Open in CoCalc
1
\documentclass[12pt]{book}
2
\textwidth=1.2\textwidth
3
\textheight=1.3\textheight
4
\hoffset=-.7in
5
\voffset=-1.28in
6
\usepackage{amsfonts}
7
\usepackage{amsmath}
8
\usepackage{amsopn}
9
\usepackage{amsthm}
10
\usepackage{amscd}
11
\usepackage{makeidx} % exec the command "makeindex ribetnote.idx"
12
\makeindex
13
\author{William Stein}
14
\title{Hecke Algebras and Modular Forms:\\
15
Notes derived from Ribet's 1996 Berkeley grad. course.}
16
17
\font\bbb=msbm10 scaled \magstep 1
18
\font\bbbs=msbm10
19
\font\german=eufm10 scaled \magstep 1
20
21
\font\script=rsfs10 scaled \magstep 1
22
23
\newcommand{\rhouniv}{\rho_{\text{univ}}}
24
\newcommand{\Set}{\mathbf{Set}}
25
\newcommand{\pr}{pr}
26
\newcommand{\ns}{N_{\Sigma}}
27
\newcommand{\rs}{R_{\Sigma}}
28
\newcommand{\ts}{\T_{\Sigma}}
29
\newcommand{\wpt}{\wp_T}
30
\newcommand{\wpr}{\wp_R}
31
\newcommand{\nd}{\!\!\not|}
32
\newcommand{\dbd}[1]{\langle#1\rangle} % make a diamond bracket d symbol
33
34
%%%% Draw a correspondence diagram.
35
\newcommand{\corr}[5]{\begin{array}
36
{ccc}&#1\\\stackrel{#4}{\swarrow}&&\stackrel{#5}{\searrow}
37
\\#2&&#3\end{array}}
38
\newcommand{\corrs}{\rightarrow\rightarrow} % a correspondence,
39
%% This should be a squiggly arrow,
40
%% but that isn't supported on some
41
%% installations.
42
43
%%%% Draw a triangular commutative diagram
44
\newcommand{\triCD}[6]{\begin{array}{ccc}
45
#1&\xrightarrow{#2}&#3\\
46
&#6\searrow&\downarrow #4\\
47
& & #5\end{array}}
48
\newcommand{\ltriCD}[6]{\begin{array}{ccc}
49
#1 \\
50
#6\downarrow&\searrow\\
51
#5&\xrightarrow{#4}&#3
52
\end{array}}
53
54
55
\newcommand{\onto}{\longrightarrow\!\!\!\!\rightarrow}
56
\newcommand{\mapdown}[1]{\Big\downarrow\rlap{$\vcenter{\hbox{$\scriptstyle#1$}}$}}
57
\newcommand{\presup}[1]{{}^{#1}\!}
58
%\newcommand{\bZ}{\mbox{\bbb Z}}
59
%\newcommand{\bF}{\mbox{\bbb F}}
60
%\newcommand{\bA}{\mbox{\bbb A}}
61
%\newcommand{\bQ}{\mbox{\bbb Q}}
62
%\newcommand{\bP}{\mbox{\bbb P}}
63
%\newcommand{\bR}{\mbox{\bbb R}}
64
%\newcommand{\bC}{\mbox{\bbb C}}
65
%\newcommand{\bsC}{\mbox{\bbbs C}}
66
%\newcommand{\bsQ}{\mbox{\bbbs Q}}
67
%\newcommand{\bsZ}{\mbox{\bbbs Z}}
68
%\newcommand{\bsR}{\mbox{\bbbs R}}
69
%\newcommand{\bT}{\mbox{\bbb T}}
70
71
\newcommand{\rholf}{\rho_{\ell,f}}
72
\renewcommand{\ss}{\text{s.s.}}
73
\newcommand{\dual}{\vee}
74
\newcommand{\uE}{\underline{E}} % an enhanced elliptic curve
75
\newcommand{\bZ}{\mathbf{Z}}
76
\newcommand{\bF}{\mathbf{F}}
77
\newcommand{\bA}{\mathbf{A}}
78
\newcommand{\bQ}{\mathbf{Q}}
79
\newcommand{\bP}{\mathbf{P}}
80
\newcommand{\bR}{\mathbf{R}}
81
\newcommand{\bC}{\mathbf{C}}
82
\newcommand{\G}{\mathbf{G}}
83
\newcommand{\bsC}{\mathbf{C}}
84
\newcommand{\bsQ}{\mathbf{Q}}
85
\newcommand{\bsZ}{\mathbf{Z}}
86
\newcommand{\bsR}{\mathbf{R}}
87
\newcommand{\bT}{\mathbf{T}}
88
\newcommand{\m}{\mathbf{m}}
89
\newcommand{\uJ}{\underline{J}}
90
\newcommand{\uV}{\underline{V}}
91
\newcommand{\uW}{\underline{W}}
92
\newcommand{\uQ}{\underline{Q}}
93
94
95
\newcommand{\sL}{\mathcal{L}}
96
\newcommand{\sH}{\mathcal{H}}
97
\newcommand{\sG}{\mathcal{G}}
98
\newcommand{\sO}{\mathcal{O}}
99
\newcommand{\sR}{\mathcal{R}}
100
\newcommand{\sF}{\mathcal{F}}
101
\newcommand{\sA}{\mathcal{A}}
102
\newcommand{\sM}{\mathcal{M}}
103
\newcommand{\sS}{\mathcal{S}}
104
\newcommand{\sC}{\mathcal{C}}
105
\newcommand{\A}{\mathbf{A}}
106
\newcommand{\soe}{\mathcal{O}_E}
107
108
\newcommand{\Mu}{\pmb{\mu}}
109
\newcommand{\Mul}{\Mu_{\ell}}
110
\newcommand{\modgp}{\sl2z}
111
\newcommand{\galq}{\Gal({\overline{\Q}/\Q})}
112
\newcommand{\galqp}{\Gal({\overline{\Qp}/\Qp})}
113
\newcommand{\Qbar}{\overline{\Q}}
114
\newcommand{\Qpbar}{\Qbar_p}
115
\newcommand{\Qlbar}{\Qbar_{\ell}}
116
\newcommand{\Z}{\bZ}
117
\newcommand{\F}{\bF}
118
\newcommand{\Fbar}{\overline{\F}}
119
\newcommand{\Fp}{\bF_p}
120
\newcommand{\Fq}{\bF_q}
121
\newcommand{\Fl}{\bF_{\ell}}
122
\newcommand{\Flbar}{\Fbar_{\ell}}
123
\newcommand{\Fln}{\bF_{\ell^{\nu}}}
124
\newcommand{\Qp}{\bQ_p}
125
\newcommand{\Zp}{\bZ_p}
126
\newcommand{\Zl}{\bZ_{\ell}}
127
\newcommand{\Ql}{\bQ_{\ell}}
128
\newcommand{\Q}{\bQ}
129
\newcommand{\R}{\bR}
130
\newcommand{\C}{\bC}
131
\renewcommand{\H}{\mathcal{H}}
132
\renewcommand{\O}{\sO}
133
\newcommand{\T}{\bT}
134
\newcommand{\Tm}{\T_{\m}}
135
\newcommand{\rmr}{R_{\m_R}}
136
\newcommand{\Af}{\mbox{\script A}_f} %% for the shimura construction 7.14
137
138
\newcommand{\isom}{\cong}
139
\newcommand{\iso}{\xrightarrow{\sim}}
140
\newcommand{\tensor}{\otimes}
141
\newcommand{\into}{\rightarrow}
142
\newcommand{\union}{\cup}
143
\newcommand{\intersect}{\cap}
144
\newcommand{\abcd}{\bigl(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\bigr)}
145
\newcommand{\bigabcd}{\Bigl(\begin{matrix}a&b\\c&d\end{matrix}\Bigr)}
146
\newcommand{\mat}[4]{\left(\begin{matrix}#1&#2\\#3&#4\end{matrix}\right)}
147
\newcommand{\smat}[4]{\left(\begin{smallmatrix}#1&#2\\#3&#4
148
\end{smallmatrix}\right)}
149
\newcommand{\cross}{\times}
150
%\newcommand{\Gal}{\mathcal{G}al}
151
\newcommand{\gal}{\mathcal{G}al}
152
\newcommand{\Tatel}{\Tate_{\ell}}
153
\newcommand{\tatel}{\Tate_{\ell}}
154
155
\newcommand{\xopn}{X_0(pN)}
156
\newcommand{\xopnp}{X_0(pN)_{\F_p}}
157
\newcommand{\xon}{X_0(N)}
158
\newcommand{\xonp}{X_0(N)_{\F_p}}
159
\newcommand{\gon}{\Gamma_0(N)}
160
\newcommand{\jon}{J_0(N)}
161
\newcommand{\jonp}{J_0(N)_{\F_p}}
162
\newcommand{\frobp}{\Frob_p}
163
\newcommand{\Frobp}{\Frob_p}
164
\newcommand{\rholam}{\rho_{\lambda}}
165
\newcommand{\rhol}{\rho_{\ell}}
166
\newcommand{\rholamf}{\rho_{f,\lambda}}
167
\newcommand{\rholamfbar}{\overline{\rho}_{\lambda,f}}
168
\newcommand{\rhom}{\rho_{\m}}
169
\newcommand{\rholambdabar}{\overline{\rho}_{\lambda}}
170
\newcommand{\pico}{\pic^0}
171
\newcommand{\fln}{\F_{\ell^{\nu}}}
172
\newcommand{\flnbar}{\overline{\fnl}}
173
\newcommand{\fl}{\F_{\ell}}
174
\newcommand{\galql}{\Gal(\Qlbar/\Ql)}
175
\newcommand{\sofl}{\sO_{f,\lambda}}
176
177
\DeclareMathOperator{\Val}{Val}
178
\DeclareMathOperator{\new}{new}
179
\DeclareMathOperator{\lcm}{lcm}
180
\DeclareMathOperator{\Gal}{Gal}
181
\DeclareMathOperator{\WD}{WD}
182
\DeclareMathOperator{\Char}{char}
183
\DeclareMathOperator{\cond}{cond}
184
\DeclareMathOperator{\Alb}{Alb}
185
\DeclareMathOperator{\Tan}{Tan}
186
\DeclareMathOperator{\Aut}{Aut}
187
\DeclareMathOperator{\End}{End}
188
\DeclareMathOperator{\ord}{ord}
189
\DeclareMathOperator{\sl2}{SL_2}
190
\DeclareMathOperator{\sl2z}{SL_2(\bZ)}
191
\DeclareMathOperator{\SL}{SL}
192
\DeclareMathOperator{\GL}{GL}
193
\DeclareMathOperator{\gl}{GL}
194
\DeclareMathOperator{\tr}{Tr}
195
\DeclareMathOperator{\trace}{trace}
196
\DeclareMathOperator{\imag}{Im} % image or imaginary part. cute, eh?
197
\DeclareMathOperator{\real}{Re} % image or imaginary part. cute, eh?
198
\DeclareMathOperator{\Hom}{Hom}
199
\DeclareMathOperator{\Div}{Div}
200
\DeclareMathOperator{\Mor}{Mor}
201
\DeclareMathOperator{\Cot}{Cot}
202
\DeclareMathOperator{\spec}{spec}
203
\DeclareMathOperator{\Sym}{Sym}
204
\DeclareMathOperator{\frob}{frob}
205
\DeclareMathOperator{\Frob}{Frob}
206
\DeclareMathOperator{\ver}{ver}
207
\DeclareMathOperator{\Ver}{Ver}
208
\DeclareMathOperator{\tors}{tors}
209
\DeclareMathOperator{\id}{id}
210
\DeclareMathOperator{\Tate}{Tate}
211
\DeclareMathOperator{\pic}{Pic}
212
\DeclareMathOperator{\rank}{rank}
213
\DeclareMathOperator{\PS}{PS}
214
\DeclareMathOperator{\ab}{ab}
215
\DeclareMathOperator{\st}{st}
216
\DeclareMathOperator{\Ind}{Ind}
217
\DeclareMathOperator{\Ann}{Ann}
218
219
\newcommand{\aut}{\Aut}
220
221
%%%% Theoremstyles
222
\theoremstyle{plain}
223
\newtheorem{thm}{Theorem}[section]
224
\newtheorem{prop}[thm]{Proposition}
225
\newtheorem{cor}[thm]{Corollary}
226
\newtheorem{lem}[thm]{Lemma}
227
\newtheorem{question}[thm]{Question}
228
229
\theoremstyle{definition}
230
\newtheorem{defn}[thm]{Definition}
231
\newtheorem{conj}[thm]{Conjecture}
232
233
\theoremstyle{remark}
234
\newtheorem{remark}[thm]{Remark}
235
\newtheorem{example}[thm]{Example}
236
\newtheorem{exercise}[thm]{Exercise}
237
238
239
240
\begin{document}
241
\frontmatter
242
\maketitle
243
\chapter{Preface}
244
{\bfseries Disclaimer: }
245
These notes record some of what I saw in Ken Ribet's course
246
on Modular Forms and Hecke Operators given at U.C. Berkeley during
247
the Spring semester 1996. They are still {\em very rough}
248
as I wrote them during my first semester of graduate school before
249
I knew any real mathematics.
250
251
The participants in the course were:
252
Amod Agashe,
253
Matt Baker,
254
Jim Borger,
255
Kevin Buzzard,
256
Bruce Caskel,
257
Robert Coleman,
258
Jan\'{o}s Csirik,
259
Annette Huber,
260
David Jones,
261
David Kohel,
262
Loic Merel,
263
David Moulton,
264
Andrew Ogg,
265
Arthur Ogus,
266
Jessica Polito,
267
Ken Ribet,
268
Saul Schleimer,
269
Lawren Smithline,
270
William Stein,
271
Takahashi,
272
Wayne Whitney, and
273
Hui Zui.
274
275
I wish to thank David Moulton, Joe Wetherall, and Kevin Buzzard who
276
helped me in preparing these notes, Arthur Ogus who asked
277
a lot of stimulating questions during the class, and of course
278
Ken Ribet who sees clearly.
279
280
William Stein, Spring 1996, Berkeley, CA, {\tt was@math.berkeley.edu}
281
\tableofcontents
282
283
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
284
\mainmatter
285
%%%%%%%%%%%%%%%%%%%%%%%%
286
%% Lecture 1, January 17
287
\chapter{Introduction}
288
289
The main objects of study in this course are:
290
\begin{itemize}
291
\item Modular Forms
292
\item Hecke Algebras
293
\item Modular Curves\index{modular curves}
294
\item Jacobians\index{Jacobian}
295
\item Abelian Varieties
296
\end{itemize}
297
298
\section{Two Dimensional Galois Representations}
299
300
\index{Galois representations}
301
The geometric objects of study are elliptic curves and more generally
302
algebraic curves of arbitrary genus. These in turn give rise via the
303
Jacobian construction to higher dimensional
304
abelian varieties. These geometric objects in turn give rise to
305
Galois representations.
