Sharedwww / artin2.texOpen in CoCalc
Author: William A. Stein
1
% artin.tex
2
3
\documentclass{article}
4
\pagestyle{myheadings}
5
\newcounter{Pagecount}
6
\markboth{}{Buzzard-Stein (\today)}%{Icosahedral Galois representations}
7
\title{Mod five approaches to modularity of icosahedral
8
Galois representations}
9
\author{Kevin Buzzard and William A. Stein}
10
11
% all of our conversations have been moved into an ``edit'' macro
12
\newcommand{\edit}[1]{\footnote{#1}}
13
%\newcommand{\edit}[1]{}
14
% Kevin speaks in ``\kevin'' whereas william speaks in ``\william''
15
\newcommand{\william}[1]{\par{}WS: {\sl #1}}
16
\newcommand{\kevin}[1]{\par{}KB: {\it #1}}
17
18
% macros
19
\usepackage{amsmath}
20
\usepackage{amssymb} % divide, not divide
21
\newcommand{\abcd}[4]{\left(
22
\begin{smallmatrix}#1&#2\\#3&#4\end{smallmatrix}\right)}
23
\newcommand{\mtwo}[4]{\left(
24
\begin{matrix}#1&#2\\#3&#4
25
\end{matrix}\right)}
26
\newcommand{\alp}{\alpha}
27
\newcommand{\bA}{\mathbf{A}}
28
\newcommand{\C}{\mathbf{C}}
29
\newcommand{\cL}{\mathcal{L}}
30
\newcommand{\cO}{\mathcal{O}}
31
\newcommand{\comment}[1]{}
32
\newcommand{\con}{\equiv}
33
\DeclareMathOperator{\cond}{cond}
34
\newcommand{\cross}{\times}
35
\newcommand{\defn}[1]{{\em #1}}
36
\newcommand{\eps}{\varepsilon}
37
\newcommand{\F}{\mathbf{F}}
38
\newcommand{\fbar}{\overline{f}}
39
\newcommand{\Fbar}{\overline{\F}}
40
\DeclareMathOperator{\Frob}{Frob}
41
\newcommand{\GQ}{G_{\Q}}
42
\DeclareMathOperator{\Gal}{Gal}
43
\newcommand{\galq}{\Gal(\Qbar/\Q)}
44
\DeclareMathOperator{\GL}{GL}
45
\DeclareMathOperator{\Hom}{Hom}
46
\newcommand{\im}{\mbox{\rm Im}}
47
\newcommand{\intersect}{\cap}
48
\newcommand{\isom}{\cong}
49
\newcommand{\magma}{{\sc Magma}}
50
\newcommand{\ncisom}{\approx} % noncanonical isomorphism
51
\DeclareMathOperator{\PGL}{PGL}
52
\DeclareMathOperator{\proj}{proj}
53
\DeclareMathOperator{\PSL}{PSL}
54
\newcommand{\ra}{\rightarrow}
55
\newcommand{\rhobar}{\overline{\rho}}
56
\newcommand{\Q}{\mathbf{Q}}
57
\newcommand{\Qbar}{\overline{\Q}}
58
\DeclareMathOperator{\SL}{SL}
59
\newcommand{\sM}{\boldsymbol{\mathcal{M}}}
60
\newcommand{\sS}{\boldsymbol{\mathcal{S}}}
61
\newcommand{\T}{\mathbf{T}}
62
\newcommand{\tensor}{\otimes}
63
\newcommand{\vphi}{\varphi}
64
\newcommand{\Z}{\mathbf{Z}}
65
66
\usepackage{amsthm}
67
\theoremstyle{plain}
68
\newtheorem{theorem}{Theorem}[section]
69
\newtheorem{proposition}[theorem]{Proposition}
70
\newtheorem{lemma}[theorem]{Lemma}
71
\theoremstyle{remark}
72
\newtheorem{remark}{Remark}[theorem]
73
74
75
% bulleted list environment
76
\newenvironment{dashlist}
77
{
78
\begin{list}
79
{---}
80
{
81
\setlength{\itemsep}{0.5ex}
82
\setlength{\parsep}{0ex}
83
\setlength{\parskip}{1ex}
84
\setlength{\topsep}{0ex}
85
}
86
}
87
{
88
\end{list}
89
}
90
%end newenvironment
91
92
% numbered list environment
93
\newcounter{listnum}
94
\newenvironment{numlist}
95
{
96
\begin{list}
97
{{\em \thelistnum.}}{
98
\usecounter{listnum}
99
\setlength{\itemsep}{.5ex}
100
\setlength{\parsep}{0ex}
101
\setlength{\parskip}{0ex}
102
\setlength{\topsep}{0ex}
103
}
104
}
105
{
106
\end{list}
107
}
108
%end newenvironment
109
110
111
\begin{document}
112
\maketitle
113
\begin{abstract}
114
Using a theorem of Buzzard and Taylor, we compute new icosahedral
115
examples of Artin's conjecture on holomorphicity of the $L$-function
116
associated to certain Galois representations. We give in detail one
117
new example of an icosahedral representation of conductor
118
${\bf 1376}=2^5\cdot 43$ that satisfies Artin's conjecture. We also
119
give\edit{\kevin{[[Still let's see what happens...]]}} seven
120
additional examples of conductors
121
${\bf 2416}=2^4\cdot 151$,
122
${\bf 3184}=2^4\cdot 199$,
123
${\bf 3556}=2^2\cdot 7\cdot 127$,
124
${\bf 3756}=2^2\cdot 3\cdot 313$,
125
${\bf 4108}=2^2\cdot 13\cdot 79$,
126
${\bf 4288}=2^6\cdot 67$, and
127
${\bf 5373}=3^3\cdot 199$.
128
\end{abstract}
129
130
\section*{Introduction}\edit{
131
\kevin{
132
behavior --$>$ behaviour,
133
analyze --$>$ analyse. William---what do we do about incompatibilities
134
in the spelling in our languages?! I once submitted a paper to MRL and
135
in the typesetting all the UK spelling was changed to US spelling so
136
in the proofs I just changed it all back again :-) But this is a joint
137
paper which we'll probably end up submitting to an American journal
138
so perhaps US spelling is more appropriate...
139
}
140
\william{
141
Let's decide based on the journal we submit to. Where should we
142
submit this paper? In a few days I can drop it in front of Ken
143
and ask him.
144
}}
145
Consider a continuous irreducible Galois representation
146
$$\rho:\galq\ra\GL_n(\C)$$
147
with $n > 1$.
148
Inspired by his reciprocity law,
149
Artin conjectured in~\cite{artin:conjecture} that
150
$L(\rho,s)$ has an analytic continuation to the whole complex plane.
151
Many of the known cases of this conjecture were obtained by
152
proving the apparently stronger assertion that~$\rho$ is \defn{automorphic},
153
in the sense that the $L$-function of~$\rho$ is equal to the $L$-function
154
of a certain automorphic representation (whose $L$-function is known to have
155
analytic continuation). In the special case where $n=2$ and $\rho$ is in
156
addition assumed to be odd, the automorphic representation in question
157
should be the one associated to a classical weight~$1$
158
modular eigenform, and in fact there is conjectured to be a
159
bijection between such~$\rho$ and the set of all weight~$1$
160
cuspidal newforms, which should
161
preserve $L$-functions. It is this bijection
162
that we are concerned with in this paper, so assume for the rest
163
of the paper that $n=2$ and~$\rho$ is odd.
164
165
In this special case, the construction
166
of~\cite{deligne-serre} shows how to construct a continuous irreducible
167
odd 2-dimensional representation from a weight~$1$ newform, and the problem
168
is to go the other way. Say that a representation is \defn{modular}
169
if it arises in this way.
170
171
If the image of~$\rho$ is solvable,
172
then~$\rho$ is known to be modular
173
\cite{langlands:basechange, tunnell:artin};
174
if the image is not solvable, then $\im(\rho)$ in $\PGL_2(\C)$
175
is isomorphic to the
176
alternating group~$A_5$, and the modularity of~$\rho$
177
is, in general, unknown. We call such a 2-dimensional representation an
178
``icosahedral representation''.
179
The published literature contains only eight examples (up to twist)
180
of odd icosahedral Galois representations that are known to satisfy Artin's
181
conjecture: one of conductor $800=2^5\cdot 5^2$
182
(see \cite{buhler:thesis}), and seven of conductors:
183
$2083,\, 2^2\cdot 487,\, 2^2\cdot 751,\,
184
2^2 \cdot 887,\, 2^2\cdot 919,\,
185
2^5\cdot 73,\,\text{ and } 2^5\cdot 193$
186
(see \cite{freyetal}).
187
188
After the first draft of this paper was written, the
189
preprint~\cite{bdsbt} appeared, which contains a general theorem that
190
yields infinitely many (up to twist) modular icosahedral representations.
191
However, we feel that our work, although much less powerful, is still
192
of some worth, because it gives an effective computational approach to
193
proving that certain mod~5 representations are modular, without
194
computing any spaces of weight~1 forms or using effective versions of
195
the Chebotarev\edit{\kevin{William, I don't have my TeXbook with me so
196
don't know how to put the accent on the C.}
197
\william{Kevin, How should we spell this?? I just checked the titles
198
of recent papers with this word in the MathSciNet database and found:
199
Chebotarev (Murty, etc.), Chebotar\"{e}v (Lenstra, however the funniness
200
of the e is dropped in the AMS review), Tschebotareff (by a Japanese author),
201
Tchebotarev (Lang), Tschebotarev (van der Ploeg),
202
Tchebotareff (a Frenchman, but translated to Chebotarev by the AMS),
203
\v{C}ebotarev (Moshe).
204
I say we go with ``Chebotarev,'' because it is seems to be the
205
most common, and simply looks nice to me. Also, it is the
206
closest to Hendrik's without using special characters. It
207
seems strange to me to transliterate into English from Russian and use special
208
characters from, I guess, French.
209
}}
210
density theorem. Finally, the
211
main theorem of~\cite{bdsbt} does not apply to any of the examples
212
considered in the present paper.
213
214
In this paper we give eight new examples that were computed
215
by applying the main theorem of~\cite{buzzard-taylor} to
216
the mod~$5$ reduction of~$\rho$.
217
\edit{In seven of the eight examples, we assume an unknown
218
congruence bound in order to keep the computations
219
manageable.\kevin{Do we? We'll have to sort this out---see
220
the lemma later on in section 1.3}
221
\william{The congruence bound is now known!}}
222
We verify modularity mod~$5$ on a case-by-case basis. Later we shall
223
explain our approach more carefully, but let us briefly summarise it here.
224
By~\cite{buzzard-taylor},
225
the problem is to show that the mod~5 reduction of~$\rho$ is modular.
226
\edit{\william{By the way, as I'm sure you've spotted, I have added this
227
para.}}
228
We do this by finding a candidate mod~5 modular form at weight~5
229
and then, using the table of icosahedral extensions of $\Q$ in~\cite{freyetal}
230
and what we know about the 5-adic representation attached to our candidate
231
form, we deduce that the mod~5 representation attached to our candidate
232
form must be the reduction of~$\rho$. In particular, this paper gives
233
a computational methods for checking the modularity of certain mod~5
234
representations whose conductors are not too large. We now give
235
more details.
236
237
In each of our examples it is easy to compute a few Hecke operators
238
and be morally convinced that a mod~$5$ representation should be modular;
239
it is far more difficult to prove this.
240
Effective variants of the Chebotarev density theorem require
241
that we check vastly more traces of Frobenius than is practical.
242
Instead we use the Local Langlands theorem for $\GL_2$, the
243
theory of companion forms, the main theorem of \cite{buzzard-taylor},
244
and Table~2 of~\cite{freyetal}, to provide proofs of modularity
245
in certain cases.
246
247
More precisely, let~$K$ be an icosahedral extension of~$\Q$ that is not
248
totally real, and consider a minimal lift $\rho:\GQ\ra \GL_2(\C)$
249
of
250
$$\GQ\ra \Gal(K/\Q)\ncisom{}A_5\subset \PGL_2(\C);$$
251
the lift is minimal in the sense that its conductor is minimal.
252
Assume that~$5$ does not ramify in~$K$, and that
253
a Frobenius element at~$5$ in $\Gal(K/\Q)$ does not have order~$1$ or~$5$.
254
Inspired by the possibility that~$\rho$ is modular,
255
we search for a mod~$5$ modular form of weight~$5$ whose existence would
256
be forced by modularity of~$\rho$. Indeed, we find
257
a candidate mod~$5$ form~$f$, and then prove that the fixed field
258
of the kernel of the projective mod~$5$ representation
259
associated to a certain twist of~$f$ must be~$K$.
260
This proves that the mod~$5$ reduction of a twist
261
of~$\rho$ is modular, and the main theorem
262
of \cite{buzzard-taylor} then implies
263
that~$\rho$ is modular.
264
We carried out this program for icosahedral representations
265
of the following conductors:
266
${\bf 1376} = 2^5\cdot 43$,
267
${\bf 2416}=2^4\cdot 151$,
268
${\bf 3184}=2^4\cdot 199$,
269
${\bf 3556}=2^2\cdot 7\cdot 127$,
270
${\bf 3756}=2^2\cdot 3\cdot 313$,
271
${\bf 4108}=2^2\cdot 13\cdot 79$,
272
${\bf 4288}=2^6\cdot 67$, and
273
${\bf 5373}=3^3\cdot 199$.
