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Author: William A. Stein
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% conjtwist.tex
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\documentclass[11pt]{article}
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\title{\Large\sc The First Newform on $\Gamma_0(N)$ such that
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$\Q(a_n)\not=\Q(\ldots,a_m,\ldots)$ for all~$n$}
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\author{William A. Stein}
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\usepackage{amsthm}
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\usepackage{amsmath}
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\theoremstyle{plain}
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\newtheorem{theorem}{Theorem}[section]
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\newtheorem{proposition}[theorem]{Proposition}
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\newtheorem{corollary}[theorem]{Corollary}
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\newtheorem{claim}[theorem]{Claim}
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\newtheorem{lemma}[theorem]{Lemma}
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\newtheorem{conjecture}[theorem]{Conjecture}
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\theoremstyle{definition}
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\newtheorem{definition}[theorem]{Definition}
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\newtheorem{algorithm}[theorem]{Algorithm}
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\newtheorem{question}[theorem]{Question}
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\theoremstyle{remark}
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\newtheorem{goal}[theorem]{Goal}
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\newtheorem{remark}[theorem]{Remark}
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\newtheorem{example}[theorem]{Example}
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\newtheorem{exercise}[theorem]{Exercise}
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\newcommand{\defn}[1]{{\em #1}}
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\newcommand{\e}{\mathbf{e}}
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\newcommand{\Gam}{\Gamma}
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\newcommand{\X}{\mathcal{X}}
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\newcommand{\E}{\mathcal{E}}
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\newcommand{\q}{\mathbf{q}}
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\newcommand{\cross}{\times}
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\newcommand{\ra}{\rightarrow}
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\newcommand{\la}{\leftarrow}
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\newcommand{\tensor}{\otimes}
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\newcommand{\eps}{\varepsilon}
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\newcommand{\vphi}{\varphi}
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\newcommand{\comment}[1]{}
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\newcommand{\Q}{\mathbf{Q}}
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\newcommand{\Qbar}{\overline{\Q}}
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\newcommand{\A}{\mathcal{A}}
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\newcommand{\p}{\mathfrak{p}}
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\renewcommand{\a}{\mathfrak{a}}
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\DeclareMathOperator{\Br}{Br}
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\DeclareMathOperator{\End}{End}
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\DeclareMathOperator{\new}{new}
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\DeclareMathOperator{\Aut}{Aut}
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\DeclareMathOperator{\Gal}{Gal}
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\begin{document}
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\maketitle
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\begin{abstract}
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We compute the minimum $N$ such that there
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is a newform $f=\sum a_n q^n\in{}S_2(\Gamma_0(N))$ with
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the odd property that
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$\Q(a_n)\not=\Q(\ldots,a_m,\ldots)$ for all~$n$.\footnote{Henri
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Cohen told me that
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he discoved this example many years ago, and Elkies ``explained'' it to him.
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This example is not in the literature, and I found
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it indepedently of Cohen and Elkies.}
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\end{abstract}
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\section{The Newform}
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Using modular symbols, we computed the first few terms of each
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newform $f=\sum a_n q^n$ on $S_2(\Gamma_0(N))$ for $N\leq 512$.
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For each newform of level $<512$, we found that there
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exists an integer~$n$ such that the field $K_f=\Q(\ldots, a_m, \ldots)$
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generated by all Fourier coefficients of~$f$ equals the field
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$\Q(a_n)$ generated by a single coefficient. This lead me
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to suspect that we always have equality.
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We do not always have equality.
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The characteristic polynomial of $T_3$ on $S_2^{\new}(\Gamma_0(512))$ is
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$
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(x^2 -6)^2(x^2 -2)^4(x^2 + 4x + 2)(x^2 -4x + 2),
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$
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and if we let $f=\sum a_n q^n$
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be one of the newforms in the four-dimensional kernel~$V$
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of $T_3^2-6$, then
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$$f = q -\sqrt{6}q^3 + 2\sqrt{3}q^5 +2\sqrt{2}q^7 + 3q^9 -\sqrt{6}q^{11}
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-2\sqrt{3}q^{13}-6\sqrt{2}q^{15} +\cdots$$
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provides an example in which $K_f\neq \Q(a_n)$ for some~$n$.
