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Author: William A. Stein
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2%% ribetofficial.tex --- Scribe notes for Ken Ribets Spring 1996 course.  %%
3%%                                                                        %%
4%% This file should compile with standard latex, the amssymb package,     %%
5%% and the diagrams package.  The diagrams package is contained in the    %%
6%% file "diagrams.tex".  Obtain it and put it in the same directory       %%
7%% as this file.                                                          %%
8%%                                                                        %%
9%% This file is being maintained by William Stein ([email protected]).%%
10%% This file was assembled by Lawren Smithline ([email protected]).%%
11%%                                                                        %%
12%% Last modification date: 9/25/96.                                       %%
13%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
14
15\documentclass[12pt]{report}
16
17%% Uncomment the following two lines if your system doesn't
18%% define the \mathfrak font.
19%\font\german=eufm10 scaled \magstep 1
20%\def\mathfrak{\german}
21
22\input diagrams
23\usepackage{amssymb}
24\marginparwidth 0pt
25\oddsidemargin  0pt
26\evensidemargin  0pt
27\marginparsep 0pt
28\topmargin   0pt
29\textwidth   6.5in
30\textheight  8.5 in
31\def\nn{{\rm N}}                    % Roman N in math mode
32\def\aa{{\cal A}}
33\def\rr{{\cal R}}
34\def\flt{F{\sc l}T}                 % Fermat's little theorem
35\def\pf{{\sc Proof.\ }}
36\def\st{\ \cdot\backepsilon\cdot\ } % such that'' symbol
37\def\vs{\vspace{4pt}}
39\def\C{{\bf C}}   % complex nos.
40\def\Z{{\bf Z}}   % integers
41\def\F{{\bf F}}   % field
42\def\P{{\bf P}}   % projective land
43
44
45\def\N{{\mathfrak N}} % index of an ideal in a number ring (norm)
46\def\H{{\mathfrak H}} % the complex upper halfplane
47\def\R{{\bf R}}   % reals
48\def\Q{{\bf Q}}   % rationals
49\def\O{{\cal O}}  % ring of integers
50\def\T{{\bf T}}   % Hecke algebra
51\def\G{{\bf G}}   % group scheme
52\def\X{{X(N)}}    % Modular curve
53\def\E{{E_j}}
54\def\K{{\overline{K}}}
55\def\ff{{\cal F}} % Modular function field
56
59
60\def\0{{\bf 0}}   % zero structure
62\def\b{\beta}
63\def\c{\gamma}
64\def\d{\delta}
65\def\e{\epsilon}
66\def\f{\zeta}
67\def\G{\Gamma}
68\def\ka{\kappa}
69\def\la{\lambda}
70\def\t{\tau}
71\def\om{\omega}
72\def\Om{\Omega}
73\def\rh{{\overline      % rho bar
74         {\rho_E}}}
75\def\rhla{{\rho_\lambda}}
76\def\th{{\theta}}
77\def\g{\gamma}
78
79\def\p{{\mathfrak p}}   % Gothic p, a prime ideal
80\def\q{{\mathfrak q}}   % Gothic q
81\def\M{{\mathfrak m}}	% maximal ideal
82\def\Tm{{\T_\M}}        % T localized at m
83\def\ia{{\mathfrak a}}
84\def\ib{{\mathfrak b}}
85\def\si{\sigma}
86\def\cent{{\bf C}}      % centralizer
87\def\qed{\hfill $\blacksquare$\smallskip}
88
89\def\tate{{\rm Ta}}                      % Tate module
90\def\tatel{{{\rm Ta}_\l}}                % l-adic Tate module
91\def\tatem{{{\rm Ta}_\M}}                % m-adic Tate module
92\def\tatels{{{\rm Ta}_\l^*}}             % contrav. l-adic T. m.
93\def\tatems{{{\rm Ta}_\M^*}}             % contrav. m-adic T. m.
94
95\def\ve{\varepsilon}    % epsilon the character
96
97\def\lan{\langle}       % angle brackets
98\def\ran{\rangle}
99\def\<{{\langle}}       % < bracket
100\def\>{{\rangle}}       % > bracket
101
102\def\l{\ell}            % script l as in l-adic.
105\def\sigmaonf{
106        {}^\sigma\!f}   % sigma acting on f, from upper left hand corner
107\def\GL{{\rm GL}}       % general linear group
108\def\SL{{\rm SL}}       % special linear group
109\def\gal{{\cal G}al}    % Galois group
110\def\GQ{\gal(\overline\Q/\Q)}
111                        % abs Galois gp of Q
112\def\GQp{\gal(\overline\Qp/\Qp)}
113                        % abs Galois gp of Qp
114\def\modgp{\SL_2(\Z)}   % modular group
115\def\abcd{\left(        % 2 x 2 matrix a b // c d
116     \begin{array}{cc}
117     a&b\\c&d\end{array}\right)}
118\def\isom{\cong}
119\def\tensor{\otimes}
120
121\def\tr{{\rm tr}}
122\def\frob{{\rm frob}}
123\def\mod{\ {\rm mod}\,}
124\def\im{{\rm im}}
125\def\ord{{\rm ord}}
126\def\rank{{\rm rank}}
127\def\aut{{\rm Aut}}
128\def\Hom{{\rm Hom}}
129\def\plim{{\displaystyle\lim_{\longleftarrow}\,}}  % Projective limit
130\def\dlim{{\displaystyle\lim_{\longrightarrow}\,}} % Direct limit
131\def\plimr{{\displaystyle\lim_{
132            \buildrel\longleftarrow\over r}\,}}    % Projective limit over r
133\def\Div{{\rm Div}}
134\def\det{{\rm det }}    % Determinant
135\def\endo{{\rm End}}
136\def\Cot{{\rm Cot}}   % contangent space
137\def\ver{{\rm ver}}   % Vershibung endmorphism
138
139\def\inj{\hookrightarrow}
140\def\into{\rightarrow}
142\def\isomap{{\buildrel \sim\over\rightarrow}}
143\newarrow{To} ---->
144\newarrow{Line} -----
145
146\newtheorem{thm}{Theorem}
147\newtheorem{dfn}{Definition}
148\newtheorem{prop}{Proposition}
149\newtheorem{lem}{Lemma}
150\newtheorem{cor}{Corollary}
151\newtheorem{res}{Result}
152\newtheorem{eg}{Example}
153\newtheorem{claim}{Claim}
154\newtheorem{exercise}{Exercise}
155\newtheorem{remark}{Remark}
156\newtheorem{conj}{Conjecture}
157\newtheorem{note}{Note}
158
159\begin{document}
160\title{Scribe notes for Ken Ribet's Math 274}
161\author{}
162\date{\null}
163\maketitle
164\section*{January 17, 1996}
165\noindent{Scribe: Lawren Smithline, \tt <lawren@math>}
166\bigskip
167
168\noindent
169Here are some topics to be discussed in this course:
170
171\begin{tabular}{l}
172Galois representations and modular forms, \\
173Hecke algebras, \\
174modular curves, and
175Jacobians (Abelian varieties).
176\end{tabular} \\
177This lecture is a brief overview of some connection between these concepts,
178and also an exercise in name-dropping.
179
180We can describe an an elliptic curve, or the Jacobian of a higher genus
181curve, or abelian variety using a lattice.  For $E$, the lattice is $L = 182H_1(E(\C),\Z) \hookrightarrow \C,$ by the map $\c \mapsto \int_\c 183\omega.$
184
185Weil considered curves over a finite field, $k$ of characteristic $p$.
186There is an algebraic definition of $L/nL$ for $n \geq 1, \ \gcd(n, p) = 1.$
187$$E[n] = \{ P \in E(\bar k) : nP = 0\} = \textstyle\frac1n L/L = L/nL.$$
188For example, let $n = \l^\nu$ for $\nu \geq 1$, and $\l$ a prime different
189from $p.$
190
191Weil further considered the limit $$E[\l^\infty] = \bigcup_{\nu = 1}^\infty 192E[\l^\nu].$$
193Tate oberserved there is a map $E[\l^\nu] \stackrel{\l}\rightarrow 194E[\l^{\nu-1}],$ and so of course this inverse limit, $$195\lim_\leftarrow E[\l^\nu] = T_\l(E)$$ is called the Tate module.
196
197As $E[n]$ is free of rank 2 over $\Z / n\Z$, so is $T_\l(E)$ free of rank 2
198over $\Z_\l$.  Also, $V_l(e) = T_\l(E) \otimes \Q_\l$ is a 2 dimensional
199vector space over $\Q_\l$.  This is the first example of $\l$-adic \'etale
200cohomology.  For topological space $X$, $X \mapsto H_{\mathaccent 19 201et}^i(X/\bar k, \Q_\l).$
202
203Here are the names of some cool folks: Taniyama Shimura Mumford Tate.
204
205Now, elliptic curve $E/\Q$ gets an action of $G = \gal(\bar \Q/\Q)$,
206and so does $E[n]$.  I.e. $\si(P+Q) = \si(P) + \si(Q).$  So we have a
207homomorphism $\rho: G \rightarrow {\rm Aut}(E[n]) = GL_2(\Z/n\Z).$
208Since we have an exact sequence $$1 \rightarrow \ker \rho \rightarrow G 209\rightarrow {\rm im}\, \rho \rightarrow 0,$$
210we get a tower of fields
211$$\bar \Q \rightarrow K \rightarrow \Q,$$ and ${\cal G}al(K/\Q) = 212{\rm im}\, \rho \subset GL_2( \Z/n\Z).$
213\smallskip
214
215And now for something completely different.  We can also get to these
216Galois representations via modular forms.  Let $k$ be the weight, such as
2172.  Let $N$ be the level.  The complex vector space $S_k(N)$ is the set of
218cusp forms on $\Gamma_1(N)$, a finite dimensional vector space, namely, the
219set of holomorphic functions $f$ on $\H$ such that
220$$f((az+b)/(cz+d)) = (cz+d)^kf(z)$$ for $$\left( \begin{array}{cc} a & b 221\\ 222c & d \end{array} \right) \in \Gamma_1(N), \ \ a,d \equiv 1, c \equiv 0 (N).$$
223Such an $f$ has a power series (or Fourier series) expansion in $q = 224\exp(2\pi i z)$: $$f(z) = 225\sum_1^\infty c_n q^n.$$
226Here is a famous example observed by Ramanujan, and proved by Mordell using
227(his) Hecke operators:
228$$q\prod_1^\infty (1 - q^n)^{24} = \sum_1^\infty \tau(n)q^n,$$
229for Ramanujan's $\tau$ function.  Now, $\tau(n)\tau(m) =\tau(nm)$ for
230$\gcd(n,m) = 1$.  Also, there is a recurrence for prime powers.
231Amusingly, the normalized basis element of $S_{12}(1)$ is $$\Delta = 232\sum_1^\infty \tau(n)\exp(2\pi i nz).$$  Even more amusingly, $$\tau(n) 233\equiv \sum_{d \mid n} d^{11} \ \ (691).$$
234\smallskip
235
236Experience and Shimura have shown that there exist $f \in S_k(N)$ such that
237$$T_n(f) =c_n \cdot f$$ for all $n \geq 1$ for some scalars $c_n$, and
238$$f = \sum c_n q^n$$ for the same $c_n$, and that these $c_n$ are algebraic
239integers in a finitely generated number field. That is, $[\Q(c_n: n\geq 1): 240\Q]$ is finite.
241
242How can we study and interpret this?  We start with the Hecke ring,
243$$\Q[T_n] \subseteq End(S_k(N)).$$
244Serre in 1968 said there should be Galois representations attached to forms
245of arbitrary weight. Deligne constructed them.  In a broad stroke, one can
246say that we get between Galois representations and modular forms via
247Frobenius elements.
248
249Next time, we continue with the semihistorical overview.
250\section*{January 19, 1996}
251\noindent{Scribe: William Stein, \tt <was@math>}
252\bigskip
253
254\noindent {\bf\large Modular Representations and Modular Curves} \smallskip
255
256\subsection*{Arithmetic of Modular Forms}
257Suppose $f=\sum_{n=1}^{\infty}a_n q^n$ is a cusp form in $S_k(N)$ which is
258an eigenform for the Hecke operators.  The Mellin transform associates
259to $f$ the $L$-function $L(f,z)=\sum_{n=1}^{\infty} n^{-s}{a_n}$.
260Let $K=\Q(a_1,a_2,\ldots)$. One can show that the $a_n$ are algebraic
261integers and $K$ is a number field. When $k=2$, $f$ is associated
262to $f$ an abelian variety
263$A_f$ over $\Q$ of dimension $[K:\Q]$, and $A_f$ has a $K$ action. (See
264Shimura,
265{\em Introduction to the Arithmetic Theory of Automorphic Functions},
266Theorem 7.14.)
267
268\begin{eg}[Modular Elliptic Curves]
269If $a_n\in\Q$ for all $n$, then $K=\Q$ and $[K:\Q]=1$.  In this case, $A_f$
270is a one dimensional abelian variety, which is an elliptic curve, since
271it has nonzero genus.
272An elliptic curve arising in this way is called modular.
273\end{eg}
274
275\begin{dfn}
276Elliptic curves $E_1$ and $E_2$ are {\em isogenous} if there is
277a morphism $E_1\into E_2$ of algebraic groups, which has a
278finite kernel.
279\end{dfn}
280
281The following conjecture motivates much of the theory.
