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%% ribetofficial.tex --- Scribe notes for Ken Ribets Spring 1996 course.  %%
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%% This file is being maintained by William Stein ([email protected]).%%  
%% This file was assembled by Lawren Smithline ([email protected]).%%
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%% Last modification date: 9/25/96.                                       %%


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\def\pf{{\sc Proof.\ }}
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\def\F{{\bf F}}   % field
\def\P{{\bf P}}   % projective land

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\def\Q{{\bf Q}}   % rationals
\def\O{{\cal O}}  % ring of integers
\def\T{{\bf T}}   % Hecke algebra
\def\G{{\bf G}}   % group scheme
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\def\ff{{\cal F}} % Modular function field

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\def\q{{\mathfrak q}}   % Gothic q
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\def\det{{\rm det }}    % Determinant
\def\endo{{\rm End}}
\def\Cot{{\rm Cot}}   % contangent space
\def\ver{{\rm ver}}   % Vershibung endmorphism

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\newarrow{To} ---->
\newarrow{Line} -----


\title{Scribe notes for Ken Ribet's Math 274}
\section*{January 17, 1996}
\noindent{Scribe: Lawren Smithline, \tt <lawren@math>}

Here are some topics to be discussed in this course:

Galois representations and modular forms, \\
Hecke algebras, \\
modular curves, and
Jacobians (Abelian varieties).
\end{tabular} \\
This lecture is a brief overview of some connection between these concepts,
and also an exercise in name-dropping.

We can describe an an elliptic curve, or the Jacobian of a higher genus
curve, or abelian variety using a lattice.  For $E$, the lattice is $L =
H_1(E(\C),\Z) \hookrightarrow \C,$ by the map $\c \mapsto \int_\c

Weil considered curves over a finite field, $k$ of characteristic $p$.
There is an algebraic definition of $L/nL$ for $n \geq 1, \ \gcd(n, p) = 1.$
$$E[n] = \{ P \in E(\bar k) : nP = 0\} = \textstyle\frac1n L/L = L/nL.$$
For example, let $n = \l^\nu$ for $\nu \geq 1$, and $\l$ a prime different
from $p.$

Weil further considered the limit $$E[\l^\infty] = \bigcup_{\nu = 1}^\infty
Tate oberserved there is a map $E[\l^\nu] \stackrel{\l}\rightarrow
E[\l^{\nu-1}],$ and so of course this inverse limit, $$
\lim_\leftarrow E[\l^\nu] = T_\l(E)$$ is called the Tate module.

As $E[n]$ is free of rank 2 over $\Z / n\Z$, so is $T_\l(E)$ free of rank 2
over $\Z_\l$.  Also, $V_l(e) = T_\l(E) \otimes \Q_\l$ is a 2 dimensional
vector space over $\Q_\l$.  This is the first example of $\l$-adic \'etale
cohomology.  For topological space $X$, $X \mapsto H_{\mathaccent 19
et}^i(X/\bar k, \Q_\l).$

Here are the names of some cool folks: Taniyama Shimura Mumford Tate.

Now, elliptic curve $E/\Q$ gets an action of $G = \gal(\bar \Q/\Q)$,
and so does $E[n]$.  I.e. $\si(P+Q) = \si(P) + \si(Q).$  So we have a
homomorphism $\rho: G \rightarrow {\rm Aut}(E[n]) = GL_2(\Z/n\Z).$
Since we have an exact sequence $$1 \rightarrow \ker \rho \rightarrow G
\rightarrow {\rm im}\, \rho \rightarrow 0,$$
we get a tower of fields
$$\bar \Q \rightarrow K \rightarrow \Q,$$ and ${\cal G}al(K/\Q) =
{\rm im}\, \rho \subset GL_2( \Z/n\Z).$

And now for something completely different.  We can also get to these
Galois representations via modular forms.  Let $k$ be the weight, such as
2.  Let $N$ be the level.  The complex vector space $S_k(N)$ is the set of
cusp forms on $\Gamma_1(N)$, a finite dimensional vector space, namely, the
set of holomorphic functions $f$ on $\H$ such that
$$ f((az+b)/(cz+d)) = (cz+d)^kf(z)$$ for $$\left( \begin{array}{cc} a & b
c & d \end{array} \right) \in \Gamma_1(N), \ \ a,d \equiv 1, c \equiv 0 (N).$$
Such an $f$ has a power series (or Fourier series) expansion in $q =
\exp(2\pi i z)$: $$f(z) =
\sum_1^\infty c_n q^n.$$
Here is a famous example observed by Ramanujan, and proved by Mordell using
(his) Hecke operators:
$$ q\prod_1^\infty (1 - q^n)^{24} = \sum_1^\infty \tau(n)q^n,$$
for Ramanujan's $\tau$ function.  Now, $\tau(n)\tau(m) =\tau(nm)$ for
$\gcd(n,m) = 1$.  Also, there is a recurrence for prime powers.
Amusingly, the normalized basis element of $S_{12}(1)$ is $$\Delta =
\sum_1^\infty \tau(n)\exp(2\pi i nz).$$  Even more amusingly, $$\tau(n)
\equiv \sum_{d \mid n} d^{11} \ \ (691).$$

Experience and Shimura have shown that there exist $f \in S_k(N)$ such that
$$T_n(f) =c_n \cdot f$$ for all $n \geq 1$ for some scalars $c_n$, and
$$f = \sum c_n q^n$$ for the same $c_n$, and that these $c_n$ are algebraic
integers in a finitely generated number field. That is, $[\Q(c_n: n\geq 1):
\Q]$ is finite.

How can we study and interpret this?  We start with the Hecke ring,
$$\Q[T_n] \subseteq End(S_k(N)).$$
Serre in 1968 said there should be Galois representations attached to forms
of arbitrary weight. Deligne constructed them.  In a broad stroke, one can
say that we get between Galois representations and modular forms via
Frobenius elements.

Next time, we continue with the semihistorical overview.
\section*{January 19, 1996}
\noindent{Scribe: William Stein, \tt <was@math>}

\noindent {\bf\large Modular Representations and Modular Curves} \smallskip

\subsection*{Arithmetic of Modular Forms}
Suppose $f=\sum_{n=1}^{\infty}a_n q^n$ is a cusp form in $S_k(N)$ which is
an eigenform for the Hecke operators.  The Mellin transform associates 
to $f$ the $L$-function $L(f,z)=\sum_{n=1}^{\infty} n^{-s}{a_n}$. 
Let $K=\Q(a_1,a_2,\ldots)$. One can show that the $a_n$ are algebraic
integers and $K$ is a number field. When $k=2$, $f$ is associated
to $f$ an abelian variety
$A_f$ over $\Q$ of dimension $[K:\Q]$, and $A_f$ has a $K$ action. (See
{\em Introduction to the Arithmetic Theory of Automorphic Functions},
Theorem 7.14.)

\begin{eg}[Modular Elliptic Curves] 
If $a_n\in\Q$ for all $n$, then $K=\Q$ and $[K:\Q]=1$.  In this case, $A_f$
is a one dimensional abelian variety, which is an elliptic curve, since
it has nonzero genus.
An elliptic curve arising in this way is called modular. 

Elliptic curves $E_1$ and $E_2$ are {\em isogenous} if there is
a morphism $E_1\into E_2$ of algebraic groups, which has a
finite kernel.

The following conjecture motivates much of the theory. 

Every elliptic curve over $\Q$ is modular, 
that is, isogenous to a curve constructed in the above way. 

For $k\geq 2$, Serre and Deligne found a way to associate to $f$ a family
of $\l$-adic representations. Let $\l$ be a prime number and $K$ be as
above. It is well known that $$K\otimes_{\Q} \Q_{\l}\isom
One can associate to $f$ a family of representations
unramified at all primes $p\not|\l N$. 
By unramified we mean that for all primes $P$ lying over $p$, 
the inertia group of the decomposition group at $P$ is contained
in the kernel of $\rho$. (The decomposition group $D_P$ at $P$ is the
set of those $g\in G$ which fix $P$ and the inertia group 
is the kernel of the map $D_P\rightarrow 
\gal(\O/P)$, where $\O$ is the ring of all algebraic integers.)

Now $I_P\subset D_P \subset \gal(\overline{\Q}/\Q)$ and
$D_P / I_P$ is cyclic, since it is isomorphic to a subgroup of the
galois group of a finite extension of finite fields.
So $D_P / I_P$
is generated by a Frobenious automorphism $\frob_p$ lying over $p$. 
We have
\tr(\rho_{\l,f}(\frob_p)) = a_p\in K \subset K\otimes \Q_{\l} $$
\det(\rho_{\l}) = \chi_{\l}^{k-1}\ve,
where $\chi_{\l}$ is the $\l$th cyclotomic character and
$\ve$ is a Dirichlet character. 

Let $f\in S_k(N)$. For all 
$\abcd \in \modgp$ with $c\cong 0 \mod{N}$ we have
f\left(\frac{az+b}{cz+d}\right) = (cz+d)^k \ve(d) f(z),
where $\ve:(\Z/n\Z)^*\rightarrow \C^*$
is a Dirichlet character mod $N$. If $f$ is an eigenform for
the diamond bracket operator $<d>$, (so that
$f|<d> = \ve(d) f$)
then $\ve$ actually takes values in $K$.

Let $\phi_n$ be the mod $n$ cyclotomic character.
The map $\phi_n: G \rightarrow (\Z/n\Z)^*$ takes $g\in G$ to
the automorphism induced by $g$ on the $n$th cyclotomic
extension $\Q(\mu_n)$ of $\Q$, where we identify
$\gal(\Q(\mu_n)/\Q)$ with $(\Z/n\Z)^*$. 
The $\ve$ appearing in (\ref{detrho})
is really the composition
 \stackrel{\ve}\longrightarrow \C^*.