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We will use the following theorem (see \cite[Prop.~3.64]{shimura}):
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\begin{theorem}[Shimura]\label{thm:shimura}
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Let $N, r, s, k>0$ be integers such that $s|N$ and let
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$M$ be the least common multiple of $N$, $r^2$, and $rs$.
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Let $\chi$ (resp. $\Psi$) be a primitive character mod $r$ (resp. $s$).
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If $f=\sum a_n q^n \in S_k(\Gamma_0(N),\Psi)$ then
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$$\sum \chi(n)a_n q^n\in S_k(\Gamma_0(M),\Psi\chi^2).$$
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\end{theorem}
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\begin{lemma}\label{preserves}
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Suppose $n\geq 6$ and let~$\eps$ be a primitive
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Dirichlet character of conductor dividing~$8$.
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Then the map $f\mapsto f\tensor\eps$
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preserves $S_k(\Gamma_0(2^n))$.
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\end{lemma}
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\begin{proof}
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Set $r=8$ in Theorem~\ref{thm:shimura}, and note that $\eps^2=1$.
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\end{proof}
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It is clear from the $q$-expansion of~$f$ that
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$\Q(\sqrt{2},\sqrt{3})\subset K_f$.
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The newform~$f$ and its companions lie inside
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of the kernel~$V$ of $T_3^2-6$. A modular symbols computation
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shows that this kernel has dimension~$4$, which proves that
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$K_f=\Q(\sqrt{2},\sqrt{3})$.
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\begin{proposition}
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There is no integer~$n$ such that $K_f\neq \Q(a_n)$.
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\end{proposition}
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\begin{proof}
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Let~$\chi$ be a character of conductor dividing~$8$. By
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Lemma~\ref{preserves}, $f\tensor\chi$ lies in $S_2(\Gamma_0(512))$.
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We check computationally that $f\tensor\chi$ is one of the four
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Galois-conjugates of~$f$, and that each conjugate is of the form
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$f\tensor\chi$ for some~$\chi$. Let~$n>2$ be a positive integer and
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let~$\sigma$ be an automorphism. Then we have just showed that
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$\sigma(a_n) = \chi(n) a_n$ for some character~$\chi$ of order at
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most~$2$, so $\chi(n)\in\{\pm 1\}$. Thus the action of
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$\Gal(\Qbar/\Q)$ on $a_n$ factors through $\{ \pm 1\}$, so
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$\Q(a_n)$ has degree at most~$2$.
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\end{proof}
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\vspace{2ex}
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\noindent{\bf Acknowledgment:} It is a pleasure to thank Kevin Buzzard and
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Ken Ribet for helpful comments.
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\begin{thebibliography}{HHHHHHH}
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%\bibitem[0]{ghate:endo} {\em Modular Endomorphism Algebras} (2002), preprint.
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\bibitem[1]{shimura} G. Shimura, {\em Introduction to the
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Arithmetic Theory of Automorphic Functions}, Princeton
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University Press, (1994).
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\end{thebibliography} \normalsize\vspace*{1 cm}
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\end{document}
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\section{Don't read this. It contains a mistake.}
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We introduce some notation in order to recall Ribet's Theorem.
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Let $E=\Q(f)=\Q(a_1,a_2,a_3,\ldots)$. Let
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$$\Gamma=\{ \gamma\in\Aut(E) : \gamma f=
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f\tensor \chi_{\gamma},\text{ some }\chi_{\gamma}\}.$$
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It is known that $\Gamma$ is an abelian subgroup
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of $\Aut(E)$. Let $F$ be the subfield of $E$ fixed by $\Gamma$.
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\begin{remark}
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What is $[F:\Q]$ in terms of $\Gamma$ and $\#\Aut(E)$?
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Embed $E$ in its normal closure $K$.
