vert \lnot \mathsf{C_{ij}^{ab}}) & = p(L_i^a \vert \lnot \mathsf{C_{ij}^{ab}}) \cdot p(R_j^b \vert \lnot \mathsf{C_{ij}^{ab}}), \label{eq:model_RPCC_02}\\
p(L_i^a \vert \mathsf{C_{ij}^{ab}}) & > p(L_i^a \vert \lnot \mathsf{C_{ij}^{ab}}),\label{eq:model_RPCC_03}\\
p(R_j^b \vert \mathsf{C_{ij}^{ab}}) & > p(R_j^b \vert \lnot \mathsf{C_{ij}^{ab}}).\label{eq:model_RPCC_04}
\end{align}
Note as well that the postulated common causes are not necessarily deterministic, i.e.~their occurrence does not necessarily entail that the corresponding outcomes will occur with certainty.
On the other hand, our friend may recall that the common causes were originally postulated for classical correlations (between the different light flashes) and that she were led to think that the same common cause would explain both the classical correlations and the corresponding quantum ones. These observations ---perhaps the latter in particular--- seem to suggest that the $\mathsf{C_{ij}^{ab}}$ are events that somehow \emph{contain} or \emph{include} the measurement operations. In other words, \emph{measurement operations are causally relevant to the common causes}. This is indeed a remarkable feature of the postulated common causes, although far from controversial. Causal relevance of measurement may be motivated through the following example.
Imagine the following fictitious situation. Suppose we have some identically looking spherical objects the nature of whose we would like to understand better. In order to do so we chose to measure something we call `elasticity'. Suppose for the sake of the argument that `elasticity' is a property that all objects have and which can only take on one of two possible values when measured, either `0' or `1'. Thus, when it comes to `elasticity' we can only find `0-elasticity' objects and `1-elasticity' objects.
We want then to find out whether our spheres have `0-elasticity' or `1-elasticity' so we set up a simple experiment to this end. As it happens, though, `elasticity' is a property which is not directly observable and needs to be inferred from other experimental facts. Thus, our experiment consists in hitting the spheres, one at a time, with a hammer ---with the same hammer and with the same strength each time. The recorded events are of two kinds: we measure the hammer's recoil after hitting the sphere as well as the sound produced. (We can suppose for simplicity that typically we will find the hammer recoiling in two modes, so to speak, a `large recoil' and a `short recoil'. Similarly, the sounds emitted when hitting the spheres with the hammer turn out to be of two kinds only, one `higher pitched' and the other `lower pitched'). Finally, suppose that the information compiled about recoil and sound modes is enough to tell whether our spheres have `0-elasticity' or `1-elasticity'. That is, if a `short' hammer recoil and a `high pitch' emitted sound are measured we can safely infer that the sphere has `0-elasticity'. Otherwise, the sphere has `1-elasticity'.
This example I think illustrates sufficiently how measurement may have a causal role in a causal explanation. In particular, our measurement operations ---the hammer hits--- are responsible, causally responsible that is, for the observed outcomes that allow us in turn to infer a property of the system at hand. This is not to say, of course, that the hammering is the cause of the sphere having a particular value of the property `elasticity', i.e.~having `0-`elasticity'' or `1-`elasticity''. But once more, what the hammering is causally relevant for is the particular sound and degree of recoil which allow us to make an inference regarding the `elasticity' of the spheres. We could perhaps say that our measurement operations \emph{reveal} what the value of the `elasticity' of the spheres is by \emph{causing} further events (related to that property).
Spin measurement experiments may be taken to some extent to be similar to `elasticity' measurements in the example above, at least empirically, i.e.~as far as empirical observations are concerned. Recall, for instance that in an EPR experiment spin components (values) of the electrons are \emph{detected} by letting the particles through an inhomogeneous magnetic field (a Stern-Gerlach magnet) and observing them flashing on the screen, either on its upper or lower side. The electron spin component is thus \emph{inferred} from the effect that the Stern-Gerlach magnet has on the particle. Thus, the magnetic field may be taken not to cause the electron to have a particular spin value but to \emph{reveal} such a spin value by causing a change of the electron trajectory (in either way).
Of course, this may be soon rejected by proponents of the orthodox interpretation of quantum mechanics on the grounds that no such property exists, really, before measurement. The orthodox's critique would surely undermine the identification I made above between the case of `elasticity' and spin to the extent that both are taken in each case to be real physical properties of the system. But it seems to me that it would leave untouched the claim that measurement settings may be taken to be causally relevant to the outcomes (whatever the underlying reality of the system is). Thus, we may still follow with such an idea in mind, regardless of the degree of realism we are committed to.
Note finally that measurement operations as considered above are not to be taken as the only relevant causal factors of the observed outcomes. In the classical case, for instance, outcomes do strongly depend on the property in question. In particular, whether the spheres have `0-`elasticity' or `1-`elasticity' seems clearly crucial for the fact that a particular sound and degree of recoil is observed. And such an influence, I would say, seems to be causal as well.
The case of spin measurements is different just for reasons similar to those concerning the orthodox critique above. For in such a view it seems difficult to maintain that a property, which is not taken to exist, is in any sense responsible for an outcome. However, there seem to be good intuitive grounds to say that if one takes it that there is some causal connection between the unmeasured quantum system and the measured outcomes, this will include some feature exclusive to the system itself. In the EPR context, in fact, it is very often assumed that the spin-singlet state itself ---or perhaps some factor closely related to it--- is causally relevant to the EPR outcomes.\footnote{This is indeed commonplace. See, for instance \cite{cartwright1987}, \cite{cartwright-jones1991}, \cite{cartwright-chang1993}, \cite{healey1992a}, \cite{vanfraassen1982a}, or more recently \cite{price1994} or \cite{suarez2007}.}
\section{A non-Factorizable Common Cause Mo\-del for EPR Correlations}
\label{sec:themodel}
The conclusions of the discussion in the previous section about measurement and the structure of the postulated common causes can be expressed formally, in terms of algebra events and probabilistic relations.
\subsection{\emph{Individual}-Common Causes and outcome independence}
\label{ssec:ccstructure}
Recall first that the common causes were each postulated to screen-off a single correlation ---these are what I called \emph{individual}-common causes. I expressed that by means of the screening-off conditions~\eqref{eq:model_RPCC_01}-\eqref{eq:model_RPCC_02}.
Now the role of measurement in the model can be explicitly accounted for if we think of the postulated \emph{individual}-common causes $\mathsf{C_{ij}^{ab}}$ as the result of a conjunction of the specific measurement operations $L_i$ and $R_j$ and some other causally relevant factor $\Lambda$. That is:
\begin{align}
\label{eq:c_in_lirilambda}
\mathsf{C_{ij}^{ab}} & \subset L_i \land R_j \land \Lambda.
\end{align}
Typically, $\Lambda$ will most probably be associated to the singlet state $\Psi_s$ itself, and perhaps to some other relevant causal factors prior to the preparation of the entangled system.\footnote{This characterisation of $\Lambda$ somehow reminds to the kind of events Cartwright considers the right common cause events for EPR. That the similarity is quite so it will become clear in a moment, since I will be requiring (or at least allowing) that the $\Lambda$ be non-screening-off events, just like in Cartwright's common cause account of EPR. See for instance~\cite{cartwright1987,cartwright-jones1991} and~\cite{cartwright-chang1993}.} In some sense $\Lambda$ seems much more deeply related to the `inner' quantum mechanical structure of the system than the postulated common causes $\mathsf{C_{ij}^{ab}}$. This observation is also supported by the fact that the $\mathsf{C_{ij}^{ab}}$ have been postulated in a purely classical context, such as that of the flashing light panels. Indeed, the model's common causes $\mathsf{C_{ij}^{ab}}$ are not to be thought of as hidden variables as such. For they are not aimed to complete the quantum description of the system in any way. Their role is much more explanatory than anything else.
We may assume furthermore that the causal factor $\Lambda$ is common to several (or even to all) correlations in the EPR experiment. It is important to note however that $\Lambda$, in contrast to the $\mathsf{C_{ij}^{ab}}$, will not in general screen-off the correlations. I shall explicitly require this in order to avoid problems concerning \emph{common}-common causes. That such problems may arise is clear, since if $\Lambda$ is a screening-off causal factor common to all the possible outcomes of the experiment, it surely is a \emph{common}-common cause of all the outcome correlations. And a \emph{common}-common cause model cannot in general be guaranteed to exist. Moreover, in the EPR context, assuming \emph{common}-common causes leads quite straightforwardly to the Bell inequalities.
Similar reasons lead us to require as well that the conjunction of $\Lambda$ with the measurement operation events, i.e.~$L_i \land R_j \land \Lambda$, be non-screening-off events. In particular, $L_i \land R_j \land \Lambda$ is assumed to be a causal factor common to all the correlations that involve these particular measurement settings.
This suggests that $L_i \land R_j \land \Lambda$ \emph{contains} all the postulated common causes $\mathsf{C_{ij}^{ab}}$ of the correlations displayed for these specific measurement settings. That is:
\begin{align}
L_i \land R_j \land \Lambda & \supseteq \mathsf{C_{ij}^{++}} \lor \mathsf{C_{ij}^{+-}} \lor \mathsf{C_{ij}^{-+}} \lor \mathsf{C_{ij}^{--}}.
\end{align}
One may however think that the above would entail that the $\mathsf{C_{ij}^{ab}}$ be \emph{common}-common causes. In order to avoid that and guarantee that $\mathsf{C_{ij}^{ab}}$ be \emph{individual}-common causes, we shall require that they be mutually exclusive, i.e.
\begin{align}
\mathsf{C_{ij}^{ab}} \land \mathsf{C_{ij}^{a'b'}} & = \emptyset,
\end{align}
\noindent where $a,b, a',b' = +, -$ and $ab \neq a'b'$.
The \emph{individual}-common cause event structure resulting from the above considerations can be seen in Figure~\ref{fig:model_structure}.
\begin{figure}
\begin{center}
\includegraphics[scale=.71]{model_structure.eps}
\end{center}
\caption{The \emph{individual}-common cause model event structure for EPR correlations with measurement settings $L_i$ and $R_j$.}\label{fig:model_structure}
\end{figure}
A remarkable feature of the resulting event structure is that our postulated common causes $\mathsf{C_{ij}^{ab}}$ satisfy a familiar condition very closely related to the derivation of the Bell inequalities, namely \emph{outcome independence} (OI):
\begin{equation}
p(L_i^a \land R_j^b \vert L_i \land R_j \land \mathsf{C_{ij}^{ab}} ) = p(L_i^a \vert
L_i \land R_j \land \mathsf{C_{ij}^{ab}} ) \cdot p(R_j^b \vert L_i \land R_j \land \mathsf{C_{ij}^{ab}} ),
\end{equation}
This is indeed what expressions~\eqref{eq:model_RPCC_01} and~\eqref{eq:model_RPCC_02} encapsulate. But note as well that, on the other hand, the event $\Lambda$ (and also $L_i \land R_j \land \Lambda$) will not in general satisfy such constraint. In particular, both $\Lambda$ and $L_i \land R_j \land \Lambda$ will in general violate \emph{outcome independence} due to the requirement that they shall not screen-off the correlations. (This was explicitly required in order to avoid the usual problems regarding \emph{common}-common causes.)
Thus the model is able to accommodate the usual interpretation of the violations of the Bell inequalities, i.e.~the violations of Bell's \emph{factorizability}. The standard view in this respect takes it that quantum mechanics violates \emph{outcome independence} ---while being compatible with so-called \emph{parameter independence}.\footnote{Such claims are normally supported by the fact that the spin-singlet state has a spherical symmetry.} Thus, the fact that the \emph{outcome independence} condition is violated when $\Lambda$ is taken as the hidden variable in our model is in complete agreement with this standard view.
On the other hand, since the postulated common causes $\mathsf{C_{ij}^{ab}}$ in the model are viewed as events which in principle are not part of the `inner' quantum description (or structure) of the system ---although they are of course clearly related to it via equation~\eqref{eq:c_in_lirilambda}---, it seems quite natural to think of them as satisfying \emph{outcome independence}.
\subsection{Common Cause Dependence is not Conspiracy}
\label{ssec:ccdependence}
A second remarkable feature of the model is that, because the dependence of the common causes on measurement, the $\mathsf{C_{ij}^{ab}}$ violate what I called \emph{measurement independence}. That is:
\begin{align}
\label{eq:model_no-cons1}
p(\mathsf{C_{ij}^{ab}} \land L_i) & \neq p(\mathsf{C_{ij}^{ab}}) \cdot p(L_i), \\
p(\mathsf{C_{ij}^{ab}} \land R_j) & \neq p(\mathsf{C_{ij}^{ab}}) \cdot p(R_j).\label{eq:model_no-cons2}
\end{align}
Recall that \emph{measurement independence} was a condition explicitly required ---usually under the name of \emph{no-conspiracy}--- on the grounds that its violation would amount to some sort of `universal conspiracy'. The question then seems quite straightforward: In the light of expressions~\eqref{eq:model_no-cons1} and~\eqref{eq:model_no-cons2}, is our model a `conspiratorial' model (in the sense above)? Alternatively, is \emph{measurement independence} a reasonable assumption for common causes of EPR correlations? The answer to both these questions, I think, is negative.
The charge of conspiracy may be avoided in different ways. Appealing to backwards in time causation is one of them. I shall briefly comment this option in the next section.
Another possibility opens up if we look closely at the role of measurement in the experiment. In fact, one of the things I pointed out when discussing the significance of \emph{measurement independence} as an assumption underlying EPR was that it seemed appropriate only if one assumed that the common causes take place prior to the measurement apparatus is set up. It is only in those cases that probabilistic dependencies between the common causes and measurement settings, i.e.~violations of \emph{measurement independence}, may be sensibly interpreted as some sort of `world conspiracy'. However, there is nothing in the notion of common cause, nor in the very structure of the EPR experiment, that forces us to stick to the idea that the common cause must take place prior to measurement (and therefore prior to the setting-up of the measurement devices). This was motivated through the flashing lights example in Section~\ref{flashing_ex}.
In fact, relations~\eqref{eq:rcc1}-\eqref{eq:rcc4} in page~\pageref{eq:rcc1} ---taken as a definition (Reichenbachian) common cause--- do not include spatio-temporal information of any sort. This is mainly due to the fact that they involve event types, which are not defined in space time. It is only in virtue of them being collections of token events that we can refer to them spatio-temporally. In this interpretation, the intuition is that common cause events in expressions~\eqref{eq:rcc1}-\eqref{eq:rcc4} are to be located in the causal past of the events which are correlated. In the particular case of the EPR experiment, we want to require that the postulated common causes and the corresponding outcome events be time-like (ensuring hence temporal priority of the common causes). But, once more, this needs not entail that such common causes be located prior to measurement operations.
So, we may perfectly allow instead that the common causes take place after the measurement devices have been set-up, and even after measurement operations have been performed. In such a case, requiring \emph{measurement independence} conditions to hold does not seem particularly appealing and, more importantly, there is nothing conspiratorial about them being violated. This is indeed the possibility the model above exploits.
\subsection{Parameter Dependence and non-factorizability}
\label{ssec:pdependence}
As a direct consequence of the violation of \emph{measurement independence}, the model turns out to be non-factorizable, i.e.~Bell's \emph{factorizability} is also violated. This soon becomes evident if we note that \emph{measurement independence} is necessary for a more general independence condition, namely \emph{parameter independence}, which is in turn necessary for \emph{factorizability}. In particular, a violation of \emph{measurement independence} entails, as we have seen in the previous section, that the common cause is not independent of \emph{both} the measurement settings. That is:
\begin{align*}
p(\mathsf{C_{ij}^{ab}} \land L_i) & \neq p(\mathsf{C_{ij}^{ab}}) \cdot p(L_i), \\
p(\mathsf{C_{ij}^{ab}} \land R_j) & \neq p(\mathsf{C_{ij}^{ab}}) \cdot p(R_j),
\end{align*}
\noindent which entails that
\begin{align*}
p(\mathsf{C_{ij}^{ab}} \land L_i \land R_j) & \neq p(\mathsf{C_{ij}^{ab}}) \cdot p(L_i \land R_j),
\end{align*}
\noindent as long as we assume $p(L_i)$ and $p(R_j)$ probabilistically independent.\footnote{This expression is nothing more that \emph{Hidden Locality} in \citet{vanfraassen1982a}.}
This in turn means that the outcomes will also statistically depend on \emph{both} measurement settings, since they obviously depend on the common cause, i.e.
\begin{align*}
p(L_i^a \vert L_i \land R_j \land \mathsf{C_{ij}^{ab}}) & \neq p(L_i^a \vert L_i \land \mathsf{C_{ij}^{ab}}), \\
p(R_j^b \vert L_i \land R_j \land \mathsf{C_{ij}^{ab}}) & \neq p(R_j^b \vert R_j \land \mathsf{C_{ij}^{ab}}).
\end{align*}
This expression is nothing more than the violation of \emph{parameter independence}. And since \emph{parameter independence} is necessary for Bell's \emph{factorizability}, the above just means that a failure of \emph{measurement independence} entails a violation of \emph{factorizability} as well.
The rejection of \emph{factorizability} has two obvious consequences, one which may be seen to support the model, the other which might be problematic for it. First, it is obvious that the Bell inequalities can not be derived for the model's event structure. This is certainly good, for it means that the model avoids the implications of Bell's theorem, and more importantly that it is perfectly compatible with the (empirically confirmed) quantum mechanical predictions for the EPR correlations.
On the other hand, the violation of Bell's \emph{factorizability} may be interpreted as a sign of non-local behaviour. I already pointed out that claims along these lines are far from being uncontroversial, especially if locality is merely associated with the requirement that there not be superluminal signalling between the two wings of the EPR experiment.
Other, more radical views retain locality intuitions by appealing, for instance, to backwards in time causation. Such views usually exploit the failure of Bell's \emph{factorizability} via violations of \emph{parameter independence} ---as opposed to standard view that quantum mechanics violates \emph{outcome independence}---, and in particular to violations of \emph{measurement independence} (just as in the model I just proposed) which are interpreted as implying backwards in time causation.
The possibility of backwards in time influences is not, by any means, new. This possibility was explored in a common cause model suggested by \citeauthor{price1994}.\footnote{Cf.~\cite{price1994}. Although the model was initially presented as a common cause model (containing backwards in time causal influences), \citeauthor{price1994} seemed later to retract from interpreting it as causal. In particular, \citet{price1996} seems to suggest that the backwards in time influences of the model be of no causal origin. An explicit causal interpretation of \citeauthor{price1994}'s model is given in \cite{suarez2007}. I shall assume, for the sake of the argument, that \citeauthor{price1994}'s model may indeed be interpreted causally.} Indeed \citeauthor{price1994}'s model needs to assume backwards in time influences in order for the postulated common causes to operate locally. In particular, \citeauthor{price1994} assumes that the common cause events take place in the overlap of the backward light-cones of the correlated EPR outcomes. Violations of \emph{measurement independence} therefore obtain due to the influences of the settings on the common causes operating backwards in time.
Note that the probabilistic event structure of \citeauthor{price1994}'s model and my own is exactly the same. Only, the interpretation of the events is different. The model I just proposed has what I think might be an advantage, even if only at the intuitive level, over that of \citeauthor{price1994}, namely that appeal to backwards in time causation is not needed. But recall that it is precisely backwards causation what allows \citeauthor{price1994}'s common causes to operate locally. This suggests that, since backwards in time causation is not a feature of our model, we might not be able to retain locality. This observation seems to be supported by the fact that our model has certain parallelisms with Bohm's quantum mechanics. They both violate, for instance, \emph{parameter independence}, while conform to \emph{outcome independence}. And, again, Bohm's quantum mechanics is explicitly non-local. So, is our model non-local as well?
A somehow qualitative and partial answer to this question might be advanced by stressing that in developing the model I have only paid attention to the significance and adequacy of \emph{measurement independence}. And recall that \emph{measurement independence} (or the equivalent \emph{no-conspiracy}) conditions are just some of the extra restrictions imposed on the idea of Reichenbachian Common Cause (RCC), alongside other further conditions, normally associated to locality requirements. What this means is that, by construction, our model is supposed to have retained some notion of locality within its event structure. So the question rather seems to be to what extent the model can be thought to be local (or equivalently non-local).
Locality issues are subtle. It is not clear, for instance, what notion of locality may best fit the (causal) structure in our model. Or whether the very concept of common cause would still stand if we are to retain local causal interactions. In addressing such matters a thorough assessment of the ontological implications of the model will quite likely be of help.
\addcontentsline{toc}{section}{Bibliography}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 474 |
Mac OS X users were panicky when an updated variant of one of the notorious Trojans started utilizing the chinks of an unpatched Java vulnerability in the OS. The vulnerability enabled the virus to hijack the system even when the user does not enter admin passwords.
The Trojan was exceptional in its character that it spread through some 600,000 Mac machines in a small span of time.
To help fight the Trojan, Apple has released two Java updates in a span of two days trying to do some firefighting for crisis management.
The zeal shown by Apple to roll out the updates is laudable, but at the same time, it should be noted that Apple has always came under fire from security experts for delaying Java updates frequently.
The delay and lethargy was the major reason that helped the Trojan spread through MAC OS X machines in blaze pace.
The first java update rolled out by the iGiant was named Java for OS X Lion 2012-001, solved the vulnerability in Java 1.6.0_29. However, there is no clue on how the second update, dubbed Java for OS X 2012-002, patches the weak spot in the system.
In the wake of recent virus attacks, Apple is planning to include a 'gatekeeper' feature in their upcoming OS X Mountain Lion version. The gatekeeper feature will verify the credibility of apps through the Mac App Store and through the identified developer program recently initiated by Apple for third party developers.
The program provides a certificate to third party developers who distribute their apps outside Mac App Store and tracks them to detect any malicious activity. | {
"redpajama_set_name": "RedPajamaC4"
} | 7,566 |
\section{Introduction}
Novae are cataclysmic variables caused by a white dwarfs in a binary systems that experience a thermonuclear runaway in a compact surface-layer of hydrogen that was accreted from the companion.
The outburst is detected as a strong sudden increase in optical brightness followed by a gradual decline over a time scale of months.
Nova light curves vary substantially from system to system \citep{Strope10}.
Some novae go through a Super Soft Source (SSS) phase, in which spectra are measured similar to those of Super Soft X-ray binary Sources \citep{Kahabka97}.
Since the advent of the X-ray grating observatories, Chandra and XMM-Newton, `high'-resolution X-ray spectra have been obtained for a number of novae in their SSS phase.
These spectra contain a wealth of detail which have provided evidence of continued mass loss during the SSS phase: blue-shifted absorption lines, the absence of sharp ionization edges, and P-Cygni line profiles \citep{Ness07}.
These data so far were interpreted with the help of hydrostatic atmosphere models \citep{Rauch10, Osborne11, Ness11}.
These models have proven to be a powerful tool for determining effective temperatures and chemical compositions for grating observations.
But in order to match observed blue-shifted absorption lines the synthetic spectra are blue-shifted in an ad-hoc fashion \citep{Nelson08, Rauch10, Ness11}, which is not physically consistent with having a velocity field in a radially extended (vs. a compact hydrostatic) atmosphere.
The effects of the velocity field on the physical conditions of the atmosphere (e.g. density and temperature stratification) can significantly change the spectrum and affect the interpretation of the observations as shown by \cite{Vanrossum10}.
\cite{Petz05} introduced expanding atmosphere models for the SSS phase calculated with the {\texttt {PHOENIX}}{} code \citep{Hauschildt92, Hauschildt99, Hauschildt04}.
Those models are similar to the {\texttt {PHOENIX}}{} models used to simulate early UV and optical nova spectra \citep{Hauschildt92} but applied to the very high temperature conditions of SSS.
Although these models naturally reproduce blue-shifted absorption lines, neither the lines nor the continuum of the synthetic spectra match the shape of the observations well.
In \cite{Vanrossum10} a wind-type (WT) expanding atmosphere model was introduced for X-ray novae using the same {\texttt {PHOENIX}}{} code.
Those models were shown to provide reasonable fits to two different SSS novae (RS Oph and V2491 Cyg), but their models were called preliminary and have since then neither been applied and examined, nor been available for the public to use for that purpose.
In this paper a large set of WT spectra is presented (and made publicly available), their properties explored, and their use demonstrated by application to the five V4743 Sgr 2003 grating observations.
V4743 Sgr 2003 was thoroughly\footnote{The abundances were tuned to individual observations.} analyzed with hydrostatic atmospheres in \cite{Rauch10}, so that the results presented here can be directly compared to previous work.
\section{The ``wind-type'' (WT) SSS Nova model atmosphere setup}
The WT model atmospheres are based on two observations.
Firstly, (at least) a part of the atmosphere is expanding.
Secondly, given the high temperatures and small effective radii typically determined for SSS spectra (e.g. \cite{Osborne11}, \cite{Ness11}, \cite{Vanrossum10}), (at least) a part of the atmosphere is very close to the compact surface of the white dwarf.
Based on these two observations the WT model atmospheres assume a spherically symmetric compound structure with a hydrostatic base and an expanding envelope.
The hydrostatic base is described with the classical parameters effective temperature $T_{\rm eff}$, gravitational acceleration $\log(g)$, and radius $R$, using the hydrostatic equilibrium condition
\begin{equation}
\frac{\mathrm{d} p}{\mathrm{d} r} = -\rho g \qquad {\rm for} \quad r\le R\;,
\end{equation}
where $r$ is the radial distance to the center of mass, $R$ the outer radius of the static core, $p$ is the gas pressure, and $\rho$ the density of the gas.
The envelope is parameterized by a constant mass-loss rate $\dot{M}$ and a beta-law velocity field (see e.g. \cite{Lamers97})
\begin{equation}
v(r>R) = v_0 + (v_\infty - v_0) \left(1 - \frac{R}{r}\right) ^\beta \;, \label{eq:EnvelopeV}
\end{equation}
with $v_\infty$ the asymptotic velocity, and $v_0$ the velocity at the base of the wind.
The parameter $\beta$ determines the length-scale in which the terminal velocity is reached.
The density of the envelope follows from the continuity equation for a constant mass loss rate
\begin{equation}
\rho(r>R) = \frac{\dot{M}}{4\pi r^2 v(r)} \;. \label{eq:EnvelopeRho}
\end{equation}
The base velocity $v_0$ is refined in the model to provide a smooth run of the density and its gradient over the transition boundary at $r = R$ (see \cite{Thesis09}).
This is not possible for any arbitrary value of $\beta$.
Experiment shows that $\beta = 1.2 \pm 0.05$ for the WT models to have a smooth transition between static core and expanding envelope\footnote{
Note that the model atmosphere, and thus it's spectrum, depends on the assumed velocity law. The effect of $\beta$ on the spectra is small within the allowed range. The effect of assuming a different type of velocity-law is potentially bigger. This is not explored within the scope of this work.}.
This is slightly higher than the value of $\beta = 0.9 \pm 0.1$ for radiation driven winds of blue giants (see e.g. \cite{Pauldrach86}).
In conclusion, the model atmosphere is defined by five parameters\footnotemark, the three classical hydrostatic parameters: $T_{\rm eff}$, $\log(g)$, $R$, and two parameters for the expanding envelope: $\dot{M}$, $v_\infty$.
\footnotetext{The chemical composition adds a large number of additional free parameters to the model, the discussion of which is deferred to next paper of this series.}
\section{Characteristics of the WT model} \label{sec:ModelCharacteristics}
With five parameters, the WT models have much more freedom than static atmospheres (which have three).
The effect of each of these parameters on the model atmosphere was examined in \cite{Thesis09}.
Here we demonstrate the effect on the observables, i.e. the spectra, focusing on four abstract properties:
\begin{enumerate} \setlength{\parskip}{0pt}
\item Luminosity (or brightness)
\item Color (or blueness)
\item Blue-shift of absorption features
\item Emission wing prominence in P Cyg profiles
\end{enumerate}
In the following, the details of the relations between model parameters and these observables is given, and an overview is shown in the Table \ref{tab:ParameterEffects}.
This table serves as a guide in the process of fitting WT spectra to data.
\begin{table}
\begin{center}
\caption{Overview on the effects of the five model parameters on four observables.} \label{tab:ParameterEffects}
\begin{tabular}{c|ccccc}
& $R$ & $T_{\rm eff}$ & $\log(g)$ & $v_\infty$ & $\dot{M}/v_\infty$ \\
\hline
Luminosity & $\Uparrow$ & $\uparrow$ & -- & -- & -- \\
Color & -- & $\Uparrow$ & $\downarrow$ & -- & $\uparrow$ \\
Blue-shift & -- & -- & -- & $\Uparrow$ & $\uparrow$ \\
Emission wings& -- & -- & -- & -- & $\Uparrow$ \\
\end{tabular}
\end{center}
{\bf Notes:}
{\footnotesize This table serves as a guide in the process of finding good spectral fits to data.
$\uparrow$ means a positive correlation, $\downarrow$ a negative correlation, and -- no correlation between the model parameter and the observable (abstract property of the spectrum).
The double arrows $\Uparrow$ denote the parameters that are most characteristic for each observable.\\
The parameter $\log(g)$ has no double arrows meaning that it is not well constrained by the observables.
The Color is affected by three parameters, which means that $T_{\rm eff}$ generally is not determined uniquely, but depends on the value for $\log(g)$.}
\end{table}
\paragraph{Radius}
The atmosphere is self-similar with variation of $R$, meaning that it does not affect the flux produced by the model.
Yet, the total luminosity
\begin{equation}
L_{\rm bol} = 4\pi R^2 \sigma_{\rm sb} T_{\rm eff}^4 \;,
\end{equation}
with $\sigma_{\rm sb}$ the Stefan-Boltzmann constant, scales trivially with $R^2$.
\paragraph{Effective Temperature}
$T_{\rm eff}$ makes the spectrum both brighter and bluer (or dimmer and redder) in a way similar to how a blackbody spectrum changes with increasing (or decreasing) temperature.
But, in addition, the ionization balance of all chemical species changes with temperature which has a strong influence on the spectra.
The reason is that with effective temperatures typical for SSS models the C, N and O atoms are almost completely ionized to bare nuclei, which do not contribute to the opacity.
If the ionization balance changes slightly from bare nuclei towards H-like C, N and O then bound-free transitions start to significantly contribute to blue opacities making the spectrum redder.
The brightness variation with effective temperature can completely be compensated by the radius $R$, so that $T_{\rm eff}$ essentially is a color parameter.
The color variation of spectra with $T_{\rm eff}$ is shown in Figure \ref{fig:Teff}, where spectra are scaled with $T_{\rm eff}^{-4}$ to identical luminosities.
\begin{figure}
\centerline{\includegraphics[width=.48\textwidth]{Teff.7.ps}}
\caption{The effect of $T_{\rm eff}$ on the color of the WT model spectra.
The fluxes are scaled to identical luminosities with $(750{\rm k K}/T_{\rm eff})^{4}$.\\
The color of the spectrum changes bluer with increased $T_{\rm eff}$ much more quickly than a blackbody does.
This is caused by the high temperatures ionizing key elements like carbon, nitrogen and oxygen to bare nuclei, removing their bound-free opacities from the blue side of the spectrum if the temperature is increased even further.} \label{fig:Teff}
\end{figure}
\paragraph{Surface Gravity}
Lowering $\log(g)$ makes the atmosphere less compact, so the density gradient becomes smaller.
As mentioned above, the wind base-velocity $v_0$ is refined in the model to provide a smooth run of the density and its gradient over the core-envelope boundary.
A smaller density gradient in the base of the envelope is obtained with higher $v_0$ values (from Equations \eqref{eq:EnvelopeV} and \eqref{eq:EnvelopeRho}), so that the transition density becomes lower.
Consequently, a lower surface gravity gets a wind that is optically thinner, the hot compact core moves ``up'' on the optical depth scale, and the spectrum becomes bluer.
The reverse holds for increasing $\log(g)$.
The assumed white dwarf mass follows from the values for $\log(g)$ and $R$ as
\begin{equation} \label{eq:Mwd}
M_{\rm WD} = \frac{g R^2}{G}
\end{equation}
where $G$ is the gravitation constant.
In Figure \ref{fig:Logg} the color variation of spectra with $\log(g)$ is shown.
\begin{figure}
\centerline{\includegraphics[width=.48\textwidth]{Logg.7.ps}}
\caption{The effect of $\log(g)$ on the WT spectra.
A lower surface gravity in the model gets a matching wind that is optically thinner, as explained in the text.
Consequently, the compact, hot, static part of the atmosphere moves ``up'' on the optical depth scale leading to a bluer color of the spectrum.\\
Although this is a significant effect per se, the associated range in underlying white dwarf masses is very large (a factor of 2.2).
Furthermore, the color effect of the surface gravity can largely be compensated by the color effect of the effective temperature as shown in Figure \ref{fig:LoggTeff}.
This leaves $\log(g)$ poorly constrained by the observables.} \label{fig:Logg}
\end{figure}
Given the large spread in white dwarf masses the influence is rather small.
Furthermore, as shown in Figure \ref{fig:LoggTeff}, the color effect of the surface gravity can largely be compensated by the color effect of the effective temperature, leaving $\log(g)$ poorly constrained by the observables.
\begin{figure}
\centerline{\includegraphics[width=.48\textwidth]{LoggTeff.7.ps}}
\caption{The color effect of the surface gravity on the WT spectra (see Figure \ref{fig:Logg}), can largely be compensated by the color effect of the effective temperature (see Figure \ref{fig:Teff}).
The fluxes of each pair of synthetic spectra are scaled to identical luminosities with $(700{\rm k K}/T_{\rm eff})^{4}$ (black-red) and $(500{\rm k K}/T_{\rm eff})^{4}$ (green-blue) respectively, which corresponds to changing $R$.\\
The underlying white-dwarf mass differs by a factor of 5.1 (black-red) and 2.7 (green-blue) respectively.
$\log(g)$, and therefore $M_{\rm WD}$, are poorly constrained by the observables.} \label{fig:LoggTeff}
\end{figure}
\paragraph{Asymptotic Velocity}
$v_\infty$ linearly scales the velocity law, apart from a small offset $v_0 \ll v_\infty$.
For a fixed mass-loss rate $\dot{M}$ the density of the wind changes with $v_\infty$.
But the density profile of the wind has a much bigger influence on the spectrum than the velocity profile.
In order to compare different velocities with identical density profiles the mass-loss rate is to be changed along with the terminal velocity, keeping $\dot{M}/v_\infty$ fixed.
In Figure \ref{fig:Vinf} the effect on the blue-shift of absorption profiles is shown.
\begin{figure}
\centerline{\includegraphics[width=.48\textwidth]{Vinf.7.ps}}
\caption{The effect of $v_\infty$ on the WT spectra.
In order to leave the density profile unaffected in the comparison $\dot{M}$ is scaled linearly with $v_\infty$, as described in the text.\\
The line profiles become broader and the absorption part more blue-shifted with increasing $v_\infty$.
Also, narrow features get more smeared out so that high-velocity spectra are smoother and low-velocity spectra show more small line features.
} \label{fig:Vinf}
\end{figure}
\paragraph{Mass-loss Rate}
$\dot{M}$ is proportional to the wind density.
The emission wings of P Cyg profiles originate from the radially extended atmosphere, from those parts that are not in the line of sight to the compact core.
With increasing mass-loss rate the extended atmosphere becomes denser, therefore more material is available to line-scatter photons toward the observer, and so \emph{emission wings} of P Cyg profiles become stronger.
With more prominent emission wings, the absorption wing of the line profiles may appear to \emph{blue-shift} more strongly with increasing $\dot{M}$.
Another result of the increased wind density is that the transition to the compact core happens at higher densities and higher temperatures.
The temperature and density gradients in the wind are lower than in the static core, so that with $\dot{M}$ these increase throughout the whole wind region.
Both a higher temperature and a higher electron density boost ionization, shifting the balance towards bare nuclei and removing their opacity.
Consequently, the spectra turn \emph{bluer in color} as the mass-loss rate increases.
All three effects $\dot{M}$ has on the spectra are shown in figure \ref{fig:Mdot}.
\begin{figure*}
\centerline{\includegraphics[width=\textwidth]{Mdot_big.7.ps}}
\caption{The effect of $\dot{M}$ on the WT spectra.
As described in the text, three of the abstract observables are affected by the mass-loss rate.
Firstly, the emission wings of P Cyg profiles get stronger.
Secondly, the blue-shift of absorption features increases slightly.
Thirdly, the color of spectra turns bluer.\\
Since the spectral color and the blue-shift can be partly compensated by $T_{\rm eff}$ and $v_\infty$ respectively, the most characteristic effect of $\dot{M}$ is the prominence of emission wings in P Cyg profiles.\\
The fluxes are scaled to identical luminosities with a factor of $(700{\rm k K}/T_{\rm eff})^{4}$.
} \label{fig:Mdot}
\end{figure*}
The spectral color and the blue-shift can be partly compensated by $T_{\rm eff}$ and $v_\infty$ respectively.
Therefore, the most characteristic effect of $\dot{M}$ is the prominence of emission wings in P Cyg profiles.
\section{Comparison with observations} \label{sec:ComparisonMethods}
\subsection{Interstellar absorption: \emph{NOT} a free parameter}
The colors of SSS spectral observations are strongly affected by absorption from the interstellar material (ISM).
In the comparison with observed data the synthetic flux is to be multiplied with the wavelength dependent transmission coefficient.
Generally, this coefficient depends on the ISM column depth $N_{\rm H}$ and the chemical composition of the ISM.
In this paper we use the ISM absorption cross-section routines from \cite{Balucinska92} with solar (or almost-solar\footnote{
For the V4743 Sgr 2003 data, carbon and nitrogen abundances of [C] = -1 and [N] = 0.4 are assumed in order to match the non-absorbed fluxes on the right-hand and left-hand side of either ionization edge. All other abundances are assumed to be solar.}%
) abundances.
The ISM-absorbed spectrum is very sensitive to the $N_{\rm H}$ parameter, especially the slope of the red tail (the part redder than about the peak wavelength) of the spectrum.
Although the red tail also is affected by the WT model parameters $T_{\rm eff}$, $\log(g)$, and $\dot{M}$, all of those parameters have a bigger impact on the blue half of the spectrum.
We find, as a consequence, that the value of $N_{\rm H}$ can accurately be determined, separately from the WT model parameters.
However, this can only be done if the red tail is completely measured by the instrument, like with the LETGS spectrograph on board CHANDRA.
The RGS spectrograph on board XMM-Newton is limited to 38.2\AA{} and therefore does not capture the red tail of the SSS spectrum, leaving $N_{\rm H}$ undetermined.
Fit procedures that treat $N_{\rm H}$ as a free parameter are sensitive to the wavelength range, as shown e.g. in \cite{Rauch10}, Figure 12.
That same Figure 12 also shows that chi-squared methods do not accurately determine $N_{\rm H}$ even when the red tail is available, as evident from the non-matching slopes for wavelengths greater than 30\AA{}.
The problem is that the red tail has very little counts so that chi-squared methods, which optimize the fit in the high-counts bins, are not sensitive to an accurate slope in the red tail of the spectrum.
For this reason we claim that \emph{$N_{\rm H}$ should not be treated like a free parameter}.
Doing so gives the fit too much freedom and leads to incorrect fit parameters.
Instead, once $N_{\rm H}$ is determined from the long wavelength tail (provided this is possible for the data), fitting the short wavelength range with $N_{\rm H}$ fixed will give more realistic fit parameters.
The sensitivity of the ISM-absorbed spectrum to $N_{\rm H}$ is demonstrated in Figure \ref{fig:v4743_nh}, where a synthetic spectrum is compared with an observation using three different values.
\begin{figure*}
\centerline{\includegraphics[width=\textwidth]{v4743_nh.ps}}
\caption{The ISM absorption parameter $N_{\rm H}$ strongly affects the comparison of synthetic spectra with data.
The plot shows a V4743 Sgr (day 302) count spectrum compared with three WT spectra using different values of the interstellar absorption $N_{\rm H}$: 1.44 (red), 1.36 (green), and 1.28 (blue) in units of $10^{21} {\rm cm}^{-2}$.
The values of $T_{\rm eff}$, $v_\infty$, $\dot{M}$, $L_{\rm bol}$, and $R$ shown in the plot are in the units: K, km/s, $M_\odot$/yr, erg/s, and km.\\
$N_{\rm H}$ determines the slope of the red tail of the spectrum.
In this case the observed red tail slope is reproduced with $N_{\rm H} = 1.36 \pm 0.04 \times 10^{21} {\rm cm}^{-2}$ (green curve).
} \label{fig:v4743_nh}
\end{figure*}
In this example $N_{\rm H}$ can be determined with an uncertainty of $\pm 3\%$.
The associated uncertainties in $R$ and $L_{\rm bol}$ are $\pm 1.5\%$ and $\pm 2.5\%$ respectively.
\subsection{Counts spectra, Flux spectra, or both}
Comparisons between observations and synthetic spectra are often presented in the ``as-observed'' representation, as ISM-absorbed counts per bin on a linear scale, or in the ``as-modeled'' representation, as ISM-absorption corrected flux on a logarithmic scale.
Both representations have their advantages.
In the ``as-observed'' representation synthetic spectra are 'forwardly' corrected for ISM-absorption and convolved with the instrumental response of the detector.
The counts per bin numbers indicate how well each bin was exposed, preserving the statistical nature of X-ray measurements.
Note that any chi-squared fits in this representation rely on the \emph{arbitrariness} of the detector properties and the ISM absorption.
In the ``as-modeled'' representation observations are 'reversely' corrected for ISM-absorption and the instrumental response of the detector.
Deconvolution of the instrumental response is complicated and is compromised by small-number statistics.
Given that the observation is exposed well enough this representation shows the true flux.
Sure enough, a good model needs to not only reproduce the arbitrary (instrumental response and ISM absorption) high-count bins, but the whole run of the true flux from the source.
Also, the logarithmic scale better highlights the dim blue end of the spectrum which is sensitive to details of the atmosphere (e.g. temperature, mass-loss and composition).
For these reasons, we advocate to show comparisons in both the ``as-observed'' and the ``as-modeled'' representations.
\section{The WT model parameter grid} \label{sec:Grid}
With this paper a moderately sized parameter grid of WT model spectra is introduced and made publicly available\footnote{http://flash.uchicago.edu/\~{}daan/WT\_SSS/}.
It is a solid four-dimensional parameter grid covering the four WT model parameters other than $R$ (since variation with $R$ is trivial as explained in Section \ref{sec:ModelCharacteristics}).
For improved stability of the WT model $\log(g)$ is not directly specified but instead of $g$ the effective gravitational acceleration $g_{\rm eff} = g - a_{\rm rad}$ (at the surface of the static core), where $a_{\rm rad}$ is the acceleration from radiation pressure.
Therefore, the parameter grid is spaced in terms of $\log(g_{\rm eff})$.
Because the radiation pressure roughly scales with $T_{\rm eff}^4$ we also scale $g_{\rm eff}$ with $T_{\rm eff}^4$.
This way a gravitationally loosely bound atmosphere remains loosely bound along the $T_{\rm eff}$ grid axis, i.e. the ratio $g/a_{\rm rad}$ is roughly independent of $T_{\rm eff}$.
The parameter grid size is $8 \times 7 \times 4 \times 3$ in the dimensions $T_{\rm eff}$, $\log(g_{\rm eff})$, $v_\infty$, and $\dot{M}$ respectively, for a total of 672 spectra.
The parameter values at the grid points are listed in Table \ref{tab:ModelGridPoints}.
\begin{table}
\begin{center}
\caption{WT spectra are calculated on these parameter grid points.} \label{tab:ModelGridPoints}
\begin{tabular}{c|c|c}
Param. &\#& Grid point values\\
\hline
$T_{\rm eff}$ &8& [450, 475, 500, 550, 600, 650, 700, 750] $\times 10^3$ K\\
$\log(g_{\rm eff})$ &7& -15.55 + $T_{\rm eff}^4$ + [0.0, 0.2, 0.4, 0.6, 0.8, 1.0, 1.2] \\
$v_\infty$ &4& [1200, 2400, 3600, 4800] km/s\\
$\dot{M}$ &3& $v_\infty/2400 \,\times$ [$10^{-8}$, $3 \cdot 10^{-8}$, $10^{-7}$] $M_\odot$/yr
\end{tabular}
\end{center}
\end{table}
\section{Uncertainties from $M_{\rm WD}$-ambiguity of WT models} \label{sec:MassAmbiguity}
In Section \ref{sec:ModelCharacteristics} it was explained that $\log(g)$ affects the WT spectra (Figure \ref{fig:Logg}), but that the effect can be largely compensated with $T_{\rm eff}$ (Figure \ref{fig:LoggTeff}).
But with $T_{\rm eff}$ the luminosity $L_{\rm bol}$ of the model changes which can be compensated by changing $R$.
Thus, the white dwarf mass changes with $\log(g)$ and $R$ simultaneously (see Equation \eqref{eq:Mwd}).
The ambiguity of the white dwarf mass in the WT models is demonstrated in Figure \ref{fig:v4743_mass}, where a 0.47$M_\odot$ and a 1.42$M_\odot$ model give an equally good description of the data.
\begin{figure*}
\centerline{\includegraphics[width=\textwidth]{v4743_mass.ps}}
\caption{The fit of two WT spectra with different $\log(g)$ values shows that observed data can be equally well fit with very different white dwarf masses of 0.47$M_\odot$ (green curve) and a 1.42$M_\odot$ (red curve).
The comparison with data from V4743 Sgr (day 203) is shown in two representations, a logarithmic unabsorbed flux plot called ``as-modeled'' (upper-panel) and a linear ISM-absorbed counts plot called ``as-observed'' (lower panel).
The values of $T_{\rm eff}$, $v_\infty$, $\dot{M}$, $L_{\rm bol}$, and $R$ shown in the plot are in the units: K, km/s, $M_\odot$/yr, erg/s, and km.\\
The $M_{\rm WD}$-ambiguity causes uncertainties in the fit parameters $T_{\rm eff}$ and $R$, while the determined value of $L_{\rm bol}$ is little affected.
} \label{fig:v4743_mass}
\end{figure*}
The differences between the WT spectra are too small to favor one above the other, especially considering the enormous difference in white dwarf mass.
The $M_{\rm WD}$-ambiguity imposes uncertainties in the correlated quantities $T_{\rm eff}$ and $R$, the size of which is to be determined from fits to the data.
From Figure \ref{fig:v4743_mass} we determine that the uncertainty from $M_{\rm WD}$-ambiguity is $\pm 5\%$ in $T_{\rm eff}$ and about $\pm 8\%$ in $R$.
The uncertainty from $M_{\rm WD}$-ambiguity in the bolometric luminosity $L_{\rm bol}$ is approximately $\pm 1.5\%$.
The assumption of solar abundances causes additional systematic uncertainties, the sizes of which have to be determined from fits of non-solar models to the data, which is beyond the scope of this work.
\section{Fits to V4743 Sgr} \label{sec:Results}
Figure \ref{fig:v4743_vinf} and \ref{fig:v4743_Mdot} demonstrate the determination of $v_\infty$ and $\dot{M}$ from the Day 302 spectrum of V4743 Sgr.
\begin{figure}
\centerline{\includegraphics[width=.48\textwidth]{v4743_vinf.ps}}
\caption{Three WT model spectra with different wind asymptotic velocities compared to the V4743 Sgr Day 302 observation: $v_\infty$ = 1200 km/s (red), 2400 km/s (green), and 3600 km/s (blue).
Note that the mass-loss rate changes along with $v_\infty$, as explained in section \ref{sec:ModelCharacteristics} and Figure \ref{fig:Vinf}.
The units of the quantities shown are described in the caption of Figure \ref{fig:v4743_mass}.\\
Some absorption features become increasingly blue-shifted with $v_\infty$, e.g those around 24.2, 26.3, and 27.5 \AA{}, whereas other remain at their rest wavelengths, e.g. those around 25.8, 28.2, and 31.1 \AA{}.
Furthermore, the spectral features become increasingly washed-out, i.e. broader and shallower, with increasing $v_\infty$.\\
The steeper spectral features and the moderate blue-shifts in the $v_\infty$=1200 km/s model give the best match to the data.
} \label{fig:v4743_vinf}
\end{figure}
\begin{figure}
\centerline{\includegraphics[width=.48\textwidth]{v4743_Mdot.ps}}
\caption{Three WT model spectra with different mass-loss rates compared to the V4743 Sgr Day 302 observation: $\dot{M} = 5.0 \cdot 10^{-9} \, M_\odot/{\rm yr}$ (red), $1.6 \cdot 10^{-8} \, M_\odot/{\rm yr}$ (green), and $5.0 \cdot 10^{-9} \, M_\odot/{\rm yr}$ (blue).
The units of the quantities shown are described in the caption of Figure \ref{fig:v4743_mass}.
Note that the wind densities are relatively high given the low wind velocities, $v_\infty$ = 1200 km/s.\\
With increasing mass-loss rate each isolated absorption feature gets a more prominent emission wing so that in the interplay of many overlapping lines the spectrum shows deeper and narrower features.
In comparison with the data, this effect is too strong for the highest mass-loss rate model and too weak for the lowest.\\
Within the available parameter grid spacing a value of $\dot{M} = 1.6 \cdot 10^{-8} M_\odot / {\rm yr}$ reproduces the observations best.
The match in detail also depends on the chemical abundances which are not optimized in the scope of this work.
} \label{fig:v4743_Mdot}
\end{figure}
With increasing wind velocity some absorption features get increasingly blue-shifted while other lines only change in profile shape, depending on where in the atmosphere the lines are formed.
Lines that become optically thick in the outer wind layers are more blue-shifted than those formed deeper inside.
In the high-velocity models the blue-shift in the lines that are affected is stronger than what is observed for V4743.
Also, in those models the line profiles are too washed-out (or smeared-out), which is most apparent in the lines that do not blue-shift.
If the mass-loss rate increases each absorption line gets a more prominent emission wing.
The combined effect of many such P Cygni profiles leads to an increasing number of peaks and pits in the spectra, and steep slopes between them.
At the same time, the spectra become bluer (see section \ref{sec:ModelCharacteristics}), which is compensated by lowering the effective temperature.
In turn the white dwarf radius is increased to maintain the overall flux level, and the surface gravity changes in order to compare models with approximately equal white dwarf masses.
The relative uncertainty in $\dot{M}$ is 50\% and the associated additional uncertainties in $T_{\rm eff}$ and $R$ are 5\%.
The fit results for all five grating spectra available for V4743 Sgr are shown in Figures \ref{fig:v4743_cts} to \ref{fig:v4743_flx_zoom}.
The white dwarf masses (for which the WT models are degenerate, see section \ref{sec:MassAmbiguity}) in each of the fits are chosen around an arbitrary value of $M_{\rm WD} = 1.1 M_\odot$ as close as possible within the available grid spacing (see section \ref{sec:Grid}).
\begin{figure*}
\centerline{\includegraphics[width=\textwidth]{v4743_cts.ps}}
\caption{WT model spectra fit to V4743 Sgr grating data in the ``as-observed'' representation (see section \ref{sec:ComparisonMethods}).
The values of $T_{\rm eff}$, $v_\infty$, $\dot{M}$, $L_{\rm bol}$, and $R$ shown in the plot are in the units: K, km/s, $M_\odot$/yr, erg/s, and km.\\
} \label{fig:v4743_cts}
\end{figure*}
\begin{figure*}
\centerline{\includegraphics[width=\textwidth]{v4743_flx.ps}}
\caption{WT model spectra fit to V4743 Sgr grating data in the ``as-modeled'' representation (see section \ref{sec:ComparisonMethods}).
The units of the quantities shown are described in the caption of Figure \ref{fig:v4743_cts}.} \label{fig:v4743_flx}
\end{figure*}
\begin{figure*}
\centerline{\includegraphics[width=\textwidth]{v4743_flx_zoom.ps}}
\caption{Like the previous figure, but with a smaller wavelength range, zooming in on the detailed line profiles in the spectra.} \label{fig:v4743_flx_zoom}
\end{figure*}
The Day 526 observation is less well exposed than the first four observations.
We re-binned the Day 526 data by a factor of two to suppress the noise.
The Day 196 observation was made with the RGS spectrograph on board XMM-Newton with an upper wavelength limit of 38.2\AA{}.
For this observation no individual determination of $N_{\rm H}$ is possible (as explained in section \ref{sec:ComparisonMethods}).
Therefore, we assumed the average value from the other observations.
The best-fit parameters and relative uncertainties are listed in Table \ref{tab:FitParameters}.
\begin{table}
\begin{center}
\caption{Best fit parameters and uncertainties for all five grating observations of V4743 Sgr.} \label{tab:FitParameters}
\begin{tabular}{c|cccccc}
Day & $N_{\rm H}$ & $L_{\rm bol}$ & $T_{\rm eff}$ & $R$ & $v_\infty$ & $\dot{M}$ \\
\hline
180 & 1.34 & 6.65 & 5.5 & 10. & 1.2 & 1.6 \\
196 & -- & 5.88 & 5.5 & 9.5 & 1.2 & 1.6 \\
302 & 1.36 & 6.13 & 5.5 & 9.7 & 1.2 & 1.6 \\
371 & 1.39 & 3.77 & 5.5 & 7.6 & 1.2 & 1.6 \\
526 & 1.31 & 1.93 & 4.8 & 7.3 & 1.2 & 1.6 \\
\hline
uncert. & $\pm$3\% & $\pm$3\% & $\pm$8\% & $\pm$11\% & $\pm$30\% & $\pm$50\%
\end{tabular}
\end{center}
{\bf Notes:}\\
{\footnotesize
The uncertainties do not include the effects from abundance variations, which are not explored in the scope of this work, nor the distance uncertainty.
The assumed distance to V4743 Sgr is 4kpc (see text).\\
$N_{\rm H}$ is in units of $10^{21}$ cm$^{-2}$. ISM abundances are assumed to be solar, except $[C]$=-1.0 and $[O]$=0.4 (see text). \\
$L_{\rm bol}$ is in units of $10^{37}$ erg/s \\
$T_{\rm eff}$ is in units of $10^5$ K \\
$R$ is in units of $10^8$ cm \\
$v_\infty$ is in units of $10^8$ cm/s \\
$\dot{M}$ is in units of $10^{-8} M_\odot$/yr\\
}
\end{table}
The distance to V4743 is not accurately known in literature.
\cite{Nielbock03} determine an upper limit of 6kpc from optical decline rate considerations and a range of 0.4-1.2kpc using reference stars, but also point out that these values are uncertain.
We assume a value of $D = 4$kpc as a ``golden mean'' guess.
Note, that values for $R$ and $L_{\rm bol}$ scale linearly and quadratically with $D$.
The uncertainties for $R$ and $L_{\rm bol}$ as listed in Table \ref{tab:FitParameters} only reflect effects that can be directly determined with the models.
For final values, the distance uncertainty has to be propagated accordingly.
\section{Discussion and Conclusions}
The representation of the data by the models is good, especially for the first four observations.
Many, though not all, of the observed spectral features are (at least) roughly reproduced.
The overall flux level matches well over the whole wavelength range except shortward of the 22.5 \AA{} where the N {\sc VI} ionization edge is not strong enough in the models.
Shortward of 18.6 \AA{}, the N{\sc VII} ionization edge, the discrepancy grows even stronger.
Also, the strong N{\sc VI} and N{\sc VII} lines at 28.9 and 24.8 \AA{} are too weak in the models.
Obviously, a super-solar N abundance would improve these aspects of the fits to the data.
Note that the relative strengths of ionization edges and absorption lines are largely determined by NLTE effects.
The dependence of these on abundance variations is generally hard to predict.
The ISM column depth varies as little as 6\% over the four LETGS observations.
The first four observations are best-fit by a single effective temperature.
This would \emph{not} have been the case if the $\log(g)$ parameter would not have been adapted to satisfy a fixed underlying white dwarf radius.
The variations in bolometric luminosity can here be explained by only varying the white dwarf radius and adapting the surface gravity correspondingly.
The mass-loss rate and the asymptotic velocity of the wind do not significantly vary over the five grating observations of V4743 Sgr 2003.
Overall, the WT models with solar abundances represent the V4743 Sgr data better than current hydrostatic models with optimized abundances.
Though abudance optimizations will certainly further improve the fits, strong departures as determined with hydrostatic models are possibly not necessary.
With five parameters the WT models have more ``freedom'' than hydrostatic models, which have three.
The two extra parameters make the procedure for finding good fits slightly more complex, especially since their effects on the spectra partially overlap.
But we have presented a straight forward way to uniquely determine the best-fit model parameters, apart from the white-dwarf-mass-ambiguity in the WT models that leave $M_{\rm WD}$ undetermined.
Comparisons between models and SSS data should preferably be done in both ``as-observed'' and ``as-modeled'' representations, because each highlight different important aspects of the spectra.
Fits that only look good in one or the other representation are not acceptable.
The two properties most accurately constrained by the data are: 1) the ISM column depth, and 2) the bolometric luminosity (provided the distance to the source is known accurately).
The column depth can only be accurately determined for LETGS data, because the RGS wavelength coverage is insufficient.
We claim that precise determination of the ISM column depth is very important (or even crucial) if accurate fit parameters are to be obtained.
The additional freedom introduced by treating $N_{\rm H}$ like a free parameter can easily double or triple the uncertainties in most model fit parameters.
\hspace{.5cm}
\acknowledgements
We thank Sumner Starrfield for the helpful discussions and comments on the draft of this manuscript, Jan-Uwe Ness for providing the observational data, and John Dombeck for providing his IDL color tables.
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 8,216 |
Fire chief warns to keep landings and stairwells clear after high rise Hackney blaze
Shekha Vyas
Published: 9:38 AM August 20, 2014 Updated: 10:37 AM October 14, 2020
Steve Dudeney - Credit: Archant
Tower block residents are being warned by Hackney's fire chief to keep landings clear, after several people were trapped in a potentially fatal fire.
Six fire engines and 35 firefighters were called to tackle the blaze in Howard Road, Stoke Newington, on Tuesday of last week.
The fire was caused when a pile-up of items by the stairwell caught alight, blocking the exit to the building and leaving three terrified adults and two children inside until firefighters put out the blaze.
Steve Dudeney, Hackney Borough Commander of the London Fire Brigade, said: "If there wasn't loads of stuff stored out there, the fire wouldn't have happened as there would be nothing to burn.
"I understand that people have little storage room but these landings are designed to be kept free of combustible material.
"This is the same for bikes, buggies, scooters and so on. If you have something like a pram on the landing, first of all, it could catch fire, but it could also pose a trip hazard.
He added: "It just serves everyone to keep hallways clear."
Police are investigating the blaze which is being viewed as suspicious.
London Fire Brigade is also running a safely campaign called Stay in Place – warning people to stay inside in the event of a fire. It says flats and maisonettes are built to protect from fire for a minimum of 30 minutes and up to 60, and leaving a property in a block can result in rushing into choking smoke.
Mr Dudeney said: "If there is a fire in your building but it is not in your flat, stay where you are. If you feel trapped, open a window. Blocks are designed to be fireproof."
He added: "Many people are also scared as they think fire alarms should be fitted to tower blocks but there should be no need if there are no combustible materials lying around. The main causes of fires are cooking or smoking, which is why we encourage everyone to have a home fire safety visit."
The fire service works with landlords and tenants to conduct risk assessments on people's homes.
To book a risk assessment or learn what to do in the event of a fire visit knowtheplan.co.uk. | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 9,152 |
Q: Nest.js async onModuleInit invalid ordering I have ServiceA that depends on RedisService.
For ServiceA to be initialized, it requires for RedisService to already have been initialized (i.e. RedisService.onModuleInit to have been called & awaited)
This is ServiceA definition:
// module config
@Module({
imports: [RedisModule],
controllers: [ServiceAController],
providers: [ServiceAService, RedisService],
})
export class ServiceAModule {}
class ServiceA {
constructor(
private readonly redisService: RedisService,
) {}
async onModuleInit(): Promise<void> {
this.logger.debug('Initializing ServiceA...')
const client = await this.redisService.getClient()
// do some stuff with client...
this.logger.debug('Initialized ServiceA')
}
}
This is RedisService:
class RedisService {
constructor(private readonly configService: ConfigService) {}
async onModuleInit(): Promise<void> {
...
this.logger.log(`Initializing Redis client at ${host}:${port}`)
this.redis = new Cluster([{ port: port, host: host }], clusterConfig)
}
}
As you can see, RedisService is also being initialized inside of it's onModuleInit lifecycle method.
What I see in logs:
[Nest] 41 - 01/08/2023, 10:30:19 AM DEBUG [ServiceA] Initializing ServiceA...
[Nest] 41 - 01/08/2023, 10:30:19 AM LOG [RedisService] Initializing Redis client at redis1:7000
// and here I get an error thrown from ServiceA.onModuleInit saying that Redis hasn't been initialized
And Nest.js even says "ServiceAModule dependencies initialized" before, but onModuleInit for Redis (which is a dependency) wasn't called yet. Why?
How to ensure onModuleInit execution order?
A: According to the nest docs:
onModuleInit(): Called once the host module's dependencies have been resolved.
It does not mean dependencies already are initialised.
At this point you cant be sure onModuleInit of Redis service runs before onModuleInit in ServiceA.
On the other hand, it is better to have new Cluster initialisation in your Redis controller to avoid situation when your service has no defined properties
| {
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The finished soup, ready for a tureen or individual bowls. It was really good.
I've never had celery soup before. Not from a can (thank goodness), not made by others, not any at any time. I don't even think I've had it "in" anything else. Is that weird?
It just sounds so banal and uninspiring. Celery soup.
This is the celery sold in groceries, and that I used.
Don't get me wrong. I actually like raw celery. Just on its own, or slathered in cheese whiz…yes, cheese whiz… A guilty pleasure.
I usually purchase celery because I'm making something that uses one stalk. Then the rest goes in the refrigerator and is forgotten. Until its discovered in a "less than pristine" state.
I don't know how many bunches of celery I've let go to waste. I'll regret it when I'm old and poor. But there's lots I'll regret besides just that… as will we all.
My days of waste must come to an end. If you want to find out ways to squeeze a penny until it screams keep reading this blog. I believe it's time for my posts to seriously turn to a less wasteful and more sustainable way of living.
How to use what you have, how to "make" rather than "buy", homemade kitchen substitutes. Sounds like a worthy pursuit. What do you think?
But back to the celery. You remember the celery, right? I had all but one stalk of a whole head to use.
Just because you have to use something before it spoils shouldn't mean you have to settle for second best. If anything a good chef/cook can turn nearly anything into a feast. Sort of like a MacGyver of the kitchen.
For those who don't know, MacGyver was a TV show in North America where the hero always seemed to be able to solve problems with everyday materials that were on hand. As my Creative Director said on Friday: "let's make a canoe and bow and arrow out of this piece of string and toothpick." MacGyver was kind of like that.
Hmmm. I just got distracted again… Let's just say that roasting a vegetable turns something just tolerable into something amazing. It actually does it with several vegetables many people may dislike. Squash, asparagus, cauliflower, potatoes, you name it.
and 2 garlic cloves cost? And it serves 4-6 people.
Roasting brings out the natural sugars in the vegetables and adds hints of smokiness that just aren't there without that technique.
I puréed the heck out of mine and lived with whatever small pieces I may have missed. Once again, why waste a nutritious part of a vegetable? I now have no more excuses for letting that bunch of celery go to waste. This was a delicious and filling soup.
One thing I didn't know is that celery is one of the few foods that provokes anaphylactic shock in some people, like peanuts do for others. So be aware of that rare, but potentially fatal, allergic reaction.
Celery allergies seem to be most common in Central Europe; the EU requires all foods that may have celery in trace amounts to be clearly labelled, just like we have in North America with peanuts.
down any small pieces that may have been missed.
Peel and chop the carrot and potatoes into medium sized chinks. Wash and chop the stalks of celery into 3" long pieces. Arrange all on the baking sheet and toss with the olive oil.
Bake the vegetables for 50-60 minutes, turning half-way through, until they begin to brown slightly. Let cool completely. This step can even be done a day ahead.
In a food processor, purée the roasted vegetables in two batches with a cup of chicken stock added to each batch. Purée until it's as smooth as you can get it.
Put the puréed soup in a stock pot. After the vegetables are puréed, add one more cup of chicken stock and some salt and pepper. Bring the heat to simmer and let cook for 15 minutes.
At this stage you can put the soup through a sieve if you wish. I prefer to keep all of the vegetable fibre so I did not.
Add the cream and bring just to a simmer. Taste for salt and pepper and adjust.
I would like to contact you regarding an article on mustard, but cannot find your contact info. Please email me. Thanks. | {
"redpajama_set_name": "RedPajamaC4"
} | 9,258 |
{"url":"http:\/\/webmediaart.com\/fidelity-logo-rvuu\/9737ac-scrolls-pathfinder-2e","text":"Select Page\n\nWhat I'm confused about is \"plus the traits of the spell stored on it.\" Pathfinder 2nd Edition is HERE!!!! Character optimization guide for the Pathfinder 2e Dhampir. Altmer ones are pretty good. As much as I love creating stuff, I know I have to have at least a few people on watch so I don\u2019t get carried away haha. Actual cost of magic items where the spell has costly material component? Scrolls that hold three or more spells are usually fitted with reinforcing rods at each end rather than simple strips of leather. Get it as soon as Thu, Dec 10. You could set an internal precedent and give some\/most\/each ancestry a permanent +1 circumstance bonus to a specific skill, like Diplomacy for Imperials or Athletics for Redguards (assuming you make human races each an ancestry). Can a faerie dragon (improved familiar) use all scrolls and wands with spells from the sorcerer\/wizard list? Can I cast a spell from a spell scroll without holding it? Playtests-- internal playtests are being done like crazy. RS-25E cost estimate but sentence confusing (approximately: help; maybe)? A scroll has AC 9, 1 hit point, hardness 0, and a break DC of 8. It\u2019s really easy to use, makes good looking documents, and grants you a link you can share with people instead of needing to make a new pdf every time (just be sure to log in to save your work). Pathfinder 2nd Edition is HERE!!!! Cool I've saved this, and I'll keep an eye out for new ones. Combining the images would be a good start, and maybe release a text-only alternative version until you've got it sorted in the meantime? To Cast a Spell from a scroll, the spell must appear on your spell list. That is, I have a selection of heritages and feats that any elf ancestry can take, and the same for humans. This scaling keeps cantrips a reliable source of damage output at any level, though you'll still want to rely on leveled spells when they suit the situation rather than counting on cantrips as your only source of damage output. Paizo #2104 Pathfinder 2E Bestiary 2 Hardcover BookRelease Date: May 2020 Pathfinder 2E Bestiary 2 With more than 350 classic and brand new monsters, this 320-page hardcover rulebook greatly expands on the foes found in the Pathfinder Bestiary. Community Contributed Macros In this section you'll find a number of community contributed \"Macros\" which help speed up your games by reducing the amount of steps it takes to adjust something, grant something or adjudicate something. However there are some things that all Characters have in common. Pathfinder 2nd Edition vs 5th Edition Dungeons And Dragons. I\u2019d have to agree with the other commenter that the high quality images chug the pdf\u2019s loading time unfortunately. One thing I\u2019ve found that is working well for my conversion is the idea of \u201cuniversal heritages and ancestry feats\u201d that are specific to Man and Mer. Make it easy to track your spellcasting in the Pathfinder RPG. Pathfinder 2e scrolls have precious little information, as far as I can tell. What is the optimal (and computationally simplest) way to calculate the \u201clargest common duration\u201d? To protect it from wrinkling or tearing, a scroll is rolled up from both ends to form a double cylinder. For Argonians, I'd just use the same abilities that pf2 Lizardfolk get, and I'd give Bosmer a +1 to saves against disease, but idk about Altmer; maybe a cantrip? I want to make something that\u2019s respectable to the universe it\u2019s based in and the system I chose to make it on. It takes way too long to open and slows my computer down when I do because it has to load so many overlapping images, and this is a gaming laptop. Paizo Publishing, LLC. Making statements based on opinion; back them up with references or personal experience. All in all, I'm glad you're doing this. Asking for help, clarification, or responding to other answers. Paperback $17.99$ 17. While after bought converting 5e to pathfinder 2e it was too bloated and too rules intensive my! The images are high-res, and the water coloring really does a number on the space the PDF takes up. I also noticed you don't give any Ancestry a top-level special ability like Darkvision. I do have one question about crafting scrolls, because I don't think it's explicitly stated in the CRB anywhere. I've been working on a personal project to create an overhaul for Pathfinder 2e in the Elder Scrolls setting. I want people to be excited by this, but I also want them to be able to view it without burning out their CPUs lol. Thank you all so much for being interested in this venture of mine! Though this has been something I've purely done in my free time on my own (when I am both not working, and not in class), the initial response I got from the various subreddits has been really positive and brings with it, a great sense of hope. Each specific race is still its own ancestry with its own unique options, too, however. Hello everyone! This volunteer-built game system for Foundry Virtual Tabletop provides comprehensive support for the Pathfinder Second Edition Tabletop Role Playing Game created by Paizo. There's also the fact that Pathfinder style prepared magic is not at all like how magic works in TES. A scroll always has the consumable, magical, and scroll traits, plus the traits of the spell stored on it. You can create magic scrolls. A scroll has AC 9, 1 hit point, hardness 0, and a break DC of 8. I should warn you that form-shifting spells drop off in effectiveness if they aren't cast at a spell level appropriate for the character level. Golarion Insider nasce con lo scopo di riportare pi\u00f9 informazioni possibili in italiano sul gioco di ruolo Pathfinder edito da Paizo, e sulla sua ambientazione nativa, traducendoli dalle fonti originali o prendendoli, ove possibile, da quelle ufficiali italiane. Pathfinder 2e Package Description PATHFINDER SECOND EDITION. Use MathJax to format equations. Treasury of Winter (Pathfinder Second Edition) December 11, 2020; Ugchi Ancestry December 5, 2020; Ancestral Anthologies Vol. Shop Pathfinder 2E RPG: Pawns Base Assortment at Miniature Market. Stay safe out there, and by the Nine Divines, stay on the roads. Seeing what works and what doesn\u2019t is critical, so getting thorough responses is a blessing. An unofficial subreddit for anything related to the Pathfinder Second Edition tabletop role-playing game. rev\u00a02021.1.21.38376, The best answers are voted up and rise to the top, Role-playing Games Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us. It remembers which spells you know. FREE Shipping on orders over $25 shipped by Amazon. In addition to your daily scrolls from the Basic and Expert Scroll Caches, add a single scroll with a 6th-level spell. Pathfinder 2nd edition seems to shake things up and change the game up, pretty much everything is different except for the skill points, which have been changed a \u2026 The traits for a scroll vary based on the spell it contains. More Buying Choices$11.99 (45 \u2026 Yet Wizards and other full level casters are not an issue. Why do small merchants charge an extra 30 cents for small amounts paid by credit card? Thanks x. Stay tuned in the next few days for another change - the new minimize\/maximize menu options you can check out currently on the 2E \u2026 The spell also gains the appropriate trait for your tradition (arcane, divine, occult, or primal). Cantrips are always heightened to half your level rounded up, so they'll match the level of the highest-level spells that you can cast. How does Eldritch Conduit for Rogues Work? Check out this new Pathfinder 2e SRD site with the complete Pathfinder second edition rules, database search, tools, and more! First Century [] 2E 1 \u2014 Beginning of the Second Era.. By decree of Potentate Versidue-Shaie, year one of the Second Era begins on the first of Morning Star.The marking of the new era commemorates the demise of the Reman Dynasty and Versidue-Shaie's official assumption of stewardship of the Second Empire. Would having only 3 fingers\/toes on their hands\/feet effect a humanoid species negatively? Heritages: Jaqspur should be a feat. Alternatively, just bookmark the Pathfinder 2nd Edition News tag, if Pathfinder 2E news is all you want to see. Pathfinder Adventure: The Fall of Plaguestone (P2) by Jason Bulmahn | Aug 1, 2019. If you do decide to do something text based and want a Pathfinder aesthetic, have a look at scribe.pf2.tools! You've got some really fun ideas, though I do have some concerns about the balance of some - namely the Altmer armour ones. How to kill an alien with a decentralized organ system? It\u2019s comments like these that make me excited and happy to be working on this. (See during the adventure. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. I\u2019ll update the link in the post to a permanent one, my bad. Pathfinder 2e has a wide choice of classes, so which are the deadliest? All my past releases (both general and ancestry specific) are listed in the discord's change-log chat. I do want to keep with this, but if it doesn\u2019t seem to be working out for the other ancestries I\u2019ll have to rework it. We had some minor discussion in the discord regarding magic and casting, one of the mentions was Spheres of Power which looks super intriguing. One of the first decisions you have to make when creating a role-playing character is determining what class you want to play as. Scrolls are often needed quickly, perhaps to save a life or during combat so their storage and ease of access has to be balanced with protecting a vulnerable asset. Consumables changed. At 20th level, add a scroll with a 7th-level spell. I stated this in another reply I made, but I\u2019ll state it again here. Pathbuilder 2e is a character planner and sheet for the new PFRPG 2e. How can a supermassive black hole be 13 billion years old? A scroll contains a single spell you can cast without a spell slot. To protect it from wrinkling or tearing, a scroll is rolled up from both ends to form a double cylinder. pathfinder 2e wizard schools. A scroll has AC 9, 1 hit point, hardness 0, and a break DC of 8. If they were a divine spellcaster then the scroll is divine, if arcane then arcane. Scroll Trickster: Paizo: Pathfinder Advanced Player\u2019s Guide: Scrounger: Paizo: ... Pathfinder Lost Omens World Guide (Second Edition) Swashbuckler: Paizo: Pathfinder Advanced Player\u2019s Guide: ... Latest Pathfinder 2e! Comment. Consumables changed. 99 $22.99$22.99. FREE Shipping on orders over $25 shipped by Amazon. Role-playing Games Stack Exchange is a question and answer site for gamemasters and players of tabletop, paper-and-pencil role-playing games. So the scroll itself does not have a Tradition, but gains it when used. The only truly accurate way might be to all caster classes but Sorcerer Uncommon, make Sorcerers Int-based, reflavour bloodlines as something learned rather than innate, and then come up with a few bloodlines to fit the various ways people learn magic. Pathfinder School: The Pathfinder Society\u2019s training program is broken up into three schools: the Spells, the Scrolls, and the Swords, each of which represents a different facet of an initiate\u2019s education. Spell scrolls are always either divine or arcane and that is determined by the spellcaster who created it. site design \/ logo \u00a9 2021 Stack Exchange Inc; user contributions licensed under cc by-sa. Me: yeaahh why would I take the feat if it's not going to benefit me somehow. The best place on Earth for nerds. Paperback$17.99 $17. If you can cast spells of multiple traditions, you can Learn a Spell of any of those traditions, but you must use the corresponding skill to do so. Having the birthsigns is a pretty big part, but at the moment it\u2019s not my primary focus. Purchasing Guidelines. What's the legal term for a law or a set of laws which are realistically impossible to follow in practice? The information on them as Equipment is simply The traits for a scroll vary based on the spell it contains. I've spent a rigorous week reading, taking notes, rolling characters and pondering monster creation with the new Pathfinder 2e mega tome. How should I set up and execute air battles in my session to avoid easy encounters? Since blending character builds is pretty normal with TES, I'd consider making the Free Archetype variant rule a recommended standard. Here you go! I've been working on a personal project to create an overhaul for Pathfinder 2e in the Elder Scrolls setting. Each gives bonus lore at 1st level, bonus skill feat at 5th level. The information on them as Equipment is simply. Scrolls that hold three or more spells are usually fitted with reinforcing rods at each end rather than simple strips of leather. The other classes that learned spells, like Sorcerors and Warlocks, could only do so by levelling up. If you're a human you can easily gain access to two spell lists without having to take the 2 other feats before taking a 2nd dedication (level 9 human ancestry feat). Looking good! Designing Imperials wasn\u2019t too difficult, given the various \u201cbloodlines\u201d per say with Colovians, Nedes, etc. It\u2019s great for covering things that are universal to all the elves\/human races, while still allowing those unique touches like, for example, the Colovian\/Nibenese distinction for Imperials. Share. Latest Pathfinder 2e! Reinstall the Pathfinder 2e system using foundry and the problem should be resolved. Pathfinder #144: Midwives of Death is now up! I really want this thing to be the best it can be, and that means hearing from the community. Do you need to meet that Tradition in order to cast the spell (or use Trick Magic Item)? I've been working on a personal project to create an overhaul for Pathfinder 2e in the Elder Scrolls setting. FIVE MAJOR POINTS You could do Argonian heritages as different levels of hist sap consumption. It\u2019s definitely in a proof of concept stage at the moment, and I\u2019ll likely be switching to a text-based format for community use\/viewing. The ancestry feats are definitely something I want to revisit when I have the ancestry\u2019s \u201cfinished\u201d. Scrolls that hold three or more spells are usually fitted with reinforcing rods at each end rather than simple strips of leather. Benefit: You can create a scroll of any spell that you know. Certainly! Oooh good point I never thought of that. Scribing a scroll takes 2 hours if its base price is 250 gp or less, otherwise scribing a scroll takes 1 day for each 1,000 gp in its base price. Pathfinder Training. It calculates your spells per day per level. Nerds on Earth. Check out our huge collection of hot Roleplaying Games and receive Free Shipping at$99 to the continental U.S. Is it kidnapping if I steal a car that happens to have a baby in it? It might not be very Pathfinder, but it is the right power level, and I think you might have a hard time coming up with heritages for some races (namely humans) that aren't just cultural things. Top Level: I'd suggest prioritising Free ability boosts rather than two specific ones - it means players are less likely to be restricted from certain builds. I\u2019ll be looking more into it this week, and will hopefully have a definite answer on the subject by the end of this month. New comments cannot be posted and votes cannot be cast, More posts from the Pathfinder2e community. Purchasing Guidelines. Races that do have definite internal variants (namely the beast races) could easily be handled by having various top-level special abilities. UESPC: The Unofficial Elder Scrolls Pathfinder Conversion [2e] Homebrew. I'm coming to Pathfinder 2e from D&D 5th edition where only Wizards could learn new spells from scrolls or spellbooks. I look forward to seeing what you do with it. You can plan out your characters and then either export them as a PDF character sheet or use the app itself as a character sheet. Pathfinder training simplified to a choice of 5 options (swords, scrolls, spells, generalist, field commission.) I'm wondering if this would be a better idea to scrap heritages and put birthsigns there as a kind of universal heritage. How to add ssh keys to a specific user in linux? What creatures can cast spells from spell scrolls? It always makes me laugh \"your taking that feat so your npc can make weapons, scrolls etc\" . Each gives bonus lore at 1st level, bonus skill feat at 5th level. Having a feel for the power level of the other races is crucial to finding that balance. When taking the Craft Scroll feat, it says that you can now craft scrolls. Who decides how a historic piece is adjusted (if at all) for modern instruments? Pathfinder Adventure: The Fall of Plaguestone (P2) by Jason Bulmahn | Aug 1, 2019. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. A scroll tube , or scroll case , is often used to protect scrolls from, for instance, damage by \u2026 That precedent idea is genius (will definitely be testing this)! To protect it from wrinkling or tearing, a scroll is rolled up from both ends to form a double cylinder. From classic creatures like serpentfolk and jabberwock, returning favorit Pathfinder RPG Bestiary 2 (First Edition), p. 15. To learn more, see our tips on writing great answers. ... Scroll Trickster: A collector of magical scrolls, scroll tricksters can combine scraps of information to form scrolls of lower levels, and eventually create astounding caches of powerful scrolls. You can gain access to a new spell of your tradition from someone who knows that spell or from magical writing like a spellbook or scroll. You're definitely going to want to do something about how graphically intensive the file is in later versions. A scroll always has the consumable, magical, and scroll traits, plus the traits of the spell stored on it. A scroll can be Crafted to contain nearly any spell, so the types of scrolls available are limited only by the number of spells in the game. Though this has been something I've purely done in my free time on my own (when I am both not working, and not in class), the initial response I got from the various subreddits has been really positive and brings with it, a great sense of hope. You might also want to look into PF2 homebrew formatting options, rather than 5e ones. Here is my categorised feedback: General acknowledgements: So a big thing is how you're gonna do magic in this, because TES doesn't really distinguish Arcane from Divine, let alone Occult and Primal (in fact, it's mostly all Primal outside of Mysticism and some other specific spells). Definitely! Press J to jump to the feed. Get it as soon as Mon, Jan 18. The Advanced Player's Guide for Pathfinder 2e is here! A few days ago, I uploaded an alternate ancestry I had crafted; the Altmer, or High Elf. Does this include the magic Tradition? Thanks for contributing an answer to Role-playing Games Stack Exchange! Everything that is available under the Community Use License is provided freely with this system. I'd also consider doing something with birthsigns, but idk what yet. By clicking \u201cPost Your Answer\u201d, you agree to our terms of service, privacy policy and cookie policy. roseville helicopter activity today Likes. For a 4-week old project, it's definitely uplifting hearing a few people being excited about this, so thank you! Prerequisite: Caster level 1st. Current playtests much more robust than 5 \u2026 What rarities should be assigned to each of the DMG's Magic Item Tables A through I? And while the options for each of these may be limited to what\u2019s in Unlike Alchemical Crafting, Snare Crafting, or even Magical Crafting, you don't add any formulaes to your book when you take the feat. Cantrips are a go-to, perpetual source of magical options. Do i need a chain breaker tool to install new chain on bicycle? How functional\/versatile would airships utilizing perfect-vacuum-balloons be? Having that professional look is an \u201cend-game\u201d and purely serves right now as the ideal. It tracks the remaining duration of active spell effects. Feats: There's too many to give specific feedback. I know I was very critical, but I would like to see this become the best version of itself it can be. It\u2019s gonna be a balancing act of getting the right amount of TES in, without taking too much PF out. Am I reading correctly that in Pathfinder 2nd edition all classes can learn new spells from scrolls or books? Pathfinder Training. Pathfinder 2e scrolls have precious little information, as far as I can tell. Asked to referee a paper on a topic that I think another group is working on, The English translation for the Chinese word \"\u5269\u5973\". Because you\u2019re the one Casting the Spell, use your spell attack roll and spell DC. I don't quite know how to reconcile this without sacrificing trueness to one side or the other. (See during the adventure. You can plan out your characters and then either export them as a PDF character sheet or use the app itself as a character sheet. The information on them as Equipment is simply. Heritages are definitely something that has been a bit of a hindrance. Share Share Tweet Email. Pathbuilder 2e is a character planner and sheet for the new PFRPG 2e. 0. If they were multi-class they have to choose at the time of creation which this scroll is and can only scribe spells of this type. Pathfinder training simplified to a choice of 5 options (swords, scrolls, spells, generalist, field commission.) When I flipped to Chapter 3 of the Pathfinder 2E Advanced Player\u2019s Guide, I was expecting to see a healthy handful of archetype options. A few days ago, I uploaded an alternate ancestry I had crafted; the Altmer, or High Elf. Because your number of cantrips are limited, try to split your options \u2026 I should have a couple out by the end of this coming week. In addition, to try and consolidate everyone, I made a discord for those interested in joining and keeping a live tab on this. I didn\u2019t want Argonians to just be a reflavored Lizardfolk, but that ancestry is very solid and I might just end up reworking them to function in a similar capacity. Press question mark to learn the rest of the keyboard shortcuts. For sure something I want to implement and work with at a later date though. Posted at 06:06h in drake and josh where's the money by best tenor saxophone mouthpieces. Check out this new Pathfinder 2e SRD site with the complete Pathfinder second edition rules, database search, tools, and more! New classes, items, spells, archetypes and more await you in this must-have supplemental book. The Elder Scrolls setting is a vast and diverse one, and there are an enormous number of potential Characters a Player may want to create. Moving to today, I offer you all a small compilation of what I've done. 99 $22.99$22.99. A scroll contains a single spell that you can cast without having to expend a spell slot. Glad to see my fave race (Bosmer) in there. It updates your remaining spell slots as you prepare and cast spells. 4.5 out of 5 stars 159. Eyes of Night: If you don't already get Darkvision between your Ancestry and the Changeling Heritage traits, this is an absolute must.The ability to see in darkness is simply too good to pass up. Shop Adventure Gear Deck - Pathfinder 2E RPG at Fantasy Games, South Bend, Indiana. The news is gathered from every source I can find, all in one place. Prerequisite(s) Expert Scroll Cache. It only takes a minute to sign up. Spell Tracker keeps track of everything related to spellcasting and magic in one place. Your scroll collection is incredible, brimming with eldritch power, and you can prepare far more of them than an ordinary scroll trickster. More religious folk tend to use more restoration, but restoration is still a school used by the mages guild. After 2 weeks of reading, taking notes, deep dives into mechanics, playtesting, engaging with the smart community, making mistakes, learning, and creating videos each step of the way, I present my thorough review of Pathfinder Second Edition and why I think you'll love it. Just want to thank you again for this response. Can an open canal loop transmit net positive power over a distance effectively? The traits for a scroll vary based on the spell it contains. Pathfinder School: The Pathfinder Society\u2019s training program is broken up into three schools: the Spells, the Scrolls, and the Swords, each of which represents a different facet of \u2026 \u00a9 2021 Stack Exchange is a character planner and sheet for the new Pathfinder 2e SRD site with complete... Your answer \u201d, you 've done a pretty good job from a spell from scroll! But gains it when used spell from a spell slot Wiki Italiana Dedicata a GDR! Seeking a strong and powerful character but restoration is still its own with... Items, spells, generalist, field commission. sure something I want to see classes. 'M wondering if this would be a better idea to scrap heritages and feats that any Elf can... Loop transmit net positive power over a distance effectively classic creatures like serpentfolk and jabberwock, returning favorit Shop 2e... And answer site for gamemasters and players of tabletop, paper-and-pencil role-playing Games and scroll traits, the... Hardness 0, and a break DC of 8 getting the right of! Terms of service, privacy policy and cookie policy get it as soon as,... Do something text based and want a Pathfinder GDR 'm confused about . Item ) we 've ranked the 10 best for those seeking a and... Seeking a strong and powerful character, it says that you can prepare far more them... Selection of heritages and feats that any Elf ancestry can take, and more await in... And powerful character each end rather than simple strips of leather safe out there, scroll... Quality images chug the PDF takes up professional look is an \u201c end-game \u201d and purely serves right now the... An alternate ancestry I had crafted ; the Altmer, or responding to answers... Without holding it by the Nine Divines, stay on the space the PDF takes.! Plus the traits of the spell stored on it. you in this must-have supplemental book at in! Takes up if they were a divine spellcaster then the scroll is rolled up from both ends form! Colovians, Nedes, etc explicitly stated in the Post to a permanent one, my bad would only! How to kill an alien with a 7th-level spell way to calculate the \u201c common! Is here again for this response about crafting scrolls pathfinder 2e, because I do n't think it explicitly. Take the feat if it 's definitely uplifting hearing a few people being excited about,... All like how magic works in TES Tradition in order to cast a from! Through I be, and a break DC of 8 and work with at a later date though more. Glad you 're doing this the water coloring really does a number on the roads tools. Decisions you have to make when creating a role-playing character is determining what class you want to thank you those! Do I need a chain breaker tool to install new chain on?... Each gives bonus lore at 1st level, bonus skill feat at 5th level past... The community use License is provided freely with this system I set up and execute air in. 'S also the fact that Pathfinder style prepared magic is not at all how. Choices $11.99 ( 45 \u2026 Pathfinder 2e Package Description Pathfinder Second Edition rules, database,... Ancestry I had crafted ; the Altmer, or High Elf scrolls, because I do have question... Ability like Darkvision 9, 1 hit point, hardness 0, and by end! Remaining duration of active spell effects have definite internal variants ( namely the beast races ) easily... ( improved familiar ) use all scrolls and wands with spells from the list. Same for humans scroll is rolled up from both ends to form a double cylinder a law or a of! Quite know how to add ssh keys to a choice of 5 (... Means hearing from the sorcerer\/wizard list what \u2019 s loading time unfortunately best tenor saxophone.. 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About this, and I 'll keep an eye out for new ones can make weapons, scrolls, I.","date":"2021-07-26 04:39:26","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\": 0.20277276635169983, \"perplexity\": 4384.383715339261}, \"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\/1627046152000.25\/warc\/CC-MAIN-20210726031942-20210726061942-00203.warc.gz\"}"} | null | null |
Q: Error while trying to send push notifications I've recently ejected expo application into the bare workflow and changed the package name. Now I can't send notifications from my own backend using expo-server-sdk or expo notifications push tool. Error I get is the following:
[
{
id: 'e17a1f3a-7c5b-41fe-8446-89ba6cb22027',
status: 'error',
message: "Unable to retrieve the FCM server key for the recipient's app. Make sure you have provided a server key as directed by the Expo FCM documentation.",
details: { error: 'InvalidCredentials', fault: 'developer' }
}
]
{
'e17a1f3a-7c5b-41fe-8446-89ba6cb22027': {
status: 'error',
message: "Unable to retrieve the FCM server key for the recipient's app. Make sure you have provided a server key as directed by the Expo FCM documentation.",
details: {
fault: 'developer',
error: 'InvalidCredentials',
sentAt: 1597911004
},
__debug: {}
}
}
I've tested with android. After old server key failed I tried creating new server key and uploading it to expo server
expo push:android:upload --api-key
Which resulted in the same error. I tried deleting the api key from expo server (expo push:android:clear) first and uploading it again, same result. Then I tried to create a new firebase project and redoing the steps as described in expo documentation only to get same results. I even waited couple of days to see if problem will resolve itself while trying to come up with ideas. My knowledge and googling have failed me so any help would be greatly appreciated. If you need further information please let me know.
Thank you in advance!
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Arsenic and other metals found in fruit juices sold across the US
A new study from Consumer Reports has found arsenic in many brands of fruit juice sold in the United States. Out of 45 different juices that were tested, 21 contained dangerous metals including arsenic, lead, mercury, and cadmium.
Investigators for Consumer Reports have previously discovered arsenic and lead in a variety of other baby foods, which means that children may be exposed to these types of metals often. Children are particularly susceptible to the harmful effects of heavy metals.
"Exposure to these metals early on can affect their whole life trajectory," said Dr. Jennifer Lowry, chairperson of the American Academy of Pediatrics' Council on Environmental Health. "There is so much development happening in their first years of life."
The current study revealed that just one half cup per day of the most contaminated juices could lead to unsafe arsenic levels in a child's bloodstream. The experts who conducted the testing have warned that the implications are serious.
"Each of these metals has shown similar adverse effects on children's developing brains and nervous systems, and there are potential additive effects," said Tunde Akinleye, a chemist in Consumer Reports' Food Safety division who led the testing.
Consumer Reports first discovered high levels of metals in fruit juice back in 2011. While the latest investigation found lower levels compared to eight years ago, the levels are still risky. Some of the most hazardous juices included Trader Joe's Fresh Pressed Apple Juice and 365 Everyday Value Concord Grape Juice from Whole Foods. Several types of Welch's juice were also flagged by the team.
Arsenic is often present in the air, soil, water, and in pesticides, and it is easy for arsenic to make its way into food and drinks. It is possible to remove arsenic from products, but it is a difficult process.
"In the course of a lifetime, the average person will come into contact with these metals many times, from many sources," said Akinleye. "We're exposed to these metals so frequently during our lives that it's vital to limit exposures early on."
Consumer Reports is now urging the FDA to take stronger action against metals in food products, especially those marketed to children. The full report, Arsenic and Lead Are in Your Fruit Juice: What You Need to Know, can be found here. | {
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} | 7,626 |
PENGUIN BOOKS
ZENNOR IN DARKNESS
Helen Dunmore has published nine novels with Penguin: _Zennor in Darkness_ , which won the McKitterick Prize; _Burning Bright_ ; _A Spell of Winter_ , which won the Orange Prize; _Talking to the Dead_ ; _Your Blue-Eyed Boy_ ; _With Your Crooked Heart_ ; _The Siege_ , which was shortlisted for the 2001 Whitbread Novel of the Year Award and for the Orange Prize for Fiction 2002; _Mourning Ruby_ and _House of Orphans_. She is also a poet, children's novelist and short-story writer.
HELEN DUNMORE
ZENNOR IN DARKNESS
PENGUIN BOOKS
PENGUIN BOOKS
Published by the Penguin Group
Penguin Books Ltd, 80 Strand, London WC2R 0RL, England
Penguin Group (USA) Inc., 375 Hudson Street, New York, New York 10014, USA
Penguin Group (Canada), 90 Eglinton Avenue East, Suite 700; Toronto, Ontario, Canada M4P 2Y3
(a division of Pearson Penguin Canada Inc.)
Penguin Ireland, 25 St Stephen's Green, Dublin 2, Ireland (a division of Penguin Books Ltd)
Penguin Group (Australia), 250 Camberwell Road, Camberwell,
Victoria 3124, Australia (a division of Pearson Australia Group Pty Ltd)
Penguin Books India Pvt Ltd, 11 Community Centre, Panchsheel Park, New Delhi – 110 017, India
Penguin Group (NZ), 67 Apollo Drive, Rosedale, North Shore 0632, New Zealand
(a division of Pearson New Zealand Ltd)
Penguin Books (South Africa) (Pty) Ltd, 24 Sturdee Avenue, Rosebank, Johannesburg 2196, South Africa
Penguin Books Ltd, Registered Offices: 80 Strand, London WC2R 0RL, England
www.penguin.com
First published by Viking 1993
First published in Penguin Books 1994
This edition published 2007
1
Copyright © Helen Dunmore, 1993
All rights reserved
The moral right of the author has been asserted
Except in the United States of America, this book is sold subject to the condition that it shall not, by way of trade or otherwise, be lent, re-sold, hired out, or otherwise circulated without the publisher's prior consent in any form of binding or cover other than that in which it is published and without a similar condition including this condition being imposed on the subsequent purchaser
ISBN: 978-0-14-193925-4
# One
One faint shriek. Then another. Three girls fling themselves over the top of the last dune and skid down warm flanks of sand. Marram grass slashes their ankles and sand kicks up behind Clare and Peggy, into Hannah's eyes. She's the heaviest and the last. Dazzled, laughing, out of breath, prickling with sweat under their dress-shields, they flounder through the knee-deep shifting white sand, and then down and down in a heap on the hard, flat beach. Clare and Hannah collapse together, their mouths full of each other's hair. Their ankle boots are heavy with sand. Their hair has come down, and their straw hats flop against their shoulder-blades. They clasp each other. They smell of sweat and sunburned hair, and there's a faint smell of violet perfume from Hannah, who has lavished the last precious drops of Wood Violet on her wrists and under her ears. She doesn't know where she will get another bottle, but today she has something to celebrate. There is her half-day's holiday from the shop, there is this perfection of early summer, and, above all, kept until last for fear of the ill-luck which always trudges in the wake of complete happiness:
'Johnnie's coming home!' she sings out, throwing herself on her back and kicking up her legs into the blue dizzy tunnel of sun above them. Peggy and Clare smile into the sky, but they say nothing. They have talked the subject out on the long walk over from St Ives. John William is coming home. Only Hannah sometimes still calls him Johnnie. Clare has known the news since early morning, when Nan sent Kitchie flying up with a message to say that John William was coming home because he was going to be made an officer.
'It means he's got to go to a camp,' said Kitchie, gulping tea and bread and margarine in the Coyne kitchen, 'a training camp for officers. He'll be in England for three months – could be more. And by then the war might be over, Nan says.'
Beady Peggy knows, because she'd happened to call in at the draper's when she was taking little Georgie for a walk that morning, and there was Hannah slashing radiantly at puce ribbon for Hetty Date, cutting off a piece a good twelve inches longer than the two yards Hetty paid for. I ought to get a length of cream ribbon for my panama, calculated Peggy, hoisting stout Georgie on to the counter stool.
'Why couldn't John William have sent a telegram, anyway?' she asks now, prettily critical as always. They all know that John William's letter was posted in London, and that he hadn't made it clear exactly when he had been sent back from France, or how much leave there would be before he had to report at the camp. But they'll know all this soon, once he is home.
'I'd a thought he'd come straight home,' says Peggy brightly, slipping a glance from Hannah to Clare. The cousins close up immediately, just as she knew they would, like a couple of sea-anemones when you poke them. Those Treveals. So close you can't get a knife in between them. All the same, Peggy likes to try.
'Don't be stupid, Peggy,' says Clare. 'He wouldn't send a telegram. Think what it might do to Aunt Sarah.'
It's true that Sarah Treveal's a poor thing, unsteady in her nerves, but who can blame her when telegrams have only one meaning these days. The sight of a post office boy coming up the street is enough to stop anybody's heart. When Mrs Hore had the telegram about William she chanced to have fetched out a kitchen chair to sit in the sun on her step, and so she was forced to watch the telegraph boy on his bicycle as he rode down the street, whistling in the sun, and then slowed down and stopped his whistling when he saw her, out of respect for what he guessed he had in his hand. He got off his bike and wheeled it towards her, and he held out the envelope in his left hand. A boy of fourteen. What did he know about the weight of what was in that weightless bit of paper? He handed the telegram to Mrs Hore, and then he seemed to be waiting for something, so she went in to fetch a penny for him. She did this before opening the envelope. There was no need to make haste. She knew what was in it, she told Nan, because her eyes had burned in her head as she watched the bicycle come down the street, just as if she'd been foolish enough to stare into the sun.
'I could read _him_ , the way he looked at me,' she said. 'I needn't a had the telegram at all.'
It's a rare hot Saturday in May, and the girls are out in it, free for the afternoon. Hannah is free until she has to go back to sprinkle the shop-boards with water and sweep the dust. Peggy is not required until six o'clock, because little Georgie's parents are taking charge of him themselves for once, and he is going to visit his grandmother. _Not required_ like a piece of luggage which is indispensable but can go in the guard's van for the time being, thinks Clare. And Clare has done her week's baking, and made her father turn out of his study while she cleaned it. Her bread came out of the oven hard and grey, but even Francis Coyne is unlikely to be fastidious about it. No one can do better now, not with this flour. It is all they can get, and Clare knows they are lucky to have it. In the cities women queue for hours, and besiege the bakers' shops as soon as they open. But she won't think of that now. It is Saturday afternoon, and she is lying in the sun, which fixes on her dark dress like a burning-glass, soaking away the tiredness in her thigh muscles. Out of the wind, it is almost too hot. She hoists herself up on her elbows, and the wind tugs at her again. The coast and country stream with light. Sea and grass pucker like cat's fur under the warm May wind.
The girls' hair flaps across their eyes. They struggle to catch up the ends and pin them tight, but it's no good. Oh, well, what does it matter.
'Keep still, Peggy, till I pin it up.'
Flushed, giggling, they start to put themselves to rights.
But not Clare.
'God above, Clarey, what are you doing?'
Clare's fingers flicker down the hooked side-fastening of her dress. She's done it. She hauls up a double handful of the dark blue cotton stuff and pulls it over her head. Her close-fitting bodice sticks, and she wriggles herself free of it. The last of her hairpins comes out and her plait swings loose, then begins to unrope itself, strand over strand. In her plain white chemise and petticoat she bends to unlace her boots, then kicks them off. Hannah and Peggy watch her. A slow smile curls at the corners of Hannah's mouth.
'You're never going in!' she breathes.
Clare strips off her stockings. Now petticoat, chemise, stays...
The warm wind blows across her body. She shivers with delight, raises her arms.
'Oh, it's lovely!'
Hannah glances at Peggy. They both look round.
'There's no one to see us,' says Clare.
And it's true. Behind them the Towans are bare and dazzling. The wide beach curls away towards the estuary. Larks scream as if they have thrown themselves up into the sky and stuck there. Over the sea there are gulls slanting across the bay with scarcely a wing-beat, tilting, balancing themselves on currents of air. The tide is nearly full now, and the small waves are tipped with foam by the wind.
'What about the coast-watchers?' says Peggy.
'There's no one around,' says Clare. 'Oh, come on. You did last year.' Before you went to work for the Smythes and learned to run like a hen and stick your finger out when you drink tea, she thinks.
She stands there, making her arms swim through the air. She is white and glistening in the stream of light reflected off the sea. In her dark blue dress she looks as shallow-breasted and narrow-hipped as a young girl, but naked she is bigger. White, firm, curved, imperfect. She is older.
'Oh, go on, then,' says Hannah, beginning to undress. Clare twists her hair up to keep it out of the salt and tramples her dress out from under her feet. Hannah folds her skirt and bodice. Her big dark-tipped breasts swing free as she pulls off her petticoat. Peggy takes off her neat pearl-grey skirt and striped blouse. Then she hesitates, eels her way out of her petticoat, but will not take off anything more. No sense in trying to persuade her. Peggy is sweetly smiling and delicate and biddable; Peggy is also as stubborn as...
'Stubborn as a cat,' Clare said once. 'She'll purr and wind herself round you until you think you can do what you like with her, but have you ever seen her do anything she didn't want?'
Now they are ready. Suddenly they're ageless, girls from anywhere, from any time. They catch hands and run towards the water. Hannah and Clare run straight and easy, seaside girls who are used to broad, flat sands, and to roads over the moors with the wind against them. Peggy has learned a more ladylike way of throwing out her legs from the knee. She runs ineffectually: perhaps she is expecting someone to help her? Peggy is the slightest and smallest of the three. Her body bows, sheltering itself.
The girls stop by the water's edge. Peggy raises the hand which isn't holding Clare's, and shades her eyes.
'Fishing-boat coming in,' she pants.
But it's so far away that they don't need to bother about it. No one will see them at that distance. It's just a black speck among the bob and glitter of the waves. Peggy's always had better eyesight than anyone else.
The water grips their ankles like an icy bracelet. They spring back, then slowly, cautiously, in again, then back as a bigger wave licks up their calves.
'Mother of God!' says Clare, hopping on one leg.
'There's only one way to do it,' says Hannah. Once again she's the guardian of their childhood, the one who always knew how cold the water was, how far out they should swim, how long it was before they had to go home. The one who packed their jam sandwiches and bottle of tea into wet cold sand under the rocks so that it would be cool at dinner-time, while Clare and John William ran off heedlessly, straight into the sea...
'One... two... three...' chants Hannah, forcing them all deeper into the water, but Clare and Peggy wriggle out of her grip.
'It's all very well for you,' complains Clare, eyeing Hannah's thighs and buttocks. 'You're subcutaneous.'
But Hannah isn't listening. Eyes shut, blind with purpose, she feels her way into deeper water. The water surges up her, splashing her thighs, pubic hair, belly. She dips her hands in the water and rubs them over her shoulders.
Then she's in and swimming. Paddling like a dog, she swings round to the shore, shouting, 'First in! First in!'
'Come on, Peg, or we'll never hear the last of this,' says Clare. The next wave rolls water up her thighs, and she gasps and shrinks away from it, standing ridiculously on tiptoe.
'Aaah! Aaah!'
Hannah wallows, laughing at Clare. She raises an arm to splash her cousin all over her tender stomach and breasts with a thousand prickling icy needles –
'No! Don't! I'm coming in.'
Clare presses her lips tight and charges, galumphing into the waves. Waves slap, water shimmies on her eyelashes. She's in. She turns like a fish, so cold that at first she can't even breathe, then she gives out a long gasp and a scream to Peggy who is still hovering in the shallows.
'I'm in! I'm in!'
It's too cold to swim. Hannah and Clare roll and bathe in waist-deep water, just where the waves break. Clare's hair straggles down her body. They call and threaten and scorn, but they can't get Peggy in. She is paddling dreamily along the edge of the water, stirring the sand into clouds. Her head is down.
'I'm getting out,' Hannah chatters. 'You'd better. You ought to mind your chest.'
'In a minute,' Clare replies.
Second in, she must be second out. And she wants the sea to herself for a minute, the noise and swell of it, her bare flesh rocking in salt water.
But it's too cold. So cold you could die of it. Slowly, legs dragging, she hauls herself out. The sea is thick and resistant. The wind doesn't feel warm any more. Hannah is lolloping along the beach, getting dry. They have no towels. Aching with cold, Clare begins to walk up the beach. Behind her the sea sighs like a disappointment.
Hannah has chocolate in her pocket. She can save a handful of sweets all week, so as to cram all her pleasures into the one afternoon that's her own. She doesn't have to go to the shop on Sundays, but she might as well. Sunday is chapel, teaching Sunday school, cooking for the family because Aunt Sarah needs to lie down after the walk back from chapel. Aunt Sarah has a female complaint which she will confide to any other female who can endure the mosquito whinge of her voice and the meaty, intimate smell of her breath. Clare shudders. Every Sunday Hannah bastes the meat, and her faint scent of violet is drowned in fumes of gravy and fat.
But even on this perfect day something is missing. Hannah has not had her letter from Sam. There is no new, limp envelope in her pocket, so new that it hasn't yet moulded itself to Hannah's curves. And it's the end of the week, so she should have had a letter by now, or at least a postcard. Perhaps he's on the move again, to another of those places he's not allowed to name. She knows other girls who'd worked out a code with their best boy before he was sent out to France, but she and Sam hadn't bothered. Sam was not much of a one for writing, and Hannah doubted privately whether he would have remembered any code had they managed to invent one. Besides, it had begun to seem as if Sam would never be sent out. He was in barracks so long, and then he caught chicken-pox and had it badly, when the rest of his draft went out. He was so bad that he was put in the isolation ward of the military hospital, and the letter she had from him there was baked brown from its fumigation. But just when they were getting used to the precarious safety of Sam being nowhere in particular, neither a soldier in training nor a soldier at the front, he was sent out with the next draft.
He's never been a good letter-writer. She shouldn't worry so much. Only a thin few lines each time, hoping this finds you as it leaves me, and always the same love at the bottom. His letters are never clean. She holds them up to her mouth and nose and breathes in the musty, terrifying smell of them. But they mean safety, another week of confidence that she's going to get him back again, brown and strange and maybe with that blind-drunk look in his eyes she's seen in other men on the train back from Bodmin Barracks, on leave. But there'll be nothing wrong with him which she can't cure, she is sure of that. No matter what, she only needs to get him back again. She shivers, thinking of the smell of him. His hands. The way he leans back and groans as if she is hurting him. She rubs her legs briskly with her petticoat and shivers again.
The chocolate is soft and warm. Peggy won't take any. She hasn't got cold, like the others, she says. And chocolate is precious. Clare lets her small piece melt under her tongue and spread out across her palate. Her body starts to tingle. She snuggles into a warm hollow of sand and shuts her eyes. She can hear the sand hissing, moving past her ears where her weight has disturbed it. When she was a child, Father warned her not to dig here in case she set the sand moving. There'd been a boy buried one July afternoon, when Clare was five. A visitor's child. They scrabbled to get at him, but it was no good. His mouth and his eyes and his ears were full of sand. His face was blue and bulging by the time they got to him, though her father did not tell her that. It was Uncle John, big and doom-laden in Nan's kitchen. And the state of the poor boy...
_He'd shat hisself all over,_ said John William secretly, when the children were kneeling by the kitchen fire in the near-dark.
She thinks now of the boy's desperate fight against the sand which feels so soft and loose and playful until it's on top of you. Maybe it began as a game. He was pretending to be buried, then suddenly the sand was too high and beginning to squeeze in on him, and he called out, though at first he thought it was part of the game and in a moment the sand would give way and turn back into that easy soft stuff he could make into castles and dams. But the sand hissed at him again and then he could not turn his head for the weight of it and he was screaming until it filled his mouth and nose and he breathed it in, dragging it down into his lungs, dry and parching, and then the sand covered his eyes. Still no one could get at him.
She breathes shallowly. The fabric of her chemise is like a clasp.
'I won't think of it. Let me not think of it any more.' So she used to pray when she was little, scrunched up at the side of the bed, bargaining for a good night as the cold seeped into her from the oilcloth. 'And let me have no bad dreams or nightmares or think of bad things in the night.' It was not the way to say her prayers, she knew that as she knelt there, chilled and guilty. She had said her proper night prayers with Father and then slid back out of bed for this illicit begging-session.
Peggy hugs her knees, looking along the coast, duty plucking at her.
'I'll have to be getting back,' she said. 'I've only got till six.'
'Can't they ever give you a proper day off?' asks Hannah. 'You didn't get home at all last Sunday, did you?'
'Oh, well, Georgie needs me,' says Peggy, vaguely, proudly. 'He won't go off to sleep without I sing to him. And he likes me to give him his supper. He won't eat an egg for his mother. She don't like it!' Peggy giggles suddenly and subversively, then the mask of responsibility drops over her face again.
'Sounds like a spoiled brat,' murmurs Clare. She tries out, briefly, this alternative life. Governess for a delicate child whose parents have come down to Cornwall for the sake of his health. They are after good air and a cheap governess, and they have got both. They call Peggy a governess, but really she is sick-nurse and nurse-maid and cook for him too. Peggy does it all. Even empties the slop-bucket, I wouldn't wonder. What would it be like, to form your life round the shape other people want? She opens her eyes a little and looks at Peggy's back and the wisps of hair at the nape of her neck. Peggy's hair is thin, but she makes the best of it, fluffing it into fair curls at her temples, piling it over a little cushion of false hair.
'Oh, well,' says Peggy complacently. 'Georgie's used to me. They get like that. I don't mind.'
Clare rolls over and pillows her face on her arm. She rubs her cheek along the silky underside of her left forearm, and smells her skin. She remembers how she and Hannah and John William would lie on the warmest patch of sand they could find, after bathing, close as they could 'to get warm' in a muddle of bare skin and tangled legs. John William's salt-stiff black hair would scrub into her face. Hannah smelled of cloveballs. John William smelled of almonds and salt when he was clean from the sea.
'What do I smell of?' Clare asked once, and John William twisted round and buried his face between her bare white chest and her arm. But he did not smell her skin. He put his mouth on the inner, whitest part of her arm and sucked. There was a sharp, pleasurable, drawing sensation on Clare's skin, then John William raised his head and saw the mark he'd made. His suck had brought up a perfect circle of reddish-purple pinpricks.
'Now look what you done to Clarey!' said Hannah.
But when Clare looked at it closely, she liked the pattern of it. Who'd have thought that just John William's mouth could do this? Like Nan's pincushion with the pins all set in it. And John William smiled proudly.
'You could a made her bleed,' said Hannah.
Could he? All three looked at each other.
'Could he really, Hannah?'
'Course I could. It's cos you're so white, see, Clarey.'
Yes, she was the whitest and frailest of them all. And she had a real dead mother, while they had only whinging Aunt Sarah. And Nan always told them they had to look after Clarey. Clare looked at John William and laughed, showing her white teeth.
'Don't roll about like that in your chemise,' says Peggy sharply. 'What if somebody comes? You ought to get yourself dressed decent.'
'This 'ud be dull compared to what they could have seen five minutes ago,' says Clare. But all the same she reaches for dress and stockings, shakes out sand, does them up. Her hair is drying. Wet, it looks almost black; dry, it is dark red. Her features are almost extinguished by the blaze of light on them. Hannah's broad, strong-planed face survives. Peggy is pretty, as always.
Clare stands and looks out to sea. There's a good view of the lighthouse from here. Water curls whitely, breaking round it. But you can't see the lighthouse-keeper's little garden from here, where the earth is seeded with salt each winter, then sweetened again with blood-and-bone and a coating of agar each spring. Now there's a boat coming round the point. Not a fishing-boat. A patrol. Lucky they'd got dressed. The patrol-boat sweeps right in here sometimes, nearly to the shore. Now, it keeps on coming black and squat through the sea. She wonders why they're round here today? They've been further down all week, off Zennor Head. Nosing blunt-headed under the cliffs, looking for something. Looking for fuel-dumps, someone told Grandad. She wonders what a U-boat looks like when it rises above the water. It must be very like a basking shark, she thinks. She has seen basking sharks from the cliffs, lying offshore in warm weather, seeming to do nothing and said to wish no harm, though they can turn a boat over if they are frightened. But the U-boat strikes, then it comes up black-shouldered and streaming with water and lies there watching to be sure of its kill, while men thresh in the oily water, and nobody comes. Next day the waves are full of tins and oil and smashed wood.
'Come on, let's go. It's getting late.'
The sun is much lower now. The shadow of the dunes covers them as they start to toil up through the sand.
# Two
Low drystone walls fill the vegetable garden with sheltering heat. A thin white-fleshed man in soft grey trousers bends over his spade, puts one foot on the lug, pauses. A trace of coconut blows past him from the gorse, then the spice of wallflowers. Gorse blazes in a long tongue of pure yellow all down the hill above the cottages. Beyond his garden wall the land slopes, glistening, down to the farm and the sea beyond. It is a landscape of irregular small fields shaped by Celtic farmers two thousand years ago. Lichened granite boulders are lodged deep into the hedges. They stand upright in the fields, a crop of stone. Lanes run tunnel-like between the furze down to the farms. Here, by the cottage, the lane dips and dampens and is lined with foxglove and hart's tongue fern and slow drops of oozing water. It is so quiet here. He has heard no human noises at all since he walked up from Lower Tregerthen, the Hockings' farm at the bottom of the lane. The men are working several fields away this afternoon, carting dung for mangels, and the wind blows their voices away from him. He shades his eyes and looks right out towards the sea. It is bare, glittering like a plate. He will walk there later, he thinks. Then a sudden uproar of lowing makes him turn. The cows three fields away to the right are staging their daily stampede, striking the hard ground with their hoofs and jostling each other. Three of them break away and canter to the field gate, lashing their tails. He laughs. He likes these small wild Zennor cows which imagine they are buffaloes in their tiny wind-swept fields. These are not the milky, huge-rumped Midland creatures he's used to, with their slow tails switching as they eat their way across a meadow. Here, the grass is sweet but the herd must scramble for it.
He hears the steady flagging of the breeze in the petticoats he hung out on the line that morning. They are hung deftly, so that they keep their shape and won't flap against the briars. They are white, soft, full petticoats, with good lace on the bodice and hem. The lace comes from his wife's first trousseau, prepared for her first marriage, and she has kept it through divorce and remarriage.
The soil is poor but healthy. William Henry has brought up five loads of dung from the farm. Lawrence has spread the dung and cleared stones and weeds. He is growing vegetables to eke out his tiny income. He earns his living by his writing, and it has shrunk close to nothing since his novel was seized by the police in November 1915 and prosecuted for obscenity. The book is shameful, say reviewers and prosecution. It is a thing _which creeps and crawls_. It dishonours the lofty sacrifices of our soldiers. All the publisher's copies have been destroyed, and the book is banned. He does not know when he will be able to publish another novel. But with a remote cottage rented at five pounds a year, and cheap rural living, he hopes that he and his wife may get through the war. He has turned his back on London, on reviewers who puff themselves up with rage and trumpet that his book _has no right to exist_ , and on influential friends whose support is lukewarm and who have trimmed their sails to the winds of censorship and war. He grows vegetables because he has no money to buy them, but also because it delights him to see bright sparks of life coming out of the earth. The soil is clean. Fragile carrot tops poke out of the earth, the broad beans are shooting, and soon his potatoes will be sprawling over the potato hill. He plans to make their household self-sufficient in fresh food. He notes that the cabbage plants are drooping: it's been a dry, brilliant month, hot for May. Here spring comes early and there are few frosts, but wild nights and salt-soaked winds kill off just as many plants. The heat on his neck is Mediterranean, glowing, scented. The sea is dark, with a broad purplish stain where the channel deepens. It would remind him of Italy, except that the land here is darker and wilder.
He will not look backwards, at the sprawl of England and the dark blotch of London, seething with rumour and propaganda. He'll work the land here. And he'll go over and help on the farm at Lower Tregerthen because he loves the life of that farm as he loved the life of The Haggs years ago. He loves to peel pickling onions and talk in the kitchen as the hens run by the open door. He loves William Henry's dark, warm smile as he pulls off his boots and sits on the settle by the fire, turning over the pages of Lawrence's _Geography of the World_ with respectful fingers. When the hay harvest comes he'll help again. Let the other farmers murmur about _free labour at Lower Tregerthen_ when they come into Zennor churchtown on a Sunday morning. They know everything that goes on, and now they are bitter with losing their own sons to the war. This bold, sculpted landscape is a watching landscape. A human being is a dab of movement in it, visible miles off in the fields. Absolutely spied-upon. Your neighbour may be listening, stone-still in the lane below your cottage, and you will never see him. But he will hear every word you say as you sit talking to your wife near an open window. He will listen to Frieda's German voice reading aloud letters from her German mother. He will hear them singing strange-vowelled Hebridean songs from the new collection a friend has sent them. He will spy on their rages and their reconciliations.
Neighbours listen, and murmur, and the murmuring spreads as quickly as flowers of gorse opening to the touch of the sun. The farmers around Zennor pass on the news. Newspaper columns tell them that it is their patriotic duty to watch, to inform, to spy for spies. There has never been such spy-fever as there is in England this summer, brewing like the unseasonal heat of May. The farmers' wives sit in their traps after church on Sundays, ready to be driven home to their dinners, and among talk of heifer calves and the price of eggs they whisper about the foreigners. The German woman. The man who speaks against the war. The farmers and farmworkers drive over the moors to market in Penzance, or down to St Ives for the Saturday market, and the talk goes with them, half jeering, half judging. Talk licks its way from one mouth to another. These two may think themselves remote, aloof, left to birdsong and the pecking of the typewriter. But they are not. Most inflammatory of all, they act as if they are alone and have to account to no one. They act as if this countryside is really the bare, shouldering granite wilderness it appears to be. This brazen couple ignores the crossed, tight webs, the drystone walls, the small signals of kinship, the spider-fine apprehensions of those who've lived there for ever once they feel a fly strumming somewhere on their web.
_He may have fooled the Hockings, but he don't fool us._
Food shortages are reaching crisis-point as U-boats down the incoming merchant ships. At last the government has been galvanized into action, and voluntary rationing is being introduced. At last land is being ploughed up to grow desperately needed food. In London and Manchester and Birmingham allotments sprout. Half a million of them have been dug within the past year. Sooty, industrial air keeps off blight and rust as effectively as Atlantic winds. In Nottinghamshire miners dig up waste ground, and keep rabbits in their backyards. He knows them so well, those miners sitting on their heels against a wall, raising their faces to the weak sun, wrung out after their long shift underground. Widows scratch earth without proper tools and with only dim childhood memories of Gran's cottage garden to guide them. The rationing system does not work and the only margarine they can get tastes like axle-grease as it congeals around their children's teeth. Bread is short. The rich eat tins of imported larks, while rumours fly that the Germans are melting down corpses for oil. At home, within government offices, there are better-kept secrets. The country is lurching towards the point where there will be only two weeks' supply of food left. Lords and ministers look eastwards to Russia, where revolution is breaking like a wave on the heels of hunger and war. The menu at the Ritz still offers five courses larded with cream. And those who can afford it must eat up all the caviare, because the poor would never swallow those clotted, salty, unfamiliar globules. To eat caviare is to perform a service to one's country.
The man leans on the lug of his spade, then eases it in. There's not much space left, but he can get a catch-crop of salad stuff just here, out of the wind. Radishes, some spring onions, a bit of lettuce. The King has just issued a proclamation urging his people to eat less bread: 'We, being persuaded that the abstention from all unnecessary consumption of grain...'
There are stray marigold plants which have self-seeded from the summer before. He bends down, works the soil loose around the stems of the plants, and lifts them out. Under the wall there's sandy soil where thrift and valerian grow. He fetches a trowel, makes space for the marigolds and dibs them in. The sharp scent of their leaves comes off on his fingers: he sniffs them. The petticoats belly out over his head, then flatten and hang limp in a sudden drop of the wind. The sun sucks out smells of turned earth, bruised onion, marigold. Larks shrill high up, so constant that he no longer attends to them. He runs his hand over the wall, liking the fit of one stone into another. But just here the wall is loose: it has been hastily repaired, hand-to-mouth. He applies a little more pressure and the top rows of stone rock gently. Perhaps he could take down the top three rows of stone and rebuild them? There will be time. It is a slow, patient job, the kind he likes. He kneels to look more closely at the base of the wall. There is a clump of white foxgloves, not yet in flower. He remembers them flowering in the long June dusks last summer, like lights. Now they show their stubby buds again. He has seen the year round in this cottage. He did not think to do so, but it is comforting in the shipwreck of the war to watch the seasons come round again here.
There is a cry from the window. She has woken up. She leans out over the sill, her broad forearms pressed against the stone. She is magnificent, frowsty with sleep, looking to him like the pulse and colour of health itself. But she is not quite well today, she says, and she went up to bed after their lunch, to sleep or perhaps to watch the light move round the white walls and to think of her children. If she thinks of them while she is with him, he always knows. There is a particular look on her, not so much sad as intensely recollecting something in which he has no part. He darkens with rage. 'Aren't I enough for you?' he asks her.
Or he is tender, mending her shawl for her, telling her what they are going to have for dinner, bringing in a handful of the first lettuce, seducing her back to the present.
'Aren't I enough for you?'
Now she leans out, feeling the sun on her arms, and waves to him. She sees row after row of little vegetables pricking the earth, more than they can possibly eat, she thinks. But it is good for him. It eases the lines of strain on his face. And that terrible look he has sometimes, after reading one more letter from a friend chiding him for his stupid obstinacy in opposing this war as he does. Does he think he is an Old Testament prophet, crying woe unto the people? He certainly behaves like one. And one must be frank, people won't stand for much more of that sort of thing. He reads through such letters and throws them on the table in disgust. Then she sees his 'white look' burning on his face. He will burn himself to nothing. So let him go to his vegetable garden, and to the farm, even if it leaves her on her own, listening to the constant sift of the wind over the roof.
She runs her hands back through her thick, rather coarse golden hair. She is putting on flesh again, even though there is almost nothing to eat. She is hungry for good things. She daren't think of cakes, of pastries and cream, of whipped hot chocolate and _torte_ after skating. He has put in some wallflowers under their window, and the heated scent of them rises to her, so that she leans out to catch more of it, to draw it up to her. Soft pansy-brown clumps of them, spiked with seed pods, with bees working away at them. He thinks _he_ is the only one who notices such things, but she loves them too.
He goes back to his work, hoeing between rows of onion-sets. She sighs. She'll get dressed and make tea for them both. There will be letters, perhaps, from their _real_ friends. Perhaps there will be one from Katherine. No one else writes such letters as she does. But they have to be careful now. They must not write what they think any more. Lorenzo is almost sure that letters are being opened and read. They are beginning to blunt some words and to sharpen others with hidden meanings. But no one is going to frighten Frieda out of writing to her mother in Germany. And she _will_ read the _Berliner Tageblatt_ , where she finds the names of men she grew up with among the lists of the dead. And their younger brothers too, who were children when she knew them. Or perhaps she will read of the exploits of her cousin whom they call the Red Baron, Manfred von Richthofen, who shot down five British planes in one day last month. But how can anyone be so stupid as to suspect her? How can anyone be so petty? After all she is a British subject. How can they call her a Hunwife? Indeed she has shown her preferences by marrying not one but two Englishmen. She has children here, even if they are not the children of the man she is married to.
'No. I must not think of the children.' It is dangerous to think of the children. It is like Lorenzo thinking of the war. _The children ran from her_. They ran from her when their governess said, 'Run, children! Run!' They saw their mother and they ran. She had come to meet them from school, and there they were with the governess. She heard their high, sweet voices, but she could not hear what they were saying. Barbara had a blue hat on, a new one which she had never seen. Then Frieda was stumbling towards her children with her arms open to seize them and hold them and breathe in the warm smell of their hair and their skin. But the governess had been told not to let the children see their mother, the wicked woman who had run away and left her husband and her children. She called out: 'Run, children! Run!' And the children ran. They did not know who I was, Frieda tells herself. They were frightened and they did not recognize me.
The light changes again as a fresh ripple of wind crosses the landscape. For a moment it looks like heaven. While it looks like this, nothing bad can happen. Let them fret and bluster with their boundaries and rules and prohibitions. Let them tamper with our mail and carry on with their surveillance which is so clumsy, so ponderous that it is comic. Surely it can never come to anything? We must continue to live as usual, to show that we have nothing to hide, that we are not afraid. These oafs of military police – they are just something one has to live with. Something decent people pretend to ignore.
Here, they are hundreds of miles from London's tumour of officialdom. Here, it is nearly all sea. Little dots of people on rock, a huddle of grey farm buildings wherever the land folds in to make shelter for them, then miles and miles of sea, stretching sheer to America.
'Aaah, how beautiful!' she says, opening her white, powerful arms. She will make tea for them both, and then they will go for a walk. She turns her back on the window and starts to pull on her stockings. Her husband bends too, rasping his hoe through the bindweed he can never quite eradicate. Each fragment is capable of rooting itself, and forming a fresh stranglehold on his young plants. It is rank green bindweed with its coarse white flowers, not the little, scented mauve-striped convolvulus which grows along the cliff-path.
Another boat noses into sight, heading for the cliff-base. It looks like nothing from here, but it is full of living sharp-eyed men, doing their job. Only doing their duty. What a world this war has made, where everyone is doing his duty, or what he conceives to be his duty. There is a rumour of a German fuel-dump near here, tucked into one of the hollow, sea-echoing chambers along the coast, ready to feed the submarines. You could never search this whole coast. And the U-boats are prowling along this coastline, sinking ship after ship after ship on the Western Approaches, leaving the British authorities helpless and raging with the pent-up secret of their helplessness. U-boats prowl, bread queues breed, allotmenteers dig up the railway embankments. If there is no food, will there be revolution here too? What if the soldiers at the Front get letters from their dear ones at home saying: _We are hungry. We are starving. You must help us_.
She dresses, he digs. Neither sees the patrol-boat swing right in, balancing itself on the swell. It is taking advantage of the calm weather to come close under the cliffs, searching for evidence. Is there a patch of oil, or a place where someone might set a signal-light? The boat rocks while its crew look and listen. Their eyes rake the rocks. Nothing. _But there must be something_. It is four o'clock. The men out in the fields straighten as they see the girl reach the gate with a cloth-covered basket and a jug of cold tea. She will hand it over the gate to them, because she does not want the dung on her skirts. Meanwhile the postman toils seven miles uphill from St Ives, with the _Berliner Tageblatt_ in his saddlebag.