282
283\begin{conj}
284Every elliptic curve over $\Q$ is modular,
285that is, isogenous to a curve constructed in the above way.
286\end{conj}
287
288For $k\geq 2$, Serre and Deligne found a way to associate to $f$ a family
289of $\l$-adic representations. Let $\l$ be a prime number and $K$ be as
290above. It is well known that $$K\otimes_{\Q} \Q_{\l}\isom 291\prod_{\la|\l}K_{\la}.$$
292One can associate to $f$ a family of representations
293$$294\rho_{\l,f}:G=\gal(\overline{\Q}/\Q) 295\rightarrow\GL(K\otimes_{\Q}\Q_{\l}) 296$$
297unramified at all primes $p\not|\l N$.
298By unramified we mean that for all primes $P$ lying over $p$,
299the inertia group of the decomposition group at $P$ is contained
300in the kernel of $\rho$. (The decomposition group $D_P$ at $P$ is the
301set of those $g\in G$ which fix $P$ and the inertia group
302is the kernel of the map $D_P\rightarrow 303\gal(\O/P)$, where $\O$ is the ring of all algebraic integers.)
304
305
306Now $I_P\subset D_P \subset \gal(\overline{\Q}/\Q)$ and
307$D_P / I_P$ is cyclic, since it is isomorphic to a subgroup of the
308galois group of a finite extension of finite fields.
309So $D_P / I_P$
310is generated by a Frobenious automorphism $\frob_p$ lying over $p$.
311We have
312$$313\tr(\rho_{\l,f}(\frob_p)) = a_p\in K \subset K\otimes \Q_{\l}$$
314and
315\begin{equation}\label{detrho}
316\det(\rho_{\l}) = \chi_{\l}^{k-1}\ve,
317\end{equation}
318where $\chi_{\l}$ is the $\l$th cyclotomic character and
319$\ve$ is a Dirichlet character.
320
321\subsection*{Characters}
322Let $f\in S_k(N)$. For all
323$\abcd \in \modgp$ with $c\cong 0 \mod{N}$ we have
324$$325f\left(\frac{az+b}{cz+d}\right) = (cz+d)^k \ve(d) f(z), 326$$
327where $\ve:(\Z/n\Z)^*\rightarrow \C^*$
328is a Dirichlet character mod $N$. If $f$ is an eigenform for
329the diamond bracket operator $<d>$, (so that
330$f|<d> = \ve(d) f$)
331then $\ve$ actually takes values in $K$.
332
333Let $\phi_n$ be the mod $n$ cyclotomic character.
334The map $\phi_n: G \rightarrow (\Z/n\Z)^*$ takes $g\in G$ to
335the automorphism induced by $g$ on the $n$th cyclotomic
336extension $\Q(\mu_n)$ of $\Q$, where we identify
337$\gal(\Q(\mu_n)/\Q)$ with $(\Z/n\Z)^*$.
338The $\ve$ appearing in (\ref{detrho})
339is really the composition
340$$341G\stackrel{\phi_n}\longrightarrow(\Z/n\Z)^* 342 \stackrel{\ve}\longrightarrow \C^*. 343$$
344
345For each positive integer $\nu$ we consider the $\l^{\nu}$th
346cyclotomic character on $G$,
347$$348\phi_{\l^{\nu}}:G\rightarrow (\Z/\l^{\nu}\Z)^*. 349$$
350Putting these together give a map
351$$352\phi_{\l^{\infty}}=\lim_{\stackrel\longleftarrow\nu} 353\phi_{\l^{\nu}}:G\stackrel{\chi_{\l}}\longrightarrow\Z_{\l}^{*}. 354$$
355
356\subsection*{Parity Conditions}
357
358Let $c\in\gal(\overline{\Q}/\Q)$ be complex conjugation.
359We have $\phi_n(c)=-1$, so $\ve(c) = \ve(-1)$ and
360$\chi_{\l}(c) = (-1)^{k-1}$. Let
361$$\abcd 362=\left(\begin{array}{cc} -1&0\\0&-1 \end{array}\right).$$
363For $f\in S_k(N)$,
364$$f(z) = (-1)^k\ve(-1)f(z),$$
365so $(-1)^k\ve(-1) = 1$. Thus,
366$$\det(\rho_{\l}(c)) = \epsilon(-1)(-1)^{k-1} = -1.$$
367The $\det$ character is odd so the representation
368$\rho_{\l}$ is odd.
369
370\begin{remark} (Vague Question) How can one recognize representations
371like $\rho_{\l,f}$ in nature''? Mazur and Fontaine have made
372relevant conjectures. The Shimura-Taniyama conjecture can be reformulated
373by saying that for any representation $\rho_{\l,E}$ comming
374from an elliptic curve $E$ there is $f$ so that
375$\rho_{\l,E}\isom \rho_{\l,f}$.
376\end{remark}
377
378\subsection*{Conjectures of Serre (mod $\l$ version)}
379Suppose $f$ is a modular form, $\l$  a rational prime,
380$\la$ a prime lying over $\l$, and the representation
381$$\rho_{\la,f}:G\rightarrow \GL_2(K_{\la})$$
382(constructed by Serre-Deligne) is irreducible.
383Then $\rho_{\la,f}$ is conjugate to a representation
384with image in $\GL_2(\O_{\la})$, where $\O_{\la}$
385is the ring of integers of $K_{\la}$.
386Reducing mod $\l$ gives a representation
387$$\overline{\rho}_{\la,f}:G\rightarrow\GL_2(\F_{\la})$$
388which has a well-defined trace and det, i.e., the det and trace
389don't depend on the choice of conjugate used to reduce mod
390$\la$. One knows from representation theory that if
391such a representation is semisimple then it is completely determined
392by its trace and det. Thus if $\overline{\rho}_{\la,f}$ is irreducible
393it is unique in the sense that it doesn't depend on the choice
394of conjugate.
395
396We have the following conjecture of Serre which remains open.
397\begin{conj}[Serre]
398All irreducible representation of
399$G$ over a finite field which are odd, i.e., $det(\sigma(c))=-1$, $c$
400complex conjugation, are of the form $\overline{\rho}_{\la,f}$
401for some representation $\rho_{\la,f}$ constructed as above.
402\end{conj}
403
404\begin{eg}
405Let $E/\Q$ be an elliptic curve and let
406$\sigma_{\l}:G\rightarrow\GL_2(\F_{\l})$ be
407the representation induced by the action of $G$
408on the $\l$-torsion of $E$. Then $\det \sigma_{\l} = \phi_{\l}$
409is odd and $\sigma_{\l}$ is usually irreducible, so Serre's conjecture
410would imply that $\sigma_{\l}$ is modular. From this one can, under Serre's
411conjecture, prove that $E$ is modular.
412\end{eg}
413
414\begin{dfn}
415Let $\sigma:G\rightarrow \GL_2(\F)$ ($\F$ is a finite field)
416be a represenation of the galois group $G$. The we say that the
417{\em representions $\sigma$ is
418modular} if there is a modular form $f$, a prime $\la$, and an embedding
419$\F\hookrightarrow \overline{\F}_{\la}$ such that
420$\sigma\isom\overline{\rho}_{\la,f}$ over
421$\overline{\F}_\la$.
422\end{dfn}
423
424\subsection*{Wile's Perspective}
425
426Suppose $E/\Q$ is an elliptic curve and
427$\rho_{\l,E}:G\rightarrow\GL_2(\Z_{\l})$
428the associated $\l$-adic representation on the
429Tate module $T_{\l}$. Then by reducing
430we obtain a mod $\l$ representation
431$$\overline{\rho}_{\l,E}=\sigma_{\l,E}:G 432\rightarrow \GL_2(\F_{\l}).$$
433If we can show this is modular for infinitely many $\l$
434then we will know that $E$ is modular.
435
436\begin{thm}[Langlands and Tunnel]
437If $\sigma_{2,E}$ and $\sigma_{3,E}$ are irreducible, then they
438are modular.
439\end{thm}
440
441This is proved by using the fact that $\GL_2(\F_2)$ and
442$\GL_2(\F_3)$ are solvable so we may apply base-change''.
443
444\begin{thm}[Wiles]
445If $\rho$ is an $\l$-adic representation which is irreducible
446and modular mod $\l$ with $\l>2$ and certain other reasonable
447hypothesis are satisfied, then $\rho$ itself is modular.
448\end{thm}
449
450%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
451
452\section*{January 22, 1996}
453\noindent{Scribe: Lawren Smithline, \tt <lawren@math>}
454(Note.  These were done about two months after the fact, since the assigned
455person didn't.  So they are terse. It was a pretty easy lecture.)
456\bigskip
457
458Today, we have a limited goal: to explain modular forms as functions on
459lattices -- or elliptic curves.  (See Serre's {\em Course in Arithmetic},
460or Katz's paper in the Proceedings of the Antwerp Conference in the
461Springer LNM series.)
462
463Let the level $N = 1$.  Consider a weight $k$ cusp form $f$.  For $\t \in 464\H$, we have $$f( a\t+b / c\t+d) = (c\t+d)^k f(\t).$$  So $f(\t+1) = 465f(\t)$.  By the map $\t \mapsto q= \exp(2\pi i \t)$, we map $\H$ to the
466punctured disc $\{ z : 0 < |z| < 1\}$.  We abuse notation and think of $f$
467as a function on this disc.  Since $f$ is a cusp form, $f$ extends to 0,
468and $f(0) = 0$.  So we have a $q$-expansion $$f = \sum _{n=1}^{\infty} a_n 469q^n.$$
470
471\subsection*{Lattices inside $\C$}
472
473Let $L = \Z\om_1 \oplus Z\om_2$.  We may assume that $\om_1 / \om_2 \in 474\H$.  Let ${\mathfrak R}$ be the set of lattices in $\C$.  $\SL_2 \Z$ acts
475on the left of $M = \{ (\om_1, \om_2) : \om_1, \om_2 \in \H\}$ by
476multiplication of the column vector.  This action fixes the lattice.
477
478Here is the relation with elliptic curves.  A lattice $L$ determines a
479complex torus $\C / L$.  There is a Weierstrass $\wp$ function on this
480torus.  Consider an elliptic curve $E$ over $\C$.  There is a lattice given
481by the inclusion $H_1(E(\C),\Z) \hookrightarrow \C$.  Choose a nonzero $\om 482\in H^0(E, \Om'_E).$  Then $\c \in H_1$ maps to $\int_\c \om \in \C$.
483
484So maybe we should think of ${\mathfrak R}$ as the set of pairs $\{ (E, 485\om) \}.$
486
487We have a map $M / \C^\times \into \H$ sending $(\om_1, \om_2)$ to $\om_1 / 488\om_2.$  Now take the quotient on the left by $\SL_2 \Z$:
489$${\mathfrak R}/\C^{times} = \SL_2 \Z \backslash M / \C^\times 490\longrightarrow \SL_2 \Z \backslash \H.$$
491But this just is the space of elliptic curves over $\C$.  So $f:\H 492\rightarrow \C$ which is a modular form and $F:M \rightarrow \C$ satisfying
493$F(\la L) = \la^{-k}F(L)$ amount to the same thing by a simple calculation.
494
495\subsection*{Hecke Operators}
496
497Let $F$ be a function on lattices.  Define the Hecke operator $T_n$ as
498$$T_n F (L) = \sum_{L' \subset L, (L:L') = n} F(L) n^{k-1}.$$
499
500The essential case on elliptic curves is for $n = \l$ a prime.  In this
501case, the $L'$ correspond to the $\l+1$ subgroups of order $\l$ of $(\Z / 502\l\Z)^ 2$.
503%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
504\section*{January 24, 1996}
505\noindent{Scribe: William Stein, \tt <was@math>}
506\bigskip
507
508\noindent {\bf \large More On Hecke Operators}\smallskip
509
510We consider modular forms $f$ on $\Gamma_1(1)=\modgp$, that
511is, holomorphic functions on $\H\cup\{\infty\}$ which satisfy
512$$f(\tau)=f(\frac{a\tau+b}{c\tau+d})(c\tau+d)^{-k}$$
513for all $\abcd\in\modgp$. Using a Fourier expansion we write
514$$f(\tau)=\sum_{n=0}^{\infty} a_ne^{2\pi i\tau n},$$
515and say $f$ is a cusp form if $a_0=0$.
516There is a correspondence between modular forms $f$ and
517lattice functions $F$ satisfying $F(\lambda L)=\lambda^{-k}F(L)$
518given by $F(\Z\tau+\Z)=f(\tau)$.
519
520\subsection*{Explicit Description of Sublattices}
521The $n$th Hecke operator $T_n$ of weight $k$ is defined by
522$$T_n(L)=n^{k-1}\sum_{{L'\subset L,\ (L:L')=n}} L'.$$
523What are the $L'$ explicitly? Note that $L/L'$ is a group of order $n$ and
524$$L'/nL\subset L/nL=(\Z/n\Z)^2.$$
525Write $L=\Z\om_1+\Z\om_2$, let $Y_2$ be the cyclic subgroup
526of $L/L'$ generated by $\om_2$ and let $d=\#Y_2$. Let
527$Y_1=(L/L')/Y_2$.  $Y_1$ is generated by the image
528of $\om_1$ so it is a cyclic group of order $a=n/d$.