306
307
When studying elliptic curves, the natural tool in the characteristic
308
zero situation is to present the elliptic curve as
309
$\bC/\sL$ for some lattice $\sL$ in $\bC$.
310
To construct $\sL$ fix a non-zero holomorphic differential $\omega$
311
of $E$ over $\bC$ and construct $\sL$ as
312
\begin{equation*}
313
\bigg\{\int_{\gamma} \omega \quad \bigg| \quad \gamma\in H_1(E(\bC),\bZ)\bigg\}.
314
\end{equation*}
315
316
\subsection{Finite Fields (Weil, Tate)}
317
In the 1940's, Weil study the analogous situation for
318
elliptic curves defined over a finite field $k$. He desperately
319
wanted to find an algebraic way to describe the above correspondence.
320
He was able to find an algebraic definition of
321
$\sL/n\sL$, where $n\geq 1$ and $(n,\Char k)=1$, which is as follows.
322
Let $E[n]=\{P\in E(\overline{k}) : nP = 0\} = (\frac{1}{n} \sL) / \sL
323
\isom \sL / n \sL$.
324
325
Now fix a prime $\ell$, we let
326
$E[\ell^\infty]=\{P\in E(\overline{k}) : \ell^{\nu}P = 0, \text{ some }
327
\nu \geq 1\} = \cup_{\nu=1}^{\infty} E[\ell^{\nu}]$.
328
Tate obtained an analogous construction by defining a rank 2 free
329
$\Zl$-module $\Tatel E:=\varprojlim E[\ell^{\nu}]$
330
(the map from $E[\ell^{\nu}]\into E[\ell^{\nu-1}]$ is multiplication
331
by $\ell$).
332
To see that the rank is $2$, check that
333
the $\bZ/\ell^{\nu}\bZ$-module structure of $E[\ell^{\nu}]$ is compatible
334
with the maps $E[\ell^{\nu}]\into E[\ell^{\nu-1}]$). See
335
\cite{silverman1} (III, 7).
336
Then $V_{\ell}(E)=T_{\ell}(E)\tensor\bQ_{\ell}$ is a two dimensional
337
vector space over $\bQ_{\ell}$. This gives the first
338
nontrivial example of $\ell$-adic \'{e}tale cohomology.
339
340
\subsection{Galois Representations (Taniyama, Shimura, Mumford-Tate)}
341
Let $E/\bQ$ be an elliptic curve and $G=\gal(\overline{\bQ}/\bQ)$.
342
Then $E[n]=\{P\in E(\overline{\bQ}) : nP = 0\} \isom (\bZ / n\bZ)^2$
343
is acted on by $G$ and this action respects the group operation
344
so we have a Galois representation
345
\begin{equation*}
346
G\xrightarrow{\rho}\aut(E[n])\isom \gl_2(\bZ/n\bZ)
347
\end{equation*}
348
Let $K$ be the fixed field of $\ker \rho$ (note that $K$ is a number
349
field), then since $\gal(K/\bQ)\isom G/\ker\rho \isom \imag \rho \subseteq
350
\gl_2(\bZ/n\bZ)$ we obtain many subgroups of $\gl_2(\bZ/n\bZ)$ as Galois
351
groups. Shimura\index{Shimura} showed that if we start with the elliptic curve
352
\begin{equation*}
353
E: \quad y^2+y = x^3-x^2
354
\end{equation*}
355
then the image of $\rho$ is often all of $\gl_2(\bZ/n\bZ)$
356
and the image is ``most'' of $\gl_2(\bZ/n\bZ)$ when $E$
357
does not have complex multiplication.
358
359
\section{Modular Forms and Galois Representations}
360
361
362
\subsection{Cusp Forms}
363
\index{cusp forms}
364
Let $S_k(N)$ denote the space of cusp forms
365
of weight $k$ and level
366
$N$ on the congruence subgroup\index{congruence subgroup} $\Gamma_1(N)=\bigl\{\bigl(\begin{smallmatrix}
367
a&b\\
368
c&d
369
\end{smallmatrix}\bigr) \in \sl2z :
370
a\equiv 1 \pmod N, c\equiv 0 \pmod N, d\equiv 1 \pmod N \bigr\}$.
371
Thus $S_k(N)$ is the finite dimensional vector space consisting of
372
all holomorphic functions $f(z)$ on $\sH=\{z\in\bC : \imag(z)>0\}$
373
vanishing at $\infty$ and satisfying
374
\begin{equation*}
375
f(\frac{az+b}{cz+d})=(cz+d)^k f(z) \quad\text{for all}
376
\quad \bigl ( \begin{smallmatrix} a&b\\c&d\end{smallmatrix}\bigr)
377
\in \Gamma_1(N).
378
\end{equation*}
379
Since, in particular, $f(z)=f(z+1)$ we can
380
expand $f(z)$ as a $q$-series (this requires rigorous justification)
381
\begin{equation*}
382
f(z) = \sum_{n=1}^{\infty} c_n q^n.
383
\end{equation*}
384
385
A famous example is
386
\begin{equation*}
387
\Delta = q\prod_{n=1}^{\infty}(1-q^n)^{24} = \sum_{n=1}^{\infty} \tau(n) q^n
388
\end{equation*}
389
$\tau$ is called the Ramanujan function.
390
One now knows that $\tau$ is multiplicative and satisfies
391
$\tau(p^{\nu})=\tau(p)\tau(p^{\nu})-p^{11}\tau(p^{\nu-1})$.
392
$\Delta$ is a normalized basis for $S_1(1)$.
393
394
\subsection{Hecke Operators (Mordell)}
395
396
Mordell defined, for $n\geq 1$, operators $T_n$ on $S_k(N)$ called
397
{\em Hecke operators}\index{Hecke operator}.
398
These proved very fruitful. The set of such
399
operators forms an ``almost'' commuting family of endomorphisms and
400
is hence ``almost'' simultaneously diagonalizable. The precise meaning
401
of ``almost'' and the actual structure of the Hecke algebra
402
$\bQ[T_1,\ldots,T_n,\ldots]$
403
will be studied in greater detail in the remainder of this course.
404
Often there will exist a basis of cusp forms $f = \sum_{n=1}^{\infty} c_n q^n
405
\in S_k(N)$ so that $f_n$ is a simultaneous eigenvector for all of the
406
Hecke operators $T_n$ and, in fact, $T_n f = c_n f$. All of
407
the $c_n$ will be algebraic integers and the field
408
$\bQ(c_1,c_2,\ldots)$ will be finite over $\bQ$.
409
410
A good claim can be made that the $c_n$ are often interesting integers
411
because they exhibit remarkable properties. For example,
412
$\tau(n) \equiv \sum_{d|n}d^{11} \pmod {691}$. How can we study the $c_n$?
413
How can we interpret the $c_n$? We can do this by studying the connection
414
between Galois representations and modular forms. In 1968 work was originally
415
begun on this by Serre\index{Serre}, Shimura, Eichler and Deligne.
416
\index{Shimura}\index{Eichler}\index{Deligne}
417
418
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
419
420
%% Lecture 2, 1/19/96
421
422
\chapter{Modular Representations and Curves}
423
424
\section{Arithmetic of Modular Forms}
425
Suppose $f=\sum_{n=1}^{\infty}a_n q^n$ is a cusp form in $S_k(N)$ which is
426
an eigenform for the Hecke operators. Then the Mellin transform associates
427
to $f$ the $L$-function $L(f,z)=\sum_{n=1}^{\infty} \frac{a_n}{n^s}$.
428
Let $K=\bQ(a_1,a_2,\ldots)$, then one can show that the $a_n$ are algebraic
429
integers and $K$ is a number field. When $k=2$ Shimura\index{Shimura}
430
associates to $f$ an abelian variety $A_f$ over $\bQ$ of
431
dimension $[K:\bQ]$ on which $K$ acts (see
432
theorem 7.14 of \cite{shimura1}).
433
434
\begin{example}[Modular Elliptic Curves]
435
436
When all of the coefficients $a_n$ of the modular form $f$ lie in $\bQ$
437
then $[K:\bQ]=1$ so $A_f$ is a one dimensional abelian variety.
438
A one dimensional abelian variety of nonzero genus is an elliptic curve.
439
An elliptic curve isogenous to one arising via this construction is
440
called {\em modular}.
441
\end{example}
442
443
\begin{defn}
444
445
Elliptic curves $E_1$ and $E_2$ are {\em isogenous} if there is
446
a morphism $E_1\into E_2$ of algebraic groups, which has a
447
finite kernel.
448
\end{defn}
449
450
The following conjecture motivates much of the theory.
451
452
\begin{conj}
453
Every elliptic curve over $\bQ$ is modular,
454
that is, isogenous to a curve constructed in the above way.
455
\end{conj}
456
457
For $k\geq 2$ Serre\index{Serre} and Deligne found a way to associate to $f$ a family
458
of $\ell$-adic representations. Let $\ell$ be a prime number and $K$ be as
459
above, then it is well known that $K\tensor_{\bQ} \bQ_{\ell}\isom
460
\prod_{\lambda|\ell}K_{\lambda}$. One can associate to $f$ a representation
461
\begin{equation*}
462
\rholf:G=\gal(\overline{\bQ}/\bQ)
463
\rightarrow\gl(K\tensor_{\bQ}\bQ_{\ell})
464
\end{equation*}
465
unramified at all primes $p\nd \ell N$.
466
For $\rholf$ to be unramified we mean that for all primes $P$ lying over $p$,
467
the inertia group of the decomposition group at $P$ is contained
468
in the kernel of $\rholf$. The decomposition group $D_P$ at $P$ is the
469
set of those $g\in G$ which fix $P$. Let $k$ be the residue
470
field $\sO/P$ where $\sO$ is the ring of all algebraic integers.
471
Then the inertia group $I_P$ is the kernel of the map $D_P\rightarrow
472
\gal(\overline{k}/k)$.
473
474
Now $I_P\subset D_P \subset \gal(\overline{\bQ}/\bQ)$ and
475
$D_P / I_P$ is cyclic (being isomorphic to a subgroup of the
476
Galois group of a finite extension of finite fields)
477
so it is generated by a Frobenious automorphism $\frob_p$ lying over $p$.
478
One has
479
\begin{align*}
480
\tr(\rholf(\frob_p))& = a_p\in K \subset K\tensor \bQ_{\ell}\\
481
&\text{and}\\
482
\det(\rholf) &= \chi_{\ell}^{k-1}\varepsilon
483
\end{align*}
484
where $\chi_{\ell}$ is the $\ell$th cyclotomic character and
485
$\varepsilon$ is the Dirichlet character associated to $f$.
486
There is an incredible amount
487
of ``abuse of notation'' packed into this statement. First, the Frobenius
488
$\frob_P$ (note $P$ not $p$) is only well defined in $\gal(K/\bQ)$
489
(so I think an unstated result is that $K$ must be Galois), and
490
then $\frob_p$ is only well defined up to conjugacy.
491
But this works out since $\rholf$ is
492
well-defined on $\gal(K/\bQ)$
493
(it kills $\gal(\overline{\bQ}/K)$) and the trace
494
is well-defined on conjugacy classes
495
($\tr(AB)=\tr(BA)$ so $\tr(ABA^{-1})=Tr(B)$).
496
497
498
\section{Characters}
499
Let $f\in S_k(N)$, then for all
500
$\bigl(\begin{smallmatrix} a&b\\c&d\end{smallmatrix}\bigr)
501
\in \sl2z$ with $c\equiv 0 \mod{N}$ we have
502
\begin{equation*}
503
f(\frac{az+b}{cz+d}) = (cz+d)^k \varepsilon(d) f(z)
504
\end{equation*}
505
where $\varepsilon:(\bZ/N\bZ)^*\rightarrow \bC^*$
506
is a Dirichlet character mod $N$. If $f$ is an eigenform for
507
the so called ``diamond-bracket operator'' $\dbd{d}$ so that
508
$f|\dbd{d} = \varepsilon(d) f$
509
then $\varepsilon$ actually takes values in $K$.
510
511
Led $\varphi_N$ be the mod $N$ cyclotomic character so that
512
$\varphi_N: G \rightarrow (\bZ/N\bZ)^*$ takes $g\in G$ to
513
the automorphism induced by $g$ on the $N$th cyclotomic
514
extension $\bQ(\Mu_N)$ of $\bQ$ (where we identify
515
$\gal(\bQ(\Mu_N)/\bQ)$ with $(\bZ/N\bZ)^*$).
516
Then what we called $\varepsilon$ above in the formula
517
$\det(\rho_{\ell})=\chi_{\ell}^{k-1}\varepsilon$
518
is really the composition
519
\begin{equation*}
520
G\xrightarrow{\varphi_N}(\bZ/N\bZ)^*\xrightarrow{\varepsilon} \bC^*.
521
\end{equation*}
522
523
For each positive integer $\nu$ we consider the $\ell^{\nu}$th
524
cyclotomic character on $G$,
525
\begin{equation*}
526
\varphi_{\ell^{\nu}}:G\rightarrow (\bZ/\ell^{\nu}\bZ)^*.
527
\end{equation*}
528
Putting these together gives the $\ell$-adic cyclotomic character
529
$$\chi_{\ell}:G\into\bZ_{\ell}^{*}.$$
530
531
\section{Parity Conditions}
532
533
Let $c\in\gal(\overline{\bQ}/\bQ)$ be complex conjugation.
534
Then $\varphi_N(c)=-1$ so $\varepsilon(c) = \varepsilon(-1)$ and
535
$\chi_{\ell}^{k-1}(c) = (-1)^{k-1}$. Now let
536
$\bigl(\begin{smallmatrix} a&b\\c&d\end{smallmatrix}\bigr)
537
=
538
\bigl(\begin{smallmatrix} -1&0\\0&-1\end{smallmatrix}\bigr)$,
539
then for $f\in S_k(N)$,
540
$$f(z) = (-1)^k\varepsilon(-1)f(z)$$
541
so $(-1)^k\varepsilon(-1) = 1$ thus
542
$$\det(\rholf(c)) = \epsilon(-1)(-1)^{k-1} = -1.$$
543
Thus the $\det$ character is odd so the representation
544
$\rholf$ is odd.
545
546
\begin{remark}[Vague Question] How can one recognize representations
547
like $\rholf$ ``in nature''? Mazur\index{Mazur} and Fontaine have made
548
relevant conjectures. The Shimura-Taniyama conjecture can be reformulated
549
by saying that for any representation $\rho_{\ell,E}$ comming
550
from an elliptic curve $E$ there is $f$ so that
551
$\rho_{\ell,E}\isom \rholf$.