# Three
Six o'clock and the house is silent, as Clare opens the porch and stands still just inside the front door. Westerly light drips through stained glass on to her hair, through the languid figure of a craggy-jawed Pre-Raphaelite beauty. Her long green robes make green blobs on the black and white tiled floor. Clare runs a finger round the outline of a big, bare foot which peeks out of the draperies. She used to reach up and huff her breath on the toes to see them cloud. She used to think the figures were angels, not second-hand Lizzie Siddals.
'Is that you, my darling?'
It is her father's fond, light, abstracted voice.
'Have you been out, Clare?'
She goes down the hall to her father's room, which smells musty after the sunshine. The room is small and full of furniture: a brown, slender-legged bureau desk, two fat brown leather armchairs, a table, a dark crimson Turkey rug. The window is open to let out the smell of his cigar. He only smokes in this one room, out of consideration for her. A narrow-skulled Burmese cat turns and glares at her from the armchair opposite Francis Coyne.
'Sheba,' he croons reprovingly.
Maps are sprawled out, open, on the floor, the chair arm, the rosewood dining-table at which they never dine. Clare's own sketchbooks are piled up on a corner of the table.
Francis Coyne is mapping the distribution of travelling plants in an area seven miles around St Ives. He is more interested in travelling species than in the close-rooted kind which has always been at home here, clinging to its habitat. Seeding by bird-droppings, accidental migrations in sailors' knapsacks, carefully harvested pods and cuttings which survive, freakishly, half-way around the planet: these interest him. When Clare was a child, the only fairy stories he told her were the stories of plants: their places of birth and their uprootings, their adventures, adaptations and survivals. At six and seven and eight she had listened, fascinated; now she's nearly twenty, and she has become impatient. Or perhaps she had never been patient, or even interested at all. She had drunk up attention, and even if it came to her filtered through a strange dialect of pedicels, racemes and bracts, she was capable of burrowing past the information to the cadences of her father's voice, his smell, his light hands that never clasped her as tightly as she wanted. Now Francis Coyne is supposed to be writing a book, and Clare is supposed to be helping him. He is a self-taught botanist, an amateur, as he will admit to anyone who asks him. No expert. But it is his enthusiasm, gentlemanly and kept well within bonds.
'Don't pretend,' says Clare. 'You know I've been out. And it's six o'clock. We went right up to Hayle and came back on the railway.'
His eyes gleam briefly, acknowledging that yes, he isn't as absent-minded as all that. He has noticed her absence, but he preferred to go on fossicking among books and maps rather than realize the time and light the kitchen stove and think about what there was to eat. It was only what she'd known and expected.
'You make tea while I wash,' says Clare. 'I'm all salt – look.' She holds out her arm, and he peers. It's true, there is a delicate rime of salt on her skin.
'Been paddling?' he asks.
'Swimming.'
'You must not take cold, Clare.'
He says it because he feels he should, but they both know that Clare chooses for herself.
'Have you anything to show me?'
'I didn't take my sketchbook.'
Clare is the one who can draw. Her pencil sketches are delightful, her father thinks, firm, yet delicate, beautifully shaded. He has taught her which details she must include for botanical drawings. But she has not taken her sketchbook.
'I'll light the stove,' he says with conscious generosity.
'No, don't do it yet. I'll boil the kettle over the kitchen fire. I asked Hannah, but Uncle Arthur hasn't any kerosene. There won't be any until the middle of next week; Wednesday, Hannah thinks.'
'Can we manage until then, with what we have?'
She shrugs. He doesn't really want to know, so why does he ask? She will have to manage.
'I shall finish this, then, while you get ready.'
And he relapses, book in his hand. She peers. It is some privately printed slender volume of poems. She hasn't seen it before. Surely he can't have been spending money on books again? Well, she will say nothing now. The cat, pleased that Clare has gone, knocks lightly against Francis Coyne's dangling hand. The hand curls and caresses the cat's long spine, which pours away under his touch.
The whole golden evening arches over the bay as Clare goes upstairs. Her calf muscles ache from scrambling the dunes: she'd like to have a bath, but there isn't enough fuel to heat the water. She pauses at the tiny landing window and looks out. They're high up here, in one of the rows of new villas built twenty years ago, already well weathered by winters of storms. The houses have stone-cladded fronts and stained-glass panels in their porches, and little gardens which are no more than backyards. But the rooms are light and airy. Her father thinks the house hideous. He has lived there since the builder signed off in the plaster of the front sitting-room, and he and Clare's mother were left to tiptoe around the raw newness of the house, planning its decoration.
Clare was two then, kept up in lodgings in London until the house was habitable. When she first came, she was frightened by the noise and the size of the sea. It was winter, and she thought their house would tumble down the hill and into the dark grey fighting waves. Her father would have preferred one of the whitewashed cottages down in the town, among narrow lanes and sudden astonishing views. But her mother wouldn't think of it. Dirty, insanitary places, full of noise and drunkenness when the boats came in after a successful catch. She loathed her husband's taste for the naïve and local. Dimly she felt that it insulted her by putting her firmly and gently within the category of the picturesque. She loathed the way he refused to know right from wrong. He could well afford to have such fancies, being what he was, coming from where he did; but she could not. She had left this town half trained as a maid by the minister's wife, made bold by a frightened ambitiousness which she revealed to nobody. She would touch fine things, smell perfumes and silks, put out bon-bons in silver filigree dishes on walnut dressing-tables. She was to be polished. The scouring away of her had worked. She was not what she had been, and yet she wasn't quite what she had been moving towards. She worked her way up to lady's maid in a secluded terrace with its own railed gardens in Kensington, not far from Brompton Oratory. Her hands were perfect, broad, with cool, tapering fingers. She had a touch that would soothe anyone, and a gift for dressing hair so lightly and easily that her lady scarcely felt the pins going in. Just her lady to dress. No daughters. Engagements to luncheon, callers, concerts. A muddle of Kensington ladies fluttering from house to house for tea and soft, murmurous, complaining conversations.
A celibacy of servitude, almost classless, had begun to shroud Susannah Treveal. She came across Francis Coyne in the course of it, and had scarcely thought of him as a man, let alone as a man with a possible sexual interest in herself. After all, he came to tea with his mother. She had dressed his mother's hair one afternoon, when Mrs Coyne had been struck down by migraine and had had to lie down upstairs in one of the cool bedrooms with wistaria tapping against the windows.
She was soon enlightened by the other staff. These Coynes were not as grand as you might think. They had the name, but Coyne Park was not what it sounded. All the land had been sold long ago to meet recusancy fines, and now there was just the place itself, with the Home Farm and a few cottages in the village of Coyne. Oh, the name might sound fine, but you wouldn't catch Ethel or Ivy working there, away down in Somerset with nowhere to go but between blank green fields. And the Coynes always had too many sons. Francis was the third, and he was no more likely to inherit than a parakeet in a cage. The eldest son was married already with a long family in the nursery. Francis's mother survived on this relation and that. He'd have a small income and he'd have to look like a gentleman on it. Wouldn't you think he'd have the gumption to go into the army? And he couldn't go into the Church. Pity, that. It was a useful place for younger sons.
'Why not? Why can't he?' asked Susannah, seeing Francis suddenly in a pulpit, long hands laced as he looked over and down at the congregation.
'They're Romans. Didn't you know? _She's_ always bobbing into the Oratory.'
'Oh!' squeaked Susannah, who had been named for the mother of all the Wesleys. This was worse than his lack of money.
Yet she felt for him. There was something about him. His long, finicking fingers: the general uselessness of him. He was a luxury item. He could make her laugh. He was light, he took nothing at its full value, she thought, not even the marriage which made her sweat and tremble at night beforehand. She said nothing to them at home. She kept dumb as a hare. Her wooden little letters didn't cease, for fear they'd suspect. For nearly two months she traipsed to the presbytery of Holy Cross every Wednesday night, with Francis, to receive instruction. She couldn't help it: the first time she set foot over the threshold she crossed her fingers behind her back, fearful that her sinning body would burst into spontaneous fire. She assented to every proposition put forward by the priest, who was kindly and tired and impatient of Francis's love of argument. Before she knew it she had _turned_. They would be able to have a nuptial mass, though they had no one to ask – Francis's family were horrified by the marriage, her own knew nothing of it, and Susannah had no friends but her fellow-servants. Francis had friends, young men who laughed too much, in her opinion. She imagined that these friends would come to an end, after their wedding.
She could avoid telling her family of her courtship but not of her marriage. It would not have been beyond her father to mount an assault upon London, in the pride of his foreignness and his honesty, should he suspect from her change of address that she had taken wrong ways. So the letter was written, and she closed off her mind to the rage and torment she could feel far down in that bony, jutting finger of land. They did not come to find her or fetch her. She wrote again when Clare was born, and received a present of a shawl from her mother, who had four huge virtuous sons and preferred girls. Susannah began her campaign to come home, away from Francis's friends and the traffic and books and rare, terrifying visits to the Café Royale and the eddy of people who either believed in a god she could not believe in or in no god at all. Sometimes she suspected Francis of belonging to both these camps at once.
Besides, she'd become delicate since giving birth to Clare. She did not put on weight. She coughed, and her breath smelled. They would be better down in Cornwall, where living was cheap, the air was pure, and Francis could write his books, since this was what he wanted to do.
And they could get a fine new house with no damp in it, no insanitary passages, no bad drains, high up above the bay, triumphantly above the piggling streets where she'd lived as a child. He objected that there was no convent school for Clare, but surely he could teach her himself – with all he knew, what better arrangement could there be?
Two years after they moved, she was dead of phthisis. A hard and ugly way to die, but he had stuck to her right through it, paying for doctors, injections, special diets, Jaeger wool. He held bowls for her and changed her sheets after she woke at three o'clock, wet with a night-sweat. By the time she died she was almost convinced that he must have really loved her, and had not married her simply in order to carry out some mazy fantasy of his own or to get his mother off his back for ever.
She wouldn't let the child come in her room. 'Don't let Clarey come near me. Send her downalong to Mum's.'
The lady's maid speech was nearly gone now: used up, like her breath. Clare went down to the insanitary cottages, and played by the harbour with her cousins and second cousins, and ran out to see the boats come in with the other children, getting in the way, getting yelled at, jeering at the blue lobsters which crawled over one another in their pots, insect-like, claws tied for the London market.
At home, the shut door. Father going in and out, smelling funny, and a doctor with legs like scissors as he got out of his trap. Covered bowls and buckets. And all night, whenever she woke, she could hear gargling and coughing from the thing that was there, behind her mother's door.
They all went to see her buried, a foreign and appalling experience for Susannah Treveal's family. Francis had insisted that her requiem be said in the little chapel at Coyne. She was his wife, and she would lie among the wives of the Coynes. Besides, there was no Catholic church in St Ives yet, though the site for it in Street-an-Pol was bought in the year of Susannah's death, and the Catholic community was growing fast. So there sat the row of Treveals, Nan and Grandad, Uncles Arthur and John and William and Stanley, their wives, their children. Not one of them had ever set foot in a Catholic church, nor had they thought ever to do so. They sat there upright and disbelieving as the coffin containing their child, their sister, was asperged and muttered over, and the censer was swung around it. It was a strange gathering of Methodists and Coynes and Francis's atheist friends from London.
And look at him now, alone with black-drenched four-year-old Clare, at the chapel door. He did not quite look as a widower should, but Nan allowed for the pinch and tuck God had put into his lips, which made him look as if he were smiling slightly. All the Treveals had seen with their own eyes how he had looked after Susannah. And now he had this child to rear. Susannah's brother Stan's wife had no children, although they'd been wed for twelve years. It was a female trouble she had. She would take the child gladly and leave Francis free for his work. Clare would be in the same street as Nan and Grandad, and just a step away from her cousins, clattering off to school with them. It would get her out of that sickly house up the hill, full in the face of the gales. So the Treveals reasoned among themselves, and the idea grew and shimmered as it passed like a blown bubble from one family branch to the next. However, there seemed to be no right time to tell Francis of it. After the funeral he stayed at Coyne with the child for a week or so, while the Treveals went back home. They came to meet him at the station when he returned to St Ives, but he preferred to walk straight home with Clare. 'She must get used to the house without her mother in it,' he said to Nan.
Francis Coyne set off uphill with the child, both of them walking soberly, but not downcast, not easy objects of pity. Then Francis bent suddenly and scooped Clare up on to his shoulders and balanced her there, a smudge of black dress and red hair bobbing against the pale tearing early winter sky. He had hold of her boots, and she gripped him firmly around his neck, spoiling his collar. Both of them leaned forward as the hill steepened, then they went out of sight together.
Clare lived half up the hill and half down the hill. She eased and flitted between the cottages and the villa, speaking two dialects, learning two pasts, instructed in two faiths. Grandad could not bear the idea of Clare 'bobbing and ducking', first in the mission up the hill, and then in the newly built church of the Sacred Heart and St Ia. So he refused to think of it, and absolutely ignored that part of her life. First Communion, Confirmation; Nan might question and remark, but Grandad never would. Instead, he told Clare of Wesley ranging the spine of Cornwall, preaching at quarries which were flooded with people as if with water. She learned the Wesleyan hymns and the legends of John Wesley's childhood. The rectory at Epworth, the savage parishioners who tried to burn the Wesley children alive, and five-year-old John plucked out of the flames by his rescuers. ' "A brand plucked from the burning." Pray that'll be true of Clare Coyne.'
Clare's education did not begin or end at fixed points. Reading was easy. She would have liked to go to school, but there were delays, bothers that were never specified. Soon there was no need for her father to explain to her why she could not: she understood. There was enough money for her French and arithmetic lessons with dull Miss Purse, and the rest her father taught her. And there were so many people around her who could teach her particular, desirable things. Her cousins taught her to swim and whistle, fish and torment crabs, handle a boat and lie. Aunt Mag taught her plain sewing. Clare liked the tick of her needle through cotton. Nan and Aunt Mag made everything: drawers and petticoats, shirts for the boys, summer frocks for Clarey and Hannah. Clare saw a sampler and longed to sew one herself, so that Nan and Grandad could put it upon the wall in its fine frame. Nan and Aunt Mag made up clothes for the draper's shop Aunt Sarah and Uncle Arthur kept, where Hannah went to work as soon as she was old enough. They worked in the front-room in good light and bad. Nan did not hold with samplers.
'Waste a time, Clarey, when you could be doing something useful. Here, thread this needle for me. Your eyes are better'n mine.'
If only Nan didn't get those sharp, shooting pains in her fingers after a day's sewing. Sometimes she could hardly hold her cup still. 'But come next season Uncle Arthur's getting yer old Nan a Singer. Think of it, Clarey.' The Singer came, black and gold, with its treadle a criss-cross weave, then a blur as Nan caught the rhythm and fed cloth across the sewing plate. Clare watched Nan's pedalling feet, so perfect in their black boots which were all out of shape at the sides because of her bunions.
'Mark that, Clarey. That un there's the tension spring. Never touch him.' If Clare touched it, the Singer would spring apart and die, and Nan and Aunt Mag would have to go back to sewing every stitch by hand, and it would be Clare's fault.
Nan taught her to cook. By the time she was ten, Clare could cook the dinner for her father and herself. Mutton stew with dumplings, spring cabbage, boiled pudding. Sarah, Arthur's wife, no longer had to go up and downhill with the Coyne dinners. Francis had paid her a consideration, but Arthur didn't like her to be so tied to another man's household. Not with the shop to run as well.
Francis thought Clare was still a child, but to the Treveals a child of ten was a worker. Her boy cousins earned money packing fish. They laid slim mackerel head to tail in salt and ice in the spring; in the summer it was hake, cod, ling, skate, halibut, and a market for all of it. In the late autumn the kipper girls came, and they teased and played rough with the boys between hours of labour. The boys were not allowed to touch the lobsters yet, for fear they'd damage a claw and spoil the price in the London market. Boys were wild and wanted watching. Their hands were raw with brine, nicked, fiery. They stank of fish, and they were proud of it. Their elbows glittered with scales. Then John William got a job after school and weekends as a grocer's errand boy. It gave him the use of the shop bicycle, and he liked it better than fish, he said, as he pushed his cycle uphill and freewheeled down, and kicked away house-dogs with his stout boots, and got given pennies. He knew all the secret drives of the big houses, with their curvy lining of blue hydrangeas and their camellias on the terraces. Hannah would go into the shop. Already she had learned the ways of it. She was quick, and deft, and she did not break things or spoil them, or make marks on cloth. By the time she was twelve she could handle the Singer better than Nan herself. She even dared to touch the tension spring, because she knew what it was for.
John William, Hannah, Harry. Kitchie, William and Mabel's only child. And Albert, Jo and George, Uncle John's boys, up on the farm and working almost since birth. They hoed potatoes, cleared furze off the crofts, leased stones, stood round while the pig was killed. The farm was small, but the land was fair to middling; with a man and three strong sons on it, it might amount to something. The farm came from their mother, who had been a Lee before she fell in love with John Treveal. She was the lucky only child of elderly parents who had nearly worked the lives out of themselves, keeping back a wilderness of gorse from their land.
The cousins grew older and worked harder, while Clare drew. Her music lessons were suspended. Her accomplishments hung in mid-air. There was tension plucking in the atmosphere, louder than piano strings: investments dipped, and her father's small income shrivelled a little smaller. Anyway, she'd hated the noise she made. Under her fingers the piano shot out notes like a machine, not a musical instrument. She'd learned enough to play songs and please Nan and Grandad with them: one good year Grandad bought a cottage piano, and he took pains to have it tuned for Clare. Her style of playing worked for Grandad's hymns. Once, when Grandad was out of the house, Hannah said, 'Clare, play one of your hymns.'
Hail QU-EEN of HEV'N,
thee oh-oh-cean STAR
crashed out Clare, then stopped suddenly, homesick for something that often didn't even feel like home.
'Is that all there is to it?' mocked Hannah.
Clare drew flowers, not because she liked to, but because her father couldn't draw and needed illustrations for his book. She preferred to draw faces. It didn't matter: she enjoyed accuracy of structure and colouring, and she was working. She was illustrating a book for her father. The shading of a corolla counted as work, just as much as lugging baskets full of stoned raisins, kitchen cheese and tapioca up the drive of Little Talland House.
'Clarey's keeping house for her father. And she's doing the pictures for his book, aren't you Clarey? Show us that one you showed Grandad. Beautiful.' Where Nan praised, no one else would disparage.
Clare is nearly twenty. She cooks quickly and off-handedly but well. She keeps the house clean, though she doesn't scrub the floors. Hat comes in to do that every Friday morning, and brings up with her the fish the Coynes will eat for their Friday dinner. But Clare does everything else. She ties up her hair in a white bandeau, puts on a long grey wrapper and polishes, scours, lifts, disinfects, bangs clean. If she could do it without being seen, she'd whiten the doorstep and clean the windows too, and save the money they pay Hat. Francis carries in the coals, because a young woman should not lift coal, for fear of doing something to her insides. He splits kindling in the back garden. For a botanist he is a poor gardener, but Clare grows nasturtiums and mignonette, candytuft, bushes of lavender and rosemary around their small parched lawn. Clare rolls up her father's Turkey carpet and takes it out to beat it over the clothes-line. Her figure bends and plies. She will leave the carpet out in the sun for the sea-air to sweeten it as it blows over the peninsula. She's working. None of the Treveals can criticize her for idleness.
The remains of the Coyne meal sprawl on the table. Plates smeared with ham fat and butter but no leftovers. Last week Uncle John sent them a pound of butter wrapped in muslin. They say the soldiers are sending back butter from France, but the Post Office is going to stop it, now the weather's getting warmer and oily butter is seeping out of its packaging. Clare stirs her tea. Her father's eyes flicker down the newspaper.
'They've moved Sam, we think. Hannah hasn't had a letter,' says Clare.
'Ah. You've seen Hannah today?'
'Yes, I told you, Father.'
'Does he say where? No, I suppose not.'
He looks down at the paper. Lists of dead, figures as high as cricketing scores. There won't be much village cricket, this year, up at Coyne. He thinks of Coyne. Mama rolling bandages and praying for the dead; Benedict and Marie-Thérèse praying as each year the war goes on and one more son becomes old enough to enlist. They have been lucky so far. Young Francis is C2 on account of his eyesight. Stephen is in a staff job. Lucky so far. May 1917. If the war goes on for five more months, it will get Gerard. Gerard is in perfect health, unfortunately. Besides, they are passing young men now who would never have got through their medical boards a year ago. So many men have been lost, and must be replaced.
'No,' says Clare, 'she hasn't had a letter this week.'
'Did you read the letter from John William?'
'Yes, Nan showed me. But it only said what Kitchie told us.'
'Ah, it's excellent news. Excellent. All the same, Sam might have sent Hannah a postcard.'
He hears his own voice, ridiculous, as if the boy is on a seaside holiday, failing to write home that he is having a good time.
'John William will be home soon.'
'We'll have to have him up here for dinner.'
But he doesn't continue. Clare will know what he means. He is your cousin. They are going to make him an officer. Even if he is a Treveal. They are hungry for officers now too, since half the product of the public schools has been scythed down already.
Francis half rises. The room is chilly now. They ought to have a fire, but Clare has stopped laying fires ready in the grate, to be lit in the evenings. He's been sitting too long. He feels cold. Better not to look at the lists. It doesn't do any good. He thinks back to Coyne, and rabbiting in the woods. There's so much blood, just from one dead thing. And the smell of it is cloying, like water from, a spring full of iron. When a creature is freshly killed and its blood spurts out, it makes steam on a winter morning. Reeking. Does Clare know that? What does his daughter think of when she reads the lists:
Missing,
Missing, believed killed in action.
Killed in action.
Perhaps she doesn't read them. Those girls, what do they talk about when they're alone?
One day last September he was walking round the rocks at low tide. He turned and saw into a cove, steep, dry-sanded, trapping the sun's late warmth. There was a girl on her back, eyes shut, mouth slack, unrecognizable. But he knew the young man straddling her, working away at her. Then he knew the stylish blue and white stripe of the girl's skirt. Goods at a discount. Hannah. He turned away to stop himself looking at her white legs, straining apart on the sand, wider and wider, to draw the man deep into her body. She knew what she was about.
Now he watches Clare's hands as she stacks the china, scoops fragments of waste food on to one plate, pours dregs of tea into the slop-bowl. Her hands are calm and capable. He thinks of all that her hands have learned. She can do so many things which he has never learned to do. She can cook, sew, wash, play the piano. How can he know what she talks about when he's not there? Futile even to want to know. But he does. He imagines the girls leaning together, confiding. He pulls back sharply from the thought, flinching at his own prurience. Again he thinks of the lists. Of Hannah's spread legs. Of Sam with the sun on his back. Now Clare feels his look on her. She glances at him and half smiles, but does not stop what she is doing.
# Four
Day after day it doesn't rain. Mid-May, blazing hot, with blackthorn fully out and bluebells bowed over, hanging their sappy stems in the heat. The air smells tauntingly of honey and salt.
When the gorse is out of blossom
Kissing is then out of fashion...
Clare can't get the rhyme out of her head. Her head has turned into a drum echoing it. And on this perfect afternoon she can't endure to be in the house, which is fusty for all her spring-cleaning, and dark, and dusty-windowed. She hurries downhill past the cemetery gleaming white and grey in the sun, each gravestone lapped with new grass. She skirts Porthmeor Beach, and on to the coast-path. The word _kiss_ clings to her like a web she has walked through.
Dry kisses. The smell of her father's hair when she lets her lips rest on his fine-grained forehead. When she bends down over his chair arm to kiss him he looks up at her with unrecognizing, dark-pupilled eyes. He is always surprised by her touch, as if for a minute he doesn't know quite who she is. Kisses. Fierce kisses for Nan when she was little. Grappling Nan's black skirts to her, snuffing Nan's apron, scrubbing her head against Nan's waist. Grandad was never much of a one for kisses, but he liked to have Clare on his knee when he smoked his pipe, and he would let her fill it with tobacco for him. Her fingers were neat at tamping down the shag. She loved Grandad's tobacco pouch too, and he would let her unroll it and breathe in the dark, rich smell which was so much more fragrant than the smell of his burning pipe. But then she had to roll it up tight and not touch it again, for fear of the expensive tobacco drying out and losing flavour.
Kissing Hannah, when they pretended to be married under the mock-orange bush in Clare's back garden, then spitting and raging at Hannah who wouldn't take her turn as bridegroom, and who pummelled their crown of blossom to brown slime rather than let Clare have it. Hannah's raking nails left red stripes in Clare's white forearms, and Clare could not retaliate because she had bitten her own nails down to the quick. No one else had skin as white as Clare's. It went with her hair, they said, and she must mind herself and not get freckled or nobody would want to marry her. But even then it would be all right, because Nan could cure freckles with lemon juice.
Kisses. One hot afternoon when she was twelve and opened the door to the grocer's boy, John William. He walked past her, slapped their order down on the kitchen-table. First time they'd sent him up here. He'd always got out of it before, given the order to one of the other boys, taken a longer round and a heavier load rather than deliver rice and yellow split peas and suet to the Coyne house. The Coyne grocery order was not impressive, and he hated the thought of being sent to serve Clare.
There stood John William, thirteen already. This was the unfair time of the year when John William would seem to be a whole year older than she was, though the difference was only seven months. His face was impassive. His tan hid everything. And there was Clare, caught out in the middle of her washing, with her hands red and chapped from washing-soda. She hid them in her wrapper and said, 'Let me make you a cup of tea, John William.' If she'd had warning she'd have gone up and rubbed cold cream into her hands. She would have combed her hair too, for it was straggly and sticking to her forehead with the steam from the copper.
'I got to get on,' he said. He was beginning to be awkward with her, not only when he was with the other boys, but even when they were alone. He had grown so much this past year. He was well past her father's shoulder, and now he wore his black, soft hair in a side-parting and smoothed it down like a man. The bones of his face were strong. People thought he was older than he was; fifteen, or sixteen even. The shape of his cheek-bones was like Hannah's, but Hannah's face was softer, and fuller and her skin was not so dark. Their bones underneath would be the same, thought Clare. They would have the same skeletons. She was not sure if she liked her own difference from them, or if she was sorry for it.
'I've made some lemonade. It's keeping cool in the larder. Shall I fetch it?'
'All right.'
She went through the scullery and undid the larder door, where flies buzzed at the fine-mesh wire. Her lemonade stood covered on the slab. She poured a glass for him, and took a couple of squashed-fly biscuits out of the tin and put them on a plate. It seemed right now to treat him like a grown-up visitor. He always used to like squashed-fly biscuits. He was still standing by the kitchen-table when she came out. She pulled out one of the deal chairs for him but he would not sit down. He drank off his lemonade in long, smooth draughts. He was always neat as a cat in his eating and drinking. 'That's good,' he said.
He was thirteen, had just left school and was working full time at the grocer's for now. Errand boy but serving behind counter too, slicing cheese and sifting dry goods from sack to blue paper bags. She had seen him cutting cheese with wire, perfectly, but he did not look as if he was in his right place there.
'Better be going,' he said. 'I've got night-school. Got to get over there by seven.'
That night-school. It had rocked the small row of cottages where Uncle Arthur and Uncle Stan lived within a few doors of one another. Nan's was just round the corner, where the alley steepened: you could tell Nan's with your eyes shut because of the pot of sweet marjoram she kept by the doorstep. Everyone rubbed their fingers on the leaves before they came into the house. Uncle William's was further down the hill, with Aunt Mabel's just-perceptible mist of sluttishness on windows and curtains and doorstep. All of them had been fetched up to Uncle Arthur's to reason with John William, who had declared his intention of going to night-school to study. And nothing useful like book-keeping or arithmetic either. He was planning to study physics, chemistry and biology, if you please. He was going to take exams. He had been writing secretly, behind their backs, to places in London, about exams and qualifications. He had been uncovering those things which were mysteries to Aunt Sarah and Uncle Arthur. He was not going to follow Uncle Arthur into the drapery, though there would be a place made for him, and a small wage which would be all he could need, living at home. The whole family must gather.
'Let's hear if you're brazen enough to tell your Grandad to his face how you're going to throw away everything we've worked for.'
He was. Mute and stubborn, playing his hand after years of keeping his cards hidden. He had watched them for a long while, weighing what it was worth, this life they led. He had been the brightest and the quickest of them all at school, doing sums for all the Treveal cousins, writing an impassioned essay on Mary Queen of Scots which earned him a bronze medal, a shilling, and the permanent mistrust of his parents. The other boys would have bullied him for his cleverness, if he had not had a reputation as a wild, unpredictable fighter, calm for months, then suddenly lashing out and fighting until the other boy writhed and screamed out, and still he kept on until three or four of the other boys had to pull him off. He never rolled and wrestled behind the school wall as the other boys did. He went with Uncle Stan to bare-knuckle fights, held in secret with a look-out posted, in quiet hollows on the moors. He saw how one fighter danced and struck, blurring in the dazed vision of the other, then came in to strike again, splitting the skin under a man's eye, sending him blind with his own blood. He would fight like that. He concentrated, sitting tense and still, not noticing Uncle Stan and the other men crouched in their circle, making noises like dogs held back and dragging their leashes as the square of uncut corn grows smaller and smaller and the rabbits double back and forth inside it with stretched panicking eyes, and the men wait for the moment to let slip the dogs.
He would stay in the grocer's. He would earn money enough to give Aunt Sarah his keep. He would go to night-school. He would obtain qualifications. Weren't there all the others to go into the farm and the shop? Weren't they glad enough to go where they were told? Look at them. He counted on his fingers: his brother Harry, Hannah, Uncle John's Albert, Jo and George, Uncle William's Kitchie.
Harry and Kitchie stood just inside the door in stockinged feet. Hannah had run down to the quay and fetched them up just as they were, raw from fooling about on the quay with the other lads. They had their boots in their hands. They looked and listened while the storm raged. Father crashed his chair back from the hearth. Grandad and Uncle Stan and Uncle William stood round him, holding him back. Mother shivered in the kitchen, secretly thrilled at the prospect of a discount on her groceries. Harry and Kitchie exchanged glances, wiped their hands on the backs of their trousers, and went back down to the quay. Best off out of it. And there'd be no shifting John William.
Clare had missed it all, at her music lesson, plonking out 'Für Elise'. The juice of it was related to her that evening by Hannah, but the great scene which would be embroidered and framed in family history over the years had an empty space where Clare ought to have been.
'Are you taking any exams this year, John William?' she asked timidly.
'I don't know enough for that yet,' he answered. 'But I shall.'
'Does Uncle Arthur know? Doesn't it cost a lot to take them? Father said it would.'
'I shall get it. I'll tell you something, Clarey. I've got money none of em know about.'
'Have you? _How_ have you? Where did you get it?'
'I get a shilling a week more in my wages than they know of. I said to old Trevithick I'd do his books for him, when I saw all his figuring was wrong. I save it for fees. And Dr Kernack gave me a guinea. And I get given money – Christmas boxes. I don't tell em.'
Clare glowed, relaxing. This is how it used to be, with John William telling her secret things he tells only her and Hannah.
'But why did Dr Kernack give you a guinea?'
John William hesitated. 'Keep a secret, now, Clarey? Don't tell a soul, not even Nan?'
'Swear to God.'
'Cos he knows what they don't know – what I'm doing it for. I want to be a doctor, Clarey. Course it's hard. I got to do better'n any of em in the exams. And I _can_ do. Cos they're lazy, see. They don't want things enough – not like you got to want em. They don't know about really wanting. I can beat em all. Don't you believe it, Clarey?'
She looked at him. She believed it. John William had set himself like an arrow on this one thing, leaving no space for anything else, and leaving no space for it to fail to happen either. She had never thought in that way herself, about wanting things. She had only thought that you had what you had, and that was all. Now she realized that she was far behind him, and that it was no longer just because of the few months between them. But there was danger in wanting anything that much, and showing that you wanted it.
If he doesn't get it, it'll finish him, thought Clare, and she believed him.
'Only you'll have to talk like me, if you want to be a doctor,' she teased him, paying him back for the hundred times they'd all jeered at her for talking like a lady. She thought he would laugh at her again, but this time he didn't. He was thinking. His forehead tensed as he took in this one more thing he would have to consider. One more part of his plan.
'You can help me with that, Clarey,' he stated. His narrow eyes gleamed with amusement. 'I can listen to you while I'm scrattin over my books.'
'Just think. You might end up rich!' she half laughed.
He looked at her and saw her believing in him. There was Clare, not doubting that he could do it. His sore, lonely self-confidence yielded to her for a moment. He ducked his head, embarrassed to let her see how it moved him.
Was he cross? Had she said too much? She hovered, twisting her hands inside the coarse grey cloth of her wrapper.
'I'll be off, then, Clarey,' he said, and in a suddenly awkward flurry he was beside her and had leaned to kiss her warm white cheek. For a second he was so close she saw something she'd never noticed before: that the iris of his eye was rayed with black streaks and then there was a thin ring of black between the iris and white. He smelled of toffee and fresh air.
Clare remembered Nan's voice. How old had they been? Six or seven. John William hadn't wanted to kiss her when she was going away to Coyne with her father. 'Give your cousin a kiss, John William. Go on, nicely now. She won't bite you, will you, Clarey?' Everyone had laughed, and John William had stood frowning in his boots, hands behind his back, not kissing Clare.
Kisses. Just today, at dinner-time, Hannah panting at their front door, her face flaring dusky red under its usual clear brown,
'I've run all the way from our Nan's. We've had another letter. John William's coming home. Coming on the London train. Will you go with us to meet him?'
'What's taken him so long?' snaps out before she can stop herself. 'Oh, I'm sorry, Hannah! You know I'm glad really. But it's days since we had that last letter. What has he been doing?'
'I expect there were things he had to see after. Uniform and such,' says Hannah vaguely. 'Never mind that. He'll be on his way already. Quick, Clare! I've got to get back to the shop. I've had no dinner and I'm late as it is and we've a new customer coming to choose curtain stuff.'
Clare feels the heat of Hannah, standing so close, her pulse throbbing in her neck after her run uphill.
'Hannah – was there anything else – any news from Sam?'
Hannah frowns. Reluctantly she says, 'Yes. I had a letter yesterday.'
'Why didn't you say? You could have told me!'
'Wasn't anything to tell.'
Her face has shut. She turns to go, then turns back. 'Clarey. Uncle Francis gets the newspaper, doesn't he?' She fumbles for the words she wants. 'Course they can't put what's really happening into the newspaper, now can they?'
Clare thinks of the newspaper reports. Movements. Advances. Salients. The same names of ridges and fields and woods, repeated over and over again. And the lists, day after day after day.
FALLEN OFFICERS. ROLL OF HONOUR: LOSSES IN THE RANKS. _Missing Believed Killed, Seriously Wounded, Missing, Missing Believed Wounded, Prisoner in Enemy Hands, Died of Wounds_... Every shade of loss has its own category. But Hannah knows that. Column after column of names, spreading out from under the classified advertisements. And then a dispatch on another page about heavy enemy casualties, and a significant advance. Clare doesn't read the dispatches any more. They are just words which mean nothing. Father talks over the war news in the porch after Mass and it sounds like the same war news he was talking about two years ago. Except that it was new then.
'The dispatches don't tell you much,' she says, looking at Hannah carefully and choosing her words. 'You know how it is. You have to guess what they really mean. They can't put everything. What did Sam say in his letter?'
Hannah stands there, looking back at Clare. Calm, shrewd Hannah waits for something Clare can't give her. The air's gone quiet: that means the schoolchildren have gone in from their dinner play-time. Hannah bites her lip, waiting.
'Only I've never had a letter like that from Sam,' she says suddenly.
'What's he say?'
'He sounds all wrong. Not like him. I don't know.'
Kisses. Hannah and Sam, locked, melting, in the shadow of the harbour wall when Clare walks by to Nan's in the dusk. Hannah's back arched, Sam's lips on hers, his hands round her buttocks. And more than that. Clare shared Hannah's week of dread last autumn, after Sam went back to France and Hannah's _visitor_ failed to arrive. Patches of crimson flaring on Hannah's cheek-bones as she slid past Clare in Nan's kitchen, hissing the good news, the relief of Mafeking.
'He sounded all wrong, Clare. And the letter – well, the letter didn't come from over there.'
'What do you mean?'
Hannah looks quickly up and down the road.
'Don't say a word.'
'I shan't.'
'It was posted in London.'
Clare is silent for a moment, slow-headed in the sunshine. Then she opens her mouth, bubbling with questions, but Hannah's turned with a switch of skirts and run off waving one hand, too far already to be called back. Already wishing she hadn't said so much. She wishes she hadn't told me, and Hannah's always told me everything.
Thank God to be away from it, away from fear and newspapers, away from letters which come too late and messages which nobody understands. Now there is not another human being in sight. Soon, John William will be here. Tomorrow morning he'll jump down at St Ives Station. He'll walk along with me and Hannah, and we'll talk. We'll be one on each of his arms. Everything will be clear again.
Clare is too hot in her plain white blouse and long dark skirt. She loosens her belt by a notch, and rolls up the sleeves of her blouse to the elbow. She has been tramping the cliff-path for over an hour. The dry turf echoes under her boots. She's turned her skirt over at the waistband to stop it from dragging in sheep and rabbit droppings. Anyway, skirts are shorter now. They have risen five inches since the beginning of the war, but not in St Ives. Even Hannah, who reads the fashion papers from London and can make up her own paper patterns, has only turned up her hems cautiously, for fear of Grandad. There are flappers in London, and all sorts of ungodly goings-on in London parks, according to Grandad. The Wesleyan Conference last year was forced to hear of some shameful goings-on in cinemas. But Hannah has other reasons for refraining from slashing off the bottoms of her skirts. You can't chop at your clothes and expect them to look right. You have to alter the whole proportion of the skirt. The girls pore over last month's _Tatler_ , passed on to the drapery by the Veryan young ladies. They scan a cloudy, radiant picture of Lady Diana Manners making her début at Blenheim. Hannah looks minutely at her face and clothes.
'Butter wouldn't melt,' she remarks, as she snaps over the page.
Clare clambers up the worn, sandy middle of the path. The sun is so strong it oozes through the straw pattern of her hat brim. She'll just get up to the boulders, she thinks, then she'll stop and rest.
But she doesn't. She tugs her blouse out of her waistband and trudges on. She doesn't look at the cliff-edge on her right side, high and sheer now, deeply cleft with gorse and bramble tangling over the drop, above dark turquoise and purple water. She changes her canvas sketching-bag to her left shoulder and watches her boots as if they were someone else's, tramping on and on up the hot path, pebbles skidding away under them, puffs of sand and dust rising. Her ears are full of her own labouring breath and the pumping of her blood. She knows she's walking too fast. If she'd any sense she'd slow down to the easy swing which takes her ten miles without tiring her, the way her father taught her to walk on these cliff-tops and hills when she was five years old. But she can't slow herself down. The throb of the sun and the throb of her heart are the same thing. Hannah's face shifts and melts into John William's. Hannah's back arches, John William closes the kitchen door and leaves Clare alone with the flies buzzing in the larder.
Clare thinks of Lady Diana Manners, wide-eyed, flaunting herself. Parties and balls. Débuts, coming-outs. That is the language of her father's past, even though he doesn't speak it any more. She remembers Hannah flicking over the page, her face expressionless in its absolute withholding of respect.
Faces. What's mine like? Sweaty and hot. This hat is too tight. I'll have a dent in my forehead. Hannah's face. Could I draw it again? Might try, if she'd sit to me. I could tell her I'd do a sketch for Sam as well. John William's coming home tomorrow. His face.
Lists of names in long columns down _The Times_ and the _Morning Post_ , as long as columns of marching men. As many names as there are grains of sand in the sea.
She stops. The sea sighs. The long, sloping Atlantic swell moves in at a diagonal to the cliffs. Today might be windless, but yesterday there was a strong south-westerly blowing up the dust in the streets of St Ives. The narrow land here is just a snag in the sea's passage. It's always trying to find a way through, thinks Clare, picturing the maze of passages under the cliff, and the sea fingering its way through from cave to cave, through membranes of rock. She knows of swimmers caught by the current, rolling in here puffed up and sodden white, bumping against the roofs of caves, splitting their swollen skins on projections of rock. It happens. Sometimes the sea wins. A boat wallows and breaks up in sight of the town. There are hymns on the edge of the sea, and a memorial stone. Nan ducks clothes in a vat of stinking, steaming black dye. Even white petticoats must be dyed black for decency. Dinners are brought in covered pails for the widow. The cost of the orphans' boots is shared among eight families. Nan lets her neighbour weep long enough, then tells her she must brace herself, have a cup of tea, think of her children.
There's a gap in the gorse here, and a patch of smooth, nibbled turf. A clump of thrift bobbles right of the lip of the cliff. Clare throws off her sketching-bag and lies face-down a few feet from the edge. She wriggles forward. Beneath her the sea booms into an undercliff cave, then sucks out. Close turf makes little pricks against her cheek. She hears the drowsy sound of a bee in clover. Clare's heart slows. She's dizzy with walking so fast. She lies under her hat in a small golden haze, hearing the sea drain from its chambers, then come in again. She can feel it right through the rock. She listens. The water shocks, withdraws, shocks, withdraws... Too hot here. Drowsy herself, she rolls over –
'You're very near the edge,' a voice remarks.
Clare shoves her hat back and scrambles up on her knees. This sudden change of posture turns her dizzy. She has to look down and clutch the turf while the world heaves under her.
The man who has spoken from the path comes over to her swiftly. She feels his shadow cross her.
'Move back a little from the edge. You're very close, did you not know?'
A foreign voice, not from round here.
'I don't mind heights,' gasps Clare.
'Oh,' says the voice. 'So you thought you'd swank by going to sleep on top of the cliff?'
That's better. The swirl of black grains in front of her eyes settles into turf, minutely studded with pimpernel and speedwell. She curls her legs round carefully and sits up, smoothing her skirt over her knees. The man drops down beside her. She turns and looks at him.
# Five
She sees lively, bright blue eyes looking straight at her. They are piercing but homely too, because they remind her of Nan. They are not the same colour as Nan's eyes, but they have her warmth and quickness. Not many people ever look directly into another person's eyes. They are frightened of what they may find there. There are so many shuttered, timid eyes.
But his beard is astonishing. It juts from his face, wiry and bright red, and then the sunlight catches it and it's all the colours she'd never have thought human hair could be: threads of orange and purple like slim flames lapping at coals. And yet the hair on his head is so mild, smooth and mousy, lying flat across his head and parted at one side. It doesn't lie very well. It doesn't look elegant, like her father's parted hair.
'Where were you walking to?'
No, he's not from here. A foreigner. His 'you' is nearly a soft, tender 'yer', but he's not Cornish. Nor London. Where then? It's not an accent she's heard before. And yet his manner, his whole bearing, is like a gentleman's. At least, it's like that of a man who does what he chooses, and she supposes that means he must be a gentleman. Uncle Stanley walks as if he cares what the world thinks of him, for all his bulk and assertiveness. There is something quite different about this man. A sense of freedom.
'I wasn't going anywhere,' she says. 'Just walking.'
'You should go on as far as the Carracks. There's rabbits sitting up, and a raven looking at 'em, wishing they were little juicy ones.'
'I know that part well. I've often walked there,' she says coolly, piqued again that this stranger should instruct her in her own landscape. How can he claim to know it as she does, when she has been walking these cliffs and moors and beaches since she was old enough to keep pace without whining at her father's heels?
'Are you staying near by?'
There. That'll show him she considers him merely a visitor. This is her own country. But he looks as if he believes the whole world belongs to him. He kneels on the close-cropped turf, perfectly at his ease. There are hollows under his flat cheekbones, but his eyes are brilliant. If he leaned forward just a few inches, their faces would touch. She ought to be nervous, but she couldn't possibly be frightened of this man.
'We're living up at Higher Tregerthen, near Eagle's Nest. D'you know it?'
She does. Only just at this moment she can't quite think exactly where it is... She screws her face up, thinking.
'Is that a part of the Hockings' farm? Lower Tregerthen?'
'It is, it's by there, just above the farm. But the cottage doesn't belong with the farm. We rent it from Captain Short, down in St Ives. I daresay you know him, too.' A little sardonic sideways look.
Oh. Yes. Now I know who you are. Or I think I do. You're the man they have all been talking about.
'Did you ever hear the like of it! Captain Short's got Germans staying up at Higher Tregerthen. They've taken his cottage. A great big German woman, by all accounts, shouting out at him in German just as if there wasn't a war on. They fight like two devils, not man and wife. And she wears red stockings, you never saw the like of them. And she's got this little bit of a husband with a big red beard. Queer-looking pair, aren't they, Mabel?'
'He looks like she found him in a Lucky Dip, that's what I say.'
Aunt Mabel and Aunt Mag exchange the smirks of women who have found themselves solid substantial husbands.
'All the same though, there are things not right up there. They say they've put different coloured curtains up. _In the same window_.'
'Why, whatever would they want to do that for?'
' _In the window looking over the sea_.'
'You mean –'
'Well, it could be, couldn't it?'
'Signals.'
'Out to sea.'
'Signalling out to sea.'
'The colours mean something. It's a code.'
'I heard –'
'U-boats. Signalling to those U-boats.'
Onlooking Stanley cuts in with authority: 'Three ships we lost in Febr'y between Land's End and St Ives, to those damned U-boats.'
'Stanley!' Nan's voice. 'We'll have no more of that language if you please, with the girls here.'
Stanley growling inside his chest. Wink from Aunt Mag.
'Lot of old nonsense anyway,' says Nan. 'Don't they say he works up at Hockings? Must be all right, then.'
'Free labour.'
'That's what they're like, you see, spies. It's in their nature. They get in with people. They get round them. If they weren't cunning-like, they couldn't _be_ spies could they?'
'It's the Unseen Hand, that's what they call it. The Unseen Hand working for the Germans. Wasn't it in your paper, Stan?'
Stan says heavily, 'They make a mockery of us, her coming here in her red stockings, singing her German songs. Captain Short should never've let the place to em. And him in the Merchant Marine.'
Nan says nothing more. She whisks her scissors under a piece of navy-blue twill, measures, nicks in the first cut.
Hannah rolls her eyes at Clare.
'For God's sake, let's get out of here,' her face suggests silently. 'This'll go on all night.'
And out they go into the wet warm evening. The wind's blowing salt and ladders of rain pulse down the main street. Hannah links arms with Clare, and they run splashing.
It's just a lot of nonsense. But there's something ugly in it all the same, when the aunts and uncles get together, faces crowded and puffed up with that queer sort of excitement which made Uncle Stanley's eyes glisten and Aunt Mag's eyes like little bright buttons nipping glances from one face to the next, not missing a crumb of it. Clare had recoiled, as if it were something that would smear her if she got too close. They are predatory and a bit frightening, even though they are her own flesh and blood. Yes, they were like gulls circling, ready to plunge. And she'd felt as if she should join in too, cawing over the spoils, to show them that she was one of the family, never an outsider on whom they could turn and rip to pieces. But now she remembers how Nan hadn't joined in. She'd let them talk themselves out, watching with that little ironic reserved way she had.
Yes, Clare knows who he is now. But what can she say to him? She can hardly tell him that his name is all over St Ives, and that his wife's stockings are providing something for Uncle Stanley to pull his whiskers and pronounce over.
'Is it really true that your wife wears red stockings?'
'Say some German words so I can hear what they sound like.'
'Don't you know they're all talking about you?'
She can't say any of it, so she'll have to leave him innocent of the gossip which perhaps he ought to know. Anyway, he doesn't look as if he knows or cares what people think of him. And in blue daylight talk of spies and warnings seems to melt away and become ridiculous. He sits on the turf, relaxed in the sunshine. He's got all the time in the world to enjoy it and her company. His fingers feel blindly and gently at a red clover head. The bee that was feeding there has stumbled on to its next flower. He ruffles the tiny lipped segments of clover, then nods at her sketching-bag.
'You paint, then?' he asks.
'These are pencil drawings. And I do water-colours.'
'What do you draw?'
'Flowers, mostly.'
'Do you? May I see?'
In her turn she nods. He is not a person you want to refuse, when he looks at you with such warm, compelling eyes, and seems so interested. She sits still, hugging her skirt round her knees as he unbuckles the canvas straps and fetches out her sketchbook. She hands him the big, public botanical sketchbook, not her own private book. He flips the pages, quick and intent. She waits, her heart beating palpably, perhaps from the heat and the walk, perhaps because it seems to matter what this man thinks of her work.
'You can draw,' he admits, as if puzzled by something. 'This, look, this is fine.' He points to the corner of one page. It's a white foxglove she saw in a lane-bank. No good to Father, but she liked the shape of the bells lighting into flower, one after the next, up the steep stem. Yes, she agreed. It wasn't a bad drawing. She'd have liked to paint the foxglove. The nubs of closed buds were lovely to draw. She'd like to paint them, now that she understood their structure through the drawing.
'Then why don't you? You _should_ paint. See, that little drawing is worth the rest of the book. Why do you draw _these_ things?' And he points at drawing after drawing, careful, shaded, exact, attentive to the detail of nodes and stipules.
'I draw them for Father. They're for a book he's writing, a botany book, so they have to be accurate.' It sounds lame, even to herself. It sounds like a child talking, a child who has never got beyond wanting to please her father. Is she still a little girl coming in from her lessons to show her father a piece of work which she knows will be praised for its neatness?
'Yes, of course they have to be accurate!' he exclaims. 'But have you ever really looked at what you are drawing? You've looked at the outside, the part everyone can see. You haven't seen what it is _really_. See that bud there. Now how can you make me believe it's got anything inside it? It's empty – just a husk. You haven't drawn it as if it's got any life inside it. I don't mean you should be sloppy. But look!' He points at the small clump of sea-pinks stirring in the sea-breeze at the edge of the cliff.
'How can you look at that without wanting to draw it as if it's alive? See how long it is! Think how the roots must grip down deep into the turf, to keep it here through the gales. And it has a little frail flower bobbing right on the edge of the cliff. See how the stem gives way to the breeze. I should think it'ud lay itself down flat before the gales; but it would spring up again, as soon as the sun shone. Don't you admire it? Isn't there something courageous about it? Look how fine it is, all the time stirring with the wind so it shan't get knocked to pieces.'
Clare smiles. When she was little she used to pull heads of sea-pinks apart. The stems were so tough they bit into the tender palms of her hands. She knows what he means.
'See, now it lies right down, to protect itself. It's tough, really, not tender. Could you draw that?'
She shrugs. She feels a devil in her, to teach him something when he thinks he has everything to teach her. 'I could draw _you_ ,' she says.
He laughs, turning to her with delight, his eyes narrowing. His blunt stubby face is nothing special but for those eyes. Can she really draw them?
'I can, if you want me to.'
'I do.'
'Then move back a little. Like that. No – stop. That's enough. Now if you look down there, at that thorn-bush, and then keep your head just as it is. Good.'
He sits perfectly still while she takes her second, smaller sketchbook out of her bag, and an H B pencil.
'Shan't you need a rubber?' he teases. 'I hear all the girls at the Slade have bits of bread to rub out their mistakes.'
'Do you know any girls at the Slade?'
'I do, some of 'em.'
'Do you? What are they like?'
'Oh, you've a long way to go if you want to be a Slade girl. You'll have to get your hair bobbed and get rid of your Christian name. The girls call each other by their surnames there. And they live in diggings – should you like that?'
'Mmm.'
She's not really listening. She'd like to know all about it, but not now.
'It's all right,' she mumbles. 'Go on talking. It's better to draw you when you're talking.'
He falls silent.
She wants to get the whole pose. The lines of face and neck flow so naturally into the body. Light, thin limbs in green corduroy trousers. Head very upright and alert, a bit bird-like. But she mustn't overdo it and turn the drawing into a caricature. The corduroy is worn and soft, so that it shows the shapes of the limbs. The face has no good bones in it, and the hair grows oddly, flat and springy at once. You could so easily make the face common, even mongrelish. Yet it is so quick and alive. Even mischievous. In spite of all its faults of colouring and structure, it's attractive.
She draws on. It's coming. But it isn't quite right. She hasn't been drawing enough lately, and her line is not as fluent as it should be. Also, she knows he is right. She is out of the habit of looking, really looking closely. Perhaps because there are things she doesn't want to see. Shuttered eyes again, only this time shuttered so that they cannot look within. She is so many people; our Clarey, my Clare, cousin Clarey, Clare Coyne, the Treveals' Clare, poor dead Susannah's daughter. But inside she is none of them. Once she stood at her window and said her own name, over and over, 'Clare, Clare, Clare, Clare', until it meant nothing.
She's lost the bold, accurate line her drawing has at its best. Has she lost it through finicking at flower drawings? She frowns, tears her first sketch off the block, scrumples it, starts again. She would need to paint him, and not in water-colour either. He ought to be painted in oils, to get the colour of him: the brilliant beard and eyes, the sunburn over his pallor. Behind him the sea, aqua and dark purple, wrinkling a little in the breeze. But she has no oils, and she doesn't understand the technique. There are so many things you can't get from books. The materials are too expensive for her, and Father would not think she needed them anyway. They would be of no use for colouring flower drawings. He likes her work to be meticulous rather than flowing and alive, as she would like to be. And like everyone she knows he is too easily impressed by her work. She knows that whatever talent she has is a curled up, deeply rooted thing. It has scarcely begun to unfold. She is not hard enough on herself, and no one is hard enough on her. There are so many things she doesn't know: technical things she tries to discover laboriously, from books. She flounders as she feels her way forward, and she wants criticism, not easy praise. But her father persists in believing that her botanical drawings represent the admirable sum of what she can achieve. This man doesn't, though. He is prepared to criticize and challenge her.
She sighs. 'It's not really finished, but I'll only spoil it if I go on.'
He stands up, his long thin legs stretching above her head where she still sits. He comes round to her side and kneels, looking at the drawing over her shoulder. He is so close that she can feel his breath on her neck. She shivers slightly and hopes that he hasn't noticed, but he is intent on the drawing. She lets the book lie open, and wonders what he'll make of it. She feels curiously unworried. It's as good as she can do.
He studies the drawing, picks up the book to look more closely. 'Will you give it to me?' he asks abruptly.
'If you like,' says Clare. Gently, she eases the page with the drawing on it away from the sketchbook's binding. 'You'll have to carry it as it is,' she says. 'I've nothing to wrap it in.'
But he's still looking down at her sketchbook. 'May I look through it?' he asks.
'No,' she says. 'This is just a private sketchbook. Drawings I do for myself.'
'I _should_ like to see one or two,' he responds, almost wheedling, almost wistful. But she glances up and catches the comic look of determination on his face. Just because she won't let him! It makes him want the more. Clare smiles.
'There's one of my cousin Hannah you can see,' she says, and opens the book and riffles through to find the sketch of Hannah sewing at the Singer. It hadn't been easy to get Hannah to pose like that, frowning, concentrated, with all the strong moulding of her face exposed. It was the best she'd ever done of Hannah. No one had liked it, though. They thought it did not do Hannah justice.
He looks up at her and smiles delightedly, as if he has suddenly recognized her as a friend. Her mouth curls into an answering smile.
'It's very good,' he says. 'Did your cousin Hannah like it?'
'She didn't. If she had, I'd have given it to her.'
'You don't make her pretty, but you make me want to know her. What does she do, your cousin? Is she a schoolteacher?'
'No, she works in my uncle's drapery. And she does dressmaking. She's very good at it. She wants to do more, but my uncle needs her in the shop.'
He looks closely at Hannah's face. 'She reminds me of a girl I used to know. Not that they're alike – but there's a look of her.'
Clare closes her book. She feels he has stared at Hannah long enough.
'I _should_ like to know her,' he repeats, with the same wistful persistence.
'I don't suppose you shall,' says Clare briskly.
'Will you sign my portrait for me? You know it adds to the value when it's signed by the artist.'
He's laughing at her, but she doesn't mind. 'Then I will,' she says. 'And when I'm famous you can sell it and make your fortune.'
'So I can.'
She takes the drawing back, about to sign it, then pauses. 'I ought to write your name too.'
'You should. Put _D. H. Lawrence_ , by...?'
' _Clare Coyne_. But I can't put D.H. It won't look right. I need your full name.'
'It's David Herbert.'
She writes carefully, _David Herbert Lawrence, by Clare Coyne, May 1917_.
'There. But what do they call you? David or Herbert?'
'My wife calls me Lorenzo. My friends call me David, or Lawrence. My sisters call me Bert. My enemies call me –'
He breaks off, his face laughing but half savage. Clare hardly notices: she's too much intrigued by the name his wife calls him.
'Does she? Why does she? I mean – she's German, isn't she, not Italian?'
A mistake, she knows at once. A bad mistake. His face closes over.
'Yes, my wife is German. Her name is Frieda. That means _peace_. Do you understand German?'
'No.'
'This foul talk,' he says. 'I suppose the town is stewing with it.'
'Not as much as that,' she says quickly, stung by the contempt in his voice when he speaks of the town. It was as if he'd seen Aunt Mag's babbling eyes himself, and heard Uncle Stan turning over the dirty business of the Lawrences in his big, raw fists.
'Because my wife is a German. Because she is a von Richthofen. And we want to live alone, to ourselves.'
She recognizes the pride in his voice when he says his wife's family name. She's familiar enough with that pride herself. Her father talking of the Coynes, affecting to laugh at what they stand for, gently mocking their enclosed, inturned world of Coyne Park and Coyne village, yet something in him always unhappy until strangers know of Coyne, and where he comes from. But surely there is nothing to be proud of in a German name? The name is familiar too. It has a newspaper ring to it. Where has she come across it?
'I'm sorry,' she says, 'I didn't mean...'
He has half turned from her, looking out over the sea. Behind him, looking north, the dark deeply folded land, the edge of the world itself, wrinkled and looking westward to America. He would like to go there. Southwards the land narrows, creaming with surf even in calm weather. There is so much more sea than land, and so much more sky than sea. The sky's perfect blue thins to lavender at the horizon. The sea breathes calmly, its back to the land, lapping, fulfilled.
It is wonderful to have your back to the land, to the whole of England: to have your back to the darkness of it, its frenzy of bureaucratic bloodshed, its cries in the night. It is heaven to have your back to the long gun-resounding coast of France, echoing against the white face of England. To have your back to this madness which finds a reason for everything: a madness of telegrams, medical examinations and popular songs; a madness of girls making shells and ferocious sentimentality. That blackness is behind him now, and he will not turn round to be trapped by it. It is a slow stain of nightmare, seeping into human minds and hearts. He refuses to let it contaminate his heart. He will keep his eyes turned to the West, towards the light and the place where the sun is going.
He falls silent. How much has he said? The girl beside him is frightened-looking, her lips parted.
'You're a writer, aren't you?' she asks.
'Yes. I am.' The words are brief, but he does not sound angry any more.
'Could you tell me... could you tell me what sort of thing you write?'
'What sort? Ah, now there's a question. It's lucky you asked me and not the men who review my books.'
Yes, the anger has really gone, just as if the sun has poured out from behind a cloud. What a face he has! She would like to draw it again, as it is now. She says boldly, 'You know what I mean. Faces or botanical drawings?'
He puts his head back and laughs, and his red beard glistens. 'Faces! Faces every time. And the likenesses are too good, so folk hate 'em, just as your family hates your portraits. How they hate them! They'd like to burn them all up, and me with them. Oh, you'll meet the book-burners yet, Clare Coyne, and the picture-burners, if you go on drawing. Some of them do it so nicely – it breaks their heart to set a match to you, but for your own sake they've got to do it, don't you understand? And all the time they're looking at you like a dog does when it wants to bite your leg but it daren't. Then there are the others who don't pretend.'
He is still laughing, but she shivers.
'It sounds horrible,' she says. 'Why have they got to be like that?'
'Don't you pretend you don't know, Clare Coyne. There's nothing you couldn't see if you wanted to.' He taps the sketchbook as if it contains his evidence. His eyes brighten, daring her.
'I must go,' she says, glancing up at the sun. 'I have to cook the dinner and the rabbit's not skinned yet.'
'Rabbit!' he exclaims. 'You'll make a rabbit stew, I suppose?'
'I will. It's a tender young'un, it won't take much cooking. It'll eat soft as butter.' Unconsciously her voice drops into the tones of her cousin Kitchie, who'd brought the rabbit that morning, banging limp and warm against his britches.
'Do you keep house for your father?'
'I have done since I was ten,' she answers.
'Coyne. That's not a Cornish name, I think?'
'It's my father's name. He's from London – Somerset, really. But my mother was Cornish.'
'Was she? You don't look it. The Cornish are soft, dark people – aren't they? Like your cousin Hannah.'
Soft, dark, stubborn Hannah.
'Yes, Hannah's pure Cornish,' she agrees. 'She's never been north of Bodmin.'
'And you have?'
'I've been to London. Twice. And to Coyne, to see my Uncle Benedict and my Aunt Marie-Thérèse.'
He scans her. Dark red hair. White skin. 'You might be Irish. Or Scots? Are you Scots?'
'My great-grandmother was Irish. I'm like her, Father says.'
He frowns. What is it now? Doesn't he like the Irish? And yet he fires up at a word against the Germans.
'So you're a lady, Miss Clare Coyne,' he says. 'A lady who skins rabbits. I must tell Frieda. She'll like that. I wish I could get Frieda to skin a rabbit, but no, she's such a swell she can't even put on her own stockings.'
An image flits into Clare's mind. She sees Lawrence kneeling, rolling Frieda's red stockings up her legs past her knees. Frieda's legs will be white and strong, more womanly than Clare's own. But there's a blank disc where Frieda's face should be.
'You'll come and visit us?' he asks suddenly. 'Come to tea. Come next Tuesday. Frieda will like it. She ought to know more women – she knows no women here.'
'Is that sufficient recommendation – the fact that I'm a woman?'
'And the drawing,' he reminds her. 'Don't forget, I've got that. So you'll come?'
She hesitates, then, 'I could call in for butter at Lower Tregerthen too,' she agrees.
He nods, and looks back over the moor. 'I've never seen anything so beautiful as the gorse this year,' he says.
She watches him walk away. He is a queer figure with his thin legs and corduroys. He is quite different from anyone else she has ever met. She has met so few people who live out in the world. Father is always saying that she meets no one, as if he regrets it. But does he mean people like Lawrence? She smiles. Next time she will get him to tell her about the girls at the Slade. And his books. She would like to know much more about his books. She shades her eyes, and looks after him. He is holding the rolled-up drawing carefully, as if he values it. He walks away fast and lightly, raises his hand to wave, then disappears into a fold of the land.
# Six
_Thursday_
_Hannah is angry with me because I did not go with them to meet John William. They did not expect Father to be there at the station. They know he will call later this evening, with a present of cigars for John William. Nan and Aunt Sarah and Aunt Mag have boiled a ham and made gooseberry tarts and cheesecakes. I did not help them. I had my baking to do here. Now Nan will be putting aside pastry crust and ham scraps for Uncle John's pigs while Aunt Mag and Hannah wash the dishes. Aunt Sarah won't be in the kitchen. They'll have shooed her out into the front-room, so that she can sit in the wing chair by the fire and watch John William. It won't matter to Aunt Sarah what he does, or whether he talks to her or not. As long as she can sit there and watch him being alive. She knows how easily it might have been us crowded into the front parlour to receive visits of condolence, with the smell of black dye bubbling from the kitchen. How thin it is now, and how easy to cross, that line between being dead and being alive. Grandad doesn't understand it like Aunt Sarah does. He's sure that the Lord will look after John William. Every night he says the Psalm of Protection for him: 'A thousand shall fall by thy side, and ten thousand at thy right hand; but it shall not come nigh thee...' I love the words, but how can they charm bullets and shells? How many people are praying them? Thousands, I should think. All hoping that others will fall on the left side, and the right side, and their own will be safe. But I do light candles for him. I would give him a badge of the Sacred Heart if I thought he would wear it._
_Grandad will sit at the other side of the fireplace, drinking in every word of John William's, so he can go down to the quay and smoke and retell it 'fresher than the newspaper and more truth in it'._
_Kitchie will interrupt with questions. Albert and Jo and George will lounge in one by one, darkening the room, waiting for John William to come out with them. Everyone knows they'll take a drink, but as long as nothing's said, nothing's questioned._
_Hannah and I stroked the cloth of his uniform when it was new and stiff, when we said goodbye and I kissed him but he wasn't looking at me. He was looking over my head at Uncle Arthur and Aunt Sarah and the boys and all the family milling on the station platform. I could not read the look on his face. Then Nan slid another packet of sandwiches into his kitbag, and I stepped back. He smiled at me then, and winked, and made us a secret world for a second, like we've always had. I wonder what his uniform is like now. He will have to have a new one if he's going to be an officer. The old one will smell like Sam's letters, muddy and frightening. When someone is killed, they send his uniform home with mud and blood still on it._
_Nan will let John William smoke in the house tonight instead of making him go out on to the street. The front-room will smell of smoke and food and elderflower cordial. In two years' time they'll call Kitchie up, and we'll never get exemption for Kitchie, even though he's the only one Aunt Mabel managed to rear._
She stops and frowns at what she's just written. It sounds coarse, like a farm diary. She raises her pen to score it through, then sighs and leaves it as it is. It's true. Draw faces, not flowers, Mr Lawrence said. A blob of ink falls on the page and spreads.
_John William is very brown, they say. Hannah was shy of him when he_ _stepped down off the train. He was like a stranger, so dark and broad. He's weathered like Albert and Jo and George now, only it's not from farm-work, and he's grown a 'tache. He's handsome now, Hannah says. I hope she didn't notice how surprised I was. Didn't she think he was handsome before? His face has changed too, Hannah says, but she couldn't tell me how._
_'You ought to have come and met him for yourself, Clarey. He asked where you were.'_
_Did he? Or is that just Hannah saying it, knowing I want her to say it?_
_'I'll go tomorrow,' I said. 'I didn't want to crowd him, that first hour. There are so many of us.'_
_'How could there be too many of us?' asks Hannah. 'You're family, aren't you?'_
_I couldn't say to Hannah that I wanted to be the only one. I wanted the train to stop, and the steam and smoke to billow round us as John William stepped off the train and folded me tight in his arms, tight, so tight I couldn't feel anything but him, so close I couldn't see anything but him._
_My cousin is back from the war. There. Do I feel something? I should. I could walk down the hill in five minutes and look through Nan's geraniums in the windows and see him. I could lift the latch and Nan would fetch me a plate of cold ham and call through, 'Here's your cousin come, John William! It's Clarey!'_
_Here he is, close enough to touch, eating gooseberry tart._
_No. He isn't really ours any more. He won't be ours as long as the war lasts. He doesn't belong to us now, he belongs to the war. Even when the men are wounded they aren't allowed to come home, not if the war still thinks it can use them. Even if they are dead they can't come home. The war takes care of it all. We never see the dead men coming back, and it's so bard to believe in what you never see. Mrs Hore told_ _me she still expects to see her William coming down the street. She thinks she hears his step in the entry. He had his own way of whistling under his breath as he fiddled with the latch-string. If only she could have seen him in his coffin, she says, she would be able to stop listening for him._
_Can they make enough coffins for them all? How many coffins, and who makes them? The chaplain wrote and told Mrs Hore that William had his rosary beads twisted around his rifle butt when they found him. Father O'Malley encouraged all the men to do that, so they would have their rosaries with them, whatever happened. They would have two essential things by them, their rifles and their rosaries. William was killed instantly, Father O'Malley wrote. Mrs Hore said it was good of him to write and tell her that. It set her mind at rest. She knew William hadn't suffered. But I wonder. Sam says they always write it. Shot through the head, or shot through the heart. Then Sam said something I can't forget: 'We pick up what we can find, enough to make a decent burial.'_
_William Hore made his First Communion with me, but he wasn't very holy._
_Who made me? Why did God make me?_
_I should not be writing this. I should keep it shuttered. I'll have that dream again._
_I won't go down to Nan's with Father. I can see John William tomorrow, after he's had a night's sleep. It'll be quieter then, and I'll see him alone._
Clare blots the page, shuts up the book, then leans forward with her elbows on her little desk. The desk is set in the bay window of her bedroom, looking out over the sea and the long flank of the cemetery. The sea is dark, wrinkled by a cool evening breeze. It is bald and empty. The war again. But long before the war, things had changed. She's heard Grandad talk of the drift-boats coming into St Ives from the herring fishery on the Irish coast. The _Agenora_ fell in with a shoal of pilchards on her way back, and brought in twenty thousand, so he heard tell, for even Grandad wasn't born then. There are no catches like that these days. The seas have run dry.
Grandad told her it was more than fish they brought back from Ireland. They'd go ashore there to buy and sell, and this time they brought back some old clothes from a woman who was selling them cheap. But she didn't tell the fishermen why she wanted to sell them. Her son had died of cholera, and the clothes were full of disease. So our boats came back not knowing what they brought with them and spread cholera to the people of Newlyn, so that a man in perfect health could eat his breakfast, say goodbye to his wife, go out to his work, and be dead by nightfall.
'This is the blue cholera I'm speaking of,' said Grandad.
And the people were so fearful that they ran away, leaving the sick on the ground to die alone. A man would leave his wife, and a wife would leave her husband. The whole country was bent down with bad news like a field of wheat under the wind.
'They would publish lists of the dead on the walls, and there were midnight funerals with hearses going through the streets at midnight to keep the disease from spreading.'
'Why was it called blue cholera?' asked Clare.
'For a good reason. It turned a man blue, see, before he died. It was the way the disease worked in a man's body, and it was a terrible thing to watch.'
'You wouldn't leave me and Nan, would you, Grandad, even if we turned blue? You wouldn't run off from us?'
'Ask God to spare us that trial, Clarey.'
That was just like Grandad. He clung too close to the truth to give her easy assurances.
'Ask God to spare us from trials and temptations, Clarey.'
Grandad did not like the Irish. But she had to like them, for she was part Irish herself. How could she dislike part of herself?
But when you think of it, cholera was nothing to the war.
'A thousand shall fall by thy side, and ten thousand at thy right hand.' A thousand thousands, and yet we don't run away. Anyway, there is nowhere to run. Everything stays so calm, as if we could go on like this for ever. We don't have burials at midnight to hide the numbers of the dead, because we don't have any funerals at all. Not here. We're not supposed to pray for the war to end. We're meant to pray for victory. The boys go out to France just as they used to go off on the drift-boats, or down the tin-mines, or to work on the farms. They all go: Medlands and Pascoes, Vivians and Trethakes, Popes and Mawgans. A whole landscape of family faces crushed into thin black lines, so dense you could not read them but that your fear and curiosity made you read them. You looked out for local names. What was cholera compared to the emptiness of the sea, the quiet streets, and the clumsy-booted young men, first time away from home, entraining and vanishing through the foggy curtain of war. We don't know what's happening behind that curtain. There's a silence which all the yammering headlines can't break.
Clare thinks of the train journey she loves, from London to Penzance. Drowsing and jogging through the heavy green fields of Devon. Then the bright red sea-cliffs at Dawlish, and waves breaking nearly on to the line. As you crossed into Cornwall you seemed to go through a curtain of light. You were sealed away from darkness and soot-smells and tall crowding buildings. Even the smoke of the train seemed to blow sideways whitely and buoyantly. You were safe and nothing could snatch you back.
But since conscription the trains had been going north and east heavily laden with boys off fishing-boats and farms and shops. It's taken a long time for the war to get down to them, but it has managed it at last. The war's long fingers can winkle a boy out of a lonely cottage on Bodmin Moor just as easily as it can pluck one out of the Manchester mills. There's nowhere safe from trains and telegrams and call-up notices. There are medical examinations held at Bodmin. Classifications. There's a class for everyone, no matter what the shape of your body or its powers and aptitudes. If you have two arms and legs, the war can make something of you for its purposes. Fit for active service. Fit for light duties. Fit for non-military duties.
IS YOUR BEST BOY IN KHAKI?
As for the dead, Clare has seen none of them. There would not be enough room for them in the graveyards, and they are as near to heaven in Picardy as they could be in the long sea-resounding graveyard of St Ives. So don't mourn because there is no funeral.
I'm one of the lucky ones. I haven't lost anyone. Well, no one close.
The fingers of the war are pulling harder. They are after Harry now. They want to reclassify him, in spite of his weak arm. _Combing out_ , it's called. Like combing lice out of hair. Kitchie is still too young for them. Albert and Jo and George were too quick for them. Before enlistment gave way to conscription Grandad had seen the way the wind was blowing. He and Uncle John conferred late into the evenings. George was a coast-watcher already, and now Albert and Jo volunteered. Food-producers, coast-watchers, patrollers, they had kept their exemptions so far. And how could the land be worked without them? The country must have food.
The names of Tribunal members were common coin to Grandad. He knew them all, weighed them, judged them. This one was envious; that one fair but hard. This one had a prejudice against farmers keeping their sons on the lands. That one might be bribable. Uncle John, Uncle Arthur and Uncle William listened to Grandad. Gifts of vegetables, butter, cheese, ham, were driven in from the farm. Uncle Stan, sonless, boomed and schemed on behalf of the rest of them. But how long could it go on now that all that mattered was more bodies for the Front?
Grandad curses himself now for not thinking so far ahead with Harry. Now they are saying that Harry has got sufficient use of his arm. And it is only his left arm. Harry must come up to Bodmin again for a more thorough examination. The war gets hungrier and hungrier, the more we give to it, thinks Clare. It is like feeding a dog which has worms.
Downstairs Francis Coyne drinks Burgundy. The wine flushes pleasantly through his body, and he reaches out to pour himself a second glass. He has spent the afternoon in Newlyn, where he goes twice a month to fuck neat plump-bodied May Foage. Tuesday or Thursday is his day: he writes her a note beforehand. He walks from Penzance Station past the harbour wall with its slopping green water and blocks of dumped limestone exposed at high tide, and then down the beach road to Newlyn. All the time he walks he thinks of fucking, loving the word for it and the anticipation of it. Under her black skirt and white petticoat she wears no drawers on the day he visits. What he likes is to be let into her white, tight little house, to look at her for a long moment, with her hair knotted smoothly at the back of her neck and her broad white forehead and cheeks unlined, placid, giving nothing away, and her two feet firmly planted on the bit of Turkey carpet he got her second-hand as a sign of her respectability. Then he sits down in her one armchair, and she perches herself on his knee, proper and neat and upright, and his left hand circles her waist while his right feels up her plump white calves (for she has taken off her stockings too) and up her thighs, which he knows have a subtle tracery of blue veins on their insides, and up to knot themselves in her mazy pubic hair. Her head hair is as smooth and brown as a salmon-river, he thinks. And on he slips into the secret moist cleft of her body, with its folds of skin that suck and hold his finger. He loves her childless tightness and springiness. She sits there through it primly, a woman in black perched on his knee, tucking in a smile.
She has no husband. She had one once who made her Mrs Foage, but he is no more. She is nothing so definite as a widow, nothing so weak as a deserted wife, but her husband is perfectly absent. By gift or charm she has no children, nor does she show any sign that she has ever wanted them. She has other visitors, but he never sees them. She is modest in what she asks for, and modest in how she asks for it. After one brisk, business-like conversation, the subject of money has never had to be reopened. They both know where they stand. She knows how investments are, these days. And long ago she was a St Ives girl and went to school with Susannah Treveal. It gives her pleasure to draw Susannah's gentleman husband into her house. It gives him pleasure to think of the two little girls with their moist clefts still as tight as almonds, pattering to school and pattering home again, not knowing that one day he would have them both.
And he has May Foage. He has her more than he ever had Susannah in her white, startled night-dresses. Sly, neat, fuckable May in her blacks. He has licked May's breasts; he has bitten her shoulders. She has rocked him, and he has heard the waves pouncing on the beach outside her cottage, as he arches over her broad, white, spread buttocks, spread open for him. How delicious she is, jouncing beneath him; how shrewd in her passivities, how nimble in her accommodations. He swirls the wine and thinks of May's cleverness. His occasional occasion of sin. He drinks the wine, and it slides along his veins, leaving them sunlit.
And here he is, home again, waiting for the tea (which he doesn't want) to be brought to him by Clare, who probably doesn't want to bring it either, since he has just heard her coming downstairs as if reluctantly, stopping on the landing, looking out, no doubt, as she does. Better not have another glass. Here she comes.
My daughter Clare. Pale face, dark dress in the doorway. Her face is mute and closed in on itself, as she puts down the tea-tray and moves off. She's incurious. She never asks me where I've been on Tuesdays or Thursdays, twice a month. I have never had to lie to Clare.
'What are you doing today, Father?'
'I'll be out most of the day, Clare. But I'll be home for dinner. I may go over to Newlyn.'
Does she smell May Foage on me? When I come home I drink wine without washing my hands of May's juices. Then I drink tea and go upstairs and pour water into the bowl in my room and soap my hands with the cheap yellow soap which is all Clare can get now.
They say the Germans are making soap out of the cadavers of fallen soldiers. They need the fat from them. _Kadaverfabriken_ have sprung up secretly to process the corpses. It was preached against at Mass. I say nothing. I know the glee of an evil imagination.
My pale daughter Clare goes out of the room. She is not pretty today. Her features are shrunken, small and tight as they were when she was angry as a child, or when she was unhappy. She drops her eyelids, makes herself absent. She has her own thoughts. Thank God that we each have our own thoughts and that we can live secretly even in this small house. Clare cannot see the marks of May's hands and teeth on me. She does not hear the cries and sweet succulent noises I hear.
'Clare!' he calls, a little too loudly, his voice like a seagull's in the small brown room. 'Clare! Will you come down with me now and see your cousin?'
A small collected pause, and her voice from the kitchen, little and purposeful, 'No, Father. Hannah has lent me a new dress pattern until tomorrow. I'm going to alter my green silk. I shall see John William tomorrow.'
Her green silk. Poor child, she doesn't need to specify the colour. She has only one silk dress. He has done nothing for her. Her hands are red with housework, and they rasp when she touches silk. While he is out, she will spread her green silk on the kitchen-table, cut it, remodel it according to Hannah's pattern, sew it again. She will be well pleased with what she achieves. All last evening she was unpicking the gathers at the waistband.
It's the war, he consoles himself. It's the same for all the girls. They have never known anything else.
He ought to have been able to buy another silk dress for his daughter, to make her beautiful. But look at Hannah, she has no silk dress at all, and she manages. Yes. Her brave blue and white stripes are upended as she opens her body to Sam. The white thighs of May Foage and Hannah Treveal flex and shine. It is May, the month of May. Mary's month. He remembers Coyne and his sisters dressing their May altars, fitting more and more flowers into tiny silver vases in front of the statue of Our Lady. And he already ironic, but still liking to see them at it, liking the flush of solemnity of their cheeks, and their smooth parted lips wondering if it was beautiful enough yet, and the way they ducked and kneeled in the chapel, adorning, adoring. His thoughts widen, covering Clare, putting a new silk dress on her, grey to match the slate of her eyes, putting her in mantilla and dark gloves, folding her hands in prayer. Her hands would be smooth and white under kid gloves, her body would be hidden and quite unimaginable.
'And a pious ejaculation to go with it!' says Francis aloud, scorning himself. He picks up the box of cigars for John William. He takes up his hat and goes out, calling goodnight to Clare, glimpsing through the kitchen door his resourceful girl nipping and tucking at her old green dress in the pale May evening light.
# Seven
Zennor church door creaks and opens. Outside, the unusual May heat simmers. It is noon and quiet. Dogs lie in the dust. Butterflies skim gravestones and the pointing finger of shadow on the sundial to the left of the church porch is sharply distinct. _The Glory of the World Passeth_ , it says. Lawrence stops to look up at it, notes the name of the sundial-maker. Paul Quick.
He enters the cool porch, and then into the church smell of wood and stone and darkness. He knows where he is going. He walks up the aisle, turns right, kneels to touch the carving of Zennor's mermaid. She floats, round-bellied, arms raised, innocent above the curve of her tail. Her fins flick and scull. He traces the carving of breast and navel. How fine she is. He is surprised that they have not put her out of the church, or gashed her into decency with axe and chisel. They have tried to hurt her, but she lifts her hands blithely above the scars. She belongs here. She is dark and smooth as the little madonnas he's seen in Bavarian wayside shrines, and she is as faceless as they are. Her body is like a sea-wave, a crest which will never break.
The church smells of its own sunlessness, as churches do. He gets up from her side and goes out, leaving the door ajar so that sweet air can blow in and ruffle the May flowers on the altar. He'll go to the farm now and help with the furze-cutting, then tonight he'll give Stanley his French lesson. He pushes down the thought of Frieda left alone, restlessly writing, restlessly reading.
*
When the wind blows hard at night, the cottage door bursts open under its pressure, and the rain slashes in across the floor, bringing darkness and the noise of the sea. I think I hear the friends of my childhood and all the young soldiers of Marienbad who marched past our garden walls to the parade-ground. My sister Nusch and I would throw apples and pears down among the marching men, to see their perfect lines falter as the soldiers scrambled for the fruit. Now they are gone, swallowed up like all the other thousands upon thousands, and I am here on the farthest outside edge of a country which hates me. I can do nothing. I am the Hunwife. The darkness is alive and wild. A south-westerly gale bangs the waves against the base of the cliffs with a sound like gunfire. I force the door shut and look around the safe, violated room. The walls are pale pink, glowing, and the furniture we got so cheaply at Benny's Sale Rooms in St Ives is so beautiful. And yet no one wanted it. But for us it makes a home. How many more homes will we have to make? Spreading out shawls on ugly chipped chests, bargaining for furniture, hanging embroidery on the wall, finding china for sixpence to put on the dresser. Such queer, beautiful bits of china, none of them matching. The bowls with peacocks on them. I tell Lorenzo that we are gypsies, and I laugh and make a joke of it, and he believes me when I say that I don't mind it.
I think about growing up in Metz. Those white dresses we had, and our stiff-sided boots, and the hats which could never crush down my hair. There was so much of it always, springing out and curling. Like a lioness, my mother said. She called me her little lioness and she said white didn't suit me. But never mind, I would wear colours when I was grown up. Then, when we went out to walk around the garrison, my big sister Else held my hand and told me to keep close to her and walk with small steps, but I could not. Something inside me made me break out and run. If I tore the lace on my petticoat, I would throw it to the floor and kick it into the corner of the room for my maid to find and mend.
Then we were at the court of the Kaiser. We were the daughters of the Baron von Richthofen, Nusch and Else and I, and so we must talk in this way, and dress in this way, and laugh in this way. So stiff – such nonsense! – I never cared for the Kaiser. But my cousins I loved, my bold cousins with their jokes and teasing and their brilliant uniforms. They would come into a room and the air would seem to dance round them. And now when I hear of the U-boats sunk and Germans killed, I think of them all, not skating or dancing or buying flowers for us any more, but men at war, as stern and stiff as even the Kaiser would have wanted.
So I read the _Berliner Tageblatt_ and scour up and down the lists of the dead, but when I have finished I hide it under a cushion or the whole day will darken for me each time I see it. I will not let the war do that to our life. So many good things come each day. I was ill – I was in such pain I could not move out of my bed. And now I am quite well again and I can walk on the cliffs, five miles, six miles, and come home and eat the pea and ham soup which Lorenzo has made. There are so many good things each day.
Only, the children. I am not fit to see my children, they say: only for half an hour in a solicitor's office with a clerk coming in to tell me when it is time for me to go. When I gave each of the children ten shillings, my husband took it from them and sent it back to me.
'You left them. You chose to leave them. You preferred to abandon your own children rather than give up _that man_. Now you must live without them.'
He would like me to die without seeing my children again. Monty and Barbara and Elsa. They have grown so much. If I held out my arms, they would not fit there as they used to do. I think of them at night, if Lorenzo is down at the farm and I'm alone. Have they gone to bed yet? Are they sleeping, or are they lying awake too, staring at the dark and imagining they see faces in it? Do they see my face? Or do they see nightmare faces and cry out for me in the dark, and I never come?
I wrap myself in the Paisley shawl Lorenzo bought and mended for me. He mended it so patiently. It was a cobweb of rents and tears, but it only cost sixpence and he saw that he could make it beautiful for me. He is late again. If he was here he would be moving around the room, quick and deft, heating some milk and chiding me for my laziness. I would play the piano, and sing, and he would sit in that chair with his head thrown back, listening to me, eyes half shut. But he is down at the farm again, laughing with William Henry and Stanley.
The Hockings are not comfortable when I come down to the farm. The talk and the laughter stop as I open the door. William Henry stands up. They do not know how to treat me, for I am a lady and a foreigner and not one of them, and yet I live in a labourer's cottage which they would not live in. Even Lorenzo looks at me across that farm kitchen as if I do not belong. So I don't stay long. I say goodnight and lift the latch and walk out into the darkness, and as I go I hear laughter and talk spurt up again, the way it always does when he is there. The yard is cool and I can smell the cattle crowding up to the field gate. I walk up to the cottage, and look down at the long slope of the fields, and the sea with the stars coming out over it.
At least there are no dark shadowy rooms leading off this one. There is only this one little pink room, with its low boarded ceiling and boxy staircase going up the wall opposite the fireplace. One warm lit room and then the window looking westward towards America. Wind and space and strangers. No Katherine next door any more. The Murrys did not like it, for it was too wild for them.
We made a home for them here. We willed them to like it, but they would not. Never shall I forget Murry squatting on the bit of rough grass outside their cottage, painting black the kitchen chairs we had bought for him, as if it were a funeral, while Katherine settled her things so that she could write. But she could not, she told me: she could not stand the way the wind blew, and the rain slashed at her ankles every time she went out. How she made me laugh! It made me feel I had never had a woman friend here in England, a real woman friend I could tell things. No one tells stories like Katherine, Lorenzo said. Before they came, he thought it was all beginning. He wrote letter after letter, talking about 'Katherine's tower'. We would be able to start the community he is always dreaming about. Well, I did not care so much about that. But I cared for Katherine: she was so fine, and small, and she made us laugh and forget the war.
It was hard for Lorenzo. Before they came here he found the cottage for them, and helped them to furnish it. He scrubbed the floors for them, and bought iron pans for their cooking. Katherine does not like to cook. She is like me, always half expecting to turn round and find a maid has slipped into the room to tell her that it is time for dinner.
Now he has shaken off his disappointment, as he does. He knows that he will never build his community here, his dream of Rananim. The Murrys could not stand it and who could blame them? This is not the place for them. Katherine talked of Bandol, and the smell of almond blossom. Here it rained all the time, and Katherine was cold, and the fires smoked and the lanes were full of farm mud. Lorenzo and I fought every day. I saw Katherine's face when we fought, when Lorenzo chased me around the table and hit me and called me a hussy and said he would make me scrub floors on my knees. I fought as hard as he did. He never got the better of me. But Katherine did not understand it; she would go away without speaking.
They would not stay. Lorenzo raged against them for weeks, but I could not. Yet it was so good to have another woman to talk to, another woman close by. Katherine was like a little mermaid herself, swimming in the cove, so slender-boned and well made in her odd, perfect clothes. Murry said it was too cold for her after Bandol, she must have the south. So they found a place at Mylor, on the south coast where it would be gentler for them, out of the storms we have here. Lorenzo says that Murry has abandoned him and betrayed the Blutsbrüderschaft between them, so now he goes down to the farm and talks with William Henry. I do not like William Henry. He looks at me as if I am nothing, not important, just a thing belonging to my husband. But who is William Henry to look at me like that? A farmer. A peasant. Always he is dragging Lorenzo down into his fields, his farm, pretending to want to learn geography and French so that Lorenzo can explain to him and teach him as he loves to do. Then Lorenzo comes home to me and tells me I do not understand William Henry, his mind is fine and quick and he is hungry to discuss ideas. I can see that he is hungry. But for what is he hungry? There are so many people like him, who want Lorenzo and what he can give to them. They see he has a gift and they want it. But I have lived with him for five years and I know they cannot capture it.
*
At Zennor the wind never quite sleeps. The typescript on the table is held down by a piece of granite, but the edge of the pages still riffle and words jerk into movement like those cinematographic images children make by drawing stick figures on the corners of a notebook and flicking the pages. Lawrence works swiftly, writing by hand in margins and between the typewritten lines, making his final revision. He is writing his article on 'The Reality of Peace'. He works with intense concentration and his handwriting flows across the page. The article is nearly finished, but he senses that there is something missing: the ideas are naked, like signposts. He stops and shuts his eyes. The blind, lost faces of the men on the special conscription train rise in his mind like the faces of drowned men. He leans over the paper and writes again, faster. Then he tucks the typescript under the stone again and goes in to scrub the potatoes.
As he goes up the path he stiffens, tense. He calls out quietly but with such urgency that she comes at once: 'Frieda! Frieda!'
She appears at the door, squinting a little against the sun.
'Come here. No, not that way – come round by me. Look! Can you see her?'
Frieda pads lightly across the hummocky grass and potato hills. She looks where he points, and her tawny eyes dilate. A little slim black and yellow adder has poured herself up from a crack in the wall and is coiling on a warm stone. Her blunt head makes soft stabbing motions into the air, butting against its new summery warmth as a lamb butts the udder playfully before it seizes the nipple.
'Ah, she is beautiful!' breathes Frieda.
He watches, and frowns. 'There may be a nest of 'em in under the wall,' he says dryly. 'Should you like that?'
'I have never seen one so close. See! There is her tongue.'
'William Henry would tell me to poke out the nest. There's a plague of adders on these hills. They bite the sheep.'
'Sheep!' exclaims Frieda scornfully. 'Often I think I should like to bite a sheep myself, they are so stupid.'
The little adder flicks her tail lightly to and fro.
'She knows we're near,' says Lawrence. 'She can sense us.'
The warm wind stirs again, carrying a dark brown spice of wallflowers.
'Smell those gillivors,' he mutters. 'They always make me think of my mother's garden. She would stop still and shut her eyes to smell them.'
He is back in the small dark-soiled garden in Eastwood, treading the cinder-path to where his mother stands stock-still and oblivious of him. Her white apron blows in the spring wind, and the noise of the pit's winding-gear grates through the piping of a robin hidden in a blackberry-bush. In a minute she'll turn and see him.
'Mother!'
But he has not spoken aloud. Frieda leans against him, watching the snake as she coils again, showing her belly, her deaf body uneasy as she picks up the vibration of their voices.
'She's going.'
The snake's body elongates like water, tilts and disappears. Frieda steps over and lays her hand on the stone where the snake has been, but there's nothing but the stone's own warmth and smoothness. She looks up at Lawrence and laughs. Her face is flushed from stooping and from the excitement of the snake, and her eyes glow.
'Lorenzo, do you know who she reminded me of, that snake?'
He smiles, and shakes his head.
'Katherine. Don't you think so?'
Her mention of Katherine presses lightly on the bruise of anger he still feels towards both the Murrys after their rejection of Zennor. But Frieda is beautiful in the sunshine.
'I suppose Katherine might be an adder. Slim, and quick, and a bit wicked.'
'And her black hair, so sleek and smooth.'
'Yes. It's true that she likes to be by herself. She hates people to touch her except when she chooses.'
'Or she will bite.'
'She can certainly bite. Frieda, you know that girl I was telling you about – Clare?'
'Yes? Why?'
'You will like her. I am sure you will like her.'
His face is warm and eager, but Frieda is cautious. Will this be another of these girls who sit at Lorenzo's feet and wonder why Frieda is not safely out of the way doing the cooking?
'She needs to meet a woman like you,' he goes on. 'It will open her eyes.'
'Oh,' mocks Frieda, opening her own fierce eyes wide. 'Is that what it will do?'
Reluctantly, he grins back. She tucks her hand into his arm and pushes back the hair which is always breaking into curls on her forehead.
'I'm glad you called me to see the snake,' she says.
The wind blows at night, but by day it's calm. The weather is extraordinary: everyone remarks on it. In London, ladies note it in their diaries. Perhaps it's something to do with the war. Or perhaps it's just that people notice things more now, when they are living in a time out of time? In the parks the wounded lie out on chairs in their blue uniform with red ties. Three men hop on their crutches, swiping at a rag ball which rolls and bounces between them. They have three legs between them. The spare leg of their blue uniform trousers is pinned neatly at the top of the thigh. They look just like crickets, thinks a nursemaid as she pulls a dense creamy fold of new wool blanket up to the lips of her charge. She pats the warm hump of his bottom. He sleeps on, whole and perfect from his bunched fists to his dwindling, purple-blotched legs. She'll take him back another way, so as not to go past the men on crutches again. There is something frightening to her in their nimbleness. And there's the way they look at her as she sways past them pushing the perambulator, her white skirt creaking faintly under its tight belt, her cheeks pinkening. There's a flat, dark look in their eyes, as if they do not see her, or as if they do not like what they see. She will go round here, and back by the pond. The extra turn will be good for baby.
# Eight
Clare snaps fully awake at quarter to six in her dark room. She's had to change her white lace curtains for chenille because of the war. No one must show a light to U-boats lying out in the Channel, sinking ship after ship after ship, creeping in at night to refuel at secret dumps along the miles of toothed coastline, fed by the Unseen Hand of German collaborators, a cancer working in the body militant, a nest to be smoked out by every observant man, woman and child. Or so they say. Clare hates the blotting out of her landscape of clouds and sea. The chenille is stuffy and faded, and, even though she aired the curtains in the sun for a full day before hanging them, a smell of must seeps out of them and taints the room.
She pads round her bed and swishes the curtains right back. There it is, her bowlful of sky and sea. The sun has risen behind the house, throwing long shadows down the hill in front of her. Everything is moist, fresh, dipped in new light. There are still some curls of mist over the sea, but they are clearing fast. It's going to be hot again. There's dew on her lettuces in the little kitchen garden over the lane, and her marigolds are already lolling open where the morning sun touches them. They look ripe, like fruit. Sheba picks her way immaculately through the wet grass, her tail twitching. She senses the movement of Clare in the window and looks up, then steps off again, lightly, as if she's just confirmed her opinion that Clare is not worth bothering about. At the horizon the sea is purple, but within the bay it is a dark, breathing blue. The tide's still going out. Clare clatters the sash window up and leans out. There's the milk-cart labouring its way up the hill, with Joss walking at Sally's head, talking to her so that the mare won't notice how steep a climb it is, how often she's done it, how tired she is. The white spots on his red neckerchief stand out like daisies on a lawn. Sally has her head down, shuddering, bracing herself. She's getting too old for this, but you can't get a decent horse for love or money, not with what he can afford, not with the war, Joss says each morning as he pours a long white tongue of milk into Clare's jug.
Clare will go out. She can't wait to be out of the house, away from this frowst of curtains and stale air. Down there to her right all the cottages hump, still sleeping. No, not quite all. The first smoke goes up from a stubby chimney wedged into a slate roof. Father's still asleep, in his dark red bedroom at the back of the house, closed in against the rising sun. He has had black shades fitted to his windows, because his eyes are sensitive to light. He won't get up for hours yet, and when he does he'll meander down to the kitchen in his Paisley silk dressing-gown (because it isn't one of Hat's days) and make tea for himself, whistling between his teeth, book in one hand. He doesn't sleep well, and Daylight Saving makes it even worse for him. She never sees him before eight, and it will be later today. She didn't hear him come in last night, even though she lay there, half dozing, half waiting, on edge. But before the door snicked she fell asleep, into heavy confused dreams which still hang about her now.
She stretches to shake away the dreams. The clean wooden boards under the window feel cool and smooth to her bare feet. She's glad she got rid of her childhood oilcloth. Now she is flooded with quick relief at simply being awake in her plain white room, with the breeze stirring just a little and the summer air flooding in. She pours water into her ewer and swooshes it over face and neck. It's silky. She soaps her arms and breasts and rinses them with her flannel while her night-dress slowly slithers past her waist and buttocks and into a heap around her feet. She catches sight of herself in the glass: her pinkish nipples are stiff from the cold water. Real red-head's nipples, Hannah calls them. Her breasts are mapped with blue veins. Clare has never liked to ask if this is ugly, or not. Her shoulders please her, and her neck, and the small delicate moulding of her collar-bones. But she is not sure about her breasts, and perhaps her arms are too thin.
Today, though, her face is at its best. Her cheeks are firm, with all the contours perfectly in place, not squashed or swollen, or somehow disfigured as they are on bad mornings. She's even got a little colour. It's only from the splash of cold water, but it suits her. What a pity no one else will see it. By the time she goes downstairs, she will be back to smooth pallor. She must remember to put Vaseline on her lashes tonight, to make them grow thick. She keeps forgetting. This is Peggy's tip: it will work wonders, she says, and it's much cheaper than Lasholene. Clare leans in close to her looking-glass and stares into her own eyes. Yes, it's a very good day. Today you could call her eyes blue. Dark blue. It's because she has some colour in her cheeks. Perhaps she ought to rouge – girls do now, and no one notices. But how lovely it is this morning to be Clare Coyne. She hugs herself tightly, crossing her legs for a faint, familiar stir of pleasure. She rubs her face against the inside of her arm where her flesh is silkiest. The smell of her skin pleases her, with its trace of night-warmth under the fresh soap. It's a shame to put on clothes on a day like this. And she has nothing right for summer. Everything is worn out, after three years of war and pinched income and rising prices. If only she could cut out like Hannah. A cream flannel skirt four inches above the ankle, with a broad scarlet leather belt and a tussore-silk blouse... But that one outfit would cost more than Clare spends on clothes in a whole season. And scarlet would clash with her hair. She scrambles into her old navy middy blouse and navy skirt and twists her hair into a knot without plaiting it.
Nothing can weigh down her lightness this morning. This old skirt fits perfectly, showing off the narrow waist which is one of her good points, making her aware at every moment of her young, firm body. She bends to pick up her hairbrush with a flourish of suppleness. Thank God for a body which is beautifully balanced and will do everything she wants. She makes her morning offering swiftly, obliquely. Should you thank God for the size of your waist? Why not? Maybe He's glad of it too. There are enough thick waists in this world and Heaven too, surely, thinks Clare.
The beach is ridged and wet with retreating sea. A black-headed gull glides, stabs at something behind the rocks, then wheels up, crying mournfully. Clare walks down towards the softly collapsing waves. Everything gleams like pearl, like the beginning of the world. Now she has her back to the land, and there's nothing between herself and the sea. The sun is warm on her back. She has eaten and drunk nothing. She is purified and ready to receive. The little waves come to the turn, bow down, come to the turn, bow down. Their lace spreads fan-like around her scuffed boots. She wants nothing but this, to stand by the slowly falling sea and watch black glistening humps of rock appear like seals' heads, streaming with weed and water.
'It's Clarey, isn't it?'
It's John William. She should have known it was bound to be him. The jolt in her heart was recognition, not shock. It's like this when you've known all along someone will come. For a moment she can't speak. He comes up alongside, very close to her, and she moves half a step away. He is huge. Even on the wide beach he crowds her.
'How are you, John William?'
He smiles at her. The points of his teeth show. He's so close that she sees how his eyes are mazed with tiny red blood-vessels all around the iris.
'I'm not so clever this morning,' he says. 'We made a night of it last night, me and Albert and Jo and George.'
'What time did you get in?'
'They set off back around midnight. To be skinned alive by Aunt Annie, I daresay. I went walking.'
'You haven't been to bed?'
No, he hasn't. Those prickling red eyes, too steady and bright. His hair's matted, not combed. And he smells of beer and smoke. He never used to drink beer. It was always cider with the boys, when they could get away from Grandad.
'I don't know where I went,' says John William. He rocks on his boots, staring, pleased with himself, stiff and rigid as a dummy in that carcass of a uniform, stinking of drink. Pleased with himself he might well be, thinks Clare, crashing his way across the countryside thinking of nothing and no one but himself. Yes, he's got that look on him. A pinched yellow look round his nose. A couple of years ago that would have meant a fight, but I suppose now he's had enough of fighting.
'You should take more care,' she says sharply. 'It's a miracle you didn't fall over the cliffs.'
John William glances round the bay. 'Here's your miracle, if you want miracles,' he says.
She knew he would feel it as she does. How often have they come down here, John William barefoot, before anyone else was awake, to fish off the rocks?
'Why should _I_ be here?' he asks, staring round him. 'Can you tell me that, Clarey?'
'Why should anyone?' she says. 'But you _are_ here.'
Still she is not sure. Is he? He doesn't seem very present. He is as temporary as a ghost.
A ghost let in, touched, talked to. Let in once a year to eat food put out by the family. There is no sense in it. How can we still have leaves and train time-tables and welcomes? How can we expect to be glad when everyone we love is going to be dragged away from us?
'I went somewhere high up, I do remember that,' says John William.
'What, last night?'
'Yes. I can hear it now. There was a noise of sea in front of me, and something else behind me. A woman's voice, singing.'
'You'd been drinking. Or else it was a mermaid.'
'No, it didn't come from the sea,' he says. 'It came from the land behind me. I couldn't hear it quite clearly. It was a thin little voice – a foreign voice.'
'Where had you got to? You ought to know that.'
He smiles. 'I do. But I thought that if I told you, you wouldn't believe me.' Now he sounds like himself again. The bragging note has gone out of his voice.
'Where?'
'Above Pendour Cove.'
She laughs. 'However in the world did you get up there?'
'Looking for something I'd lost.'
'Were you? What was that, John William?'
'Why didn't you come down last night, Clarey? I was looking out for my little cousin, and then you never came?'
His voice blurs sentimentally. She looks at him and realizes he's still slightly drunk, even now after what must have been a fourteen-mile walk. That's why he's talking so much. Perhaps it's why he came to talk to her in the first place. That's why he went to Pendour Cove to look for the mermaid.
'You ought to get to bed, John William. You need some sleep.'
'I can't sleep,' he says. His eyes are suddenly empty of life. He sways a little. A small wave rushes around their boots, the last try of an ebbing tide.
'Waste of time to sleep anyway,' he goes on. 'I've only got forty-eight hours at home. Let's sit down by the rocks.'
The rocks are warm already. They sit facing the sea, with a pool below them where pale red anemones pulse and fray in the current of falling water. The sun falls deliciously on Clare's face. She'll freckle without her hat. Never mind. She shuts her eyes. This is just like when they were little, stealing out of Nan's with the boys, to hunt in the rock-pools at low tide when she was supposed to be going to bed.
But John William is restless beside her. He hoists himself down to the pool and kneels beside it to cup sea-water in his hands and sluice his face and the back of his neck. His hair is cut brutally short, bristling and stubby around the nape, a stubble through which she can see his scalp. All his beautiful soft black hair chopped to nothing. And all its shine is gone. It looks dead. He grins up at her.
'That's better!'
'They cut your hair too short,' she says. 'It's a shame.'
'You wouldn't say that, if you saw the lice we have out there. When you wash your clothes, you have to press all along the seams – like this – so it kills them.'
'How horrible!'
He laughs. 'It's just the way it is. Better'n getting typhoid.'
She looks down on him, hunkered by the pool, splashing himself and puffing out sea-water. He is busy and blind as a badger. The spread of his back and haunches looks scarcely human. He's undone his collar and rolled up his sleeves. His arms are brown, with fine black hair on the backs of them glistening with water, and his neck is pale where the deep line of tan stops. He's darker than she's ever seen him, but it's not the fine-grained salt and air tan he gets from working up at the farm with Albert and Jo and George on summer Sundays, helping with the hay harvest. Now his colour is mahogany, rich and somehow crude, as if it's a boot-polish he's rubbed into himself. His clear Red Indian look is blunted by the filling out of his face. His head looks small somehow, and again, animalish, quick and a bit frightening on his thick shoulders. She would not want to touch him now. His white teeth show as he calls up, 'Feels a treat!'
'Your hair will dry full of salt,' she observes.
He flicks his head and showers of drops fly off, to spatter against Clare's skirt. They make dark blotches on fading navy cloth. Then he bounds back up the rock to sit by Clare, so close she smells not only beer and smoke and salt but the smell of John William himself, unaltered underneath it all, a dry, almondy smell which is him and will always be him, a smell which even the war can't change, though it's changed everything else. He's still there. Her body gives and warms on the side where he almost touches her.
He stares out at the horizon. 'You don't get anything like this in the trenches,' he remarks.
She shifts in irritation. Why is he talking like this, with a crudity and banality which would suit Sam or one of the other lads, but is quite foreign to John William? It's as if he wants to pretend he's only got the same thoughts everyone else has. And John William's never had those, nor wanted them. He is different from the others, and she can't bear him acting the same.
'You surprise me,' she says tartly. 'I should have thought the trenches would be just like Porthmeor Beach. Not that I know anything about it, of course.'
'Good old Blighty,' he says, looking at her mockingly, appreciatively. 'I'm glad I saw you this morning, Clarey. I need your opinion.'
Her heart quickens. Something's wrong. He's going to tell her –
'Is it true what the Bishop of London's saying?'
'What on earth do you mean?'
His eyes yellow and glitter. 'The curse of lust and sin has fallen on London, according to him. Girls abandoning the decent ways they were brought up to. Is there much of that going on in St Ives?'
She slaps down her skirts and stands up. 'You're drunk. You don't know what you're saying. You'd better go home,' she says.
But he gets hold of her hem. 'Don't go, Clarey, don't rush off like that. I was only asking.'
Now he looks like himself, sharp and funny and a bit pleading. Well, he's been through a bad time. She mustn't take him up like that. He was only joking. And those articles in the paper Grandad reads are always about war-babies and sin and shame in cinemas. She sits down again. And isn't it the way he always talks, saying things other people wouldn't say?
'It's just the same here. Hellfire from Grandad if Hannah shows more than a couple of inches of leg.'
'Oh, Hannah...' He looks thoughtful.
'Have you seen Sam?'
He hesitates, picks up her hand which is lying in her lap, runs her fingers lightly against his palm. She's glad she filed her nails to smooth moony ovals.
'Yes. Sam's in London.'
So Hannah was right; the letter did come from London.
'But how did he get there?'
'Sick-leave.'
'But Sam's not sick. We'd have heard.'
'No,' he agrees. 'But Sam's not stupid either. Our Sam saw which way the wind was blowing. He had dysentery – well, that's nothing. Half the men have got it. But then he got blood in his stools, trust Sam for that, so they had to take him back to base camp for treatment. Course they'd a spotted he was faking it in five minutes. They're sharp enough, those army medical officers, and they know what their job is. So the next thing we know Sam is rolling around and shrieking out because he's suffering from shell-shock.'
'How do you know all this?'
'He told me himself. He's not ashamed, don't think it. He's set on living, that's all, and who am I to say he shouldn't be? He hadn't got one chance in ten, the section he was in, not with an advance due. So he gets sent back to hospital in London and the first chance he has he deserts. He made a fool of one of the VADs, I daresay.'
'But where is he now? Does Hannah know?'
'He's stopping in London. He met a girl there.'
She stares into his inscrutable face, tight on itself. He's lidded his eyes, the way he does when he doesn't want to look at someone.
'But he can't just stay there! I mean – what about Hannah? She's worried to death over him. Can't you do something?'
'No.'
'But – but he loves her. He loves Hannah. You know he does.'
John William moves his shoulders slightly. It's not a shrug, nothing as crude as that, but it's a gesture which repudiates everything she's said, and everything on which her words are founded.
'You won't say anything to Hannah,' he says quietly.
'Of course I won't,' says Clare, thinking of the letter Hannah couldn't understand, of Hannah's suspended nightmare of waiting last autumn, of the kisses she has seen and the love-making she can't imagine.
'He's going to see a new show called _Zig-Zag_ tonight. Good old Sam,' remarks John William surprisingly.
'Good old Sam! I don't see what's good about it. Don't you care about Hannah at all?' snaps Clare.
'He'll come back. As much as anyone's likely to come back. You don't know what it's like out there.'
'Is it so awful?' she asks.
'You can get through it as long as you don't think about it too much. I wish I could draw like you, though, Clarey. I've seen some queer sights. Flowers in the middle of everything. You'd like the flowers: you know I thought of you a couple of weeks ago. We were in an apple orchard, not far from Arras. Where Fritz moved back a couple of months ago. All the farm people had gone, but they hadn't been able to take anything with them except what they could get on a handcart, so our officers billeted themselves in the farmhouse and we slept in the orchard. Oh, the smell of it in the night, Clarey. All those trees flowering. You've no idea. We killed a hen for breakfast.' He gives a crack of laughter. 'One of the officers wanted to shoot it. I wish you'd seen him. He got down on his belly and went stalking the hen through the long grass, as high as your waist and wet. And the old hen didn't take a blind bit of notice of him. She kept on pecking away the grain we'd thrown for her. We thought we'd fatten her up a bit first. She was a stringy dud of a creature.'
'Did he shoot her?'
'No, his batman came up and wrung her neck. She'd have been blown to shreds if the silly bugger'd shot her. No good to us like that.'
'And you all had some?'
He laughs in real amusement. 'Luckily, as it happened, she turned out too tough for the officers. So I kindly took her off their hands and made soup for the men. They quite thought it had been their own idea.'
'Officers' training camp,' she says. 'You must have been pleased.'
The same movement of the shoulders again. Another rebuff. The weakness and silliness of what she has just said echoes in her own ears.
'You could say pleased,' he says. 'It's like being born. You can feel your blood pouring through you like it's for the first time in your life. When they told me, we'd been three days in the reserve trench, waiting for orders. There's big manoeuvres on, but you never know the whole picture, only your own bit of it. I suppose I'll have to know more now. But we knew it would be bad. Everyone was on edge. Then I was sent for by our colonel. I walked up the farm lane, out of range of the guns, noticing every flower in the ditch as I walked past it. It was warm. There was a sparrow having a dustbath. Things seem to burn into your mind sometimes. I went in and saluted him, thinking what now, was there something I'd done wrong. And he looked up from his paperwork and asked me if I'd like to go in for a commission. Just like that. I was to go to headquarters to be interviewed a week later. It poured through me so I could hardly stand, and all the time I was looking at the yellow in the white of his eyes, thinking he had something wrong with his liver, or else he drank too much.
'That was the worst week. I thought I was sure to be killed, just when I'd seen my way out of it. I knew I'd be sent back to Blighty for training, and I'd been told it would be three months. The war could be over by then. All the boys were round me, telling me what a lucky bugger I was. You'd a thought they'd grudge me my luck, but they didn't. I was going right away to Blighty, and they knew there was going to be another big push coming up soon. And they knew their chances.'
He is lost to her. He is a thousand miles away, hearing the guns, seeing the ring of faces round him and knowing their chances.
'I must get back, I haven't lit the stove yet,' she says quickly. Perhaps he wants to be on his own.
'Stay a bit. It's nice here. Listen, Clarey, you mustn't worry over Hannah and Sam. Sam's all right.'
Easy, glib words again. Anybody's words. But she wants to believe them. Hannah's all right, and Sam's all right, and John William's all right. Sam's letters stink of death and he'd rather hide with a prostitute in London than trust them here at home. But it will be all right, once the war's over.
She doesn't even try to believe it. She listens to the sea and the screaming of the gulls. They're gutting the catch round by the harbour. The gulls dive with slimy trails hanging from their beaks.
John William shifts position, frees his arm and puts it around her waist. It circles her there, warm and definite. She breathes in quickly, unevenly. He must feel her breathing. And her heart, beating so fast. She is sitting with John William's arm around her. She looks down and there is the back of his hand, curled over the light mound of her stomach, the black fine hairs on it lying flat now, the hand blunt and strong, the colour of the skin like that of another kind of creature from herself and her own paleness. The underside of her left breast just skims the top of his wrist as she breathes out. She sits as still as still. If she says nothing and doesn't move, they will stay together like this. Excitement trembles in her, and peace too. He is here, with her. No one else can touch him. For now he does not belong to the war. No matter how close he may be to the boys, they are not here. She is.
His hand begins to stroke her lightly, as he would stroke the belly of a cat. The tips of her breasts burn under warm cotton.
'There's a Red Cross concert on tonight, at the Drill Hall,' he says. 'Had you thought of going? I'd like to hear a woman singing.'
'I don't know – I hadn't considered it...'
No, but she's seen the poster. A Belgian woman singing in aid of the Red Cross. Opera-trained, with her piano accompanist, it said. What was her name exactly? Not Elaine, but something like it. Eliane, that was it. A preposterous bosom and a background of gunfire over a distant shore. But people are getting tired of Belgians.
'How are your Belgian atrocities?' they joke.
She can picture the concert, and the singer, her eyes wide with unshed tears against the crossed flags of Britain and Belgium. A faulty, shuddering soprano... But it is no use criticizing. Clare had flared in passionate pity with the rest of them, three years back at the invasion of Belgium. Now it is all different. People cope by making jokes and going on blindly, heads down, without flourishes. Her father would raise his eyebrows if she said she was going to the Saturday night concert.
'A predictable type of evening, I'd have thought, Clare.'
A woman's voice. John William must hear nothing but men's voices out there, except for the VADs. Only men, day after day, living together and dying together. No wonder they talk about each other the way they do, as if the other men in the trenches are more real than all of us here. The way even Sam talked about Billy when he came back on leave after Billy was killed. Tender. You'd never imagine Sam could talk like that.
'The singing I heard last night,' goes on John William. 'That was a woman. Or a seal maybe.'
'Or a mermaid.'
'That's right. Will you come?'
'With you?'
'Yes. We could go for a walk after. Like I did last night. Tell Uncle Francis you're going with me and Hannah and he won't trouble himself.'
'Yes.' Her voice is muffled. Surely now he can hear her heart thudding through it.
Yes, we could go for a walk. Like Hannah and Sam. Just John William and me and no one else in the world. He'll talk to me...
His hand on her waist tightens.
'You'd better sleep this afternoon, or you won't enjoy the concert,' she says.
His face shutters against her. 'I shan't sleep.'
There is a long pause, with the sand sucking as the tide drains out through it. A bee flies past them, out of its element and dangerously low over the water. The pouncing tip of a wave nearly gets it. How long would it struggle to fly again, weighed down by its wet fur and salt-sodden wings? His arm feels less warm and she is stiff with sitting still on the rock. She must go and cook breakfast for her father. She twists to free herself and he lets her go with disconcerting promptness.
'Goodbye, then,' she says.
'I'll call for you, Clarey. Seven o'clock.'
She nods, and turns to go. He hunches forward, clasping his knees. She's going, so he'll hold on to himself, smelling his own khaki, watching his boots glitter. The others are up already, working. Essential workers they are, Jo and Albert and George, and they are there to be seen working essentially by anyone prowling or nosing up at the farm. Coast-watchers, military police, tribunal members, let em all come. The boys will be blinding their way across the yard to swill down the pig-sheds now, hung-over and breakfastless.
She doesn't want to leave John William like this, worn out and half sober. After all this is her cousin. He's neither slept nor eaten. He isn't like himself. All the intelligence which scares the others like a fire which might singe them is sleeping somewhere way down in the middle of him. Like the red core of an overnight fire, piled up with slack. Almost sleeping. And when he put his arm around her, was that really him, or was it just part of his being a soldier, part of the strangeness? Is it something which has woken up in him because of war and death, which needs to be coaxed back into cousinly sleep? Or is it just that she was there? He doesn't really need her, not Clare Coyne herself. What he needs is a girl in London, and the music-hall, and a few drinks inside him so the lights grow muzzy while an old man plays _Goodby-eee_ on his violin and the audience roars its approval and John William's girl smiles at him the way she smiles at all the boys. A noble sense of renunciation swells inside Clare. She is family. Cousin Clarey. She ought to look after him.
'Nan'll make you a bacon sandwich if you come on up,' she calls to him across the sand.
But her words come to nothing. He remains sitting on the rock, looking out to sea, not wanting her.
# Nine
Twenty-five to six. The confessional curtain opens behind her, and Francis Coyne walks past her, calm and everyday. He walks the length of the church, dips into the pew in front of the altar, crosses himself, bows his head.
Good, thinks Clare. He won't take long now. She feels uneasy and inhibited as long as her father is here in the church. The fruits of a bad conscience. She's deep enough in the sticky web of family as it is. The confessional queue shuffles up by one, and now she joins it. She wanted to be sure that her father would be well out of the Church by the time she made her own confession. She kneels again, sighs, pulls off her gloves. The thumb seam is splitting and she's not going to be able to mend it again. The left palm is worn too. She will have to buy new gloves. She twists her fingers together, then covers her face with her hands to blot out the milky sea-light filtering through stained glass, and the row of patient backs, bowed over in the pews. The familiar church smell comforts her as she takes a deep breath, calms herself, spins out a prayer which she can follow without panic: 'Oh Guardian Angel, sent to watch over me during my life, be with me now in sorrow for my sins. May my holy patron, Saint Clare, whose name I bear...'
Father's still kneeling there. He's taking a long time over his penance. I wish he'd go. What if he's decided to wait for me?
No, it's all right. She sees her father's back stirring, his head lifting. Francis Coyne crosses himself again, unfolds his long legs awkwardly, genuflects as he leaves his pew and walks out into the sunshine, putting on his hat. An obscure sense of propriety stops him from acknowledging Clare as he passes her.
That's that. The spice of May Foage has been cleansed from his flesh and his spirit. The child looks miserable today. I wonder what's wrong? Better not ask her. It'll be some quarrel that's blown up between her and Hannah. And now she's exaggerated it into sin in her own mind, no doubt. He smiles fondly at the thought of his daughter's innocent conscience. But he'd felt sorry for her, kneeling there, looking small and young and extinguished. She isn't pretty at all at the moment. Slate shadows round slate eyes, her face narrow and pinched, with spots of colour on the cheek-bones. It takes him back to the sight of her when she was ten or eleven, with all that fierce hair strained back into tight plaits. She used to kneel beside him during Mass and hold up her fingers in front of her eyes to make patterns of light through them. She didn't have her mother to make the church a proper home to her, that was the root of it. Even if she'd lived, Susannah wouldn't have known how to do it. She was stiff and not a quick learner. She would never have called in easily with Clare on her way back from the shops, to light a candle, pray for an anniversary or remember a small local saint. Besides, this church wasn't built then, and Susannah would have been even more uneasy in the over-intimate atmosphere of the Mission. And he too had failed. Perhaps it was because he was a man. Somehow he did not have the gift of making her faith both immediate and mysterious to her, in the way it had been to his sisters. He had spent too long explaining her faith to her. Clare fidgeted, he became impatient. If she wanted familiarity, she went down to Nan's.
Now there's this business of the Red Cross concert. I suppose she's going because John William's home on leave. They've always got on well. The two of them talking, talking, thin clever faces across the kitchen-table. But that was a long time ago now. Ah, well, she'll be safe enough with Hannah and Harry and John William, listening to the warblings of some lachrymose Belgian. All the cousins are going, apparently.
Clare hears the church door thud shut behind her father. She presses her face deeper into her warm hands and gabbles through her prayers of preparation: 'I desire now to confess sincerely all my sins to you and your priest, and for this purpose I wish to know myself and call myself to account by a diligent examination of my conscience...'
Her mind scutters through prayer after prayer. The worn phrases click and whirr. But it isn't true. I don't desire it. I don't want to know all my sins and call myself to account. So what am I doing here? John William's grin. John William's warm arm around her waist, his hand on her stomach, the spreading tingle in her breasts, the sweet heaviness in her groin. That's why I'm here. Because all that is sinful.
'Grant me, Father, this change of heart...' But a song is hurdy-gurdying round inside her all the while.
We're here because we're here because we're here
because we're here because...
There's no beginning or ending to this song. It goes on for ever, picked up by one voice as soon as it's dropped by another, just as the army lives on while the individual men die, one by one. It's a song for marching. It goes with the lift and crash of thousands of boots, and with the weary stumble of men who can scarcely put one foot in front of another any more. The army goes on for ever to the noise of boots and trench songs:
Send out the boys of the Old Brigade
Who made Old England free,
Send out my mother, my sister and my brother,
But for Gawd's sake don't send me.
Sam sang the song to Hannah and Clare. At first they were shocked, though they wouldn't show it to Sam. Later, talking in Clare's bedroom, Hannah said, 'But you have to look at it their way. When you think of it, all the time they're singing they're getting closer to the Front.'
Sam has got a girl up in London. They'll go to the music-hall together. The last number is always a throbbing romantic song, and there'll be swaying rows of soldiers up on the balcony with their best girls, the girls they're treating tonight. Then it's over and the lights go up and they'll spill out into the dusk, art silk rubbing against khaki, pressed close together and not giggling any more but solemn, drifting away down the streets, away from lights and music and the smell of wine and beer on warm breath. They'll go into the parks where there's a smell of jasmine and rank laurel and the night policeman turns a blind eye to movements in the shadows.
She sighs and shifts her weight to her left knee.
'I wish Hannah was there for Sam.'
She is being stupid anyway. Sam will not dare to go to a public place for fear of the military police. John William must have been joking when he said Sam was going to a show.
'But what will it avail to know my sins if you do not also give me the grace of sorrow and repentance?'
Sam is sinning: Hannah is sinning: I am sinning. Clare presses down a bubbling giggle. She's reminded irresistibly of the conjugation of French verbs. The perfect, the pluperfect, the imperfect. I shall have sinned, he will have sinned, we shall all have sinned...
Mrs Hore sighs enormously and creaks off her knees and into a box. She is so heavily in mourning she seems to drip black behind her. You look for a slow stain of black spreading out on the floor where she's been kneeling. Muttering begins again from the confessional. There's no one before me now. She glances sideways. It's the Driscoll girl who works as a maid up at Middleton House. Bouncy, red-faced Driscoll girl, not bouncy any more. Things have gone badly for her. Her brother, Caddie Driscoll – what was it they said about him? Poor soul, he was not right in his mind any more, not since he came home from the war. And the Driscolls had been going about boasting that they'd got him back without even a Blighty one, before they realized what it meant and that they'd got him back without the spirit in him any more. He'd never have been sent back unless he was fit for nothing but to be spat out by the war. Shell-shock was what he had. Clare had seen him bobbing and trembling in the back pew at Benediction. But he had to go out of the church because he could not bear the noise, and the smoke from the censer frightened him. His face was as white as a wax candle.
She turns and whispers to Annie Driscoll: 'Should you like to go before me?'
'Are you not ready?'
'Not yet.'
No, I'm not ready. More prayers. Can't leave now when there's six behind me, all murmuring, knowing, watching. And a quizzical face with a blazing beard questioning in front of my eyes.
O purify me, then I shall be clean
O wash me, I shall be whiter than snow...
She's loved those words so much, but today they don't work. She doesn't want them to: she'd rather cling to the sweetness of John William. The Driscoll girl ducks her head as she comes out, flushed, with her curls bobbing against her cheeks. A pause while the queue waits to see if Clare's piety will lead her into a yet more scrupulously lengthy examination of conscience. The pressure of their waiting defeats her and she goes in.
She drops to her knees, rubs her fingers against the splintery wood of the ledge below the grille, and speaks. In the close, dark silence her panic dissolves. The statue of Our Lady floats in her niche above Clare's head. The cool syllables of the priest's listening voice fall across her restless conscience as they have always fallen, quietening it. She accuses herself of impurity of thought and deed and thanks God for the phrases which shawl her meaning even from herself as her words leave her lips. It will not happen again. In the dark she detests her sins. For the first time that day she can think of John William without a prickle in her conscience or her breasts. Now, with the cool web of the Church embracing her, it seems impossible that she has burned all day long, her heart thudding suddenly each time she realizes that she's one hour nearer to night, to seeing him again. It seems impossible that she, Clare Coyne, has spent her whole day planning, intending, longing to commit the sin of fornication with her cousin John William. How easy it is to resist sins once they are put into ugly words. It is all so much easier than she'd thought it was going to be.
The mild implacability of the priest's voice warms after absolution, as he blesses her. Here she kneels in the confessional's wooden heart as she's done from the age of seven, confessing to greed over sweets, to disobedience, to anger, to stealing a red hair-ribbon which belonged to her cousin Hannah and which she never wore but hid away in her underclothes drawer. It wouldn't have suited her anyway, she knows that now. Gorgeous on dark-haired Hannah, infinitely desirable, but a flat glare of colour against her own complexion. Here she kneels, empty and grateful. She is still Clare Coyne, the same Clare Coyne, brushing dust off her skirts as she stands, her cheeks pale now that the rare high colour which has flared in them all day has gone. Her hips sway composedly as she taps her way across the floor and into a pew to say her penance, which is going to take her some considerable time.
Clare threads her way quickly down Tregenna Hill, up Gabriel Street and through Wesley Place. On and up she goes, up Windsor Hill with her quick, elastic, virtuous step, on and up to crest the hill above Barnoon Cemetery. Her head is down. She is not so much praying as telling over her prayers like beads.
A shrill whistle. She looks up.
'Clarey.'
She looks wildly left and right, and there's Kitchie sitting on the cemetery wall, his head framed by a bent Scotch pine. He lounges, looking at her satirically. He knows the Coyne religious routines, and has his own Treveal reservations about them.
'Been to get it all off your chest?' he inquires.
How long his legs are now. He's nearly a man. Kitchie's cap's always been too big for him – that's just Kitchie. But now it fits him. And his hair springs less wildly than it did. He is like her mother, they say, but she can't remember her mother as a distinct face. Only the shape and smell of her. Kitchie puts oil on his hair now, she supposes.
She grins back. It is always good to see Kitchie.
'Don't you wish you might do as much, you cheeky monkey. I'm glad I haven't got your conscience. What are you doing here, anyway?'
'Waiting for you. Knew you'd come up this way.'
'Did you! Then next time I'll go round by the beach, just to spite you.'
'Then you'll spite yourself with a longer walk.'