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Let $H\subset G =\Gal(K/\Q)$ be the subgroup of automorphisms
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which preserve $E$. Since every automorphism extends,
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$\Aut(E)$ is a quotient of $H$ with kernel $J$, the subgroup
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of $G$ corresponding to $E$ via Galois theory.
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Note that $J$ is normal in $H$ but need not be normal in $G$.
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The inverse image $\Gamma'$ of
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$\Gamma$ in $H$ is a subgroup of the same index as the index
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of $\Gamma$ in $\Aut(E)$.
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Unless I make a mistake computing indices, we get
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$$[F:\Q] = [G:\Gamma'+J] = \frac{[G:J]}{[\Gamma'+J:J]}
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= \frac{[E:\Q]}{\#\Gamma}$$
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which is just what we would expect.
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\end{remark}
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For a primitive character $\vphi$ of conductor $c$ define
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$$g(\vphi)=\sum_{u=1}^{c}\vphi(u)e^{2\pi i u/c}.$$
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For $\gamma,\delta\in\Gamma$, let
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$$c(\gamma,\delta)=\frac{g(\chi_{\gamma}^{-1})
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g(\gamma(\chi_{\delta}^{-1}))}
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{g(\chi_{\gamma\delta}^{-1})}.$$
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In \cite{ribet} there is an explicit description of $\End(A_f)\tensor\Q$
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in terms of generators and relations.
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Let $\X$ be the $E$-vector space
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$$\X = \sum_{\gamma\in\Gamma} E\cdot X_{\gamma}$$
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where the $X_{\gamma}$ are formal symbols. By imposing
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on the $X_{\gamma}$ the rules
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\begin{eqnarray*}
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X_{\gamma}\cdot e &=& \gamma(e)X_{\gamma},
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\quad\text{for $e\in E$ and $\gamma\in\Gamma$}\\
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X_{\gamma}X_{\delta} &=&c(\gamma,\delta) X_{\gamma \delta}
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\end{eqnarray*}
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we make $\X$ into an associative algebra.
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\begin{theorem}[Ribet\cite{ribet}]\label{ribet}
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There algebra $\X$ is a central simple
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algebra over $F$ which is isomorphic to $(\End A)\tensor\Q$.
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Furthermore, the $2$-cocycle $c$ corresponds to the class of
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$(\End A)\tensor \Q$ in $\Br(F)=H^2(F,\overline{F}^{\star})$.
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\end{theorem}
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\section{Example in which the endomorphism ring has order $2$ in the Brauer group}
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\comment{%PARI code
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? nf=nfinit(subst(E[1],x,t));
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? nffactor(nf,x^2-2)
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? a=-1/12*t^2 + 3/2;
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? nffactor(nf,x^2-3)
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? b=-1/24*t^3 + 7/4*t;
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? c=lift(Mod(a*b,t^4-36*t^2+36))
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? d=1;
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? v=subst(E[2],x,t);
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? N=mattranspose([3/2,0,-1/12,0;0,7/4,0,-1/24;0,5/2,0,-1/12;1,0,0,0])^(-1);
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? w(f)=vector(4,i,polcoeff(f,i-1))~;;
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? N*w(v[5])
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? N*w(v[7])
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}
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\begin{theorem}
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The endomorphism ring of $A_f$ is $M_2(\Q)$.
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\end{theorem}
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\begin{proof}
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{\bf NOTE, added July 2002: } The proof below is wrong, as Pete
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Clark has pointed out. I recommend ignoring it, and looking at
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\cite{ghate:endo} instead (see their Example 4.3.1).
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Write $G=\Gal(E/\Q)=\{\gamma_1=1,\gamma_2,\gamma_3,\gamma_6\}$
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where $\gamma_d$ fixes $\sqrt{d}$, $d=2,3,6$.
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Let $\chi_d$, $d=\pm 1, \pm 2$ be the four Dirichlet characters of
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conductor dividing $8$, where $\chi_d$ corresponds to the field
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$\Q(\sqrt{d})$.