552
\end{remark}
553
554
\section{Conjectures of Serre (mod $\ell$ version)}
555
\index{Serre}
556
Suppose $f$ is a modular form, $\ell\in\Z$ prime,
557
$\lambda$ a prime lying over $\ell$, and the representation
558
$$\rho_{\lambda,f}:G\rightarrow \gl_2(K_{\lambda})$$
559
(constructed by Serre-Deligne) is irreducible.
560
Then $\rho_{\lambda,f}$ is conjugate to a representation
561
with image in $\gl_2(\sO_{\lambda})$, where $\sO_{\lambda}$
562
is the ring of integers of $K_{\lambda}$.
563
Reducing mod $\lambda$ gives a representation
564
$$\overline{\rho}_{\lambda,f}:G\rightarrow\gl_2(\bF_{\lambda})$$
565
which has a well-defined trace and det, i.e., the det and trace
566
don't depend on the choice of conjugate representation used to
567
obtain the reduced representation.
568
One knows from representation theory that if
569
such a representation is semisimple then it is completely determined
570
by its trace and det (more precisely, the characteristic polynomials
571
of all of its elements -- see chapter ??).
572
Thus if $\overline{\rho}_{\lambda,f}$ is irreducible (and hence semisimple)
573
then it is unique in the sense that it does not depend on the choice
574
of conjugate.
575
576
\section{General remarks on mod $p$ Galois representations}
577
\index{Galois representations}
578
579
%% By Joe Wetherall
580
[[This section was written by Joseph Loebach Wetherell.]]
581
582
First, what are semi-simple and irreducible representations? Remember
583
that a representation $\rho$ is a map from a group $G$ to the endomorphisms of
584
some vector space $W$ (or a free module $M$ if we are working over a ring
585
instead of a field, but let's not worry about that for now). A subspace $W'$
586
of $W$ is said to be invariant under $\rho$ if $\rho$ takes $W'$ back into itself.
587
(The point is that if $W'$ is invariant, then $\rho$ induces representations on
588
both $W'$ and $W/W'$.) An irreducible representation is one where the only
589
invariant subspaces are ${0}$ and $W$. A semi-simple representation is one
590
where for every invariant subspace $W'$ there is a complementary invariant
591
subspace $W''$ -- that is, you can write $\rho$ as the direct sum of $\rho|_{W'}$
592
and $\rho|_{W''}$.
593
594
Another way to say this is that if $W'$ is an invariant subspace then we get a
595
short exact sequence $$0\into\rho|_{W/W'}\into\rho\into\rho|_{W'}\into 0.$$
596
Furthermore $\rho$ is
597
semi-simple if and only if every such sequence splits.
598
599
Note that irreducible representations are semi-simple.
600
601
One other fact is that semi-simple Galois representations
602
are uniquely determined (up to isomorphism class) by their trace
603
and determinant.
604
605
Now, since in the case we are doing, $G = \galq$ is
606
compact, it follows that the image of any Galois representation $\rho$ into
607
$\gl_2(K_{\lambda})$ is compact. Thus we can conjugate it into
608
$\gl_2(\sO_{\lambda})$. Irreducibility is not needed for this.
609
610
Now that we have a representation into $\gl_2(\sO_{\lambda})$, we can reduce
611
to get a representation $\overline{\rho}$ to $\gl_2(\bF_{\lambda})$. This
612
reduced representation is not uniquely determined by $\rho$, since we had a
613
choice of conjugators. However, the trace and determinant are invariant
614
under conjugation, so the trace and determinant of the reduced
615
representation are uniquely determined by $\rho$.
616
617
So we know the trace and determinant of the reduced representation. If we
618
also knew that it was semi-simple, then we would know its isomorphism class,
619
and we would be done. So we would be happy if the reduced representation is
620
irreducible. And in fact, it is easy to see that if the reduced
621
representation is irreducible, then $\rho$ must also be irreducible. Now, it
622
turns out that all $\rho$ of interest to us will be irreducible;
623
unfortunately, we can't go the other way and claim that $\rho$ irreducible
624
implies the reduction is irreducible.
625
626
\section{Serre's Conjecture}
627
\index{Serre}
628
Serre has made the following conjecture which is still open at
629
the time of this writing.
630
\begin{conj}[Serre]
631
All irreducible representation of
632
$G$ over a finite field which are odd, i.e., $det(\sigma(c))=-1$, $c$
633
complex conjugation, are of the form $\overline{\rho}_{\lambda,f}$
634
for some representation $\rho_{\lambda,f}$ constructed as above.
635
\end{conj}
636
637
\begin{example}
638
Let $E/\bQ$ be an elliptic curve and let
639
$\sigma_{\ell}:G\rightarrow\gl_2(\bF_{\ell})$ be
640
the representation induced by the action of $G$
641
on the $\ell$-torsion of $E$. Then $\det \sigma_{\ell} = \varphi_{\ell}$
642
is odd and $\sigma_{\ell}$ is usually irreducible, so
643
Serre's conjecture\index{Serre's conjecture}
644
would imply that $\sigma_{\ell}$ is modular. From this one can, assuming
645
Serre's conjecture, prove that $E$ is modular.
646
\end{example}
647
648
\begin{defn}
649
Let $\sigma:G\rightarrow \gl_2(\bF)$ ($\bF$ is a finite field)
650
be a represenation of the Galois group $G$. The we say that the
651
{\em representions $\sigma$ is
652
modular} if there is a modular form $f$, a prime $\lambda$, and an embedding
653
$\bF\hookrightarrow \overline{\bF}_{\lambda}$ such that
654
$\sigma\isom\overline{\rho}_{\lambda,f}$ over
655
$\overline{\bF}_\lambda$.
656
\end{defn}
657
658
\section{Wiles' Perspective}
659
660
Suppose $E/\bQ$ is an elliptic curve and
661
$\rho_{\ell,E}:G\rightarrow\gl_2(\bZ_{\ell})$
662
the associated $\ell$-adic representation on the
663
Tate module $T_{\ell}$. Then by reducing
664
we obtain a mod $\ell$ representation
665
$$\overline{\rho}_{\ell,E}=\sigma_{\ell,E}:G
666
\rightarrow \gl_2(\bF_{\ell}).$$
667
If we can show this representation is modular for infinitely many $\ell$
668
then we will know that $E$ is modular.
669
670
\begin{thm}[Langland's and Tunnel]
671
If $\sigma_{2,E}$ and $\sigma_{3,E}$ are irreducible, then they
672
are modular.
673
\end{thm}
674
675
This is proved by using the fact that $\gl_2(\bF_2)$ and
676
$\gl_2(\bF_3)$ are solvable so we may apply ``base-change''.
677
678
\begin{thm}[Wiles]
679
If $\rho$ is an $\ell$-adic representation which is irreducible
680
and modular mod $\ell$ with $\ell>2$ and certain other reasonable
681
hypothesis are satisfied, then $\rho$ itself is modular.
682
\end{thm}
683
684
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
685
686
%% Lecture 3, 1/21/96
687
688
\chapter{Modular Forms}
689
690
Our goal is to explain modular forms
691
as functions of lattices
692
or of elliptic curves. Good references are Serre \cite{serre2}
693
\index{Serre}
694
and Katz \cite{antwerp}.
695
696
697
698
\section{Cusp Forms}
699
First suppose $N=1$, then we must define $S_k=S_k(1)$. Let
700
$\Gamma_1(1)=\sl2z$, then $S_k$ consists of all functions
701
$f$ holomorphic on the upper half plane $\sH$ and such that for all
702
$\abcd\in\sl2z$
703
one has
704
$$f(\frac{a\tau+b}{c\tau+d})=(c\tau+d)^k f(\tau),$$
705
and $f$ vanishes at infinity.
706
Thus, in particular, $f(\tau+1)=f(\tau)$ and so $f$ passes
707
to a well defined function of $q=e^{2\pi i\tau}$. So $f(q)$
708
is a function on $\{z:0<|z|<1\}$ and the condition that $f(\tau)$
709
vanishes at infinity is that $f(q)$ extends to a holomorphic
710
function on $\{z:|z|<1\}$ and $f(0)=0$. In this case, we may
711
write $f(q)=\sum_{n=1}^{\infty}a_n q^n$.
712
713
\section{Lattices}
714
A lattice $L\subset \bC$ is a subring $L=\bZ\omega_1 +\bZ\omega_2$
715
for which $\omega_1, \omega_2\in \bC$ are lineary independent over $\bR$.
716
Without loss, we may assume that $\omega_1 / \omega_2 \in \sH$.
717
718
Let
719
$$\sR=\{\text{lattices in $\bC$}\}=\{(E,\omega):
720
\text{ $E$ is an elliptic curve, } \omega\in\Omega_E^{1}\}$$
721
722
and
723
$$M=\{(\omega_1,\omega_2):\omega_1,\omega_2\in\bC,
724
\imag(\omega_1/\omega_2)>0\}.$$
725
There is a left action of $\sl2z$ on $M$
726
$$\abcd:(\omega_1,\omega_2)\mapsto
727
(a\omega_1+b\omega_2,c\omega_1+d\omega_2)$$
728
and $\sl2z\backslash M\isom \sR$.
729
730
\section{Relationship With Elliptic Curves}
731
732
There is a map $L\mapsto \bC/L$ from lattices to
733
complex tori which, by Weierstrass theory, correspond
734
to elliptic curves defined over $\bC$ along with
735
a distinguished differential $\omega=dz$. % huh?
736
737
Conversely, if $E/\bC$ is an elliptic curve, we can
738
obtain the corresponding lattice by fixing a differential
739
$\omega$ and taking the lattice to be the image of the map
740
$$H_1(E(\bC),\bZ)\xrightarrow{\text{integration}}\bC$$
741
which takes $\gamma\in H_1$ to $\int_{\gamma}\omega\in\bC$.
742
743
There is a map $M/\bC\into\sH$ defined by $(\omega_1,\omega_2)
744
\mapsto \omega_1/\omega_2$. This gives an isomorphism
745
$$\sR/\bC^*=(\sl2z\backslash M)/\bC^* \xrightarrow{ ~ } \sl2z\backslash\sH$$
746
and
747
$$\sR/\bC^*=\{\text{ elliptic curves /$\bC$ (without differentials)}\}.$$
748
749
750
If $f:\sH\into\bC$ we define $F:M\into\bC$ by
751
$F(\omega_1,\omega_2)=f(\omega_1/\omega_2)$. Suppose now
752
that $F$ is a lattice function and sattisfies the homogeneity % (spelling!?)
753
condition $F(\lambda L)=\lambda^{-k} F(L)$.
754
Then
755
\begin{align*}
756
f(\frac{a\tau+b}{c\tau+d})&=F(\bZ\frac{a\tau+b}{c\tau+d}+\bZ)\\
757
&= F((c\tau+d)^{-1}(\bZ(a\tau+b)+\bZ(c\tau+d)))\\
758
&= (c\tau+d)^k F(\bZ(a\tau+b)+\bZ(c\tau+d))\\
759
&= (c\tau+d)^k F(\bZ+\tau\bZ)\\
760
&= (c\tau+d)^k f(\tau)
761
\end{align*}
762
so functions of lattices with the homogeneity condition come
763
from functions $f\in M_k$. Thus, if $f\in M_k$ and $F$ is the corresponding
764
lattice function then
765
$$F(\bZ\omega_1+\bZ\omega_2)=F(\omega_2(\bZ+\bZ \frac{\omega_1}{\omega_2}))
766
=\omega_2^{-k}F(\bZ+\bZ
767
\frac{\omega_1}{\omega_2})
768
=\omega_2^{-k}f(
769
\frac{\omega_1}{\omega_2}),
770
$$
771
so we can recover $F$ from $f$.
772
773
%% I don't understand the stuff with H^0(E,\Omega_E^{1}) being a
774
%% 1 dimensional vector space...
775
776
\section{Hecke Operators}
777
778
Define a map $T_n$ from the free abelian group generated by all
779
$\bC$-lattices
780
into itself by
781
$$T_n(L)=
782
\sum_{(L:L')=n} L'.$$
783
Then if $F$ is a function on lattices define $T_nF$ by
784
$$(T_nF)(L)=n^{k-1}\sum_{(L:L')=n}F(L').$$
785
786
Since $(n,m)=1$ implies $T_nT_m=T_{nm}$ and $T_{p^k}$ is a polynomial
787
in $\bZ[T_p]$ the essential case to consider is $n$ prime.
788
789
Suppose $L'\subset L$ with $(L:L')=n$, then $L/L'$ is killed by $n$
790
so $nL\subset L'\subset L$ and
791
$$L'/nL\subset L/nL\isom (\bZ/n\bZ)^2.$$
792
Thus the subgroups of $(\bZ/n\bZ)^2$ of order $n$ correspond to
793
the sublattices $L'$ of $L$ of index $n$. When $n=\ell$ is prime
794
there are $\ell+1$ such subgroups. (The subgroups correspond to
795
nonzero vectors in $\bF_{\ell}$ modulo scalar equivalence and
796
there are $\frac{\ell^2-1}{\ell-1}$ of them.)
797
798
Suppose $L'\subset L$ is a sublattice of index $\ell$ and let
799
$L''=\ell^{-1}L'$. Note that $\ell L\subset L'$ so
800
$L \subset \ell^{-1} L'=L''$ and $L$ is a sublattice of $L''$ of
801
index $\ell$. Thus, assuming $F$ satisfies the homogeneity condition
802
$F(\lambda L)=\lambda^{-k}F(L)$,
803
$$\ell^{k-1}\sum_{L'}F(L') = \frac{1}{\ell}\sum_{L''}F(L'')$$
804
which helps explain the extra factor of $n^{k-1}$ in our
805
definition of $T_n F$ -- we are ``averaging'' over the sublattices
806
(note that there are $\ell+1$ terms yet we divide by $\ell$ so
807
we aren't exactly averaging).
808
809
We now give a geometric description of the $\ell$th Hecke operator\index{Hecke operator}.
810
Let $L\subset L''$ be lattices with $(L'':L)=\ell$ and let
811
$E=\bC/L$, $E''=\bC/L''$ be the elliptic curves corresponding
812
to $L$, $L''$, respectively. Then $E[\ell]=\frac{1}{\ell}L/L$
813
contains $H=L''/L$ which may be thought of as a line
814
[Ed: I don't know why!]. Then the Hecke operator is
815
$$E\mapsto \frac{1}{\ell}\sum_{\text{lines }H\subset E[\ell]} E/H.$$
816
Let $\hat{\pi}$ be the isogeny dual to $\pi:E\into E/H$.