Yes, he's nearly a man. He has thickened up this past year. He is still the noisiest of them all, and he never seems to know what he's going to say until he's said it, but he's a good worker up at the farm with his cousins. He has the kind of hands which can repair anything. Quick and instinctive, they feel for the right place to slot a newly oiled bolt, or a strip of leather he's cut and stitched to repair the harness. But however good he is, however valuable he is to Uncle John and Aunt Annie, they won't keep him on the farm once he reaches call-up age, not with Albert and George and Jo, their own sons, precariously holding on to their exemptions. One more application might be the straw that breaks the camel's back, and all the Treveal boys would whirl together into the mouth of the war. That must never happen. It's agreed. Kitchie's mother knows the war's got to be over by the time Kitchie's eighteen. How can they take her Kitchie from her, when he's the only one left to her out of the four children born to her? Dead Florrie, dead Rebecca, dead little William. Clare remembers their blue wizened faces in their small rough coffins which Aunt Mabel lined each time with the shawl she crocheted for each baby as she waited for it to be born. A waste, Nan muttered. There were plenty would be glad of the warmth of those shawls. But she helped Mabel to fold in the shawls around the small skulls with the deep dip of the fontanelle in them. Nobody would dare say anything to Mabel's white, frantic face.
They can't take Kitchie from Aunt Mabel. Why, it's impossible.
But Clare knows that they can, and that nothing is impossible any more. There is no language to describe the world she lives in now, where lists of thousands of dead are published in the newspapers each morning as routinely as the small advertisements. She reads the newspaper. Not in _The Times_ , but in Grandad's _Daily Mail_ there are stories of heroic mothers who give up eight sons, or ten sons, to conscription. Stories of families who have eighteen adult males fighting at the front – sons, cousins, husbands. And all we have is John William. Of course they'll take Kitchie. Her heart tightens as she looks at him in his best white shirt and a waistcoat she recognizes as an old one of John William's. He's brushed his hair down flat for the concert, and his Sunday boots are carefully cleaned and polished. He's ready and she's not. Quick, she must get changed – John William will be here at seven.
'Hannah sent this up for you,' says Kitchie. Messenger-boy Kitchie – now she sees that there's a flat brown paper parcel on the wall beside him, beautifully tied with string. A draper's parcel. Kitchie's grin widens as he hands it to her.
'Go on, then – open it. I know you want to.'
'Don't be silly, Kitchie, not out here. But thank Hannah for me and tell her I'll see her down at the hall with the rest of you. She is coming, isn't she?'
'She wasn't going to, but John William brought her round. She's thinking of that bugger Sam.'
'Kitchie,' Clare murmurs, automatically reproving. But he has never minded his language in front of her. And she doesn't mind either. It's part of their shared escape from Nan and Grandad.
'Aren't you going to ask me in?' he reminds her.
She starts. How stupid of her. She should have remembered how much Kitchie likes coming to the Coyne house.
'You'll come in and have a cup of tea with us? Then I'll have to get dressed.'
'I don't mind,' he says, and swings himself off the wall and into step beside her as they cross the road up from Porthmeor Beach and into Clare's row of villas.
Kitchie scans the sea to their right. 'They say those U-boats come skulking in round Zennor Head Wednesday night. I heard tell yesterday they've got a cache in under Tregerthen cliff. Fuel and food. The coast-watchers've been up and around there, Jim Bossinney told me.'
'They're always up and down there. It doesn't mean anything,' says Clare.
'How can you say that when they've sunk so many good ships?' says Kitchie in passionate reproof of her automatic coolness and good sense.
And Kitchie doesn't speak alone. As their footsteps clip along the pavement to Clare's door, he is joined by innumerable, swelling voices. The air is thick with them. The war is not going well. After three years it is bloated and invalidish. And each month it grows trickier to handle. Generals shift and scramble and make stratagems and bury their mistakes. U-boats nose impudently all along the Western Approaches, drowning men, sinking ships full of supplies while bread queues lengthen. Rusty, split tins of food are washed up on the beaches. Three ships sunk between Land's End and St Ives a couple of months ago. And in January farmers had to watch from the cliffs above Zennor as a Norwegian vessel wallowed and her crew drowned in front of their eyes. Lords and politicians quietly panic inside the fastnesses of the War Office.
And now we need more enemies. Even the Germans are not enough any more. So many men are gone, so many are wounded. So many have their minds and spirits destroyed, and the news of this leaks from family to neighbour to acquaintance and spills into newsprint. Look at this group photograph of wounded veterans. Seven of them. Count the limbs. Between the seven of them they have one leg, and even that one leg lacks a foot. Their faces stare unreadably at the camera. Their hair is mostly parted to the side, but one has an exquisite, knife-sharp central parting. Two are moustached, five clean-shaven. Their round collars are white and immaculate. Behind them dense, heavy summer foliage stirs a little in a breeze which also ruffles the hair of the one man who looks upward, away from the camera. He is perhaps the oldest of them, with a bony, vivid face. He must be thirty at least. He looks like a man standing on a cliff in the face of a sea-breeze, wondering if the weather is about to change. He cannot bear to look into the camera. On his far right there are two handsome legless boys, one with his hands folded in his lap, the other with his trunk perched on a high stool between wheelchairs, balancing himself with an arm round the shoulders of the men on his right hand and on his left hand.
Up and down, even and then uneven, runs the row of heads.
The faces are young, fully moulded and perfect. There are no lines on them but for the tight line of each mouth holding back what it is never going to say, not now, not here. The youngest of them is only three or four years older than Kitchie. No wonder he cries out.
In the dull anger of late May 1917 Kitchie's not alone. Shops open and close, newspapers are printed, girls make their summer dresses and make themselves beautiful even though they have more best boys among the dead than among the living. An animal anger growls and murmurs. The war is full grown, lolling over its attendants, sprawled like a giant child which still won't fend for itself. The War Office scuttles to shovel it full of living food, for who knows what will happen if it gets hungry? And we've got Lloyd George now, not Squiffy Asquith. Lloyd George knows how to talk to people. He understands from deep inside himself the sway of public passions, and how to point them out and away from himself. He is the master of the jabbing finger, the whipped-up passion, the sweet permission to hate. Revenge is good and revenge is justified. As for enemies, let us also look within.
'Come on in, Kitchie,' says Clare, holding the front door open as she counts over the contents of her larder in her mind and wonders what she can offer him. She's prepared a cold supper for Father and herself. Hard-boiled eggs from the farm, lettuce, new potatoes, a bit of gooseberry tart Nan sent up. That'll do. She wouldn't have time to eat hers anyway.
'Will you have a bit of supper here, Kitchie?'
He will. He can always eat.
She leaves him breaking open white, fresh potatoes, like little eggs themselves, and picks up Hannah's parcel as she goes through the hall and up to her bedroom to change. Her room is warm with trapped heat, even though she's left the window open at the top. She drops the parcel on to her bed and tears it open. A blue and white striped fold of cloth ripples out. Hannah's new summer skirt. And a note: 'You can wear this to the concert if you like. Wear your white Sunday blouse with it, and your white belt.'
Hannah can't be going to the concert, if she's lending me her skirt. Clare strokes the material. It still smells of newness to her, although Hannah has worn it and washed it. It's a heavy French cotton, pre-war. If Sam had come home, Hannah would be wearing it and they'd be out, strolling along the beach or down to the harbour, having a quiet hour together after tea at Nan's.
But, in spite of her scruples, Clare can't bear not to wear it. She strips off her grey dress and washes quickly. Her white petticoat is clean, and her white cotton stockings. She steps into the skirt and shakes it out around her. It hangs beautifully, but the waist is too big – she'll have to thread the loops through her belt and pull it in evenly all the way round. There. She glances back at the brown paper and string on the bed and sees something small and round tucked into the corner of the parcel. She picks it up and examines it. Rouge-papers. She rubs her cheeks carefully, over the bone, then damps her lips and rubs the paper over them too. In the glass her eyes are big and hectic. The rouge changes her face into a narrow, inviting muzzle. She runs her hands carefully back through her hair to loosen it, so that it will cloud a little around her temples. The blouse is dull, but it will do. The skirt says everything. She turns, making it switch at her hips, then she kicks her grey dress under the chair and goes out, running lightly down the stairs to show Kitchie.
'My eye!' he exclaims gratifyingly; then, 'That's Hannah's skirt you've got on.'
'I know. She's lent it to me for the concert. Do you like it?'
She can't resist twirling, even for Kitchie. If only she could cut out patterns like Hannah! – or pay to have someone like Hannah cut out for her.
'Where's Uncle Francis?' asks Kitchie through a mouthful of gooseberry tart. 'This tart Nan's?'
'Mine's as good and you know it. Father was in church – I don't know where he went. He's not coming to the concert.'
'Well, he never does, does he?'
Sensing criticism, Clare says quickly, 'He always sends in money to the Red Cross. He doesn't like the crowds.'
'Oh, no, it wouldn't suit Uncle Francis,' agrees Kitchie tolerantly. 'You got any more of this tart, Clarey?'
She goes over to cut him another slice, but at that moment Francis Coyne's clock strikes the three quarters. Clare snatches Kitchie's plate from under him, makes a dive to brush off his crumbs with the tea-cloth, and chivvies him out of his chair.
'Come _on_ , Kitchie! There's no time for that now. Here, I'll wrap it in a bit of paper and you can put it in your pocket.'
But he's aggrieved. He's a young man in his best, off to a concert, not a kid with his pockets full of sweet-stuff.
'And I never had my tea,' he grumbles, moving towards the door reluctantly. 'What's the rush? It's only a concert.'
'I know, I know! Never mind the plate, Kitchie – I've just got to –' and she flurries to the back door. She'll pick a flower. But which? Not a marigold. Hannah'd never forgive her.
'You ought to of asked Nan for one of her roses.'
He's watching her fly about, noticing the colour rise in her cheeks again.
'Go _on_ , Kitchie!'
The door slams.
# Ten
The woman is bent over, sobbing, shaking out sparks of tears and rage. The solidly beautiful golden flesh of her shoulders is chased over by cloud shadows through the bedroom window. Her face is broad, wet and furious. Her hair spills down in a rough bright rope, over her collar-bone and her breasts.
'And as for you, Lorenzo – ha! I spit on you, you and your dirty little games, your fools of farmers. Just because they will listen to you and think you are God Almighty you run to them every day and leave me here alone. I am nothing! I am the Hunwife! Leave her, she is stupid, she will not understand. Go to Gurnard's Head to your Heseltines, your Grays, go down to the farm to talk with your William Henrys. I am only your wife! See if I will weep for you!'
Tears splatter across her face as she feels with a sure hand for a painted Italian plate on her dressing-table, swipes it up and sends it singing through the air at her husband's head. He steps sideways and the plate thuds against the wall and breaks. But for once he won't take up the challenge. They are both so tired, and he is beginning to realize how much confidence it takes to fight as they have fought.
' _Nu_ , Frieda,' he begins quietly, watchfully.
'They are all spying on me!' she screams out. 'This hateful spying everywhere. Because I am a German they must hide in ditches and under our windows to hear what I am saying. Fools, nothing but fools. We should laugh in their faces.'
'We should do no such thing. You know that. We should lie as still as hares in the field, and let them forget us.'
'Ha! That is fine from you, you who go around telling the farm-people the war is evil and that they are going down the slope to Hell. Tell me you don't talk to William Henry and Stanley about the war! I have heard you. And you dare to tell me that I must be silent.'
He frowns. She is right, and yet not right. Things are shifting so fast now, and so dangerously. The immense apparatus of censorship and surveillance is not new-born and timid any more; it flexes its muscles and looks about it for new opportunities. Three years ago we had freedom, such freedom that we did not even know we had it, and now it is gone.
He has seen which way the wind is blowing now, how hard it is and how cold. He no longer hopes to sway, to convince, to stand in the shallows and hold back the tide of the war. He is holding tightly, light-footedly now, clinging on to the rocks while the war lashes at his feet, hungry to tear him away and smash him to pulp. The militarism he has spoken against and written against for three years is now as powerful as he foresaw it would become. It can swat him, humiliate him, destroy him. It can force him on to a train to the barracks at Bodmin, it can strip him naked and part his buttocks to stare up his anus as part of a 'medical examination'. It can jeer at him. It can open his letters and question his friends. It can treat his wife as a creature less than human, to be watched and judged. It can talk over his head, civilly, smoothly, so that his writing is stoppered and even his friends feel entitled to juggle opinions about him.
_He is extreme. He is half mad. He is doing himself no good, and he makes enemies everywhere. If only he would keep quiet. Who is he to preach and lecture to us? Can't he understand how people take it? You_ _would think he'd be more careful; after all, he has a German wife. That's the root of it. Frieda has changed him. She is ruining him. All that brilliance – do you remember? Now she has crammed him full of her German ideas, this semi-mystical philosophical rubbish which no one wants to read. And how charming he used to be. It is the fault of that German woman..._
'We must keep ourselves apart, in our souls,' he says, frowning, disliking the words even as they come out of his mouth. There are too many words. He has spoken too many, written too many. Just now, just this summer, he has had enough of words. He prefers to walk down to the farm and help with the hay harvest, and collect eggs, and eat fried potatoes with the Hockings in the farm kitchen, and teach Stanley French. It reminds him of his days up at The Haggs, years ago when he was a boy, teaching Jessie French, peeling a basketful of pickling onions for Mrs Chambers and going up with the men to work in the hay-fields at Greasley.
The war wants to crush him, he knows that. And he knows now that he can be crushed as easily as a snail is battered to bits by a thrush which does not even want to eat it. He is no good to the war. Any military doctor can tell it at a glance. He is white, and thin, and when they listen to his lungs they can hear the harsh noise of his breathing. He has had pleurisy twice, and double pneumonia, but he can manage his body if he is left alone. It is serviceable to him. He cannot bear the way they look him up and down, these doctors, sneering and dismissing him. They would like to play with him like cats with a mouse. They would like to pass him 'Fit for Non-military Duties', so that they would have him to play with until he was broken. They want to force him to clean out latrines. He must be cleverer then they are, he must be still and watchful and keep out of their snares. He will be as wise as a serpent, as silent as an adder. He will write careful letters to his friends. And he will cry out against Frieda if she dances in the wind with her scarf flying above her like a banner. She dances for pure joy, but the war does not recognize that kind of dancing. It knows that she's twirling her scarf in a prearranged signal to the U-boats lying out offshore, waiting. And how shocked she is when he shouts at her for a fool, because, for all her passionate antagonism to the coldly watchful times in which she finds herself, Frieda does not yet realize how much she is hated. They would be glad to knock her over the cliff, and him too, if they could get away with it. But they do not quite dare; not yet.
He has a quick, bright vision of his vegetable patch. His onions are already swelling on the surface of the soil, strongly fattening themselves, half secret and half exposed. Water drains away quickly here, and the ground is dry. His hoe scratches between the onion rows and lettuces. Perhaps they'll have rain soon. The bare, blowing slopes between Higher Tregerthen and the sea are enough for him now. He will go no farther. Besides, the return fare to London is three pounds, ten shillings – money they haven't got. They have their cottage, where they can live cheaply. They have even bought a cottage piano, as well as their rosewood table and the chairs he has mended and painted.
Why isn't it enough for Frieda? A year ago he would have chased her around the kitchen-table and given as good as he had got for the throwing of that plate, but now he hasn't the heart for it. Besides, she is very strong. A powerful woman, Frieda von Richthofen, he thinks, looking at her admiringly, ruefully. A woman who would crack a plate across your head when your back was turned. And indeed has done so.
She is not crying any more. She gets up wearily, pads to the window, stares down.
'Why can't they leave us alone here?' she asks. 'We need so little. Just this house. Why can't they let us have our home in peace?'
But it's not enough any more to have few wants and try to hide away from the war in the hollow of an empty landscape. There aren't any empty landscapes, though you think there are when you first arrive, full of pure naïvety and hope. It won't work. Ordinary things are dangerous. They must not show a light, they must not tar their chimney, they must not have curtains of different colours hanging in the same window, they must not sing German folk songs on Christmas Eve in the tower with the Hockings. They must not try out Hebridean lullabies in case the outlandish sounds are taken for coded German. A block of salt in a bag may be a spy's camera. Oh, it's not the civil police – they are decent enough men. They don't like what they are doing. They aren't yet in the habit of believing that a man's life is not his own affair. It is the military police, tumbling over themselves to produce reports, to magnify gossip, to lurk shamelessly under windows and garden walls. They are ridiculous, really, but they don't care that they are ridiculous. They are only doing their job, and their job has to be done.
Yes, it's the military police, thinks Frieda. Never would she have believed it possible, but now she has seen it with her own eyes. She should laugh in their faces with derision, but by now even Lorenzo has realized that this is dangerous. But her anger comes up – she cannot help herself. They are opening letters from her mother in Germany, she knows it, and letters from Nusch. And there's the way even the decent people look at you now, so that she scarcely ever goes into the town any more. The Cornish people are still human, still individuals, but the war is winning. People don't meet her eyes, or they go past with an embarrassed, regretful half-nod. They have read about Hunwives in the newspapers, and how they must be on their guard.
_Shame it has to be like this._
_We didn't want it, you know._
_But still. While there's the war on..._
_This isn't the right place for you._
And the others, the ones who hate nakedly. Poison seeps out of their eyes as they watch Frieda lugging her shopping up the long, lonely road to the cottage.
_Big bag she's got there._
_Makes you wonder what she's got in it._
_It does._
_They ought to do something about it. It's not right._
They are foreigners twice over, he with his red beard, she with her red stockings. There she goes with her loud voice, unashamedly German, with her big scandalous body vivid with life when so many good men have died.
_Letters, she gets_.
_And sends them_.
_We know._
_Who is not for me, is against me. Who does not love the war, will lose it._
He sighs. He cannot fight with Frieda now. The vigour of it's gone, and they are unsafe and exposed enough as it is.
'Put your shawl on, Frieda, and come out for a walk.'
'I suppose you have hidden away my white scarf,' she says bitterly.
He watches her as she twists up her hair again. She never looks in the mirror save for one confident narrowing of the eyes when she is finished. Her hands seem incapable of making a wrong move. Her body cannot be awkward. Now she ducks her head round and looks at him, her face bathed in sudden radiance. 'It looks nice, I think, my new blouse?'
They walk up the lane side by side. They will walk into Zennor, proudly, her blouse glowing. Frieda walks quickly because, although she is not angry any more, there is still anger in her body. They link arms and he feels the pace and heat of her movements, and knows why she is walking so fast. No matter how furious she makes him, he loves her perfect physical honesty. If she is angry she spits and throws things. If she is happy she runs and dances. If she is sad her whole body is heavy with it. She does not lie or cover things up. He has never met another woman like her.
The lane turns and they see William Henry sauntering down from the high road, bare-headed, the collie bitch at his heels. He walks slowly towards them, nods to Frieda. The bitch growls but a sharp word silences her.
'Good afternoon, Mrs Lawrence. Shall you come down to see us tonight?' he asks, turning to Lawrence, who hesitates before he answers,
'I have some new music from London – you know, Frieda, the songs Dollie Radford sent. Perhaps Stanley would like to learn them? I'll bring them down and show them to you after supper.'
'Will you come, Mrs Lawrence?' asks William Henry.
His eyes are so dark that they are quite unreadable. Lawrence is right, there is something uncanny about him, as if he scarcely belongs to this century at all. Once, in a rare moment of expansiveness with Frieda, he told her that his fields were still the same shape as they had been in Celtic times, when the Celtic farmers had cleared the ground and made granite hedges of the boulders. He spoke as if those farmers were as close to him as his own grandparents, as if he could turn and see the shadows of them working his fields.
'Thank you, not tonight,' she replies.
No, he did not want her to come. He nods quickly at her refusal.
'You been walking over to St Ives by the coast,' he remarks to Lawrence. It is not a question.
'Yesterday, you mean? No, not all the way. But I walked along the cliff-path.'
'That's what I heard. They say you met Miss Coyne – the Treveals' Clare.'
Lawrence's nostrils tighten. Frieda knows how such gossip annoys him, as much the sly satisfaction with which it is brought out as what is actually said.
'I did. We spoke about painting. We hope she will come here and meet Frieda.'
'Oh. Yess,' says William Henry, drawing out the long soft Cornish _s_. 'That'ud be the way of it. I made sure it'ud be something like that. But you know how folks are here in Zennor. They'd say she didn't belong to be roaming off by herself, since she's half a lady.'
'Can't a girl do something as harmless as go for a walk without people watching and tattling about her?' bursts out Frieda in disgust and quick sympathy with Clare.
'Oh, she _can_ ,' says William Henry, watching Frieda. 'She _can_ do what she like. But folks'll talk, for she's Francis Coyne's daughter.'
Frieda shrugs. 'Well, I'm sure _we_ shall take no notice of what they say,' she remarks with an hauteur which goes straight back to the court of Kaiser Wilhelm.
'It's getting late,' says Lawrence. 'If we want our walk, we must go on now.'
They say goodbye to William Henry, and he lounges off down the lane towards the farm, the dog pattering behind him.
'What foul gossip,' says Frieda.
'William Henry is my friend,' remarks Lawrence. 'So you can imagine what the rest of 'em are saying. It was good of him to warn us.'
Frieda gives a slight, impatient toss of her head, and walks on. After a while she asks, 'Is your Clare like the Hockings?' He glances sideways at the word 'your', but there is no weight behind it. She is offended by William Henry and the whole population of Zennor and St Ives, not by Clare or Lawrence.
'Why, she's not my Clare! You know that. I hope she'll be yours, really. She's so quick and bright, yet I'm sure she's never met a woman who thinks for herself. She knows scarcely any educated people. You could show her so much.'
A breath of Frieda's pioneer spirit stirs in her at the thought of a girl who is gifted and eager and new to everything. But a breath of suspicion counters it. Is he offering her Clare so that he can have William Henry? She brushes the thought away. I am tired, she thinks, and I am letting myself grow weak and afraid. I must not be afraid. We started from nothing and look, here we are, in this place which can be more beautiful than anywhere in the world when it wants. We have our little house, and I am not ill any more. And Lorenzo says the sea-air is making him stronger. We will not have another winter such as we had last year. Let this Clare come and see what we have.
# Eleven
Clare sees John William coming along on the other side of the road from where she stands tucked right in at one side of her window, with her hand on the curtain to twitch it over if he looks up. How strangely he walks. This is not John William, who walks with easy grace and never seems to hurry. This is someone holding himself painfully upright against a storm. Yet the evening is perfectly still. He comes closer, and she sees him glance up at the house. Suddenly her blouse feels dreary and everyday. But there's no time to do anything about it, so she hoists in her belt another notch, ducks down to the mirror which plays back leaping lozenges of light and tells her nothing, then skims across her room, across the landing, and down the worn stair-carpet to wait by the newel-post, her heart thumping. She isn't going to be caught just inside the front door, where he can see her through the stained glass. But what if he goes round the back? No, he won't. Even when he was the grocer's errand boy, bringing up her pound and a half of lentils and yellow soap, he was too proud ever to go round the back entry. And here he is, fine and soft and clean in his uniform. She's flustered at the suddenness of him and the way he fills up the hall.
'Come in – come in and have something, John William. Will you have a glass of Father's sherry?'
'We'd much better get down along,' he says easily. 'Concert starts in ten minutes.'
And it does, she knows that, for hasn't she been watching each quivering touch of the long hand on to the minute all day long? But now she looks behind her and dismisses the face of the clock airily. 'Oh, is it so late? I hadn't realized.'
'Not that they ever start on time. You got your wrap – or something?'
She shakes her head.
'It's warm enough. I don't need anything.'
'You might be needing it later. It gets cool.' He's smiling.
She darts into her father's room and snatches up the Indian shawl he keeps draped over a chair. It is dark and gorgeous, though it smells of her father and his tobacco and his closed room.
'That's pretty,' says John William. 'Silk.'
He rubs the fine shawl between his fingers approvingly. He has always liked the things about her which are different from Hannah and the cousins. They stand in the empty, dim hall with evening sunlight filtering on to them through coloured panes of glass. The house is luxurious, suddenly, as she sees it through his senses, with its space and smell of polish. The polish has been rubbed in by Clare herself, but if he liked he could imagine that she has sat in a room playing the piano and asking some Annie or Edith to do the banisters again, because they are smeared. Behind them the clock which came from Coyne ticks, measuring out time. John William glances at it, watches the second hand go round at the top of the dial. It never stops; Francis Coyne says that it has been going since before Clare was born. She watches John William watching it. Time sweeping away, time which will sweep him away from her. She touches his arm, and he turns away from the clock and looks at her. Here is Clare with an Indian shawl gliding over her arms. Her workaday blouse is hidden, and the broad flirtatious stripes of Hannah's skirt. She moves slightly, reservedly, feeling herself to be desirable. This time John William doesn't try to touch her. He points at the undulating stained-glass maiden with her heavy knot of red hair and her rapt expression which Clare once loved and now mocks.
'She looks like you, Clarey,' he says.
She thinks he is teasing her, but no. He is serious as he traces the outline of hair, cheek-bone, neck on the stained glass. She knows how the glass feels under his finger.
'She looks like you look sometimes. Always in a dream. Maybe that's a good way to be.' He turns round to her and smiles.
Clare stills her lips and looks down. This is important. This is what John William wants, so it can't be ridiculous.
They open the door, and he folds her arm through his as they meet the flood of evening light and step out past the little staring row of tiptoe houses.
But unlike the Pre-Raphaelite beauty on her front door, Clare can't keep her mouth shut. She starts by making conversation, but soon she's accelerating through the void of his lack of response. And she can't be silent, as she has always been able to be with John William. Even though she hears herself gabbling and wants to stop herself, she can't. Words fall out, and they are all the wrong words.
'How is Sam _really_? Will you see him at the end of your leave? What will you say to him? Don't you think someone ought to say something – after all, there's Hannah...'
When? Why? What has come over him, how can he, how can he do this to Hannah? I don't understand.
John William's arm stiffens to wood. Clare ceases to be his young woman and becomes a canary clinging desperately to its perch as they reach the steepest part of the hill. She nags on, her voice high and shrill. Sam and Hannah. His own sister. What's changed Sam so much?
And then, unforgivably at last, what she really wants to know. She burns at her own words, driven on by a curiosity which feels more shameful than the curiosity which led them to lift skirts and unbutton knickerbockers when they were all young together and knew nothing but this town and one another. She asks about the war. About being _out there_. What is it like? What it is _really_ like. The war. Your friends around you, dying. Billy. No one tells us anything. But you know. Tell me, John William...
Tell me. Let me sympathize. Let me be the one.
You can talk to me.
You can spill yourself on me.
She doesn't even need to say it. He knows it. She is warm and sticky and eager. But he wants none of it. He cannot expose his exhaustion to Clare. It is all right for her to try to draw his soul out of him, but he has got to go back into the world of the war. He cannot look to the left of him or to the right of him.
She looks up and sees the tight skin around his eyes and the set of his mouth. It is only his heavy tan that makes him look healthy, she realizes. Under it his colour is bad. How they used to swing along together, arm in arm, she and Hannah and John William. They could walk for miles. But it hurts her to walk with him now. They hurt each other.
And here's the Drill Hall and so he doesn't have to say anything. And here's Pretty Peggy, free for the evening, dancing up to them. Little Georgie's in bed, and her employers were only too pleased to let her go to a Red Cross concert.
Peggy's eyes dart over Clare and John William as they barge forward together, out of step with one another. Clare bangs the back of her hand against the rough wall. She reddens slightly with the effort of suppressing pain. Peggy's little, off-centre face glows at John William. She is really lovely tonight, dressed in something he doesn't notice – white, perhaps. Cool. She teases him.
'I wouldn't have known you in that uniform, John William. He's grown so handsome, hasn't he, Clare?'
Now, appealing to Clare, she becomes the privileged friend, licensed to flirtatious sisterliness with John William.
Clare says nothing. Somehow she and John William have dropped arms. Her own arm feels numb where he held it, and she'd like to rub it. Instead she pulls the shawl round her more closely, because he's right, it is getting colder already.
Peggy puts up a little, light hand to marvel at the shortness of John William's hair. She caresses the bristle at the back of his neck. And he just stands there like a prize bull, thinks Clare. Her offended body recoils farther from his.
'Such a shame there's no dancing tonight,' says Peggy. 'I love dancing, Clare can tell you. But we haven't had any for ages, with all you boys gone.'
'You _should_ dance, in that dress,' says John William. The breeze stirs a wisp of it and the skirts flatten against Peggy's slender legs.
'Oh, well,' she says. She looks down, then up again, under lasholened eyelids.
'We ought to get our tickets,' says Clare. Her voice feels clumsy in her mouth. If only Hannah were here. She always knew how to deal with Peggy. When they were children, it was Clare who snapped at Peggy's provocations, and raged and pulled hair and got a whipping from Nan for treating poor little Peggy like that. Hannah sized up situations and then deflated them.
'In a minute,' says John William. The crowd bulges around the entrance, under the bunting, the home-made red, white and blue decorations and the skewed portrait of the King. There are farmers, shop-people, fishermen and swarms of kids without tickets getting their caps knocked off as they dive under people's elbows to see through the door. Not many in khaki. They'd all make way for John William. The Belgian singer has arrived. She's in the back, resting her voice. She's been sucking glycerine lozenges since noon, they say, because this is the last leg of a tour which has taken her from Exeter all down Cornwall. Travelling through all that dust, no wonder she's hoarse. But she's good. Oh, yes, there's no doubt about that. Your Uncle Arthur spoke to someone at Penzance market who heard her up at Plymouth and he did say...
Uncle Arthur and Aunt Sarah. Even Aunt Mabel who never comes to things and looks as washed out and wobbly as ever, with her fingers tense on Uncle William's sleeve. Kitchie ducks through the crowd towards Clare, lighting up at the sight of John William. Two gulls swing over the crowd, louder than the milling voices below. They poise on a current of air and then dive-bomb Mrs Cocking's feathered Sunday hat. There are shrieks of laughter, and she too decides she'd better make light of it in spite of her annoyance at the damned thing and the sight of her mangled, claw-struck best straw. Besides, one of the gulls has shat on it in fright. There's the black shape of Mrs Hore. The crowd makes space for her, respecting her inky trail of grief.
'Hello, John William,' says Kitchie. 'Hello, Peggy.'
For they stand together. They are a couple, suddenly, her white dress hazing his uniform, her laughter drawing in Kitchie too. Her eyes flit over Clare without looking at her. But tomorrow they'll be friends again.
'You looked _lovely_ in that skirt of Hannah's last night, Clarey,' she'll say tomorrow, widening her eyes. Because Peggy will never let herself know anything. What is there to know? If you said anything, she'd answer, 'But what's the matter with you? We were only having a bit of fun.' All she's doing is having a bit of fun and a lovely time. Everybody's having a lovely time. The sky's clear blue, striped with mackerel, the gulls are raving over fish-heads, Peggy's got the evening off and everybody loves her. Even little Georgie wouldn't go to sleep tonight without a kiss from his Peggy.
Peggy's flittering eyes have spotted the skirt Clare's wearing and identified it instantly, but she won't say anything about it now, because she's not that sort of a girl.
'He's so _brown_ , isn't he, Clare?' she coos. 'Even the back of his neck!' And she just touches it.
'It goes down farther'n that,' says John William, with his straight, unreadable look, and Peggy flushes deliciously and shrinks towards him.
'I expect you'd like to take a look, wouldn't you, Peg?' a voice remarks. 'I'm sure we could arrange it for you, couldn't we, Clare?' It's Hannah, slipping in suddenly beside Kitchie, wearing just her plain navy skirt and white blouse. Peggy frowns. Her native prudishness peeks out as she says, 'Really, Hannah, what a thing to suggest.'
But everyone is looking at Hannah now, not at Peggy. You can't keep a good story hidden. Everyone knows Sam's got leave but he hasn't come home. How will Hannah take it? Shame on him. You've got to feel sorry for the girl.
But you don't and you can't feel sorry for her, not when Hannah is there before you. Her eyes are steady, and after a few moments the furtive glances die down like a breeze-eddy on the bay.
She smiles warmly at Clare. 'You look lovely, Clare,' she says. 'Doesn't she, Johnnie?' Only Hannah calls him that, and hardly ever. But she does so now, asserting that _she's_ John William's sister as she moves in close to him, leaving no space for Peggy's insinuating sisterliness. As Hannah moves forward, she seems to draw Clare closer too.
'You and Clarey better be going on in,' she tells her brother. 'You don't want to be stuck in those nasty seats at the back, not with our Clarey, do you?'
She winks at the couple. They have all heard Grandad fulminating against 'those nasty seats at the back of the cinema', and the evils they hold for the young girls of Penzance when the boys are on leave. There's a ripple of laughter. Kitchie guffaws, then glances quickly at John William. Only Peggy does not see the joke.
Clare's breathing steadies. She is _their_ Clarey.
'All right, let's get our tickets, shall we?' says John William. 'Three, is it, Hannah?'
'I've got mine. I'm over there with Nan. Come on, Peggy, Nan was wanting to have a word with you. Come and sit with us.' And she tows off the reluctant Peggy, who is by now rather chilly in her white dress, and glad enough to be in with the press of warm, breathing people.
Some instinct keeps Clare very still. She draws her shawl close, and looks down.
'There's such a crowd,' she demurs. Hannah would be proud of her. John William takes her arm, and thrusts forward with the other shoulder. As she's expected, the people around them make way. All the farmers struggling for exemptions for their sons want to be seen here tonight at the Red Cross concert. They make way for John William and murmur approving nothings to him as he goes by. Clare and John William are drawn through the narrow central doorway and into the hall.
It is true enough about the glycerine lozenges. In the small side room Eliane leans into a mirror which has been propped ready for her. Her practised fingers twist up the last scraps of her reddish hair over her false hairpiece and jab in the pins. She hennas her hair and she had better do it again soon, because a narrow strip of grey is worming its way to the surface at her parting. However, no one will see it on stage. She damps her index finger with spit and combs the hair a little looser, clouding, hiding. Once she's back in London she'll go to Emile to have her hair cleaned, then she'll henna it again. But her costume is perfect. Dove-grey, demure but piquant, trimmed with black velvet. Eliane despises singers who trail about in flowing white with patriotic sashes, particularly those who continue to do so once they are over thirty. Who do they imagine they are deceiving? Besides, they have no elegance. Do they wish to pretend they are still at their First Communion? Then they should put a veil over their heads and have done with it.
Exeter, Plymouth, Bodmin, Liskeard, Truro, Redruth. St Ives tonight, Penzance tomorrow, and there's an end of it, for this tour at least. Back to London on what passes for a fast train these days, crowded with soldiers who will give up two seats if she asks them, so that she can keep her hatboxes beside her. Then she can draw her veil over her face and turn to the window and sleep. There is a place for veils, certainly: Eliane knows that no woman is at her most charming when she is sleeping. No woman over thirty, that is sure, when her unguarded jaw sags a little and a bubble of spittle comes out at the corner of her mouth. It's necessary to give thought to such things.
She finishes putting kohl around her eyes. The art lies in wiping nearly all of it away again, so that there is only a suggestion of shadowiness around the pure violet which is her best feature – really, she knows, her only feature. Her nose is little but poorly shaped; her mouth elegant, nothing more. And another week of these train-smuts will ruin her complexion for good. But her eyes are unassailable. She turns, opening them slowly to their most direct, most shadowy regard. Her face vanishes and there is a garden at dusk, a bowl of hyacinths, the sea at Naxos. Everything has been said already and sometimes she wonders if she will ever hear it again, when she wakes up in the train and for the hundredth time adjusts the tired, crumpled flesh of her face into a vase for her eyes.
They are terrible, these little towns. Smells of sweat and dust, great red faces lolling on the other side of the stage, greedy for her but always holding themselves back, never really enthralled. She knows quite well that even as her husky, hesitant voice casts out its spell and trawls among them a good third are still thinking of their dinner, or whether the position of their seat adequately reflects their social position, or how much money they need to put into the collection. If they are women, they are wondering about Eliane. Is she married or not married, is she – you know – people on the stage... They sit there preening themselves a little beside the bulwark of their Sunday-best husbands, who are bolt upright, hands on thighs, straining the cloth of their good Sunday suits. What they are thinking glides under the surface of their eyes as they devour her body and her clothes, and are satisfied that after all, there's nothing so special about her. But they can never quite satisfy themselves. Afterwards, if there is a reception and she must talk to the mayor or the freemen of the town or the local vicar, she will feel the prickle of eyes upon her still. What is it about her? Her dress is quite ordinary. The eyes cost it. Plain fine grey wool. Velvet trimmings, one and six-three the yard. No jewels. You would think she would be more – well, dressed up.
And yet the mayor and the curate kindle in spite of themselves. Warmth rises; one, more daring or simply more human than the others, presses her hand as he assures her again that it was magnificent, deeply moving. It's there again, that curious ethereal element which is called her success. It glows around her as she smiles faintly at the joke of one, or acknowledges the bow of another. And the more it grows, the more the faces smile and the crowd thickens around her, the more she longs to be in the next dirty train, with the window open and the cold night air coming in, on her way to her next engagement.
She isn't sleeping here tonight; she will go on to Penzance after the concert and spend the next morning resting her face and her voice in her hotel. There have been too many concerts and that faint, attractive huskiness is becoming too pronounced.
She has rubbed off most of the rouge too. Time to go on. She puts down her handkerchief.
Clare and John William are in the middle of their row and near the front. They both sit stiffly upright, conscious of their position among Chellews and Stephens and isn't that the Canon there, on the right? But then John William turns and gives Clare a lightning wink and suddenly this is their own concert, their night out together, and no one has more right here than they have. Clare glows and relaxes for the first time since morning.
Eliane steps out on to the stage, and stands there in front of them, 'quite simply', in a way which is so accomplished that it looks perfectly natural, her hands clasped at her waist. Her hands are extraordinarily white and fine, but that may be the light on them, thinks Clare. Eliane gives a little smile and lets her blue gaze cross the audience.
'I am going to sing for you first – a little song of my country – where some of you may have been,' and she gives a small, acknowledging nod towards the nearest man in khaki, who happens to be John William.
Clare's French is just about up to the song.
Little pigeon on my sweetheart's roof
Little pigeon with grey feathers
Fly to my sweetheart's soft breast
And tell him of my love...
John William does not understand, thinks Clare, glancing at him. But from the look on his face she wonders if he might have learned some French over there. And everyone knows the word for love.
The songs follow one another tenderly, hesitantly, humorously. Little villages sleep in the sun, girls are married, lovers run away, old mothers long for their sons to come home. Eliane sketches out the meaning of each song to them before she sings it. A red cockerel is deceived by a sly brown hen. A young girl sings in the early morning as she tills her vegetable garden. Then there's a funny one with a chorus about a little sailor who runs back home to his mother. Eliane translates for them and coaches them in the French chorus, and they sing along, first shyly and then with gusto as she draws them in with her white fingers. One last time, the little sailor runs away to sea and does not like the look of the dark green waves. His little boat shakes and makes him afraid. He wants to be home by the fireside, does he, drinking his soup? But he has to go, there's no way out of it. The contingent from downalong roars its approval. They know the feeling.
Then there is a change of tone. The piano accompanist drops his hands to his lap. Eliane steps forward almost to the edge of the stage. She closes her eyes and her eye-sockets became huge and shadowy but remain violet, as if the colour is still shining through her closed lids. She cannot sing this one with her eyes open. The audience leans forward. She says, 'This is a song I wrote myself, about my country. I am going to sing it to you now in English.'
In my country there grows an apple tree
Full of white blossom
The blossom is blowing
In my country there grows an apple tree
Full of green apples
The apples are ripening
In my country there grows an apple tree
Full of red apples
The apples are falling
In my country there grows an apple tree
Black and empty
The wind is blowing...
It works. It always works. The slender, intimate thread of Eliane's voice swells and dies down. The tree is empty, the orchard has been broken open, and the fields are ploughed up by guns and shells. There will never be any apples any more.
Eliane stands very still, her hands folded. She opens her eyes and looks along the rows. Each man thinks she is scanning the row to find his face. Clare sits back in her seat. Father should have come – this wasn't at all the usual Belgian thing. Not a mention of priests, churchbells, atrocities, flags, vengeance or valorous resistance. And people are standing up in their seats, applauding Eliane. Waves of sound beat around her grey dress and her small bowed head. Clare looks around at the audience, because for some reason she does not want to look at Eliane any more. Their faces glisten as they clap and cheer; only a few matrons sit it out, with reserved, sceptical looks on their faces. One or two are already striding purposefully towards the collecting boxes, ready to position themselves at the exit. Eliane remains still, and it is obvious that she's not going to sing again. She bows once and leaves the stage, and the steady clapping dissolves at once into hubbub as each person turns to a neighbour to say the same things about the singer. They are quickly said, and then it's time to scan the audience. Who's here, who's unaccountably absent, who is with whom, what is she wearing, _she_ looks very bad poor soul, is that John William Treveal, I wouldn't have known him.
John William doesn't say anything. He is still looking at the stage, though Eliane is no longer on it. He is grey with fatigue, and he is unnaturally still, leaning forward slightly, as if he hears something Clare cannot hear. There is a light film of sweat on his forehead. He does not look well, but perhaps that is the heat of the hall and the noise of the people? Clare touches his arm, and he turns towards her like someone waking up.
'That apple song,' says Clare. 'It reminds me of the orchard where you caught the hen.'
'The orchard she was singing about,' he says. 'You want to find it every day when you wake up. You would give the rest of your life to wake up in that orchard, with one blackbird singing. All grey and cool with wet grass, Clarey, and so quiet you can hear the leaves moving. Only you never get there. You start hearing the guns.'
'Do you want to stay on?' she asks, feeling her own generosity. 'Do you want to speak to her? After all _you've_ been over there...'
'Me and how many more? No. She's got nothing more to say to me than she's said already, and anyone who thinks otherwise is fooling himself.' But he speaks harshly, as if he's crushing down the person who might think otherwise and fool himself.
'We'll go, then, will we?' says Clare, and they stand and he takes her arm and they wait while the press of people thins. Then she sees something at the other side of the room. Red, flaming, jutting. A beard. Well, she would have never thought to see him here. This would have been the very last place. But the crowd is sweeping her closer to him. If he doesn't move, they're going to pass right by him. Will he speak? Should she? Will other people know who he is?
He turns and sees her and his face dissolves into a smile of such warmth that she smiles back at once, openly, the way a baby does to its mother, without even thinking about it. He reaches and takes her hand and pulls her out of the stream of people. She catches a curious look from Aunt Mabel. She's going to hear about this later, for sure. John William still has hold of her arm and suddenly there they are, the three of them forming an island among the people.
Lawrence speaks only to Clare, quickly and intimately: 'What did you think? Did you like her?'
She hesitates, then tells the truth.
'Not really. I felt as if I were _meant_ to be feeling something, but I couldn't feel it.'
He laughs. 'It's true, she hasn't a good voice. But _can't_ she put it over!'
Clare smiles back at him, delighted. 'You didn't like her either!'
'Nevertheless she knows her stuff. Look at 'em!' He gestures at the buoyant, glistening, departing crowd, fumbling in its pockets, smiling gratefully at the collectors as if glad to be given the chance to express its feelings in coin.
Beside them, John William darkens so perceptibly that Clare remembers him with a start.
'This is my cousin, John William Treveal. John William, this is Mr Lawrence. He's staying up at Higher Tregerthen.'
The men acknowledge one another. John William's eyes make the rapid, automatic scan of the conscripted soldier. A man of military age. Civilian clothes. He notes the oddness of the clothes, but doesn't bother to think about it yet. Not called up. Why not? Health, probably. Rejected. Look at his colour. And his thinness.
'You are on leave, I think, Mr Treveal?' inquires Lawrence, and in spite of himself John William finds himself responding.
'I'm due at the training camp tomorrow night,' he says.
'And you liked Eliane?'
'I did.'
Clare looks from one face to the other. John William's is alive now, and intelligent. He has lost the small-headed animal look he had on him in the morning, down by the sea. He looks like what he is, her cousin with his books and his night-school and his carefully guarded plans, and his wild fights under the hedges.
'I'm glad. Your time is so short, you should enjoy it.'
John William says, 'There's a woman who knows what it's like over there. No one here does.'
He is unanswerable. He has played his card and the two of them are silenced.
'And if I'd a wanted _that_ sort of enjoyment I'd a stayed in London,' continues John William, who is clearly responding to some argument which has been going on inside himself, rather than to what is actually being said.
What sort of enjoyment does he mean, thinks Clare? Oh. He thinks Lawrence is telling him to have a good time, the way soldiers are supposed to. He's offended. He imagines Lawrence thinks he is not clever, that he is just a common soldier. But I'm sure he is mistaken. Lawrence is watching John William attentively. The words between them have been prickly, but the two men don't now feel unsympathetic to one another. Suddenly Clare can imagine them walking together, talking. But now there are people jostling at their backs – they are in the way here.
'I must go,' says Lawrence. 'Frieda doesn't like to be alone late at night. She's expecting you on Tuesday, Miss Coyne.'
'How will you get back – you won't walk, surely?' For it's late, and dark, and six and a half miles to Higher Tregerthen.
'I shall walk. I shall like to walk. There'll be a moon.'
He must be stronger than he looks, thinks John William. He'll walk through the moonlight, on the high road up to Zennor. The coolness and freshness of that walk through the night sweeps over him suddenly like the most desirable thing in the world. He must go too – but there's Clarey, holding on to his arm. But Clare won't mind. She will know how he feels.
'If you like,' he says to Lawrence, 'I'll walk my cousin home, then give you company up to Zennor.'
Lawrence must have caught the shock of disappointment that went through Clare, or the look in her eyes before she drops her lids. He says quickly, 'I don't want to take you out of your way. I s'll enjoy the walk.'
'I may as well walk,' says John William. 'I walked all last night. I don't sleep here.'
'You slept this afternoon, Kitchie said,' Clare reminds him.
He hitches his shoulders. 'That was easy. Nan was talking.'
'John William!'
'I don't mean it like that, though. She was talking to Aunt Mag, not to me. Only, it felt good to sleep while she was in the room.'
'Yes, you sleep best in the presence of someone who loves you,' says Lawrence. 'But walking may do it too. You can fall asleep on your feet. _I_ shan't stop you.'
'I've seen columns of men marching, all of them asleep,' says John William. 'And you should hear the sound of their boots when they go out of step. It's like waves breaking on Treen Cove in a south-westerly. D'you know Treen Cove?'
Lawrence nods.
'We walk all along – all around the country,' he says.
'Ah, then you're a real Zennor goat!' says John William.
Clare says nothing as they laugh. She is very tired suddenly and knows that she looks pale and shrunken. Now she has swallowed her first disappointment, she wants to be rid of them both. She hasn't got close to John William at all the whole evening. First Peggy, then Eliane, and now her own friend. And yet under it all there's some small, secret part of her which is relieved. What might he want of her? They are not children any more. She begins to see how much John William might want of her now, and how much he could hurt her. If he asked something of her, she could never help giving it. If she could make herself into an orchard for him, she would be bound to do it.
They set out past the Wesley Chapel and turn up Wesley Place. As they come to Windsor Hill, the men each take one of Clare's arms, though she is supple and light and used to any hill. She has the red-bearded one on her left, moving quickly, though she hears the strain in his breathing, and her dark cousin on her right, drawing her arm close within his. The moon is up, just as Lawrence said, and there's honeysuckle in one of the gardens. She smells it briefly and tantalizingly as they go past its garden. In a few more minutes they are up on the top, with the cemetery beneath them. Clare looks along the row of houses and sees the light in her porch.
'Father will be waiting up for me,' she says. Let John William know that she never intended to do anything but accompany him to the concert and then home at perhaps just a slightly slower pace than the one they've taken. She is wanted, waited for, shielded by Father and this house and God's Holy Church and her own good sense. She is not like Hannah, she thinks betrayingly, then catches her hand on Hannah's skirt and flushes in the darkness, as if she's smirched Hannah's quick loyalty to her. How Hannah had spoken to Peggy – _that_ was what Hannah was like. Hannah's brother stops in front of Clare's house.
'My train leaves at eleven,' he says. 'You'll be there, Clarey? You won't stay away this time?'
She nods. She'll come.
'I'll see you then.'
He leans over her and draws her shawl close over her shoulders, although the night's not cold and she's about to go indoors. His bare hand touches her neck and she shivers. He leaves his hand lying there, heavy and solid. He is not a ghost. He is real and present, but tomorrow he'll be gone. She tilts her head so that her cheek touches the back of his hand.
'You go on in now, Clarey. You don't want to catch cold.'
'Goodnight, John William.'
And she wants to cry out to Lawrence to be careful of him, not to let him roam back along the coast-path but to make him go home by the road. For it's night, and the moon'll set, and he's in the mood to think he hears singing from down in one of the coves. _Women's voices_... Don't leave him on his own up there, she wants to say, and what she sees is Eliane, gleaming on a black, salt-soaked rock with just enough moon to show her eyes and her small half-smile turned to the land, beckoning.
'Goodnight,' she says again and stretches up and brushes his cheek quickly, lightly, with her own. It's not really a kiss.
'Goodnight, Mr Lawrence,' she says into the dark where she can barely see his outline, and he answers, 'Be sure and come on Tuesday. Frieda expects you.'
She opens the house-door and there is the smell of her father's tobacco and a glint of Sheba disappearing down the passage.
# Twelve
Father's study door opens as she crosses the hall, and he comes out, stooped and blinking. How old he looks. She's never noticed before that the space between his eyes is webbed with small lines. He'll have been straining his eyes, reading late in bad light, and tomorrow he'll have a headache.
'Ah! Clare. Did you enjoy the concert?'
'Yes, Father. It was very good. Have you had your Bengers? Shall I make some for us?'
He smiles gratefully. She is looking more like herself, he thinks. The concert has done her good. She ought to go out more. It's a pity there are so few people here for her to get to know. She should be going to dances, concerts, tea with friends, weekends in the country. Look at her now, all dressed up just to go out to a Red Cross concert with her cousins.
'I am glad you wore your grandmother's shawl,' he says.
'Oh!' She'd forgotten. She'd meant to slip the shawl back without his noticing. 'You don't mind?'
'Not at all, you wear it, you wear it. She would have liked you to have it. It was made to be wrapped around beautiful young ladies.'
She wishes he wouldn't talk like that. The role of creaking complimentary Papa doesn't suit him at all. She much prefers him bawling absent-mindedly for his cup of tea while she's in the middle of peeling potatoes. That's the way they've always lived together. He peers at Clare's skirt. That seems familiar too. Where has he seen it? He can't quite recall but it brings a disturbing association into his mind, one which doesn't belong with his Clare.
'I hope John William brought you home?' For these cousins can be careless. They forget that his Clare is not a child any more, to knock about the streets with them and find her own way home.
'Yes; he was going to go for a walk with a friend, so he brought me home first.'
A friend, thinks Francis Coyne. Clare is being discreet, or is she just innocent? John William is on leave, sandwiched between war and preparation for war. There's bound to be a woman somewhere who'll let him bury himself in her tonight. Someone he'll pay, or perhaps a girl who's always cared for him and won't wait now he may be going off again for ever. It happens all the time. Perhaps Clare even knows the girl. She was always closest to John William, out of all her boy cousins. Does he confide in her? Does he tell her things she ought not to know?
'How is your cousin?' he asks abruptly.
Clare moves off towards the kitchen, fidgeting with her shawl.
'Oh, well, you know John William. He hasn't said much. He looks well, though, doesn't he? Isn't he brown? Didn't you think so when you saw him?'
'He looks older. I'd never have thought the boy would grow up so fast.'
She rounds on him, blazing, her eyes dark in a white violent face.
'He isn't a boy any more, Father, he's a man.'
He is shocked. His Clare sounds as if she hates him.
'It was just a manner of speaking, Clare. Of course we are all proud of him. He's done great things, great things –'
'Or if he's a boy, then why don't the men go? The ones who are always telling us how much they'd like to go if they could? It makes me sick when Uncle John says he wishes he was twenty years younger. He knows it's a lie, and we know it's a lie. But we have to pretend to be grateful to him for what he'd never had done anyway.'
'But Clare – you know your Uncle John. He's not an educated man. He talks as he hears others talking.'
'Yes, why don't they all go,' she repeats bitterly. 'Why not Uncle Arthur and Uncle Stan and Uncle William and Uncle John, since they know so much about the war? They're always so proud of telling us what they've read in the _Daily Mail_.'
'I think you're being a little unfair. Look how hard Uncle John's worked to get the boys exempted.'
'He's tried for Albert, you mean. And Jo, and George. But not for Kitchie, has he? What about Kitchie when his time comes up? What about Harry? What's he going to do for Harry when they say that Harry's bad arm doesn't count any more, that's fine, he can still fight with it? Because they _will_ say that, you know they will. And what about John William?' Her breath runs out, and her voice squeaks as she says John William's name. She stares at her father.
'Clare. Clare. Come now. I'm sure if you asked John William, he would tell you that he understands why we have to fight. And think how well John William's done. His commission will stand him in good stead after the war.'
He has to say it, and even six months ago he could just about have believed it. He could have believed in what they were fighting for, and in the evil of what they were fighting against. But now he can't believe in it any more. He is reminded of the deadly frustration he used to feel when his mother called him into her bedroom because she had not seen him at the altar rails for months. She would talk to him of her own 'implicit faith', her God-given freedom from doubt or difficulty in matters of religion, and how she only longed to share this with him. He can see now her blind, ecstatic face as she told him she would pray night and day for God to grant him this same faith.
To cease to believe in the war feels like that loss of faith; it is something he has kept just as secret. All he can see is confusion, and newspaper columns full of deaths and explanations which mask still more confusion. We're frantic with fear of what we've already done, and what we're doing now, and what we're about to do, he thinks. But we have to go on pretending that we know what we are doing. Newspaper justifications. We are like children whose game has gone terribly wrong, all gathered round the one who's got hurt, begging him to say it's all right, it doesn't matter, the blood can be wiped off and we can all go home.
But we can't let out the truth, not now that so many have died. It would be like killing them again. We can't say how bloody and pointless it's become, and that we don't know what to do any more. These are our children who are dying, rows of them in uniform like the children they were a few years ago in their school photographs. Our sons, dying for us. It's the wrong way round. People die for their children, that's the way it has always been. Now we send them out fresh from school. Everything is confusion.
'Clare,' he goes on diffidently, 'has John William been talking to you about the war? Has it distressed you?'
She laughs, then her eyes fill with tears. Her face quivers and she hides it behind her hands, then sinks down on to the carpet, kneeling, obliterating herself, shaken with sobs.
Now he knows something must have happened. After the concert perhaps. How he dislikes these patriotic concerts. They send people out into the streets swollen with enjoyable sentimentalities which they confuse with feeling. This whole country is in love with the idea of death, as long as the death happens elsewhere.
He has a sudden vision of seven-year-old Clare, beside herself with rage, drumming the carpet with her heels. It's best for her if I let her cry, he tells himself, and the thought makes him easier. After a few moments, hovering, he says, 'I'll go and make our Bengers.'
A hot drink will do her good. He bustles gratefully to the kitchen, feels along the cupboard-top for the matches – isn't that where she keeps them? – and lights the stove. A soft blue bubble of flame comes up. He watches it, thinking of May. He tries to remember whether May has ever talked about the war. He doesn't think so – or if she has, it's only to say the same things everybody says. While he is with her, there is no world outside her cottage, her bed, the white, blue-hazed circle of her thighs.
He scrapes the bottom of the Bengers tin, stirs the powder into hot milk, fills two striped mugs and carries them back to Clare. Her hand slips out and takes one of the mugs. He sits down on the shabby carpet beside her and puts his arm around her. She is still shaking, but she seems calmer. She's over the worst of it. How she used to cry! He smiles, remembering. He'd think she would choke herself, or scream herself into a fit. He had nearly forgotten. When her grandmother used to scold her or Hannah wouldn't play or the others had gone off without her. And now she's grown up and he doesn't know what she cries about any more. Who does she talk to now?
'Clare,' he suggests, 'perhaps we should have the rosary tonight? With a special intention for John William?'
He braces himself for the sting of her refusal, but after a minute she nods. He was right not to talk about the war any more. They both need to recollect themselves. He goes and fetches their beads, and they kneel on the hall carpet together, with the cheap cord pressing into their flesh. He makes the sign of the Cross and leads the prayers, feeling his way forward through Clare's silence, until her voice responds in the third Hail Mary. He sighs in relief and plunges on through the decade of the Resurrection. This is putting things right! Somehow they've got out of the way of the evening rosary – he can't quite remember how it happened. He had always resolved to keep it going for Clare, no matter how he felt himself. Religion is more important to women. They are beautiful within it, he knows, thinking of his shy, plain sisters.
To Clare, the hall is as cold as the bottom of the sea, as cold as the current that tugs past the Island, taking dead things with it. She's stood there often in early spring and watched the mackerel luggers buck as they come out of the harbour and set sail south-west. From here you can look down on the men bending over their gear, with their backs to the land and their legs braced against the swing of the decks.
Prayers slip and drift behind her; her cold little beads trail through her fingers. She shuts her eyes. Now, she's leaning over the back of Uncle Arthur's rowing boat. The boys are rowing; Harry at the back because of his arm, John William pulling away strongly. The water is navy-blue and the sun blazes around the white circle of her cotton sun-bonnet. She is ten years old. She kneels in the stern and holds her beads down in the water, into the bubbling stream of the wake, and then slowly, slowly she uncurls her fingers and lets the sea slip the rosary right off her hand. There, it's gone. She looks over her shoulder at John William.
'An owl,' says John William. The two men look up and see its white spread wings ghost across the road in the moonlight.
'A little owl, isn't it?'
'Yes, that's what we get commonly round here. Sometimes you'll see a pair of short-eared owls hunting over the moor when we've a plague of voles. They'll raise two broods in a season then, and take hundreds of voles.'
They walk on in silence, as they have walked most of the last three miles, brushed by moths and night breezes. The moon is up, sheeted by a thin layer of cloud. Its light is grey and wan, like dawn. It has none of the glamour of unclouded moonlight and the two men's moon-shadows are vague splotches in front of them. Far off on their right the long Atlantic swell rolls in against the cliffs.
Their feet crunch loose stones. They are both wearing evening shoes, more suitable for the concert than for miles of country road. Lawrence had a lift into St Ives in the Hockings' trap earlier, and would have ridden home in it but for meeting Clare and John William. But it is good to walk. It feels like walking out of the world entirely, into a moonscape of sky and moorlands which were here long before the first people came to live in the hollows and valleys. Lawrence and John William walk on past streams rushing off the moor, through the smell of dung spread on fields, through knotty, complicated shadows of thorn-trees. Granite boulders loom, balancing on top of one another as they have done for more than two thousand years. They seem to pivot over the little fields like blind men. Surely they are moving? But when you stop, and blink, it is the effect of the moonlight. You walk on, and just out of the corner of your eye you see the loggans stir...
'Zennor witches,' says John William.
'This is a country for witches, isn't it? I've heard stories from the Hockings.'
'Oh, you'll hear stories from everyone here.'
Lawrence hears the other man's soft laugh. There is no humour in it. He glances at John William's face and sees that it is set hard. The moon draws out odd resemblances. John William's bulky shoulders and blunt, cropped head make him seem like a cousin of those standing stones. But how ill he looks, suddenly. The wash of moonlight shows a sheen of sweat on his face.
'So, you are going to study medicine after the war?' Lawrence prompts him, wanting to resume the conversation they'd started as they left behind the last few houses of St Ives.
John William lowers his head, draws in a deep breath of exhaustion. 'Yes,' he says vaguely.
'You are tired. I should not have brought you so far,' says Lawrence with quick, warm sympathy. 'Should you like to stay at the cottage with us? We have only two rooms, but we can make up a comfortable bed for you. It is very quiet – only Frieda and myself. You could walk back to St Ives in the morning.'
John William shakes his head slowly, as if to clear it. His footsteps drag. His shoe scuffs against gravel, then he stands stock-still.
'Listen,' he whispers.
The soughing of the wind, the sea's slow suck and heave, the thousand sounds of the countryside stirring at night. A small agonized squeak, cut off as soon as it forms. Around the moon the clouds thin and race north.
'Listen,' whispers John William, 'listen. Can't you hear them?'
'No, I hear nothing,' says Lawrence.
John William sinks with a slow staggering motion to his knees. His fists clench, his face turns up to the moon, he scans the thorn-bushes. _There it is_. He freezes, staring, then, 'Get down, Hawker! Get down! Get your fucking head down!'
A pause. The breeze lifts Lawrence's hair as John William slowly lowers his own head and knuckles his fists against his eyes, hiding what they have seen. In the quiet Lawrence hears his short, harsh breathing. He waits, not daring to touch or speak to the kneeling man. Slowly the shoulders relax and sag. Lawrence reaches out and puts his hand on John William's arm. He shudders, but does not strike out at the other man.
'Come on,' says Lawrence. 'Best come to the cottage. You can rest there.'
John William takes his hands from his face, shivers slightly, and looks up. His face is blank for a moment or two, then it relaxes. He says in the same easy tones in which he discussed his career in medicine, 'I ought to get back now.'
'Ought you?' asks Lawrence carefully.
'There's not much time,' explains John William.
'No, you are going back to camp tomorrow, of course. You will want to be with your family tonight.'
'You know Clarey's my cousin?'
'Yes. Yes, I did know.'
'Cousin Clarey,' repeats John William thoughtfully. His smile is real now, and he gets up lithely, sweeping the dirt off his knees as if it's an immense joke to find himself kneeling in the road.
'But you should not go back so soon. You aren't well.'
'Me?' asks John William. He sounds astonished. 'Why, I'm well enough. Fit for an officer.' He laughs again, looks around, says, 'We used to walk up here, you know. All the way to Zennor we'd walk, Clare and Hannah and me. The whole day up on the moor with our dinner basket Nan gave us. When we were tired, Hannah made house for us under a thorn-tree. Clare always wanted to go and see the little mermaid.'
'And did she?'
'Oh, yes. We got as far as the church one time. Then we had milk at one of the farms, warm out of the cow. The farm-people knew Nan. They wanted Clare to drink buttermilk because she was so pale, but she pressed her mouth shut. She didn't like it.'
His smile is pure reminiscent affection. Lawrence sees that he remembers nothing of Hawker now. John William has gone back to where Clare is a stubborn child in a grass-stained petticoat, lips pinched to a thin line, and Hawker is an unknown boy of seven somewhere in England, wandering out with his pail to pick blackberries or blubbing in a corner because he has had the strap.
'So you'll turn here?'
'I shall,' says John William. The two men are very close. Far off, a chained dog barks, but they feel themselves to be the only living creatures in the landscape. The cloud is dissolving now, blowing into flakes over the moon, and the light is strong. They shake hands, feeling the living flesh behind the transfer of warmth. They move apart.
'Come and visit us when you are next on leave,' says Lawrence.
'I shall,' says John William. He smiles, turns and walks swiftly back along the road to St Ives while Lawrence watches. John William doesn't hesitate or look back; he looks neither to the right of him nor to the left of him.
# Thirteen
He comes whistling into her dreams. The sea tosses; she stirs uneasily. Don't go so close in to the rocks, she begs him, in small desperate whispers. But he pulls harder, hauling at the sweat-worn oars. They are toiling out into the bay – she can see Clodgy Point with white foam smothering the rocks, then the waves roll back and the rock streams black and clear. The sea heaves as it roughens, but there is no sound of wind or water, only thin whistling, on and on.
And she wakes. The whistling goes on like bad luck. What had Nan told her? 'It's ill-luck to whistle, for you never know what you're calling to come to you. Don't let the men hear you.' The boys knew they must never whistle at night. The seven whistlers mean death when you hear them. But the night is still and there's someone out in it, whistling for Clare.
She slithers out of her bed. Her night-gown sticks to her, damp with nightmare. She lifts aside an inch of the curtain, and looks down. Dark shadows and a skim of moonlight show on the sea, and she still hears that thin whistling, insistent, on one note, then fluking up, then down to its home note again. Ah, but she knows that whistle, the one he was whipped for. And there's his dark shape too, and she's looking down on a strange new map of him in the moonlight, all face and dwindling legs. He is staring whitely up at her window. Cautiously, she pushes up the sash, and the whistling stops. They look at one another, neither knowing how much the other can see. Clare scrapes the window up a little farther and leans out and whispers, pointing down the road.
'Go round the back! Go round the entry. I'll come down.'
He nods, and disappears.
No time for anything. She throws on her dressing-gown and shakes her hair back, and runs on her bare feet down the stairs, through kitchen and scullery, to fumble with the bolts. The top one's rusty – everything rusts if you don't keep oiling it. And there he is already, close and sudden, making her catch her breath and shrink back. Even now, knowing who it is, she's afraid of him. It is blind dark in the kitchen, and she feels for the matches, but: 'Don't light the candle, Clarey! Come on out.'
'But I'm only in my night-dress. I can't.'
'You'll be warm enough. It's still as still out here. The sea's flat as a plate but for the swell. Come on, Clarey, come down and see it with me. There's no one about to look at you. All the tabbies have been in their beds for hours.'
He sounds as young and eager as Kitchie. She thrusts out a bare white foot and says, 'Wait a moment.'
Her old boots are in the cupboard under the stairs. She knows them well enough to lace them in the dark. There. She draws her dressing-gown sash tight and runs back to him. They listen, but the house is quite silent.
'Father never wakes,' says Clare. But he sleeps at the back of the house, and they are right under him.
They slip through the door and the little moonlit garden with white shells glowing at the path edge. Clare knocks her foot at the gate, then they are down the alley, over the ash and clinker, past the grumble of sleepy Kaiser at the Lees' ('I call him Kaiser so I can say "Down, Kaiser", see'), and then out on to the flat whitish top road.
'We'll go down through the cemetery,' says John William.
'Where are we going?'
'To the sea,' he says, as if there's no other answer.
They take the diagonal path across the steepness of the cemetery where she's so often played and hidden. For a moment she can feel all the sleepers moan and turn over beneath her. All their poor bones, some of them scarcely laid down, she thinks, and now here we are disturbing them. Our footprints are too light. We ought to be sad and heavy, carrying water and flowers. They'll know we're not thinking of them. But John William catches her waist and she laughs aloud. If anyone hears it, they'll only stop their ears and pray to God to deliver them from seeing a ghost that night. Her dressing-gown sash is loose and her white night-dress billows and glimmers as they flit down the slopes between the stones. And down they go out of the cemetery, past St Ia's holy well and along the top of the beach, towards the rocks. He is heading for the coast-path, and she'll come with him, no matter where. Tonight her feet are light and sure and she cannot lose her footing. The grass is tussocky and cold around the tops of her boots and dew wipes against her calves. They used to go out like this just as it got light, mushrooming. Hannah, John William, little Kitchie, Clare. Albert would meet them up at the leys. The air smells sweetly of salt and gorse, and though it is so still they begin to feel the thud and shock of the swell coming against the rocks, coming in under the rocks to its secret caverns.
Now they must climb. John William goes first, takes her hand, and swings her. She makes a long sure step and she's beside him. They are up on the rocks, with black sea in front and to their sides, and the land behind. There, on the left, the swell comes in under a jumble of rocks. They hear the surface noise of the water, and under it something deeper and darker, like a kept caged lion waiting for food which comes too late and never enough.
Here they are. There is always wind here, enough to flap her white skirts and strew her hair over her face. He is fully dressed and they have cut his hair so close that the wind can do nothing but comb through it. They stand side by side on the rock, facing out to sea. They are hidden from land here. Even spies would see nothing of them. Clare's heard how people used to spy on lovers on the Island, in their private places where they thought they were alone. Some of the watchers even took a telescope and claimed they were watching the boats go out. Then there would be scandal, hot and delicious, licking up narrow streets until it reached the watched girl's home in a thunderous explosion. The wind blows John William and Clare. It feels strong and warm, even though she knows it cannot be warm, for this is only May and long past midnight.
Her dressing-gown peels back, catching on her arms. They are side by side and then John William turns to her and suddenly, quickly, he ducks and with one movement sweeps his bare hands up under the night-dress from her calves to her thighs to her buttocks and round to her stomach and her breasts, sweeping so firmly that it is hardly a caress, rucking up her night-dress so that her bare white legs and stomach are exposed and she stands there and laughs at him, feeling the wind blowing all over her and his hands all over her. He pulls her close to him, his hands under her armpits, and all sensation goes as she rubs against his buttons and rough cloth and his stubbled cheeks. They buckle together and lie down. They are on the black, glistening edge of the rock, but they are safely wedged. Now, with her head on the rock, she hears the underground lion's cough and then the pause and shock of the wave beneath her. John William is pulling off his clothes. He bites his lip as he unbuttons his trousers, the way he always does when he's concentrating. He looks up and gives her his sudden, intimate, white-toothed, cousinly smile.
'Lay down your dressing-gown,' he says. 'It'll be better'n lying on the rock.'
She kneels up and her night-dress drops down, covering her body while she spreads out the dressing-gown. It's like laying out the rugs to air on the back lawn, after she's beaten them. The thought makes her pause. But already he's lying down, and he pulls his cousin on top of him. What kind of game is this? Is this how you play it? She's always thought –
He draws her night-dress up to her neck again, and she lies white and spread on top of him, her back to the moon.
'Open your legs, Clarey!' he says, and because he has always known how to do things she opens them and he runs his hands up the inside of her thighs and along the cleft of her buttocks and up to the moist surface of her vulva. And again. And again. He makes velvet of her. So this is what you do. She jolts and laughs and rubs her face against his.
'You touch me, Clarey. You touch me.' His voice sounds as if she is hurting him.
But she doesn't know how, or where to touch him, and when he guides her fingers she is astonished, for where can such a thing have come from or grown in the few years since she's seen John William naked, swimming around the boat? And it's hard. Is this what is supposed to go inside her? How can it? But all the same she touches him, first tentatively, then boldly exploring, caressing him as he has caressed. He groans and twists his face into the angle of her neck. She feels his tongue lick her skin, and she relaxes. Now she feels older than him and sure of herself. She is doing it right.
He raises himself and twists them both over so that she is underneath. He rests on his elbows, his face above hers so she can't see it in the shadow.
'You never done it, have you, Clarey?'
'No.'
'I knew you wouldn't've. Here. Lie like this. Put your legs up. I won't hurt you.'
And he doesn't. She's wet and ready for him. Maybe all that swimming and running and those pretend weddings with Hannah have made her ready. He is too big, she thinks, and for a desperate minute it won't work, it's all impossible and he strains while she tries to arch back from him and hits solid rock. Then with a sharp burn he's in her and it's done and they fit as she widens and has him. He's so heavy, but she has the rhythm now and then she's leading them. There's a rise in the rock under her back, just where it needs to be. Her body knows what to do now – she spreads herself wider and wider and then just like a sea-anemone when the tide goes out she folds in and laces her legs across his back and now they are one rocking thing, sweating and crying out with the thump of the sea and their own blood equal in their ears.
The wind sifts against Clare's naked left side. She lies triumphant, taking John William's weight. There is cold air blowing around the soles of her feet. She doesn't care. She lies still for as long as she can, then the pressure of rock under her wins and she shifts minutely.
'Mind out,' says John William. 'We're near the edge.' The words echo.
'Not too near,' she says.
'No, not too near.'
She moves another quarter-inch and her lips brush his shoulder.
'Here, keep still,' he says, easing himself off her. 'I've made you messy.'
He reaches for his cotton singlet and wipes down her stomach and the inside of her thighs. She wriggles to free her night-dress, and he smooths it down over her legs.
'There. That's it. You're fine,' he says. 'I didn't hurt you, did I?'
'No,' she says. 'It didn't hurt at all.'
Does he remember that she never used to cry at falls and fights? Only at words. He picked her up once, after she'd fallen running down Smeaton's Pier and cut her knee open so blood splashed on the cobbles, and he rode her home to Nan's on his back. There should be blood, she knows, since she's a virgin. But now she's like Hannah, not a virgin any more.
'There'll be rain tomorrow,' he says, looking at the little clouds shadowing the moon.
'Is it tomorrow now – or today?'
'Today, I reckon.'
She says nothing more. If it's today, then it's already the morning when John William leaves on the eleven o'clock train.
'Hours to go,' she says. 'The moon's not even set.'
She sits up on the rock and looks out to sea. It winks and glitters – maybe there's a shoal of mackerel running. Like a shoal-place in a moonlit field.
'Take off your things,' says John William. 'You don't know how nice you are without your clothes.'
'Do you remember,' she starts, then trails off. She feels herself blushing through the dark.
'What?'
'You know. When we were little. That game we used to play. You and me and Hannah.'
Silence for a moment, then he laughs softly. 'I'd a thought you'd forgotten all about that, Clarey.'
Does he see what she sees?
Two solid, bare little girls caper in the cove. They are Hannah and Clare, prancing in warm, shallow water. John William has his back to them, because he is digging the deepest hole in the world. His skin is dark brown all over but for the shallower tan over his buttocks. He is slicing through cold, hard, water-seeping sand towards Australia.
'Let's play our game,' says Hannah.
They sidle up to John William. Now he's standing aside from his hole. He is very busy, mending gear. Bits of driftwood crisscross importantly on the sand.
'Please, sir,' says Hannah. 'I found a mermaid.'
'Oh a mermaid,' says John William in a deep, growling voice. 'S'pose I better take a look at her.'
'Here she is, sir,' says Hannah, leading up Clare.
'You lay her down on the rock and I'll take a look and tell you if she's a true mermaid. Hide your eyes and stand over there till I tell you.'
Hannah arranges Clare on a flat rock, legs together, arms straight by her sides.
'Shut your eyes. Remember you can't talk cos you're a mermaid.'
Then she leaves Clare and stands in the cold shadows at the back of the cove, digging her feet into wet sand, knuckling her eyes until red spots come.
John William walks over to Clare with his hands folded behind his back. He marches up and down beside the rock on which she lies. Her legs are in the shape of a tail, and her eyes are screwed tight shut. After a minute she peeps at him through the slit of her eyelid. His face is solemn as he parts her knees with a smooth bit of driftwood and looks at her bare little vulva. Now she pretends to wake up. She sighs and opens her eyes.
'Are you a human boy?' and her blind, straying hand touches his little coiled penis. She shuts her eyes.
He turns and calls to Hannah, 'This is a true mermaid you found! Now take her back to the sea before she dies.'
He bends down, digging, while Hannah leads Clare into the water. But suddenly they see a jellyfish and Clare shrieks and the game is forgotten. It goes down and down to the bottom of their minds and they never even think of it until next time.
How many times did they play it? She can't remember. All these years she's never once thought of it. The next summer Hannah and Clare wore their bathing-dresses and screeched if anyone even caught a glimpse of them changing. John William swam with the boys then.
Now Clare sits naked on the rock for John William. They're quiet for a long while and they don't touch. It's getting darker, and the moon is going down. There's not much time – soon dawn will start to streak in the sky behind them, and there'll be people about, and voices, and the milk-cart moving, and the boats ready to go out, and the coal that's lying heaped in the sidings will be stoked into the train John William will catch. But she'll keep her back to all of it – she'll only look at the sea. Or she'll shut her eyes and when she opens them it'll be blazing July and they'll be six years old again, sent out for the day by Nan with their pasties and a bottle of cold tea and a handful of cherries. 'And don't come in till you hear five o'clock on the church bell, for I've all my washing to do. And you, Hannah, you're the eldest, you watch out for our Clarey. Mind she don't run too much and start up her coughing.' For Hannah is seven, and responsible for them.
'I'll watch Clare,' says John William.
She opens her eyes and it's dark. The waves shoot in under the rock with a sound like gunfire. They say in the south of England you can hear the guns from France.
'Come here,' says John William. 'S'pose I better make sure if I found a true mermaid, before I take you home.'
# Fourteen
John William was right about the rain. By ten to eleven on Grandad's watch big drops of rain are falling, pocking the sand of Porthminster Beach below the railway station. The wind has whipped up coldly, and everyone's clothes are quickly spattered. They stand in a tight family bunch, smelling of damp cloth and hair. Nan is in her best black, standing with Aunt Annie and Aunt Mag. Aunt Mabel isn't here yet – perhaps she's not coming. The women surround Aunt Sarah, who is struggling to keep back tears the way she managed to do last autumn, when John William went out first time. She held them back until the train was past the platform then. Heroic, uncharacteristic effort; but it's plain she won't equal it today. Uncle Arthur stands aside, talking to Uncle John. They've heard this morning that Harry has been called up for re-examination. Clare watches Uncle Arthur fumble the letter out of his pocket and show it to his brother.
The station is full of family and well-wishers, for three other young men are going out. One has come in on a farm trap from Zennor. He has a basket of eggs with him, hard-boiled, and his mother is fussing over two which have cracked, for fear the dirt will get into them.
Nobody fusses John William. He has his greaseproof packet of ham sandwiches in his pocket, and his cigarettes, and his kitbag. He stands with Hannah and Clare. The train is in already, hissing in the wet, but nobody gets on to it. They stand still, and the rain begins to fall more heavily, in long vertical streaks. Clare looks right through the train windows and sees a mist of rain settling on the sea.
'I better say goodbye to Mother now,' John William says to Hannah.
'She looks very bad. Do it as quick as you can.'
He pushes his way through the aunts to Aunt Sarah. She turns and sees him, and she can't hold back any more: she clings to his arms and his shoulders, terrible jerking sobs come out of her, her hat falls off, and she would fall herself if he wasn't holding her.
'There now, Sarey, come away now, come away,' says Nan, gripping her daughter-in-law's arm. Sarah shudders all over, and peels herself off from her son. They grip her, Uncle Arthur on one side, Nan on the other. The aunts flurry their goodbyes. If only the train'll go now, and it'll be over and they can get Sarah home. She'd've been all right, if it hadn't been for that letter about Harry coming this morning.
'Johnnie,' says Hannah. She hugs him, but she does not cling. She is giving him her strength, thinks Clare, not draining his courage out of him. And I shall do the same.
'There's my girl,' says John William.
'Write, mind.'
Now it's Clare. And it's really time to go now – people are bundling on to the train. Some of them are going on ordinary errands, Sunday visits from which they'll come back safe tonight with eggs or butter in their baskets. Back home, back safe. Shielded from the rest of his family by Hannah, John William takes Clare fully into his arms. It feels strange. They have no public history together. They know each other naked, not clothed. They've never held each other like this with their clothes on. Everything has been the wrong way round. Pudding before meat, Nan would say. His cheeks are smooth – he must have shaved. She feels the shape of him through the clothes – his arms tight, his chest flat and hard, her breast bruising against him. She breathes in the smell of him.
'Look out for me now,' he says, then there's nothing against her and he's gone. He pulls down the window strap and the window is full of him, looking out over her head at Aunt Sarah, who cannot look at him because she has turned faint and they are forcing her head down. The guard walks swiftly past with his two flags in his hand, then waves down the green one and steps on to the train. It's moving. Steam hisses and the engine pours down a rain of smuts on them. Kitchie begins to run along the platform, waving his cap in his hand. The rain slashes.
'Get in, get in!' shouts Hannah, seeing her brother take the rain full in his face. And he's gone.
Clare sees her father on the edge of the family group, looking down the line after the train. She hadn't noticed him there. The rest are still gathered around Sarah, who is smelling Nan's salts and coughing.
'I'll walk up along with you,' says Hannah to Clare.
'No, let's go by the harbour.'
'All right, if you don't mind the rain.'
Clare doesn't.
'No Peggy this morning,' remarks Hannah dryly.
'She'll be at chapel with little Georgie, I daresay.'
'I thought they were church?'
'Yes, I suppose they are,' says Clare dully. She hasn't the heart for malice just yet. She yawns hugely.
'Johnnie was late home last night, after the concert,' remarks Hannah.
'Yes,' says Clare.
'I was glad of it.'
No need to say anything. Hannah knows. It has made her glad to think of John William being with Clare, on his last night. The last night of his leave, that is, she corrects herself quickly.
But what about Sam? She hasn't asked Hannah – she hasn't helped Hannah. She's only thought of herself. And yet Hannah looks happier than she did the day before. Has she heard something? Has she had a letter too?
'Hannah. Have you heard from Sam?'
'I got another letter.' Her voice is low; she glances around. Nothing but seagulls and a thicket of masts and sails humping up and down in the choppy harbour water. The cobbles gleam with rain.
'What's it say?' asks Clare.
'He's going to stop in London for a while.'
So she knows. How much does she know? I must pretend it is new to me.
'What do you mean – stop in London? He can't. He's in the army.'
'London's a big place, isn't it? You've been there, Clarey. You should know. People can lose themselves there, easy.'
'But not if you're in the army. They'll come looking for him. There's the military police, you know. Then he'll be court-martialled.'
'There's a girl,' says Hannah with difficulty. 'He'll be staying with her. He won't need to go out. Nobody'll know.'
Clare is silent.
'But Hannah –'
'Why should I mind?' says Hannah fiercely. 'She'll keep him alive, won't she? If he can get through this war alive, what've I got to mind for? When it's over, he'll come home.'
'Do you think he'll be able to?'
'He'll have to,' says Hannah simply. 'I know Sam. He won't be able to keep himself from coming home. He'll be lonely for it.'
'And you don't mind – if he's a deserter?'
'Once the war's over, they won't shoot them for it any more, John William says. He might have to go to prison, but he'll come back. It won't be for ever. Not like it'd be for ever, if he's killed.'
'A deserter,' says Clare, trying the word over in her mind.
'And I wish they _would_ desert, all of them, every last one,' says Hannah. 'But they won't. Our Johnnie won't, so there's no use ever thinking of it. But now he's to get his commission, maybe that'll keep him safe.'
There's a couple of kids playing out, barefoot, their hair streaking down over their faces. Hannah knows the mother, who drinks. She'll be snoring without so much as a fire lit in the house. The kids play out forlornly in the rain, with no one to shout at them or to chivvy them back home.
We'll talk about it later, thinks Clare. She is so tired now that she cannot help Hannah. And Hannah is braver than she is. Clare's missed first Mass now, and second Mass. Anyway, she isn't going to go. She'll be no worse off committing two mortal sins than one.
'I'm going home to sleep,' says Clare. 'My head aches.'
'You might as well. A day like this, there's nothing to stay awake for.'
Better to sleep. She might dream. His touch. The smell of him is still on her. The inner muscles of her thighs ache from gripping his back, and she's sore, bruised, tender. She'll tell Father she's ill, and climb the stairs to her bed and lie in her cool sheets and go back in time, holding on to the night before. The wind's rising – she'll hear the noise of it roaming over the bay from her bedroom, and the noise of the waves around the rocks where she lay with John William.
But the sheets are cold, and her room is not a refuge. Rain skitters down the window. The wind is veering west, and rough gusts blow up off the sea. Her window always bangs when the wind is in this direction. She ought to get up and wedge it with newspaper, but she's too tired and cold and sore to bother. In the time it takes thinking about it, and being irritated by the irregular banging of window against frame, she could have been up and mended it six times.
Downstairs her father clatters about. It's not so much that he wants to wake Clare, as that he finds himself doing the things which will stop her from sleeping. He goes to open the front door – she can't think why. Then he must have gone back into the kitchen without shutting the door properly because she hears wind sucking and whining down the hall, gathering force until the door leaps open and crashes against the porch wall. She nearly springs out of bed, thinking of the glass panels, but stops herself. Then he leaves the kitchen door open so she can hear him banging about with the saucepans, until he finds one in which he can boil his solitary, pathetic Sunday egg. What a meal for a man. No wonder he has to clash her careful nest of saucepans until the enamel chips. He is angry with Clare in the way children are angry with mothers who suddenly stop their perpetual motion in kitchen and backyard and shops, and take off their aprons and say in flat voices which the children have never heard before: _I can't be doing with it any more. Go on out with_ _this (she bands out doorsteps of bread with no butter or jam) and stay out till your father comes home._
Clare knows all about this. She knows how she ought to be downstairs, slapping slices of cold pork on to plates; it's the least she can do when there's no Sunday dinner made, and he has come back from church cold and wet and out of temper. And she has not gone with him. And he has had no sponge-cake for his Sunday tea for weeks. He likes it dredged with sugar, but they can't get enough sugar to whiten the surface the way she used to, or to make patterns by shaking the sugar through a paper cut-out. How quickly her father would brighten if she went down now. He would flap about, making her a cup of tea while she sliced bread and butter with a small martyrish smile.
She twists in the bed, fighting off thoughts which elbow their way through. Hannah has gone home to cook and keep house; she will not slummock in bed indulging herself like Clare. Besides, there's no room for it. The house will echo like a drum to Aunt Sarah's weeping. Nan will be helping with her, and no doubt Aunt Mabel and Aunt Annie and Aunt Mag too, gathered round the steaming broth of Aunt Sarah's emotions. Oh, yes, they'll be comforting her all right, thinks Clare, and Aunt Sarah won't know about the avidity in their faces when they talk about her in the kitchen as they bolt down the meal which decorum insists they don't eat in front of grieving Sarah. Then Grandad'll lead them all in prayer. Clare shudders.
They will pray to God for John William's safe homecoming, their faces red and relieved because they're doing something for him. Even Aunt Sarah will feel the better for it as she whimpers over the best tea-cups. Poor Aunt Sarah with her red eyes and her red worn hands. And everyone telling her they'll never take Harry, it won't be allowed.
Her bed is as uncomfortable as the bottom of a boat in an uneasy sea. Sam has deserted. Does anyone know but John William and Hannah and me? Hannah thinks I'm safe with a secret. She doesn't know how I feel about Sam rolling up his big body like a hedgehog in a hole in London, to crouch there behind some girl for the duration. Why should he? Why should it be easy for him?
I don't think that John William would desert like Sam, even if the idea had occurred to him. Not because he's braver than Sam. Sam's brave enough, when it's wanted, but he'd say he won't be brave for nothing.
'Where's the sense of it?' that's what Sam would say. 'All of us getting killed like rabbits.'
Perhaps it's that Sam doesn't care what people think of him. He doesn't mind if he loses his place among 'all the fine young men we've sent over there'. He would rather that people looked sidelong at him for the rest of his life than go back to fight – that is if Hannah's right and they won't shoot him for desertion once the war's over.
'I'm not having any more of it. Let the rest of em make fools of theirselves. Not me.'
No. That's not Sam either. What the rest of them think and do has always been important to Sam. He was proud to swing Hannah on his arm because all the boys were after her and she could sew so fine and earn her own living. He'd got what the rest of them wanted.
Clare turns her face into the pillow and grips it. Hannah hinted at something in Sam's letter; something which had happened to Sam, and changed him. There must be things I don't know and can't begin to imagine. Experiences which no one in England can share. But now I know how Hannah feels after Sam has gone.
John William won't ever desert. I know that – I am sure he won't. He will never give people that chance to scorn him. He'll call himself a fool, but he'll rather die than give anyone else the right to name him one. He might mock them for making him an officer, but he's glad that it forces fools to respect him. And Sam will survive in his burrow in London, doing what any creature in its senses does when it hears the guns. Doing what you get shot for doing now. But John William's on the train, laughing maybe and talking to men whom I don't know and who will never hear of me – not from him. They'll ask if he had a good time at home and he'll smile and say something which will satisfy them but give away nothing. John William is good at smiling and giving nothing away. He never got into rages. If you lose your temper, you lose yourself. Other people learn what you want. Now I must hold myself still, like him, thinks Clare.
'My cousin Clarey.' He'll hold a letter from her, creased and rumpled from being reread as the guns boom and the men wait, tense, to go over the top. Or he'll tuck a bit of her hair under his tunic.
You fool, you fool, she tells herself, thumping the pillow.
'Men don't want spoiled goods,' says Nan. 'A man'll say anything on God's earth to get you to go with him, and then you'll never see his face again – you'll see nothing but the back of his britches when he sees you coming.'
It was advice given to Hannah and meant for her and Clare. Clare might be half a lady but all girls need watching, in Nan's experience. There's nothing worse than a fool who's a fool to herself.
Clare turns and lies on her back. Runnels of rain run down straight now, breaking their channels, sluicing the window.
What did Mr Lawrence say to John William, all that time they were walking together? What did John William tell him, that he couldn't tell me?
# Fifteen
_I was on Tom Stevens's croft._
_And that's where you saw em?_
_That's it. Brazen as you like, sat there on the cliff-top_. He _didn't give tuppence who saw em, nor by the look of it did she neither_.
_That's one in the eye for the Treveals_.
_It is_.
_Well, now, who'd a thought it. Clare Coyne. Go on. What did you see?_
_I'd stopped by with Tom. He was cutting furze with the two lads and I stopped to inquire after Ettie. He was telling me that brendy cow of his isn't doing. Well, we were standing there like I said; you know how Tom Stevens's croft run down to the cliff-path?_
_I do._
_And we see em there, the two of em at it in God's good daylight_.
_And Tom Stevens being strong Chapel too. Must a bin a shock for him._
_He is. A powerful voice for the Lord, Tom Stevens._
_Course the Coyne girl's a Roman._
_That's so. Not that I've anything against the Romans_...
_No._
_They have their ways and we have ours._
_That's just so. You hit it. And coddling and kissing with foreigners in brazen daylight isn't our way_.
_Mind, I'm not saying there was anything in it_.
_Oh, no!_
_But a young girl and a married man_.
_That's just it. No hiding theirselves either. You can see that red beard of his six miles off_.
_You can._
_And the girl with her red hair too_.
_Hot_
_blood_
_wonder if it's the same red down below_
_like a couple of foxes getting together_
_stink_
_smell of my Susan when she's got her visitor_
_So what did you do then?_
_Went back. Tom had a couple of pigs to alter_.
_She wants to get_ him _altered, that wife of his does. Keep him and his red beard safe at home_.
_Then after he left her he went on over to the Hockings._
_Teaching Stanley Hocking French, it is now. And piano. I heard William Henry tell it at Penzance market_.
_Piano. I wonder the Hockings don't throw him out of their house. Sits there in the kitchen like he belongs there, talking against the war_.
_What's Tom getting for his pigs now?_
_Ten and eleven-three a score, last I heard_.
_Is he now?_
_Selling em up to Redruth bacon factory. Ask him yourself at St Ives Saturday market_.
_And then the girl went home?_
_Back along the cliff-path._
_And he don't know nothing, her father?_
_Too busy going over Newlyn himself, I reckon. But that's an old story_.
_Same as May Foage's an old story_.
_Ah, well that's just the beauty of it for the Romans. It don't matter what they do all week just so long as they go along into the box and tell the priest all about it on a Saturday. Then there they are, ready to start up again on Saturday night. Just like magic._
_But_ he's _not a Roman, is he? Red-beard?_
_He's purely Godforsaken. Said as bold as brass to Rector he didn't hold with Church nor Chapel. Said a man should take care of his soul for hisself_.
_That'll've pleased Rector_.
_It did. He called Red-beard a snake in the grass, outside church Sunday before last. Tom had it from Jo Quick. Said we should remember how the serpent spoke smooth to tempt Eve._
The Rector sits at his desk. A blob of ink wavers on the end of his nib, but the flat creamy sheets of paper on which he writes his sermon are still blank. Words run through his mind:
The serpent was more subtil than any beast of the field which the Lord God had made. Dear people, my dear people, consider what these words mean. None of us, alas, can hope to see the Garden of Eden with our earthly eyes. With the hope given to us by God's infinite mercy we may trust that one day our eyes will be opened. Our eyes are clouded by sin. Our minds are dark and wretched, but even in our darkness we struggle to turn towards the glorious light of our Redeemer. But there are some who love the darkness. When they see the light of God they hide their faces and hate it. They love their own sins and abominations more than the light. They cling to them. Their Prince is the Prince of Darkness. They will come to you with words like honey; they will speak to you with as smooth a tongue as the serpent's when he spoke to Eve in the Garden of Eden, and drove the children of this world to damnation. They will speak with the tongue of the Great Deceiver who is their Lord and Master and if you listen they will tempt you.
He imagines the upturned faces of his parishioners, shrewd, subtle, earthly. 'Rector give it us hot this mornin.'
And then there's his insolence. The way he looks at me. One day I met him in the churchyard. What was he doing? Looking at the sundial, he said. That beard of his. It ought to be shaved off. His clothes. He's a figure of fun. And his wife's a madwoman or worse. Consider a serpent with the head of a man. Here I am holding back darkness. My people are never far from it. They speak of witches and loggans and mermaids and when I come by they fall silent. Once I thought I would write down the old stories and kill off their magic, but the stories still slip from mouth to ear, born in darkness and told by lamplight. There was a witch up here who worked her magic in the shape of a hare, and even when four strong men carried her up to God's holy churchyard in her coffin they couldn't hold her. The coffin shivered and turned over like an eel in their arms and there she went fleeting across the grass..
He dips his pen and begins to write. The first few words stick as his pen digs into the paper, then his fingers relax and lines of black firm writing fly down the page.
Evil goes fleeting across the earth, making a mock of all that good men hold dear. It will mock your wives and your daughters. It will mock your country and all we are fighting for. All those fine young men who have answered the call of King and Country, soberly, advisedly, knowing what is asked of them – are they to be made a mockery? Evil will mock at them and make a bonfire of their sacrifices. It will hiss and whisper in your ears. There is evil everywhere, prowling our countryside, waiting for the breach through which it can enter. It will settle here and make a home for itself. It will work through our land like a snake, and make rotten where it touches.
He stops writing and chews his pen. He knows that he will never say these words aloud; not as they stand. They are to be dropped, one by one, a word here and a murmur there.
He blots the paper and folds it over, stamping the crease with the side of his fist. These are notes to himself. They will keep his purpose firm.
Lawrence isn't writing. He has his flower and vegetable gardens to keep up, both the little plot which goes with the cottage at a shilling a year, and the two plots he has rented from the Hockings. The land is full of granite stones, standing upright or leaning over so that you think in a year or two they'll fall. But they don't fall. In a hundred years they might add a fraction of an inch to their angle. Here and there shoal-places break the surface of the fields. The stones are thick as fish under thin soil. He has learned to cultivate around the stones without driving his spade or hoe against them. The stones have been here longer than anyone knows. Long ago the first Celtic farmers shaped their small fields around granite boulders, and their granite hedges remain. Druids left the stones, as thick a crop of them as you get anywhere in this world. The work of the farms goes on around them so that you might think no one thinks of them or looks at them. But how can these stones fail to enter you somewhere? There are the shadows of them falling on your back as you earth up potatoes, or spray soapy water on blackfly-ridden runner beans. Or they stand at the base of the hedge in a froth of montbretia and wild fuchsia, their grey sides licked by purple and scarlet and flame. As you go down the little deep lane to the Hockings, tall stones stand in the field on your right, peering over the hedge at you. You might believe that at night when there's no one to watch them, they walk.
Walking stones. Lawrence thinks of the moon-blank loggans the night before, and the sweat on John William's face. Hawker. Who was Hawker? What was it that John William saw when he dropped to his knees on the road to Zennor?
The night-owl was flying. If a man believed in souls, how many of them would be flying now, looking for home? Thousands of them released every evening into the low grey skies above the Flanders plains, hungry for the lives they had been torn from, wanting to tell someone. Like souls in the Inferno they would press forward, naked and shivering, drifting like brown leaves on the banks of the Styx, squeaking. Behind them there crowd the next battalions, more and more until they are beyond counting. He thinks of the sigh that would go up from them.
He finds he is standing still, holding his hoe, too tired to grub up the weeds between another row of beans. He has some rare things planted. There are scorzonera and salsify, vegetables new to William Henry, who touches them with curious fingers and asks what they are good for. But where is Frieda? She left a note that she had gone out walking. She is so careless. She never says where she is going, though there are cliffs and dangerous places all around. She might stumble and turn her ankle, then call for hours. He smiles. He cannot really imagine Frieda calling for hours, and nobody coming. All her life people have come. She is as irresistible as wind or sun. Besides, she is not really careless. She knows that there are risks but she still takes them, just as she did when she gave up house and husband and children and threw in her life with Lawrence after knowing him for six weeks.
How quiet the cottage is without her. He potters through the kitchen, washing pearly scarlet and white radishes, tapping the bottom of the loaves he made earlier. They are perfectly baked. He lays out china on the little table, and sets their black-bottomed kettle on the fire. Soon she will be home, bursting open the cottage door, dropping a bunch of bluebells on a chair, unpinning her hat. The room will flare with life and vigour.
'Ah, Lorenzo!' she will exclaim, shedding gloves as she claps her hands for joy at the sight of the tea laid ready. 'Such a walk! You cannot imagine...'
And she will sit down opposite him and make a festival of the meal.
But it is late and she is not home. He cannot get the sound of John William's voice out of his head. For once, briefly, he allows himself to think how much he needs his wife.
# Sixteen
'Father.'
Francis flails his _Times_ open and hits the dish of rhubarb jam. There is no marmalade; there never seems to be any these days. Every day starts wrong for him as he bites through a greasy layer of margarine and then tastes this jam which leaves a metallic coating on his teeth for the rest of the morning. Why hasn't Clare got any butter? Surely to God, she ought to be able to manage it, with all her family slipping each other parcels of this and that the whole day long. And the damned paper won't fold in the middle either.
Clare pokes a spoon critically into the jam dish. Yes, she thought as much. There wasn't enough sugar when she made it and now the jam is beginning to ferment. There'll be mould frilling the jars in a few more days. Perhaps if she boils up the jam again with a bit more of next week's sugar which she's hidden in a blue canister marked 'salt'? But it takes a lot of fuel to boil jam. Nothing on the table is as it should be. The bread is grey. The margarine tastes like axle-grease, Father says, though how he knows what axle-grease tastes like she can't imagine.
Francis Coyne thrashes the newspaper. Pinkish brown streaks of rhubarb fibre cling to the corner of the front page.
'Father,' says Clare.
'Mmm? What?'
'I'm going up to Zennor today, to get some eggs and butter.'
'Zennor! Why would you go all the way up there for eggs and butter? Haven't you asked your uncle?'
'I can't keep asking Uncle John, Father. He can get much more for butter at the market than we pay him.'
It's true, but it's a slander to Uncle John all the same. He takes pride in himself as provider for the Treveals. She knows that if she asked him Albert or Jo would be sent down that same evening with a plump pound of butter wrapped in a damp muslin cloth. 'The sweetest butter in all Cornwall,' Uncle John called it, when he was feeling sentimental, and, 'It's all due to the hand at the churn,' looking at Aunt Annie, who bridled at that because she had a girl to help her now, and it was the girl's hand at the churn.
'But really,' Clare goes on quickly, 'I want to take my sketching things. I want to draw the white Himalayan balsam in the churchtown. We haven't an illustration of it yet.'
But he frowns and twitches another piece of toast on to his plate.
'Don't be ridiculous, Clare, it won't be in flower for a good two months. Sometimes I think I've taught you nothing.'
'And I'm going to visit Mr Lawrence and his wife. They have asked me to tea with them.'
'Mr Lawrence? Mr Lawrence? Who is he? I seem to know the name, but I can't place him. What sort of people are they?'
'He's a writer, Father. They've taken a cottage at Higher Tregerthen.'
'And how did you meet them?'
'At the concert, with my cousins.'
It sounds all right. She is margarining a slice of bread, rubbing the grease carefully into the crumb. There's something tricky here which he can't put his finger on. Something's been up with the child since – when? Last Friday? Saturday? And then her cousin leaving on Sunday will have upset her, though she's said nothing. He hopes they aren't going to have any more upsets like the one they had after the concert. This is where a girl misses her mother. The trouble is that Clare reads the newspapers and she knows what is going on. It would be easier for her if she were stupid. And now these new people – surely he's heard that name somewhere? Lawrence. Of course. The writer-fellow there's been all the trouble about. A German wife or something. And they won't like that round here. It's quite bad enough to be a Catholic from London; Francis Coyne is well aware that he is a foreigner and will remain a foreigner long after he has been buried in Coyne Chapel. Let her go. It will do her good to see new people, and a different way of life. A writer ought to have something to talk about beyond the price of pork and pilchards. And if his wife is German, perhaps she is a Catholic? He remembers Bavaria, and his walking tour there during the Long. People walked miles over the mountains to Mass, women in their costumes, men in black hats. And that greeting of theirs: _'Grüss Gott, Grüss Gott.'_ God's greeting. A true Catholic country; it had touched him more than he expected. He could open any church door and find himself at home. All the beautiful churches were still ours. And now we're fighting them. What a country it was. Those lakes we bathed in – as cold as death in the height of summer, cold like iron bracelets around wrists and ankles when we swam, weighing us down. We must have walked fifteen miles a day through the forests. The smell of resin, and the silence of the birds. White-haired children herding goats who made the sign of the evil eye as we went past. Those children will be old enough to fight now, I daresay.
'How will you get there? Is somebody going up in a trap?'
'I shall walk. It will do me good,' she says firmly.
'Higher Tregerthen; that's this side of the churchtown. Still, it must be six miles.'
'Oh, Father, you know six miles is nothing. And look what a perfect day it is! And it'll be downhill coming home. I'll take my sketchbook in case I see anything interesting.'
She will too. You need a salting of truth in the broth of a lie. She wants to sketch Frieda. Frieda will be beautiful, she is sure of it, even though Mr Lawrence is not beautiful at all. But he is a person whose wife would not be like anybody else's. You would find yourself looking longest at Mr Lawrence in any crowded room. He'd seem insignificant at a first glance, but your eye would be drawn back to him. He looks so alive.
The day is alive too, and glorious, even the golden slice of it which is all they can see through their window as they breakfast. Francis looks at his daughter affectionately, reaches over and pats her hand. His strong girl.
'Will you be going over to Newlyn?' she asks, collecting tea-cups. Her face is averted, her voice casual.
'Why, no – you know I only went there last Thursday. I have no business there today. Why do you ask?'
'I only just wondered – I wasn't sure.'
A very slight tremor in the hand which holds the cups, then she's gone, whisking up from the table with a pile of crockery. Could she have been laughing? Could she have been teasing him? Does she know – has she guessed? Could someone have spoken to her? One of her uncles, letting something slip, trusting the words not to mean anything to her. They don't know his Clare's intelligence.
But no. If she is intelligent, she is also innocent. It's impossible that she should think or suspect any such thing of her own father.
'I shall go to my study now, Clare.'
'I'm going down to Nan's to help with her ironing. Then we'll have lunch early, at twelve. Pickled herring,' she adds, knowing that her father likes to know but hates to appear greedy. 'And I have some spring onions.'
'Have you!' he brightens. 'Do you know, Clare, I think I shan't work on my classifications today.'
'Shan't you?'
'No. I have something else in mind.' He pauses, shyly, importantly.
'What is it, Father?'
'I thought I'd like to attempt a little sketch of my Oxford days. Nothing significant. I don't pretend to be in the same league as your Mr Lawrence,' he disclaims, wondering what Mr Lawrence _does_ write. He must try to find out. He could ask the Rector of Zennor when he next meets him in town. The Rector is an educated man, and must know these Lawrences. It would be interesting to have his opinion of them. And the Rector might be interested in Francis's memories of Oxford too?
'A small volume, Clare. It could be printed privately, for circulation among friends.' Francis Coyne looks at his daughter expectantly.
'Oh, yes, Father, that does sound interesting.'
'When we were all young together.' He puts irony into his voice but there's something else there too, something which she hears far more rarely from him. Eagerness.
'Stacey – Montfort – all of them. They were wonderful fellows. I made sure Montfort would make his name as a poet – we all did.'
He sees dripping sun; tall glasses of hock and seltzer. Montfort has just finished reading his new poem. They sprawl in a circle and he stands, his back to the chimney-piece. One of his arms lies easily along the stone shelf, but his left hand grips the paper tight, with the words on it which he's worked at from tipsy midnight through to morning. He stares round at his friends. For all his nervousness he can't help knowing, _knowing_ , that they must find it good, as he does. Stacey, Coyne, Chatterne. What will they say?
'Well, what do you say?' He trusts them enough to ask, to risk it. Tender, humorous, mockingly affectionate, their words swarm in the sunlit room.
'You've really done it this time, old fellow!'
'Stupendous.'
'That line about the lily.'
'Let's hear it again – I thought I detected a false quantity in line 47.'
They all laugh.
'God, I feel – I can't tell you!'
'Sit on that man, someone.'
A cushion is hurled, a wine-glass topples, a bubbling pool of wine and water runs over the carpet.
His Clare is looking at him. If only he could show her – make her see – but all that comes out is a repeated 'They were wonderful fellows'.
Clare smiles, and straightens the table-cloth.
Stacey. Montfort. Chatterne. And Eliot, and Stillington too, and all of them. Their sons ought to be at Oxford now, drinking wine and feeling as if it were the morning of the world. Getting drunk and writing poetry and travelling to Prague in the Long. Stacey went to Rome and wrote back, 'It is all just as we imagined it.' He fell in love there – we talked of it all Michaelmas term.
But Oxford's empty now. All the young men are in France, buried or still breathing, turning up their faces to the sun to feel its warmth. Ranks of schoolboys come up behind to replace them. We have not been able to give our sons anything. Not even one golden year. Gashed and splattered...
He fumbles _The Times_ in his hands. But he has his Clare, and here she is beside him, solid and warm.
'It was just a thought,' he apologizes. 'I don't suppose I shall do it.'
She is not interested, he thinks, shutting the study door. He does not guess that she will think of him all the way up the long steep pull of hill out of St Ives, and keep on thinking of him as the road bends and rises and she walks at a good easy pace the six miles to Higher Tregerthen. She thinks of his loneliness. She has always known about it, all her life, because it is half hers. If it weren't for Nan, and Grandad, and Hannah and Aunt Mag and all of them, and John William and Kitchie and the boys, she would live in the same globe of loneliness as he does. Her mother's family have a family liking for her father and they respect him, but she knows that they cannot really include him in their lives. None of them can be themselves while he is there. When he comes stooping through Nan's low doorway the atmosphere changes. Clare and Hannah are chivvied up from the table where they have been sprawling over the trimming of Hannah's new petticoat, threading ribbons at bodice and hem. Nan talks differently. Grandad leaves off bellowing hymns. Nobody can be himself or herself, even though Francis Coyne must have spent hundreds of hours with them over the years.
He spends less time now. Now she comes to think of it, she realizes that he hardly ever goes down to Nan's. He seems to think that now Clare is grown up, and there is no obvious reason for him to call to fetch her or to ask Nan's advice over a cough or a bad tooth, he has no place there. He has done his duty. That particular tangle of family doesn't require his presence any more. Nan likes him, but either he doesn't know it or it wouldn't mean anything to him. Nan says to Clare: 'Your father has beautiful teeth.' That's so like Nan. She fancies the idea of a son-in-law with beautiful teeth, instead of her broad, lummocking sons. Oh, yes, she likes his teeth and his way of eating and his fineness, but what good's that if you can't be happy in the same room together?
And he's not at ease with new people. It's just the same in the Church. He hates meetings, shrinks from comfortable intimacy with 'Father'. He couldn't do a good work to save his life, thinks Clare, though he gives enough to charity to stop mouths against him.
Yes, he's lonely. Better not think about it. It's like a bruise, and the day is magnificent. You could sing aloud, glorying in it. You could understand that the Magnificat was once a wild and unstoppable song of triumph, not a delicate lacework of church voices. Little complicated fields glitter. The sun is high and hot, but a breeze keeps the temperature perfect. She's walking up into wildness, up the little road with rising land on her left. The bracken's uncurling, white and hooked over at the top of each frond. Rough boulders glitter and the larks' song melts into the flare of gorse. On her right the sea shines like a shield. She shades her eyes and picks out farms, set in where a hollow will protect them from sea-gales, reached by thick-tunnelled lanes full of hart's tongue ferns, foxgloves, yarrow. She has been walking for two hours, and soon she should come to the lane for Higher Tregerthen and the farm.
And there it is, sandy and open at first, and then it turns and is coolly shaded by the high bank on the left.
'Just a little way down,' he'd said, and soon she sees it. But not one cottage – there are two or three, with the long front of the second one facing the track, just before another turn in the lane. This must be the place. Her boots and stockings and her skirt hem are thick with dust. She stops, and bangs some of it off.
The door's open. A snatch of voice – a woman's voice, quite deep and rich. Then a burst of song without recognizable words. Is it German? No, surely not. It sounds wilder than German. Someone begins to play the piano, not very well. About as well as Clare plays herself.
Clare walks quietly to the deep-set open door, which is slightly below the level of the lane. She's looking straight into the one room – and straight through it, for there's the window through on the other side, looking out over fields and down to the sea. It's like looking straight through the train John William left on. The room is dark. There she is, a big yellow-haired woman in a white blouse at a cottage piano.
'Hello,' says Clare.
The hands drop from the keys and snatch up a Paisley shawl. The woman looks round at her. A broad face, tense and watchful. She does not want anybody here.
'Who are you? What do you want?'
'I'm Clare Coyne. I've come for tea – have I got the day right? It _was_ Tuesday, wasn't it?'
The face changes completely. The woman jumps up and comes to the doorway and holds out her hands, both hands. Direct, warm, impulsive. This is her real face, thinks Clare, the other one was what she has learned to put on. Why?
Frieda takes Clare's hand and holds it between hers.
'Of course, I know now. You will think I am so stupid but I am playing the piano and dreaming and did not know the time. And Lorenzo is not here to tell me.'
'Oh. Isn't he at home?'
'He is only in his garden. His everlasting garden I call it, but I should not complain for he grows us many good things to eat. Radishes and lettuce and carrots and so. You will see. I will call him.'
She goes to the back and sings out in a full-throated way which reminds Clare more of the women downalong than any lady: 'Lor-enzo-o! Come!' Then she turns to Clare and announces confidently, 'He will come. And he will make tea for us and we have some little cakes too, I think.'
Gracious, how peculiar, thinks Clare. Isn't she going to make the tea herself?
'I am lazy, I know,' confides Frieda. 'But I have been ill – very ill. Neuritis so bad I could not walk.'
'I'm sorry. Is it better now?'
'Oh, it's nothing now,' shrugs Frieda, losing interest in the topic. 'Illness is boring, eh? Especially the illnesses of other people. It was so cold, so long a winter here. So much wind and rain – did you not think so?'
'I like the wind – or perhaps I'm used to it. And I think it is wilder up here than in St Ives. You are more exposed.'
'Oh, yes, it's wild. Why, at night, when Lorenzo was out' – she breaks off, glances out the back to see if he is coming, then continues – 'when he was out, I would be sitting here and the door would bang open – so – just of itself – and all the wind would fly around our house and the curtains would flap and these cups would shake so I think they would fall off. It would be like a wild animal in the house. A wolf perhaps.'
'Have you ever seen a wolf?' asks Clare eagerly. Germany, after all...
'No – oo, no wolves! When I was a girl we would hear of wolves coming down off the mountains, in the winter, when it had frozen hard for weeks. But I never saw it. We did not live in those parts. But such frosts! – you cannot imagine them. My sister and I would wrap ourselves up in furs, right to our noses.'
Clare looks around the room. The walls are washed pink – it is not so very dark, after all. The little boxed staircase goes straight up on the right-hand side, opposite the deep black fireplace. And the floor-stones are reddish too, scrubbed very clean. It is a queer, definite little room, quite unlike any she has seen before. No farm-worker's wife would think of making her cottage like this.
'I like your furniture,' she says, looking at the round rosewood table. 'Did it come with the cottage?'
She has said the right thing. Frieda glows, and reaches out to stroke the lovely, responsive wood.
'No, no, we bought it in St Ives for sixpence!' cries Frieda. 'They are so stupid there that nobody wants it. They must have new things. Everything must be new and ugly. What fools! So we buy all this, and bring it up on a cart and make our home. Even the piano.'
'You didn't get that for sixpence!' says Clare, laughing.
'No, that is quite true, it was very extravagant, it cost us five pounds. We only bought it this spring, but now we play it all the time, and sing.'
'And your china too – it's so pretty.' For Frieda has arranged it, none of it matching, bright, curious, delicate. One vase is stuffed with bluebells. They keel over lightly, beautifully, dipping their heads, bringing their smoky hyacinthine scent into the room.
'Ah! Now he comes!'
He is at the back, taking off his soil-clodded boots. He nods at Clare as he comes in and goes straight to the stone sink to wash his hands, which he does carefully, turning one hand over the other to examine his nails. All his movements are neat and exact. Unlike Frieda, he looks as if he belongs in a place like this. And yet he's a gentleman, isn't he? A writer? Or perhaps he's a half-and-half, like her.
'I'll put the kettle on,' he says. 'She's given you nothing to eat or drink, I'll be bound.'
'Why should I, when I know you will come and do it so _beautifully_ ,' mocks Frieda. 'You see, Clare, I am nothing here – just a useless wife.'
'She's like one of those lambs out there, frisking about in the sunshine, looking as if butter wouldn't melt in her mouth. "Oh, aren't they _perfect_ , aren't they _heaven_?" Yes, but they've eaten my broad beans, for all that.'
Frieda laughs. Clare thinks she relishes the imitation – it is of some other woman, affectedly ladylike. He is a good mimic. And Frieda appears not to resent the comparison of herself with the lambs.
Clare has never heard husband and wife talk to one another like this. In fact she is not used to hearing husband and wife talk to each other very much at all. Nan and the aunts talk together; Grandad with the uncles. An intimate, personal conversation between a married couple is something she is not used to and she finds it both exciting and embarrassing.
The house is spotless – as clean as Nan's. Lawrence builds up the fire with turves, and sets the kettle to boil.
'We'll have a proper tea, Clare,' he announces. 'Should you like that? I've some eggs, laid this morning, and bread and butter, and lettuce.'
'Such good things!' says Frieda.
Clare is hungry after the walk, and says so. They both approve.
'Clare's no lady, you see,' remarks Lawrence. 'She walks from St Ives to have tea with us. Now that is what I call friendship.'
Clare beams. His snub, ugly face charms her. Someone like this to talk to all the time! How lucky Frieda is.
Frieda looks carefully sidelong at Clare. No. This one is all right. She is no moth sucked to the flame of Lorenzo's vitality, convinced that she and only she can make a life worth living for him; once Frieda is out of the way. There have been several of them, and there will be more, Frieda knows. It's inevitable. But this girl, though friendly and eager, is shielded by something. Her eyes are not hungry, even though she responds to his jokes and his quick, warm way of looking round over his shoulder to answer her, as he sets out the tea-things and puts the eggs to boil. He likes her. Probably she is already in love, thinks Frieda, to whom such a solution will always be the first and most natural. And she is intelligent. Not the kind of girl you would expect to find down here. 'Half a peasant and half a lady,' Lorenzo had described her. 'And she can draw too. She has the energy to do something, but she doesn't yet know what it is.'
'I liked your drawing of Lorenzo very much,' says Frieda now. 'See, there it is, on the wall.' And there it is, not badly framed, although the frame is a little too heavy for a pencil drawing.
'Lorenzo found the frame, at the market. There was only rubbish in it. The wife of a farmer – oh so miserable, she was! So we took her out, and Lorenzo cleaned the frame, and made a new back for it. See! Doesn't your drawing look nice in it!'
'I'm glad you like it.' The girl flushes a little with pleasure, then frowns. 'But it wasn't quite right. It would have been better if I had drawn him here, in this room. The light's right – and the shape of things around him. I ought to draw him now, just as he is. But I wanted to draw _you_ today, if you'd let me. I brought my sketchbook. But perhaps there won't be time.'
Frieda lies back in her armchair. She is fully herself: supple, radiant, beautiful in a way Clare has not thought of women being beautiful before. She is pleased, but not surprised, that someone should wish to draw her. It must have happened many times before, thinks Clare, when she was young. Her confidence fills the room like sunlight. Lawrence and Clare lean forward looking at her. Clare measures the curves of flesh over strong bone, the spring of hair, the white swell of the forearm, the vitality of the pose. Lawrence looks at his wife.
'I should like that,' he says quietly.
The tea is fresh and fragrant, the eggs well cooked, the bread and butter good. Clare praises the bread and is told it is home-made.
'And the butter's from the farm – Lower Tregerthen. Do you know it?' asks Lawrence.
'I haven't been there, but I know the people – the Hockings. They come into market. And my cousins know William Henry.'
'I work down there sometimes,' says Lawrence.
'Do you?' She's surprised. What would the Hockings be doing with a man like Lawrence on their land? All this gardening he does too. And yet he doesn't look strong. He has that look on him, indefinable but familiar. It's the look of a sick man, for all his vitality.
'Yes. I like it.'
'It's heavy work, though, this time of the year,' goes on Clare, thinking of what Uncle John and the boys say. 'Leasing stones and cutting the furze.'
'I was hoeing mangels with Stanley this morning,' he says. 'But the work goes like steam, you know, when you are talking. We only stopped for croust. And I used to work on a farm when I was a boy. I know the way of it.'
Clearly this is important to him. How extraordinary. You would think a _writer_... but then I don't know much about writers. Frieda doesn't look too pleased about it. I wonder why? Doesn't she like them at the farm? Perhaps she thinks he ought not to waste his time there.
'Has your cousin gone back?' Lawrence asks abruptly.
The small neat room darkens. Clare puts down her bread and butter. There are too many faces.
'Yes,' she says. 'He went on the eleven o'clock train on Sunday morning.'
Her heart beats hard, right up in her throat, so that her voice is forced to squeak past the pressure of it.
'He walked back with me all the way after the concert – did he tell you?'
'Yes. At least – he said he was going to. There wasn't much time to talk at the station, not with all the family there.'
And we would have had much better things to talk about anyway, she thinks. Things about ourselves. But they hadn't talked about themselves. They hadn't said any of those things.
'He talked about medicine,' says Lawrence. 'But I wish he had talked about the war. I think he needed to talk. There was something he said...'
'What?'
'About a man called Hawker. I think he was shot. Do you know who he means? Was he a friend of your cousin? Did he talk to you about it?'
'No.'
Clare feels as if he has hit her. Six miles; two hours of talking. Telling Lawrence things he would not tell her. Sharing the life she has not shared. But she won't ask questions; she won't hear any of it second-hand.
Lawrence passes her cup to her. He looks into her pale face, closed in on its distress. He's seen that look before. Clare and John William are more like one another than he had realized.
'It can't go on like this. Things must be different by the time your cousin has finished his training,' exclaims Frieda, her sympathy quickened towards Clare.
'Now the French have got rid of Nivelle,' says Lawrence, 'their army's in a queer state. Pétain'll have to handle 'em with kid gloves if he wants to avoid a mutiny. Everything will fall on our army this summer. And Haig won't mind that, because it'll give him a free hand.'
Tactics. Clare's used to that. The maps, and pins, and names of little woods and hills suddenly on everybody's lips. And at first nobody knows how to pronounce the names, then the more that are killed there, the more we get used to saying the names of the places where they died. You think they are huge places, but the men say, no, they are just little woods and orchards, like those we have here.
'And what do you think?' she asks. 'Will it succeed – will they advance?'
She's never seen anyone's face change so fast. A bitter, jeering black look comes down on it.
'Oh – they'll advance. Have you ever heard men singing – a mass of men swaying together? They sing ragtime, your cousin says. Can't you hear it? Death in the air, everything dark, nobody daring to show a light. Can't you feel that there's murder in the air, even here? It is not just death but hatred of life, a desire that life _should not be_. And they sing ragtime on the troop trains.'
'Lorenzo,' says Frieda, her voice uninflected, warning.
'I can't write any more,' he says. His voice is thin and exhausted. 'How is it possible to write in the middle of this? The only way is simply _not to think_ , to work in the fields and in the gardens and slip back and forget it all. Let your soul go into abeyance and think of nothing. After all, there are living things...' His hands sketch softly, subtly in the air. She seems to see something growing there.
'So I plant my potatoes and my broad beans.'
'And the lambs eat them,' remarks Frieda.
'You are quite right. But at least the lambs have no evil intentions. They are just following their nature. I can curse them and chase them out of my field plot, and in a minute they'll whisk up their tails and forget it. They don't bear a grudge against me – they just look for another occasion to eat my broad beans.'
'But John William isn't like that,' insists Clare, brushing aside this talk of lambs and beans.
'How can you say that? On his own, he is not. Here, he's single and separate. Even aloof. But once they are there, they are not men, they are machines. They belong to the military, and it can do as it likes with them. And they are glad of it, somewhere in them. They let go of their wills, and lapse, and accept the will of the military.'
'All the same, you're wrong. I know him. He's my cousin, I've known him all my life.'
And besides – but she doesn't say it. She holds it back.
And besides, he came to me. Where were you when we were lying on the rocks? _That_ was John William, not the man you talked to. He'd let you say anything you wanted to. He wouldn't argue with you – that's not his way. John William keeps quiet for a long time, and then he turns...
... to fighting with teeth and fists and nails under the schoolyard wall, or to making love.
She is proud. She has her secret. She's got something of John William and let no one dare take it away from her – not even this man.
Frieda watches the taut, vixenish face. They must be careful. Lorenzo has promised her that he will not talk to the Cornish about the war any more. He will not tell them that what is in the newspapers is a pack of lies, and that Lloyd George is a little trickster, slippery as butter, a corner-shop grocer who will say anything. But Lorenzo likes this girl, and so he will talk to her. Doesn't he realize that what he is telling her is that the young man she loves has gone off to a war which is nothing but pointless destruction? That he will die for nothing? How can she accept that? She must deny it. And then he wonders that the people here don't love him. In some part of himself he is always quite simply surprised when people don't love him. Because so often they do, and then they turn against him, and they hate him more than if they had never loved him. He is always telling people things they don't want to hear. And for all his cleverness, he doesn't understand soldiers, as she does. When she was a girl, living near the barracks at Metz, she knew soldiers. And he should keep quiet now, in front of this girl.
'Will you draw me now?' she asks Clare. 'And Lorenzo will make us some more tea.'
'No, first she must see my garden,' he cries. The dark cloud slips away. Clare is swept out along with him to look at his vegetable and flower plots, stone-walled, high above farmland and sea. The Hockings have let him cultivate a corner of one of their fields, as well as his own cottage garden. The gardens are warm and well kept, the vegetables free of pests or mildew. Better than her own. He has manure from the farm, he tells her, and the soil is good – see –
'There are lapwings round here,' he says. 'And a goldcrest, once.' She smiles at his funny, possessive way of speaking of the land. But she has to admit that he seems to understand it. He is getting good crops out of it, and he knows the flowers and the birds better than she does. But she is sick of knowing about flowers. How restful it is just to notice vaguely that they are white, or yellow, or that they smell sweet. The small noise of bees makes the garden intimate.
'Do they keep stocks, down at the farm?' she asks. He looks puzzled, and she realizes he doesn't understand her.
'I mean, beehives.'
'They do, I believe. We shall have honey later on. Frieda will like that.'
'I must go in. I should like to draw her now.' She touches the sun-heated stone wall.
'I like this,' she says, noting valerian and foxglove. 'It's a good wall. Our house is new – we have nothing like this.'
'Folk seem to find it convenient,' he replies, dryly.
'Why? How do you mean?'
'The Cornish seem to think nothing of lying under a wall to spy on their neighbours.'
'Spy? On you?'
'Or of lying in a ditch either,' he goes on, his eyes amused but hard, fixed on her face. 'They seem to find our conversation interesting.'
She turns her head away. She is angry, not ashamed. What is he suggesting?
'I'm very sure they do not!' she flashes back at him. Then she remembers the click of insinuations she has heard in Nan's front-room. And where has that information come from?
'You have to understand people,' she says. 'They've always lived here. You're very strange to them.'
'I feel as if the whole of England's a stranger to me now,' he says harshly. 'And those who are destroying it are the first to claim that they are defending it.'
'They _do_ say things...' she begins, and then hesitates.
'What?'
'It is too stupid. I can't say it. The colour of your wife's stockings – they think they are a signal.'
He barks with laughter.
'And our curtains. You'll have heard of that. And the way we tar our chimney. They are fools enough to think anything – or to pretend that they think it. Poor Frieda is terrified of showing a light. She would rather sit in the dark some evenings than risk lighting a chill.'
'Is she afraid? She doesn't seem so.'
'I think it's harder for a woman. She should have friends – woman friends. She needs other women about her. Someone to talk to. But here there is no one. I thought, perhaps you might come up here sometimes?'
'Let's go in now,' she says.
The drawing doesn't go well. The sunlight has gone out of the room, and Frieda changes so quickly. An hour ago Clare could have drawn a sure line around that contained energy and form, but now Frieda is restless. Her restlessness reminds Clare of Sheba padding from room to room with a kitten in her mouth, looking for a new nest after Hat stupidly tipped out the safe home she had made for her litter in the boot-box. One of the kittens died. Sheba could not look after them properly any more. Why is Frieda so uneasy? Once Clare sees her eyes fill with tears, but, rather than let them fall, she tips her head slightly back so that they are absorbed again, swallowed without acknowledgement. How hard it is when people won't tell you what is wrong with them. Was it because they talked about the war? It is difficult to keep on remembering that Frieda is German.
Frieda sits there with her hands in her lap. Now she is the grand lady, abrupt and distant. The daughter of a baron, Lawrence has told Clare. She quite defeats Clare, who pencils on laboriously, knowing that there is no rhythm in the drawing. The portrait is a disappointment, and Frieda does not hide it. Yet she has resisted being drawn as hard as she could. Now she turns over the pages of Clare's sketchbook. Once again she is eager and appreciative, praising drawings of Kitchie and Hannah. She looks for a long while at an old sketch of Nan.
'Have it,' cries Clare, tearing out the page and giving it to Frieda. There. How odd – she never does things like that.
'It reminds me a little of my mother,' says Frieda. 'It is a long time since I have seen her.'
'Where is she?'
'She is in Germany. Things are terrible for her there. They have nothing. I am so worried about her. She cannot get food and firewood, and old people must keep warm. It is much worse than here.'
Clare snaps a band round her sketchbook, and stands up. She does not know what to say to Frieda. It was so lovely earlier on, and now the war is here too. It seeps everywhere, like dark-brown London fog. It's time to go – besides, it's getting late. Well past six o'clock, and although Clare has left a cold supper ready, her father will worry if she is not back by nine. Now she looks forward to the walk home, in the long, light, sweet May evening.
Frieda comes to the door with her. Lawrence will see her to the lane-end. She must often stay there alone, thinks Clare, remembering the conversation, watching the light drain out over the fields towards evening. She looks sad. Behind her there is the little, bravely washed room, the five-pound cottage piano, the quiet fields. The air smells of turned soil, cattle and salt. Where does a woman like Frieda belong? Not here, surely. And the war has caught her. Suddenly she is no longer interestingly exotic. She is the enemy.
