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