817
Then in terms of pairs $(E,\omega)$ we have
818
$$(E,\omega)\mapsto
819
\frac{1}{\ell} \sum_{H\subset E[\ell], \#H=\ell}(E/H,\pi_{*}\omega)
820
=\ell^{k-1}\sum_{H\subset E[\ell]} (E/H,\hat{\pi}^{*}(\omega)).$$
821
822
% Lecture 4, 1/24/96
823
824
We consider modular forms $f$ on $\Gamma_1(1)=\sl2z$, that
825
is, holomorphic functions on $\sH\cup\{\infty\}$ which satisfy
826
$$f(\tau)=f(\frac{a\tau+b}{c\tau+d})(c\tau+d)^{-k}$$
827
for all $\abcd\in\sl2z$. Using a Fourier expansion we write
828
$$f(\tau)=\sum_{n=0}^{\infty} a_ne^{2\pi i\tau n},$$
829
and say $f$ is a cusp form if $a_0=0$.
830
There is a correspondence between modular forms $f$ and
831
lattice functions $F$ satisfying $F(\lambda L)=\lambda^{-k}F(L)$
832
given by $F(\bZ\tau+\bZ)=f(\tau)$.
833
834
\section{Explicit Description of Sublattices}
835
The $n$th Hecke operator $T_n$ of weight $k$ is defined by
836
$$T_n(L)=n^{k-1}\sum_{\substack{L'\subset L\\(L:L')=n}} L'.$$
837
What are the $L'$ explicitly? Note that $L/L'$ is a group of order $n$ and
838
$$L'/nL\subset L/nL=(\bZ/n\bZ)^2.$$
839
Write $L=\bZ\omega_1+\bZ\omega_2$, let $Y_2$ be the cyclic subgroup
840
of $L/L'$ generated by $\omega_2$ and let $d=\#Y_2$. Let
841
$Y_1=(L/L')/Y_2$, then $Y_1$ is generated by the image
842
of $\omega_1$ so it is a cyclic group of order $a=n/d$.
843
We want to exhibit a basis of $L'$. Let
844
$\omega_2'=d\omega_2\in L'$ and use the fact that $Y_1$ is
845
generated by $\omega_1$ to write $a\omega_1=\omega_1'+b\omega_2$
846
for some integer $b$ and some $\omega_1'\in L'$. Since $b$ is only
847
well-defined modulo $d$ we may assume $0\leq b\leq d-1$.
848
Thus
849
$$
850
\Bigl(\begin{matrix}\omega_1'\\ \omega_2'\end{matrix}\Bigr)
851
=
852
\Bigl(\begin{matrix}a&b\\0&d\end{matrix}\Bigr)
853
\Bigl(\begin{matrix}\omega_1\\ \omega_2\end{matrix}\Bigr)
854
$$
855
and the change of basis matrix has determinent $ad=n$.
856
Since
857
$$\bZ\omega_1'+\bZ\omega_2'\subset L' \subset L=\bZ\omega_1+\bZ\omega_2$$
858
and $(L:\bZ\omega_1'+\bZ\omega_2')=n$ (since the change of basis matrix has
859
determinent $n$) and $(L:L')=n$ we see that $L'=\bZ\omega_1'+\bZ\omega_2'$.
860
861
Thus there is a one-to-one correspondence between sublattices $L'\subset L$
862
of index $n$ and matrices
863
$\bigl(\begin{smallmatrix}a&b\\0&d\end{smallmatrix}\bigr)$
864
with $ad=n$ and $0\leq b\leq d-1$.
865
In particular, when $n=p$ is prime there $p+1$ of these. In general, the
866
number of such sublattices equals the sum of the positive divisors
867
of $n$.
868
869
\section{Action of Hecke Operators on Modular Forms}
870
Now assume $f(\tau)=\sum_{m=0}^{\infty} c_m q^m$ is a modular
871
form with corresponding lattice function $F$. How can we describe the
872
action of the Hecke operator $T_n$ on $f(\tau)$? We have
873
\begin{align*}
874
T_nF(\bZ\tau+\bZ) & = n^{k-1}\sum_{\substack{a,b,d\\ ab=n\\ 0\leq b<d}}
875
F((a\tau+b)\bZ + d\bZ)\\
876
& = n^{k-1}\sum d^{-k} F(\frac{a\tau+b}{d}\bZ+\bZ)\\
877
& = n^{k-1}\sum d^{-k} f(\frac{a\tau+b}{d})\\
878
& = n^{k-1}\sum_{a,d,b,m} d^{-k}c_m e^{2\pi i(\frac{a\tau+b}{d})m}\\
879
& = n^{k-1}\sum_{a,d,m} d^{1-k}c_m e^{\frac{2\pi i a m \tau}{d}}
880
\frac{1}{d}\sum_{b=0}^{d-1} (e^{\frac{2\pi i m}{d}})^b\\
881
& = n^{k-1}\sum_{\substack{ad=n\\m'\geq 0}}d^{1-k} c_{dm'}e^{2\pi i a m' \tau}\\
882
& = \sum_{\substack{ad=n\\m'\geq 0}} a^{k-1} c_{dm'}q^{am'}.
883
\end{align*}
884
In the second to the last expression we
885
let $m=dm'$, $m'\geq 0$, then used the fact that the
886
sum
887
$\frac{1}{d}\sum_{b=0}^{d-1} (e^{\frac{2\pi i m}{d}})^b$
888
is only nonzero if $d|m$.
889
890
Thus
891
$$T_nf(q)=\sum_{\substack{ad=n\\m\geq 0}} a^{k-1}c_{dm} q^{am}$$
892
and if $\mu\geq 0$ then the coefficient of $q^{\mu}$ is
893
$$\sum_{\substack{a|n\\ a|\mu}}a^{k-1}c_{\frac{n\mu}{a^2}}.$$
894
895
\begin{remark}
896
When $k\geq 1$ the coefficients of $q^{\mu}$ for all $\mu$ belong
897
to the $\bZ$-module generated by the $c_m$.
898
\end{remark}
899
900
\begin{remark}
901
Setting $\mu=0$ gives the constant coefficient of $T_n f$ which is
902
$$\sum_{a|n}a^{k-1}c_0 = \sigma_{k-1}(n)c_0.$$
903
Thus if $f$ is a cusp form so is $T_nf$. ($T_nf$ is holomorphic
904
since its original definition is as a finite sum of holomorphic
905
functions.)
906
\end{remark}
907
908
\begin{remark}
909
Setting $\mu=1$ shows that the coefficient of $q$ in $T_n f$ is
910
$\sum_{a|1}1^{k-1}c_n=c_n$. As an immediate corollary we have the
911
following important result.
912
\end{remark}
913
914
\begin{cor}
915
Suppose $f$ is a cusp form for which $T_n f$ has 0 as coefficient
916
of $q$ for all $n\geq 1$, then $f=0$.
917
\end{cor}
918
919
\begin{remark}
920
When $n=p$ is prime we get an interesting formula for the
921
action of $T_p$ on the $q$-expansion of $f$.
922
One has
923
$$T_p f = \sum_{\mu\geq 0} \sum_{\substack{a|n\\a|\mu}}a^{k-1}
924
c_{\frac{n\mu}{a^2}} q^{\mu}. $$
925
Since $n=p$ is prime either $a=1$ or $a=p$. When
926
$a=1$, $c_{p\mu}$ occurs in the coefficient of $q^{\mu}$
927
and when $a=p$, we can write $\mu=p\lambda$ and we get
928
terms $p^{k-1}c_{\lambda}$ in $q^{\lambda p}$.
929
Thus
930
$$T_n f = \sum_{\mu\geq 0}c_{p\mu}q^{\mu}+
931
p^{k-1}\sum_{\lambda\geq 0} c_{\lambda}q^{p\lambda}.$$
932
\end{remark}
933
934
%% Lecture 5, 1/26/96
935
936
\chapter{Embedding Hecke Operators in the Dual}
937
938
\section{The Space of Modular Forms}
939
Let $\Gamma=\Gamma_1(1)=\sl2z$ and for $k\geq 0$
940
let
941
\begin{align*}
942
M_k&=\{f=\sum_{n=0}^{\infty}a_n q^n : \text{$f$ is a modular form
943
for $\Gamma$}\}\\
944
&\subset S_k=\{f=\sum_{n=1}^{\infty}a_n q^n\}\end{align*}
945
These are finite dimensional $\bC$-vector spaces whose dimensions
946
are easily computed. Furthermore, they are generated by familiar elements
947
(see Serre \cite{serre2} or Lang \cite{lang1}.)
948
\index{Serre}
949
The main tool is the formula
950
$$\sum_{p\in D\union\{\infty\}} \frac{1}{e(p)}\ord_p(f) = \frac{k}{12}$$
951
where $D$ is the fundamental domain for $\Gamma$ and
952
$$e(p)=\begin{cases} 1&\text{otherwise}\\
953
2&\text{if $p=i$}\\
954
3&\text{if $p=\rho$}
955
\end{cases}$$
956
One can alternatively define $e(p)$
957
as follows. If $p=\tau$ and $E=\bC/(\bZ\tau+\bZ)$
958
then $e(p)=\frac{1}{2}\#\aut(E)$.
959
960
For $k\geq 4$ we define the {\em Eisenstein series}\index{Eisenstein series} $G_k$ by
961
$$G_k(q)=\frac{1}{2}\zeta(1-k)+\sum_{n=1}^{\infty}\sigma_{k-1}(n)q^n,$$
962
then the map
963
$$\tau\mapsto\sum_{\substack{(m,n)\neq(0,0)\\m,n\in\bZ}}
964
\frac{1}{(m\tau+n)^k}$$
965
differs from $G_k$ by a constant (no proof).
966
Also, $\zeta(1-k)\in\bQ$ and one may say, {\em symbolically} at least,
967
``$\displaystyle \zeta(1-k)=\sum_{d=1}^{\infty} d^{k-1} =
968
\sigma_{k-1}(0)$.''
969
The {\em $n$th Bernoulli number $B_n$} is defined by the equation
970
$$\frac{x}{e^x-1}=\sum_{n=0}^{\infty} \frac{B_nx^n}{n!}.$$
971
One can show that $\zeta(1-k)=-\frac{B_k}{k}$ so the constant
972
%% How?? Reference???
973
coefficient of $G_k$ is $-\frac{B_k}{2k}$ which is rational.
974
975
\section{Inner Product}\index{inner product}
976
In what follows we assume $k\geq 2$ to avoid trivialities..
977
The Hecke operators $T_n$ acts on the space $M_k$. Fix a
978
subspace $V\subset M_k$ which is stable under the action
979
of the $T_n$. Let $\bT(V)$ be the $\bC$-algebra generated by
980
the endomorphism $T_n$ acting on $V$ and note that $\bT(V)$
981
is actually a finite dimensional $\bC$-vector space since it
982
is a subspace of $End(V)$ and $V$ is finite dimensional. Recall
983
that $\bT$ is commutative.
984
985
There is a bilinear form
986
\begin{align*}
987
\bT\times V &\into \bC \\
988
\langle T,f\rangle & \mapsto a_1(f|T)
989
\end{align*}
990
where $f|T=\sum_{n=0}^{\infty}a_n(f|T)q^n$.
991
We thus get maps
992
\begin{align*}
993
V\into\Hom(\bT,\bC)=\bT^{*}\\
994
\bT\into \Hom(V,\bC)=V^{*}.
995
\end{align*}
996
997
\begin{thm}
998
The above maps are isomorphisms.
999
\end{thm}
1000
\begin{proof}
1001
It just remains to show each map is injective.
1002
Then since a finite dimensional
1003
vector space and its dual have the same dimension the result follows.
1004
First suppose $f\mapsto 0\in\Hom(\bT,\bC)$, then
1005
$a_1(f|T)=0$ for all $T\in\bT$ so, in particular,
1006
$a_n=a_1(f|T_n)=0$ for all $n\geq 1$. Thus $f$ is a constant,
1007
but since $k\geq 2$ this implies $f=0$ (otherwise $f$ wouldn't
1008
transform correctly with respect to the action of the modular group).
1009
1010
Next suppose $T\mapsto 0\in \Hom(V,\bC)$, then
1011
$a_1(f|T)=0$ for all $f\in V$. Substiting $f|T_n$ for
1012
$f$ and using the commutativity of $\bT$ we have
1013
\begin{align*}
1014
a_1((f|T_n)|T)&=0 && \text{for all $f$, $n\geq 1$}\\
1015
a_1((f|T)|T_n)&=0 && \text{by commutativity}\\
1016
a_n(f|T)&=0 && \text{$n\geq 1$}\\
1017
f|T&=0 && \text{since $k\geq 2$, as above}
1018
\end{align*}
1019
Thus $T=0$ which completes the proof.
1020
\end{proof}
1021
1022
\begin{remark}
1023
The above isomorphisms are {\em $\bT$-equivariant}.
1024
$\Hom(\bT,\bC)$ is a $\bT$-module if we let $T\in\bT$ act
1025
on $\varphi\in\Hom(\bT,\bC)$ by
1026
$(T\cdot\varphi)(T')=\varphi(TT')$. If $\alpha:V\into\Hom(\bT,\bC)$
1027
is the above isomorphism
1028
(so $\alpha:f\mapsto\varphi_f:=(T'\mapsto a_1(f|T'))$)
1029
then equivariance is the statement that $\alpha(Tf)=T\alpha(f).$
1030
This follows since
1031
\begin{align*}
1032
\alpha(Tf)(T')&=\varphi_{Tf}(T')=a_1(Tf|T')=a_1(f|T'T)\\
1033
&=\varphi_{f}(T'T)=T\varphi(T')=T\alpha(f)(T').
1034
\end{align*}
1035
\end{remark}
1036
1037
\section{Eigenforms}
1038
\index{eigenforms}
1039
We continue to assume that $k\geq 2$.
1040
\begin{defn}
1041
A modular form $f\in M_k$ is an {\em eigenform for $\bT$} if
1042
$f|T_n=\lambda_n f$ for all $n\geq 1$ and some complex numbers $\lambda_n$.
1043
\end{defn}
1044
Let $f$ be an eigenform, then
1045
$a_n(f)=a_1(f|T_n)=\lambda_n a_1(f)$
1046
so if $a_1(f)=0$ then $a_n(f)=0$ for all $n\geq 1$ so
1047
since $k\geq 2$ this would imply $f=0$. Thus $a_1(f)\neq 0$
1048
and we may as well divide through by $a_1(f)$ to obtain
1049
the {\em normalized eigenform} $\frac{1}{a_1(f)}f$. We thus
1050
assume that $a_1(f)=1$, then the formula becomes $a_n(f)=\lambda_n$
1051
and so $f|T_n = a_n(f) f$, for all $n\geq 1$.
