CoCalc Shared Fileswww / papers / twist / twistold.tex
Author: William A. Stein
1% conjtwist.tex
2\documentclass[11pt]{article}
3\include{macros}
4\title{A newform with inner twists at level~$512$}
5\author{K.\thinspace{}M. Buzzard and W.\thinspace{}A. Stein}
6\begin{document}
7\maketitle
8
9\section{Introduction}
10
11Doi, Koike, Ribet, Shimura, and Yamauchi have studied newforms $f$
12which admit an {\em inner twist}.  That is, for which there
13exists a nontrivial primitive character $\chi$ so that
14$f\tensor\chi=\sigma f$ for some $\sigma\in\Gal(\Qbar/\Q)$.
15Let $A_f$ be the abelian variety associated to $f$ (considered
16only up to isogeny). Then $A_f$ is simple over $\Q$.
17Inner twists of $f$ are of interest because they can give rise to
18nontrivial decompositions of $A_f$ over a finite extension of $\Q$.
19For example, Koike used the theory to see that the
20$2$ dimensional abelian variety $J_0(81)^{\new}$ is isogeneous over
21$\Q(\sqrt{-3})$ to $E\cross E^{\tau}$ where $E$ is an explicitely
22determined elliptic curve over $\Q(\sqrt{-3})$ and $\tau$ is the
23nontrivial automorphism.
24
25In this paper we work out a single example.  It is hoped that
26the ideas used in working out the example can be used to both
27find more examples and prove a converse to some of Ribet's theorems.
28
29\section{Theorems of Shimura and Ribet}
30\begin{theorem}[Shimura]\label{shimura}
31Let $N, r, s, k>0$ be integers such that $s|N$ and let
32$M$ be the least common multiple of $N$, $r^2$, and $rs$.
33Let $\chi$ (resp. $\Psi$) be a primitive character mod $r$ (resp. $s$).
34If $f=\sum a_n q^n \in S_k(\Gamma_0(N),\Psi)$ then
35$$f\tensor\chi:=\sum \chi(n)a_n q^n\in S_k(\Gamma_0(M),\Psi\chi^2).$$
36\end{theorem}
37\begin{proof}
38Proposition 3.64 of \cite{shimura}.
39\end{proof}
40
41As a special case we have
42\begin{corollary}
43Let $2^n\geq 64$ and let $\chi$
44be a primitive character of conductor dividing $8$.
45Then twisting by $\chi$'' (i.e., the
46map $f\mapsto f\tensor\chi$) preserves $S_k(\Gamma_0(2^n),\Psi)$.
47\end{corollary}
48\begin{proof}
49Put $r=8$ in the theorem and note that $\chi^2=1$.
50\end{proof}
51
52We introduce some notation in order to recall Ribet's Theorem.
53Let $E=\Q(f)=\Q(a_1,a_2,a_3,\ldots)$.  Let
54$$\Gamma=\{ \gamma\in\Aut(E) : \gamma f= 55 f\tensor \chi_{\gamma},\text{ some }\chi_{\gamma}\}.$$
56It is known that $\Gamma$ is an abelian subgroup
57of $\Aut(E)$.  Let $F$ be the subfield of $E$ fixed by $\Gamma$.
58
59\begin{remark}
60What is $[F:\Q]$ in terms of $\Gamma$ and $\#\Aut(E)$?
61Embed $E$ in its normal closure $K$.
62Let $H\subset G =\Gal(K/\Q)$ be the subgroup of automorphisms
63which preserve $E$.  Since every automorphism extends,
64$\Aut(E)$ is a quotient of $H$ with kernel $J$, the subgroup
65of $G$ corresponding to $E$ via Galois theory.
66Note that $J$ is normal in $H$ but need not be normal in $G$.
67The inverse image $\Gamma'$ of
68$\Gamma$ in $H$ is a subgroup of the same index as the index
69of $\Gamma$ in $\Aut(E)$.
70Unless I make a mistake computing indices, we get
71$$[F:\Q] = [G:\Gamma'+J] = \frac{[G:J]}{[\Gamma'+J:J]} 72 = \frac{[E:\Q]}{\#\Gamma}$$
73which is just what we would expect.
74\end{remark}
75
76For a primitive character $\vphi$ of conductor $c$ define
77$$g(\vphi)=\sum_{u=1}^{c}\vphi(u)e^{2\pi i u/c}.$$
78For $\gamma,\delta\in\Gamma$, let
79$$c(\gamma,\delta)=\frac{g(\chi_{\gamma}^{-1}) 80 g(\gamma(\chi_{\delta}^{-1}))} 81 {g(\chi_{\gamma\delta}^{-1})}.$$
82
83In \cite{ribet} there is an explicit description of $\End(A_f)\tensor\Q$
84in terms of generators and relations.