529We want to exhibit a basis of $L'$. Let
530$\om_2'=d\om_2\in L'$ and use the fact that $Y_1$ is
531generated by $\om_1$ to write $a\om_1=\om_1'+b\om_2$
532for $b\in\Z$ and $\om_1'\in L'$. Since $b$ is only
533well-defined modulo $d$ we may assume $0\leq b\leq d-1$.
534Thus
535$$536\left(\begin{array}{c}\om_1'\\ \om_2'\end{array}\right) 537= 538\left(\begin{array}{cc}a&b\\0&d\end{array}\right) 539\left(\begin{array}{cc}\om_1\\ \om_2\end{array}\right) 540$$
541and the change of basis matrix has determinent $ad=n$.
542Since
543$$\Z\om_1'+\Z\om_2'\subset L' \subset L=\Z\om_1+\Z\om_2$$
544and $(L:\Z\om_1'+\Z\om_2')=n$ (since the change of basis matrix has
545determinent $n$) and $(L:L')=n$ we see that $L'=\Z\om_1'+\Z\om_2'$.
546
547Thus there is a one-to-one correspondence between sublattices $L'\subset L$
548of index $n$ and matrices
549$\bigl(\begin{array}{cc}a&b\\0&d\end{array}\bigr)$
550with $ad=n$ and $0\leq b\leq d-1$.
551In particular, when $n=p$ is prime there $p+1$ of these. In general, the
552number of such sublattices equals the sum of the positive divisors
553of $n$.
554
555\subsection*{Action of Hecke Operators on Modular Forms}
556Now assume $f(\tau)=\sum_{m=0}^{\infty} c_m q^m$ is a modular
557form with corresponding lattice function $F$. How can we describe the
558action of the Hecke operator $T_n$ on $f(\tau)$? We have
559$$\begin{array}{ll} 560T_nF(\Z\tau+\Z) & = n^{k-1}\displaystyle\sum_{ 561\stackrel{\stackrel{\stackrel{a,b,d}{ab=n}}{0\leq b<d}}\null 562 } 563F((a\tau+b)\Z + d\Z)\smallskip \\ 564& = n^{k-1}\displaystyle\sum d^{-k} F(\frac{a\tau+b}{d}\Z+\Z)\smallskip \\ 565& = n^{k-1}\displaystyle\sum d^{-k} f(\frac{a\tau+b}{d})\smallskip \\ 566& = n^{k-1}\displaystyle\sum_{a,d,b,m} d^{-k}c_m e^{2\pi i(\frac{a\tau+b}{d})m}\smallskip \\ 567& = n^{k-1}\displaystyle\sum_{a,d,m} d^{1-k}c_m e^{\frac{2\pi i a m \tau}{d}} 568\frac{1}{d}\displaystyle\sum_{b=0}^{d-1} (e^{\frac{2\pi i m}{d}})^b\smallskip \\ 569& = n^{k-1}\displaystyle\sum_{\stackrel{ad=n}{m'\geq 0}\null} 570d^{1-k} c_{dm'}e^{2\pi i a m' 571\tau}\smallskip \\ 572& = \displaystyle\sum_{{ad=n, \ m'\geq 0}} a^{k-1} c_{dm'}q^{am'}. 573\end{array}$$
574In the second to the last expression we
575 let $m=dm'$, $m'\geq 0$, then used the fact that the
576 sum
577$\frac{1}{d}\sum_{b=0}^{d-1} (e^{\frac{2\pi i m}{d}})^b$
578is only nonzero if $d|m$.
579
580Thus
581$$T_nf(q)=\sum_{{ad=n, \ m\geq 0}} a^{k-1}c_{dm} q^{am}$$
582and if $\mu\geq 0$ then the coefficient of $q^{\mu}$ is
583$$\sum_{{a|n, \ a|\mu}}a^{k-1}c_{\frac{n\mu}{a^2}}.$$
584
585\begin{remark}
586When $k\geq 1$ the coefficients of $q^{\mu}$ for all $\mu$ belong
587to the $\Z$-module generated by the $c_m$.
588\end{remark}
589
590\begin{remark}
591Setting $\mu=0$ gives the constant coefficient of $T_n f$ which is
592$$\sum_{a|n}a^{k-1}c_0 = \sigma_{k-1}(n)c_0.$$
593Thus if $f$ is a cusp form so is $T_nf$. ($T_nf$ is holomorphic
594since its original definition is as a finite sum of holomorphic
595functions.)
596\end{remark}
597
598\begin{remark}
599Setting $\mu=1$ shows that the coefficient of $q$ in $T_n f$ is
600just $c_n$. As an immediate corollary we have the
601following important result.
602\end{remark}
603
604\begin{cor}
605Suppose $f$ is a cusp form for which $T_n f$ has 0 as coefficient
606of $q$ for all $n\geq 1$, then $f=0$.
607\end{cor}
608
609\begin{remark}
610When $n=p$ is prime we get an interesting formula for the
611action of $T_p$ on the $q$-expansion of $f$.
612One has
613$$T_p f = \sum_{\mu\geq 0} \sum_{{a|n, \ a|\mu}}a^{k-1} 614 c_{\frac{n\mu}{a^2}} q^{\mu}.$$
615Since $n=p$ is prime either $a=1$ or $a=p$. When
616$a=1$, $c_{p\mu}$ occurs in the coefficient of $q^{\mu}$
617and when $a=p$, we can write $\mu=p\lambda$ and we get
618terms $p^{k-1}c_{\lambda}$ in $q^{\lambda p}$.
619Thus
620$$T_n f = \sum_{\mu\geq 0}c_{p\mu}q^{\mu}+ 621 p^{k-1}\sum_{\lambda\geq 0} c_{\lambda}q^{p\lambda}.$$
622\end{remark}
623
624
625
626
627
628\section*{January 26, 1996}
629\noindent{Scribe: Amod Agashe, \tt <amod@math>}
630\bigskip
631
632Following the notation of the last few lectures, let $M_k$ denote
633the space of modular forms of weight $k$ for $SL_2(\Z)$ and $S_k$
634denote the subspace of cusp forms.
635
636Then we have:
637
638\begin{prop}. $M_k$ is a finite dimensional $\C$-vector space and is generated
639by modular forms having the coefficients of their Fourier expansion in $\Q$.
640\end{prop}
641{\sc Sketch of Proof}. (For details, refer Serre's A course in
642Arithmetic'' or Lang's Introduction to modular forms''.) \smallskip
643
644\noindent
645The key ingredient that goes into proving finite dimensionality is
646the following result, which can be obtained by contour integration:
647
648Let $f\in M_k$ and let $D=\{z\in \C :Im(z)>0, \mid z\mid \geq 1, 649\mid Re(z)\mid \leq 1/2\}$ be the fundamental domain for $SL_2(\Z)$. Then
650$$\sum_{p \in D\cup\infty} \frac{1}{e_p} ord_p(f) = \frac{k}{12}$$
651where
652$$e_p = \frac{1}{2} \# \{\gamma \in SL_2(\Z) : \gamma p =p \} 653 = \frac{1}{2} \# Aut(E_p)$$
654Here, the latter equality follows from the observation that the
655category of elliptic curves over $\C$ with isogenies is the same
656as the category of lattices in $\C$ upto homothety with maps being
657multiplication by elemets of $\C$. One can
658show that the invertible maps that preserve the lattice $\Z \oplus 659\Z p$ are in one-to-one correspondence with the set
660$\{\gamma \in SL_2(\Z) : \gamma p =p \}$ and hence the latter equality.
661
662In particular,
663$$e_p = \left\{ \begin{array}{ll} 664 2 & \mbox{if p=i} \\ 665 3 & \mbox{if p=\sqrt[3]{-1}} \\ 666 1 & \mbox{otherwise} 667 \end{array} 668 \right.$$
669
670Using this formula and relating the dimensions of $M_k$ and
671$S_k$, one can show that $M_k$ is finite dimensional and also
672explicitly calculate its dimension.
673
674To get a basis with Fourier coefficients in $\Q$, first observe
675that $M_k$ is generated by the set of Eisenstein series $G_k$ for all $k$.
676As a function on the complex upper half plane,
677the Eisenstein series $G_k$ for $k \in \Z$ and $k>1$ is given by
678$$G_k(\tau) = \sum_{(m,n) \in \Z^2 \backslash (0,0) 679 } 680 \frac{1}{(m\tau+n)^k}$$
681One can then show that
682$$G_k(\tau) = \frac{1}{2}\zeta(1-k) + \sum_{k=1}^\infty \sigma_{k-1}(n)q^n$$
683where $q=e^{2\pi i \tau}$, $\zeta$ is the Riemann zeta function and
684$$\sigma_k(n) = \sum_{d\mid n}d^k$$
685There is a theorem due to Euler which states that $\zeta(1-k)= -\frac{b_k}{k}$
686where $b_k$ are the Bernoulli numbers defined by the following power series
687expansion:
688$$\frac{x}{e^x-1} = \sum_{k=0}^\infty \frac{b_kx^k}{k!}$$
689The constant term of $G_k$ is thus $-b_k/{2k}$, which is rational.
690Thus the Fourier expansion of $G_k$ has rational coefficients and thus
691we have found a basis with rational coefficients.
692\bigskip
693
694Next, let $V$ be a subspace of $M_k$ which is stable under the
695action of all the Hecke operators $T_n$. For example, observing
696that $T_n(G_k)=\sigma_{k-1}(n)G_k$, we see that $V=\C(G_k)$ is
697one such subspace.
698
699Let $\T=\T(V)$ = $\C$-algebra generated by the $T_n$'s inside $End(V)$
700	= $\C$-vector space generated by the $T_n$'s inside $End(V)$.
701The latter equality of sets follows because the product of two Hecke
702operators can be expressed as a linear combination
703of finitely many Hecke operators.
704
705For $k>0$, we define a bilinear map
706$\T \times V \rightarrow \C$ by
707$$(T,f) \mapsto a_1(f\mid T).$$
708
709\begin{prop}. The induced maps $\T \rightarrow Hom(V,\C)$ and
710$V \rightarrow Hom(\T, \C)$ are isomorphisms.
711\end{prop}
712\pf
713We first show that the maps are injective.\\
714Injectivity of the second map:
715$$\begin{array}{l} 716f \in V \mapsto 0 \\ 717\Rightarrow a_1(f \mid T) = 0 \ \forall T\in \T \\ 718\Rightarrow a_1(f \mid T_n) = 0 \ \forall n \\ 719\Rightarrow a_n(f) = 0 \ \forall n \geq 1 \\ 720\Rightarrow f\mbox{ is constant} \\ 721\Rightarrow f=0\mbox{ if }k>0 722\end{array}$$
723Injectivity of the first map:
724$$\begin{array}{l} 725T \in \T \mapsto 0 \\ 726\Rightarrow a_1(f \mid T)=0 \ \forall f\in V \\ 727\Rightarrow a_1((f\mid T_n)\mid T) = 0 \ \forall f\in V \\ 728\Rightarrow a_1((f\mid T)\mid T_n) = 0 \ \forall f \in V \\ 729\Rightarrow a_n(f\mid T) = 0 \ \forall n>0, \forall f\in V \\ 730\Rightarrow f\mid T = 0 \ \forall k>0, \forall f\in V \\ 731\Rightarrow T = 0 732\end{array}$$
733
734In the fourth line, $f$ is replaced by $f \mid T_n$.
735Next observe that $V$, being a subspace of $M_k$, is finite dimensional.
736Hence we have from the injectivities of both the above maps that each map
737is actually an isomorphism. \smallskip
738
739A map $\phi: M \rightarrow N$ of $\T$-modules is said to be $\T$-equivariant
740if $\phi (Tm) = T\phi(m) \ \forall m\in M$.
741
742\begin{prop}.
743The isomorphisms $\T \cong Hom(V,\C)$
744and $V \cong Hom(\T, \C)$
745as defined above are $\T$-equivariant.
746\end{prop}
747\pf
748Consider the first map.
749Here is the $\T$-module structure on $Hom(V,\C)$.
750Given $\psi \in Hom(V,\C)$, i.e. $\psi: V \rightarrow \C$,
751define $T\psi: V \rightarrow \C$ by $(T\psi) (f)=\psi (f\mid T)$.
752Let $\beta$ denote the map $\T \rightarrow Hom(V, \C)$.
753Then given $T'\in \T$ and $T \in \T$,
754we have to show that $\beta (T(T'))=T(\beta (T'))$. Let $f \in V$.
755Now $(\beta (TT'))(f) = a_1(f\mid TT')$,
756while $(T(\beta T'))(f) = \beta (T') (f\mid T) = a_1((f\mid T) \mid T') 757= a_1(f\mid TT')$.
758Thus $(\beta (TT'))(f)=(T(\beta (T')))(f) \ \forall f\in V$ and we
759are done.
760
761\noindent Next consider the second map.
762We define the $\T$-module structure on $Hom(\T,\C)$.
763Given $\phi \in Hom(\T,\C)$, i.e. $\phi: \T \rightarrow \C$,
764define $T\phi: \T \rightarrow \C$ by $(T\phi) (T')=\phi (TT')$.
765Let $\alpha$ denote the map $V \rightarrow Hom(\T, \C)$.
766Then given $f\in V$ and $T \in \T$,
767we have to show that $\alpha (T(f))=T(\alpha (f))$. Let $T' \in \T$.
