CoCalc Public Fileswww / papers / endo_alg / endring.tex
Author: William A. Stein
Compute Environment: Ubuntu 18.04 (Deprecated)
1
\documentclass{article}
2
\hoffset=-0.03\textwidth
3
\textwidth=1.06\textwidth
4
\voffset=-0.03\textheight
5
\textheight=1.06\textheight
6
\bibliographystyle{amsalpha}
7
%\usepackage[hyperindex,pdfmark]{hyperref}
8
\usepackage[hypertex]{hyperref}
9
10
\include{macros}
11
\title{Explicitly Computing the Endomorphism Rings of Modular Abelian Varieties}
12
\author{William Stein}
13
\setcounter{section}{1}
14
\begin{document}
15
\maketitle
16
17
Let $\Gamma$ be a subgroup of $\SL_2(\Z)$ that contains $\Gamma_1(N)$
18
for some $N$, and let $J=\Jac(X_\Gamma)$ be the Jacobian of the
19
corresponding modular curve. The abelian variety~$J$ is defined
20
over~$\Q$ and has dimension $g=\dim S_2(\Gamma)$. It is isogenous to
21
a product (with multiplicities) of simple abelian subvarieties
22
$A_f\subset J$ attached to newforms of level dividing~$N$.
23
When $f$ is a newform (of level~$N$), the Hecke algebra
24
$$25 \T=\T_\Gamma = \Z[\ldots, T_n, \ldots] \subset \End(J) 26$$
27
preserves $A_f$, so there is a restriction homomorphism
28
$\T\to \End(A_f)$. This homomorphism is almost surjective,
29
in the following sense.
30
\begin{proposition}
31
The homomorphism $\T\to \End(A_f)$ has finite cokernel.
32
\end{proposition}
33
\begin{proof}
34
By \cite{shimura:factors}, the image of
35
$T \tensor \Q$ in $\End(A_f/\Q) \tensor \Q$ is a
36
field of degree $\dim A_f$. But~$A_f$ is simple by
37
\cite[Cor.~4.2]{ribet:twistsendoalg}, so
38
\cite[Thm.~2.1]{ribet:abvars} implies that
39
$\End(A_f/\Q) \tensor \Q$ also has dimension $\dim(A_f)$.
40
Thus $\T\tensor\Q$ surjects onto $\End(A_f/\Q) \tensor \Q$,
41
which proves the claim.
42
\end{proof}
43
44
When $\Gamma=\Gamma_0(N)$, we have computed all cokernels
45
$\T\to\End(A_f)$ for all $N\leq 332$. See
46
Example~\ref{ex:tim} below.
47
48
\begin{remark}
49
Note that $\End(A_f/\Qbar)$ is sometimes larger than $\End(A_f/\Q)$.
50
The ring $\End(A_f/\Qbar)$ can also be computed via methods similar to the ones
51
described here, but in addition to Hecke operators one has
52
to compute endomorphisms on modular symbols attached
53
to Shimura's inner twist operators. I have not found an
54
{\em efficient} way to compute these inner twist operators on
55
modular symbols.
56
Motivation: If
57
one could make computation of $\End(A_f/\Qbar)$ efficient, then
58
combined with a characteristic zero meataxe'' (the presumed
59
subject of Allan Steele's Ph.D. research), one would have a general
60
algorithm for computing all $\Q$-curves of given level.
61
Note also that Ribet has studied the abstract structure of
62
$\End(A_f/\Qbar)$ in detail (see \cite{ribet:twistsendoalg}).
