CoCalc Shared Fileswww / papers / twist / twistold_2.tex
Author: William A. Stein
1% conjtwist.tex
2\documentclass[11pt]{article}
3%\include{macros}
4\title{\Large\sc The First Newform on $\Gamma_0(N)$ such that
5$\Q(a_n)\not=\Q(\ldots,a_m,\ldots)$ for all~$n$}
6\author{William A. Stein}
7\usepackage{amsthm}
8\usepackage{amsmath}
9\theoremstyle{plain}
10\newtheorem{theorem}{Theorem}[section]
11\newtheorem{proposition}[theorem]{Proposition}
12\newtheorem{corollary}[theorem]{Corollary}
13\newtheorem{claim}[theorem]{Claim}
14\newtheorem{lemma}[theorem]{Lemma}
15\newtheorem{conjecture}[theorem]{Conjecture}
16\theoremstyle{definition}
17\newtheorem{definition}[theorem]{Definition}
18\newtheorem{algorithm}[theorem]{Algorithm}
19\newtheorem{question}[theorem]{Question}
20
21\theoremstyle{remark}
22\newtheorem{goal}[theorem]{Goal}
23\newtheorem{remark}[theorem]{Remark}
24\newtheorem{example}[theorem]{Example}
25\newtheorem{exercise}[theorem]{Exercise}
26
27\newcommand{\defn}[1]{{\em #1}}
28\newcommand{\e}{\mathbf{e}}
29\newcommand{\Gam}{\Gamma}
30\newcommand{\X}{\mathcal{X}}
31\newcommand{\E}{\mathcal{E}}
32\newcommand{\q}{\mathbf{q}}
33\newcommand{\cross}{\times}
34\newcommand{\ra}{\rightarrow}
35\newcommand{\la}{\leftarrow}
36\newcommand{\tensor}{\otimes}
37\newcommand{\eps}{\varepsilon}
38\newcommand{\vphi}{\varphi}
39\newcommand{\comment}[1]{}
40\newcommand{\Q}{\mathbf{Q}}
41\newcommand{\Qbar}{\overline{\Q}}
42\newcommand{\A}{\mathcal{A}}
43\newcommand{\p}{\mathfrak{p}}
44\newcommand{\m}{\mathfrak{m}}
45\renewcommand{\L}{\mathcal{L}}
46\renewcommand{\l}{\ell}
47\renewcommand{\t}{\tau}
48\renewcommand{\star}{\times}
49\renewcommand{\P}{\mathbf{P}}
50\renewcommand{\a}{\mathfrak{a}}
51\DeclareMathOperator{\Br}{Br}
52\DeclareMathOperator{\End}{End}
53\DeclareMathOperator{\new}{new}
54\DeclareMathOperator{\Aut}{Aut}
55\DeclareMathOperator{\Gal}{Gal}
56
57\begin{document}
58\maketitle
59
60\begin{abstract}
61We compute the minimum $N$ such that there
62is a newform $f=\sum a_n q^n\in{}S_2(\Gamma_0(N))$ with
63the odd property that
64$\Q(a_n)\not=\Q(\ldots,a_m,\ldots)$ for all~$n$.\footnote{Henri
65Cohen told me that
66phe discoved this example many years ago, and Elkies explained'' it to him.
67This example is not in the literature, and I found
68it indepedently of Cohen and Elkies.}
69%We use inner twists,
70%and compute the endomorphism ring of the associated abelian variety~$A_f$.
71\end{abstract}
72
73\section{The Newform}
74
75%A weight~$2$ newform~$f=\sum{}a_n q^n$
76%admits an \defn{inner twist} by a primitive Dirichlet
77%character~$\eps$ if there exists $\sigma\in\Gal(\Qbar/\Q)$
78%such that $a_p\eps(p) = \sigma(a_p)$ for almost all~$p$.
79
80
81%Let $A_f$ be the optimal quotient of $J_0(N)$ attached
82%to a modular form~$f$.
83%Shimura proved that inner twists
84%give rise to extra $\Qbar$-endomorphisms of~$A_f$.
85%For example, Koike showed that
86%the two-dimensional abelian variety $J_0(81)^{\new}$ is
87%isogeneous, over $\Q(\sqrt{-3})$,
88%to $E\cross E^{\tau}$, where~$E$ is an
89%elliptic curve over $\Q(\sqrt{-3})$ and~$\tau$ generates
90%$\Gal(\Q(\sqrt{-3})/\Q)$.