768Now $$(\alpha (Tf))(T') = a_1((f\mid T) \mid T') = a_1(f\mid TT'),$$
769while $$(T(\alpha f))(T') = \alpha (f) (TT') = a_1(f\mid TT').$$
770Thus $$(\alpha (Tf))(T')=(T(\alpha f))(T') \ \forall T'\in \T$$
771and we are done.
772
773\begin{dfn}.
774An element $f$ of $M_k$ is said to be an eigenform if it is an eigenfunction
775for all the Hecke operators.\\
776i.e. $f\mid T_n = \lambda_n f$ for some $\lambda_n \in \C \ \forall n\geq 1$.
777\end{dfn}
778Let $f$ is an eigenform with eigenvalues $\lambda_n$. Then
779$a_n(f) = a_1(f\mid T_n) = a_1(\lambda_n f) = \lambda_n a_1(f) \ \forall n\geq 1$. \\
780Thus if $a_1(f) = 0$, then $a_n(f) = 0 \ \forall n\geq 1$.  \\
781If $k>0$ then this implies $f=0$.
782Hence if $k>0$, then if $f\neq 0$, we can normalize $f$ to
783$\frac{1}{a_1(f)}f$.
784
785\begin{dfn}.
786An eigenform $f$ is said to be normalized if $a_1(f) = 1$.
787\end{dfn}
788
789If $f$ is a normalized eigenform with eigenvalues $\lambda_n$,
790then $a_n(f)=\lambda_n$ and $f\mid T_n = \lambda_n f = a_n(f)f$.
791
792% Given $f \in M_k$ define $\phi_f: \T \rightarrow \C$ by
793% $\phi_f(T) = a_1(f\mid T)$. Note that if $\alpha$ denotes the map
794% $V \rightarrow Hom(\T, \C)$ induced by the bilinear pairing mentioned
795% before, then $\phi_f$ is just $\alhpa(f)$.
796% Then we have: \\
797
798Again, let $\alpha$ denote the map $V \rightarrow Hom(\T, \C)$
799induced by the bilinear pairing mentioned earlier. Then if $f \in V$,
800we have the map $\alpha(f): \T \rightarrow \C$.
801
802\begin{prop}.
803Let $f$ be an eigenform.
804Then $f$ is normalized $\Leftrightarrow$ $\alpha(f)$ is a ring homomorphism.
805\end{prop}
806\pf
807If $f=0$ then the statement is trivial. So assume $f\neq 0$.
808Then as discussed above, $a_1(f)\neq 0$. Also recall
809from the same discussion that if $f\neq 0$ is an eigenform, then
810$$f\mid T_n = \frac{a_n(f)}{a_1(f)}f.$$ For ease of notation, let
811$\alpha_f$ denote the map $\alpha(f)$. So
812$$\begin{array}{l} 813\alpha_f(T_nT_m) \\ = a_1(f\mid T_nT_m) \\ = a_1((f\mid T_n)T_m) 814\\ = a_m(f\mid T_n) \\ = a_m((a_n(f)/a_1(f))f) \\ = a_m(f)a_n(f)/a_1(f). 815\end{array}$$
816We have $$\alpha_f(T_n)\alpha_f(T_m) = a_1(f\mid T_n) a_1(f\mid T_m) 817= a_n(f) a_m(f).$$
818The following are equivalent:
819
820\begin{tabular}{l}
821$\alpha_f$ is a ring homomorphism, \\
822$\alpha_f(T_nT_m) = \alpha_f(T_n)\alpha_f(T_m) \ \forall T_n,T_m$, \\
823$a_1(f) = 1$, and \\
824$f$ is normalized.
825\end{tabular} \\
826The first implication follows because $\T$ is generated by the $T_n$'s.
827
828% We will only prove the $\Rightarrow$'' part and leave the reverse
829% implication as an exercise. \\
830% Assuming that $f$ is normalized, we have to show that
831% $(\alpha(f))(T_nT_m) = (\alpha(f))(T_n)\phi_f(T_m) \ \forall n,m \geq 1$.
832% Now $(\alpha(f))(T_nT_m) = a_1(f\mid T_nT_m) = a_1((f\mid T_n)T_m) 833% = a_m(f\mid T_n) = a_m(a_n(f)f) = a_m(f)a_n(f)$. \\
834% While $(\alpha(f))(T_n)\phi_f(T_m) = a_1(f\mid T_n) a_1(f\mid T_m) 835% = a_n(f) a_m(f)$. \\
836% Thus the two are equal and we are done.
837
838
839\section*{January 29, 1996}
840\noindent{Scribe: J\'anos Csirik, \tt <janos@math>}
841\bigskip
842
843Today we'll consider questions of rationality and integrality.
844(References: Serre: {\em A Course in Arithmetic} and Lang: {\em
845Introduction to Modular Forms}.)  Let
846$S=S_k$ be the space of cusp forms of weight $k$.  Let
847$S(\Q) = S_k \cap \Q [[ q ]]$
848and
849$S(\Q) \supseteq S(\Z) = S_k \cap \Z [[ q ]].$
850The following fact is easy to prove using explicit formul\ae: $S_k$
851has a $\C$-basis consisting of forms with integral coefficients (see
852Victor Miller's construction below).
853
854Recall that for all even $k\geq4$, there is an Eisenstein series
855$G_k = \frac{-b_k}{2k} + 856 \sum^\infty_{n=1} \left ( \sum_{d|n} d^{k-1} \right) q^n,$
857which is a modular form of weight $k$.  Renormalize this to obtain
858$E_k = \frac{2k}{-b_k}\cdot G_k = 1 + \cdots.$
859The first few Bernoulli numbers of even positive index are $b_2=1/6$,
860$b_4=-1/30$, $b_6=1/42$, $b_8=-1/30$, $b_{10}=5/66$, $b_{12}=-691/2730$.
861The fact that the first four of these have numerator 1 is closely
862related to the arithmetic of cyclotomic fields.
863
864The modular forms $E_4$ and $E_6$ have $q$-expansions with constant
865terms equal to $1$, and all coefficients in $\Z$.  The functions
866$E_4^aE_6^b$ with $4a+6b=k$ form a basis for $M_k$.
867It is easy to see that they are modular forms.
868From the formula that the (weighted) number of zeros of any modular form
869of weight $k$ is $k/12$, we deduce that $E_4$ has a simple zero at
870$\rho$, and $E_6$ hasn't got a zero at $\rho$.  Hence $E_4^aE_6^b$ has a
871simple zero of order $a$ at $\rho$ so these expressions are linearly
872independent over $\C$.
873To
874show that they span $M_k$, consider the following modular form of weight
875$12$:
876$\Delta = (E_4^3 - E_6^2)/1728.$
877Here the coefficient of $q$ is a simple number: $1$.  $\Delta$ has a
878simple zero at $\infty$.  Since a cusp form of weight $k$ has $k/12$
879zeros, it follows that (since
880the weighting $e_\infty$ is $1$), that $\Delta$ does not vanish
881anywhere on $\H$.  Therefore $S_{k+12}=\Delta\cdot M_k$.
882Since $E_k(i\infty)=1\neq0$, it follows that
883$M_k=E_k\cdot\C\oplus S_k=E_k\cdot\C\oplus\Delta\cdot M_{k-12}$.  Hence
884${\rm dim}\,M_k={\rm dim}\,M_{k-12}+1$.  Again using the fact that $f\in 885M_k$ has $k/12$ zeros, we quickly deduce that the dimensions of $M_0$,
886$M_2$, $M_4$, $M_6$, $M_8$, $M_{10}$ are $1$,$0$,$1$,$1$,$1$,$1$
887respectively. (e.g., for $k=4$ any modular form must have just a simple
888zero at $\rho$.  So for any $f\in M_4$, it is the case that
889 $f(\tau)-f(i\infty)E_4(\tau)$ vanishes at
890$\tau=i\infty$ and hence is identically zero.  So $E_4$ spans $M_4$.)  Thus
891we have determined ${\rm dim}\,M_k$ for all $k\geq0$, and it is easy to
892see that this number is equal to the number of solutions to $4a+6b=k$
893for $a,b\geq0$.  Hence the $E_4^aE_6^b$ span $M_k$ and therefore we have
894proved that they form a basis.
895
896The following construction comes from the first page of Victor Miller's
897thesis.  Let $d={\rm dim}_\C\, S_k$.  Then there exist $f_1,\ldots,f_d\in 898S(\Z)$ such that $a_i(f_j)=\delta_{ij}$ for $1\leq i,j\leq d$.
899To show this, recall that $E_4\in M_4$ and $E_6\in M_6$ have
900$q$-expansions with coefficients in $\Z$.  $\Delta\in S_{12}$ has
901constant coefficient $0$, and the coefficient of $q$ is $1$.  Also
902$\Delta\in S_{12}(\Z)$, as can be seen for example from the formula
903$\Delta = q\prod_{n=1}^\infty (1-q^n)^{24}.$
904Now pick $a,b\geq0$ so that $14\geq 4a+6b\equiv k \pmod{12}$, with
905$a=b=0$ when $k\equiv0\pmod{12}$.  Note that then $12d+6a+4b=k$ by our
906previous result on the dimension of $M_k$ (and the fact that the
907dimension of $S_k$ is one less than that for $k\geq12$).  Hence the
908functions
909$g_j = \Delta^j E_6^{2(d-j)+a}E_4^b$
910for $1\leq j\leq d$ will be cusp forms of weight $k$.  By our previous
911remarks on the coefficients of $\Delta$, $E_6$, $E_4$, we have $g_j\in 912S_k(\Z)$ and
913$a_i(g_j) = \delta_{ij}$
914for $i\leq j$.  A straightforward elimination now yields the
915$f_1,\ldots,f_d$ with the stated properties.
916It is clear that these $f_1,\ldots,f_d$ are
917linearly independent over $\C$, hence they form a basis of $S_k$.
918
919If you take $T_1,\ldots,T_d\in\T=\T(S_k)$, they are also linearly
920independent: for given any linear relation
921$\sum_{i=1}^d c_iT_i=0,$
922apply this to $f_j$ and look at the first coefficient
923$0 = a_1\left(f_j\left|\sum_{i=1}^d c_iT_i\right.\right) 924 = \sum_{i=1}^d c_ia_i(f_j) = \sum_{i=1}^d c_i\delta_{ij} = c_j,$
925hence the linear relation given in the first place was trivial, so
926$T_1,\ldots,T_d$ form a basis for $\T(S_k)$, since they are linearly
927independent, and ${\rm dim}_\C\,\T={\rm dim}_\C\, V=d$.
928
929Let ${\cal R}=\Z[\ldots T_n\ldots]\subseteq {\rm End}(S_k)$.
930\begin{claim}
931${\cal R}=\bigoplus_{i=1}^d\Z T_i.$
932\end{claim}
933
934\pf Since the $T_i$ form a basis of $\T$,
935we have $T_n =\sum_{i=1}^d c_{n_i}T_i$ with $c_{n_i}\in\C$.  We need
936to check that $c_{n_i}\in\Z$.  With the $f_j$ as above, consider
937$$938\begin{array}{lll} 939 a_n(f_j) & = a_1(f_j|T_n) 940\\ & = a_1\left(f_j\left|\sum_{i=1}^d c_{n_i}T_i\right.\right) 941\\ & = \sum_{i=1}^d c_{n_i}a_1(f_j|T_i) 942\\ & = \sum_{i=1}^d c_{n_i}a_i(f_j) 943\\ & = \sum_{i=1}^d c_{n_i}\delta_{ij} & = c_{n_j}. 944\end{array}$$
945Hence $c_{n_i}\in\Z$.  \qed
946
947${\cal R}$ is called the {\em  integral Hecke algebra}.  It is a finite
948$\Z$-module of rank $d$.  We still have (from the last lecture) a
949pairing
950\begin{eqnarray*}
951   S(\Z)\times{\cal R}  &  \rightarrow  &  \Z             \\
952   (f,T)                &  \mapsto      &  a_1(f|T).
953\end{eqnarray*}
954Now $S(\Z)\hookrightarrow{\rm Hom}({\cal R},\Z)\cong\Z^d$ by the
955argument given before.  Therefore $S(\Z)$ is a free $\Z$-module of
956finite rank.  But it also contains the $f_i$, so $S(\Z)\cong\Z^d$.
957
958What is $S(\Z)$ as an ${\cal R}$-module?
959
960\begin{exercise}
961The map $S(\Z)\hookrightarrow{\rm Hom}({\cal R},\Z)$ is in fact an
962isomorphism of $\T$-modules.
963\end{exercise}
964
965{\sc Hint.\ }  The cokernel is a torsion (in fact finite) group.
966So if we show it torsion free, we are done.
967
968\begin{thm}
969The $T_n$ are all diagonalizable on $S_k$.
970\end{thm}
971
972$S_n$ supports a Hermitian non-degenerate inner product, the Petersson
973inner product
974$(f,g)\mapsto \langle f,g\rangle\in\C.$
975We have $\langle f,f\rangle\geq0$, with equality iff $f=0$.  Furthermore
976$\langle f|T_n,g\rangle = \langle f,g|T_n\rangle,$
977i.e., $T_n$ is self-adjoint with respect to the given inner product.
978
979An operator $T$ is {\em normal} if it commutes with $T^*$ (which denotes
980its Hermitian transpose).  Normal operators are diagonalizable (for a
981proof, refer to Math H110).  In our case, $T_n^*=T_n$, so this fact
982applies.  It is also true (same proof) that a commuting family of
983semi-simple (i.e., diagonalizable) operators is simultaneously
984diagonalizable.