63
\end{remark}
64
65
Using modular symbols, one can describe $\H_1(J,\Z)\ncisom \Z^{2g}$
66
explicitly as a module over the Hecke algebra $\T=\Z[\ldots, T_n, 67 \ldots]$. Also, for each newform~$f$ there is an injection $A_f\hra 68 J$, which induces an inclusion $$69 \Lambda_f = \H_1(A_f,\Z) \hra \H_1(J,\Z)$$
70
with
71
saturated image, i.e., torsion free cokernel. Given~$f$, the image of
72
$\H_1(A_f,\Z)$ in $\H_1(J,\Z)$ can be explicitly computed just using
73
Hecke operators. We view $A_f$ as a complex torus given by this
74
image, which defines a subtorus of the complex torus $$75 J(\C) \isom 76 \H_1(J,\R)/\H_1(J,\Z).$$
77
78
79
\begin{algorithm}{Saturate}\label{alg:saturate}
80
Given a subspace $V$ of $\Q^n$, this algorithm computes
81
the intersection $L=V \cap \Z^n$.
82
\begin{steps}
83
\item{} [Echelon Form] Using the reduced row echelon form of a basis
84
matrix of~$V$, find a matrix $A \in M_n(\Z)$ whose integer kernel is~$L$.
85
(In practice, $V$ will be presented by giving the reduced row echelon
86
form of a basis matrix.)
87
\item{} [Kernel] Compute $\Ker(A)$, e.g., as described in \cite[\S2.7.1]{cohen:course_ant},
88
using the LLL lattice reduction algorithm.
89
Since $L$ is saturated in $\Z^n$, we have $L=\Ker(A)$.
90
\end{steps}
91
\end{algorithm}
92
93
94
\begin{lemma}\label{lem:gal}
95
Let $K$ be a number field.
96
If an element $x\in \C$ is fixed by every element of $\Aut(\C/K)$,
97
then $x\in K$.
98
\end{lemma}
99
\begin{proof}
100
If $x\in\Kbar$, this is standard Galois theory. If $x\not\in 101 \Kbar$, then $x$ is transcendental.
102
Since $x+1$ is also transcendental,
103
the fields $\Kbar(x)$ and $\Kbar(x+1)$ are isomorphic via a map $\sigma$
104
sending $x$ to $x+1$.
105
Every automorphism of a subfield of $\C$
106
extends to $\C$, so $\sigma$ extends to an automorphism of $\C$
107
that does not fix~$x$.
108
\end{proof}
109
110
\begin{proposition}\label{prop:end}
111
If $A$ is a simple abelian variety over a number field $K$, then
112
$$\End(A/K) = (\End(A/K)\tensor \Q)\cap \End(\Lambda_A),$$
113
where
114
$\Lambda_A = \H_1(A,\Z)$ and we implicitly embed $\End(A/K)$ in
115
$\End(\Lambda_A)$, so the intersection takes place in
116
$\End(\Lambda_A)\tensor\Q$.
117
\end{proposition}
118
\begin{proof}
119
120
The inclusion of $\End(A/K)$ in the right hand side is obvious,
121
so suppose $\vphi \in (\End(A/K)\tensor \Q)\cap \End(\Lambda_A)$.
122
Then there is a positive integer $n$ such that $n\vphi\in \End(A/K)$.
123
Thus $n\vphi$ induces a complex-linear endomorphism of $\Tan_0(A_\C)$.
124
Hence $\vphi$ induces a complex-linear endomorphism of $\Tan_0(A_\C)$,
125
and by hypothesis $\vphi$ preserves $\Lambda_A$.
126
An element of $\End(A/\C)$ is a complex linear map on $\Tan_0(A_\C)$
127
that preserves $\Lambda_A$, so $\vphi\in\End(A/\C)$.
128
129
There is an action of $\Gal(\C/K)$ on $\End(A/\C)$, which
130
we extend $\Q$-linearly to an action on $\End(A/\C)\tensor\Q$.
131
Since $n\vphi \in \End(A/K) \subset \End(A/\C)$ is
132
defined over $K$, for any $\sigma\in\Gal(\C/K)$,
133
we have $\sigma(n\vphi) = n\vphi$.
134
But
135
$$\sigma(n\vphi) = \sigma([n])\sigma(\vphi) = [n]\sigma(\vphi),$$
136
so
137
$$[n](\sigma(\vphi) - \vphi) = 0,$$
138
which implies $\sigma(\vphi)=\vphi$, since the kernel of $[n]$
139
is finite and image of $\sigma(\vphi)-\vphi$ is either infinite or $0$.