274
275
276
We choose an icosahedral field~$K$ and representation~$\rho$,
277
then proceed as follows:
278
\vspace{.5ex}
279
\begin{numlist}
280
\item Search for a form~$f \in S_5(N,\eps;\Fbar_5)$ whose
281
associated mod~$5$ Galois representation looks like
282
it is the mod~$5$ reduction of~$\rho$.
283
\item Twist~$f$ to obtain an eigenform~$g$ with coefficients in~$\F_5$.
284
\item Prove that~$\rho_g$ is unramified at~$5$ by finding a companion form.
285
\item Prove that the image of $\proj\rho_g$ is~$A_5$ by ruling out all
286
other possibilities.
287
\item Prove that the fixed field~$L$ of $\proj\rho_g$ has
288
root field of discriminant at most $2083^2$,
289
so~$L$ is in Table~2 of~\cite{freyetal}; deduce that~$L=K$.
290
\item Apply the main theorem of~\cite{buzzard-taylor}
291
to a lift of $\rhobar=\rho_g$
292
to conclude that~$\rho$ is modular.
293
\end{numlist}
294
295
296
297
\section{Modularity of an icosahedral representation of
298
conductor~$1376=2^5\cdot 43$}\label{sec:1376}
299
In this section we prove the following theorem.
300
\begin{theorem}\label{thm:1376}
301
The icosahedral representations whose corresponding
302
icosahedral extension
303
is the splitting field of $x^5 + 2x^4+6x^3+8x^2+10x+8$
304
are modular.
305
\end{theorem}
306
307
Let~$K$ be the splitting field of $h=x^5 + 2x^4+6x^3+8x^2+10x+8$.
308
The Galois group of~$K$ is~$A_5$, so we obtain a homomorphism
309
$G_\Q\ra{}A_5\subset \PGL_2(\C)$;
310
let $\rho:G_\Q\ra\GL_2(\C)$ be a minimal lift, minimal
311
in the sense that the Artin conductor of~$\rho$ is minimal.
312
By Table~$A_5$ of~\cite{buhler:thesis}, the conductor of~$\rho$
313
is $N=1376=2^5\cdot 43$. Since
314
$h\con (x-1)(x^2-x+1)(x^2-x+2)\pmod{5}$,
315
and ${\rm disc}(h)$ is coprime to~$5$,
316
any Frobenius element at~$5$ in $\Gal(K/\Q)$ has order~$2$.
317
318
We use the notation of Tables 3.1 and 3.2 of~\cite[pg. 46]{buhler:thesis};
319
from Table 3.2 we see that the type of~$\rho$ at~$2$
320
is~$17$ and the type at~$43$ is~$2$.
321
The mod~$N$ Dirichlet character~$\eps=\det(\rho)$
322
factors as~$\eps=\eps_2\cdot \eps_{43}$ where~$\eps_2$ is
323
a character mod~$2^5$ and~$\eps_{43}$ is a character mod~$43$.
324
Corresponding to each type in Buhler's table, there is a character,
325
and fortunately Buhler's level $800$ example also was of type~$17$ at~$2$
326
(see the first line of~\cite[Table~3.2]{buhler:thesis}).
327
By~\cite[pg.~80]{buhler:thesis} $\eps_2$ is the unique
328
character\edit{
329
\kevin{William: we should
330
check that Buhler's example turned out to be a weight~$1$ form
331
whose character at~$2$ was $\eps_2$.}
332
\william{\em Yep, by Table 3.2.}
333
\kevin{I didn't
334
mean this---but after reading Buhler I found that in fact the answer
335
to the question I meant to ask was ``Yep, by the beginning of chapter 6''
336
(and this is probably the p80 you mention above). So we can kill
337
this footnote. By the way, how long does
338
it take you to check the ``fact'' near the beginning of Chapter 6, that
339
Buhler talks philosophically about? (not that it matters)}
340
\william{
341
On a PII 350, it takes a total of 258 seconds total
342
for my program to compute a basis of newforms
343
for each space of level dividing $800$, and from this to then
344
compute a Gauss-reduced basis of $q$-expansions, up to precision $360$
345
for the space $S_2(800;\C)$. I didn't check how long it
346
takes to verify that the $h_{i,d}$ lie in this space, but
347
this should be completely trivial because my basis is
348
Gauss-reduced.}}
349
of conductor~$4$ and order~$2$.
350
A local computation shows that the image
351
of~$\eps_{43}$ has order~$3$.
352
353
If~$\rho$ is modular, then there is a weight~$1$
354
newform $f_?\in S_1(N,\eps;\Qbar)$ that gives rise to~$\rho$.
355
Suppose for the moment that~$\rho$ is modular, so that~$f_?$ exists.
356
Choose a prime of~$\overline{\Z}$ lying
357
over~$5$, and denote by~$\fbar_?$ the reduction
358
of $f_?$ modulo this prime. The Eisenstein series
359
$E_4\in M_4(1;\F_5)$ is congruent to~$1$ modulo~$5$, so
360
$E_4\cdot{}\fbar_?\in S_5(N,\eps;\Fbar_5)$ has the same $q$-expansion
361
as $\fbar_?$. Using a computer, we can search for a
362
form $f\in S_5(N,\eps,\Fbar_5)$ that has the same
363
$q$-expansion as the conjectural form $E_4\cdot{}\fbar_?$.
364
365
Instead of multiplying $\fbar_?$ by~$E_4$, we could have multiplied
366
it by an Eisenstein series of weight~$1$, level~$5$, and character $\eps'$.
367
We used $E_4$ because the dimension of $S_5(N,\eps;\Fbar_5)$
368
is~$696$ whereas the dimension of the relevant space
369
$S_2(5\cdot 1376, \eps_{43})$ of weight~$2$ cusp forms is~$1040$.
370
371
\subsection{Searching for the newform~$f$}
372
Using modular symbols (see Section~\ref{sec:modsym}) we
373
compute (at least up to semi-simplification) the space
374
$S_5(1376,\eps;\F_{25})$. Note that there is injective map
375
from the image of~$\eps$ into $\F_{25}^*$. By computing
376
the kernels of various Hecke operators on this space, we find~$f$.
377
In the following computations, we represent nonzero elements of~$\F_{25}$
378
as powers of a generator~$\alp$ of~$\F_{25}^*$, which satisfies
379
$$\alp^2 + 4\alp + 2=0.$$
380
Our character $\eps_{43}$ was represented
381
as the map sending $(1,3)\in(\Z/2^5\Z)^*\cross(\Z/43\Z)^*$ to
382
$2\alp+1$. Note that~3 is a primitive root mod~43, and that $2\alp+1$
383
has order~3.
384
385
If the least common multiple of the degrees of the factors of
386
the polynomial~$h$ modulo an
387
unramified prime~$p$ is~$2$, then $\Frob_p\in\Gal(K/\Q)$
388
has order~$2$. The minimal polynomial of $\rho(\Frob_p)\in\GL_2(\C)$
389
is then $x^2-1$, so $\rho(\Frob_p)$ has trace~$0$.
390
The first three primes $p \nmid 5\cdot 1376$ such
391
that $\rho(\Frob_p)$ has order~$2$ are $p= 19,31,97$.
392
We computed the mod~$5$ reduction $\sS_5(1376,\eps;\F_{25})^{+}$
393
of the $\Z_5[\zeta_3]$-lattice of
394
modular symbols of level~$1376$ and
395
character~$\tilde{\eps}$ where
396
complex conjugation acts as $+1$.
397
Here~$\tilde{\eps}$ denotes the Teichm\"uller lift of~$\eps$.
398
399
Let~$V$ be the intersection of the kernels of $T_{19}$, $T_{31}$, and
400
$T_{97}$ inside of the space $\sS_5(1376,\eps;\F_{25})^{+}$ of mod~5
401
modular symbols.
402
The space~$V$ is $8$-dimensional,
403
and no doubt all the eigenforms in this space give rise to~$\rho$ or one
404
of its twists. One of the eigenvalues of~$T_3$ on this space
405
is~$\alp^{16}$, and the kernel $V_1$ of $T_3-\alp^{16}$ is $2$-dimensional
406
over $\F_{25}$. The Hecke operator~$T_5$ acted as a diagonalisable matrix on
407
$V_1$, with eigenvalues $\alp^{10}$ and $\alp^{22}$, so the corresponding
408
two systems of eigenvalues must correspond to mod~$5$ modular eigenforms,
409
and furthermore we must have found all mod~$5$ modular eigenforms
410
of this level, weight and character,
411
such that $a_{19}=a_{31}=0$ and $a_3=\alp^{16}$.
412
413
\begin{remark}
414
The careful reader might wonder how we know that the
415
systems of mod~$5$ eigenvalues really do correspond to mod~$5$ modular
416
forms, and not to perhaps some strange mod~$5$ torsion in the space of
417
modular symbols. However, we eliminated this possibility by
418
computing the dimension of the full space of mod~$5$ modular
419
symbols where complex conjugation acts as~$+1$, and checking that it
420
equals $696$, the dimension of $S_5(1376,\tilde{\eps},\C)$, which we
421
computed using the formula in \cite{cohen-oesterle}.
422
\edit{
423
\kevin{I still have never seen this
424
paper, so can't give a precise reference. Did you check it all in magma?
425
My pari port gives 696 :-) :-) :-)}
426
\william{I added the biblio reference.
427
I have carefully re-checked all of these numbers in Magma, using
428
Cohen-Oesterle. See the end of this tex file for the Magma code.}
429
}
430
\end{remark}
431
432
Let~$f$ be the eigenform in~$V_1$ that satisfies
433
$a_5=\alp^{22}$; the $q$-expansion of~$f$ begins
434
$$f=q + \alp^{16}q^3 + \alp^{22}q^5 + \alp^{14}q^7
435
+ \alp^{14}q^9 + 4q^{11}+\cdots.$$
436
Further eigenvalues are given in Table~\ref{table:1376} below.
437
The primes~$p$ in the table such that~$a_p=0$ are
438
exactly those
439
predicted by considering the splitting behavior of~$h$.
440
This is strong evidence that~$\rho$ is modular,
441
and also that our modular symbols algorithm have been correctly
442
implemented.
