% conjtwist.tex1\documentclass[11pt]{article}2\include{macros}3\title{A newform with inner twists at level~$512$}4\author{K.\thinspace{}M. Buzzard and W.\thinspace{}A. Stein}5\begin{document}6\maketitle78\section{Introduction}910Doi, Koike, Ribet, Shimura, and Yamauchi have studied newforms $f$11which admit an {\em inner twist}. That is, for which there12exists a nontrivial primitive character $\chi$ so that13$f\tensor\chi=\sigma f$ for some $\sigma\in\Gal(\Qbar/\Q)$.14Let $A_f$ be the abelian variety associated to $f$ (considered15only up to isogeny). Then $A_f$ is simple over $\Q$.16Inner twists of $f$ are of interest because they can give rise to17nontrivial decompositions of $A_f$ over a finite extension of $\Q$.18For example, Koike used the theory to see that the19$2$ dimensional abelian variety $J_0(81)^{\new}$ is isogeneous over20$\Q(\sqrt{-3})$ to $E\cross E^{\tau}$ where $E$ is an explicitely21determined elliptic curve over $\Q(\sqrt{-3})$ and $\tau$ is the22nontrivial automorphism.2324In this paper we work out a single example. It is hoped that25the ideas used in working out the example can be used to both26find more examples and prove a converse to some of Ribet's theorems.2728\section{Theorems of Shimura and Ribet}29\begin{theorem}[Shimura]\label{shimura}30Let $N, r, s, k>0$ be integers such that $s|N$ and let31$M$ be the least common multiple of $N$, $r^2$, and $rs$.32Let $\chi$ (resp. $\Psi$) be a primitive character mod $r$ (resp. $s$).33If $f=\sum a_n q^n \in S_k(\Gamma_0(N),\Psi)$ then34$$f\tensor\chi:=\sum \chi(n)a_n q^n\in S_k(\Gamma_0(M),\Psi\chi^2).$$35\end{theorem}36\begin{proof}37Proposition 3.64 of \cite{shimura}.38\end{proof}3940As a special case we have41\begin{corollary}42Let $2^n\geq 64$ and let $\chi$43be a primitive character of conductor dividing $8$.44Then ``twisting by $\chi$'' (i.e., the45map $f\mapsto f\tensor\chi$) preserves $S_k(\Gamma_0(2^n),\Psi)$.46\end{corollary}47\begin{proof}48Put $r=8$ in the theorem and note that $\chi^2=1$.49\end{proof}5051We introduce some notation in order to recall Ribet's Theorem.52Let $E=\Q(f)=\Q(a_1,a_2,a_3,\ldots)$. Let53$$\Gamma=\{ \gamma\in\Aut(E) : \gamma f=54f\tensor \chi_{\gamma},\text{ some }\chi_{\gamma}\}.$$55It is known that $\Gamma$ is an abelian subgroup56of $\Aut(E)$. Let $F$ be the subfield of $E$ fixed by $\Gamma$.5758\begin{remark}59What is $[F:\Q]$ in terms of $\Gamma$ and $\#\Aut(E)$?60Embed $E$ in its normal closure $K$.61Let $H\subset G =\Gal(K/\Q)$ be the subgroup of automorphisms62which preserve $E$. Since every automorphism extends,63$\Aut(E)$ is a quotient of $H$ with kernel $J$, the subgroup64of $G$ corresponding to $E$ via Galois theory.65Note that $J$ is normal in $H$ but need not be normal in $G$.66The inverse image $\Gamma'$ of67$\Gamma$ in $H$ is a subgroup of the same index as the index68of $\Gamma$ in $\Aut(E)$.69Unless I make a mistake computing indices, we get70$$[F:\Q] = [G:\Gamma'+J] = \frac{[G:J]}{[\Gamma'+J:J]}71= \frac{[E:\Q]}{\#\Gamma}$$72which is just what we would expect.73\end{remark}7475For a primitive character $\vphi$ of conductor $c$ define76$$g(\vphi)=\sum_{u=1}^{c}\vphi(u)e^{2\pi i u/c}.