85Let $\X$ be the $E$-vector space
86$$\X = \sum_{\gamma\in\Gamma} E\cdot X_{\gamma}$$
87where the $X_{\gamma}$ are formal symbols.  By imposing
88on the $X_{\gamma}$ the rules
89\begin{eqnarray*}
90X_{\gamma}\cdot e &=& \gamma(e)X_{\gamma},
91         \quad\text{for $e\in E$ and $\gamma\in\Gamma$}\\
92X_{\gamma}X_{\delta} &=&c(\gamma,\delta) X_{\gamma \delta}
93\end{eqnarray*}
94we make $\X$ into an associative algebra.
95
96\begin{theorem}[Ribet\cite{ribet}]\label{ribet}
97There algebra $\X$ is a central simple
98algebra over $F$ which is isomorphic to $(\End A)\tensor\Q$.
99Furthermore, the $2$-cocycle $c$ corresponds to the class of
100$(\End A)\tensor \Q$ in $\Br(F)=H^2(F,\overline{F}^{\star})$.
101\end{theorem}
102
103\section{Example in which the endomorphism ring has order $2$ in the Brauer group}
104
105The characteristic polynomial of $T_3$ on $S_2^{\new}(\Gamma_0(512))$ is
106$$(x^2 -6)^2(x^2 -2)^4(x^2 + 4x + 2)(x^2 -4x + 2)$$
107Let $f=\sum a_n q^n$ be one of the newforms in the four dimensional kernel
108$V$ of $(T_3^2-6)^2$.
109and let $E=\Q(f)=\Q(a_1,a_2,a_3,a_4,\ldots)$.
110
111\begin{proposition}
112We have $E=\Q(\sqrt{2},\sqrt{3})$
113and the compa-nions of $f$ coincide
114with the twists of $f$ by the four Dirichlet characters
115of conductor dividing $8$.  In particular, $\Gamma=\Gal(E/\Q)$.
116\end{proposition}
117\begin{proof}
118The newform $f$ and its companions lie inside
119of $V$.  By multiplicity one'', $V$ has dimension
120$4$ so $[\Q(f):\Q]\leq 4$.
121The characteristic polynomial of $T_7$ on $V$ is
122$(x^2-8)^2$ so $a_3 = \pm \sqrt{6}$ and
123$a_7 = \pm 2\sqrt{2}$ so
124$E=\Q(\sqrt{2},\sqrt{3})$.
125
126Let $\chi$ be a character of conductor dividing $8$.
127By Theorem~\ref{shimura}, $f\tensor\chi$ lies in
128$S_2(\Gamma_0(512))$.  In fact, it must lie
129in $S_2^{\new}(\Gamma_0(512))$ because if $f\tensor\chi$ lies in
130$S_2(\Gamma_0(2^n))$ for $n\geq 6$ then Theorem~\ref{shimura}
131implies that $f=(f\tensor\chi)\tensor\chi$ lies in $S_2(\Gamma_0(2^n))$,
132and certainly we can take $n\geq 6$.  So we're done.
133
134Alternatively, since the image of $\chi$ is
135contained in $\{\pm 1\}$, the $T_3$ eigenvalue of
136$f$ equal $\pm\sqrt{6}$.
137The factor $(x^2-6)^2$ exactly divides
138the characteristic polynomial of $T_3$ on $S_2(\Gamma_0(512))$
139so $f\tensor\chi$ must be a companion of $f$.
140\end{proof}
141
142The $q$-expansion of $f$ can be computed to be
143$$f = q -\sqrt{6}q^3 + 2\sqrt{3}q^5 +2\sqrt{2}q^7 + 3q^9 -\sqrt{6}q^{11} 144 -2\sqrt{3}q^{13}-6\sqrt{2}q^{15} + 4q^{17}+\cdots$$
145
146\begin{remark} $f$ is the first example of a weight $2$ newform
147having trivial character for which $\Q(a_n)$ is a {\em proper}
148subfield of $E$ for all $n$.