1052
1053
\begin{thm}
1054
Let $f\in V$ and let $\psi$ be the image of
1055
$f$ in $\Hom(\bT,\bC)$, thus $\psi(T)=a_1(f|T)$.
1056
Then $f$ is a normalized eigenform iff $\psi$ is a
1057
1058
ring homomorphism.
1059
\end{thm}
1060
1061
\begin{proof}
1062
First suppose $f$ is a normalized eigenform so $f|T_n=a_n(f)f$.
1063
Then
1064
\begin{align*}
1065
\psi(T_nT_m) &=a_1(f|T_nT_m)=a_m(f|T_n)\\
1066
&=a_m(a_n(f)f)=a_m(f)a_n(f)\\
1067
&=\psi(T_n)\psi(T_m),
1068
\end{align*}
1069
so $\psi$ is a homomorphism.
1070
1071
Conversely, assume $\psi$ is a homomorphism. Then
1072
$f|T_n=\sum a_m(f|T_n)q^m$, so to show that $f|T_n=a_n(f)f$
1073
we must show that $a_m(f|T_n)=a_n(f)a_m(f)$. Recall that %% remark 3.3
1074
$\psi(T_n)=a_1(f|T_n)=a_n$, thus
1075
\begin{align*}
1076
a_n(f)a_m(f) &= a_1(f|T_n)a_1(f|T_m) = \psi(T_n)\psi(T_m) \\
1077
& = \psi(T_n T_m) = a_1(f|T_n|T_m) \\
1078
& = a_m(f| T_n)
1079
\end{align*}
1080
as desired.
1081
1082
\end{proof}
1083
1084
%% Lecture 6, 1/29/96
1085
1086
\chapter{Rationality and Integrality Questions}
1087
1088
\section{Review}
1089
In the previous lecture we looked at subspaces
1090
$V \subset M_k \subset \bC[[q]]$, $(k\geq 4)$, and considered the
1091
space $\bT=\bT(V)=\bC[\ldots,T_n,\ldots]\subset\End_{\bsC}V$
1092
of Hecke operators on $V$. We defined a pairing
1093
$\bT\cross V\into \bC$ by $(T,f)\mapsto a_1(f|T)$ and
1094
showed this pairing is nondegenerate and that
1095
it induces isomorphisms
1096
$\bT\isom\Hom(V,\bC)$ and $V\isom\Hom(\bT,\bC)$.
1097
1098
\section{Integrality}
1099
Fix $k\geq 4$ and let $S=S_k$ be the space of weight $k$
1100
cusp forms with respect to the action of $\sl2z$. Let
1101
\begin{align*}
1102
S(\bQ)& =S_k\intersect \bQ[[q]]\\
1103
S(\bZ)& = S_k\intersect \bZ[[q]].
1104
\end{align*}
1105
1106
\begin{thm}
1107
There is a $\bC$-basis of $M_k$ consisting of forms
1108
with integral coefficients.
1109
\end{thm}
1110
\begin{proof}
1111
This is seen by
1112
exhibiting a basis. Recall that for all $k\geq 4$
1113
$$G_k=-\frac{b_k}{2k}+\sum_{k=1}^{\infty}\sum_{d|k}d^{k-1}q^n$$
1114
is the $k$th Eisenstein series\index{Eisenstein series} which is a modular form of weight $k$
1115
and
1116
$$E_k=-\frac{2k}{b_k}\cdot G_k=1+\cdots$$
1117
is its normalization. Since the Bernoulli numbers $b_2,\ldots,b_8$
1118
have $1$ as numerator (this isn't always the case,
1119
$b_{10}=\frac{5}{66}$) we see that $E_4$ and $E_6$ have coefficients
1120
in $\bZ$ and constant term $1$. Furthermore one shows by dimension and
1121
independence arguments that the forms
1122
$$\{E_4^aE_6^b|4a+6b=k\}$$
1123
form a basis for $M_k$.
1124
\end{proof}
1125
1126
\section{Victor Miller's Thesis}
1127
Let $d=\dim_{\bsC}S_k$, then Victor Miller showed in his thesis (see
1128
\cite{lang2}, ch. X, theorem 4.4) that there exists
1129
$$f_1,\ldots,f_d\in S_k(\bZ) \quad\text{such that}\quad a_i(f_j)=\delta_{ij}$$
1130
for $1\leq i,j\leq d$. The $f_i$ clearly form a basis.
1131
\begin{prop}
1132
Let $R=\bZ[\ldots,T_n,\ldots]\subset End(S_k)$, then
1133
$R=\bigoplus_{i=1}^{d} \bZ T_i$.
1134
\end{prop}
1135
1136
\begin{proof}
1137
To see that $T_1,\cdots,T_d\in \bT=\bT(S_k)$
1138
are linearly independent over $\bC$ suppose
1139
$\sum_{i=1}^{d} c_i T_i = 0$, then
1140
$$0=a_1(f_j|\sum c_i T_i)=\sum_{i}c_i a_i(f_j) =
1141
\sum_{i} c_i \delta_{ij} = c_j.$$ From the isomorphism
1142
$\bT\isom\Hom(S_k,\bC)$ we know that $\dim_{\bsC}\bT=d$,
1143
so we can write any $T_n$ as a $\bC$-linear combination
1144
$$T_n=\sum_{i=1}^{d}c_{n_i}T_i,\quad c_{n_i}\in\bC.$$
1145
But
1146
$$\bZ\ni a_n(f_j)=a_1(f_j|T_n)=\sum_{i=1}^{d}c_{n_i}a_1(f_j|T_i)
1147
=\sum_{i=1}^{d}c_{n_i}a_i(f_j) = c_{n_j}$$
1148
so the $c_{n_i}$ all lie in $\bZ$ which completes the proof.
1149
\end{proof}
1150
1151
Thus $R$ is an integral Hecke algebra of finite rank $d$ over $\bZ$.
1152
We have a map
1153
\begin{align*}
1154
S(\bZ)\cross R & \into \bZ \\
1155
(f,T) & \mapsto a_1(f|T)
1156
\end{align*}
1157
which induces an embedding
1158
$$S(\bZ)\hookrightarrow\Hom(R,\bZ)\isom \bZ^d.$$
1159
1160
\begin{exercise}
1161
Prove that the map $S(\bZ)\hookrightarrow\Hom(R,\bZ)$ is
1162
in fact an isomorphism of $\bT$-modules.
1163
[Hint: Show the cokernel is torsion free.]
1164
\end{exercise}
1165
1166
\section{Petersson Inner Product}\index{inner product}
1167
1168
The main theorem is
1169
\begin{thm}
1170
The $T_n\in\bT(S_k)$ are all diagonalizable over $\bC$.
1171
\end{thm}
1172
1173
To prove this we note that $S_k$ supports a non-degenerate positive definite
1174
Hermitean inner product (the Petersson inner product)
1175
$$(f,g)\mapsto\langle f,g\rangle\in\bC$$
1176
such that $\langle f|T_n,g\rangle =\langle f,g|T_n\rangle$.
1177
We need some background facts.
1178
1179
\begin{defn}
1180
An operator $T$ is {\em normal} if it commutes with its adjoint, thus
1181
$TT^{*}=T^{*}T$.
1182
\end{defn}
1183
$T_n$ is clearly normal since $T_n^{*}=T_n$,
1184
\begin{thm}
1185
A normal operator is diagonalizable.
1186
\end{thm}
1187
Thus each $T_n$ is diagonalizable.
1188
\begin{thm}
1189
A commuting family of semisimple (=diagonalizable) operators
1190
can be simultaneously diagonalized.
1191
\end{thm}
1192
Since the $T_n$ commute this implies $S_k$ has a basis consisting
1193
of normalized eigenforms $f$. Their eigenvalues are real since
1194
\begin{align*}
1195
a_n(f)\langle f,f\rangle &=\langle a_n(f)f,f\rangle =\langle f|T_n,f\rangle\\
1196
&=\langle f,a_n(f)f\rangle =\overline{a_n(f)}\langle f,f\rangle.
1197
\end{align*}
1198
\begin{exercise}
1199
The coefficients $a_n$ of the eigenforms are totally real algebraic integers.
1200
[Hint: The space $S_k$ is stable under the action of $\aut(\bC)$ on
1201
coefficients: if $f=\sum_{n=1}^{\infty}c_n q^n\in S_k$ and
1202
$\sigma\in\aut(\bC)$ then $\sigma(f)=\sum_{n=1}^{\infty}\sigma(c_n)q^n$
1203
is again in $S_k$ (check this by writing $f$ in terms of a basis
1204
$f_1,\ldots,f_d\in S(\bZ)$). Next use the fact that $f$ is an eigenform
1205
iff $\sigma(f)$ is an eigenform.]
1206
%%% I have absolutely no idea how to do this!!!
1207
\end{exercise}
1208
1209
Let
1210
$$\sH=\{x+iy : x, y\in \bR, \text{ and } y>0\}$$
1211
be the upper half plane. Then the volume form
1212
$\frac{dx\wedge dy}{y^2}$ is invariant under the action of
1213
$$\gl_2^{+}(\bR)=\{M\in\gl_2(\bR) | \det(M)>0\}.$$
1214
If $\alpha=\abcd\in\gl_2^{+}(\bR)$ then $\abcd$ acts on $\sH$ by
1215
$$\bigabcd:\quad z\mapsto\frac{az+b}{cz+d}$$
1216
and one has
1217
$$\imag(\frac{az+b}{cz+d})=\frac{\det(\alpha)}{|cz+d|^2}y.$$
1218
Differentiating $\frac{az+b}{cz+d}$ gives
1219
\begin{align*}
1220
d(\frac{az+b}{cz+d})&=
1221
\frac{a(cz+d)dz-c(az+b)dz}{(cz+d)^2}\\
1222
&=\frac{(ad-bc)dz}{(cz+d)^2}\\
1223
&= \frac{det(\alpha)}{(cz+d)^2}dz
1224
\end{align*}
1225
Thus, under the action of $\alpha$, $dz\wedge d{\overline z}$
1226
takes on a factor of
1227
$$\frac{\det(\alpha)^2}{(cz+d)^2(c\overline{z}+d)^2}
1228
=\Bigl(\frac{\det(\alpha)}{|cz+d|^2}\Bigr)^2.$$
1229
1230
\begin{defn}
1231
The {\em Petersson inner product} of forms $f,g\in S_k$ is defined by
1232
$$<f,g>=\int_{\Gamma\backslash\sH}(f(z)\overline{g(z)}y^k)
1233
\frac{dx\wedge dy}{y^2},$$
1234
where $\Gamma=\sl2z$.
1235
\end{defn}
1236
1237
Integrating over $\Gamma\backslash\sH$ can be taken to mean integrating
1238
over a fundamental domain for the action of $\sH$. Showing that the
1239
operators $T_n$ are self-adjoint with respect to the Petersson inner
1240
product is a harder computation than Serre \cite{serre2}
1241
\index{Serre}
1242
might lead one to believe --- it takes a bit of thought.
1243
1244
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
1245
%% Lecture 7, 1/31/96
1246
1247
\chapter{Modular Curves}
1248
\index{modular curves}
1249
% define various gammas
1250
\section{Cusp Forms}
1251
Recall that if $N$ is a positive integer we define the congruence
1252
subgroups
1253
$\Gamma(N)\subset\Gamma_1(N)\subset\Gamma_0(N)$ by
1254
\begin{align*}
1255
\Gamma_0(N) & = \{\abcd \in \modgp : c\equiv 0 \pmod{N}\}\\
1256
\Gamma_1(N) & = \{\abcd \in \modgp : a\equiv d\equiv 1, c\equiv 0 \pmod{N}\}\\
1257
\Gamma(N) & = \{\abcd \in \modgp : \abcd \equiv
1258
\bigl(\begin{smallmatrix}1&0\\0&1\end{smallmatrix}\bigr) \pmod{N}\}.
1259
\end{align*}
1260
1261
Let $\Gamma$ be one of the above subgroups.
1262
One can give a construction of the space $S_k(\Gamma)$ of cusp forms
1263
of weight $k$ for the action of $\Gamma$ using the language of
1264
algebraic geometry.
1265
Let $X_{\Gamma}=\overline{\Gamma\backslash\H^{*}}$
1266
be the compactifaction of the upper half plane (union the cusps)
1267
modulo the action of $\Gamma$. Then $X_{\Gamma}$ can be given the structure
1268
of Riemann surface and
1269
$S_2(\Gamma)=H^0(X_{\Gamma},\Omega^1)$ where
1270
$\Omega^1$ is the sheaf of differential 1-forms on $X_{\Gamma}$.
1271
This works since an element of $H^0(X_{\Gamma},\Omega^1)$
1272
is a differential form $f(z)dz$, holomorphic on $\H$ and
1273
the cusps, which is invariant with respect to the action
1274
of $\Gamma$. If $\gamma=\abcd\in\Gamma$ then
1275
$$d(\gamma(z))/dz=(cz+d)^{-2}$$
1276
so
1277
$$f(\gamma(z))d(\gamma(z))=f(z)dz$$
1278
iff $f$ satisfies the modular condition
1279
$$f(\gamma(z))=(cz+d)^{2}f(z).$$
1280
1281
There is a similar construction of $S_k$ for $k>2$.
1282
1283
\section{Modular Curves}\index{modular curves}
1284
One knows that $\sl2z\backslash\sH$ parameterizes isomorphism
1285
classes of elliptic curves. The other congruence subgroups also
1286
give rise to similar parameterizations.
1287
Thus $\Gamma_0(N)\backslash\sH$ parameterizes pairs $(E,C)$ where
1288
$E$ is an elliptic curve and $C$ is a cyclic subgroup of order
1289
$N$, and $\Gamma_1(N)\backslash\H$ parameterizes pairs $(E,P)$ where
1290
$E$ is an elliptic curve and $P$ is a point of exact order $N$.
1291
Note that one can also give a point of exact order $N$ by giving
1292
an injection $\bZ/N\bZ\hookrightarrow E[N]$
1293
or equivalently an injection $\Mu_N\hookrightarrow E[N]$
1294
where $\Mu_N$ denotes the $N$th roots of unity.
1295
$\Gamma(N)\backslash\sH$ parameterizes pairs $(E,\{\alpha,\beta\})$
1296
where $\{\alpha,\beta\}$ is a basis for
1297
$E[N]\isom(\bZ/N\bZ)^2$.