'Goodbye, Clare. Come again and see us. We had no singing – or playing. I meant us to have singing. And we have some new songs – Hebridean lullabies. A friend has sent them to us. They are beautiful, are they not, Lorenzo?'
But he hums another tune:
Schatz, mein Schatz, reite nicht so weit...
Frieda smiles.
'I shall think of your cousin,' she says. 'You know I have cousins of my own at the war. And you must come again. It is so good to have friends near us.'
How magnificent she is in the dark doorway. That lovely, sure outline as she stands with her arms folded in the quickly changing May light. Irresolutely, Clare takes half a step towards her. Their cheeks brush softly. For a moment Clare is in the world of Frieda's skin and scent and rough golden hair, as if a wave has lifted her there.
'Goodbye.'
'Goodbye.'
It will be dusk, and big moths will be flying across the white road, by the time she gets home.
'Come again.'
'Come on, Clare,' says Lawrence, touching her elbow.
# Seventeen
He says very little as they walk up the lane side by side. The silence is cool spring water after so much talk. And such talk! – she can't absorb it all at once. She'll have to take it in slowly, while she wipes the plates and sets the table for tomorrow's breakfast and talks to her father of spring onions and Sheba. Lawrence points to a clump of white foxgloves, like the ones she drew. They are flowering early this year, forced on by the heat of May. The bracken has unfurled to fresh silvery points. She catches one between her fingers, and it breaks off with a dull, tender snap. Puffs of dust kick up under their boots. Clare is tired: the day's been so long, and she's swung far out of the orbit of home and Nan. Yet she's excited too. Suddenly it seems much too tame just to turn left and walk down the quiet high road home, towards her father and the smells of polish and pickled herring. They are nearly at the top of the lane, and the sweet evening wind is blowing from the sea. The light's golden here, glowing, almost incandescent. If she were to paint someone in this light... Lawrence, say... to paint him out of pure light and show the way his solid form makes the light jump and dance... And yet she's got to go home, just when she feels so alive. She could stay up all night, walk all night, paint as she's never painted in her life.
'Why do we live in houses?' she demands and stands still, reaching up into the warm sky, bathing her hands in the breeze, laughing at the feel of it, laughing back at Lawrence.
'What do you want to do – pull down the sky?' asks Lawrence, but she looks at his face and knows that the same restless energy is licking through him, filling his limbs, his hands, his voice. She stares out at the big sun as it drops quickly westward.
'I don't want to go home,' she says.
'No. Why should you?' he responds. But he's waiting for her to take the lead. She's in charge.
'Let's go the other way,' she says.
'What, towards the churchtown? Do you still want to go there?'
'How do you mean – still?'
'Something your cousin said. That you always wanted to go to the church, to see the mermaid.'
'Did he say that? Why, I'd forgotten. I haven't thought of that for years. How strange...'
It's certainly strange. She doesn't forget much to do with John William. She fumbles for the memory but it won't come; there's a blur of sensation, nothing more. Never mind, it'll come back if she doesn't force it. Everything to do with John William comes back in the end.
'We'll go there now,' she announces. Lawrence will come. He's bound to do what she wants. Just at this moment she knows that everything is with her; the warm expectant evening, the long _hush_ of the sea, the constant secret presence of John William. 'Are you coming?' she challenges.
'Look, there's the moon,' he answers. There it hangs, pale and insignificant. Daylight takes a long time to die here.
'Frieda will be all right, won't she?' asks Clare.
'Oh, she's used to it. I often go out in the evenings. I can't stay cooped in one place, thinking. She knows that, really.'
'You go down to the farm, don't you?'
'Yes, or I go walking with William Henry. Like us, now. It's the best time. You feel you can do anything: run over the fields to the sea, turn round and find the farm gone and nothing there but little dark Cornish farmers from a thousand years ago, looking at you... have you ever noticed the way the Cornish look at you? No, of course you haven't. This is your place, isn't it, Clare Coyne? But you haven't got those Cornish eyes – soft and dark and a bit inscrutable. Your cousin Hannah has those eyes. Do you know William Henry?'
'Yes, he comes into market, doesn't he – and my uncle knows him. My Uncle John, who has the farm.'
'Everyone knows everyone here. I'm a fool to forget it. But William Henry – he's not like a common farmer. Those eyes of his – they draw all your thoughts out of you. Yet there's something else there – something a bit sceptical – even jeering – as if he'd like to do you down for getting close to him.'
Where's she heard that tone before? Wistful, puzzling, eager? Why, in herself. That's what she hears when she turns over and over the secret of John William, like turning a coin in her pocket at full moon, and the secret quickens into life, flames up, dies down. When you love someone, you just want to say his name. That's what Nan told her years ago, when Clare asked how it was that she always knew when a girl liked a boy long before anything was ever said. No courtship, no marriage, no pregnancy had ever surprised Nan.
'It's from watching, Clarey. Most folk walk around with their eyes shut. And there's other sure signs: say, if a girl always brings a man's name into her conversation where it doesn't belong, then you know she's saying it for the pure pleasure of having his name in her mouth. Names are magical things, Clarey, never forget that.'
But how can Lawrence feel anything like that about another man? She must have misunderstood, that's all.
He's walking with her and the last bees swing past them, heavily laden, down to the farm stocks. They fly close to the earth now, late in the evening, going home. Frieda will be waiting. No she won't, thinks Clare quickly. She's not like the women of the town, pulling the potatoes off the fire before they boil to rags, keeping her man's plate covered and warm at the bottom of the range for fear of feeling the back of his hand when he comes home late and truculent to a spoiled dinner. Frieda does not cook or sweep or spin. Frieda sits in her chair like a queen. Clare smiles, making it up, but her powers of invention falter as the road opens to the right and reveals the long, changeful, violet-shadowed slope to the sea.
The graveyard is full of people Clare has known. She kneels, examining names, tracing family trees, stories, allegiances, old hates and likings, while Lawrence sits back on his heels and urges her on. She matches each name to its own patch of earth: to the one it tilled before it came here, or to the restless acres of the sea. There are Celtic crosses, trefoils and rough, potent granite stones too hard for a name to be worn into them. They rise like a wave towards the sweep of the moor. She stops, exhausted and looks up at him.
'That's a funny way of sitting,' she says.
'It's very comfortable. You ought to try it. It's the way all the miners sit at home, squatting against the walls after the shift. I used to pass them on my way home from school. My father'ud be there too, making them all laugh.'
'Did he?'
'Oh, yes. He was famous through the whole pit. He could mimic anyone: the men loved him.'
The bitterness in his voice is so much at odds with his words that she searches his face, puzzled.
'Didn't you like that?'
There's a pause. Carefully Lawrence selects a blade of grass from a long tangled clump, breaks it off, raises it to his lips and blows. There's a feeble fluttering squeak, then he blows again and a keen whistle parts the air. She smiles: it's ages since she thought of doing that. He grins back at her.
'My father taught me,' he says. 'He taught me lots of things like that. Trouble was, I never thought they were worth anything. Not compared to real learning, the kind of thing they taught us at school. And my mother agreed with me.'
'That was sad.'
'Yes,' he agrees. 'It was, though at the time I though it was just what I wanted. Once I went back home, and an old man who'd known my father stopped me in the street. I can see him now, in his clean Sunday clothes, a bit bent and having to walk carefully. He'd danced with my father when they were young... but now he was old and tired and angry. He said that I'd shamed a finer man than I'd ever be, writing of him in my books. He stood there telling me this, and I couldn't even remember his name. My father knew every man in the pit, and they all knew him. But my father's friends never came to our house. He would meet them in the pub. When I was little I'd think what a place it must be, that all the men wanted to go there so badly. But my mother said it was a nasty place where men went to drink away the money that might have made decent-looking women of their wives and chances for their children. He used to run a dancing-class...'
'Your father?'
'The best dancer for miles around, they said he was, before he married. But he'd long stopped dancing by the time I knew him. And my mother never danced with him once she was married. He was the best dancer, and the handsomest – so quick and always the kind of man you warmed to – you couldn't help yourself. Unless you'd grown to hate him...'
There he goes, brooding over his father just like he brooded over William Henry, with the same helpless regret and longing. Why can't he stop thinking of them both and sit in the last bit of sun and whistle with the grass? He ought to have seen Francis Coyne and Susannah Treveal together. That would have given him something to think about.
'They were as different as cream from herring,' said Nan, when Clare asked for stories of her parents. But Nan would never say that cream was better than herring; each was good in its kind. And you would sicken of the one in time, and want the other.
Lawrence smiles suddenly, unfolds himself and stands over her. He reaches out his hand.
'Come on. We'll go and look inside the church.'
She takes his hand, smiling back, and yields to let him pull her up.
Neither of them sees that they are seen. At that moment the Rector goes down the path from the church door, head down. He catches a movement to his right and turns and there they are: two figures balanced like figures in a dance, one lifting, one lifted, both smiling, the warm evening sun catching and lighting them. Unmistakably they flame in consecrated ground: the daughter of his friend with her rare flag of red hair, and his outlandish red-bearded enemy. There they are. They have not seen him. He steps forward as if to speak, then moves back. He will watch. He will let them betray themselves. Even from here he can see that the girl is flushed, and her hair has come loose around her face. There she is lolling on a gravestone, on the bones of Christ's chosen. He moves back further, into the shadows, and watches. They stand, and after a long interval their parted hands separate. The girl brushes herself down, then turns as if to display herself. The man leans forward and brushes grass from her back, her skirt. They pick their way through the gravestones, faces animated and bent together, noticing nothing. They turn towards the church porch, and the Rector wants to cry out against sacrilege, but he holds back the cry. Let them eat and drink to their own damnation. They melt into the porch and he hears the click and groan of the church door, then they are gone. Out of their own mouths they have condemned themselves. What I have seen I have seen, he mutters quoting wildly from his own scriptures as he backs out and confronts the Tinner's Arms, glaring.
'Is it this way?'
They blunder down the aisle. The church is dark and cool, a vessel of stone-chilled air. It feels so cold that Clare shivers. She begins to remember something – some dim recollection of stumbling out of the blaze of a summer day, the church air washing over a prickle of sunburn, the sudden silence.
'It's down here.'
Lawrence's voice sounds frighteningly loud, but he is only talking as he usually does.
'Do you always whisper in churches?' he mocks. 'Who do you think is listening?'
No one at all. There's no steady red glow and dip of the sanctuary lamp, no candles flickering with the prayers of those who have already walked out into the sunlight, no murmur of voices from the confessionals. Only this ancient, untenanted silence. You couldn't say a prayer here. It would be soaked up at once by these granite walls, and blotted out.
'It's cold in here,' she whispers.
But he's not listening. 'Here she is,' he says, and kneels. The little mermaid, carved into dark wood. And Clare knows her. Her fingers reach out for the familiar curves of that body which drew her again and again in her childhood. How could she have forgotten.
'Little mermaid,' Clare greets her.
She has got no more wounds since Clare last touched her. Those gashes are polished with age now, and if she is lucky no one will add to them. Will they throw her out of the church again, as they have done before? She smiles and raises her arms, and will not answer. Does she even know what she can stir in those who crouch beside her to pray or to draw out a knife? 'Virgin most pure, star of the sea,' Clare mutters. That was it. That was what she used to sing, blurring this beautiful thing in her heart with the Virgin she was taught to love. And Hannah laughed once and they fought blindly and silently, scuffling on the cold floor for fear of waking the echoes. And John William traced the mermaid's scales and her round breasts and said nothing.
'She doesn't belong here,' says Lawrence.
'Why, of course she does!' says Clare, far more loudly than she intended, surprised by the force of her own anger. 'It's only bigots who say she doesn't, and try to drive her out.'
'No,' says Lawrence slowly. 'She doesn't belong. She's half and half. They laugh at her on land, and hate her. And when she's in the sea, she can't breathe. So what's left for her? She can only sit on her rock, neither on land nor in the sea, and wait, and drive men mad. So they hate her all the more, all the time they're pretending to love her and want her –'
'But why? Why do they have to hate her? I don't understand.'
'Because she makes people think that there's something more – something they haven't been told about. And they'll never have it. Look at her face.'
'It's worn to nothing,' says Clare.
'Only because so many people have touched it. It makes her more beautiful.'
Clare does not want to hear any more. His words are taking the little mermaid away from her.
'There's another way of looking at it,' she says. 'She can swim and she can breathe – and she loves both. She has both. Why should there be anything sad about it?'
He turns to her and smiles. It's a smile fully for herself, acknowledging, appreciative.
'Long may you believe that, Clare Coyne,' he answers.
# Eighteen
Two weeks later, out in her backyard in the morning sun, Clare thinks she would like to go and see Frieda again. Lugging the heavy wash-basket of wet clothes into the garden has made her think of Frieda. What does she do on washing-day? No doubt she laughs and says Lorenzo must do it, he does it so beautifully. Once in her life, Frieda scrubbed a floor, or so Lawrence said. But it was only once. She wasn't brought up to do such work – not like us.
Not like us, indeed, thinks Clare grimly, as she hauls out the first twisted sheet from her pile, ready to put it through the mangle. Prices are going up all the time, but their income has dipped again. She's decided that they'll have to stop sending the sheets to the laundry. She has always done the clothes-washing at home, but it was her luxury not to wash the sheets and to get them back from the laundry in a brown-paper parcel, starched so stiff so that they creaked when she unfolded them. It is heavy work to fold over the wet sheets and feed them little by little through the mangle, and into the zinc bucket on the other side. The mangle-rollers ride roughly over four thicknesses of sheet. Her hands are red and raw from washing-soda, and her hair sticks to her forehead in tight curls from the steam. She's sweating so much that her overall sticks to her. She wriggles inside it. The sun beats on the back of the house, and bounces against the whitewashed garden wall. They have even mentioned _this_ _unusual spell of fine weather_ in the newspaper. But other news is crowding for space. There was a bad air-raid in London yesterday, one of the worst of the war. Here the sky stays empty and blue. What must it be like to live in London and see bombs falling out of the skies? But here the talk is of U-boats, not aeroplanes. Everyone is asking what Jellicoe's going to do about the latest sinkings. It can't go on like this, with the Germans stealing in under our very noses. Those U-boats do just as they like. They pick off our supply-ships one by one.
Oh, there's good news too. There's always good news. The country has been on the point of a decisive victory for three years now. Now the papers are full of General Plumer's triumph at the Messines Ridge. It's better to write about the million tons of TNT it took to blow up the ridge than of the 600,000 tons of shipping lost to the U-boats every month.
'Haig's chance has come,' said her father that morning. 'Now they'll have to give him a free hand.'
He sounds eager, as if the news has refreshed him. He has started to order the _Morning Post_ , as well as _The Times_ , to find out what Gwynne's got to say.
'He's after Lloyd George to "take charge" in France. Thinks the French would move aside for our Welsh Wizard. Mark you, Clare, Gwynne's got no more time for Lloyd George than he has for any other politician. What he wants is for the generals to get a free hand.'
Clare thinks that the cost of two newspapers a day would go a long way to paying her laundry bills.
How they long for a final onslaught, no matter what it costs. If only it could all be over, no matter what. Last night, slicing a pair of raw kidneys, she was dizzy, seeing how the newly sharpened knife parted the caul and exposed a delicate tracery of ducts. All so exquisitely made, and so easy to spoil. Those kidneys would begin to stink within the hour in this hot weather if she left the meat-dish out in the sun. Then the blowflies would come. She shuddered, covered the kidneys with a plate and put them in the meat-safe until supper-time. The third sheet. Only this one, and one more. Then the table-cloths, and her boiled glass-cloths and dish-rags. She'd rest for a minute before pegging up the sheets. She dreads doing it today; the line is so heavy once all the clothes are on it and it has to be hoisted up into the sunshine by the pulley. She shivers. She must have got over-heated. Just this one sheet now.
And there's Hannah, a blur of light at the back-entry gate. Good. That means I'll have to stop for a bit. Hannah wavers on, indistinct in the blinding light, fumbling with the gate-latch. It always sticks, but Hannah knows the way of it – why is she taking so long?
'Lift the gate – it's got stuck,' Clare calls. Hannah's face is averted under the brim of her straw hat. Then she pushes the gate open, and comes on down the path.
The path is twenty feet long, no more. Hannah takes two steps, and Clare lets the sheet fall. It will get grass stains on it, it will all have to be done again, she thinks with one part of her mind. She sees her own hand gripping the handle of the mangle.
'Hannah,' she cries out. 'Go away.'
But Hannah comes on. Her face is a colour Clare has never seen it. Her yellow skin sticks to her bones. Her eyes are startled, wide-open and staring. She doesn't say anything. She comes up and lays both her hands on top of the mangle, and looks at Clare.
'Mind your fingers,' says Clare in a quick, frightened voice.
'Clare,' says Hannah.
'I know what you're going to say. Don't say it.'
'You got to come downalong to Nan's, Clarey.'
'No, I can't, I'm in the middle of my washing.'
'Bugger your washing,' says Hannah. Now she is round the other side of the mangle, with Clare. She puts her arm around Clare's waist. Clare is sweating. Hannah can smell her frightened sweat.
The two girls go into the scullery, and Hannah pours two glasses of water from the big jug Clare keeps on a stone slab behind the door, covered.
'But it can't be,' says Clare. 'He was safe. He was in camp. He was training to be an officer.'
'It was an accident,' says Hannah, her mouth shaking. 'It was an accident in training. He was killed instantly, it said.'
The two girls look at each other. They both know that is what telegrams always say.
'But it can't be true,' says Clare. 'He'd got out of it. He was safe. He can't have been killed in an officers' training camp. Not after all that long time in France.'
Surely this is some awful joke. Perhaps it's a test. Perhaps it's a test to see if they're fit to be an officer's family. But then she looks at Hannah's face and sees that John William is already dead there.
'Do you know when it was?'
'Saturday.'
'When we were digging up the end of the vegetable plot, then.'
For Clare had thought she would plant cauliflower, and winter cabbage, and Hannah had come up to help her after work. They had dibbed in the young plants, and watered them in the early evening –
'And then we walked over to meet Peggy – didn't we?'
Hannah nods. That's what they did, and some time in between their planting the winter cabbages and hustling Peggy out for an hour's evening walk, John William had died. But the cabbages would live, because Clare had been watering them every day, carrying the water carefully across to the plot.
'I shall stop watering them,' she says aloud.
'What?'
'The cabbages. I shall stop watering them. It's not right that they should keep growing when he's dead.'
She has said it now. When he's dead. The words go down through her like stones falling through deep water. It feels as if those words have opened up some new place in her, which has never been touched before. But she will have to say them again, and again, because it will never not be true now.
They finish drinking the water, and Clare says she will go upstairs to take off her overall.
'But Hannah! What can I do? I've no mourning.'
'Bring down your silk. Nan'll dye it. She's dyeing our dresses tonight.'
'When did the telegram come?'
'An hour ago. I was at work. Kitchie was round at Nan's, and he fetched me home. He wanted to come up here with me, but I didn't let him.'
Yes, Hannah has her black work dress on. What had she been doing when Kitchie came into the shop? Trying to get rid of a tiresome customer who came in for a packet of needles and stayed to talk for twenty minutes? Pulling down the blinds because the sun was threatening to fade a roll of cretonne at 1/9d a yard? Hannah moving. The way she would reach up with the blind-pole and hook the blind down. Supple and full of life. Drat the thing, it always catches. Wants oiling. John William's smell of carbolic and, under it, almonds. They could never kill the smell of his skin.
Those kidneys. Don't think of them.
'Hannah, I'm going to be sick.'
She gets to the privy in time and vomits, while Hannah holds her head and strokes her forehead.
'It's all right now, Clarey. It's all right.' Now Hannah is crying.
The privy stinks of vomit and sweat as Clare struggles up. She is dizzy, and there are red sparks coming and going in front of her eyes.
You have to keep on doing things; you can't stop, even when this happens. Now I have to rinse out my mouth or else it will smell. I must wash, and comb my hair, and put on different clothes. I have to do that, but I can leave the washing. I needn't think about that now, but I shall have to think of it later, for there's no one else who will.
Hannah is crying. Hannah must sit down at the table, and I'll make us both tea. It won't matter if we don't go straight down to Nan's. Hannah cries. She has forgotten Clare now. She sits at the kitchen table with her head on her arms, her back shaking, curled in on herself and private. She can't cry at home, thinks Clare, because of Aunt Sarah. Everybody will be round Aunt Sarah, the way they were at the railway station. They will expect Hannah to be strong.
For a moment Clare hates Aunt Sarah – weak, greedy Aunt Sarah, who will feebly gulp down a macaroon in the middle of a fit of hysterics. Licensed mourner, chief in suffering. And I am just his cousin.
Clare puts on her grey dress and packs her Sunday dress into a brown-paper parcel. How strange: for once she is the competent one, the one who does things, while Hannah sits at the table. She looks out of the window and the sun is just the same, falling on the white tangle of sheet on the grass, and the loaded line with the other sheets nearly touching the worn grass. I shall go out and peg up the rest of it, and raise the line. But not yet. Hannah will think it isn't seemly, even though it doesn't make any difference if I do it now or tonight. John William is dead. He will always be dead.
They go out of the house and creep down the hill close together, dark, dull blotches on the brilliant morning. The cemetery looks beautiful, warm in the sun, curled up asleep on the curve of the hill. When they were children and they went out mackerel fishing, they would look back and see the cemetery, bigger than the town itself from the sea. Only John William wouldn't look up. He frowned with concentration as he baited the lines.
'Isn't that a shoal of mackerel there?' she cries.
'No,' says Hannah. 'Just a bit of wind getting up.'
As they come into Nan's, they're engulfed by a close, suffocating smell of black dye. Nan's decided not to wait; she has begun on her vat of mourning. Curtains are drawn, rooms are in shadow. There's steam in the kitchen, snaking up from the vat of dye where deep bubbles pock and the clothes writhe slowly to the surface to be poked down again by Nan's long wooden spoon. The kettle steams too, and the biggest tea-pot is drawing at the side of the range. Nan slices bread, and one of Uncle John's hams sits at her right hand. The flesh of it is dark and rosy, exquisitely marbled with white fat. A basin of brown eggs waits to be hard-boiled.
'That poor useless Ellen-creature brought them. She had them from her sister. Not that I would wish to eat out of Ellen's kitchen, but there's little enough she can do to an egg.'
Nan taps the eggs and tells Clare to get them hard-boiled. Her face is tired and severe, but there are no marks of tears on it. Like Clare, she hasn't cried yet. Sarah is lying down in the back-bedroom. She took on terribly, but she's fallen into a drowse now. Hannah must go and sit with her mother.
'But where's Uncle Arthur?' asks Clare.
Nan shrimps up her lips.
'Your Uncle Arthur's gone out cursing God and all Creation – that's the way it took him.'
How can Nan be like this? Her hands slice ham unerringly.
'Fetch the mustard, Clarey, and there's a last big jar of those pickled onions I put up in the winter. He's better that way,' says Nan. 'Sometimes a man's better on his own.'
'Nan,' says Clare. 'Why don't you sit down? I can do this.'
'There's plenty for all of us to do,' says Nan. 'Better to be doing than thinking. You might just turn those clothes, case they scorch on the bottom.'
The clothes smell ugly. The fume of the dye catches at the back of Clare's throat and brings tears to her eyes.
'I'll have to see about the cards,' continues Nan, running down her thoughts aloud.
The cards. They will have to be in all the shop windows by tomorrow morning.
WILL FRIENDS PLEASE ACCEPT THIS THE ONLY INTIMATION
How often she had seen a new card in a window as she went by, stopped to read it and thought vaguely of the gone dead one, Stevens or Rosewall or Date. And deep down in the impulse of sympathy there had been guilty gladness, as if to read of another's death left her twice as alive. When they were children they used to play in the cemetery. They didn't mind the dead. They would pick daisies and yarrow and make tiny wreaths for the graves. Those dead ones were another race. They had never really been alive. Caught, weak and silent: they could not run in the sunshine, scream with laughter, nibble the fat juicy ends of summer grasses. Death was something that was never going to happen to us.
Clare can picture the people stopping by the shop windows to read the cards. The cards will have John William's name on them. How strange and impossible that seems, when only a little over a fortnight before he had been walking in these streets, buying Grandad a present of tobacco in the very shop that will bear his name in its window now. One person will stop, then two, then a little bunch. If they haven't heard already, they will catch in their breath and say how terrible it must be for the Treveals, just when they were so proud of their John William going for an officer. If they already know, they will find their knowledge stamped there in black, black-bordered letters. They will linger, tracing over each letter. Then they will walk on in the sunshine, twice as rich as before. The boys John William fought with, the ones who would have liked to jeer at his ambitions but did not dare. For he was going to do it, she thinks hotly. He would have shown them all.
'Don't put the cards in, Nan,' begs Clare.
'Why, Clarey, what can you be thinking of? Don't you know what's right?'
Silence in the kitchen. They'll all be here soon, snatched out of their daily lives, wreathed in shawls. Aunt Mag's come already, and she is padding about upstairs, tending to Aunt Sarah. And your father, Clare? He doesn't know. He was out when Hannah came.
Ah, yes. Out. That secret place that men go to and women don't know about.
'He'll be back soon, Clarey. He'll be down, soon as he hears.'
Nan is trying to comfort her, and Clare can't say that her father will be no help to her now.
The door bangs hard. Clare jumps and gives a small ridiculous shriek. Now there is a blundering of boots at the door, quickly stilled by Nan. For a terrible moment she thinks they are bringing in John William's coffin. But no – it can't be. And at the door there are three lads, grown huge in the crowded space. A Harte and two Lees. Jack, Sammie and Esau. There they stand in their boots, with a white, mauled-looking Harry, barely on his feet between them.
'There's bin an accident,' blurts Jack Harte. 'It's is arm. Is bad arm's smashed up.'
They will not come into the house out of respect for the dead, for they are in their working clothes. So they know about John William already, thinks Clare. So quick: the cards not in the windows yet. They stand at the cottage doorway nervously, gaping with respect for its time of sorrow.
'When did this happen?' asks Nan, whisking over to white-faced Harry. She bends and examines his arm without touching it.
'Help him in, then. Don't just let him stand there,' she orders, and they tramp over the threshold and deposit Harry with clumsy gentleness on the kitchen settle, then stand around uneasily, sniffing the dye and meat-smells that mean death in the house. A thread of spittle comes out of Harry's mouth as he sags down and shuts his eyes. Esau Lee wipes his hands against his trousers.
'When did this happen?' asks Nan again.
'Not long after your Kitchie come up and told us about John William. Bit a gear we got wanted fixing – it went an caught Harry's arm.' The Harte boy looks beyond Nan, at the wall.
'Don't let our Sarah hear em,' says Nan sharply to Clare. 'Now, Kitchie, run for Dr Kernack.'
'It's bad,' says Jack Harte. What's that in his voice? Something queer that ought not be there. You might almost say... satisfaction? Nan nods. She looks Jack straight in the eye and she nods slowly, acknowledging that same strange _something_. It's bad enough. It'll do. Clare has the feeling they are speaking another language, one which she is not quick enough to catch. But they've scarcely said anything, so how can that be?
'I'll tell Mother,' says Hannah, who has been drawn downstairs by the noise of boots and voices.
'You will not,' says Nan. 'They say one sorrow drives out another, but Sarah needs rest.'
'She's not resting,' says Hannah.
Nan thinks for a minute. Sarah's a poor thing, but she's sharp enough. She'll cotton on.
'Maybe you're right. Maybe it's best she thinks of Harry rather than our Johnnie. Now wait there, Jack, Esau, Samuel.'
Sweat and tears are gathering on Harry's face, but Nan takes no notice of them yet. She goes upstairs, and they hear her feet creak across the bedroom boards, and then the lock of her trunk snick, and the lid squeaking back. When she comes down, she has in her hand a screw of paper with something hidden in it. She closes Jack Harte's fingers around it and pats his hand. But he pulls back. He won't take it from her. She will need it – specially at a time like this.
'You take it,' she says. 'Take it and don't add to my troubles. You're good boys, all of you.'
'How much did you give em?' asks Hannah, in quite her usual voice, the minute they are out of the house and clumping off up the street, back to work.
'That's for you to ask and me to know,' says Nan, as if Hannah were six years old again.
'Well, they can't call him up now!' exclaims Clare, suddenly realizing.
Nan and Hannah exchange slow, ironic glances. 'Quick, isn't she,' remarks Hannah.
'Get a bucket for Harry,' says Nan, watching his colour.
'Don't let the doctor put his hands on me, Nan, don't let the doctor touch me,' he begs, his courage all drained now the lads have gone. 'You do it, you can set a bone, Nan, you do it.'
'I can't handle an injury like this one, my boy,' she says. 'You should of thought a that before you did it.'
'Don't mind, Harry, Dr Kernack will give you chloroform,' says Clare.
'Dear God, how's all this to be paid for,' fusses Nan.
And there's Sarah in the doorway, her face white as a halibut's belly, gaping at them.
'Our Harry's hurt his arm, Mother,' says Hannah distinctly. 'So bad they won't be able to send him to the war.'
Nothing from Sarah.
'Did you hear me, Mother? Harry'll be all right. They won't be able to send Harry.'
Sarah's face crumbles. She totters towards Harry and crouches by his feet, stroking his leg. Then she gets hold of his whole, living hand and kisses it over and over, covering it with tears. Clare looks away. In spite of herself, she is disgusted by Aunt Sarah, greedily crouching over Harry. Hannah shrugs imperceptibly.
'There now,' says Nan. 'She'll be all right. Mind, Hannah, your mother's to be the one that looks after Harry. Don't you go doing nothing. Leave her to do it.'
Kitchie comes flying through the door.
'Doctor's coming soon as he can!' he announces.
I've got to get out of here, thinks Clare. Blood and chloroform and bits of bone. And Nan too; I know it's her way to turn to what's living and what needs to be done, but I can't bear it.
'Clare. Open your legs.'
'Please, sir, I found a mermaid.'
'Are you a human boy?'
'Are you a true mermaid?'
Nan, I'm going out. But she doesn't say it. She is ensnared by Aunt Sarah's snorting sobs of relief over her Harry, by the rolling boil of the black clothes, by the click of Nan's knife over the meat.
At tea-time they are all still there, all gathered: John William's flesh and blood. Clare's father has arrived, and he is in the front-room with Uncle Arthur. Both of them smell of brandy. Uncle Arthur doesn't like Francis Coyne, but now he is spilling out his life to him in raw gobbets of bewilderment and sorrow. Father says nothing. He has a hip-flask which he offers to Uncle Arthur from time to time. Occasionally he murmurs that John William was a fine boy, a fine boy.
For one hour the story of Uncle Arthur and his boy rambles, takes shape and becomes beautiful. He makes Francis see the boy standing in front of him, outfacing him, dark and insolent. Any father would chastise him – wouldn't you? Speak the truth, now. I tell you, Francis, I wanted to knock it out of him. It was my duty as his father. I thought he was a fool to himself, always after what he could never have.
'But then they wanted him for an officer,' says Uncle Arthur, in the voice of someone slapping down his full hand on the table. He counts them off on his fingers: 'He'd been Lance-Corporal. Corporal. Sergeant. And then he'd a had his commission, not a doubt of it. Learning was nothing to John William.'
No one can take that away from him: not now. Harry may be Sarah's boy, but, dead, John William is his father's.
In the kitchen there is still a pile of ham sandwiches by the tea-pot. Nan has been cutting and replenishing all afternoon. Uncle John reaches out and weighs his sandwich in his hand before biting into it. He smells the meat. It is moist and succulent, just as it ought to be. He bred the pig, and the sow which farrowed it. He looks around, holding the sandwich ostentatiously. Sure enough, Nan says, 'If you hadn't a made us a present of that ham, John, I don't know what we should a done, not with all these that've been in here today.'
'Well, I don't know,' he says consideringly, turning the sandwich. 'Good enough. Good enough, eh, Kitchie.'
Kitchie, scarlet, bolts his sandwich. That was rough of Uncle John. Kitchie can't help being greedy, not even today. But at least he doesn't pretend about it. He doesn't say in a faint voice that he couldn't eat a morsel, like Aunt Sarah, then gnaw her way through three sandwiches. If John William was here, he'd spot that. And maybe he'd give Clare a smile so quick nobody else would see it, and one of his split-second winks. How she envied him for that when she was little – he could wink, and whistle through his teeth, and imitate a dog barking. He tried to teach her, but she never learned. Once she thought she'd got it, but he made her look in the glass and she saw that all she was doing was to screw up both eyes and look foolish.
The men are talking about the funeral. Will John William's body be brought home for burial here? Uncle John is blustering, sounding important but knowing nothing. Her father says quietly, 'I don't think that they will release his body for burial here.'
Clare looks at him, startled. She has hardly ever heard such finality in her father's voice. She cannot think why he says it, for surely he knows nothing more than any of them? But the family defers to him. He is a gentleman, and he knows the way of the world and the ways of officers.
'They'll give him a military funeral, with full honours,' says Father.
'I don't know about that,' begins Aunt Mag, eyes darting around for support. 'Surely he did ought to be brought home? It's only right –'
Unexpectedly Nan quells her. 'He's a soldier, Mag. He got to have what soldiers have.'
Surely Nan would want John William brought home, so he could rest here in his coffin and the neighbours could come in and pay their respects, and look at him before the coffin was closed. If he'd been killed in France, we would know that couldn't happen. But he's here, in England. Why doesn't Nan want it? Why is she agreeing with Father?
'There will be a letter, from the camp commander, to notify us,' says Father.
Nan nods. A look passes between the two of them: Nan in her black, for she always wears black and needs to dye nothing; Father with his fine, thin face turned to her. Something is between the two of them, as if they are alone in the room. Now Nan's face is slack and heavy with sorrow. She looks old, and afraid. But Nan isn't afraid of death, Clare knows. Clare cries out silently to the two of them, 'Don't let them bring him here. Take it away from me.'
She sees a day of blazing sun, and the coffin bobbing down the steep path through Barnoon Cemetery, and the people of the town following it with faces like hungry gulls.
'Nan, take it away from me. Make it not happen.'
Nan sighs. She is very tired, but she will not give way yet. There are things to be taken care of first, and this is something they are taking care of between the two of them, she and Francis Coyne.
# Nineteen
At first light the next day Clare hears her father creak past on the landing. Thank God, it's morning, even though the room's still shadowy because of its heavy curtains. Last night she went to bed thinking that she'd never be able to get to sleep, but she slept heavily, scarcely moving in the bed, plummeting so far into solid waveless sleep that she was beyond dreaming until the last minutes of the night. Then, in her dream, the church of the Sacred Heart and St Ia was being built again, this time with a smooth forest of slender marble columns supporting a roof so fragile that Clare could not tell if it were made of glass or air. The old altar had vanished, and a new altar was being built into the side of the church. It was made of marble, the colour of sweet butter and scrolled with carving.
In her dream Clare advanced down the quiet church, stepping on sheets spattered with white and gold, past stepladders and chisels and fine brushes for applying gold-leaf. All the tools looked as if they had just been put down a moment before. One drop of gold paint hung on the end of a brush, a globe ready to topple. There was no one to be seen, but she must find the builders and tell them they were making the altar in the wrong place. The altar must be in the front of the church. But just as she reached the altar someone took her right hand. The grip was no more than a band of warmth around her little finger. She did not see the person who touched her, because in the dream she knew she was not allowed to look sideways. She had to look ahead of her. Suddenly she wasn't worried about anything any more. Let them build the altar where they liked. It was beautiful. She laughed as she touched a coil of marble, not caring about the sound of laughter in the empty church. Then the building was suddenly full of people, talking and laughing and making more noise than she had ever heard in a church before. They were hanging branches of cherry and bunches of grapes around the altar, so that leaves, tendrils and fruit curved over the swell of the marble. Now its marble hollows were flushed with crimson from the colour of the cherry-flesh. Slowly Clare felt the grip on her finger thin away like mist. She was released. She turned and saw that it was John William walking swiftly away towards the door. He did not look at her. She knew when he opened the door that he would walk out into the sunlight.
Clare could not be sad to see him go. He had to leave. She could tell that from the set of his back as he turned, and the way he looked as if he were hurrying towards something she could not see. At that same moment she saw that someone had planted a naked baby in one of the hollows of marble. It lay there cupped, on its back, looking up through leaves and fruit. Then, as Clare watched, a big man with a battered face knelt on the step of the altar and reached up to the child in its stone cradle. Clare knew his face: he was a herring fisherman. He held out his finger, calloused from handling the nets, and the child grasped it. There they remained. Now there was no one else in the church, and Clare woke.
'Clare. Clare.'
'Yes, I'm awake.'
He pushes open her door. His voice is tired. Has he slept? He never gets up this early. Clare hoists herself up on her elbows and says, 'Draw the curtains, Father, I can't see you.'
But he stays in the doorway.
'Clare, I'm taking the half past seven train. I am going to make arrangements at the officers' training camp. I have been thinking about it, and I must go. I won't be home tonight, so you must get Hannah to come up and sleep with you, if your Aunt Sarah can spare her.'
'But why are you going, Father? What about Uncle Arthur?'
'Arthur is in no fit state. It's better that I go. And Arthur would get nothing out of the people in charge. They would know they could get away with telling him anything, and he would take it gratefully, no matter what the truth of it.'
The truth? Why is he talking like this? There's no truth except that John William is dead. Or does he think – can he possibly think –
'Father! Could it have been a mistake? Do you think it's possible? Is that why you're going?'
'No, no, Clare, not that. There can be no mistake, not here in England. If it had happened in France, perhaps: though the communications are excellent, I believe. But we ought to know what kind of accident it was.'
'I see,' she says. She had not really thought it could be a mistake. She could not wake up and believe John William was alive. She could not be like Mrs Hore who still said that she would hark as she sat by her fire in the evening and think she heard her William's footstep. There was not a trace of John William left in the world. Only his body; an unthinkable thing, slumped, never moving any more. Even in her dream he was walking away; but he had touched her before he left her, and that dream-touch was still as real as anything that had happened between her and John William.
'Wait, Father, I'll get up and make your breakfast. You mustn't go without eating anything. You know what the trains are like: it will take hours.'
'Go back to sleep. You need to sleep. I only put my head round the door to tell you I was going.'
'I don't mind. I shan't sleep now.'
She throws on her dressing-gown and stumbles downstairs to boil an egg and cut bread and butter for her father. At least there is butter today. For once her father wants to talk to her. He wants to share this with her – his fears, whatever they are. Is she going to let him? Or will she cook for him and tidy up after him and let him go without asking questions?
The way I always do.
Francis Coyne sits there, grey and old with a shaving cut on his right cheek and a rusty bubble of blood oozing over the lint he has put on it. He chips off the top of his egg and the kitchen fills with the murky smell of it. Yet he lifts his spoon absently. He's going to eat it. She sweeps the plate away from him.
'That egg is bad. Can't you smell it, Father? Eat your bread and wait five minutes while I cook you another. You have twenty minutes yet.' Reckless of their bare larder, she takes another egg off the shelf and boils it. Fresh tea for them both. Her head is thick and aching. She has just these few minutes before it boils. Heart thumping, she sits down opposite him, elbows on the table, and asks, 'Tell me why are you going. Truly.'
He sighs. 'It would be wrong to say anything before I am sure.'
'But you can say it to me. You can tell me what it is you are afraid of.'
'I want to find out more about the accident. How it happened. Clare, John William has been two years in France. How many came safe through that? But he did. I know they used to say he was wild, but I always thought he was a careful enough boy, when it mattered. He knew his job, or they would never have thought of making him an officer. He was used to taking charge of men. He had his platoon. They will not tell your Uncle Arthur. Probably they will not tell me the truth either, but I shall be able to piece it out from what they don't tell me.'
Yes, her father is right. He knew John William better than she thought. Her father goes on, 'They said it was a miracle he came through two years in France, but John William knew it wasn't. You weren't there when he was talking down at your grandmother's. We were all in the parlour – the men. Your Aunt Sarah had gone to bed, so I suppose he felt freer to talk.
'No, it wasn't a miracle. There was luck in it, perhaps nine tenths of it was luck, but without the tenth part all the luck in the world wouldn't have got him through it. And I am sure that the tenth part was his carefulness, Clare. He never let up on himself. He said he always had to think ahead, even when the noise of shells made them all stupid, and the starlights bursting over their heads knocked their senses out of them. What were those other things he spoke of? Whizz-bangs. We can't imagine it. Now I see that I have never let myself imagine it. Barbed-wire entanglements, and shell-craters everywhere, and smoke so thick that you can't see the man next to you. And always the fear that it would be gas next time, not smoke. You can tell gas by its colour, but by the time it is near enough for you to tell, it is too late. The men cough until they cough their own lungs up. He told us in some sections of the Front there wasn't a blade of grass left, though it had all been farmland and woodland. There were only a few trees left sticking up out of the mud with all their branches splintered, and rags hanging down from them where a man's clothes had been blown off him in a shell explosion. Or his flesh. Everywhere there were pieces of equipment the men had abandoned in retreat, or when they'd been killed. You didn't always know if the stuff was British or German, for the line swayed back and forth over the same land like a tug-of-war. Tin hats. Bully-beef tins. They had to walk on duck-boards over the shell-holes. The holes filled up with water, you see. Men drowned in them – can you imagine it, Clare? Wounded men, unable to get their mouths out of the mud, drowning. But even in the middle of all this your cousin was careful of himself. And he was careful of his men once they gave him men to look after. He would have made a good officer, Clare. I'm sure he was a damned fine sergeant.'
Her cool father swallows.
'He was just as careful over the men. He knew which ones had given up already long before they knew that they had. They had a look on them, he said. He forced all his men to take care of their feet. Even when they were too tired to stand, he'd have them take off their boots and dry their feet when they could. They'd curse him for it, but they didn't get trench-foot. On hot days when they were behind the lines he'd tell them to strip off their shirts and hang them over their rifle butts – bayonets fixed and the whole thing stuck into the ground – and then the sun would draw out the lice from the seams of their shirts. Clever notion, whoever thought of it. You could kill the lice with a lighted match once they came out. Lice weaken a man, you see, Clare, quite apart from the disease they might bring. And a lice-ridden man takes less care of himself than a clean one.'
He talks of lice as he might talk of stamens, or pollination. He says, 'It made me feel ashamed, because I didn't know what he was talking about. He had to explain it all.'
But you asked, thinks Clare, and he answered. Mr Lawrence was right. He did need to talk. He could have talked to me too, but perhaps I did not ask the right questions. But Father is right about his carefulness. John William was always careful when it mattered. If we were fishing with maggots, he would hook them on just right, through the loose skin at the top, so that none of their inner juices would spurt out. And he could bring in a rowing-boat through thickets of masts and hulls when the herring-fleet was in, and never once graze another boat. But the thing she sees most vividly is his hands pouring an exact pound of pearl barley into a blue paper bag, then sealing the top. Not a grain spilt.
'You will lose your train,' she says. 'And you haven't had your egg.'
'Leave it now. You eat it later,' says Francis Coyne. 'All those young men, learning all those things we never knew. And then he's thrown away in an accident after two years of the war.'
They will tell Father, Clare thinks, because he is one of them. A gentleman – like John William would have been. And because he is angry. They won't know why, but there's a force in him she has scarcely ever sensed before.
Francis Coyne picks up hat and overcoat and his polished London walking-stick and kisses her. He has sixteen minutes before his train, and he will do it easily. There'll be nobody about, he says.
'What shall I tell Hannah?'
'Just that I've gone to make the funeral arrangements.'
She smells the sharp, new morning as she lets him out, and he raises his stick and walks swiftly away. He looks purposeful: not like Father at all.
Inside, she sets the kettle on again. The kitchen looks cluttered and dismal. She shovels her father's crusts into the compost bin, and, as she does so, she sees the sheets still in the bucket and on the grass. The smell of the bad egg is sickening, and her father smelled old this morning. She had had to turn her face aside when he kissed her. Then he said, 'Make sure you pray for your cousin.'
Now I'm alone, with the washing, and some dried cod to soak for tomorrow, and a message to send down to Hat to tell her not to come today, for I can't bear to have her clattering around the house, sympathizing with those big greedy eyes. I shall scrub the floor myself. I shall wash those sheets again, and water my vegetables, for we're going to need them. What will Father's train fare be? I ought to have asked him. But he will find it from somewhere, I suppose. Perhaps he will sell more books, although last time he told me it was not worth the journey to Truro to sell them. He ought to have sold them in London, but we cannot go there.
Kitchie will come up. Another day of everyone crowding together. And I cannot bear any of it. I would like the day to be over already, and the night here. And then tomorrow to be over, and another night.
Hannah sleeps, worn out after a day with her mother and Harry. Harry's arm had kept them awake the whole of the previous night, fetching drinks and wiping sweat from him and easing pillows under his shoulder to keep pressure off his arm. It is a bad break, and now he is feverish. And Harry is a terrible patient. He will not keep still. He tosses to rid himself of pain, and then the tossing jars his arm, and he cries out. He was confused after the chloroform Dr Kernack gave him when he set the bone.
'I don't know where I am! I don't know where I am!' he cried, clinging on to Hannah, his eyes wide and black and horror-struck.
'And who's to pay the doctor's bill, I don't know,' said Hannah, as she drank the cocoa Clare made for her. 'Nan had to fetch him out again today, Harry was so bad.'
'What about your mother? Can't she nurse him?'
Hannah made a quick, dismissive motion of her hand. She was tired, and she didn't want to talk.
'Mother can't. But I have to be back at work tomorrow. Father can't manage the shop on his own. But Nan'll be there, and Aunt Mag. You have to mind him. If he sets up an inflammation...'
'You should let your mother nurse him. You should _make_ her do it, Hannah. You're too soft on her.'
'It's not in her nature, Clare, and that's all there is to it. Besides, she's thinking of Johnnie –'
Or you want to believe she is, thought Clare. No. Aunt Sarah's shallow whirlpool of feeling was something Clare refused to enter.
The house makes settling night noises around the two girls, one sleeping, one wide awake. Hannah has the best half of Clare's bed and her warmth fills it. She lies round and peaceful and still, drinking up sleep, snuffling against the pillow, her dark night-plait coiled against her spine. But Clare lies on the cold edge of the bed, so that her body will not touch Hannah's. She does not want to touch anybody. She lies on her back, her arms behind her head like a declaration of wakefulness, and she stares into the dark. It is not so very late: it is somewhere between quarter to twelve and midnight. Soon the clock in the hall will strike, because her father is not here to muffle it for the night. It will go on striking each quarter until morning. But Clare prefers not to tamper with it. That clock is one of the few things in the house which her father takes care of, and Clare was told too often as a child that its delicate coilings could be wrecked for ever by one interfering finger.
If I could hear him calling out in the street, and lift the window, and go down through the graves again. This time I wouldn't be frightened. I would know what it all meant. Afterwards I would make him sit on the rock and I would draw him and talk to him and I would have him for ever.
'For ever? A man who didn't want you enough to spend more than an hour with you before he went?'
'He did. It was the war. He had to get away from people.'
'People? If you love someone, they're not "people".'
'He needed to be on his own. He needed to walk it out of himself.'
'Oh, yes? And the way he looked at Peggy? But he knew he wouldn't get up Peggy's skirt without marrying her, because Peggy's a good girl. So, since it wasn't worth wasting time on Peggy, he had a quick look round and there was Cousin Clare to hand.'
'It wasn't like that.'
'It never is, is it? But all the same, he's just like any other man, your precious John William. As soon as he's got what he wants, he'll show you the back of his britches.'
'You're wrong. He didn't turn his back on me. When we were at the station saying goodbye, he kissed me in front of all of the family.'
'In front of all of them. Are you so sure about that? The way I remember it, Hannah was standing in front of you, blocking the sight of the pair of you. Anyway, he could afford to kiss you. He was going off to the war, wasn't he? Anyone can kiss a girl then, especially if she's his cousin. And any sensible girl would forget about it right away.'
'No. You've got it all wrong. He was going off to the training camp. He was going to get his commission. He would have been safe for three months, at least. That's what he told me. He'd have written to me. He might even have had leave again.'
'But it wasn't safe, was it. He's dead.'
'He didn't know he was going to die! He didn't know it when he did those things with me out on the rocks. He thought he was going to live. That's why he felt so bad, because he was going to live and all his friends were dying. They'd picked him out to be an officer and a gentleman.'
'Gentleman! Very much the gentleman he looked that night, to be sure. I wish the pair of you could have seen yourselves.'
'It wasn't like that. You can't tell what love feels like from the outside. And he would have come back to me, I know he would. He would have come back and married me. Cousins can marry.'
'They can; yes, they can. _If_ they want to. But it's not going to happen, is it? He went and died, didn't he?'
'He didn't know he was going to die.'
'Didn't he?'
'No, he didn't. He didn't. How could he? How can anybody know he's going to die? That's all rubbish and superstition and I'm not going to listen to another word of it.'
'Ah. You're not listening anyway, are you? You don't want to, in case it spoils your pretty pictures. But you will. You'll have to. You don't believe it now, but you'll want to.'
# Twenty
The train creaks into Lelant Saltings, and lets off steam. Let it stay here, thinks Francis Coyne, looking out of the window at the wet spits of the saltings lying ready for the next tide. The window is open at the top, and a smell of salt and rain blows in. Let it stay here and never move again. It is grey today and the wind is blowing, so that when we come round to the sea there won't be that shockingly beautiful vision of dark blue water and purple water and white sands. Porth Kidney Sands and Carbis Bay. Coming home. Susannah showed them to me first. Could I have loved them if they hadn't belonged to her? She would not share them. She kept this landscape locked tight inside herself and never said whether she found it beautiful or not. It was just there – hers. No doubt she couldn't have lived without it, after so many hungry years in London, but then she didn't live long with it either. She fussed with plant-stands and lace mats on polished tables and frilled pinafores for Clare. Only once I came into the bedroom and saw her by the open window, with the cold air streaming over her, looking out at the sea. I heard the thud of waves hitting rock. She did not seem to hear it, or feel the wind. I told her to shut her window, for the cold would hurt her, and then she turned to me but she still said nothing. She put up her hand to her mouth to stifle the cough that was bubbling there. She had no shawl. Her face was white and red in patches, the skin stretched over her cheek-bones. She could not stop coughing. I took hold of her and felt her vibrate with the cough, as if there were something alive there tearing its way out of her. I held her while she coughed, and I closed the window, and drew the curtains across so that she might sleep in the darkened room.
She's left me nothing to love here, only my Clare. I can see the beauty everywhere, and it pierces me, but I am never at home. It is a beauty which disturbs me without ever offering the comfort of familiarity or possession. I live with my back turned to everything I know, and no matter how long I live here it will be the same. I would find it easier if the place were less beautiful.
The glass in the train window rattles. Cold for June, they'll say, just as they said before that it was hot for May. There is strangeness in the weather these days. It is as if the days have misplaced themselves. Sometimes the wind tastes as if it has blown here across a desert. Heat swells until it seems as if the spirits-of-wine will burst through the top of the thermometer. The sky is brassy and frighteningly indifferent. Why shouldn't it roast us if it wants? We believe it will not do so, because it has never done so before. This is our world, and it will not turn on us and rend us. But it's the same sun that glitters over the trenches on perfect June mornings before the obliteration of thousands in a couple of hours. Men who shave themselves in the morning, and put tissue-paper over a shaving cut are blown to rags by noon. Phoebus Apollo, flame-crowned. What are those lines?
Stand in the trench, Achilles,
flame-capped, and shout for me...
Yes. It sounds glorious. It does not sound like a man trying to scrabble his way up the greased yellow walls of a flooded shell-hole. It does not sound like a man trying to force his wet, squirming, glistening bowels back into the hole in his stomach. Dying in a lather of blood and excrement, yapping like a puppy. No words, no tears, no prayers. _You wouldn't think the insides of a man would a been such a colour_. We sat in our seats, and we listened as the women made the food in the kitchen. We could not relish our ham and gooseberry tart after that. John William broke the compact: soldiers are not supposed to tell us about such things.
How does the poem begin? Now I have it:
I saw a man this morning
Who did not wish to die
I ask but cannot answer
If otherwise wish I...
I see them. Young men like we were when we travelled through Europe. Young men in an _estaminet_ , drunk on fear and cheap wine and exhaustion. Young men in orchards which won't fruit this year, because they'll be blown to bits before the blossom is off the branches. Rags of flesh the colour of cherries before they brown and harden.
... who did not wish to die...
The train jerks forward. The fat guard hangs dreamily from the window, looking over the estuary to Hayle Towans. He lifts his hand – who can he be waving to? Francis Coyne can see no one. That's where Clare bathes. Not too near the estuary, I hope. Dangerous water.
... who did not wish...
And the wheels go rumptitump, rumptitump, _wish to die, wish to die_.
They said he should not see John William's body and he had not insisted. Coward. It would serve no purpose they said. There was no question of identification.
The Colonel who had come in from somewhere was confidential in the smoky small room. From time to time a young man entered, saluted, took away a file.
'If I am not disturbing you, sir.'
Perfect manners. What were they – adjutants?
'It's the calibre of a man which matters to us now. Not his background. That's all old hat,' said the Colonel.
You mean you are desperate for new men, thought Francis Coyne, watching the young officer's small, perfect salute as he withdrew.
'John William Treveal. Terrible business. A moment's carelessness,' he told Francis Coyne. His fire bubbled in the corner. 'Splendid report came with him from his platoon officer.' He twitched another file over the desk. 'Yes. Here.' He read it through, but did not offer to show it across the desk.
'You can be proud of him. Be sure of that. A terrible thing, but it's happened before and it'll happen again. The best man can get careless cleaning his rifle. Why, at home –' But he broke off without relating his anecdote of a safe, domestic shooting accident.
He got careless. No, there is no purpose to be served by your seeing the body. A drink? No? As you wish. The funeral arrangements. Yes.
Francis Coyne walked through the camp on his way back to the world. He was offered an escort, but no, he preferred to be alone. They understood. Of course. The Colonel stood, and Francis stood, and there were handshakes. They trusted him to be stupid and to ask no questions. He was given a packet containing ten pounds in cash, which he signed for, and John William's watch. The other effects would be sent on.
He paused to watch a group of soldiers playing football. They were not officers yet – or were they? They leapt and yelled and there was a little mild horseplay of the kind that is only to be expected of young men filled with adrenalin and bully-beef. They had all been out there. Some of them would have known John William. The ball shot past him – he kicked at it too late, ineffectually civilian, but the man who pounded after it didn't mind. His boots thumped the dry turf. He was big, muscled and tanned, panting with the exercise. He stopped by Francis Coyne.
'Looking for someone?' he asked.
'No. No. I am here to see after something. Someone. My nephew. John William Treveal.'
The man's face changed. He turned and threw the ball back into the field and shouted, 'Get Simcox in. I'm wanted here a minute.'
The pair of them walked away from the field.
'Who have you seen?' asked the man.
'Colonel Lacey.'
'What did he tell you?'
'About the accident. He said John William was cleaning his rifle. But...' He looked carefully into the soldier's face. 'Did you know my nephew? Did you know John William?'
'Yes, I knew him. We were over from Boulogne on the same boat. Right away to Blighty! I still couldn't believe it, and I don't think he could either. I can't yet. I had a twenty-franc note in my pocket and the sun was shining and we were out of the war. We were leaning on the rails, looking down on the men throwing the ropes off down on the quayside, making a lot of racket. Just like the French.'
'Did you see him? Before the accident?'
The man paused and glanced round. He was very close, so close that Francis Coyne could smell his sweat. He spoke quietly.
'When he came back, he wasn't the same.'
'After his leave.'
'Yes. Course it takes a man all ways, seeing his home again. I'd allow for that. But with John William it was different. He got into a wild rage the first night here. Something was said – I thought he should a killed a chap. But we all got hold of him, asked him did he want to get himself RTU.'
'What do you mean?'
'Returned To Unit. That's the penalty – worst thing that can happen.'
'What did he say?'
The man frowned. 'Seemed like he didn't care any more. The others thought he had drink taken, but I knew he hadn't. He wasn't one for making a beast of himself.'
'So what was it?'
'Later, when I was asleep, I woke up cos I heard him moving about. He was feeling for something under his bed, it looked like. Sort of groping. Somehow I didn't care for it. I called to him, soft-like. I don't know why but he didn't look right to me. Not like a man should.'
'I suppose you get used to spotting that, out in the trenches.'
'You do. You have to. So I motion him to come outside the hut and he does, quite meek he was. Not like himself. And I said to him, "What's up?" and then I saw he was trembling. Well, I felt it, really.'
'What did you do?'
'I put my arm around him.' He flashed a look at Francis Coyne. 'And he didn't shake me off neither.'
'No.'
'He was crying. I asked what was up with him, and he said he had done wrong.'
'What was it? Did he say?'
'I thought it might a been a girl, seeing as he had been home. I didn't get it clear, sir, but it was that he thought he had left someone. He thought he had abandoned them. He ought to be out in France still, that's what it was. He thought he had let the men down, coming here to be an officer when they were over there in Hell still. Course it was rubbish, and I told him so, and he must a known it. I was surprised at him. He struck me as level-headed enough most of the time, John William, for all there was a wild streak in him.'
'And was he better?'
'I thought he was. Then, as we were going in again, he freezes up and he says, _"Hammond!"_ (that's my name, sir) and then _"Can't you hear them?"_ and I say, what, there's nothing to hear, trying to calm him down, like, and he says again just like this, _"Can't you hear them? Those 5.9s? Don't tell me you can't hear them_?" '
'So what did you do?'
'Well, I knew of other men it had taken that way. I thought I had better go along with it, to humour him. But it was a queer thing – I could feel gooseflesh creeping out all along my arms, though it wasn't a cold night. And as I listened I was half thinking I would hear them too, and maybe voices singing out the way they do in the trenches. So I was humouring him, and not humouring him, if you see what I mean.'
'Yes, I do. Go on.'
'And so then I said, _"Come on. Come along out of it and get some kip,"_ like I was relieving him from duty, and he came with me. He was stiff under my hands, and he still had a listening look on him. He wasn't right. So I thought I would go and have a word with the medical orderly the next day, not naming his name, just to find out what might be helpful to him. For I know men who've been taken like that. I've seen it. It's in any man, if he's tired enough, or if he's just seen his pal killed in front of him. But in the morning he was fine again, shaving hisself with a steady hand, for I took note of it, and eating his breakfast. One moment, he was laughing at something Simcox said. I can see him now – soap all over his face, laughing. So I wasn't worried. When I got a quiet minute I said to him, "Everything all right now?" He looked at me as if he didn't know what I was talking about. So I said no more, for it felt awkward. Strange things happen at night, and a man'ull say things he wouldn't want remembered in cold daylight. And that was on the Saturday. We went on parade, then there was drill, and map-work, and then he wasn't with us. I wondered for a minute, then it went out of my mind. The next we heard was Lucas howling out from the huts.'
'Howling?'
'Well, he'd seen dead men before; lived with them, you might say. But he didn't expect to find one back in Blighty; not in the officers' training camp.'
'Hammond, I am going to ask you a question. You needn't say a word if you prefer not: just signify to me, yes or no. Would it be in order for a man to clean his rifle where they say that John William was cleaning his when it went off?'
'I can answer that,' said Hammond readily. 'It would not.'
'Let me ask you something more. Did you or any of the other men see John William's body – apart from Lucas?'
Slowly, Hammond shook his head. 'We did not. Our hut was put out of bounds to us, except for the detail which took away his body and cleaned the place.'
'You didn't happen – did you? – to speak to any of that detail?'
'I did,' said Hammond. His eyes were pleading. 'I made it my business to speak to them. I had to satisfy myself of what had happened to him.'
'You did right. And that is why I am here too. What did they say?'
'They said that he had shot hisself.'
The two men had stopped, but now they walked on, close together, taking pace after pace, heads down. Francis Coyne said, 'Thank you. Don't be afraid – you are not telling me anything I hadn't feared.'
'It's a mortal sin, they say,' ventured Hammond, turning over the words.
'Are you a Catholic?'
'I was brought up to it. But out there – it changes you. If there is damnation, you don't have to look far for it. And I can't believe any God would damn a man for not being able to get those guns out of his head – can you?'
'No. No, I can't believe that.'
'Or if He does, then I want none of Him,' said Hammond.
'I shall tell no one of what you have told me. Only my daughter. It's best his family doesn't know.'
'They'd stop being so proud of him then, would they?' said Hammond. How they must despise us, thought Francis Coyne, safe at home, judging their courage. He would like to give this man something. Would he take it?
'They have given me John William's watch. Will you have it?' he asked. 'As a friend?'
'There'll be plenty closer who should have it, surely. He'd no son, a course, but maybe there was a girl?'
'Not many close when he needed us,' said Francis, putting the watch into Hammond's hand.
Now he is coming home with ten pounds, and a story no one is going to hear. He thinks of May's eyes glazing over briefly, masking her irritation that he should try to harrow her with it. Isn't there enough to think of in life without wallowing in other people's troubles?
Old Mrs Treveal knows. He's sure of it. He doesn't think that she will ever allude to it, or ask questions of him, but he has seen that she knows. And perhaps she won't ever need, to tell or to talk about it. That's strength, he supposes. She also knows what people are made of, and that Sarah and Arthur won't be able to stand knowing what has happened to their son. How quick she was to assure them that Francis must go to the camp, not them: they have fixed it up together, he and Nan, like a couple of conspirators. She will protect the family, even from its own child, even though there's no doubting that she loved John William purely, wanting nothing from him but that he should flourish and grow away from them. She'll forgo walking behind his coffin; she'll never be able to visit his grave; she'll silence the murmurs of townspeople who ask why the Treveals couldn't have brought home their John William.
And now for Clare. The weight of it: can she stand it? She will have to stand it. He will tell Clare what has happened to her cousin. They will share the weight of it. They will share everything, and there will be no secrets between them any more. They have drifted so far apart; he had his own life, and thought that by having it he did not injure his daughter. But it is not so, he thinks now. The Church is right; our sins are small seeds which grow and flourish until they are great trees with shadowing branches and sucking roots, drawing the light and life out of the house. He will stop going to see May. It will be his sacrifice.
All that is over, he tells himself, exalted. He will never again see that closed look on his Clare's face. She will turn to him, her face cleft with grief, and he will hold her. There'll be an end to the two of them in the house, picking their way over silences. They have grown too far apart, and it is dangerous. He has let her smooth out his life for him and serve him, but he has not watched over her. He has let her go her own way, and the world is a wilder place than she knows. Even down here in St Ives we are blown by the desperate breath of those men struggling with their feet lodged in mud and foul water up to their lips.
Her cousin is dead. And he, her father, has been half asleep. He has let his Clare form an acquaintance with those Lawrences without bothering to find out if they were suitable people for her to know. Now he finds that they have roused the hatred of half the county. Now he hears that the Rector of Zennor fears the corruption of his parishioners.
God knows if there is any truth in what people say of them, but Francis Coyne knows the power of local voices. He knows the tentacle-grip of gossip. He ignores it, usually, and lets it flow past him without ever discerning the words. Gentlemen do not listen to gossip. It is less troublesome that way. But he came alive at the hint Sarah let drop, half fearful, half daring. Coupling the Red-beard's name with his Clare. The weak look of triumph on Sarah's face, as she taunted him with something she knew and he did not. He did not want to question her, for it would seem to give some credence to what she suggested. Anyway she had taken fright and hidden behind her easy tears, her fog of denials. He went home with sharp, prickling alarm in his breast. And then the Rector's curt, disgusted repudiation of Lawrence, and behind it a hint subtler than Sarah's, man to man, gentleman to gentleman.
But I did not know, Francis Coyne tells himself. How could I have stopped her meeting him?
You did not choose to know. You did not want to know. You left her unprotected, says the voice he is accustomed to calling his conscience.
It will end now. He will give up May. He'll give up hearing the cluck of water against the harbour wall at Newlyn, and the little intimate creak of May's door going backward and forward in the draught. He will not say her name softly any more, as he steps into her room. He will not trace the greeny branches of her veins, or lay his cheek against the mound of her belly and listen to the sea-like movement of food and blood. He will not run his tongue around the knobs of her nipples as they stiffen in the cold air of her bedroom. She lies on her back, eyes shut, mouth askew, breath quick.
John William lies on his hard wooden board, bandaged into shape. His mouth and nostrils are icy as the entrance to caves. A cold foul air sifts in and out of them. He is hard all over, his head cocked and propped so that it makes him look as if he is still listening.
Francis Coyne imagines, and blames himself. He ought to have listened; he ought to have watched. He has abandoned John William just as he has neglected his Clare. No eye has looked on John William's body with love. No one has lit a candle at his head or at his foot, or twisted flowers into his hands, or kept watch over him through the night.
But he will tell Clare how her cousin died. She will understand the weight of it. She will cry out with her soft lips parted, and she will turn to him.
'Clare,' rehearses Francis Coyne. 'I am telling you this because...'
# Twenty-one
Clare opens the door.
And there he is, Mr Lawrence in his worn green corduroy jacket, with a paper parcel under his arm which is neatly tied but which sifts out a thin stream of black soil from its corner.
'May I come in?'
She glances up the street and says, 'My father is away.'
'Surely you don't care for such things? And there is no one about.'
She laughs. 'At least six people will tell another six people each that you have been here. This is not London, you know.'
'And I am not a Londoner, tha knowst,' he says, teasing out the soft Midland dialect as he steps through her doorway.
In the hall both their smiles disappear.
'I heard in the town that your cousin was dead,' he says.
She waits for the conventional expressions of sympathy, but they do not come.
'I wrote once,' he says slowly. 'It was to another woman; very beautiful she was, in the English way, you know? With her little air of being somewhere else even while she was talking to you. Her brother was killed. I wrote to her that I would rather put out my eyes than stand as a witness to this deliberate horror. And I believed it. But now I would not put out my eyes. I need them to look on other things – flowers, and beasts, and a little hut in the mountains. So I set myself to _keep separate_ –' He breaks off.
'He was killed in an accident,' she says. 'When we thought he was home safe.'
'Yes, yes,' he says quickly, looking intently at her face as if willing her to follow his train of thought. 'You know that I was with your cousin after we left you? We walked up to Zennor that night after the concert.'
'Yes, when I came to your cottage for tea you told me you'd talked to John William that night.'
What did they say, she wonders? Did John William mention it? But she can't remember anything now, except, 'You don't know how nice you are without your clothes.' She feels her skin warming.
'He was sick to his soul too, though there wasn't any wound to see,' says Lawrence, looking at her out of his taut, white face. 'Perhaps that's what is coming – a time when men will go clean out of their minds. Except a man who can sit apart in his own soul and watch the foxgloves come out.'
'We can't all do that,' says Clare sharply.
'Why not? Why not, if it'll keep us living – _really_ living, in the middle of all this madness?'
'John William wasn't mad,' says Clare. 'He knew what he was doing.'
'No one knows what he is doing any more. He is joined to a machine. He is not free to act as a man any longer. He is part of a machine of colossal stupidity. And I think your cousin knew it. He was a brave man, but he wasn't blessed with stupidity. It would have been easier for him if he had been able to sink into a state of mindlessness and forget that he was a man with a soul of his own for which he was responsible.'
'You talk of soldiers as if they were doing wrong – as if they were animals!'
'No; they are not animals. Perhaps it would be better for them if they were. An animal does only what is in its nature to do. It cannot kill its own instincts. Think of a fox when you come upon him suddenly. He looks straight at you, and knows you, and then he trots away, quite self-possessed. He knows whether you mean him harm or not. You can't fool him. But you can fool a man's soul out of him, if you set yourself to do it. The war has fooled England's soul out of her.'
He must have talked to you too, the way he talked to Father, thinks Clare. He showed you things about the war. But he wasn't like that with me. It was like when we were children. Why? Was it because he thought I wouldn't understand if he told me about men dying? Because he thought he had to protect me? Or perhaps he didn't choose at all. She remembers the lion roar of water under their bodies that night. John William's face, wiped clean and astonished back into the naked lines of childhood. He did not look like that the next day. The line of his body was stiff and burdened then.
She looks at Lawrence's face, clenched with marks of pain. There is no light in it any more. The vitality which would draw a friend to him across a crowded room has sunk down deep inside him. It is not a clear flame now, but something that smokes and smoulders and throws out long, frightening shadows. He talked to you, she thinks. And what did you say to him? What answer did you find?
'How did your cousin die?' asks Lawrence.
'It was an accident. We do not know any more yet. My father has gone to the camp, to talk to the commander.'
'An accident. Yet he was two years in France, I believe?'
He looks down, frowning. She hasn't offered him so much as a cup of tea, she thinks: and yet he and Frieda made her tea, and cake, and had flowers in the vases. Her boiled egg was set perfectly, its deep yellow yolk steaming as she cracked the top off it. Lawrence had kept cutting her more bread and butter, for he could not bear girls finicking at their food, and she had had a long walk. She had eaten five slices and he had wanted to wrap up a loaf for her to take home in greased paper, since she liked it so much. But today she stands in the hall with him, tense and fearful. She does not want to draw him deeper into her house. He makes sweat prickle on her palms and her breath come quickly. She is afraid he will tell her something about John William which she does not want to know.
'I brought you these new carrots, from my garden,' he says, unwrapping the parcel a little to show slender, clean roots. More soil falls from the paper.
'Come into the kitchen with them. I will make us tea.'
'Frieda eats them raw,' he says, following her and sitting down at the kitchen-table. 'She has fine teeth. I should leave them in the earth a little longer, to fatten, but they are so good like this.'
He unpacks the fronded bundle of carrots, and some spring onions.
'Thank you,' she says, thinking of Frieda's white teeth. She is alive, eating Cornish carrots with her white German teeth, looking out over our sea and exclaiming how beautiful it is, while John William is dead. There they live in their cottage, untouched, growing their vegetables, singing their songs. A pang of horror sweeps over her like sickness. John William has got to stay in the same place for ever. She won't ever see him swing round the corner towards her, frowning against the sun, lightening as he sees her.
'I am glad you like them,' says Lawrence. 'I shall send more down for you, on the farm trap. William Henry will bring them for you, when he comes into market. Shall you like that?'
'It is very good of you.'
'It's good for us to make a new friend. Frieda is lonely; I have the farm-people, but she has no one.'
He's told her this before – does he think she cares so much for his wife? He looks at her with simple pleasure in his face. He changes so quickly! – she cannot keep up with him. His face is warm again, because he has forgotten about the war for a moment. Let him forget it. It is all talk; all words. He did not know John William and John William's death is nothing to him.
He does not know me either. All I am is a girl who can draw, who might make a friend for his wife. She is lonely, is she? Let her get used to it. She likes me, does she? More's the pity. Why does he sit here looking like that, making me think?
'Perhaps you ought to go back to London,' she says lightly, pretending innocence, pretending that it scarcely matters if the Lawrences live in London or in Zennor. They think they are at home here. She hands him his cup. 'Perhaps you ought to think of leaving, if Frieda is so lonely. It is hard to make friends with Cornish people, they say.'
He looks at her, surprised, as if he is readjusting his ideas about her – as if she has proved less intelligent, or less perceptive, than he had thought. And she wants to cry out and reassure him. But he smiles, and explains:
'No, we are much better here. We have lived in so many places, and this cottage is ours, you know; we found it for ourselves. We pay the rent, and it is only five pounds a year. We can afford to stay here. And London is vile now; it makes me wretched to think of it. I should go mad if I had to live there now. But in Zennor we have a wild place of our own, where no one bothers us.'
'Don't they?' she asks.
'Oh – they try. They would like to. They would like to get their hands on the life we have and smear dirt all over it and make it vile too. But they do not quite dare. I am not speaking of the ordinary people, you understand; we have nothing to fear from them. It is the little officials of the war. Some of them are decent men, and they don't like what they are doing. But there are others who have wanted this war all their lives without knowing it. They have wanted the power it gives them. They would like to destroy us, but we don't let them. We lie low.'
She laughs, imagining Frieda lying low.
'Why do you laugh?'
'Because you are wrong. Because everyone sees everything here,' she says, meaning to disturb him, to hurt him. 'Just as they will be wondering why I should have you in my house, with your German wife, when my cousin has just been killed.'
But he does not take offence, as she expects.
'They _would_ think that. It's only to be expected,' he says, his face hardening again with contempt. 'But you know better.'
You take too much for granted, she thinks from a cold hard space inside herself. You don't stop to think that I too might be wondering what I am doing.
The front door clicks, opens, closes gently.
'My father,' says Clare, jumping up as he calls her name from the hall.
'I had better leave you,' he says quickly. 'You will want to talk to him.'
But Francis Coyne is down the hall and into the kitchen. The two men will have to meet now, and as she realizes this Clare also realizes it would have been much better to keep them apart. Clare sees the scene as her father will see it: the cups of tea half drunk, carrots spilling across the table, her own hair pinned loosely and her sleeves rolled up, ready for housework. He hates other people to see her like this. Lawrence stands at one side of the table. His red beard flames and his deliberately self-possessed thin figure seems combative. Clare sees that her father is exhausted. His cheeks are sunk and white under his cheek-bones, and there is a light film of sweat on his forehead. He knuckles his eyes for a moment, bewildered. He cannot take in what this man is doing here.
'Mr Lawrence,' he repeats.
'Mr Lawrence has taken a cottage at Zennor – Higher Tregerthen. You remember, Father, I visited them.'
'Ah, yes. I have heard about you,' says her father dryly. He straightens himself and scans Lawrence with cool, open dislike. Is it his clothes? Or his beard? What is it about Lawrence that her father dislikes so much? He did not react like this when she first told him she had met the Lawrences. And he is so polite, usually. Perhaps he is angry because Mr Lawrence has come to visit when she is alone in the house. Clare blunders on, 'Mr Lawrence came because he heard of John William's death. He came to offer his sympathy.'
'Very friendly of you,' says Francis Coyne. 'Now, Clare, if you will excuse me.'
He has not even offered Lawrence his hand. He mounts the stairs, thud, thud, thud, to his bedroom. The door shuts.
'I'm sorry,' says Clare. 'I'm so sorry. He is distressed – and the journey – '
But now she wants only to be rid of Lawrence. Something has happened – something terrible. Is it something her father has found out? What have they told him about John William, to drain the colour out of her father's face and make him look as if he does not know where he is? She must go to him. She gives her hand to Lawrence to hurry him out of the house. He bows, a small bow. His hand is warm and hard and he does not let go of hers; he looks into her face. He wants something more from her, some warmth to heal her father's coldness. But she has nothing to give, and he is too quick not to spot this at once. He looks at her consideringly for a minute, holding her hand in his, then he says, 'Goodbye, Clare. Remember Frieda – won't you?'
She cannot answer him. She feels an absurd desire to raise his warm hand to her cheek, and pillow her face against it, and sleep. To be in the darkness where none of this is happening. Then he smiles and he's gone.
He's gone. She flies up the stairs and knocks at her father's door. He is sitting on his bed, taking off his boots.
'I don't want that man in my house ever again,' he says. 'What were you thinking of to ask him here, Clare?'
'But, Father – you knew I had met him. You knew I had been to their house. And I didn't ask him; he heard of John William's death in the town, and called in.'
'I hadn't realized. I hadn't thought. I have been remiss towards you, Clare. Your Mr Lawrence is a dangerous man. There's talk in the town, and it isn't just the usual gossip, I believe. You may not have heard it, or understood what it means. You can't understand what it might mean to have your name coupled with a man like that. He has a reputation. And his wife's a German. They are not fit people for you to know.'
He sounds like one of her uncles. What can have happened? Father never talks like this. Dull red mottles his neck as he struggles with his left boot.
'Let me.'
'No, I have it. There.'
'Father, what happened?'
She is terrified by the bitterness in his voice. The flat ungivingness of it. Everyone is changing – even her father is no longer the same.
'Tell me.'
'You must give me your word that what I tell you will never be repeated.'
'Aren't you going to tell Aunt Sarah – or Nan?'
'It must never be repeated.'
'All right. I promise.'
And he tells her. The day is dark already, so it does not darken. The grey sky stays just as grey, and the swiftly moving clouds do not stop, or turn backward in their course. The noise of the sea goes on and on. The war continues at Passchendaele in the Ypres sector. The noise of guns.
'Can't you hear it? Can't you hear it?'
Francis Coyne can hear it now, and the voices singing out, and the cry of wounded men as they hold their entrails together and beg for water which cannot come until dark when a stretcher-party will try to reach them. The faces of the stretcher-bearers are no longer appalled by anything, but their hands are still gentle as they refuse to put in the bullet the wounded man screams for. They are men, not beasts.
John William had a revolver. He put it into his mouth, pointed it upwards, and squeezed the trigger. He was used to shooting. He did not make any stupid mistake like trying to shoot himself through the heart, and he made no error in the angle of the barrel. All the same it was lucky he could not foresee the heart-stopping spray of his brains and bone over the whitewashed walls of the hut, or the way he spoiled the clean laundry for eight men. And someone had to pick it all up, and clean it away. One of the detail said, 'Yer'd a thought the bugger could've done it out in the woods.' For that was the way they talked, to make it all bearable. And perhaps the bugger could have, or perhaps he'd had enough of men dying in mud and dirt holes where no one had to clear up after them. Maybe he thought he would spoil one white room in Blighty, and make them clean it, and know that a man had died there.
There is Clare with her white, white face, his own daughter shrinking back from him.
'Did you see him?' she whispers at last.
He shakes his head.
She raises both her clenched fists and bangs them against the side of her head, at the temples.
'Why did you tell me!' she screams. 'I shall never get it out of my mind.'
And she bangs and batters, as if she will fight her way into her own skull and beat out the knowledge which he has put in there. He seizes her hands, and holds them down. He is just stronger than her.
'Clare. Clare. Clare, listen to me...'
And she collapses against him, burrowing her head into his chest. It is frightful, he thinks. He would never have thought she'd take it like this. Could there have been something between them – some understanding he didn't know about? No, it couldn't be. The rest of the family would have known about it. Nan would have told him. And Clare is half a child still. It is the shock.
'Clare, Clare,' he remonstrates, rocking her. But he is fiercely glad. She is his, his own daughter, his own Clare. His little red-haired one. Let Nan and Hannah and Mag and all the rest of them know it. They are outside, and he is inside. He will be the one who comforts her.
'Nan,' she hiccups. 'Nan.'
'What, my darling?'
'Don't tell her.'
'I won't tell her. I won't tell anyone. We shall be the only ones who know.'
'Don't light the candle, Clarey! Come on out.'
'But I'm only in my night-dress. I can't.'
'You'll be warm enough.'
_We shall be the only ones who know._
'Stop that whistling, John William! You'll whistle up ill-luck for yourself.'
_We shall be the only ones who know._
'Where are we going?'
'Down to the sea.'
'Which way are we going?'
'Down through the cemetery.'
'Are you a mermaid?'
'Are you a human boy?'
'Come here and let me hold you.'
_We shall be the only ones who know._
# Twenty-two
Little Clarey in her black dress, a smudge on the front steps, in high-walled steep alleys. The houses are white cliffs to Clare, and the women who lean out of their windows to screech across at one another and hang out their washing are seagulls roosting fiercely in the cliff crevices. They are big women with brawny arms and hard hands who will slap Clare if she goes near their washing baskets and meddles. Nan has put her out for the morning. All the others are at school, and Clare isn't to have her lessons today because Miss Purse's mother has had one of her turns. (Gagging, reeling Mrs Purse with her face like a white and red crumpled cloth at the door of the poky parlour where Clare is taught.) But Nan and Aunt Mag have got the morning-room curtains to finish for Carrack House, and if they are not delivered on time a shilling must be deducted from the charge.
Clare rumples up her petticoat and sits on the thickness of it, because the stone is cold. Her black bonnet drags at the back of her neck by the strings which she has nearly chewed through. Her red hair straggles over her shoulders. A grey day. No Nan and no cousins. Grandad is not there for her to smell his tobacco and play round his feet. Clare has a thick piece of bread with lardy bacon in it for her dinner. She must mind and play and be a good girl and not run off.
She stands up and pats the warm nibbled lump of bread in her pocket. And down the cobbles she goes, stepping from one to another, pretending that she is crossing a river, pretending to herself that she does not notice that she is getting farther and farther from the cottage door. Now she is out of Nan's hearing. Now she is under the arch, and she skips across the deep launder, hopscotch, back and forth, draggling her skirt in the dirty running water. A big seagull voice cawks at her, and she dodges off, farther again, slipping along by the high steps on one side of the houses. And down, and down, until the sheer white light of the harbour bursts all over her and she blinks and screws up her eyes. She will watch the carts and horses drawing off the pilchards.
Everyone's working. No one notices Clare except to shout her out of the way once, when she comes close to the hoofs of a straining horse and an arm reaches out and half lifts, half knocks her out of its path. For a second the horse's huge, hairy hoof is by her temples. It strikes out, struggling for a grip on the fish-slimed cobbles. She tumbles out of its way and makes off up the harbour wall with a curse after her. The sails are the colour of blood setting on a scab, she thinks, and remembers her knee, and picks at it slowly, deliciously, easing the big brown crust off the shiny pink surface of new skin underneath. Half off. She will eat it. Already she can taste its dense chewiness. But the middle of the scab won't come. She pulls again and there it is, with blood on it, and a new dent of blood where the scab was, slowly filling up. She eats the scab and tastes fresh salty blood on it. The air wheels with screaming seagulls, plummeting and rising high into the air again, maddened by the dense silver mountains of pilchards. She looks down and is giddy with pilchards. Are they alive still? Oh, the man with his hands plunged into them, his hands all scaled and rich with silver pilchards, and then he lets them fall as if he were lifting them like silver money for the pure joy of having and letting go.
Clare shies her way back up the alley, face bland, eyes watchful. How long has she been gone? She is not properly sure if she's been asleep or not. It was warm by the bollard, and she ate her bread and bacon in mouse-bites, hiding it inside her hand against seagulls and bad boys who hadn't gone to school. She made herself small and put her apron over her head to dull the horses and carts and men and women and screaming gulls. Beat after beat of voice and noises. Dark. Sharp. Dark. Sharp. Clouds going over the sun when Clare's eyes are shut.
I'm dawdling, she thinks now, tasting the word as she comes up the cobbles, battering her stout little boots the way she's been told never to do, for it wastes shoe-leather. Nan's always telling her to run and not dawdle. The bread and bacon is not solid inside her any more. It must be a long, long time since she ate it, and perhaps it's nearly night now. She looks up at the sky and can't tell. Night has a way of slamming down on her, pouring out stars when she is playing out wildly, ready to play for ever.
A hand snatches her from behind.
'You young divil!'
It's Grandad.
'Don't yer know yer Nan 'n' Aunt Mag've had to run all over the town looking for you stead a getting their work done? Where've you been to?'
But she stands brazenly looking up at him, straight at him without a tear or a blush. He sees she has the devil in her today. Five years old and fit to lie her head off like a woman of fifty. Satan has got half his work done for him here. Grandad roars out, and Nan and Aunt Mag come to the door. A flock of seagulls from the opposite cottages leans out of the windows and caws to the Treveals.
'Thank the Lord you've found her!' cries Nan.
'Found her! She's found herself! Run off for pure devilment, that's all it was!'
'You bad girl,' says Nan, but the relief in her voice is stronger than the anger. Clare smiles, and makes to run and snuggle against her, but this is more than Grandad can bear. He catches her back, switches up her dress and petticoats and pulls down her drawers. Clare struggles to run, but with her drawers round her ankles she cannot. Instead she screeches shrilly, 'Bog! Bog!' which is all she dares to say of the word 'bugger'.
However, it is enough for Grandad, and for the amusement of the watching women. In a second he has hauled her over his knee. In a second she has twisted round and bitten the side of his left hand, which is holding her down across the back and shoulders. She tastes the familiar smell of blood, spits and shrieks again.
'Bog! Bog! Bog, Grandad!'
'Tell she's got red hair without lookin at er,' remarks one woman to another in the next doorway.
Grandad smacks Clare so hard that Nan darts forward to catch hold of his arm. But before she can do so, he shoves the child off his knee and stamps into the house, leaving Clare upended on the cobbles, still screeching. But she stops at once as Nan picks her up.
'Pull yer drawers up, Clarey. What a sight! And everyone looking at yer.'
Clare is silent at once. She settles her own clothes as coolly as if she were getting dressed in the morning in her own bedroom. She will not let Nan help her. Then she takes Nan's hand and tucks her thumb inside Nan's in their 'special' way, and walks into the cottage, her head well up. Mute, and stubborn, she is put to bed until her father should fetch her. Rumblings downstairs. Nan's voice, then Grandad's.
Clare sleeps again. When she wakes, her father has come for her, but Nan and Aunt Mag are still working. Nan gets up from her chair, blinking as she looks up from the fine stuff she's sewing, putting one hand to her tired back. She kisses Clare just as she always does, and says, 'Go and kiss Grandad.'
But Clare will not. She pretends to be sleepy, and clings to her father's hand, and stumbles out past Grandad without looking at him. When she has gone, he laughs. He is tickled by her small fierce face, her deliberate aloofness.
'Did you see that, Alice? Er won't come to me! Er won't come to me!'
'She'll come,' murmurs Nan, frowning over the gathers.
'Just because I tan her backside for her, our Clarey won't as much as look at me!' he marvels, almost appreciatively. Nan smiles. Thank the Lord the child's devilment's caught his fancy.
Mag's lips narrow. She's heard enough from Stan of thrashings handed out to the boys when they were young. Lucky Stan's not here now. Nothing makes him angrier than Grandad's softness to Hannah and Clare.
Nan sews. Her thoughts are like stitches, each one tiny and precise and not much in itself, but making up the strong seam. She thinks about Clare, and her responsibility for Clare. She weighs over Francis Coyne in her mind, and what can be expected from him. But she cannot judge the weight of him. Is he going to be able to rear that child as she should be reared? In her mind she feels around the edges of what Francis Coyne lacks. It is hard to put into words, and she does not need to try. These thoughts are for herself alone. Clare's red hair and her white, wild face. Say what you like, but some things are born, not bred, and you can't change them. A good child seven days out of the seven, but she's good because she wants to be, not because it's in her nature. Nan sighs and smiles, thinking of the turn of Clare's thumb under her own, the secret sign of love. How she watches us all. Well, we can do nothing but pray to the Lord and watch her. Keep her with her cousins. She's safe enough along with them. Better than with Miss Purse. Nan's face wrinkles in distaste for the Purse family's fits and vapours.
Fifteen years later Nan sews and thinks of Clare. She treadles the Singer in its placid, slightly irregular rhythm, and yards of plain seaming spill down over her sewing-table. Her eyes are not good tonight. It has strained them to make Sarah's mourning. Sarah would not have her Sunday dress dyed like the rest of them, and Arthur was weak and gave in to her. You would never think how black would grow to dazzle your eyes as you sat sewing it. Still, these sheets are white. Good, seemly white sheets. Nan feels the quality of the cotton twill. Not like the rubbish we get now, since the war.
Well, Sarah and Harry are back in their own place now. Nan doesn't like the look of Harry's arm. It wasn't more than a week before she knew it was going to take bad ways. And she'd been right. Dr Kernack had gone quiet the last time he looked at it, then he had spoken to her privately in the kitchen and told her that Harry would lose the use of it entirely. The nerve had been severed. But there was no use troubling Sarah. Let her come to that when she must. She was a poor thing at the best of times and what sense she had was driven out of her by John William's death. It was as well she had a daughter like Hannah.
Clare. Paler than ever, and she never had but a pinch of colour at the best of times. Now she's like whey. No one'd look at her.
Nan treadles, and pushes the cloth smoothly under the machine-foot. Thoughts of Susannah. Push them away, don't let them form. They do nothing but harm. Besides, Susannah had a good colour right up to the day the doctor told them there was no hope for her. Clare is a Coyne through and through to look at her. There's not a red hair among all the Treveals, nor one face the colour of Clare's.
Nan treadles on. The things they think they hide from us, these girls. Take Hannah now. Hannah tearing out her heart over Sam and saying nothing. But I could tell her, he'll be back, you can be sure of it. Because he needs you, Hannah, because your Sam is a poor thing for all he's so handsome and six foot in his stockings. But there's no telling you that. There's trouble between you now, I know, but your trouble will pass. Or you will think it's passed, though to my mind it will be only just beginning.
But Clare. My Clarey. Hovering in the doorway, with her shawl wrapped around her like a smudge on a ghost.
'I think I won't stop, Nan. I ought to get Father's supper. And I'm colouring a drawing for him.'
Your father's supper has never been started at four o'clock since this world began. I want to tell you to come on in, girl, and get your things off and sit down with your Nan, but I don't say anything. For fear of scaring you. You don't want to be with us, do you, my lovey? You don't want us to share your grieving. You don't want your Nan.
Oh, if I could just look up now and see the pair of you laughing over your nonsense. But Hannah's taken up entirely with Harry, because that's the way she wants to be. To stop her thinking. And what's my Clarey taken up with?
Nan treadles on. A quiet house for once, with Sarah back in her own kitchen, and Hannah hanging over her brother, and Mag visiting John and Annie purely for the sake of coming back snappish and discontented. There's nothing worse in God's creation than a discontented woman. But there's no Clarey coming in through the doorway on a swirl of wind, and John bellowing out to the girls to keep the door shut. Or maybe I should go up there and see her? But it'll seem strange to her. Better let her think she's able to hide her troubles.
Clare crouches on the middle of the kitchen-table in her stocking-feet. She shifts her weight cautiously to the edge of the table. It creaks, but does not tip. A bit farther. Then she braces her calf and thigh muscles, and springs out towards the hard stone floor. Her heels strike with a sickening jar which goes right up through her spine, making her gag. Again. If only the kitchen ceiling was higher and she could stand up and jump properly. That would be sure to work better. She crouches again, shuts her eyes and springs. But this time she misjudges it so badly that she falls backwards, striking the knob of her spine against the table-leg and biting her tongue. Once more. I must do it once more, then I'll let myself go and lie down. Her next jump is soft and useless. She despises herself, but she can't bring herself to scramble back on to that table. She limps upstairs and lies down on her bed.
It's not easy to say why suspicion and fear and curiosity and excitement and disbelief should suddenly vanish, leaving nothing but hard certainty. But it's very easy to say when it happened. It was at quarter to five last Friday evening, between peeling the potatoes and dropping them into a panful of cold salted water. Before, when she fetched the potatoes out of the cellar, she was not pregnant. Her _visitor_ was late: but then it had been late before. Eight days; well, perhaps not quite as late before. Fourteen days. It is the shock of John William. I'm quite well and not any fatter. Peer in the mirror: my face is thinner, if anything. Look at my eyes. Seventeen days. A hot, insistent beat of panic behind everything, all day long. Twenty-one days. A surge of confidence. I am stupid to worry. It's perfectly all right. It is simply that I'm going to miss out one month, and then everything will go on as usual.
And then, suddenly, twenty-two days late and I'm peeling the potatoes and I know I'm pregnant. There's no need to think and guess any more. My breasts haven't ever felt like this before, so sore that I wince when I put my clothes on in the mornings. I don't want to eat anything until I've drunk cup after cup of boiling hot weak tea. And the sight of Father's egg makes me leave the table, pretending I've got to fetch something from the scullery or the larder. And most of all the way nothing feels close or real any more. Not Nan, not Father, not Hannah. Even John William is behind me, outside the circle, tapping on my shoulder but I don't turn round to him.
I'm surprised by how cold and practical I feel inside. No one would think it who has seen me jumping off that table, but I know exactly what I am doing. I know it can't work that easily. I have to do it, but I don't have to hope that it will work. If it was so easy to shake a baby out of the womb, there wouldn't be a single girl brought to shame anywhere. They say the Baptists put a black hood on a girl who has fallen, and make her stand before the deacons to be banished from the congregation. But no one is going to touch me. And when I've finished jumping off the table, I'll go up and lie on my bed and curl my arms around myself and shut out all the light, and hold myself so tight that I feel the blood bumping in the big vein in my arm. Bump, bump, bump, bump. The sound of it is fast and perfect, as if my blood knows exactly what it is doing.
I know what I am going to tell them. I will not let them send me away, though Father will want to. He will want to pretend that I am married, one of those hasty war marriages, and send me up to Coyne. My aunt and uncle might even believe him, because they think we have such strange ways down here that it would be quite possible for me to get married with no announcement, no presents and no ceremony.
But I will not go there. I will not creep off with John William's baby like an ant rolling a crumb which is too big for it into its hole. The baby belongs here. It is more Treveal than I am. It is nearly pure Treveal. Nan will see that as soon as she looks at it. She will see and she'll say nothing, just as I'll say nothing. I won't have to explain anything to Nan. She loved John William.
I will tell Hannah, and she will say to Aunt Sarah and Uncle Arthur that John William and I were secretly engaged, and we meant to marry. It happens. Even in the newspaper you can read about war babies now. There are so many they can't hush it up any more. I will say that we were going to marry. Aunt Sarah will scream and snivel, and the uncles will bluster, and Father will look white and thin and appalled, and I will not care for any of them. I will not let them upset me, for now I have to think of my baby. Mine and John William's. It is more important than any of them.
If I have to, I can do anything. Mr Lawrence pays five pounds a year for his cottage. And what do they spend beside that? He earns everything they live on by his writing. Frieda told me she has not a penny of her own. They don't get their money from investments which are always going down. They don't need to keep up appearances: they scrub their own floors, and the doorstep too. It doesn't matter what people see. And they have friends who live the same way: writers and painters. They get little cottages and they do their own work and they get their furniture from places like Benny's Sale Rooms. And they have children too, some of them. One of their friends is going to have a child, and she's writing a book too.
And in the end Father will say nothing. He will want something of John William to live. He will outface them all by not even seeming to notice what they think, just as he must have done over marrying Mother; or if he doesn't, I shall make him. He will never turn me out of this house. 'All those young men dead, leaving nothing behind them.' That is what he said, when he came back from the camp where they killed John William. And let no one dare say to me that they did not kill him; let no one dare say to me that he wanted to die. They made it impossible for him to want to live any more. It may have been his hand that pointed the gun, but who forced it there? They owe John William his child.
I don't think about that night when we were together now. I think about when we were little, before any of this happened, when people meant something different if they said the word 'war'. I think about John William and Hannah and me, and our games, and the way the tide came in and we ran and screamed and kicked down our own sandcastles rather than let the water break them. I think about the way we would cling on to one another when the waves were rough and we swam in the breakers, and then how we'd huddle together when we got out and the wind was cold and we would hop about drying ourselves on my flannel petticoat. We all had to share it, because I was the only one who had flannel petticoats. Nan thought my chest might be weak, like my mother's. And we would giggle when John William's penis went stiff as we dried it, and he would look proud. I wonder if people ever saw us? If they did, they left us alone. We were always hungry. What did Nan give us? Jam sandwiches, and sometimes it was meat-paste for a treat. We would share them out, every bit, then John William would run and get water for us to drink from the well, because he had his tin water-bottle.
So I am not going to Coyne, and I am not going to disappear up to London so that no one will know me, and I will not bring shame on the Coynes and the Treveals. I am staying here. Later, I shall go to Zennor again, to talk to Frieda and Mr Lawrence. I shall write and tell them first. They do what no one approves of, but they survive. They have made a life for themselves. And it is not their fault that John William died. I shall talk to Hannah in the morning, and then I shall tell Father. Let him ask what questions he wants: I know the answers I shall give. And some of the questions I shall never answer.
# Twenty-three
Francis Coyne wakes to the drum of rain out on the roof. His room is dark, with only a trace of grey light leaking over the top of the blind. The gutter rushes with water. He remembers that it needs repairing, and that he had put it off early in spring because he had not enough money to pay the bill. The water will run down the side of the house and damp will spread across the inside wall. He sighs, turns over in bed and swims down into a drowse again. No sound of Clare moving yet. She is getting up late these mornings. She yawns, and rubs bleary eyes, and drinks tea endlessly, and eats nothing. He asks her what is the matter, is she ill, and she says there is nothing wrong. She is just tired. His heart tightens briefly in fear: for years he has watched over her for signs of the consumption which ate away her mother's lungs and turned her into a dry skeleton with red-splashed cheeks who put up her arms to him at the last and asked him to hold her. 'I can't feel myself, Francis. I don't know where I am.'
But Clare is not getting thinner. There are splotches of shadow under her eyes, but her cheeks are rounder if anything, and there is a new fullness under her jaw. Francis Coyne humps the bedclothes tight round him and pulls the counterpane over his ears to block out the scream of the gulls.
He is back at Coyne. He is thirteen, a tall, gangling boy with a tutor who is too old for him. He is growing so fast that his wrists bulge from his cuffs and his knuckles are always skinned and he hunches himself awkwardly when he sits down. He never seems quite to know how much of him there is to be folded away. His features have grown too big for his face and they are topped incongruously by a cloud of fine, soft, downy hair, child's hair still, which will not lie down over his collar no matter how much he stabs at it with his sponge in the mornings. He looks down when he is spoken to and people say he is growing sullen. As soon as his lessons are over, he is off across the fields to shoot in Pesthouse Wood. He shoots rabbits, mostly. His brother Benedict, who inclines towards the Franciscan, shudders when he sees his own Francis come home mud-clotted and blood-streaked, with a couple of bundles of limp fur dangling from a belt the keeper has given him. He skins his rabbits too, quickly and skilfully, but the family will not eat them. The scullery-maid tucks them under her cloak and runs with them down to her mother's cottage in the village. He looks down so he will not have to meet her timid, grateful eyes.
The water-supply at Coyne is erratic: the well does not give enough water for the house, and washing-day is becoming a nightmare. A new well is to be sunk. A man is coming from far away, down in Cornwall, to find the site. There have been arguments about it, and Father McLennan has been consulted in case water-divining should be against the teachings of the Church. But Father McLennan is a practical man, who has to live at Coyne and bath in an inch of muddy water. He will bless the new well: God wishes us to use all His gifts.
The diviner comes from a family of diviners. It has been their everyday job for generations. They travel around Cornwall and Devon, up to Dorset and Somerset, around the outlying farms and homesteads where the well has dried or the spring is tainted. The family's reputation goes before them, and in remote parts their coming is as good as a festival. They marry among themselves, it's said, or else they pick a girl who carries the gift.
Francis does not go to the woods with his gun on the day the diviner is due to arrive. He is more curious than he wants to admit, so he fiddles with cleaning a gun-barrel, then lounges around the paddocks with his bitch Maisie fawning before him. There is a little crowd gathered by the time the diviner's cart arrives. It has a broad blue stripe painted on it, and the pony is a tough little cob which drops its head and starts to tear at the rich grass with its yellow teeth as soon as the cart comes to a halt. Francis does not go forward with the others. He hangs back, watching the diviner. He is not going to gape like Adie there with her mouth open. The diviner is a strongly built man with a tanned, outdoor face. He does not look at all magical. Before he starts his work he has to come into the house and eat and drink: it is part of the ritual. He must eat the food of the house. It's the custom, though very likely it has only grown into a custom because any man is naturally hungry and thirsty after a dawn start and a cross-country journey from the last farm. And he has come a long way, he says. Since June he has been in three counties. The scullery-maid and the garden-boy marvel: they have never been more than eight miles from Coyne.
When the diviner comes out of the kitchen, wiping his mouth after his bread and cheese and pickles and his pint of cider, Francis is still waiting. Maisie whines at his heel, shivering, for she is afraid of the diviner's dog. The little group walks down the drive to the soft, spongy grass at the side of the first paddock, where the washing-lines are hung on Mondays. It is quite hidden from the house. The grass is bright green and succulent here, and in winter it keeps its colour and quakes under your boots. The diviner stops. He walks around, quartering the area of grass, stops again. Seems to listen, to smell the air. He touches the grass briefly, and stands up again, hitching his trousers. He nods to Francis, to the servants, to Benedict, who is watching with his face pursed in disapproval. The girls are there too, fluttering, excited and half afraid. The diviner stumps off to his cart.
Francis thought he would have a wand, a light, flexible twig which would yield to each tiny twitch of the finger so that it could be said to be moving with the flow of the water. He frowns in surprise when he sees the diviner returning with a strong, forked, polished-looking branch. It looks used and ancient. Black sweat from generations of hands has put that sheen on it. It is not the only fork he has, the scullery-maid whispers. He has three others, but he has set his dog to watch over them. And the teeth on the animal – you durst never go near the cart. That band of blue paint is magic too, to watch over his wands.
The diviner shoves his hat farther back on his head, rolls up his sleeves and steps lightly over the grass as if he were walking over the body of some huge animal – a whale, perhaps, cast up and sleeping. He treads his path back over the way he walked before. Then without any show he stops and holds out the rod of hazel, a fork in each fist. Nothing. He steps on. Francis watches, eyes narrowed now, waiting for the trickery. As if an arm has reached up from the ground and grabbed it, the hazel rod jolts down. The muscles in the man's forearms bulge.
'Iss strong here,' he comments.
He is workman-like. It might be the weight of any common thing he is judging. Yet he moves as if he is aware of something infinitely stronger than himself, just under the ground. He moves again, tests. The rod jerks less powerfully. He steps back, presenting the little patch of bright green grass to them all.
'This yer's the place,' he remarks.
Francis finds he has moved to the front of the little crowd. The diviner catches his eye as he straightens with the rod in his hand.
'Like ter feel it?' he asks.
Francis is astonished. He was sure the diviner would keep his magic to himself, but he goes forward.
'Will I be able?'
The diviner ignores the question. He places Francis's hands on the forks of the hazel branch, and turns his body round a little. There is force in the diviner's hands. I wouldn't want to have to fight him, thinks Francis. He angles the boy so that he is standing by the place where the water was strongest. Francis balances the hazel branch nervously.
'Not like that. Yer got ter get a grip on im. The water's strong when it comes. It'll tear that wand out a yer ands.'
For a moment it feels to Francis as if he is trying to turn a key in a lock which doesn't fit. Nothing happens but the grate and click of a false turn. The door remains blank, immovable. Then he grips the rod more tightly and moves forward a little. There is a pulse through the length of it. It twists violently and is nearly pulled out of his hands. He is breathless. He finds he has jumped back in fear and in wonder. He had thought it would be a little twitch, something gossamer, something you could fool yourself about and pretend you had felt when you had not felt anything. Not this real thing, as dangerous as the kick of a horse.
The diviner's weather-seamed face has creaked into a smile.
'Yer felt it, then?'
Francis nods.
'Iss not the water yer feeling – iss what the water's carryin.'
'How far down is it?'
The diviner shrugs. 'Thirty foot. An there's enough water down there ter wash an army clean.'
Francis turns in his bed again. He remembers how the new water tasted, once they had bored the well. It was full of iron, and you could not leave white clothes to soak in it or they would be stained with orange. The taste of metal in it made your mouth pucker, if it was the first thing you drank in the morning. Next day the cart with its blue stripe had gone. The dog looked over the side of it, still showing his teeth.
But still no sound of Clare. What can ail her? Suddenly anxious, he hauls himself out of the bed's warm frowst and ties on his dressing-gown. The house is much too quiet, apart from the lick and spatter of rain. He will make her a cup of tea and take it into her.
He treads lightly along the landing, not wanting to disturb her. But by her door he pauses. A tiny sound. Is it a moan? Is she crying? Or ill, hurt? Her thick door is firmly closed, and he can hear nothing more. But he lingers. Should he leave her be? No, he will just look in and see that she is all right. She has not been herself since her cousin's death.
He opens the door to a pale flood of rainy light. He blinks. The curtains are pulled back, the window is open, and chill air sucks out the curtains as the draught from the door reaches them. Clare turns at the sound of the door opening. She is standing naked by her wash-stand, with the ewer in her hand, pouring a thin stream of water into her bowl. The window squeaks as the wind blows it. That is the moaning sound he heard.
He has not seen his daughter naked since she was a child. Why does he not back out, babble apologies and shut the door? She stands quite still, her level grey eyes finding his. She has ceased to pour out the water. Surely she must be cold, standing there? She is a blur, blue-veined, branched, a spreading map of flesh. Her head looks small because she has twisted up her hair and knotted it tight. Or else it looks small because her body is so vast. It seems to fill the room with its own light. The sight of it goes through him like electricity, like the jolt of the hazel branch in his hand. His daughter's white body is alive with something deep inside it. The heaviness under her jaw, the startling blueness of veins over her breasts, the swelling knot of them around her dark-tipped nipples, the blurring lines of her stomach: his daughter is pregnant.
She continues to meet his eyes without shielding her body. She puts down the ewer.
'Shut the window, Clare, you will get cold,' he says, and goes out of her room.
I will not think. I will not think. I will not try to understand it. There is water underneath us wherever we walk. We walk on the crust of it. We jump and balance. How old she looked. She is older than I ever imagined. My daughter.
Does she know? Does she understand what is happening to her? Is she afraid?
I left her unprotected. We have left all our children unprotected, to scramble their way out of shell-holes, or crawl across no-man's-land with blood in their mouths.
When I looked at her naked body, I was afraid that she would arouse me. But she did not. I felt something immense for her, something deep inside me, alive. I could not move for pity of her.
Her cousin killed himself. He blew his head off because this world had become a place of horror to him. That must never happen to Clare.
# Twenty-four
Three girls walk gently on the cliff-path. There is not room for them all to walk abreast, so two walk ahead and one a little behind. They are going so slowly that they can still talk quite comfortably between the three of them. Sometimes the red-haired and the fair-haired girls form a couple, sometimes the red-head and the dark girl. They talk over their shoulders to the one behind, but their talk is quiet, as if they do not want to disturb the peace they are walking through. At the moment Peggy is ahead, scuffing up dry earth into dust with her light canvas boots. For the whole summer has been hot, and even now in early September the sun is low but warm. There are vipers on these hills. More have been seen this summer than ever before, and they have been brought into the St Ives pubs hanging crushed and limp from the end of sticks. Clare walks on the inside, carefully, for here the path narrows and the cliff falls away sheer only a foot or so from their feet. She has seen no adders, but there are butterflies everywhere. She has never seen so many butterflies sunning themselves on this path. Their wings flirt up at the vibration of the girls' boots. They tilt off into the hazy sunshine, over a drop of a hundred feet to the water, or brushing among yellow-tipped bracken. Far below, bladder wrack and oar-weed stir with the tide in water which is so clear you cannot tell its depth.
Peggy calls back, 'Are you all right, Clare? Are we walking too fast for you?'
'No, I'm fine,' says Clare. It is too much to hope that Peggy's rather greedy sympathy can be satisfied without constant references to Clare's pregnancy. She wants every detail. Sickness, tightness of clothes, names for the child. Above all Peggy wants to ask the unanswerable question: 'But what does it feel like, Clare? What is it really like?' Clare can only shrug and say that she doesn't feel so very different, not yet. The baby hasn't quickened, but Clare is beginning to show now, even though Nan has altered her waistbands cleverly, letting in a panel of matching fabric and taking out tucks. This skirt will do her for another couple of months.
'Where are we going, anyway?' asks Hannah, but no one knows. Just as far as they can, and then perhaps they'll pick blackberries on their way back, on the open slope of rough land which catches the sun most of the day and always ripens the sweetest blackberries. But the blackberries are small this year because there has been so little rain.
In Flanders the struggle for the Passchendaele Ridge continues. The poppy-blowing fields are ploughed by German and English guns, and sown with a litter of lost equipment, a seeding of blood and bone. Soon it will be autumn there too, and heavy northern rains will fall. Men will be listed _missing, presumed drowned_ – a new classification for the lists in the newspaper. They are presumed drowned in the mud in which they live and often die. The men who came 'right away to Blighty' with John William will return to Flanders with their new commissions soon. Their training lasts only three months, and then they are wanted back at the Front. Hammond will die on a mission described to him by a senior officer as 'rather a tricky bit of patrol-work'. His body will not be found. Simcox, a dozen feet to the left of him, will survive.
'Clare, I really think you ought to rest for a while. Let's all sit down here.'
Peggy's sharp blue eyes never seem to leave Clare. The girls sit on hummocks of grass and look out to sea, idly at first, and then they watch another patrol-boat coming close in below them. Patrols have been stepped up again.
'U-boats about,' says Peggy knowledgeably.
Hannah sighs in exasperation. Peggy is always the same. If she tells you that the sun's shining, she manages to make it sound like a piece of privileged information.
'Aren't they always?'
Hannah is a little thinner, but nothing can diminish her, thinks Clare, looking at her cousin's profile. Hannah's broad brow is exposed by her new way of dressing her hair, without a parting and drawn back. Her lower lip is caught in by two of her bottom teeth. These days Hannah looks puzzled when her face is in repose. The slight line she has always had between her eyebrows has deepened. She has not heard from Sam, but then she couldn't expect to, Clare reassures her. It would be dangerous for him. Sam. They are looking for him: they have even visited Hannah's house, to ask questions, thinking perhaps that she was concealing him in the wash-house, says Hannah scornfully. For where is there to hide in a place like St Ives, where everybody knows everyone else's business? How would they feed him, even, without the shop-people noting the extra quarter of a pound of sago, the extra quarter pound of butter?
'You got a visitor, then, Hannah?'
'No one need know. After all, you might get food from your uncle's farm. You Treveals could be feeding an army,' suggests Peggy helpfully.
But there are no more secrets. Hannah is not hiding Sam romantically in wash-house or sea-cove. Sam remains in London; in Wapping to be exact, where he has melted into a shadowy city-life which accepts what comes and asks no questions. They know; maybe. People in cities are as quick-guessing and inquisitive as people anywhere, but Sam's girl is the daughter and sister of prize-fighters, and Sam is her best boy, and as such he is to be left alone. So what if he lounges in the courtyard's one scrap of dirty sunlight, yawning and scratching himself, never doing a stroke of work while his girl pounds clothes at the laundry where she works a ten-hour day for 2/7d? She would be better off in munitions, but her father has sworn no girl of his will ever get a yellow face as a munitionette, no matter what. And she goes along with it all, placidly. She likes the laundry: there is a bit of life there with the other girls, and one day, after the war, when Sam can work, things will be different... For she is not a prostitute, as John William supposed, and as Clare and Hannah now believe. She is not a girl who has picked up Sam lightly out of the whirlpool of war-time London, when girls have money and men have leave and desperation. She is not one of the girls the Bishop of London thundered against; she is not even particularly pretty. But she has a room, and in that room she has Sam.
Clare's body is rounding daily, but it does not matter now. Everyone knows. Gossip is like a furze fire: it burns hard, then the wind changes and it flares off elsewhere, racing and crackling. The story of Red-beard and Clare Coyne has died down, though it is not quite dead yet. Some red embers sleep in the white ash, but the flame and the fire are on another hillside now, eagerly consuming the story of Clare Coyne and her dead cousin.
Nan has a way of putting things which is almost magically effective in silencing vicious tongues before they can begin to cut. Nan knows exactly what gossip she wants to seed. She must have thought hard and well in the long night after Francis Coyne came down to her cottage and talked to her alone. She must have known that Treveal secrecy would not serve the family now. The Treveals must bend if Clare was not to be broken; or they must appear to bend. Nan knew in whose ear to drop her flattering confidences, in the sure knowledge that they would be passed on with an insider's pride in her source of information. Soon everyone in St Ives would know that Clare Coyne had had an understanding with her dead cousin, and that the two would have married on his next leave if he had not died for his country. The Treveals had kept it close, but didn't they always? But now secrecy was beyond them. They have broken the habit of generations and made the town privy to their family secrets.
'They do keep themselves to themselves, the Treveals, but Mrs Treveal said to me that at times like this you have to go outside the family – and since we're old neighbours she came to me...'
Clare is drawn into the curious embrace of the town as her story passes like a delicious spice from lip to lip. She is one of their own again, shared among them. And wasn't John William a soldier, going for an officer? The mothers and fathers of exempted sons know better than to open their mouths against John William and Clare Coyne: And it has to be admitted that these things do happen. How many young people of the town anticipate their weddings, and then draw a quick veil of respectability over a too-short pregnancy? And are we to blame Clare more than all those girls, just because her young man died before he could marry her? And he would have married her. Everyone knows the Treveal closeness. They would never bring shame on one of their own. Whatever you may think of first cousins marrying, there is no doubt that a wedding would have taken place.
Nan watches, bird-sharp and apparently indifferent. Nan listens with satisfaction to the coupling of her grandchildren's names. The breath of open scandal has licked at Clare, and receded. She is safe for now.
Even the Church has hesitated to condemn Clare. That first Sunday after her pregnancy became talked of in the town, she came up with her father to first Mass. She had been seen going to confession the evening before. Her father gave her his arm, his face impassive as the two of them walked into the church. Now his long-cultivated distance and indifference worked in his favour. There had never been such a show of style in St Ives. The daughter did not blush or look aside from the devouring stares; the father bowed slightly to one or two acquaintances, and showed no consciousness of any change in his own position. Almost against their will, his fellow-parishioners have found themselves treating Clare more as a young widow to be commiserated with, than as a fornicatress who is likely to burn in Hell if she should happen not to survive childbirth. And fortunately the priest who heard Clare's confession is likely to remain silent upon the subject. A little of her father's indifference has rubbed off on Clare. Shielded by the child, she does not care so very much what people think of her now. As long as they believe that John William loved her, she does not care how sinful they think she is. And since she does not care, they find themselves chatting to her for a few moments in the porch, on ordinary topics, quite as usual. And then the chance to scorn Clare Coyne has gone, if they ever wanted it. Everywhere there is news of death, and more death. Perhaps that is why the news of a birth does not seem so terrible.
'It just goes to show,' says the less charitable Methodists of the town. 'How those Romans do think nothing of their sins.'
But they do not say it very loud, for Nan has vanquished them. One or two strong Chapel women are even to be seen knitting for the little Coyne baby. Their fingers, hypnotized by Nan, click over their white-woolled needles.
Peggy looks down at the sea nostalgically.
'No more swimming this year,' she says. 'Remember that swim we had at Hayle, Clarey? May, wasn't it? Ooh! I could've died of the cold!' The other two exchange glances and smile.
'I made sure I'd catch pneumonia and die,' continues Peggy, then stops abruptly, for it is not right to talk about death with Clare in her present condition. 'But I didn't,' she adds lamely.
'Scarcely surprising, Peg, considering you were never in the water,' remarks Hannah.
'Yes; I remember it,' says Clare, answering not Hannah but some train of thought of her own. And next year, she thinks, the baby will be three months old. Old enough to have his toes dipped in the water.
'How old were we when we started to swim?' she asks Hannah, who shrugs.
'I don't know. We were always swimming. Johnnie – ' She's said his name. She doesn't often. 'Johnnie was the one for swimming.'
'Yes.'
A beach. Cold footprints in cold incoming sea. Bare skin tingling with sunburn, but beginning to chill.
'Quick, get your clothes on.'
'I can't find my drawers!'
'Why, there they are, Clarey. You weren't looking.'
'Where's John William?'
'He's still out there. Can't you see him? Look. Out past the rocks. He's gone too far out again. Johnnie! Johnnie! You shout too, Clarey. Nan'll kill me if we're not back in time.'
'John Will-i-am! John Will-i-am!'
He turns, his black head sleek with water. He turns and dives to tease them. The water grows smooth over his head. He's gone.
Francis Coyne watches from the window of Clare's bedroom as the girls set off on their walk. He has the afternoon sun in his eyes, but he can still see the three figures distinctly: Hannah, Peggy and his Clare between them. The slight breeze blows back their skirts over their legs. Three young girls against the softly glowing sea. They look mysterious now, the way people you love look when they walk away from you and disappear into a landscape, their colours melting into pleats of light and shadow.
He sighs, and turns his back to the window, and cracks his finger-joints in the way he is trying not to do when Clare is with him. It's a new habit, and one which maddens her. He goes down to his study, which is dim and cool. There is no work on his desk. He has not touched his book for weeks; he has had other things to think of. Money. They must have more money. He knows now that prices will never come down, not even after the war is over, and his investments will never right themselves. He cracks his knuckles and thinks of money and how he can get it. Clare is quite sure. Her confidence astonishes him. Sometimes she frightens him, for she seems so changed. She goes here and there and she makes inquiries and she comes home and tells him what she has done. She is going to take up her portrait-sketching seriously, and sell sketches of summer visitors. She is sure they will pay a good price, especially for sketches of their children. She has been to talk to a lady at the St Ives Arts Club who liked Clare's drawings and thought her plans perfectly possible. She is going to give Clare some introductions. She has heard Clare's story, though she does not tell her this, and thinks it is romantic. Very Cornish, she thinks. And she finds it quite moving, she tells her friends from London, the way the people of the town accept Clare and look after her. She is one of their own really. You would not find _that_ spirit in Lewisham, she laughs.
Francis will continue to sell whatever he has of value, though there is not much left. He will swallow his pride – or that's what he'll call it, though he will neither feel nor seem in the least humble – and he will ask his brother for money. He has never asked for himself, but he will ask for Clare. He will give Benedict and Marie-Thérèse to understand that she had actually married her cousin, and it will suit them very well to accept what he tells them.
And no one will ever know the truth. He is amazed at Clare's silence and how she has kept it up. She has never so much as hinted at where the real blame lies. She sticks to her story, and it is a good one since the only man who might have contradicted it is dead. She astonishes him. He had thought a girl in her position would melt with shame and grief and sob out her accusations against the man who had seduced and then betrayed her. For she has been betrayed. What story can that man have told her? He must have taken her by surprise, in her innocence. She will not tell anyone, and it seems she is too proud even to tell her father, who would defend her against the whole world, he feels now with a rush of emotion he has not known since he took her away from the Treveals after her mother's funeral. She is his Clare, his daughter, part of him. And they are so alike that she cannot keep the truth from him. He knows that her cousin's part in this baby is nothing but a convenient fiction, but he does not blame her for using it. It is not such a slur on John William's memory. There's a sense of gladness in the Treveals, even, now that they are used to the idea.
He had gone to the cottage that night in desperation, ready to confide all his fears in Nan, but he did not need to. When he met her blue gaze, even he could believe for a moment that the baby could only be John William's. Her eyes presented no other possibility. Yet in the depths of their directness secrets swam. He could never fathom Clare's grandmother, but they could work together. He has watched with admiration the skill with which Nan has handled the gossips of St Ives and has helped Francis Coyne to protect his child.
Clare's cousin is buried now and can never deny the child she is fathering on him. Surely John William would want to help Clare? They were always close. It cannot really harm him to be thought the father of her child. Clare has put her child first, not her own sense of betrayal or her desire for revenge. She must desire it. She is his daughter after all. But however bitter it feels he must not punish his daughter's seducer, if by doing so he exposes Clare. There'd be no place for her in the town once the truth was known.
There was gossip in the town and he heard it. It flared up and he saw the danger, but with Nan's help it has died down. He has seen idle malice in street-corner faces as he passed them, but he thinks Clare never knew of it. She never knew that she had been seen with that man on the cliffs. That she had been seen talking alone with him in his cottage garden. That he had called her across the concert room to him, and she had gone, dragging her cousin with her. That they were watched as they laughed and lolled on gravestones in Zennor churchyard. All this people saw, and her father knows. They say Lawrence visited her in her father's house, in her father's absence.
The voices have been silenced and the ash of gossip looks cool. But can Lawrence be trusted to keep quiet? A man who writes about the most sacred and intimate aspects of life as if he were a dog in the gutter, the Rector told Francis. What if Lawrence should take it into his head to notice the gossip and speak of Clare? What if the Rector were to say what he saw and what he knows? So far the rumours have done nothing but deepen local hatred of the Lawrences, but fire can spurt up and run with the wind again, Francis knows.
Francis knows the truth. He has seen Lawrence with his own eyes, elbows on the table in the Coyne kitchen, familiar. The man must have known her father was away and Clare was without protection. He had seized the opportunity of John William's death. In Francis Coyne's house he sat alone with Clare. God knows how long he had been there, or what he had done to her. And then he got up and went out with his red beard jutting in front of him. Now everything is clear. Clare's weeping, her anger, her wild ideas. Everything that had puzzled her father. All of it had come from that man. He would have made her hate her own father if he could, and her home, and those who loved her.
The gossip has only died down because a better rumour killed it: Clare Coyne carrying her cousin's child. No one couples Red-beard's name with his daughter's now. Perhaps they think there was nothing in it.
Thank God, thinks Francis Coyne, that my daughter has red hair. There will be nothing remarkable in her having a red-haired child. There will be no link with him.
He is a little in awe of Clare's silence. Sometimes he wishes he could go to her and tell her that he knows, he understands, he will punish the guilty. She was too innocent. She had never been anywhere, and she knew nothing. For a man of that type she must have been easy prey. Anger against Lawrence swells in him again.
Anger against Lawrence and his German wife swells like a tide through the countryside, filling hungry lips. Now, with the third battle of Ypres floundering into mud, the couple is an offence in its very existence. The Lawrences must be watched. It is an honourable duty to spy upon them. They are not people like us. They have no place here. Young military police visit them in groups of three or four to keep each other from bamboozlement and contamination. Reports are gathered and sent to Southern Command in Salisbury. Military men, official men, take a serious interest in this case. It is swelling like a boil upon the body of this suffering country, on the fragile and exposed flank of England.
Hundreds of thousands of men have died, thinks Francis Coyne. They've died hearing the sound of the guns, like John William.
'Can't you hear em? Those 5.9s? Don't tell me you can't hear em?'
He would have died hearing that surf-noise of guns, wilder than any storm we have here and more pitiless. Surely he shot himself because he could not get the noise of the guns out of his head.
Francis Coyne thinks of the football field and strong boots battering the turf. Men who had come through two years of the war, like John William. And this man, with his beard waggling and his eyes staring at you so that you felt you could not bear to look at him, sat talking to his daughter in the kitchen while John William lay in the mortuary with half his head blown off.
Such men are dangerous. They are injurious to our peace. They do not belong here. Whatever wind blew those two to Zennor, let it blow them away again.
Once they are gone, thinks Francis, we shall be safe. Red-beard will be forgotten and no one will measure his features against the child's.
Francis Coyne picks up his pen. He cannot go to this man and tell him he has injured his daughter, but the war has given him another way. She will be walking by the sea now. She mustn't tire herself. But Hannah is a sensible girl, and she will make Clare rest. They will not be back for an hour at least.
He has a sheet of his best cream-laid paper in front of him. Nearly the last of the stock, and he won't be able to buy any more. No more cream-laid paper; no more May Foage. He puts the paper up to his lips and tastes its cool smoothness. But he has Clare, and he will have her child. It will be Clare's child only, like its mother with its red hair. That other will be blotted out. He is going to disappear and it will be as if he had never been. For the first time in all these years Francis Coyne feels a tremor of Cornishness in him as he thinks of the landscape around Zennor empty of Red-beard and his German wife. A Prussian, they say she is. Poor brute, one might even feel sorry for him.
He dips his pen, and begins to write.
Dear Sir,
As a resident of St Ives, and a patriotic subject of his Majesty the King, I feel it is my unpleasant duty to inform you in strictest confidence...
He grimaces fastidiously. The language is distressingly crude. It curdles as he writes it. But he continues, and the long elegant lines of his black writing unreel across the surface of the paper to the luxurious scratching of his nib. He scarcely has to think. He signs the letter, blots it, folds it into an envelope. He leans his fist on the envelope, presses down with all his weight. Yes, the thing is abominable. But he believes it may work.
# Twenty-five
A cool, still October morning in Zennor. It is early, not quite yet eight o'clock. Four men turn off the high road which leads from St Ives to Zennor, and tramp down the farm lane towards Higher Tregerthen. Pods of Himalayan balsam burst with sudden explosions as the men brush past them. They know their way; they have been here before. They know every inch of the Lawrences' bare little cottage with its bits of fabric draped over furniture and its pink-washed walls. They have flipped over Frieda's embroideries and left them lying skew-whiffed. They would not be caught dead living in a place like this themselves. They are contemptuous of the Lawrences' meagre possessions. And yet this Lawrence is an educated man – surely it only confirms the suspicion that he must have some hidden motive for living in such an out-of-the-way place, high above the coast where a light can be seen for miles out to sea? They have rifled the cottage once, and it will open its doors to them easily this time. The man is as frail as a dandelion puff, so there is no fear of a fight there. And if need be they can put the fear of God into the German woman. A few words about the risk of internment should silence her.
If the cottage ever had that virginity of lostness and secrecy which Lawrence once thought it possessed, it is gone now. The red floor is printed over with clumsy bootmarks from yesterday's search. The searchers did not care what traces they left. They wanted the Lawrences to know that their lives had been stripped bare and pawed over. Drawers have been pulled open, small belongings tipped out and searched. Letters and manuscripts have been taken.
The Lawrences were not at home when the men came yesterday. The first search is over, and nothing that follows it can shock them as sharply. Frieda came home humming to herself, pushed her door open absently, thinking of something else, and found her home broken open like an egg.
But this morning they are in their cottage when the army officer, the two detectives, and the police sergeant from St Ives knock on their door. Frieda has got up first, and she is drinking her tea standing up by the fireplace. She does not want to sit down in any of the chairs, or touch her own things. She shrinks from them as if they have been contaminated. Even her clothes smell wrong, as her children do now on the rare occasions when she is able to see them and take them into her arms. They smell of the other people who have taken possession of them. Last night she could scarcely sleep in her bed for the knowledge that her sheets and blankets had been switched back by the searchers as they made their way quietly and methodically through the cottage. There was a small stain of blood on her undersheet. They would have seen it. But she will not feel humiliated: she _will not_. It is for them to feel shame, after what they have done. Let them tell her the cause for what they have done, if they dare.
By eleven o'clock the second search is over and the men have gone. They have left behind the smear of the oldest message in the world: 'We can do what we like to you, and you cannot stop us, for we have the power and you are powerless.' She has writhed and wept under the weight of it, and her own helplessness, but now she is silent. In the solicitor's office she learned the cost of leaving her husband for Lawrence. Now she learns that the cost of living with him still has to be paid, over and over. She will pay it.
Lawrence fingers the official notice they have left, as if he would like to tear it up. It orders them to leave Cornwall by Monday, to live subsequently within an unprohibited area, and to report to the police. Its curt, chopped words bristle off the page at him. He smiles slightly, involuntarily, as the style of it reaches him. He could write it so much better himself.