985
986\pf  Put together the above facts.   \qed
987
988We can also prove that the eigenvalues are real.  This depends on the
989following trick.  For $f\neq0$ consider
990$a_n\langle f,f\rangle = \langle a_nf,f\rangle 991 = \langle f|T_n,f\rangle = \langle f,f|T_n\rangle 992 = \langle f,a_nf\rangle = \bar{a_n}\langle f,f\rangle.$
993$a_n\in\R$ now follows since $f\neq0$ implies
994 $\langle f,f\rangle\neq0$.
995
996\begin{exercise}
997The $a_n$ are totally real algebraic integers.
998\end{exercise}
999
1000{\sc Hint.\ }  The space $S_k$ is stable under the action of
1001${\rm Aut}(\C)$ on the coefficients''.  Given a cusp form
1002$f=\sum_{n=1}^\infty c_nq^n$
1003and some $\sigma\in{\rm Aut}(\C)$, define
1004$\sigmaonf=\sum_{n=1}^\infty \sigma(c_n)q^n$.  This function is in
1005$S_k$, since $S_k$ has a basis in $S(\Q)$, which is fixed by $\sigma$.
1006Then $f$ is an eigenform iff $\sigmaonf$ is an eigenform.
1007
1008We'll use a lame definition of the {\em Petersson inner product} for
1009this section.  Let $z=x+iy\in\H$.  Then we have a volume form
1010$y^{-2}{dx\,dy}$ which is invariant under $GL^+_2(\R)$ (the subgroup
1011of the general linear group of matrices of positive determinant).
1012To prove this, note that
1013$dz\wedge d\bar z = (dx+i\,dy)\wedge(dx-i\,dy) = -2i(dx\wedge dy)$,
1014and hence
1015$dx\,dy = dx\wedge dy = \frac{-1}{2i} dz\wedge d\bar z$
1016Then for any
1017$\alpha = \left( \begin{array}{cc} a & b \\ c & d \end{array} \right) 1018 \in GL^+_2(\R)$ we can consider the usual action
1019$\alpha: z \mapsto \frac{az+b}{cz+d} 1020 = \frac{(az+b)(c\bar z+d)}{|cz+d|^2}$
1021where the imaginary part of the result is
1022$|cz+d|^{-2}(ad-bc){\rm Im}\, z = y|cz+d|^{-2}\det(\alpha).$
1023As for differentials,
1024$d\left(\frac{az+b}{cz+d}\right) 1025 = \frac{a(cz+d)\,dz-(az+b)c\,dz}{(cz+d)^2} 1026 = \frac{ad-bc}{(cz+d)^2}\,dz$
1027hence under application of $\alpha$, $dz\wedge d\bar z$ takes on a
1028factor of
1029$$\frac{\det(\alpha)}{(cz+d)^2}\frac{\det(\alpha)}{(c\bar z+d)^2} 1030 = \left( \frac{\det(\alpha)}{|cz+d|^2} \right)^2.$$
1031This finally proves that the differential $y^{-2}{dx\,dy}$
1032is invariant under the action of $GL^+_2(\R)$.
1033
1034The formula for the {\em Petersson inner product} is
1035$\langle f,g\rangle= 1036 \int_{SL_2(\Z)\setminus\H}\left(f(z)\overline{g(z)} 1037 y^k\right)\,\frac{dx\,dy}{y^2}.$
1038
1039This could be considered either an integral over the fundamental region
1040or, noting that the integrand is invariant under $SL_2(\Z)$, an integral
1041over the quotient space.  One then checks that for $f,g$ cusp forms, the
1042integral will converge, since they go down exponentially as $z$ tends to
1043infinity.  It is then clear that this inner product is Hermitian.
1044
1045It is not immediately clear that the Hecke operators are self-adjoint.
1046\section*{January 31, 1996}
1047\noindent{Scribe: William Stein, \tt <was@math>}
1048\bigskip
1049
1050{\bf \large Modular Curves}\smallskip
1051
1052
1053
1054\subsection{Cusp Forms}
1055Recall that if $N$ is a positive integer we define the congruence
1056subgroups
1057$\Gamma(N)\subset\Gamma_1(N)\subset\Gamma_0(N)$ by
1058$$1059\begin{array}{cl} 1060\Gamma_0(N) & = \{\abcd \in \modgp : c\equiv 0 \pmod{N}\}\\ 1061\Gamma_1(N) & = \{\abcd \in \modgp : a\equiv d\equiv 1, c\equiv 0 \pmod{N}\}\\ 1062\Gamma(N) & = \{\abcd \in \modgp : \abcd \equiv 1063 \bigl(\begin{array}{cc}1&0\\0&1\end{array}\bigr) \pmod{N}\} 1064\end{array} 1065$$
1066
1067Let $\Gamma$ be one of the above subgroups.
1068One can give a construction of the space $S_k(\Gamma)$ of cusp forms
1069of weight $k$ for the action of $\Gamma$ using the language of
1070algebraic geometry.
1071Let $X_{\Gamma}=\Gamma\backslash\H^{*}$
1072be the upper half plane (union the cusps)
1073modulo the action of $\Gamma$. Then $X_{\Gamma}$ can be given the structure
1074of Riemann surface. Furthermore,
1075$S_2(\Gamma)=H^0(X_{\Gamma},\Omega^1)$ where
1076$\Omega^1$ is the sheaf of differential 1-forms on $X_{\Gamma}$.
1077This works since an element of $H^0(X_{\Gamma},\Omega^1)$
1078is a differential form $f(z)dz$, holomorphic on $\H$ and
1079the cusps, which is invariant with respect to the action
1080of $\Gamma$. If $\gamma=\abcd\in\Gamma$ then
1081$$d(\gamma(z))/dz=(cz+d)^{-2}$$
1082so
1083$$f(\gamma(z))d(\gamma(z))=f(z)dz$$
1084iff $f$ satisfies the modular condition
1085$$f(\gamma(z))=(cz+d)^{2}f(z).$$
1086
1087There is a similiar construction when $k>2$.
1088
1089\subsection{Modular Curves}
1090$\modgp\backslash\H$ parametrizes isomorphism
1091classes of elliptic curves. The other congruence subgroups also
1092give rise to similiar parametrizations.
1093$\Gamma_0(N)\backslash\H$ parametrizes pairs $(E,C)$ where
1094$E$ is an elliptic curve and $C$ is a cyclic subgroup of order
1095$N$, and $\Gamma_1(N)\backslash\H$ parametrizes pairs $(E,P)$ where
1096$E$ is an elliptic curve and $P$ is a point of exact order $N$.
1097Note that one can also give a point of exact order $N$ by giving
1098an injection $\Z/N\Z\hookrightarrow E[N]$
1099or equivalently an injection $\mu_N\hookrightarrow E[N]$
1100where $\mu_N$ denotes the $N$th roots of unity.
1101$\Gamma(N)\backslash\H$ parametrizes pairs $(E,\{\alpha,\beta\})$
1102where $\{\alpha,\beta\}$ is a basis for
1103$E[N]\isom(\Z/N\Z)^2$.
1104
1105The above quotients spaces are {\em moduli spaces} for the
1106{\em moduli problem} of determining equivalence classes of
1107pairs ($E +$ extra structure).
1108
1109\subsection{Classifying $\Gamma(N)$-structures}
1110\begin{dfn}
1111Let $S$ be an arbitrary scheme. An {\bfseries elliptic curve}
1112$E/S$ is a proper smooth curve
1113$$\begin{array}{c} E \\ \bigl| \\ S\end{array}$$
1114with geometrically connected fibers all of genus one, together with a
1115section 0''.
1116\end{dfn}
1117
1118Loosely speaking, proper generalizes the notion of projective,
1119and smooth generalizes nonsingularity. See Chapter III, section 10 of
1120Hartshorne's {\em Algebraic Geometry} for the precise definitions.
1121
1122\begin{dfn}
1123Let $S$ be any scheme and $E/S$ an elliptic curve.
1124A {\bfseries $\Gamma(N)$-structure} on $E/S$ is
1125a group homomorphism
1126$$\varphi:(\Z/N\Z)^2\into E[N](S)$$
1127whose image generates'' $E[N](S)$.
1128\end{dfn}
1129
1130See Katz and Mazur, {\em Arithmetic Moduli of
1131Elliptic Curves}, 1985, Princeton University Press, especially
1132chapter 3.
1133
1134Define a functor from the category of $\Q$-schemes to the
1135category of sets by sending a scheme $S$ to the
1136set of isomorphism classes of pairs
1137 $$(E, \Gamma(N)\mbox{\rm -structure})$$
1138where $E$ is an elliptic curve defined over $S$ and
1139isomorphisms (preserving the $\Gamma(N)$-structure) are taken
1140over $S$. An isomorphism preserves the $\Gamma(N)$-structure
1141if it takes the two distinguished generators to the two
1142distinguished generators in the image (in the correct order).
1143
1144\begin{thm}
1145For $N\geq 4$ the functor defined above is representable and
1146the object representing it is the modular curve $X(N)$ corresponding
1147to $\Gamma(N)$.
1148\end{thm}
1149
1150What this means is that given a $\Q$-scheme $S$, the
1151set $X(S)=Mor_{\Q\mbox{\rm -schemes}}(S,X)$ is isomorphic to
1152the image of the functor's value on $S$.
1153
1154There is a natural way to map a pair $(E,\Gamma(N)\mbox{\rm -structure})$
1155to an $N$th root of unity.
1156If $P,Q$ are the distinguished basis of $E[N]$ we send
1157the pair $(E,\Gamma(N)\mbox{\rm -structure})$ to
1158 $$e_N(P,Q)\in\mu_N$$
1159where $$e_N:E[N]\times E[N]\into \mu_N$$ is the Weil pairing. For
1160the definition of this pairing see chapter III, section 8 of
1161Silverman's {\em The Arithmetic of Elliptic Curves}. The Weil pairing
1162is bilinear, alternating, non-degenerate, galois invariant, and
1163maps surjectively onto $\mu_N$.
1164\section*{February 2, 1996}
1165\noindent{Scribe: Lawren Smithline, \tt <lawren@math>}
1166(Note.  These were done about a month after the fact, since the assigned
1167person dropped the course.  So they are terse.)
1168\bigskip
1169
1170Earlier, we looked at V. Miller's construction for eigenforms.  (See, for
1171instance, Lang X \S4 in the course references.  This is a special miracle for
1172$\SL_2(\Z)$.)  Over $\Z$, $\T \cong \Z^d$ and $S(\Z)$ are dual, where $d = 1173\dim S_k(\C)$.
1174
1175Here is Shimura's explanation (Lang III \S5, VIII).  The Hecke operator $\T_n$
1176maps $S_k$ to itself.  Let $A\subset \C$ be a subring and
1177$$\T_A = A[\T_n : n > 0] \subseteq {\rm End}_\C S_k.$$
1178Denote by $\T$, $\T_\Z$.  There is a natural tensor product $\T_A \otimes_A 1179\C \twoheadrightarrow \T_\C$.
1180
1181There is also a complex conjugation automorphism of $S_k$ by
1182$f \mapsto \overline{f(-\bar\tau)}$.  This map is conjugate linear.
1183The map $\tau \mapsto \exp(2\pi i \tau)$ becomes
1184$\tau \mapsto \overline{\exp(-2\pi i \bar\tau)} = \exp(2 \pi i \tau)$.
1185Say $f = \sum a_n q^n.$  Its conjugate
1186$g = \sum \bar a_n q^n.$  If you know $S_k(\C) = \C \otimes_\Q S_k(\Q)$,
1187then you know that modular forms can be conjugated in this sense.
1188
1189There is an isomorphism $\T_\R \otimes_\R \C \stackrel\sim\longrightarrow 1190\T_\C$ since the map is surjective and the complex dimensions on each side
1191are equal.
1192
1193Shimura (1959) exhibited the (Eichler-)Shimura isomorphism
1194$$S_k(\C) \cong H^1(X_\Gamma,\R).$$
1195We have $S_k(\C) = H^0(X_\Gamma, \Omega^1)$ and a map
1196$H_1(X_\Gamma,\Z) \times S_k(\C) \rightarrow \C$.
1197Now, $H_1(X,\Z) \cong \Z^{2d}$ embeds in $\Hom_\C(S_k(\C),\C) \cong 1198\C^{2d}$ as a lattice.
1199
1200So we have $S_k(\C) \rightarrow \Hom(H_1(X,\Z),\C))$ and
1201$S_k(\C) \stackrel\sim\rightarrow \Hom(H_1(X,\Z),\R))$ as real vector
1202spaces.  By the Shimura isomorphism, this is isomorphic to $H^1(X,\R) \sim 1203H^1_p(\Gamma,\R)$, the parabolic cohomology of $\Gamma$.