$$77For $\gamma,\delta\in\Gamma$, let78$$c(\gamma,\delta)=\frac{g(\chi_{\gamma}^{-1})79g(\gamma(\chi_{\delta}^{-1}))}80{g(\chi_{\gamma\delta}^{-1})}.$$8182In \cite{ribet} there is an explicit description of $\End(A_f)\tensor\Q$83in terms of generators and relations.84Let $\X$ be the $E$-vector space85$$\X = \sum_{\gamma\in\Gamma} E\cdot X_{\gamma}$$86where the $X_{\gamma}$ are formal symbols. By imposing87on the $X_{\gamma}$ the rules88\begin{eqnarray*}89X_{\gamma}\cdot e &=& \gamma(e)X_{\gamma},90\quad\text{for $e\in E$ and $\gamma\in\Gamma$}\\91X_{\gamma}X_{\delta} &=&c(\gamma,\delta) X_{\gamma \delta}92\end{eqnarray*}93we make $\X$ into an associative algebra.9495\begin{theorem}[Ribet\cite{ribet}]\label{ribet}96There algebra $\X$ is a central simple97algebra over $F$ which is isomorphic to $(\End A)\tensor\Q$.98Furthermore, the $2$-cocycle $c$ corresponds to the class of99$(\End A)\tensor \Q$ in $\Br(F)=H^2(F,\overline{F}^{\star})$.100\end{theorem}101102\section{Example in which the endomorphism ring has order $2$ in the Brauer group}103104The characteristic polynomial of $T_3$ on $S_2^{\new}(\Gamma_0(512))$ is105$$(x^2 -6)^2(x^2 -2)^4(x^2 + 4x + 2)(x^2 -4x + 2)$$106Let $f=\sum a_n q^n$ be one of the newforms in the four dimensional kernel107$V$ of $(T_3^2-6)^2$.108and let $E=\Q(f)=\Q(a_1,a_2,a_3,a_4,\ldots)$.109110\begin{proposition}111We have $E=\Q(\sqrt{2},\sqrt{3})$112and the compa-nions of $f$ coincide113with the twists of $f$ by the four Dirichlet characters114of conductor dividing $8$. In particular, $\Gamma=\Gal(E/\Q)$.115\end{proposition}116\begin{proof}117The newform $f$ and its companions lie inside118of $V$. By ``multiplicity one'', $V$ has dimension119$4$ so $[\Q(f):\Q]\leq 4$.120The characteristic polynomial of $T_7$ on $V$ is121$(x^2-8)^2$ so $a_3 = \pm \sqrt{6}$ and122$a_7 = \pm 2\sqrt{2}$ so123$E=\Q(\sqrt{2},\sqrt{3})$.124125Let $\chi$ be a character of conductor dividing $8$.126By Theorem~\ref{shimura}, $f\tensor\chi$ lies in127$S_2(\Gamma_0(512))$. In fact, it must lie128in $S_2^{\new}(\Gamma_0(512))$ because if $f\tensor\chi$ lies in129$S_2(\Gamma_0(2^n))$ for $n\geq 6$ then Theorem~\ref{shimura}130implies that $f=(f\tensor\chi)\tensor\chi$ lies in $S_2(\Gamma_0(2^n))$,131and certainly we can take $n\geq 6$. So we're done.132133Alternatively, since the image of $\chi$ is134contained in $\{\pm 1\}$, the $T_3$ eigenvalue of135$f$ equal $\pm\sqrt{6}$.136The factor $(x^2-6)^2$ exactly divides137the characteristic polynomial of $T_3$ on $S_2(\Gamma_0(512))$138so $f\tensor\chi$ must be a companion of $f$.139\end{proof}140141The $q$-expansion of $f$ can be computed to be142$$f = q -\sqrt{6}q^3 + 2\sqrt{3}q^5 +2\sqrt{2}q^7 + 3q^9 -\sqrt{6}q^{11}143-2\sqrt{3}q^{13}-6\sqrt{2}q^{15} + 4q^{17}+\cdots $$144145\begin{remark} $f$ is the first example of a weight $2$ newform146having trivial character for which $\Q(a_n)$ is a {\em proper}147subfield of $E$ for all $n$.148\end{remark}149150\comment{%PARI code151? nf=nfinit(subst(E[1],x,t));152? nffactor(nf,x^2-2)153? a=-1/12*t^2 + 3/2;154? nffactor(nf,x^2-3)155? b=-1/24*t^3 + 7/4*t;156? c=lift(Mod(a*b,t^4-36*t^2+36))157? d=1;158? v=subst(E[2],x,t);159? 