Yet he is white with shock. For two hours the four men have thrashed through the cottage, touching everything, emptying pails and buckets, opening cupboards, feeling in jacket pockets. They have found nothing new – the cottage is so small that they can have missed nothing during their first search yesterday. But they needed to hunt and paw again in front of the Lawrences. They needed to make the couple watch while they read letters, combed through address books, opened underclothes-drawers. For where is the point of a demonstration of power if there is no one there to be crushed by it?
Well, they have had their satisfaction. Frieda's face is snail-tracked with tears. He wishes she had not wept in front of them. He will not let his own rage rise in him yet. If he does, it will shake him apart.
He looks down at the pile of papers smashed into a heap on the rosewood table. These are the ones the officials discarded as not worth bothering about. They had found so little to take, really, for all their frowning and leafing – a few letters from Frieda's mother, some notes.
'Shameless little article, wasn't he – the ferrety one?' he remarks.
She smiles faintly. 'Lorenzo, you know he was not at all like a ferret. A bit more like...' She thinks. 'A bit more like a warthog!'
'Blessed if I'd know one of 'em if I saw 'em. But let him be one, if it pleases you.'
She bends down to pick up a piece of red and gentian embroidery which has slithered off the back of the table. He riffles through the notes and letters.
There is another knock at the door. Frieda jumps, looks at Lawrence with suddenly darkened eyes, then goes to open it in response to his almost imperceptible nod. There stands one of the little boys Frieda knows to smile to, a farm-worker's child. He is out of breath, puffing with excitement and importance. But, on seeing Frieda, he suddenly turns shy, and thrusts out his hand with a bit of paper in it.
'Why, what's that?' asks Lawrence.
Frieda looks at it. In pencil, on the outside of the letter, someone has written in neat, ticked-off letters: 'Am returning this as a private communication.'
'Must be one of our letters,' says Lawrence. 'What's he up to, sending it back?' And he frowns, suspecting another trap. 'Did an army man give this to you?' he asks the child, but the boy backs out of the doorway, giving only a sudden, violent nod before he pelts off down the lane.
'What can he mean, a "private communication"?' puzzles Frieda. 'He has only just taken it from my drawer. He must know that it is private! He did not let that stop him before. They have taken all my mother's letters.'
Lawrence takes the letter from her and unfolds the paper. The large, strongly curved, slightly childish lines are instantly familiar. They have read over this letter many times, and discussed it late into the night. But they have not yet answered it, for they are waiting for replies from friends in London who may be interested in Clare's work. The letter is from Clare, and it is addressed to Lawrence:
There is something I must tell you which you may have heard already in Zennor or in St Ives. We both know how news travels here. And how lies travel too. So I thought I would write to you and tell you that this time it is not lies, or gossip, but the truth. I am expecting a child. I shall continue to live with my father, and I am hoping to make my living as you do, by my own work. Already I have met a woman in a gallery who thinks she can sell my things. I draw every day – I should like to show you. My own drawings, not flowers any more. I enclose a sketch so that you can see. Perhaps, if you have any friends who would like to have their portraits drawn, you might mention me to them? I believe you liked my work.
Perhaps I should not write to you and tell you this. But I had never met anyone like you before. And you knew John William.
Tell Frieda I would like to see her, if she still wants to see me. I want to draw her properly: she is beautiful.
Yours truly
Clare Coyne
He pinches the stilted little letter along its crease. Stilted, but with a rush of confidence in the middle of it: 'My own drawings, not flowers any more.'
'Did they take the sketch?' he asks.
'No, I have it. I put it into the frame, behind the one she drew of you, so that it would not crease. You had not fastened the back of the frame.'
Lawrence lifts the picture off the wall, removes the back of the frame, eases out the two drawings.
'Be careful! They are stuck together – they will tear,' says Frieda.
'It's the damp,' says Lawrence, unpeeling the top sketch with skilful fingers. 'There's no damage. Lucky it's not charcoal.'
He spreads the two portraits on the table, side by side. The two faces look at one another. 'David Herbert Lawrence, by Clare Coyne.' 'Francis Coyne, by Clare Coyne.'
'The portrait of her father is very good,' observes Frieda.
High forehead, finely moulded temples, dark receding hair. The eyes are deeply set and the eyelids curve down, half hiding the eyes. Shy, or remote? But this face has its own fierceness. It would not be entirely out of place in the von Richthofen album, thinks Frieda. The hooded eyes stare at the portrait of her husband, but they seem not to see it or to take note of it.
The portrait of Lawrence is less confident. It is only a few months since Clare drew it, but how much she has improved since then, to judge by the recent sketch of her father. Frieda bends over her husband's pencilled face. The eyes meet hers. Their startling directness is a promise which can surely never quite be fulfilled? But he is not really looking at her. This is how he looked at Clare as she drew him. The rest of the face is not quite as good, though she has caught a likeness. Who could ever capture that vitality? As soon as you think you have grasped it, it moves on. Frieda looks up from the drawing to the living man.
'We have not answered her letter,' says Lawrence.
'No,' laments Frieda, as if this is the last, unbearable disappointment of these bitter two days. 'And now we shall not see her. We shall never see her again.'
But Lawrence rereads Clare's letter, his face alight with amused sympathy.
'She will be all right,' he says. 'See, even now she thinks ahead. She thinks that we may be useful to her. And we can write to her from London. She might come to see us there – you know she told me she had visited London before. I am sure I shall hear something from Lady Cynthia. She has so many rich friends – some of 'em _must_ want their portraits done. Or their children's – such women always "adore" their children.'
Frieda pulls a quick face at Lady Cynthia's name.
'Let her help _you_ first!' she exclaims. 'You have so many friends, yet you cannot get your own books published!'
It is too true. He will have to pack up thick wads of the novel which he may never publish. And yet it is good: he knows it. But they will stop this new novel too. They will put their censor's boot down on it, just as they did on _The Rainbow_. He looks around the room. It is empty for him now. They have two days to pack up their things and get out of Cornwall. But he has not the heart for preserving this small temporary home so that they can start again and re-create it in yet another cheap cottage found for them by friends. Perhaps one day they will come back to the only home they've found and paid for themselves. The war cannot go on for ever. Let the piano and the rosewood table stay here, and Frieda's bits and pieces. There are onions and leeks left in the soil, and winter cabbages. Let them stand there until the salt or the frost rots them. Or the Hockings will come up from the farm and harvest them. They will be sorry to lose his company, the stories in the kitchen, Stanley's French lessons and piano lessons, the jokes, the political discussions while they picked over the peas or skinned pickling onions. But regret will not stop William Henry from eating the vegetables Lawrence has planted. His dark, wary face is the face of a survivor whose family has held on to their land through centuries of storms. And why not? So much waste, everywhere. Why shouldn't something be saved?
The marigolds will seed themselves and the white foxglove will come out again at the bottom of the wall, like a ghost in the summer light, and there will be fine flowers on the hydrangea. He has been feeding the soil around it with spent tea-leaves all year. How many times he's stepped out into the earth-scented, pearl-coloured evenings and tipped the tea-pot out around its roots. It takes time to build up the soil. Stanley carted up five loads of manure from the farm midden, for the vegetable gardens.
But he will not think of that – not now. It would drive him mad. This is the place where he once thought he would build his ideal community: his Rananim. There will never be Rananim at Zennor now. He thinks of the Murrys, leaving a few weeks after they arrived. He had such blazing hopes for life together here, the four of them. And yet – he smiles without meaning to, without knowing he was going to smile. How funny they had looked, Jack and Katherine packing up their cart with their things, Katherine so meticulous, so dark and neat, with her face set mute and purposeful towards the south, away from the Lawrences. There was no changing her mind. And now he and Frieda are to be kicked out of Cornwall. He would like to shake off this country entirely and go westward to America, but the authorities will not let him. They will not endorse his passport. Again, they give the excuse of war.
Now he and Frieda must go. Another cart full of furniture, another couple of bedraggled wanderers balancing perilously on top of their possessions. Back up the spine of Cornwall towards the seething madness of London with its newsboys in the streets, its Zeppelin raids, its crowds of feverish soldiers on leave spilling out of the pubs and theatres. If only they would all kneel down on the London pavements as John William had knelt on the white Zennor road, and cry out for the dead men who were walking at their shoulders.
'Perhaps Clare will bring her baby here, when we are gone,' says Frieda suddenly. 'We have paid the rent until next summer. It is good for babies to sunbathe. She could put him in the garden to kick in the sun on a blanket. The air is so pure here. I have never tasted air like it.'
She has gone into the scullery and is wrapping cups in newspaper. He glances at her face quickly, suspiciously. Is she thinking of her own children again? But no, that look of mingled grief and inwardness is absent.
'Not she,' says Lawrence. 'What would there be for Clare in Zennor? She has her own life.'
'And her own place,' says Frieda. She has moved back to the main room and is folding up the table-cloth. And he, usually so deft and handy, does nothing to help her. He just stands there.
'I used to think, when we got here,' continues Frieda. 'When we looked out of our window on a clear night, it was like a door into heaven. So many stars! Nothing but stars between us and the sea.'
'Stars!' he exclaims. 'I'm glad you think so well of 'em, for they're all we've got, or likely to have.'
He will not think now. His skin prickles, receiving the impression of the tumbled room, his tear-streaked wife packing away their linen, the cold smell of autumn air seeping in through the door neither of them has bothered to close, for there's no safety or privacy left here to guard. Soon the Hockings will come up from the farm, wondering, knowing, guessing. Stanley will help to pile their books into boxes with his big clumsy hands. And William Henry will want to talk. He will want to tease out what has happened in his slow, taunting Cornish voice.
He must not think, he must keep still and let it all happen to him without resisting it. Clare is going to have a child. He thinks of her, ruffled and sweaty with wind and sun, drawing his portrait on the edge of the cliff. How she frowned as she concentrated. Her tense little fox-face was not pretty at all, and she did not care. She thought of nothing but her drawing. He smiles.
'It is good that he left a child,' says Frieda, kneeling to roll up a rug.
'What?'
'That young man. Her cousin. It is good that he left something behind him that she will love.'
'Oh – love,' he says, tasting the word as if it comes from a foreign language. 'Yes, I daresay she'll do that.'
They look at one another, then she stoops and packs on, indefatigable. Her crown of rich rough hair glistens, though the day is sunless. She has rolled up her sleeves for work and there are shadows in the creamy hollows inside her elbows. She has stripped off her rings, and her bare fingers move confidently, filling up space in the boxes. He watches her for a while, his face smoothing, relaxing. Suddenly it sharpens, attentive as the muzzle of a fox scenting down the wind. The scent blows sharp, then thins to nothing, leaving an itch against his senses. The packing-cases vanish; Frieda blurs to gold. He goes to the table, shoves the muddle of their things aside, pulls paper and pen towards him, and begins to write.
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Birkalands välfärdsområde (finska: Pirkanmaan hyvinvointialue) är ett av de 21 välfärdsområdena i Finland. Välfärdsområdet grundades som en del av reformen som berör social- och hälsovården och räddningsväsendet i Finland, och det omfattar samma område som landskapet Birkaland. Birkalands välfärdsområde är befolkningsmässigt ett av Finlands största välfärdsområden.
Kommuner
Birkalands välfärdsområde består av 23 kommuner varav 12 är städer.
Ackas stad
Birkala kommun
Ikalis stad
Juupajoki kommun
Kangasala stad
Kihniö kommun
Kuhmois kommun
Lembois kommun
Mänttä-Filpula stad
Nokia stad
Orivesi stad
Parkano stad
Pungalaitio kommun
Pälkäne kommun
Ruovesi kommun
Sastamala stad
Tammerfors stad
Tavastkyro kommun
Urdiala kommun
Valkeakoski stad
Vesilax kommun
Virdois stad
Ylöjärvi stad
I april 2022 fanns det invånare i området.
Tjänster
Från och med 1 januari 2023 överförs ansvaret för social- och hälsovårdstjänster och räddningsväsende från kommuner och samkommuner till välfärdsområdena. Enligt lag ska kommuners och samkommuners gällande avtal överföras till välfärdsområden.
Sjukvård
Alla Birkalands kommuner förutom Pungalaitio tillhör Birkalands sjukvårdsdistrikt. Områdets centralsjukhus är Tammerfors centralsjukhus inom vilket också Tammerfors universitetssjukhus är verksamt. Områdets andra sjukhus är Hatanpää sjukhus, Pitkäniemi sjukhus, Sastamala sjukhus och Valkeakoski sjukhus. Specialsjukvård tillhandshålls av Tammerfors universitetssjukhus.
Räddningsverk
Birkalands räddningsverk är verksamt i området.
Beslutsfattande
Välfärdsområdesvalet
Vid välfärdsområdesval utses välfärdsområdesfullmäktige för välfärdsområdena, som ansvarar för ordnandet av social- och hälsovården och räddningsväsendet från och med den 1 januari 2023. Välfärdsområdena har självstyre och den högsta beslutanderätten utövas av välfärdsområdesfullmäktige. Välfärdsområdesval förrättas samtidigt med kommunalval.
Det första välfärdsområdesvalet hölls den 23 januari 2022. Då valdes 79 personer till välfärdsområdesfullmäktige.
Välfärdsområdesfullmäktige
Välfärdsområdesfullmäktige ansvarar för välfärdsområdets verksamhet och ekonomi. Fullmäktige fattar beslut om årsbudgeten, godkänner bokslutet och ansvarar för strategiska linjer.
Partier
Partier och antalet platser i fullmäktige (valet 2022):
Källor
Birkaland
Välfärdsområden i Finland | {
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1298 () fue un año común comenzado en miércoles del calendario juliano.
Acontecimientos
2 de julio: Batalla de Falkirk, por la independencia de Escocia.
1 de diciembre: en Poggio Bustone, Spoleto, Rieti (Umbría, Italia) a la madrugada (o en la noche del 30 de noviembre) se registra un terremoto de 6,6 grados en la escala sismológica de Richter (intensidad de 9-10), que deja un saldo de «muchos» muertos.
Se produce el sitio de Siracusa.
Nacimientos
12 de diciembre: Alberto II, duque austriaco.
Fallecimientos
2 de julio: Adolfo de Nassau, rey alemán.
Enlaces externos | {
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{"url":"http:\/\/stackoverflow.com\/questions\/6417092\/svn-switch-what-does-relocate-do","text":"# svn switch , what does --relocate do?\n\nI tried reading the man page on this a few times but fail to comprehend clearly.\n\nWhen one does svn switch <source> <target>, I understand that it switches the working directory to target path.\n\nI would like to understand what --relocate is used for. I thought this was only used when the url of repository had changed and one wanted to update working copy url to point to new svn server location. But it seems Relocate can only change the repository part of an URL.\n\nSo my question is, what does the --relocate switch do? In particular, can it be used to integrate to user branch a vendor branch that has many differences since previous version, ie\n\nsvn co path\\to\\vendor\\lib\n\nsvn --relocate lib url\\path\\to\\my\\branch\n\nsvn commit -m \"Updated my branch with vendor code without using cp, this way _I think_ enables deleting folders and files that don't exist in new version of vendor lib.\n\nAm I getting any of this right?\n\nThanks SO.\n\n-\n\nSVN Relocate is for when those pesky admins move the path to your repository.\n\nFor example original:\n\nhttp:\/\/mydomain.com\/svn\/repo\/Project\/MyApplication\/trunk\n\nAdmin gets bored and changes it to:\n\nhttp:\/\/mydomain.com\/svn\/MyApplication\/trunk\n\nYou would need to do svn relocate to tell svn that the path to your project has changed.\n\nAs for your provided example it can not be used to reintegrate branches, the command is to let the svn client update all of its paths to the new location of your repository. You need to svn merge them.\n\n-\nYes, but is the usage scenario in my example valid? \u2013\u00a0 stefgosselin Jun 20 '11 at 20:36\nDoes my last edit answer your question ? \u2013\u00a0 Nix Jun 20 '11 at 20:40\n\nI thought this was only used when the url of repository had changed and one wanted to update working copy url to point to new svn server location.\n\nThat is exactly what it does. If your the subversion server has changed its location, use \"switch --relocate\", it will just contact the server and chack if its still there, and then rewrite the URLs in your working copy to the new URL. Your working copy isn't updated with any differences from the server (until you run svn update again)\n\nsvn switch, without the --relocate switches your working copy to a new location (typically a branch) and makes whatever changes needed to your working copy to transform it into that branch It's pretty much the same as just checking out that branch, but usually faster and more convenient.\n\nIn particular, can it be used to integrate to user branch a vendor branch that has many differences since previous version, ie\n\nNo, that's what svn merge is for.\n\nAlthough svn switch could allow you to test out that vendor branch in your working copy first. You keep your main project as it used to, but you switch the subdirectory where the vendor stuff is to the new vendor branch.\n\nNote that subversion does not (yet) support any merging\/brancing between different subversion servers, so your vendor branch have to be imported into the server where your \"user branch\" is first. Then you could merge that new vendor branch with your user branch.\n\nYou could use svn:external for this too, it's common to keep vendor code outside your project path, and \"import\" them in a working copy using svn:external. e.g. you have \/myproject\/trunk\/ as your project and \/3party\/libraryFoo\/version1\/\n\nWhen working with myproject you want to have it under \/myproject\/trunk\/libraryFoo so you just add an svn:external on \/myproject\/trunk\/ to e.g. libraryFoo http:\/\/myserverver\/\/3party\/libraryFoo\/version1\/ . When you want to update to a new version ,you just change the svn:external to e.g. libraryFoo http:\/\/myserverver\/\/3party\/libraryFoo\/version2 . This allows you to pull in stuff from another subversion server as well, e.g. directly to the 3.party subersion server - though that's not very useful if you intend to make local changes.\n\n-","date":"2014-09-17 09:09:39","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\": 0.4880455434322357, \"perplexity\": 3139.6517499146707}, \"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-2014-41\/segments\/1410657121798.11\/warc\/CC-MAIN-20140914011201-00147-ip-10-196-40-205.us-west-1.compute.internal.warc.gz\"}"} | null | null |
Q: Styling a result set Perhaps this is not the proper place to ask this question, if that is the case, please direct me to the correct venue.
I'm looking for research, guides, any kind of information pertaining to the structuring and styling of results sets; data which comes back from a search, or when looking at content in a list view.
Links would be appreciated, but opinion and commentary are also valued.
A: *
*http://icant.co.uk/csstablegallery/
*http://www.smashingmagazine.com/2008/08/13/top-10-css-table-designs/
*http://www.noupe.com/css/21-fresh-ajax-css-tables.html
A: The CSS table gallery may prove useful.
| {
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{"url":"https:\/\/www.alloprof.qc.ca\/helpzone\/discussion\/36237\/question","text":"# Help Zone\n\n### Student Question\n\n>:C What is the thecnique to calculate 10^10, 10^100 pls help me ( If anyone is here that is ;-; )\n\nMathematics\n\n## Explanations (1)\n\n\u2022 Explanation from Alloprof\n\nExplanation from Alloprof\n\nThis Explanation was submitted by a member of the Alloprof team.\n\nOptions\nTeam Alloprof \u2022 3mo.\n\nHi !\n\nIt is relatively simple to do so. First, you sount the number of zeros of each number. This value is the nomber of zeros on your final answer. And then, you put your number 1 in front of the zeros.\n\nIn the cas of $$10\\times10$$, you can count two zeros. This means that the answer will have two zeros and a 1 in front of them :\n\n$$10\\times10=100$$\n\nYou can try it for yourself and if you have any other questions, feel free to ask them ! Also, this link might useful :","date":"2022-10-07 19:29:03","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 2, \"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\": 0.33047130703926086, \"perplexity\": 988.3269220609753}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2022-40\/segments\/1664030338244.64\/warc\/CC-MAIN-20221007175237-20221007205237-00508.warc.gz\"}"} | null | null |
{"url":"https:\/\/www.groundai.com\/project\/lower-bounds-to-the-accuracy-of-inference-on-heavy-tails\/","text":"Lower bounds to the accuracy of inference on heavy tails\n\n# Lower bounds to the accuracy of inference on heavy tails\n\n\\fnmsS.Y.\u00a0\\snmNovak\\correflabel=e1]S.Novak@mdx.ac.uk [ Middlesex University, The Burroughs, London NW44BT, UK. \\printeade1\n\\smonth6 \\syear2012\\smonth1 \\syear2013\n\\smonth6 \\syear2012\\smonth1 \\syear2013\n\\smonth6 \\syear2012\\smonth1 \\syear2013\n###### Abstract\n\nThe paper suggests a simple method of deriving minimax lower bounds to the accuracy of statistical inference on heavy tails. A well-known result by Hall and Welsh (Ann. Statist. 12 (1984) 1079\u20131084) states that if is an estimator of the tail index and is a sequence of positive numbers such that , where is a certain class of heavy-tailed distributions, then . The paper presents a non-asymptotic lower bound to the probabilities . We also establish non-uniform lower bounds to the accuracy of tail constant and extreme quantiles estimation. The results reveal that normalising sequences of robust estimators should depend in a specific way on the tail index and the tail constant.\n\n\\kwd\n\\aid\n\n0 \\volume20 \\issue2 2014 \\firstpage979 \\lastpage989 \\doi10.3150\/13-BEJ512\n\n\\runtitle\n\nLower bounds\n\n{aug}\n\nheavy-tailed distribution \\kwdlower bounds\n\n## 1 Introduction\n\nA growing number of publications is devoted to the problem of statistical inference on heavy-tailed distributions. Such distributions naturally appear in finance, meteorology, hydrology, teletraffic engineering, etc. EKM (), R97 (). In particular, it is widely accepted that frequent financial data (e.g., daily and hourly log-returns of share prices, stock indexes and currency exchange rates) often exhibits heavy tails FR (), EKM (), M63 (), N09 (), while less frequent financial data is typically light-tailed. The heaviness of a tail of the distribution appears to be responsible for extreme movements of stock indexes and share prices. The tail index indicates how heavy the tail is; extreme quantiles are used as measures of financial risk EKM (), N09 (). The need to evaluate the tail index and extreme quantiles stimulated research on methods of statistical inference on heavy-tailed data.\n\nThe distribution of a random variable (r.v.) is said to have a heavy right tail if\n\n P(X\u2265x)=L(x)x\u2212\u03b1(\u03b1>0), (1)\n\nwhere the (unknown) function is slowly varying at infinity:\n\n limx\u2192\u221eL(xt)\/L(x)=1(\u2200t>0).\n\nWe denote by the non-parametric class of distributions obeying (1).\n\nThe tail index is the main characteristic describing the tail of a distribution. If then is called the tail constant.\n\nLet denote the distribution function (d.f.). Obviously, the tail index is a functional of the distribution function:\n\n \u03b1F\u2261\u03b1P=\u2212limx\u2192\u221elnP(X\u2265x)lnx. (2)\n\nIf tends to a constant (say, ) as then the tail constant is also a functional of :\n\n cF\u2261cP=limx\u2192\u221ex\u03b1FP(X\u2265x).\n\nThe statistical inference on a heavy-tailed distribution is straightforward if the class of unknown distributions is assumed to be a regular parametric family. The drawback of the parametric approach is that one usually cannot reliably check whether the unknown distribution belongs to a chosen parametric family.\n\nA lot of attention during the past three decades has been given to the problem of reliable inference on heavy tails without parametric assumptions. The advantage of the non-parametric approach is that a class of unknown distributions, is so large that the problem of testing the hypothesis that the unknown distribution belongs to does not arise. The disadvantage of the non-parametric approach is that virtually no question concerning inference on heavy tails can be given a simple answer. In particular, the problem of establishing a lower bound to the accuracy of tail index estimation remained open for decades.\n\nA lower bound to the accuracy of statistical inference sets a benchmark against which the accuracy of any particular estimator can be compared. When looking for an estimator of a quantity of interest, where is the unknown distribution, is the class of distributions and is a functional of one often would like to choose an estimator that minimises a loss function uniformly in (e.g., where is a particular loss function). A\u00a0lower bound to follows if one can establish a lower bound to\n\n supP\u2208PP(|^an\u2212aP|\u2265u)(u>0).\n\nThe first step towards establishing a lower bound to the accuracy of tail index estimation was made by Hall and Welsh HW84 (), who proved the following result. Note that the class of heavy-tailed distributions is too \u201crich\u201d for meaningful inference, and one usually deals with a subclass of imposing certain restrictions on the asymptotics of . Hall and Welsh dealt with the class of distributions on with densities\n\n f(x)=c\u03b1x\u2212\u03b1\u22121(1+u(x)), (3)\n\nwhere . Note that the range of possible values of the tail index is restricted to interval . Let\n\n ^\u03b1n\u2261^\u03b1n(X1,\u2026,Xn)\n\nbe an arbitrary tail index estimator, where are independent and identically distributed (i.i.d.) random variables, and let be a sequence of positive numbers. If\n\n limn\u2192\u221esupF\u2208Db,APF(|^\u03b1n\u2212\u03b1F|\u2265zn)=0(\u2200A>0), (4)\n\nthen\n\n zn\u226bn\u2212b\/(2b+1)(n\u2192\u221e)\n\n(to be precise, Hall and Welsh HW84 () dealt with the random variables where are distributed according to (3)).\n\nBeirlant et al. BBW06 () have a similar result for a larger class of distributions but require the estimators are uniformly consistent in . Pfanzagl Pf () has established a lower bound in terms of a modulus of continuity related to the total variation distance . Let be the class of distributions with densities (3) such that , and set\n\n sn(\u03b5,P0)=supP\u2208Pn,\u03b5|\u03b1P\u2212\u03b1P0|,\n\nwhere is the tail index of distribution and is a neighborhood of Pfanzagl has showed that neither estimator can converge to uniformly in with the rate better than and\n\n inf0<\u03b5<1\u03b5\u22122b\/(1+2b)liminfn\u2192\u221enb\/(1+2b)sn(\u03b5,P0)>0.\n\nDonoho and Liu DL () present a lower bound to the accuracy of tail index estimation in terms of a modulus of continuity . However, they do not calculate . The claim that a particular heavy-tailed distribution is stochastically dominant over all heavy-tailed distributions with the same tail index appears without proof. Assuming that the range of possible values of the tail index is restricted to an interval of fixed length, Drees Drees2001 () derives the asymptotic minimax risk for affine estimators of the tail index and indicates an approach to numerical computation of the asymptotic minimax risk for non-affine ones.\n\nThe paper presents a simple method of deriving minimax lower bounds to the accuracy of non-parametric inference on heavy-tailed distributions. The results are non-asymptotic, the constants in the bounds are shown explicitly, the range of possible values of the tail index is not restricted to an interval of fixed length. The information functional seems to be found for the first time, as well as the lower bound to the accuracy of extreme quantiles estimation.\n\nThe results indicate that the traditional minimax approach may require revising. The classical approach suggests looking for an estimator that minimises, say,\n\n supP\u2208PEP|^an\u2212aP|\n\n(cf. Hu97 (), IH81 (), Tsy ()), while our results suggest looking for an estimator that minimises\n\nwhere is the \u201cinformation functional\u201d (an analogue of Fisher\u2019s information). Theorems 14 reveal the information functionals and indicate that the normalising sequence of a robust estimator should depend in a specific way on the characteristics of the unknown distribution.\n\n## 2 Results\n\nIn the sequel, we deal with the non-parametric class\n\n H(b)={P\u2208H\\dvtsupx>K\u2217(P)\u2223\u2223c\u22121Fx\u03b1FP(X\u2265x)\u22121\u2223\u2223xb\u03b1F<\u221e} (5)\n\nof distributions on , where and is the left end-point of the distribution. If then\n\n P(X\u2265x)=cFx\u2212\u03b1F(1+O(x\u2212b\u03b1F))(x\u2192\u221e).\n\nThe class is larger than ; the range of possible values of the tail index is not restricted to an interval of fixed length. Below, given a distribution function (d.f.) , we put\n\n aFi=1\/\u03b1Fi,r=b\/(1+2b),\n\nmeans the mathematical expectation with respect to and is the corresponding distribution. We set .\n\n###### Theorem 1\n\nFor any any tail index estimator and any estimator of index there exist d.f.s such that and\n\n maxi\u2208{0;1}Pi(|^\u03b1n\/\u03b1Fi\u22121|\u03b1r\/bFicrFinr\u2265v\/2) \u2265 (1\u2212v1\/r\/8n)2n\/4, (6) maxi\u2208{0;1}Pi(|^an\/aFi\u22121|a\u2212r\/bFicrFinr\u2265v\/2) \u2265 (1\u2212v1\/r\/8n)2n\/4 (7)\n\nas and .\n\nNote that if as then for any we have for all large enough , yielding . Thus, the Hall\u2013Welsh result follows from\u00a0(6).\n\nTheorem 1 shows that the natural normalising sequence for is The information functional plays here the same role as Fisher\u2019s information function in the Fr\u00e9chet\u2013Rao\u2013Cram\u00e9r inequality.\n\nTheorem 1 yields also minimax lower bounds to the moments of In particular, there holds\n\n###### Corollary 2\n\nFor any there exist distribution functions such that and for any tail index estimator\n\n maxi\u2208{0;1}\u03b1r\/bFicrFiEFi|^\u03b1n\/\u03b1Fi\u22121|nr\u22654rr\u0393(r)\/8+o(1). (8)\n\nThe result holds if in (8) is replaced with\n\nLet be a class of d.f.s such that as . Then for any estimator\n\n supF\u2208Hn(b)\u03b1r\/bFcrFEF|^\u03b1n\/\u03b1F\u22121|nr\u22654rr\u0393(r)\/8+o(1). (???\u2217)\n\nA lower bound to seems to be established for the first time.\n\nThe presence of the information functional makes the bound non-uniform. Note that a uniform lower bound would be meaningless: as the range of possible values of is not restricted to an interval of fixed length, it follows from (???) that\n\n supF\u2208Hn(b)EF|^\u03b1n\/\u03b1F\u22121|\u2192\u221e(n\u2192\u221e).\n\nMore generally, may tend to as if .\n\nLet be an arbitrary tail constant estimator. The next theorem presents a lower bound to the probabilities .\n\n###### Theorem 3\n\nLet be an arbitrary tail constant estimator. For any and there exist distribution functions such that and for all large enough , as ,\n\n maxi\u2208{0;1}Pi(|^cn\/cFi\u22121|\u03b1r\/bFicrFi\u2265rvrn\u2212rln(n\/lnn)tn\/2b)\u2265(1\u2212v\/8n)2n\/4, (9)\n\nwhere\n\nSimilarly to (8) Theorem 3 yields lower bounds to the moments of In particular, (9) entails\n\n maxi\u2208{0;1}\u03b1r\/bFicrFiEFi|^cn\/cFi\u22121|\u2265(lnn)n\u2212rr24r\u22121\u0393(r)\/(2b+o(1)). (???\u2217)\n\nAccording to Hall and Welsh HW84 (),\n\n zn\u226b(lnn)n\u2212b\/(2b+1)\n\nif . This fact can be obtained as a consequence to Theorem 3: if as then for any we have for all large enough , hence .\n\nWe now present a lower bound to the accuracy of estimating extreme upper quantiles. We call an upper quantile of level \u201cextreme\u201d if tends to 0 as grows. In financial applications (see, e.g., EKM (), N09 ()), an upper quantile of the level as high as 0.05 can be considered extreme as the empirical quantile estimator appears unreliable. Of course, there is an infinite variety of possible rates of decay of Theorem 4 presents lower bounds in the case where is bounded away from and .\n\nSet We denote the upper quantile of level by\n\n xF,n=\u00afF\u22121(qn).\n\nLet be an arbitrary estimator of . Denote .\n\n###### Theorem 4\n\nFor any there exist distribution functions such that and for all large enough and\n\n maxi\u2208{0;1}Pi(|^xn\/xFi,n\u22121|\u03b12(1\u2212r)FicrFi\/wFit\u22c6i,n\u2265un\u2212r\/2b) \u2265 (1\u2212u1\/r\/8n)2n\/4, (10) maxi\u2208{0;1}Pi(|xFi,n\/^xn\u22121|\u03b12(1\u2212r)FicrFi\/wFit\u22c6i,n\u2265un\u2212r\/2b) \u2265 (1\u2212u1\/r\/8n)2n\/4, (11)\n\nwhere as .\n\n## 3 Proofs\n\nOur approach to establishing lower bounds requires constructing two distribution functions and where is a Pareto d.f. and is a \u201cdisturbed\u201d version of . We then apply Lemma 5 that provides a non-asymptotic lower bound to the accuracy of estimation when choosing between two close alternatives.\n\nThe problem of estimating the tail index, the tail constant and from is equivalent to the problem of estimating and quantiles from a sample of i.i.d. positive r.v.s with the distribution\n\n F(y)\u2261P(Y\u2264y)=y\u03b1\u2113(y)(y>0), (12)\n\nwhere function slowly varies at the origin.\n\nWe denote by the class of distributions obeying (12). Note that if and only if Obviously, a tail index estimator can be considered an estimator of index from the sample , and vice versa. The tradition of dealing with this equivalent problem stems from H82 (). We proceed with this equivalent formulation.\n\nA counterpart to is the following non-parametric class of d.f.s on :\n\n F(b)={F\u2208F\\dvtsupy\n\nwhere and is the right end-point of . A d.f. obeys\n\n F(y)=cFy\u03b1F(1+O(yb\u03b1F))(y\u21920),\n\nwhere and\n\n{pf*}\n\nProof of Theorem 1 Let , and denote\n\n \u03b10=\u03b1,\u03b11=\u03b1+\u03b3,\u03b3=h\u03b1b.\n\nWe will employ the distribution functions and where\n\n F0(y) = (y\/c)\u03b1\\mathbh1{0\n\nThe counterparts to these distributions are\n\n P0(X>x) = (cx)\u2212\u03b1\\mathbh1{x\u22651\/c}, P1(X>x) = (cx)\u2212\u03b1\\mathbh1{1\/c\u2264x<1\/h}+c\u2212\u03b1h\u2212\u03b3x\u2212\u03b11\\mathbh1{x\u22651\/h}.\n\nIt is easy to see that and\n\n \u03b1F0=\u03b1,\u03b1F1=\u03b11,cF0=c\u2212\u03b1,cF1=c\u2212\u03b1h\u2212\u03b3. (14)\n\nObviously, We now check that\n\nSince\n\n c\u22121F1y\u2212\u03b11F1(y)=y\u2212\u03b3h\u03b3(h\n\nwe have\n\n sup0\n\nThe right-hand side of (15) takes on its maximum at ; the supremum is bounded by . Note that\n\nLet denote the Hellinger distance. It is easy to check that\n\n d2H(F0;F1)\u2264\u03b31\/r\/8\u03b12c\u03b1. (16)\n\nAccording to Lemma 5 below,\n\n maxi\u2208{0;1}Pi(|^\u03b1n\u2212\u03b1Fi|\u2265\u03b3\/2)\u2265(1\u2212\u03b31\/r\/8\u03b12c\u03b1)2n\/4. (17)\n\nLet where\n\n \u03b3n\u2261\u03b3n(\u03b1,c,v)=v(\u03b12c\u03b1\/n)r.\n\nNote that as From (17),\n\n maxi\u2208{0;1}Pi(|^\u03b1n\/\u03b1Fi\u22121|\u03b1r\/bFicrFinr\u2265vtn,i\/2)\u2265(1\u2212v1\/r\/8n)2n\/4, (18)\n\nwhere and Note that as . Hence, as and (6) follows.\n\nLet be an arbitrary estimator of index . Denote . Since Lemma 5 yields\n\n maxi\u2208{0;1}Pi(|^an\u2212aFi|\u2265\u03b3aa1\/2)\u2265(1\u2212\u03b31\/r\/8\u03b12c\u03b1)2n\/4.\n\nWith the left-hand side of this inequality is\n\n maxi\u2208{0;1}Pi(|^an\u2212aFi|\u2265vn\u2212ra1\u22122ra1\/2crF0)=maxi\u2208{0;1}Pi(|^an\/aFi\u22121|a\u2212r\/bFicrFinr\u2265vt+n,i\/2),\n\nwhere and , leading to (7).\n\n{pf*}\n\nProof of Corollary 2 Note that\n\n E\u03be=\u222b\u221e0P(\u03be\u2265x)dx (19)\n\nfor any non-negative r.v. Since\n\n \u222bzn0(1\u2212v1\/r\/8n)2ndv=4rr\u0393(r)+o(1)(n\u2192\u221e)\n\nas (6) and (19) entail (8).\n\n{pf*}\n\nProof of Theorem 3 With and defined as above, we have\n\n cF1\u2212cF0=c\u2212\u03b1(\u03b3\u2212\u03b3\/\u03b1b\u22121)\u2265c\u2212\u03b1\u03b3|ln\u03b3|\/\u03b1b.\n\nUsing this inequality, (17) and Lemma 5, we derive\n\n maxi\u2208{0;1}Pi(|^cn\u2212cFi|\u2265c\u2212\u03b1\u03b3|ln\u03b3|\/2\u03b1b)\u2265(1\u2212\u03b31\/r\/8\u03b12c\u03b1)2n\/4.\n\nLet . Then\n\n maxi\u2208{0;1}Pi(|^cn\u2212cFi|\u2265cF0(v\u03b12c\u03b1\/n)rrln(n\/lnn)\/2\u03b1b)\u2265(1\u2212v\/8n)2n\/4.\n\nNote that as . The result follows.\n\n{pf*}\n\nProof of Theorem 4 Denote\n\n xi\u2261xFi,n,yi=1\/xi.\n\nObviously, is the quantile of . We find convenient dealing with the equivalent problem of estimating quantiles of the distribution of a random variable .\n\nWith functions defined as above, it is easy to see that\n\n y0=cq1\/\u03b1n=c\u03bah,y1=c\u03b1\/\u03b11q1\/\u03b11nh\u03b3\/\u03b11=y0(c\u03ba)\u2212\u03b3\/\u03b11, (20)\n\nwhere we put Note that Hence if ().\n\nDenote\n\n \u03b3\u2261\u03b3n(\u03b1,b,c)=u(\u03b12c\u03b1\/n)r. (21)\n\nThen and\n\n c\u03ba=u\u22121\/\u03b1bs1\/\u03b1c2r\u03b1\u22122r\/\u03b1b<1 (22)\n\nby the assumption.\n\nUsing the facts that and as , we derive\n\n y1\u2212y0 = y0((c\u03ba)\u2212\u03b3\/\u03b11\u22121) \u2265 \u03b31+1\/\u03b1b(c\u03ba)1\u2212\u03b3\/2\u03b11|lnc\u03ba|\/\u03b11.\n\nHence, and . By Lemma 5,\n\n maxi\u2208{0;1}Pi(|^yn\u2212yi|\u2265\u03b31+1\/\u03b1b(c\u03ba)1\u2212\u03b3\/2\u03b11|lnc\u03ba|\/2\u03b11)\u2265(1\u2212\u03b31\/r\/8\u03b12c\u03b1)2n\/4\n\nfor any estimator . Thus,\n\nwhere and Taking into account (21) and (22), we derive\n\n maxi\u2208{0;1}Pi(|^yn\/yi\u22121|\u2265u\u03b12(r\u22121)Fic\u2212rFin\u2212rln(u\u03b12r\/sbc2r\u03b1b)t\u22c6n,i\/2b)\u2265(1\u2212u1\/r\/8n)2n\/4.\n\nRecall that . From (20),\n\n |x1\u2212x0|=|y1\u2212y0|\/y0y1\u2265\u03b31\u22121\/\u03b1b(c\u03ba)\u22121+\u03b3\/2\u03b11|lnc\u03ba|\/\u03b11.\n\nBy Lemma 5,\n\n maxi\u2208{0;1}Pi(|^xn\u2212xi|\u2265\u03b31\u22121\/\u03b1b(c\u03ba)\u22121+\u03b3\/2\u03b11|lnc\u03ba|\/2\u03b11)\u2265(1\u2212\u03b31\/r\/8\u03b12c\u03b1)2n\/4.\n\nHence,\n\n maxi\u2208{0;1}Pi(|^xn\/xi\u22121|\u2265u\u03b12(r\u22121)Fic\u2212rFin\u2212r\u2223\u2223ln(sbc2r\u03b1b\/\u03b12ru)\u2223\u2223~tn,i\/2b)\u2265(1\u2212u1\/r\/8n)2n\/4,\n\nwhere and . The proof is complete.\n\nThe next lemma presents a lower bound to the accuracy of choosing between two \u201cclose\u201d alternatives.\n\nLet be an arbitrary class of distributions, and assume that the quantity of interest, is an element of a metric space . An estimator of is a measurable function of taking values in a subspace of the metric space .\n\nExamples of functionals include (a) where is a parametric family of distributions (); (b) where is the density of with respect to a particular measure; (c) . A minimax lower bound over follows from a lower bound to where\n\nDenote","date":"2020-10-27 11:51:20","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\": 0.9099385142326355, \"perplexity\": 833.7139406473262}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2020-45\/segments\/1603107894175.55\/warc\/CC-MAIN-20201027111346-20201027141346-00620.warc.gz\"}"} | null | null |
#ifndef ANDROID_VOLUME_SHAPER_H
#define ANDROID_VOLUME_SHAPER_H
#include <cmath>
#include <list>
#include <math.h>
#include <sstream>
#include <binder/Parcel.h>
#include <media/Interpolator.h>
#include <utils/Mutex.h>
#include <utils/RefBase.h>
#if defined(_MSC_VER) || defined(__linux__) // M3E:
#include <functional>
#include <random>
#endif
#pragma push_macro("LOG_TAG")
#undef LOG_TAG
#define LOG_TAG "VolumeShaper"
// turn on VolumeShaper logging
#define VS_LOGGING 0
#define VS_LOG(...) ALOGD_IF(VS_LOGGING, __VA_ARGS__)
namespace android {
namespace media {
#if defined(__linux__) // M3E:
# if defined(Status)
# undef Status
# endif
#endif
// The native VolumeShaper class mirrors the java VolumeShaper class;
// in addition, the native class contains implementation for actual operation.
//
// VolumeShaper methods are not safe for multiple thread access.
// Use VolumeHandler for thread-safe encapsulation of multiple VolumeShapers.
//
// Classes below written are to avoid naked pointers so there are no
// explicit destructors required.
class VolumeShaper {
public:
// S and T are like template typenames (matching the Interpolator<S, T>)
using S = float; // time type
using T = float; // volume type
// Curve and dimension information
// TODO: member static const or constexpr float initialization not permitted in C++11
#define MIN_CURVE_TIME 0.f // type S: start of VolumeShaper curve (normalized)
#define MAX_CURVE_TIME 1.f // type S: end of VolumeShaper curve (normalized)
#define MIN_LINEAR_VOLUME 0.f // type T: silence / mute audio
#define MAX_LINEAR_VOLUME 1.f // type T: max volume, unity gain
#define MAX_LOG_VOLUME 0.f // type T: max volume, unity gain in dBFS
/* kSystemVolumeShapersMax is the maximum number of system VolumeShapers.
* Each system VolumeShapers has a predefined Id, which ranges from 0
* to kSystemVolumeShapersMax - 1 and is unique for its usage.
*
* "1" is reserved for system ducking.
*/
static const int kSystemVolumeShapersMax = 16;
/* kUserVolumeShapersMax is the maximum number of application
* VolumeShapers for a player/track. Application VolumeShapers are
* assigned on creation by the client, and have Ids ranging
* from kSystemVolumeShapersMax to INT32_MAX.
*
* The number of user/application volume shapers is independent to the
* system volume shapers. If an application tries to create more than
* kUserVolumeShapersMax to a player, then the apply() will fail.
* This prevents exhausting server side resources by a potentially malicious
* application.
*/
static const int kUserVolumeShapersMax = 16;
/* VolumeShaper::Status is equivalent to status_t if negative
* but if non-negative represents the id operated on.
* It must be expressible as an int32_t for binder purposes.
*/
using Status = status_t;
// Local definition for clamp as std::clamp is included in C++17 only.
// TODO: use the std::clamp version when Android build uses C++17.
template<typename R>
static constexpr const R &clamp(const R &v, const R &lo, const R &hi) {
return (v < lo) ? lo : (hi < v) ? hi : v;
}
/* VolumeShaper.Configuration derives from the Interpolator class and adds
* parameters relating to the volume shape.
*
* This parallels the Java implementation and the enums must match.
* See "frameworks/base/media/java/android/media/VolumeShaper.java" for
* details on the Java implementation.
*/
class Configuration : public Interpolator<S, T>, public RefBase, public Parcelable {
public:
// Must match with VolumeShaper.java in frameworks/base.
enum Type : int32_t {
TYPE_ID,
TYPE_SCALE,
};
// Must match with VolumeShaper.java in frameworks/base.
enum OptionFlag : int32_t {
OPTION_FLAG_NONE = 0,
OPTION_FLAG_VOLUME_IN_DBFS = (1 << 0),
OPTION_FLAG_CLOCK_TIME = (1 << 1),
OPTION_FLAG_ALL = (OPTION_FLAG_VOLUME_IN_DBFS | OPTION_FLAG_CLOCK_TIME),
};
// Bring from base class; must match with VolumeShaper.java in frameworks/base.
using InterpolatorType = Interpolator<S, T>::InterpolatorType;
Configuration()
: Interpolator<S, T>()
, RefBase()
, mType(TYPE_SCALE)
, mId(-1)
, mOptionFlags(OPTION_FLAG_NONE)
, mDurationMs(1000.) {
}
explicit Configuration(const Configuration &configuration)
: Interpolator<S, T>(*static_cast<const Interpolator<S, T> *>(&configuration))
, RefBase()
, mType(configuration.mType)
, mId(configuration.mId)
, mOptionFlags(configuration.mOptionFlags)
, mDurationMs(configuration.mDurationMs) {
}
Type getType() const {
return mType;
}
status_t setType(Type type) {
switch (type) {
case TYPE_ID:
case TYPE_SCALE:
mType = type;
return NO_ERROR;
default:
ALOGE("invalid Type: %d", type);
return BAD_VALUE;
}
}
OptionFlag getOptionFlags() const {
return mOptionFlags;
}
status_t setOptionFlags(OptionFlag optionFlags) {
if ((optionFlags & ~OPTION_FLAG_ALL) != 0) {
ALOGE("optionFlags has invalid bits: %#x", optionFlags);
return BAD_VALUE;
}
mOptionFlags = optionFlags;
return NO_ERROR;
}
double getDurationMs() const {
return mDurationMs;
}
status_t setDurationMs(double durationMs) {
if (durationMs > 0.) {
mDurationMs = durationMs;
return NO_ERROR;
}
// zero, negative, or nan. These values not possible from Java.
return BAD_VALUE;
}
int32_t getId() const {
return mId;
}
void setId(int32_t id) {
// We permit a negative id here (representing invalid).
mId = id;
}
/* Adjust the volume to be in linear range from MIN_LINEAR_VOLUME to MAX_LINEAR_VOLUME
* and compensate for log dbFS volume as needed.
*/
T adjustVolume(T volume) const {
if ((getOptionFlags() & OPTION_FLAG_VOLUME_IN_DBFS) != 0) {
const T out = powf(10.f, volume / 10.f);
VS_LOG("in: %f out: %f", volume, out);
volume = out;
}
return clamp(volume, MIN_LINEAR_VOLUME /* lo */, MAX_LINEAR_VOLUME /* hi */);
}
/* Check if the existing curve is valid.
*/
status_t checkCurve() const {
if (mType == TYPE_ID) return NO_ERROR;
if (this->size() < 2) {
ALOGE("curve must have at least 2 points");
return BAD_VALUE;
}
if (first().first != MIN_CURVE_TIME || last().first != MAX_CURVE_TIME) {
ALOGE("curve must start at MIN_CURVE_TIME and end at MAX_CURVE_TIME");
return BAD_VALUE;
}
if ((getOptionFlags() & OPTION_FLAG_VOLUME_IN_DBFS) != 0) {
for (const auto &pt : *this) {
if (!(pt.second <= MAX_LOG_VOLUME) /* handle nan */) {
ALOGE("positive volume dbFS");
return BAD_VALUE;
}
}
} else {
for (const auto &pt : *this) {
if (!(pt.second >= MIN_LINEAR_VOLUME)
|| !(pt.second <= MAX_LINEAR_VOLUME) /* handle nan */) {
ALOGE("volume < MIN_LINEAR_VOLUME or > MAX_LINEAR_VOLUME");
return BAD_VALUE;
}
}
}
return NO_ERROR;
}
/* Clamps the volume curve in the configuration to
* the valid range for log or linear scale.
*/
void clampVolume() {
if ((mOptionFlags & OPTION_FLAG_VOLUME_IN_DBFS) != 0) {
for (auto it = this->begin(); it != this->end(); ++it) {
if (!(it->second <= MAX_LOG_VOLUME) /* handle nan */) {
it->second = MAX_LOG_VOLUME;
}
}
} else {
for (auto it = this->begin(); it != this->end(); ++it) {
if (!(it->second >= MIN_LINEAR_VOLUME) /* handle nan */) {
it->second = MIN_LINEAR_VOLUME;
} else if (!(it->second <= MAX_LINEAR_VOLUME)) {
it->second = MAX_LINEAR_VOLUME;
}
}
}
}
/* scaleToStartVolume() is used to set the start volume of a
* new VolumeShaper curve, when replacing one VolumeShaper
* with another using the "join" (volume match) option.
*
* It works best for monotonic volume ramps or ducks.
*/
void scaleToStartVolume(T volume) {
if (this->size() < 2) {
return;
}
const T startVolume = first().second;
const T endVolume = last().second;
if (endVolume == startVolume) {
// match with linear ramp
const T offset = volume - startVolume;
static const T scale = 1.f / (MAX_CURVE_TIME - MIN_CURVE_TIME); // nominally 1.f
for (auto it = this->begin(); it != this->end(); ++it) {
it->second = it->second + offset * (MAX_CURVE_TIME - it->first) * scale;
}
} else {
const T scale = (volume - endVolume) / (startVolume - endVolume);
for (auto it = this->begin(); it != this->end(); ++it) {
it->second = scale * (it->second - endVolume) + endVolume;
}
}
clampVolume();
}
// The parcel layout must match VolumeShaper.java
status_t writeToParcel(Parcel *parcel) const override {
if (parcel == nullptr) return BAD_VALUE;
return parcel->writeInt32((int32_t)mType)
? -1 : parcel->writeInt32(mId) // M3E:
? -2 : mType == TYPE_ID
? NO_ERROR
: parcel->writeInt32((int32_t)mOptionFlags)
? -3 : parcel->writeDouble(mDurationMs) // M3E:
? -4 : Interpolator<S, T>::writeToParcel(parcel);
}
status_t readFromParcel(const Parcel *parcel) override {
int32_t type, optionFlags;
return parcel->readInt32(&type)
? -1 : setType((Type)type) // M3E:
? -2 : parcel->readInt32(&mId)
? -3 : mType == TYPE_ID
? NO_ERROR
: parcel->readInt32(&optionFlags)
? -4 : setOptionFlags((OptionFlag)optionFlags) // M3E:
? -5 : parcel->readDouble(&mDurationMs)
? -6 : Interpolator<S, T>::readFromParcel(*parcel)
? -7 : checkCurve();
}
// Returns a string for debug printing.
std::string toString() const {
std::stringstream ss;
ss << "VolumeShaper::Configuration{mType=" << static_cast<int32_t>(mType);
ss << ", mId=" << mId;
if (mType != TYPE_ID) {
ss << ", mOptionFlags=" << static_cast<int32_t>(mOptionFlags);
ss << ", mDurationMs=" << mDurationMs;
ss << ", " << Interpolator<S, T>::toString().c_str();
}
ss << "}";
return ss.str();
}
private:
Type mType; // type of configuration
int32_t mId; // A valid id is >= 0.
OptionFlag mOptionFlags; // option flags for the configuration.
double mDurationMs; // duration, must be > 0; default is 1000 ms.
}; // Configuration
/* VolumeShaper::Operation expresses an operation to perform on the
* configuration (either explicitly specified or an id).
*
* This parallels the Java implementation and the enums must match.
* See "frameworks/base/media/java/android/media/VolumeShaper.java" for
* details on the Java implementation.
*/
class Operation : public RefBase, public Parcelable {
public:
// Must match with VolumeShaper.java.
enum Flag : int32_t {
FLAG_NONE = 0,
FLAG_REVERSE = (1 << 0), // the absence of this indicates "play"
FLAG_TERMINATE = (1 << 1),
FLAG_JOIN = (1 << 2),
FLAG_DELAY = (1 << 3),
FLAG_CREATE_IF_NECESSARY = (1 << 4),
FLAG_ALL = (FLAG_REVERSE | FLAG_TERMINATE | FLAG_JOIN | FLAG_DELAY
| FLAG_CREATE_IF_NECESSARY),
};
Operation()
: Operation(FLAG_NONE, -1 /* replaceId */) {
}
Operation(Flag flags, int replaceId)
: Operation(flags, replaceId, std::numeric_limits<S>::quiet_NaN() /* xOffset */) {
}
explicit Operation(const Operation &operation)
: Operation(operation.mFlags, operation.mReplaceId, operation.mXOffset) {
}
explicit Operation(const sp<Operation> &operation)
: Operation(*operation.get()) {
}
Operation(Flag flags, int replaceId, S xOffset)
: mFlags(flags)
, mReplaceId(replaceId)
, mXOffset(xOffset) {
}
int32_t getReplaceId() const {
return mReplaceId;
}
void setReplaceId(int32_t replaceId) {
mReplaceId = replaceId;
}
S getXOffset() const {
return mXOffset;
}
void setXOffset(S xOffset) {
mXOffset = clamp(xOffset, MIN_CURVE_TIME /* lo */, MAX_CURVE_TIME /* hi */);
}
Flag getFlags() const {
return mFlags;
}
/* xOffset is the position on the volume curve and may go backwards
* if you are in reverse mode. This must be in the range from
* [MIN_CURVE_TIME, MAX_CURVE_TIME].
*
* normalizedTime always increases as time or framecount increases.
* normalizedTime is nominally from MIN_CURVE_TIME to MAX_CURVE_TIME when
* running through the curve, but could be outside this range afterwards.
* If you are reversing, this means the position on the curve, or xOffset,
* is computed as MAX_CURVE_TIME - normalizedTime, clamped to
* [MIN_CURVE_TIME, MAX_CURVE_TIME].
*/
void setNormalizedTime(S normalizedTime) {
setXOffset((mFlags & FLAG_REVERSE) != 0
? MAX_CURVE_TIME - normalizedTime : normalizedTime);
}
status_t setFlags(Flag flags) {
if ((flags & ~FLAG_ALL) != 0) {
ALOGE("flags has invalid bits: %#x", flags);
return BAD_VALUE;
}
mFlags = flags;
return NO_ERROR;
}
status_t writeToParcel(Parcel *parcel) const override {
if (parcel == nullptr) return BAD_VALUE;
return parcel->writeInt32((int32_t)mFlags)
? -1 : parcel->writeInt32(mReplaceId) // M3E:
? -2 : parcel->writeFloat(mXOffset);
}
status_t readFromParcel(const Parcel *parcel) override {
int32_t flags;
return parcel->readInt32(&flags)
? -1 : parcel->readInt32(&mReplaceId) // M3E:
? -2 : parcel->readFloat(&mXOffset)
? -3 : setFlags((Flag)flags);
}
std::string toString() const {
std::stringstream ss;
ss << "VolumeShaper::Operation{mFlags=" << static_cast<int32_t>(mFlags) ;
ss << ", mReplaceId=" << mReplaceId;
ss << ", mXOffset=" << mXOffset;
ss << "}";
return ss.str();
}
private:
Flag mFlags; // operation to do
int32_t mReplaceId; // if >= 0 the id to remove in a replace operation.
S mXOffset; // position in the curve to set if a valid number (not nan)
}; // Operation
/* VolumeShaper.State is returned when requesting the last
* state of the VolumeShaper.
*
* This parallels the Java implementation.
* See "frameworks/base/media/java/android/media/VolumeShaper.java" for
* details on the Java implementation.
*/
class State : public RefBase, public Parcelable {
public:
State(T volume, S xOffset)
: mVolume(volume)
, mXOffset(xOffset) {
}
State()
: State(NAN, NAN) { }
T getVolume() const {
return mVolume;
}
void setVolume(T volume) {
mVolume = volume;
}
S getXOffset() const {
return mXOffset;
}
void setXOffset(S xOffset) {
mXOffset = xOffset;
}
status_t writeToParcel(Parcel *parcel) const override {
if (parcel == nullptr) return BAD_VALUE;
return parcel->writeFloat(mVolume)
? -1 : parcel->writeFloat(mXOffset); // M3E:
}
status_t readFromParcel(const Parcel *parcel) override {
return parcel->readFloat(&mVolume)
? -1 : parcel->readFloat(&mXOffset); // M3E:
}
std::string toString() const {
std::stringstream ss;
ss << "VolumeShaper::State{mVolume=" << mVolume;
ss << ", mXOffset=" << mXOffset;
ss << "}";
return ss.str();
}
private:
T mVolume; // linear volume in the range MIN_LINEAR_VOLUME to MAX_LINEAR_VOLUME
S mXOffset; // position on curve expressed from MIN_CURVE_TIME to MAX_CURVE_TIME
}; // State
// Internal helper class to do an affine transform for time and amplitude scaling.
template <typename R>
class Translate {
public:
Translate()
: mOffset(0)
, mScale(1) {
}
R getOffset() const {
return mOffset;
}
void setOffset(R offset) {
mOffset = offset;
}
R getScale() const {
return mScale;
}
void setScale(R scale) {
mScale = scale;
}
R operator()(R in) const {
return mScale * (in - mOffset);
}
std::string toString() const {
std::stringstream ss;
ss << "VolumeShaper::Translate{mOffset=" << mOffset;
ss << ", mScale=" << mScale;
ss << "}";
return ss.str();
}
private:
R mOffset;
R mScale;
}; // Translate
static int64_t convertTimespecToUs(const struct timespec &tv)
{
return tv.tv_sec * 1000000ll + tv.tv_nsec / 1000;
}
// current monotonic time in microseconds.
static int64_t getNowUs()
{
struct timespec tv;
if (clock_gettime(CLOCK_MONOTONIC, &tv) != 0) {
return 0; // system is really sick, just return 0 for consistency.
}
return convertTimespecToUs(tv);
}
/* Native implementation of VolumeShaper. This is NOT mirrored
* on the Java side, so we don't need to mimic Java side layout
* and data; furthermore, this isn't refcounted as a "RefBase" object.
*
* Since we pass configuration and operation as shared pointers (like
* Java) there is a potential risk that the caller may modify
* these after delivery.
*/
VolumeShaper(
const sp<VolumeShaper::Configuration> &configuration,
const sp<VolumeShaper::Operation> &operation)
: mConfiguration(configuration) // we do not make a copy
, mOperation(operation) // ditto
, mStartFrame(-1)
, mLastVolume(T(1))
, mLastXOffset(MIN_CURVE_TIME)
, mDelayXOffset(MIN_CURVE_TIME) {
if (configuration.get() != nullptr
&& (getFlags() & VolumeShaper::Operation::FLAG_DELAY) == 0) {
mLastVolume = configuration->first().second;
}
}
// We allow a null operation here, though VolumeHandler always provides one.
VolumeShaper::Operation::Flag getFlags() const {
return mOperation == nullptr
? VolumeShaper::Operation::FLAG_NONE : mOperation->getFlags();
}
/* Returns the last volume and xoffset reported to the AudioFlinger.
* If the VolumeShaper has not been started, compute what the volume
* should be based on the initial offset specified.
*/
sp<VolumeShaper::State> getState() const {
if (!isStarted()) {
const T volume = computeVolumeFromXOffset(mDelayXOffset);
VS_LOG("delayed VolumeShaper, using cached offset:%f for volume:%f",
mDelayXOffset, volume);
return new VolumeShaper::State(volume, mDelayXOffset);
} else {
return new VolumeShaper::State(mLastVolume, mLastXOffset);
}
}
S getDelayXOffset() const {
return mDelayXOffset;
}
void setDelayXOffset(S xOffset) {
mDelayXOffset = clamp(xOffset, MIN_CURVE_TIME /* lo */, MAX_CURVE_TIME /* hi */);
}
bool isStarted() const {
return mStartFrame >= 0;
}
/* getVolume() updates the last volume/xoffset state so it is not
* const, even though logically it may be viewed as const.
*/
std::pair<T /* volume */, bool /* active */> getVolume(
int64_t trackFrameCount, double trackSampleRate) {
if ((getFlags() & VolumeShaper::Operation::FLAG_DELAY) != 0) {
// We haven't had PLAY called yet, so just return the value
// as if PLAY were called just now.
VS_LOG("delayed VolumeShaper, using cached offset %f", mDelayXOffset);
const T volume = computeVolumeFromXOffset(mDelayXOffset);
return std::make_pair(volume, false);
}
const bool clockTime = (mConfiguration->getOptionFlags()
& VolumeShaper::Configuration::OPTION_FLAG_CLOCK_TIME) != 0;
const int64_t frameCount = clockTime ? getNowUs() : trackFrameCount;
const double sampleRate = clockTime ? 1000000 : trackSampleRate;
if (mStartFrame < 0) {
updatePosition(frameCount, sampleRate, mDelayXOffset);
mStartFrame = frameCount;
}
VS_LOG("frameCount: %lld", (long long)frameCount);
const S x = mXTranslate((T)frameCount);
VS_LOG("translation to normalized time: %f", x);
std::tuple<T /* volume */, S /* position */, bool /* active */> vt =
computeStateFromNormalizedTime(x);
mLastVolume = std::get<0>(vt);
mLastXOffset = std::get<1>(vt);
const bool active = std::get<2>(vt);
VS_LOG("rescaled time:%f volume:%f xOffset:%f active:%s",
x, mLastVolume, mLastXOffset, active ? "true" : "false");
return std::make_pair(mLastVolume, active);
}
std::string toString() const {
std::stringstream ss;
ss << "VolumeShaper{mStartFrame=" << mStartFrame;
ss << ", mXTranslate=" << mXTranslate.toString().c_str();
ss << ", mConfiguration=" <<
(mConfiguration.get() == nullptr
? "nullptr" : mConfiguration->toString().c_str());
ss << ", mOperation=" <<
(mOperation.get() == nullptr
? "nullptr" : mOperation->toString().c_str());
ss << "}";
return ss.str();
}
Translate<S> mXTranslate; // translation from frames (usec for clock time) to normalized time.
sp<VolumeShaper::Configuration> mConfiguration;
sp<VolumeShaper::Operation> mOperation;
private:
int64_t mStartFrame; // starting frame, non-negative when started (in usec for clock time)
T mLastVolume; // last computed interpolated volume (y-axis)
S mLastXOffset; // last computed interpolated xOffset/time (x-axis)
S mDelayXOffset; // xOffset to use for first invocation of VolumeShaper.
// Called internally to adjust mXTranslate for first time start.
void updatePosition(int64_t startFrame, double sampleRate, S xOffset) {
double scale = (mConfiguration->last().first - mConfiguration->first().first)
/ (mConfiguration->getDurationMs() * 0.001 * sampleRate);
const double minScale = 1. / static_cast<double>(INT64_MAX);
scale = std::max(scale, minScale);
VS_LOG("update position: scale %lf frameCount:%lld, sampleRate:%lf, xOffset:%f",
scale, (long long) startFrame, sampleRate, xOffset);
S normalizedTime = (getFlags() & VolumeShaper::Operation::FLAG_REVERSE) != 0 ?
MAX_CURVE_TIME - xOffset : xOffset;
mXTranslate.setOffset(static_cast<float>(static_cast<double>(startFrame)
- static_cast<double>(normalizedTime) / scale));
mXTranslate.setScale(static_cast<float>(scale));
VS_LOG("translate: %s", mXTranslate.toString().c_str());
}
T computeVolumeFromXOffset(S xOffset) const {
const T unscaledVolume = mConfiguration->findY(xOffset);
const T volume = mConfiguration->adjustVolume(unscaledVolume); // handle log scale
VS_LOG("computeVolumeFromXOffset %f -> %f -> %f", xOffset, unscaledVolume, volume);
return volume;
}
std::tuple<T /* volume */, S /* position */, bool /* active */>
computeStateFromNormalizedTime(S x) const {
bool active = true;
// handle reversal of position
if (getFlags() & VolumeShaper::Operation::FLAG_REVERSE) {
x = MAX_CURVE_TIME - x;
VS_LOG("reversing to %f", x);
if (x < MIN_CURVE_TIME) {
x = MIN_CURVE_TIME;
active = false; // at the end
} else if (x > MAX_CURVE_TIME) {
x = MAX_CURVE_TIME; //early
}
} else {
if (x < MIN_CURVE_TIME) {
x = MIN_CURVE_TIME; // early
} else if (x > MAX_CURVE_TIME) {
x = MAX_CURVE_TIME;
active = false; // at end
}
}
const S xOffset = x;
const T volume = computeVolumeFromXOffset(xOffset);
return std::make_tuple(volume, xOffset, active);
}
}; // VolumeShaper
/* VolumeHandler combines the volume factors of multiple VolumeShapers associated
* with a player. It is thread safe by synchronizing all public methods.
*
* This is a native-only implementation.
*
* The server side VolumeHandler is used to maintain a list of volume handlers,
* keep state, and obtain volume.
*
* The client side VolumeHandler is used to maintain a list of volume handlers,
* keep some partial state, and restore if the server dies.
*/
class VolumeHandler : public RefBase {
public:
using S = float;
using T = float;
// A volume handler which just keeps track of active VolumeShapers does not need sampleRate.
VolumeHandler()
: VolumeHandler(0 /* sampleRate */) {
}
explicit VolumeHandler(uint32_t sampleRate)
: mSampleRate((double)sampleRate)
, mLastFrame(0)
, mVolumeShaperIdCounter(VolumeShaper::kSystemVolumeShapersMax)
, mLastVolume(1.f, false) {
}
VolumeShaper::Status applyVolumeShaper(
const sp<VolumeShaper::Configuration> &configuration,
const sp<VolumeShaper::Operation> &operation_in) {
// make a local copy of operation, as we modify it.
sp<VolumeShaper::Operation> operation(new VolumeShaper::Operation(operation_in));
VS_LOG("applyVolumeShaper:configuration: %s", configuration->toString().c_str());
VS_LOG("applyVolumeShaper:operation: %s", operation->toString().c_str());
AutoMutex _l(mLock);
if (configuration == nullptr) {
ALOGE("null configuration");
return VolumeShaper::Status(BAD_VALUE);
}
if (operation == nullptr) {
ALOGE("null operation");
return VolumeShaper::Status(BAD_VALUE);
}
const int32_t id = configuration->getId();
if (id < 0) {
ALOGE("negative id: %d", id);
return VolumeShaper::Status(BAD_VALUE);
}
VS_LOG("applyVolumeShaper id: %d", id);
switch (configuration->getType()) {
case VolumeShaper::Configuration::TYPE_SCALE: {
const int replaceId = operation->getReplaceId();
if (replaceId >= 0) {
VS_LOG("replacing %d", replaceId);
auto replaceIt = findId_l(replaceId);
if (replaceIt == mVolumeShapers.end()) {
ALOGW("cannot find replace id: %d", replaceId);
} else {
if ((operation->getFlags() & VolumeShaper::Operation::FLAG_JOIN) != 0) {
// For join, we scale the start volume of the current configuration
// to match the last-used volume of the replacing VolumeShaper.
auto state = replaceIt->getState();
ALOGD("join: state:%s", state->toString().c_str());
if (state->getXOffset() >= 0) { // valid
const T volume = state->getVolume();
ALOGD("join: scaling start volume to %f", volume);
configuration->scaleToStartVolume(volume);
}
}
(void)mVolumeShapers.erase(replaceIt);
}
operation->setReplaceId(-1);
}
// check if we have another of the same id.
auto oldIt = findId_l(id);
if (oldIt != mVolumeShapers.end()) {
if ((operation->getFlags()
& VolumeShaper::Operation::FLAG_CREATE_IF_NECESSARY) != 0) {
// TODO: move the case to a separate function.
goto HANDLE_TYPE_ID; // no need to create, take over existing id.
}
ALOGW("duplicate id, removing old %d", id);
(void)mVolumeShapers.erase(oldIt);
}
/* Check if too many application VolumeShapers (with id >= kSystemVolumeShapersMax).
* We check on the server side to ensure synchronization and robustness.
*
* This shouldn't fail on a replace command unless the replaced id is
* already invalid (which *should* be checked in the Java layer).
*/
if (id >= VolumeShaper::kSystemVolumeShapersMax
&& numberOfUserVolumeShapers_l() >= VolumeShaper::kUserVolumeShapersMax) {
ALOGW("Too many app VolumeShapers, cannot add to VolumeHandler");
return VolumeShaper::Status(INVALID_OPERATION);
}
// create new VolumeShaper with default behavior.
mVolumeShapers.emplace_back(configuration, new VolumeShaper::Operation());
VS_LOG("after adding, number of volumeShapers:%zu", mVolumeShapers.size());
}
// fall through to handle the operation
HANDLE_TYPE_ID:
case VolumeShaper::Configuration::TYPE_ID: {
VS_LOG("trying to find id: %d", id);
auto it = findId_l(id);
if (it == mVolumeShapers.end()) {
VS_LOG("couldn't find id: %d", id);
return VolumeShaper::Status(INVALID_OPERATION);
}
if ((operation->getFlags() & VolumeShaper::Operation::FLAG_TERMINATE) != 0) {
VS_LOG("terminate id: %d", id);
mVolumeShapers.erase(it);
break;
}
const bool clockTime = (it->mConfiguration->getOptionFlags()
& VolumeShaper::Configuration::OPTION_FLAG_CLOCK_TIME) != 0;
if ((it->getFlags() & VolumeShaper::Operation::FLAG_REVERSE) !=
(operation->getFlags() & VolumeShaper::Operation::FLAG_REVERSE)) {
if (it->isStarted()) {
const int64_t frameCount = clockTime ? VolumeShaper::getNowUs() : mLastFrame;
const S x = it->mXTranslate((T)frameCount);
VS_LOG("reverse normalizedTime: %f", x);
// reflect position
S target = MAX_CURVE_TIME - x;
if (target < MIN_CURVE_TIME) {
VS_LOG("clamp to start - begin immediately");
target = MIN_CURVE_TIME;
}
VS_LOG("reverse normalizedTime target: %f", target);
it->mXTranslate.setOffset(it->mXTranslate.getOffset()
+ (x - target) / it->mXTranslate.getScale());
}
// if not started, the delay offset doesn't change.
}
const S xOffset = operation->getXOffset();
if (!std::isnan(xOffset)) {
if (it->isStarted()) {
const int64_t frameCount = clockTime ? VolumeShaper::getNowUs() : mLastFrame;
const S x = it->mXTranslate((T)frameCount);
VS_LOG("normalizedTime translation: %f", x);
const S target =
(operation->getFlags() & VolumeShaper::Operation::FLAG_REVERSE) != 0 ?
MAX_CURVE_TIME - xOffset : xOffset;
VS_LOG("normalizedTime target x offset: %f", target);
it->mXTranslate.setOffset(it->mXTranslate.getOffset()
+ (x - target) / it->mXTranslate.getScale());
} else {
it->setDelayXOffset(xOffset);
}
}
it->mOperation = operation; // replace the operation
} break;
}
return VolumeShaper::Status(id);
}
sp<VolumeShaper::State> getVolumeShaperState(int id) {
AutoMutex _l(mLock);
auto it = findId_l(id);
if (it == mVolumeShapers.end()) {
VS_LOG("cannot find state for id: %d", id);
return nullptr;
}
return it->getState();
}
/* getVolume() is not const, as it updates internal state.
* Once called, any VolumeShapers not already started begin running.
*/
std::pair<T /* volume */, bool /* active */> getVolume(int64_t trackFrameCount) {
AutoMutex _l(mLock);
mLastFrame = trackFrameCount;
T volume(1);
size_t activeCount = 0;
for (auto it = mVolumeShapers.begin(); it != mVolumeShapers.end();) {
const std::pair<T, bool> shaperVolume =
it->getVolume(trackFrameCount, mSampleRate);
volume *= shaperVolume.first;
activeCount += shaperVolume.second;
++it;
}
mLastVolume = std::make_pair(volume, activeCount != 0);
VS_LOG("getVolume: <%f, %s>", mLastVolume.first, mLastVolume.second ? "true" : "false");
return mLastVolume;
}
/* Used by a client side VolumeHandler to ensure all the VolumeShapers
* indicate that they have been started. Upon a change in audioserver
* output sink, this information is used for restoration of the server side
* VolumeHandler.
*/
void setStarted() {
(void)getVolume(mLastFrame); // getVolume() will start the individual VolumeShapers.
}
std::pair<T /* volume */, bool /* active */> getLastVolume() const {
AutoMutex _l(mLock);
return mLastVolume;
}
std::string toString() const {
AutoMutex _l(mLock);
std::stringstream ss;
ss << "VolumeHandler{mSampleRate=" << mSampleRate;
ss << ", mLastFrame=" << mLastFrame;
ss << ", mVolumeShapers={";
bool first = true;
for (const auto &shaper : mVolumeShapers) {
if (first) {
first = false;
} else {
ss << ", ";
}
ss << shaper.toString().c_str();
}
ss << "}}";
return ss.str();
}
void forall(const std::function<VolumeShaper::Status (const VolumeShaper &)> &lambda) {
AutoMutex _l(mLock);
VS_LOG("forall: mVolumeShapers.size() %zu", mVolumeShapers.size());
for (const auto &shaper : mVolumeShapers) {
VolumeShaper::Status status = lambda(shaper);
VS_LOG("forall applying lambda on shaper (%p): %d", &shaper, (int)status);
}
}
void reset() {
AutoMutex _l(mLock);
mVolumeShapers.clear();
mLastFrame = 0;
// keep mVolumeShaperIdCounter as is.
}
/* Sets the configuration id if necessary - This is based on the counter
* internal to the VolumeHandler.
*/
void setIdIfNecessary(const sp<VolumeShaper::Configuration> &configuration) {
if (configuration->getType() == VolumeShaper::Configuration::TYPE_SCALE) {
const int id = configuration->getId();
if (id == -1) {
// Reassign to a unique id, skipping system ids.
AutoMutex _l(mLock);
while (true) {
if (mVolumeShaperIdCounter == INT32_MAX) {
mVolumeShaperIdCounter = VolumeShaper::kSystemVolumeShapersMax;
} else {
++mVolumeShaperIdCounter;
}
if (findId_l(mVolumeShaperIdCounter) != mVolumeShapers.end()) {
continue; // collision with an existing id.
}
configuration->setId(mVolumeShaperIdCounter);
ALOGD("setting id to %d", mVolumeShaperIdCounter);
break;
}
}
}
}
private:
std::list<VolumeShaper>::iterator findId_l(int32_t id) {
std::list<VolumeShaper>::iterator it = mVolumeShapers.begin();
for (; it != mVolumeShapers.end(); ++it) {
if (it->mConfiguration->getId() == id) {
break;
}
}
return it;
}
size_t numberOfUserVolumeShapers_l() const {
size_t count = 0;
for (const auto &shaper : mVolumeShapers) {
count += (shaper.mConfiguration->getId() >= VolumeShaper::kSystemVolumeShapersMax);
}
return count;
}
mutable Mutex mLock;
double mSampleRate; // in samples (frames) per second
int64_t mLastFrame; // logging purpose only, 0 on start
int32_t mVolumeShaperIdCounter; // a counter to return a unique volume shaper id.
std::pair<T /* volume */, bool /* active */> mLastVolume;
std::list<VolumeShaper> mVolumeShapers; // list provides stable iterators on erase
}; // VolumeHandler
} // namespace media
} // namespace android
#pragma pop_macro("LOG_TAG")
#endif // ANDROID_VOLUME_SHAPER_H
| {
"redpajama_set_name": "RedPajamaGithub"
} | 3,202 |
KNOW YOUR DECK ,HAVE IT INSPECTED EVERY COPLE OF YEARS !!
AFTER READING NICK GROMICKO, Founder, International Association of Certified Home Inspectors (InterNACHI) AND A MASTER CERTIFIED INSPECTOR'S FACTS ABOUT DECK'S ? WHY WOULDN'T YOU HIRE L.A.C HOMEINSPECTIONS WHO'S IS CERTIFIED BY (InterNACHI) EXPLAIN THE SAFTY AND CONDITION'S OF YOUR DECK. LETS FACE IT, MOST FAMILY'S SPEND ALOT OF TIME ON THEM, ALL TOGETHER IN THE SUMMER MONTHS.
The North American Deck and Railing Association: Improper Fastening of Deck Ledgers to Houses !
Building a deck is the ideal DIY project — it's fairly straightforward and materials are simple.
But a recent spate of deck failures reveals that many decks fail where the deck ledger fastens to the house — one of the more technically challenging steps of deck-building.
Improper (or missing) flashing to keep water from seeping behind the ledger where it can soften and rot out wood.
Using old fastening methods, such as plain nails, to secure the ledger to the house.
It's a good idea to have your deck inspected for safety and proper construction techniques when you buy it and every couple of years after. | {
"redpajama_set_name": "RedPajamaC4"
} | 1,783 |
Q: Definir a prioridade de threads em C++11 No programa que estou desenvolvendo tenho dois std::threads que estão
sempre ativos durante toda a vida do programa. No entanto, considero
que a função de um deles é de menor importância e gostaria de alterar
a prioridade deles.
Dei uma olhada na documentação e não achei nenhuma função que defina
a prioridade do std::thread.
Minha duvida é: como defino a prioridade de um std::thread? É possível ou
o próprio sistema operacional se encarrega de definir isso em tempo de execução?
Obs: O programa só rodará em Linux (Debian), por isso, não há necessidade de
portabilidade com Windows.
A: Não há nenhuma forma de alterar a prioridade de um std::thread no C++11 ou no C++14. A única forma de fazer isso seria pelo uso de funções do linux (não portável). Obtenha um identificador nativo com o std::thread::native_handle() e o use com a função pthread_setschedparam. Um exemplo (retirado da primeira referência):
#include <thread>
#include <iostream>
#include <chrono>
#include <cstring>
#include <pthread.h>
std::mutex iomutex;
void f(int num)
{
std::this_thread::sleep_for(std::chrono::seconds(1));
sched_param sch;
int policy;
pthread_getschedparam(pthread_self(), &policy, &sch);
std::lock_guard<std::mutex> lk(iomutex);
std::cout << "Thread " << num << " executando na prioridade "
<< sch.sched_priority << '\n';
}
int main()
{
std::thread t1(f, 1), t2(f, 2);
sched_param sch;
int policy;
pthread_getschedparam(t1.native_handle(), &policy, &sch);
sch.sched_priority = 20;
if (pthread_setschedparam(t1.native_handle(), SCHED_FIFO, &sch)) {
std::cout << "setschedparam falhou: " << std::strerror(errno) << '\n';
}
t1.join(); t2.join();
}
No Windows a mesma ideia pode ser aplicada com a SetThreadPriority.
A: Segundo essa resposta a biblioteca padrão do C++11 não presta suporte padrão para o controle de prioridade de threads. (O autor ainda acredita que não isso não mudará no C++14)
Nessa mesma resposta ele cita um comando que funciona em sistemas que sigam as normas POSIX:
pthread_setschedparam(thread.native_handle(), politica, {prioridade});
Como você quer só para linux, esse método deve resolver seu problema. Existem ainda alguns pontos relevantes a serem levados em conta.
A política padrão de threads do Linux tem prioridade dinâmica
Geralmente, quando você inicia uma thread, o Linux coloca a política SCHED_OTHER, como é visto nessa resposta da SOEN.
Nessa mesma resposta, ele coloca os tipos de política que podem ser adotadas para o sistema de thread e qual a prioridade mínima e a máxima:
SCHED_FIFO: Esquema de fila, primeira a entrar, é o primeiro a sair. (1/99)
SCHED_RR: Esquema de política round-robin. (1/99)
Onde a prioridade está da seguinte forma (min/máx). Eu optei por colocar as políticas que tinha prioridade. Apesar de eu ler nos comentários que o SCHED_OTHER pode oferecer um certo nível de controle de prioridade, ele é definido pelo próprio sistema conforme o comportamento da thread, o que você pode fazer é dar "uma dica da importância da thread", setando sua prioridade como muito alta (-20) ou muito baixa (19).
Políticas para troca de threads
Threads com as políticas SCHED_RR ou SCHED_FIFO serão trocadas se um dos dois eventos acontecerem, ainda segundo esse link:
*
*Uma thread é posta para dormir (sleep) ou passa a esperar um evento
*Uma thread de tempo real de prioridade maior está pronta para rodar
Esses pontos devem ser levados em conta quando você for implementar as suas threads.
Dito isso, vamos ao nosso exemplo:
Exemplo retirado do cpp reference:
#include <thread>
#include <mutex>
#include <iostream>
#include <chrono>
#include <cstring>
#include <pthread.h>
std::mutex iomutex;
void f(int num) {
std::this_thread::sleep_for(std::chrono::seconds(1));
sched_param sch;
int policy;
pthread_getschedparam(pthread_self(), &policy, &sch);
std::lock_guard<std::mutex> lk(iomutex);
std::cout << "Thread " << num << " esta executando com prioridade "
<< sch.sched_priority << '\n';
}
int main(){
//A thread 2 será uma thread padrão
std::thread t1(f, 1), t2(f, 2);
sched_param sch;
int policy;
pthread_getschedparam(t1.native_handle(), &policy, &sch);
sch.sched_priority = 20;
//Nessa linha ele seta a política e a prioridade da thread 1
if(pthread_setschedparam(t1.native_handle(), SCHED_FIFO, &sch)) {
std::cout << "Falha para utilizar setschedparam: " << std::strerror(errno) << '\n';
}
t1.join(); t2.join();
}
Apesar de tudo, eu fiquei com uma dúvida, talvez pelo meu desconhecimento de como sistemas POSIX tratam as prioridades de threads, mas, pelo que o exemplo coloca como saída, threads com valores mais altos de prioridade, tem prioridade menor.
Saída do exemplo:
Thread 2 esta executando com prioridade 0
Thread 1 esta executando com prioridade 20
A: A forma de fazer isso envolve a obtenção de uma handle para a implementação nativa da thread. Um exemplo de uso da função std::thread::native_handle está descrito em http://pt.cppreference.com/w/cpp/thread/thread/native_handle.
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 4,337 |
The appearance was related to the State Library of Tasmania, a long-time LibraryThing for Libraries member, adding our "Reviews" enhancement, and public talk John is giving on Wednesday at the State Library in Hobart tomorrow, Wednesday at 4:00pm.
More on the talk here.
An example book at the State Library, with reviews, here.
The kindest interpretation of statements like "the future is mobile" or "the future of reference is SMS" or "the future is librarians in pods" or whatever is that the librarians are trying to create that future by speaking it. The incantation will somehow make it so…. The less kind interpretation is that the authors of such statements are reductionist promoters, reducing a complex field to whatever marginal utility they're focused on and claiming that this is the future, while simultaneously promoting themselves as seers.
The obvious and most likely statement is that nothing is the future, as in no thing is the future, period. Anyone who tells you different is just plain wrong. With technology, it should be clear to anyone who bothers to see past their obsessions that formats and tools die hard. Some people like to imply that if librarians don't take up every new trend they'll become like buggy whip makers. I should point out that there are still people who make buggy whips. Buggy whips aren't as popular as they once were, but they're still around. There are even buggies to accompany them.
Though a purveyor of "Web 2.0″ ideas—I founded LibraryThing, what can I say?—I think it's a great post.
It's depressing, but it's not unique to library technology. You see it in other trends, like "green libraries" (they're the future, didn't you get the memo?). It's in the dynamics of communication. Your post is a good corrective to it.
At the same time, you're missing something. I don't know if you're missing it for real, or just in this focused expression. But there's a powerful "yes but" here, and it needs saying—shouting even!—lest people take the wrong thing from your post.
For all the nonsense and hype, librares are subject to an extraordinary and rapid cultural change. They have already changed drastically—especially if "libraries" means what libraries mean to culture generally, and people who don't work in them.
Libraries are in the "information business" and this business is in one of the most profound transformations in human history. This isn't buggies vs. Stanley Steamers—different ways of getting to the habberdasher. It's horse-and-buggy culture vs. everything the car has brought—mass production, suburban living, the Blitzkreig, the global economy, global warming and the sexual revolution. Certainly, as you say, carriges continue to exist as objects that convey people, but their meaning has been utterly transformed. If libraries end up as a way for rich people to indulge children on a visit to a big city—what carriages mean today—well, crap! How did that happen?!
The world is changing, and for all the noise about this or that technology, I don't think libraries are dealing with it squarely. (Forget Web 2.0; libraries haven't really ingested Web 1.0 yet.) "The future is X" isn't the best response to that change, but it's a response.
I expect your post will get wide circulation. It says something that hasn't been said before as well. But if it prompts librarians to dismiss technology's impact on the future of libraries, it will do great harm. Instead, I hope people use your essay as a way to "kick it up a notch" intellectually, get past the small stuff and confront the very real changes ahead.
PS: By the way, LibraryThing is releasing a universal mobile catalog. It's the future. No, really!
Shelf Browse—which we announced last week—is now live in High Plains Library District's catalog. As we mentioned in our brief ALA announcement, Shelf Browse lets you browse your library's shelves visually, just as you would do in the physical library.
Shelf Browse lets your patrons see where a book sits on your actual shelves, and what's near it. It includes a "mini-browser" that sits on your detail pages, and a full-screen version, launched from the detail page.
Scroll back and forth, serendipitously browsing through the shelves. If lists are more your speed, in the full-screen version, you can switch between shelf and list mode.
For ordering information contact Peder Christensen at Bowker—toll-free at 877-340-2400 or email Peder.Christensen@bowker.com. | {
"redpajama_set_name": "RedPajamaC4"
} | 3,011 |
\section{Introduction}
\label{sec:introduction}
Bioinformaticians define the $k$th-order de Bruijn graph for a string or set of strings to be the directed graph whose nodes are the distinct $k$-tuples in those strings and in which there is an edge from $u$ to $v$ if there is a \((k + 1)\)-tuple somewhere in those strings whose prefix of length $k$ is $u$ and whose suffix of length $k$ is $v$.\footnote{An alternative definition, which our data structure can be made to handle but which we do not consider in this paper, has an edge from $u$ to $v$ whenever both nodes are in the graph.} These graphs have many uses in bioinformatics, including {\it de novo\/} assembly~\cite{zerbino2008velvet}, read correction~\cite{DBLP:journals/bioinformatics/SalmelaR14} and pan-genomics~\cite{siren2014indexing}. The datasets in these applications are massive and the graphs can be even larger, however, so pointer-based implementations are impractical. Researchers have suggested several approaches to representing de Bruijn graphs compactly, the two most popular of which are based on Bloom filters~\cite{wabi,cascading} and the Burrows-Wheeler Transform~\cite{bowe2012succinct,boucher2015variable,belazzougui2016bidirectional}, respectively. In this paper we describe a new approach, based on minimal perfect hash functions~\cite{mehlhorn1982program}, that is similar to that using Bloom filters but has better theoretical bounds when the number of connected components in the graph is small, and is fully dynamic: i.e., we can both insert and delete nodes and edges efficiently, whereas implementations based on Bloom filters are usually semi-dynamic and support only insertions. We also show how to modify our implementation to support, e.g., jumbled pattern matching~\cite{BCFL12} with fixed-length patterns.
Our data structure is based on a combination of Karp-Rabin hashing~\cite{KR87} and minimal perfect hashing, which we will describe in the full version of this paper and which we summarize for now with the following technical lemmas:
\begin{lemma}
\label{lem:static}
Given a static set $N$ of $n$ $k$-tuples over an alphabet $\Sigma$ of size $\sigma$, with high probability in $O(kn)$ expected time we can build a function \(f : \Sigma^k \rightarrow \{0, \ldots, n - 1\}\) with the following properties:
\begin{itemize}
\item when its domain is restricted to $N$, $f$ is bijective;
\item we can store $f$ in $O(n + \log k+\log\sigma)$ bits;
\item given a $k$-tuple $v$, we can compute \(f (v)\) in $\Oh{k}$ time;
\item given $u$ and $v$ such that the suffix of $u$ of length \(k - 1\) is the prefix of $v$ of length \(k - 1\), or vice versa, if we have already computed \(f (u)\) then we can compute \(f (v)\) in $\Oh{1}$ time.
\end{itemize}
\end{lemma}
\begin{lemma}
\label{lem:dynamic}
If $N$ is dynamic then we can maintain a function $f$ as described in Lemma~\ref{lem:static} except that:\
\begin{itemize}
\item the range of $f$ becomes \(\{0, \ldots, 3 n - 1\}\);
\item when its domain is restricted to $N$, $f$ is injective;
\item our space bound for $f$ is $\Oh{n (\log \log n + \log \log \sigma)}$ bits with high probability;
\item insertions and deletions take $\Oh{k}$ amortized expected time.
\item the data structure may work incorrectly with very low probability
(inversely polynomial in $n$).
\end{itemize}
\end{lemma}
Suppose $N$ is the node-set of a de Bruijn graph. In Section~\ref{sec:static} we show how we can store $\Oh{n \sigma}$ more bits than Lemma~\ref{lem:static} such that, given a pair of $k$-tuples $u$ and $v$ of which at least one is in $N$, we can check whether the edge \((u, v)\) is in the graph. This means that, if we start with a $k$-tuple in $N$, then we can explore the entire connected component containing that $k$-tuple in the underlying undirected graph. On the other hand, if we start with a $k$-tuple not in $N$, then we will learn that fact as soon as we try to cross an edge to a $k$-tuple that is in $N$. To deal with the possibility that we never try to cross such an edge, however --- i.e.,\@\xspace that our encoding as described so far is consistent with a graph containing a connected component disjoint from $N$ --- we cover the vertices with a forest of shallow rooted trees. We store each root as a $k$-tuple, and for each other node we store \(1 + \lg \sigma\) bits indicating which of its incident edges leads to its parent. To verify that a $k$-tuple we are considering is indeed in the graph, we ascend to the root of the tree that contains it and check that $k$-tuple is what we expect. The main challenge for making our representation dynamic with Lemma~\ref{lem:dynamic} is updating the covering forest. In Section~\ref{sec:dynamic} how we can do this efficiently while maintaining our depth and size invariants. Finally, in Section~\ref{sec:jumbled} we observe that our representation can be easily modified for other applications by replacing the Karp-Rabin hash function by other kinds of hash functions. To support jumbled pattern matching with fixed-length patterns, for example, we hash the histograms indicating the characters' frequencies in the $k$-tuples.
\section{Static de Bruijn Graphs}
\label{sec:static}
Let \(G\) be a de Bruijn graph of order \(k\), let \(N = \{v_0, \ldots, v_{n-1}\}\) be the set of its nodes, and let \(E = \{a_0, \ldots, a_{e-1}\}\) be the set of its edges. We call each \(v_i\) either a node or a \(k\)-tuple, using interchangeably the two terms since there is a one-to-one correspondence between nodes and labels.
We maintain the structure of \(G\) by storing two binary matrices, \IN and \OUT, of size \(n \times \sigma\). For each node, the former represents its incoming edges whereas the latter represents its outgoing edges. In particular, for each \(k\)-tuple \(v_x = c_1 c_2 \ldots c_{k-1} a\), the former stores a row of length \(\sigma\) such that, if there exists another \(k\)-tuple \(v_y = b c_1 c_2 \ldots c_{k-1}\) and an edge from \(v_y\) to \(v_x\), then the position indexed by \(b\) of such row is set to \TRUE. Similarly, \OUT contains a row for \(v_y\) and the position indexed by \(a\) is set to \TRUE. As previously stated, each \(k\)-tuple is uniquely mapped to a value between \(0\) and \(n-1\) by \(f\), where $f$ is as defined in Lemma~\ref{lem:static}, and therefore we can use these values as indices for the rows of the matrices \IN and \OUT, i.e.,\@\xspace in the previous example the values of \(\IN[f(v_x)][b]\) and \(\OUT[f(v_y)][a]\) are set to \TRUE. We note that, e.g., the SPAdes assembler~\cite{Ban12} also uses such matrices.
Suppose we want to check whether there is an edge from \(b X\) to \(X a\).
Letting \(f(b X) = i\) and \(f(X a) = j\), we first assume \(b X\) is in \(G\) and check the values of \(\OUT [i] [a] \) and \( \IN [j] [b]\). If both values are \TRUE, we report that the edge is present and we say that the edge is \emph{confirmed} by \IN and \OUT; otherwise, if any of the two values is \FALSE, we report that the edge is absent. Moreover, note that if \(b X\) is in \(G\) and \(\OUT [i] [a] = \TRUE\), then \(X a\) is in \(G\) as well. Symmetrically, if \(X a\) is in \(G\) and \(\IN [j] [b] = \TRUE\), then \(b X\) is in \(G\) as well. Therefore, if \(\OUT [i] [a] = \IN [j] [b] = \TRUE\), then \(b X\) is in \(G\) if and only if \(X a\) is. This means that, if we have a path \(P\) and if all the edges in \(P\) are confirmed by \IN and \OUT, then either all the nodes touched by \(P\) are in \(G\) or none of them is.
We now focus on detecting false positives in our data structure maintaining a reasonable memory usage. Our strategy is to sample a subset of nodes for which we store the plain-text \(k\)-tuple and connect all the unsampled nodes to the sampled ones. More precisely, we partition nodes in the undirected graph \(G^\prime\) underlying \(G\) into a forest of rooted trees of height at least \(k \lg \sigma \) and at most \(3 k \lg \sigma\). For each node we store a pointer to its parent in the tree, which takes \(1 + \lg \sigma\) bits per node, and we sample the \(k\)-mer at the root of such tree. We allow a tree to have height smaller than \(k \lg \sigma\) when necessary,
e.g., if it covers a connected component. Figure~\ref{fig:trees} shows an illustration of this idea.
\begin{figure}[t!]
\begin{center}
\includegraphics[width=\textwidth]{trees.pdf}
\caption{Given a de Bruijn graph (left), we cover the underlying undirected graph with a forest of rooted trees of height at most \(3 k \lg \sigma\) (center). The roots are shown as filled nodes, and parent pointers are shown as arrows; notice that the directions of the arrows in our forest are not related to the edges' directions in the original de Bruijn graph. We sample the $k$-tuples at the roots so that, starting at a node we think is in the graph, we can verify its presence by finding the root of its tree and checking its label in $\Oh{k \log \sigma}$ time. The most complicated kind of update (right) is adding an edge between a node $u$ in a small connected component to a node $v$ in a large one, $v$'s depth is more than \(2 k \lg \sigma\) in its tree. We re-orient the parent pointers in $u$'s tree to make $u$ the temporary root, then make $u$ point to $v$. We ascend \(k \lg \sigma\) steps from $v$, then delete the parent pointer $e$ of the node $w$ we reach, making $w$ a new root. (To keep this figure reasonably small, some distances in this example are smaller than prescribed by our formulas.)}
\label{fig:trees}
\end{center}
\end{figure}
We can therefore check whether a given node \(v_x\) is in \(G\) by first computing \(f(v_x)\) and then checking and ascending at most \(3 k \lg \sigma\) edges, updating \(v_x\) and \(f(v_x)\) as we go. Once we reach the root of the tree we can compare the resulting \(k\)-tuple with the one sampled to check if \(v_x\) is in the graph. This procedure requires \Oh{k \lg \sigma} time since computing the first value of \(f(v_x)\) requires \Oh{k}, ascending the tree requires constant time per edge, and comparing the \(k\)-tuples requires \Oh{k}.
We now describe a Las Vegas algorithm for the construction of this data structure that requires, with high probability, \Oh{kn + n\sigma} expected time.
We recall that \(N\) is the set of input nodes of size \(n\).
We first select a function \(f\) and construct bitvector \(B\) of size \(n\) initialized with all its elements set to \FALSE.
For each elements \(v_x\) of \(N\) we compute \(f(v_x) = i\) and check the value of \(B[i]\).
If this value is \FALSE we set it to \TRUE and proceed with the next element in \(N\), if it is already set to \TRUE, we reset \(B\), select a different function \(f\), and restart the procedure from the first element in \(N\).
Once we finish this procedure --- i.e.,\@\xspace we found that \(f\) do not produces collisions when applied to \(N\) --- we store \(f\) and proceed to initialize \IN and \OUT correctly.
This procedure requires with high probability \Oh{kn} expected time for constructing \(f\) and \Oh{n\sigma} time for computing \IN and \OUT.
Notice that if \(N\) is the set of \(k\)-tuples of a single text sorted by their starting position in the text, each \(f(v_x)\) can be computed in constant time from \(f(v_{x-1})\) except for \(f(v_0)\) that still requires \Oh{k}.
More generally, if \(N\) is the set of \(k\)-tuples of \(t\) texts sorted by their initial position, we can compute \(n - t\) values of the function \(f(v_x)\) in constant time from \(f(v_{x-1})\) and the remaining in \Oh{k}.
We will explain how to build the forest in the full version of this paper.
In this case the construction requires, with high probability, \(\Oh{kt + n + n\sigma} = \Oh{kt + n\sigma}\) expected time.
Combining our forest with Lemma~\ref{lem:static}, we can summarize our static data structure in the following theorem:
\begin{theorem}
\label{thm:static}
Given a static $\sigma$-ary $k$th-order de Bruijn graph $G$ with $n$ nodes, with high probability in $\Oh{k n + n \sigma}$
expected time we can store $G$ in $\Oh{\sigma n}$ bits plus $\Oh{k \log \sigma}$ bits for each connected component in the
underlying undirected graph, such that checking whether a node is in $G$ takes $\Oh{k \log \sigma}$ time, listing the edges incident to a node we are visiting takes $\Oh{\sigma}$ time, and crossing an edge takes $\Oh{1}$ time.
\end{theorem}
In the full version we will show how to use monotone minimal perfect hashing~\cite{BBPV09}
to reduce the space to $(2+\epsilon)n\sigma$ bits of space
(for any constant $\epsilon>0$). We will also show how to reduce the time to list the
edges incident to a node of degree $d$ to $O(d)$, and the time
to check whether a node is in $G$ to $\Oh{k}$. We note that the obtained space and query times are
both optimal up to constant factors, which is unlike previous methods which have additional
factor(s) depending on $k$ and/or $\sigma$ in space and/or time.
\section{Dynamic de Bruijn Graphs}
\label{sec:dynamic}
In the previous section we presented a static representation of de Buijn graphs, we now present how we can make this data structure dynamic. In particular, we will show how we can insert and remove edges and nodes and that updating the graph reduces to managing the covering forest over \(G\).
In this section, when we refer to $f$ we mean the function defined in Lemma~\ref{lem:dynamic}.
We first show how to add or remove an edge in the graph and will later describe how to add or remove a node in it.
The updates must maintain the following invariant: any tree must have size at least $k\log\sigma$
and height at most $3k\log\sigma$ except when the tree covers (all nodes in) a connected component
of size at most $k\log\sigma$.
Let \(v_x\) and \(v_y\) be two nodes in \(G\), \(e = (v_x, v_y)\) be an edge in \(G\), and let \(f(v_x) = i\) and \(f(v_y) = j\).
Suppose we want to add \(e\) to \(G\). First, we set to \TRUE the values of \(\OUT[i][a]\) and \(\IN[j][b]\) in constant time.
We then check whether \(v_x\) or \(v_y\) are in different components of size less than \(k \lg \sigma\) in \Oh{k \lg \sigma} time for each node.
If both components have size greater than \(k \lg \sigma\) we do not have to proceed further since the trees will not change. If both connected components have size less than \(k \lg \sigma\) we merge their trees in \Oh{k \lg \sigma} time by traversing both trees and switching the orientation of the edges in them, discarding the samples at the roots of the old trees and sampling the new root in \Oh{k} time.
If only one of the two connected components has size greater than \(k \lg \sigma\) we select it and perform a tree traversal to check whether the depth of the node is less than \(2 k \lg \sigma\).
If it is, we connect the two trees as in the previous case. If it is not, we traverse the tree in the bigger components upwards for \(k \lg \sigma\) steps, we delete the edge pointing to the parent of the node we reached creating a new tree, and merge it with the smaller one.
This procedure requires \Oh{k \lg \sigma} time since deleting the edge pointing to the parent in the tree requires \Oh{1} time, i.e.,\@\xspace we have to reset the pointer to the parent in only one node.
Suppose now that we want to remove \(e\) from \(G\).
First we set to \FALSE the values of \(\OUT[i][a]\) and \(\IN[j][b]\) in constant time.
Then, we check in \Oh{k} time whether \(e\) is an edge in some tree by computing \(f(v_x)\) and \(f(v_y)\) checking for each node if that edge is the one that points to their parent.
If \(e\) is not in any tree we do not have to proceed further whereas if it is we check the size of each tree in which \(v_x\) and \(v_y\) are.
If any of the two trees is small (i.e.,\@\xspace if it has fewer than \(k \lg \sigma\) elements) we search any outgoing edge from the tree that connects it to some other tree.
If such an edge is not found we conclude that we are in a small connected component that is covered by the current tree and we sample a node in the tree as a root and switch directions of some edges if necessary.
If such an edge is found, we merge the small tree with the bigger one by adding the edge and switch the direction of some edges originating from the small tree if necessary. Finally if the height of the new tree exceeds $3k\log\sigma$, we traverse the tree upwards from the deepest node in the tree (which was necessarily a node in the smaller tree before the merger) for \(2k \lg \sigma\) steps, delete the edge pointing to the parent of the reached node, creating a new tree.
This procedure requires
$\Oh{k \lg \sigma}$ since the number of nodes traversed is at most \(O(k \lg \sigma)\)
and the number of changes to the data structures is also at most \(O(k \lg \sigma)\)
with each change taking expected constant time.
It is clear that the insertion and deletion algorithms will maintain the invariant on the tree sizes.
It is also clear that the invariant implies that the number of sampled nodes is $O(n/(k\log\sigma))$ plus the
number of connected components.
We now show how to add and remove a node from the graph.
Adding a node is trivial since it will not have any edge connecting it to any other node.
Therefore adding a node reduces to modify the function \(f\) and requires \Oh{k} amortized expected time.
When we want to remove a node, we first remove all its edges one by one and, once the node is isolated from the graph, we remove it by updating the function \(f\).
Since a node will have at most \(\sigma\) edges and updating \(f\) requires \Oh{k} amortized expected time, the amortized expected time complexity of this procedure is
$\Oh{\sigma k\lg \sigma+ k}$.
Combining these techniques for updating our forest with Lemma~\ref{lem:dynamic}, we can summarize our dynamic data structure in the following theorem:
\begin{theorem}
\label{thm:dynamic}
We can maintain a $\sigma$-ary $k$th-order de Bruijn graph $G$ with $n$ nodes that is fully dynamic (i.e., supporting node and edge insertions and deletions) in $\Oh{n (\log \log n + \sigma)}$ bits (plus $\Oh{k \log \sigma}$ bits for each connected component) with high probability, such that we can add or remove an edge in expected \Oh{k\lg\sigma} time, add a node in expected \Oh{k+\sigma} time, and remove a node in expected \Oh{\sigma k\lg \sigma} time, and queries have the same time bounds as in Theorem~\ref{thm:static}. The data structure may work incorrectly with very low probability (inversely polynomial in $n$).
\end{theorem}
\section{Jumbled Pattern Matching}
\label{sec:jumbled}
Karp-Rabin hash functions implicitly divide their domain into equivalence classes --- i.e., subsets in which the elements hash to the same value. In this paper we have chosen Karp-Rabin hash functions such that each equivalence class contains only one $k$-tuple in the graph. Most of our efforts have gone into being able, given a $k$-tuple and a hash value, to determine whether that $k$-tuple is the unique element of its equivalence class in the graph. In some sense, therefore, we have treated the equivalence relation induced by our hash functions as a necessary evil, useful for space-efficiency but otherwise an obstacle to be overcome. For some applications, however --- e.g., parameterized pattern matching, circular pattern matching or jumbled pattern matching --- we are given an interesting equivalence relation on strings and asked to preprocess a text such that later, given a pattern, we can determine whether any substrings of the text are in the same equivalence class as the pattern. We can modify our data structure for some of these applications by replacing the Karp-Rabin hash function by other kinds of hash functions.
For indexed jumbled pattern matching~\cite{BCFL12,KRR13,ACLL14} we are asked to pre-process a text such that later, given a pattern, we can determine quickly whether any substring of the text consists of exactly the same multiset of characters in the pattern. Consider fixed-length jumbled pattern matching, when the length of the patterns is fixed at pre-processing time. If we modify Lemmas~\ref{lem:static} and~\ref{lem:dynamic} so that, instead of using Karp-Rabin hashes in the definition of the function $f$, we use a hash function on the histograms of characters' frequencies in $k$-tuples, our function $f$ will map all permutations of a $k$-tuple to the same value. The rest of our implementation stays the same, but now the nodes of our graph are multisets of characters of size $k$ and there is an edge between two nodes $u$ and $v$ if it is possible to replace an element of $u$ and obtain $v$. If we build our graph for the multisets of characters in $k$-tuples in a string $S$, then our process for checking whether a node is in the graph tells us whether there is a jumbled match in $S$ for a pattern of length $k$. If we build a tree in which the root is a graph for all of $S$, the left and right children of the root are graphs for the first and second halves of $S$, etc., as described by Gagie et al.~\cite{GHLW15}, then we increase the space by a logarithmic factor but we can return the locations of all matches quickly.
\begin{theorem}
\label{thm:jumbled}
Given a string \(S [1..n]\) over an alphabet of size $\sigma$ and a length $k \ll n$, with high probability in $\Oh{k n + n \sigma}$ expected time we can store \((2n \log \sigma)(1+o(1))\) bits such that later we can determine in $\Oh{k \log \sigma}$ time if a pattern of length $k$ has a jumbled match in $S$.
\end{theorem}
\section*{Acknowledgements}
Many thanks to Rayan Chikhi and the anonymous reviewers for their comments.
\bibliographystyle{splncs03}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 5,873 |
{"url":"https:\/\/johncarlosbaez.wordpress.com\/page\/2\/","text":"## Saving Climate Data (Part\u00a02)\n\n16 December, 2016\n\nI want to get you involved in the Azimuth Environmental Data Backup Project, so click on that for more. But first some background.\n\nA few days ago, many scientists, librarians, archivists, computer geeks and environmental activists started to make backups of US government environmental databases. We\u2019re trying to beat the January 20th deadline just in case.\n\nBacking up data is always a good thing, so there\u2019s no point in arguing about politics or the likelihood that these backups are needed. The present situation is just a nice reason to hurry up and do some things we should have been doing anyway.\n\nAs of 2 days ago the story looked like this:\n\nSaving climate data (Part 1), Azimuth, 13 December 2016.\n\nA lot has happened since then, but much more needs to be done. Right now you can see a list of 90 databases to be backed up here:\n\nGov. Climate Datasets (Archive). (Click on the tiny word \u201cDatasets\u201d at the bottom of the page!)\n\nDespite the word \u2018climate\u2019, the scope includes other environmental databases, and rightly so. Here is a list of databases that have been backed up:\n\nBy going here and clicking \u201cStart Here to Help\u201d:\n\nyou can nominate a dataset for rescue, claim a dataset to rescue, let everyone know about a data rescue event, or help in some other way (which you must specify). There\u2019s also other useful information on this page, which was set up by Nick Santos.\n\nThe overall effort is being organized by the Penn Program in the Environmental Humanities, or \u2018PPEHLab\u2019 for short, headed by Bethany Wiggin. If you want to know what\u2019s going on, it helps to look at their blog:\n\nHowever, the people organizing the project are currently overwhelmed with offers of help! People worldwide are proceeding to take action in a decentralzed way! So, everything is a bit chaotic, and nobody has an overall view of what\u2019s going on.\n\nI can\u2019t overstate this: if you think that \u2018they\u2019 have a plan and \u2018they\u2019 know what\u2019s going on, you\u2019re wrong. \u2018They\u2019 is us. Act accordingly.\n\nHere\u2019s a list of news articles, a list of \u2018data rescue events\u2019 where people get together with lots of computers and do stuff, and a bit about archives and archivists.\n\n### News\n\nHere are some things to read:\n\n\u2022 Jason Koebler, Researchers are preparing for Trump to delete government science from the web, Vice, 13 December 2016.\n\n\u2022 Brady Dennis, Scientists frantically copying U.S. climate data, fearing it might vanish under Trump, Washington Post, 13 December, 2016. (Also at the Chicago Tribune.)\n\n\u2022 Eric Holthaus, Why I\u2019m trying to preserve federal climate data before Trump takes office, Washington Post, 13 December 2016.\n\n\u2022 Nicole Mortillaro, U of T heads \u2018guerrilla archiving event\u2019 to preserve climate data ahead of Trump presidency, CBC News, 14 December 2016.\n\n\u2022 Audie Kornish and Eric Holthaus, Scientists race to preserve climate change data before Trump takes office, All Things Considered, National Public Radio, 14 December 2016.\n\n### Data rescue events\n\nThere\u2019s one in Toronto:\n\nGuerrilla archiving event, 10 am \u2013 4 pm EST, Saturday 17 December 2016. Location: Bissell Building, 4th Floor, 140 St. George St. University of Toronto.\n\nThere will be one in Philadelphia:\n\nDataRescuePenn Data Harvesting, Friday\u2013Saturday 13\u201314 January 2017. Location: not determined yet, probably somewhere at the University of Pennsylvania, Philadelphia.\n\nI hear there will also be events in New York City and Los Angeles, but I don\u2019t know details. If you do, please tell me!\n\n### Archives and archivists\n\nToday I helped catalyze a phone conversation between Bethany Wiggin, who heads the PPEHLab, and Nancy Beaumont, head of the Society of American Archivists. Digital archivists have a lot of expertise in saving information, so their skills are crucial here. Big wads of disorganized data are not very useful.\n\nIn this conversation I learned that some people are already in contact with the Internet Archive. This archive always tries to save US government websites and databases at the end of each presidential term. Their efforts are not limited to environmental data, and they save not only webpages but entire databases, e.g. data in ftp sites. You can nominate sites to be saved here:\n\n\u2022 Internet Archive, End of Presidential Term Harvest 2016.\n\n\u2022 Internet Archive blog, Preserving U.S. Government Websites and Data as the Obama Term Ends, 15 December 2016.\n\n## Globular\n\n14 December, 2016\n\nOne of my goals is to turn category theory, and even higher category theory, into a practical tool for science. For this we need good scientific ideas\u2014but we also need good software.\n\nMy friend Jamie Vicary has been developing some of this software, together with Aleks Kissinger and Krzysztof Bar and others. Jamie demonstrated it at the Simons Institute workshop on compositionality. You can watch his demonstration here:\n\nBut since Globular runs on a web browser, you can also try it out yourself here:\n\nGlobular.\n\nYou can see his talk slides:\n\n\u2022 Jamie Vicary, Data structures for quasistrict higher categories. (Talk slides here.)\n\nAbstract. Higher category theory is one of the most general approaches to compositionality, with broad and striking applications across computer science, mathematics and physics. We present a new, simple way to define higher categories, in which many important compositional properties emerge as theorems, rather than axioms. Our approach is amenable to computer implementation, and we present a new proof assistant we have developed, with a powerful graphical calculus. In particular, we will outline a substantial new proof we have developed in our setting.\n\nAnd in December 2015, he wrote an article about this software on the n-Category Caf\u00e9. It\u2019s been improved since then, but it can\u2019t hurt to read what he wrote\u2014so I append it here!\n\n### Globular: the basic idea\n\nWhen you\u2019re trying to prove something in a monoidal category, or a higher category, string diagrams are a really useful technique, especially when you\u2019re trying to get an intuition for what you\u2019re doing. But when it comes to writing up your results, the problems start to mount. For a complex proof, it\u2019s hard to be sure your result is correct\u2014a slip of the pen could lead to a false proof, and an error that\u2019s hard to find. And writing up your results can be a huge time-sink, requiring weeks or months using a graphics package, all just for some nice pictures that tell you little about the correctness of the proof, and become useless if you decide to change your approach. Computers should be able help with all these things, in the way that proof assistants like Coq and Agda are allowing us to work with traditional syntactic proofs in a more sophisticated way.\n\nThe purpose of this post is to introduce Globular, a new proof assistant for working with higher-categorical proofs using string diagrams. It\u2019s available at http:\/\/globular.science, with documentation on the nab. It\u2019s web-based, so everything happens right in your browser: build formal proofs, visualize and step through them; keep your proofs private, share them with collaborators, or make them publicly available.\n\nBefore we get into the technical details, here\u2019s a screenshot of Globular in action:\n\nThe main part of the screen shows a diagram, which in this case is 2-dimensional. It represents a composite 2-cell in a finitely-presented 2-category, with the blue and red regions representing objects, the lines representing 1-cells, and the vertices representing 2-cells. In fact, this 2d diagram is just an intermediate state of a 3d proof, through which we\u2019re navigating with the \u2018Slice\u2019 controls in the top-right. The proof itself has been built up by composing the generators listed in the signature, down the left-hand side of the screen. (If you want to take a look at this proof yourself, you can go straight there; in the top-right, set \u2018Project\u2019 to 0, then increment the second \u2018Slice\u2019 counter to scroll through the proof.)\n\nGlobular has been developed so far in the Quantum Group in\nthe Oxford Computer Science department, by Krzysztof\nBar\n, Katherine Casey, Aleks Kissinger, Jamie Vicary and Caspar Wylie. We haven\u2019t quite got around to it yet, but Globular will be open-source, and we\u2019re really keen for people to get involved and help build the software\u2014there\u2019s a huge amount to do! If you want to help out, get in touch.\n\n### Mathematical foundations\n\nGlobular is based on the theory of finitely-presented semistrict n-categories; at the moment, it works up to the level of 3-categories, with an extension to 4-categories actively in development. (You can build cells of any dimension, but from 4-cells and up, some structures are missing.)\n\nDefinitions of n-category vary in how strict they are; a definition is semistrict when it\u2019s as strict as possible, while still having the property that every weak n-category satisfies it, up to equivalence. Definitions of semistrict n-category are not unique: in dimension 3, Gray categories put all the weak structure in the interchangers, while Simpson snucategories put it all in the unitors. Globular implements the axioms of a Gray category, because this is the most appropriate for the graphical calculus: the interchangers can be seen graphically, as changes in height of the components of the diagram. By the theory of k-tuply monoidal n-categories, this also lets you build proofs in a monoidal category, or a braided monoidal category, or a monoidal 2-category.\n\nThe only things that Globular understands are $k$-cells, for some value of $k$. So if you want to build an n-category where an equation $f=g$ holds between n-cells, you have to do it by adding $(n+1)$-cells $a:f \\to g$ and $b:g \\to f$. If you then build some composite $C(f)$ involving $f$, you can apply the cell $a$ to obtain $C(g)$, and we interpret this as the equation $C(f) = C(g)$. In a slogan, this is equality via rewriting. This is consistent with the basic premise of homotopy type theory: treat your proofs as first-order structures, which can in turn be reasoned about themselves.\n\nGlobular can also handle invertibility in a nice way. For a cell $F:A \\to B$ to be invertible, indicated by ticking a box in the signature, means that there also exists an invertible cell $F^{-1}: B \\to A$, and invertible cells $\\text{id}_A \\to F . F^{-1}$ and $\\text{id}_B \\to F^{-1} . F$. This is a coinductive definition (see Mike Shulman\u2019s nice post on this topic), since we\u2019re defining the notion of invertibility in terms of itself in a higher dimension. This sort of a definition is great for proof assistants to work with, as it allows a lot of structure to be generated from a single compact definition.\n\n### How it works\n\nFor a lot more details, take a look at the nLab page. Everything that happens in Globular involves in interaction between the signature on the left-hand side, and the diagram in the main part of the screen. The signature stores the \u2018library\u2019 of cells you have available, and the diagram is a particular composite of cells that you have constructed.\n\nTo construct a new diagram, clear whatever is currently displayed by clicking the \u2018Clear\u2019 button on the right, or pressing \u2018c\u2019. Then start by clicking the icon of a n-cell in your signature, which will make a diagram consisting just of that cell. Clicking on the icons of other $k$-generators for $0 < k \\leq n$ will display a list of ways the cell can be attached, and when you choose one of these ways, the attachment will be performed, growing your n-diagram. (If you\u2019re starting with a blank workspace you will only have a single 0-cell available, so you won\u2019t be able to do this yet!) Clicking an $(n+1)$-cell $G$ displays a list of ways that your n-diagram $D$ can be rewritten, by identifying the source of $G$ as a subdiagram of $D$. Selecting one of these ways will implement the rewrite, by \u2018cutting out\u2019 the chosen subdiagram of $D$, and replacing it with the target of $G$.\n\nAnother way to modify the diagram is to click directly on it. Clicking near the edge of the diagram performs an attachment, while clicking in the interior of the diagram performs a rewrite. If more than one attachment or rewrite is consistent with your click, a little menu will pop up for you to choose what you want to do. When you move your mouse pointer over the diagram, a little label pops up to show you what your cursor is hovering over, which is helpful when choosing where to click.\n\nYou can also click-and-drag on the diagram. This will attach or rewrite with an interchanger, or naturality for an interchanger, or invertibility for an interchanger, depending on where you have clicked and the direction of the drag. Clicking and dragging is designed to work as if you were really \u2018touching\u2019 the strings. So if you want to braid one strand over another, click the strand to \u2018grab\u2019 it, and \u2018pull\u2019 it to one side. If you want to pull a vertex through a braiding, click the vertex to \u2018grab\u2019 it, and \u2018pull\u2019 it up or down through its adjacent braiding. Of course, Globular will only carry out the command if the move you are attempting to make is actually valid in that location.\n\n### Example theorems\n\nHere are four worked examples of nontrivial proofs in Globular:\n\nFrobenius implies associative: http:\/\/globular.science\/1512.004. In a monoidal category, if multiplication and comultiplication morphisms are unital, counital and Frobenius, then they are associative and coassociative.\n\nStrengthening an equivalence: http:\/\/globular.science\/1512.007. In a 2-category, an equivalence gives rise to an adjoint equivalence, satisfying the snake equations.\n\nSwallowtail comes for free: http:\/\/globular.science\/1512.006. In a monoidal 2-category, a weakly-dual pair of objects gives rise to a strongly-dual pair, satisfying the swallowtail equations.\n\nPentagon and triangle implies $\\lambda_I = \\rho_I$: http:\/\/globular.science\/1512.002. In a monoidal 2-category, if a pseudomonoid object satisfies pentagon and triangle equations, then it satisfies $\\lambda_I = \\rho_I$.\n\nWe\u2019ll focus on the second example project \u201cStrengthening an equivalence\u201d listed above, and see how it was constructed. This project investigates the factthat every equivalence in a 2-category gives rise to an adjoint equivalence. To start, we therefore need the basic data that exhibits an equivalence in a 2-category: two 0-cells $A$ and $B$, and an invertible 1-cell $F:A \\to B$, by the weak definition of \u2018invertible\u2019 discussed above. This gives us the following signature:\n\nThe 2-cells that witness invertibility of $F$ look like cups and caps in the graphical calculus, but they won\u2019t satisfy the snake equations that define an adjoint equivalence. The idea of this proof is to define a new cap, built out of the invertible structure of $F$, which does satisfy the snake equations with the existing cup.\n\nBy starting with a diagram consisting of $F$ alone, pressing \u2018i\u2019 to take the identity diagram, and clicking-and-dragging, we build the following 2-diagram, out of the invertible structure associated to $F$:\n\nThis is our candidate for our redefined cup. Its source is the identity on $A$, and its target is $F$ composed with $F^{-1}$. It looks a bit like the curved end of a hockey stick.\n\nTo store it for later use, we now click the \u2018Theorem\u2019 button. Writing $D$ for the 2-diagram we have constructed, this does two things. First, it creates a 2-cell generator that we call \u201cNew Cup\u201d, whose source is $s(D)$, and whose target is $t(D)$. This is the redefined cup that we can use in future expressions. Second, it creates an invertible 3-cell generator that we call \u201cNew Cup Definition\u201d, with source given by \u201cNew Cup\u201d, and with target given by our hockey-stick diagram $D$. This says what \u201cNew Cup\u201d means in terms of our original structure. This adds the following cells to our signature:\n\nBecause \u201cNew Cup Definition\u201d is a 3-cell, by default we see two little icons: one for its source, and one for its target. See how its source is a little picture of \u201cNew Cup\u201d, and its target is a little picture of the hockey stick, just as we defined it.\n\nWe\u2019re now ready to prove one of the snake equations. We start by building the snake composite, using \u201cNew Cup\u201d for the cup, and the invertible structure of $F$ for the cap:\n\nNow have to prove that this equals the identity. Since equality is implemented by rewriting, we must construct a 3-diagram whose source is this snake composite, and whose target is the identity on $F$. To start, we click the \u2018Identity\u2019 button to convert our diagram into an identity 3-diagram. The only apparent effect this has is to add a number scroller to the \u2018Slice\u2019 area of the controls in the top-right. At the moment we can set this to \u20180\u2019 and \u20181\u2019, representing the source and target of our identity 3-diagram respectively. We set it to \u20181\u2019, because we want to compose things to the target.\n\nWe now build up our proof. First, we click on the pink vertex that represents \u201cNew Cup\u201d. This will attach our 3-cell \u201cNew Cup Definition\u201d, replacing \u201cNew Cup\u201d with our hockey-stick picture. By clicking-and-dragging on the diagram, we obtain the following sequence\nof pictures:\n\nPictures 3 to 10 were created by attaching interchangers, and pictures 11 to 15 were created by attaching higher structure generated by the invertibility of $F$. In all cases, this structure was attached just by clicking-and-dragging on the appropriate vertices of the diagram. We\u2019ve turned the snake into the identity, so we\u2019ve finished our proof, which required 14 3-cells. By using the \u2018Slice\u2019 control in the top-right, we can navigate through the 15 slices that make up our proof, and review what we just did. Even better, turning the \u2018Project\u2019 control to the value \u20181\u2019 tells Globular to project out the lowest dimension. This means that our entire 3-diagram proof can be viewed as a single 2-dimensional diagram, as follows:\n\nThis is just like the Morse singularity graphics used by topologists to study the structure of higher-dimensional manifolds. In this picture, the vertices are 3-cells, the lines are 2-cells, and the regions are 1-cells (in fact, every region is the 1-cell $F$.) By moving your mouse pointer over the different parts of the diagram, you can see what the different components are. Interchangers are represented in this projection by braidings.\n\nNow we can do something cool: we can modify our proof, by clicking-and-dragging on the Morse projection. For example, just to the right of centre, there is a crossing, out of which emerge two long vertical lines that travel up a long way before annihilating with one another. Our proof would be much simpler if these two lines just annihilated with each other right after the interchanger. So, we click the vertex at the top of the lines, and drag it down repeatedly, until it gets to where we want it:\n\nWe\u2019ve simplified our proof. By clicking-and-dragging some more, you can change the proof in lots of different ways, although you probably won\u2019t get it much simpler than this. Putting the \u2018Project\u2019 control back to \u20180\u2019, and navigating through the stages of the proof with the \u2018Slice\u2019 control as we were doing before, we can see that our proof has indeed been modified.\n\nThis project has been in development for about 18 months, so it feels great to finally launch. We hope the whole community will get clicking-and-dragging, and let us know how easy it is to use, and what other features would be useful. There are certain to still be bugs, so let us know about them too, and we\u2019ll get right on them.\n\n## Saving Climate Data (Part\u00a01)\n\n13 December, 2016\n\nI try to stay out of politics on this website. This post is not mainly about politics. It\u2019s a call to action. We\u2019re trying to do something rather simple and clearly worthwhile. We\u2019re trying to create backups of US government climate data.\n\nThe background is, of course, political. Many signs point to a dramatic change in US climate policy:\n\n\u2022 Oliver Milman, Trump\u2019s transition: sceptics guide every agency dealing with climate change, The Guardian, 12 December 2016.\n\nSo, scientists are now backing up large amounts of climate data, just in case the Trump administration tries to delete it after he takes office on January 20th:\n\n\u2022 Brady Dennis, Scientists are frantically copying U.S. climate data, fearing it might vanish under Trump, Washington Post, 13 December 2016.\n\nOf course saving the data publicly available on US government sites is not nearly as good as keeping climate programs fully funded! New data is coming in all the time from satellites and other sources. We need it\u2014and we need the experts who understand it.\n\nAlso, it\u2019s possible that the Trump administration won\u2019t go so far as trying to delete big climate science databases. Still, I think it can\u2019t be a bad thing to have backups. Or as my mother always said: better safe than sorry!\n\nQuoting the Washington Post article:\n\nAlarmed that decades of crucial climate measurements could vanish under a hostile Trump administration, scientists have begun a feverish attempt to copy reams of government data onto independent servers in hopes of safeguarding it from any political interference.\n\nThe efforts include a \u201cguerrilla archiving\u201d event in Toronto, where experts will copy irreplaceable public data, meetings at the University of Pennsylvania focused on how to download as much federal data as possible in the coming weeks, and a collaboration of scientists and database experts who are compiling an online site to harbor scientific information.\n\n\u201cSomething that seemed a little paranoid to me before all of a sudden seems potentially realistic, or at least something you\u2019d want to hedge against,\u201d said Nick Santos, an environmental researcher at the University of California at Davis, who over the weekend began copying government climate data onto a nongovernment server, where it will remain available to the public. \u201cDoing this can only be a good thing. Hopefully they leave everything in place. But if not, we\u2019re planning for that.\u201d\n\n[\u2026]\n\n\u201cWhat are the most important .gov climate assets?\u201d Eric Holthaus, a meteorologist and self-proclaimed \u201cclimate hawk,\u201d tweeted from his Arizona home Saturday evening. \u201cScientists: Do you have a US .gov climate database that you don\u2019t want to see disappear?\u201d\n\nWithin hours, responses flooded in from around the country. Scientists added links to dozens of government databases to a Google spreadsheet. Investors offered to help fund efforts to copy and safeguard key climate data. Lawyers offered pro bono legal help. Database experts offered to help organize mountains of data and to house it with free server space. In California, Santos began building an online repository to \u201cmake sure these data sets remain freely and broadly accessible.\u201d\n\nIn Philadelphia, researchers at the University of Pennsylvania, along with members of groups such as Open Data Philly and the software company Azavea, have been meeting to figure out ways to harvest and store important data sets.\n\nAt the University of Toronto this weekend, researchers are holding what they call a \u201cguerrilla archiving\u201d event to catalogue key federal environmental data ahead of Trump\u2019s inauguration. The event \u201cis focused on preserving information and data from the Environmental Protection Agency, which has programs and data at high risk of being removed from online public access or even deleted,\u201d the organizers said. \u201cThis includes climate change, water, air, toxics programs.\u201d\n\nThe event is part of a broader effort to help San Francisco-based Internet Archive with its End of Term 2016 project, an effort by university, government and nonprofit officials to find and archive valuable pages on federal websites. The project has existed through several presidential transitions.\n\nI hope that small \u201cguerilla archiving\u201d efforts will be dwarfed by more systematic work, because it\u2019s crucial that databases be copied along with all relevant metadata\u2014and some sort of cryptographic certificate of authenticity, if possible. However, getting lots of people involved is bound to be a good thing, politically speaking.\n\nIf you have good computer skills, good understanding of databases, or lots of storage space, please get involved. Efforts are being coordinated by Barbara Wiggin and others at the Data Refuge Project:\n\n\u2022 PPEHLab (Penn Program in the Environmental Humanities), DataRefuge.\n\nYou can contact them at DataRefuge@ppehlab.org. Nick Santos is also involved, and if you want to get \u201cmore plugged into the project\u201d you can contact him here. They are trying to build a climate database mirror website here:\n\nAt the help form on this website you can nominate a dataset for rescue, claim a dataset to rescue, let them know about a data rescue event, or help in some other way (which you must specify).\n\nPPEHLab and Penn Libraries are organizing a data rescue event this Thursday:\n\n\u2022 PPEHLab, DataRefuge meeting, 14 December 2016.\n\nAt the American Geophysical Union meeting in San Francisco, where more than 20,000 earth and climate scientists gather from around the world, there was a public demonstration today starting at 1:30 PST:\n\nRally to stand up for science, 13 December 2016.\n\nAnd the \u201cguerilla archiving\u201d hackathon in Toronto is this Saturday\u2014see below. If you know people with good computer skills in Toronto, get them to check it out!\n\n### Guerrilla archiving in Toronto\n\nHere are details on this:\n\nGuerrilla Archiving Hackathon\n\nDate: 10am-4pm, December 17, 2016\n\nLocation: Bissell Building, 4th Floor, 140 St. George St. University of Toronto\n\nRSVP and up-to-date information: Guerilla archiving: saving environmental data from Trump.\n\nBring: laptops, power bars, and snacks. Coffee and pizza provided.\n\nThis event collaborates with the Internet Archive\u2019s End of Term 2016 project, which seeks to archive the federal online pages and data that are in danger of disappearing during the Trump administration. Our event is focused on preserving information and data from the Environmental Protection Agency, which has programs and data at high risk of being removed from online public access or even deleted. This includes climate change, water, air, toxics programs. This project is urgent because the Trump transition team has identified the EPA and other environmental programs as priorities for the chopping block.\n\nThe Internet Archive is a San Francisco-based nonprofit digital library which aims at preserving and making universally accessible knowledge. Its End of Term web archive captures and saves U.S. Government websites that are at risk of changing or disappearing altogether during government transitions. The Internet Archive has asked volunteers to help select and organize information that will be preserved before the Trump transition.\n\nEnd of Term web archive: http:\/\/eotarchive.cdlib.org\/2016.html\n\nNew York Times article: \u201cHarvesting Government History, One Web Page at a Time\n\nActivities:\n\nIdentifying endangered programs and data\n\nSeeding the End of Term webcrawler with priority URLs\n\nIdentifying and mapping the location of inaccessible environmental databases\n\nHacking scripts to make accessible to the webcrawler hard to reach databases.\n\nBuilding a toolkit so that other groups can hold similar events\n\nSkills needed: We need all kinds of people \u2014 and that means you!\n\nPeople who can locate relevant webpages for the Internet Archive\u2019s webcrawler\n\nPeople who can identify data targeted for deletion by the Trump transition team and the organizations they work with\n\nPeople with knowledge of government websites and information, including the EPA\n\nPeople with library and archive skills\n\nPeople who are good at navigating databases\n\nPeople interested in mapping where inaccessible data is located at the EPA\n\nHackers to figure out how to extract data and URLs from databases (in a way that Internet Archive can use)\n\nPeople with good organization and communication skills\n\nPeople interested in creating a toolkit for reproducing similar events\n\nContacts: michelle.murphy@utoronto.ca, p.keilty@utoronto.ca\n\n## Modelling Interconnected Systems with Decorated\u00a0Corelations\n\n9 December, 2016\n\nHere at the Simons Institute workshop on compositionality, my talk on network theory explained how to use \u2018decorated cospans\u2019 as a general model of open systems. These were invented by Brendan Fong, and are nicely explained in his thesis:\n\n\u2022 Brendan Fong, The Algebra of Open and Interconnected Systems. (Blog article here.)\n\nBut he went further: to understand the externally observable behavior of an open system we often want to simplify a decorated cospan and get another sort of structure, which he calls a \u2018decorated corelation\u2019.\n\nIn this talk, Brendan explained decorated corelations and what they\u2019re good for:\n\n\u2022 Brendan Fong, Modelling interconnected systems with decorated corelations. (Talk slides here.)\n\nAbstract. Hypergraph categories are monoidal categories in which every object is equipped with a special commutative Frobenius monoid. Morphisms in a hypergraph category can hence be represented by string diagrams in which strings can branch and split: diagrams that are reminiscent of electrical circuit diagrams. As such they provide a framework for formalising the syntax and semantics of circuit-type diagrammatic languages. In this talk I will introduce decorated corelations as a tool for building hypergraph categories and hypergraph functors, drawing examples from linear algebra and dynamical systems.\n\n## Semantics for Physicists\n\n7 December, 2016\n\nI once complained that my student Brendan Fong said \u2018semantics\u2019 too much. You see, I\u2019m in a math department, but he was actually in the computer science department at Oxford: I was his informal supervisor. Theoretical computer scientists love talking about syntax versus semantics\u2014that is, written expressions versus what those expressions actually mean, or programs versus what those programs actually do. So Brendan was very comfortable with that distinction. But my other grad students, coming from a math department didn\u2019t understand it\u2026 and he was mentioning it in practically ever other sentence.\n\nIn 1963, in his PhD thesis, Bill Lawvere figured out a way to talk about syntax versus semantics that even mathematicians\u2014well, even category theorists\u2014could understand. It\u2019s called \u2018functorial semantics\u2019. The idea is that things you write are morphisms in some category $X,$ while their meanings are morphisms in some other category $Y.$ There\u2019s a functor $F \\colon X \\to Y$ which sends things you write to their meanings. This functor sends syntax to semantics!\n\nBut physicists may not enjoy this idea unless they see it at work in physics. In physics, too, the distinction is important! But it takes a while to understand. I hope Prakash Panangaden\u2019s talk at the start of the Simons Institute workshop on compositionality is helpful. Check it out:\n\n## Compositionality in Network\u00a0Theory\n\n29 November, 2016\n\nI gave a talk at the workshop on compositionality at the Simons Institute for the Theory of Computing next week. I spoke about some new work with Blake Pollard. You can see the slides here:\n\n\u2022 John Baez, Compositionality in network theory, 6 December 2016.\n\nand a video here:\n\nAbstract. To describe systems composed of interacting parts, scientists and engineers draw diagrams of networks: flow charts, Petri nets, electrical circuit diagrams, signal-flow graphs, chemical reaction networks, Feynman diagrams and the like. In principle all these different diagrams fit into a common framework: the mathematics of symmetric monoidal categories. This has been known for some time. However, the details are more challenging, and ultimately more rewarding, than this basic insight. Two complementary approaches are presentations of symmetric monoidal categories using generators and relations (which are more algebraic in flavor) and decorated cospan categories (which are more geometrical). In this talk we focus on the latter.\n\nThis talk assumes considerable familiarity with category theory. For a much gentler talk on the same theme, see:\n\n## Compositional Frameworks for Open\u00a0Systems\n\n27 November, 2016\n\nHere are the slides of Blake Pollard\u2019s talk at the Santa Fe Institute workshop on Statistical Physics, Information Processing and Biology:\n\n\u2022 Blake Pollard, Compositional frameworks for open systems, 17 November 2016.\n\nHe gave a really nice introduction to how we can use categories to study open systems, with his main example being \u2018open Markov processes\u2019, where probability can flow in and out of the set of states. People liked it a lot!","date":"2017-01-20 05:40:54","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\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 47, \"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\": 0.41666102409362793, \"perplexity\": 1622.0193597835796}, \"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-2017-04\/segments\/1484560280791.35\/warc\/CC-MAIN-20170116095120-00047-ip-10-171-10-70.ec2.internal.warc.gz\"}"} | null | null |
Home White House Trump touts Brexit in Ireland, compares border with Britain to U.S.-Mexico border,...