443
\newpage
444
445
\begin{center}
446
\label{table:1376}
447
{\bf Table~\ref{table:1376}}
448
$$\begin{array}{|rl|}\hline
449
2&0\\
450
3&\alpha^{16}\\
451
5&\alpha^{22}\\
452
7&\alpha^{14}\\
453
11&4\\
454
13&\alpha^{14}\\
455
17&\alpha^{14}\\
456
19&0\\
457
23&\alpha^{16}\\
458
29&\alpha^{8}\\
459
31&0\\
460
37&\alpha^{10}\\
461
41&1\\
462
43&\alpha^{10}\\
463
47&1\\
464
53&\alpha^{22}\\
465
59&4\\
466
61&\alpha^{14}\\
467
67&\alpha^{4}\\
468
71&\alpha^{20}\\
469
73&\alpha^{2}\\
470
\hline\end{array}
471
\begin{array}{|rl|}\hline
472
79&\alpha^{20}\\
473
83&\alpha^{4}\\
474
89&\alpha^{10}\\
475
97&0\\
476
101&\alpha^{8}\\
477
103&\alpha^{14}\\
478
107&0\\
479
109&\alpha^{10}\\
480
113&2\\
481
127&0\\
482
131&2\\
483
137&0\\
484
139&\alpha^{22}\\
485
149&\alpha^{4}\\
486
151&1\\
487
157&\alpha^{14}\\
488
163&0\\
489
167&\alpha^{22}\\
490
173&4\\
491
179&\alpha^{2}\\
492
181&\alpha^{14}\\
493
\hline\end{array}
494
\begin{array}{|rl|}\hline
495
191&\alpha^{10}\\
496
193&4\\
497
197&0\\
498
199&3\\
499
211&0\\
500
223&0\\
501
227&\alpha^{10}\\
502
229&0\\
503
233&\alpha^{14}\\
504
239&0\\
505
241&\alpha^{2}\\
506
251&\alpha^{2}\\
507
257&3\\
508
263&\alpha^{16}\\
509
269&2\\
510
271&\alpha^{8}\\
511
277&0\\
512
281&\alpha^{16}\\
513
283&0\\
514
293&3\\
515
307&\alpha^{4}\\
516
\hline\end{array}
517
\begin{array}{|rl|}\hline
518
311&\alpha^{22}\\
519
313&0\\
520
317&0\\
521
331&\alpha^{14}\\
522
337&0\\
523
347&\alpha^{16}\\
524
349&\alpha^{4}\\
525
353&0\\
526
359&0\\
527
367&\alpha^{22}\\
528
373&0\\
529
379&3\\
530
383&3\\
531
389&1\\
532
397&\alpha^{16}\\
533
401&0\\
534
409&2\\
535
419&3\\
536
421&\alpha^{20}\\
537
431&4\\
538
433&\alpha^{4}\\
539
\hline\end{array}
540
\begin{array}{|rl|}\hline
541
439&\alpha^{20}\\
542
443&0\\
543
449&0\\
544
457&0\\
545
461&0\\
546
463&\alpha^{10}\\
547
467&0\\
548
479&0\\
549
487&\alpha^{8}\\
550
491&\alpha^{2}\\
551
499&\alpha^{20}\\
552
503&\alpha^{2}\\
553
509&\alpha^{8}\\
554
521&\alpha^{10}\\
555
523&\alpha^{14}\\
556
541&\alpha^{20}\\
557
547&\alpha^{22}\\
558
557&3\\
559
563&1\\
560
569&\alpha^{16}\\
561
571&\alpha^{22}\\
562
\hline\end{array}
563
\begin{array}{|rl|}\hline
564
577&\alpha^{14}\\
565
587&\alpha^{20}\\
566
593&0\\
567
599&\alpha^{22}\\
568
601&0\\
569
607&\alpha^{16}\\
570
613&2\\
571
617&0\\
572
619&\alpha^{20}\\
573
631&\alpha^{20}\\
574
641&4\\
575
643&1\\
576
647&4\\
577
653&1\\
578
659&\alpha^{14}\\
579
661&2\\
580
673&\alpha^{8}\\
581
677&4\\
582
683&0\\
583
691&\alpha^{16}\\
584
701&\alpha^{14}\\
585
\hline\end{array}
586
\begin{array}{|rl|}\hline
587
709&4\\
588
719&\alpha^{4}\\
589
727&4\\
590
733&0\\
591
739&2\\
592
743&\alpha^{22}\\
593
751&\alpha^{4}\\
594
757&\alpha^{2}\\
595
761&\alpha^{2}\\
596
769&0\\
597
773&0\\
598
787&\alpha^{20}\\
599
797&\alpha^{16}\\
600
809&3\\
601
811&\alpha^{16}\\
602
821&2\\
603
823&\alpha^{10}\\
604
827&\alpha^{10}\\
605
829&\alpha^{22}\\
606
839&0\\
607
853&\alpha^{14}\\
608
\hline\end{array}
609
\comment{
610
\begin{array}{|rl|}\hline
611
857&0\\
612
859&0\\
613
863&\alpha^{4}\\
614
877&\alpha^{8}\\
615
881&1\\
616
883&0\\
617
887&2\\
618
907&0\\
619
911&1\\
620
919&1\\
621
929&0\\
622
937&\alpha^{2}\\
623
941&\alpha^{4}\\
624
947&2\\
625
953&\alpha^{8}\\
626
967&4\\
627
971&\alpha^{2}\\
628
977&0\\
629
983&0\\
630
991&3\\
631
997&3\\
632
\hline\end{array}}
633
$$
634
\end{center}
635
636
\subsection{Twisting into $\GL(2,\F_5)$}
637
Although there is a representation
638
$\rho_f:\GQ\ra\GL(2,\F_{25})$ attached to $f$,
639
it is difficult to say anything about its image without further
640
work. We use a trick to show that the image of $\rho_f$ is small.
641
Firstly, for a character~$\chi:\GQ\to\Fbar_5$, let~$\tilde\chi$
642
denote its Teichm\"uller lift to~$\Qbar_5$. By a result of Carayol,
643
there is a characteristic 0 eigenform
644
$\tilde{f}\in S_5(N,\tilde{\eps};\Qbar_5)$ lifting $f$.
645
The twist $\tilde{g}=\tilde{f} \tensor \tilde{\eps}_{43}$ is, by
646
\cite[Prop. 3.64]{shimura:intro}, an eigenform in
647
$S_5(43N, \tilde{\eps}_2; \Qbar_5)$, and its reduction is
648
a form $g\in S_5(43N,\eps_2,\F_{25})$.
649
The eigenvalues $a_p(g) = a_p(f) \eps_{43}(p)$, for
650
$p\nmid 5N$, are given in Table~\ref{table:1376twist}.
651
\begin{center}\label{table:1376twist}
652
{\bf Table~\ref{table:1376twist}}
653
$$
654
\begin{array}{|rl|}\hline
655
2&*\\%0\\
656
3&1\\
657
5&*\\%3\\
658
7&2\\
659
11&4\\
660
13&2\\
661
17&2\\
662
19&0\\
663
23&1\\
664
29&1\\
665
31&0\\
666
37&3\\
667
41&1\\
668
43&*\\%0\\
669
47&1\\
670
53&2\\
671
\hline\end{array}
672
\begin{array}{|rl|}\hline
673
59&4\\
674
61&2\\
675
67&4\\
676
71&4\\
677
73&3\\
678
79&4\\
679
83&4\\
680
89&3\\
681
97&0\\
682
101&1\\
683
103&2\\
684
107&0\\
685
109&3\\
686
113&2\\
687
127&0\\
688
131&2\\
689
\hline\end{array}
690
\begin{array}{|rl|}\hline
691
137&0\\
692
139&2\\
693
149&4\\
694
151&1\\
695
157&2\\
696
163&0\\
697
167&2\\
698
173&4\\
699
179&3\\
700
181&2\\
701
191&3\\
702
193&4\\
703
197&0\\
704
199&3\\
705
211&0\\
706
223&0\\
707
\hline\end{array}
708
\begin{array}{|rl|}\hline
709
227&3\\
710
229&0\\
711
233&2\\
712
239&0\\
713
241&3\\
714
251&3\\
715
257&3\\
716
263&1\\
717
269&2\\
718
271&1\\
719
277&0\\
720
281&1\\
721
283&0\\
722
293&3\\
723
307&4\\
724
311&2\\
725
\hline\end{array}
726
\begin{array}{|rl|}\hline
727
313&0\\
728
317&0\\
729
331&2\\
730
337&0\\
731
347&1\\
732
349&4\\
733
353&0\\
734
359&0\\
735
367&2\\
736
373&0\\
737
379&3\\
738
383&3\\
739
389&1\\
740
397&1\\
741
401&0\\
742
409&2\\
743
\hline\end{array}
744
\begin{array}{|rl|}\hline
745
419&3\\
746
421&4\\
747
431&4\\
748
433&4\\
749
439&4\\
750
443&0\\
751
449&0\\
752
457&0\\
753
461&0\\
754
463&3\\
755
467&0\\
756
479&0\\
757
487&1\\
758
491&3\\
759
499&4\\
760
503&3\\
761
\hline\end{array}
762
\begin{array}{|rl|}\hline
763
509&1\\
764
521&3\\
765
523&2\\
766
541&4\\
767
547&2\\
768
557&3\\
769
563&1\\
770
569&1\\
771
571&2\\
772
577&2\\
773
587&4\\
774
593&0\\
775
599&2\\
776
601&0\\
777
607&1\\
778
613&2\\
779
\hline\end{array}
780
\begin{array}{|rl|}\hline
781
617&0\\
782
619&4\\
783
631&4\\
784
641&4\\
785
643&1\\
786
647&4\\
787
653&1\\
788
659&2\\
789
661&2\\
790
673&1\\
791
677&4\\
792
683&0\\
793
691&1\\
794
701&2\\
795
709&4\\
796
719&4\\
797
\hline\end{array}
798
\comment{
799
\begin{array}{|rl|}\hline
800
727&4\\
801
733&0\\
802
739&2\\
803
743&2\\
804
751&4\\
805
757&3\\
806
761&3\\
807
769&0\\
808
773&0\\
809
787&4\\
810
797&1\\
811
809&3\\
812
811&1\\
813
821&2\\
814
823&3\\
815
827&3\\
816
\hline\end{array}
817
\begin{array}{|rl|}\hline
818
829&2\\
819
839&0\\
820
853&2\\
821
857&0\\
822
859&0\\
823
863&4\\
824
877&1\\
825
881&1\\
826
883&0\\
827
887&2\\
828
907&0\\
829
911&1\\
830
919&1\\
831
929&0\\
832
937&3\\
833
941&4\\
834
\hline\end{array}
835
\begin{array}{|rl|}\hline
836
947&2\\
837
953&1\\
838
967&4\\
839
971&3\\
840
977&0\\
841
983&0\\
842
991&3\\
843
997&3\\
844
&\\
845
&\\
846
&\\
847
&\\
848
&\\
849
&\\
850
&\\
851
&\\
852
\hline\end{array}}
853
$$
854
\end{center}
855
856
\begin{proposition}\label{prop:1376-g}
857
Let $g=f\tensor \eps_{43}$. Then $a_p(g)\in \F_5$
858
for all~$p\nmid \ell N$.
859
\end{proposition}
860
\begin{proof}
861
Consider an eigenform $\tilde{f}\in S_5(N,\tilde{\eps};\Qbar_5)$
862
lifting~$f$ as above.
863
Associated to~$\tilde{f}$ there is an automorphic
864
representation~$\pi=\tensor_v'\pi_v$ of $\GL(2,\bA)$, where~$\bA$
865
is the ad\`{e}le ring of~$\Q$.
866
Because $43\mid\mid N$, and~$43$ divides the conductor
867
of $\eps$, we see that the local component $\pi_{43}$ of $\pi$ at
868
$43$ must be ramified principal series. By Carayol's theorem,
869
$\rho_{\tilde{f}}|_{D_{43}} \sim
870
\abcd{\Psi_1}{0}{0}{\Psi_2}$
871
with, without loss of generality,~$\Psi_2$ unramified. We have
872
$(\Psi_1\cdot \Psi_2)|_{I_{43}}=\tilde{\eps}|_{I_{43}}=\tilde{\eps}_{43}$,
873
therefore, $\rho_{\tilde{f}}|_{I_{43}} \sim
874
\abcd{\tilde{\eps}_{43}}{0}{0}{1}$.
875
876
Now twist~$\tilde{f}$ by $\tilde{\eps}_{43}^{-1}$; we find that
877
$\rho_{\tilde{f}\tensor\tilde{\eps}_{43}^{-1}}|_{I_{43}} \sim
878
\abcd{1}{0}{0}{\tilde{\eps}^{-1}_{43}}$.
879
In particular, there is an
880
eigenform~$\tilde{f}'\in S_5(N,\tilde{\eps}_2\tilde{\eps}^{-1}_{43},\Qbar_5)$
881
whose associated Galois representation is the twist by $\tilde{\eps}^{-1}_{43}$
882
of that of $\tilde{f}$ (recall that $N=1376$ and so~$43$ divides~$N$
883
exactly once). Let~$f'$ denote the mod~$5$ reduction of~$\tilde{f}'$. Then
884
one checks easily that $f'\in S_5(N,\eps_2\eps^{-1}_{43},\F_{25})=S_5(N,\eps^5,\F_{25})$.
885
886
For all primes $p\nmid5N$ we have $a_p(f')=\eps_{43}(p)^{-1}a_p(f)$.
887
In particular, we have $a_p(f')=0$ for
888
$p=19,31$.
889
Also, $\eps_{43}(3)=\alp^8$ and $\eps_{43}(5) =\alp^8$, so
890
$$a_3(f')=\alp^{16}/\alp^8 = \alp^8 = (\alp^{16})^5$$
891
$$a_5(f')=\alp^{22}/\alp^8 = \alp^{14} = (\alp^{22})^5.$$
892
Now if $\sigma$ is the non-trivial automorphism of $\F_{25}$,
893
then $\sigma(f')$ and $f$ both lie in
894
$S_5(1376,\eps;\F_{25})$ and have same~$a_p$ for
895
$p=3,5,19,31$, so they are equal because we found~$f$
896
by computing the unique eigenform with given~$a_p$ for $p=3,5,19,31$.
897
So $g = f\tensor\eps_{43} = \sigma(f)\tensor\eps_{43}^2$.
898
Thus for all $p\nmid 5N$, we see that
899
$a_p(g) = a_p(f)^5 \eps_{43}^2$ has fifth power
900
$a_p(g)^5 = a_p(f)^{25} \eps_{43}^{10}
901
= a_p(f) \eps_{43} = a_p(g)$.
902
\end{proof}
903
904
\subsection{Proof that~$\rho_g$ is unramified at~$5$}
905
We begin with a generalisation of~\cite{sturm:cong}.\edit{\kevin{Are you happy
906
to put this lemma here? I expect you're just relieved to see the
907
lemma at all :-)}
908
\william{It fits naturally here. And {\em yes}, I am very relieved
909
to see the lemma at all!}}
910
Let $M>4$ be an integer, and let $h=\sum_{n\geq1}c_nq^n$ be a
911
normalised cuspidal eigenform
912
of some weight~$k\geq1$, level~$M$ and character~$\chi$, defined over some
913
field of characteristic not dividing~$M$.