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We have
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\begin{eqnarray*}
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f & = & f\tensor\chi_1\\
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\gamma_2 f = q +\sqrt{6}q^3 - 2\sqrt{3}q^5 +2\sqrt{2}q^7 + \cdots &=& f\tensor\chi_{2}\\
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\gamma_3 f = q +\sqrt{6}q^3 + 2\sqrt{3}q^5 -2\sqrt{2}q^7 + \cdots
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&=& f\tensor\chi_{-1}\\
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\gamma_6 f = q -\sqrt{6}q^3 - 2\sqrt{3}q^5 -2\sqrt{2}q^7 + \cdots
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&=& f\tensor\chi_{-2}
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\end{eqnarray*}
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The sums are
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\begin{eqnarray*}
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g(\chi_1) &=& 1\\
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g(\chi_{-1}) &=& e^{2\pi i/4} - e^{2\pi i 3/4} = i - (-i) = 2i\\
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g(\chi_{2}) &=&
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e^{2\pi i/8} - e^{2 \pi i 3/8} - e^{2\pi i 5/8} + e^{2\pi i 7/8}
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= 2\sqrt{2}\\
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g(\chi_{-2}) &=& e^{2\pi i/8} + e^{2 \pi i 3/8}
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- e^{2\pi i 5/8} - e^{2\pi i 7/8}
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= 2i\sqrt{2}.
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\end{eqnarray*}
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Thus we can compute
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$$c(\gamma_i,\gamma_j) = \frac{g(\chi_{\gamma_i}) g(\chi_{\gamma_j})}
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{g(\chi_{\gamma_i\gamma_j})}.$$
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For example,
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$$c(\gamma_2,\gamma_3) = \frac{2\sqrt{2} \cdot 2 i}{2i\sqrt{2}} = 2$$
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By Theorem~\ref{ribet} we obtain a presentation of
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$(\End A)\tensor\Q$.
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Note that $c(\gamma_i,\gamma_j)=c(\gamma_j,\gamma_i)$ so
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that the generators $X_{\gamma}$ commute. The endomorphism
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ring is not commutative as $E$ does not commute with $X_{\gamma}$
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for nontrivial $\gamma$.
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The $2$-cocycle $c$ represents the element of the Brauer group $\Br(\Q)$
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corresponding to $(\End A)\tensor\Q$. We ask, does this element have
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order $1$ or order $2$? Let $K=\Q_2(\sqrt{2},\sqrt{3})$.
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Then, by inf-res, $c$ arises from an
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element of
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$$H^2(\Gal(K/\Q_2),K^{\star})\ra \Br(\Q).$$
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Note that inf {\em is} injective because of Hilbert's
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Theorem 90; see Proposition 6 on page 156 of \cite{serre}.
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I think that $c$ must have order dividing $2$ and
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$$H^2(\Gal(\Q_2(\sqrt{2})/\Q_2),\Q_2(\sqrt{2})^{\star})\subset H^2(\Gal(K/\Q_2),K^{\star})$$
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is the unique subgroup of order $2$ of $\Br(\Q)$, hence $c$ must lie in it.
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Using the Local symbol in Chapter XIV of we can write down the
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nonzero element of $H^2(\Gal(\Q_2(\sqrt{2})/\Q_2),\Q_2(\sqrt{2})^{\star})$.
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Let $G'=\Gal(\Q_2(\sqrt{2})/\Q_2)=\{ 1, \tau\}$. What is
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$c$ as a map $G'\cross G' \ra \Q_2(\sqrt{2})^\star$?
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We have:
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$$c(1,1)=1, \quad c(1,\tau) = c(\tau,1) = 1, \quad
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c(\tau,\tau) = -4.$$
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This is precisely the $2$-cocycle given by the local symbol $(2,-4)$;
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see Washington's article \cite{washington}. Thus this cocycle is
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trivial iff $-4$ is a norm from $\Q_2(\sqrt{2})$.
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If $-4$ is a norm then so is $-1$ and hence $-1\in\Q_2(\sqrt{2})$.