Trump touts Brexit in Ireland, compares border with Britain to U.S.-Mexico border, Irish leader says he wishes to avoid a crisis, Trump says it will work out well
By Douglas Christian
President Donald Trump leaves Capitol Hill after attending after a "Friends of Ireland" lunch an celebrating Irish American politicians on the eve of St. Patrick's Day. With him are, from left, former House Speaker Paul Ryan, Irish Taoiseach Enda Kenny, Rep. Peter King (R-NY), and Vice President Mike Pence in March 2017. The president and First Lady Melania Trump visited Ireland on Wednesday. (Photo ©2018 Doug Christian/TMN)
(Audio: Trump with Irish Taoiseacht (Tea-shock) Leo Varadkar says "I think it will all work out very well, and also for you with your wall, your border")
THE WHITE HOUSE – President Donald Trump waded into Irish politics minutes after he arrived in Ireland, comparing the possible result of the United Kingdom's Brexit resolution cutting off Northern Ireland from Britain, to building a wall along the border between the United States and Mexico.
Sitting with Irish Defense Minister Taoiseach Leo Varadkar at Shannon Airport, Trump said:
(Audio: "I mean, we have a border situation in the United States, and you have one over here. But I hear it's going to work out very well here.")
Varadkar interjected, saying:
(Audio: Varadkar says that Ireland wished to avoid a border or a wall)
Trump interjected back, saying:
(Audio: "I think you do, I think you do. The way it works now is good, you want to try and to keep it that way. I know that's a big point of contention with respect to Brexit. I'm sure it's going to work out very well. I know they're focused very heavily on it.")
Doug Christian, The White House
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Boris Johnson emerges as frontrunner to replace Theresa May
Trump sees cheering crowds, not protesters; tariffs likely, Republicans back him; 94% GOP approval rating
Trump visits UK as a deeply unpopular US president, criticizes PM May and supports her rival Boris Johnson | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 7,701 |
Canadians' average saving rate 5% too low
A BMO Household Savings Report shows that Canadians plan to save on average $9,859 in 2013.
By Warren S. Hersch | February 07, 2013 at 08:38 AM
Canadians must increase their average savings rate to close to 9 percent from the current 4 percent to meet their long-term financial goals, according to new research.
BMO Financial Group, Toronto, released this finding in The BMO Economics Report. The study follows a BMO Household Savings Report, which shows that Canadians plan to save on average $9,859 in 2013.
"The prospect of a prolonged period of subdued investment returns after the recent rebound suggests that Canadian personal savings trends, on average, are on the light side for adequate retirement purposes," says Douglas Porter, Chief Economist, BMO Capital Markets. "While the most straightforward way for policymakers to address this over the longer term is to get interest rates back to neutral levels, the higher-interest-rate cavalry is nowhere on the horizon."
Porter adds that, while debt has risen to 165 percent of disposable income, financial assets have rebounded to 453 percent of disposable income; this likely rose further in the fourth quarter, probably surpassing the record high of 454 percent in 2007, just before the financial crisis began.
"It's true that the ratio of net financial assets to income—285 percent in Q3 of 2012— remains below pre-financial-crisis highs, but it has moved back above the 10-year average, suggesting Canadian household finances are fully under repair."
Households continue to increase their risk in search of higher yields, with equities and high-yield bonds benefitting from the rising demand. This shift has lifted household assets to a record high, but cannot be counted on to continue closing the savings gap.
Porter notes that Canadians will likely need about 18 times their desired retirement income (exceeding payments expected from the Canada Pension Plan and Old Age Security) to retire comfortably.
"While every case is unique, a typical and reasonable retirement income goal is often about 60 percent of pre-retirement earnings," says Porter. "Given that the CPP and OAS are designed to replace about 25 percent of pre-retirement income for the average Canadian, that leaves a target of about 35 percent of pre-retirement income to be funded from individual savings."
Family Caregiving Is a Long-Term Job: Report
Marlene Satter | July 11, 2019
The average caregiver for a parent or spouse provides five years of care, the GAO finds.
5 House Districts With the Lowest Ratios of Older People to Working-Age People
Allison Bell | July 11, 2019
For lawmakers in these districts, the Silver Tsunami may look like a ripple.
5 House Districts With the Highest Ratios of Older People to 'Working-Age' People
Lawmakers in these districts are already living in the demographic future. | {
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} | 8,892 |
\section{Introduction}\label{intro_sec}
\renewcommand{\thefootnote}{\fnsymbol{footnote}}
NGC~205 is a low-luminosity, early-type, dwarf galaxy in the Local Group
and is one of the brightest nearby examples of a dwarf elliptical (dE)
galaxy. A close satellite of the large spiral galaxy Andromeda (M31),
NGC~205 has a projected separation to M31 of merely $40'$ or 8\,kpc
(Fig.~\ref{fig_masks}); the physical separation between NGC~205 and
M31 is estimated to be $\sim 40$\,kpc \citep{dem03,mcc05}. This
separation distance is intermediate between that of the Milky Way and
its dwarf satellite galaxies Sagittarius \citep[24\,kpc;][]{iba95} and
the Large Magellanic Cloud \citep[50\,kpc;][]{fre01}. Observational
evidence such as isophotal twisting at large radii
\citep*{hod73,ken87,cho02}, recent star formation \citep{pel93,cap99},
and a steadily increasing velocity dispersion with radius
\citep*{ben91,sim02} suggests that NGC~205 is tidally
interacting with its parent galaxy. However, inside the effective
radius, $r_{\rm eff} = 2.5' (0.6$\,kpc), NGC~205 exhibits regular
surface brightness and kinematic profiles typical of a normal dE
galaxy. NGC~205 therefore offers a unique opportunity to study both
the internal dynamics of a nearby dE galaxy while providing a detailed
view of a satellite undergoing disruption.
A crucial question for current galaxy formation theories is to what
extent the accretion of dwarf satellite galaxies contributes to the
growth of massive galactic halos \citep{bul00,fon05}. The M31 galaxy
and its associated dwarf satellite system are an excellent laboratory
to study this question in detail: M31 is the nearest massive galaxy
external to the Milky Way in which current observing techniques are
able to resolve individual stars. \citet{iba01} and \citet{fer02}
first detected evidence for tidal streams in the halo of M31 based on
wide-area star-count maps. \citet{guh04} and \citet{iba04} provided
confirmation that the stars in these streams are dynamically related
and have a small enough velocity dispersion to have feasibly been
stripped from a dwarf-sized galaxy. \citet{mey01} have suggested that
the globular cluster G1, one of the brightest globular clusters
belonging to M31, is the remnant core of a nucleated dwarf galaxy
based on its internal dynamics and stellar populations (however see
\citet{rei04}). From these observations, it is clear that dwarf
satellite disruption must contribute to the build-up of M31's halo;
detailed study of current, on-going interactions between M31 and its
satellites should help quantify what fraction of halo material is
attributed to this process.
Galaxy formation scenarios for the origin of dE galaxies lie broadly
in two very distinct categories: (1)~dEs are old, primordial objects,
and (2)~dEs have recently evolved or transformed from a progenitor
galaxy population \citep{dek86,moo98,may01,mas04}. Dwarf elliptical
galaxies are highly clustered \citep{bin88}-- the majority of dEs in
the local Universe are found in dense galaxy clusters. Kinematic
studies of dEs in the Virgo and Fornax Clusters suggest a large range
in their kinematic properties, from rotationally-supported galaxies to
dEs with no detectable major axis rotation \citep{geh03,der01,van04}.
These results, in particular the abundance of dE galaxies with a
complete lack of rotation, have so far been difficult to explain in
the context of current formation models. The three bright Local Group
dE satellites NGC~205, NGC~147, and NGC~185 provide an excellent
opportunity to study dE kinematics in greater detail and to larger
radii than their cluster counterparts. In this paper, we present
kinematics for the brightest of the Local Group dEs, NGC~205.
The absolute magnitude of NGC~205 is $M_V = -16.5$. It has a nearly
exponential surface brightness profile \citep{cho02}, characteristic
of dE galaxies. As a result, NGC~205 has a far more diffuse
appearance than the other bright M31 satellite galaxy, M32, which has
a similar absolute magnitude but is classified as a compact
elliptical. Near the center of this galaxy, inside a radius of $1'$
(0.24\,kpc), NGC~205 contains dust \citep{haa98}, atomic and molecular
gas \citep{wel98,you97}, and a number of OB associations implying a
recent episode of star formation in the past 100 Myr
\citep{pel93,lee96}. Beyond this radius, NGC~205 is gas and dust-free
and is composed of an intermediate stellar population \citep{dem03},
typical for similarly-sized dEs in the Virgo cluster \citep{geh03}.
Early dynamical studies suggested that NGC~205 had negligible internal
rotation velocity. Dynamical measurements by \citet{ben91} placed an
upper limit on the rotation velocity of $1.5\pm0.8$\,\,km~s$^{-1}$\ at $\sim r_{\rm
eff}$. Combined with observations of two other Local Group dEs, this
established the paradigm that dE galaxies are supported primarily by
anisotropic velocity dispersions. However, \citet{sim02} more
recently measured the major-axis rotation of NGC~205 out to twice the
effective radius, measuring a rotation speed $13\pm 2\,$\,km~s$^{-1}$. The
central velocity dispersion of NGC~205 has a value of $16\pm4\,$\,km~s$^{-1}$\,
\citep{pet93,car90}, increasing with radius to 42\,\,km~s$^{-1}$\ at $2'$
\citep{ben91} and 50\,\,km~s$^{-1}$\ at $5'$ \citep{sim02}. Assuming an isotropic
spheroid model, \citet{ben91} inferred a mass-to-light ratio of
$M/L\sim7$ for NGC~205, similar to the mass-to-light ratios inferred
for a sample of Virgo and Fornax Cluster dE galaxies based on more
sophisticated dynamical modeling \citep{geh02,der01}. These ratios
are consistent with that of a modestly old stellar population with no
dark matter. This lack of dark matter in the inner region is in
stark contrast to the much larger mass-to-light ratios inferred for
fainter dwarf spheroidal (dSph) galaxies in the Local Group, such as
the Draco and Ursa Minor dSph, which appear to be
dark-matter-dominated even in their central region \citep{kle03}.
Integrated-light spectroscopy cannot probe the kinematics of dE
galaxies much beyond the effective radius due to their characteristic
low surface brightness ($\mu_{V,\rm eff} = 23$ mag arcsec$^{-1}$).
Unlike dwarf irregular galaxies, dEs are devoid of gas at large radii
so their outer kinematics cannot be studied via HI gas dynamics.
Other kinematical probes exist at large radius, such as globular
clusters \citep{bea05} and planetary nebulae \citep{cor05}, but their
small numbers make these poor kinematic tracers. The proximity of
NGC~205 presents a unique opportunity to study the dynamics of a dE
galaxy out to much larger radii than possible in more distant systems:
via spectroscopy of individual red giant branch (RGB) stars.
In this paper, we present accurate radial velocities for 725~RGB
stars in the Local Group dE galaxy NGC~205 measured via Keck/DEIMOS
multi-slit spectroscopy. We present dynamics of this galaxy well
beyond the inferred tidal radius, twice the radial extent of previous
measurements. This paper is organized as follows: in
\S\,\ref{sec_data} we discuss target selection for our DEIMOS
slitmasks, the observing procedure and data reduction. In
\S\,\ref{sec_dyn}, a detailed kinematic analysis of NGC~205 is
presented along with a maximum-likelihood estimate of the kinematic
contamination from stars in M31. Finally, in \S\,\ref{sec_orbit},
we discuss the implications of our kinematic observations for the tidal
interaction between NGC~205 and M31.
Throughout this paper we adopt a distance modulus to NGC~205 of $(m
-M)_0 = 24.58 \pm 0.07$, i.e.,~a distance of $824\pm 27$\,kpc, as
determined by \citet{mcc05} via the tip of the RGB
method; this places NGC~205 40\,kpc further than its parent galaxy M31.
\section{Data}\label{sec_data}
\subsection{Target Selection}\label{subsec_targets}
Stars were selected for spectroscopy according to their probability of
being a RGB star in NGC~205. Stars were selected
based on Canada-France-Hawaii Telescope CFH12K mosaic imaging in the $R$
and $I$ bands kindly provided by \citet*[][private
communication]{dem03}. The CCD mosaic covers a $42'\times28'$ region
centered on NGC~205 with $0.206''$ pixels. The images were obtained in
sub-arcsecond seeing conditions. Stellar photometry from these images
extends 2~mag
below the tip of the RGB. The color-magnitude diagram for this region
is shown in Figure~\ref{fig_cmd}. A broad RGB is seen extending to
very red colors ($R-I > 3.0$); the reddest of these stars are likely to be
metal-rich/reddened M31 disk RGB contaminants \citep{dem03}. The vertical
ridge near $(R-I) = 0.2$ is due to foreground Milky Way stars. The
spatial distribution of stars is shown in Figure~\ref{fig_imvel}. The
gaps between CFH12K CCD chips are $\sim7''$ wide. The apparent hole in
stellar density map near the center of NGC~205 is due to crowding; the
stellar density increase in the lower left corner is due to the
presence of field RGB stars belonging to M31.
The absolute magnitude of the RGB is $M_I\sim -4$. At the assumed
distance of NGC~205, the apparent magnitude of this stellar population
is $I \sim 21$. Since metallicity variations within NGC~205 will
cause a spread in the colors of its RGB stars, we select spectroscopic
targets based primarily on apparent magnitude. Highest priority in
the spectroscopic target list was assigned to stars between $20.5 \le
I \le 21.0$. Second priority was given to stars between $20.0 \le I <
20.5$ and $21.0 < I \le 21.5$; lowest priority was assigned to objects
$I > 21.5$. To minimize Galactic foreground contamination, targets
were required to have $(R-I) > 0.2$. Stars with photometric errors
larger than $\sigma_{I} > 0.1$ were rejected, as were stars having
neighbors of equal or greater brightness within a radius $r< 4''$. The
photometric criteria used for spectroscopic target selection are shown
in Figure~\ref{fig_cmd}.
\subsection{DEIMOS Multi-Slit Mask Design}\label{ssec_maskdesign}
Four multislit masks were observed on the nights of 2003 September 30 and
October 1 with the Keck~II 10-m telescope and the DEIMOS
multi-object spectrograph \citep{fab03}. Three of the four masks
(N205-1, N205-2, N205-3) were designed to be observed in a
conventional mode, while a fourth mask was designed to have multiple
tiers of spectra as described in \S\,\ref{sssec_multitier}. Science
exposures for all four masks were $3\times1200$s per mask; the average
seeing FWHM during the spectroscopic observations was $0.7''$.
In Figure~\ref{fig_masks}, the placement of the four DEIMOS slit-masks
is shown relative to a Digitized Palomar Optical Sky Survey image of
NGC~205. Each DEIMOS multislit mask covers a rectangular area of
$\approx16'\times4'$. Two slit-masks (N205-1 and N205-4) were
positioned on the center on NGC~205 with the long axis of the mask
along the major axis of the galaxy. The centers of masks N205-2 and
N205-3 were placed $10'$ to the South-East and North-West of NGC~205,
respectively. The long axis of these two masks were roughly placed
along the direction of tidal distortion. The direction of tidal
distortion was determined from surface brightness ellipse fitting by
\citet{cho02} based on wide-field $B$-band photometry. The locus of
semi-major axes of the ellipse fitting forms a gentle 'S' curve which
we plot in Figure~\ref{fig_masks} and refer to as the major-axis
of NGC~205 (see \S\,\ref{sec_dyn}). Slitlets in each mask have the
same the position angle as the overall mask---i.e.~we did not use
tilted slits. In Table~2, we list the observing details for each
mask.
\subsubsection{Conventional Masks}\label{sssec_conventional}
The three conventional DEIMOS slit-masks (N205-1, N205-2, N205-3) were
designed to be observed with the 1200~line~mm$^{-1}$ grating covering
a wavelength region $6400-9100\mbox{\AA}$. The spectral dispersion of
this setup is $0.33\mbox{\AA}$, and the resulting spectral resolution,
taking into account the anamorphic distortion factor of 0.706, is
$1.37\mbox{\AA}$ (FWHM). This wavelength range includes the Ca\,{\smcap ii}\
triplet lines which are expected to be strong in absorption for our
targeted RGB stars. Slitlets were $0.7''$ wide to match the typical
seeing conditions while maintaining good wavelength resolution. The
spatial scale is $0.12''$~per pixel and the spectral dispersion of
$0.33\mbox{\AA}$ per pixel. To allow adequate sky subtraction, the
minimum slit length was $5''$; the minimum spatial separation between
slit ends was $0.4''$ (three pixels).
Using the above input parameters and the target list described in
\S\,\ref{subsec_targets}, slitmasks were created using the DEIMOS {\tt
dsimulator}\footnote{Available at
http://www.ucolick.org/$\sim$phillips/deimos\_ref/masks.html} slitmask
design software. For each slitmask the software starts with the
highest priority input targets and automatically fills in the mask
area to the extent possible and filling in the remaining space on the
slitmask with lower priority targets. An average of nearly 200
slitlets were placed on each mask (see Table~2). A handful of targets were
deliberately included on multiple masks. These overlapping targets
are used to quantify measurement errors in \S\,\ref{subsec_rvel}.
\subsubsection{Multi-Tier Masks}\label{sssec_multitier}
The DEIMOS slitmask N205-4 was designed for use with a blocking filter
centered on the Ca\,{\smcap ii}\ triplet region. The blocking filter allows
multiple tiers of slitlets to placed on a single mask increasing the
observing efficiency by a factor of 2--3 over conventional DEIMOS
masks. The Ca\,{\smcap ii}\ blocking filter has a central wavelength of
$8550\mbox{\AA}$ and is $350\mbox{\AA}$ wide. In combination with the
831~line~mm$^{-1}$ grating, the spectral dispersion of this
setup is $0.47\mbox{\AA}$, and the resulting spectral resolution is
$1.96\mbox{\AA}$ (FWHM). Two and a half tiers of spectra are
possible without significant overlap in the wavelength region $8450 -
8850\mbox{\AA}$. Three full tiers of spectra are not possible because
the DEIMOS field-of-view is narrower in the middle as compared to the
ends. The width and typical length of the slitlets are the same as
the conventional masks above. This mask was designed by running {\tt
dsimulator} three times over the mask. In each pass, slitlets were
placed on targets in a row running along the spatial axis of the mask
to maximize the number of slitlets. For each pass of {\tt
dsimulator}, the region of the CCD detector plane occupied by the spectra
of all the slitlets was calculated, the spectroscopic targets in this region
excluded, and the resulting target list fed into the next pass. A total
of 332~slitlets were placed on the N205-4 mask. The tier pattern of
slitlets can be seen in the spatial distribution of targets in
Figure~\ref{fig_imvel}.
\subsection{Data Reduction}\label{subsec_redux}
Spectra from the three conventional DEIMOS multi-slit masks were
reduced using the {\tt spec2d} software pipeline (version~1.1.4)
developed by the DEEP2 team at the University of California-Berkeley
for that survey. A detailed description of the reductions can be
found in \citet{coo06}. Briefly, the flat-field exposures are used to
rectify the curved raw spectra into rectangular arrays by applying
small shifts and interpolating in the spatial direction. A
one-dimensional slit function correction and two-dimensional
flat-field and fringing correction are applied to each slitlet. Using
the DEIMOS optical model as a starting point, a two-dimensional
wavelength solution is determined from the arc lamp exposures with
residuals of order $\rm0.01\mbox{\AA}$. Each slitlet is then
sky-subtracted exposure by exposure using a B-spline model for the
sky. The individual exposures of the slitlet are averaged with
cosmic-ray rejection and inverse-variance weighting. Finally one
dimensional spectra are extracted for all science targets using the
optimal scheme of \citet{hor86} and are rebinned into logarithmic
wavelength bins with 15\,\,km~s$^{-1}$\ per pixel. Sample one-dimensional
spectra are shown in Figure~\ref{fig_spec}.
Spectra from the multi-tier mask NGC~205-4 were reduced using a
combination of IRAF multi- and long-slit tasks similar to the method
described in \citet{geh02}. The flat-field exposure for this mask was
used to trace the ends of each slitlet. The APALL task was used in
``strip'' mode to extract and rectify two-dimensional rectangular
strips for each slitlet; a similar extraction was applied to the arc
lamp calibration and science frames. Data reduction proceeded on these
rectified strips. Each strip was divided by its corresponding
normalized flat-field image. Individual science exposures were
cleaned of cosmic rays and combined. A wavelength solution was
determined for each slitlet from the combined Kr/Ar/Ne/Xe arc lamp
spectrum and was applied to the data. One-dimensional spectra were
then extracted from each strip and rebinned into logarithmic
wavelength bins with 20\,\,km~s$^{-1}$\ per pixel.
\subsection{Radial Velocities}\label{subsec_rvel}
Radial velocities were measured for spectra extracted from the four
DEIMOS masks by cross-correlating the observed spectra with a series
of high signal-to-noise stellar templates originally created for the
Sloan Digital Sky Survey covering a wide range of stellar spectral
type \citep{coo06}. These templates were rebinned to match the final
wavelength resolution of the observations of 15 and 20\,\,km~s$^{-1}$\ per
pixel for the conventional and multi-tier masks, respectively. The
science and template spectra were continuum-subtracted; the template
was then shifted and scaled to minimize the reduced-$\chi^2$. A
heliocentric correction was applied to all the measured radial
velocities. The fitted velocities were visually inspected and
assigned a quality code to indicate the reliability of the measured
redshift and the overall quality of the spectrum.
Radial velocities for individual stars were successfully measured for
769 of the 869 extracted spectra. Of the 100 spectra for which we did
not measure a redshift, 51 spectra were unusable due to bad columns or
vignetting and 49 spectra had insufficient signal-to-noise ratios to
determine a redshift. Of the spectra with measured velocities, 3 were
background galaxies and 41 were duplicate measurements. Duplicate
measurements are used to estimate our velocity error bars as discussed
below. The final sample consists of 725~unique stellar radial
velocity measurements.
We estimate the accuracy of our radial velocity measurements via
repeat measurements of stars across the four observed masks.
Twenty-nine duplicate stars were observed across the three
conventional mask. Twelve stars on the multi-tier mask were
duplicated on the conventional masks. The root-mean-square (rms)
radial velocity difference between pairs of measurements for stars on
the conventional masks is 16.3~km~s$^{-1}$. Assuming the measurement
uncertainty is the same for each member of the pair, the radial
velocity error for an individual measurement is $\sqrt{2}$ times
smaller than the rms of the difference, or $11.5$~km~s$^{-1}$. The
rms difference between the conventional and multi-tier mask is
21.7\,\,km~s$^{-1}$; this rms difference is larger than the conventional masks due to
lower spectral resolution and the smaller available wavelength region
in the multi-tier design. Assuming velocity errors between the
conventional and multi-tier mask add in quadrature, the radial
velocity error on individual multi-tier measurements is 18.4\,\,km~s$^{-1}$.
Velocity measurements for individual stars are listed in Table~3.
\section{Results}\label{sec_dyn}
The measured velocities of individual stars allow us to probe the
dynamics of NGC~205 to much larger radius than possible via
integrated-light spectroscopy. In Figure~\ref{fig_vhist}, we show the
distribution of radial velocities of individual RGB stars in the four
Keck/DEIMOS slitmasks. The distribution of velocities in slitmasks
N205-1 and N205-4 is centered on the systemic velocity of NGC~205.
The systemic velocity of NGC~205 is $v_{\rm sys} = -246 \pm 1$\,\,km~s$^{-1}$\
based on the median velocity of stars with semi-major axis distances
less than $5'$. The two slitmasks off-center from NGC~205 show skewed
velocity distributions. The mask N205-2, placed to the NW of NGC~205
on the farside from M31, is skewed toward more positive velocities
than the NGC~205 systemic velocity; mask N205-3, to the SE of NGC~205
nearer to M31, has more negative velocities relative to the NGC~205
systemic. To illustrate this observation, we fit a Gaussian profile
to the combined distribution of velocities (Figure~\ref{fig_vhist},
bottom panel). The fitted Gaussian has a central velocity of
$-246$\,km~s$^{-1}$\, in agreement with the systemic velocity of NGC~205
measured above, and a velocity width of $42$\,km~s$^{-1}$\, in agreement with
the average velocity dispersion of NGC~205 measured by \citet{sim02}.
In the four top panels of Figure~\ref{fig_vhist} we plot the scaled
Gaussian profile; the excess of stars in mask N205-2 and N205-3 is
clearly visible to the right and left, respectively of the main
velocity peak. In all four masks, there are a small number of stars
with velocities less negative than NGC~205 due to foreground
contamination, while the tail of stars to more negative velocities are
primarily M31 halo stars whose systemic velocity is $-300$\,\,km~s$^{-1}$. We
estimate M31 contamination fractions in our sample in
\S\,\ref{subsec_m31}.
A semi-major axis distance is assigned to each slitlet based on its
minimum distance to the major axis of NGC~205. The major-axis of
NGC~205 is shown in Figure~\ref{fig_imvel} and was determined from
surface brightness ellipse fitting by \citet{cho02} based on
wide-field $B$-band images of NGC~205, corrected for the underlying
light of M31. The major-axis of NGC~205 forms a gentle 'S' curve as
plotted in Figure~\ref{fig_imvel}. For each slitlet, we determine its
minimum distance to the major-axis and assign the semi-major axis
length (a) at the point. Although there are alternative methods to
assign this distance (e.g.~the semi-major distance corresponding to
the nearest elliptical isophote), this method most closely resembles
integrated-light spectroscopy of more distant dEs to which we will
compare our results.
\subsection{The Velocity Profile of NGC~205}\label{vp_prof}
In Figure~\ref{fig_vp}, we present the major-axis velocity profile for
NGC~205 determined from the combined velocity measurements of RGB
stars. Individual stellar velocity measurements are shown in the top
panel of this figure. To determine the ensemble major-axis velocity
profile, individual measurements were binned into minimum $1'$ radial
bins with 25 or more stars per bin. The velocity in each bin was
determined by simultaneously fitting a double Gaussian profile to the
distribution of stars in that bin. To account for contamination from
M31 stars, one of the two fitted Gaussian profiles had a fixed mean
and velocity width of -300 \,km~s$^{-1}$\ and 150\,km~s$^{-1}$, respectively,
corresponding to best estimates for M31's halo (see
\S\,\ref{subsec_m31}). We fit for the width and mean of the second
Gaussian profile to determine both the velocity and velocity
dispersion of NGC~205 in each radial bin. The relative height of
these two profiles is fixed based on the M31 contamination fraction
determined by maximum likelihood fitting in \S\,\ref{subsec_m31}. We
have also run fits in which the contamination fraction is a free
parameter which does not change the resulting profile appreciably.
The resulting velocity and velocity dispersion profile as a function
of radius is shown in the bottom two panels of Figure~\ref{fig_vp}.
The velocity dispersion profile is determined using a coarser binning
scheme of 50 or more stars per radial bins. Error bars were computed
based on number statistics in each radial bin which dominate over the
radial velocity measurement errors of individual stars. We list the
velocity and velocity dispersion as a function of semi-major axis
distance, right ascension and declination in Table~4.
We determine the maximum major-axis rotation velocity, $v_{\rm max}$,
for NGC~205 by differencing the maximum/minimum velocity in the
upper-left quadrant and lower-right quadrant of Figure~\ref{fig_vp},
and dividing this number by two. The maximum rotation velocity for
NGC~205 is $v_{\rm max} = 11 \pm 5$\,\,km~s$^{-1}$. The peak rotation velocity
occurs at slightly different radii on either side of the galaxy, with
an average radius of $r_{\rm max} = 4.5'~(1.1$\,kpc). Our value is
somewhat smaller, but within the 1-sigma error bars, of the value
measured by \citet{sim02} of $13\pm2$\,\,km~s$^{-1}$\ at $r=4'$ based on
integrated-light measurements. The Simien \& Prugniel data hint at a
flattening in the velocity profile, but the integrated-light data is
noisy at these large radii. We observe this turnover conclusively
and note that $v_{\rm max}$ is the physical, rather than a
observational, maximum rotation velocity. Previous observations
of dE galaxies have not reached sufficient radii to observe a
definite turnover in the rotation curve; this is the first dE galaxy
in which the {\it maximum} rotation velocity has been measured.
We plot the ratio of the maximum rotational velocity to the average
velocity dispersion ($v_{\rm max}/\sigma$) versus ellipticity in
Figure~\ref{fig_vsigma}. We assume an ellipticity for NGC~205 of
$\epsilon = 0.43$ \citep{cho02}, and a velocity dispersion of $\sigma
= 35\pm5$\,\,km~s$^{-1}$ determined from our profile excluding data
beyond the tidal radius. The observed ratio for NGC~205 is $v_{\rm
max}/\sigma = 0.21$. At the ellipticity of NGC~205, the expected
ratio for an oblate, isotropic, rotationally-flattened body seen
edge-on is slightly more than unity \citep{bin87}. NGC~205 lies
midway between a rotationally supported and anisotropic object.
Interestingly, it was early kinematic observations of NGC~205 which
established the paradigm that dE galaxies are supported by anisotropic
velocity dispersions in contrast to rotationally supported normal
ellipticals \citep{ben90}. In Figure~\ref{fig_vsigma}, we compare
NGC~205 to a sample of similar luminosity dEs in the Virgo Cluster
taken from \citet{geh03}. The Virgo dEs naturally fall into
``rotating'' and ``non-rotating'' categories. NGC~205 does not fall
into either category having a mixture of both rotational and
anisotropic support.
At a radius of $r\sim4.5'$, the velocity profile of NGC~205 turns
over; stars beyond this radius rotate in a direction opposite that of
the main galaxy body. This turnover is in the sense that stars on the
side of NGC~205 closest to M31 move with more negative velocities
(approaching the systemic velocity of M31) and stars on the farside
move with less negative velocities. The turnover radius is coincident
with the onset of isophotal twisting as observed by \citet{cho02},
suggesting that the outer dynamics of NGC~205 have been affected by
the processes of tidal stripping. We discuss further evidence for
tidal interactions and its significance in \S\,\ref{sec_orbit}.
\subsection{Contamination from M31 Stars and Maximum-Likelihood Analysis}\label{subsec_m31}
NGC~205 lies merely $40'$ (8\,kpc) in projection from the center of
M31 and our kinematic sample is likely contaminated with stars
associated with M31. We consider two sources for contamination, the
M31 disk and halo. We estimate that contamination from M31's disk is
negligible: NGC~205 lies behind M31 and its line-of-sight intersects
the M31 disk at a point 37\,kpc (7 scale lengths) along the disk from
the center of M31 (assuming an inclination angle of the M31 disk of
$12.5^{\circ}$ and a disk scale length of 5.3\,kpc \citep{wal88}). At
this scale length, the M31 disk surface brightness is $\mu_B =
29.2$\,mag arcsec$^{-2}$. In comparison, at the average radial
distance of our spectroscopic sample ($r = 5'$), the surface
brightness of NGC~205 is $\mu_B = 25.4$\,mag arcsec$^{-2}$. Thus, our
kinematic sample is unlikely to contain stars from the disk of M31.
We estimate the contamination fraction from stars in the M31 halo
based on maximum likelihood fitting. We assume the M31 halo is
spherically symmetric with a power-law surface brightness profile. We
normalize the surface brightness (SB) of the halo at the center of NGC~205,
defining the free parameter $f_0$ to be the fractional contamination
at the center of NGC~205. The M31 halo surface brightness at the
position of a given slitlet $i$ is then:
\begin{equation}
{{\rm SB}_i}^{\rm M31} = f_0 ~ {\rm SB}^{\rm N205}_0 ~\Big{(}\frac{r^{\rm M31}_i}
{r^{\rm M31-N205}}\Big{)}^\alpha
\end{equation}
\noindent
where the angular distance between M31 and NGC~205 is $r^{M31-N205} =
40'$, the angular distance from slitlet $i$ to the center of M31 is
$r^{\rm M31}_i$ , and the free parameter $\alpha$ is the power-law
index of the halo profile. The surface brightness profile of NGC~205
is based on a high-order fit to the photometry of \citet{cho02} with a
central $B$-band surface brightness of $SB^{\rm N205}_0 = 19.2$~mag
arcsec$^{-2}$. We assume a Gaussian velocity distribution for the M31
halo with a velocity dispersion of width $\sigma^{\rm M31}$ and a
systemic velocity of $v^{\rm M31}_{\rm sys} = -300$\,\,km~s$^{-1}$\
\citep{guh04}. Given a measured velocity of $v_i$ for slitlet $i$,
the probability that it was drawn from the velocity distribution
of the M31 halo is:
\begin{equation}
P_i^{\rm M31} = \frac{1}{\sqrt{2\pi} \sigma^{\rm M31}}
exp\Big{[}-\frac{1}{2}\Big{(}\frac{v_i - v^{\rm M31}_{\rm
sys}}{\sigma^{\rm M31}}\Big{)}^2\Big{]}
\end{equation}
\noindent
Similarly, the probability that a given velocity is drawn from
velocity distribution of NGC~205, $P_i^{\rm N205}$, is the same as
Eqn.~(2), however $\sigma^{\rm N205}$ and $v^{\rm N205}_{\rm sys}$ are
the measured quantities as a function of radius. The final probability
function for each slitlet $i$ is:
\begin{equation}
P_i = C_i [{\rm SB}^{\rm N205}_i P^{\rm N205}_i + {\rm SB}^{\rm M31}_i P^{\rm
M31}_i]
\end{equation}
\noindent
The normalization constant $C_i \equiv 1/(SB^{N205}_i + SB^{M31}_i)$
is defined such that the integral over $P_i$ is equal to unity. The
probability $P_i$ is evaluated for each slitlet and the natural
logarithm of this quantity is summed over all slits. We do a gridded
parameter search over the free parameters $f_0$, $\sigma^{M31}$, and
$\alpha$ and maximize the function $M = \sum_i {\rm ln}(P_i)$. As
shown in the left panel of Figure~\ref{fig_maxlike}, the best fitting
parameters which maximize the quantity $M$ are: $f_0$ = 0.065,
$\sigma_{M31} = 150$\,\,km~s$^{-1}$\ and $\alpha=-2.25$. We note that these best
fitting M31 halo parameters are in good agreement with the measured
M31 halo velocity dispersion by \citet{guh04} and an $r^{-2}$ halo
surface brightness profile.
In the right panel of Figure~\ref{fig_maxlike}, we plot the fractional
contamination in our kinematic sample from the M31 halo as a function
of distance from the center of NGC~205 using the best fit parameters
determined above. At the center of NGC~205, the best estimate of the
contamination from M31 is 6\%; this corresponds to 1.5 out of 25 stars
per radial bin. On the side of NGC~205 closest to M31 the
contamination is 20\%, while on the farside the contamination is less
than 3\%. The average halo contamination fraction for our sample is
6.5\%.
\section{The Tidal Disruption of NGC~205}\label{sec_orbit}
The outer dynamics of NGC~205 place significant constraint on the
tidal interaction between NGC~205 and its parent galaxy M31. While
the main body of NGC~205 is rotating with a maximum velocity of
11\,\,km~s$^{-1}$\,, the major-axis velocity profile turns over abruptly at a
radius of $4.5'~(1$\,kpc); stars at larger radii are moving in the
opposite direction than the main galaxy body. The turnover radius
coincides with the onset of isophotal twisting \citep{hod73,ken87} and
a downward break in the surface brightness profile \citep{cho02}.
Both features have been demonstrated by Choi~et~al.~to be hallmarks of
tidal stripping. We make a simple estimate of the tidal radius of
NGC~205, assuming NGC~205 and M31 to be point masses. We adopt a
physical separation distance between these two galaxies of
$r_{N205-M31} = 40$\,kpc, the mass of M31 inside this distance of
$M_{\rm M31} = 1\times10^{11}\mbox{\,$M_{\odot}$}$ \citep{gee05} and the total mass
of NGC~205 $M_{\rm N205} = 2\times10^{9} \mbox{\,$M_{\odot}$}$ (calculated from the
maximum rotation velocity and velocity dispersion at the turnover
radius). The tidal radius is calculated to be $r_{\rm tidal} \sim 4'$
(1.0\,kpc). The observed velocity turnover in NGC~205, at roughly
this distance, is therefore due to tidal stripping from M31; stars
beyond this radius are no longer bound to NGC~205.
The stellar motion beyond the tidal radius of NGC~205 suggests that it
is on a prograde encounter with its parent galaxy: the spin angular
momentum vector and the orbital angular momentum vector are
parallel. The eventual destruction of a satellite on a
prograde orbit is expected to proceed more slowly and create less
spectacular tidal tails than one on a retrograde orbit
\citep[e.g.,~compare Figs.~1 and 2 of][]{too72}. We calculate the
timescale for the destruction of NGC~205 based on the dynamical
friction timescale. Using Eqn.~(7.26) in \citet{bin87}, we assume
the orbital velocity of NGC~205 is related the velocity difference
between NGC~205 and M31 ($=\sqrt{3} \times 60$\,\,km~s$^{-1}$) and calculate
$t_{\rm fric} \sim 3 \times 10^9$ years. Detailed dynamical modeling
is required to determine if features in the M31 halo, such as stellar
substructure \citep{mcc04}, or HI clouds \citep{thi04} are indeed
associated with the orbit of NGC~205.
\section{Summary}\label{sec_disc}
We present the stellar kinematics of NGC~205 out to large radii based
on Keck/DEIMOS multislit spectroscopic observations of 725 individual
red giant branch stars. NGC~205 is one of the closest examples of a
dwarf elliptical galaxy and the prototype of this galaxy class. Early
kinematic observations of NGC~205 established the paradigm that dE
galaxies are supported primarily by anisotropic velocity dispersions
\citet{ben91}. Although kinematic study of dEs outside the Local
Group suggest that a fraction of dEs are indeed supported by
anisotropic velocity dispersion alone \citep{geh03,der01}, NGC~205
itself has significant rotation. We measure a maximum major-axis
rotation speed for the body of NGC~205 of $11\pm5$\,\,km~s$^{-1}$, implying
that this galaxy is supported by a mixture of both rotational and
anisotropic velocities.
The velocity profile of NGC~205 turns over at a major-axis distance of
$4.5'~(1$\,kpc). This turnover is due to gravitational interaction
between NGC~205 and its parent galaxy M31. The motion of stars beyond
the tidal radius suggest that NGC~205 is on a prograde encounter
with M31. Detailed dynamical modeling should clarify what, if any,
substructure in the M31 halo is associated with disruption of NGC~205
and provide insight into the on-going interactions between these two
galaxies.
\acknowledgments
We thank S.~Demers and P.~Battinelli for kindly providing their
photometric catalogs. We also thank P.~Choi, G.~Laughlin, D.~Kelson
and R.~P.~van der Marel for productive and enjoyable conversations.
M.~G.~is supported by NASA through Hubble Fellowship grant
HF-01159.01-A awarded by the Space Telescope Science Institute, which
is operated by the Association of Universities for Research in
Astronomy. P.G. acknowledges support from NSG grant AST-0307966 and
NASA/STScI grant GO-10265.02. R.M.R. acknowledges funding by grant
AST-0307931.
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 5,173 |
Protecting biodiversity in the European Union: National barriers and European opportunities?
J Fairbrass, A Jordan
Tyndall Centre for Climate Change Research
The European Union (EU) is an evolving system of multi-level governance (MLG). For scholars of the EU, a critical question is which level of governance has the most decisive influence on the integration process? Some studies of EU regional policy claim that subnational actors, using channels of interest representation that bypass national officials, interact directly with EU policy-makers generating outcomes that are neither desired nor intended by national executives. This article examines the development of EU biodiversity policy over a thirty-year period (c. 1970-2000) and finds that environmental groups, who were generally marginalized at the national level in Britain, have learnt to use EU opportunities to outflank the government, resulting in policy outcomes that they would be unlikely to secure through national channels of representation. However, the evidence presented suggests that supranational actors were the major cause of these unintended consequences, not environmental groups.
Journal of European Public Policy
Fairbrass, J., & Jordan, A. (2001). Protecting biodiversity in the European Union: National barriers and European opportunities? Journal of European Public Policy, 8(4), 499-518. https://doi.org/10.1080/13501760110064366
Fairbrass, J ; Jordan, A. / Protecting biodiversity in the European Union: National barriers and European opportunities?. In: Journal of European Public Policy. 2001 ; Vol. 8, No. 4. pp. 499-518.
@article{f4bca317f0ca489fad89c7327886702a,
title = "Protecting biodiversity in the European Union: National barriers and European opportunities?",
abstract = "The European Union (EU) is an evolving system of multi-level governance (MLG). For scholars of the EU, a critical question is which level of governance has the most decisive influence on the integration process? Some studies of EU regional policy claim that subnational actors, using channels of interest representation that bypass national officials, interact directly with EU policy-makers generating outcomes that are neither desired nor intended by national executives. This article examines the development of EU biodiversity policy over a thirty-year period (c. 1970-2000) and finds that environmental groups, who were generally marginalized at the national level in Britain, have learnt to use EU opportunities to outflank the government, resulting in policy outcomes that they would be unlikely to secure through national channels of representation. However, the evidence presented suggests that supranational actors were the major cause of these unintended consequences, not environmental groups.",
author = "J Fairbrass and A Jordan",
journal = "Journal of European Public Policy",
Fairbrass, J & Jordan, A 2001, 'Protecting biodiversity in the European Union: National barriers and European opportunities?', Journal of European Public Policy, vol. 8, no. 4, pp. 499-518. https://doi.org/10.1080/13501760110064366
Protecting biodiversity in the European Union: National barriers and European opportunities? / Fairbrass, J; Jordan, A.
In: Journal of European Public Policy, Vol. 8, No. 4, 2001, p. 499-518.
T1 - Protecting biodiversity in the European Union: National barriers and European opportunities?
AU - Fairbrass, J
AU - Jordan, A
N2 - The European Union (EU) is an evolving system of multi-level governance (MLG). For scholars of the EU, a critical question is which level of governance has the most decisive influence on the integration process? Some studies of EU regional policy claim that subnational actors, using channels of interest representation that bypass national officials, interact directly with EU policy-makers generating outcomes that are neither desired nor intended by national executives. This article examines the development of EU biodiversity policy over a thirty-year period (c. 1970-2000) and finds that environmental groups, who were generally marginalized at the national level in Britain, have learnt to use EU opportunities to outflank the government, resulting in policy outcomes that they would be unlikely to secure through national channels of representation. However, the evidence presented suggests that supranational actors were the major cause of these unintended consequences, not environmental groups.
AB - The European Union (EU) is an evolving system of multi-level governance (MLG). For scholars of the EU, a critical question is which level of governance has the most decisive influence on the integration process? Some studies of EU regional policy claim that subnational actors, using channels of interest representation that bypass national officials, interact directly with EU policy-makers generating outcomes that are neither desired nor intended by national executives. This article examines the development of EU biodiversity policy over a thirty-year period (c. 1970-2000) and finds that environmental groups, who were generally marginalized at the national level in Britain, have learnt to use EU opportunities to outflank the government, resulting in policy outcomes that they would be unlikely to secure through national channels of representation. However, the evidence presented suggests that supranational actors were the major cause of these unintended consequences, not environmental groups.
JO - Journal of European Public Policy
JF - Journal of European Public Policy
Fairbrass J, Jordan A. Protecting biodiversity in the European Union: National barriers and European opportunities? Journal of European Public Policy. 2001;8(4):499-518. https://doi.org/10.1080/13501760110064366 | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 6,436 |
Q: find substrings between two string I have a string like this:
string = r'''<img height="233" src="monline/" title="email example" width="500" ..
title="second example title" width="600"...
title="one more title"...> '''
I am trying to get anything that appears as title (title="Anything here")
I have already tried this but it does not work correctly.
re.findall(r'title=\"(.*)\"',string)
A: I think your Regex is too Greedy. You can try something like this
re.findall(r'title=\"(?P<title>[\w\s]+)\"', string)
As @Austin and @Plato77 said in the comments, there is a better way to parse HTML in python. See other SO Answers for more context. There are a few common tools for this like:
*
*https://docs.python.org/3/library/html.parser.html
*https://www.simplifiedpython.net/parsing-html-in-python/
*https://github.com/psf/requests-html / Get html using Python requests?
If you would like to read more on performance testing of different python HTML parsers you can learn more here
A: As @Austin and @Plato77 said in the comments, there is a better way to parse HTML in python. I stand by this too, but if you want to get it done through regex this may help
c = re.finditer(r'title=[\"]([a-zA-Z0-9\s]+)[\" ]', string)
for i in c:
print(i.group(1))
A: The problem here is that the next " symbol is parsed as a character and is considered part of the (.*) of your RE. For your usecase, you can use only letters and numbers.
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 8,098 |
{"url":"https:\/\/moderndive.netlify.app\/8-2-resampling-simulation.html","text":"8.2 Computer simulation of resampling\n\nLet\u2019s now mimic our tactile resampling activity virtually with a computer.\n\n8.2.1 Virtually resampling once\n\nFirst, let\u2019s perform the virtual analog of resampling once. Recall that the pennies_sample data frame included in the moderndive package contains the years of our original sample of 50 pennies from the bank. Furthermore, recall in Chapter 7 on sampling that we used the rep_sample_n() function as a virtual shovel to sample balls from our virtual bowl of 2400 balls as follows:\n\nvirtual_shovel <- bowl %>%\nrep_sample_n(size = 50)\n\nLet\u2019s modify this code to perform the resampling with replacement of the 50 slips of paper representing our original sample 50 pennies:\n\nvirtual_resample <- pennies_sample %>%\nrep_sample_n(size = 50, replace = TRUE)\n\nObserve how we explicitly set the replace argument to TRUE in order to tell rep_sample_n() that we would like to sample pennies with replacement. Had we not set replace = TRUE, the function would\u2019ve assumed the default value of FALSE and hence done resampling without replacement. Additionally, since we didn\u2019t specify the number of replicates via the reps argument, the function assumes the default of one replicate reps = 1. Lastly, observe also that the size argument is set to match the original sample size of 50 pennies.\n\nLet\u2019s look at only the first 10 out of 50 rows of virtual_resample:\n\nvirtual_resample\n# A tibble: 50 x 3\n# Groups: replicate [1]\nreplicate ID year\n<int> <int> <dbl>\n1 1 37 1962\n2 1 1 2002\n3 1 45 1997\n4 1 28 2006\n5 1 50 2017\n6 1 10 2000\n7 1 16 2015\n8 1 47 1982\n9 1 23 1998\n10 1 44 2015\n# \u2026 with 40 more rows\n\nThe replicate variable only takes on the value of 1 corresponding to us only having reps = 1, the ID variable indicates which of the 50 pennies from pennies_sample was resampled, and year denotes the year of minting. Let\u2019s now compute the mean year in our virtual resample of size 50 using data wrangling functions included in the dplyr package:\n\nvirtual_resample %>%\nsummarize(resample_mean = mean(year))\n# A tibble: 1 x 2\nreplicate resample_mean\n<int> <dbl>\n1 1 1996\n\nAs we saw when we did our tactile resampling exercise, the resulting mean year is different than the mean year of our 50 originally sampled pennies of 1995.44.\n\n8.2.2 Virtually resampling 35 times\n\nLet\u2019s now perform the virtual analog of our 35 friends\u2019 resampling. Using these results, we\u2019ll be able to study the variability in the sample means from 35 resamples of size 50. Let\u2019s first add a reps = 35 argument to rep_sample_n() to indicate we would like 35 replicates. Thus, we want to repeat the resampling with the replacement of 50 pennies 35 times.\n\nvirtual_resamples <- pennies_sample %>%\nrep_sample_n(size = 50, replace = TRUE, reps = 35)\nvirtual_resamples\n# A tibble: 1,750 x 3\n# Groups: replicate [35]\nreplicate ID year\n<int> <int> <dbl>\n1 1 21 1981\n2 1 34 1985\n3 1 4 1988\n4 1 11 1994\n5 1 26 1979\n6 1 8 1996\n7 1 19 1983\n8 1 21 1981\n9 1 49 2006\n10 1 2 1986\n# \u2026 with 1,740 more rows\n\nThe resulting virtual_resamples data frame has 35 $$\\cdot$$ 50 = 1750 rows corresponding to 35 resamples of 50 pennies. Let\u2019s now compute the resulting 35 sample means using the same dplyr code as we did in the previous section, but this time adding a group_by(replicate):\n\nvirtual_resampled_means <- virtual_resamples %>%\ngroup_by(replicate) %>%\nsummarize(mean_year = mean(year))\nvirtual_resampled_means\n# A tibble: 35 x 2\nreplicate mean_year\n<int> <dbl>\n1 1 1995.58\n2 2 1999.74\n3 3 1993.7\n4 4 1997.1\n5 5 1999.42\n6 6 1995.12\n7 7 1994.94\n8 8 1997.78\n9 9 1991.26\n10 10 1996.88\n# \u2026 with 25 more rows\n\nObserve that virtual_resampled_means has 35 rows, corresponding to the 35 resampled means. Furthermore, observe that the values of mean_year vary. Let\u2019s visualize this variation using a histogram in Figure 8.12.\n\nggplot(virtual_resampled_means, aes(x = mean_year)) +\ngeom_histogram(binwidth = 1, color = \"white\", boundary = 1990) +\nlabs(x = \"Resample mean year\")\n\nLet\u2019s compare our virtually constructed bootstrap distribution with the one our 35 friends constructed via our tactile resampling exercise in Figure 8.13. Observe how they are somewhat similar, but not identical.\n\nRecall that in the \u201cresampling with replacement\u201d scenario we are illustrating here, both of these histograms have a special name: the bootstrap distribution of the sample mean. Furthermore, recall they are an approximation to the sampling distribution of the sample mean, a concept you saw in Chapter 7 on sampling. These distributions allow us to study the effect of sampling variation on our estimates of the true population mean, in this case the true mean year for all US pennies. However, unlike in Chapter 7 where we took multiple samples (something one would never do in practice), bootstrap distributions are constructed by taking multiple resamples from a single sample: in this case, the 50 original pennies from the bank.\n\n8.2.3 Virtually resampling 1000 times\n\nRemember that one of the goals of resampling with replacement is to construct the bootstrap distribution, which is an approximation of the sampling distribution. However, the bootstrap distribution in Figure 8.12 is based only on 35 resamples and hence looks a little coarse. Let\u2019s increase the number of resamples to 1000, so that we can hopefully better see the shape and the variability between different resamples.\n\n# Repeat resampling 1000 times\nvirtual_resamples <- pennies_sample %>%\nrep_sample_n(size = 50, replace = TRUE, reps = 1000)\n\n# Compute 1000 sample means\nvirtual_resampled_means <- virtual_resamples %>%\ngroup_by(replicate) %>%\nsummarize(mean_year = mean(year))\n\nHowever, in the interest of brevity, going forward let\u2019s combine these two operations into a single chain of pipe (%>%) operators:\n\nvirtual_resampled_means <- pennies_sample %>%\nrep_sample_n(size = 50, replace = TRUE, reps = 1000) %>%\ngroup_by(replicate) %>%\nsummarize(mean_year = mean(year))\nvirtual_resampled_means\n# A tibble: 1,000 x 2\nreplicate mean_year\n<int> <dbl>\n1 1 1992.6\n2 2 1994.78\n3 3 1994.74\n4 4 1997.88\n5 5 1990\n6 6 1999.48\n7 7 1990.26\n8 8 1993.2\n9 9 1994.88\n10 10 1996.3\n# \u2026 with 990 more rows\n\nIn Figure 8.14 let\u2019s visualize the bootstrap distribution of these 1000 means based on 1000 virtual resamples:\n\nggplot(virtual_resampled_means, aes(x = mean_year)) +\ngeom_histogram(binwidth = 1, color = \"white\", boundary = 1990) +\nlabs(x = \"sample mean\")\n\nNote here that the bell shape is starting to become much more apparent. We now have a general sense for the range of values that the sample mean may take on. But where is this histogram centered? Let\u2019s compute the mean of the 1000 resample means:\n\nvirtual_resampled_means %>%\nsummarize(mean_of_means = mean(mean_year))\n# A tibble: 1 x 1\nmean_of_means\n<dbl>\n1 1995.36\n\nThe mean of these 1000 means is 1995.36, which is quite close to the mean of our original sample of 50 pennies of 1995.44. This is the case since each of the 1000 resamples is based on the original sample of 50 pennies.\n\nCongratulations! You\u2019ve just constructed your first bootstrap distribution! In the next section, you\u2019ll see how to use this bootstrap distribution to construct confidence intervals.\n\nLearning check\n\n(LC8.1) What is the chief difference between a bootstrap distribution and a sampling distribution?\n\n(LC8.2) Looking at the bootstrap distribution for the sample mean in Figure 8.14, between what two values would you say most values lie?","date":"2020-12-03 01:25:47","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 1, \"mathjax_asciimath\": 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\": 0.5415066480636597, \"perplexity\": 1938.2946549435073}, \"config\": {\"markdown_headings\": false, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2020-50\/segments\/1606141717601.66\/warc\/CC-MAIN-20201203000447-20201203030447-00521.warc.gz\"}"} | null | null |
\section*{}
\section{Introduction}
One of the fundamental problems of modern pulsar astrophysics is the generation of electromagnetic waves in the high energy domain. It is believed that pulsar's radiation is produced mainly by means of the synchrotron mechanism \citep{pacini,shklovsky} and the inverse Compton scattering \citep{blandf}. On the other hand, one can straightforwardly show that typical timescales of synchrotron emission is so small compared to the escape timescale (which is of the order of rotation period of a pulsar) that almost from the very beginning of motion the particles go to the ground Landau state and after that slide along the magnetic field lines. This means that the overall dynamics of particles is described by a one-dimensional picture.
It is observationally evident that some pulsars exhibit efficient radiation in the X-ray band \citep{x1,x2} and some of them usually belong to binary systems \citep{x3,x4}. It is clear that generation of radiation requires efficient acceleration of particles. In general it is thought that particles might accelerate from a nearby region of the polar caps (polar cap acceleration) \citep{polar1,polar2,polar3} as well as from the outer magnetosphere, by means of the so called outer gap mechanism \citep{outer1,outer2} or centrifugal acceleration \citep{or1,or2,or3}.
Already accelerated particles might generate electromagnetic radiation via several mechanisms: the synchrotron process, the inverse Compton scattering and the curvature radiation. In the magnetospheres of pulsars the synchrotron radiation power is so efficient that if the pitch angles are not very small the particles very soon transit to the ground Landau state and slide almost along the magnetic field lines, making dynamics almost one-dimensional. If the particles follow the so-called open field lines, the magnetic field decreases and the synchrotron process is strongly suppressed. Unlike the mentioned scenario, the particles which stay on closed field lines might turn back to the star's surface, potentially leading to a very interesting result. Due to the conservation of the adiabatic invariant, $I = 3cp_{\perp}^2/2eB$, where $c$ is the speed of light, $p_{\perp}$ is the transversal momentum of electron, $e$ is its charge and $B$ is the magnetic field induction, the value of $p_{\perp}$ \citep{landau} must inevitably increase if the particle slides along the field line and penetrates the region with strong magnetic field. On the other hand, the synchrotron emission is strongly dependent on the transversal component of momentum and as a result the corresponding emission mechanism will be revived inside the magnetosphere.
If the particles are in frozen-in condition, they co-rotate with the field lines and consequently the force responsible for co-rotation is significant. On the other hand, the force responsible for radiation will be important if it exceeds the co-rotation force.
The paper is organised in the following way: in Sec. 2, we develop the model, and in Sec. 3
we summarise our results for typical parameters of pulsars.
\section[]{Main approach}
By assuming that the particles are in frozen-in condition, they follow the magnetic field lines and thus dynamics of particles is described by a bead on the wire approximation. In the framework of the paper we consider the dipolar field lines when the vector of angular velocity of rotation, $\vec{\omega}$, and the dipolar magnetic momentum are perpendicular (see Fig. 1). Then, if one combines the prescribed dipolar channels, $\phi = \varphi (r)+\omega t$, with the Minkowskian metric defined in the polar coordinates
\begin{equation}
\label{metr1}
ds^2 = -c^2dt^2+d\ell^2+\ell^2d\phi^2+dz^2,
\end{equation}
after some boring but straightforward algebra one obtains the metric on the co-rotating trajectories
\begin{equation}
\label{metr2}
ds^2 = g_{00}(dx^0)^2+2g_{01}dx^0dx^1+g_{11}(dx^1)^2,
\end{equation}
\begin{equation}
\
g_{\alpha\beta}=
\begin{bmatrix}
-1+\frac{\omega^2r^2}{c^2}m(r) & \frac{\omega r^2\varphi'(r)\sin\delta}{c} \\
\\
\frac{\omega r^2\varphi'(r)\sin\delta}{c} & g^2(r)+k^2(r)+r^2h^2(r)m(r)
\end{bmatrix}
\
\end{equation}
where
$$\beta=\{0,1\}
$$
$$m(r) =1-\frac{r}{L}\cos^2\delta$$
$$g(r) = \frac{1-\frac{3r}{2L}\cos^2\delta}{\sqrt{1-\frac{r}{L}\cos^2\delta}}$$
$$k(r) = \frac{3}{2}\sqrt{\frac{r}{L}}\cos\delta$$
$$h(r) = \frac{\varphi'(r)\sin\delta}{1-\frac{r}{L}\cos^2\delta}$$
$\varphi (r)$ describes a shape of the dipolar magnetic field line in the proper plane ($r = L\sin^2\theta$), $L$ is the corresponding lengthscale, $\delta$ is the inclination angle of a particular field line's plane with resect to $\vec{\omega}$ and $x^0 = ct$, $x^1 = r$.
As we have already mentioned we consider acceleration of particles close to the light cylinder (LC) surface (a hypothetical area where the linear velocity of rotation equals the speed of light). On the other hand, different planes of field lines, corresponding to different values of $\delta$ intersect the LC with different distances from the pulsar. It is clear that higher Lorentz factors will correspond to stronger magnetic fields, which is the case with $\delta = \pi/2$. For this reason we consider this particular case and make the relevant calculations.
By setting $\delta = \pi/2$ the polar radial coordinate, $\ell$, coincides with the proper radial coordinate of a particular field line, $r$, and consequently the aforementioned metric tensor reduces to
\begin{equation}
\
g_{\alpha\beta}=
\begin{bmatrix}
-1+\frac{\omega^2r^2}{c^2} & \frac{\omega r^2\varphi'(r)}{c} \\
\\
\frac{\omega r^2\varphi'(r)}{c} & 1+r^2(\varphi'(r))^2
\end{bmatrix}
\
\end{equation}
For this case the Lagrangian
\begin{equation}
\label{lag}
L = -mc^2\left(-g_{00}-2g_{01}\frac{\upsilon}{c}-g_{11}\frac{\upsilon^2}{c^2}\right)^{1/2}
\end{equation}
does not depend on time explicitly, therefore, the Hamiltonian
\begin{equation}
\label{ham}
\mathcal{H} = \upsilon\frac{\partial L}{\partial\upsilon}-L = \gamma mc^2\left(-g_{00}-g_{01}\frac{\upsilon}{c}\right) = E
\end{equation}
is a constant of motion and consequently the Lorentz factor behaves as \citep{or2}
\begin{equation}
\label{gamma}
\gamma = \gamma_0\frac{1-\omega r_0^2\Omega_0/c^2}{1-\omega r^2\Omega/c^2},
\end{equation}
where $\upsilon = dr/dt$ is the radial velocity, $\Omega = \omega+\varphi'(r)\upsilon$ is the effective angular velocity of rotation and $\gamma_0$, $r_0$ and $\Omega_0 $ are the initial values of the Lorentz factor, the radial coordinate and effective angular velocity of rotation respectively.
From Eq. (\ref{ham}) one can straightforwardly show that the radial velocity behaves with $r$ as follows \citep{arsen}
$$v = c \frac{\sqrt{g_{11}+E^2}}{g_{12}^2+E^2g_{22}}\times$$
\begin{equation}
\times\left(-g_{12}\sqrt{g_{11}+E^2}\pm E\sqrt{g_{12}^2-g_{11}g_{22}} \right).
\end{equation}
Components of the effective force acting on the particle responsible for co-rotation is expressed as \citep{grg}
\begin{equation}
\label{Fr}
F_r = \frac{d}{dt}\left(\gamma m\upsilon\right)-\gamma m\Omega^2r,
\end{equation}
\begin{equation}
\label{Ff}
F_{\varphi} = \frac{d}{dt}\left(\gamma m\Omega r\right)+\gamma m\Omega\upsilon,
\end{equation}
where $\gamma =\left(-g_{00}-2g_{01}\frac{\upsilon}{c}-g_{11}\frac{\upsilon^2}{c^2}\right)^{-1/2}$ is the relativistic factor of the particle and the time dependence is governed by the value of the radial acceleration \citep{grg}
\begin{equation}
\label{dv}
\frac{d\upsilon}{dt} = \frac{r\omega\Omega-\gamma^2r\upsilon\left(\varphi'+\omega\upsilon/c^2\right)\left(\Omega+r\varphi''\upsilon\right)}{\gamma^2\left(1-\omega^2r^2/c^2+r^2(\varphi')^2\right)}.
\end{equation}
Apart from the aforementioned force the charged particles also undergo the force providing the conservation of the adiabatic invariant
\begin{equation}
\label{Gpp}
G_{\perp} = -\frac{c}{\rho}p_{\perp},
\end{equation}
\begin{equation}
\label{Gpr}
G_{\parallel} = \gamma mc^2 \;\frac{p_{\perp}^2}{p_{\parallel}^2},
\end{equation}
where $p_{\perp}$ and $p_{\parallel}$ are respectively the transversal and longitudinal components of momentum and $\rho$ represents the curvature radius of the field lines. This force contributes in a very important process of maintenance of synchrotron emission despite the efficient energy losses. In particular, by means of the so called cyclotron instability the quasi linear diffusion leads to the following kinetic equation \citep{QLD}
\begin{equation}
\label{kin}
\frac{\partial f}{\partial t}+\frac{1}{p_{\perp}}\frac{\partial}{\partial p_{\perp}}\left(p_{\perp}G_{\perp}f\right) =
\frac{1}{p_{\perp}}\frac{\partial}{\partial p_{\perp}}\left(p_{\perp} D_{\perp,\perp}\frac{\partial f}{\partial p_{\perp}} \right),
\end{equation}
where
\begin{equation}
\label{D}
D_{\perp,\perp} \approx \frac{e^2mc^2n\gamma}{16\omega_B},
\end{equation}
is the corresponding diffusion coefficient, $n$ is the relativistic particles' number density and $\omega_B$ is the cyclotron frequency.
By considering the quasi stationary case $\partial/\partial t = 0$, one can straightforwardly obtain
\begin{equation}
\label{f}
f(p_{\perp}) = C \exp\left(\int\frac{G_{\perp}}{D_{\perp,\perp}}dp_{\perp}\right) = Ce^{-\left(\frac{p_{\perp}}{p_{\perp,0}}\right)^2},
\end{equation}
where
\begin{equation}
\label{pp0}
p_{\perp,0} = \sqrt{\frac{2\rho D_{\perp,\perp}}{c}}.
\end{equation}
After averaging the value of $p_{\perp}$ one derives an expression for the residual pitch angle, $\psi_0 = p_{\perp,av}/(\gamma mc)$,
\begin{equation}
\label{pitch}
\psi_0 = \frac{1}{\sqrt{\pi}}\frac{p_{\perp,0}}{\gamma mc}.
\end{equation}
Therefore, if the particles are accelerated close to the LC, the perpendicular momentum rapidly almost completely vanishes and a residual pitch angle defined by Eq. (\ref{pitch}) plays a role of the initial pitch angle. Usually these values are very small and consequently the synchrotron mechanism is strongly suppressed.
According to the standard approach, the particles accelerated on the open field lines, leave the magnetosphere and contribute in emission processes generated on the LC scales and beyond. On the other hand, there is a certain fraction of relativistic particles moving along the closed magnetic field lines. They sooner or later will reach the inner areas of pulsar's magnetosphere and as a result, by means of the adiabatic invariant, the corresponding transversal component (pitch angle) will inevitably increase turning the synchrotron process back into the game. This in turn means that the synchrotron radiation reaction force having the following components \citep{landau}
\begin{equation}
\label{Hpp}
H_{\perp} =-\alpha\;\frac{p_{\perp}}{p_{\parallel}}\left(1+\frac{p_{\perp}^2}{m^2c^2}\right),
\end{equation}
\begin{equation}
\label{Hpr}
H_{\parallel} =-\alpha\;\frac{p_{\perp}^2}{m^2c^2}
\end{equation}
with $\alpha = 2e^2\omega_B^2/(3c^2)$ will become significant.
The major difference from the standard emission mechanisms is that we consider the trapped particles, accelerating in the outer gap region, but which are going back towards the neutron star's surface emitting from higher and higher magnetic field regions. The analysis of the aforementioned forces might reveal an area in the pulsar's magnetosphere, where the synchrotron mechanism will be efficient. This particular problem we consider in the following section.
\begin{figure}
\centering {\includegraphics[width=8cm]{magnetosphere.eps}}
\caption{Here we schematically show the pulsar's magnetosphere. The arrow indicates a direction of the vector of the angular velocity of rotation which is perpendicular to the magnetic dipole momentum.}\label{fig1}
\end{figure}
\begin{figure}
\centering {\includegraphics[width=8cm]{fig1.eps}}
\caption{Here we plot the forces, F (black), G (blue) and H (red) versus distance from the star's centre (normalised by the star's radius). The set of parameters is $P = 1$sec, $\dot{P} = 10^{-15}$ss$^{-1}$, $L = 0.9 R_{lc}$, $R_{st} = 10^6$cm, $\rho \simeq L$, $r_0\simeq L$ and $\gamma_0 = 10^6$.}\label{fig2}
\end{figure}
\begin{figure}
\centering {\includegraphics[width=8cm]{fig2.eps}}
\caption{In this plot we show the dependence of emission energy on distance (normalised by the neutron star's radius) is. The set of parameters is the same as in Fig. 1.}\label{fig3}
\end{figure}
\section{Discusion}
Here we examine normal pulsars with typical periods of rotation $P\simeq 1$sec, for which the magnetic field and the particle number density (the Goldreich-Julian density) are respectively given by
\begin{equation}
\label{B}
B \simeq 10^{12}\times\left(\frac{P}{1s}\times\frac{\dot{P}}{10^{-15}erg/s}\right)^{1/2}\times\left(\frac{R_{st}}{r}\right)^3 \;G,
\end{equation}
and
\begin{equation}
\label{gj}
n_{GJ} = \frac{\omega B}{2\pi ec}\frac{1}{1-\frac{\omega^2r^2}{c^2}}.
\end{equation}
where $\dot{P}\equiv dP/dt$, $R_{st} \simeq10km$ represents the neutron star's radius.
In this paper we consider the dipolar configuration of the field lines in polar coordinates $r = L\sin\varphi$, where $L$ is the scale parameter. In the framework of the approach we assume that particles are accelerated in the outer gap zone \citep{outer1,outer2}, where usually the Lorentz factors are of the order of $\gamma = 10^{6-7}$. In general it is believed that the mentioned area is very close to the LC, therefore the choice $L = 0.9 R_{lc}$ ($R_{lc}=c/\Omega$ is the light cylinder radius) might be considered as to be a reasonable parameter.
On the other hand, we are interested in particles, following the field lines. The frozen-in condition works if the plasma energy density is small compared to the magnetic field energy density
\begin{equation}
\label{cond}
\gamma mc^2n_{GJ} \leq\frac{B^2}{8\pi},
\end{equation}
leading to the following Lorentz factor
\begin{equation}
\label{gcor}
\gamma \leq\frac{eB}{4\omega mc}\left(1-\frac{\omega^2r^2}{c^2}\right).
\end{equation}
By combining this expression with Eq. (\ref{B}), one can show that the highest value of the initial Lorentz factor still providing the frozen-in condition is of the order of $2\times 10^6$.
As a first example we consider $\gamma_0 = 10^6$, which gives the initial residual pitch angle of the order of $8\times 10^{-4}$. One can straightforwardly check that the force responsible for co-rotation exceeds all other forces, therefore, on the initial stage synchrotron emission is negligible. The electrons, sliding along the field lines, will move toward the star's surface, entering the higher magnetic field regions. As a result, by means of the adiabatic invariant the perpendicular momentum will increase
\begin{equation}
\label{pper}
p_{\perp} = \frac{p_{\perp,0}}{\sqrt{\pi}}\left(\frac{r_0}{r}\right)^{3/2},
\end{equation}
leading to the increase of the value of $G$ (see Eqs. (\ref{Gpp},\ref{Gpr})). It is worth noting that if initially the plasma particles are frozen in strong magnetic fields this condition will be maintained in the whole course of motion. Indeed, when particles move along the field lines back to the neutron star's surface, by means of the centrifugal force it will decelerate, leading to the decrease of the Lorentz factor. On the other hand, the regions closer to the star have stronger magnetic fields. Therefore, if initially the particles follow the field lines, they will always stay on them.
In Fig. 2 we show the behavioour of the aforementioned forces, $F = \sqrt{F_{r}^2+F_{\varphi}^2}$ (black), $G = \sqrt{G_{\perp}^2+G_{\parallel}^2}$ (blue) and $H = \sqrt{H_{\perp}^2+H_{\parallel}^2}$ (red) versus the distance from the pulsar's centre. The distance is normalised by the neutron star's radius. The set of parameters is $P = 1$sec, $\dot{P} = 10^{-15}$ss$^{-1}$, $L = 0.9 R_{lc}$, $R_{st} = 10^6$cm, $\rho \simeq L$, $r_0\simeq L$ and $\gamma_0 = 10^6$. As it is clear from the plots, on the distance of the order of $500 R_{st}$ the force responsible for the adiabatic invariant, $G$, becomes comparable to $F$ and consequently the synchrotron emission becomes important from this area. On the distance of the order of $150 R_{st}$ the synchrotron emission becomes most efficient because as it is evident from the figure the radiation reaction force becomes of the same order of magnitude as the force responsible for co-rotation. The corresponding gamma ray energy of synchrotron emission is given by \citep{rybicki}
\begin{equation}
\label{eps}
\epsilon = \frac{3\gamma^2 eB}{4h\pi mc}\sin\psi,
\end{equation}
where $h$ is the Plank'c constant and the pitch angle behaves with distance as (see Eq. (\ref{pper}))
\begin{equation}
\label{psi}
\psi = \psi_0\left(\frac{r_0}{r}\right)^{3/2},
\end{equation}
In Fig 3 we show the behaviour of synchrotron frequency vs distance (normalised by the neutron star's radius). The set of parameters is the same as in Fig. 1. As it is clear from the plot, from $150 R_{st}$ to $500 R_{st}$ the synchrotron emission provides energies in the interval $(0.1-10)$MeV. The energy is a continuously decreasing function of distance, which is a natural result of Eqs. (\ref{eps},\ref{psi}). In particular, by increasing the distance the magnetic field induction as well as the pitch angle decrease leading to the following behaviour.
It is worth noting that a similar mechanism has been examined in the context of generation of infrared radiation in the binary system \citep{mst}. In particular, the authors have shown that the cyclotron instability by means of the quasi-linear difussion can explain emission of a neutron star's accretion flow in the frequency domain $\omega\simeq 10^{14}$Hz.
\section{Conclusion}
For studying the generation of high energy emission by means of the synchrotron mechanism we examine particles already accelerated in the outer magnetosphere. By means of the so-called quasi linear diffusion it has been shown that if the particles have Lorentz factors of the order of $10^6$, the pitch angle becomes very small $8\times 10^{-4}$.
In due course of motion, as the particles reach the stars' surface the pitch angle increases. We have shown that initially the force responsible for co-rotation is higher compared to other two forces. At the distance $500 R_{st}$ the value of $G$ becomes of the order of $F$, which means that the pitch angle becomes significant and the synchrotron emission comes into the game.
By reaching the surface the pitch angles increase even more and at $150 R_{st}$ the radiation reaction force becomes comparable to $F$, implying that at the mentioned location the emission is extremely efficient.
We have found that in the interval $150-500$ stellar radii the synchrotron emission might provide energies from $0.1$MeV to $\sim 10$MeV.
Generally speaking, the considered process might be significant in millisecond pulsars as well as magnetars and therefore, a certain extension of the present work could be done.
\section*{Acknowledgments}
The research was supported by the Shota Rustaveli National Science Foundation grant (NFR17-587). ZO and VB also acknowledge hospitality of the Department of Advanced Energy at the University of Tokyo during their visit in 2019.
\bibliographystyle{spr-mp-nameyear-cnd}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 7,306 |
\section{Introduction}
\label{intro}
The top hat spherical collapse formalism dates back to Gunn and
Gott \cite{gg}. In addition to its beautiful simplicity, it has
proven to be a powerful tool for understanding and analysing the
growth of inhomogeneities and bound systems in the Universe. It
describes how a small spherical patch of homogeneous overdensity
decouples from the expansion of the Universe, slows down, and
eventually turns around and collapses. One assumes that the
collapse is not completed into a singularity, but that the system
eventually virializes and stabilizes at a finite size. The
definition of the moment of virialization depends on energy
considerations. The top hat spherical collapse is incorporated,
for example, in the Press-Schecter \cite{ps} formalism. It is
therefore widely used in present day interpretation of data sets.
For an Einstein-de Sitter Universe (EdS), i.e.~a Universe with
$\Omega_m=1$ and $\Omega_{\Lambda} = 0$ that is composed strictly
of non relativistic dust, there is an analytical solution for the
spherical collapse. The ratio of the final, virialized radius to
the maximal size at turnaround of the bound object is
$R_{vir}/R_{ta}=\frac{1}{2}$. The situation becomes more
complicated and parameter dependent when one considers a component
of dark energy. This was a subject of numerous studies
\cite{llpr,ws,is,wk,bw,zg,mb}. Lahav {\it et al} (LLPR)
\cite{llpr} generalized the formalism to a Universe composed of
ordinary matter and a cosmological constant. In this case the
cosmological constant is `passive', and only the matter
virializes. This leads to $R_{vir}/R_{ta}< \frac{1}{2}$. This
scenario also corresponds to the dynamics implemented in $N$-body
simulations for the concordance $\Lambda CDM$ model, i.e. the
cosmological constant only affects the time evolution of the scale
factor of the background Universe. Wang and Steinhardt (WS)
\cite{ws} included quintessence with a constant or slowly varying
equation of state. Battye and Weller (BW) \cite{bw} included
quintessence in a different manner to WS, taking into account its
pressure. Mota and Van de Bruck (MB) \cite{mb} considered
spherical collapse for different potentials of the quintessence
field, and checked what happens when one relaxes the common
assumption that the
quintessence field does not cluster on the relevant scales. \\
In this work we wish to review the inclusion of a cosmological
constant and quintessence into the formalism of spherical
collapse. Adding dark energy creates a system with two components
- the matter and the additional fluid. Most existing works
\cite{llpr,ws,bw,zg} look at the virialization of the matter
component (luminous and dark), which feels an additional potential
due to the presence of the dark energy. With this procedure, one
implicitly assumes that the dark energy either does not virialize,
or does so separately from the matter component. MB on the other
hand, included the additional fluid in the virialization
equations, the assumption here being that all of the system's
components virialize together. However, they did not remark either
on the difference in physical understanding of the system between
their approach and the one common in the literature, or on the
case of the cosmological constant. Our aim here is to critically
contrast the two approaches - the assumption that the dark matter
component virializes separately (as in LLPR, WS, and BW), and the
assumption that the whole system virializes as a whole (as in MB).
We wish to consider the meaning of including or not including the
additional fluid into the virialization, and point out a few
puzzles.
A second issue which we will address here is the use of energy
conservation in order to find the condition of virialization.
Assuming that the quintessence field does not collapse with the
mass perturbation but stays homogeneous as the background means
that the system must lose energy as it collapses. Yet, energy
conservation between turnaround and virialization is assumed. This
inconsistency is for quintessence fields with equation of state
$w\neq -1$. Energy is conserved with a cosmological constant
($w=-1$), for reasons that will be discussed later on. We will
propose a way to take into account this energy loss for
quintessence, and introduce a correction to the equation that
defines the final virialized radius of the system. \\
The inclusion of dark energy in the virialization process changes
the results in a fundamental manner. As we will show, the ratio of
final to maximal size of the spherical perturbation is
{\it{larger}} if the dark energy is part of the virialization.
While the results we will show are of the cosmological constant or
quintessence with a constant $w$, the methods we use are
applicable to a time dependent equation of state, as well as to
models in which quintessence is coupled to matter \cite{qc}. We
limit the discussion here to $w\geq-1$. While an equation of state
which is more negative than $-1$ is observationally interesting,
the physical interpretation of it is unclear, and beyond the scope
of this work. We assume throughout that the background is
described by a flat, FRW metric, with two energy components - the
matter and the dark energy.
The paper is organized as follows. In section \ref{1cs} we give
the general picture of how one calculates the point of
virialization of a single component system, and define the
relevant fundamental quantities. In section \ref{2cs} we consider
the case of a two component system. Section \ref{cq} reviews the
case of a clustered quintessence. In section \ref{g} we examine
the transition from clustered to homogeneous quintessence and in
section \ref{cc} we examine the transition toward $w=-1$. We
summarize and conclude in section \ref{conclusions}.
\section{Virialization of a single component system}
\label{1cs} The spherical collapse provides a mathematical
description of how an initial inhomogeneity decouples from the
general evolution of the Universe, and expands in a slower
fashion, until it reaches the point of turnaround and collapses on
itself. The mathematical solution gives a point singularity as the
final state. Physically though, we know that objects go through a
virialization process, and stabilize towards a finite size.
Since virialization is not `built in' into the spherical collapse
model (see though \cite{pad}), the common practice is to
{\it{define}} the virialization radius as the radius at which the
virial theorem holds, and the kinetic energy $T$ is related to the
potential energy $U$ by $T_{vir}=\frac{1}{2}(R ~\partial
U/\partial R)_{vir}$. Using energy conservation between
virialization and turnaround (where $T_{ta}=0$) gives
\begin{eqnarray}
\left[ U+\frac{R}{2}\frac{\partial U}{\partial R}
\right]_{vir} & = & U_{ta} \label{ec} ~.
\end{eqnarray}
Equation (\ref{ec}) defines $R_{vir}$. Thus in order to calculate
the final size of a bound object, we need to know how to calculate
the potential energy of the spherical perturbation, and to use
energy conservation between turnaround and the time of
virialization. We discuss later the case where energy is not
conserved, and how to account for it. For an EdS Universe,
$U=-\frac{3}{5}GM/R$ ($M$ is the conserved mass within the
spherical perturbation) and $T_{vir}=-\frac{1}{2}U$, so the ratio
of final to maximal
radii of the system is $x=R_{vir}/R_{ta}=\frac{1}{2}$. \\
The virial equation, which at equilibrium gives the above results,
is usually derived from the Euler Equation for particles. It is
worth noting here that one can derive the virial equation for a
cosmological fluid. The starting point is the continuity equation
of a perfect fluid (derived from $T^{0\nu}~_{;\nu}=0$), with
equation of state (the ratio of pressure to energy density)
$w=p/\rho$:
\begin{eqnarray}
\dot\rho+3\left( 1+w \right) \frac{\dot r}{r}\rho &=& 0 ~.
\label{vir1}
\end{eqnarray}
Multiplying equation (\ref{vir1}) by $r^2$, taking the time
derivative and integrating over a sphere of radius $R$ gives
\begin{eqnarray}
\int\dot G dV+\frac{1}{2}\left(1+3w \right)\left[
\int\rho\dot r^2 dV+\int r\frac{d}{dt}\left(\rho \dot r
\right)dV \right] &=& 0 ~,
\label{vir2}
\end{eqnarray}
where $G=(d/dt)\left(\frac{1}{2} \rho r^2\right)$. In the
classical analogy, $\dot G=\ddot I$ is the second derivative of
the inertia tensor. In a state of equilibrium, $\dot G=0$. The
quantity $\int\rho\dot r^2 dV$ is twice the kinetic energy, and
$\int r(d/dt)\left(\rho \dot r \right)dV $ is $R~ \partial
U/\partial R$. As equation (\ref{vir2}) shows, the value of $w$
factorizes out when one is looking for the equilibrium condition.
In the case where the fluid does not conserve energy, the right
hand side of of equation (\ref{vir1}) will be equal to some
function $\Gamma$. In that case, the virial equation (\ref{vir2})
will have an additional surface term. In equilibrium, the surface
term and
$\dot G$ should vanish. \\
This is the non-relativistic version of the scalar virial theorem.
Hence, it is not applicable for a fluid with a relativistic
equation of state, $w\rightarrow \frac{1}{3}$. It can be shown
that when writing the relativistic version, the energy of the
radiation field drops out of the virial equation \cite{ll}.
\section{A two - component system}
\label{2cs}
When adding a component to the system, there are three questions
to be asked - {\bf{(a)}} how does the potential induced by the new
component affect the system? {\bf{(b)}} does the new component
participate in the virialization? and {\bf{(c)}} does the new
component cluster, or does it stay homogeneous? These questions
should be addressed in the framework of a fundamental theory for
dark energy. Here we work out the consequences of virialization
and clustering of dark energy, if they do take place. We
shall try to address each of these questions separately. \\
\vspace{0.50cm}\\
{\bf{(a)}} In the case where the new component does not cluster or
virialize, its sole effect is contributing to the potential energy
of component 1. LLPR \cite{llpr} calculated this contribution to
the potential energy using the Tolman-Bondi equation. Their result
was generalized for quintessence by WS \cite{ws} and, in a
different manner, by BW \cite{bw}. We follow LLPR and WS in our
numerical calculations.
\vspace{0.50cm}\\
{\bf{(b)}} Any energy component with non vanishing kinetic energy
is capable of virializing, but the question is: on what time
scale? If one imagines that the full system virializes, then the
virial theorem should relate the {\it{full}} kinetic and potential
energies of the system,
\begin{eqnarray}
U &=& U_{11}+U_{12}+U_{21}+U_{22}=
\frac{1}{2}\int\left(\rho_1+\rho_2\right)
\left(\Phi_1+\Phi_2 \right)dV \label{uf} ~,
\end{eqnarray}
where the potential $\Phi_x$ induced by each energy component in a
spherical homogeneous configuration is
\begin{eqnarray}
\Phi_x(r) &=& -2\pi G (1+3w_x)\rho_x\left(R^2-\frac{r^2}{3}
\right)~.
\end{eqnarray}
The kinetic energy at virialization is then
\begin{eqnarray}
T_{tot} &=& \frac{1}{2}R\frac{\partial}{\partial R}
\left(U_{11}+U_{12}+U_{21}+U_{22}\right) ~.
\label{vf}
\end{eqnarray}
The expression above is the full potential energy of the system.
As we will show, the addition of these new terms to the virial
theorem makes a fundamental difference in the final state of the
system, so the question of whether dark energy participates in
the virialization is crucial. \\
\vspace{0.50cm}\\
{\bf{(c)}} Every positive energy component other than the
cosmological constant is capable of clustering. Even though
Caldwell {\it{et al}} \cite{c} have shown that quintessence cannot
be perfectly smooth, it is assumed that the clustering is
negligible on scales less than $100~Mpc$. It is therefore common
practice to keep the quintessence homogeneous during the evolution
of the system. The effects of relaxing this assumption were
explored in MB. We will consider both cases here.
The continuity equation for a Q component which is kept
homogeneous is
\begin{eqnarray}
\dot\rho_{Qc}+3(1+w)\frac{\dot a}{a}\rho_{Qc} & = & 0 ~,
\end{eqnarray}
and for clustering Q is
\begin{eqnarray}
\dot\rho_{Qc}+3(1+w)\frac{\dot r}{r}\rho_{Qc} & = & 0
\end{eqnarray}
($a$ and $r$ are the global and local scale factors,
respectively). One can ask what happens if one slowly `turns on'
and enables the possibility of clustering for the Q component. To
enable a slow continuous `turn on' of the clustering, one can
write
\begin{eqnarray}
&& \dot\rho_{Qc}+3(1+w)\left(\frac{\dot r}{r}\right)\rho_{Qc} = \gamma \Gamma
\label{13}\\
&& \Gamma = 3(1+w)\left(\frac{\dot r}{r}-
\frac{\dot a}{a}\right)\rho_{Qc} \\
&& 0\leq \gamma \leq 1 ~,
\end{eqnarray}
where $\rho_{Qc}$ is the dark energy's density within the cluster.
The notation $\Gamma$ follows MB. $\gamma$ is the `clustering
parameter', $\gamma=0$ gives clustering behaviour and $\gamma=1$
gives homogeneous behaviour. In the case of $\gamma=1$, the dark
energy inside the spherical region and the background dark energy
$\rho_Q$, behave in similar ways: $\rho_{Qc}=\rho_Q$. A point to
bear in mind is that for the case of homogeneous quintessence, the
system does not conserve energy as it collapses from turnaround to
virialized
state. \\
Putting all this together, the equations governing the dynamics of
the spherical perturbation are
\begin{eqnarray}
\left(\frac{\ddot r}{r} \right) & = &-\frac{4\pi G}{3}
\left(\frac{}{}\rho_{mc}+\left(1+3w \right)\rho_{Qc}\right) \label{qr2} \\
\dot\rho_{mc} & + & 3\left(\frac{\dot r}{r}\right)\rho_{mc} = 0\\
\dot \rho_{Qc} & + & 3\left(1+w \right)\left(\frac{\dot r}{r}\right)
\rho_{Qc} = \gamma \Gamma \label{qc} ~.
\end{eqnarray}
Our results are going to be presented as a function of $q$, which
is defined as the ratio of the system's dark energy to matter's
densities at the time of turnaround,
$q(z_{ta})\equiv\rho_{Qc}(z_{ta})/\rho_{mc}(z_{ta})$. The system's
matter density $\rho_{mc}$ at turnaround is
$\rho_{mc}(z_{ta})=\zeta(z_{ta})\rho_m(z=0)(1+z_{ta})^3$,
expressed in terms of the background matter density
$\rho_m(z_{ta})$ and the density contrast at turnaround,
$\zeta(z_{ta})$. In order to estimate which values of $q$ are of
interest,we plotted, in figure \ref{qz}, the dependence of $q$ on
the turnaround redshift $z_{ta}$, for various values of $\Omega_m$
and $\Omega_{\Lambda}$. As can be seen, $q$ takes typical values
no larger than $0.3$.
\begin{figure}
\begin{center}
\epsfig{file=qz.eps,height=80mm}
\end{center}
\caption{$q=\rho_{\Lambda}/\rho_{mc}$ as a function of
the turnaround redshift $z_{ta}$, for various
values of $\Omega_m$ and $\Omega_{\Lambda}$.
\label{qz}}
\end{figure}
\section{A clustered quintessence}
\label{cq}
In the case of fully clustering quintessence, $\gamma=0$, the
quintessence field responds to the infall in the same way as
matter, the only difference being its equation of state which
dictates a different energy conservation (in the general
relativity sense). However, energy is conserved, and since the
quintessence is active in the dynamics of the system, it is quite
reasonable to imagine that it takes part in the virialization. We
therefore present here the calculation assuming the whole system
virializes, matter and dark energy. \\
Following equation (\ref{uf}), the potential energy of the full
system is
\begin{eqnarray}
U &=& -\frac{3}{5}\frac{GM^2}{R}-(2+3w)\frac{4\pi G}{5}M\rho_{Qc} R^2
-(1+3w)\frac{16\pi^2 G}{15}\rho_{Qc}^2R^5 ~. \label{qec}
\end{eqnarray}
Once the potential energy has been calculated, virialization is
found with the use of equation (\ref{ec}). Expressing it in term
of $q=\rho_{Qc}/\rho_{mc}$ at turnaround and $x=R_{vir}/R_{ta}$
gives
\begin{eqnarray}
&&\left[1+(2+3w)q+(1+3w)q^2\right]x \nonumber \\
&&-\frac{1}{2}(2+3w)(1-3w)qx^{-3w}-
\frac{1}{2}(1+3w)(1-6w)q^2x^{-6} = \frac{1}{2} ~.
\label{gen0}
\end{eqnarray}
Equation (\ref{gen0}) is valid under the assumption that the whole
system virializes. \\
If, on the other hand, only the matter virializes, then the
equation defining $x$ is
\begin{eqnarray}
&&\left(1+q \right)x-\frac{q}{2}\left(1-3w \right)x^{-3w}
=\frac{1}{2} ~.
\label{ws0}
\end{eqnarray}
\begin{figure}
\begin{center}
\epsfig{file=q0.eps,height=80mm}
\end{center}
\caption{The ratio of final to turnaround radii as a function
of $q=\rho_{Qc}/\rho_{mc}$ at turnaround, for quintessence with
a constant equation of state, $w=-0.8$, which fully clusters
($\gamma=0$). The dotted line is the ratio in the case where
only matter virializes. The solid line
is the ratio when the whole system,
including the dark energy component, has virialized.
\label{wc08}}
\end{figure}
In figure \ref{wc08} we show $x$ as a function of $q$, for a fluid
with $w=-0.8$ that fully clusters, $\gamma=0$. The dotted line is
the ratio in the case where only matter virializes, equation
(\ref{ws0}). The solid line is the ratio when the whole system,
including the dark energy component, have virialized, equation
(\ref{gen0}). As can be seen, there is a fundamental difference of
the solutions: if only the matter virializes then the final ratio
is smaller than the EdS $\frac{1}{2}$ value, while when the whole
system virializes, the final ratio is larger than $\frac{1}{2}$.
\section{Turning off the clustering}
\label{g}
It is the usual practice to neglect spatial perturbations of the
quintessence field, and to keep it homogeneous \cite{c}. With our
generalized notation of the `clustering parameter' $\gamma$, one
can also allow a small but non-zero amount of clustering for the
quintessence. \\
For any $\gamma\neq 0$, the quintessence field within the system
does not conserve energy. As equation (\ref{ec}) assumes energy
conservation, the problem with not allowing the quintessence to
fully cluster is how to find the radius of virialization. We will
now propose a correction to equation (\ref{ec}), that will take
into account the loss of energy.
We denote the potential energy at turnaround as $U_{ta}$, and at
virialization as $U_{vir}$. We also define a function $\tilde U$
as the system's potential energy {\it had} it conserved energy.
Thus by construction the energy that the system lost is
\begin{eqnarray}
\Delta U & \equiv & \tilde U-U ~.
\end{eqnarray}
Equation (\ref{ec}) which describes energy conservation between
turnaround and virialization now needs to be corrected. Accounting
for the lost energy gives
\begin{eqnarray}
\left[ U+\frac{R}{2}\frac{\partial U}{\partial R}
\right]_{vir}+\Delta U_{vir} =
\left[\tilde U+\frac{R}{2}\frac{\partial U}{\partial
R}\right]_{vir}
= U_{ta} ~. \label{new}
\end{eqnarray}
We are now set to calculate $\Delta U$. \\
Looking at equation (\ref{qec}), one can treat $U$ as
$U(\rho_x,R)$ ($\rho_x$ being the various energy density
components). In order to calculate $\tilde U(\rho_x,R)$, one needs
to replace $\rho_x$ with $\tilde \rho_x$ in the expression for
$U$, which has $\gamma=0$ and conserves energy. The continuity
equation for $\tilde \rho_x$ is then
\begin{eqnarray}
\dot{\tilde \rho_x}+3\left(\frac{\dot R}{R} \right)
\left(1+w_x\right)\tilde\rho_x=0 ~, \label{tilde}
\end{eqnarray}
and we impose boundary conditions such that
$\tilde\rho_x(a_{ta})=\rho_x(a_{ta})$. \\
For a constant equation of state, this gives
\begin{eqnarray}
\tilde\rho_x & = & \tilde \rho_x(a_{ta})
\left(\frac{R_{ta}}{R_{vir}} \right)^{3(1+w_x)}
=\rho_x(a_{ta})
\left(\frac{R_{ta}}{R_{vir}} \right)^{3(1+w_x)} ~.
\end{eqnarray}
For a time dependent $w$ one needs to use the integral
expression for $\tilde \rho$. \\
We therefore have
\begin{eqnarray}
\tilde U(\rho_x,R) & = & U(\tilde \rho_x, R) ~.
\end{eqnarray}
Equation (\ref{new}) is now a function of $R_{vir}$ and values
determined at turnaround time (such as $U_{ta}$ and
$\rho_x(a_{ta})$), and defines $R_{vir}$ in the same manner as
equation (\ref{ec}) did. With the definitions of $q$, $x$,
$y=(a_{vir}/a_{ta})^{1+w}$ and $p=x/y$, the final form of equation
(\ref{ec}) for a quintessence with a general value of $\gamma$ is
then
\begin{eqnarray}
&&
\frac{q^2}{2}\left(1+3w\right)
\left[\frac{}{}1+6w-6\gamma\left(1+w\right)\right]
\left[\frac{}{}\left(1-\gamma \right)x^{-3w}+
\gamma p^3\right]^2
\nonumber \\
&+&
\frac{q}{2}\left(2+3w\right)
\left[\frac{}{}1+3w-3\gamma\left(1+w\right)\right]
\left[\frac{}{}\left(1-\gamma \right)x^{-3w}+\gamma p^3 \right]
\nonumber \\
&+&
\left[\frac{}{}1+\left(2+3w \right)q+\left(1+3w\right)q^2 \right]x
-\left(2+3w \right)qx^{-3w}- \left(1+3w\right)q^2x^{-6w}
\nonumber \\
&=& \frac{1}{2} ~.
\label{general}
\end{eqnarray}
For the case (common in literature) of a completely homogeneous
quintessence, $\gamma=1$, the virialization condition
(\ref{general}) is reduced to
\begin{eqnarray}
&& \left[1+(2+3w)q+(1+3w)q^2\right]x- \nonumber \\
&& (2+3w)q\left(p^3+x^{-3w}\right)-
(1+3w)q^2\left(\frac{5}{2}p^6+x^{-6w}\right)= \frac{1}{2} ~.
\label{gen1}
\end{eqnarray}
\\
The equation for $x$ when only the matter virializes does not need
to be corrected for energy conservation, as it counts only the
energy associated with the matter. Its general form is
\begin{eqnarray}
&& \left(1+q\right)x-\frac{q}{2}
\left[\frac{}{}1-3w+3\gamma\left(1+w\right)\right]
\left[\left(1-\gamma \right) x^{-3w}+\gamma p^3 \right]
=\frac{1}{2} ~,
\label{wsg}
\end{eqnarray}
and for $\gamma=1$ it reproduces the solution of WS,
\begin{eqnarray}
&& \left(1+q\right)x-2qp^3 =\frac{1}{2} ~,
\label{ws1}
\end{eqnarray}
\\
which is a generalization of LLPR's results (to be discussed
later on, see equation (\ref{llpr})).
One should consider which of the solutions is more plausible.
Ultimately the choice between the two solutions should be dictated
by the theory with which the dark energy is modelled. As we are
looking at an effective description of the dark energy as a
perfect fluid and not with a fundamental theory, this information
is lost.
For the case of $\gamma=1$, when one keeps the evolution of the
dark energy in the system identical to that of the background, it
is reasonable to assume that it does not participate in the local
processes that lead to virialization. This gives credibility to
the solution of equation (\ref{ws1}), allowing only the matter to
virialize. However, this presents a question of continuity,
presented in figure \ref{gamma}. The figure shows the solutions of
$x$ as a function of $\gamma$, with fixed $w=-0.8$ and $q=0.2$.
The circle on the right is the WS's result when the quintessence
is kept completely homogeneous. The square on the left is the
result when both the matter and the quintessence virialize, for
the fully clustering case. The `clustering parameter' allows us to
think of a continuous transition between the two cases. One would
expect the transition in the behavior of the system along $\gamma$
to be smooth. Allowing the dark energy to virialize for the
clustering case, $\gamma=0$, and keeping it out of the
virialization process when $\gamma=1$, raises the question of how
one should extrapolate smoothly between the two cases. As figure
\ref{gamma} suggests, there will be a
discontinuity. \\
\begin{figure}
\begin{center}
\epsfig{file=gamma.eps,height=80mm}
\end{center}
\caption{$R_{vir}/R_{ta}$ as a function of $\gamma$, for
$w=-0.8$ and $q=0.2$. $\gamma=0$ describes the case of a
fully clustering $Q$
field, and $\gamma=1$ is the case of a homogeneous $Q$,
allowing only the matter component to cluster. For
$\gamma=0$, taking the dark energy into the virialization is
highly plausible, (see square on left).
If one assumes that only the matter component virializes
for $\gamma=1$ (see circle on right), it is unclear how to
extrapolate in a smooth way between the two cases.
This will produce a discontinuity in the transition
from the `clustering' to the `non-clustering' behaviour.
\label{gamma}}
\end{figure}
\begin{figure}
\begin{center}
\epsfig{file=q1.eps,height=80mm}
\end{center}
\caption{The ratio of final to turnaround radii as a function
of $q=\rho_{Qc}/\rho_{mc}$ at turnaround, for quintessence with
a constant equation of state, $w=-0.8$, which stays homogeneous
($\gamma=1$). The dotted line follows
WS's calculation, assuming only the matter component virializes.
The dashed line is the ratio when the whole system,
including the dark energy component, has virialized. The solid
line takes into account the loss of energy between turnaround
and virialization.
\label{r08}}
\end{figure}
In figure \ref{r08} we compare the solution of equation
(\ref{ws1}) and (\ref{gen1}) for $w=-0.8$ and $\gamma=1$, and show
the effect of the energy correction. As can be seen, taking into
account the loss of energy produces a small quantitative
correction, but keeps the general feature of enlarging the final
size of the system if the dark energy is allowed to virialize.
\section{The limit of a cosmological constant}
\label{cc}
Equations (\ref{general}) and (\ref{wsg}) are valid for any
constant $w$. As $w$ approaches $-1$, we get that $p\rightarrow
x$, and the dependency on $\gamma$ vanishes. The reason that
$\gamma$ plays no role in the limit of $w\rightarrow -1$ is that
the question whether such a fluid is allowed to cluster
($\gamma=0$) or not ($\gamma=1$) is rather abstract. It stays
homogeneous in any case, because of its equation of state,
$w_{\Lambda}=-1$ (which leads to
$\Gamma=0$). Accordingly, energy is automatically conserved. \\
In that limit, equation (\ref{general}) which assumes that the
whole system to virialize, is simplified into
\begin{eqnarray}
7 q^2 x^6 + 2 q x^3 + \left(1-q-2q^2\right)x & = & \frac{1}{2} ~.
\label{gencc}
\end{eqnarray}
Taking the same limit for equation (\ref{wsg}) yields the familiar
result of LLPR:
\begin{eqnarray}
\left(1+q\right)x-2qx^3 & = & \frac{1}{2} ~~~~~(LLPR) ~,
\label{llpr}
\end{eqnarray}
which is valid under the assumption that the matter component
alone virializes \footnote{One can look at a test particle feeling
an inverse square force and an additional repulsive $\Lambda$
force. Consider two possible orbits of the particle: one circular,
and one in which its kinetic energy can vanish. The ratio of the
circular radius to the radius of zero kinetic energy
(`turnaround') is described exactly by equation (\ref{llpr}). This
assumes that the test particle does not contribute to the forces
of the system. We thanks Martin Rees for pointing out this
similarity.} (notice that our definition of $q$ differs by a
factor of $2$ from the definition of $\eta$ of LLPR,
$q=\frac{1}{2}\eta$). \\
Again, we wish to consider the plausibility of the two solutions.
If one considers the cosmological constant as a true constant of
Nature, $\rho_{\Lambda}=\Lambda/(8 \pi G)$, it is hard to imagine
it participating in the dynamics that lead to virialization, as it
is a true constant. In this case, one could categorically say that
the right procedure is to look at the virialization of the matter
fluid only, and follow LLPR's solution, equation (\ref{llpr}). The
sole effect of the cosmological constant, then, is to modify the
potential that the matter fluid feels.
If, on the other hand, one considers the origin of a perfect fluid
with $w\approx -1$ as a special case of quintessence, which is
indistinguishable from a cosmological constant, it is reasonable
to expect continuity in the behaviour of the system as one slowly
changes the value of $w$ toward $-1$. In other words, if the
physical interpretation of the fluid with $w\approx -1$ is of a
dynamical field that {\it{mimics}} a constant, the idea of
including it in the dynamics of the system has a physical meaning.
\begin{figure}
\begin{center}
\epsfig{file=w0.eps,height=80mm}
\end{center}
\caption{$R_{vir}/R_{ta}$ as a function of $w$, for
$q=0.2$ and $\gamma=0$. The dotted line is the ratio when
the matter alone virializes, and the solid is for the case where
the whole system virializes. The circle on the left is LLPR's
solution for the cosmological constant. The square on the right
is an example of a clustered quintessence, where we expect to
take into account the whole system in the virialization. The
figure suggests we should expect a discontinuity in the
behaviour of quintessence fields and a true cosmological
constant.
\label{w0}}
\end{figure}
The result, then, is that we possibly have a signature
differentiating between a cosmological constant which is a true
constant, and something else which {\it{mimics}} a constant. This
point is shown in figure \ref{w0}. The figure shows $x$ as a
function of $w$, with $q=0.2$ and $\gamma=0$. The dotted line
follows the solution of equation (\ref{wsg}), with the matter
alone virializing. The circle on the left is LLPR's solution for
the cosmological constant. The solid line follows the solution of
equation (\ref{general}). The square on the right is an example of
a clustered quintessence, where we expect to take into account the
whole system in the virialization. As with figure \ref{gamma},
there is a suggested discontinuity, but here one can associate the
discontinuity with a clear physical meaning: a true cosmological
constant is not on the continuum of perfect fluids with general
$w$, as its physical behaviour is different. \\
An observational detection of virialized objects with
$R_{vir}>\frac{1}{2}R_{ta}$ would be a strong evidence against a
cosmological constant which is a true constant, regardless of the
measured value of the equation of state.
\section{Conclusions}
\label{conclusions}
In this work, we have reconsidered the inclusion of a dark energy
component into the formalism of spherical collapse. We compared
existing results (such as those of LLPR and WS) which implicitly
assume that only the dark matter virializes, to the case where the
whole system's energy is taken into account for virialization,
implying that the dark energy component also participates in the
process (MB). While previous studies allow the dark energy
component either to fully cluster or keep completely homogeneous,
we generalized and allowed a smooth transition between the two
cases. Additionally, we addressed the issue of energy
non-conservation when the dark energy is kept homogeneous.
Our main conclusions are:
\begin{itemize}
\item
In the case of a true cosmological constant only the matter
component virlializes and the LLPR solution is valid.
\item
If both components of the system virialize, two additional terms
to the potential energy appear. These are the self - energy of the
additional energy source, and its reaction to the presence of the
matter.
\item
The inclusion of these terms results in a fundamentally different
behaviour of the system. If only dark matter virializes, the final
size of the system is {\it{smaller}} than half of its maximal
size. When the whole system virializes, its final size is
{\it{bigger}} than half of its maximal size.
\item
It is hard to understand the physical meaning of a cosmological
constant `virializing', if it is a true constant. Accordingly,
observational evidence for $R_{vir}>\frac{1}{2}R_{ta}$ would be
strong evidence in favour of a dynamical field for the dark
energy, regardless of the measured value of the equation of state.
On the other hand, $R_{vir}<\frac{1}{2}R_{ta}$ is compatible with
both a true constant and a field mimicking the cosmological
constant.
\item
Keeping the dark energy component homogeneous implies that the
overdense region does not conserve energy. The exception here is
the case of the cosmological constant, for which the
non-clustering behaviour is exact and not an approximation. The
equation defining virialization needs to be corrected, in order to
account for the energy lost by the Q field between turnaround and
virialization. It should read
\begin{eqnarray}
\left[\tilde U+\frac{R}{2}\frac{\partial U}{\partial
R}\right]_{vir} & = & U_{ta} ~. \nonumber
\end{eqnarray}
This introduces a small quantitative correction.
\end{itemize}
Table \ref{table} gives a summary of the relevant solution for the
different cases that we considered in this work.
\begin{table}
\begin{center}
\begin{tabular}{||c||c|c||}
\hline \hline
&{\bf{$\rho_{mc}$ virializes}}
&{\bf{$\rho_{mc}$ and $\rho_{Qc}$ virialize}} \\
\hline \hline
{\bf{general case}} & (\ref{wsg}) & (\ref{general}) \\
\hline
{\bf{$\gamma=0$}} & (\ref{ws0}) & (\ref{gen0}) \\
\hline
{\bf{$\gamma=1$}} & (\ref{ws1}) & (\ref{gen1}) \\
\hline
{\bf{$w\rightarrow -1$}} & (\ref{llpr})& (\ref{gencc}) \\
\hline
{\bf{Cosmological Constant}}& (\ref{llpr})& -- \\
\hline \hline
\end{tabular}
\end{center}
\caption{A summary of the relevant equations defining $x$ for
the various cases that we considered.
\label{table}}
\end{table}
Our work has consequences for the linear theory as well. As we
have not altered any of the equations governing the evolution of
the system, the linear equation of growth will not be altered
either. Nonetheless, reconsidering the energy budget changed the
time in which we perceive virialization to happen, and as a
consequence the linear contrast at virialization ($1.686$ for the
EdS case) will change. In practice though, the numerical change is
rather small. We find that for the cosmological constant, the
maximal deviation from the EdS value is a rise of about $3\%$.
A future work would be to incorporate the possible virialization
of dark energy into numerical simulations and into analyses of
observations. It would be particularly interesting to see how it
affects cluster abundances, and which approach provides a better
fit to the observations. Several works are pursuing such
directions \cite{additional}.
For various models of coupled quintessence \cite{qc}, it is very
likely that the dark energy component clusters and virializes. For
models in which the dark energy doesn't cluster, one could ask how
plausible the scenario of the dark energy participating in the
virialization is. Of course, should one argue that the dark energy
virializes and not just the dark matter component, a mechanism of
how it physically happens would be needed. An additional direction
to pursue is the actual mechanism of virialization, which at the
moment is still rather obscure. Understanding what the physical
process by which the system virializes is will hopefully give us a
clue as to whether we should include the dark energy or not.
\ack We would like to thank Jacob Bekenstein, Carsten van de
Bruck, Ramy Brustein, Uri Keshet, Donald Lynden-Bell, David Mota,
Amos Ori, Martin Rees and Jochen Weller for useful discussions. OL
acknowledges a PPARC Senior Research Fellowship. IM acknowledges
the support from the Leverhulme Quantitative Cosmology grant. Part
of this investigation was carried out while one of us (IM) was
visiting the Weizmann Institute for Science. IM wishes to thank
Micha Berkuz and Yosi Nir for their kind invitation and warm
hospitality.
\section*{References}
| {
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Truck crashes through East Mount Airy building causing collapse
By Jeff Cole
Published July 1
FOX 29 Philadelphia
Crews begin demolition of East Mount Airy home after partial collapse
Crews began to knock down a home after a truck crashed into it on Monday.
PHILADELPHIA - A home in Philadelphia's East Mount Airy neighborhood has collapsed after a driver struck the front of the structure Monday morning.
The crash occurred on the unit block of Pleasant Street before 7:30 a.m.
A home in Philadelphia's East Mount Airy neighborhood has collapsed after a driver struck the front of the structure Monday morning.
Video from the scene shows a truck inside the front of the corner building before the second floor of the building collapsed onto the vehicle.
The homeowner was inside and asleep at the time of the crash. She lives in the home with three adult sons, two of whom were out when the incident occurred. The homeowner says she was asleep in the bedroom above where the truck crashed into the home. She and her son were able to get out of the house safely.
It is not yet clear if the neighboring rowhomes are affected by the collapse.
The driver of the truck was evaluated by paramedics, and the extent of his injuries is unknown at this time.
Philadelphia Licenses and Inspections says the building is dangerous and must be knocked down.
A GoFundMe page has been set up in order to help the family. For more information, click here.
Virginia Gov. Ralph Northam yearbook photo shows one man in blackface, another in KKK regalia
Teen hospitalized after man dives into her tube
Code Red information for Philadelphia | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 9,332 |
layout: page
title: Poole Reef Tech Executive Retreat
date: 2016-05-24
author: Judy Lee
tags: weekly links, java
status: published
summary: Curabitur nec volutpat turpis, non.
banner: images/banner/leisure-02.jpg
booking:
startDate: 08/18/2016
endDate: 08/22/2016
ctyhocn: YEGSPHX
groupCode: PRTER
published: true
---
Cras lacinia volutpat urna, a blandit est tincidunt eu. Phasellus eu posuere est. Mauris sollicitudin odio a dignissim tincidunt. Fusce a elit vehicula, condimentum nulla eu, elementum ligula. Nam vitae sapien feugiat, malesuada sapien sed, dapibus lacus. Vestibulum est augue, rhoncus ac bibendum ut, rhoncus eget felis. Nunc sit amet mi fringilla, vehicula purus in, interdum mi. Nunc a tortor risus. Morbi vitae eros non elit fringilla ornare sit amet ac tellus. Vivamus auctor tortor eget nulla luctus, ac fermentum risus imperdiet. Aliquam erat volutpat. Praesent scelerisque, erat in ullamcorper semper, nulla orci interdum erat, a porta enim metus id lorem. Maecenas tellus ipsum, posuere eget enim at, faucibus accumsan tortor. Etiam at est a purus fringilla vestibulum. Vestibulum vel eros erat.
* Nunc in sem sed ante semper rutrum in et enim
* Phasellus non dui sollicitudin, rhoncus tellus id, consequat nibh.