140
By Lemma~\ref{lem:gal}, $\vphi\in \End(A/K)$.
141
\end{proof}
142
143
\begin{algorithm}{Compute $\End(A_f)$}\label{alg:endaf}
144
Let $f\in S_2(\Gamma)$ be a newform.
145
This algorithm computes $\End(A_f)$.
146
\begin{enumerate}
147
\item{} [Initialize] Let $d=\dim(A_f)$, let $n=2$, and let $V=\Q I$ be the
148
subspace of $\End(A_f/\Q)\tensor\Q \subset \End(\H_1(A_f,\Q))$
149
spanned by the identity matrix.
150
\item{} [Compute Hecke Operator]
151
Using modular symbols, compute the restriction $T_n|_{A_f}$ of the Hecke operator
152
$T_n$ to $\H_1(A_f,\Q)$.
153
\item{} [Increase $V$]\label{step:increase}
154
Replace $V$ by $V+W$, where $W$ is the span of the powers
155
$T_n|_{A_f}^i$, for $i=1,2,\ldots, d$.
156
\item{} [Finished?]\label{step:finished}
157
If $\dim(V) < d$, go to step 2.
158
\item{} [Saturate] Compute $\End(A_f/\Q) = 159 V \cap \End(\Lambda_{A_f})$ using Algorithm~\ref{alg:saturate}.
160
\end{enumerate}
161
\end{algorithm}
162
163
\begin{remark}
164
Often we find $A_f$ by finding a Hecke operator $T$, which is a
165
random linear combination of $T_n$ with~$n$ small, such that the
166
characteristic polynomial of~$T$ on $S_2(\Gamma)_{\new}$ is square
167
free. Then the image of~$T$ in $\End(A_f/\Q)$ will generate
168
$\End(A_f/\Q)\tensor\Q$ as an algebra.
169
If we use $T$ instead of $T_n$ in Step~\ref{step:increase},
170
we will always be finished in Step~\ref{step:finished}.
171
\end{remark}
172
173
\begin{remark}
174
Algorithm~\ref{alg:endaf} can be extended to arbitrary
175
explicit factors~$A$ of $\Jac(X_\Gamma)$ by
176
decomposing~$A$ up to isogeny as a product of $A_f$'s.
177
\end{remark}
178
179
\begin{remark}
180
Everything we have done formally make sense with~$f$ replaced
181
by a newform of weight $k>2$. What is the importance of $\End(A_f(\C))$
182
and the cokernel of $\T \to \End(A_f(\C))$ in this case?
183
\end{remark}
184
185
186
One can also compute the image of $\T$ in $\End(A_f)$ as a subring,
187
since $T_1,\ldots, T_r$, generate $\T$ as a $\Z$-module,
188
where $r=\frac{km}{12}$ with $m=[\SL_2(\Z):\Gamma]$
189
(see \cite[App.]{lario-schoof}).
190
191
\begin{example}
192
We compute $\End(J_0(23))$ and note that it equals the Hecke
193
algebra.
194
\begin{verbatim}
195
> J := JZero(23);
196
> R := End(J);
197
> Basis(R);
198
[ Homomorphism from JZero(23) to JZero(23) given by:
199
(Identity Matrix)
200
Homomorphism from JZero(23) to JZero(23) given by:
201
[ 0 1 -1 0]
202
[ 0 1 -1 1]
203
[-1 2 -2 1]
204
[-1 1 0 -1]]
205
> T := HeckeAlgebra(J);
206
> Index(R,T);
207
1
208
\end{verbatim}
209
\end{example}
210
211
% \begin{example}
212
% {\small
213
% \begin{verbatim}
214
% > J := JZero(389); A := Factorization(J)[5][1]; A;
215
% Modular abelian variety 389E of dimension 20, level 389 and
216
% conductor 389^20 over Q
217
% > time Index(End(A),HeckeAlgebra(A));
218
% 1
219
% Time: 8.470 seconds (on my laptop)
220
% > R := BaseRing(Newform(A)); O := MaximalOrder(R);
221
% > factor(Discriminant(End(A)) / Discriminant(O));
222
% [ <2, 24> ]
223
% \end{verbatim}
224
% }
225
% \end{example}
226
227
\begin{example}
228
Let $A_f\subset J_0(559)$ be the newform abelian variety of dimension~$15$.