91
92
93%\section{Ribet and Shimura}
94
95Using modular symbols, we computed the first few terms of each
96newform $f=\sum a_n q^n$ on $S_2(\Gamma_0(N))$ for $N\leq 512$.
97For every single newform of level $N<512$, we found that there
98is an integer~$n$ such that the field $K_f=\Q(\ldots, a_m, \ldots)$
99generated by all Fourier coefficients of~$f$ equals the field
100$\Q(a_n)$, for some~$n$. However,
101the characteristic polynomial of $T_3$ on $S_2^{\new}(\Gamma_0(512))$ is
102$$103 (x^2 -6)^2(x^2 -2)^4(x^2 + 4x + 2)(x^2 -4x + 2), 104$$
105and if we let $f=\sum a_n q^n$
106be one of the newforms in the four-dimensional kernel~$V$
107of $(T_3^2-6)^2$, then
108$$f = q -\sqrt{6}q^3 + 2\sqrt{3}q^5 +2\sqrt{2}q^7 + 3q^9 -\sqrt{6}q^{11} 109 -2\sqrt{3}q^{13}-6\sqrt{2}q^{15} +\cdots$$
110provides an example in which $K_f\neq \Q(a_n)$ for some~$n$.
111We use the following theorem of Shimura (see \cite[Prop.~3.64]{shimura})
112to prove this assertion about~$f$:
113
114\begin{theorem}[Shimura]\label{shimura}
115Let $N, r, s, k>0$ be integers such that $s|N$ and let
116$M$ be the least common multiple of $N$, $r^2$, and $rs$.
117Let $\chi$ (resp. $\Psi$) be a primitive character mod $r$ (resp. $s$).
118If $f=\sum a_n q^n \in S_k(\Gamma_0(N),\Psi)$ then
119$$f\tensor\chi:=\sum \chi(n)a_n q^n\in S_k(\Gamma_0(M),\Psi\chi^2).$$
120\end{theorem}
121
122\begin{proposition}
123Suppose $n\geq 6$ and let~$\eps$ be a primitive
124character of conductor dividing~$8$.
125Then the map $f\mapsto f\tensor\eps$
126preserves $S_k(\Gamma_0(2^n))$.
127\end{proposition}
128\begin{proof}
129Set $r=8$ in Theorem~\ref{shimura}, and note that $\eps^2=1$.
130\end{proof}
131
132
133
134\begin{proposition}
135We have $K_f=\Q(\sqrt{2},\sqrt{3})$
136and the companions of $f$ coincide
137with the twists of~$f$ by the four Dirichlet characters
138of conductor dividing~$8$.
139\end{proposition}
140\begin{proof}
141The newform~$f$ and its companions lie inside
142of~$V$.
143By multiplicity one'', $V$ has dimension~$4$
144so $[\Q(f):\Q]\leq 4$.
145The characteristic polynomial of $T_7$ on $V$ is
146$(x^2-8)^2$ so $a_3 = \pm \sqrt{6}$ and
147$a_7 = \pm 2\sqrt{2}$ so
148$E=\Q(\sqrt{2},\sqrt{3})$.
149
150Let $\chi$ be a character of conductor dividing $8$.
151By Theorem~\ref{shimura}, $f\tensor\chi$ lies in
152$S_2(\Gamma_0(512))$.  In fact, it must lie
153in $S_2^{\new}(\Gamma_0(512))$ because if $f\tensor\chi$ lies in
154$S_2(\Gamma_0(2^n))$ for $n\geq 6$ then Theorem~\ref{shimura}
155implies that $f=(f\tensor\chi)\tensor\chi$ lies in $S_2(\Gamma_0(2^n))$,
156and certainly we can take $n\geq 6$.  So we're done.
157
158Alternatively, since the image of $\chi$ is
159contained in $\{\pm 1\}$, the $T_3$ eigenvalue of
160$f$ equal $\pm\sqrt{6}$.
161The factor $(x^2-6)^2$ exactly divides
162the characteristic polynomial of $T_3$ on $S_2(\Gamma_0(512))$
163so $f\tensor\chi$ must be a companion of $f$.
164\end{proof}
165
166
167\begin{remark} $f$ is the first example of a weight $2$ newform
168having trivial character for which $\Q(a_n)$ is a {\em proper}
169subfield of $E$ for all $n$.