1204
1205So $S_k(\C) \cong H^1_p(\Gamma, V_k)$, for a certain $d-1$ dimensional
1206subspace $V_k$.  (Let $W = \R \oplus \R$. $\Gamma$ acts by linear fractional
1207transformation.  Let $V_k = {\rm Sym}^{k-2} W$.  There is a lattice in
1208$S_k(\C)$
1209corresponding to $H^1_p(\Gamma, {\rm Sym}^{k-2} \Z^2)$)  We have an action
1210of
1211$\Gamma$ by $$f\cdot \c \mapsto \int_{\tau_0}^{\c(\tau_0)} 1212f(\tau)\tau^{k-1}d\tau.$$
1213
1214Recall $\T = \T_\Z$ is a set of endomorphisms of a lattice $L$, and $\T$
1215has finite rank over $\Z$.  We have the inclusion
1216$S_k(\Z) \hookrightarrow S_k(\C)$, or equivalently, $$\Hom_\Z(\T,\Z) 1217\hookrightarrow \Hom_\Z(\T,\C) = \Hom(\T_\C,\C) = S_k(\C) = S_k(\Z) 1218\otimes_\Z \C.$$
1219
1220Here is a nifty inner product (the Petersson innner product) on $S_k(\C)$.
1221For $f,g \in S_k(\C)$, let
1222$$\lan f,g \ran = \int_{\Gamma \backslash \H} f(\tau)g(\tau) y^k 1223\frac{dx\,dy}{y^2}.$$
1224The Hecke operators are self-adjoint for $(p, N) = 1$:
1225$$\lan f | T_p, g \ran = \lan f, g | T_p \ran.$$
1226Indeed, for $\a \in \GL_2^+ (\R)$,
1227$$\lan f | \a , g | \a \ran = \lan f,g \ran.$$
1228
1229\section*{February 5, 1996}
1230\noindent{Scribe: Shuzo Takahashi, \tt <shuzo@math>}
1231\bigskip
1232
1233We have studied actions of $\T$ on
1234$S_k(\C)$, $S_k(\Q)$ and $S_k(\Z)$ where
1235$\Gamma = SL_2(\Z)$.  What we know so far is
1236$$S_k(\Z) \simeq {\rm Hom}_{\Z}(\T,\Z).$$
1237Also we have studied the Peterson product. It is Hermitian, i.e.,
1238$$\lan f|T_n, g\ran = \lan f,g|T_n\ran$$
1239for $T_n \in \T$ and for $f,g \in S_k(\C)$.
1240
1241\noindent
1242Note: $T_n$ defined on $S_k(\C)$ preserves $S_k(\Z)$.
1243
1244Today we study when $\Gamma = \G_1(N)$ or $\G_0(N)$ for
1245$N \geq 1$.
1246
1247\medbreak
1248\noindent
1249{\bf\large 1. The Diamond Operator and the Decomposition of
1250$S_k(\G_1(N))$}
1251
1252\begin{thm}
1253$\G_1(N)$ is a normal subgroup of $\G_0(N)$ and we have
1254$\G_0(N)/\G_1(N) \simeq (\Z/n\Z)^*$.
1255\end{thm}
1256
1257\begin{dfn}
1258The diamond operator $< >$ is defined as follows: for
1259$\pmatrix{a & b \cr c & d \cr} \in \G_0(N)$, the map on $S_k(\G_1(N))$
1260$$f \rightarrow f | \pmatrix{a & b \cr c & d \cr}$$
1261defines an endmorphism $< d>$ of $S_k(\G_1(N))$ which depends only
1262on $d \ {\rm mod}\ N$. Thus we get an action $< >$ of
1263$(\Z/n\Z)^*$ on $S_k(\G_1(N))$.
1264\end{dfn}
1265
1266Since $(\Z/n\Z)^*$ is a finite group, we have the following decomposition
1267theorem:
1268
1269\begin{thm}
1270$$S_k(\G_1(N)) = \bigoplus_{\e} S_k(\G_0(N),\e)$$
1271where $\e$ runs over the set of characters $(\Z/n\Z)^* \rightarrow \C^*$
1272and $S_k(\G_0(N),\e)$ is defined as:
1273$$S_k(\G_0(N),\e) = \{ f \in S_k(\G_1(N)) : f | < d> = \e(d) f\}.$$
1274We have  $S_k(\G_0(N),\e) = 0$ unless $\e(-1) = (-1)^k$.
1275\end{thm}
1276
1277\medbreak
1278\noindent
1279{\bf\large 2. The Hecke Operators on $S_k(\G_1(N))$}
1280
1281For $n \geq 1$, we have the operation on $S_k(\G_1(N))$ of the $n$th Hecke
1282operator $T_n$. The following are basic properties:
1283
1284\begin{thm}
1285
1286(1) $T_n$'s commute each other and with $< d>$.
1287
1288(2) $T_n$'s preserve $S_k(\G_0(N),\e)$.
1289
1290(3) if $(n,N) = 1$, then $\lan f | T_n,g\ran = 1291\lan f,g|{< n>}^{-1}T_n\ran$.
1292
1293(4) $\lan f | < d> ,g\ran = 1294\lan f,g|{< d>}^{-1}\ran$.
1295
1296(5) if $(n,N) = 1$,  then $T_n$ is diagonalizable.
1297
1298(6) if $(n,N) \neq 1$,  then $T_n$ is not diagonalizable.
1299
1300\end{thm}
1301
1302The action of $T_n$ is described in the following theorem:
1303
1304\begin{thm}
1305Let $f = \sum_{n = 1}^\infty a_n q^n \in S_k(\G_0(N),\e)$. Then
1306
1307%$$f | T_p = 1308%\cases 1309%\sum_{n=1}^{\infty} a_{pn} q^n + p^{k-1}\e(p) \sum_{n=1}^{\infty} a_n 1310%q^{pn} 1311%&\text{if p \not| N;} \\ 1312%\sum_{n=1}^{\infty} a_{pn} q^n 1313%&\text{if p | N.} 1314%\endcases$$
1315
1316(1) $f | T_p = \sum_{n=1}^{\infty} a_{pn} q^n + p^{k-1}\e(p) 1317\sum_{n=1}^{\infty} a_n q^{pn}$.
1318(Note: when $p|N$, $\e(p)$ is considered to be $0$ and $U_p$ is used instead
1319of $T_p$ which is called Atkin-Lehner operator.)
1320
1321(2) if $(n,m)=1$, then $T_{nm} = T_n T_m$.
1322
1323(3) if $p \not| N$, then $T_{p^l} = T_{p^{l-1}}T_p - p^{k-1}< p> 1324T_{p^{l-2}}$.
1325
1326(4) if $p | N$, then $T_{p^l} = (T_p)^l$.
1327\end{thm}
1328
1329The last formula in the theorem can be proved by comparing the coefficients
1330of $q^{p^{l-1}}$ in both sides of the following formal identity:
1331$$\left(\sum_{n=1}^{\infty} T_n q^n\right)|T_p = 1332\sum_{n=1}^{\infty} T_{pn} q^n + p^{k-1}< p> 1333\sum_{n=1}^{\infty} T_n q^{pn}.$$
1334For example, the coefficient of $q^p$ in the LHS is $T_p T_p$. On the other
1335hand, the coefficient of $q^p$ in the RHS is
1336$T_{p^2} + p^{k-1}< p> T_1$. Thus, we have
1337$$T_{p^2} = (T_p)^2 - p^{k-1}< p> Id$$
1338where $< p>$ should be considered to be a null map if $p | N$.
1339
1340\medbreak
1341\noindent
1342{\bf\large 3. The Old Forms}
1343
1344Suppose $M|N$. Let $f \in S_k(\Gamma_1(M))$. Then for $d$ such that
1345$d | \frac{N}{M}$, $f(d\tau) \in S_k(\G_1(N))$.
1346Thus we have a map
1347$$\phi_M : \bigoplus_{d | \frac{N}{M}} S_k(\Gamma_1(M)) \rightarrow 1348S_k(\G_1(N)).$$
1349The old part of $S_k(\G_1(N))$ is defined as the subspace generated by the
1350images of $\phi_M$ for $M | N$, $M \neq N$.
1351
1352\begin{eg} $\phi_M$ is not injective. Consider the case that $k = 12$,
1353$M = p$ and $N = p^2$. $S_k(\Gamma_1(p))$ contains $\Delta(\tau)$ and
1354$\Delta(p\tau)$. But $\phi_p$ maps both of them to $\Delta(p\tau)$ in
1355$S_k(\Gamma_1(p^2))$.
1356\end{eg}
1357
1358\begin{thm}
1359Suppose $p \nmid N$. Consider $f,g \in S_k(\G_1(N))$.  Then $f$ and
1360$g(p\tau)$ are both in $S_k(\Gamma_1(Np))$.  Then we have
1361$$f | U_p = (f | T_p) - p^{k-1}\e(p)(f(p\tau))$$
1362and
1363$$g(p\tau) | U_p = g(\tau)$$
1364where $f | T_p$ is considered in $S_k(\G_1(N))$.
1365\end{thm}
1366\noindent
1367{\bf Proof.} Let $f = \sum_{n=1}^{\infty} a_n q^n$. Then, considering in
1368$S_k(\G_1(N))$, we have
1369$$1370f | T_p = 1371\sum_{n=1}^{\infty} a_{pn} q^n + p^{k-1}\e(p) \sum_{n=1}^{\infty} a_n 1372q^{pn}.$$
1373Also, considering in $S_k(\Gamma_1(Np))$, we have
1374$$f | U_p = \sum_{n=1}^{\infty} a_{pn} q^n.$$
1375Thus, we have
1376$$f | U_p = (f | T_p) - p^{k-1}\e(p)(f(p\tau)).$$
1377Now, let $g = \sum_{n=1}^{\infty} b_n q^n$. Then
1378$$g(p\tau) | U_p = 1379\left(\sum_{n=1}^{\infty} b_{n/p} q^n \right) | U_p = g(\tau)$$
1380where $b_{n/p} = 0$ unless $p | n$.
1381
1382
1383
1384\section*{February 7, 1996}
1385\noindent{Scribe: Amod Agashe, \tt <amod@math>}
1386\bigskip
1387
1388We are in the process of showing that the Hecke operators $T_p$ acting
1389on the space of cusp forms $S_k(\Gamma_1(N))$ are not necessarily
1390semisimple if $p\mid N$.
1391
1392Recall from last time that if $M \mid N$ then for every divisor $d$
1393of $M/N$, we had a map $S_k(\Gamma_1(M)) \rightarrow S_k(\Gamma_1(N))$
1394given by $f(\tau) \mapsto f(d\tau)$.
1395
1396Note that the various $f(d\tau)$'s are linearly independent over $\C$,
1397because the Fourier expansion of $f(d\tau)$ starts with $q^d$.
1398
1399Let $f$ be an eigenfuntion for all the Hecke operators $T_n$ in
1400$S_k(\Gamma_1(M))$. Let $p$ be a prime not dividing $M$.
1401So $f\mid T_p = af$ where $a=a_p(f)$ and $f \mid <p>=\epsilon (p)f$
1402where $\epsilon (p)$ is the character associated to the modular
1403form $f$. Note that one can prove that if $f$ is an eigenfunction
1404for the $T_n$'s then it is an eigenfunction for the diamond operators
1405also (or alternatively, make it part of the definition of eigenform).
1406Let $N=p^\alpha M$ with $\alpha \geq 1$. We will look at the action of
1407the $p^{th}$ Hecke operator $U_p$ in $S_k(\Gamma_1(N))$ on the
1408images of $f$ under the maps described above. Let $f_i(\tau) = f(p^i \tau)$
1409for $0\leq i\leq \alpha$. As we showed earlier, \\
1410$$f\mid T_p = \sum a_{np}q^n + \epsilon (p)p^{k-1}\sum a_nq^{pn}.$$
1411So $$af = f_0\mid U_p + \epsilon (p)p^{k-1}f_1.$$
1412Thus, $$f_0\mid U_p = af_0 -\epsilon(p)p^{k-1}f_1.$$
1413From last time, we have $f_1\mid U_p = f_0.$ In fact, in general,
1414one can see easily that $f_i\mid U_p=f_{i-1}$ for $i\geq 1$.
1415
1416So $U_p$ preserves the 2-dimensional space spanned by $f_0$ and $f_1$.
1417The matrix of $U_p$ (acting on the right) with this basis is given
1418(from the equations above) by: \smallskip
1419$\left( \begin{array}{cc} 1420 a & 1 \\ 1421 -\epsilon(p)p^{k-1} & 0 1422 \end{array} 1423 \right) 1424$
1425The characteristic polynomial of this matrix is $x^2-ax+p^{k-1}\epsilon(p)$.
1426
1427There is the following striking coincidence:
1428Let $E$ be the number field generated over $\Q$ by the coefficients of the
1429Fourier series expansion of $f$ and let $\la$ be a prime ideal
1430of $\O_E$ lying over some rational prime $l$.
1431Then we have a Galois representation \smallskip
1432$$\rho_\la: Gal(\overline{\Q}/\Q) \rightarrow \GL_2(E_\la)$$
1433If $p\not|Nl$ then $\rho_\la$ is unramified and also
1434$det\ \rho_\la(Frob_p)=\epsilon(p)p^{k-1}$ and
1435$tr\ \rho_\la(Frob_p)=a_p(f)=a$. Thus the characteristic
1436polynomial of $\rho_\la(Frob_p)$ is $x^2-ax+p^{k-1}\epsilon(p)$,
1437the same as that of the matrix of $U_p$!
1438
1439A question one can ask is: Is $U_p$ semisimple on the space spanned by
1440$f_0$ and $f_1$? The answer is yes if the eigenvalues of $U_p$ are
1441different.
1442
1443Now, the eigenvalues are the same iff the
1444discriminant of the characteristic polynomial is zero i.e.