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);160? w(f)=vector(4,i,polcoeff(f,i-1))~;;161? N*w(v[5])162? N*w(v[7])163}164165166\begin{theorem}167The endomorphism ring of $A_f$ is a $2\times 2$ matrix ring168over a quaternion division algebra with center $\Q$.169\end{theorem}170\begin{proof}171Write $G=\Gal(E/\Q)=\{\gamma_1=1,\gamma_2,\gamma_3,\gamma_6\}$172where $\gamma_d$ fixes $\sqrt{d}$, $d=2,3,6$.173Let $\chi_d$, $d=\pm 1, \pm 2$ be the four Dirichlet characters of174conductor dividing $8$, where $\chi_d$ corresponds to the field175$\Q(\sqrt{d})$.176We have177\begin{eqnarray*}178f & = & f\tensor\chi_1\\179\gamma_2 f = q +\sqrt{6}q^3 - 2\sqrt{3}q^5 +2\sqrt{2}q^7 + \cdots &=& f\tensor\chi_{2}\\180\gamma_3 f = q +\sqrt{6}q^3 + 2\sqrt{3}q^5 -2\sqrt{2}q^7 + \cdots181&=& f\tensor\chi_{-1}\\182\gamma_6 f = q -\sqrt{6}q^3 - 2\sqrt{3}q^5 -2\sqrt{2}q^7 + \cdots183&=& f\tensor\chi_{-2}184\end{eqnarray*}185186The sums are187\begin{eqnarray*}188g(\chi_1) &=& 1\\189g(\chi_{-1}) &=& e^{2\pi i/4} - e^{2\pi i 3/4} = i - (-i) = 2i\\190g(\chi_{2}) &=&191e^{2\pi i/8} - e^{2 \pi i 3/8} - e^{2\pi i 5/8} + e^{2\pi i 7/8}192= 2\sqrt{2}\\193g(\chi_{-2}) &=& e^{2\pi i/8} + e^{2 \pi i 3/8}194- e^{2\pi i 5/8} - e^{2\pi i 7/8}195= 2i\sqrt{2}.196\end{eqnarray*}197Thus we can compute198$$c(\gamma_i,\gamma_j) = \frac{g(\chi_{\gamma_i}) g(\chi_{\gamma_j})}199{g(\chi_{\gamma_i\gamma_j})}.$$200For example,201$$c(\gamma_2,\gamma_3) = \frac{2\sqrt{2} \cdot 2 i}{2i\sqrt{2}} = 2$$202By Theorem~\ref{ribet} we obtain a presentation of203$(\End A)\tensor\Q$.204Note that $c(\gamma_i,\gamma_j)=c(\gamma_j,\gamma_i)$ so205that the generators $X_{\gamma}$ commute. The endomorphism206ring is not commutative as $E$ does not commute with $X_{\gamma}$207for nontrivial $\gamma$.208209The $2$-cocycle $c$ represents the element of the Brauer group $\Br(\Q)$210corresponding to $(\End A)\tensor\Q$. We ask, does this element have211order $1$ or order $2$? Let $K=\Q_2(\sqrt{2},\sqrt{3})$.212Then, by inf-res, $c$ arises from an213element of214$$H^2(\Gal(K/\Q_2),K^{\star})\ra \Br(\Q).$$215Note that inf {\em is} injective because of Hilbert's216Theorem 90; see Proposition 6 on page 156 of \cite{serre}.217218I think that $c$ must have order dividing $2$ and219$$H^2(\Gal(\Q_2(\sqrt{2})/\Q_2),\Q_2(\sqrt{2})^{\star})\subset H^2(\Gal(K/\Q_2),K^{\star})$$220is the unique subgroup of order $2$ of $\Br(\Q)$, hence $c$ must lie in it.221Using the Local symbol in Chapter XIV of we can write down the222nonzero element of $H^2(\Gal(\Q_2(\sqrt{2})/\Q_2),\Q_2(\sqrt{2})^{\star})$.223224Let $G'=\Gal(\Q_2(\sqrt{2})/\Q_2)=\{ 1, \tau\}$. What is225$c$ as a map $G'\cross G' \ra \Q_2(\sqrt{2})^\star$?226We have:227$$c(1,1)=1, \quad c(1,\tau) = c(\tau,1) = 1, \quad228c(\tau,\tau) = -4.$$229230This is precisely the $2$-cocycle given by the local symbol $(2,-4)$;231see Washington's article \cite{washington}. Thus this cocycle is232trivial iff $-4$ is a norm from $\Q_2(\sqrt{2})$.233If $-4$ is a norm then so is $-1$ and hence $-1\in\Q_2(\sqrt{2})$.234But this is impossible as the ramification degree of235$2$ in $\Q(\sqrt{2},i)$ is $e=4$, so that the local degree of $2$236is $4$. Thus we finally conclude that $c$ is nontrivial and hence237obtain the theorem.238\end{proof}239240241\begin{thebibliography}{HHHHHHH}242\bibitem[D]{doi} K. Doi, M. Yamauchi, {\em On the Hecke operators243for $\Gamma_0(N)$ and class fields over quadratic number fields},244J. Math. Soc. Japan, {\bf 25} (1973), 629--643.245\bibitem[K]{koike} M. Koike, {\em On certain abelian varieties246obtained from new forms of weight $2$ on247$\Gamma_0(3^4)$ and $\Gamma_0(3^5)$.