For each positive integer $\nu$ we consider the $\l^{\nu}$th
cyclotomic character on $G$, 
\phi_{\l^{\nu}}:G\rightarrow (\Z/\l^{\nu}\Z)^*.
Putting these together give a map

\subsection*{Parity Conditions}

Let $c\in\gal(\overline{\Q}/\Q)$ be complex conjugation.
We have $\phi_n(c)=-1$, so $\ve(c) = \ve(-1)$ and
$\chi_{\l}(c) = (-1)^{k-1}$. Let 
=\left(\begin{array}{cc} -1&0\\0&-1 \end{array}\right).$$
For $f\in S_k(N)$,
$$f(z) = (-1)^k\ve(-1)f(z),$$ 
so $(-1)^k\ve(-1) = 1$. Thus,
$$\det(\rho_{\l}(c)) = \epsilon(-1)(-1)^{k-1} = -1.$$
The $\det$ character is odd so the representation 
$\rho_{\l}$ is odd.

\begin{remark} (Vague Question) How can one recognize representations
like $\rho_{\l,f}$ ``in nature''? Mazur and Fontaine have made
relevant conjectures. The Shimura-Taniyama conjecture can be reformulated
by saying that for any representation $\rho_{\l,E}$ comming
from an elliptic curve $E$ there is $f$ so that 
$\rho_{\l,E}\isom \rho_{\l,f}$.

\subsection*{Conjectures of Serre (mod $\l$ version)}
Suppose $f$ is a modular form, $\l$  a rational prime,
$\la$ a prime lying over $\l$, and the representation
$$\rho_{\la,f}:G\rightarrow \GL_2(K_{\la})$$ 
(constructed by Serre-Deligne) is irreducible. 
Then $\rho_{\la,f}$ is conjugate to a representation
with image in $\GL_2(\O_{\la})$, where $\O_{\la}$
is the ring of integers of $K_{\la}$. 
Reducing mod $\l$ gives a representation
which has a well-defined trace and det, i.e., the det and trace
don't depend on the choice of conjugate used to reduce mod
$\la$. One knows from representation theory that if
such a representation is semisimple then it is completely determined
by its trace and det. Thus if $\overline{\rho}_{\la,f}$ is irreducible
it is unique in the sense that it doesn't depend on the choice
of conjugate.  

We have the following conjecture of Serre which remains open.
All irreducible representation of 
$G$ over a finite field which are odd, i.e., $det(\sigma(c))=-1$, $c$
complex conjugation, are of the form $\overline{\rho}_{\la,f}$
for some representation $\rho_{\la,f}$ constructed as above. 

Let $E/\Q$ be an elliptic curve and let 
$\sigma_{\l}:G\rightarrow\GL_2(\F_{\l})$ be
the representation induced by the action of $G$
on the $\l$-torsion of $E$. Then $\det \sigma_{\l} = \phi_{\l}$
is odd and $\sigma_{\l}$ is usually irreducible, so Serre's conjecture
would imply that $\sigma_{\l}$ is modular. From this one can, under Serre's
conjecture, prove that $E$ is modular. 

Let $\sigma:G\rightarrow \GL_2(\F)$ ($\F$ is a finite field) 
be a represenation of the galois group $G$. The we say that the
{\em representions $\sigma$ is
modular} if there is a modular form $f$, a prime $\la$, and an embedding
$\F\hookrightarrow \overline{\F}_{\la}$ such that 
$\sigma\isom\overline{\rho}_{\la,f}$ over 

\subsection*{Wile's Perspective}

Suppose $E/\Q$ is an elliptic curve and
the associated $\l$-adic representation on the
Tate module $T_{\l}$. Then by reducing 
we obtain a mod $\l$ representation
\rightarrow \GL_2(\F_{\l}).$$
If we can show this is modular for infinitely many $\l$
then we will know that $E$ is modular.

\begin{thm}[Langlands and Tunnel]
If $\sigma_{2,E}$ and $\sigma_{3,E}$ are irreducible, then they
are modular. 

This is proved by using the fact that $\GL_2(\F_2)$ and
$\GL_2(\F_3)$ are solvable so we may apply ``base-change''. 

If $\rho$ is an $\l$-adic representation which is irreducible
and modular mod $\l$ with $\l>2$ and certain other reasonable 
hypothesis are satisfied, then $\rho$ itself is modular.


\section*{January 22, 1996}
\noindent{Scribe: Lawren Smithline, \tt <lawren@math>}
(Note.  These were done about two months after the fact, since the assigned
person didn't.  So they are terse. It was a pretty easy lecture.)

Today, we have a limited goal: to explain modular forms as functions on
lattices -- or elliptic curves.  (See Serre's {\em Course in Arithmetic},
or Katz's paper in the Proceedings of the Antwerp Conference in the
Springer LNM series.)

Let the level $N = 1$.  Consider a weight $k$ cusp form $f$.  For $\t \in
\H$, we have $$f( a\t+b / c\t+d) = (c\t+d)^k f(\t).$$  So $f(\t+1) =
f(\t)$.  By the map $\t \mapsto q= \exp(2\pi i \t)$, we map $\H$ to the
punctured disc $ \{ z : 0 < |z| < 1\}$.  We abuse notation and think of $f$
as a function on this disc.  Since $f$ is a cusp form, $f$ extends to 0,
and $f(0) = 0$.  So we have a $q$-expansion $$ f = \sum _{n=1}^{\infty} a_n
q^n. $$

\subsection*{Lattices inside $\C$}

Let $L = \Z\om_1 \oplus Z\om_2$.  We may assume that $\om_1 / \om_2 \in
\H$.  Let ${\mathfrak R}$ be the set of lattices in $\C$.  $\SL_2 \Z$ acts
on the left of $M = \{ (\om_1, \om_2) : \om_1, \om_2 \in \H\}$ by
multiplication of the column vector.  This action fixes the lattice.

Here is the relation with elliptic curves.  A lattice $L$ determines a
complex torus $\C / L$.  There is a Weierstrass $\wp$ function on this
torus.  Consider an elliptic curve $E$ over $\C$.  There is a lattice given
by the inclusion $H_1(E(\C),\Z) \hookrightarrow \C$.  Choose a nonzero $\om
\in H^0(E, \Om'_E).$  Then $\c \in H_1$ maps to $\int_\c \om \in \C$.

So maybe we should think of ${\mathfrak R}$ as the set of pairs $\{ (E,
\om) \}.$

We have a map $M / \C^\times \into \H$ sending $(\om_1, \om_2)$ to $\om_1 /
\om_2.$  Now take the quotient on the left by $\SL_2 \Z$:
$$ {\mathfrak R}/\C^{times} = \SL_2 \Z \backslash M / \C^\times
\longrightarrow \SL_2 \Z \backslash \H.$$
But this just is the space of elliptic curves over $\C$.  So $f:\H
\rightarrow \C$ which is a modular form and $F:M \rightarrow \C$ satisfying
$F(\la L) = \la^{-k}F(L)$ amount to the same thing by a simple calculation.

\subsection*{Hecke Operators}

Let $F$ be a function on lattices.  Define the Hecke operator $T_n$ as
$$T_n F (L) = \sum_{L' \subset L, (L:L') = n} F(L) n^{k-1}.$$

The essential case on elliptic curves is for $n = \l$ a prime.  In this
case, the $L'$ correspond to the $\l+1$ subgroups of order $\l$ of $(\Z /
\l\Z)^ 2$.
\section*{January 24, 1996}
\noindent{Scribe: William Stein, \tt <was@math>}

\noindent {\bf \large More On Hecke Operators}\smallskip

We consider modular forms $f$ on $\Gamma_1(1)=\modgp$, that
is, holomorphic functions on $\H\cup\{\infty\}$ which satisfy
for all $\abcd\in\modgp$. Using a Fourier expansion we write
$$f(\tau)=\sum_{n=0}^{\infty} a_ne^{2\pi i\tau n},$$ 
and say $f$ is a cusp form if $a_0=0$.
There is a correspondence between modular forms $f$ and 
lattice functions $F$ satisfying $F(\lambda L)=\lambda^{-k}F(L)$
given by $F(\Z\tau+\Z)=f(\tau)$. 

\subsection*{Explicit Description of Sublattices}
The $n$th Hecke operator $T_n$ of weight $k$ is defined by
$$T_n(L)=n^{k-1}\sum_{{L'\subset L,\ (L:L')=n}} L'.$$
What are the $L'$ explicitly? Note that $L/L'$ is a group of order $n$ and 
$$L'/nL\subset L/nL=(\Z/n\Z)^2.$$
Write $L=\Z\om_1+\Z\om_2$, let $Y_2$ be the cyclic subgroup
of $L/L'$ generated by $\om_2$ and let $d=\#Y_2$. Let 
$Y_1=(L/L')/Y_2$.  $Y_1$ is generated by the image
of $\om_1$ so it is a cyclic group of order $a=n/d$. 
We want to exhibit a basis of $L'$. Let
$\om_2'=d\om_2\in L'$ and use the fact that $Y_1$ is
generated by $\om_1$ to write $a\om_1=\om_1'+b\om_2$
for $b\in\Z$ and $\om_1'\in L'$. Since $b$ is only
well-defined modulo $d$ we may assume $0\leq b\leq d-1$. 
\left(\begin{array}{c}\om_1'\\ \om_2'\end{array}\right)
\left(\begin{array}{cc}\om_1\\ \om_2\end{array}\right)
and the change of basis matrix has determinent $ad=n$.
$$\Z\om_1'+\Z\om_2'\subset L' \subset L=\Z\om_1+\Z\om_2$$
and $(L:\Z\om_1'+\Z\om_2')=n$ (since the change of basis matrix has
determinent $n$) and $(L:L')=n$ we see that $L'=\Z\om_1'+\Z\om_2'$.   

Thus there is a one-to-one correspondence between sublattices $L'\subset L$
of index $n$ and matrices  
with $ad=n$ and $0\leq b\leq d-1$.
In particular, when $n=p$ is prime there $p+1$ of these. In general, the
number of such sublattices equals the sum of the positive divisors 
of $n$. 