1445$a^2=4\epsilon(p)p^{k-1}$ i.e. $a=2p^{\frac{k-1}{2}}\zeta$ where
1446$\zeta$ is some square root of $\epsilon(p)$.
1447Here is a curious fact: the Ramanujan-Petersson conjecture proved by Deligne
1448says $|a|\leq 2p^{\frac{k-1}{2}}$; thus the above equality is allowed
1449by it, so we do not get any conclusion about the semisimplicity of
1450$U_p$.
1451
1452Let us now specialize to $k=2$. Weil has shown that $\rho_\la(Frob_p)$
1453is semisimple. Thus if the eigenvalues of $U_p$ are equal, then
1454$\rho_\la(Frob_p)$ is a scalar. Edixhoven proved that it is not.
1455So the eigenvalues of $U_p$ are different and hence $U_p$ is semisimple
1456in this case. So this example (for k=2) does not give us an example
1457of $U_p$ being not semisimple.
1458
1459There is the following example given by Shimura which shows that the Hecke
1460operator $U_p$ need not be semisimple. Let $W$ denote the space spanned
1461by $f_0, f_1$ and $V$ denote the space spanned by $f_0,f_1,f_2,f_3$.
1462$U_p$ preserves both spaces $W$ and $V$, so it acts on $V/W$. The action
1463is given by $\overline{f_2}\mapsto \overline{f_1}=0$ and
1464$\overline{f_3}\mapsto \overline{f_2}$ where the bar denotes the image
1465in $V/W$. Thus the matrix of $U_p$ on the space $V/W$ is
1466$\left( \begin{array}{cc} 1467 0 & 1 \\ 1468 0 & 0 1469 \end{array} 1470 \right) 1471$
1472which is nilpotent, and in particular not semisimple. If $U_p$ were
1473semisimple on $V$ then it would be semisimple on $V/W$ also; but we
1474have just shown that it is not. Thus $U_p$ is not semisimple
1475on $V$, and hence not on $S_2(\Gamma_1(M))$ (because $V$ is invariant
1476under $U_p$).
1477
1478\bigskip
1479
1480We next discuss the structure of the $\C$-algebra $\T =\T_\C$ generated
1481by the Hecke and diamond operators and the structure of $S_k(\Gamma_1(N))$
1482as a $\T$-module.
1483
1484First we consider the case of level $1$ i.e. $N=1$.
1485Then $\Gamma_1(1)=SL_2(\Z)$. All the $T_n$'s are diagonalizable.
1486$S_k=S_k(\Gamma_1(N))$ has a basis of $f_1,....,f_d$ of normalized eigenforms
1487where $d=dim(S_k)$. Thus $S_k\cong \C^d$ as a $\C$-vector space.
1488Then we have the $\C$-algebra homomorphism  $\T\rightarrow \C^d$ given by
1489$T\mapsto (\la_1,....,\la_d$) where $f_i\mid T=\la_i f_i$.
1490It is injective because if the image of $T$ is zero, then it kills all
1491$f_i$ i.e. all of $S_k$ i.e. it is the zero operator. The map is
1492surjective because $\T$ has dimension $d$.
1493%and hence it is surjective
1494%when we consider both sides as $\C$-vector spaces.
1495Thus as a $\C$-algbebra, $\T\cong \C^d$.
1496Next, we claim that the modular form $v=f_1+...+f_d$ generates
1497$S_k$ as a $\T$-module. This follows because under the
1498map $S_k\cong \C^d, v\mapsto (1,....,1)$ and our
1499statement is just the trivial fact that $(1,....,1)$ generates
1500$\C^d$ as a $\C^d$-module (acting component-wise).
1501
1502Thus $S_k$ is free of rank $1$ as a $\T$-module.
1503We already know that $S_k\cong Hom(\T,\C)$ as $\T$-modules.
1504Thus $\T\cong Hom(\T,\C)$ as $\T$-modules. In fact the isomorphism
1505is canonical since the $f_i$'s are normalized.
1506We remark that $v$ in fact lies in $S_k(\Q)$.
1507
1508Next, we deal with the general case where the level is not necessarily $1$.
1509
1510First we need to talk about newforms. Recall the maps
1511$S_k(\Gamma_1(M)) \rightarrow S_k(\Gamma_1(N))$
1512for every divisor $d$ of $M/N$ mentioned at the beginning of this lecture.
1513The old part of $S_k(\Gamma_1(N))$ is defined as the space generated by
1514all the images of $S_k(\Gamma_1(M))$ for all $M\mid N, M\neq N$
1515under these maps.
1516The new part of $S_k(\Gamma_1(N))$ can be defined in two different ways.
1517Firstly we can define it as the orthogonal complement of the old part
1518with respect to the Petersson inner product.
1519There is also an algbraic definition
1520as follows. There are certain maps going the other way:
1521$S_k(\Gamma_1(N)) \rightarrow S_k(\Gamma_1(M))$ for $M\mid N, M\neq N$.
1522The new part is the space killed under all these maps. The space
1523of newforms, denoted $S_k(\Gamma_1(N))_{new}$ is like $S_k(\Gamma(1))$ in the
1524sense that all the $T_n$'s (including $U_p$) are semisimple and there
1525is a basis consisting of newforms. A form of level $N$ is said to
1526be new of level $N$ if it is in $S_k(\Gamma_1(N))_{new}$.
1527
1528Next, one can show that the map
1529$\bigoplus_{M\mid N,M\leq N} S_k(\Gamma_1(M))_{new} \rightarrow 1530S_k(\Gamma_1(N))$ given by $f(\tau)\mapsto f(d\tau)$ for $d\mid \frac{N}{M}$
1531is injective (See W.-C. W.Li, Newforms and functional equations,
1532Math. Annalen, 212(1975), 285-315).
1533Note that an eigenform in one of the subspaces of the source
1534need not be an eigenfuntion for all the operators in the image.
1535If $f$ is a newform, then let $M_f$ denote its level (i.e. $f$
1536is new of level $M_f$). Let $S$ be the set of newforms of weight $k$
1537and some level dividing $N$. Let
1538$$v=\sum_{f\in S} f(\frac{N}{M_f}\tau).$$
1539Then one can show that $S_k(\Gamma_1(N))$ is free of rank $1$ over
1540$\T_\C$ with $v$ as the basis element. Also one can show that
1541$v$ has rational coefficients.
1542
1543\section*{February 9, 1996}
1544\noindent{Scribe: J\'anos Csirik, \tt <janos@math>}
1545\vspace{.2in}
1546
1548
1549Recall that for the case $\Gamma=SL(2,\Z)$, if we set $f_1,\ldots,f_d$
1550to be the normalized eigenforms (newforms of level 1), then they have
1551possibly complex coefficients but in any case $\{f_i\}$ is finite and
1552stable under automorphisms of $\C$ and all the coefficients
1553$a_n(f_i)$ lie in some number field.  Furthermore this field is totally
1554real: to show this we used that since the set is stable under
1555conjugation, it suffices to show that all the $a_n(f_i)$ are real, which
1556followed by remarking that they are eigenvalues of the operators $T_n$
1558
1559More generally, let $f\in S(\Gamma_1(N))$ be a normalized eigenform of
1560character $\varepsilon$.  Then $a_n=\varepsilon(n)\overline{a_n}$.
1561
1562Note that the algebra $\T_\Q$ generated by the $T_i$ over $Q$ contains
1563the diamond bracket operators: the formula relating $T_{p^2}$ and
1564$(T_p)^2$ tells us that the difference is $p^{k-1}\varepsilon(p)$, so
1565$\varepsilon(p)\in\T_\Q$.  Using Dirichlet's Theorem on primes in
1566arithmetic progressions, for any $d$ relatively prime to $N$ we can find
1567a prime $p\equiv d\pmod{N}$, so $\varepsilon(d)=\varepsilon(p)\in\T_\Q$.
1568
1569If the space of modular forms has dimension 1, then it is spanned by a
1570(normalized) eigenform with rational coefficients, so the eigenvalues
1571are all in $\Q$.
1572
1573The next simplest example is $k=24$, which is the smallest weight such
1574that the dimension is more than one.  There are two eigenforms, which
1575are conjugate to each other.  If $f=\sum a_nq^n$ is an eigenform, then
1576$\Q(\ldots a_n\ldots)=\Q(\sqrt{144169})$.  In fact $S_{24}$ is spanned
1577by $\Delta^2$ and $\Delta^2|T_2$.  (Note that $\Delta^2$ is definitely
1578not an eigenform since its $q$-coefficient is 0.)  The action of $T_2$
1579on $S_{24}$ with respect to this basis is described by a two by two
1580matrix of trace $1080$ and determinant $-2^{10}3^2 2221$.  For high $k$,
1581eigenforms tend to form a single orbit; however, no proof
1582is known for this.
1583
1584For every newform $f$, let $E_f$ be the number field generated by its
1585coefficients.  Let $\Sigma$ be a set of representatives for $f$'s modulo
1586$Gal(\bar\Q/\Q)$.  Define
1587\begin{eqnarray*}
1588   \T_\Q   &   \rightarrow   &   E_f             \\
1589   T       &  \mapsto        &  \lambda_T,
1590\end{eqnarray*}
1591with $f|T=\lambda_Tf$.  Thus $T_n\mapsto a_n$ so this map is surjective.
1592
1593In fact the induced
1594$\T_\Q \to \prod_{f\in\Sigma} E_f$
1595is an isomorphism of $\Q$-algebras: it is injective since if $T$ dies on
1596the image then it acts as zero on $f\in\Sigma$ and so on all of f, by
1597the rationality of Hecke operators: ${}^\sigma\!(g|T)={}^\sigma\!g|T$.
1598(And if an operator acts as zero on everything, then it is zero.)
1599
1600Here we used the fact that there were no oldforms around.
1601
1602For example, consider $S_2(\Gamma(N))$ for $N$ prime.  Then
1603$S_2(\Gamma(1))$ is empty, hence we get an isomorphism
1604$1605 \T_\Q\cong E_1\times\ldots\times E_t, 1606$
1607 with the right hand side a
1608product of totally real number fields.  $t>1$ is possible, e.g. for
1609$N=37$, $\T_\Q=\Q\times\Q$.
1610
1611In general, oldforms complicate the situation.
1612
1613
1614
1616
1617We'll only treat the case $k=2$.  Then (for $\Gamma$ a congruence
1618subgroup),
1619$S_2(\Gamma)=H^0(X_\Gamma,\Omega^1)$
1620where $X_\Gamma=\Gamma\setminus\H\cup\Gamma\setminus\P^1(\Q)$, with
1621$\Gamma\setminus\P^1(\Q)$ being the set of cusps that we need to adjoin
1622to make it a compact Riemann surface.
1623
1624Therefore
1625${\rm dim}\,S_2(\Gamma)= g(X_\Gamma).$
1626
1627{\sc Example.\ } $SL(2,\Z)\setminus\P^1(\Q)$ has just one point.  The
1628proof involves the Euclidean algorithm: any element of $\P^1(\Q)$ can be
1629written as $\left( \begin{array}{cc} x\\y \end{array} \right)$ with $x$
1630and $y$ relatively prime integers.  By the Euclidean Algorithm, we can
1631find $a$ and $b$ integers such that $ax+by=1$.  Then
1632$1633\left(\begin{array}{cc} a & b \\ -y & x \end{array} \right) 1634\left( \begin{array}{cc} x\\y \end{array} \right) = 1635\left( \begin{array}{cc} 1\\0 \end{array} \right) 1636$
1637
1638To calculate $g(X_\Gamma)$, use the following covering (recall that
1639$X_\Gamma$ is the compactification of $\Gamma\setminus\H$ we obtain by
1641$1642 X_\Gamma \rightarrow X_{\Gamma(1)}, 1643$
1644keeping in mind the isomorphism
1645\begin{eqnarray*}
1646     j:X_{\Gamma(1)}   &   \rightarrow   &  \P^1(\C)             \\
1647    (i,\rho,\infty) &  \mapsto        &  (1728,0,\infty)
1648\end{eqnarray*}
1649The only ramification in our covering
1650occurs above the points $0,1728,\infty$.
1651
1652{\sc Example.\ } Let $\Gamma=\Gamma_0(N)$.  The degree of the covering
1653is $(PSL(2,\Z):\Gamma_0(N)/\pm1)$ which is the number of cyclic subgroups
1654of order $N$ in $SL(2,\Z/N\Z)$.  We have a covering
1655$Y_0(N)\to Y_{\Gamma(1)}$ where $Y_0(N)$ parametrizes elliptic curves
1656$E$ with a cyclic subgroup of order $N$,
1657$C\subseteq E[N]\cong\Z/N\Z\times\Z/n\Z$ up to isomorphism; and
1658$Y_{\Gamma(1)}$ parametrises elliptic curves.  The isomorphism
1659$(E,C_1)\cong(E,C_2)$ is an automorphism $\alpha$ of $E$ with
1660$\alpha:C_1\mapsto C_2$.  Usually $\alpha=\pm1$, unless
1661$j=1728,0,\infty$.