229
In this example we verify that $\End(A_f/\Q)$ is generated by the Hecke
230
operators, but note that the index of $\End(A_f/\Q)$ in its normalization
231
is $7$. Note that the discriminant of the endomorphism ring
232
is the discriminant of the trace pairing matrix acting on homology,
233
not of left multiplication on itself, so the discriminant
234
is $2^{\dim(A)}$ times as big as it would be otherwise.
235
\begin{verbatim}
236
> J := JZero(559);
237
> [Dimension(D[1]) : D in Factorization(J)];
238
[ 3, 4, 7, 14, 15, 1, 2 ]
239
> A := Factorization(J)[5][1]; A;
240
Modular abelian variety 559E of dimension 15, level 13*43 and
241
conductor 13^15*43^15 over Q
242
> Index(End(A),HeckeAlgebra(A));
243
1
244
> Discriminant(End(A));
245
2747410093977522170045665218920448
246
> R := BaseRing(Newform(A)); O := MaximalOrder(R);
247
> Discriminant(O);
248
1711108207844339282005880064
249
> Factorization(Discriminant(End(A)) div Discriminant(O));
250
[ <2, 15>, <7, 2> ]
251
> Factorization(Discriminant(O));
252
[ <2, 8>, <3, 1>, <29, 1>, <37, 1>, <97, 1>, <4802947, 1>,
253
<4456942220789, 1> ]
254
> Factorization(Discriminant(R));
255
[ <2, 8>, <3, 1>, <7, 2>, <29, 1>, <37, 1>, <97, 1>,
256
<4802947, 1>, <4456942220789, 1> ]
257
\end{verbatim}
258
\end{example}
259
260
\begin{example}\label{ex:tim}
261
If $A_f\subset J_0(N)$ is a newform abelian subvariety, then
262
the cokernel of the restriction map
263
$264 \T \to \End(A_f) 265$
266
is relevant to questions involving congruences
267
between modular forms.
268
For $N\leq 338$, the map $\T\to \End(A_f)$ is surjective,
269
except in a few cases.
270
The cokernel has order~$2$ for the following $A_f$:
271
$$94B, 125C, 160C, 166B, 196C, 199C, 224C, 272 224D, 227B, 227E, 233C, 249E,$$
273
$$250B, 250C, 256E, 259B, 277D, 278D, 295D, 299G, 274 303E, 307F, 320G, 326D,$$
275
$$329D, 331C, 332C, 338G, 338H.$$
276
Note that $199$ and several other of these levels are prime.
277
The cokernel has order~$3$ for
278
$321D$ and $335D$. Note that $335=5\cdot 67$ is coprime to~$3$.
279
\end{example}
280
\begin{question}
281
Is the cokernel of
282
$283 \T \to \End(A_f) 284$
285
a power of $2$ times a power of $3$?
286
\end{question}
287
288
\bibliography{biblio}
289
\end{document}
290
291
We used the following program to
292
verify the above statement.
293
\begin{verbatim}
294
> procedure f(N)
295
D := Factorization(NewSubvariety(JZero(N)));
296
for i in [1..#D] do
297
n := Index(End(D[i][1]),HeckeAlgebra(D[i][1]));
298
if n gt 1 then
299
print N, i, n;
300
end if;
301
end for;
302
end procedure;
303
> for N in [1..1000] do f(N); end for;
304
94 2 2
305
125 3 2
306
307
\end{verbatim}
308