914
Let~$I$ be the set of primes~$p$
915
dividing~$M$ such that~$h$ is $p$-new, and
916
such that either~$p$ does not divide $M/\cond(\chi)$, or~$p$
917
divides~$M$ exactly once. Let~$C$ denote
918
the orbit of the cusp~$\infty$ in $X_1(M)$
919
under the action of the group generated by $w_p$ for $p\in I$, and
920
the diamond operators $\langle d\rangle_M$. By Corollary~4.6.18
921
of \cite{miyake}\edit{\kevin{don't know the publisher/year and don't have
922
it with me, sorry. Working offline and at home is hard!}\william{Fixed.}},
923
we see that the first~$t$ terms of the $q$-expansion
924
of~$h$ at any cusp in~$C$ are determined by~$M$,~$k$, $\chi$, $c_p$
925
for~$p$ in~$I$, and $c_n$ for $1\leq n\leq t$. The size of~$C$ is
926
$\phi(M).2^{|I|-1}.$ The usefulness of this result is that
927
if $h_1$ and $h_2$ are two normalised eigenforms of the same level,
928
weight and character as above, both new at all primes in~$I$,
929
and the coefficients of $q^n$ in the
930
$q$-expansions of $h_1$ and $h_2$ agree for $n\in I$ and $n\leq t$,
931
then $h_1-h_2$ has a zero of order at least $t+1$ at all cusps in $C$,
932
and in particular if $\phi(M).2^{|I|-1}(t+1)>k/12[\SL_2(\Z):\Gamma_1(M)]$
933
then $h_1=h_2$. Using the fact that $[\Gamma_0(M):\Gamma_1(M)]=\phi(M)/2$,
934
we deduce
935
\begin{lemma}\label{lem:bound}
936
Let $h_1$ and $h_2$ be two normalised eigenforms as above.
937
If the coefficients of $q^n$ in the $q$-expansions of $h_1$ and $h_2$ agree
938
for all primes in $I$ and for all
939
$n\leq\frac{k}{12}[\SL_2(\Z):\Gamma_0(M)]/2^{|I|}$ then $h_1=h_2$.
940
\end{lemma}\edit{\kevin{\bf William: I have been a
941
bit vague here---but tell me whether this
942
result (a) looks right and (b) looks useful, and then we can go from
943
there. I could strengthen it a little in the case when $p$ is in $I$
944
and $p^2$ divides $M$ but I don't know whether this is necessary. Note that
945
it gives exactly what Sturm says in the squarefree case. Note also that
946
it's easy to check that e.g. our first form is new at 43 because the
947
char is non-trivial! I could give details of Miyake's argument (which
948
is in fact over $\C$ but which I believe I can generalise to an arbitrary
949
field if necessary) if you want.}
950
\william{
951
I think that both (a) and (b) are true.
952
I would like to see details of Miyake's argument over an arbitrary field.
953
Even if we don't put them in the paper, I need to understand them
954
because surely somebody will grill me about this... Anyway, I can
955
also use this result to improve my MAGMA software.
956
957
Am I mistaken in believing that ``multiplicity one'' implies that
958
the forms we consider are all new?
959
}}
960
961
We now go back to the explicit situation we are concerned with.
962
Although~$g$ is an eigenform of level $59168=2^5\cdot 43^2$,
963
we can still consider the corresponding representation
964
$\rho_g :\GQ\ra \GL(2,\F_5)$, and then directly analyze
965
its ramification.
966
\begin{proposition}
967
The representation~$\rho_g$ is unramified at~$5$.
968
\end{proposition}
969
\begin{proof}
970
Continuing the modular symbols computations as above,
971
we find that~$V_1$ is spanned by the two eigenforms
972
\begin{align*}
973
f\,\,&=q + \alp^{16}q^3 + \alp^{22}q^5 + \alp^{14}q^7
974
+ \alp^{14}q^9 + 4q^{11}+\cdots\\
975
f_1&=q + \alp^{16}q^3 + \alp^{10}q^5 + \alp^{14}q^7
976
+ \alp^{14}q^9 + 4q^{11}+\cdots.
977
\end{align*}
978
For $p\neq 5$ and $p\leq 997$, we have $a_p(f_1)=a_p(f)$.
979
To check that $a_p(f) = a_p(f_1)$ for all $p\neq 5$,
980
it suffices to show that the difference~$f-f_1$ has
981
$q$-expansion involving only powers of~$q^5$;
982
for this we use the $\theta$-operator
983
$q\frac{d}{dq}:S_5(1376,\eps,\F_{25})\ra S_{11}(1376,\eps;\F_{25})$.
984
Since~$\theta$ sends normalized eigenforms to normalized eigenforms,
985
it suffices to check that the subspace of
986
$S_{11}(1376,\eps;\F_{25})$ generated by~$\theta(f)$
987
and~$\theta(f_1)$ has dimension~$1$.
988
Lemma~\ref{lem:bound} implies that it suffices to verify that the
989
coefficients $a_p(\theta(f))$ and $a_p(\theta(f_1))$ are equal for all
990
$$p \leq \frac{11}{12}\cdot [\SL_2(\Z):\Gamma_0(1376)]\cdot \frac{1}{4}
991
= 484.$$
992
The eigenform~$f$ must be new because we computed it by finding
993
the intersections of the kernels of Hecke operators $T_p$ with
994
$p\nmid 1376$; if~$f$ were an oldform then the intersection of the
995
kernels of these Hecke operators
996
would necessarily have dimension greater than~$1$.
997
Because it takes less than a second
998
to compute each $a_p(\theta(f))$, we were easily able to verify that the
999
space generated by $\theta(f)$ and $\theta(f_1)$ has dimension~$1$.
1000
1001
\begin{remark}
1002
It is possible to avoid appealing to Lemma~\ref{lem:bound} by using
1003
one of the following two alternative methods:
1004
\begin{enumerate}
1005
\item Define~$\theta$ directly on modular symbols and compute it.
1006
\item Compute the intersection
1007
$$\bigcap_{p\geq 2} \ker(T_p - pa_p(f))
1008
\subset S_{11}(1376,\eps;\F_{25}).$$
1009
Since~$\theta(f)$ and~$\theta(f_1)$ both lie
1010
in the intersection, the moment the dimension
1011
of a partial intersection is~$1$, it follows
1012
that $\theta(f-f_1)=0$.
1013
\end{enumerate}
1014
We successfully carried out both alternatives.
1015
For the first, we showed that~$\theta$ on modular symbols is
1016
induced by multiplication by
1017
$X^5Y - Y^5X$.
1018
For the second, we find that after intersecting
1019
kernels for $p\leq 11$, the dimension is already~$1$.
1020
The first of these two methods took much less
1021
time than the second.
1022
\end{remark}
1023
\edit{\kevin{Have I done this? That is, is my result strong enough?
1024
The point is that for arbitrary forms on $\Gamma_1(N)$ you have to
1025
check a long way but for 2 forms which are eigenforms for the diamond
1026
ops with the same character you can check
1027
they're the same just by using the $\Gamma_0(N)$ bounds, because if the
1028
q-exps agree at one cusp then they agree at lots of cusps
1029
($\phi(N)/2$ cusps, to be precise, the orbit of infinity under the
1030
diamond operators) Is this all you needed? It's kind
1031
of like why you don't need to deal with a huge huge space when working
1032
with modular symbols for $\Gamma_1(N)$ with a character. Anyway, if my
1033
result is strong enough then perhaps this section needs mild rewriting.}
1034
}
1035
1036
Next we use that $\theta(f-f_1)=0$ to show that $\rho_g$ is unramified,
1037
thus finishing the proof of the proposition.
1038
Since~$f$ is ordinary, Deligne's theorem (see~\cite[\S12]{gross:tameness})
1039
implies that
1040
$$\rho_f|_{D_5}\sim
1041
\mtwo{\alp}{*}{0}{\beta}\qquad\text{over $\Fbar_5$}$$
1042
with both~$\alp, \beta$ unramified,
1043
$\alp(\Frob_5)=\eps(5)/a_5=\alp^8/\alp^{22}=\alp^{10}$, and
1044
$\beta(\Frob_5)=\alp^{22}$.
1045
Since $a_p(f_1)=a_p(f)$, for $p\neq 5$, we have
1046
$${\rho_f}|_{D_5}\sim {\rho_{f_1}}|_{D_5} \sim \mtwo{\alp'}{*}{0}{\beta'}$$
1047
with
1048
$\alp'(\Frob_5)=\alp^8/\alp^{10}=\alp^{22}$ and
1049
$\beta'(\Frob_5)=\alp^{10}$;
1050
in particular, $\alp'=\beta$.
1051
Thus $\rho_f|_{D_5}$ contains $\alp\oplus \beta$, so
1052
$\rho_f|_{D_5}\sim\alp\oplus\beta$ and hence there is a choice
1053
of basis so that $*=0$.
1054
1055
\end{proof}
1056
1057
1058
1059
\subsection{The image of $\proj \rho_g$}
1060
\begin{proposition}
1061
The image of $\proj \rho_g$ is $A_5$.
1062
\end{proposition}
1063
\begin{proof}
1064
The image~$H$ of $\proj \rho_g$ in $\PGL_2(\F_5)$ is easily checked to
1065
lie in $\PSL_2(\F_5)\cong A_5$ because of what we know about the determinant
1066
of $\rho_g$. Hence $H$ is a subgroup of $A_5$ that
1067
contains an element of order~$2$ (complex conjugation)
1068
and an element of order~$3$ (for example, $\rho_g(\Frob_7)$ has
1069
characteristic polynomial $x^2-2x-1$).
1070
This proves that~$H$ is isomorphic to either~$S_3$,~$A_4$, or~$A_5$.
1071
Let $L$ be the number field cut out by~$H$.
1072
If~$L$ were an $S_3$ extension, then there would be a
1073
quadratic extension contained in it which is unramified
1074
outside $2\cdot 5\cdot 43$; it is furthermore unramified at~$5$ by
1075
the previous section and unramified at $43$ because $I_{43}$
1076
has order~$3$. Thus it is one of the three quadratic
1077
fields unramified outside~$2$. In particular, the trace of $\Frob_p$
1078
would be zero for all primes in a certain congruence class
1079
modulo~8.
1080
However, there are primes~$p$ congruent to $3$, $5$, and $7$
1081
mod $8$ such that $a_p(g)\neq 0$, e.g., $3$, $7$, and $13$.
1082
1083
If $H$ were isomorphic to $A_4$,
1084
then let~$M$ denote the cyclic extension of degree~3 over~$\Q$ contained
1085
in~$L$. Now~$M$ is unramfied at~2 and~5, and hence is the subfield
1086
of $\Q(\zeta_{43})$ of degree~3.
1087
Choose $p\nmid 1376\cdot 5$ that is inert in~$M$. The
1088
order of $\rho_g(\Frob_p)$ in $\GL_2(\F_5)$ must be divisible
1089
by~$3$. However, a quick check using
1090
Table~\ref{table:1376twist} shows that this is
1091
usually not case, even for $p=3$.
1092
\end{proof}
1093
1094
1095
\subsection{Bounding the ramification at~$2$ and~$43$}
1096
Let~$L$ be the fixed field of $\ker(\proj(\rho_g))$. We have just
1097
shown that $\Gal(L/\Q)$ is isomorphic to $A_5$.
1098
By a root field for~$L$, we mean
1099
a non-Galois extension of $\Q$ of degree~5 whose Galois closure is~$L$.
1100
\begin{proposition}
1101
The discriminant of a root
1102
field for~$L$ divides $(43\cdot 8)^2=344^2$, and
1103
in particular,~$L$ must be mentioned in Table~1
1104
of \cite[pg 122]{freyetal}.\end{proposition}
1105
\begin{proof}
1106
The analysis of the local behavior of~$\rho_f$ at~$43$ given in
1107
Proposition~\ref{prop:1376-g}
1108
shows that the inertia group at~$43$ in $\Gal(L/\Q)$ has order~$3$. Using
1109
Table~3.1 of~\cite{buhler:thesis}, we see that if
1110
$\Gal(L/\Q)\isom A_5$
1111
then it must be ``type~$2$'' at 43, and hence the discriminant of a root
1112
field of~$L$, that is, of a non-Galois extension of~$\Q$ of degree~$5$
1113
whose Galois closure is~$L$, must be $43^2$ at~$43$.
1114
1115
At~$2$ the behavior of~$\rho$ is more subtle and we shall not analyze
1116
it fully. But we can say that, because~$\rho$ has arisen from
1117
a form of level $1376=2^5.43$, we must be either of type~$5$
1118
or one of types~$14$--$17$. In particular, the discriminant at~$2$ of a root
1119
field for~$L$ will be at most~$2^6$.
1120
1121
Finally,~$L$ is unramified at all other primes, because~$\rho$ is.
1122
Hence the discriminant of a root field for~$L$, assuming that
1123
$\Gal(L/\Q)\cong A_5$, divides $(43.8)^2=344^2$.
1124
\end{proof}
1125
1126
We know that~$L$ is an icosahedral extension of~$\Q$ with
1127
discriminant dividing $43^2\cdot 2^6$. Table~1 of \cite[pg 122]{freyetal}
1128
contains all icosahedral extensions, such that the discriminant
1129
of a root field is bounded by $2083^2$. The table
1130
must contain~$L$; there is only one icosahedral extension with
1131
discriminant dividing $43^2\cdot 2^6$, so $L=K$.