\subsection*{Action of Hecke Operators on Modular Forms}
Now assume $f(\tau)=\sum_{m=0}^{\infty} c_m q^m$ is a modular
form with corresponding lattice function $F$. How can we describe the 
action of the Hecke operator $T_n$ on $f(\tau)$? We have
T_nF(\Z\tau+\Z) & =  n^{k-1}\displaystyle\sum_{
\stackrel{\stackrel{\stackrel{a,b,d}{ab=n}}{0\leq b<d}}\null
F((a\tau+b)\Z + d\Z)\smallskip \\
& = n^{k-1}\displaystyle\sum d^{-k} F(\frac{a\tau+b}{d}\Z+\Z)\smallskip \\
& = n^{k-1}\displaystyle\sum d^{-k} f(\frac{a\tau+b}{d})\smallskip \\
& = n^{k-1}\displaystyle\sum_{a,d,b,m} d^{-k}c_m e^{2\pi i(\frac{a\tau+b}{d})m}\smallskip \\
& = n^{k-1}\displaystyle\sum_{a,d,m} d^{1-k}c_m e^{\frac{2\pi i a m \tau}{d}}
\frac{1}{d}\displaystyle\sum_{b=0}^{d-1} (e^{\frac{2\pi i m}{d}})^b\smallskip \\
& = n^{k-1}\displaystyle\sum_{\stackrel{ad=n}{m'\geq 0}\null}
d^{1-k} c_{dm'}e^{2\pi i a m'
\tau}\smallskip \\
& = \displaystyle\sum_{{ad=n, \ m'\geq 0}} a^{k-1} c_{dm'}q^{am'}.
In the second to the last expression we
 let $m=dm'$, $m'\geq 0$, then used the fact that the 
$\frac{1}{d}\sum_{b=0}^{d-1} (e^{\frac{2\pi i m}{d}})^b$
is only nonzero if $d|m$. 

$$T_nf(q)=\sum_{{ad=n, \ m\geq 0}} a^{k-1}c_{dm} q^{am}$$
and if $\mu\geq 0$ then the coefficient of $q^{\mu}$ is
$$\sum_{{a|n, \  a|\mu}}a^{k-1}c_{\frac{n\mu}{a^2}}.$$

When $k\geq 1$ the coefficients of $q^{\mu}$ for all $\mu$ belong
to the $\Z$-module generated by the $c_m$.

Setting $\mu=0$ gives the constant coefficient of $T_n f$ which is
$$\sum_{a|n}a^{k-1}c_0 = \sigma_{k-1}(n)c_0.$$ 
Thus if $f$ is a cusp form so is $T_nf$. ($T_nf$ is holomorphic
since its original definition is as a finite sum of holomorphic 

Setting $\mu=1$ shows that the coefficient of $q$ in $T_n f$ is 
just $c_n$. As an immediate corollary we have the
following important result.

Suppose $f$ is a cusp form for which $T_n f$ has 0 as coefficient
of $q$ for all $n\geq 1$, then $f=0$. 

When $n=p$ is prime we get an interesting formula for the
action of $T_p$ on the $q$-expansion of $f$. 
One has
$$T_p f = \sum_{\mu\geq 0} \sum_{{a|n, \ 	a|\mu}}a^{k-1}
                         c_{\frac{n\mu}{a^2}} q^{\mu}. $$
Since $n=p$ is prime either $a=1$ or $a=p$. When
$a=1$, $c_{p\mu}$ occurs in the coefficient of $q^{\mu}$
and when $a=p$, we can write $\mu=p\lambda$ and we get
terms $p^{k-1}c_{\lambda}$ in $q^{\lambda p}$. 
$$T_n f = \sum_{\mu\geq 0}c_{p\mu}q^{\mu}+
          p^{k-1}\sum_{\lambda\geq 0} c_{\lambda}q^{p\lambda}.$$

\section*{January 26, 1996}
\noindent{Scribe: Amod Agashe, \tt <amod@math>}

Following the notation of the last few lectures, let $M_k$ denote
the space of modular forms of weight $k$ for $SL_2(\Z)$ and $S_k$ 
denote the subspace of cusp forms. 

Then we have:

\begin{prop}. $M_k$ is a finite dimensional $\C$-vector space and is generated
by modular forms having the coefficients of their Fourier expansion in $\Q$.
{\sc Sketch of Proof}. (For details, refer Serre's ``A course in
Arithmetic'' or Lang's ``Introduction to modular forms''.) \smallskip

The key ingredient that goes into proving finite dimensionality is
the following result, which can be obtained by contour integration:

Let $f\in M_k$ and let $D=\{z\in \C :Im(z)>0, \mid z\mid \geq 1, 
\mid Re(z)\mid \leq 1/2\}$ be the fundamental domain for $SL_2(\Z)$. Then
$$\sum_{p \in D\cup\infty} \frac{1}{e_p} ord_p(f) = \frac{k}{12} $$
$$ e_p = \frac{1}{2} \# \{\gamma \in SL_2(\Z) : \gamma p =p \}
       = \frac{1}{2} \# Aut(E_p)$$
Here, the latter equality follows from the observation that the 
category of elliptic curves over $\C$ with isogenies is the same
as the category of lattices in $\C$ upto homothety with maps being
multiplication by elemets of $\C$. One can
show that the invertible maps that preserve the lattice $\Z \oplus
\Z p$ are in one-to-one correspondence with the set 
$\{\gamma \in SL_2(\Z) : \gamma p =p \}$ and hence the latter equality.

In particular,
$$ e_p = \left\{ \begin{array}{ll}
			2 & \mbox{if $p=i$} \\
			3 & \mbox{if $p=\sqrt[3]{-1}$} \\
			1 & \mbox{otherwise}
	 \right. $$

Using this formula and relating the dimensions of $M_k$ and
$S_k$, one can show that $M_k$ is finite dimensional and also
explicitly calculate its dimension. 

To get a basis with Fourier coefficients in $\Q$, first observe
that $M_k$ is generated by the set of Eisenstein series $G_k$ for all $k$.
As a function on the complex upper half plane, 
the Eisenstein series $G_k$ for $k \in \Z$ and $k>1$ is given by
$$G_k(\tau) = \sum_{(m,n) \in \Z^2 \backslash (0,0)
One can then show that
$$G_k(\tau) = \frac{1}{2}\zeta(1-k) + \sum_{k=1}^\infty \sigma_{k-1}(n)q^n$$
where $q=e^{2\pi i \tau}$, $\zeta$ is the Riemann zeta function and
$$\sigma_k(n) = \sum_{d\mid n}d^k$$
There is a theorem due to Euler which states that $\zeta(1-k)= -\frac{b_k}{k}$
where $b_k$ are the Bernoulli numbers defined by the following power series
$$\frac{x}{e^x-1} = \sum_{k=0}^\infty \frac{b_kx^k}{k!} $$
The constant term of $G_k$ is thus $-b_k/{2k}$, which is rational.
Thus the Fourier expansion of $G_k$ has rational coefficients and thus
we have found a basis with rational coefficients.

Next, let $V$ be a subspace of $M_k$ which is stable under the 
action of all the Hecke operators $T_n$. For example, observing 
that $T_n(G_k)=\sigma_{k-1}(n)G_k$, we see that $V=\C(G_k)$ is
one such subspace.

Let $\T=\T(V)$ = $\C$-algebra generated by the $T_n$'s inside $End(V)$
	= $\C$-vector space generated by the $T_n$'s inside $End(V)$.
The latter equality of sets follows because the product of two Hecke
operators can be expressed as a linear combination
of finitely many Hecke operators.

For $k>0$, we define a bilinear map
$\T \times V \rightarrow \C$ by
$$(T,f) \mapsto a_1(f\mid T).$$

\begin{prop}. The induced maps $\T \rightarrow Hom(V,\C)$ and
$V \rightarrow Hom(\T, \C)$ are isomorphisms.
We first show that the maps are injective.\\
Injectivity of the second map:
f \in V \mapsto 0  \\
\Rightarrow a_1(f \mid T) = 0 \ \forall T\in \T \\
\Rightarrow a_1(f \mid T_n) = 0 \ \forall n  \\
\Rightarrow a_n(f) = 0 \ \forall n \geq 1 \\
\Rightarrow f\mbox{ is constant} \\
\Rightarrow f=0\mbox{ if }k>0  
Injectivity of the first map:
T \in \T \mapsto 0  \\
\Rightarrow a_1(f \mid T)=0 \ \forall f\in V \\
\Rightarrow a_1((f\mid T_n)\mid T) = 0 \ \forall f\in V \\
\Rightarrow a_1((f\mid T)\mid T_n) = 0 \ \forall f \in V \\
\Rightarrow a_n(f\mid T) = 0 \ \forall n>0, \forall f\in V \\
\Rightarrow f\mid T = 0 \ \forall k>0, \forall f\in V \\
\Rightarrow T = 0

In the fourth line, $f$ is replaced by $f \mid T_n$.
Next observe that $V$, being a subspace of $M_k$, is finite dimensional.
Hence we have from the injectivities of both the above maps that each map
is actually an isomorphism. \smallskip

A map $\phi: M \rightarrow N$ of $\T$-modules is said to be $\T$-equivariant
if $\phi (Tm) = T\phi(m) \ \forall m\in M$.

The isomorphisms $\T \cong Hom(V,\C)$
and $V \cong Hom(\T, \C)$
as defined above are $\T$-equivariant.
Consider the first map. 
Here is the $\T$-module structure on $Hom(V,\C)$.
Given $\psi \in Hom(V,\C)$, i.e. $\psi: V \rightarrow \C$,
define $T\psi: V \rightarrow \C$ by $(T\psi) (f)=\psi (f\mid T)$. 
Let $\beta$ denote the map $\T \rightarrow Hom(V, \C)$.
Then given $T'\in \T$ and $T \in \T$,
we have to show that $\beta (T(T'))=T(\beta (T'))$. Let $f \in V$.
Now $(\beta (TT'))(f) = a_1(f\mid TT')$, 
while $(T(\beta T'))(f) = \beta (T') (f\mid T) = a_1((f\mid T) \mid T') 
= a_1(f\mid TT')$.
Thus $(\beta (TT'))(f)=(T(\beta (T')))(f) \ \forall f\in V$ and we
are done.

\noindent Next consider the second map. 
We define the $\T$-module structure on $Hom(\T,\C)$.
Given $\phi \in Hom(\T,\C)$, i.e. $\phi: \T \rightarrow \C$,
define $T\phi: \T \rightarrow \C$ by $(T\phi) (T')=\phi (TT')$. 
Let $\alpha$ denote the map $V \rightarrow Hom(\T, \C)$.
Then given $f\in V$ and $T \in \T$,
we have to show that $\alpha (T(f))=T(\alpha (f))$. Let $T' \in \T$. 
Now $$(\alpha (Tf))(T') = a_1((f\mid T) \mid T') = a_1(f\mid TT'),$$ 
while $$(T(\alpha f))(T') = \alpha (f) (TT') = a_1(f\mid TT').$$
Thus $$(\alpha (Tf))(T')=(T(\alpha f))(T') \ \forall T'\in \T$$
and we are done.