1298
1299
The above quotients spaces are called {\em moduli spaces} for the
1300
{\em moduli problem} of determining equivalence classes of
1301
pairs ($E + $ extra structure). \index{$\Gamma(N)$-structures}
1302
1303
\section{Classifying $\Gamma(N)$-structures}
1304
\begin{defn}
1305
Let $S$ be an arbitrary scheme. An {\bfseries elliptic curve}
1306
$E/S$ is a proper smooth curve
1307
$$\begin{CD} E \\ @VfVV \\ S\end{CD}$$
1308
with geometrically connected fibers all of genus one, give with a
1309
section ``0''.
1310
\end{defn}
1311
1312
Loosely speaking, proper is a generalization of projective
1313
and smooth generalizes nonsingularity. See
1314
Hartshorne \cite{hartshorne}, chapter III, section 10,
1315
for the precise definitions.
1316
1317
\begin{defn}
1318
Let $S$ be any scheme and $E/S$ an elliptic curve.
1319
A {\bfseries $\Gamma(N)$-structure} on $E/S$ is
1320
a group homomorphism
1321
$$\varphi:(\bZ/N\bZ)^2\into E[N](S)$$
1322
whose image ``generates'' $E[N](S)$.
1323
\end{defn}
1324
1325
A good reference is chapter 3 of Katz and Mazur \cite{katzmazur}.
1326
1327
Define a functor from the category of $\Q$-schemes to the
1328
category of sets by sending a scheme $S$ to the
1329
set of isomorphism classes of pairs
1330
$$(E, \Gamma(N)\text{-structure})$$\index{$\Gamma(N)$-structures}
1331
where $E$ is an elliptic curve defined over $S$ and
1332
isomorphisms (preserving the $\Gamma(N)$-structure) are taken
1333
over $S$. An isomorphism preserves the $\Gamma(N)$-structure
1334
if it takes the two distinguished generators to the two
1335
distinguished generators in the image (in the correct order).
1336
1337
\begin{thm}
1338
For $N\geq 4$ the functor defined above is representable and
1339
the object representing it is the modular curve $X$ corresponding
1340
to $\Gamma(N)$.
1341
\end{thm}
1342
1343
What this means is that given a $\Q$-scheme $S$, the
1344
set $X(S)=\Mor_{\Q\text{-schemes}}(S,X)$ is isomorphic to
1345
the image of the functor's value on $S$.
1346
1347
There is a natural way to map a pair $(E,\Gamma(N)\text{-structure})$
1348
\index{$\Gamma(N)$-structures}
1349
to an $N$th root of unity.
1350
If $P,Q$ are the distinguished basis of $E[N]$ we send
1351
the pair $(E,\Gamma(N)\text{-structure})$ to
1352
$$e_N(P,Q)\in\Mu_N$$
1353
where $e_N:E[N]\cross E[N]\into \Mu_N$ is the Weil pairing. For
1354
the definition of this pairing see chapter III, section 8 of
1355
Silverman \cite{silverman1}. The Weil pairing
1356
is bilinear, alternating, non-degenerate, Galois invariant, and
1357
maps surjectively onto $\Mu_N$.
1358
1359
%% 2/2/96
1360
\section{More on Integral Hecke Operators}
1361
1362
We are considering the algebra of integral Hecke operators $\T=\T_{\Z}$
1363
on the space of cusp forms $S_k(\C)$ with respect to the action
1364
of the full modular group $\sl2z$. Our goal is to see why
1365
$\T\isom\Z^d$ where $d=\dim_{\bsC}S_k(\C)$.
1366
1367
Suppose $A\subset\C$ is any {\em subring} of $\C$ and recall that
1368
$$\T_A=A[\ldots,T_n,\ldots]\subset \End_{\C}S_k.$$
1369
We have a natural map
1370
$$\T_A\tensor_A\C\into\T_C$$
1371
but we do not yet know that it is
1372
an isomorphism.
1373
1374
\section{Complex Conjugation}
1375
We have a conjugate linear map on functions
1376
$$f(\tau)\mapsto \overline{f(-\overline{\tau})}.$$
1377
Since $\overline{(e^{-2\pi i\overline{\tau}})}=e^{2\pi i \tau}$,
1378
it follows that
1379
$$\sum_{n=1}^{\infty} a_n q^n \mapsto
1380
\sum_{n=1}^{\infty}\overline{a_n}q^n$$
1381
so it is reasonable to call this map ``complex conjugation''. Furthermore,
1382
if we know that
1383
$$S_k(\C)=\C\tensor_{\bsQ}S_k(\Q)$$
1384
then it follows that complex conjugation
1385
takes $S_k(\C)$ into $S_k(\C)$. To see this note that if
1386
we have the above equality then every element of
1387
$S_k(\C)$ is a $\C$-linear combination of elements of
1388
$S_k(\Q)$ and conversely, and it is clear that the set of such
1389
$\C$-linear combinations is invariant under the action
1390
of complex conjugation.
1391
1392
\section{Isomorphism in the Real Case}
1393
1394
\begin{prop} $\T_{\R}\tensor_{\bsR}\C \isom \T_{\C}$, as $\C$-vector spaces.
1395
\end{prop}
1396
\begin{proof}
1397
Since $S_k(\R)=S_k(\C)\intersect\R[[q]]$ and since
1398
theorem 5.1 assures us that there is a $\C$-basis of
1399
$S_k(\C)$ consisting of forms with integral coefficients,
1400
we see that $S_k(\R)\isom \R^d$ where $d=\dim_{\bsC}S_k(\C)$.
1401
(Any element of $S_k(\R)$ is a $\C$-linear combination
1402
of the integral basis, hence equating real and imaginary
1403
parts, an $\R$-linear combination of the integral basis,
1404
and the integral basis stays independent over $\R$.)
1405
By considering the explicit formula for the action
1406
of the Hecke operators $T_n$ on $S_k$ (see section 3)
1407
we see that $\T_{\R}$ leaves $S_k(\R)$ invariant, thus
1408
$$\T_{\R}=\R[\ldots,T_d,\ldots]\subset \End_{\R}S_k(\R).$$
1409
In section 4 we defined a ``perfect'' pairing
1410
$$\T_{\C}\times S_k(\C)\into \C$$
1411
which allowed us to show that
1412
$\T_{\C}\isom S_k(\C).$ By restricting
1413
to $\R$ we again get a perfect pairing so we see that
1414
$\T_{\R}\isom S_k(\R) \isom {\R}^d$ which
1415
implies that
1416
$$\T_R\tensor_{\bsR}\C \xrightarrow{\sim}\T_{\C}.$$
1417
\end{proof}
1418
1419
This also shows that $S_k(\C)\isom \C\tensor_{\bsR} S_k(\R)$
1420
so we have complex conjugation over $\R$.
1421
%% For some reason I feel something circular is going on here, but
1422
%% i can't put my finger on it... :(
1423
1424
\section{The Eichler-Shimura Isomorphism}
1425
\index{Eichler-Shimura}
1426
1427
Our goal in this section is to outline a homological interpretation
1428
of $S_k$. For details see chapter 6 of Lang \cite{lang2},
1429
the original paper of Shimura \cite{shimura2},
1430
or chapter VIII of Shimura \cite{shimura1}.
1431
1432
How is $S_k(\C)$ sort of isomorphic to $H^1(X_{\Gamma},\R)$?
1433
Suppose $k=2$ and $\Gamma\subset\sl2z$ is a congruence subgroup,
1434
let $X_{\Gamma}=\overline{\Gamma\backslash\H}$ be the
1435
Riemann surface obtained by compactifying the upper half plane modulo
1436
the action of $\Gamma$. Then $S_k(\C)=H^0(X_{\Gamma},\Omega^{1})$ so
1437
we have a pairing
1438
$$H_1(X_{\Gamma},\Z)\cross S_k(\C)\into \C$$
1439
given by integration
1440
$$(\gamma,\omega)\mapsto \int_{\gamma}\omega.$$
1441
This gives an embedding
1442
$$\Z^{2d}\isom{}H_1(X_{\Gamma},\Z)\hookrightarrow
1443
\Hom_{\C}(S_k(\C),\C)\isom {\C}^d$$
1444
of a ``lattice'' in $\C^d$. (We say ``lattice'' since there
1445
were some comments by Ribet that $Z^{2d}$ isn't a lattice because the rank
1446
might be too small since a subring of $\C^d$ having $\Z$-rank $2d$
1447
might not spans $\C^d$ over $\C$).
1448
Passing to the quotient (and compactifying) gives a complex torus
1449
called the Jacobian\index{Jacobian} of $X_{\Gamma}.$
1450
Again using the above pairing we get an embedding
1451
$${\C}^d\isom S_k(\C)\hookrightarrow
1452
\Hom(H_1(X_{\Gamma},\Z),\C)\isom\C^{2d}$$
1453
which, upon taking the real part, gives
1454
$$
1455
S_k(\C) \into \Hom(H_1(X_{\Gamma},\Z),\R)
1456
\isom H^1(X_{\Gamma},\R) \isom H^1_p(\Gamma,\R)
1457
$$
1458
where $H^1_p(\Gamma,\R)$ denotes the {\em parabolic} group cohomology
1459
of $\Gamma$ with respect to the trivial action. It is this result, that we
1460
may view $S_k(\C)$ as the cohomology group $H^1_p(\Gamma,\R)$, that was
1461
alluded to above.
1462
1463
Shimura\index{Shimura} generalized this for arbitrary $k\geq 2$ so that
1464
$$S_k(\C)\isom{}H_p^1(\Gamma,V_k)$$
1465
where $V_k$ is a $k-1$ dimensional $\R$-vector space.
1466
The isomorphism is (approximately) the following:
1467
$f\in S_k(\C)$ is sent to the map
1468
$$\gamma\mapsto \real\int_{\tau_0}^{\gamma\tau_0}f(\tau){\tau}^{i}d\tau,\quad
1469
i=0,\ldots,k-2.$$
1470
Let $W=\R\oplus\R$, then $\Gamma$ acts on $W$ by
1471
$$\begin{pmatrix}a&b\\c&d\end{pmatrix}:
1472
\begin{pmatrix}x\\y\end{pmatrix}\mapsto
1473
\begin{pmatrix}ax+by\\cx+dy\end{pmatrix}
1474
$$
1475
so $\Gamma$ acts on
1476
$$V_k=\Sym^{k-2}W=W^{\otimes k-2}/S_{k-2}$$
1477
where $S_{k-2}$ is the symmetric group on $k-2$ symbols
1478
(note that $\dim V_k = k-1$).
1479
Let $$L=H_p^1(\Gamma,\Sym^{k-2}(\Z\oplus\Z))$$
1480
then under the isomorphism $$S_k(\C)\isom H^1_p(\Gamma,\R)$$
1481
$L$ is a sublattice of $S_k(\C)$ of $\Z$-rank 2 which
1482
is $T_n$-stable for all $n$. Thus we have an embedding
1483
$$\T_{\Z} = \T \hookrightarrow \End L$$
1484
and so
1485
$\T_{\R}\subset\End_{\bsR}(L\tensor\R)$
1486
and $\T_{\Z}\tensor_{\bsZ}\R\isom \T_{\R}$ which has rank $d$.
1487
1488
%% What is the point of this last bit??
1489
1490
1491
\section{The Petterson Inner Product is Hecke Compatible}\index{inner product}
1492
1493
\begin{thm}
1494
Let $\Gamma=\sl2z$, let $f,g\in S_k(\C)$, and let
1495
$$\langle f,g\rangle=\int_{\Gamma\backslash\H} f(\tau)\overline{g(\tau)}y^{k}
1496
\frac{dx dy}{y^2}.$$
1497
Then this integral is well-defined and Hecke compatible, that is,
1498
$\langle f|T_n,g\rangle=\langle f,g|T_n\rangle$ for all $n$.
1499
\end{thm}
1500
\begin{proof}
1501
See Chapter 3 of Lang \cite{lang2}.
1502
\end{proof}
1503
1504
%% 2/5/96
1505
1506
\chapter{Higher Weight Modular Forms}
1507
1508
We are considering the spaces $S_k(\C)$, $S_k(\Q)$ and $S_k(\Z)$
1509
which all have rank $d$. Each is acted upon by the Hecke algebra $\T$.
1510
We defined a Hecke compatible inner product (the Peterrson product)
1511
and used it to show that $$S_k(\Z)\isom\Hom_{\bsZ}(\T,\Z).$$
1512
1513
\section{Definitions of $\T$}
1514
1515
\begin{quote}
1516
``I may be asking you to explain something we have already discussed,
1517
but have we intrinsically defined the Hecke operators yet?''
1518
1519
-- Saul Schliemer
1520
\end{quote}
1521
1522
The $T_n$ are defined as operators on $S_k(\C)$ by defining their action
1523
on modular forms and noting from explicit formulas that $S_k(\C)$ is
1524
preserved. But the $T_n$ can be thought of in other ways, for example,
1525
since $$S_k(\C)\isom H_p^1(\Gamma,\Sym^{k-2}(\R\oplus\R))$$
1526
one may give an {\em explicit} description of the action of $T_n$ on
1527
$H_p^1$.
1528
1529
\section{Double Cosets}
1530
Let $p$ be a prime and
1531
$$\sM_p=\{\alpha\in M_2(\Z) : \det(\alpha)=p\}.$$
1532
Let $F:\sL\into\C$ be a function on the free abelian group of lattices
1533
and recall that $T_p$ acts on $F$ by
1534
$$(T_{p}F)(L)=p^{k-1}\sum_{\substack{L'\subset L\\(L:L')=p}}F(L').$$
1535
One can write $\sM_p$ as a disjoint union of left cosets,
1536
$$\sM_p=
1537
\sl2z \begin{pmatrix}p&0\\0&1\end{pmatrix} \sl2z
1538
= \bigcup_{\substack{ad=p\\0\leq b<d}}
1539
\begin{pmatrix}a&b\\0&d\end{pmatrix} \sl2z.$$
1540
Then $\sum_{(L:L')=p}L'$ may be thought of as the sum of the images
1541
of $L$ under the action of the left cosets of $\sM_p$.
1542
For a complete exposition, in greater generality, see Shimura\index{Shimura}
1543
\cite{shimura1}, especially chapter 3.
1544
1545
\section{More General Congruence Subgroups}
1546
\begin{defn}
1547
A {\bfseries Dirichlet character} mod $N$ is a homomorphism
1548
$$\varepsilon:(\Z/n\Z)^{*}\into\C^{*}$$
1549
extended to $\Z/N\Z$ by putting $\varepsilon(m)=0$ if
1550
$(m,N)\neq 1$.
1551
\end{defn}
1552
1553
Fix integers $k\geq 0$ and $N\geq 1$. In this section we consider the spaces
1554
$$S_k(\Gamma_1(N),\epsilon)$$
1555
for Dirichlet characters $\varepsilon$ mod $N$
1556
and explicitly describe the action of the Hecke operators
1557
$$\begin{cases}
1558
T_n,&n\geq{}1\\
1559
\dbd{d},&d\in(\Z/N\Z)^{*}
1560
\end{cases}
1561
$$
1562
on these spaces.