170\end{remark}
171\vspace{2ex}
172
173\noindent{\bf Acknowledgment:} It is a pleasure to thank Kevin Buzzard and
175
176\begin{thebibliography}{HHHHHHH}
177\bibitem[1]{shimura} G. Shimura, {\em Introduction to the
178Arithmetic Theory of Automorphic Functions}, Princeton
179University Press, (1994).
180\end{thebibliography} \normalsize\vspace*{1 cm}
181
182\end{document}
183
184
185We introduce some notation in order to recall Ribet's Theorem.
186Let $E=\Q(f)=\Q(a_1,a_2,a_3,\ldots)$.  Let
187$$\Gamma=\{ \gamma\in\Aut(E) : \gamma f= 188 f\tensor \chi_{\gamma},\text{ some }\chi_{\gamma}\}.$$
189It is known that $\Gamma$ is an abelian subgroup
190of $\Aut(E)$.  Let $F$ be the subfield of $E$ fixed by $\Gamma$.
191
192\begin{remark}
193What is $[F:\Q]$ in terms of $\Gamma$ and $\#\Aut(E)$?
194Embed $E$ in its normal closure $K$.
195Let $H\subset G =\Gal(K/\Q)$ be the subgroup of automorphisms
196which preserve $E$.  Since every automorphism extends,
197$\Aut(E)$ is a quotient of $H$ with kernel $J$, the subgroup
198of $G$ corresponding to $E$ via Galois theory.
199Note that $J$ is normal in $H$ but need not be normal in $G$.
200The inverse image $\Gamma'$ of
201$\Gamma$ in $H$ is a subgroup of the same index as the index
202of $\Gamma$ in $\Aut(E)$.
203Unless I make a mistake computing indices, we get
204$$[F:\Q] = [G:\Gamma'+J] = \frac{[G:J]}{[\Gamma'+J:J]} 205 = \frac{[E:\Q]}{\#\Gamma}$$
206which is just what we would expect.
207\end{remark}
208
209For a primitive character $\vphi$ of conductor $c$ define
210$$g(\vphi)=\sum_{u=1}^{c}\vphi(u)e^{2\pi i u/c}.$$
211For $\gamma,\delta\in\Gamma$, let
212$$c(\gamma,\delta)=\frac{g(\chi_{\gamma}^{-1}) 213 g(\gamma(\chi_{\delta}^{-1}))} 214 {g(\chi_{\gamma\delta}^{-1})}.$$
215
216In \cite{ribet} there is an explicit description of $\End(A_f)\tensor\Q$
217in terms of generators and relations.
218Let $\X$ be the $E$-vector space
219$$\X = \sum_{\gamma\in\Gamma} E\cdot X_{\gamma}$$
220where the $X_{\gamma}$ are formal symbols.  By imposing
221on the $X_{\gamma}$ the rules
222\begin{eqnarray*}
223X_{\gamma}\cdot e &=& \gamma(e)X_{\gamma},
224         \quad\text{for $e\in E$ and $\gamma\in\Gamma$}\\
225X_{\gamma}X_{\delta} &=&c(\gamma,\delta) X_{\gamma \delta}
226\end{eqnarray*}
227we make $\X$ into an associative algebra.
228
229\begin{theorem}[Ribet\cite{ribet}]\label{ribet}
230There algebra $\X$ is a central simple
231algebra over $F$ which is isomorphic to $(\End A)\tensor\Q$.
232Furthermore, the $2$-cocycle $c$ corresponds to the class of
233$(\End A)\tensor \Q$ in $\Br(F)=H^2(F,\overline{F}^{\star})$.
234\end{theorem}
235
236
237\section{Example in which the endomorphism ring has order $2$ in the Brauer group}
238
239\comment{%PARI code
240? nf=nfinit(subst(E[1],x,t));
241? nffactor(nf,x^2-2)
242? a=-1/12*t^2 + 3/2;
243? nffactor(nf,x^2-3)
244? b=-1/24*t^3 + 7/4*t;
245? c=lift(Mod(a*b,t^4-36*t^2+36))
246? d=1;
247? v=subst(E[2],x,t);
248? 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);
249? w(f)=vector(4,i,polcoeff(f,i-1))~;;
250? N*w(v[5])
251? N*w(v[7])
252}
253
254
255\begin{theorem}
256The endomorphism ring of $A_f$ is a $2\times 2$ matrix ring
257over a quaternion division algebra with center $\Q$.