An element $f$ of $M_k$ is said to be an eigenform if it is an eigenfunction
for all the Hecke operators.\\
i.e. $f\mid T_n = \lambda_n f$ for some $\lambda_n \in \C \ \forall n\geq 1$.
Let $f$ is an eigenform with eigenvalues $\lambda_n$. Then 
$a_n(f) =  a_1(f\mid T_n) = a_1(\lambda_n f) = \lambda_n a_1(f) \ \forall n\geq 1$. \\
Thus if $a_1(f) = 0$, then $a_n(f) = 0 \ \forall n\geq 1$.  \\
If $k>0$ then this implies $f=0$.
Hence if $k>0$, then if $f\neq 0$, we can normalize $f$ to 

An eigenform $f$ is said to be normalized if $a_1(f) = 1$.

If $f$ is a normalized eigenform with eigenvalues $\lambda_n$, 
then $a_n(f)=\lambda_n$ and $f\mid T_n = \lambda_n f = a_n(f)f$.

% Given $f \in M_k$ define $\phi_f: \T \rightarrow \C$ by 
% $\phi_f(T) = a_1(f\mid T)$. Note that if $\alpha$ denotes the map 
% $V \rightarrow Hom(\T, \C)$ induced by the bilinear pairing mentioned
% before, then $\phi_f$ is just $\alhpa(f)$.
% Then we have: \\

Again, let $\alpha$ denote the map $V \rightarrow Hom(\T, \C)$ 
induced by the bilinear pairing mentioned earlier. Then if $f \in V$, 
we have the map $\alpha(f): \T \rightarrow \C$.

Let $f$ be an eigenform. 
Then $f$ is normalized $\Leftrightarrow$ $\alpha(f)$ is a ring homomorphism.
If $f=0$ then the statement is trivial. So assume $f\neq 0$.
Then as discussed above, $a_1(f)\neq 0$. Also recall
from the same discussion that if $f\neq 0$ is an eigenform, then 
$$f\mid T_n = \frac{a_n(f)}{a_1(f)}f.$$ For ease of notation, let
$\alpha_f$ denote the map $\alpha(f)$. So 
\alpha_f(T_nT_m) \\ = a_1(f\mid T_nT_m) \\ = a_1((f\mid T_n)T_m)
\\ = a_m(f\mid T_n) \\ = a_m((a_n(f)/a_1(f))f) \\ = a_m(f)a_n(f)/a_1(f).
We have $$\alpha_f(T_n)\alpha_f(T_m) = a_1(f\mid T_n) a_1(f\mid T_m)
= a_n(f) a_m(f).$$ 
The following are equivalent:

$\alpha_f$ is a ring homomorphism, \\
$\alpha_f(T_nT_m) = \alpha_f(T_n)\alpha_f(T_m) \ \forall T_n,T_m$, \\
$a_1(f) = 1$, and \\
$f$ is normalized.
\end{tabular} \\
The first implication follows because $\T$ is generated by the $T_n$'s.

% We will only prove the ``$\Rightarrow$'' part and leave the reverse
% implication as an exercise. \\
% Assuming that $f$ is normalized, we have to show that 
% $(\alpha(f))(T_nT_m) = (\alpha(f))(T_n)\phi_f(T_m) \ \forall n,m \geq 1$.
% Now $(\alpha(f))(T_nT_m) = a_1(f\mid T_nT_m) = a_1((f\mid T_n)T_m) 
% = a_m(f\mid T_n) = a_m(a_n(f)f) = a_m(f)a_n(f)$. \\
% While $(\alpha(f))(T_n)\phi_f(T_m) = a_1(f\mid T_n) a_1(f\mid T_m) 
% = a_n(f) a_m(f)$. \\
% Thus the two are equal and we are done.

\section*{January 29, 1996}
\noindent{Scribe: J\'anos Csirik, \tt <janos@math>}

Today we'll consider questions of rationality and integrality.
(References: Serre: {\em A Course in Arithmetic} and Lang: {\em
Introduction to Modular Forms}.)  Let
$S=S_k$ be the space of cusp forms of weight $k$.  Let
\[ S(\Q) = S_k \cap \Q [[ q ]] \]
\[ S(\Q) \supseteq S(\Z) = S_k \cap \Z [[ q ]]. \]
The following fact is easy to prove using explicit formul\ae: $S_k$
has a $\C$-basis consisting of forms with integral coefficients (see
Victor Miller's construction below).

Recall that for all even $k\geq4$, there is an Eisenstein series 
\[ G_k = \frac{-b_k}{2k} + 
         \sum^\infty_{n=1} \left ( \sum_{d|n} d^{k-1} \right) q^n, \]
which is a modular form of weight $k$.  Renormalize this to obtain
\[ E_k = \frac{2k}{-b_k}\cdot G_k = 1 + \cdots. \]
The first few Bernoulli numbers of even positive index are $b_2=1/6$,
$b_4=-1/30$, $b_6=1/42$, $b_8=-1/30$, $b_{10}=5/66$, $b_{12}=-691/2730$.
The fact that the first four of these have numerator 1 is closely
related to the arithmetic of cyclotomic fields.

The modular forms $E_4$ and $E_6$ have $q$-expansions with constant
terms equal to $1$, and all coefficients in $\Z$.  The functions
$E_4^aE_6^b$ with $4a+6b=k$ form a basis for $M_k$.  
It is easy to see that they are modular forms.
From the formula that the (weighted) number of zeros of any modular form
of weight $k$ is $k/12$, we deduce that $E_4$ has a simple zero at
$\rho$, and $E_6$ hasn't got a zero at $\rho$.  Hence $E_4^aE_6^b$ has a
simple zero of order $a$ at $\rho$ so these expressions are linearly
independent over $\C$.
show that they span $M_k$, consider the following modular form of weight
\[ \Delta = (E_4^3 - E_6^2)/1728.  \]
Here the coefficient of $q$ is a simple number: $1$.  $\Delta$ has a
simple zero at $\infty$.  Since a cusp form of weight $k$ has $k/12$
zeros, it follows that (since
the weighting $e_\infty$ is $1$), that $\Delta$ does not vanish
anywhere on $\H$.  Therefore $S_{k+12}=\Delta\cdot M_k$.  
Since $E_k(i\infty)=1\neq0$, it follows that 
$M_k=E_k\cdot\C\oplus S_k=E_k\cdot\C\oplus\Delta\cdot M_{k-12}$.  Hence 
${\rm dim}\,M_k={\rm dim}\,M_{k-12}+1$.  Again using the fact that $f\in
M_k$ has $k/12$ zeros, we quickly deduce that the dimensions of $M_0$, 
$M_2$, $M_4$, $M_6$, $M_8$, $M_{10}$ are $1$,$0$,$1$,$1$,$1$,$1$
respectively. (e.g., for $k=4$ any modular form must have just a simple
zero at $\rho$.  So for any $f\in M_4$, it is the case that 
 $f(\tau)-f(i\infty)E_4(\tau)$ vanishes at
$\tau=i\infty$ and hence is identically zero.  So $E_4$ spans $M_4$.)  Thus
we have determined ${\rm dim}\,M_k$ for all $k\geq0$, and it is easy to
see that this number is equal to the number of solutions to $4a+6b=k$
for $a,b\geq0$.  Hence the $E_4^aE_6^b$ span $M_k$ and therefore we have
proved that they form a basis.

The following construction comes from the first page of Victor Miller's
thesis.  Let $d={\rm dim}_\C\, S_k$.  Then there exist $f_1,\ldots,f_d\in
S(\Z)$ such that $a_i(f_j)=\delta_{ij}$ for $1\leq i,j\leq d$.  
To show this, recall that $E_4\in M_4$ and $E_6\in M_6$ have
$q$-expansions with coefficients in $\Z$.  $\Delta\in S_{12}$ has
constant coefficient $0$, and the coefficient of $q$ is $1$.  Also
$\Delta\in S_{12}(\Z)$, as can be seen for example from the formula
\[ \Delta = q\prod_{n=1}^\infty (1-q^n)^{24}.  \]
Now pick $a,b\geq0$ so that $14\geq 4a+6b\equiv k \pmod{12}$, with
$a=b=0$ when $k\equiv0\pmod{12}$.  Note that then $12d+6a+4b=k$ by our
previous result on the dimension of $M_k$ (and the fact that the
dimension of $S_k$ is one less than that for $k\geq12$).  Hence the
\[ g_j = \Delta^j E_6^{2(d-j)+a}E_4^b  \]
for $1\leq j\leq d$ will be cusp forms of weight $k$.  By our previous
remarks on the coefficients of $\Delta$, $E_6$, $E_4$, we have $g_j\in
S_k(\Z)$ and 
\[ a_i(g_j) = \delta_{ij} \]
for $i\leq j$.  A straightforward elimination now yields the
$f_1,\ldots,f_d$ with the stated properties.
It is clear that these $f_1,\ldots,f_d$ are
linearly independent over $\C$, hence they form a basis of $S_k$.

If you take $T_1,\ldots,T_d\in\T=\T(S_k)$, they are also linearly
independent: for given any linear relation 
\[ \sum_{i=1}^d c_iT_i=0, \]
apply this to $f_j$ and look at the first coefficient
\[ 0 = a_1\left(f_j\left|\sum_{i=1}^d c_iT_i\right.\right) 
     = \sum_{i=1}^d c_ia_i(f_j) = \sum_{i=1}^d c_i\delta_{ij} = c_j, \] 
hence the linear relation given in the first place was trivial, so
$T_1,\ldots,T_d$ form a basis for $\T(S_k)$, since they are linearly
independent, and ${\rm dim}_\C\,\T={\rm dim}_\C\, V=d$.