149\end{remark}
150
151\comment{%PARI code
152? nf=nfinit(subst(E[1],x,t));
153? nffactor(nf,x^2-2)
154? a=-1/12*t^2 + 3/2;
155? nffactor(nf,x^2-3)
156? b=-1/24*t^3 + 7/4*t;
157? c=lift(Mod(a*b,t^4-36*t^2+36))
158? d=1;
159? v=subst(E[2],x,t);
160? 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);
161? w(f)=vector(4,i,polcoeff(f,i-1))~;;
162? N*w(v[5])
163? N*w(v[7])
164}
165
166
167\begin{theorem}
168The endomorphism ring of $A_f$ is a $2\times 2$ matrix ring
169over a quaternion division algebra with center $\Q$.
170\end{theorem}
171\begin{proof}
172Write $G=\Gal(E/\Q)=\{\gamma_1=1,\gamma_2,\gamma_3,\gamma_6\}$
173where $\gamma_d$ fixes $\sqrt{d}$, $d=2,3,6$.
174Let $\chi_d$, $d=\pm 1, \pm 2$ be the four Dirichlet characters of
175conductor dividing $8$, where $\chi_d$ corresponds to the field
176$\Q(\sqrt{d})$.
177We have
178\begin{eqnarray*}
179  f & = & f\tensor\chi_1\\
180  \gamma_2 f  = q +\sqrt{6}q^3 - 2\sqrt{3}q^5 +2\sqrt{2}q^7 + \cdots &=& f\tensor\chi_{2}\\
181  \gamma_3 f  = q +\sqrt{6}q^3 + 2\sqrt{3}q^5 -2\sqrt{2}q^7 + \cdots
182     &=& f\tensor\chi_{-1}\\
183  \gamma_6 f  = q -\sqrt{6}q^3 - 2\sqrt{3}q^5 -2\sqrt{2}q^7 + \cdots
184     &=& f\tensor\chi_{-2}
185\end{eqnarray*}
186
187The sums are
188\begin{eqnarray*}
189  g(\chi_1) &=& 1\\
190  g(\chi_{-1}) &=& e^{2\pi i/4} - e^{2\pi i 3/4} = i - (-i) = 2i\\
191  g(\chi_{2}) &=&
192    e^{2\pi i/8} - e^{2 \pi i 3/8} - e^{2\pi i 5/8} + e^{2\pi i 7/8}
193     = 2\sqrt{2}\\
194  g(\chi_{-2}) &=& e^{2\pi i/8} + e^{2 \pi i 3/8}
195      - e^{2\pi i 5/8} - e^{2\pi i 7/8}
196     = 2i\sqrt{2}.
197\end{eqnarray*}
198Thus we can compute
199$$c(\gamma_i,\gamma_j) = \frac{g(\chi_{\gamma_i}) g(\chi_{\gamma_j})} 200 {g(\chi_{\gamma_i\gamma_j})}.$$
201For example,
202$$c(\gamma_2,\gamma_3) = \frac{2\sqrt{2} \cdot 2 i}{2i\sqrt{2}} = 2$$
203By Theorem~\ref{ribet} we obtain a presentation of
204$(\End A)\tensor\Q$.
205Note that $c(\gamma_i,\gamma_j)=c(\gamma_j,\gamma_i)$ so
206that the generators $X_{\gamma}$ commute.  The endomorphism
207ring is not commutative as $E$ does not commute with $X_{\gamma}$
208for nontrivial $\gamma$.
209
210The $2$-cocycle $c$ represents the element of the Brauer group $\Br(\Q)$
211corresponding to $(\End A)\tensor\Q$.  We ask, does this element have
212order $1$ or order $2$?  Let $K=\Q_2(\sqrt{2},\sqrt{3})$.
213Then, by inf-res, $c$ arises from an
214element of
215  $$H^2(\Gal(K/\Q_2),K^{\star})\ra \Br(\Q).$$
216Note that inf {\em is} injective because of Hilbert's
217Theorem 90; see Proposition 6 on page 156 of \cite{serre}.
218
219I think that $c$ must have order dividing $2$ and
220$$H^2(\Gal(\Q_2(\sqrt{2})/\Q_2),\Q_2(\sqrt{2})^{\star})\subset H^2(\Gal(K/\Q_2),K^{\star})$$
221is the unique subgroup of order $2$ of $\Br(\Q)$, hence $c$ must lie in it.
222Using the Local symbol in Chapter XIV of we can write down the
223nonzero element of $H^2(\Gal(\Q_2(\sqrt{2})/\Q_2),\Q_2(\sqrt{2})^{\star})$.
224
225Let $G'=\Gal(\Q_2(\sqrt{2})/\Q_2)=\{ 1, \tau\}$.  What is
226$c$ as a map $G'\cross G' \ra \Q_2(\sqrt{2})^\star$?