}, Nagoya Math. J.,248{\bf 62} (1976), 29--39.249\bibitem[R]{ribet} K. Ribet, {\em Twists of Modular Forms and250Endomorphisms of Abelian Varieties}, Math. Ann. {\bf 253},251(1980), 43--62.252\bibitem[SE]{serre} J.P. Serre, {\em Local Fields}, Springer-Verlag,253(1979).254\bibitem[S]{shimura} G. Shimura, {\em Introduction to the255Arithmetic Theory of Automorphic Functions}, Princeton256University Press, (1994).257\bibitem[W]{washington} L.C. Washington, {\em Galois Cohomology},258In ``Modular Forms and Fermat's Last Theorem'', Ed.'s259Cornell-Silverman-Stevens, (1997).260\end{thebibliography} \normalsize\vspace*{1 cm}261262\end{document}263264265\comment{%Letter to Luiz266267THe endomorphism algebra of A_f (over Q) is a central simple algebra over268F_f which contains Q_f as a maximal conmutative subfield. Its degree over269Q is [Q_f : Q]*[Q_f : F_f] .270271Moreover, the central simple algebra has order either one or two272in the Brauer group of F_f. Thus273274/- matrix algebra over F_f275(End A_f) tensor Q = or276\- matrix algebra over quaternion algebra over F_f.277278Knowing which one gives the exact isogeny decomposition of A_f over Qbar.279280This is proved in : K. Ribet: "Twists of Modular Forms and Endomorphisms281of Abelian Varieties", Math. Ann. 253, 43-62 (1980)282283There is also some discussion in284Ken's: ``Endomorphism algebras of abelian varieties attached285to newforms of weight 2''286287Have you tried to work out the isogeny structure (over Qbar) of the288abelian variety A_f attached to your level 8192 form f? I suspect289that f has inner twists corresponding to all of the Dirichlet characters290of conductor dividing 8, i.e., corresponding to the quadratic291subfields of Q(zeta_8)=Q(i,sqrt(2)). I doubt that f has any292inner twists by characters of degree > 2 or ramified outside of 2.293I wonder what the order of (End A_f)tensor Q is in the Brauer group.294Is it one or two?295296297298Best,299William.300301302Ken,303304Thanks for giving me a copy of your paper on extra twists.305Let D be the endomorphism ring (tensor Q) of the abelian306variety A_f corresponding to the newform in f in307S_2(Gamma_0(512)) that we've been discussing (the one308for which Q(a_n) never equals Q(f)). In this case, the subfield309of Q(f) fixed by the group which you call Gamma is the rational310numbers Q. Thus D is a central simple Q-algebra. I think I can use311the theorem in your paper to show that D = M_2(K) where K is a division312quaternion algebra with center Q, i.e., D has order two in the313Brauer group of Q. This example may be interesting in314light of your remark on page 60 (of the 1980 Math. Annalen paper)315that it does not seem easy to determine the order of D316by ``pure thought''. You showed it was 1 in many cases, but317I don't think you gave conditions which imply that it must be 2.318319Thanks again for the paper,320William321}322323