Let ${\cal R}=\Z[\ldots T_n\ldots]\subseteq {\rm End}(S_k)$.  
\[ {\cal R}=\bigoplus_{i=1}^d\Z T_i. \]

\pf Since the $T_i$ form a basis of $\T$,
we have $T_n =\sum_{i=1}^d c_{n_i}T_i$ with $c_{n_i}\in\C$.  We need
to check that $c_{n_i}\in\Z$.  With the $f_j$ as above, consider
 a_n(f_j) &  = a_1(f_j|T_n) 
\\ &            = a_1\left(f_j\left|\sum_{i=1}^d c_{n_i}T_i\right.\right)
\\ &            = \sum_{i=1}^d c_{n_i}a_1(f_j|T_i)
\\ &            = \sum_{i=1}^d c_{n_i}a_i(f_j)
\\ &            = \sum_{i=1}^d c_{n_i}\delta_{ij}  & = c_{n_j}.
Hence $c_{n_i}\in\Z$.  \qed
${\cal R}$ is called the {\em  integral Hecke algebra}.  It is a finite
$\Z$-module of rank $d$.  We still have (from the last lecture) a
   S(\Z)\times{\cal R}  &  \rightarrow  &  \Z             \\
   (f,T)                &  \mapsto      &  a_1(f|T).
Now $S(\Z)\hookrightarrow{\rm Hom}({\cal R},\Z)\cong\Z^d$ by the
argument given before.  Therefore $S(\Z)$ is a free $\Z$-module of
finite rank.  But it also contains the $f_i$, so $S(\Z)\cong\Z^d$.

What is $S(\Z)$ as an ${\cal R}$-module?

The map $S(\Z)\hookrightarrow{\rm Hom}({\cal R},\Z)$ is in fact an
isomorphism of $\T$-modules.

{\sc Hint.\ }  The cokernel is a torsion (in fact finite) group.
So if we show it torsion free, we are done.

The $T_n$ are all diagonalizable on $S_k$.

$S_n$ supports a Hermitian non-degenerate inner product, the Petersson
inner product
\[ (f,g)\mapsto \langle f,g\rangle\in\C. \]
We have $\langle f,f\rangle\geq0$, with equality iff $f=0$.  Furthermore
\[ \langle f|T_n,g\rangle = \langle f,g|T_n\rangle, \]
i.e., $T_n$ is self-adjoint with respect to the given inner product.

An operator $T$ is {\em normal} if it commutes with $T^*$ (which denotes
its Hermitian transpose).  Normal operators are diagonalizable (for a
proof, refer to Math H110).  In our case, $T_n^*=T_n$, so this fact
applies.  It is also true (same proof) that a commuting family of
semi-simple (i.e., diagonalizable) operators is simultaneously

\pf  Put together the above facts.   \qed

We can also prove that the eigenvalues are real.  This depends on the
following trick.  For $f\neq0$ consider
\[ a_n\langle f,f\rangle = \langle a_nf,f\rangle
    = \langle f|T_n,f\rangle = \langle f,f|T_n\rangle 
    = \langle f,a_nf\rangle = \bar{a_n}\langle f,f\rangle.  \]
$a_n\in\R$ now follows since $f\neq0$ implies 
 $\langle f,f\rangle\neq0$.  

The $a_n$ are totally real algebraic integers.

{\sc Hint.\ }  The space $S_k$ is stable under the action of 
${\rm Aut}(\C)$ ``on the coefficients''.  Given a cusp form 
\[f=\sum_{n=1}^\infty c_nq^n\]
and some $\sigma\in{\rm Aut}(\C)$, define 
$\sigmaonf=\sum_{n=1}^\infty \sigma(c_n)q^n$.  This function is in
$S_k$, since $S_k$ has a basis in $S(\Q)$, which is fixed by $\sigma$.
Then $f$ is an eigenform iff $\sigmaonf$ is an eigenform.

We'll use a lame definition of the {\em Petersson inner product} for
this section.  Let $z=x+iy\in\H$.  Then we have a volume form
$y^{-2}{dx\,dy}$ which is invariant under $GL^+_2(\R)$ (the subgroup
of the general linear group of matrices of positive determinant).
To prove this, note that 
$dz\wedge d\bar z = (dx+i\,dy)\wedge(dx-i\,dy) = -2i(dx\wedge dy)$,
and hence
\[ dx\,dy = dx\wedge dy = \frac{-1}{2i} dz\wedge d\bar z \]
Then for any 
$\alpha = \left( \begin{array}{cc} a & b \\ c & d \end{array} \right)
    \in GL^+_2(\R)$ we can consider the usual action
\[ \alpha: z \mapsto \frac{az+b}{cz+d}
      = \frac{(az+b)(c\bar z+d)}{|cz+d|^2} \]
where the imaginary part of the result is 
$|cz+d|^{-2}(ad-bc){\rm Im}\, z = y|cz+d|^{-2}\det(\alpha).$
As for differentials, 
\[ d\left(\frac{az+b}{cz+d}\right) 
    = \frac{a(cz+d)\,dz-(az+b)c\,dz}{(cz+d)^2}
    = \frac{ad-bc}{(cz+d)^2}\,dz \]
hence under application of $\alpha$, $dz\wedge d\bar z$ takes on a
factor of 
$$\frac{\det(\alpha)}{(cz+d)^2}\frac{\det(\alpha)}{(c\bar z+d)^2} 
   = \left( \frac{\det(\alpha)}{|cz+d|^2} \right)^2.$$
This finally proves that the differential $y^{-2}{dx\,dy}$
is invariant under the action of $GL^+_2(\R)$.

The formula for the {\em Petersson inner product} is 
\[ \langle f,g\rangle= 

This could be considered either an integral over the fundamental region
or, noting that the integrand is invariant under $SL_2(\Z)$, an integral
over the quotient space.  One then checks that for $f,g$ cusp forms, the
integral will converge, since they go down exponentially as $z$ tends to
infinity.  It is then clear that this inner product is Hermitian.

It is not immediately clear that the Hecke operators are self-adjoint.
\section*{January 31, 1996}
\noindent{Scribe: William Stein, \tt <was@math>}

{\bf \large Modular Curves}\smallskip

\subsection{Cusp Forms}
Recall that if $N$ is a positive integer we define the congruence
$\Gamma(N)\subset\Gamma_1(N)\subset\Gamma_0(N)$ by
\Gamma_0(N) & = \{\abcd \in \modgp : c\equiv 0 \pmod{N}\}\\
\Gamma_1(N) & = \{\abcd \in \modgp : a\equiv d\equiv 1, c\equiv 0 \pmod{N}\}\\
\Gamma(N) & = \{\abcd \in \modgp : \abcd \equiv
             \bigl(\begin{array}{cc}1&0\\0&1\end{array}\bigr) \pmod{N}\}

Let $\Gamma$ be one of the above subgroups.
One can give a construction of the space $S_k(\Gamma)$ of cusp forms
of weight $k$ for the action of $\Gamma$ using the language of
algebraic geometry.
Let $X_{\Gamma}=\Gamma\backslash\H^{*}$
be the upper half plane (union the cusps)
modulo the action of $\Gamma$. Then $X_{\Gamma}$ can be given the structure
of Riemann surface. Furthermore, 
$S_2(\Gamma)=H^0(X_{\Gamma},\Omega^1)$ where
$\Omega^1$ is the sheaf of differential 1-forms on $X_{\Gamma}$.
This works since an element of $H^0(X_{\Gamma},\Omega^1)$
is a differential form $f(z)dz$, holomorphic on $\H$ and
the cusps, which is invariant with respect to the action
of $\Gamma$. If $\gamma=\abcd\in\Gamma$ then
iff $f$ satisfies the modular condition

There is a similiar construction when $k>2$.

\subsection{Modular Curves}
$\modgp\backslash\H$ parametrizes isomorphism
classes of elliptic curves. The other congruence subgroups also
give rise to similiar parametrizations.
$\Gamma_0(N)\backslash\H$ parametrizes pairs $(E,C)$ where
$E$ is an elliptic curve and $C$ is a cyclic subgroup of order
$N$, and $\Gamma_1(N)\backslash\H$ parametrizes pairs $(E,P)$ where
$E$ is an elliptic curve and $P$ is a point of exact order $N$.
Note that one can also give a point of exact order $N$ by giving
an injection $\Z/N\Z\hookrightarrow E[N]$
or equivalently an injection $\mu_N\hookrightarrow E[N]$
where $\mu_N$ denotes the $N$th roots of unity.
$\Gamma(N)\backslash\H$ parametrizes pairs $(E,\{\alpha,\beta\})$
where $\{\alpha,\beta\}$ is a basis for

The above quotients spaces are {\em moduli spaces} for the
{\em moduli problem} of determining equivalence classes of
pairs ($E + $ extra structure).

\subsection{Classifying $\Gamma(N)$-structures}
Let $S$ be an arbitrary scheme. An {\bfseries elliptic curve}
$E/S$ is a proper smooth curve
$$\begin{array}{c} E \\ \bigl| \\ S\end{array}$$
with geometrically connected fibers all of genus one, together with a
section ``0''.

Loosely speaking, proper generalizes the notion of projective,
and smooth generalizes nonsingularity. See Chapter III, section 10 of
Hartshorne's {\em Algebraic Geometry} for the precise definitions.

Let $S$ be any scheme and $E/S$ an elliptic curve.
A {\bfseries $\Gamma(N)$-structure} on $E/S$ is
a group homomorphism
$$\varphi:(\Z/N\Z)^2\into E[N](S)$$
whose image ``generates'' $E[N](S)$.

See Katz and Mazur, {\em Arithmetic Moduli of
Elliptic Curves}, 1985, Princeton University Press, especially
chapter 3.

Define a functor from the category of $\Q$-schemes to the
category of sets by sending a scheme $S$ to the
set of isomorphism classes of pairs
 $$(E, \Gamma(N)\mbox{\rm -structure})$$
where $E$ is an elliptic curve defined over $S$ and
isomorphisms (preserving the $\Gamma(N)$-structure) are taken
over $S$. An isomorphism preserves the $\Gamma(N)$-structure
if it takes the two distinguished generators to the two
distinguished generators in the image (in the correct order).

For $N\geq 4$ the functor defined above is representable and
the object representing it is the modular curve $X(N)$ corresponding
to $\Gamma(N)$.

What this means is that given a $\Q$-scheme $S$, the
set $X(S)=Mor_{\Q\mbox{\rm -schemes}}(S,X)$ is isomorphic to
the image of the functor's value on $S$.