Ut mattis nisi dui, eu iaculis quam blandit vitae. Nullam sagittis sed lacus ac facilisis. Donec dictum, nulla a semper porttitor, leo libero tincidunt nunc, a ultrices mi lorem vitae mauris. Maecenas odio mi, suscipit a sapien eu, consequat dictum nulla. Aenean varius rhoncus neque. Aliquam vel posuere massa. Nulla ultrices cursus viverra.
| {
"redpajama_set_name": "RedPajamaGithub"
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{"url":"https:\/\/www.nature.com\/articles\/s41467-020-18170-9?error=cookies_not_supported&code=b22abc9c-96b5-422f-90ee-42328f90d4c2","text":"# Identification of molecular quantum states using phase-sensitive forces\n\n## Abstract\n\nQuantum-logic techniques used to manipulate quantum systems are now increasingly being applied to molecules. Previous experiments on single trapped diatomic species have enabled state detection with excellent fidelities and highly precise spectroscopic measurements. However, for complex molecules with a dense energy-level structure improved methods are necessary. Here, we demonstrate an enhanced quantum protocol for molecular state detection using state-dependent forces. Our approach is based on interfering a reference and a signal force applied to a single atomic and molecular ion. By changing the relative phase of the forces, we identify states embedded in a dense molecular energy-level structure and monitor state-to-state inelastic scattering processes. This method can also be used to exclude a large number of states in a single measurement when the initial state preparation is imperfect and information on the molecular properties is incomplete. While the present experiments focus on N$${}_{2}^{+}$$, the method is general and is expected to be of particular benefit for polyatomic systems.\n\n## Introduction\n\nThe quantum control of isolated particles forms the basis for recent advancements in quantum computation1,2,3, for precise time and frequency measurements4,5, for searches for new physics beyond the standard model6,7, for controlled chemistry8,9,10 and for quantum communication11,12.\n\nRecent progress in the quantum control of single trapped molecules13,14,15,16,17 has enabled the detection13,15, coherent manipulation14,16 and entanglement17 of molecular quantum states on the single-particle level. These advances aim at encoding qubits in rotational and vibrational molecular energy states18,19 and at performing precise spectroscopic measurements20 with applications including the development of mid-infra-red frequency standards21 and testing a possible variation of fundamental constants such as the proton-to-electron-mass ratio22,23,24,25,26.\n\nPrevious studies13,14,15,16,17 built on prior knowledge of the molecular energy-level structure to engineer level subspaces that enable a precise control on the quantum level. These approaches relied on the entanglement of the internal molecular states with the external motion of a molecular-ion (MI) and atomic-ion (AI) Coulomb crystal27,28. The motion was generated by, e.g., applying state-dependent forces on the MI13,15. The molecular state was then inferred from the excitation amplitude of the MI\u2013AI crystal read out on the AI.\n\nHowever, for larger and more complex molecules such prior knowledge of the molecular energy-level structure is often not available. Moreover, situations occur, even in comparatively simple diatomic molecules, in which several states that are potentially populated result in similar excitation amplitudes generated by the state-dependent forces such that it is not readily possible to unambiguously distinguish them. Indeed, it can be expected that this is a typical scenario given that molecular-state preparation down to the hyperfine or Zeeman level is currently only feasible for the simplest systems14,29.\n\nHere, we enhanced the capabilities of a recently developed method for the quantum-non-demolition readout of molecular states15,18. This was achieved by interfering a reference force applied on the AI with the state-dependent force experienced by the MI. By changing the relative phase of the forces, additional information on the molecular state could be obtained, which is not available from only measuring the amplitude of the forces.\n\nUsing this method, we experimentally identified specific hyperfine-Zeeman states of a single 14N$${}_{2}^{+}$$ MI in different excited rotational states of its electronic and vibrational ground state, thus also probabilistically preparing the MI in a specific hyperfine state. We also showed how this method could potentially be used to track the initial and final states during state-changing processes like chemical reactions and inelastic collisions. Finally, we demonstrated a protocol to simultaneously exclude a large subset of rotational levels. This protocol is useful when probing an unknown molecular state within a complex and only partially known energy level structure.\n\n## Results\n\n### Phase-sensitive forces\n\nConsider a string of two ions, an AI and an MI confined in a linear radiofrequency ion trap (Fig.\u00a01a). The two-ion string exhibits two fundamental motional modes along the trap axis, an in-phase (IP) mode and an out-of-phase mode30,31. The ions are superimposed with a running-wave optical lattice generated by two overlapping linearly polarized counterpropagating laser beams with frequency difference \u0394f. The lattice induces a modulated ac-Stark shift on the ions which results in a state-dependent optical-dipole force (ODF) oscillating at the frequency \u0394f. When the oscillation frequency of the ODF matches the one of a motional mode (e.g., \u0394f\u2009=\u2009fIP for the in-phase mode), the ODF resonantly excites motion in the mode15,18,32. This motion can be detected on the AI through sideband thermometry15,33,34.\n\nIn addition to the ac-Stark shift on the MI, the lattice may induce a non-negligible ac-Stark shift on the AI. Since two ODFs are applied on the crystal simultaneously, the relative phase of the forces will influence the total degree of motional excitation. This phase depends on both the sign of the detuning of the lattice wavelength from spectroscopic resonances in the ions as well as the distance of the ions from each other. The former determines the sign of the ac-Stark shift experienced by the AI and the MI while the latter determines the relative phase of the optical lattice at the positions of the AI and MI.\n\nFor the AI, the sign and amplitude of the ac-Stark shift were constant in our experiments because the lattice beams were far detuned (\u2212\u0394a) from any atomic resonance (Fig. 2a). Here, the minus sign indicates that the lattice beams were red detuned from the nearest transition. On the other hand, for the MI, the amplitude and sign of the induced ac-Stark shift depended on the specific rovibronic, hyperfine and Zeeman state of the MI. Within the frequency range of the optical lattice used in the present experiments, a range of different spectroscopic transitions were accessible (Fig.\u00a02b) leading to different detunings (\u00b1\u0394m) depending on the exact state of the MI.\n\nFigure\u00a01b shows a schematic of two different configurations of forces on the two ions used in our experiment. Since the AI was always red detuned, it always experienced a force in the direction of higher lattice intensity (high-field seeking configuration). For the MI, however, both high- and low-field seeking (blue detuning) scenarios were possible (upper and lower panels in Fig.\u00a01b). The relative phase of the forces was set by controlling the distance between the two ions, d. By placing the two ions at a separation that corresponded to an integer multiple of lattice nodes, d\u2009=\u2009n\u03bb\/2 ($$n\\in {\\mathbb{N}}$$ is an integer number and \u03bb is the wavelength of the lattice lasers), the ions experienced the same intensity gradient from the lattice. We denote this lattice configuration as \u201csame phase\u201d (SP). In the SP scenario, the forces were in the same (opposite) direction when the MI was red (blue) detuned (SPR and SPB in Fig.\u00a01b). By changing the two-ion distance by half a lattice spacing, \u0394d\u2009=\u2009\u00b1\u2009\u03bb\/4, the ions experienced an opposite gradient of the lattice field, i.e., the opposite phase (OP) of the lattice. In the OP lattice configuration, the forces were in the same (opposite) direction when the MI was blue (red) detuned (OPR and OPB in Fig.\u00a01b).\n\nIn the case when the forces interfered constructively (same directions), the resulting motional-excitation signal was stronger than in the case of destructive interference (opposite directions). This by itself did not yield information about the sign of the detuning since the ac-Stark shift varied greatly between different states such that a priori the strength of the signal could not be predicted. In order to disentangle the strength and sign of the ac-Stark shift, the SP and OP configurations were compared by varying the two-ion distance and thus obtained both the magnitude and sign of the ac-Stark shift. As the detuning (\u0394m) changes across a resonance in the molecule, the direction of the induced ac-Stark shift (and hence the dipole force) changes while the direction of the shift (and hence the force) on the AI remains constant (Fig.\u00a01c). As will be shown below, this information proves extremely valuable for distinguishing states in a complex energy level structure.\n\nThe signed value of the ac-Stark shift, \u0394Ej, is related to the dynamic molecular polarizability, \u03b1j(\u03c9), via $$\\Delta {E}^{j}=-\\frac{{\\alpha }^{j}(\\omega )}{2{\\varepsilon }_{0}c}I$$ (see Methods). Here, the index j specifies the quantum state of the molecule, \u03b50 is the vacuum permittivity, c is the speed of light and I\u2009\u2248\u200911.5\u2009W\u2009mm\u22122 is the intensity of a single lattice beam. The AI was used as a probe for the lattice intensity at the position of the AI\u2013MI crystal (see Methods).\n\n### Experiment\n\nA detailed description of our experimental apparatus is given in refs. 15,18, a schematic representation is shown in Fig.\u00a01a and a brief overview is given in the Methods section.\n\nThe relevant energy levels of 40Ca+ and 14N$${}_{2}^{+}$$ are shown in Fig.\u00a02. With lattice beams linearly polarized and parallel to the magnetic-field direction, the lattice mainly interacted with high-lying (nf)F7\/2(m\u2009=\u2009\u22125\/2) and (nf)F5\/2(m\u2009=\u2009\u22125\/2) levels in Ca+ since the AI was prepared in the metastable (3d)2D5\/2(m\u2009=\u2009\u22125\/2) state. Here, m denotes the magnetic quantum number. Even though the lattice was highly red-detuned from these AI levels (\u0394a >250\u2009THz), their contribution to the ac-Stark shift was not negligible (see Methods). For N$${}_{2}^{+}$$, the lattice was nearly resonant with the $${A}^{2}{\\Pi }_{u}(v^{\\prime} =2)\\leftarrow {X}^{2}{\\Sigma }_{g}^{+}(v^{\\prime\\prime} =0)$$ vibronic transition where\u00a0$$v^{\\prime}(v^{\\prime\\prime})$$ denotes the quantum number of the\u00a0upper (lower)\u00a0vibrational level of the transition. The detuning was dependent on the exact rotational, fine, hyperfine and Zeeman state of the MI. Figure\u00a02b shows the detuning of the lattice on the P(J), Q(J), and R(J) lines initiating from the two spin\u2013rotation components of the vibronic ground state. Here, J is the quantum number of the total molecular angular momentum without nuclear spin of the MI in its $${X}^{2}{\\Sigma }_{g}^{+}(v^{\\prime\\prime} =0,N^{\\prime\\prime} ,\\;J^{\\prime\\prime} =N^{\\prime\\prime} \\pm 1\/2)$$ ground state and P,Q,R stand for spectroscopic branches with $$\\Delta J=J^{\\prime} -J^{\\prime\\prime} =-1,0,1$$.\n\nAn ODF pulse was applied to detect the state of the molecule. In this pulse, the lattice lasers were turned on for a duration of 3\u2009ms to excite motion in the crystal. The motional state of the IP mode was subsequently probed on the AI by Rabi sideband thermometry on a blue sideband of the D5\/2(m\u2009=\u2009\u22125\/2)\u2009\u2192\u2009S1\/2(m\u2009=\u2009\u22121\/2) transition33,34. The frequency and contrast of the resulting Rabi-oscillation signal was proportional to the ODF strength (Methods).\n\nIn order to obtain both the sign and the magnitude of the ac-Stark shift experienced by the MI, the Rabi-oscillation experiment was performed twice. Once with an ion distance of d\u2009=\u2009n\u03bb\/2 where n\u2009=\u200919 and \u03bb\u2009\u2248\u2009789\u2009nm corresponding to the SP configuration ($${f}_{{\\rm{IP}}}^{{\\rm{SP}}}=695$$\u2009kHz) and a second time with the ion distance shifted by \u0394d\u2009=\u2009+\u03bb\/4 corresponding to the OP configuration ($${f}_{{\\rm{IP}}}^{{\\rm{OP}}}=668$$\u2009kHz). The configurations were changed by adjusting the voltages on the ion-trap end-cap electrodes. The relative uncertainty in determining the ion\u2013ion distance was estimated to be \u03b4d\/d\u2009=\u200910\u22123 (Methods).\n\nExample of Rabi-oscillation signals from two SP\/OP experiments are shown in Fig.\u00a03. The blue-sideband-pulse duration, t729, was scanned and averaged 20 times for each data point to retrieve two Rabi-oscillation signals in each experiment, one for the SP (blue) and the other one for the OP (purple) configurations. The frequency and contrast of the Rabi signals were observed to be higher for the SP configuration compared to the OP configuration in Fig.\u00a03a, implying that the ODF and overall motional excitation was stronger for the SP configuration. This indicated that the detuning of the lattice frequency from a molecular resonance was to the red in this experiment (Fig.\u00a01b,\u00a0upper panel). The resulting Rabi oscillations were fitted to a phenomenological function which was used to extract the amplitude of the ac-Stark shift (Methods). Based on the obtained values for the phase and amplitude, the molecular state could unambiguously be identified as N\u2033\u2009=\u20096, J\u2033\u2009=\u200911\/2, as illustrated in the following section.\n\nIn Fig.\u00a03b, the opposite situation occurred for the same settings of the lattice laser beams. Here, the observed frequency and contrast of the Rabi signals were higher for the OP configuration. This implied that the lattice frequency was blue-detuned from the closest molecular resonance. Below it is shown how based on this observation only, all molecular states with N\u2009\u2264\u20094 can be excluded.\n\n### Results\n\nOur method applied to the identification of specific spin\u2013rotation and Zeeman levels of N$${}_{2}^{+}$$ molecules in the $${X}^{2}{\\Sigma }_{g}^{+}(v^{\\prime\\prime} =0)$$ electronic and vibrational ground state is exemplified in Fig.\u00a04 with the N\u2033\u2009=\u20094 and N\u2033\u2009=\u20096 states. By tuning the lattice-laser wavelength near 789\u2009nm, the detection was sensitive to states originating from the N\u2033\u2009=\u20094, J\u2033\u2009=\u20097\/2 spin\u2013rotation manifold and induced a significant ac-Stark shift through the Q12(7\/2) transition at 788.624\u2009nm (red lines in Fig.\u00a04). For these states, the lattice frequency was red detuned with respect to the transition frequency. At the same wavelength, the detection was also sensitive to states originating from the N\u2033\u2009=\u20096, J\u2033\u2009=\u200911\/2 spin\u2013rotation manifold. These states induced a significant ac-Stark shift through the Q12(11\/2) transition at 789.1872\u2009nm (blue lines in Fig.\u00a04). For these states, the lattice frequency was blue detuned with respect to the transition frequency.\n\nIn the present experiments, molecules in the electronic and vibrational ground state were primarily created in rotational states with N\u2033\u2009\u2264\u20098 by a rotationally unselective photoionization scheme. In each experiment, the molecules were randomly initialized in one of the 540 corresponding hyperfine Zeeman states. The detection protocol was then applied to extract the amplitude and phase of the ODF signal. In many experiments, the signal was very low, indicating that the state of the molecule was not within the subspace of states which could effectively be traced at the chosen lattice-laser wavelength (examples are shown by the red triangles in the bottom of Fig.\u00a04). However, for some experiments (labelled 1\u20133 in Fig.\u00a04), the ODF amplitude was large enough indicating that the molecule was either in the N\u2033\u2009=\u20096, J\u2033\u2009=\u200911\/2 or the N\u2033\u2009=\u20094, J\u2033\u2009=\u20097\/2 spin\u2013rotational states. The added sign information enabled us to distinguish between these two spin\u2013rotation manifolds (blue and red triangles, respectively). The black square exemplifies an experiment in which the sign information was lacking and, therefore, an unambiguous identification of the molecular state was not possible.\n\nFor the experiment marked 1 in Fig.\u00a04, the state was identified as N\u2033\u2009=\u20094, J\u2033\u2009=\u20097\/2. The measured and the predicted ac-Stark shifts agree within 1\u03c3 with only six hyperfine Zeeman states (Table\u00a01). Thus, 534 out of 540 possible states can be excluded for a molecule with N\u2009\u2264\u20098. Even if we assume a discrepancy of 2\u03c3 between theory and our measured ac-Stark shifts, 528 of the 540 possible states of the molecule can be excluded (Table\u00a01). For the experiment marked 2 in Fig.\u00a04, the state was identified as N\u2033\u2009=\u20096, J\u2033\u2009=\u200911\/2. Here again, only 6 (12) hyperfine Zeeman states agree with theory within 1\u03c3 (2\u03c3), thus excluding 99% (98%) of the possible states (Table\u00a01).\n\nThe experiments marked 3 and 4 in Fig.\u00a04 also show\u00a0opposite signs at a comparable amplitude. While in experiment 3, the state was identified to be 1 out of 4 (6) hyperfine Zeeman states of the N\u2033\u2009=\u20096, J\u2033\u2009=\u200911\/2 spin\u2013rotation manifold, experiment 4 was associated with 22 (42) hyperfine Zeeman states of the N\u2033\u2009=\u20094, J\u2033\u2009=\u2009\u00a07\/2 and N\u2033\u2009=\u20098, J\u2033\u2009=\u200917\/2 spin\u2013rotation manifolds (Table\u00a01). These experiments exemplified that using the phase information, it is possible to exclude more than 50% (here \u00a0~85%) of the states compared to when only the amplitude information is available.\n\nTo further illustrate the capabilities of the present state-detection protocol, a state-sensitive chemical reaction experiment was implemented to observe how N$${}_{2}^{+}$$ molecules in a determined state react with H2 background gas molecules to form N2H+. Figure\u00a05 shows the results of such an experiment. An N$${}_{2}^{+}$$ molecule was identified to be in the N\u2009=\u20094, J\u2033\u2009=\u20097\/2 spin-rotational state (Fig. 5a). The molecule eventually reacted with background gas and turned into N2H+. The change of the Rabi-oscillation signal from panel a to panel b indicated a possible change of the chemical composition of the molecule, which was then verified by mass spectrometry (Methods). Due to the mass change of the crystal in the reaction, the frequency difference of the lattice beams, \u0394f, had to be adjusted to match the new crystal frequency $${f}_{{\\rm{IP}}}^{{\\rm{SP}}}({\\mathrm{{{N}}}}_{2}{\\mathrm{{{H}}}}^{+})=690$$\u2009kHz (Methods). The energy-level structure of N2H+ is not sufficiently known to be able to clearly identify the molecular quantum state generated in the reaction. However, the phase information indicated that the lattice beams were red detuned from the relevant resonance in the N2H+ molecule. This experiment exemplifies the basic possibility to perform state-to-state chemical reaction experiments involving polyatomic species. The ability to perform such experiments depends on the lifetime of the product state compared to the time required for state detection.\n\nFollowing the same principle, inelastic, i.e., state-changing, processes of a single molecule can also be traced. In panels c and d of Fig.\u00a05, the results of such an experiment are shown. An N$${}_{2}^{+}$$ ion in the N\u2033\u2009=\u20096, J\u2033\u2009=\u200911\/2 spin\u2013rotational state underwent a quantum-jump to a different rotational state as can be seen by the change in the amplitude and phase of the Rabi-oscillation signal. The change of state could have been caused by either an inelastic collision with a background-gas molecule or by the scattering of a photon from the lattice laser15. Prior to the quantum jump, the OP configuration showed a stronger signal than the SP configuration while after the collision, the OP configuration showed a weaker signal than the SP configuration. This suggests that the molecule underwent a rotational-state change, most probably to the N\u2033\u2009=\u20094 state though the low amplitude of the signal does not allow us to unambiguously exclude other rotational or vibrational states. The state of the molecule after the quantum jump could have been tested with higher fidelity by changing the lattice laser frequency to a value at which the assumed state would result in an increased signal amplitude.\n\nIn this experiment, the interrogation time for state identification was on the order of few seconds. This relatively long interrogation time posed no problem since all states of homonuclear diatomics in the electronic ground state are long-lived. However, this timescale could set a limitation for the state identification of polar molecules in which the rovibrational states typically exhibit shorter radiative lifetimes. This can become important, for example, in a state-to-state reaction experiment in which the product states are short lived or if interaction with black-body radiation scrambles the state populations. In this case, the interrogation time can be reduced by carefully choosing the lattice and AI parameters to maximize the signal and minimize the averaging time. Moving the experiment to a cryogenic environment should also\u00a0reduce the limitation on the interrogation time.\n\nThe experiments presented so far relied on prior knowledge of the strengths and positions of molecular transitions to positively identify specific states of a molecule by comparing the experimentally obtained ac-Stark shift to theory. In a variety of molecules, this information will not be completely available. For the common situation in which only the frequencies of the transitions are known, e.g., from a prediction based on known spectroscopic constants, we propose and demonstrate here an adaption of the present method for partial state readout35 that can be used to exclude a large subset of molecular states simultaneously and provide a non-destructive spectroscopic signal.\n\nAn example is given for the present case of N$${}_{2}^{+}$$. The positions of all spectral lines belonging to the $${A}^{2}{\\Pi }_{u}(v^{\\prime} =2)\\leftarrow {X}^{2}{\\Sigma }_{g}^{+}(v^{\\prime\\prime} =0)$$ transitions up to N\u2033\u2009=\u20096 colour coded by rotational state are shown in Fig.\u00a06. As can be seen in the figure, a lattice-laser wavelength larger than 789.4\u00a0nm (dashed line) is red-detuned with respect to all transitions with N\u2009\u2264\u20094. Therefore, detecting a blue detuning in an SP\/OP experiment excludes this entire manifold regardless of its substructure or strengths. In Fig.\u00a03b, a demonstration for this partial state readout is shown. Here, a lattice wavelength of 789.71\u2009nm was used and a stronger OP than SP signal was detected which implied that the lattice was blue detuned to the molecular transition, thus enabling to exclude all states with rotational quantum number N\u2009\u2264\u20094 for this molecule.\n\nConversely, the present protocol can be used as a non-destructive readout for spectroscopic excitations when exciting from a partially known initial state, e.g., N\u2009\u2264\u20094 (v\u2033\u2009=\u20090) to a long-lived excited state with a known detuning with respect to the lattice such as the $$v^{\\prime} =1$$ state of the $${X}^{2}{\\Sigma }_{g}^{+}$$ ground state. For instance, at around 804\u2009nm the lattice laser is blue-detuned from the closest transitions out of $$v =1$$ and one can detect a successful spectroscopic excitation from low rotational states\u00a0in the vibrational ground state to the first excited vibrational state\u00a0as a change in the sign of the ac-Stark shift even with no available information about the hyperfine or Zeeman structure of either upper or lower states or the transition strengths. Such an approach will greatly enhance the possibilities for non-destructive spectroscopic experiments in complex molecules.\n\nTo conclude, we have demonstrated an experimental protocol for molecular-state detection incorporating both the amplitude and sign of the ac-Stark shift. Using this protocol, Zeeman levels of rotational states of N$${}_{2}^{+}$$ were identified in a region where information about only the amplitude of the ac-Stark shift leads to ambiguity in the state identification. The state identification of hyperfine-spin-rotational levels can also be regarded as a probabilistic state preparation. The hyperfine-state preparation of molecular ions has been a long-standing open problem in molecular physics29,36 for which the present scheme offers a complementary solution.\n\nThe present method was also demonstrated to trace reactive and state-changing collisions of single molecules with partial state selectively. We also showed how the sign information provided by the present method can be used for a partial state determination in situations in which only incomplete spectroscopic information is available on the molecule. The present work thus introduces important new tools towards non-destructive state identification and spectroscopy of complex molecular systems.\n\n## Methods\n\n### Experimental setup\n\nA molecular-beam machine was coupled to a linear radio-frequency ion trap in which\u00a0a 40Ca+ AI and a 14N$${}_{2}^{+}$$ MI were trapped simultaneously (Fig.\u00a01 of the main text). A small magnetic field of 4.6\u2009G defined the quantization axis along the viewing direction of the camera, orthogonal to the trap axis. A continuous-wave (CW) laser beam at 789\u2009nm was split into two paths which were then superimposed on the trapping region in a counter-propagating configuration to form an optical lattice. The polarization of the beams was chosen parallel to the magnetic-field vector. The frequency difference, \u0394f, between the beam paths was matched to the frequency of the IP motional mode using acousto-optic modulators. Two pulsed dye-laser beams at 202 and 375\u2009nm were used to produce 14N$${}_{2}^{+}$$ ions by resonance-enhanced multi-photon ionization (REMPI) from the pulsed molecular beam of neutral 14N2 molecules37,38. The ionization scheme was chosen to create 14N$${}_{2}^{+}$$ in the lowest rotational levels of the I\u2009=\u20090 or I\u2009=\u20092 nuclear-spin isomers (corresponding to levels with even rotational angular momentum quantum numbers N\u2009=\u20090,\u00a02,\u00a04,\u00a0\u2026) in the electronic and vibrational ground state. We achieved this isomeric selectivity by using a [$$2+1^{\\prime}$$] REMPI scheme consisting of two 202\u2009nm photons at 49,426\u2009cm\u22121 on resonance with the S(0) transition from the rovibronic ground-state (N\u2009=\u20090) in neutral 14N2. Therefore, the I\u2009=\u20091 isomer, with its lowest rotational state N\u2009=\u20091 was excluded from the ionization. Two CW lasers at 397 and 866\u00a0nm were used for Doppler laser cooling of Ca+, and another two laser beams at 729 and 854\u00a0nm were used for resolved-sideband cooling and coherent state manipulation.\n\nOur experimental procedure was described in detail in refs. 15,18. Approximately ten Ca+ ions were loaded into the trap and Doppler cooled on the (4s)2S1\/2\u2009\u2194\u2009(4p)2P1\/2\u2009\u2194\u2009(3d)2D3\/2 closed optical cycling transitions to form a string of Coulomb-crystallized ions. A single N$${}_{2}^{+}$$ ion was then loaded into the trap using REMPI. The appearance of a dark ion in the string signalled the sympathetic cooling of a molecule. By lowering the trap depth, Ca+ ions were successively ejected from the trap until a Ca+\u2013N$${}_{2}^{+}$$ two-ion string remained. The ions were cooled to the motional ground state of the IP motional mode by resolved sideband cooling on the (3d)2D5\/2(m\u2009=\u2009\u22125\/2)\u2009\u2190\u2009\u00a0(4s)2S1\/2(m\u2009=\u2009\u22121\/2) transition of Ca+ followed by quenching the D5\/2 state to the S1\/2 level through the (4p)2P3\/2 state and optical pumping back to the S1\/2(m\u2009=\u2009\u22121\/2) state. To prepare the Ca+ ion\u00a0in the metastable D5\/2(m\u2009=\u2009\u22125\/2) state, we used a \u03c0-pulse on the narrow D5\/2(m\u2009=\u2009\u22125\/2)\u2009\u2190\u2009S1\/2(m\u2009=\u2009\u22121\/2) transition followed by a state-purification pulse14,15. The state-purification pulse projected the Ca+ either to the D5\/2(m\u2009=\u2009\u22125\/2) or the S1\/2 state with a heralded signal (photon scattering) that allowed us to exclude experiments with improper state preparation. We chose the D5\/2(m\u2009=\u2009\u22125\/2) level\u00a0due to its reduced ac-Stark shift compared to the S1\/2 state which would have overwhelmed the molecular signal.\n\n### Phase-sensitive forces\n\nIn ref. 15, we used a similar type of force-spectroscopic method to the one presented here. In that work, we employed a smaller lattice-laser detuning from a specific, well known resonance in N$${}_{2}^{+}$$ prepared in a specific quantum state and consequently required a smaller lattice-laser intensity and duration compared to the present study. Under these conditions, small effects such as the contribution of the ODF on Ca+ to the motional excitation, the effects of far-detuned molecular resonances including other electronic transitions and the influence of the polarizability of the N$${}_{2}^{+}$$ core electrons on the excitation strength could be neglected.\n\nFor the objectives of the present work, i.e., the determination of an initially unknown quantum state of the molecule, larger detunings from molecular resonances and consequently higher laser intensity and duration were necessary. Here, we give a more complete account of phase-sensitive force spectroscopy that includes the effect of the lattice on both the atomic and the molecular ions.\n\n### Normal modes of the system\n\nThe derivation of the normal modes of a Coulomb crystal of two ions with unequal masses in a harmonic trap is described in refs. 30,31. Here, we only give the final results and definitions which will be used later in the extraction of the molecular ac-Stark shift from the experimental signal.\n\nThe present system is composed of two ions, a molecular ion with mass m1\u2009=\u200928\u00a0a.u. and an atomic ion with mass m2\u2009=\u200940\u00a0a.u. The ions are confined in a harmonic potential characterized by a spring constant, u0, such that the frequency of a single particle is given by $${\\omega }_{i}=\\sqrt{{u}_{0}\/{m}_{i}}$$. Here, the subscripts i\u2009=\u20091,\u00a02 refer to the molecular ion and the atomic ion, respectively.\n\nWhen both ions are trapped together, they form a crystal due to the balance between their Coulomb repulsion, $${F}_{\\text{c}}=-\\frac{{e}^{2}\/4\\pi {\\varepsilon }_{0}}{{r}_{12}^{2}}$$, and the harmonic confinement of the trap. Here, e is the electron charge, \u03b50 is the vacuum permittivity and r12 is the separation between the two ions. The distance between the equilibrium positions, $${x}_{i}^{0}$$, of the ions in the crystal is given by\n\n$$d=| {x}_{2}^{0}-{x}_{1}^{0}| =\\root{3}\\of {\\frac{{e}^{2}}{4\\pi {\\varepsilon }_{0}}\\frac{2}{{u}_{0}}}.$$\n(1)\n\nBy measuring the trap-oscillation frequency of a single atomic ion, \u03c92, with relative uncertainty \u03b4\u03c92\/\u03c92\u2009<\u200910\u22123, the distance d can be calculated with a similar accuracy.\n\nAt the motional-excitation amplitudes reached in the present experiments, the ions performed only small oscillations around their equilibrium positions, $${q}_{i}={x}_{i}-{x}_{i}^{0}$$, such that it is safe to keep only harmonic terms in an expansion of the potential. The displacements of the individual particles, however, do not correspond to the normal modes of the system. To transform to a normal-mode basis, scaling and rotation transformations given by\n\n$$\\left(\\begin{array}{l}{\\beta }_{+}\\\\ {\\beta }_{-}\\end{array}\\right)=\\left(\\begin{array}{ll}\\cos (\\theta )\/\\sqrt{\\mu }&-\\sin (\\theta )\\\\ \\sin (\\theta )\/\\sqrt{\\mu }&\\cos (\\theta )\\end{array}\\right)\\left(\\begin{array}{l}{q}_{1}\\\\ {q}_{2}\\end{array}\\right).$$\n(2)\n\nwere performed. Here, \u03b2\u00b1 are the in-phase (\u2212) and out-of-phase (+) normal-mode displacements, \u03b8 is a rotation angle where $$\\tan (\\theta )=1\/\\sqrt{\\mu }-\\sqrt{\\mu }+\\sqrt{1\/\\mu +\\mu -1}$$ and \u03bc\u2009=\u2009m2\/m1. In the normal-mode picture, the system is equivalent to an ion of mass m2 trapped in a 2D harmonic potential with two uncoupled mode frequencies,\n\n$${\\Omega }_{\\pm }={\\omega }_{2}\\sqrt{1+\\mu \\pm \\sqrt{1+{\\mu }^{2}-\\mu }}.$$\n(3)\n\nNote that a change of one unit in the mass of the molecular ion (as in the case of the chemical reaction described in Fig.\u00a05 of the main text) corresponds to a relative change of the in-phase mode frequency of \u03b4\u03a9\u2009\u2248\u20096\u2009\u00d7\u200910\u22123, which can readily be detected in the present experiments.\n\n### Lattice excitation of the normal modes\n\nThe optical lattice induces a modulated ac-Stark shift on both the atomic and molecular ion,\n\n$$\\Delta {E}_{i}=2\\Delta {E}_{i}^{0}\\left(1+\\cos (2k{q}_{i}-\\Delta\\omega t+{\\phi }_{i}^{0})\\right).$$\n(4)\n\nHere, $$\\Delta {E}_{i}^{0}$$ is the ac-Stark shift induced by a single lattice beam, \u0394\u03c9\u2009=\u20092\u03c0\u0394f\u00a0is the difference in\u00a0angular frequency\u00a0of the lattice-laser beams, k\u2009=\u20092\u03c0\/\u03bb is the lattice-laser k-vector and \u03bb\u2009\u2248\u2009789\u00a0nm is the lattice-laser wavelength. The phase, $${\\phi }_{i}^{0}=2k{x}_{i}^{0}$$, depends on the equilibrium positions of the particles. In the SP configuration, the phase difference between the two particles, $${\\phi }_{21}^{0}={\\phi }_{2}^{0}-{\\phi }_{1}^{0}=2k{d}_{{\\rm{SP}}}=2\\pi n$$ (n an integer), is such that the phase of the lattice is equal for both particles. In the OP configuration, $${\\phi }_{21}^{0}=2k{d}_{{\\rm{OP}}}=2\\pi (n+1\/2)$$, and the particles experience opposite phases\u00a0of the lattice intensity gradient (Fig.\u00a01b of the main text). An additional phase difference is attributed to the sign of the ac-Stark shift, $$\\pm | \\Delta {E}_{i}^{0}|$$, which depends on the detuning of the lattice-laser frequency with respect to a dominant resonance in the relevant particle.\n\nTo get an intuition on how the ODF on the molecular ion and the atomic ion affect the excitation of the in-phase mode of the crystal, we first expand Eq. (4) in a Taylor series around the equilibrium positions and neglect constant terms of the potential which do not exert any force,\n\n$$\\Delta {E}_{i}\\approx -4k\\Delta {E}_{i}^{0}{q}_{i}\\sin (\\Delta\\omega t-{\\phi }_{i}^{0}).$$\n(5)\n\nIn this approximation, the lattice exerts an oscillating spatially homogeneous force with an amplitude $${F}_{i}=4k\\Delta {E}_{i}^{0}$$ on particle i. In the case of a single particle, this force will lead to coherent excitation of motion in the trap. In the case of two ions with unequal masses, the ac-Stark shift resulting in the driving of the in-phase mode can be obtained using Eq. (2),\n\n$$\\Delta {E}_{-} =-4k\\left(\\Delta {E}_{1}^{0}\\sqrt{\\mu }\\sin (\\theta )\\pm \\Delta {E}_{2}^{0}\\cos (\\theta )\\right){\\beta }_{-}\\sin (\\Delta\\omega t)\\\\ \\equiv -4k\\Delta {E}_{-}^{0}{\\beta }_{-}\\sin (\\Delta\\omega t).$$\n(6)\n\nHere, the \u00a0\u00b1\u00a0 sign corresponds to the SP and OP configurations, respectively, and $$\\Delta {E}_{-}^{0}$$ defines the single-beam ac-Stark shift of the in-phase mode. Thus, the two forces exerted by the lattice on the two ions are combined to a single effective force on the in-phase mode with an amplitude, $${F}_{-}=4k\\Delta {E}_{-}^{0}$$. This force will lead to coherent excitation of the in-phase mode of the two-ion crystal. In this analytical derivation, higher order terms of the lattice potential were neglected, which lead to squeezing of the motional states and to mixing of the in-phase and out-of-phase modes. In the data analysis, we used a classical simulation of the two-ion system to account for high-order terms in the lattice excitation15,18.\n\nBy measuring the in-phase motional excitation amplitude both in the SP ($$| \\Delta {E}_{-}^{0}({\\rm{SP}})|$$) and in the OP configuration ($$| \\Delta {E}_{-}^{0}({\\rm{OP}})|$$), the amplitude of the ac-Stark shift, $$| \\Delta {E}_{\\,\\text{m}\\,}^{0}|$$, exerted by the lattice on the molecule can be determined,\n\n$$| \\Delta {E}_{\\,\\text{m}\\,}^{0}| \\equiv | \\Delta {E}_{1}^{0}| =\\frac{| \\Delta {E}_{-}^{0}({\\rm{SP}})| +| \\Delta {E}_{-}^{0}({\\rm{OP}})| }{2\\sqrt{\\mu }\\sin (\\theta )},$$\n(7)\n\nunder the assumption that $$| \\Delta {E}_{\\,\\text{m}}^{0}| \\,> \\,| \\Delta {E}_{\\text{a}\\,}^{0}|$$. The phase of the ac-Stark shift is inferred from the relative amplitudes obtained in the SP and OP measurements assuming that the sign of the ac-Stark shift on the atomic ion is negative.\n\n### Ac-Stark shift of Ca+\n\nThe ac-Stark shift, \u0394Ej, of a level j in Ca+ is given by\n\n$$\\Delta {E}^{j}=-\\frac{{\\alpha }^{j}(\\omega )}{2{\\varepsilon }_{0}c}I,$$\n(8)\n\nwhere \u03b50 is the vacuum permittivity, c is the speed of light, I the laser intensity, and \u03b1j(\u03c9) the dynamic polarizability of level j which is given by\n\n$${\\alpha }^{j}(\\omega )={\\alpha }_{s}^{j}(\\omega )+\\left(\\frac{3{\\cos }^{2}\\Theta -1}{2}\\right)\\frac{3{m}_{j}^{2}-{J}_{j}({J}_{j}+1)}{{J}_{j}(2{J}_{j}-1)}{\\alpha }_{t}^{j}(\\omega ).$$\n(9)\n\nHere, $${\\alpha }_{s}^{j}(\\omega )$$ and $${\\alpha }_{t}^{j}(\\omega )$$ are the scalar and tensor dynamic polarizabilities of level j, \u0398 is the angle of linear polarization (\u0398\u2009=\u20090 for \u03c0-polarized light) and Jj and mj are the quantum numbers of the total angular momentum and its projection.\n\nThe (4s)2S1\/2 state of Ca+ has only a scalar contribution to the polarizability which results in \u03b1S(\u03c9)\u2009=\u200997.5 a.u.39 for a lattice laser wavelength of 789.0\u2009nm.\n\nThe (3d)2D5\/2(m\u2009=\u2009\u22125\/2) state of Ca+ has both scalar and tensor contributions to the polarizability. With linearly \u03c0-polarized light, this state interacts only with high lying F states. The contribution of these states to the polarizability decays rather slowly such that all states up to the continuum need to be taken into account39. The polarizability of this state results in, \u03b1D(\u03c9)\u2009=\u20094.44 a.u.39 at a lattice-laser wavelength of 789.0\u2009nm. Small uncertainties in the linear lattice polarization at the level of 1 amount to \u00a0~1% error in the D-state polarizability.\n\nThe low polarizability of the (3d)2D5\/2(m\u2009=\u2009\u22125\/2) state with linearly polarized light is comparable to the polarizability of the core electrons, $${\\alpha }^{D,{\\rm{core}}}=3.03$$\u2009a.u.39 for this state. The core contribution is almost identical in the (3d)2D5\/2 and the (4s)2S1\/2 states ($${\\alpha }^{S,{\\rm{core}}}=3.134$$\u2009a.u.39 in the S state). Therefore, in a spectroscopic experiment on the 2D5\/2(m\u2009=\u2009\u2212\u20095\/2)\u2009\u2190\u20092S1\/2(m\u2009=\u2009\u22121\/2) transition, the measured ac-Stark shift has contributions mostly from the polarizability of the valence electron in both states,\n\n$$\\Delta {E}^{D\\leftarrow S} =\\left(\\Delta {E}^{D}+\\Delta {E}^{D,{\\rm{core}}}\\right)-\\left(\\Delta {E}^{S}+\\Delta {E}^{S,{\\rm{core}}}\\right)\\\\ \\approx \\Delta {E}^{D}-\\Delta {E}^{S}.$$\n(10)\n\nHowever, in a lattice-excitation experiment, the contribution to the ac-Stark shift from the atomic ion is due to both the core and valence electrons in the relevant state,\n\n$$\\Delta {E}_{2}^{0}\\equiv \\Delta {E}^{j,{\\rm{lattice}}}=\\Delta {E}^{j}+\\Delta {E}^{j,{\\rm{core}}}.$$\n(11)\n\n### Ac-Stark shift calibration\n\nThe amplitude of the ac-Stark shift was determined from the Rabi-oscillation data (see, e.g., Fig.\u00a03 of the main text) by comparison to a calibration experiment. In this calibration, the Ca+\u2013N$${}_{2}^{+}$$ Rabi-oscillation signal was simulated by exciting motion on a two-ion Ca+\u2013N2H+ string with a well-defined ac-Stark shift amplitude applied on the atomic ion15. We used an N2H+ molecular ion for convenience due to its longer chemical lifetime. The mass difference of 1 u between N$${}_{2}^{+}$$ and N2H+ amounts to only a small correction compared to other experimental errors.\n\nIn the calibration experiment, Ca+ was prepared in the $${(4s)}^{2}{S}_{1\/2}\\left(m=-1\/2\\right)$$ state due to its large polarizability that allowed us to tune the desired ac-Stark shift. The ac-Stark shift was calibrated by a spectroscopic measurement on the $$(3d){}^{2}{D}_{5\/2}\\left(m=-5\/2\\right)\\leftarrow {}^{2}{S}_{1\/2}\\left(m=-1\/2\\right)$$ transition of Ca+ and varied by changing the lattice-laser power. Six different ac-Stark shifts in the range 0.8\u22124.6\u2009kHz were applied for generating the calibration data. The spectroscopically measured shifts were the combined shifts of both spectroscopic levels (Eq. (10)). However, since the lattice excitations were preformed in the $${}^{2}{S}_{1\/2}\\left(m=-1\/2\\right)$$ state only, the $${}^{2}{D}_{5\/2}\\left(m=-5\/2\\right)$$ contribution (Eq. (11)) was subtracted.\n\nSubsequently, a lattice-excitation experiment was performed in which the ODF was applied for 3\u2009ms with well-defined shifts on the Ca+\u2013N2H+ system so that a Rabi-oscillation signal was obtained for each applied ac-Stark shift (Fig.\u00a07). We used these Rabi-oscillation signals as a reference for the ac-Stark shift amplitudes generated in the experiments described in the main text. In order to extract the amplitude from Rabi-oscillation data with unknown ac-Stark shift, an interpolation between the six calibrated ac-Stark shifts was performed using a single parameter (the ac-Stark shift) fitting function (see details in the supplementary materials of ref. 15).\n\nAnother advantage of using N2H+ as opposed to N$${}_{2}^{+}$$ for the calibration experiment besides the longer lifetime is that it has no strong transitions near 789\u2009nm that could interfere with the calibration. There is, however, a residual force on the N2H+ due to highly detuned transitions in the molecule that must be taken into account (compare Fig.\u00a05b and the associated discussion in the main text). Since the ac-Stark shift on N2H+, $$\\Delta {E}_{{{\\rm{N}}}_{2}{{\\rm{H}}}^{+}}$$ is small compared to the Ca+ shift, $$\\Delta {E}_{{{\\rm{Ca}}}^{+}}$$, in the (4s)2S1\/2(m\u2009=\u2009\u22121\/2) state, it was neglected to first order. This zero-order fitting function was then used to extract the ac-Stark shift of N2H+ (Fig.\u00a05). An approximate value of $$\\Delta {E}_{{{\\rm{N}}}_{2}{{\\rm{H}}}^{+}}=-0.81$$\u2009kHz was obtained corresponding to \u00a0~15.8% of $$\\Delta {E}_{{{\\rm{Ca}}}^{+}}=5.41$$\u2009kHz measured with the same power. This shift was then added to the calibration data for a more realistic fitting function which now includes the total shift from both Ca+ and N2H+. This procedure was iterated for a first-order estimate of the ac-Stark shift of N2H+. The first iteration changed the measured shift to $$\\Delta {E}_{{{\\rm{N}}}_{2}{{\\rm{H}}}^{+}}=-0.93$$\u2009kHz corresponding to \u00a0~18.1% of $$\\Delta {E}_{{{\\rm{Ca}}}^{+}}$$. The second iteration yielded $$\\Delta {E}_{{{\\rm{N}}}_{2}{{\\rm{H}}}^{+}}=-0.99$$\u2009kHz corresponding to \u00a0~18.5% of $$\\Delta {E}_{{{\\rm{Ca}}}^{+}}$$. Thus, any residual error in the fitting function after the second iteration was neglected.\n\nFigure\u00a07 shows the results of the calibration experiments accounting for the effect of N2H+ excitation. The Ca+ and N2H+ shifts were added together to produce the effective shift as indicated in the figure legend since the lattice was red detuned for both the atomic and molecular ions and the calibration was preformed in the SPR configuration (Fig.\u00a01b of the main text).\n\n### Ac-Stark shift of N$${}_{2}^{+}$$ including hyperfine structure\n\nThe lattice-laser beam induced an ac-Stark shift on the molecular ion according to Eq. (8). This shift was calculated by summing up the contributions of the different transitions in the ro-vibrational band $${A}^{2}{\\Pi }_{u}(v^{\\prime} =2)\\leftarrow {X}^{2}{\\Sigma }_{g}^{+}(v^{\\prime\\prime} =0)$$,\n\n$${\\alpha }^{j}(\\omega )=\\sum _{k}\\frac{2}{\\hslash }\\frac{{\\omega }_{jk}}{{\\omega }^{2}-{{\\omega }_{jk}}^{2}}{\\left|\\langle k| {\\boldsymbol{\\mu }}| j\\rangle \\right|}^{2}.$$\n(12)\n\nHere, is the reduced Planck constant, \u03c9jk are the transition angular frequencies40 and k\u03bcj2 is the squared transition-dipole matrix element calculated using a spherical-tensor-algebra approach41 using the value of the vibronic Einstein A coefficient taken from ref. 42. Here, the ground state j is the $${X}^{2}{\\Sigma }_{g}^{+}(v^{\\prime\\prime} =0)$$ vibronic state of N$${}_{2}^{+}$$ which is adequately described by a Hund\u2019s case (b) coupling scheme. The excited state k is the $${A}^{2}{\\Pi }_{u}(v^{\\prime} =2)$$ vibronic state described by an intermediate Hund\u2019s case (a)\/Hund\u2019s case (b) coupling scheme43. Hyperfine effects are included in the calculation of the dipole matrix element. The effect of mixing of states with different total angular momentum quantum number J by hyperfine interactions44 should only have a small effect on the ac-Stark shifts and was therefore not included in the calculations.\n\nThe contribution of additional vibrational bands in the $${A}^{2}{\\Pi }_{u}(v^{\\prime} \\,\\ne\\, 2)$$ excited state and of the $${B}^{2}{\\Sigma }_{u}^{+}(v^{\\prime} )$$ electronic state were included. Here, due to the large detuning, only the rotationless transition frequencies were used45. The corresponding Einstein A coefficients were also taken from ref. 45.\n\nThe core polarizability of N$${}_{2}^{+}$$ was estimated from a calculation of the polarizability of N$${}_{2}^{2+}$$. The calculation was performed using Gaussian 09 (ref. 46) at the CCSD\/aug-cc-pVQZ level of theory. The value of the polarizability was found to be $${\\alpha }^{{{\\rm{N}}}_{2}^{+},{\\rm{core}}}\\approx {\\alpha }^{{{\\rm{N}}}_{2}^{2+}}=7.23$$ a.u. which corresponds to an ac-Stark shift of \u2212390\u2009Hz at our experimental parameters.\n\n## Data availability\n\nThe data that support the findings of this study have been deposited in Zenodo with the identifier DOI: https:\/\/doi.org\/10.5281\/zenodo.3898047.\n\n## References\n\n1. 1.\n\nWright, K. et al. Benchmarking an 11-qubit quantum computer. Nat. Commun. 10, 1\u20136 (2019).\n\n2. 2.\n\nFriis, N. et al. Observation of entangled states of a fully controlled 20-qubit system. Phys. Rev. X 8, 021012 (2018).\n\n3. 3.\n\nArute, F. et al. Quantum supremacy using a programmable superconducting processor. Nature 574, 505\u2013510 (2019).\n\n4. 4.\n\nBrewer, S. 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Commun. 10, 5429 (2019).\n\n11. 11.\n\nKimble, H. J. The quantum internet. Nature 453, 1023 (2008).\n\n12. 12.\n\nNorthup, T. & Blatt, R. Quantum information transfer using photons. Nat. Photon. 8, 356 (2014).\n\n13. 13.\n\nWolf, F. et al. Non-destructive state detection for quantum logic spectroscopy of molecular ions. Nature 530, 457 (2016).\n\n14. 14.\n\nChou, C. W. et al. Preparation and coherent manipulation of pure quantum states of a single molecular ion. Nature 545, 203 (2017).\n\n15. 15.\n\nSinhal, M., Meir, Z., Najafian, K., Hegi, G. & Willitsch, S. Quantum-nondemolition state detection and spectroscopy of single trapped molecules. Science 367, 1213\u20131218 (2020).\n\n16. 16.\n\nChou, C. et al. Frequency-comb spectroscopy on pure quantum states of a single molecular ion. Science 367, 1458\u20131461 (2020).\n\n17. 17.\n\nLin, Y., Leibrandt, D. R., Leibfried, D. & Chou, C.-w Quantum entanglement between an atom and a molecule. Nature 581, 273\u2013277 (2020).\n\n18. 18.\n\nMeir, Z., Hegi, G., Najafian, K., Sinhal, M. & Willitsch, S. State-selective coherent motional excitation as a new approach for the manipulation, spectroscopy and state-to-state chemistry of single molecular ions. Faraday Discuss. 217, 561\u2013583 (2019).\n\n19. 19.\n\nNajafian, K., Meir, Z. & Willitsch, S. From megahertz to terahertz qubits encoded in molecular ions: theoretical analysis of dipole-forbidden spectroscopic transitions in $${\\,\\text{N}\\,}_{2}^{+}$$. Preprint at arXiv\u00a0https:\/\/arxiv.org\/abs\/2007.11097 (2020).\n\n20. 20.\n\nBiesheuvel, J. et al. Probing QED and fundamental constants through laser spectroscopy of vibrational transitions in HD+. Nat. Commun. 7, 10385 (2016).\n\n21. 21.\n\nSchiller, S., Bakalov, D. & Korobov, V. I. Simplest molecules as candidates for precise optical clocks. Phys. Rev. Lett. 113, 023004 (2014).\n\n22. 22.\n\nSchiller, S. & Korobov, V. 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Blackbody radiation shift, multipole polarizabilities, oscillator strengths, lifetimes, hyperfine constants, and excitation energies in Ca+. Phys. Rev. A 83, 012503 (2011).\n\n40. 40.\n\nWu, Y.-D. et al. Study of (2, 0) band of A \u2212 X system of $${\\,\\text{N}\\,}_{2}^{+}$$ by optical heterodyne detected velocity modulation spectroscopy. Chin. J. Chem. Phys. 20, 285 (2007).\n\n41. 41.\n\nZare, R. N. Angular Momentum (Wiley, 1988).\n\n42. 42.\n\nLanghoff, S. R., Bauschlicher Jr, C. W. & Partridge, H. Theoretical study of the $${\\,\\text{N}\\,}_{2}^{+}$$\u00a0Meinel system. J. Chem. Phys. 87, 4716\u20134721 (1987).\n\n43. 43.\n\nBennett, R. H\u00f6nl\u2013London factors for doublet transitions in diatomic molecules. Mon. Not. R. Astron. Soc. 147, 35\u201346 (1970).\n\n44. 44.\n\nGermann, M. Dipole-Forbidden Vibrational Transitions in Molecular Ions. (Thesis, Univ. Basel, 2016).\n\n45. 45.\n\nGilmore, F. R., Laher, R. R. & Espy, P. J. Franck\u2013Condon factors, r-centroids, electronic transition moments, and Einstein coefficients for many nitrogen and oxygen band systems. J. Phys. Chem. Ref. Data 21, 1005\u20131107 (1992).\n\n46. 46.\n\nFrisch, M. J. et al. Gaussian 09 Revision D.01 (Gaussian Inc., Wallingford, CT, 2009).\n\n## Acknowledgements\n\nWe thank M. S. Safronova for providing accurate values for the Ca+ polarizability and P. Stra\u0148\u00e1k for calculating the polarizability of N$${}_{2}^{2+}$$. This work has been supported by the Swiss National Science Foundation as part of the National Centre of Competence in Research, Quantum Science and Technology (NCCR-QSIT), grant nr. CRSII5_183579, and by the University of Basel.\n\n## Author information\n\nAuthors\n\n### Contributions\n\nK.N. and Z.M. performed the experiments and the analysis of the data. M.S. developed parts of the theory underlying the analysis. S.W. conceived and supervised the project. All authors contributed to writing the manuscript.\n\n### Corresponding author\n\nCorrespondence to Stefan Willitsch.\n\n## Ethics declarations\n\n### Competing interests\n\nThe authors declare no competing interests.\n\nPeer review information Nature Communications thanks the anonymous reviewers for their contribution to the peer review of this work.\n\nPublisher\u2019s note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.\n\n## Rights and permissions\n\nReprints and Permissions\n\nNajafian, K., Meir, Z., Sinhal, M. et al. Identification of molecular quantum states using phase-sensitive forces. Nat Commun 11, 4470 (2020). https:\/\/doi.org\/10.1038\/s41467-020-18170-9\n\n\u2022 Accepted:\n\n\u2022 Published:\n\n\u2022 ### From megahertz to terahertz qubits encoded in molecular ions: theoretical analysis of dipole-forbidden spectroscopic transitions in N2+\n\n\u2022 Kaveh Najafian\n\u2022 , Ziv Meir\n\u2022 \u00a0&\u00a0Stefan Willitsch\n\nPhysical Chemistry Chemical Physics (2020)\n\nBy submitting a comment you agree to abide by our Terms and Community Guidelines. 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\section{Introduction}
Let $F$ be a number field and $G$ be a reductive algebraic group over $F$. The unitary irreducible representations of the adelic group $G(\mathbb{A}_F)$ are of great interest as the automorphic representation side of the global Langlands correspondence.
It is then natural to consider the Plancherel measure on the equivalence classes of these representations, i.e., the unitary dual $\widehat{G(\mathbb{A}_F)}$.
It can be shown that the dual space $\widehat{G(\mathbb{A}_F)}$ with the Plancherel measure $\nu_{G(\mathbb{A}_F)}$ is the restricted product of its local dual $\widehat{G(F_v)}$ with respect to the unramified representations (see Theorem \ref{tPm}).
Meanwhile, the study of the von Neumann algebra $\mathcal{L}(\Gamma)$ of a countable discrete group $\Gamma$ is one of the most challenging topics in operator algebras.
Very little is known of how $\mathcal{L}(\Gamma)$ depends on the group $\Gamma$.
Especially, a well known conjecture asks whether $\mathcal{L}(\Gamma_1)\cong \mathcal{L}(\Gamma_2)$ implies $\Gamma_1\cong \Gamma_2$ provided $
\Gamma_1,\Gamma_2$ are free groups or lattices in a real Lie group \cite{J00ten}.
In this paper, we consider the group von Neumann algebra $\mathcal{L}(G(F))$ of the countable group $G(F)$ defined over the number field $F$.
The modules over $\mathcal{L}(G(F))$ can be described by a canonical dimension function $\dim_{\mathcal{L}(G(F))}$, which determines the isomorphic class of an $\mathcal{L}(G(F))$-module if $G(F)$ is an infinite conjugacy class (ICC) group (or equivalently, $\mathcal{L}(G(F))$ is a factor of type $\text{II}_1$).
We show the coincidence of the Plancherel measures $\nu_{G(\mathbb{A}_F)}$ of the adelic group $G(\mathbb{A}_F)$ and the dimensions over the von Neumann algebra $\mathcal{L}(G(F))$ of the group $G(F)$.
More precisely, let $X$ be a subset of $\widehat{G(\mathbb{A}_F)}$ with a finite Plancherel measure, i.e., $\nu_{G(\mathbb{A}_F)}(X)<\infty$.
Denoting the underlying Hilbert space of an irreducible representation $\pi$ by $H_{\pi}$, we let $H_X$ be the direct integral of the spaces $H_{\pi}$'s such that the isomorphism class $[\pi] \in X$, which is $H_X=\int_X^{\oplus} H_{\pi}d\nu_{G(\mathbb{A}_F)}([\pi])$.
The following result (Theorem \ref{tmain1}) is proved :
\begin{theorem*}
Let $G$ be a simply connected semisimple algebraic group defined over a number field $F$.
Then, for a subset $X$ of $\widehat{G(\mathbb{A}_F)}$ with finite Plancherel measure, we have
\begin{center}
$\dim_{\mathcal{L}(G(F))}H_X=\nu_{G(\mathbb{A}_F)}(X)$.
\end{center}
\end{theorem*}
In another word, one can tell the Placherel measure of the adelic group $G(\mathbb{A}_F)$ just by treating the representations as modules over the von Neumann algebra $\mathcal{L}(G(F))$ of the discrete subgroup $G(F)$.
In Section \ref{sPadele}, we give an explicit description of the Plancherel measure of an adelic group.
In Section \ref{svndim}, we construct the Hilbert space from the direct integral of irreducible representations of $G(\mathbb{A}_F)$ and discuss its von Neumann over $\mathcal{L}(G(F))$.
In Section \ref{sprod}, we reach a product formula involving the covolume of a lattice.
An example is also provided.
In Section \ref{smain}, we prove the theorem above.
{\it Acknowledgements}
The author would like to thank V. Jones for encouraging me to work on these problems.
The author is grateful for the detailed comments from D. Bisch, F. Radulescu, and many fruitful conversations with A. Jaffe, B. Mazur, D. Vogan, and W. Schmid.
\section{The Plancherel measure of an adelic group}\label{sPadele}
Let $G$ be a locally compact Hausdorff group with a Haar measure $\mu$.
Let $\widehat{G}$ be the unitary dual of $G$, i.e., the set of equivalence classes $[\pi]$ of irreducible unitary representations $(\pi,H_{\pi})$ of $G$.
For convenience, we will not distinguish a representation $\pi$ and its equivalence class $[\pi]$
In the following sections, we will let $\pi$ denote its equivalence class $[\pi]$ or also a representative in it.
We first give a brief review of type I groups following \cite{DiCalg,Fo2}.
A unitary representation $(\pi,H)$ of $G$ is called {\it primary} if the von Neumann algebra $\mathcal{A}(\pi)\subset B(H)$ generated by $\{\pi(g)|g\in G\}$ is a factor, i.e., its center $Z(A(\pi))$ are scalar multiples of the identity.
If $\pi$ is a direct sum of irreducible representations, then $\pi$ is primary if and only if all these irreducible summands are unitary equivalent.
A group $G$ is called {\it type I} if every primary representation of $G$ is a direct sum of copies of some irreducible representation.
That is to say, $\mathcal{A}(\pi)$ is a type I factor for every primary representation $\pi$.
It is known that compact groups, connected semisimple Lie groups, connected real algebraic groups,
reductive $p$-adic groups, and
also the adelic group $G(\mathbb{A})$ of a connected reductive group $G$ defined over a number field are type I (see \cite{Bnst74,Clz07,Fo2,Kiri76}).
\iffalse
\begin{enumerate}
\item Compact groups \cite{Kiri76};
\item Connected semisimple Lie groups, connected nilpotent Lie groups, and connected real algebraic groups \cite{Fo2};
\item Reductive $p$-adic groups \cite{Bnst74};
\item The adelic group $G(\mathbb{A})$ of a connected reductive group $G$ defined over a number field \cite{Clz07}.
\end{enumerate}
\fi
Let us consider the regular representations of a type I locally compact group $G$ and the Fourier transform.
Let $\rho,\lambda$ be the right and left regular representation of $G$ on $L^2(G)$:
\begin{center}
$(\rho(g)f)(x)=R_{g}f(x)=f(xg)$, $(\lambda(g)f)(x)=L_{g}f(x)=f(g^{-1}x)$
\end{center}
for $f\in L^2(G)$.
This gives us the {\it two sided regular representation} $\tau$ of $G\times G$ on $L^2(G)$:
$(\tau(g,h)f)(x)=R_gL_hf(x)=L_hR_gf(x)=f(h^{-1}xg)$.
If $f\in L^1(G)$, the {\it Fourier transform} of $f$ is defined to be the measurable field of operators over $\widehat{G}$ given by
\begin{center}
$\hat{f}(\pi)=\int_G f(x)\pi(x^{-1})d\mu(x)$,
\end{center}
with a representative of a equivalence class $[\pi]\in \widehat{G}$.
We have the convolution $\widehat{f_1 * f_2}=\widehat{f_1}\cdot\widehat{f_2}$ \cite{Fo2,Kiri04}.
Now we let
\begin{center}
$\mathcal{J}^1=L^1(G)\cap L^2(G)$, $\mathcal{J}^2=\spn\{f*g|f,g\in \mathcal{J}^1\}$.
\end{center}
For a unitary representation $(\pi,H_{\pi})$ of $G$, we denote its contragradient by $(\overline{\pi},H_{\overline{\pi}})$.
We let $\Tr$ denote the usual trace defined on the trace-class operators.
\begin{theorem}[The Plancherel Theorem]\label{tplancherel}
Suppose $G$ is a second-countable, unimodular, type I group.
There is a measure $\nu$ on $\widehat{G}$, uniquely determined once the Haar measure $\mu$ on $G$ is fixed, with the following properties.
\begin{enumerate}
\item The Fourier transform $f\mapsto \hat{f}$ maps $\mathcal{J}^1$ into $\int_{G}^{\oplus}H_{\pi}\otimes H_{\overline{\pi}}d\nu(\pi)$ and it extends to a unitary map from $L^2(G)$ onto $\int_{G}^{\oplus}H_{\pi}\otimes H_{\overline{\pi}}d\nu(\pi)$ that intertwines the two-sided regular representation $\tau$ with $\int_{G}^{\oplus}{\pi}\otimes {\overline{\pi}}d\nu(\pi)$.
\item For $f_1,f_2\in \mathcal{J}^1$, one has the Parseval formula
\begin{center}
$\int_G f_1(x)\overline{f_2(x)}d\mu(x)=\int_{\widehat{G}}\Tr[\hat{f_1}(\pi)\hat{f_2}(\pi)^*]d\nu(\pi)$.
\end{center}
\item For $f\in \mathcal{J}^2$, one has the Fourier inversion formula
\begin{center}
$f(x)=\int_{\widehat{G}}\Tr[{\pi}(x)^*\hat{f}(\pi)]d\nu(\pi)$, $x\in G$.
\end{center}
\end{enumerate}
\end{theorem}
Indeed, the Plancherel measure can be determined by either the second or third property described above.
This theorem also implies the decomposition of right and left regular representation of $G$:
\begin{center}
$\rho\cong \int_{\widehat{G}}^{\oplus}\pi\otimes {\id_{\overline{H_\pi}}}~d\nu(\pi)$, $\lambda\cong \int_{\widehat{G}}^{\oplus}{\id_{H_\pi}}\otimes\overline{\pi}~d\nu(\pi)$.
\end{center}
The measure $\nu$ above is called the {\it Plancherel measure} on $\widehat{G}$.
Note if $G$ is abelian, the Plancherel measure is a Haar measure;
if $G$ is compact, the Plancherel measure is given by $\nu(E)=\sum_{\pi \in E}\dim_{\mathbb{C}}{H_\pi}$ for $E\subset \widehat{G}$.
The support of $\nu$ is not always all of $\widehat{G}$.
Indeed, ${\rm supp}(\nu)=\widehat{G}$ if and only if $G$ is amenable (see \cite{DiCalg}).
\begin{remark}\label{rtenHS}
For a Hilbert space $H$, let $H^*$ be its dual space, which is isomorphic to the conjugate space $\overline{H}$.
We may identify the tensor Hilbert space $H\otimes H^*$ with the Hilbert-Schmidt operators on $H$, denoted by $B(H)_{\rm HS}$, which is equipped with the inner product $\langle x,y\rangle_{B(H)_{\rm HS}}=\Tr(xy^*)$.
Taking an orthonormal basis $\{e_i\}_{i\geq 1}$ of $H$, there is an isometric isomorphism $\Psi\colon B(H)_{\rm HS}\to H\otimes H^*$ given as
\begin{center}
$\Psi(T)=\sum_{i,j}\langle Te_j,e_i\rangle\cdot e_i\otimes e_j^*$, $T \in B(H)_{\rm HS}$,
\end{center}
such that $\Tr(TT^*)=\|\Psi(T)\|_{H\otimes H^*}^2$.
\end{remark}
Now let $F$ be a number field and $G$ be a reductive linear algebraic group defined over $F$.
We let $V$ ($V_f$ and $V_{\infty}$, respectively) denote the set of equivalence classes of places (finite places and infinite places, respectively) of $F$ and $F_v$ denotes the local fields at $v\in V$.
Let $G_v=G(F_v)$ and $G(\mathbb{A}_F)=\prod_{v\in V_{\infty}}G(F_v)\times \prod'_{v\in V_f}G(F_v)$, the group over the adele $\mathbb{A}_F$ of $F$.
For each $v\in V_f$, let $K_v$ be a special open maximal compact subgroup of $G_v$.
Indeed, for almost all $v$, $K_v$ can be taken to be $G(\mathcal{O}_v)$, the integral points of $G_v$.
For each finite place $v\in V_f$, we fix a Haar measure $\mu_v$ on the local group $G_v=G(F_v)$.
We may assume it is normalized in the sense that $\mu_v(K_v)=1$.
Let $\nu_v$ be the Planacherel measure on $\widehat{G_v}$ determined by $\mu_v$.
Denote an irreducible unitary representation of $G_v$ by $(\pi_v,H_{\pi_v})$, which will also stand for the isomorphism class in the unitary dual $\widehat{G_v}$.
Hence we have $L^2(G_v)=\int_{\widehat{G_v}}H_{\pi_{v}}\otimes H_{\pi_v}^* d\nu_v(\pi_v)$ by the Plancherel Theorem \ref{tplancherel}.
Consider the characteristic function $\chi_v$ of $K_v$, i.e.,
\begin{equation*}
\chi_v(x)=
\begin{cases}
1, & \text{if}\ x\in K_v; \\
0, & \text{otherwise}.
\end{cases}
\end{equation*}
Observe $\|\chi_v\|_{L^2(G_v)}=\mu_v(K_v)=1$.
Now we take an irreducible representation $(\pi_v,H_v)$ from $\widehat{G_v}$ and consider the Fourier transform $\widehat{\chi_v}$ of $\chi_{v}$ at $\pi_v$, which is given by
\begin{center}
$\widehat{\chi_v}(\pi_{v})=\int_{x\in G_v}\chi_v(x)\pi_{v}(x^{-1})d\mu_v(x)$.
\end{center}
Please note $\widehat{\chi_v}(\pi_{v})$ is a Hilbert-Schmidt operator, or equivalently, a vector in $H_{\pi_{v}}\otimes H_{\overline{\pi_{v}}}$ almost everywhere on $\widehat{G_v}$ (see \ref{rtenHS}).
\begin{lemma}\label{lunivec}
If $\widehat{\chi_v}(\pi_{v})$ is nonzero, we have
$\|\widehat{\chi_v}(\pi_{v})\|_{H_{\pi_{v}}\otimes H_{\overline{\pi_{v}}}}=1$.
\end{lemma}
\begin{proof}
Since the group $K_v$ is compact, the restriction $\pi_{v}|_{K_v}$ can be written as a direct sum of irreducible subrepresentations of $K_v$, i.e.,
$\pi_{v}|_{K_v}=\bigoplus_{i\in I}\rho_i$.
Then the Fourier transform of the characteristic function $\chi_v$ can be obtained as follows:
\begin{equation*}
\begin{aligned}
\widehat{\chi_v}(\pi_{v})&=\int_{G_v}\chi_v(x)\pi_{v}(x^{-1})d\mu_v(x)=\int_{K_v}\pi_{v}(x^{-1})d\mu_v(x)\\
&=\bigoplus_{i\in I}\int_{x\in K_v}\rho_i(x^{-1})d\mu_v(x).
\end{aligned}
\end{equation*}
If $\rho_i$ is a nontrivial irreducible representation, we have $\int_{ K_v}\rho(x^{-1})d\mu_v(x)=0$.
But for the trivial representation ${\rm 1}_{K_v}$, $\int_{ K_v}{\rm 1}_{K_v}(x^{-1})d\mu_v(x)=1$.
Therefore $\widehat{\chi_v}(\pi_{v})$ is the multiplicity of ${\rm 1}_{K_v}$ in $\pi_v$.
It is well-known that an irreducible unitary representation of $G(F_v)$ contains ${\rm 1}_{K_v}$ at most once (which are the unramified ones, see \cite{Flath77} Theorem 2).
Hence $\widehat{\chi_v}(\pi_{v})$ is $0$ or a projection on $B(H_{\pi_{v}})$ with rank $1$, or equivalently, the zero vector or a unit vector in the tensor space $H_{\pi_{v}}\otimes H_{\overline{\pi_{v}}}$.
\end{proof}
Let $X_v\subset \widehat{G_v}$ be the subset of unramified representations of $G_v$.
\begin{corollary}\label{cmeassph}
We have $\nu_v(X_v)=1$, the Plancherel measure of unramified representations is $1$.
\end{corollary}
\begin{proof}
It follows the fact that
\begin{center}
$1=\|\chi_v\|_{L^2(G_v)}^2=\int_{\widehat{G_v}}\|\widehat{\chi_v}(\pi_{v})\|^2d\nu_v(\pi_{v})$,
\end{center}
where $\|\widehat{\chi_v}(\pi_{v})\|=1$ for unramified representations and $\|\widehat{\chi_v}(\pi_{v})\|=0$ otherwise.
\end{proof}
\begin{remark}[The restricted product of measure spaces \cite{Blkd77}]
Let $\{(X_i,\mu_i):i\in I\}$ be a set of measure spaces with subsets $Y_i\subset X_i$ such that $\mu_i(Y_i)=1$ for all but finitely many $i\in I$.
Write $(X_S,\nu_S)=(\prod_{i\in S}X_i)\times (\prod_{i\notin F}Y_i)$ for a finite subset $S\subset I$.
Note for $S'\subset S$, there is a natural embedding of measure spaces $X_{S'}\subset X_S$.
The {\it restricted product of $\{(X_i,\mu_i):i\in I\}$ with respect to $\{Y_i:i\in I\}$} is defined by
\begin{center}
$\prod'_{i\in I} (X_i,Y_i,\mu_i)=\varinjlim_S X_S=\{(x_i)_{i\in I}\in \prod X_i|x_i\in Y_i\text{~for~almost~all~}i\}$,
\end{center}
which will also be denoted simply by $(X,\nu)$.
A subset $Z\subset X$ is measurable if and only if $Z\cap X_S$ is measurable for all finite $S$ and its measure is given by $\mu(S)=\sup_{S}\nu_S(Z\cap X_S)$.
\end{remark}
Now we take the restricted product of the measures $\{\mu_v\}_{v\in V}$ as a Haar measure on the adelic group $G(\mathbb{A})$, denoted by $\mu_{G_{\mathbb{A}}}$, or simply $\mu$.
Let $\nu_{G(\mathbb{A})}$ be the Plancherel measure of $\widehat{G_{\mathbb{A}}}$ determined by $\mu$.
\begin{theorem}\label{tPm}
$(\widehat{G_{\mathbb{A}}},\nu_{G(\mathbb{A})})=\prod'_{v\in V}(\widehat{G_v},X_v,\nu_v)$.
\end{theorem}
\begin{proof}
We have $L^2(G_{\mathbb{A}},\mu_{\mathbb{A}})=\bigotimes'_{v\in V}(L^2(G_v),\chi_v)$ by \cite{Blkd77} Corollary 2.3, where the restriced product is taken with respect to the unit vectors $\widehat{\chi_v}$.
This can be rewritten as
\begin{center}
$L^2(G_{\mathbb{A}})=\bigotimes'_v\left(\int_{\widehat{G_v}}H_{\pi_{v}}\otimes H_{\overline{\pi_{v}}}d\nu_v(\pi_{v})\right)$.
\end{center}
As shown in \cite{Flath77}, an irreducible representation $\pi$ of $G(\mathbb{A})$ can be written as $\pi\cong \bigotimes \pi_{v}$, where each $\pi_v$ is an irreducible unitary representation of $G_v$.
Almost all $\pi_v$ are unramified, i.e., $\pi_v\in X_v$ for all but finitely many $v$.
Take a finite set $S\subset V$ and consider the function $h_S=(\otimes_{v\in S}~h_v)\otimes (\otimes_{v\in V\backslash S}~\chi_{v})$ with $h_v\in \mathcal{J}^2_{G_v}$.
For any $x=(x_v)_{v\in V}\in G_{\mathbb{A}}$, we let $S_x=\{v \in V|x_v\notin K_v\}$, the finite set of the places $v$ whose local coordinate $x_v$ is outside of the compact group $K_v$.
Then the Fourier inversion formula of $h$ at $x$ is given by
{\small
\begin{equation}\label{eadeleplan1}
\begin{aligned}
h((x_v)_{v\in V})
=&\int_{\widehat{G_{\mathbb{A}}}}\Tr(\widehat{h}(\pi))\pi((x_v)_{v\in V})^{*}d\nu_{\mathbb{A}}(\pi)\\
=&\int_{\widehat{G_{\mathbb{A}}}}\left(\prod_{v\in S}\Tr(\widehat{h_v}(\pi_v))\pi_v(x_v)^{*})\cdot \prod_{v\in V\backslash S}\Tr(\widehat{\chi_{K_v}}(\pi_v)\pi_v(x_v)^{*})\right)d\nu_{\mathbb{A}}(\otimes \pi_v)\\
=&\int_{\widehat{G_{\mathbb{A}}}}\left(\prod_{v\in S}\Tr(\widehat{h_v}(\pi_v))\pi_v(x_v)^{*})\cdot \prod_{v\in S_x\backslash S}\Tr(\widehat{\chi_{K_v}}(\pi_v)\pi_v(x_v)^{*})\right)\cdot \\
&~~~~~~\left(\prod_{v\in V\backslash (S\cup S_x)}\delta_{X_v}(\pi_v)\right)d\nu_{\mathbb{A}}(\otimes'\pi_v)
\end{aligned}
\end{equation}
}
Observing that for $x_v\in K_v$, $\widehat{\chi_{v}}(\pi_v)\circ\pi_v(x_v)=\int_{K_v}\pi_v(y^{-1}x_v)d\mu_v(y)$ depends only on $\pi_v$,
we have $\delta_{X_v}(\pi_v)=1$ or $0$ depends on whether $\pi_v$ is unramified or not.
On the other hand, we have
{\small
\begin{equation}\label{eadeleplan2}
\begin{aligned}
h((x_v)_{v\in V})
=&\prod_{v\in S}h_v(x_v)\cdot \prod_{v\in V\backslash S}\chi_{K_v}(x_v)\\
=&\prod_{v\in S}\int_{\widehat{G_v}}\Tr(\widehat{h_v}(\pi_v)\pi_v(x_v)^{*})d\nu_v(\pi_v)\cdot \prod_{v\in V\backslash S}\int_{\widehat{G_v}}\Tr(\widehat{\chi_{K_v}}(\pi_v)\pi_v(x_v)^{*})d\nu_v(\pi_v)\\
=&\prod_{v\in S}\int_{\widehat{G_v}}\Tr(\widehat{h_v}(\pi_v)\pi_v(x_v)^{*})d\nu_v(\pi_v)\cdot \prod_{v\in S_x\backslash S}\int_{\widehat{G_v}}\Tr(\widehat{\chi_{K_v}}(\pi_v)\pi_v(x_v)^{*})d\nu_v(\pi_v)\\
&~~~~~~\cdot \prod_{v\in V\backslash (S\cup S_x)}\int_{\widehat{G_v}}\delta_{X_v}(\pi_v)d\nu_v(\pi_v),\\
\end{aligned}
\end{equation}
}
where the last product is $1$ by Corollary \ref{cmeassph}.
Observe the Plancherel measure is uniquely determined by the Fourier inversion formula.
The functions of the same form as $h$ above span a dense subset of the $L^2$-space.
Comparing the equation (\ref{eadeleplan1}) with (\ref{eadeleplan2}), we conclude $(\widehat{G_{\mathbb{A}}},\nu_{\mathbb{A}})=\prod'(\widehat{G_v},X_v,\nu_v)$.
\end{proof}
Note an irreducible unitary representation $\pi=\otimes \pi_v$ is a discrete series representation (an irreducible direct summand in the $L^2$-space) if and only if $\nu_{\mathbb{A}}(\{\pi\})>0$.
But $\nu_v(\pi_v)=0$ for almost all $v$.
Hence $\nu_{\mathbb{A}}(\pi)=\prod'\nu_{p}(\pi_v)=0$ and we obtain the following known result:
\begin{corollary}
$G(\mathbb{A})$ has no discrete series representations.
\end{corollary}
\iffalse
By the Plancherel Theorem \ref{tplancherel}, the two-sided regular representation of $G$ can be written as a direct integral such that
\begin{equation*}
\begin{aligned}
L^2(G,\mu)&\cong\int_{\widehat{G}}H_{\pi}\otimes H_{\overline{\pi}}d\nu(\pi)\\
f~~~&\mapsto \{\widehat{f}\colon\pi\mapsto \int_{G}f(x)\pi(x)^{-1}d\mu(x)\}
\end{aligned}
\end{equation*}
as a $G$-$G$ bimodule while the right regular representation $\rho$ and the right regular representation $\lambda$ are decomposed as
\begin{equation}\label{edecompregrep}
\rho(g)\cong \int_{\widehat{G}}\pi(g)\otimes \id_{H_{\overline{\pi}}}d\nu(\pi),~~~~ \lambda(g)\cong \int_{\widehat{G}}\id_{H_{\pi}}\otimes \overline{\pi}(g)d\nu(\pi)
\end{equation}
\fi
\section{The von Neumann dimensions of direct integrals}\label{svndim}
Let $\Gamma$ be a countable group with the counting measure.
Let $\{\delta_{\gamma}\}_{\gamma\in \Gamma}$ be the usual orthonormal basis of $l^2(\Gamma)$.
We also let $\lambda$ and $\rho$ be the left and right regular representations of $\Gamma$ on $l^2(\Gamma)$ respectively.
For all $\gamma,\gamma'\in \Gamma$, we have
$\lambda(\gamma')\delta_{\gamma}=\delta_{\gamma'\gamma}$ and $\rho(\gamma')\delta_{\gamma}=\delta_{\gamma\gamma'^{-1}}$.
Let $\mathcal{L}(\Gamma)$ be the strong operator closure of the complex linear span of $\lambda(\gamma)$'s (or equivalently, $\rho(\gamma)$'s).
This is the {\it group von Neumann algebra of $\Gamma$}.
There is a canonical faithful normal tracial state $\tau_{\Gamma}$, or simply $\tau$, on $\mathcal{L}(\Gamma)$, which is given by
\begin{center}
$\tau(x)=\langle x\delta_e,\delta_e\rangle_{l^2(\Gamma)}$, $x\in \mathcal{L}(\Gamma)$.
\end{center}
Hence $\mathcal{L}(\Gamma)$ is a finite von Neumann algebra (hence it is of type $\text{I}$ or $\text{II}_1$).
More generally, for a tracial von Neumann algebra $M$ with the trace $\tau$, we consider the GNS representation of $M$ on the Hilbert space constructed from the completion of $M$ with respect to the inner product $\langle x,y\rangle_{\tau}=\tau(xy^*)$.
The underlying space will be denoted by $L^2(M,\tau)$, or simply $L^2(M)$.
Consider a normal unital representation $\pi\colon M\to B(H)$ with both $M$ and $H$ separable.
There exists an isometry $u\colon H\to L^2(M)\otimes l^2(\mathbb{N})$, which commutes with the actions of $M$:
\begin{center}
$u\circ\pi(x)=(\lambda(x)\otimes\id_{l^2(\mathbb{N})} )\circ u$, $\forall x\in M$,
\end{center}
where $\lambda\colon M\mapsto L^2(M)$ denotes the left action.
Then $p=uu^*$ is a projection in $B(L^2(M)\otimes l^2(\mathbb{N}))$ such that $H\cong p( L^2(M)\otimes l^2(\mathbb{N}))$.
We have the following result (see \cite{APintrII1} Chapter 8).
\begin{proposition}\label{ptrdim}
The correspondence $H\mapsto p$ above defines a bijection between the set of equivalence classes of left $M$-modules and the set of equivalence classes of projections in $(M'\cap B(L^2(M)))\otimes B(l^2(\mathbb{N}))$.
\end{proposition}
The {\it von Neumann dimension} of the $M$-module $H$ are defined to be $(\tau\otimes \Tr)(p)$ and denoted by $\dim_M(H)$, which takes its value in $[0,\infty]$.
We have:
\begin{enumerate}
\item $\dim_M(\oplus_i H_i)=\sum_i \dim_M(H_i)$.
\item $\dim_M(L^2(M))=1$.
\end{enumerate}
Note $\dim_M(H)$ depends on the trace $\tau$.
If $M$ is a finite factor, i.e., $Z(M)\cong\mathbb{C}$, there is a unique normal tracial state (see \cite{J83,MvN36}) and we further have:
\begin{enumerate}
\setcounter{enumi}{2}
\item $\dim_M(H)=\dim_M(H')$ if and only if $H$ and $H'$ are isomorphic as $M$-modules (provided $M$ is a factor).
\end{enumerate}
When $M$ is not a factor, there is a $Z(M)$-valued trace which determines the isomorphism class of an $M$-module (see \cite{Bek04}).
In the following sections, we will consider the group von Neumann algebra $\mathcal{L}(\Gamma)$ with the canonical trace $tr(x)=\langle x\delta_e,\delta_e \rangle$.
Hence the von Neumann dimension of $\mathcal{L}(\Gamma)$ is the one uniquely determined by this trace.
Note a discrete group $\Gamma$ is called an infinite conjugacy class (ICC) group if every nontrivial conjugacy class $C_{\gamma}=\{g\gamma g^{-1}|g\in \Gamma\}$, $\gamma\neq e$, is infinite.
It is well-known that $\mathcal{L}(\Gamma)$ is a $\rm{II}_1$ factor if and only if $\Gamma$ is a nontrivial ICC group.
\iffalse
\begin{lemma}
The group von Neumann algebra $\mathcal{L}(\Gamma)$ is a $\rm{II}_1$ factor if and only if $\Gamma$ is a nontrivial ICC group.
\end{lemma}
\fi
Now we consider the case that $\Gamma$ is a discrete subgroup of a locally compact unimodular type I group $G$.
Let $\mu$ be a Haar measure of $G$. A measurable set $D\subset G$ is called a {\it fundamental domain} for $\Gamma$ if $D$ satisfies $\mu(G\backslash \cup_{\gamma\in\Gamma}\gamma D)=0$ and
$\mu(\gamma_1 D\cap \gamma_2 D)=0$ if $\gamma_1\neq \gamma_2$ in $\Gamma$.
In this section, we always assume $\Gamma$ is a lattice, i.e., $\mu(D)<\infty$.
The measure $\mu(D)$ is called {\it covolume} of $\Gamma$ and will be denoted by $\covol(\Gamma)$.
Note the covolume depends on the Haar measure $\mu$ (see Remark \ref{rcovol}).
There is a natural isomorphism $L^2(G)\cong l^2(\Gamma)\otimes L^2(D,\mu)$ given by
\begin{center}
$\phi\mapsto \sum_{\gamma\in\Gamma}\delta_{\gamma}\otimes \phi_{\gamma}$ with $\phi_{\gamma}(z)=\phi(\gamma\cdot z)$,
\end{center}
where $z\in D$ and $\gamma\in \Gamma$.
The restriction representation $\rho_G|_{\Gamma}$ of $\Gamma$ is the tensor product of $\rho_{\Gamma}$ on $l^2(\Gamma)$ and the identity operator $\id$ on $L^2(D,\mu)$.
Hence the von Neumann algebra $\rho_G(\Gamma)''\cong \mathcal{L}(\Gamma)\otimes \mathbb{C}=\mathcal{L}(\Gamma)$, which will be denoted by $M$ throughout this section.
Please note $L^2(M)=l^2(\Gamma)$.
Suppose $X$ is a measurable subset of
$\widehat{G}$ with the Plancherel measure $\nu(X)<\infty$.
Define
\begin{center}
$H_X=\int_{X}^{\oplus}H_{\pi}d\nu(\pi)$,
\end{center}
which is the direct integral of the spaces $H_{\pi}$ with $\pi\in X$.
Suppose $\{e_k(\pi)\}_{k\geq 1}$ is an orthonormal basis of its underlying Hilbert space $H_{\pi}$.
We have the following natural $G$-equivariant isometric isomorphism from $H_X$ to a subspace of $L^2(G)$:
\begin{equation}\label{eHXsub}
\begin{aligned}
H_X~~~~~~~~~&\cong ~~~\int_{X}^{\oplus}H_{\pi}\otimes e_1(\pi)^*d\nu(\pi)\\
v=\int_{X}^{\oplus}v(\pi)d\nu(\pi)~~&\mapsto~~~ \int_{X}v(\pi)\otimes e_1(\pi)d\nu(\pi),
\end{aligned}
\end{equation}
Therefore we will not distinguish these two spaces and denote them both by $H_X$.
Consider the projection $P_X\colon L^2(G)\to H_X$ defined on a dense subspace of $L^2(G)$ as follows:
\begin{center}
{\small
$\int_{\widehat{G}}^{\oplus}\left(\sum\limits_{i,j\geq 1}a_{i,j}(\pi)e_j(\pi)\otimes e_i(\pi)^*\right)d\nu(\pi)\mapsto \int_{\widehat{X}}^{\oplus}\left(\sum\limits_{j\geq 1}a_{1,j}(\pi)e_j(\pi)\otimes e_1(\pi)^*\right)d\nu(\pi)$
},
\end{center}
where all but finite $a_{i,j}(\pi)\in \mathbb{C}$ are zero for each $\pi$.
Observe that $P_X$ commutes with the right regular representation $\rho$, i.e., $\rho\circ P_X=P_X\circ \rho$.
We have a unitary representation of $G$ on $H_X$ and denote it by $\rho_X$.
Given two vectors $v=\int_{X}^{\oplus}v(\pi)d\nu(\pi)$ and $w=\int_{X}^{\oplus}w(\pi)d\nu(\pi)$ in $H_X$ with $v(\pi),w(\pi)\in H_{\pi}$, we have $v(\pi)\otimes w(\pi)^{*}\in H_{\pi}\otimes H_{\pi}^*$.
The corresponding Hilbert-Schmidt operator will also be denoted by $v(\pi)\otimes w(\pi)^{*}$ (see Remark \ref{rtenHS}).
We define a function on $G$ by
\begin{center}
$C_{v,w}(g)=\langle \rho_X(g^{-1})v,w\rangle_{H_X}=\int_{X}\Tr(\pi(g)^{*}v(\pi)\otimes w(\pi)^*)d\nu(\pi)$.
\end{center}
\begin{lemma}
For $v,w\in H_X$, we have $C_{v,w}\in L^2(G)$.
Moreover,
\begin{center}
$\langle C_{v_1,w_1},C_{v_2,w_2}\rangle_{L^2(G)}=\int_{X}\Tr(v_1(\pi)\otimes w_1(\pi)^{*}\cdot w_2(\pi)\otimes v_2(\pi)^{*})d\nu(\pi)$,
\end{center}
for any vectors $v_1,v_2,w_1,w_2\in H_X$.
\end{lemma}
\begin{proof}
Since $v_i(\pi)\otimes w_i(\pi)^*$ is Hilbert-Schmidt, $v_1(\pi)\otimes w_1(\pi)^{*}\cdot w_2(\pi)\otimes v_2(\pi)^{*}$ is a trace-class operator.
The integral is bounded by the norm of $v,w$ in $H_X$ and hence well-defined.
As $v(\pi)=w(\pi)=0$ for $\pi \in \widehat{G}- X$, we have
\begin{center}
$C_{v,w}(g)=\int_{X}\Tr(\pi(g)^*v(\pi)\otimes w(\pi)^*)d\nu(\pi)=\int_{\widehat{G}}\Tr(\pi(g)^*v(\pi)\otimes w(\pi)^*)d\nu(\pi)$.
\end{center}
By Theorem \ref{tplancherel}, we conclude that $v\otimes w^*=\int_{X}v(\pi)\otimes w(\pi)^*d\nu(\pi)=\widehat{C_{v,w}}$, the Fourier transform of the function $C_{v,w}$ on $G$.
Thus we obtain
\begin{equation*}
\begin{aligned}
\langle C_{v_1,w_1},C_{v_2,w_2}\rangle_{L^2(G)}&=\int_{\widehat{G}}\Tr \left(\widehat{C_{v_1,w_1}}(\pi)\widehat{C_{v_1,w_1}}(\pi)^*\right)d\nu(\pi)\\
&=\int_{X}\Tr \left(\widehat{C_{v_1,w_1}}(\pi)\widehat{C_{v_1,w_1}}(\pi)^*\right)d\nu(\pi)\\
&=\int_{\widehat{X}}\Tr \left(v_1(\pi)\otimes w_1(\pi)^{*}\cdot w_2(\pi)\otimes v_2(\pi)^{*}\right)d\nu(\pi)\\
\end{aligned}
\end{equation*}
Then it is easy to get $C_{v,w}\in L^2(G)$.
\end{proof}
Let $\{d_k\}_{k\geq 1}$ be an orthonormal basis of $L^2(D)$.
Note the restricted representation $\rho_X|_{\Gamma}$ makes the Hilbert space $H_X$ an $M$-module, whose von Neumann dimension can be obtained as follows.
\begin{lemma}\label{ldimsumnorm}
With the assumption above, we have
\begin{center}
$\dim_M(H_X)=\sum_{k\geq 1}\|Pd_k\|_{H_X}^2$.
\end{center}
\end{lemma}
\begin{proof}
Let $u$ be the inclusion $H_X\to L^2(G)$.
We have $u^*u={\rm id}_{H_X}$ and $uu^*=P_X$.
Note $L^2(G)\cong L^2(M)\otimes L^2(D,dg)$, where $L^2(M)$ is the standard $M$-module and $L^2(D,dg)$ is regarded as a trivial $M$-module.
Thus, by definition (see Proposition \ref{ptrdim}), we know
\begin{center}
$\dim_M(H_X)=\Tr_{M'\cap B(L^2(G))}(P)$,
\end{center}
where $M'\cap B(L^2(G))=\{T\in B(L^2(G))|Tx=xT,~\forall x\in M\}$, the commutant of $M$ on $L^2(G)$.
On the right-hand side,
\begin{center}
$\Tr_{M'\cap B(L^2(G))}=\tr_{M'\cap B(L^2(M))}\otimes \Tr_{B(L^2(D))}$
\end{center}
is the natural trace on $M'$.
The commutant $M'$ is generated by the finite sums of the form
\begin{center}
$x=\sum_{\gamma\in \Gamma}\rho_{\gamma}\otimes a_{\gamma}$,
\end{center}
where $\rho_{\gamma}=J\lambda(\gamma)J\in M'\cap L^2(M)$ (here $J\colon L^2(M)\to L^2(M)$ is the conjugate linear isometry extended from $x\mapsto x^*$) and $a_{\gamma}$ is a finite rank operator in $B(L^2(D))$.
Let $d_m^*\otimes d_n$ denotes the operator $\xi \mapsto \langle d_m,\xi\rangle\cdot d_n$ on $L^2(D)$.
Then each $a_{\gamma}$ can be written as $a_{\gamma}=\sum_{m,n\geq 1}a_{\gamma,m,n}d_m^*\otimes d_n$ with $a_{\gamma,m,n}\in \mathbb{C}$ and all but finite terms of $a_{m,n}$ are trivial.
Thus we obtain
\begin{center}
$\Tr_{M'}(\rho_{\gamma}\otimes a_{\gamma})=\tr_M(\lambda_{\gamma})\sum_{m\geq 1}a_{\gamma,m,m}=\delta_{e}(\lambda)\Tr_{L^2(D)}(a_\gamma)$.
\end{center}
This is equivalent to say
\begin{center}
$\Tr_{M'}(x)=\Tr_{L^2(D)}(a_e)$.
\end{center}
Let $Q$ be the projection of $L^2(G)$ onto $L^2(D)\cong \mathbb{C}\delta_e\otimes L^2(D)$.
Then $\Tr_{L^2(D)}=\Tr_{L^2(G)}(QyQ)$ for $y\in B(L^2(D))$.
We have
\begin{equation}\label{eTrQxQ}
\Tr_{M'}(x)=\Tr_{L^2(D)}(a_e)=\Tr_{L^2(G)}(Qa_eQ)=\Tr_{L^2(G)}(QxQ)
\end{equation}
As $P$ is a strong limit of elements that have the same form as $x$ above and the traces are normal, formula (\ref{eTrQxQ}) holds for $x=P$ and we obtain
\begin{equation*}
\begin{aligned}
\dim_M(H_X)&=\Tr_{M'}(P)=\Tr_{L^2(G,dg)}(QPQ)\\
&=\sum\nolimits_{k\geq 1}\langle QPQd_k,d_k\rangle_{L^2(G)}=\sum\nolimits_{k\geq 1}\langle Qd_k,PQd_k\rangle_{L^2(G)}\\
&=\sum\nolimits_{k\geq 1}\langle d_k,Pd_k\rangle_{L^2(G)}=\sum\nolimits_{k\geq 1}\langle Pd_k,Pd_k\rangle_{L^2(G)}\\
&=\sum\nolimits_{k\geq 1}\langle Pd_k,Pd_k\rangle_{H_X}=\sum\nolimits_{n\geq 1}\|Pd_k\|_{H_X}^2
\end{aligned}
\end{equation*}
\end{proof}
\section{A product formula}\label{sprod}
\begin{theorem}\label{tdimmeas}
Let $G$ be a locally compact unimodular type I group with Haar measure $\mu$.
Let $\nu$ be the Plancherel measure on the unitary dual $\widehat{G}$ of $G$.
Suppose $\Gamma$ is a lattice in $G$ and $\mathcal{L}(\Gamma)$ is the group von Neumann algebra of $\Gamma$.
Let $X\subset\widehat{G}$ such that $\nu(X)<\infty$ and $H_X=\int_X^{\oplus} H_{\pi}d\nu(\pi)$.
We have
\begin{center}
$\dim_{\mathcal{L}(\Gamma)}(H_X)=\covol(\Gamma)\cdot \nu(X)$.
\end{center}
\end{theorem}
\begin{proof}
We take a vector $\eta=\int_{X}^{\oplus}\eta(\pi)d\nu(\pi)$ in $H_X$ such that $\|\eta(\pi)\|_{H_{\pi}}^2=\frac{1}{\nu(X)}$ almost everywhere in $X$.
Then $\eta$ is a unit vector in $H_X$ and also in $L^2(G)$.
Observe $\{\delta_{\gamma}\otimes d_k\}_{\gamma\in \Gamma,n\geq 1}$ is an orthogonal basis of $L^2(G,\mu)$ via the isomorphism $L^2(G)\cong l^2(\Gamma)\otimes L^2(D,\mu)$.
Note $\{d_k\}_{k\geq 1}$ can be regarded as functions in $L^2(G)$ with support in $D$. We have $\delta_{\gamma}\otimes d_k=\rho_G(\gamma)d_k$ for $k\geq 1$ and $\gamma\in \Gamma$.
Hence we have
\begin{center}
$1=\|\rho_X(g)\eta\|_{H_X}^2=\|\rho(g)\eta\|_{L^2(G)}^2=\sum_{\gamma\in \Gamma,k\geq 1}|\langle \rho(g)\eta,\rho(\lambda)d_k\rangle_{L^2(G)}|^2$.
\end{center}
Consequently, we obtain:
\begin{equation*}
\begin{aligned}
{\rm covol}(\Gamma)&=\int_{D}1d\mu(g)=\int_{D}\sum_{\gamma\in \Gamma,k\geq 1}|\langle \rho(\lambda)^{*}\rho(g)\eta,d_k\rangle|^2d\mu(g)\\
&=\sum_{k\geq 1}\int_{G}|\langle P\rho(g)\eta,d_k\rangle_{L^2(G)}|^2d\mu(g)=\sum_{k\geq 1}\int_{G}|\langle \rho(g)\eta,Pd_k\rangle_{H_X}|^2d\mu(g)\\
&=\sum_{k\geq 1}\int_{G}|\langle \rho(g)^*Pd_k,\eta\rangle|^2d\mu(g)=\sum_{k\geq 1}\langle C_{Pd_k,\eta},C_{Pd_k,\eta}\rangle_{L^2(G)}\\
&=\sum_{k\geq 1}\int_X \Tr( (Pd_n)(\pi)\otimes \eta(\pi)^*\cdot (\eta(\pi)\otimes(Pd_k)(\pi)^*)d\nu(\pi)\\
&=\sum_{k\geq 1}\int_X \langle (Pd_n)(\pi)\otimes \eta(\pi)^*,(Pd_n)(\pi)\otimes \eta(\pi)^*\rangle_{H_{\pi}\otimes H_{\pi}^*}d\nu(\pi)\\
&=\sum_{k\geq 1}\int_{X}\|\eta(\pi)\|_{H_\pi}^2\cdot \|(Pd_k)(\pi)\|_{H_{\pi}}^2d\nu(\pi)\\
&=\frac{1}{\nu(X)}\sum_{n\geq 1}\|Pd_k\|_{H_X}^2,
\end{aligned}
\end{equation*}
which is $\dim_M(H_X)\cdot \nu(X)^{-1}$ by Lemma \ref{ldimsumnorm}.
Hence we get $\dim_M(H_X)=\covol(\Gamma)\cdot \nu(X)$.
\end{proof}
\begin{remark}\label{rcovol}
\begin{enumerate}
\item If $\mu'=k\cdot \mu$ is another Haar measure on $G$ for some $k>0$, the covolumes are related by $\covol'(\Gamma)=\mu'(G/\Gamma)=k'\cdot\mu(G/\Gamma)=k\cdot\covol(\Gamma)$.
But the induced Plancherel measure $\nu'=k^{-1} \cdot\nu$ and the dependencies cancel out in the formula above.
\item The conditions of $G$ above are satisfied for all connected semisimple Lie groups and connected reductive algebraic groups over local fields and adelic rings (see Section \ref{sPadele}).
\end{enumerate}
\end{remark}
If $\pi$ is an atom in $\widehat{G}$, i.e., $\nu(\{\pi\})>0$,
we can show $\pi$ is a discrete series representation and $\nu(\{\pi\})$ is just the formal dimension of $\pi$ \cite{DiCalg,Ro}.
Under this assumption, if $G$ is a real Lie group that has discrete series and $\Gamma$ is an ICC group, the theorem reduces to the special case of a single representation (see \cite{GHJ} Theorem 3.3.2)
\begin{center}
$\dim_{\mathcal{L}(\Gamma)}(H_{\pi})=\covol(\Gamma)\cdot d_{\pi}$.
\end{center}
This is motivated by the study of discrete series of Lie groups by M. Atiyah and W. Schmid \cite{ASds77}.
(See also \cite{Ruthprod19} for a discussion of the $p$-adic $\GL(2)$ together with its cuspidal and Steinberg representations.)
It is known that a Lie group $G$ has discrete series if and only if $\rk G=\rk K$ for a maximal compact subgroup $K$.
Thus it excludes a large family of Lie groups, e.g, $\SL_n(\mathbb{R})$ ($n\geq 2$), $\SO(n,\mathbb{C})$, $\GL_n(\mathbb{C})$.
A natural question is what is the analog for other Lie groups or other irreducible representations that are not discrete series.
This is one of the motivations for this result.
Before closing this section, I would like to introduce an example of $\SL(2,\mathbb{R})$ as an application of Theorem \ref{tdimmeas}.
\begin{example}
Consider $G=\SL(2,\mathbb{R})$ and $\mathcal{H}=\{x+iy|x,y\in \mathbb{R},y>0\}$, the Poincar\'{e} upper-half plane with a $\SL(2,\mathbb{R})$-invariant measure $y^{-2}dxdy$.
Observe that $\PSL(2,\mathbb{R})/\{\SO(2)/\{\pm 1\}\}\cong \mathcal{H}$.
We fix a Haar measure on $\SO(2)/\{\pm 1\}$ with total measure $1$ and then take the Haar measure $\mu$ on $\PSL(2,\mathbb{R})$ to be the product measure on $\mathcal{H}\times \{\SO(2)/\{\pm 1\}\}$.
Let $\nu$ be the Plancherel measure given by $\mu$.
The classification of irreducible unitary representations of $\SL(2,\mathbb{R})$ is well-known (see \cite{Gel75}).
However, the support of the Plancherel measure $\nu$ on the unitary dual $\widehat{\SL(2,\mathbb{R})}$ contains only the following two families of irreducible representations (see \cite{Fo2,Kn}).
\begin{enumerate}
\item {\it Discrete series representations}: $\{\pi_n^{\pm}|n\geq 2\}$.
The underlying space of $\pi_n^+$ is given as
\begin{center}
$H_n^+=\{f\colon \mathcal{H}\to\mathbb{C}\text{ holomorphic}~|\int_{\mathcal{H}}|f(x+iy)|^2y^{n-2}dxdy<\infty\}$.
\end{center}
For $g=\begin{pmatrix}
a & b\\
c & d
\end{pmatrix}\in \SL(2,\mathbb{R})$ and $f\in H_n^+$, the action is given by $(\pi_k^+(g)f)(z)=(bz+d)^{-k}f\left(\frac{az+b}{cz+d}\right)$.
The representation $\pi_n^-$ is defined to be the contragradient of $\pi_n^+$.
We have $\nu(\{\pi_n^+\})=\nu(\{\pi_n^-\})=\frac{n-1}{4\pi}$ for $n\geq 2$.
\item {\it Principal series representations}: $\{\pi^{\pm}_{it}|t>0\}$.
The spaces $\{H_{it}^{\pm}\}_{t>0}$ for them are all $L^2(\mathbb{R})$ with distinct actions given by
\begin{center}
$\pi^{+}_{it}\begin{pmatrix}
a & b\\
c & d
\end{pmatrix}f(x)=|bx+d|^{-1-it}f\left(\frac{ax+c}{bx+d}\right)$,\\
$\pi^{-}_{it}\begin{pmatrix}
a & b\\
c & d
\end{pmatrix}f(x)={\rm sgn}(bx+d)|bx+d|^{-1-it}f\left(\frac{ax+c}{bx+d}\right)$.
\end{center}
Let $P=\{g=\begin{pmatrix}
a & b\\
0 & a^{-1}
\end{pmatrix}|a,b\in \mathbb{R}, a\neq 0\}$ with characters given by
$\chi_t^{\pm}\begin{pmatrix}
a & b\\
0 & a^{-1}
\end{pmatrix}=\varepsilon^{\pm}(a)|a|^{it}$,
where $\varepsilon^+(a)=1$ and $\varepsilon^{-}(a)=\rm{sgn}(a)$.
One can show $\pi^{\pm}_{it}=\Ind_P^G(\chi_t^{\pm})$.
We have $d\nu(\pi_t^+)=\frac{t}{8\pi}\tanh{\frac{\pi t}{2}}dt$ and $d\nu(\pi_t^-)=\frac{t}{8\pi}\coth{\frac{\pi t}{2}}dt$.
\end{enumerate}
For the lattice $\PSL(2,\mathbb{Z})$, we have $\covol(\Gamma)=\frac{\pi}{3}$.
By Theorem \ref{tdimmeas}, we list the dimensions of some modules over $\mathcal{L}(\PSL(2,\mathbb{Z}))$ (which is a factor of type ${\rm II}_1$).
\begin{enumerate}
\item (Discrete part) Let $X=\{\pi_{n_1}^{+},\dots,\pi_{n_k}^{+}\}$. We have $H_X=\oplus_{1\leq i\leq k}\pi^+_{n_i}$.
Hence
\centerline{ $\dim_{\mathcal{L}(\PSL(2,\mathbb{Z}))}H_X=\frac{\sum_{i=1}^{k}n_i}{12}-\frac{k}{12}$. }
Note the dimension above remains the same if we replace any $\pi_{n_i}^{+}$ by its dual $\pi_{n_i}^{-}$.
\item (Continuous part) Let $Y=\{\pi^{+}_{it}|t\in [a,b]\}$ for some $0<a<b$.
We have
$H_{Y}=\int_{[a,b]}^{\oplus}H_{\pi_{it}^{+}}d\nu(\pi_{it}^{+})$.
Hence
\centerline{$\dim_{\mathcal{L}(\PSL(2,\mathbb{Z}))}H_{Y}=\frac{1}{24}\int_a^b t\tanh{\frac{\pi t}{2}}dt$.}
\iffalse
We have
\begin{center}
$\dim_{\mathcal{L}(\Gamma)}H_{(a,b)}^+=x\ln(e^{2x}+1)-\frac{3x^2}{2}-\frac{\Li_2(-e^{-2x})}{2}|^b_a$, and\\
$\dim_{\mathcal{L}(\Gamma)}H_{(a,b)}^-=x\ln(|e^{2x}-1|)-\frac{3x^2}{2}-\frac{\Li_2(e^{-2x})}{2}|^b_a$.~~~~~~~
\end{center}
\fi
\end{enumerate}
\end{example}
\section{The dimension and the adelic Plancherel measure}\label{smain}
Let $G$ be a reductive group defined over a number field $F$.
Assume $G$ is of dimension $d$ over $F$ so that we are able to take a nonzero left-invariant $d$-form $\omega$ on $G$.
For each infinite place $v\in V_{\infty}$, $\omega$ determines a left-invariant different form $\omega_v$ on the smooth manifold $G(F_v)$.
This in turn gives a Haar measure $\mu_v$ of $G(F_v)$.
For each finite place $v\in V_f$, let $p_v\in \mathbb{Z}$ be the prime number lying under $v$.
Let $\mu_v$ be the Haar measure on $G(F_v)$, which is uniquely determined by the corresponding $d$-form $\omega_v$ on the $p_v$-adic analytic manifold $G(F_v)$.
The {\it Tamagawa measure}, denoted by $\mu_{G(\mathbb{A}_F)}$, or simply $\mu$, is defined to be the restricted product of the measures $\{\mu_v\}_{v\in V}$ on the adelic group $G(\mathbb{A}_F)=\prod_{v\in V_{\infty}}G(F_v)\times \prod'_{v\in V_f}G(F_v)$, which is independent of the choice of $\omega$ \cite{PR94}.
The volume of $G(\mathbb{A}_F)/G(F)$ (if it exists) is called the {\it Tamagawa number} of $G$ and denoted by $\tau_F(G)$, or simply $\tau(G)$.
In fact, for any finite extension of number fields $L/F$, we can prove $\tau_L(G)=\tau_F(G)$ \cite{Osttmgw84}.
It suffices to consider the case $F=\mathbb{Q}$.
The following result was conjectured by A. Weil and proved by R. Langlands \cite{Llds66Tama}, K. Lai \cite{Lai80Tama}, R. Kottwitz \cite{Kott88Tama} and V. Chernousov \cite{CnsvE8}.
\begin{theorem
\label{tWeilTama}
If $G$ is a simply connected semisimple group, then $\tau(G)=1$.
\end{theorem}
By fixing the Tamagama measure $\mu$ on $G(\mathbb{A}_F)$ as a Haar measure and the corresponding Plancherel measure $\nu$ on $\widehat{G(\mathbb{A}_F)}$, we have the following result.
\begin{theorem}\label{tmain1}
Let $G$ be a simply connected semisimple algebraic group over a number field $F$.
Let $X\subset \widehat{G(\mathbb{A}_F)}$ such that $\nu(X)<\infty$ and $H_X=\int_{\pi\in X}^{\oplus}H_{\pi}d\nu_{G(\mathbb{A}_F)}(\pi)$.
We have
\begin{center}
$\dim_{\mathcal{L}(G(F))}H_X=\nu_{G(\mathbb{A}_F)}(X)$.
\end{center}
\end{theorem}
\begin{proof}
By Theorem \ref{tdimmeas}, we have
\begin{center}
$\dim_{\mathcal{L}(G(F))}H_X=\covol(G(F))\cdot\nu_{G(\mathbb{A}_F)}(X)$.
\end{center}
Then it follows Theorem \ref{tWeilTama}.
\end{proof}
Here we give a criterion for the case when $\mathcal{L}(G(F))$ is a type $\text{II}_1$ factor, which is related to the local groups at some infinite places.
\begin{proposition}
If there is an infinite place $v$ such that $G(F_v)$ is center-free and connected,
$G(F)$ is an ICC group and hence $\mathcal{L}(G(F))$ is a $\rm{II}_1$ factor.
\end{proposition}
\begin{proof}
First, we observe $G(F)$ is dense in each local group $G(F_v)$ if the place $v$ is archimedean ($v\in V_{\infty}$).
Take $\gamma\in G(F)$ and consider its conjugacy class $C_\gamma=\{g\gamma g^{-1}|g\in G(F)\}$ in $G(F)$.
The map $\gamma\to C_{\gamma}$ defined on $G(F)$ can be extended to the group $G(F_v)$ by continuity.
If $C_{\gamma}$ is finite, $C_{\gamma}=\overline{C_\gamma}=\{g\gamma g^{-1}|g\in G(F_v)\}$, which can be denoted as $Y=\{\gamma,g_{1}\gamma g_{1}^{-1},\dots,g_{N}\gamma g_{N}^{-1}\}$.
Let $G(F_v)$ act on the set $Y$ by conjugation.
The stabilizer of $\gamma$ is $Z_\gamma(G(F_v))=\{g\in G(F_v)|g\gamma g^{-1}=\gamma\}$.
Note $N+1=|Y|=|G(F_v)x|=|G(F_v)/Z_\gamma(G(F_v))|$.
That is to say, $Z_\gamma(G(F_v))$ is a closed subgroup of $G(F_v)$ of finite index.
Therefore $Z_\gamma(G(F_v))=G(F_v)$ by the connectedness.
As $G(F_v)$ is center-free, $\gamma=e$ and $G(F)$ is an ICC group.
\end{proof}
\iffalse
\begin{example}[$\SL_2$]\label{esl2}
As shown in \cite{PR94} Chapter 5.3, it suffices to consider the differential form
\begin{center}
$\omega=\frac{1}{x}dx\wedge dy\wedge dt$, for $g=\begin{pmatrix}
x & y\\
z & t
\end{pmatrix}\in \SL_2$.
\end{center}
Let us denote the corresponding Haar measure on $\SL_2(\mathbb{Q}_p)$ by $\omega_p$.
Then $\omega_p(\SL_2(\mathbb{Z}_p))=1-p^{-2}$.
Let $\Sigma$ be the fundamental domain of $\SL_2(\mathbb{R})$ relative to $\SL_2(\mathbb{Z})$.
We can show $\omega_{\infty}(\Sigma)=\frac{\pi^2}{6}$.
Note the set $\Omega=\Sigma\times \prod \SL_2(\mathbb{Z}_p)$ is a fundamental domain in $\SL_2(\mathbb{A}_{\mathbb{Q}})$ relative to $\SL_2(\mathbb{Q})$. (\cite{PR94}, P163)
Therefore, $\tau(\SL_2)=\omega(\Sigma)\otimes \prod \omega_p(\SL_2(\mathbb{Z}_p))=1$.
\end{example}
\fi
\bibliographystyle{abbrv}
\typeout{}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 2,771 |
\section{Introduction}
Semi-abelian categories \cite{JMT02} extend the validity of classical homological algebras to non-abelian categories such as groups, Lie algebras, etc. Torsion theories, originally introduced for abelian categories, have been studied in semi-abelian categories (and other more general non-abelian settings) in \cite{BouGrn06} and \cite{CDT06}.
The category $Grpd(\mathbb{X})$ of internal groupoids in $\mathbb{X}$, which is semi-abelian when $\mathbb{X}$ is so, exhibits two examples of torsion theories given by the pairs:
\[ (Ab(\mathbb{X}) , Eq(\mathbb{X}))\quad \mbox{and} \quad (ConnGrpd(\mathbb{X}), Dis(\mathbb{X}))\] where $Ab(\mathbb{X})$, $Eq(\mathbb{X})$, $ConnGrpd(\mathbb{X})$ and $Dis(\mathbb{X})$ are, respectively, the subcategories of $Grpd(\mathbb{X})$ of internal abelian groups, internal equivalence relations, connected groupoids and discrete internal groupoids (\cite{BouGrn06}, \cite{EveGrn10}).
For the case of $\mathbb{X}=Grp$ the category of groups, the category of internal groupoids is equivalent to the category $\mathbb{X} Mod$ of Whitehead's crossed modules, thus the torsion theories of internal groupoids correspond to the torsion theories in crossed modules: (abelian groups, inclusion of normal monomorphisms) and (central extension in groups, groups as discrete crossed modules).
In \cite{Lop22b}, the examples of torsion theories of internal groupoids in groups have been expanded to a linearly ordered lattice $\mu(Grp)$ of torsion theories in the category $Simp(Grp)$ of simplicial groups. In such a way that the categories $Dis(Grp)$, $Eq(Grp)$ and $Grpd(Grp)$ are torsion-free subcategories of $Simp(Grp)$. On the other hand, the categories $Ab(Grp)$ and $ConnGrpd(Grp)$ are not torsion subcategories of $Simp(Grp)$. In this work we will show that when we restrict the torsion theories of $\mu(Grp)$ to the subcategory $\mathcal{M}_{n \geq}$ of simplicial groups with truncated Moore complex above $n$. First, we obtain that the corresponding lattice $\mu(\mathcal{M}_{n \geq})$ has a minimal element where the torsion category is equivalent to $Ab$ the category of abelian groups. Second, that the following (the second smallest) torsion category can be characterized using central extensions in groups, in a similar way as the torsion category $ConnGrpd(Grp)$ of $Grpd(Grp)$ is equivalent to surjective crossed modules.
The subcategory $\mathcal{M}_{n \geq}$ is a Birkhoff subcategory of $Simp(Grp)$, this means a regular epireflective subcategory closed under subobjects and quotients in $Simp(Grp)$. Birkhoff subcategories of a semi-abelian category are again semi-abelian. As a second case study, we prove that the subcategory of Dakin's $T$-group complexes is also a Birkhoff subcategory of $Simp(Grp)$, and hence, it is semi-abelian. In order to study torsion thoeries in $T$-group complexes, it will be easier to work to the equivalent category $Crs(Grp)$ of Ashley's reduced crossed complexes. Our interest in reduced crossed complexes lies in their `similar behaviour' as chain complexes. In particular, torsion theories in $Crs(Grp)$ present examples of TTF theories in a weak sense. Used mainly in categories of modules over rings (\cite{jans65}), a torsion torsion-free theory (or TTF theory for short) in an abelian category $\mathbb{X}$ is a triplet of full subcategories $(\mathcal{C}, \mathcal{T}, \mathcal{F})$ such that $(\mathcal{C}, \mathcal{T})$ and $(\mathcal{T}, \mathcal{F})$ in $\mathbb{X}$ are torsion theories in $\mathbb{X}$.
In section 2, we recall basic facts of torsion theories in semi-abelian categories, as well as, the definition of the lattice $\mu(Grp)$ of torsion theories in $Simp(Grp)$. Section 3 and 4 studies torsion theories in the subcategories $\mathcal{M}_{n \geq}$ and $Crs(Grp)$, respectively. Section 5, studies how reduced crossed complexes present examples of TTF theories in a weak sense. To this end we introduce, on one hand, CTF-theories, subcategories that are torsion-free and mono-coreflective. On the other hand, $\mathcal{E}$-torsion theories, torsion theories relative to a particular class of objects $\mathcal{E}$ of $\mathbb{X}$.
\section{Torsion theories in simplicial groups with truncated Moore complex}
A category $\mathbb{X}$ is called semi-abelian \cite{JMT02} if it is pointed with binary coproducts, Barr exact and Bourn protomodular.
\begin{defi} A torsion theory in a semi-abelian category $\mathbb{X}$ is a pair $(\mathcal{T}, \mathcal{F})$ of full and replete subcategories of $\mathbb{X}$ such that:
\begin{enumerate}
\item[TT1] A morphism $f:T \to F$ with $T$ in $\mathcal{T}$ and $F$ in $\mathcal{F}$ is a zero morphism.
\item[TT2] For any object $X$ in $\mathbb{X}$ there is a short exact sequence:
\begin{equation}\label{sestt}
\begin{tikzcd} 0\ar[r] &T_X\ar[r, "\epsilon_X"] & X \ar[r, "\eta_X"] & F_X \ar[r]& 0\end{tikzcd}
\end{equation}
with $T_X$ in $\mathcal{T}$ and $F_X$ in $\mathcal{F}$.
\end{enumerate}
\end{defi}
In a torsion theory $(\mathcal{T}, \mathcal{F})$, $\mathcal{T}$ is called the torsion category and its objects are called torsion objects, in a similar way, $\mathcal{F}$ is the torsion-free category. Any subcategory is called torsion (torsion-free resp.) if it is the torsion (torsion-free resp.) category of a torsion theory. The torsion category $\mathcal{T}$ is a normal mono-coreflective subcategory of $\mathbb{X}$, i.e., the inclusion $J:\mathcal{T} \to \mathbb{X}$ has a right adjoint $T: \mathbb{X} \to \mathcal{T}$ and each component of the counit $\epsilon_X: JT(X) \to X$ is a normal monomorphism (a kernel of some arrow in $\mathbb{X}$). Similarly, $\mathcal{F}$ is a normal epi-reflective subcategory of $\mathbb{X}$, i.e., the inclusion $I$ has a left adjoint $F$ and each component $\eta_X:X \to IF(X)$ is a normal epimorphism:
\begin{equation}\label{ttadj}
\begin{tikzcd}[column sep=large] \mathcal{T} \ar[r, bend left, "J"]\ar[r, phantom, "\perp"] & \mathbb{X} \ar[r, bend left, "F"]\ar[l, bend left,"T"]\ar[r, phantom, "\perp"] & \mathcal{F} \ar[l, bend left, "I"] \end{tikzcd}.
\end{equation}
Then, it is easy to observe that $\mathcal{T}$ is closed under colimts in $\mathbb{X}$ (those that exist in $\mathbb{X}$) and $\mathcal{F}$ is closed under limits in $\mathbb{X}$. Both $\mathcal{T}$ and $\mathcal{F}$ are closed under extensions in $\mathbb{X}$, this means that given a short exact sequence in $\mathbb{X}$:
\[\begin{tikzcd} 0\ar[r] & A\ar[r] & X\ar[r] & B\ar[r]& 0 \end{tikzcd}\]
with $A$ and $B$ in $\mathcal{T}$ (resp. in $\mathcal{F}$) then $X$ is in $\mathcal{T}$ (resp. in $\mathcal{F}$). A torsion theory $(\mathcal{T}, \mathcal{F})$ is called \textit{hereditary} if $\mathcal{T}$ is closed under subobjects in $\mathbb{X}$, i.e., given a monomorphism $m:M \to T$ with $T$ in $\mathcal{T}$ then $M$ is also torsion. Conversely, a torsion theory is called \textit{cohereditary} if $\mathcal{F}$ is closed under quotients in $\mathbb{X}$, i.e., given a normal epimorphism $q: F \to Q$ with $F$ in $\mathcal{F}$ then $Q$ is also in $\mathcal{F}$. It is useful to recall the following result.
\begin{prop}\label{semiab} \cite{Lop22b} Let $(\mathcal{T}, \mathcal{F})$ be a torsion theory of a semi-abelian category $\mathbb{X}$, if $(\mathcal{T},\mathcal{F})$ is hereditary then $\mathcal{T}$ is a semi-abelian category. Accordingly, if $(\mathcal{T}, \mathcal{F})$ is cohereditary then $\mathcal{F}$ is a semi-abelian category.
\end{prop}
Given two torsion theories $(\mathcal{T}, \mathcal{F})$ and $(\mathcal{S}, \mathcal{G})$ in $\mathbb{X}$, we have that $\mathcal{T} \subseteq \mathcal{S}$ if and only if $\mathcal{G} \subseteq \mathcal{F}$. This allows to introduce a partial order in the class of torsion theories in a category $\mathbb{X}$ by $(\mathcal{T}, \mathcal{F}) \leq (\mathcal{S}, \mathcal{G})$ if $\mathcal{T} \subseteq \mathcal{S}$. There is a bottom element and a top element given by the trivial torsion theories $0:= (0, \mathbb{X})$ and $\mathbb{X}:=(\mathbb{X},0)$.
The simplicial category $\Delta$ has, as objects, finite ordinals $[n]=\{0,1,\dots n\}$ and as morphisms monotone functions. A simplicial group is a functor $X: \Delta^{op} \to Grp$, we will denote $Simp(Grp)$ for the category of simplicial groups. A simplicial group $X$ can be equivalently defined by the following data:
a family of objects $\{X_n\}_{n\in \mathbb{N}}$ in $\mathbb{X}$, the \textit{face morphisms} $d_i: X_n\to X_{n-1}$ and the \textit{degeneracies morphisms} $s_i: X_n \to X_{n+1}$:
\[X= \begin{tikzcd}
\dots X_n \ar[r, "\vdots", phantom, shift left]
\ar[r, "d_0"', shift right=8]
\ar[r, "d_n", shift left=8]
& \dots \ar[r, "\vdots", phantom, shift left]
\ar[r, "d_0"', shift right=8]
\ar[r, "d_4", shift left=8]
\ar[l, "s_{n-1}"', shift right=4, shorten=1.5mm ]
\ar[l,"s_0", shift left=4, shorten=1.5mm ]
& X_3 \ar[r,"d_0"' ,shift right=12]
\ar[r, , "d_1"', shift right=4]
\ar[r, "d_2"' ,shift left=4]
\ar[r, "d_3",shift left=12]
\ar[l, "s_3"', shift right=4, shorten=1.5mm ]
\ar[l,"s_0", shift left=4, shorten=1.5mm ]
& X_2 \ar[r, "d_1"']
\ar[r, "d_2", shift left=8]
\ar[r, "d_0"', shift right=8]
\ar[l, "s_1", shorten=1.5mm]
\ar[l, "s_2"', shorten=1.5mm,shift right=8]
\ar[l, "s_0", shorten=1.5mm, shift left=8]
& X_1 \ar[r, "d_1", shift left=4]
\ar[r, "d_0"', shift right=4]
\ar[l,"s_1"', shift right=4, shorten=1.5mm ]
\ar[l,"s_0", shorten=1.5mm ,shift left=4 ]
& X_0 \ar[l, "s_0", shorten=1.5mm]
\end{tikzcd}\]
satisfying the \textit{simplicial identities}:
\begin{align*}
d_i d_j &= d_{j-1} d_i \quad \mbox{if} \quad i<j \\
s_i s_j &= s_{j+1} s_i \quad \mbox{if} \quad i\leq j \\
d_i s_j &=
\left\{\begin{array}{cc}
s_{j-1} d_i & \mbox{if} \quad i<j \\
1 & \mbox{if} \quad i=j\, \mbox{or}\,\ i=j+1 \\
s_j d_{i-1} & \mbox{if} \quad i>j+1\, .
\end{array}\right.
\end{align*}
Since $Simp(Grp)$ is a functor category with codomain a semi-abelian category, it is semi-abelian.
\begin{strc} An $n$-truncated simplicial group $X$ is a simplicial group where the objects $X_i$, the face morphisms $d_i$ and degeneracies morphisms $s_i$ are defined up to level $n$ and they satisfy the simplicial identities (those that make sense). Let $Simp_n(Grp)$ be the category of $n$-truncated simplicial groups then there is a truncation functor:
\[tr_n: \begin{tikzcd} Simp(Grp)\ar[r] & Simp_n(Grp) \end{tikzcd}\]
which simply forgets everything above degree $n$. For all $n$, the functor $tr_n$ has a left adjoint $sk_n$ called the $n$-skeleton, and a right adjoint $cosk_n$ named the $n$-coskeleton: $sk_n \dashv tr_n \dashv cosk_n$. We will write $Sk_n=sk_ntr_n$ and $Cosk_n=cosk_n tr_n$.
In addition, for $n=0$, the functor $sk_0$ has a left adjoint $\pi_0$ where $\pi_0(X)=coeq(d_0,d_1)$ is the coequalizer of the face morphisms $d_0,d_1:X_1\to X_0$. Moreover, the adjunctions $\pi_0 \dashv sk_0 \dashv tr_0 \dashv cosk_n$ correspond to the functors $\pi_0 \dashv Dis \dashv ()_0 \dashv Ind$. For a group $G$, $Dis(G)$ is the discrete simplicial group of $G$
\[\begin{tikzcd}
Dis(G)= & \dots\ar[r, shift left=3, "1"]\ar[r, shift right=3, "1"']\ar[r, phantom, "\vdots", shift left]
& G\ar[r, shift left=3, "1"]\ar[r, shift right=3, "1"']\ar[r, phantom, "\vdots", shift left]
& G\ar[r, shift left=3, "1"]\ar[r, shift right=3, "1"'] \ar[r, phantom, "\vdots", shift left]
& G\ar[r, shift left=3, "1"]\ar[r, shift right=3, "1"']
& G\ar[l, "1"] \end{tikzcd}\]
and $Ind(G)$ is the indiscrete simplicial group of $G$
\[\begin{tikzcd}
Ind(G)= & \dots\ar[r, shift left=3, "p_0"]\ar[r, shift right=3, "p_4"']\ar[r, phantom, "\vdots", shift left]
& G^4\ar[r, shift left=3, "p_0"]\ar[r, shift right=3, "p_3"']\ar[r, phantom, "\vdots", shift left]
& G^3\ar[r, shift left=3, "p_0"]\ar[r, shift right=3, "p_2"'] \ar[r, phantom, "\vdots", shift left]
& G^2\ar[r, shift left=3, "p_0"]\ar[r, shift right=3, "p_1"']
& G\ar[l, "s_0"] \end{tikzcd}\]
where $G^n$ is the $n$-fold product, $(X)_0=X_0$ and the morphisms $p_i$ are induced by product projections.
\end{strc}
A chain complex $M$ is a family of morphisms $\{\delta_n: M_n \to M_{n-1}\}_{n \in \mathbb{N}}$ such that $\delta_n \delta_{n+1}=0$ for all $n$. A chain complex $M$ is \textit{proper} if for each differential $\delta_n$ the monomorphism $m_n$ of the normal epi/mono factorization $(e_n, m_n)$ of $\delta_n$ is a normal monomorphism:
\[\begin{tikzcd} \dots\ar[r] & M_{n+1}\ar[r, "\delta_{n+1}"]& M_n\ar[rr, "\delta_n"]\ar[rd, "e_n"', two heads] & & M_{n-1}\ar[r, "\delta_{n-1}"]& \dots\, . \\ &&& \delta_n(M_n)\ar[ru, "m_n"', hook]&& \end{tikzcd}\]
The category of chain complexes and the subcategory of proper chain complexes in groups will be denoted as $chn(Grp)$ and $pch(Grp)$.
\begin{defi} The Moore normalization functor $N: Simp(Grp) \to chn(Grp)$ is defined as follows. Let $X$ be a simplicial group then $N(X)$ is the group chain complex
\[\begin{tikzcd} \dots\ar[r] & N(X)_n\ar[r, "\delta_n"] & N(X)_{n-1}\ar[r]& \dots\ar[r] & N(X)_1 \ar[r, "\delta_1"] & N(X)_0 \end{tikzcd} \]
such that $N(X)_0=X_0$ and
\[N(X)_n = \bigcap_{i=0}^{n-1} ker(d_i: X_n \to X_{x-1})\]
and differentials $\delta_n= d_n \circ \cap_i ker(d_i): N(X)_n \to N(X)_{n-1}$ for $n \geq 1$.
The Moore chain complex $N(X)$ of a simplicial group is proper. It is known that the functor $N$ preserves finite limits and normal epimorphisms, and also it is conservative.
We will write $\mathcal{M}_{n\geq}$ for the subcategory of simplicial groups with trivial Moore complex above degree $n$. Similarly, $\mathcal{M}_{\geq n}$ is the subcategory of simplicial groups with trivial Moore complex below degree $n$.
\end{defi}
In \cite{Lop22b} it is proved that $\mathcal{M}_{n \geq}$ is a torsion-free subcategory of $Simp(Grp)$, and respectively, $\mathcal{M}_{\geq n}$ is a torsion subcategory for all $n$. Indeed, the corresponding torsion theories form a linearly order lattice $\mu(Grp)$ in $Simp(Grp)$. We recall how these are defined and useful properties.
\begin{teo}\label{teolop} \cite{Lop22b} There is a linear order lattice $\mu(Grp)$ of torsion theories in $Simp(Grp)$:
\begin{multline*}
0 \leq \dots \leq \; \mu_{ n+1\geq} \; \leq \; \mu_{\geq n+1} \; \leq \; \mu_{n\geq} \; \leq \; \mu_{\geq n} \; \leq \dots \\ \dots \leq \; \mu_{\geq 2} \; \leq \; \mu_{1\geq} \; \leq \; \mu_{\geq 1} \; \leq \; \mu_{0\geq} \; \leq Simp(Grp)\, .
\end{multline*}
where
\begin{enumerate}
\item The torsion theory $\mu_{n \geq}$ is given by the pair $(Ker(Cot_n), \mathcal{M}_{n \geq})$, where $Cot_n:Simp(Grp) \to \mathcal{M}_{n \geq}$ (introduced in \cite{Por93}) is the left adjoint of the inclusion $i:\mathcal{M}_{n \geq} \to Simp(Grp)$ and $Ker(Cot_n)$ is the full subcategory of simplicial groups $X$ such that $Cot_n(X)=0$.
\item The torsion theory $\mu_{\geq n}$ is given by the pair $(\mathcal{M}_{\geq n}, \mathcal{F}_{tr_n})$, where $\mathcal{F}_{tr_n}$ is the subcategory of simplicial groups $X$ with $\eta_X$ monic where $\eta$ is the unit of the adjunction $tr_n \dashv cosk_n$.
\item Let $X$ be a simplicial group and $M$ its Moore complex. For $X$ the associated short exact sequence of $\mu_{n \geq}$ under normalization is the short exact sequence in chain complexes (written vertically):
\begin{equation}\label{sescok}
\begin{tikzcd}
\hdots \ar[r]
& M_{n+2} \ar[r, "\delta_{n+2}"] \ar[d, "id"]
& M_{n+1} \ar[r, "e_{n+1}", two heads] \ar[d, "id"]
& \delta_{n+1}(X_{n+1}) \ar[r] \ar[d, "m_{n+1}", hook]
& 0 \ar[r] \ar[d]
& \hdots
\\
\hdots \ar[r]
& M_{n+2} \ar[r, "\delta_{n+2}"] \ar[d]
& M_{n+1} \ar[r, "\delta_{n+1}"] \ar[d]
& M_{n} \ar[r, "\delta_{n}"] \ar[d, "cok(\delta_{n+1})", two heads]
& M_{n-1} \ar[r] \ar[d, "id"]
& \hdots
\\
\hdots \ar[r]
& 0 \ar[r]
& 0 \ar[r]
& Cok(\delta_{n+1}) \ar[r]
& M_{n-1} \ar[r]
& \hdots
\end{tikzcd}
\end{equation}
\item Let $X$ be a simplicial group and $M$ its Moore complex. For $X$ the associated short exact sequence of $\mu_{ \geq n}$ under normalization is the short exact sequence in chain complexes (written vertically):
\begin{equation}\label{sesker}
\begin{tikzcd}
\hdots \ar[r]
& M_{n+1} \ar[r] \ar[d, "id"]
& Ker(\delta_n) \ar[r] \ar[d, "ker(\delta_n)", hook]
& 0 \ar[r] \ar[d]
& 0 \ar[r] \ar[d]
& \hdots
\\
\hdots \ar[r]
& M_{n+1}\ar[r, "\delta_{n+1}"] \ar[d]
& M_n \ar[r, "\delta_n"] \ar[d, "e_n", two heads]
& M_{n-1} \ar[r, "\delta_{n-1}"] \ar[d, "id"]
& M_{n-2} \ar[r] \ar[d, "id"]
& \hdots
\\
\hdots \ar[r]
& 0 \ar[r]
& \delta(M_n)\ar[r, "m_n"]
& M_{n-1} \ar[r]
& M_{n-2} \ar[r]
& \hdots
\end{tikzcd}
\end{equation}
\end{enumerate}
\end{teo}
As shown form diagram (\ref{sescok}) and (\ref{sesker}) the coreflector of $\mathcal{M}_{\geq n}$ and respectively the reflector of $\mathcal{M}_{n\geq}$ behave, at the level of Moore complexes, as the truncation functors introduced by L. Illusie in \cite{ill71}. As a consequence, torsion/torsion-free objects can be characterized by their Moore complex as follows.
\begin{coro}\label{cotnsurj} \cite{Lop22b} Let $X$ be a simplicial group with Moore complex $M$, then
\begin{enumerate}
\item $X$ belongs in $Ker(Cot_n)$ if and only if $M$ is of the form
\[\begin{tikzcd}M=\dots\ar[r]& M_{n+2}\ar[r]& M_{n+1}\ar[r, "\delta_n", two heads] & M_n\ar[r]& 0\ar[r]& 0\ar[r]& \dots \end{tikzcd}\]
with $\delta_n$ a normal epimorphism.