1563
1564
\begin{remark}
1565
Let $n$ be a positive integer. If $(n,N)=1$, then the $T_n$ behave
1566
like they do for $\sl2z$. In fact, the $T_n$ and $\dbd{d}$ commute and
1567
$$(f|T_n,g)=(f,g|\dbd{n}^{-1}T_n)$$
1568
$$(f|\dbd{d},g)=(f,g|\dbd{d}^{-1})$$
1569
so the $T_n$ (for $n$ prime to $N$) and $\dbd{d}$ are
1570
simultaneously diagonalizable. But if
1571
$(n,N)\neq{}1$ then $T_n$ may not be diagonalizable.
1572
\end{remark}
1573
1574
\begin{defn}
1575
Let
1576
$$S_k(\Gamma_1(N)) = \{ f : f(\gamma \tau) = (c\tau+d)^{k}f(\tau) \text{ all }
1577
\gamma \in \Gamma_1(N) \}$$
1578
where the $f$ are assumed holomorphic on $\sH\union\{\text{cusps}\}$.
1579
For each Dirichlet character $\varepsilon$ mod $N$ let
1580
$$S_k(\Gamma_1(N),\varepsilon)=\{ f :
1581
f(\gamma\tau)(c\tau+d)^{-k} = \varepsilon(d) f
1582
\text{ all } \gamma=\abcd \in \Gamma_0(N) \}.$$
1583
\end{defn}
1584
When $\varepsilon\neq 0$ and $f\in{}S_k(\Gamma_1(N),\varepsilon)$
1585
one calls $\varepsilon$ the {\bfseries nebentypus} of $f$.
1586
1587
Let $d\in(\Z/N\Z)^{*}$ and let $f\in S_k(\Gamma_1(N))$. Let
1588
$\gamma=\bigl(\begin{smallmatrix}a&b\\0&d\end{smallmatrix}\bigr)
1589
\in\Gamma_1(N)$ be a matrix
1590
whose lower right entry is congruent to $d$ mod $N$. Then we define
1591
$$f(\tau)|\dbd{d} = f(\gamma\tau)(c\tau+d)^{-k}.$$
1592
1593
Since $f|\dbd{d}=\varepsilon(d)f$, $S_k(\Gamma_1(N),\varepsilon)$ is
1594
the $\varepsilon(d)$ eigenspace of $\dbd{d}$ and $\dbd{d}$ is diagonalizable
1595
so one has a direct sum decomposition
1596
$$S_k(\Gamma_1(N))=\bigoplus_{\varepsilon:(\bsZ/N\bsZ)^{*}\into\bsC^{*}}
1597
S_k(\Gamma_1(N),\varepsilon).$$
1598
If $f\in{}S_k(\Gamma_1(N),\varepsilon)$ then
1599
$$\begin{pmatrix}-1&0\\0&-1\end{pmatrix}\in\Gamma_0(N)$$
1600
so $$f(-\tau)(-1)^{-k}=\varepsilon(-1)f(\tau)$$
1601
so that $S_k(\Gamma_1(N),\varepsilon)=0$ unless $\varepsilon(-1)=(-1)^{k}$.
1602
Thus about half of the direct summands vanish.
1603
1604
\section{Explicit Formulas}
1605
Let $$f=\sum_{n=1}^{\infty}a_n q^n \in S_k(\Gamma_1(N),\varepsilon)$$
1606
and let $p$ be a prime, then
1607
$$f|T_p = \begin{cases}
1608
\displaystyle
1609
\sum_{n=1}^{\infty} a_{np}q^n + p^{k-1}\varepsilon(p)
1610
\sum_{n=1}^{\infty} a_n{}q^{pn}, &p\nd N\\
1611
\displaystyle
1612
\sum_{n=1}^{\infty} a_{np}q^n + 0, &p|N
1613
\end{cases}
1614
$$
1615
When $p|N$, $T_p$ is often denoted $U_p$ and called an Atkin-Lehner
1616
operator.
1617
1618
We have the relations
1619
\begin{align*}
1620
T_mT_n&=T_{mn},\quad (m,n)=1\\
1621
T_{p^k}&=\begin{cases}
1622
(T_p)^k, & p|N\\
1623
?, & p\nd N
1624
\end{cases}
1625
\end{align*}
1626
1627
\section{Old and New Forms}
1628
\index{newform}
1629
{\bfseries Warning:} $T_p$ is not necessarily diagonalizable if $p|N$.
1630
There is an example due to Shimura\index{Shimura}, to present it we must first
1631
introduce old and new forms.
1632
1633
Let $M$ and $N$ be positive integers such that $M|N$ and let $d|\frac{N}{M}$.
1634
If $f(\tau)\in S_k(\Gamma_1(M))$ then $f(d\tau)\in{}S_k(\Gamma_1(N))$.
1635
We thus have a map $S_k(\Gamma_1(M))\into{}S_k(\Gamma_1(N))$ for each
1636
$d|\frac{N}{M}$. Combining these gives a map
1637
$$\varphi_M:\bigoplus_{d|\frac{N}{M}}S_k(\Gamma_1(d))\into S_k(\Gamma_1(N)).$$
1638
1639
\begin{defn} The {\bfseries old part} of $S_k(\Gamma_1(N))$ is the subspace
1640
generated by the images of the $\varphi_M$ for $M|N$, $M\neq N$.
1641
\end{defn}
1642
1643
We remark that the {\bfseries new part} of $S_k(\Gamma_1(N))$ is the
1644
orthogonal complement of the old part with respect to the Petersson inner
1645
product.
1646
1647
%% NOte; he goes on to map S_k(\Gamma_1(N))^2\into S_k(\Gamma_1(p))
1648
%% which makes no sense to me in this framework -- see the written notes
1649
%% if it makes sense wednesday.
1650
1651
%% 2/7/96
1652
1653
\chapter{New Forms}
1654
1655
Today we discuss how the Hecke operators $T_n$ on $S_k(\Gamma_1(N))$
1656
can fail to be diagonalizable. Let $N$ be a positive integer and
1657
$M$ a divisor of $N$. For each $d|\frac{N}{M}$ we define a map
1658
$$\alpha_{d}:S_k(\Gamma_1(M))\into S_k(\Gamma_1(N)):\quad{}
1659
f(\tau)\mapsto{}f(d\tau).$$
1660
Note that when $T_p$ acts on the image space $S_k(\Gamma_1(N))$ we
1661
will often denote it by $U_p$.
1662
We must check that $f(d\tau)\in{}S_k(\Gamma_1(N))$. Define for
1663
$\gamma=\abcd$,
1664
$$(f|[\gamma]_k)(\tau)=\det(\gamma)^{k-1}(cz+d)^{-k}f(\gamma(\tau)).$$
1665
Thus $f\in S_k(\Gamma_1(N))$ iff $f|[\gamma]_k(\tau)=f(\tau)$ (and
1666
$f$ is holomorphic). Now let $f(\tau)\in\Gamma_1(M)$ and let
1667
$\iota_d=\bigl(\begin{smallmatrix}d&0\\0&1\end{smallmatrix}\bigr)$. Then
1668
$f|[\iota_d]_k(\tau)=d^{k-1}f(d\tau)$ is a modular form on
1669
$\Gamma_1(N)$ since $\iota_d^{-1}\Gamma_1(M)\iota_d$ contains
1670
$\Gamma_1(N)$ (check this directly by conjugating an element
1671
of $\Gamma_1(N)$ by $\iota_d$).
1672
Moreover if $f$ is a cusp form then so is $f|[\iota_d]_k$.
1673
If $f\in S_k(\Gamma_1(M))$ is nonzero, then as $d$ varies over divisors
1674
of $\frac{N}{M}$, the various $f(d\tau)$ are linearly
1675
independent.
1676
1677
Suppose $f\in S_k(\Gamma_1(M))$ is a normalized eigenform for
1678
all of the Hecke operators $T_n$ and $\dbd{n}$, and $p$ is a prime
1679
not dividiing $M$. Then
1680
$$f|T_p=a f \quad \text{and} \quad f|\dbd{p}=\varepsilon(p)f.$$
1681
Assume $N=p^{r}M$ where $r$ is an integer $\geq 1$.
1682
Let $$f_i(\tau)=f(p^i\tau),$$
1683
so $f_0,\ldots,f_r$ are the images of $f$ under the maps
1684
defined above and $f=f_0$. Consider the action of $U_p$ on the $f_i$.
1685
From previous work we have
1686
\begin{align*}
1687
f|T_p & = \sum_{n\geq 1} a_{np}q^n+\varepsilon(p)p^{k-1}\sum a_n{}q^{pn}\\
1688
& = f_0|U_p + \varepsilon(p)p^{k-1} f_1
1689
\end{align*}
1690
so
1691
$$f_0|U_p = f|T_p - \varepsilon(p)p^{k-}f_1
1692
= af_0 - \varepsilon(p)p^{k-1}f_1.$$
1693
Also
1694
$$f_1|U_p = (\sum a_n q^{pn}) | U_p = \sum a_n q^n = f_0.$$
1695
More generally one can show that $f_i|U_p = f_{i-1}$.
1696
1697
$U_p$ preserves the two dimensional vector space spanned by
1698
$f_0$ and $f_1$. The matrix of $U_p$ is
1699
$$A=\Bigl(\begin{matrix}a&1\\-\varepsilon(p)p^{k-1}&0\end{matrix}\Bigr)$$
1700
which has characteristic polynomial
1701
$$\chi_A(X)=X^2 - aX + p^{k-1}\varepsilon(p).$$
1702
1703
\section{Connection With Galois Representations}
1704
\index{Galois representations}
1705
This leads to a striking connection with Galois representations.
1706
Let $f$ be a modular form and $E$ be the field generated over $\Q$
1707
by the coefficients of $f$. Let $\ell$ be a prime and $\lambda$
1708
a prime lying over $\ell$. Then one constructs a representation
1709
$$\rho_{\lambda}:\gal(\overline{\Q}/\Q)\into\GL(2,E_{\lambda}).$$
1710
If $p\nd N\ell$, then $\rho_{\lambda}$ is unramified at $p$,
1711
so there is a Frobenious element $\frob_p\in\gal(\overline{\Q}/\Q)$.
1712
One can show that
1713
\begin{align*}
1714
\det(\rho_{\lambda}(\frob_p)) &= p^{k-1}\varepsilon(p) \\
1715
\tr(\rho_{\lambda}(\frob_p)) & = a_p = a,
1716
\end{align*}
1717
so the characteristic polynomial of $\rho_{\lambda}(\frob_p)\in\GL_2(E_{\lambda})$
1718
is $$X^2-a_p X + p^{k-1}\varepsilon(p).$$
1719
1720
\section{Semisimplicity of $U_p$}
1721
\begin{question}
1722
Is $U_p$ semisimple on the span of $f_0$ and $f_1$?
1723
\end{question}
1724
1725
If the eigenvalues are distinct the answer is clearly yes.
1726
If the eigenvalues are the same, then $X^2-aX+p^{k-1}\varepsilon(p)$
1727
has discriminant zero, that is, $4\varepsilon(p)p^{k-1}=a^2$ so
1728
$$a=2p^{\frac{k-1}{2}}\sqrt{\varepsilon(p)}.$$
1729
Is this possible? The answer is still {\em unknown}, although it is
1730
a curious fact that the Ramanujan conjectures (proved by Delign in 1973)
1731
imply that $|a|\leq 2p^{\frac{k-1}{2}}$, so the above equality remains
1732
taunting.
1733
1734
When $k=2$ Weil showed that $\rho_{\lambda}(\frob_p)$ is semisimple
1735
so if the eigenvalues of $U_p$ are equal then $\rho_{\lambda}(\frob_p)$
1736
is a scalar. But Edixhoven and Coleman \cite{edixcole}
1737
show that it is not a scalar by looking at the abelian
1738
variety attached to $f$.
1739
1740
\section{Shimura's Example of Nonsemisimple $U_p$}
1741
\index{Shimura}
1742
1743
Let $W$ be the space spanned by $f_0, f_1$ and let
1744
$V$ be the space spanned by $f_0, f_1, f_2, f_3$.
1745
$U_p$ acts on $V/W$ by $\overline{f_2}\mapsto 0$
1746
and $\overline{f_3}\mapsto \overline{f_2}$. Thus the matrix of the
1747
action of $U_p$ on $V/W$ is
1748
$\bigl(\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\bigr)$
1749
which is nonzero and nilpotent hence not semisimple.
1750
Since $W$ is invariant under $U_p$ this shows that
1751
$U_p$ is not semisimple on $V$.
1752
%% I CAN'T SEE THIS! It seems like basic representation theory...
1753
1754
\section{An Interesting Duality}
1755
1756
Now suppose $N=1$ thus $\Gamma_1(N)=\sl2z$. Because of the
1757
Petersson product all the $T_n$ are diagonalizable, so
1758
$S_k=S_k(\Gamma_1(1))$ has a basis
1759
$$f_1,\ldots,f_d$$
1760
of normalized eigenforms where $d=\dim S_k$. Let $\T=\T_{\C}$, then
1761
there is a {\em canonical} map
1762
$$\T_{\C}\hookrightarrow{}\C^d: \quad T\mapsto(\lambda_1,\ldots,\lambda_d)$$
1763
where $f_i|T=\lambda_{i}f_i$. This map is clearly injective and we
1764
know by previous arguments that $\dim\T_{\C}=d$ so the map is an isomorphism
1765
of $\C$-vector spaces.
1766
1767
The form
1768
$$v=f_1+\cdots+f_n$$ generates $S_k$ as a $\T$-module.
1769
Since $v$ corresponds to the vector $(1,\ldots,1)$ and
1770
$\T\isom\C^d$ acts on $S_k\isom\C^d$ componentwise this
1771
is just the statement that $\C^d$ is generated by
1772
$(1,\ldots,1)$ as a $\C^d$-module -- which is clear.
1773
Thus we have simultaneously:
1774
1775
1) $S_k$ is free of rank 1 over $\T$, and
1776
1777
2) $S_k=\Hom_{\C}(\T,\C)$ as $\T$-modules, thus
1778
1779
$$\T\isom\Hom_{\C}(\T,\C).$$ The isomorphism sends
1780
an element of $T\in \T$ to $Tv\in S_k$. Since
1781
the identification $S_k=\Hom_{\C}(\T,\C)$ was
1782
constructed using the Petersson product it is canonical
1783
and since the choice of a normalized eigenbasis $f_1,\ldots,f_d$
1784
is canonical we see that the isomorphism $T\isom\Hom_{\C}(\T,\C)$
1785
is canonical.