There is a natural way to map a pair $(E,\Gamma(N)\mbox{\rm -structure})$
to an $N$th root of unity.
If $P,Q$ are the distinguished basis of $E[N]$ we send
the pair $(E,\Gamma(N)\mbox{\rm -structure})$ to
where $$e_N:E[N]\times E[N]\into \mu_N$$ is the Weil pairing. For
the definition of this pairing see chapter III, section 8 of
Silverman's {\em The Arithmetic of Elliptic Curves}. The Weil pairing
is bilinear, alternating, non-degenerate, galois invariant, and
maps surjectively onto $\mu_N$.
\section*{February 2, 1996}
\noindent{Scribe: Lawren Smithline, \tt <lawren@math>}
(Note.  These were done about a month after the fact, since the assigned
person dropped the course.  So they are terse.)

Earlier, we looked at V. Miller's construction for eigenforms.  (See, for
instance, Lang X \S4 in the course references.  This is a special miracle for
$\SL_2(\Z)$.)  Over $\Z$, $\T \cong \Z^d$ and $S(\Z)$ are dual, where $d =
\dim S_k(\C)$.

Here is Shimura's explanation (Lang III \S5, VIII).  The Hecke operator $\T_n$
maps $S_k$ to itself.  Let $A\subset \C$ be a subring and
$$\T_A = A[\T_n : n > 0] \subseteq {\rm End}_\C S_k.$$
Denote by $\T$, $\T_\Z$.  There is a natural tensor product $\T_A \otimes_A
\C \twoheadrightarrow \T_\C$.

There is also a complex conjugation automorphism of $S_k$ by
$f \mapsto \overline{f(-\bar\tau)}$.  This map is conjugate linear.
The map $\tau \mapsto \exp(2\pi i \tau)$ becomes
$\tau \mapsto \overline{\exp(-2\pi i \bar\tau)} = \exp(2 \pi i \tau)$.
Say $f = \sum a_n q^n.$  Its conjugate
$g = \sum \bar a_n q^n.$  If you know $S_k(\C) = \C \otimes_\Q S_k(\Q)$,
then you know that modular forms can be conjugated in this sense.

There is an isomorphism $\T_\R \otimes_\R \C \stackrel\sim\longrightarrow
\T_\C$ since the map is surjective and the complex dimensions on each side
are equal.

Shimura (1959) exhibited the (Eichler-)Shimura isomorphism
$$S_k(\C) \cong H^1(X_\Gamma,\R).$$
We have $S_k(\C) = H^0(X_\Gamma, \Omega^1)$ and a map
$H_1(X_\Gamma,\Z) \times S_k(\C) \rightarrow \C$.
Now, $H_1(X,\Z) \cong \Z^{2d}$ embeds in $\Hom_\C(S_k(\C),\C) \cong
\C^{2d}$ as a lattice.

So we have $S_k(\C) \rightarrow \Hom(H_1(X,\Z),\C))$ and
$S_k(\C) \stackrel\sim\rightarrow \Hom(H_1(X,\Z),\R))$ as real vector
spaces.  By the Shimura isomorphism, this is isomorphic to $H^1(X,\R) \sim
H^1_p(\Gamma,\R)$, the parabolic cohomology of $\Gamma$.

So $S_k(\C) \cong H^1_p(\Gamma, V_k)$, for a certain $d-1$ dimensional
subspace $V_k$.  (Let $W = \R \oplus \R$. $\Gamma$ acts by linear fractional
transformation.  Let $V_k = {\rm Sym}^{k-2} W$.  There is a lattice in
corresponding to $H^1_p(\Gamma, {\rm Sym}^{k-2} \Z^2)$)  We have an action
$\Gamma$ by $$f\cdot \c \mapsto \int_{\tau_0}^{\c(\tau_0)}

Recall $\T = \T_\Z$ is a set of endomorphisms of a lattice $L$, and $\T$
has finite rank over $\Z$.  We have the inclusion
$S_k(\Z) \hookrightarrow S_k(\C)$, or equivalently, $$\Hom_\Z(\T,\Z)
\hookrightarrow \Hom_\Z(\T,\C) = \Hom(\T_\C,\C) = S_k(\C) = S_k(\Z)
\otimes_\Z \C.$$

Here is a nifty inner product (the Petersson innner product) on $S_k(\C)$.
For $f,g \in S_k(\C)$, let 
$$\lan f,g \ran = \int_{\Gamma \backslash \H} f(\tau)g(\tau) y^k
The Hecke operators are self-adjoint for $(p, N) = 1$:
$$\lan f | T_p, g \ran = \lan f, g | T_p \ran.$$
Indeed, for $\a \in \GL_2^+ (\R)$,
$$\lan f | \a , g | \a \ran = \lan f,g \ran.$$

\section*{February 5, 1996}
\noindent{Scribe: Shuzo Takahashi, \tt <shuzo@math>}

We have studied actions of $\T$ on 
$S_k(\C)$, $S_k(\Q)$ and $S_k(\Z)$ where 
$\Gamma = SL_2(\Z)$.  What we know so far is
$$S_k(\Z) \simeq {\rm Hom}_{\Z}(\T,\Z).$$
Also we have studied the Peterson product. It is Hermitian, i.e.,
$$\lan f|T_n, g\ran  = \lan f,g|T_n\ran $$
for $T_n \in \T$ and for $f,g \in S_k(\C)$.

Note: $T_n$ defined on $S_k(\C)$ preserves $S_k(\Z)$.

Today we study when $\Gamma = \G_1(N)$ or $\G_0(N)$ for
$N \geq 1$.

{\bf\large 1. The Diamond Operator and the Decomposition of 

$\G_1(N)$ is a normal subgroup of $\G_0(N)$ and we have
$\G_0(N)/\G_1(N) \simeq (\Z/n\Z)^*$.

The diamond operator $< > $ is defined as follows: for 
$\pmatrix{a & b \cr c & d \cr} \in \G_0(N)$, the map on $S_k(\G_1(N))$
$$f \rightarrow f | \pmatrix{a & b \cr c & d \cr} $$
defines an endmorphism $< d> $ of $S_k(\G_1(N))$ which depends only
on $d \ {\rm mod}\  N$. Thus we get an action $< > $ of 
$(\Z/n\Z)^*$ on $S_k(\G_1(N))$.  

Since $(\Z/n\Z)^*$ is a finite group, we have the following decomposition 

$$S_k(\G_1(N)) = \bigoplus_{\e} S_k(\G_0(N),\e)$$
where $\e$ runs over the set of characters $(\Z/n\Z)^* \rightarrow \C^*$ 
and $S_k(\G_0(N),\e)$ is defined as:
$$S_k(\G_0(N),\e) = \{ f \in S_k(\G_1(N)) : f | < d>  = \e(d) f\}.$$
We have  $S_k(\G_0(N),\e) = 0$ unless $\e(-1) = (-1)^k$.  

{\bf\large 2. The Hecke Operators on $S_k(\G_1(N))$} 

For $n \geq 1$, we have the operation on $S_k(\G_1(N))$ of the $n$th Hecke 
operator $T_n$. The following are basic properties:


(1) $T_n$'s commute each other and with $< d> $.

(2) $T_n$'s preserve $S_k(\G_0(N),\e)$.

(3) if $(n,N) = 1$, then $\lan f | T_n,g\ran = 
\lan f,g|{< n>}^{-1}T_n\ran $.

(4) $\lan f | < d> ,g\ran = 
\lan f,g|{< d>}^{-1}\ran $.

(5) if $(n,N) = 1$,  then $T_n$ is diagonalizable.

(6) if $(n,N) \neq 1$,  then $T_n$ is not diagonalizable.


The action of $T_n$ is described in the following theorem:

Let $f = \sum_{n = 1}^\infty a_n q^n \in S_k(\G_0(N),\e)$. Then

%$$f | T_p = 
%\sum_{n=1}^{\infty} a_{pn} q^n + p^{k-1}\e(p) \sum_{n=1}^{\infty} a_n 
%&\text{if $p \not| N$;}  \\
%\sum_{n=1}^{\infty} a_{pn} q^n
%&\text{if $p | N$.}

(1) $f | T_p = \sum_{n=1}^{\infty} a_{pn} q^n + p^{k-1}\e(p) 
\sum_{n=1}^{\infty} a_n q^{pn}$.
(Note: when $p|N$, $\e(p)$ is considered to be $0$ and $U_p$ is used instead 
of $T_p$ which is called Atkin-Lehner operator.)

(2) if $(n,m)=1$, then $T_{nm} = T_n T_m$.

(3) if $p \not| N$, then $T_{p^l} = T_{p^{l-1}}T_p - p^{k-1}< p> 

(4) if $p | N$, then $T_{p^l} = (T_p)^l$.

The last formula in the theorem can be proved by comparing the coefficients 
of $q^{p^{l-1}}$ in both sides of the following formal identity:
$$\left(\sum_{n=1}^{\infty} T_n q^n\right)|T_p =
\sum_{n=1}^{\infty} T_{pn} q^n + p^{k-1}< p> 
\sum_{n=1}^{\infty} T_n q^{pn}.$$
For example, the coefficient of $q^p$ in the LHS is $T_p T_p$. On the other
hand, the coefficient of $q^p$ in the RHS is 
$T_{p^2} + p^{k-1}< p> T_1$. Thus, we have
$$T_{p^2} = (T_p)^2 - p^{k-1}< p> Id$$
where $< p> $ should be considered to be a null map if $p | N$.