\item $X$ belongs in $\mathcal{F}_{tr_n}$ if and only if $M$ is of the form
\[\begin{tikzcd}M=\dots\ar[r] & 0\ar[r] & 0\ar[r]& M_{n+1}\ar[r, "\delta_{n+1}", hook]& M_n\ar[r]& M_{n-1}\ar[r] &\dots \end{tikzcd}\]
with $\delta_{n+1}$ a normal monomorphism.
\end{enumerate}
\end{coro}
\section{Torsion subcategories of $\mathcal{M}_{n \geq}$}
The lattice $\mu(Grp)$
\[ 0 \leq \quad \dots \leq \quad \mu_{\geq 2} \quad \leq \quad \mu_{1\geq} \quad \leq \quad \mu_{\geq 1} \quad \leq \quad \mu_{0\geq} \quad \leq Simp(Grp)\, . \]
extends the torsion theories in internal groupoids $Grpd(Grp)$
\[0=(0, Grpd(Grp)) \leq \; (Ab,Eq(Grp) ) \; \leq \; (ConnGrpd(Grp) , Dis(Grp)) \; \leq \; Grpd(Grp)\]
in the sense that the torsion-free categories of the first 3 largest non-trivial torsion theories, $ \mu_{1 \geq} \leq \mu_{\geq 1} \leq \mu_{0 \geq}$, in $Simp(Grp)$ are the torsion-free categories ($Grpd(Grp)$, $Eq(Grp)$ and $Dis(Grp)$, respectively) of $Grpd(Grp)$.
On the other hand, the torsion categories of $Simp(Grp)$ are not as easily described as in the case of $Grpd(Grp)$. However, as in the case $Grpd(Grp)= \mathcal{M}_{1 \geq}$, some similarities arise when we restrict the lattice $\mu(Grp)$ to subcategories of the form $\mathcal{M}_{n \geq}$. To this end, we recall how simplicial groups are equivalent to chain complexes with operations following the work of Loday in \cite{Lod82}, D. Conduch\'e in \cite{Con84} and, P. Carrasco and A. M. Cegarra in \cite{CaCe91}.
\subsection{The case of $\mathcal{M}_{1 \geq}$ and crossed modules}
For completeness sake we quickly recall how torsion theories in internal groupoids correspond to torsion theories in Whitehead's crossed modules.
We will write group actions acting on the left ${}^b(a)$ and each group $G$ is consider to act on itself by conjugation as ${}^g(g')=gg'g^{-1}$.
\begin{defi} A \textit{crossed module} (in groups) is a morphism of groups $\delta: A \to B$ with a groups action $B \to Aut(A)$ such that:
\begin{enumerate}
\item $\delta({}^b(a))=b\delta(a)b^{-1} $ ($\delta$ is equivariant).
\item $ {}^{\delta(a)a'}=aa'a^{-1}$ (Peiffer identity).
\end{enumerate}
If $\delta$ only satisfy axiom 1 then it is called a \textit{precrossed module}. We will denote the category of crossed modules and of precrossed modules as $\mathbb{X} Mod$ and $P \mathbb{X} Mod$, respectively.
\end{defi}
From \cite{Lod82}, the following categories are equivalent:
\begin{enumerate}
\item The category $\mathcal{M}_{1 \geq}$ of simplicial groups with trivial Moore complex above degree 1.
\item The category $Grpd(Grp)$ of internal groupoids in groups.
\item The category $\mathbb{X} Mod$ of crossed modules in groups.
\item The category of \textit{Cat-1-groups}.
\end{enumerate}
Indeed, given a simplicial group $X$ with Moore complex $M=N(X)$, the differential $\delta_1: M_1 \to M_0$ is a precrossed module with the action given by conjugation with $s_0$. Furthermore, $\delta_1$ is a crossed module if and only if $M_i=0$ for $i>1$.
We will write $\mu(\mathcal{M}_{1 \geq})$ for the lattice given by the torsion theories of $\mu(Grp)$ restricted to $\mathcal{M}_{1 \geq}$, i.e., we consider the torsion theory $(\mathcal{T} \cap \mathcal{M}_{1 \geq},\mathcal{F}\cap \mathcal{M}_{1 \geq})$ for each element $(\mathcal{T}, \mathcal{F})$ of $\mu(Grp)$. We will write $\mu'_{n \geq}$ and $\mu'_{\geq n}$ for the corresponding restrictions of $\mu_{n \geq}$ and $\mu_{\geq n}$.
\begin{prop}\label{latm1} The lattice $\mu(\mathcal{M}_{1\geq})$ is given by
\[0 \quad \leq \quad \mu'_{\geq 1} \quad \leq \quad \mu'_{0 \geq} \quad \leq \quad \mathcal{M}_{1\geq} \, . \]
\end{prop}
\begin{proof} Recall that $\mu(Grp)$ is given by
\begin{equation}\label{mugrpdiag}
\begin{tikzcd}[column sep=scriptsize]
Simp(Grp)= & Simp(Grp) \ar[r, shift left=2.5]\ar[r, phantom, "\perp"]
& Simp(Grp) \ar[r, shift left=2.5]\ar[l, shift left=2]\ar[r, phantom, "\perp"]
& 0 \ar[l, shift left=2] \ar[d, hook] \\
\mu_{0\geq}= & Ker(Cot_0) \ar[r, shift left=2.5]\ar[r, phantom, "\perp"]\ar[u, hook]
& Simp(Grp) \ar[r, shift left=2.5]\ar[l, shift left=2.5]\ar[r, phantom, "\perp"]
& \mathcal{M}_{0 \geq} \cong Grp \ar[l, shift left=2.5] \ar[d, hook] \\
\mu_{\geq 1}= & \mathcal{M}_{\geq 1} \ar[r, shift left=2.5]\ar[r, phantom, "\perp"] \ar[u, hook]
& Simp(Grp) \ar[r, shift left=2.5]\ar[l, shift left=2.5]\ar[r, phantom, "\perp"]
& \mathcal{F}_{tr_0} \cong Eq(Grp) \ar[l, shift left=2.5] \ar[d, hook]\\
\mu_{1\geq}= & Ker(Cot_1) \ar[r, shift left=2.5]\ar[r, phantom, "\perp"]\ar[u, hook]
& Simp(Grp) \ar[r, shift left=2.5]\ar[l, shift left=2.5]\ar[r, phantom, "\perp"]
& \mathcal{M}_{1 \geq} \cong Grpd(Grp) \ar[l, shift left=2.5] \ar[d, hook] \\
\mu_{\geq 2}= & \mathcal{M}_{\geq 2} \ar[r, shift left=2.5]\ar[r, phantom, "\perp"] \ar[u, hook]
& Simp(Grp) \ar[r, shift left=2.5]\ar[l, shift left=2.5]\ar[r, phantom, "\perp"]
& \mathcal{F}_{tr_1} \ar[l, shift left=2.5] \ar[d, hook] \\
\mu_{2\geq}= & Ker(Cot_2) \ar[r, shift left=2.5]\ar[r, phantom, "\perp"]\ar[u, hook]
& Simp(Grp) \ar[r, shift left=2.5]\ar[l, shift left=2.5]\ar[r, phantom, "\perp"]
& \mathcal{M}_{2 \geq} \ar[l, shift left=2.5] \\
\dots & \dots & \dots & \dots
\end{tikzcd}
\end{equation}
Since $\mathcal{M}_{1 \geq}$ is itself a torsion-free subcategory comprised in the lattice $\mu(Grp)$, the restriction of each torsion theory below $\mu_{1 \geq}$ is the trivial torsion theory $(0, \mathcal{M}_{1 \geq})$ in $\mathcal{M}_{1 \geq}$.
\end{proof}
It is easy to observe that the torsion-free categories $Eq(Grp)$ and $Dis(Grp)$ in $Simp(Grp)$ are also torsion-free categories of $\mathcal{M}_{1 \geq}$. We can conclude the following.
\begin{prop} The lattice $\mu(\mathcal{M}_{1 \geq})$ corresponds to
\[0 \quad \leq \quad (Ab(Grp), Eq(Grp)) \quad \leq \quad (ConnGrpd(Grp), Dis(Grp)) \quad \leq \quad \mathcal{M}_{1 \geq}. \]
Moreover, under the Moore normalization this lattice corresponds to the lattice of torsion theories in $\mathbb{X} Mod$
\[0 \quad \leq \quad (Ab, Mono) \quad \leq \quad (Cext, Dis) \quad \leq \quad \mathbb{X} Mod\, ,\]
where
\begin{enumerate}
\item $Ab$ is the category of crossed modules of the form $A \to 0$ for an abelian group $A$.
\item $Mono$ is the category of crossed modules given by the inclusion of a normal subgroup $i: N \to G$.
\item $Cext$ is the category of crossed modules given by central extensions, epimorphisms $p: G \to Q$ with a $ker(p) \leq Z(G)$.
\item $Dis$ is the category of crossed modules given of the form $0 \to G$ for a group $G$.
\end{enumerate}
\end{prop}
\begin{proof} In a torsion theory the torsion and the torsion-free category determine each other uniquely, so if $Eq(Grp)$ and $Dis(Grp)$ are torsion-free categories of $\mu'_{0 \geq}$ and $\mu'_{\geq 1}$ the torsion categories must correspond accordingly to $Ab(Grp)$ and $ConnGprd(Grp)$.
The second statement is well-known, for instance the equivalences are mentioned in \cite{BouGrn06}, \cite{EveGrn10} and \cite{Man15}.
\end{proof}
\subsection{The case of $\mathcal{M}_{n \geq}$}
Following the previous case, we will write $\mu(\mathcal{M}_{n \geq})$ for the restriction of $\mu(Grp)$ to $\mathcal{M}_{n \geq}$ and $\mu'_{i \geq}$, $\mu'_{\geq i}$ for the restrictions of the torsion theories $\mu_{ i \geq}$, $\mu_{\geq i}$.
Similar to Proposition \ref{latm1}, also from the diagram (\ref{mugrpdiag}) we have the following result.
\begin{prop}\label{latmn} The lattice $\mu(\mathcal{M}_{n\geq})$ is given by
\[0 \quad \leq \quad \mu'_{\geq n} \quad \leq \quad \mu'_{n-1 \geq} \quad \leq \quad \mu'_{\geq n-1} \quad \leq \dots \leq \quad \mu'_{\geq 1} \quad \leq \quad \mu'_{0 \geq} \quad \leq \quad \mathcal{M}_{n\geq}. \]
\end{prop}
Just as in $\mathcal{M}_{1 \geq}$ the subcategories $Dis(Grp)$, $Eq(Grp)$ and $Grpd(Grp)$ are torsion-free subcategories of $\mathcal{M}_{n \geq}$. In order to characterise the torsion categories with recall some categories introduced by D. Conduch\'e.
\begin{defi} \cite{Con84} A group chain complex
\[\begin{tikzcd} L\ar[r, "\delta_2"] & M\ar[r, "\delta_1"] & N \end{tikzcd} \]
is called a 2-\textit{crossed module} if $N$ acts on $L$ and $M$ and the differentials $\delta_2, \delta_1$ are equivariant ($N$ acts over itself with conjugation), and there is a mapping
\[\{\ , \ \}: \begin{tikzcd} M \times M \ar[r]& L \end{tikzcd}\]
satisfying:
\begin{itemize}
\item[2XM1] $\delta_2\{m_0 , m_1\}= m_0m_1m_0^{-1} \ ^{\delta_1(m_0)}(m_1^{-1})$;
\item[2XM2] $\{ \delta_2(l_0), \delta_2(l_1) \}=[l_0 , l_1]$;
\item[2XM3] $\{\delta_2(l), m\}\{m, \delta_2(l)\}=l\ ^{\delta_1(m)}l^{-1}$;
\item[2XM4] $\{m_0,m_1m_2\}=\{m_0, m_1\}\{m_0,m_2\}\{\delta_2\{m_0,m_2\}^{-1},\ ^{\delta_1(m_0)}m_1 \}$;
\item[2XM5] $\{m_0m_1,m_2 \}=\{m_0,m_1m_2m_1^{-1}\}\ ^{\delta_1(m_0)}\{m_1,m_2\}; $
\item[2XM6] $ {}^n\{m_0,m_1\}= \{ {}^n m_0, {}^nm_1\}$.
\end{itemize}
The map $\{\ , \ \}$ is called the \textit{Peiffer lifting}.
A morphism of 2-crossed modules is a morphism of chain complexes that preserves the group actions and the Peiffer lifting. The category of 2-crossed complexes will be denoted as $_2\mathbb{X} Mod$.
\end{defi}
In a 2-crossed module the Peiffer lifting defines an action of $M$ over $L$ as ${}^m(l)=l\{\delta_2(l)^{-1} ,m\}$, so $\delta_2$ is indeed a crossed module. On the other hand, $\delta_1$ is only a precrossed module. A crossed module is a 2-crossed module by setting $L=0$. If a 2-crossed module has $N=0$ then the equations get simplify, thus we obtain a reduced 2-crossed module as follows.
\begin{defi} \cite{Con84}
A \textit{reduced 2-crossed module} is a group morphism $\delta: L \to M$ with a map $\{\ , \ \}: M\times M \to L$ satisfying:
\begin{enumerate}
\item $\delta \{m_0,m_1\}=[m_0, m_1]$,
\item $\{\delta(l_0), \delta(l_1) \}= [l_0, l_1]$,
\item $\{ \delta(l), m \}\{m, \delta(l)\}=1$,
\item $\{ m_0, m_1m_2 \}=\{m_0,m_1 \}\{m_0,m_2 \}\{ [m_2,m_0], m_1\}$,
\item $\{ m_0m_1, m_2 \}=\{m_0, m_1m_2m_1^{-1} \}\{m_1,m_2\}$.
\end{enumerate}
The category of reduced 2-crossed modules will be denoted as $R_2\mathbb{X} Mod$.
\end{defi}
\begin{defi} \cite{Con84}
A \textit{stable crossed module} is a group morphism $\delta: L \to M$ with a map $\{\ , \ \}: M\times M \to L$ satisfying:
\begin{enumerate}
\item $\delta \{m_0,m_1\}=[m_0,m_1]$,
\item $\{\delta(l_0), \delta(l_1)\}=[l_0,l_1]$,
\item $\{m_1, m_0 \}=\{m_0, m_1 \}^{-1}$,
\item $\{m_0m_1,m_2 \}=\{m_0m_1m_0^{-1} , m_0m_2m_0^{-1} \}\{m_0,m_2\}$.
\end{enumerate}
We will denote $St\mathbb{X} Mod$ for the category of stable crossed modules.
\end{defi}
The underlying morphism $\delta:L \to M$ of reduced 2-crossed module or of a stable crossed module is in fact a crossed module. Similar to the case of crossed modules/internal groupoids, these categories characterise simplicial groups via the normalization functor.
\begin{teo}\label{con} \cite{Con84}
\begin{enumerate}
\item The category $\mathcal{M}_{2 \geq}$ of simplicial groups with trivial Moore complex above degree 2 is equivalent to the category ${_2 \mathbb{X} Mod}$ of 2-crossed modules.
\item The category $\mathcal{M}_{1,2}$ of simplicial groups with trivial Moore complex except at degrees 1,2 is equivalent to the category $R_2 \mathbb{X} Mod$ of reduced 2-crossed modules.
\item The category $\mathcal{M}_{p,p+1}$ of simplicial groups with trivial Moore complex except at degrees $p,p+1$ for $p\geq 2$ is equivalent to the category $St_2 \mathbb{X} Mod$ of stable crossed modules.
\end{enumerate}
\end{teo}
The bottom torsion categories of $\mathcal{M}_{n \geq}$ can be characterized in a similar way as the torsion categories of internal groupoids.
\begin{teo}\label{teom2} For $n=2$, consider the lattice of torsion theories $\mu(\mathcal{M}_{2 \geq})$:
\[ 0 \quad \leq \quad \mu_{ \geq 2}' \quad \leq\ \quad \mu_{1 \geq}' \quad \leq \quad \mu'_{\geq 1} \quad \leq \quad \mu_{0 \geq}' \quad \leq \quad \mathcal{M}_{2 \geq}\]
For the bottom torsion categories we have the equivalences:
\begin{enumerate}
\item the torsion category $\mathcal{M}_{ \geq 1}\cap \mathcal{M}_{2 \geq}= \mathcal{M}_{1,2}$ of $\mu_{ \geq 1}'$ is equivalent to the category $R_2\mathbb{X} Mod$ of reduced 2-crossed modules;
\item the torsion category $Ker(Cot_1)\cap \mathcal{M}_{2 \geq}$ of $\mu_{1 \geq }'$ is equivalent to the category $R_2\mathbb{X} Mod\cap Cext$ of reduced 2-crossed modules $\delta: L \to M$ with $\delta$ a central extension;
\item the torsion category $\mathcal{M}_{\geq 2}\cap \mathcal{M}_{2 \geq}$ of $\mu_{ \geq 2}'$ is equivalent to the category of 2-crossed modules of the form $A \to 0 \to 0$ and hence, it is also equivalent to the category $Ab$ of abelian groups.
\end{enumerate}
\end{teo}
\begin{proof}
1) It follows immediately from the fact that $\mathcal{M}_{ \geq 1}\cap \mathcal{M}_{2 \geq}$ is equivalent by definition to the category $\mathcal{M}_{1,2}$ of simplicial groups with trivial Moore complex except at degrees 1,2. and Theorem \ref{con}.
2) A simplicial group $X$ belongs to $Ker(Cot_1)\cap \mathcal{M}_{2 \geq}$ if and only if its Moore complex is trivial except $\delta_2:M_2 \to M_1$ which is, in addition, surjective (form Corollary \ref{cotnsurj}). This happen if and only if $\delta_2$ is a central extension, since a 2-reduced crossed module is in particular a crossed module.
3) The category $\mathcal{M}_{\geq 2}\cap \mathcal{M}_{2 \geq}$ is equivalent to the category of simplicial groups with trivial except at degree 2, so it corresponds to a 2-crossed module of the form $L \to 0\to 0$. Since, in a 2-crossed module the morphism $\delta_2$ is a crossed module then $L$ must be an abelian group.
\end{proof}
\begin{teo}\label{teomn} For $n> 2$, in $\mu(\mathcal{M}_{n \geq})$ consider the bottom torsion theories:
\[ 0 \quad \leq \quad \mu_{ \geq n}' \quad \leq\ \quad \mu_{n-1 \geq}' \quad \leq \quad \mu'_{\geq n-1} \quad \dots \]
Then for the torsion categories we have the equivalences:
\begin{enumerate}
\item the torsion category $\mathcal{M}_{ \geq n-1}\cap \mathcal{M}_{n \geq}= \mathcal{M}_{n,n-1}$ of $\mu_{ \geq n-1}'$ is equivalent to the category $St \mathbb{X} Mod$ of stable crossed modules;
\item the torsion category $Ker(Cot_{n-1})\cap \mathcal{M}_{n \geq}$ of $\mu_{n-1 \geq }'$ is equivalent to the category $St\mathbb{X} Mod\cap Cext$ of stable crossed modules $\delta: L \to M$ with $\delta$ a central extension;
\item the torsion category $\mathcal{M}_{\geq n}\cap \mathcal{M}_{n \geq}$ of $\mu_{ \geq n}'$ is equivalent to the category $Ab$ of abelian groups.
\end{enumerate}
\end{teo}
\begin{proof} It follows from 3) of Theorem \ref{con}, the proof is similar to Theorem \ref{teom2}.
\end{proof}
\begin{coro} A simplicial group $X$ belong to the torsion category $Ker(Cot_{n-1})\cap \mathcal{M}_{n \geq}$ of $\mathcal{M}_{n \geq}$ if and only if its Moore complex $M$ is a central extension in groups.
\end{coro}
\begin{coro} The categories ${}_2\mathbb{X} Mod$ of 2-crossed modules, $R_2\mathbb{X} Mod$ of reduced 2-crossed modules and $St \mathbb{X} Mod$ of stable crossed modules are semi-abelian.
\end{coro}
\begin{proof} From Proposition \ref{semiab}, the categories $\mathcal{M}_{n \geq}$ are semi-abelian. In particular, for $n=2$ the category of 2-crossed modules is semi-abelian. Similarly, $R_2\mathbb{X} Mod$ and $St \mathbb{X} Mod$ are semi-abelian from Theorems \ref{teom2}, \ref{teomn} and Proposition \ref{semiab}.
\end{proof}
\subsection{The case of $\mathcal{M}_{\geq n}$}
A dual behaviour can be noticed when we work with the torsion categories $\mathcal{M}_{\geq n}$, which are also semi-abelian (again form Proposition \ref{semiab}). When considering $\mu(\mathcal{M}_{\geq n})$, the restriction of $\mu(Grp)$ to $\mathcal{M}_{\geq n}$, we first obtain that the torsion theories above $\mu_{n\geq}$ are trivialized and second, the upper torsion-free categories are equivalent to abelian groups and the different kinds of crossed modules.
\begin{prop} The lattice $\mu(\mathcal{M}_{\geq n})$ is given by:
\[0 \quad \leq \quad \dots \quad \leq \quad \mu_{n+1\geq}' \quad \leq \quad \mu_{\geq n+1}' \quad \leq \quad \mu_{n \geq}' \quad \leq \quad \mathcal{M}_{\geq n} \]
where $\mu_{i \geq}'$, $\mu_{\geq i}'$ are the restriction of the torsion theories of $\mu(Grp)$.
\end{prop}
\begin{teo}
For the category $\mathcal{M}_{\geq 1}$ we have
\begin{enumerate}
\item the torsion-free category of $\mu_{1 \geq}'$ given by $\mathcal{M}_{1 \geq}\cap\mathcal{M}_{\geq 1}$ is equivalent to the category of abelian groups;
\item the torsion-free category of $\mu_{2 \geq}'$ given by $\mathcal{M}_{2 \geq}\cap\mathcal{M}_{\geq 1}$ is equivalent to the category $R_2 \mathbb{X} Mod$ of Reduced 2-crossed modules.
\end{enumerate}
For the category $\mathcal{M}_{\geq n}$ and $n \leq 2$ we have
\begin{enumerate}
\item the torsion-free category of $\mu_{n \geq}'$ given by $\mathcal{M}_{n \geq}\cap\mathcal{M}_{\geq n}$ is equivalent to the category of abelian groups;
\item the torsion-free category of $\mu_{n+1 \geq}'$ given by $\mathcal{M}_{n+1 \geq}\cap\mathcal{M}_{\geq n}$ is equivalent to the category $St \mathbb{X} Mod$ of stable crossed modules.
\end{enumerate}
\end{teo}
\begin{proof} The proof is similar to the case of torsion categories in $\mathcal{M}_{n \geq}$.
\end{proof}
\section{Torsion theories in reduced crossed complexes}
Introduced by M. K. Dakin \cite{Dak77}, a T-complex is a Kan simplicial object that admits a canonical filler for horns, for example internal groupoids have this property. Following the work of N. Ashley in \cite{Ash78} and the observations in \cite{CaCe91} a group T-complex (a T-complex in simplicial groups) can be defined as simplicial group $X$ with $M_n \cap D_n=0$ where $M$ is the Moore complex of $X$ and $D$ is the graded subgroup of $X$ generated by the degenerated elements of $X$.
\begin{defi}\cite{Ash78}
A \textit{reduced crossed complex} $M$, or a crossed complex in groups, is a proper chain complex
\[M= \begin{tikzcd} \dots \ar[r]& M_n \ar[r, "\delta_n"]& M_{n-1}\ar[r]& \dots\ar[r] & M_2\ar[r, "\delta_2"] & M_1\ar[r, "\delta_1"] & M_0\end{tikzcd} \]
where
\begin{enumerate}
\item $M_n$ is abelian for $n \geq 2$;
\item $M_0$ acts on $M_n$ for $n \geq 1$ and the restriction to $\delta_1(M_1)$ acts trivially on $M_n$ for $n \geq 2$;
\item $\delta_n$ preserves the action of $M_0$ and $\delta_1:M_1 \to M_0$ is a crossed module.
\end{enumerate}
A morphism of reduced crossed complexes is a chain complex morphism that preserves all actions. We will write $Crs(Grp)$ for the category of reduced crossed complexes.
\end{defi}
It is proved in \cite{Ash78} that the category of group T-complexes is equivalent to the category of reduced crossed complexes via the Moore normalization. And in \cite{EhPo97} it is shown that the category of reduced crossed complexes is an epi-reflective subcategory of simplicial groups.
\begin{coro} The category of Dakin's group $T$-complexes is semi-abelian. Hence, the category $Crs(Grp)$ of reduced crossed complexes is also semi-abelian.
\end{coro}
\begin{proof} From \cite{EhPo97}, we have that group $T$-complexes is a normal epi-reflective subcategory of $Simp(Grp)$. In fact, it is a Birkhoff subcategory so it is semi-abelian. Indeed, we just need to prove that it is closed under regular epimorphism in $Simp(Grp)$. So, let $X$ be a group $T$-complex, a simplicial group with $M^X_n\cap D^X_n=0$ and $f:X \to Y$ a regular epimorphism where $M^X_n,M^Y_n$ are the Moore subobjects and $D^X_n, D^Y_n$ are the graded subgroups generated by degenerated elements of $X$ and $Y$ respectively.
Since the Moore normalization preserves regular epimorphisms, then the restriction $f_M: M_n^X \to M^Y_n$ is also a regular epimorphism. Now the restriction $f_D: D_n^X \to D^Y_n$ is surjective, since we have a commutative diagram:
\[\begin{tikzcd} X_n \ar[d, "f"] & X_m\ar[l, "s_\mathbf{i}"]\ar[d,"f"] \\ Y_n & Y_m \ar[l, "s_\mathbf{i}"] \end{tikzcd}\]
so for a degenerate element $s_\mathbf{i}(y_m)$ in $Y_n$ ($s_\mathbf{i}$ a composition of degenerate morphisms), there is $x_m$ in $X_M$ such that $s_\mathbf{i}f(x_m)=fs_\mathbf{i}(x_m)$.
Finally, since both $f_M$ and $f_D$ are regular epimorphisms the morphism:
\[f_M \times f_D: \begin{tikzcd} M^X_n \times_{X_n} D^X_n \ar[r] & M^Y_n \times_{Y_n} D^Y_n \end{tikzcd}\]
is also a regular epimorphism, so if $M^X_n \times_{X_n} D^X_n=M^X_n \cap D^X_n=0$, then so is $M^Y_n \cap D^Y_n=0$.
\end{proof}
Our interest in reduced crossed complexes lies in their similar behaviour to chain complexes, thus torsion theories of simplicial groups can be easily studied when restricted to subcategory of $Crs(Grp)$. To this end, we recall some properties of reduced crossed complexes, we refer the reader to \cite{BrHiSi10} for the details and proofs.
\begin{strc}\label{adj}
An $n$-reduced crossed complex is an $n$-truncated chain complex $M$:
\[M= \begin{tikzcd} M_n \ar[r, "\delta_n"]& M_{n-1}\ar[r]& \dots\ar[r] & M_2\ar[r, "\delta_2"] & M_1\ar[r, "\delta_1"] & M_0\end{tikzcd} \]
satisfying all the axioms of a reduced crossed complex (those that make sense), thus we have a category $Crs(Grp)_{n \geq}$ of $n$-truncated crossed complexes. Clearly, a 1-truncated reduced crossed complex is nothing but a crossed module, thus $ Crs(Grp)_{1 \geq} = \mathbb{X} Mod$.
Moreover, we have the functors for all $n\in \mathbb{N}$:
\begin{itemize}
\item the truncation functor $\mathbf{tr}_n: Crs(Grp) \to Crs(Grp)_{n \geq}$
\[\mathbf{tr}_n(M)=\begin{tikzcd} M_n \ar[r] & M_{n-1}\ar[r] & M_{n-2} \ar[r] & \dots \end{tikzcd};\]
\item the skeleton functor (or natural embedding) $\mathbf{sk}_n: Crs(Grp)_{n \geq} \to Crs(Grp)$
\[\mathbf{sk}_n(M)= \begin{tikzcd} \dots\ar[r] & 0\ar[r]&0\ar[r] &M_n\ar[r] &M_{n-1}\ar[r] & \dots \end{tikzcd}; \]
\item the coskeleton functor $\mathbf{cosk}_n: Crs(Grp)_{n \geq} \to Crs(Grp)$
\[\mathbf{cosk}_n(M)= \begin{tikzcd} \dots\ar[r] & 0\ar[r]&ker(\delta_n)\ar[r] &M_n\ar[r] &M_{n-1}\ar[r] & \dots \end{tikzcd}; \]
\item the cotruncation $\mathbf{cot}_n: Crs(Grp) \to Crs(Grp)_{n \geq}$
\[\mathbf{cot}_n(M)=\begin{tikzcd} M_n/\delta_{n+1}(M_{n+1}) \ar[r] & M_{n-1}\ar[r] & M_{n-2} \ar[r] & \dots \end{tikzcd}.\]
\end{itemize}
We will write $\mathbf{Sk}_n=\mathbf{sk}_n\mathbf{tr}_n$, $\mathbf{Cosk}_n=\mathbf{cosk}_n\mathbf{tr_n}$ and $\mathbf{Cot}_n=\mathbf{sk}_n\mathbf{cot}_n$. These functors give a string of adjunctions
\[\mathbf{cot}_n \dashv \mathbf{sk}_n \dashv \mathbf{tr}_n \dashv \mathbf{cosk}_n:
\begin{tikzcd}[row sep=large]
Crs(Grp)
\ar[d, bend left, shift right=.5] \ar[d, bend right, shift right=4] \ar[d, phantom, "\dashv"] \ar[d,phantom,"\dashv", shift right=5.5 ] \ar[d,phantom, "\dashv", shift left=5.5 ] \\
Crs(Grp)_{n \geq }
\ar[u, bend left, shift right=.5] \ar[u, bend right, shift right=4]
\end{tikzcd} \, .\]
It is worth mentioning that the adjunctions of simplicial groups $sk_n \dashv tr_n \dashv cosk_n$ restricted to crossed complexes correspond (under Moore normalization) to $\mathbf{sk}_n \dashv \mathbf{tr}_n \dashv \mathbf{cosk}_n$ and Porter's cotruncation $Cot_n$ corresponds to $\mathbf{Cot}_n$. In addition, unlike the general case of simplicial groups, the adjunction $\mathbf{cot}_n\dashv \mathbf{sk}_n$ for reduced crossed complexes.
On the other hand, it will be useful to write $Crs(Grp)_{\geq n}$ for the subcategory of reduced crossed complexes with $M_i=0$ for $n>i$. For all $n\geq 1$, $Crs(Grp)_{\geq n}$ equivalent to the category of $chn(Ab)_{\geq n}$ chain complexes of abelian groups. Moreover, we can easily define the dual functors:
\begin{itemize}
\item $\mathbf{tr}'_n:Crs(Grp) \to Crs(Grp)_{\geq n}$;
\item $\mathbf{sk}'_n:Crs(Grp)_{\geq n} \to Crs(Grp)$;
\item $\mathbf{cot}'_n:Crs(Grp) \to Crs(Grp)_{\geq n}$.
\end{itemize}
However, only the adjunction $\mathbf{sk}'_n \dashv \mathbf{cot}'_n$ holds.
\end{strc}
\begin{prop}\label{ttcrs} Let $\mu(Crs(Grp))$ be the lattice given by the restriction of torsion theories in $\mu(Grp)$ to $Crs(Grp)$:
\[ \mu(Crs(Grp)) = \quad \dots \; \leq \quad \mu_{\geq 2}' \quad \leq \quad \mu_{1 \geq}' \quad \leq \quad \mu_{\geq 1}' \quad \leq \quad \mu_{0 \geq}' \quad \leq \quad Crs(Grp)\, . \]
Then, the torsion theories $\mu'_{n \geq}$ and $\mu'_{ \geq n}$ can be expressed with the functors $\mathbf{cot}_n \dashv \mathbf{sk}_n \dashv \mathbf{tr}_n \dashv \mathbf{cosk}_n$:
\[\mu'_{n \geq}=(Ker(\mathbf{cot}_n) ,Crs(Grp)_{n \geq}), \quad \mu'_{\geq n}=(Crs(Grp)_{\geq n} , \mathcal{F}_{\mathbf{tr}_n} )\, .\]
\end{prop}
\begin{proof} It follows from the fact that the equivalence between group $T$-complexes and reduced crossed complexes is given by the Moore normalization, and also from the short exact sequences in Theorem \ref{teolop} and the previous observations \ref{adj}.
\end{proof}
\begin{teo} For $n>1$, let $\mu(Crs(Grp)_{n \geq})$ be the lattice given by restriction of $\mu(Grp)$ to $Crs(Grp)_{n \geq}$:
\[\mu(Crs(Grp)_{n \geq})= 0 \leq \quad \mu_{\geq n}' \quad \leq \quad \mu_{n-1 \geq}' \quad \leq \; \dots \; \leq \quad \mu_{\geq 1}' \quad \leq \quad \mu_{0 \geq}' \quad \leq \quad Crs(Grp)_{n \geq}\, . \]
Then for the torsion categories we have the equivalences:
\begin{enumerate}
\item $Ker(\mathbf{cot}_{n-1})\cap Crs(Grp)_{n \geq}$, the torsion category of $\mu'_{n-1 \geq}$, is equivalent to the category $CExt(Ab)$ of central extension in abelian groups, i.e., surjective morphisms.
\item $Crs(Grp)_{\geq n} \cap Crs(Grp)_{n \geq}$, the torsion category of $\mu'_{\geq n}$, is equivalent to the category $Ab$ of abelian groups.
\end{enumerate}
\end{teo}
\begin{proof} The proof is similar to Theorem \ref{teom2}.
\end{proof}
\section{Torsion torsion-free theories}
\begin{defi} A \textit{torsion torsion-free theory} in $\mathbb{X}$, or TTF theory, is a triplet $(\mathcal{C}, \mathcal{T}, \mathcal{F})$ of full subcategories of $\mathbb{X}$ such that $(\mathcal{C}, \mathcal{T})$ and $(\mathcal{T}, \mathcal{F})$ are torsion theories in $\mathbb{X}$. A subcategory $\mathcal{T}$ of $\mathbb{X}$ is called a torsion torsion-free category (or a TTF category) if there are subcategories $\mathcal{C}$ and $\mathcal{F}$ such that $(\mathcal{C}, \mathcal{T}, \mathcal{F})$ is a TTF theory.
\end{defi}
TTF theories were introduced in \cite{jans65} with applications mainly to categories of modules over rings, in particular they are used to study when any object of a category is a joint of torsion objects for different torsion theories. TTF theories have been studied in different non-abelian settings, for example in triangulated categories in \cite{BeRe07}. In our context of simplicial groups, we present examples of TTF theories in a weak sense.
\subsection{TTF theories in chain complexes}
Let $\mathbb{X}$ be a semi-abelian category and $chn(\mathbb{X})$ and $pch(\mathbb{X})$ the categories of chain complexes and proper chain complexes, respectively. We will write $chn(\mathbb{X})_{n \geq}$ and $chn(\mathbb{X})_{\geq n}$ for the category of chain complexes bounded above/below $n$.
We have the adjunctions:
\[ \mathbf{cot}_n \dashv \mathbf{sk}_n \dashv \mathbf{tr}_n \dashv \mathbf{cosk}_n:
\begin{tikzcd}[row sep=large]
ch(\mathbb{X})
\ar[d, bend left, shift right=.5] \ar[d, bend right, shift right=4] \ar[d, phantom, "\dashv"] \ar[d,phantom,"\dashv", shift right=5.5 ] \ar[d,phantom, "\dashv", shift left=5.5 ] \\
ch(\mathbb{X})_{n \geq }
\ar[u, bend left, shift right=.5] \ar[u, bend right, shift right=4]
\end{tikzcd} ;\]
as well as their duals:
\[ \mathbf{cosk}'_n \dashv \mathbf{tr}'_n \dashv \mathbf{sk}'_n \dashv \mathbf{cot}'_n:
\begin{tikzcd}[row sep=large]
ch(\mathbb{X})_{ \geq n}
\ar[d, bend left, shift right=.5] \ar[d, bend right, shift right=4] \ar[d, phantom, "\dashv"] \ar[d,phantom,"\dashv", shift right=5.5 ] \ar[d,phantom, "\dashv", shift left=5.5 ] \\
ch(\mathbb{X})
\ar[u, bend left, shift right=.5] \ar[u, bend right, shift right=4]
\end{tikzcd} .
\]
defined similarly as in \ref{adj}. These adjunction can be restricted to adjuntions between $pch(\mathbb{X})$ and $pch(\mathbb{X})_{n \geq}$ (or $pch(\mathbb{X})_{\geq n}$ respectively).
Similar to simplicial groups, torsion theories are given by the cotruncation functors $\mathbf{cot}$ and $\mathbf{cot}'$ as follows.
\begin{teo}\label{ttchcompl} \cite{Lop22b} Let $\mathbb{X}$ be a semi-abelian category, then we have:
\begin{enumerate}
\item in $chn(\mathbb{X})$ we have a torsion theory $(chn(\mathbb{X})_{\geq n}, \mathcal{F}_{\mathbf{tr}_{n-1}})$:
\[\begin{tikzcd}[column sep= large, row sep=huge] ch(\mathbb{X})_{\geq n}\ar[r, bend left, "\mathbf{sk}_n'" above]\ar[r, phantom, "\perp"] & ch(\mathbb{X}) \ar[l, bend left, "\mathbf{cot}_n'" below]\ar[r, bend left]\ar[r,phantom, "\perp"] & \mathcal{F}_{\mathbf{tr_{n-1}}} \ar[l, bend left] \end{tikzcd} ; \]
\item the previous torsion theory is restricted to a torsion theory in $pch(\mathbb{X})$ as $(pch(\mathbb{X})_{\geq n}, \mathcal{MN}_n)$:
\[\begin{tikzcd}[column sep= large, row sep=huge] pch(\mathbb{X})_{ \geq n}\ar[r, bend left, "\mathbf{sk}_n'" above]\ar[r, phantom, "\perp"] & pch(\mathbb{X}) \ar[l, bend left, "\mathbf{cot}_n'" below]\ar[r, bend left]\ar[r,phantom, "\perp"] & \mathcal{MN}_n \ar[l, bend left] \end{tikzcd} ; \]
\item the category $chn(\mathbb{X})_{n \geq}$ is a normal epi-reflective subcategory of $chn(\mathbb{X})$, but it is not a torsion-free subcategory ;
\item in $pch(\mathbb{X})$ we have a torsion theory $(\mathcal{EP}_n, pch(\mathbb{X})_{n-1 \geq})$:
\[\begin{tikzcd}[column sep= large, row sep=huge] \mathcal{EP}_n\ar[r, bend left]\ar[r, phantom, "\perp"] & pch(\mathbb{X}) \ar[l, bend left ]\ar[r, bend left, "\mathbf{cot}_{n-1}" above]\ar[r,phantom, "\perp"] & pch(\mathbb{X})_{n-1\geq } \ar[l, bend left, "\mathbf{sk}_{n-1}" below] \end{tikzcd} .\]
\end{enumerate}
where
\begin{itemize}
\item $\mathcal{F}_{\mathbf{tr}_{n-1}}$ is the subcategory of chain complexes such that their component of the unit of $\mathbf{tr}_{n-1}\dashv \mathbf{cosk}_{n-1}$ is monic.
\item $\mathcal{MN}_n$ is the subcategory of proper chain complexes such that $M_i=0$ for $i>n$ and the differential $\delta_n$ is monic.
\item $\mathcal{EP}_n$ is the subcategory of proper chain complexes such that $M_i=0$ for $n-1>i$ and the differential $\delta_n$ is epic.
\end{itemize}
The short exact sequences of the torsion theories in 2) and 3) are given, for a proper chain complex $M$, as in diagrams (3) and (4) in Theorem \ref{teolop}, respectively.
\end{teo}
In addition to this torsion theories, we can add the following.
\begin{teo} Let $\mathbb{X}$ be a semi-abelian category. For each $n \in \mathbb{Z}$ the pair $(ch(\mathbb{X})_{n-1 \geq}, ch(\mathbb{X})_{\geq n})$ is a hereditary cohereditary torsion theory in $ch(\mathbb{X})$. Moreover, the reflector and coreflector are given by $\mathbf{sk}_{n-1} \dashv \mathbf{tr}_{n-1}$ and $\mathbf{tr}'_n \dashv \mathbf{sk}'_n$:
\[\begin{tikzcd}[column sep= large, row sep=huge]
ch(\mathbb{X})_{n-1 \geq}\ar[r, bend left, "\mathbf{sk}_{n-1}" above]\ar[r, phantom, "\perp"]
& ch(\mathbb{X}) \ar[l, bend left, "\mathbf{tr}_{n-1}" below]\ar[r, bend left, "\mathbf{tr}'_n"]\ar[r,phantom, "\perp"]
& ch(\mathbb{X})_{\geq n} \ar[l, bend left, "\mathbf{sk}'_n"] \end{tikzcd} . \]
\end{teo}
\begin{proof} For $X$ in $ch(\mathbb{X})_{n-1 \geq}$ and $Y$ in $ch(\mathbb{X})_{\geq n}$ it is clear that a morphism $\mathbf{sk}_{n-1}(X) \to \mathbf{sk}'_n(Y)$ must be trivial:
\[\begin{tikzcd}[column sep=scriptsize, row sep=scriptsize]
\mathbf{sk}_{n-1}(X)\ar[d]= & \dots\ar[r] & 0\ar[r] \ar[d]& 0 \ar[r] \ar[d]& X_{n-1}\ar[r]\ar[d] & X_{n-2} \ar[d]
\\
\mathbf{sk}'_n(Y)= & \dots \ar[r] & Y_{n+1}\ar[r] & Y_n \ar[r] & 0 \ar[r]& 0
\end{tikzcd} \]
Since limits and colimits are computed component-wise in $ch(\mathbb{X})$, the short exact sequence of the torsion theory for a chain complex $X$ in $ch(\mathbb{X})$ is given by:
\[ \begin{tikzcd}[column sep=scriptsize, row sep=scriptsize]
\dots \ar[r] & 0\ar[r] \ar[d]& 0 \ar[r] \ar[d]& X_{n-1}\ar[r]\ar[d] & X_{n-2} \ar[d]\ar[r] & \dots
\\
\dots \ar[r] & X_{n+1}\ar[r] \ar[d] & X_n \ar[r]\ar[d] & X_{n-1} \ar[r] \ar[d]& X_{n-2}\ar[r]\ar[d] & \dots
\\
\dots \ar[r] & X_{n+1}\ar[r] & X_n \ar[r] & 0 \ar[r] & 0 \ar[r]& \dots
\end{tikzcd} \]
\end{proof}
\begin{coro}\label{ttfpch} Let $\mathbb{X}$ be semi-abelian category. For each $n \in \mathbb{Z}$ the triplets of full subcategories
\[(ch(\mathbb{X})_{n-1 \geq}, ch(\mathbb{X})_{\geq n}, \mathcal{F}_{\mathbf{tr}_{n-1}})\] are TTF theories in $ch(\mathbb{X})$. Moreover, by restriction these determine the TTF theories in $pch(\mathbb{X})$
\[(pch(\mathbb{X})_{n-1 \geq}, pch(\mathbb{X})_{\geq n}, \mathcal{MN}_n).\]
Similarly, the triplet of subcategories in $ch(\mathbb{X})$
\[(Ker(\mathbf{cot}_{n-1 \geq}),ch(\mathbb{X})_{n-1 \geq} , ch(\mathbb{X})_{\geq n} )\]
determines the TTF theories in $pch(\mathbb{X})$
\[ (\mathcal{EP}_n, pch(\mathbb{X})_{n-1 \geq}, pch(\mathbb{X})_{ \geq n}).\]
\end{coro}
\subsection{Weak TTF theories in chain complexes with operations}
Through this work we have mention the similarities between torsion theories in simplicial groups and in chain complexes, this happens since they are both defined with similar set of adjuntions ($sk_n\dashv tr_n \dashv cosk_n$ and $\mathbf{cot}_n \dashv \mathbf{sk}_n \dashv \mathbf{tr}_n \dashv \mathbf{cosk}_n$). However, the TTF theories in chain complexes cannot be easily adapted to the simplicial case, not even in the initial example of internal groupoids. For instance, in $Grpd(Grp)$ the subcategory of discrete groupoids $Dis(Grp) \cong \mathcal{M}_{0 \geq}$ is a torsion-free subcategory (with reflector $\pi_0\dashv Dis$) and also mono-coreflective (with the adjunction $Dis \dashv tr_0$) but not normal mono-coreflective, so it is not a torsion subcategory. In general, the subcategories $\mathcal{M}_{ n \geq}$ are only torsion-free.
Two different kinds of torsion theories are introduced.
\begin{defi}
For a class of objects $\mathcal{E}$ of $\mathbb{X}$, a pair $(\mathcal{T}, \mathcal{F})$ of full subcategories of $\mathbb{X}$ will be called a $\mathcal{E}$-torsion theory or a torsion theory relative to the class $\mathcal{E}$ if:
\begin{itemize}
\item[TT1] for all $X \in \mathcal{T}$ and $Y \in \mathcal{F}$, every morphism $f:X \to Y$ is zero;
\item[TT2'] for every object $X \in \mathcal{E}$ exists a short exact sequence
\[\begin{tikzcd} 0\ar[r] & T_X\ar[r, "t_X"] & X\ar[r, "f_X"] & F_X\ar[r] &0 \end{tikzcd}\]
with $T_X \in \mathcal{T}$ and $F_X \in \mathcal{F}$.
\end{itemize}
\end{defi}
As a first example of a $\mathcal{E}$-torsion theory we have the pair $(Ker(\mathbf{cot}_n), ch(\mathbb{X})_{n \geq})$ in $chn(\mathbb{X})$ as in Theorem \ref{ttchcompl}.
\begin{lemma} In $ch(\mathbb{X})$ the category of chain complexes the pair
\[(Ker(\mathbf{cot}_n), ch(\mathbb{X})_{n \geq})\] and $\mathcal{E}$ the class of proper chain complexes $pch(\mathbb{X})$ is a $\mathcal{E}$-torsion theory in $ch(\mathbb{X})$.
\end{lemma}
\begin{proof} The objects in $Ker(\mathbf{cot}_n)$ are the chain complexes $X$ such that $X_i=0$ for $n> i$ and the differential $\delta_{n+1}$ has a trivial cokernel. Thus, to verify TT1 it suffices to notice that given a commutative diagram:
\[\begin{tikzcd} X_{n+1}\ar[r, "\delta_{n+1}"]\ar[d, "f_{n+1}"] & X_n\ar[d, "f_n"] \\ 0\ar[r] & Y_n \end{tikzcd}\]
with $\delta_{n+1}$ a morphism with trivial cokernel then the morphism $f$ must be trivial.
TT2' holds since it has been established that the restriction of the pair $(Ker(\mathbf{cot}_n), ch(\mathbb{X})_{n \geq})$ to proper chain gives a torsion theory $(\mathcal{EP}_n, pch(\mathbb{X})_{n \geq})$ in $pch(\mathbb{X})$. In addition, for a proper chain complex $M$ the short exact sequence is given by diagram (3) in Theorem \ref{teolop}.
\end{proof}
Any normal epireflective subcategory yield a $\mathcal{E}$-torsion theory. Let $F \dashv I: \mathbb{X} \to \mathbb{A}$ be a normal epireflective subcategory of $\mathbb{X}$ with unit $\eta$. Following \cite{BouGrn06}, we consider two subcategories of $\mathbb{X}$ :
\[\mathcal{T}_\mathcal{F}=\{T \mid T \cong ker(\eta_X) \; \mbox{for some} \; X\} \]
and
\[Ker(F)= \{X \mid F(X)=0 \}.\]
Clearly, we have $Ker(F) \subseteq \mathcal{T}_\mathcal{F}$. The pair $(\mathcal{T}_\mathcal{F} , \mathcal{F})$ satisfies axiom TT2 of a torsion theory while the pair $(Ker(F), \mathcal{F})$ satisfy axiom TT1. Indeed, if $Ker(F)= \mathcal{T}_\mathcal{F}$ we have a torsion theory. Clearly, in the relative case we have the following:
\begin{lemma} Let $F \dashv I: \mathbb{X} \to \mathbb{A}$ be a normal epireflective subcategory of $\mathbb{X}$ with unit $\eta$. If $\mathcal{E}=\{X \mid F(ker(\eta_X))=0 \}$ then the pair $(Ker(F), \mathcal{F})$ is a $\mathcal{E}$-torsion theory.
\end{lemma}
\begin{exmp} The category of $Ab$ of abelian groups is a normal epireflective subcategory of $Grp$ (in fact, it is a Birkhoff subcategory) where the reflector is the abelianization functor $ab(G)=G/G'$ where $G'$ is the commutator subgroup. Hence, if $Pr$ is the category of perfect groups, groups $G$ such that $G'=G$, then the pair $(Pr, Ab)$ is a $\mathcal{E}$-torsion theory with respect the class $\mathcal{E}$ of groups such that $(G')'=G'$ or equivalently, groups with a perfect commutator.
\end{exmp}
We introduce our main definition.
\begin{defi}
Let $(\mathcal{T}, \mathcal{F})$ be a torsion theory in $\mathbb{X}$, if $\mathcal{F}$ is a mono-coreflective category of $\mathbb{X}$ we will call $(\mathcal{T}, \mathcal{F})$ a \textit{CTF theory}. This means, that the embedding $I$ of $\mathcal{F}$ into $\mathbb{X}$ has both a right and left adjoint, $F\dashv I \dashv C$:
\[\begin{tikzcd}[column sep=large] \mathcal{T} \ar[r, bend left, "J"]\ar[r, phantom, "\perp"] & \mathbb{X} \ar[r, bend left, "F"]\ar[l, bend left,"T"]\ar[r, phantom, "\perp"]\ar[r, bend right, "C"', shift right=6]\ar[r, phantom, "\perp", shift right=6] & \mathcal{F} \ar[l, bend left, "I", near start] \end{tikzcd} .\]
\end{defi}
Clearly, in a TTF theory $(\mathcal{C}, \mathcal{T}, \mathcal{F})$ the pair $(\mathcal{C}, \mathcal{T})$ is a CTF-theory. In \cite{CDT06}, it is proved that a normal mono-coreflective subcategory closed under extension is in fact a torsion category, so a CTF theory with $\mathcal{F}$ a normal-monocoreflective subcategory is a TTF theory.
The subcategory of discrete crossed modules $Dis$ in $\mathbb{X} Mod$ behaves almost as a torsion torsion-free subcategory, it presents an example of a CTF theory as well as a relative $\mathcal{E}$-torsion theory. To this end, recall that a monomorphism in $\mathbb{X} Mod$ is given by an injective crossed module morphism $f=(f_1,f_0)$:
\[\begin{tikzcd} A\ar[r]\ar[d, "f_1"] & B\ar[d, "f_0"] \\ X\ar[r] & Y\, . \end{tikzcd}\]
In addition, $f$ is a normal subcrossed module if and only if $A$, $B$ are normal subgroups of $X$, $Y$, and the conditions ${}^{y}(a)\in A$ and ${}^b(x)x^{-1}\in X$ hold for all $a \in A$, $b \in B$, $y\in Y$ and $x\in X$.
\begin{prop}\label{ctfxmod} In $\mathbb{X} Mod$ consider the triplet of subcategories:
\[(CExt, Dis, Ab )\]
then:
\begin{enumerate}
\item the pair $(CExt ,Dis)$ is a CTF theory in $\mathbb{X} Mod$;
\item the pair $(Dis, Ab)$ is an $\mathcal{E}$-torsion theory where $\mathcal{E}$ is the class of crossed modules $\delta:A \to B$ where the action $B \to Aut(A)$ is trivial.
\end{enumerate}
\end{prop}
\begin{proof} 1) The discrete functor $D$ has right adjoint $(\, )_0$ where the component of the counit for a crossed module $\delta: A \to B$ is given by horizontal arrows in the diagram:
\begin{equation}\label{counitdis0}
\begin{tikzcd} 0\ar[r,"0"]\ar[d] & A\ar[d,"\delta"] \\ B\ar[r, "1"] & B \end{tikzcd}
\end{equation}
which is a monomorphism since the pair $(0,1)$ are injective morphisms.
2) It is clear that the pair $(Dis, Ab)$ satisfies TT1 since in a commutative diagram
\[\begin{tikzcd} 0\ar[r, "f_1"]\ar[d] & A\ar[d] \\ G\ar[r, "f_0"]& 0 \end{tikzcd}\]
the morphism $f=(f_1,f_0)$ is zero. For TT2', recall that the unit Diagram (\ref{counitdis0}) is a normal monomorphism in $\mathbb{X} Mod$ if and only if $^b(a)a^{-1}=0$, i.e., the action of $B$ over $A$ is trivial. From the Peiffer identity $^{\delta(a)}(a')=aa'a^{-1}$, a crossed module with trivial action also has $A$ as an abelian group then we have the short exact sequence in $\mathbb{X} Mod$:
\[\begin{tikzcd} 0\ar[r]&0\ar[r]\ar[d]&A\ar[r]\ar[d, "\delta"]&A\ar[r]\ar[d]&0 \\ 0\ar[r]&B\ar[r]&B\ar[r] &0\ar[r]&0 \, . \end{tikzcd} \]
\end{proof}
\begin{rmk} The pair of subcategories $Dis$ and $Ab$ of $\mathbb{X} Mod$ present another example of an $\mathcal{E}$-torsion theory in $\mathbb{X} Mod$. This time as $(Ab, Dis)$ and $\mathcal{E}$ as the subcategory $Mod$ of modules of groups. For a module we mean a pair $(A, G)$ such that $G$ is a group and $A$ is an abelian groups with a group action $G \to Aut(A)$. Then a module $(A,G)$ is a crossed modules as $\delta=0:A\to G$. In fact, the subcategory $Mod$ is a Birkhoff subcategory of $\mathbb{X} Mod$. The associated short exact sequence of the $\mathcal{E}
$-torsion theory for a module $(A, G)$ is
\[\begin{tikzcd} 0\ar[r]&A\ar[r]\ar[d]&A\ar[r]\ar[d, "\delta=0"]&0\ar[r]\ar[d]&0 \\ 0\ar[r]&0\ar[r]&G\ar[r] &G\ar[r]&0 \, . \end{tikzcd} \]
\end{rmk}
The category of reduced crossed complexes present a similar example as Proposition \ref{ctfxmod}.
\begin{strc} In $Crs(Grp)$ for each $n\geq 0$ consider the full subcategory $Crs(Grp)_{\geq n}$ of reduced crossed complexes $M$ who are trivial in degrees below $n$:
\[M= \begin{tikzcd}\dots\ar[r] & M_{n+1} \ar[r] & M_n \ar[r] & 0\ar[r] & 0\ar[r] & \dots\, ; \end{tikzcd} \]
for all $n>0$ the category $Crs(Grp)_{\geq n}$ is equivalent to the category $ch(Ab)_{\geq n}\cong ch(Ab)$ of chain complexes in abelian groups.
Thus, for $n \geq 2$ we have a functor $\mathbf{tr'}_n: Crs(Grp) \to Crs(Grp)_{\geq n}$ defined for a crossed complex $M$ by
\[\mathbf{tr'}_n (M)= \begin{tikzcd}\dots\ar[r] & M_{n+1} \ar[r] & M_n \ar[r] & 0\ar[r] & \dots \end{tikzcd}. \]
The natural chain complex morphism $f:M \to \mathbf{tr'}_n(M)$:
\[ \begin{tikzcd}\dots\ar[r] & M_{n+1}\ar[d, "1"] \ar[r] & M_n \ar[r]\ar[d, "1"] & M_{n-1}\ar[r]\ar[d, "0"] & \dots \ar[r] & M_1\ar[r]\ar[d, "0"] & M_0\ar[d, "0"] \\ \dots\ar[r] & M_{n+1} \ar[r] & M_n \ar[r] & 0\ar[r] & \dots\ar[r] & 0\ar[r] & 0 \end{tikzcd} \]
is a morphism in $Crs(Grp)$ if and only if all the actions $M_0\to Aut(M_i)$ are trivial for $i \geq n$. Indeed, $M$ should satisfy $ ^{m_0}m_n= f_n(^{m_0}m_n)= ^{f_0(m_0)}f_n(m_n)=m_n$ for all $m_0 \in M_0$ and $m_n \in M_n$. In particular, this condition holds if $\delta_1: M_1 \to M_0$ is a central extension, since in a crossed complex the restrictions of the actions $\delta_1(M_1) \to Aut(M_0)$ are trivial.
\end{strc}
\begin{prop} For $n \geq 2$, consider the triplet of subcategories:
\[ (Ker(\mathbf{cot}_{n-1}) , Crs(Grp)_{n-1 \geq}, Crs(Grp)_{ \geq n} )\, . \]
in $Crs(Grp)$. Then:
\begin{enumerate}
\item the pair $(Ker(\mathbf{cot}_{n-1}), Crs(Grp)_{n-1 \geq})$ is CTF theory, i.e., the subcategory $Crs(Grp)_{n-1 \geq}$ is mono-coreflective;
\item the pair $(Crs(Grp)_{n-1 \geq}, Crs(Grp)_{ \geq n})$ is an $\mathcal{E}$-torsion theory where $\mathcal{E}$ is the class of crossed complexes $M$ with all actions $M_0 \to Aut(M_i)$ trivial for $i \geq n$;
\item if $M$ is a crossed complex with $\delta_1:M_1 \to M_0$ a crossed module central extension then $M$ belongs to $\mathcal{E}$.
\end{enumerate}
In particular, for $n=2$ this holds for the triplet:
\[ (Ker(\mathbf{cot}_{1}) , \mathbb{X} Mod, chn(Ab)_{\geq 2})\, . \]
\end{prop}
\begin{proof} 1) From Proposition \ref{ttcrs}, $\mu'_{n-1 \geq}=(Ker(\mathbf{cot}_{n-1}) ,Crs(Grp)_{n-1 \geq})$ is a torsion theory. It suffices to notice that the counit of $\mathbf{sk}_{n-1} \dashv \mathbf{tr}_{n-1}$ given by
\[\begin{tikzcd} \dots\ar[r] & 0\ar[r]\ar[d] & M_{n-1}\ar[r]\ar[d] & M_{n-2}\ar[r] \ar[d] & \dots \\ \dots\ar[r] & M_n\ar[r] & M_{n-1}\ar[r] & M_{n-2}\ar[r] & \dots \end{tikzcd} \]
is monic since each component is an injective morphism.
2) It is clear that the pair $(Crs(Grp)_{n-1 \geq}, Crs(Grp)_{ \geq n})$ satisfies TT1 of the definition of a $\mathcal{E}$-torsion theory. Now, let $M$ be a crossed complex with trivial actions $M_0 \to Aut(M_i)$ and consider the morphisms $\mathbf{tr}_{n-1}(M) \to M \to \mathbf{tr'}_n(M)$ in $Crs(Grp)$:
\[\begin{tikzcd}
\dots\ar[r] & 0\ar[r]\ar[d] & 0\ar[r]\ar[d] & M_{n-1}\ar[r]\ar[d] & M_{n-2}\ar[r] \ar[d] & \dots
\\ \dots\ar[r] & M_{n+1}\ar[d] \ar[r] & M_n\ar[r]\ar[d] & M_{n-1}\ar[r]\ar[d] & M_{n-2}\ar[r]\ar[d] & \dots
\\ \dots\ar[r] & M_{n+1} \ar[r] & M_n\ar[r] & 0\ar[r] & 0\ar[r] & \dots
\end{tikzcd} \]
recall that the morphism $M \to \mathbf{tr'}_n(M)$ is indeed a morphism in $Crs(Grp)$ since the actions are trivial. It is a short exact sequence in $Crs(Grp)$ since it is a short exact sequence as chain complexes and the forgetful functor is conservative.
3) It follows from the definition of crossed complex that if $\delta_1$ is surjective the actions $\delta_1(M_1)=M_0\to Aut(M_i)$ are trivial.
\end{proof}
\subsection{A semi-abelian splitting CTF theory}
In an abelian category a torsion theory $(\mathcal{T}, \mathcal{F})$ is called splitting if the torsion subobject $t(X)$ of $X$ is a direct summand. In a semi-abelian category we will call a torsion theory $(\mathcal{T}, \mathcal{F})$ \textit{splitting} if for every object $X$ the associated exact sequence splits:
\[\begin{tikzcd} 0\ar[r] & t(X)\ar[r] & X\ar[r, shift right] & X/t(X)\ar[r]\ar[l, shift right]&0 \end{tikzcd}. \]
In $RMod$ the category of modules over the ring $R$, a central idempotent element of $R$ induces a splitting torsion theory $(\mathcal{T}, \mathcal{F})$ (also called centrally splitting), and even yields a TTF theory $(\mathcal{F}, \mathcal{T}, \mathcal{F})$. Connections of splitting torsion theories and TTF theories are studied in \cite{jans65}.
\begin{exmp} Let $KHopf_{coc}$ the category of cocommutative Hopf alegras over the field $K$ of characteristic 0. In \cite{GKV16}, the category $KHopf_{coc}$ is proved that it is a semi-abelian category and have a torsion theory $(KLie, Grp)$ where the $K$-Lie algebras are consider as the primitive Hopf algebras and $Grp$ is consider as the category of group Hopf algebras. Indeed, the associated short exact sequence is given by the Cartier-Gabriel-Moore-Milnor-Kostant theorem: for every $K$-algebra $H$ there is a split short exact sequence
\[\begin{tikzcd} 0\ar[r] & U(L_H) \ar[r] & H \ar[r, shift right] & K[G_H]\ar[l, shift right] \ar[r] &0 \end{tikzcd} \]
where $K[G_H]$ is the group algebra of the group-like elements $G_H$ of $H$ and $U(L_H)$ is the enveloping algebra of the primitive elements $L_H$ of $H$.
In \cite{GKV18}, it is proved that the funtor $\mathcal{G}:KHopf_{coc}\to Grp$ that takes the group-like elements and $K[\_]: Grp \to KHopf_{coc}$ yield the adjunctions $\mathcal{G}\dashv K[\_]$ and $K[\_]\dashv \mathcal{G}$. So finally, $(KLie, Grp)$ is a splitting CTF theory.
\end{exmp}
\section*{Acknowledgements}
This work is part of the author's Ph.D. thesis \cite{Lop22}. The author would like to thank his advisor Marino Gran for his guidance and suggestions, specially for mentioning the example in Section 5.3.
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 9,097 |
Q: Как можно спозиционировать данные типы картинок с помощью css? Есть два типа фото, с которыми у меня постоянно проблемы:
*
*Когда фото большое (1920px x 1080px) , но сам элемент картинки находится в центре и занимает максимум 1170px, а по бокам уши какого-либо цвета(например, белого). Можно ли как-то с помощью css обрезать эти уши по бокам, что бы width: 100%; отрабатывало корректно?
Пример фото для 1го комментария:
*Тоже самое с фото, только если активный элемент находится слева. Можно ли как-то дать "фокус" на левую сторону картинки?
Пример фото для 2го комментария:
Фото вставляются тегом img
Спасибо
A: Например, так:
*
*завернуть картинку в блок;
*картинке дать отрицательные боковые отступы, чтобы сместить её в нужную сторону;
*увеличить ширину картинки, чтобы её хватило на 100% блока и на добавленные отступы;
*блоку дать overflow: hidden;, чтобы прятал всё, что высовывается за края.
.crop-both-sides,
.crop-right-side {
overflow: hidden;
}
.crop-both-sides > img {
width: 160%;
margin: 0 -30%;
}
.crop-right-side > img {
width: 200%;
margin: 0 -80% 0 -20%;
}
<div class="crop-both-sides"><img alt="" src="https://i.stack.imgur.com/icsDL.jpg"></div>
<div class="crop-right-side"><img alt="" src="https://i.stack.imgur.com/VEPot.jpg"></div>
Еще можно задать картинку фоном и управлять ею с помощью background-size и background-position.
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 7,250 |
{"url":"https:\/\/tex.stackexchange.com\/questions\/157436\/nested-enumerate-with-multicols","text":"Nested enumerate with multicols\n\nI am currently working with nested enumerate environments in which the second level is typeset using a multicols, as follows :\n\n\\documentclass[12pt, letterpaper]{article}\n\n\\usepackage{amsmath}\n\\usepackage{amsfonts}\n\\RequirePackage{amssymb}\n\n\\usepackage[T1]{fontenc}\n\\usepackage[french]{babel}\n\n\\usepackage{comment}\n\\usepackage{enumitem}\n\\usepackage{lmodern}\n\\usepackage{multicol}\n\\usepackage[usenames, dvipsnames, svgnames, table]{xcolor}\n\n\\begin{document}\n\n\\setlength{\\columnsep}{2em}\n\\setlength{\\columnseprule}{0pt}\n\\begin{enumerate}\n\\item \\begin{multicols}{5}\n\\raggedcolumns\n\\begin{enumerate}\n\\item $0$\n\\item $0$\n\\item $-1$\n\\item $-\\infty$\n\\item $\\infty$\n\\item $1$\n\\item $2$\n\\item $2$\n\\item $1$\n\\item $1$\n\\item $2$\n\\item $1$\n\\item $0$\n\\item $\\textcolor {red}\\nexists$\n\\item $1$\n\\item $\\textcolor {red}\\nexists$\n\\end{enumerate}\n\\end{multicols}\n\\end{enumerate}\n\n\\end{document}\n\n\nAlthough the columns are created properly, the last column remains empty. Frank's balancing algorithm seems to provide a 4+4+4+4+0 balancing solution for this five (5) column layout. I would like the layout to use all five columns, given that there are more than a total of five items.\n\nMy questions are:\n\n1. Is this really due to multicols' balancing algorithm, or is it something specific in or missing from my code?\n2. Main question : is there a way to balance the layout using ALL available columns, for example 4+3+3+3+3, that does not require the use of \\columnbreak?\n3. I tried using \\usepackage[balancingshow]{multicol} to see what was happening with the balancing algorithm, but the tracing output is only showing badness for columns 1-4. Is this normal?\n\nNote that if you comment the last \\item, the balancing solution is 3+3+3+3+3, which uses all five (5) columns.\n\nMy motivation for wanting this is that I'm creating a very personalized question\/answer-type package for my colleagues and the answers should be typeset using the \"least possible amount of space\". The presentation does not seem very optimal\/natural when the last column remains empty.\n\nI am aware that multicols' balancing algorithm cannot be optimal for all given situations and that it does a very good job in most cases, along with the fact that the package was created to typeset text.\n\nHere is a solution with theshortlst package and a small patch in order to be able to choose the number of columns. It requires using shortlst, of course, setspace to adjust interlining and xkeyval. As shortlst in not in TeXLive nor MiKTeX (due to problems with its license, as far I know), you'll have to download from CTAN and install it in your local texmf directory.\n\n\\documentclass[12pt, letterpaper]{article}\n\\usepackage[utf8]{inputenc}\n\\usepackage[T1]{fontenc}\n\n\\usepackage{amsmath,amsfonts,amssymb}\n\\usepackage{lmodern}\n\\usepackage[dvipsnames,svgnames,table]{xcolor}\n\n\\usepackage{shortlst,setspace,xkeyval}%\n\\makeatletter\n\\newcounter{ncol}\n\\define@key{lex}{nc}[5]{\\setcounter{ncol}{#1}}%% 5 columns by default\n\\define@key{lex}{il}[1.5]{\\def\\@intln{#1}}% interlining![enter image description here][1]\n\\setkeys{lex}{nc,il,#1}\n\\settowidth{\\labelwidth}{\\mbox{(m)}}\n\\setlength{\\leftmargini}{\\dimexpr\\labelwidth+\\labelsep\\relax}%[1][3]\n\\setlength{\\shortitemwidth}{\\dimexpr\\textwidth\/\\value{ncol}-\\labelwidth-2\\labelsep\\relax}%\n\\renewcommand{\\labelenumi}{\\ensuremath{(\\alph{enumi})}}\n\\setstretch{\\@intln}\n\\begin{shortenumerate}}%\n{\\end{shortenumerate}\n}%\n\\makeatother\n\n\\begin{document}\n\n\\begin{enumerate}\n\\item%\n\\item $0$\n\\item $0$\n\\item $-1$\n\\item $-\\infty$\n\\item $\\infty$\n\\item $1$\n\\item $2$\n\\item $2$\n\\item $1$\n\\item $1$\n\\item $2$\n\\item $1$\n\\item $0$\n\\item $\\textcolor {red}\\nexists$\n\\item $1$\n\\item $\\textcolor {red}\\nexists$\n\\end{enumerate}\n\n\\end{document}\n\n\nwhich results in:\n\n\u2022 Interesting solution! I've used shortlst before, but I ended up switching to a combination of enumitem and multicols for two reasons: 1. I found shortlst not to be easily customizable with regards to spacing, columns, etc. I see you've countered this by \"patching\" its functionalities, which brings me to the second point. 2. For the moment, I'm not sure the use of shortlst is a good choice on the long run, especially given that my entire team is using MiKTeK. I'd much prefer using \"standard\" packages that won't disappear or require \"patches\/hacks\" every now and then. \u2013\u00a0feculededentier Feb 5 '14 at 13:57\n\u2022 @feculededentier: For me the main point is that it works as I want. It's the only package, along with tablists(version 0.0e!) that has these functionalities: putting items in columns while retaining a simple list-like synta. Multienum can do a similar formatting, but with a tabular-like syntax. And it 's the only one that uses 2 (or more) columns for an item automatically. Unfortunately I am not sufficiently aware of the internals of (La)TeX to make a valuable successor to this package that would incorporate my patches\u2026 \u2013\u00a0Bernard Feb 5 '14 at 14:17\n\nedited (2017): no more use of xintfrac as computations can be done with \\numexpr. Only need to load xinttools.\n\nMy two cents guess about the situation creating this stress to multicol is that it arises when you want D columns, have N items (which will occupy the same vertical space, imagine short words to simplify), when N\\leq ceil(N\/D)*(D-1).\n\nFor example N=16, D=5 gives 4*4=16. Bad. D=7 gives 3*6=18. Bad. D=4 gives 4*3=12 good, D=6 gives 3*5=15 good, D=3 gives 6*2=12 good.\n\n\\input xinttools.sty % \\xintApply, \\xintListWithSep, \\\u00bbintSeq\n\\def\\GoodDivisions #1{For #1:\n\\xintListWithSep{, }{\\xintApply{\\TestGoodness {#1}}{\\xintSeq {1}{#1}}}}\n\n\\def\\Ceil #1#2{\\numexpr(#1+#2\/2-1)\/#2\\relax}\n\n\\catcode@ 11\n\\long\\def\\@firstoftwo#1#2{#1}\n\\long\\def\\@secondoftwo#1#2{#2}\n\n\\def\\TestGoodness #1#2{#2 is\n\\ifnum #1>\\numexpr\\Ceil{#1}{#2}*(#2-1)\\relax\n\\expandafter\\@firstoftwo\n\\else\n\\expandafter\\@secondoftwo\n\\fi\n\\catcode@ 12\n\n\\GoodDivisions {16}\n\n\\GoodDivisions {20}\n\n\\GoodDivisions {32}\n\n\\GoodDivisions {23}\n\n\\nopagenumbers\n\n\\bye\n\n\nThese predictions are untested! (works for 16 and 20 items at least.\n\nHere is now a table giving for each number of items up to 60 the compatible choices for a multicol with D columns.\n\n\\documentclass[a4paper]{article}\n\\usepackage[vscale=0.9]{geometry}\n\\usepackage{color}\n\n\\usepackage{xinttools}\n\n\\newcommand*\\TestColumns[1]{%\n\\xintListWithSep{&}{\\xintApply{\\TestGoodness {#1}}{123456789{10}}}%\n}\n\n\\makeatletter\n\\newcommand*\\TestGoodness[2]{%\n\\ifnum #1>\\numexpr\\Ceil{#1}{#2}*(#2-1)\\relax\n\\expandafter\\@firstoftwo\n\\else\n\\expandafter\\@secondoftwo\n\\fi G{\\textcolor{red}{B}}%\n}\n\\makeatother\n\n\\newcommand*\\Ceil[2]{\\numexpr(#1+#2\/2-1)\/#2\\relax}\n\n\\begin{document}\\thispagestyle{empty}\n\n\\begin{tabular}{c*{10}c}\n&1&2&3&4&5&6&7&8&9&10\\\\\n\\hline\n\\xintFor* #1 in {\\xintSeq {1}{60}}\n\\do {#1&\\TestColumns {#1}\\\\}\n\\hline\n\\end{tabular}\n\n\\end{document}\n\n\u2022 my predictions for 20 items seem validated! the 9 columns case is interesting, multicol does then only 7 columns, 20=6x3+2. \u2013\u00a0user4686 Feb 4 '14 at 19:04\n\u2022 so far everything I have tested (N=16, N=19, N=20, N=48, N=49 etc..) is compatible with the predictions from the table... \u2013\u00a0user4686 Feb 4 '14 at 19:51\n\u2022 Interesting analysis! I'd be intrigued to know whether this analysis is accurate, because it implies that the solution in which every column has the same amount of equally sized elements, except for the first column which contains an extra element is always bad. Even for displaying text, I don't see how this output is \"bad\" or undesirable. \u2013\u00a0feculededentier Feb 5 '14 at 13:44\n\u2022 @feculededentier the first case is with 4 lines we want to have over 3 columns:\\documentclass{article} \\usepackage{multicol} \\begin{document} \\begin{multicols}{3} first line\\newline second line\\newline third line\\newline fourth line \\end{multicols} \\end{document} multicol understandably excludes doing 1+1+2 but does not seem to consider making 2+1+1, and ends up doing 2+2, thus leaving an empty third column. \u2013\u00a0user4686 Feb 5 '14 at 14:10\n\u2022 @feculededentier from the table above using 3 columns is safe if the text ends up on at least 5 lines, and using 4 columns is safe for text ending up on at least 10 lines, and using 5 columns for text ending up on at least 17` lines. Naturally, regular text might have enough stretch to move a word to an additional line and thus the stress on multicols is more easily seen when doing things like in your post, with one-liners, rather than full paragraphs allowing reflowing of text. \u2013\u00a0user4686 Feb 5 '14 at 14:15","date":"2020-05-25 11:59:48","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\": 0.99315345287323, \"perplexity\": 1926.0697962052361}, \"config\": {\"markdown_headings\": false, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2020-24\/segments\/1590347388427.15\/warc\/CC-MAIN-20200525095005-20200525125005-00523.warc.gz\"}"} | null | null |
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Is there a way to play Legend of Zelda: Majora's Mask on the PC? | {
"redpajama_set_name": "RedPajamaC4"
} | 3,011 |
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Learn more about how the LawStudio cloud-based collaboration platform can save you time and money. | {
"redpajama_set_name": "RedPajamaC4"
} | 274 |
Het Provinciaal Domein Dommelhof is een instelling van de provincie Belgisch Limburg voor cultuur, podiumkunsten en sport in de gemeente Pelt. Dommelhof werd geopend in 1967. De instelling was het eerste cultureel centrum in Vlaanderen.
Geschiedenis
In 1962 nam de provincieraad van Belgisch Limburg het besluit om Dommelhof op te richten. Toenmalig minister voor cultuur Frans Van Mechelen besliste om de bouw van de culturele infrastructuur te subsidiëren. Deze infrastructuur naar een ontwerp van Alfons Hoppenbrouwers omvatte een theater-, vormings- en verblijfsgebouw. In 1967 was de eerste schouwburgvoorstelling een feit. Tot de jaren '80 werd inhoudelijk vooral ingezet op sociaal-cultureel vormingswerk en een experimentele programmatie in de schouwburg. In 1979 werd de site uitgebreid met een sportcentrum.
In de jaren '80 werd een internationaal poppenfestival uitgebouwd en werd gestart met een zomerfestival voor openluchttheater. Het poppenfestival is intussen verdwenen en het festival voor openluchttheater groeide in de volgende jaren uit tot het internationaal festival voor openlucht-, locatie- en circustheater Theater op de Markt. Ook het platform voor arthousecinema in Limburg, Zebracinema, ontstond in Dommelhof.
Werking
De huidige activiteiten van Dommelhof omvatten:
Theater op de Markt: spreiding en creatie van circus- en openluchttheater via een zomerfestival en een herfstfestival, en een productiecel voor (inter)nationale creatie-opdrachten en residenties
C-TAKT: ondersteuning van transdisciplinair talent i.s.m. o.a. C-mine Cultuurcentrum Genk
Sportcentrum: faciliteren van bovenregionale sportactiviteiten
G-Sportcentrum: sportactiviteiten voor mensen met een beperking aanbieden
Verblijfsgebouw: faciliteren van meerdaagse vorming voor het socio-cultureel veld, alsook voor G-sportkampen
Partners
Dommelhof werkt structureel samen met regionale en Vlaamse culturele partners:
Musica: impulscentrum voor muziek, en beheerder van het klankenbos op de Dommelhof-site
Zebracinema: arthouse-cinema in Belgisch Limburg
Circuscentrum: Vlaams ankerpunt voor circuskunsten
JazzCase: jazzplatform Noord-Limburg
Zie ook
Lijst van provinciedomeinen in Limburg (België)
Klankenbos
Externe links
Provinciaal Domein Dommelhof
Theater op de Markt
TAKT
Zebracinema
Musica
Circuscentrum
JazzCase
Theaterzaal in Vlaanderen
Neerpelt
Cultuur in Limburg (België)
Cultuurcentrum in Vlaanderen
Provinciaal domein in Limburg (België) | {
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Fuzzy Notes!
Fuzzy Notes 87 - UnRambled
Harkening back to the olden days of Fuzzy Notes (...like...2013?) this is a short, bite-sized...dare I say it 'B-Sides' like episode with three tracks by artists we've heard from before! Also...devoid of Roo Rambles (send praise or complaints to @Potoroo on Twitter). Hear from Jerms, Tristyn and Eternal Infamy in today's smidgeny episode! Fuzzy Notes is a podcast featuring the music made by members of the furry fandom, with chatter and rambles from it's host Potoroo (or Roo for short!) It started in 2012 and has featured over a hundred artists and nearly three hundred songs. Currently, Fuzzy Notes is being moved to a new home where you will be able to hear the entire series from Episode One. Please keep an eye and ear out for the big move in early 2015. THE MUSIC: Jerms - Gotta Truckin' Go * Track: https://soundcloud.com/jeremy-allen-15/gotta-truckin-go * Soundcloud: https://soundcloud.com/jeremy-allen-15/ Tristyn - Collapsing Flow * Track: http://www.furaffinity.net/view/15625097/ * FA: http://www.furaffinity.net/user/tristyn/ Eternal Infamy - The Writings On The Wall * Track: http://www.furaffinity.net/view/15706325/ * FA: http://www.furaffinity.net/user/syrieltiger/ * Web: http://eternalinfamy.com --- HOUSEKEEPING: FIND ROO TO GIVE HIM MUSIC! * Twitter: http://twitter.com/Potoroo * Facebook: http://www.facebook.com/FuzzyNotes * Fuzzy Notes Dropbox: http://soundcloud.com/fuzzy-notes/dropbox HEY! Please give Fuzzy Notes a review on iTunes or Podomatic...it helps us get furry musicians to more ears. Which is the whole point, isn't it?? * Android/RSS Feed: http://fuzzynotes.podomatic.com/rss2.xml * iTunes: https://itunes.apple.com/us/podcast/id581600769 * Podomatic: http://fuzzynotes.podomatic.com Also find Fuzzy Notes on Stitcher Smart Radio!
Fuzzy Notes 86 - There You Go, Now You Know
There you go! Now you know! In this very swallowy episode (Roo apologizes...for some reason his mouth was very watery this episode) we hear a wide variety of music from some great furry artists including newcomers Palm Village and DJ Limination, and returning alum PartyWolf17, NIIC and ODIN's Disciple. Oh...and Roo plays one of his old band's tracks...randomly in the episode... Also, you get to hear Roo go FULL TIME MACHINE at one point. See if you can catch it...he seems to think he's in high school again o.O Fuzzy Notes is a podcast featuring the music made by members of the furry fandom, with chatter and rambles from it's host Potoroo (or Roo for short!) It started in 2012 and has featured over a hundred artists and nearly three hundred songs. Currently, Fuzzy Notes is being moved to a new home where you will be able to hear the entire series from Episode One. Please keep an eye and ear out for the big move in early 2015. THE MUSIC: Palm Village - I've Got Bigger Things to Worry About Than You (Irony's Sway) * Track: http://www.furaffinity.net/view/13451346/ * FA: http://www.furaffinity.net/user/leslichu/ DJ Limination - Synergy * Track: https://soundcloud.com/limination-1/synergy-1 * Soundcloud: https://soundcloud.com/limination-1 * FA: http://www.furaffinity.net/view/15190441/ * Google+: https://plus.google.com/+StaffordOtter/ PartyWolf17 (aka. Sonny B) - Misunderstood * Track: http://www.furaffinity.net/view/15467759/ * FA: http://www.furaffinity.net/user/partywolf17/ NIIC - It's How We Do * Soundcloud: https://soundcloud.com/niicmusic * FA: http://www.furaffinity.net/user/NIIC * Instinct (iTunes): https://itunes.apple.com/us/album/instinct/id941004746?ign-mpt=uo%3D4 * Instinct (Bandcamp): https://niic.bandcamp.com/album/instinct Brillohead - Duct Tape (Roo's Band from 1994) Odin's Disciple - Law of Fives * Track: https://soundcloud.com/odins-disciple/law-of-fives * Soundcloud: https://soundcloud.com/odins-disciple/ DaSilva - True Love * Track: https://soundcloud.com/dasilva_music/true-love * Soundcloud: https://soundcloud.com/dasilva_music/ Two Bonus Tracks: Jess Moskaluke - Cheap Wine and Cigarettes **Gold single by one of Roo's friends. She was just nominated for two Juno Awards: 'Country Album of the Year' and 'Breakthrough Artist of the Year' Brillohead - Bamboo Shoots **Not very well recorded, but compare this song to 'Duct Tape' played earlier in the episode. Roo included 'Duct Tape' as an example of why you should hold off releasing a song right after you record it, especially early in your songwriting/recording life. Some of your early work may sound awesome at the time, but sound rough when compared to your later songs. 'Bamboo Shoots' is EXACTLY the same band - Roo and his high school band Brillohead. This song was written and recorded about two years later. Unlike 'Duct Tape,' 'Bamboo Shoots' is a song being played by a tight band, the chords progression and instrumentation is far more complicated, and if you ignore the fact that Roo wrecked his voice recording the vocals are fare more interesting. Yet, in that last moment where you hear the whole band screaming out the names of canned food, the same silly-goofy nature from 'Duct Tape' carried through even as the band improved. ...I guess what I'm saying is, love your early stuff and learn from it, but give yourself the chance to decide what you share with the world and what you keep to yourself. And if you want to share something with the world you don't have to do it right away: once you have that finished product in your hands you now have all the power in the world to decide how to share it, when to share it. If I didn't play 'Duct Tape' on this show, it quite literally would have only sat as an MP3 on two people's computers (unless the other person had deleted it). I chose to release it here, now, about 22 years later. But if 'then was now' I might have put it up on my Bandcamp two years ago when I first wrote it, and now when I'm releasing my 'Bamboo Shoots' songs it would be out in the world. ...damn, 'Bloggy Show Notes'! --- HOUSEKEEPING: FIND ROO TO GIVE HIM MUSIC! * Twitter: http://twitter.com/Potoroo * Facebook: http://www.facebook.com/FuzzyNotes * Fuzzy Notes Dropbox: http://soundcloud.com/fuzzy-notes/dropbox HEY! Please give Fuzzy Notes a review on iTunes or Podomatic...it helps us get furry musicians to more ears. Which is the whole point, isn't it?? * Android/RSS Feed: http://fuzzynotes.podomatic.com/rss2.xml * iTunes: https://itunes.apple.com/us/podcast/id581600769 * Podomatic: http://fuzzynotes.podomatic.com Also find Fuzzy Notes on Stitcher Smart Radio!
Fuzzy Notes 85 - Tricentennial Twenty Fifteen
Technological chaos, throat-destroying sickness, travels to the Sooner State and holiday obligations kept Fuzzy Notes off the air for way too long...but we're back with more music made by furs! Join Fuzzy Notes alum Bunnies for Real and Asid Rayne, and new comers Skootums, Kenko Sabre Wolf, FeoranPride and Zackwell in the first episode of 2015! Fuzzy Notes is a podcast featuring the music made by members of the furry fandom, with chatter and rambles from it's host Potoroo (or Roo for short!) It started in 2012 and has featured over a hundred artists and nearly three hundred songs. Currently, Fuzzy Notes is being moved to a new home where you will be able to hear the entire series from Episode One. Please keep an eye and ear out for the big move in early 2015. NOTE: The song 'Elephant Song' by Bunnies for Real aka. Flipp The Bunny is ACTUALLY the real 'Elephant Song'. In episode 83, Roo accidentally played the song '43v3r' but called it 'Elephant Song'. He admits that he was dumb. We apologize for the mistake!! THE MUSIC: Bunnies for Real - Elephant Song * Soundcloud: https://soundcloud.com/bunniesforreal/ * FA: http://www.furaffinity.net/user/flippthebunny * Twitter: https://twitter.com/flippthebunny Namo - One Man In The Red Room * Track: http://www.furaffinity.net/view/14911665/ * Soundcloud: https://soundcloud.com/namooo FeoranPride - November * Track: http://www.furaffinity.net/view/14928799/ * FA: http://www.furaffinity.net/user/feoranpride/ * Animation on Vimeo: http://vimeo.com/user13747140 Zackwell (as T3G)- That's A Feeling * Track: http://www.furaffinity.net/view/14921797/ * FA: http://www.furaffinity.net/user/zackwelldaesther/ * T3G on FB: https://www.facebook.com/Tee.Three.Gee Kenko Sabre Wolf Discussing his music: * a short audio doc...I sent Kenko some questions, he sent me these answers. Note, I cut out the parts where he repeated the questions I asked, and edited it together to be a bit more like a short documentary, but his answers are very VERY awesome so you'll get it. I'm hoping to do more little 'Mini Documentaries' here and there, so please let me know what you think about this and if you want to hear more from furry musicians in episodes of Fuzzy Notes! Kenko - Worms Armageddon * FA: http://www.furaffinity.net/user/kenkosaberwolf/ Asid Rayne - Miss Murder (Cover) * Track: http://www.furaffinity.net/view/15401226/ * FA: http://www.furaffinity.net/user/tundra-gevin/ * Bandcamp: https://asidrayne.bandcamp.com HOUSEKEEPING: FIND ROO TO GIVE HIM MUSIC! * Twitter: http://twitter.com/Potoroo * Facebook: http://www.facebook.com/FuzzyNotes * Fuzzy Notes Dropbox: http://soundcloud.com/fuzzy-notes/dropbox HEY! Please give Fuzzy Notes a review on iTunes or Podomatic...it helps us get furry musicians to more ears. Which is the whole point, isn't it?? * Android/RSS Feed: http://fuzzynotes.podomatic.com/rss2.xml * iTunes: https://itunes.apple.com/us/podcast/id581600769 * Podomatic: http://fuzzynotes.podomatic.com Also find Fuzzy Notes on Stitcher Smart Radio!
Episode 84 - CODA11: Husky In Denial (Part 1)
The most confusing title in the history of Fuzzy Notes: It is the 84th episode total, it is also another one of our special CODA episodes (the 11th) and it's being split into two parts (making it Part 1). So, really...all you need to know is that Roo is talking to HUSKY IN DENIAL!!! Don't worry about all the numbery stuff! Roo chats with Husky In Denial about music, music, music! Part 2 of the podcast will be released soon, continuing the conversation about music, music, music...and furry. FIND ROO TO GIVE HIM MUSIC! * Twitter: http://twitter.com/Potoroo * Facebook: http://www.facebook.com/FuzzyNotes * Fuzzy Notes Dropbox: http://soundcloud.com/fuzzy-notes/dropbox FIND HUSKY IN DENIAL ONLINE! * Bandcamp (Kung-Fu EP out NOW): http://huskyindenial.bandcamp.com * Twitter: http://twitter.com/HuskyInDenial * Facebook: http://facebook.com/HuskyInDenial HEY! Please give Fuzzy Notes a review on iTunes or Podomatic...it helps us get furry musicians to more ears. Which is the whole point, isn't it?? * Android/RSS Feed: http://fuzzynotes.podomatic.com/rss2.xml * iTunes: https://itunes.apple.com/us/podcast/id581600769 * Podomatic: http://fuzzynotes.podomatic.com
Episode 83 - Ramble Rouser
We're BACK! It's Roo, the host with the most (pink) with another multi-genre collection of music from electronic to a live recording of a piano concerto to some hard hitting dance tracks to...well...Roo rambling! And ooooooooooh does he ramble today! Expect tracks from Zoa, Draks, Flipp The Bunny, DJ Ear remixing Bandit, Vixavian and a return to form with Roo's ramble about the importance of context, clarity and converstaion when criticizing creativity! (...FIVE C's!) FIND ROO TO GIVE HIM MUSIC! * Twitter: http://twitter.com/Potoroo * Facebook: http://www.facebook.com/FuzzyNotes * Fuzzy Notes Dropbox: http://soundcloud.com/fuzzy-notes/dropbox HEY! Please give Fuzzy Notes a review on iTunes or Podomatic...it helps us get furry musicians to more ears. Which is the whole point, isn't it?? * Android/RSS Feed: http://fuzzynotes.podomatic.com/rss2.xml * iTunes: https://itunes.apple.com/us/podcast/id581600769 * Podomatic: http://fuzzynotes.podomatic.com Also find Fuzzy Notes on Stitcher Smart Radio! THE MUSIC: Zoa - Molto Allegro * FA: http://www.furaffinity.net/user/zoasounds/ * Track: http://www.furaffinity.net/view/14831624/ Draks - Piano Concerto in Bb Major (Live) * FA: http://www.furaffinity.net/user/draks/ * Track: http://www.furaffinity.net/view/14619550/ Bunnies For Real - 43v3r NOTE: In the episode, Roo refers to this song as 'Elephant Song' but played the wrong track!! The ACTUAL Elephant Song appears on Fuzzy Notes Episode 85. We apologize for the mistake!! * Soundcloud: https://soundcloud.com/bunniesforreal/ * FA: http://www.furaffinity.net/user/flippthebunny * Twitter: https://twitter.com/flippthebunny Bandit - Simulation (EARs Remix) * Track: https://bandittheraccoon.bandcamp.com/track/simulation-dj-ear-remix-2 * DJ EAR'S TWITTER: https://twitter.com/huepow00 * BANDIT'S TWITTER: https://twitter.com/Bandit_Raccoon Vixavian - Stripes * Track: http://www.furaffinity.net/view/14913952/ * FA: http://www.furaffinity.net/user/proceleon/ Chíl - 27 (with intro from Chíl!) * Track: http://www.furaffinity.net/view/15074917/ * FA: http://www.furaffinity.net/user/goja-/ | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 4,720 |
Ralph Alexander Raphael (1 January 1921 – 27 April 1998) was a British organic chemist, well known for his use of acteylene derivatives in the synthesis of natural products with biological activity.
Early life and education
Ralph Raphael was born in Croydon, London on New Year's Day 1921, the son of master tailor Jacob ("Jack") Raphael (1889-1978) and his wife, Lily (née Woolf; 1892-1956). He attended secondary school at Wesley College, Dublin and then Tottenham County School, where a chemistry master, Edgar Ware, introduced him to the subject that would become Raphael's lifetime passion. In 1939 he won scholarships to study at Imperial College, graduating BSc with a first-class degree in 1941 and winning the Hofmann Prize for practical chemistry.
During the Second World War both the undergraduate and PhD courses at Imperial College were of two year's duration and Raphael completed the latter in 1943. His doctoral work, aimed at the synthesis of vitamin A, was published in five collaborative papers on the chemistry of acetylenes and that topic became a hallmark of his subsequent research career.
Career
As a new PhD, Raphael was allocated to the wartime effort on the antibiotic penicillin, working from 1943 to 1946 at the May & Baker laboratories. After the war, he obtained an ICI fellowship (for 1946–1949) that allowed him to return to Imperial College to pursue independent research: an early highlight was his synthesises of penicillic acid, the major product of acid degradation of penicillin (although not containing its characteristic β-lactam substructure). Another was his collaboration with Franz Sondheimer on natural products including an insecticide extracted from Zanthoxylum clava-herculis (a diene then called herculin, now systematically named as (2E,8E)-N-isobutyl-2,8-dodecadienamide); work which led to Raphael's award of the Meldola Medal in 1948.
In 1949, Raphael was appointed to his first permanent job, as a lecturer at Glasgow University. During this period he developed his teaching skills and his prodigious work rate can be judged by the fact that he also completed nine chapters in one volume of what would become a classic chemistry text. In 1954, Raphael moved to Queen's University, Belfast as its first Professor of Organic Chemistry. There he published an important book on acetylene chemistry, building on his broad experience of these compounds.
In 1957, Raphael returned to the University of Glasgow as the Regius Professor of Chemistry,. In 1960 he finished work on a text-book for undergraduates, which was updated and re-issued several times. In 1972 Raphael became head of the Department of Organic, Inorganic and Theoretical Chemistry at Cambridge University. This post had been made vacant by the retirement, on ill-health grounds, of Lord Todd, the previous holder of the 1702 Chair in Organic Chemistry. Raphael also became a Fellow of Christ's College. On retirement in 1988 he was granted emeritus status within his college and department, reflecting his distinguished service.
Teaching and research
Despite having a slight stammer, Professor Raphael was an inspiring lecturer who engaged his undergraduate students with up-to-date material on organic chemistry, based on his extensive knowledge of the current literature. He had an excellent sense of humour, illustrated by Dudley Williams's report that"he delivered spoof lectures. One on the synthesis of catenanes began with serious chemistry and gradually — imperceptibly — became less credible; it culminated in the description of their absorption spectra in the audible region" The output of Raphael's own work and that of his research group of postgraduate and postdoctoral students was published in over 150 peer-reviewed articles.
Raphael was funded by external grants, including those from the SERC, NRC Canada, Glaxo Smith Kline, Hoffmann-La Roche and ICI, for whom he was a retained consultant. He also consulted for Beecham Group, Chiroscience and Fisons. His consultancy and other work led to a number of patent filings.
Synthesis of natural products
Raphael studied many natural products, especially of the type that were biologically active and which would provide a challenge for synthesis but might be the realistic target of a single PhD student's thesis. He and his students published syntheses of 2-deoxyribose, aaptamine, aphidicolin, apiose, arachidonic acid, arcyriaflavin B, baikiain, bullatenone, chrysanthemic acid, clovene, cordycepose, cuparene, erythrulose, exaltolide, farnesiferol C, geiparvarin, gibberone, histamine, linoleic acid, linolenic acid, lipoic acid, pseudomonic acid, pyrenophorin, Queen bee acid, shikimic acid, staurosporinone, strigol, steganacin, steganone, trichodermin and virantmicin. Raphael also investigated the composition of the wax coating of plant leaves, describing the hydrocarbons of which they are composed. In another intriguing publication in Nature, Raphael collaborated with David Rubio to identify components used in the surface treatment of the wood of stringed instruments made by Stradivarius in Cremona and showed that a version of these substances could be used to improve the tone of modern instruments.
Molecules of theoretical interest
Raphael was interested in molecules of theoretical, as well as practical, interest. In 1951, co-worked and co-authored with J. W. Cook and A. I. Scott, he published the first synthesis of the quasi-aromatic compound tropolone and the thujaplicin natural products which contained this unusual ring system. His interest in acetylenes led him to study macrocyclic compounds containing this functional group, and bridged ring systems that could be derived from them. Before the first synthesis by Ralph Raphael, thujaplicins had been naturally isolated from Chamaecyparis taiwanensis by Tetsuo Nozoe in 1936 (the β-isomer; hinokitiol), and from Thuja plicata independently by Holger Erdtman in 1948 (all three isomers; α-, β- and γ-thujaplicins).
Honours, awards and service to the scientific community
In 1958 Raphael was elected a Fellow of the Royal Society of Edinburgh. His proposers were John Monteath Robertson, James Norman Davidson, Robert Campbell Garry, and Guido Pontecorvo. In 1962 he was also elected a Fellow of the Royal Society of London; he was the Davy Medalist for the latter in 1981. He was appointed a CBE in the Honours list of June 1982.
Raphael received Honorary Doctorates from his alma mater Imperial College, in 1991, Stirling University in 1982, the University of East Anglia in 1986, and Queen's University Belfast in 1989,
Among his many awards and service to learned bodies were:
Meldola Medal and Prize (1948)
Tilden Medal (1960)
Vice-President of the Chemical Society, London (1967-1970)
Chemistry Committee Member of SRC (1968)
Pedler Award (1973)
Ciba-Geigy Sponsored Award for Synthetic Chemistry (1975)
Member of the Council of the Royal Society (1975-1977)
President of the Perkin Division of the Royal Society of Chemistry (1979-1981)
Visiting Professor, Hebrew University of Jerusalem (1981)
Honorary membership of the Royal Irish Academy (1987)
Visiting Professor, University of Hong Kong (1989)
Visiting Professor, Université of Haute Alsace (1990)
Personal life
In 1944 Raphael married Prudence Maguerire Anne née Gaffikin, who was a professional violin and viola player. They had a son, Tony, and a daughter, Sonia. By 1998 there were two grandchildren and a great-granddaughter. Ralph Raphael was keenly interested in the visual and performing arts, becoming a member of the Scientific Advisory Committee of the National Gallery in 1986; his favourite pastime was contract bridge. Raphael died of ischaemic heart disease, in Cambridge on 27 April 1998.
Further reading
References
1921 births
1998 deaths
British chemists
Organic chemists
Fellows of the Royal Society
Fellows of the Royal Society of Edinburgh
British Jews
Jewish scientists
Academics of the University of Glasgow
Professors of chemistry (Cambridge, 1702)
Fellows of Christ's College, Cambridge
People educated at Wesley College, Dublin | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 8,045 |
from __future__ import unicode_literals
from django.db import migrations, models
class Migration(migrations.Migration):
dependencies = [
('portal', '0003_auto_20170617_1222'),
]
operations = [
migrations.AddField(
model_name='newsandupdate',
name='slug',
field=models.SlugField(default=1, unique=True),
preserve_default=False,
),
migrations.AlterField(
model_name='newsandupdate',
name='body',
field=models.TextField(blank=True),
),
migrations.AlterField(
model_name='newsandupdate',
name='category',
field=models.CharField(choices=[('realese', 'New release'), ('bug', 'Bug fix'), ('announcement', 'Announcement'), ('other', 'Other')], max_length=150),
),
]
| {
"redpajama_set_name": "RedPajamaGithub"
} | 9,021 |
<?php
/**
* W3 Config object
*/
/**
* Class W3_ConfigBase
*/
class W3_ConfigBase {
/**
* Array of config values
*
* @var array
*/
protected $_data = array();
/**
* Constructor
*/
function __construct($data) {
$this->_data = $data;
}
/**
* Returns string value
*
* @param string $key
* @param string $default
* @param boolean $trim
* @return string
*/
function get_string($key, $default = '', $trim = true) {
$value = (string)$this->get($key, $default);
return ($trim ? trim($value) : $value);
}
/**
* Returns integer value
*
* @param string $key
* @param integer $default
* @return integer
*/
function get_integer($key, $default = 0) {
return (integer)$this->get($key, $default);
}
/**
* Returns boolean value
*
* @param string $key
* @param boolean $default
* @return boolean
*/
function get_boolean($key, $default = false) {
return (boolean)$this->get($key, $default);
}
/**
* Returns array value
*
* @param string $key
* @param array $default
* @return array
*/
function get_array($key, $default = array()) {
return (array)$this->get($key, $default);
}
/**
* Sets config value.
* Method to override
*
* @param string $key
* @param string $value
* @return value set
*/
function set($key, $value) {
$this->_data[$key] = $value;
return $value;
}
/**
* Returns config value. Implementation for overriding
*
* @param string $key
* @param mixed $default
* @return mixed
*/
function get($key, $default = null) {
if (array_key_exists($key, $this->_data)) {
$v = $this->_data[$key];
return $v;
}
return $default;
}
}
| {
"redpajama_set_name": "RedPajamaGithub"
} | 1,618 |
Q: Optional navigation property is not being loaded I can't for the life of me figure out why Entity Framework (4.1, Code First) is not materializing an optional navigation property.
POCOs:
public class PlanMember {
public Int64 Id { get; set; }
public String FirstName { get; set; }
public virtual SystemUser CaseManager { get; set; }
public String LastName { get; set; }
public virtual PlanMemberStatus Status { get; set; }
}
public class SystemUser {
public String EMail { get; set; }
public String FirstName { get; set; }
public String Id { get; set; }
public String LastName { get; set; }
}
public class PlanMemberStatus {
public Int32 Id { get; set; }
public String Code { get; set; }
}
Configuration Classes:
public class ForPlanMemberEntities:EntityTypeConfiguration<PlanMember> {
public ForPlanMemberEntities(String schemaName) {
this.HasOptional(e => e.CaseManager)
.WithMany()
.Map(m => m.MapKey("CaseManagerID"));
this.HasRequired(e => e.Status)
.WithMany()
.Map(m => m.MapKey("StatusID"));
this.ToTable("PlanMember", schemaName);
}
}
public class ForSystemUserEntities:EntityTypeConfiguration<SystemUser> {
public ForSystemUserEntities(String schemaName) {
this.ToTable("SystemUser", schemaName);
}
}
public class ForPlanMemberStatusEntities:EntityTypeConfiguration<PlanMemberStatus> {
public ForPlanMemberStatusEntities(String schemaName) {
this.ToTable("PlanMemberStatus", schemaName);
}
}
Context:
public class Context:DbContext {
public DbSet<PlanMember> PlanMemberDbSet { get; set; }
protected override void OnModelCreating(DbModelBuilder modelBuilder) {
base.OnModelCreating(modelBuilder);
var schemaName = Properties.Settings.Default.SchemaName;
modelBuilder.Configurations
.Add(new Configuration.ForPlanMemberEntities(schemaName))
.Add(new Configuration.ForPlanMemberStatusEntities(schemaName))
.Add(new Configuration.ForSystemUserEntities(schemaName))
;
}
}
Based on that configuration, if I run a query like the following it will not populate the CaseManager property. The PlanMemberStatus property loads just fine.
var test = (from member in PlanMemberDbSet
where member.CaseManager != null
select member).First();
var cm = test.CaseManager;
In my example cm is always null, despite the fact that - logically speaking - it should be impossible. Any thoughts?
EDIT - If I do PlanMemberDbSet.Include("CaseManager") then it works as expected. Why is this necessary? In other scenarios I haven't had to do this.
A: My guess is that the CaseManagerID foreign key column value in the PlanMember table is different to the Id primary key column value in table SystemUser - different for .NET but not for SQL Server (with standard sorting system), for example different in casing: "a" in CaseManagerID but "A" in Id.
Querying for != null (translates into IS NOT NULL in SQL) works, so you get indeed only the PlanMember entities which have a SystemUser assigned in the database. Also the query by the CaseManager (triggered by lazy loading) is correctly executed in the database (INNER JOIN) and transmitted to the client. Even the object materialization works (there are 2 objects in the context). But EF fails to relate the loaded PlanMember with the loaded CaseManager because the keys are not identical (for example with respect to capital/small letters). As a result your navigation property is null.
This doesn't occur with the Status navigation property because the key is an Int32.
The problem looks closely related to this one: EF 4.1 Code First bug in Find/Update. Any workaround? Should it be reported?
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 9,944 |
I was very impressed with Hito Steyerl when I listened to her lecture in Former West. Mischievous, bold and uncanny, her delivery was impeccable and she made no compromise, neither personal nor theoretical. Known for her writing and her video work, this is an artist who is as much a thinker as a doer. Why should there be a distinction? Today we think through making…but then again, when was art ever different?
Poor images are the contemporary Wretched of the Screen, the debris of audiovisual production, the trash that washes up on the digital economies' shores. They testify to the violent dislocation, transferrals, and displacement of images—their acceleration and circulation within the vicious cycles of audiovisual capitalism. Poor images are dragged around the globe as commodities or their effigies, as gifts or as bounty. They spread pleasure or death threats, conspiracy theories or bootlegs, resistance or stultification. Poor images show the rare, the obvious, and the unbelievable—that is, if we can still manage to decipher it. | {
"redpajama_set_name": "RedPajamaC4"
} | 251 |
describe("Dependency injector", function() {
var api = require('../../index.js')();
it("should have the injector API available", function(done) {
expect(api.di).toBeDefined();
expect(api.di).not.toBe(null);
done();
});
it("should define a dependency and use it", function(done) {
api.di.register("AwesomeModule", function() {
return 42;
});
var doSomething = api.di.resolve(function(AwesomeModule) {
expect(AwesomeModule()).toBe(42);
done();
});
doSomething();
});
it("should use other parameters", function(done) {
api.di.flush().register("AwesomeModule", function() {
return 30;
});
var doSomething = api.di.resolve(function(a, AwesomeModule, b) {
expect(a).toBe(2);
expect(b).toBe(10);
expect(AwesomeModule() + a + b).toBe(42);
done();
});
doSomething(2, 10);
});
it("should protect from minification", function(done) {
api.di.flush()
.register("AwesomeModule", function() {
return 30;
})
.register("UselessModule", function() {
return 'absolutely nothing';
});
var doSomething = api.di.resolve(',AwesomeModule,,UselessModule', function(a, AwesomeModule, b, UselessModule) {
expect(a).toBe(2);
expect(b).toBe(10);
expect(AwesomeModule() + a + b).toBe(42);
expect(UselessModule()).toBe('absolutely nothing');
done();
});
doSomething(2, 10);
});
it("should keep the scope", function(done) {
api.di.flush().register("AwesomeModule", function() {
return 42;
});
var component = {
answer: 0,
print: function() {
return "The answer is " + this.answer;
},
doSomething: function(AwesomeModule) {
this.answer = AwesomeModule();
expect(this.print()).toBe("The answer is 42");
done();
}
}
component.doSomething = api.di.resolve(component.doSomething, component);
component.doSomething();
});
it("should resolve an object", function(done) {
api.di.flush().register("AwesomeModule", function() {
return 42;
});
var component = {
answer: 0,
print: function() {
return "The answer is " + this.answer;
},
doSomething: function(AwesomeModule) {
this.answer = AwesomeModule();
expect(this.print()).toBe("The answer is 42");
done();
}
}
api.di.resolveObject(component);
component.doSomething();
});
it("should resolve an object protected from minification", function(done) {
api.di.flush()
.register("AwesomeModule", function() {
return 42;
})
.register("UselessModule", function() {
return 'absolutely nothing';
});
var component = {
answer: 0,
print: function() {
return "The answer is " + this.answer;
},
doSomething: ['AwesomeModule, UselessModule', function(AwesomeModule, UselessModule) {
this.answer = AwesomeModule();
expect(this.print()).toBe("The answer is 42");
expect(UselessModule()).toBe("absolutely nothing");
done();
}]
}
api.di.resolveObject(component);
component.doSomething();
});
it("should use same function with different parameters", function(done) {
api.di.flush().register("service", function(value) {
return value;
});
var App = {
init: [',service,', function(a, service, b) {
if(!this.tested) {
this.tested = true;
expect(service(42)).toBeDefined(42);
expect(a.value).toBe("A");
expect(b.value).toBe("B");
} else {
expect(a.value).toBe(10);
expect(b.value).toBe(20);
done();
}
return this;
}]
};
api.di.resolveObject(App);
App.init(
{ value: "A" },
{ value: "B" }
).init(
{ value: 10 },
{ value: 20 }
);
});
it("should be able to pass a boolean", function(done) {
api.di.register("BooleanValue", false);
var doSomething = api.di.resolve(function(BooleanValue) {
expect(BooleanValue).toBeDefined();
expect(typeof BooleanValue).toBe('boolean');
expect(BooleanValue).toBe(false);
done();
});
doSomething();
});
it("should use the host", function(done) {
var ExternalService = {
gogo: function() {
return this.host.name;
}
}
api.di.register("es", ExternalService);
var doSomething = api.di.resolve(function(es) {
this.name = 42;
expect(es.gogo()).toBe(42);
done();
});
doSomething();
});
it("should use the host while a function is injected", function(done) {
var ExternalService = function() {
return { name: this.host.name };
}
api.di.register("es", ExternalService);
var doSomething = api.di.resolve(function(es) {
this.name = 42;
expect((new es()).name).toBe(42);
done();
});
doSomething();
});
}); | {
"redpajama_set_name": "RedPajamaGithub"
} | 9,374 |
The Carrier, Personnel Half-track, was an American armored personnel carrier from World War II. It was produced in the United States by International Harvester. It was supplied to Allied nations (the British Commonwealth, France, and the Soviet Union) under the Lend-Lease. After the war it was leased to many countries in the NATO. | {
"redpajama_set_name": "RedPajamaC4"
} | 7,786 |
package com.hcd.jbox2d.game.view;
import java.util.ArrayList;
import org.jbox2d.collision.shapes.EdgeShape;
import org.jbox2d.common.Vec2;
import org.jbox2d.dynamics.Body;
import org.jbox2d.dynamics.BodyDef;
import org.jbox2d.dynamics.World;
import com.hcd.jbox2d.game.activity.LevelActivity;
import com.hcd.jbox2d.game.activity.Stage2Activity;
import com.hcd.jbox2d.game.obj.Line;
import com.hcd.jbox2d.game.obj.Platform;
import com.hcd.jbox2d.game.obj.Polygon;
import com.hcd.jbox2d.game.utils.CutPolygonUtils;
import com.hcd.jbox2d.game.utils.GrahamScanUtils;
import com.hcd.jbox2d.game.utils.PolygonCenterUtils;
import android.content.Context;
import android.graphics.Canvas;
import android.graphics.Color;
import android.graphics.Paint;
import android.graphics.Path;
import android.graphics.Typeface;
import android.os.Handler;
import android.util.AttributeSet;
import android.util.Log;
import android.view.MotionEvent;
import android.view.View;
public class Stage2View extends View {
private Stage2View gameView;
//¹ý¹Ø°Ù·Ö±È
private static float PASSSCORE = 0.8f;
//ÎïÀíÆÁÄ»ÓëÎïÀíÊÀ½çµÄ±ÈÀýpx/m
private final static int RATE = 60;
private World world;
//Ä£ÄâµÄƵÂÊ
private float timeStep;
//µü´úÔ½´ó£¬Ä£ÄâÔ½¾«È·£¬µ«ÐÔÄÜÔ½µÍ
private int iterations;
private Handler mHandler;
//ÆÁÄ»µÄ¿í¶ÈÓë¸ß¶È
private float screenWidth, screenHeight;
private Polygon polygon2;
private ArrayList<Polygon> polygons;
private Line line = new Line(0.0f, 0.0f, 0.0f, 0.0f);
private boolean inScreen, outScreen;
private static int lineNum;
private Platform platform;
private ArrayList<Platform> platforms;
//ÅжÏËùÓÐÎïÌåÊÇ·ñ¶¼¾²Ö¹
private boolean isSleeping;
//ÓÎÏ·ËùµÃ·ÖÊýµÄ°Ù·Ö±È
private int removed;
//³õ'ÎïÌåµÄÖÊÁ¿
private float initArea;
private float cutArea;
public boolean gameOver;
public boolean gameSuccess;
public boolean haveWriteDb;
private Canvas canvas;
private Paint paint;
public Stage2View(Context context, AttributeSet attrs) {
super(context, attrs);
paint = new Paint();
paint.setAntiAlias(true);
initGame();
}
private void drawPlatform(Platform platform) {
paint.setAntiAlias(true);
paint.setColor(Color.DKGRAY);
canvas.drawRect(platform.getX1() * RATE, platform.getY1() * RATE, platform.getX2() * RATE, platform.getY2() * RATE, paint);
}
private void drawLine(float x1, float y1, float x2, float y2) {
paint.setAntiAlias(true);
paint.setColor(Color.BLACK);
canvas.drawLine(x1, y1, x2, y2, paint);
}
private void drawPolygon(Polygon polygon) {
paint.setAntiAlias(true);
paint.setColor(Color.rgb(111, 100, 10));
Path path = new Path();
Vec2[] vecs = polygon.getVecs();
Vec2[] tmp = new Vec2[polygon.getEdge()];
//¸ù¾ÝÐýתµÄ½Ç¶È»ñµÃ¸÷¸öµãµÄ×ø±ê
for (int i = 0; i < polygon.getEdge(); i++) {
float vecX = (float) (vecs[i].x * Math.cos(polygon.getAngle()) - vecs[i].y
* Math.sin(polygon.getAngle()));
float vecY = (float) (vecs[i].x * Math.sin(polygon.getAngle()) + vecs[i].y
* Math.cos(polygon.getAngle()));
tmp[i] = new Vec2(vecX, vecY);
}
Vec2 centerPoint = new Vec2(0, 0);//PolygonCenterUtils.getPolygonCenter(vecs);
path.moveTo((polygon.getX() + tmp[0].x - centerPoint.x) * RATE,
(polygon.getY() + tmp[0].y - centerPoint.y) * RATE);// ´ËµãΪ¶à±ßÐεÄÆðµã
for (int i = 1; i < polygon.getEdge(); i++) {
path.lineTo((polygon.getX() + tmp[i].x - centerPoint.x) * RATE,
(polygon.getY() + tmp[i].y - centerPoint.y) * RATE);
}
path.close(); // ʹÕâЩµã¹¹³É·â±ÕµÄ¶à±ßÐÎ
canvas.drawPath(path, paint);
}
private void setScore() {
paint.setTypeface(Typeface.SANS_SERIF);
paint.setTextSize(20);
canvas.drawText("Removed:" + removed+"%", 10 , screenHeight - 10, paint);
canvas.drawText("Target:" + (int)Math.round(PASSSCORE*100)+"%" , 150, screenHeight - 10, paint);
if (gameOver) {
if (gameSuccess) {
canvas.drawText("SUCCESS!", screenWidth - 100, 30, paint);
//ÓÎÏ·¹ý¹Øºó³õ'»¯ÏÂÒ»¹ØÊý¾Ý
if (!haveWriteDb) {
if (LevelActivity.lvManager.getStageByLevel(3).size() == 0){
LevelActivity.lvManager.insertLevelInfo(3, 0, 0);
}else if (LevelActivity.lvManager.getStageByLevel(3).size() == 1) {
//ÏÂÒ»¹ØÓÐÊý¾Ý²»Óóõ'»¯
//LevelActivity.lvManager.updateLevelInfo(3, 0, 0);
} else {
Log.i("Erro Message", "²é³öÊý¾Ý²»ÊÇΨһµÄ");
}
}
} else {
canvas.drawText("FAILED!", screenWidth - 80, 30, paint);
}
if (!haveWriteDb) {
//ÓÎÏ·½áÊøºó±£´æÊý¾Ý
if (LevelActivity.lvManager.getStageByLevel(2).size() == 0){
LevelActivity.lvManager.insertLevelInfo(2, removed, 1);
}else if (LevelActivity.lvManager.getStageByLevel(2).size() == 1) {
int oldScore = LevelActivity.lvManager.getStageByLevel(2).get(0).getScore();
LevelActivity.lvManager.updateLevelInfo(2, oldScore > removed ? oldScore : removed, 1);
} else {
Log.i("Erro Message", "²é³öÊý¾Ý²»ÊÇΨһµÄ");
}
haveWriteDb = true;
}
} else {
if (lineNum == 3) {
canvas.drawLine(screenWidth - 20, 10, screenWidth - 30, 30, paint);
canvas.drawLine(screenWidth - 30, 10, screenWidth - 40, 30, paint);
canvas.drawLine(screenWidth - 40, 10, screenWidth - 50, 30, paint);
} else if (lineNum == 2) {
canvas.drawLine(screenWidth - 20, 10, screenWidth - 30, 30, paint);
canvas.drawLine(screenWidth - 30, 10, screenWidth - 40, 30, paint);
} else if(lineNum == 1) {
canvas.drawLine(screenWidth - 20, 10, screenWidth - 30, 30, paint);
}
}
}
@Override
protected void onDraw(Canvas canvas) {
super.onDraw(canvas);
this.canvas = canvas;
setBackgroundColor(Color.WHITE);
for (int i = 0; i < platforms.size(); i++) {
drawPlatform(platforms.get(i));
}
for (int i = 0; i < polygons.size(); i++){
drawPolygon(polygons.get(i));
}
if (line != null) {
drawLine(line.getV1().x, line.getV1().y, line.getV2().x, line.getV2().y);
}
setScore();
}
@Override
public boolean onTouchEvent(MotionEvent event) {
if (lineNum > 0) {
if (event.getAction() == MotionEvent.ACTION_MOVE) {// Èç¹ûÍ϶¯
line.setV2(new Vec2(event.getX(), event.getY()));
invalidate();
Log.d("´¥Ãþ'þ", "Òƶ¯");
}
if (event.getAction() == MotionEvent.ACTION_DOWN) {// Èç¹ûµã»÷
line.setV1(new Vec2(event.getX(), event.getY()));
line.setV2(new Vec2(event.getX(), event.getY()));
invalidate();
inScreen = true;
Log.d("´¥Ãþ'þ", "µãÏÂÆÁÄ»");
}
if (event.getAction() == MotionEvent.ACTION_UP) {
line.setV2(new Vec2(event.getX(), event.getY()));
outScreen = true;
Log.d("´¥Ãþ'þ", "À뿪ÆÁÄ»");
}
}
return true;
}
/**
* ³õ'»¯ÓÎÏ·
*/
private void initGame() {
polygons = new ArrayList<Polygon>();
platforms = new ArrayList<Platform>();
lineNum = 3;
gameOver = false;
gameSuccess = false;
haveWriteDb = false;
gameView = this;
removed = 0;
inScreen = false;
outScreen = false;
isSleeping = true;
cutArea = 0.0f;
screenWidth = Stage2Activity.screenWidth;
screenHeight = Stage2Activity.screenHeight;
Vec2 gravity = new Vec2(0.0f, 10.0f); // ÏòÁ¿£¬ÓÃÀ´±êʾµ±Ç°ÊÀ½çµÄÖØÁ¦·½Ïò£¬µÚÒ»¸ö²ÎÊýΪˮƽ·½Ïò£¬¸ºÊýΪ×ö£¬ÕýÊýΪÓÒ¡£µÚ¶þ¸ö²ÎÊý±íʾ´¹Ö±·½Ïò
world = new World(gravity);
createPlatform();
createBorder(false, false, false, false);
createPolygon();
timeStep = 1.0f / 60.0f; // ¶¨ÒåƵÂÊ
iterations = 10; // ¶¨Òåµü´ú
mHandler = new Handler();
mHandler.post(update);
mHandler.post(cutloop);
}
private Runnable update = new Runnable() {
@Override
public void run() {
if (!gameOver){
world.step(timeStep, iterations, iterations);
gameView.invalidate();
mHandler.postDelayed(update, (long) timeStep * 1000);
}
}
};
private Runnable cutloop = new Runnable() {
@Override
public void run() {
if(inScreen && outScreen && !gameOver) {
boolean isCut = false;
ArrayList<Polygon> polytemp = new ArrayList<Polygon>();
for (int i = 0; i < polygons.size(); i++) {
Vec2[] temp = new Vec2[2];
temp[0] = new Vec2(line.getV1().x / RATE, line.getV1().y / RATE);
temp[1] = new Vec2(line.getV2().x / RATE, line.getV2().y / RATE);
if ((CutPolygonUtils.getLeftCutPolygon(polygons.get(i).getNowVecs(), temp) != null) &&
(CutPolygonUtils.getRightCutPolygon(polygons.get(i).getNowVecs(), temp) != null)) {
isCut = true;
//ÎïÌå±»ÇгÉÁ½¿éºó£¬Ó¦¸ÃÏÈÈ¥³ý֮ǰ´´½¨ÔÚÊÀ½çÖеÄÎïÌå
world.destroyBody(polygons.get(i).getBody());
Vec2[] left = CutPolygonUtils.getLeftCutPolygon(polygons.get(i).getNowVecs(), temp);
float x = PolygonCenterUtils.getPolygonCenter(left).x;
float y = PolygonCenterUtils.getPolygonCenter(left).y;
left = PolygonCenterUtils.getStandardPolygon(left);
Polygon pleft = new Polygon(world, x, y, left, left.length, 0.1f, 0.1f, 1.0f, 0.0f);
Vec2[] right = CutPolygonUtils.getRightCutPolygon(polygons.get(i).getNowVecs(), temp);
x = PolygonCenterUtils.getPolygonCenter(right).x;
y = PolygonCenterUtils.getPolygonCenter(right).y;
right = PolygonCenterUtils.getStandardPolygon(right);
Polygon pright = new Polygon(world, x, y, right, right.length, 0.1f, 0.1f, 1.0f, 0.0f);
polytemp.add(pright);
polytemp.add(pleft);
} else {
polytemp.add(polygons.get(i));
}
}
polygons.clear();
polygons = polytemp;
inScreen = false;
outScreen = false;
line = new Line(0.0f, 0.0f, 0.0f, 0.0f);
if (isCut) lineNum--;
}
//ÅжÏÎïÌåÊÇ·ñµ÷³öÆÁÄ»ÍâÃæÒÔ¼°ÎïÌåÊÇ·ñÍ£Ö¹Ô˶¯
isSleeping = true;
{
ArrayList<Polygon> temp_poly = new ArrayList<Polygon>();
for (int i = 0; i < polygons.size(); i++) {
if (polygons.get(i).getBody().isAwake()) {
isSleeping = false;
}
Vec2[] vecs = polygons.get(i).getNowVecs();
boolean out_screen = true;
for (int j = 0; j < vecs.length; j++) {
if (vecs[j].x * RATE < screenWidth && vecs[j].y * RATE < screenHeight) {
out_screen = false;
}
}
if (!out_screen) {
temp_poly.add(polygons.get(i));
} else {
cutArea += polygons.get(i).getMass();
Log.i("µô³öÆÁÄ»ÁË","cutArea / initArea=" + cutArea / initArea);
removed = (int) Math.rint((cutArea / initArea) * 100);
}
}
polygons.clear();
polygons = temp_poly;
}
if (isSleeping && lineNum <= 0) {
gameOver = true;
Log.i("ÓÎϷ״̬", "ÓÎÏ·½áÊø" + cutArea / initArea);
if (PASSSCORE <= cutArea / initArea) {
//ÓÎÏ·¹ý¹Ø
gameSuccess = true;
Log.i("ÓÎÏ·½á¹û", "¹ý¹Ø");
}else {
gameSuccess = false;
}
}
if (0.99 <= cutArea / initArea) {
gameOver = true;
gameSuccess = true;
}
mHandler.postDelayed(cutloop, (long) timeStep * 1000);
}
};
private void createPlatform() {
platform = new Platform(world, screenWidth / 8 / RATE, screenHeight * 3 / 4 / RATE, screenWidth / 2 / RATE, screenHeight * 13 / 16 / RATE);
platforms.add(platform);
platform = new Platform(world, screenWidth * 2 / 3 / RATE, screenHeight * 3 / 4 / RATE, screenWidth * 43 / 48 / RATE, screenHeight * 13 / 16 / RATE);
platforms.add(platform);
platform = new Platform(world, screenWidth / 16 / RATE, screenHeight * 1 / 2 / RATE, screenWidth / 8 / RATE, screenHeight * 15 / 16 / RATE);
platforms.add(platform);
platform = new Platform(world, screenWidth * 5 / 6 / RATE, screenHeight * 5 / 13 / RATE, screenWidth * 43 / 48 / RATE, screenHeight * 13 / 16 / RATE);
platforms.add(platform);
}
private void createBorder(boolean top,boolean left, boolean bottom, boolean right) {
BodyDef bd = new BodyDef();
Body border = world.createBody(bd);
EdgeShape es = new EdgeShape();
if (left) {
es.set(new Vec2(0, 0), new Vec2(0, screenHeight / RATE));
border.createFixture(es, 0.0f);
}
if (bottom) {
es.set(new Vec2(screenWidth / RATE, 0), new Vec2(screenWidth / RATE,
screenHeight / RATE));
border.createFixture(es, 0.0f);
}
if (right) {
es.set(new Vec2(0, screenHeight / RATE), new Vec2(screenWidth / RATE,
screenHeight / RATE));
border.createFixture(es, 0.0f);
}
if (top) {
es.set(new Vec2(0, 0), new Vec2(screenWidth / RATE, 0));
border.createFixture(es, 0.0f);
}
}
private void createPolygon() {
//ÕýÁù±ßÐÎÎïÌå
Vec2[] vecs = new Vec2[6];
// vecs[0] = new Vec2(screenWidth / 8 / RATE, (screenHeight) / 4 / RATE);
float startX = (float) (((screenWidth) / 7 / RATE) * Math.cos(Math.PI / 6));
float startY = (float) (((screenWidth) / 7 / RATE) * Math.sin(Math.PI / 6));
int edge = vecs.length;
for (int i = 0; i < edge; i++) {
float vecX = (float) (startX * Math.cos(2 * i * Math.PI / edge) - startY * Math.sin(2 * i * Math.PI / edge));
float vecY = (float) (startX * Math.sin(2 * i * Math.PI / edge) + startY * Math.cos(2 * i * Math.PI / edge));
vecs[i] = new Vec2(vecX, vecY);
}
Vec2[] vecst = GrahamScanUtils.getGrahamScan(vecs);
polygon2 = new Polygon(world, screenWidth * 7 / 12 / RATE, (screenHeight) / 2 / RATE, vecst, vecst.length, 0.0f,
0.5f, 1.0f, 0.0f);
initArea = polygon2.getMass();
polygons.add(polygon2);
}
}
| {
"redpajama_set_name": "RedPajamaGithub"
} | 6,377 |
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