1662
1663If we understand ramification, we can use the {\em Riemann-Hurwitz
1664formula}.  The following mnemonic way of thinking about it is due to
1665N. Katz.  The Euler characteristic (alternating sum of the dimensions of
1666cohomology groups) should be thought of as totally additive:
1667$\chi(A\coprod B) = \chi(A) + \chi(B).$
1668
1669If $X$ is a Riemann surface of genus $G$, $\chi(X)=2-2G$.  A single
1670point has Euler characteristic $\chi(P)=1$.  Hence
1671$\chi(X\setminus\{P_1,\ldots,P_n\})=2-2G-n.$
1672Therefore
1673$\chi(X_{\Gamma(1)}\setminus\{0,1728,\infty\})=-1$
1674and
1675$\chi(X_\Gamma\setminus\{\mbox{points over 1728,0,\infty}\})=2-2g-n$
1676if $n$ points lie over ${0,1728,\infty}$.
1677If the covering map $X_\Gamma\to X_{\Gamma(1)}$ has degree $d$, then we
1678can think of the top space as $d$ copies of the bottom space, so
1679$d \cdot (-1) = 2-2g-n$
1680and therefore
1681$2g-2 = d-n = d-n_0 - n_{1728} -n_\infty.$
1682
1683{\sc Example.\ }  Let $\Gamma=\Gamma(N)$.  What happens over $j=0$?
1684This corresponds to an elliptic curve $E$ with an automorphism $\alpha$
1685of order three.  Let $N>3$.
1686
1687For any $(E,P,Q)$, we have $(E,\alpha P,\alpha Q)$ and
1688$(E,\alpha^2 P,\alpha^2 Q)$ which are isomorphic to it and hence they
1689are the same point on $X_\Gamma$.  So $E$ only has one-third the usual
1690number of points lying over it, $n_0=d/3$.  (Except if the above three
1691points are equal, i.e., $\alpha$ fixes $E[N]$.  This can't happen since
1692$N>3$.)
1693
1694Similarly we get $n_{1728}=d/2$.
1695
1696To determine the degree $d$, fix $E$ and count the points lying over it:
1697these are all of the form $(\C/\Z\oplus\tau\Z,1/n,\tau/N)$ with Weil
1698pairing $e^{2\pi i/N}$ and all such occur, so we need to
1699count the number of $P,Q\in E[N]$ which form a basis of
1700$E[N]$ and $e_N\langle P,Q\rangle=e^{2\pi i/N}$.  This gives us the
1701order of $SL(2,\Z/n\Z)$.  However, (since $N\neq2$) we have to take into
1702account that $(E,P,Q)\cong(E,-P,-Q)$ but they are not equal
1703so the degree of the covering is
1704$\#(SL(2,\Z/N\Z))/2$.
1705
1706We also have $d=(PSL(2,\Z):\Gamma(N)/(\Gamma(N)\cap\pm1))$ since
1707$\Gamma(N)\setminus SL(2,\Z)\cong SL(2,\Z/N\Z)$.
1708
1709So we have established that
1710$2g-2 = d-n_0 - n_{1728} -n_\infty = d/6 - n_\infty.$
1711To determine $n_\infty$, note that $SL(2,\Z)$ acts on
1712$\P^1(\Q)\ni\infty= 1713\left( \begin{array}{cc} 1 \\ 0 \end{array} \right)$.
1714The stabilizer $\Gamma(N)_\infty$
1715of $\left( \begin{array}{cc} 1 \\ 0 \end{array} \right)$
1716is $U=\pm\left( \begin{array}{cc} 1 & * \\ 0 & 1 \end{array} \right)$
1717So the index
1718$((\modgp/\pm1)_\infty:\Gamma(N)_\infty)=N$ is the ramification degree
1719of a point over $\infty$, so $n_\infty=N/d$.
1720
1721Hence $2g(X(N))-2=d/6-d/N$.
1722\section*{February 12, 1996}
1723\noindent{Scribe: J\'anos Csirik, \tt <janos@math>}
1724\vspace{.2in}
1725
1726\def\T{{\bf T}}   % Hecke algebra
1727\def\qed{\hfill $\blacksquare$}
1728
1729Plug: A useful reference for the next lecture is Andrew Ogg: {\em Rational
1730points on certain elliptic modular curves} (1972).
1731
1732From last lecture's results on easily deduces that
1733$g(X(N)) = 1 + \frac{d}{12N}(N-6).$
1734
1735{\sc Example.}  Let $N=5,7$.  If $N$ is prime then the degree of the
1736covering is $\frac{(N^2-1)(N^2-N)}{N-1}$.  Therefore $N=5$ gives $d=60$
1737and $g=0$ (the Galois group of the covering in this case is $A_5$.)
1738Similarly, $N=7$ yields $g=3$ and $d=168$.  (Remark: For $p\geq5$ prime, the
1739group $SL(2,\Z/p\Z)/\pm1=L_2(p)$ is simple.)
1740
1741{\sc Example.}  What is the genus of $X_0(N)$, for $N$ a prime?
1742
1743It is an exercise to show that there are two cusps:
1744$\left( \begin{array}{cc} 1 \\ 0 \end{array} \right)=\infty$
1745and
1746$\left( \begin{array}{cc} 0 \\ 1 \end{array} \right)=0$.
1747The covering $X_0(N)\to X(1)$ is easily seen to have degree $N+1$, since
1748an elliptic curve has $N+1$ cyclic subgroups of order $N$.  Here
1749$\infty$ is unramified and $0$ has ramification index $N$.
1750Hence
1751$2g-2 = N+1 -n_\infty-n_{1728}-n_0$
1752with $n_\infty=2$, $n_{1728}$ approximately $d/2$ and
1753$n_0$ approximately $d/3$.
1754
1755To calculate $n_0$, we need the number of isomorphism classes $(E,C)$
1756with $E$ fixed with $End(E)=\Z[\mu_6]$.  $\mu_6$ acts on the
1757set of $C$'s.  Since $\pm1$ is acting trivially, we really have an
1758action of $\mu_3$ with $(E,C)\cong(E,\zeta C)\cong(E,\zeta^2C)$ with
1759$\zeta$ some third root of unity.
1760
1761If we consider $E=\C/\O$ with $\O=\Z[(-1+i\sqrt3)/2]$, then
1762$E[N]=\O/N\O$ as an $\O$-module.  In fact
1763$\O/N\O = 1764 \left\{ \begin{array}{ll} 1765 \F_N\oplus\F_N & \mbox{if N splits in \O} \\ 1766 \F_{N^2} & \mbox{if N does not split in \O.} 1767 \end{array} 1768 \right. 1769$
1770The above covers all possibilities, because $N>3$ can't ramify in $\O$.
1771By Kummer's theorem, $N$ splits in $\O$ iff
1772$\left(\frac{-3}{N}\right)=1$, and
1773there are exactly two $\O$-stable submodules of $\O/N\O$.  In the second
1774case, which happens iff $\left(\frac{-3}{N}\right)=-1$, $\O/N\O$ has no
1775$\O$-stable submodules.  Therefore
1776$n_0 = 1777 \left\{ \begin{array}{ll} 1778 \frac{N-1}{3} + 2 & 1779 \mbox{if N splits in \O, i.e., N\equiv1\pmod3.}\\ 1780 \frac{N+1}{3} & 1781 \mbox{if N does not split in \O, i.e., N\equiv2\pmod3} 1782 \end{array} 1783 \right. 1784$
1785In the first case, note that the two $\O$-stable submodules have to be
1786counted separately.
1787
1788Similarly we obtain
1789$n_{1728} = 1790 \left\{ \begin{array}{ll} 1791 \frac{N-1}{2}+2 & \mbox{if N\equiv1\pmod4} \\ 1792 \frac{N+1}{2} & \mbox{if N\equiv3\pmod4} 1793 \end{array} 1794 \right. 1795$
1796depending on whether$N$ splits in $\Q(i)$.
1797
1798{\sc Examples.}  For $N=37$ the genus is $2$.  Using the formula, we get
1799$2g-2=36-(2+18)-14=2$ (so it works).
1800
1801Using the formula we can also obtain $g(X_0(13))=0$ and $g(X_0(11))=1$.
1802
1803It is therefore clear that the genus of $X_0(N)$ is approximately
1804$N/12$.  In the article by Serre in the Lecture Notes in Mathematics 349
1805(Antwerp), we find the following table.  Write $N=12a+b$ with $0\leq 1806b\leq11$.  Then
1807\begin{center}
1808\begin{tabular}{c|cccc}
1809  $b$  &  $1$  &  $5$  &  $7$  &  $11$  \\ \hline
1810  $g$  & $a-1$ &  $a$  &  $a$  & $a+1$.  \\
1811\end{tabular}
1812\end{center}
1813Hence
1814$1815 1 + g(X_0(p)) = {\rm dim}\,M_{p+1}(\modgp). 1816$
1817
1818It was Serre's idea to think of modular forms mod $p$'', for some
1819congruence subgroup
1820$\Gamma\ni\left(\begin{array}{cc}1&1\\0&1\end{array}\right)$, like
1821$\Gamma_0(N)$ or $\Gamma_1(N)$.  We could use our moduli theoretic
1823$1824 M_k(\Gamma,\F_p) \subseteq \F_p[[q]]. 1825$
1826
1827By Shimura's cohomology trick, we know that $M_k(\Gamma,\Z)$ is a
1828lattice in $M_k(\Gamma,\C)$.  Hence we can set
1829$1830 M_k(\Gamma,\F_p) = M_k(\Gamma,\Z) \otimes_{\Z} \F_p. 1831$
1832Then {\em Serre's equality} states that for a prime $p$,
1833$1834 M_{p+1}(\modgp,\F_p) = M_2(\Gamma_0(p),\F_p) 1835$
1836in $\F_p[[q]]$.  The philosophy is {\em mod $p$ forms with $p$ in the
1837level can be taken to mod $p$ formswith no $p$ in the level, but of a
1838higher weight}.  So for example
1839$M_k(\Gamma_1(p^\alpha N),\F_p)$ is a subset of
1840$M_?(\Gamma_1(N),\F_p)$ of forms of some higher level.
1841
1842Finally, consider the map from the right hand side to the left hand
1843side in Serre's equality.  Recall that
1844$G_k = \frac{-B_k}{2k} + \sum^\infty_{n=1}\sigma_k(n) q^n 1845 \in M_k(\modgp).$
1846By Kummer, ${\rm ord}_p(B_{p-1}) = -1$, so
1847$1848 E_{p-1}=1+\frac{-2(p-1)}{B_{p-1}}\sum\sigma_{p-1}(n)q^n \equiv 1 \pmod{p}. 1849$
1850Hence we got the map from the right to the left:
1851multiply by $E_{p-1}$ to get to $M_{p+1}(\Gamma_0(p),\F_p)$.  Then take
1852the trace to get to $M_{p+1}(\modgp,\F_p)$.  The trace map is dual to
1853the inclusion and is expressed by
1854$1855 \tr(f) = \sum_{i=1}^{p+1} f | \gamma_i 1856 \qquad 1857 \gamma_i\in\Gamma_0(p)\setminus\modgp. 1858$
1859\section*{February 14 and 16, 1996}
1860\noindent{Scribe: Jessica Polito, \tt <polito@math>}
1861\bigskip
1862
1863
1864Our goal, for these two days, is to define the modular curves $\X$ over
1865$\Q(\mu_N)$, and
1866$X_1(N)$, and $X_0(N)$ over $\Q$.
1867These notes will spell out the construction of $\X$, with some
1868discussion of the construction of the other two types of curves.
1869The idea comes from Shimura.
1870
1871Let $\Q(t)$ be the function field of $\P^1/\Q$, and pick an elliptic
1872curve $\E/\Q(t)$ with $j$-invariant $t$:
1873$$\E: y^2 = 4x^3 - \frac{27t}{t-1728}x - \frac{27t}{t-1728}$$
1874(Note that the general formula for the $j$-invariant of a curve $y^2 = 18754x^3 - g_2x -g_3$ is $j = \frac{1728g_2^3}{g_2^3 - 27g_3^2}$, and
1876that $j(E)$ determines the isomorphism class of the given curve $E$
1877over the algebraic closure of the field of definition.)
1878
1879By substituting in a given value $j$ for $t$, we would get a formula
1880for an elliptic curve over $\Q(j)$ with $j$-invariant $j$; for $j= 0$
1881or 1728, we could pick a diferent formula for $\E$ which would give an
1882isomorphic curve over $\Q(t)$, for which that substitution would make
1883sense.
1884
1885Notice that, in general, if we have an elliptic curve $E/K$, with $K$
1886some field of charictaristic prime to $N$, then we can consider
1887$E[N](\K)$, the set of all $N$-torsion point of $E$ defined over $\K$,
1888which is isomorphic to $(\Z/N\Z)^2$.  Then we let $K(E[N])$ be the
1889smallest extension of $K$ over which all the points of $E[N]$ are
1890defined.  Notice also that $E/K$ and $N$ together define a
1891representation of the Galois groups $\gal(\K/K)$ into the
1892automorphisms of the $N$-torsion points of $E$, as they are defined by
1893polymial equations with coefficients in $E$.  We get
1894$$\rho_{E,N}: \gal(\K/K) \into \aut(E[N]) \isom \GL_2(\Z/N\Z)$$
1895with $\gal(\K/K(E[N])) = \ker(\rho_{E,N})$.  Unsurprisingly, we will
1896frequently leave out the $N$, writing simply $\rho_E$.
1897
1898Now, to return to our construction, where $K=\Q(t)$.  We will show
1899that $\Q(t)(E[N])$ is the function field of a curve defined over