{\bf\large 3. The Old Forms} 

Suppose $M|N$. Let $f \in S_k(\Gamma_1(M))$. Then for $d$ such that
$d | \frac{N}{M}$, $f(d\tau) \in S_k(\G_1(N))$.
Thus we have a map
$$\phi_M : \bigoplus_{d | \frac{N}{M}} S_k(\Gamma_1(M)) \rightarrow
The old part of $S_k(\G_1(N))$ is defined as the subspace generated by the 
images of $\phi_M$ for $M | N$, $M \neq N$.

\begin{eg} $\phi_M$ is not injective. Consider the case that $k = 12$, 
$M = p$ and $N = p^2$. $S_k(\Gamma_1(p))$ contains $\Delta(\tau)$ and
$\Delta(p\tau)$. But $\phi_p$ maps both of them to $\Delta(p\tau)$ in

Suppose $p \nmid N$. Consider $f,g \in S_k(\G_1(N))$.  Then $f$ and 
$g(p\tau)$ are both in $S_k(\Gamma_1(Np))$.  Then we have
$$f | U_p = (f | T_p) - p^{k-1}\e(p)(f(p\tau))$$
$$g(p\tau) | U_p = g(\tau)$$
where $f | T_p$ is considered in $S_k(\G_1(N))$.
{\bf Proof.} Let $f = \sum_{n=1}^{\infty} a_n q^n$. Then, considering in 
$S_k(\G_1(N))$, we have
f | T_p = 
\sum_{n=1}^{\infty} a_{pn} q^n + p^{k-1}\e(p) \sum_{n=1}^{\infty} a_n 
Also, considering in $S_k(\Gamma_1(Np))$, we have
$$f | U_p = \sum_{n=1}^{\infty} a_{pn} q^n.$$
Thus, we have
$$f | U_p = (f | T_p) - p^{k-1}\e(p)(f(p\tau)).$$
Now, let $g = \sum_{n=1}^{\infty} b_n q^n$. Then
$$g(p\tau)  | U_p = 
\left(\sum_{n=1}^{\infty} b_{n/p} q^n \right) | U_p = g(\tau)$$
where $b_{n/p} = 0$ unless $p | n$.

\section*{February 7, 1996}
\noindent{Scribe: Amod Agashe, \tt <amod@math>}

We are in the process of showing that the Hecke operators $T_p$ acting
on the space of cusp forms $S_k(\Gamma_1(N))$ are not necessarily
semisimple if $p\mid N$.

Recall from last time that if $M \mid N$ then for every divisor $d$
of $M/N$, we had a map $S_k(\Gamma_1(M)) \rightarrow S_k(\Gamma_1(N))$
given by $f(\tau) \mapsto f(d\tau)$. 

Note that the various $f(d\tau)$'s are linearly independent over $\C$,
because the Fourier expansion of $f(d\tau)$ starts with $q^d$.

Let $f$ be an eigenfuntion for all the Hecke operators $T_n$ in
$S_k(\Gamma_1(M))$. Let $p$ be a prime not dividing $M$. 
So $f\mid T_p = af$ where $a=a_p(f)$ and $f \mid <p>=\epsilon (p)f$
where $\epsilon (p)$ is the character associated to the modular 
form $f$. Note that one can prove that if $f$ is an eigenfunction
for the $T_n$'s then it is an eigenfunction for the diamond operators
also (or alternatively, make it part of the definition of eigenform).
Let $N=p^\alpha M$ with $\alpha \geq 1$. We will look at the action of
the $p^{th}$ Hecke operator $U_p$ in $S_k(\Gamma_1(N))$ on the
images of $f$ under the maps described above. Let $f_i(\tau) = f(p^i \tau)$
for $0\leq i\leq \alpha$. As we showed earlier, \\
$$f\mid T_p = \sum a_{np}q^n + \epsilon (p)p^{k-1}\sum a_nq^{pn}.$$
So $$af = f_0\mid U_p + \epsilon (p)p^{k-1}f_1.$$
Thus, $$f_0\mid U_p = af_0 -\epsilon(p)p^{k-1}f_1.$$
From last time, we have $f_1\mid U_p = f_0.$ In fact, in general,
one can see easily that $f_i\mid U_p=f_{i-1}$ for $i\geq 1$.

So $U_p$ preserves the 2-dimensional space spanned by $f_0$ and $f_1$.
The matrix of $U_p$ (acting on the right) with this basis is given 
(from the equations above) by: \smallskip
\[ \left( \begin{array}{cc}
  	     a & 1 \\
	     -\epsilon(p)p^{k-1} & 0
The characteristic polynomial of this matrix is $x^2-ax+p^{k-1}\epsilon(p)$.

There is the following striking coincidence:
Let $E$ be the number field generated over $\Q$ by the coefficients of the
Fourier series expansion of $f$ and let $\la$ be a prime ideal
of $\O_E$ lying over some rational prime $l$. 
Then we have a Galois representation \smallskip
$$\rho_\la: Gal(\overline{\Q}/\Q) \rightarrow \GL_2(E_\la)$$
If $p\not|Nl$ then $\rho_\la$ is unramified and also
$det\ \rho_\la(Frob_p)=\epsilon(p)p^{k-1}$ and 
$tr\ \rho_\la(Frob_p)=a_p(f)=a$. Thus the characteristic
polynomial of $\rho_\la(Frob_p)$ is $x^2-ax+p^{k-1}\epsilon(p)$,
the same as that of the matrix of $U_p$!

A question one can ask is: Is $U_p$ semisimple on the space spanned by
$f_0$ and $f_1$? The answer is yes if the eigenvalues of $U_p$ are

Now, the eigenvalues are the same iff the
discriminant of the characteristic polynomial is zero i.e.
$a^2=4\epsilon(p)p^{k-1}$ i.e. $a=2p^{\frac{k-1}{2}}\zeta$ where
$\zeta$ is some square root of $\epsilon(p)$.
Here is a curious fact: the Ramanujan-Petersson conjecture proved by Deligne
says $|a|\leq 2p^{\frac{k-1}{2}}$; thus the above equality is allowed
by it, so we do not get any conclusion about the semisimplicity of

Let us now specialize to $k=2$. Weil has shown that $\rho_\la(Frob_p)$
is semisimple. Thus if the eigenvalues of $U_p$ are equal, then
$\rho_\la(Frob_p)$ is a scalar. Edixhoven proved that it is not.
So the eigenvalues of $U_p$ are different and hence $U_p$ is semisimple
in this case. So this example (for k=2) does not give us an example
of $U_p$ being not semisimple.

There is the following example given by Shimura which shows that the Hecke
operator $U_p$ need not be semisimple. Let $W$ denote the space spanned
by $f_0, f_1$ and $V$ denote the space spanned by $f_0,f_1,f_2,f_3$.
$U_p$ preserves both spaces $W$ and $V$, so it acts on $V/W$. The action
is given by $\overline{f_2}\mapsto \overline{f_1}=0$ and 
$\overline{f_3}\mapsto \overline{f_2}$ where the bar denotes the image
in $V/W$. Thus the matrix of $U_p$ on the space $V/W$ is
\[ \left( \begin{array}{cc}
             0 & 1 \\
             0 & 0
which is nilpotent, and in particular not semisimple. If $U_p$ were
semisimple on $V$ then it would be semisimple on $V/W$ also; but we
have just shown that it is not. Thus $U_p$ is not semisimple 
on $V$, and hence not on $S_2(\Gamma_1(M))$ (because $V$ is invariant
under $U_p$).


We next discuss the structure of the $\C$-algebra $\T =\T_\C$ generated
by the Hecke and diamond operators and the structure of $S_k(\Gamma_1(N))$
as a $\T$-module.

First we consider the case of level $1$ i.e. $N=1$. 
Then $\Gamma_1(1)=SL_2(\Z)$. All the $T_n$'s are diagonalizable.
$S_k=S_k(\Gamma_1(N))$ has a basis of $f_1,....,f_d$ of normalized eigenforms
where $d=dim(S_k)$. Thus $S_k\cong \C^d$ as a $\C$-vector space.
Then we have the $\C$-algebra homomorphism  $\T\rightarrow \C^d$ given by
$T\mapsto (\la_1,....,\la_d$) where $f_i\mid T=\la_i f_i$.
It is injective because if the image of $T$ is zero, then it kills all
$f_i$ i.e. all of $S_k$ i.e. it is the zero operator. The map is
surjective because $\T$ has dimension $d$.
%and hence it is surjective
%when we consider both sides as $\C$-vector spaces. 
Thus as a $\C$-algbebra, $\T\cong \C^d$.
Next, we claim that the modular form $v=f_1+...+f_d$ generates
$S_k$ as a $\T$-module. This follows because under the
map $S_k\cong \C^d, v\mapsto (1,....,1)$ and our
statement is just the trivial fact that $(1,....,1)$ generates
$\C^d$ as a $\C^d$-module (acting component-wise).

Thus $S_k$ is free of rank $1$ as a $\T$-module. 
We already know that $S_k\cong Hom(\T,\C)$ as $\T$-modules.
Thus $\T\cong Hom(\T,\C)$ as $\T$-modules. In fact the isomorphism
is canonical since the $f_i$'s are normalized.
We remark that $v$ in fact lies in $S_k(\Q)$.

Next, we deal with the general case where the level is not necessarily $1$.

First we need to talk about newforms. Recall the maps
$S_k(\Gamma_1(M)) \rightarrow S_k(\Gamma_1(N))$
for every divisor $d$ of $M/N$ mentioned at the beginning of this lecture.
The old part of $S_k(\Gamma_1(N))$ is defined as the space generated by
all the images of $S_k(\Gamma_1(M))$ for all $M\mid N, M\neq N$
under these maps.
The new part of $S_k(\Gamma_1(N))$ can be defined in two different ways.
Firstly we can define it as the orthogonal complement of the old part
with respect to the Petersson inner product. 
There is also an algbraic definition
as follows. There are certain maps going the other way: 
$S_k(\Gamma_1(N)) \rightarrow S_k(\Gamma_1(M))$ for $M\mid N, M\neq N$.
The new part is the space killed under all these maps. The space
of newforms, denoted $S_k(\Gamma_1(N))_{new}$ is like $S_k(\Gamma(1))$ in the
sense that all the $T_n$'s (including $U_p$) are semisimple and there
is a basis consisting of newforms. A form of level $N$ is said to
be new of level $N$ if it is in $S_k(\Gamma_1(N))_{new}$.

Next, one can show that the map 
$\bigoplus_{M\mid N,M\leq N} S_k(\Gamma_1(M))_{new} \rightarrow 
S_k(\Gamma_1(N))$ given by $f(\tau)\mapsto f(d\tau)$ for $d\mid \frac{N}{M}$
is injective (See W.-C. W.Li, Newforms and functional equations,
Math. Annalen, 212(1975), 285-315).
Note that an eigenform in one of the subspaces of the source
need not be an eigenfuntion for all the operators in the image.
If $f$ is a newform, then let $M_f$ denote its level (i.e. $f$
is new of level $M_f$). Let $S$ be the set of newforms of weight $k$
and some level dividing $N$. Let
$$v=\sum_{f\in S} f(\frac{N}{M_f}\tau).$$
Then one can show that $S_k(\Gamma_1(N))$ is free of rank $1$ over
$\T_\C$ with $v$ as the basis element. Also one can show that
$v$ has rational coefficients.