227We have:
228$$c(1,1)=1, \quad c(1,\tau) = c(\tau,1) = 1, \quad 229 c(\tau,\tau) = -4.$$
230
231This is precisely the $2$-cocycle given by the local symbol $(2,-4)$;
232see Washington's article \cite{washington}.  Thus this cocycle is
233trivial iff $-4$ is a norm from $\Q_2(\sqrt{2})$.
234If $-4$ is a norm then so is $-1$ and hence $-1\in\Q_2(\sqrt{2})$.
235But this is impossible as the ramification degree of
236$2$ in $\Q(\sqrt{2},i)$ is $e=4$, so that the local degree of $2$
237is $4$.  Thus we finally conclude that $c$ is nontrivial and hence
238obtain the theorem.
239\end{proof}
240
241
242\begin{thebibliography}{HHHHHHH}
243\bibitem[D]{doi} K. Doi, M. Yamauchi, {\em On the Hecke operators
244for $\Gamma_0(N)$ and class fields over quadratic number fields},
245J. Math. Soc. Japan, {\bf 25} (1973), 629--643.
246\bibitem[K]{koike} M. Koike, {\em On certain abelian varieties
247obtained from new forms of weight $2$ on
248$\Gamma_0(3^4)$ and $\Gamma_0(3^5)$.}, Nagoya Math. J.,
249{\bf 62} (1976), 29--39.
250\bibitem[R]{ribet} K. Ribet, {\em Twists of Modular Forms and
251Endomorphisms of Abelian Varieties}, Math. Ann. {\bf 253},
252(1980), 43--62.
253\bibitem[SE]{serre} J.P. Serre, {\em Local Fields}, Springer-Verlag,
254(1979).
255\bibitem[S]{shimura} G. Shimura, {\em Introduction to the
256Arithmetic Theory of Automorphic Functions}, Princeton
257University Press, (1994).
258\bibitem[W]{washington} L.C. Washington, {\em Galois Cohomology},
259In Modular Forms and Fermat's Last Theorem'', Ed.'s
260Cornell-Silverman-Stevens, (1997).
261\end{thebibliography} \normalsize\vspace*{1 cm}
262
263\end{document}
264
265
266\comment{%Letter to Luiz
267
268     THe endomorphism algebra of A_f (over Q) is a central simple algebra over
269     F_f which contains Q_f as a maximal conmutative subfield. Its degree over
270     Q is [Q_f : Q]*[Q_f : F_f] .
271
272Moreover, the central simple algebra has order either one or two
273in the Brauer group of F_f.  Thus
274
275                          /- matrix algebra over F_f
276     (End A_f) tensor Q =  or
277                          \- matrix algebra over quaternion algebra over F_f.
278
279Knowing which one gives the exact isogeny decomposition of A_f over Qbar.
280
281     This is proved in : K. Ribet: "Twists of Modular Forms and Endomorphisms
282     of Abelian Varieties", Math. Ann. 253, 43-62 (1980)
283
284There is also some discussion in
285Ken's: Endomorphism algebras of abelian varieties attached
286         to newforms of weight 2''
287
288Have you tried to work out the isogeny structure (over Qbar) of the
289abelian variety A_f attached to your level 8192 form f?  I suspect
290that f has inner twists corresponding to all of the Dirichlet characters
291of conductor dividing 8, i.e., corresponding to the quadratic
292subfields of Q(zeta_8)=Q(i,sqrt(2)). I doubt that f has any
293inner twists by characters of degree > 2 or ramified outside of 2.
294I wonder what the order of (End A_f)tensor Q is in the Brauer group.
295Is it one or two?
296
297
298
299 Best,
300 William.
301
302
303Ken,
304
305Thanks for giving me a copy of your paper on extra twists.
306Let D be the endomorphism ring (tensor Q) of the abelian
307variety A_f corresponding to the newform in f in
308S_2(Gamma_0(512)) that we've been discussing (the one
309for which Q(a_n) never equals Q(f)).  In this case, the subfield
310of Q(f) fixed by the group which you call Gamma is the rational
311numbers Q. Thus D is a central simple Q-algebra.  I think I can use
312the theorem in your paper to show that D = M_2(K) where K is a division
313quaternion algebra with center Q, i.e., D has order two in the
314Brauer group of Q.  This example may be interesting in
315light of your remark on page 60 (of the 1980 Math. Annalen paper)
316that it does not seem easy to determine the order of D
317by pure thought''.  You showed it was 1 in many cases, but
318I don't think you gave conditions which imply that it must be 2.
319
320  Thanks again for the paper,
321  William
322}
323