258\end{theorem}
259\begin{proof}
260Write $G=\Gal(E/\Q)=\{\gamma_1=1,\gamma_2,\gamma_3,\gamma_6\}$
261where $\gamma_d$ fixes $\sqrt{d}$, $d=2,3,6$.
262Let $\chi_d$, $d=\pm 1, \pm 2$ be the four Dirichlet characters of
263conductor dividing $8$, where $\chi_d$ corresponds to the field
264$\Q(\sqrt{d})$.
265We have
266\begin{eqnarray*}
267  f & = & f\tensor\chi_1\\
268  \gamma_2 f  = q +\sqrt{6}q^3 - 2\sqrt{3}q^5 +2\sqrt{2}q^7 + \cdots &=& f\tensor\chi_{2}\\
269  \gamma_3 f  = q +\sqrt{6}q^3 + 2\sqrt{3}q^5 -2\sqrt{2}q^7 + \cdots
270     &=& f\tensor\chi_{-1}\\
271  \gamma_6 f  = q -\sqrt{6}q^3 - 2\sqrt{3}q^5 -2\sqrt{2}q^7 + \cdots
272     &=& f\tensor\chi_{-2}
273\end{eqnarray*}
274
275The sums are
276\begin{eqnarray*}
277  g(\chi_1) &=& 1\\
278  g(\chi_{-1}) &=& e^{2\pi i/4} - e^{2\pi i 3/4} = i - (-i) = 2i\\
279  g(\chi_{2}) &=&
280    e^{2\pi i/8} - e^{2 \pi i 3/8} - e^{2\pi i 5/8} + e^{2\pi i 7/8}
281     = 2\sqrt{2}\\
282  g(\chi_{-2}) &=& e^{2\pi i/8} + e^{2 \pi i 3/8}
283      - e^{2\pi i 5/8} - e^{2\pi i 7/8}
284     = 2i\sqrt{2}.
285\end{eqnarray*}
286Thus we can compute
287$$c(\gamma_i,\gamma_j) = \frac{g(\chi_{\gamma_i}) g(\chi_{\gamma_j})} 288 {g(\chi_{\gamma_i\gamma_j})}.$$
289For example,
290$$c(\gamma_2,\gamma_3) = \frac{2\sqrt{2} \cdot 2 i}{2i\sqrt{2}} = 2$$
291By Theorem~\ref{ribet} we obtain a presentation of
292$(\End A)\tensor\Q$.
293Note that $c(\gamma_i,\gamma_j)=c(\gamma_j,\gamma_i)$ so
294that the generators $X_{\gamma}$ commute.  The endomorphism
295ring is not commutative as $E$ does not commute with $X_{\gamma}$
296for nontrivial $\gamma$.
297
298The $2$-cocycle $c$ represents the element of the Brauer group $\Br(\Q)$
299corresponding to $(\End A)\tensor\Q$.  We ask, does this element have
300order $1$ or order $2$?  Let $K=\Q_2(\sqrt{2},\sqrt{3})$.
301Then, by inf-res, $c$ arises from an
302element of
303  $$H^2(\Gal(K/\Q_2),K^{\star})\ra \Br(\Q).$$
304Note that inf {\em is} injective because of Hilbert's
305Theorem 90; see Proposition 6 on page 156 of \cite{serre}.
306
307I think that $c$ must have order dividing $2$ and
308$$H^2(\Gal(\Q_2(\sqrt{2})/\Q_2),\Q_2(\sqrt{2})^{\star})\subset H^2(\Gal(K/\Q_2),K^{\star})$$
309is the unique subgroup of order $2$ of $\Br(\Q)$, hence $c$ must lie in it.
310Using the Local symbol in Chapter XIV of we can write down the
311nonzero element of $H^2(\Gal(\Q_2(\sqrt{2})/\Q_2),\Q_2(\sqrt{2})^{\star})$.
312
313Let $G'=\Gal(\Q_2(\sqrt{2})/\Q_2)=\{ 1, \tau\}$.  What is
314$c$ as a map $G'\cross G' \ra \Q_2(\sqrt{2})^\star$?
315We have:
316$$c(1,1)=1, \quad c(1,\tau) = c(\tau,1) = 1, \quad 317 c(\tau,\tau) = -4.$$
318
319This is precisely the $2$-cocycle given by the local symbol $(2,-4)$;
320see Washington's article \cite{washington}.  Thus this cocycle is
321trivial iff $-4$ is a norm from $\Q_2(\sqrt{2})$.