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But this is impossible as the ramification degree of
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$2$ in $\Q(\sqrt{2},i)$ is $e=4$, so that the local degree of $2$
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is $4$. Thus we finally conclude that $c$ is nontrivial and hence
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obtain the theorem. {\bf No, it is trivial.}
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\end{proof}
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\begin{thebibliography}{HHHHHHH}
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\bibitem[0]{ghate:endo} {\em Modular Endomorphism Algebras} (2002), preprint.
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\bibitem[1]{doi} K. Doi, M. Yamauchi, {\em On the Hecke operators
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for $\Gamma_0(N)$ and class fields over quadratic number fields},
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J. Math. Soc. Japan, {\bf 25} (1973), 629--643.
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\bibitem[2]{koike} M. Koike, {\em On certain abelian varieties
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obtained from new forms of weight $2$ on
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$\Gamma_0(3^4)$ and $\Gamma_0(3^5)$.}, Nagoya Math. J.,
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{\bf 62} (1976), 29--39.
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\bibitem[3]{ribet} K. Ribet, {\em Twists of Modular Forms and
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Endomorphisms of Abelian Varieties}, Math. Ann. {\bf 253},
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(1980), 43--62.
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\bibitem[4]{serre} J-P. Serre, {\em Local Fields}, Springer-Verlag,
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(1979).
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\bibitem[5]{shimura} G. Shimura, {\em Introduction to the
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Arithmetic Theory of Automorphic Functions}, Princeton
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University Press, (1994).
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\bibitem[6]{washington} L.C. Washington, {\em Galois Cohomology},
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In ``Modular Forms and Fermat's Last Theorem'', Ed.'s
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Cornell-Silverman-Stevens, (1997).
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\end{thebibliography} \normalsize\vspace*{1 cm}
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\end{document}
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\comment{%Letter to Luiz
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THe endomorphism algebra of A_f (over Q) is a central simple algebra over
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F_f which contains Q_f as a maximal conmutative subfield. Its degree over
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Q is [Q_f : Q]*[Q_f : F_f] .
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Moreover, the central simple algebra has order either one or two
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in the Brauer group of F_f. Thus
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/- matrix algebra over F_f
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(End A_f) tensor Q = or
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\- matrix algebra over quaternion algebra over F_f.
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Knowing which one gives the exact isogeny decomposition of A_f over Qbar.
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This is proved in : K. Ribet: "Twists of Modular Forms and Endomorphisms
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of Abelian Varieties", Math. Ann. 253, 43-62 (1980)
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There is also some discussion in the (not so nicely typeset) paper of
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Ken's: ``Endomorphism algebras of abelian varieties attached
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to newforms of weight 2''
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Have you tried to work out the isogeny structure (over Qbar) of the
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abelian variety A_f attached to your level 8192 form f? I suspect
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that f has inner twists corresponding to all of the Dirichlet characters
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of conductor dividing 8, i.e., corresponding to the quadratic
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subfields of Q(zeta_8)=Q(i,sqrt(2)). I doubt that f has any
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inner twists by characters of degree > 2 or ramified outside of 2.
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I wonder what the order of (End A_f)tensor Q is in the Brauer group.
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Is it one or two?
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Best,
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William.
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Ken,
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Thanks for giving me a copy of your paper on extra twists.
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Let D be the endomorphism ring (tensor Q) of the abelian
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variety A_f corresponding to the newform in f in
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S_2(Gamma_0(512)) that we've been discussing (the one
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for which Q(a_n) never equals Q(f)). In this case, the subfield
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of Q(f) fixed by the group which you call Gamma is the rational
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numbers Q. Thus D is a central simple Q-algebra. I think I can use
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the theorem in your paper to show that D = M_2(K) where K is a division
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quaternion algebra with center Q, i.e., D has order two in the
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Brauer group of Q. This example may be interesting in
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light of your remark on page 60 (of the 1980 Math. Annalen paper)
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that it does not seem easy to determine the order of D
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by ``pure thought''. You showed it was 1 in many cases, but
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I don't think you gave conditions which imply that it must be 2.
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Thanks again for the paper,
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William
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}
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