1786
1787
\begin{prop}
1788
$v\in S_k(\Q)$
1789
\end{prop}
1790
\begin{proof}
1791
Let $\sigma\in\gal(\overline{\Q}/\Q)$, then if $f_i$ is
1792
a normalized eigenform so is $\sigma(f_i)$ (from the explicit
1793
formula). Thus $\sigma(f_1+\cdots+f_n)=f_1+\cdots+f_n$ for
1794
all $\sigma$ as desired.
1795
\end{proof}
1796
1797
Now we consider the case for general $N$.
1798
Recall that we have defined maps
1799
$$S_k(\Gamma_1(M))\into S_k(\Gamma_1(N))$$
1800
for all $M$ dividing $N$ and all divisors $d$ of $\frac{N}{M}$.
1801
1802
\begin{defn}
1803
The {\bfseries old part} of $S_k(\Gamma_1(N))$ is the space
1804
generated by all images of these maps with $M|N$ but $M\neq N$.
1805
The {\bfseries new part} is the orthogonal complement of the
1806
old part with respect to the Petersson product.
1807
\end{defn}
1808
1809
There is an algebraic definition of the new part. One defines
1810
certain trace maps
1811
$$S_k(\Gamma_1(N))\into S_k(\Gamma_1(M))$$
1812
for all $M<N$, $M|N$
1813
which are the adjoints to the above maps (w.r.t Petersson product).
1814
Then $f$ is in the new part of $S_k(\Gamma_1(N))$ iff $f$
1815
is killed by all of these maps.
1816
1817
It can be shown that the $T_n$ act semisimply on
1818
$S_k(\Gamma_1(M))_{\text{new}}$ for all $M\geq 1$.
1819
Thus $S_k(\Gamma_1(M))_{\text{new}}$ has a basis of eigenforms.
1820
We have a natural map
1821
$$\bigoplus_{M|N} S_k(\Gamma_1(M))_{\text{new}}\hookrightarrow
1822
S_k(\Gamma_1(N)).$$
1823
The image in $S_k(\Gamma_1(N))$ of an eigenform $f$ for some
1824
$S_k(\Gamma_1(M))_{\text{new}}$ is called a {\bfseries newform}
1825
of level $M_f=M$. Note that a newform is not necessarily
1826
an eigenform for the Hecke operators acting on $S_k(\Gamma_1(N))$.
1827
Let
1828
$$v=\sum_{f} f(q^{\frac{N}{M_f}})\in S_k(\Gamma_1(N))$$
1829
where the sum is taken over all newforms $f$ of weight
1830
$k$ and some level $M|N$.
1831
This generalizes the $v$ constructed above when $N=1$
1832
and has many of the same good properties. For example,
1833
$S_k(\Gamma_1(N))$ is free of rank $1$ over $\T$ with
1834
basis element $v$. The coefficients of $v$ lie in $\Q$,
1835
but to show this we need to know the new part of $S_k(\Gamma_1(N))$
1836
is stable under the action of the Galois group of $\Q$.
1837
This is not easy since the new part is defined in terms of
1838
the Petersson product which is an analytic construction.
1839
Serre\index{Serre} circumvents this problem by giving an alternative
1840
definition in terms of trace maps going the other way.
1841
1842
1843
%% 2/9/96
1844
\section{Observations on $T_n$}
1845
Let $\T_{\Q}=\Q[\cdots,T_n,\cdots]$ and
1846
$\Gamma=\Gamma(1)=\modgp$. Let $f_1,\ldots,f_d$ be a basis of $\Gamma$
1847
consisting of normalized eigenforms.
1848
\begin{prop}
1849
The coefficients of the $f_i$ are totally real algebraic integers.
1850
\end{prop}
1851
\begin{proof}
1852
$\gal(\C/\Q)$ acts on $f_i$ by acting on the coefficients
1853
of its $q$-expansion. From the explicit formula in section 3.2
1854
one sees that the set $\{f_1,\ldots,f_d\}$ is
1855
stable under the action of $\gal(\C/\Q)$.
1856
For any $i$, $a_n(f_i)$ is an eigenvalue of $T_n$ since
1857
$f_i|T_n = a_n(f_i)f_i$, and $T_n$ is self-adjoint so
1858
$a_n(f_i)$ must be real. Thus all conjugates of $a_n(f_i)$
1859
are real and there are only finitely many since a conjugate
1860
of $a_n(f_i)$ must be $a_n(f_j)$ for some $j$, $1\leq j\leq d$.
1861
\end{proof}
1862
1863
\begin{prop}
1864
The operators $\dbd{d}$ on $S_k(\Gamma_1(N))$
1865
lie in $\Z[\ldots,T_n,\ldots]$.
1866
\end{prop}
1867
\begin{proof}
1868
It is enough to show $\dbd{p}\in\Z[\ldots,T_n,\ldots]$ for
1869
There is a formula relating $\dbd{p}$ and $T_p$,
1870
$$p^{k-1}\dbd{p}={T_p}^2-T_{p^2}.$$
1871
By Dirichlet's theorem on prime's in arithmetic progression,
1872
see VIII.4 of Lang \cite{lang1},
1873
there is another prime $q$ congruent to $p$ mod $N$.
1874
Since $p^{k-1}$ and $q^{k-1}$ are relatively prime there
1875
exist integers $a$ and $b$ so that
1876
$a p^{k-1} + b q^{k-1} = 1$. Then
1877
$$
1878
\dbd{p}=\dbd{p}(a p^{k-1} + b q^{k-1})
1879
= a({T_p}^2-T_{p^2}) + b({T_q}^2-T_{q^2}).
1880
$$
1881
\end{proof}
1882
1883
Let $\Sigma$ be a set of representatives of
1884
$\{f_1,\ldots,f_d\} \backslash \gal(\C/\Q)$. It
1885
is unknown whether or not $\#\Sigma$ can be
1886
larger than one, that is, whether the eigenforms
1887
are all conjugate under the action of Galois.
1888
Let $K_f = \Q(\ldots,a_n(f),\ldots)$ and defined a homomorphism
1889
of $\Q$-algebras
1890
$$T_{\bsQ}\into K_f : T_n\mapsto \lambda
1891
\text{ where }T_n f = \lambda f$$
1892
Taking the product over a set of representatives of the $f_i$
1893
yields a map
1894
$$\T_{\bsQ}\xrightarrow{\sim}\prod_{f\in\Sigma}K_f$$
1895
which one can show is an isomorphism of $\Q$-algebras.
1896
1897
\begin{example}
1898
Consider $S_2(\Gamma_0(N))$ with $N$ prime, then
1899
$$\T_{\bsQ}\isom E_1\times\cdots E_t$$
1900
with the $E_i$ totally real fields. When $N=37$,
1901
that $\T_{\bsQ}\isom \Q\cross\Q$.
1902
\end{example}
1903
1904
\chapter{Some Explicit Genus Computations}
1905
\section{Computing the Dimension of $S_k(\Gamma)$}
1906
Let $k=2$ unless otherwise noted, and let
1907
$\Gamma\subset\modgp$ be a congruence subgroup.
1908
Then $$S_2(\Gamma)=H^{0}(X_{\Gamma},\Omega^1)$$
1909
where $$X_{\Gamma}=(\Gamma\backslash\H)\union
1910
(\Gamma\backslash\bP^1(\Q)).$$
1911
By definition $\dim H^{0}(X_{\Gamma},\Omega^1)$
1912
is the genus of $X_{\Gamma}$.
1913
\begin{exercise}
1914
Prove that when $\Gamma=\modgp$ then $\Gamma\backslash\bP^1(\Q)$
1915
is a point.
1916
\end{exercise}
1917
1918
Since $\Gamma\subset\Gamma(1)$ there is a covering
1919
$$\begin{CD}
1920
\Gamma\backslash\H @>>> X_{\Gamma} \\
1921
@VVV @VVV \\
1922
\Gamma(1)\backslash\H @>>> X_{\Gamma(1)} @>j>> \bP^1(\C)
1923
\end{CD}$$
1924
which is only ramified at points above $0, 1728, \infty$
1925
($0$ corresponds to $i$ and $\rho$ to $1728$ under $j$).
1926
1927
\begin{example} Suppose $\Gamma=\Gamma_0(N)$, then the degree
1928
of the covering is the index $(\modgp/\{\pm 1\}:\Gamma_0(N)/\{\pm 1\})$.
1929
A point on $Y_{\Gamma(1)}$ corresponds to an elliptic curve, whereas
1930
a points on $Y_{0}(N)$ correspond to a pair consisting of an
1931
elliptic curve and a subgroup of order $N$.
1932
\end{example}
1933
1934
\section{Application of Riemann-Hurwitz}
1935
Now we compute the genus of $X_{\Gamma}$ by applying the
1936
Riemann-Hurwitz formula.
1937
Intuitively the Euler charcteristic
1938
should be totally additive, that is, if $A$ and $B$ are
1939
disjoint spaces then
1940
$$\chi(A\union B)=\chi(A)+\chi(B).$$
1941
Let $X$ be a compact Riemann surface of genus $g$, then
1942
$\chi(X)=2-2g$. Since $\chi(\{\text{point}\})=1$ we should
1943
have that
1944
$$\chi(X-\{p_1,\ldots,p_n\})=\chi(X)-n\chi(1)=(2-2g)-n.$$
1945
If we have an umramified covering $X\into Y$
1946
of degree $d$ then $\chi(X)=d\cdot\chi(Y)$.
1947
Consider the covering
1948
$$\begin{CD}X_{\Gamma}-\{\text{points over $0,1728,\infty$}\}\\
1949
@VVV
1950
\\ X_{\Gamma(1)}-\{0,1728,\infty\}
1951
\end{CD}$$
1952
Since $X_{\Gamma(1)}$ has genus $0$, $X_{\Gamma(1)}-\{0,1728,\infty\}$
1953
has Euler characteristic $2-3=-1$. If we let $g=\chi(X_{\Gamma})$ then
1954
$\chi(X_{\Gamma}-\{\text{points over $0,1728,\infty$}\}
1955
= 2-2g -n_{0} - n_{1728} - n_{\infty}$,
1956
where $n_p$ denotes the number of points lying over $p$.
1957
Thus $-d=2-2g-n_0-n_{1728}-n_{\infty}$ whence
1958
$$2g-2 = d - n_0 -n_{1728}-n_{\infty}.$$
1959
1960
Suppose $\Gamma=\Gamma_0(N)$ with $N>3$, then $n_0=d/3$ and $n_{1728}=d/2$
1961
(I'm not sure why).
1962
The degree $d$ of the covering is equal to the number of unordered
1963
ordered basis of $E[N]$, thus
1964
$$d=\#\SL_2(\Z/N\Z)/2.$$
1965
We still need to compute $n_{\infty}$. $\modgp$ acts on $\bP^1(\Q)$
1966
if we view $\bP^1(\Q)$ as all pairs $(a,b)$ of relatively prime integers
1967
and suppose $\infty$ corresponds to $(1,0)$. The stabilizer of $(1,0)$ is
1968
the sugroup $\{\abcd \in \modgp : c=0 \}$ of upper triangular matrices.
1969
Since the points lying over $\infty$ are all conjugate by the Galois
1970
group of the covering (which is $\SL_2(\Z/N\Z)/\{\pm 1\}$),
1971
$$\text{number of cusps}=\frac{\text{order of $\SL_2(\Z/N\Z)/\{\pm 1\}$}}
1972
{\text{order of stabilizer of $\infty$}}.$$
1973
We thus have
1974
$$2g(X(N))-2=\frac{d}{6}-\frac{d}{N}$$
1975
where $\frac{d}{N}$ is the number of cusps.
1976
1977
%% 2/12/96
1978
\section{Explicit Genus Computations}
1979
1980
Let $N>3$ and consider the modular curve $X=X(N)$.
1981
There is a natural covering map $X\into{}X(1)\xrightarrow{j}\C$.
1982
Let $d$ be the degree, then
1983
$$2g-2=d-m_0-m_{1728}-m_{\infty}$$
1984
where $g$ is the genus of $X$ and
1985
$m_x$ is the number of points lying over $x$.
1986
Since $m_0$ is approximately $\frac{d}{3}$ and $m_{1728}$
1987
is approximately $\frac{d}{2}$,
1988
$$2g-2=\frac{d}{6}-m_{\infty}\pm\text{ small correction factor}.$$
1989
1990
\section{The Genus of $X(N)$}
1991
Now we count the number of cusps of $X(N)$, that is, the
1992
size of $\Gamma(N)\backslash\bP^1(\Q)$. There is a surjective map
1993
from $\modgp$ to $\bP^1(\Q)$ given by
1994
$$\bigabcd\mapsto \bigabcd\Bigl(\begin{matrix}1\\0\end{matrix}\Bigr).$$
1995
Let $U$ be the kernel, thus $U$ is the stabilizer of
1996
$\infty=\bigl(\begin{smallmatrix}1\\0\end{smallmatrix}\bigr)$,
1997
so $U=\{\pm\bigl(\begin{smallmatrix}1&a\\0&1\end{smallmatrix}\bigr):a\in\Z\}$.
1998
Then the cusps of $X(N)$ are the elements of
1999
$$\Gamma(N)\backslash(\modgp/U)=(\Gamma(N)\backslash\modgp)/U=\SL_2(\Z/N\Z)/U$$
2000
which has order $$\frac{\#\SL_2(\Z/N\Z)}{2N}=\frac{d}{N}.$$
2001
2002
Substituting this into the above formula gives
2003
$$2g-2=\frac{d}{6}-\frac{d}{N}=\frac{d}{6N}(N-6)$$
2004
so $$g=1+\frac{d}{12N}(N-6).$$
2005
When $N$ is prime
2006
$$d=\frac{1}{2}\#\SL_2(\Z/N\Z)=\frac{1}{2}\cdot\frac{(N^2-1)(N^2-N)}{N-1}.$$
2007
Thus when $N=5$, $d=60$ so $g=0$, and when $N=7$, $d=168$ so $g=3$.
2008
2009
\section{The Genus of $X_0(N)$}
2010
Suppose $N>3$ and $N$ is prime. The covering map
2011
$X_0(N)\into X(1)$ is of degree $N+1$ since a point of $X_0(N)$
2012
corresponds to an elliptic curve along with a subgroup of order $N$
2013
and there are $N+1$ such subgroups because $N$ is prime.
2014
\begin{exercise}
2015
$X_0(N)$ has two cusps; they are the orbit of $\infty$ which
2016
is unramified and $0$ which is ramified of order $N$.
2017
\end{exercise}
2018
Thus
2019
$$2g-2=N+1-2-n_