1132
1133
\subsection{Obtaining a classical weight one form}
1134
We have shown that a twist of the icosahedral
1135
representation $\rho:\GQ\ra\GL(2,\C)$,
1136
obtained by lifting $\GQ\ra \Gal(K/\Q)\ncisom A_5$,
1137
has a mod~$5$ reduction $\rho_g:\GQ\ra \GL_2(\F_5)$ that
1138
is modular. Since~$\rho$ ramifies at only finitely many primes,
1139
and~$\rho$ can be chosen to be unramified at~$5$ with distinct eigenvalues,
1140
\cite{buzzard-taylor} implies that~$\rho$ arises from
1141
a classical weight~$1$ newform.\edit{\kevin{You write ``say something
1142
about choice of iso $\overline{\Q}_5=\C$'' but I think that we can just
1143
say nothing.}\william{I'm happy with saying nothing, as the reader
1144
will know what needs to be said upon looking at [BT].}}
1145
1146
1147
1148
1149
\section{More examples}
1150
\subsection{Conductor~$2416=2^4\cdot 151$}
1151
\label{sec:2416}
1152
Consider the icosahedral extension~$K$ defined by $h=x^5-2x^3+2x^2+5x+6$.
1153
A Frobenius element at~$5$ has order~$2$.
1154
The first three primes~$p$ such that $a_p=0$ are
1155
$53$, $97$, and $127$.
1156
The type at~$2$ is~16, and the type at~$151$ is~2.
1157
The order of $\eps_{151}$ is~$3$, and a local computation would tell us
1158
what $\eps_2$ was but we preferred to guess---if we guessed wrong then
1159
we would almost certainly find no forms and we could just guess again.
1160
We guessed that $\eps_2$ was the character of conductor~$4$, and this
1161
turned out to be correct.
1162
1163
Using \cite{cohen-oesterle}\edit{\kevin{Computing in general is currently
1164
a real pain for me. I am currently spending as much time as possible
1165
at home (Joel is too young to be put in a creche at the minute)
1166
and hence am (a) using Windows 98 and (b) usually offline (it's
1167
expensive in France to use the net) so access to magma is a pain! Added
1168
to this is the intense irritation caused by the fact that my office in
1169
London is being used by someone who has rebooted the computer into
1170
Windows so I can't telnet into it, and you can see why doing any computations
1171
at all is a real pain! Fortunately I have gp and I have ported all the
1172
DimensionCuspForms stuff (it's a pain because ``evaluate'' typically gives
1173
complex numbers to 30dps!!) and indeed it's 1210. What's the precise
1174
reference to [CO] by the way?}} we find $\dim S_5(2416,\eps)=1210$; this
1175
is the same as the dimension obtained using modular symbols, so
1176
there is no spurious $5$-torsion.
1177
We compute the space of modular symbols corresponding to
1178
$S_5(2416,\eps;\F_{25})$, and then the $8$-dimensional
1179
intersection~$V$ of the kernels of~$T_{53}$, $T_{97}$, and~$T_{127}$.
1180
The characteristic polynomial of~$T_3$ is
1181
$(x+2)^4(x+3)^4$. The kernel~$V_1$ of $T_3+2$ has dimension~$4$.
1182
The characteristic polynomial of~$T_7$ on~$V_1$
1183
is $(x+\alp^4)^4$; the kernel~$V_2\subset V_1$ of $T_7+\alp^4$ has
1184
dimension~$2$.
1185
The characteristic polynomial of~$T_5$ on~$V_2$ is
1186
$(x+\alp^{10})(x+\alp^{22})$.
1187
Let~$f$ be the newform such that $a_5+\alp^{10}=0$.
1188
We have
1189
$$ f = q +3q^3 + \alp^{22}q^5 + \alp^{16}q^7 + \alp^{4} q^{11} + \cdots.$$
1190
1191
There is one other newform $f'$ in~$V_2$, and it is a companion of~$f$.
1192
We verified this using Lemma~\ref{lem:bound} by
1193
checking that $a_p(f)=a_p(f')$ for those primes~$p\neq 5$ such that
1194
$$p \leq
1195
\frac{11}{12}\cdot
1196
[\SL_2(\Z):\Gamma_0(2416)]\cdot \frac{1}{4} = 836.$$
1197
1198
1199
Let $g=f\tensor\eps_{151}$.
1200
We have $a_p(f\tensor\eps_{151}^{-1})=0$ for $p=53,97,127$.
1201
Also,
1202
$\eps_{151}(3)=1$,
1203
$\eps_{151}(5)=\alp^8$,
1204
$\eps_{151}(7) =\alp^{20}$, so
1205
$a_3(f\tensor\eps_{151}^{-1})=3/1=3=3^5$,
1206
$a_5(f\tensor\eps_{151}^{-1})= \alp^{22}/\alp^8 = \alp^{14}=(\alp^{22})^5$,
1207
$a_7(f\tensor\eps_{151}^{-1})= \alp^{16}/\alp^{8} = \alp^{8} = (\alp^{16})^5$.
1208
Since $f\tensor\eps_{151}^{-1}$ and~$\sigma(f)$ both lie
1209
in $S_5(2416,\eps_2\eps_{151}^2;\F_{25})$ and have same~$a_p$ for
1210
$p=3,5,7,53,97,127$, they are equal.
1211
Exactly as in Proposition~\ref{prop:1376-g} we see
1212
that $g\in S_5(151N,\eps_2;\F_5)$.
1213
1214
We have not yet bounded ramification or checked that the image
1215
of~$\rho_g$ is~$A_5$.\edit{\kevin{This shouldn't take too long but I didn't
1216
want to waste time on it until you confirmed that my bound was good enough.
1217
By the way, do you want to move the lemma to a different section or
1218
anything?}\william{{\bf If I understand it correctly,
1219
you're bound is definitely good enough.}}}
1220
This computation can be done with the
1221
help of the following table:
1222
{\arraycolsep=.36em
1223
$$\begin{array}{|rl|}\hline
1224
2&0\\
1225
3&3\\
1226
5&\alp^{22}\\
1227
7&\alp^{16}\\
1228
11&\alp^{4}\\
1229
13&\alp^{2}\\
1230
17&\alp^{22}\\
1231
19&3\\
1232
\hline\end{array}
1233
\begin{array}{|rl|}\hline
1234
23&\alp^{22}\\
1235
29&3\\
1236
31&\alp^{16}\\
1237
37&\alp^{22}\\
1238
41&2\\
1239
43&\alp^{8}\\
1240
47&\alp^{8}\\
1241
53&0\\
1242
\hline\end{array}
1243
\begin{array}{|rl|}\hline
1244
59&1\\
1245
61&\alp^{8}\\
1246
67&3\\
1247
71&\alp^{8}\\
1248
73&2\\
1249
79&2\\
1250
83&2\\
1251
89&\alp^{20}\\
1252
\hline\end{array}
1253
\begin{array}{|rl|}\hline
1254
97&0\\
1255
101&2\\
1256
103&\alp^{16}\\
1257
107&1\\
1258
109&\alp^{20}\\
1259
113&\alp^{22}\\
1260
127&0\\
1261
131&4\\
1262
\hline\end{array}
1263
\begin{array}{|rl|}\hline
1264
137&\alp^{20}\\
1265
139&0\\
1266
149&\alp^{22}\\
1267
151&2\\
1268
157&\alp^{22}\\
1269
163&\alp^{8}\\
1270
167&\alp^{10}\\
1271
173&0\\
1272
\hline\end{array}
1273
\begin{array}{|rl|}\hline
1274
179&2\\
1275
181&\alp^{2}\\
1276
191&\alp^{10}\\
1277
193&0\\
1278
197&\alp^{14}\\
1279
199&\alp^{16}\\
1280
211&0\\
1281
223&0\\
1282
\hline\end{array}
1283
\begin{array}{|rl|}\hline
1284
227&\alp^{20}\\
1285
229&3\\
1286
233&\alp^{22}\\
1287
239&\alp^{10}\\
1288
241&\alp^{2}\\
1289
251&\alp^{4}\\
1290
257&0\\
1291
263&\alp^{2}\\
1292
\hline\end{array}
1293
\begin{array}{|rl|}\hline
1294
269&0\\
1295
271&\alp^{10}\\
1296
277&\alp^{20}\\
1297
281&\alp^{16}\\
1298
283&0\\
1299
293&2\\
1300
307&\alp^{4}\\
1301
311&2\\
1302
\hline\end{array}$$}
1303
1304
1305
\subsection{Conductor~$3184=2^4\cdot 199$}
1306
Consider the icosahedral extension~$K$ defined by
1307
$h=x^5+5x^4+8x^3-20x^2-21x-5$.
1308
A Frobenius element at~$5$ has order~$2$.
1309
The first~$3$ primes~$p$ such that $a_p=0$ are
1310
$31$, $89$, and $97$.
1311
The type at~$2$ is~$16$ and the type at~$199$ is~$2$.
1312
As in Section~\ref{sec:2416}, $\eps_2$ is of
1313
conductor~$4$ and order~$2$, and the order of~$\eps_{199}$ is~$3$.
1314
\edit{\kevin{We don't need to
1315
guess any more---I think that the calculation in the last section
1316
has convinced us that type 16 is associated to char non-trivial of
1317
conductor~4. Probably I should think of some way of explaining this
1318
that makes us look less like idiots.}\william{I fixed this.}}
1319
1320
Using \cite{cohen-oesterle}
1321
we find $\dim S_5(3184,\eps)=1594$,
1322
which agrees with the dimension computed using modular symbols.
1323
We compute $S_5(3184,\eps;\F_{25})$ and then the
1324
$8$-dimensional intersection~$V$ of the kernels of
1325
$T_{31}$, $T_{89}$, and $T_{97}$.
1326
The characteristic polynomial of~$T_3$ on~$V$ is
1327
$(x+\alp^4)^4(x+\alp^{16})^4$. The dimension of
1328
$V_1=\ker(T_3+\alp^4)$ is~$2$. It appears that all
1329
Hecke operators~$T_p$ act diagonally on $V_1$, except~$T_5$
1330
which has the distinct eigenvalues~$2$ and~$3$.
1331
We thus isolate a one-dimensional eigenspace, spanned by
1332
a form~$f$ whose Hecke eigenvalues are:
1333
{\arraycolsep=.4em
1334
$$
1335
\begin{array}{|rl|}\hline
1336
2&0\\
1337
3&\alpha^{16}\\
1338
5&3\\
1339
7&\alpha^{22}\\
1340
11&3\\
1341
13&\alpha^{22}\\
1342
17&3\\
1343
19&\alpha^{16}\\
1344
\hline\end{array}
1345
\begin{array}{|rl|}\hline
1346
23&\alpha^{4}\\
1347
29&\alpha^{16}\\
1348
31&0\\
1349
37&\alpha^{8}\\
1350
41&\alpha^{2}\\
1351
43&\alpha^{10}\\
1352
47&\alpha^{4}\\
1353
53&\alpha^{14}\\
1354
\hline\end{array}
1355
\begin{array}{|rl|}\hline
1356
59&3\\
1357
61&1\\
1358
67&4\\
1359
71&\alpha^{22}\\
1360
73&\alpha^{10}\\
1361
79&\alpha^{16}\\
1362
83&3\\
1363
89&0\\
1364
\hline\end{array}
1365
\begin{array}{|rl|}\hline
1366
97&0\\
1367
101&2\\
1368
103&2\\
1369
107&3\\
1370
109&0\\
1371
113&\alpha^{2}\\
1372
127&0\\
1373
131&\alpha^{14}\\
1374
\hline\end{array}
1375
\begin{array}{|rl|}\hline
1376
137&1\\
1377
139&1\\
1378
149&\alpha^{22}\\
1379
151&\alpha^{2}\\
1380
157&0\\
1381
163&\alpha^{16}\\
1382
167&0\\
1383
173&0\\
1384
\hline\end{array}
1385
\begin{array}{|rl|}\hline
1386
179&\alpha^{8}\\
1387
181&2\\
1388
191&3\\
1389
193&\alpha^{14}\\
1390
197&0\\
1391
199&2\\
1392
211&4\\
1393
223&0\\
1394
\hline\end{array}
1395
\begin{array}{|rl|}\hline
1396
227&1\\
1397
229&\alpha^{2}\\
1398
233&0\\
1399
239&3\\
1400
241&2\\
1401
251&0\\
1402
257&0\\
1403
263&1\\
1404
\hline\end{array}
1405
\begin{array}{|rl|}\hline
1406
269&\alpha^{2}\\
1407
271&\alpha^{2}\\
1408
277&0\\
1409
281&1\\
1410
283&\alpha^{10}\\
1411
293&\alpha^{8}\\
1412
307&\alpha^{20}\\
1413
311&\alpha^{20}\\
1414
\hline\end{array}
1415
$$}
1416
1417
The bound from Lemma~\ref{lem:bound} is
1418
$$\frac{11}{12}\cdot [\SL_2(\Z):\Gamma_0(3184)]\cdot
1419
\frac{1}{4} = 1100.$$
1420
1421
1422
\subsection{Conductor~$3556=2^2\cdot 7\cdot 127$}
1423
Consider the icosahedral extension~$K$ defined by
1424
$h=x^5+3x^4+9x^3-6x^2-4x-40$.
1425
A Frobenius element at~$5$ has order~$3$.
1426
The first~$3$ unramified~$p$ such that $a_p=0$ are~$19$,~$29$, and~$89$.
1427
The type at~$2$ is~5; the type at~$7$ is~3; the type at~$127$ is~2.