322If $-4$ is a norm then so is $-1$ and hence $-1\in\Q_2(\sqrt{2})$.
323But this is impossible as the ramification degree of
324$2$ in $\Q(\sqrt{2},i)$ is $e=4$, so that the local degree of $2$
325is $4$.  Thus we finally conclude that $c$ is nontrivial and hence
326obtain the theorem.
327\footnote{\bf Woops! In fact, the cocycle is trivial, as Pete Clark pointed
328out.}
329\end{proof}
330
331
332\begin{thebibliography}{HHHHHHH}
333\bibitem[1]{doi} K. Doi, M. Yamauchi, {\em On the Hecke operators
334for $\Gamma_0(N)$ and class fields over quadratic number fields},
335J. Math. Soc. Japan, {\bf 25} (1973), 629--643.
336\bibitem[2]{koike} M. Koike, {\em On certain abelian varieties
337obtained from new forms of weight $2$ on
338$\Gamma_0(3^4)$ and $\Gamma_0(3^5)$.}, Nagoya Math. J.,
339{\bf 62} (1976), 29--39.
340\bibitem[3]{ribet} K. Ribet, {\em Twists of Modular Forms and
341Endomorphisms of Abelian Varieties}, Math. Ann. {\bf 253},
342(1980), 43--62.
343\bibitem[4]{serre} J-P. Serre, {\em Local Fields}, Springer-Verlag,
344(1979).
345\bibitem[5]{shimura} G. Shimura, {\em Introduction to the
346Arithmetic Theory of Automorphic Functions}, Princeton
347University Press, (1994).
348\bibitem[6]{washington} L.C. Washington, {\em Galois Cohomology},
349In Modular Forms and Fermat's Last Theorem'', Ed.'s
350Cornell-Silverman-Stevens, (1997).
351\end{thebibliography} \normalsize\vspace*{1 cm}
352
353\end{document}
354
355
356\comment{%Letter to Luiz
357
358     THe endomorphism algebra of A_f (over Q) is a central simple algebra over
359     F_f which contains Q_f as a maximal conmutative subfield. Its degree over
360     Q is [Q_f : Q]*[Q_f : F_f] .
361
362Moreover, the central simple algebra has order either one or two
363in the Brauer group of F_f.  Thus
364
365                          /- matrix algebra over F_f
366     (End A_f) tensor Q =  or
367                          \- matrix algebra over quaternion algebra over F_f.
368
369Knowing which one gives the exact isogeny decomposition of A_f over Qbar.
370
371     This is proved in : K. Ribet: "Twists of Modular Forms and Endomorphisms
372     of Abelian Varieties", Math. Ann. 253, 43-62 (1980)
373
374There is also some discussion in the (not so nicely typeset) paper of
375Ken's: Endomorphism algebras of abelian varieties attached
376         to newforms of weight 2''
377
378Have you tried to work out the isogeny structure (over Qbar) of the
379abelian variety A_f attached to your level 8192 form f?  I suspect
380that f has inner twists corresponding to all of the Dirichlet characters
381of conductor dividing 8, i.e., corresponding to the quadratic
382subfields of Q(zeta_8)=Q(i,sqrt(2)). I doubt that f has any
383inner twists by characters of degree > 2 or ramified outside of 2.
384I wonder what the order of (End A_f)tensor Q is in the Brauer group.
385Is it one or two?
386
387
388
389 Best,
390 William.
391
392
393Ken,
394
395Thanks for giving me a copy of your paper on extra twists.
396Let D be the endomorphism ring (tensor Q) of the abelian
397variety A_f corresponding to the newform in f in
398S_2(Gamma_0(512)) that we've been discussing (the one
399for which Q(a_n) never equals Q(f)).  In this case, the subfield
400of Q(f) fixed by the group which you call Gamma is the rational
401numbers Q. Thus D is a central simple Q-algebra.  I think I can use
402the theorem in your paper to show that D = M_2(K) where K is a division
403quaternion algebra with center Q, i.e., D has order two in the
404Brauer group of Q.  This example may be interesting in
405light of your remark on page 60 (of the 1980 Math. Annalen paper)
406that it does not seem easy to determine the order of D
407by pure thought''.  You showed it was 1 in many cases, but
408I don't think you gave conditions which imply that it must be 2.
409
410  Thanks again for the paper,
411  William
412}
413