1428
The order of~$\eps_{127}$ is~$3$.
1429
A local analysis shows that the order
1430
of~$\eps_{2}$ is~$1$ and of~$\eps_7$ is~$2$.\edit{\kevin{William: I took
1431
out some bad guesses here! If the type is $\leq15$ and not 6 then
1432
I think I can work out the character, and indeed I can verify that your
1433
first guess is wrong!}}
1434
The formula of~\cite{cohen-oesterle}
1435
gives $\dim S_5(3556,\eps)=2042$, which agrees
1436
with the dimension of the corresponding space of modular symbols.
1437
1438
The intersection~$V$ of the kernels of~$T_{19}$,~$T_{29}$, and~$T_{89}$
1439
has dimension~$8$.
1440
The characteristic polynomial of~$T_{3}$ on~$V$
1441
is $(x+\alp^4)^4(x+\alp^{16})^4$,
1442
and the kernel~$V_1$ of $T_3+\alp^4$ is of dimension~$2$.
1443
It appears that all Hecke operators~$T_p$ act
1444
diagonally on~$V_1$ except~$T_5$, whose eigenvalues
1445
\edit{\kevin{How come it says $\alpha^{14}$ in
1446
the table then? Are we out by a sign here?}\william{I gave the
1447
signs incorrectly in this particular sentence. The table is fine,
1448
and the sentence is now fixed.}}
1449
are~$\alp^{14}$ and~$\alp^{22}$. The eigenvalues of the
1450
newform with $a_5 = \alp^{14}$ are given in the following table.
1451
{\arraycolsep=.4em
1452
$$
1453
\begin{array}{|rl|}\hline
1454
2&0\\
1455
3&\alpha^{16}\\
1456
5&\alpha^{14}\\
1457
7&\alpha^{10}\\
1458
11&\alpha^{2}\\
1459
13&\alpha^{22}\\
1460
17&\alpha^{14}\\
1461
19&0\\
1462
\hline\end{array}
1463
\begin{array}{|rl|}\hline
1464
23&\alpha^{10}\\
1465
29&0\\
1466
31&\alpha^{16}\\
1467
37&\alpha^{20}\\
1468
41&\alpha^{14}\\
1469
43&\alpha^{2}\\
1470
47&1\\
1471
53&\alpha^{2}\\
1472
\hline\end{array}
1473
\begin{array}{|rl|}\hline
1474
59&\alpha^{8}\\
1475
61&4\\
1476
67&\alpha^{10}\\
1477
71&\alpha^{4}\\
1478
73&4\\
1479
79&\alpha^{16}\\
1480
83&\alpha^{8}\\
1481
89&0\\
1482
\hline\end{array}
1483
\begin{array}{|rl|}\hline
1484
97&\alpha^{16}\\
1485
101&\alpha^{4}\\
1486
103&\alpha^{4}\\
1487
107&0\\
1488
109&\alpha^{14}\\
1489
113&\alpha^{4}\\
1490
127&\alpha^{20}\\
1491
131&1\\
1492
\hline\end{array}
1493
\begin{array}{|rl|}\hline
1494
137&2\\
1495
139&\alpha^{22}\\
1496
149&\alpha^{2}\\
1497
151&0\\
1498
157&0\\
1499
163&\alpha^{14}\\
1500
167&0\\
1501
173&\alpha^{10}\\
1502
\hline\end{array}
1503
\begin{array}{|rl|}\hline
1504
179&\alpha^{4}\\
1505
181&1\\
1506
191&1\\
1507
193&0\\
1508
197&\alpha^{22}\\
1509
199&\alpha^{2}\\
1510
211&\alpha^{8}\\
1511
223&0\\
1512
\hline\end{array}
1513
\begin{array}{|rl|}\hline
1514
227&3\\
1515
229&0\\
1516
233&0\\
1517
239&0\\
1518
241&\alpha^{10}\\
1519
251&0\\
1520
257&\alpha^{16}\\
1521
263&\alpha^{14}\\
1522
\hline\end{array}
1523
\begin{array}{|rl|}\hline
1524
269&\alpha^{4}\\
1525
271&\alpha^{20}\\
1526
277&\alpha^{16}\\
1527
281&0\\
1528
283&\alpha^{20}\\
1529
293&\alpha^{14}\\
1530
307&\alpha^{8}\\
1531
311&0\\
1532
\hline\end{array}
1533
$$
1534
}
1535
By checking that the Hecke operators $T_p$ act diagonally
1536
on $V_1$, for $p\neq 5$ with
1537
$$p\leq\frac{11}{12}\cdot [\SL_2(\Z):\Gamma_0(3556)]\cdot
1538
\frac{1}{8} = 704,$$
1539
we see that the other newform in~$V_1$ is a companion of~$f$.
1540
1541
Let $g=f\tensor\eps_{127}$.
1542
We have $a_p(f\tensor\eps_{127}^{-1})=0$ for $p=19, 29, 89.$
1543
We have $\eps_{127}(3)=\alp^8$ and $\eps_{127}(5)=1$, so
1544
$a_3(f\tensor\eps_{127}^{-1})= \alp^{16}/\alp^8 = \alp^{8}$,
1545
and $a_{5}(f\tensor\eps_{127}^{-1})= \alp^{14}/1=\alp^{14}$.
1546
At first glance this form looks like the Galois conjugate
1547
of~$f$; and if it were the Galois conjugate then the form
1548
twisted by the character at~$127$ will, as in
1549
Proposition~\ref{prop:1376-g}, have all~$a_p$ in~$\F_5$
1550
for $p \nmid \ell N$ and hence~$\rho_g$
1551
will go into $\GL(2,\F_5)$. Unfortunately the
1552
newform associated to the twisted form is not
1553
quite the conjugate of our original form; the coefficient
1554
of~$a_5$ is wrong! The conjugate form has
1555
$a_5=(\alp^{14})^5 = \alp^{22}$.
1556
The other newform $f_1\in V_1$, which is a companion
1557
of~$f$, has $a_5(f_1)=\alp^{22}$, so $a_5(f_1)^5 = \alp^{14}$.
1558
This proves that~$g=\sigma(f_1)$.
1559
An argument as in
1560
Proposition~\ref{prop:1376-g} shows that $a_p(g)\in\F_5$
1561
for all $p\nmid \ell N$.
1562
1563
\edit{\william{[[Kevin has an argument that
1564
uses that~$\Frob_5$ has order~$3$
1565
in the~$A_5$ field to show that~$\rho_f$ is unramified at~$5$.
1566
Unfortunately his argument uses that the $a_p(g)$ are in~$\F_5$,
1567
for $p\nmid \ell N$!
1568
It might be possible to appeal to the Gross and Coleman-Voloch theorems
1569
to prove that~$f_1$ is indeed a companion form. This would
1570
avoid any further computation.]]}
1571
\kevin{I'm not sure I believe you.
1572
You see: if we want to go down this path, we have to check that the
1573
repn is unram at 5. But we don't know a darn thing about the repn yet.
1574
Am I right in thinking that we are stuck in this case?}
1575
\william{Given the lemma 1.3, I think we can just use the standard
1576
technique and not go down this path!?}}
1577
\comment{
1578
\edit{\william{This shouldn't be needed.}}
1579
Now $\Frob_5$ has order~$3$ in the~$A_5$
1580
extension in Buhler so I think what should happen in your computations is
1581
that the form of level $3556\cdot 127$
1582
given by twisting the form of level~$3556$
1583
will have all~$a_p$ in $\Z/5\Z$ apart from $a_5$ which will be
1584
in~$\F_{25}$ and nonzero. The image of a
1585
decomposition group at~$5$ can then be shown
1586
to be upper-triangularisable (Deligne) and the chars on the diagonal
1587
are unramified of order~$3$. So the image of a decomp group at~$5$
1588
will have order $3\cdot\text{(power of $5$)}$. But there
1589
are no subgroups of ``the subgroup of $\GL(2,\Z/5\Z)$ consisting of elements
1590
of det $\pm 1$'' of order~$15$ because their image in
1591
$\PSL(2,\Z/5\Z)$ would also
1592
have order~$15$ and~$A_5$ has no such subgroups. So no companion form
1593
argument is needed---the repn will be unramified at~$5$.
1594
}
1595
1596
1597
\subsection{Conductor~$3756=2^2\cdot 3\cdot 313$}
1598
Consider the icosahedral extension~$K$ defined by
1599
$h=x^5 - 3x^3 + 10x^2 + 30x - 18$.
1600
A Frobenius element at~$5$ has order~$3$.
1601
The first~$3$ unramified~$p$ such that $a_p=0$ are $17, 61, 67$.
1602
The type at~$2$ is~$5$; the type at~$3$ is~$3$; the type
1603
at~$127$ is~$2$.
1604
The order of~$\eps_2$ is~$1$,
1605
the order of~$\eps_3$ is~$2$, and
1606
the order of~$\eps_{127}$ is~$3$;
1607
1608
Using \cite{cohen-oesterle} we find $\dim S_5(3756,\eps)=2506$;
1609
this is the same as the dimension computed using modular symbols,
1610
so there is no spurious $5$-torsion.
1611
We find that~$V$ has dimension~$8$,
1612
and the characteristic polynomial of~$T_{11}$ is
1613
$(x+\alp^{4})^4(x+\alp^{16})^4$
1614
(we do not use $T_3$ since $3\mid N$).
1615
The kernel~$V_1$ of $T_{11}+\alp^{4}$ has dimension~$2$.
1616
The characteristic polynomial of~$T_5$ on~$V_1$
1617
is $(x+\alp^2)(x+\alp^{10})$.
1618
The first few~$a_p$ of the newform~$f$
1619
in~$V_2$ such that $a_5+\alp^2=0$ are given in
1620
the following table.
1621
{\arraycolsep=.4em
1622
$$\begin{array}{|rl|}\hline
1623
2&0\\
1624
3&\alp^{14}\\
1625
5&\alp^{14}\\
1626
7&3\\
1627
11&\alp^{16}\\
1628
13&\alp^{10}\\
1629
17&0\\
1630
19&3\\
1631
\hline\end{array}
1632
\begin{array}{|rl|}\hline
1633
23&3\\
1634
29&\alp^{2}\\
1635
31&\alp^{22}\\
1636
37&\alp^{22}\\
1637
41&\alp^{20}\\
1638
43&\alp^{16}\\
1639
47&\alp^{4}\\
1640
53&4\\
1641
\hline\end{array}
1642
\begin{array}{|rl|}\hline
1643
59&\alp^{8}\\
1644
61&0\\
1645
67&0\\
1646
71&0\\
1647
73&0\\
1648
79&3\\
1649
83&\alp^{10}\\
1650
89&\alp^{20}\\
1651
\hline\end{array}
1652
\begin{array}{|rl|}\hline
1653
97&4\\
1654
101&0\\
1655
103&2\\
1656
107&\alp^{4}\\
1657
109&\alp^{20}\\
1658
113&2\\
1659
127&\alp^{10}\\
1660
131&\alp^{14}\\
1661
\hline\end{array}
1662
\begin{array}{|rl|}\hline
1663
137&0\\
1664
139&0\\
1665
149&0\\
1666
151&2\\
1667
157&\alp^{20}\\
1668
163&1\\
1669
167&0\\
1670
173&\alp^{20}\\
1671
\hline\end{array}
1672
\begin{array}{|rl|}\hline
1673
179&0\\
1674
181&\alp^{14}\\
1675
191&0\\
1676
193&\alp^{20}\\
1677
197&\alp^{22}\\
1678
199&0\\
1679
211&\alp^{10}\\
1680
223&0\\
1681
\hline\end{array}
1682
\begin{array}{|rl|}\hline
1683
227&\alp^{8}\\
1684
229&\alp^{16}\\
1685
233&\alp^{4}\\
1686
239&\alp^{20}\\
1687
241&\alp^{10}\\
1688
251&\alp^{20}\\
1689
257&1\\
1690
263&\alp^{10}\\
1691
\hline\end{array}
1692
\begin{array}{|rl|}\hline
1693
269&2\\
1694
271&1\\
1695
277&3\\
1696
281&\alp^{20}\\
1697
283&2\\
1698
293&\alp^{20}\\
1699
307&1\\
1700
311&\alp^{16}\\
1701
\hline\end{array}
1702
$$}
1703
1704
It appears that the other newform in~$V_1$ is a companion of~$f$,
1705
but we have not proved this.
1706
1707
Let $g=f\tensor\eps_{313}$.
1708
We have $a_p(f\tensor\eps_{313}^{-1})=0$ for $p=17, 61, 67.$
1709
We have $\eps_{313}(5)=1$ and $\eps_{313}(11)=\alp^8$, so
1710
$a_5(f\tensor\eps_{313}^{-1})= \alp^{14}/1 = \alp^{14}$,
1711
and $a_{11}(f\tensor\eps_{313}^{-1})= \alp^{16}/\alp^{8}=\alp^8$.
1712
At first glance this form looks like the Galois conjugate
1713
of~$f$; and if it were the Galois conjugate then the form
1714
twisted by the character at~$313$ will, as in
1715
Proposition~\ref{prop:1376-g}, have all~$a_p$ in~$\F_5$
1716
for $p \nmid \ell N$ and hence~$\rho_g$
1717
will go into $\GL(2,\F_5)$. Unfortunately the
1718
newform associated to the twisted form is not
1719
quite the conjugate of our original form; the coefficient
1720
of~$a_5$ is wrong! The conjugate form has
1721
$a_5=(\alp^{14})^5 = \alp^{22}$.
1722
The other newform $f_1\in V_1$, which appears to be a companion
1723
of~$f$, has $a_5(f_1)=\alp^{22}$, so $a_5(f_1)^5 = \alp^{14}$.
1724
This proves that $g=\sigma(f_1)$.
1725
The bound from Lemma~\ref{lem:bound} is
1726
$$\frac{11}{12}\cdot [\SL_2(\Z):\Gamma_0(3756)]\cdot
1727
\frac{1}{8} = 863.5,$$
1728
and we are easily able to verify that $f_1$ is a companion.
1729
An argument as in
1730
Proposition~\ref{prop:1376-g} then shows that $a_p(g)\in\F_5$
1731
for all $p\nmid \ell N$.
1732
1733
For this example it remains to check
1734
the discriminant bound, and surjectivity onto~$A_5$.
1735
1736
\subsection{Conductor~$4108=2^2\cdot 13\cdot 79$}
1737
Consider the icosahedral extension~$K$ defined by
1738
$h=x^5+4x^4+3x^3+9x^2+4x+5$.
1739
A Frobenius element at~$5$ has order~$3$.
1740
The first~$4$ unramified~$p$ such that
1741
$a_p=0$ are $17$, $23$, $31$, and $89$.
1742
The type at~$2$ is~5, the type at~$13$ is~2,
1743
and the type at~$79$ is~3.
1744
The order of~$\eps_2$ is~$1$,
1745
the order of~$\eps_{13}$ is~$3$,
1746
and the order of~$\eps_{79}$ is~$2$.
1747
1748
Using~\cite{cohen-oesterle} we find $\dim S_5(4108,\eps)= 2234 $;
1749
1750
The intersection~$V$ of the kernels of $T_{17}$, $T_{23}$,
1751
$T_{31}$, and $T_{89}$ has dimension~$8$.
1752
The characteristic polynomial of~$T_3$ on~$V$ is
1753
$(x+\alp^4)^4(x+\alp^{16})^4$.
1754
The dimension of $V_1=\ker(T_3+\alp^4)$ is~$2$.
1755
The characteristic polynomial of~$T_5$ on~$V_1$ is
1756
$(x+\alp^4)(x+\alp^{20})$ and all of the other~$T_p$ appear
1757
to act as scalars.
1758
1759
{\arraycolsep=.4em
1760
$$
1761
\begin{array}{|rl|}\hline
1762
2&0\\
1763
3&\alpha^{16}\\
1764
5&\alpha^{11}\\
1765
7&\alpha^{20}\\
1766
11&\alpha^{10}\\
1767
13&4\\
1768
17&0\\
1769
19&\alpha^{14}\\
1770
\hline\end{array}
1771
\begin{array}{|rl|}\hline
1772
23&0\\
1773
29&\alpha^{22}\\
1774
31&0\\
1775
37&\alpha^{22}\\
1776
41&\alpha^{10}\\
1777
43&\alpha^{2}\\
1778
47&3\\
1779
53&4\\
1780
\hline\end{array}
1781
\begin{array}{|rl|}\hline
1782
59&\alpha^{14}\\
1783
61&\alpha^{2}\\
1784
67&\alpha^{10}\\
1785
71&\alpha^{20}\\
1786
73&2\\
1787
79&3\\
1788
83&3\\
1789
89&0\\
1790
\hline\end{array}
1791
\begin{array}{|rl|}\hline
1792
97&\alpha^{2}\\
1793
101&\alpha^{22}\\
1794
103&2\\
1795
107&\alpha^{16}\\
1796
109&1\\
1797
113&\alpha^{8}\\
1798
127&0\\
1799
131&4\\
1800
\hline\end{array}
1801
\begin{array}{|rl|}\hline
1802
137&\alpha^{8}\\
1803
139&0\\
1804
149&\alpha^{8}\\
1805
151&0\\
1806
157&4\\
1807
163&\alpha^{20}\\
1808
167&0\\
1809
173&\alpha^{14}\\
1810
\hline\end{array}
1811
\begin{array}{|rl|}\hline
1812
179&\alpha^{4}\\
1813
181&2\\
1814
191&\alpha^{20}\\
1815
193&0\\
1816
197&0\\
1817
199&\alpha^{20}\\
1818
211&\alpha^{4}\\
1819
223&\alpha^{4}\\
1820
\hline\end{array}
1821
\begin{array}{|rl|}\hline
1822
227&0\\
1823
229&0\\
1824
233&3\\
1825
239&2\\
1826
241&\alpha^{8}\\
1827
251&0\\
1828
257&\alpha^{10}\\
1829
263&\alpha^{10}\\
1830
\hline\end{array}
1831
\begin{array}{|rl|}\hline
1832
269&0\\
1833
271&\alpha^{10}\\
1834
277&0\\
1835
281&2\\
1836
283&\alpha^{4}\\
1837
293&\alpha^{14}\\
1838
307&4\\
1839
311&0\\
1840
\hline\end{array}
1841
$$}
1842
1843
The bound of Lemma~\ref{lem:bound} is
1844
\edit{\kevin{Before we go any further, which forms can
1845
we deal with rigorously now?}\william{All of them?}}
1846
$$\frac{11}{12}\cdot [\SL_2(\Z):\Gamma_0(4108)]\cdot
1847
\frac{1}{8} = 770.$$
1848
1849
\subsection{Conductor~$4288=2^6\cdot 67$}
1850
Consider the icosahedral extension~$K$ defined by
1851
$h=x^5+4x^4+5x^3+8x^2+3x+2$.
1852
A Frobenius element at~$5$ has order~$3$.
1853
The first~$3$ unramified~$p$ such that $a_p=0$ are $19$,
1854
$23$, $47$.
1855
The type at~$2$ is~18, and the type at~$67$ is~2.
1856
The order of~$\eps_{67}$ is~$3$, so in order
1857
for~$\eps$ to be odd, the order of~$\eps_2$
1858
must be~$2$. First we try letting $\eps_2$ be the character
1859
of conductor~$4$.\edit{\kevin{I don't know anything
1860
about ``type 18'' and your statement ``order 2'' is not enough
1861
to tell me the character, so you should perhaps say what it is.
1862
Oh! I've just noticed that this is wrong anyway!}}
1863
1864
Using \cite{cohen-oesterle} we find that\edit{\kevin{These
1865
sections are getting really tedious, aren't they. I think we should
1866
perhaps consider instead summarising things more succinctly,
1867
e.g. saying ``we tried our approach for 4 other forms and it worked/didn't
1868
work.}
1869
\william{
1870
You may be right. However, there is a lot of information in each
1871
situation. One wants to know how many Hecke operators are needed
1872
to cut out the space, what the character is, and so on. All of
1873
this information would have to be summarized.}}
1874
$\dim S_5(4288,\eps)=2164$.
1875
the intersection~$V$ of the kernels of $T_{19}$, $T_{23}$, and
1876
$T_{47}$ has dimension~$0$ --- we must have chosen~$\eps_2$
1877
incorrectly. Next we try with $\eps_2$ one of the characters
1878
of conductor~$8$.\edit{\william{The
1879
character $\eps$ is [1,2,2].}} \edit{\kevin{William: this character
1880
also has order 2!}} The dimension is also~$2164$. This time we are
1881
able to compute the intersection~$V$, and it has dimension~$16$.
1882
%There is a quadratic twist of
1883
%conductor~$8$ that preserves the level, so the dimension should
1884
%be~$16$.
1885
\edit{\kevin{William---this is a very optimistic statement!
1886
All these spaces have Hecke acting non-semi-simply so I wouldn't dare
1887
to guess the dimension of the space, just the number of
1888
eigenforms\ldots}\william{I removed the remark about the dimension
1889
being~$16$.}}
1890
1891
The characteristic polynomial of~$T_3$ on~$V$ is
1892
$(x+2)^8(x+3)^8$. The dimension of the kernel~$V_1$
1893
of $T_3|_V+2$ is~$4$. The characteristic polynomial
1894
of~$T_7$ on~$V_1$ is $(x+\alp^8)^2(x+\alp^{20})^2$.
1895
The dimension of the kernel~$V_2$
1896
of $T_3|_{V_1}+\alp^8$ is~$2$.
1897
The characteristic polynomial of~$T_5$ on~$V_2$ is
1898
$(x+\alp^2)(x+\alp^{10})$.
1899
By checking up to the bound
1900
$$\frac{11}{12}\cdot [\SL_2(\Z):\Gamma_0(4288)]\cdot
1901
\frac{1}{4} = 1496$$
1902
of Lemma~\ref{lem:bound}
1903
we see that all $T_p$, with $p\neq 5$,
1904
act as scalars on~$V_2$.
1905
1906
{\arraycolsep=.4em
1907
$$
1908
\begin{array}{|rl|}\hline
1909
2&0\\
1910
3&3\\
1911
5&\alpha^{14}\\
1912
7&\alpha^{20}\\
1913
11&\alpha^{20}\\
1914
13&\alpha^{16}\\
1915
17&\alpha^{16}\\
1916
19&0\\
1917
\hline\end{array}
1918
\begin{array}{|rl|}\hline
1919
23&0\\
1920
29&\alpha^{14}\\
1921
31&\alpha^{20}\\
1922
37&\alpha^{4}\\
1923
41&\alpha^{8}\\
1924
43&2\\
1925
47&0\\
1926
53&4\\
1927
\hline\end{array}
1928
\begin{array}{|rl|}\hline
1929
59&1\\
1930
61&\alpha^{16}\\
1931
67&\alpha^{4}\\
1932
71&\alpha^{20}\\
1933
73&0\\
1934
79&\alpha^{20}\\
1935
83&0\\
1936
89&2\\
1937
\hline\end{array}
1938
\begin{array}{|rl|}\hline
1939
97&\alpha^{10}\\
1940
101&\alpha^{14}\\
1941
103&\alpha^{20}\\
1942
107&2\\
1943
109&0\\
1944
113&\alpha^{2}\\
1945
127&\alpha^{2}\\
1946
131&1\\
1947
\hline\end{array}
1948
\begin{array}{|rl|}\hline
1949
137&1\\
1950
139&2\\
1951
149&4\\
1952
151&\alpha^{4}\\
1953
157&\alpha^{22}\\
1954
163&0\\
1955
167&\alpha^{14}\\
1956
173&0\\
1957
\hline\end{array}
1958
\begin{array}{|rl|}\hline
1959
179&1\\
1960
181&\alpha^{14}\\
1961
191&\alpha^{4}\\
1962
193&3\\
1963
197&0\\
1964
199&0\\
1965
211&\alpha^{16}\\
1966
223&2\\
1967
\hline\end{array}
1968
\begin{array}{|rl|}\hline
1969
227&\alpha^{14}\\
1970
229&\alpha^{4}\\
1971
233&\alpha^{2}\\
1972
239&\alpha^{20}\\
1973
241&3\\
1974
251&\alpha^{4}\\
1975
257&\alpha^{2}\\
1976
263&4\\
1977
\hline\end{array}
1978
\begin{array}{|rl|}\hline
1979
269&0\\
1980
271&0\\
1981
277&2\\
1982
281&\alpha^{10}\\
1983
283&3\\
1984
293&1\\
1985
307&\alpha^{22}\\
1986
311&3\\
1987
\hline\end{array}
1988
$$}
1989
1990
1991
\subsection{Conductor~$5373=3^3\cdot 199$}
1992
There are two $A_5$ entries in Frey's
1993
Table~2 of conductor 5373. Though neither 2 nor 5 ramify,
1994
the hypothesis of~\cite{bdsbt} do not apply
1995
to the example below because $\Frob_2$ has order~$5$.
1996
This is the only example of odd level treated in this paper.
1997
1998
Consider the icosahedral extension~$K$ defined by
1999
$h=x^5+2x^4+x^3+7x^2+23x-11$.
2000
A Frobenius element at~$5$ has order~$2$.
2001
The first few $p\nmid \ell N$ such that
2002
$a_p=0$ are $7$, $23$, $37$, $79$, $89$.
2003
The type at~$3$ is~11, and the type at~$199$ is~2.
2004
The order of~$\eps_3$ is~$2$, and
2005
the order of~$\eps_{199}$ is~$3$.
2006
2007
Using~\cite{cohen-oesterle} we find $\dim S_5(5373,\eps)= 2394$;
2008
this agrees with the dimension of the space of modular symbols.
2009
2010
The intersection~$V$ of the kernels of
2011
$T_7$, $T_{23}$, $T_{37}$, $T_{79}$, and $T_{89}$
2012
is of dimension~$8$.
2013
The characteristic polynomial of~$T_2$ on~$V$ is
2014
$(x+\alp^4)^4(x+\alp^{16})^4$.
2015
The kernel~$V_1$ of $T_2|_V+\alp^4$
2016
is of dimension~$2$. It appears that all~$T_p$ act
2017
as scalars on~$V_1$ except~$T_5$ which has
2018
characteristic polynomial $(x+1)(x+4)$.
2019
Let~$f$ be the form with $a_5=-1$; some eigenvalues of~$f$ are
2020
given in the following table.
2021
{\arraycolsep=.4em
2022
$$\begin{array}{|rl|}\hline
2023
2&\alpha^{16}\\
2024
3&0\\
2025
5&4\\
2026
7&0\\
2027
11&2\\
2028
13&\alpha^{22}\\
2029
17&1\\
2030
19&\alpha^{16}\\
2031
\hline\end{array}
2032
\begin{array}{|rl|}\hline
2033
23&0\\
2034
29&\alpha^{4}\\
2035
31&\alpha^{14}\\
2036
37&0\\
2037
41&\alpha^{14}\\
2038
43&\alpha^{10}\\
2039
47&\alpha^{4}\\
2040
53&\alpha^{8}\\
2041
\hline\end{array}
2042
\begin{array}{|rl|}\hline
2043
59&3\\
2044
61&3\\
2045
67&4\\
2046
71&\alpha^{22}\\
2047
73&\alpha^{10}\\
2048
79&0\\
2049
83&3\\
2050
89&0\\
2051
\hline\end{array}
2052
\begin{array}{|rl|}\hline
2053
97&\alpha^{14}\\
2054
101&3\\
2055
103&0\\
2056
107&2\\
2057
109&3\\
2058
113&\alpha^{20}\\
2059
127&\alpha^{14}\\
2060
131&\alpha^{20}\\
2061
\hline\end{array}
2062
\begin{array}{|rl|}\hline
2063
137&2\\
2064
139&2\\
2065
149&\alpha^{4}\\
2066
151&0\\
2067
157&4\\
2068
163&\alpha^{16}\\
2069
167&\alpha^{8}\\
2070
173&\alpha^{8}\\
2071
\hline\end{array}
2072
\begin{array}{|rl|}\hline
2073
179&\alpha^{8}\\
2074
181&1\\
2075
191&3\\
2076
193&\alpha^{20}\\
2077
197&0\\
2078
199&1\\
2079
211&2\\
2080
223&0\\
2081
\hline\end{array}
2082
\begin{array}{|rl|}\hline
2083
227&1\\
2084
229&\alpha^{20}\\
2085
233&\alpha^{4}\\
2086
239&2\\
2087
241&1\\
2088
251&4\\
2089
257&\alpha^{20}\\
2090
263&4\\
2091
\hline\end{array}
2092
\begin{array}{|rl|}\hline
2093
269&\alpha^{2}\\
2094
271&\alpha^{20}\\
2095
277&3\\
2096
281&1\\
2097
283&\alpha^{22}\\
2098
293&\alpha^{14}\\
2099
307&0\\
2100
311&\alpha^{2}\\
2101
\hline\end{array}
2102
$$
2103
}
2104
2105
The bound of Lemma~\ref{lem:bound} is
2106
$$\frac{11}{12}\cdot [\SL_2(\Z):\Gamma_0(5373)]\cdot
2107
\frac{1}{4} = 1650.$$
2108
2109
\comment{
2110
\subsection{Conductor $N=1825=5^2\cdot 73$ and $\ell=7$}
2111
In this example, the method does not apply because $5\mid N$.
2112
Nonetheless, we proceed using~$\ell=7$ instead.\edit{\kevin{We can't
2113
\emph{hope} to finish though, because we'll never show the thing is $A_5$
2114
and not something like $GL_2(\F_7)$. So probably we should kill this
2115
example completely. Somehow our paper gives a method for dealing
2116
with $\GL_2(\F_5)$ (and it would also perhaps work with $\GL_2(\F_4)$)
2117
but doesn't have a chance of working wiht $GL_2(\F_7)$, I shouldn't think.}
2118
\william{It is completely dead.}}
2119
2120
Consider the icosahedral extension~$K$ defined by
2121
$h=x^5+4x^4-x^3-21x^2-x-7$.
2122
A Frobenius element at~$7$ has order~$3$, so
2123
distinct eigenvalues mod~$7$ (I think).
2124
The first three unramified~$p$ such that
2125
$a_p=0$ are $17$, $59$, $67$, $89$.
2126
The type at~$5$ is~6, and the type at~$73$ is~2.
2127
The order of~$\eps_5$ is~$4$ (I think), and
2128
the order of~$\eps_{73}$ is~$3$.
2129
2130
Using~\cite{cohen-oesterle} we find $\dim S_8(1825,\eps)= 1104$, and
2131
this agrees with modular symbols.
2132
The intersection~$V$ of the kernels of $T_{17}$, $T_{59}$,
2133
and~$T_{67}$ has dimension~$8$.
2134
The operator~$T_{89}$ also vanishes on~$V$.
2135
The character takes values in $\F_{7^2}$ which we represent
2136
using a generator~$\alp$ of the multiplicative group such that
2137
$\alp^2-\alp+3=0$.
2138
The characteristic polynomial of~$T_2$ on~$V$ is
2139
$(x+\alp)^2(x+\alp^{19})^2(x+\alp^{25})^2(x+\alp^{43})^2$.
2140
Let $V_1$ be the kernel of $T_2+\alp$; then $V_1$ is spanned
2141
by two newforms~$f$ and~$f_1$:
2142
\begin{align*}
2143
f &= q+\alp^{25}q^2 + \alp^{45}q^3 + \alp^{23}q^4+ \alp^{22}q^6+\alp^{46}q^7+\cdots\\
2144
f_1 &= q+\alp^{25}q^2 + \alp^{45}q^3 + \alp^{23}q^4+ \alp^{22}q^6+\alp^{14}q^7+\cdots
2145
\end{align*}
2146
It appears that~$f$ and~$f_1$ are companions; and undoubtedly this can
2147
be checked.
2148
2149
It doesn't seem possible to twist and obtain a representation
2150
into $\GL(2,\F_7)$. The $\eps_{73}$ part of the character
2151
is $\F_7$-rational. The $\eps_5$ part has order~$4$, and is
2152
hence not $\F_7$-rational. Twisting by~$\eps_5$ (or its inverse)
2153
only replaces $\eps_5$ by $\eps_5^3$, which is still not $\F_7$-rational.
2154
2155
\begin{dashlist}
2156
\item Even if we could twist into $\GL(2,\F_7)$,
2157
the order~$60$ of $A_5$ doesn't divide the order of $\GL(2,\F_7)$.
2158
\item In this $N=1825$ example, we found a modular representation
2159
$\rho_g : \GQ \ra \GL(2,\F_{7^2})$
2160
that (morally) is the modulo a prime-over-$7$ reduction of
2161
the Artin representation $\rho : \GQ \ra \GL(2,\cO)$.
2162
Here~$\cO$ is the ring of integers in a finite extension of~$\Q_7$.
2163
If someone ever proves modularity of~$\rho$ they will
2164
have proved that~$\rho_g$ admits a lift to a characteristic zero
2165
representation whose image is an extension of~$A_5$;
2166
perhaps we can prove this. We could then try and analyze the
2167
ramification and conclude that the lift of~$\rho_g$ is (a twist) of~$\rho$.
2168
Then~\cite{buzzard-taylor} implies modularity of~$\rho$.
2169
\end{dashlist}
2170
}
2171
2172
2173
\section{Computations}
2174
\subsection{Higher weight modular symbols}
2175
\label{sec:modsym}
2176
The second author developed software that computes the space of
2177
weight~$k$ modular symbols $\sS_k(N,\eps)$, for $k\geq 2$ and
2178
arbitrary~$\eps$.
2179
See~\cite{merel:1585} for the standard facts about higher weight
2180
modular symbols, and~\cite{stein:phd} for a description of
2181
how to efficiently compute with them.
2182
2183
Let $K=\Q(\eps)$ be the field generated by the values of~$\eps$.
2184
The cuspidal modular symbols $\sS_k(N,\eps)$ are a
2185
finite dimensional vector space over~$K$, which is generated by all
2186
linear combinations of higher weight modular symbols
2187
$$X^i Y^{k-2-i}\{\alp,\beta\}$$
2188
that lie in the kernel of an appropriate boundary map. There is an
2189
involution~$*$ that acts on $\sS_k(N,\eps)$, and
2190
$\sS_k(N,\eps)^+\tensor_K\C$ is isomorphic, as a module over the Hecke
2191
algebra, to the space $S_k(N,\eps;\C)$ of cusp forms.
2192
2193
Fix $k=5$. In each case considered in this paper,
2194
there is a prime ideal~$\lambda$
2195
of the ring of integers $\mathcal{O}$ of~$K$
2196
such that $\mathcal{O}/\lambda\isom \F_{25}$.
2197
Let~$\cL$ be the $\mathcal{O}$-module generated by all modular
2198
symbols of the form $X^iY^{3-i}\{\alp,\beta\}$,
2199
and let
2200
$$\sS_5(N,\eps;\F_{25})=(\cL\tensor_{\mathcal{O}}\F_{25})\cap \sS_5(N,\eps).$$
2201
This is the space that we computed.
2202
The Hecke algebra acts on $\sS_5(N,\eps;\F_{25})$, so when
2203
we find an eigenform we find a maximal ideal of the Hecke algebra.
2204
2205
As an extra check on our computation of
2206
$\sS_5(N,\eps;\F_{25})$, we computed the dimension
2207
of $S_5(N,\eps;\C)$ using both the formula of~\cite{cohen-oesterle}
2208
and the Hijikata trace formula (see~\cite{hijikata:trace})
2209
applied to the identity Hecke operator.
2210
2211
2212
\comment{%it's all in my thesis and it's not that relevant.
2213
The Manin symbols are
2214
$[i, (c,d)]$ where $0\leq i\leq k-2=3$ and
2215
$(c,d)$ vary over points in the projective plane.
2216
The Manin symbol $[i,(c',d')]$ corresponds to the
2217
modular symbol $(g.X^iY^{3-i})\{g(0),g(\infty)\}$
2218
where $g=\abcd{a}{b}{c}{d}\in\SL_2(\Z)$ is a matrix whose lower
2219
two entries are congruent to $(c',d')$ modulo $N$,
2220
and $g.X^iY^{3-i} := (dX-bY)^i(-cX+aY)^{3-i}$.
2221
Let $\sigma=\abcd{0}{-1}{1}{0}$, $\tau={0}{-1}{1}{-1}$
2222
and for $\gamma\in\SL_2(\Z)$, let
2223
$[i,(c,d)]\gamma = [\gamma.X^iY^{3-i}, (c,d)\gamma]$.
2224
Since there are only finitely many
2225
Manin symbols, we can
2226
compute $\sS_5(N,\eps)$ as the quotient of the $\F$-vector
2227
space generated by Manin symbols modulo
2228
the following relations:
2229
\begin{align*}
2230
{[i,(c,d)] + [i,(c,d)]\sigma} &= 0\\
2231
{[i,(c,d)] + [i,(c,d)]\tau + [i,(c,d)]\tau^2} &= 0\\
2232
{[i,(n c,n d)]}&=\eps(n)[i,(c,d)]
2233
\qquad n\in (\Z/N\Z)^*
2234
\end{align*}
2235
The quotient was computed by using a fast ``hashing'' function
2236
to quotient out by the $2$-term relations. The quotient
2237
by the $3$-term relations was then computed using sparse
2238
Gauss elimination. One important subtlety is that, e.g., $\sigma$
2239
and~$\tau$ do not commute so, after modding out by
2240
the~$\sigma$ relations, it is important to mod out by~$3$
2241
term relations coming both from~$\tau$ and~$\sigma\tau$.
2242
2243
The main result of~\cite{merel:1585} gives
2244
a way to compute the action of $T_p$ directly
2245
on the Manin symbols.
2246
Suppose $f\in\sS_5(N,\eps;\F_{25})$ is an eigenvector; to
2247
naively compute the action of~$T_p$ on~$f$ requires computing
2248
the action of~$T_p$ on each Manin symbol involved in~$f$,
2249
and then summing the result. This requires roughly
2250
$\dim\sS$ times as long as computing~$T_p$ on a single
2251
Manin symbol.
2252
In order to quickly compute a large number of
2253
Hecke eigenvalues we use the following projection trick.
2254
Let $\vphi\in\Hom(\sS_5(N,\eps;\F_{25}),\F_{25})$ be a (left) eigenvector for all
2255
Hecke operators~$T_p$ having the same eigenvalues as~$f$.
2256
Choose a Manin symbol $x=[i,(c,d)]$ such
2257
that $\vphi(x)\neq 0$. Since~$x$ is of a very simple form,
2258
it is easy to compute~$T_p(x)$ quickly. We have
2259
$\vphi(T_p(x)) = (T_p(\vphi))(x) = a_p \vphi(x)$,
2260
so since $\vphi(x)\neq 0$ we divide and find
2261
$a_p = \vphi(T_p(x))/\vphi(x)$.
2262
In fact, we use a generalization of this trick to
2263
quickly compute the action of~$T_p$ on any Hecke stable subspace
2264
$V\subset \Hom(\sS(N,\eps;\F_{25}),\F_{25})$.
2265
}
2266
2267
2268
\subsection