\section*{February 9, 1996}
\noindent{Scribe: J\'anos Csirik, \tt <janos@math>}

\subsubsection*{Final comments about Hecke algebras}

Recall that for the case $\Gamma=SL(2,\Z)$, if we set $f_1,\ldots,f_d$
to be the normalized eigenforms (newforms of level 1), then they have
possibly complex coefficients but in any case $\{f_i\}$ is finite and
stable under automorphisms of $\C$ and all the coefficients
$a_n(f_i)$ lie in some number field.  Furthermore this field is totally
real: to show this we used that since the set is stable under
conjugation, it suffices to show that all the $a_n(f_i)$ are real, which
followed by remarking that they are eigenvalues of the operators $T_n$
which are self-adjoint with respect to the Petersson inner product.

More generally, let $f\in S(\Gamma_1(N))$ be a normalized eigenform of
character $\varepsilon$.  Then $a_n=\varepsilon(n)\overline{a_n}$.

Note that the algebra $\T_\Q$ generated by the $T_i$ over $Q$ contains
the diamond bracket operators: the formula relating $T_{p^2}$ and
$(T_p)^2$ tells us that the difference is $p^{k-1}\varepsilon(p)$, so
$\varepsilon(p)\in\T_\Q$.  Using Dirichlet's Theorem on primes in
arithmetic progressions, for any $d$ relatively prime to $N$ we can find
a prime $p\equiv d\pmod{N}$, so $\varepsilon(d)=\varepsilon(p)\in\T_\Q$.

If the space of modular forms has dimension 1, then it is spanned by a
(normalized) eigenform with rational coefficients, so the eigenvalues
are all in $\Q$.

The next simplest example is $k=24$, which is the smallest weight such
that the dimension is more than one.  There are two eigenforms, which
are conjugate to each other.  If $f=\sum a_nq^n$ is an eigenform, then 
$\Q(\ldots a_n\ldots)=\Q(\sqrt{144169})$.  In fact $S_{24}$ is spanned
by $\Delta^2$ and $\Delta^2|T_2$.  (Note that $\Delta^2$ is definitely
not an eigenform since its $q$-coefficient is 0.)  The action of $T_2$
on $S_{24}$ with respect to this basis is described by a two by two
matrix of trace $1080$ and determinant $-2^{10}3^2 2221$.  For high $k$,
eigenforms tend to form a single orbit; however, no proof
is known for this.

For every newform $f$, let $E_f$ be the number field generated by its
coefficients.  Let $\Sigma$ be a set of representatives for $f$'s modulo
$Gal(\bar\Q/\Q)$.  Define
   \T_\Q   &   \rightarrow   &   E_f             \\
   T       &  \mapsto        &  \lambda_T,
with $f|T=\lambda_Tf$.  Thus $T_n\mapsto a_n$ so this map is surjective.

In fact the induced
\[  \T_\Q \to \prod_{f\in\Sigma} E_f   \]
is an isomorphism of $\Q$-algebras: it is injective since if $T$ dies on
the image then it acts as zero on $f\in\Sigma$ and so on all of f, by
the rationality of Hecke operators: ${}^\sigma\!(g|T)={}^\sigma\!g|T$.
(And if an operator acts as zero on everything, then it is zero.)

Here we used the fact that there were no oldforms around.

For example, consider $S_2(\Gamma(N))$ for $N$ prime.  Then
$S_2(\Gamma(1))$ is empty, hence we get an isomorphism 
  \T_\Q\cong E_1\times\ldots\times E_t,
 with the right hand side a
product of totally real number fields.  $t>1$ is possible, e.g. for
$N=37$, $\T_\Q=\Q\times\Q$.

In general, oldforms complicate the situation.

\subsubsection*{Final comments about Hecke algebras}

We'll only treat the case $k=2$.  Then (for $\Gamma$ a congruence
\[ S_2(\Gamma)=H^0(X_\Gamma,\Omega^1) \]
where $X_\Gamma=\Gamma\setminus\H\cup\Gamma\setminus\P^1(\Q)$, with 
$\Gamma\setminus\P^1(\Q)$ being the set of cusps that we need to adjoin
to make it a compact Riemann surface.

\[ {\rm dim}\,S_2(\Gamma)= g(X_\Gamma). \]

{\sc Example.\ } $SL(2,\Z)\setminus\P^1(\Q)$ has just one point.  The
proof involves the Euclidean algorithm: any element of $\P^1(\Q)$ can be
written as $\left( \begin{array}{cc} x\\y \end{array} \right)$ with $x$
and $y$ relatively prime integers.  By the Euclidean Algorithm, we can
find $a$ and $b$ integers such that $ax+by=1$.  Then 
\left(\begin{array}{cc} a & b \\ -y & x \end{array} \right)
\left( \begin{array}{cc} x\\y \end{array} \right) =
\left( \begin{array}{cc} 1\\0 \end{array} \right)

To calculate $g(X_\Gamma)$, use the following covering (recall that
$X_\Gamma$ is the compactification of $\Gamma\setminus\H$ we obtain by
adjoining the cusps)
  X_\Gamma \rightarrow X_{\Gamma(1)},
keeping in mind the isomorphism
     j:X_{\Gamma(1)}   &   \rightarrow   &  \P^1(\C)             \\
    (i,\rho,\infty) &  \mapsto        &  (1728,0,\infty)
The only ramification in our covering 
occurs above the points $0,1728,\infty$.

{\sc Example.\ } Let $\Gamma=\Gamma_0(N)$.  The degree of the covering
is $(PSL(2,\Z):\Gamma_0(N)/\pm1)$ which is the number of cyclic subgroups
of order $N$ in $SL(2,\Z/N\Z)$.  We have a covering 
$Y_0(N)\to Y_{\Gamma(1)}$ where $Y_0(N)$ parametrizes elliptic curves
$E$ with a cyclic subgroup of order $N$, 
$C\subseteq E[N]\cong\Z/N\Z\times\Z/n\Z$ up to isomorphism; and
$Y_{\Gamma(1)}$ parametrises elliptic curves.  The isomorphism 
$(E,C_1)\cong(E,C_2)$ is an automorphism $\alpha$ of $E$ with 
$\alpha:C_1\mapsto C_2$.  Usually $\alpha=\pm1$, unless

If we understand ramification, we can use the {\em Riemann-Hurwitz
formula}.  The following mnemonic way of thinking about it is due to
N. Katz.  The Euler characteristic (alternating sum of the dimensions of
cohomology groups) should be thought of as totally additive:
\[ \chi(A\coprod B) = \chi(A) + \chi(B).  \]

If $X$ is a Riemann surface of genus $G$, $\chi(X)=2-2G$.  A single
point has Euler characteristic $\chi(P)=1$.  Hence
\[ \chi(X\setminus\{P_1,\ldots,P_n\})=2-2G-n.  \]
\[ \chi(X_{\Gamma(1)}\setminus\{0,1728,\infty\})=-1  \]
\[ \chi(X_\Gamma\setminus\{\mbox{points over $1728,0,\infty$}\})=2-2g-n  \]
if $n$ points lie over ${0,1728,\infty}$.
If the covering map $X_\Gamma\to X_{\Gamma(1)}$ has degree $d$, then we
can think of the top space as $d$ copies of the bottom space, so 
\[ d \cdot (-1) = 2-2g-n \]
and therefore
\[ 2g-2 = d-n = d-n_0 - n_{1728} -n_\infty.\]

{\sc Example.\ }  Let $\Gamma=\Gamma(N)$.  What happens over $j=0$?
This corresponds to an elliptic curve $E$ with an automorphism $\alpha$ 
of order three.  Let $N>3$.

For any $(E,P,Q)$, we have $(E,\alpha P,\alpha Q)$ and 
$(E,\alpha^2 P,\alpha^2 Q)$ which are isomorphic to it and hence they
are the same point on $X_\Gamma$.  So $E$ only has one-third the usual
number of points lying over it, $n_0=d/3$.  (Except if the above three
points are equal, i.e., $\alpha$ fixes $E[N]$.  This can't happen since

Similarly we get $n_{1728}=d/2$.

To determine the degree $d$, fix $E$ and count the points lying over it:
these are all of the form $(\C/\Z\oplus\tau\Z,1/n,\tau/N)$ with Weil
pairing $e^{2\pi i/N}$ and all such occur, so we need to
count the number of $P,Q\in E[N]$ which form a basis of
$E[N]$ and $e_N\langle P,Q\rangle=e^{2\pi i/N}$.  This gives us the
order of $SL(2,\Z/n\Z)$.  However, (since $N\neq2$) we have to take into
account that $(E,P,Q)\cong(E,-P,-Q)$ but they are not equal
so the degree of the covering is 

We also have $d=(PSL(2,\Z):\Gamma(N)/(\Gamma(N)\cap\pm1))$ since
$\Gamma(N)\setminus SL(2,\Z)\cong SL(2,\Z/N\Z)$.

So we have established that 
\[ 2g-2 = d-n_0 - n_{1728} -n_\infty = d/6 - n_\infty. \]
To determine $n_\infty$, note that $SL(2,\Z)$ acts on 
\left( \begin{array}{cc} 1 \\ 0 \end{array} \right)$.
The stabilizer $\Gamma(N)_\infty$
of $\left( \begin{array}{cc} 1 \\ 0 \end{array} \right)$ 
is $U=\pm\left( \begin{array}{cc} 1 & * \\ 0 & 1 \end{array} \right)$
So the index 
$((\modgp/\pm1)_\infty:\Gamma(N)_\infty)=N$ is the ramification degree
of a point over $\infty$, so $n_\infty=N/d$. 

Hence $2g(X(N))-2=d/6-d/N$.
\section*{February 12, 1996}
\noindent{Scribe: J\'anos Csirik, \tt <janos@math>}

\def\T{{\bf T}}   % Hecke algebra
\def\qed{\hfill $\blacksquare$}

Plug: A useful reference for the next lecture is Andrew Ogg: {\em Rational
points on certain elliptic modular curves} (1972).  

From last lecture's results on easily deduces that
\[  g(X(N)) = 1 + \frac{d}{12N}(N-6). \]

{\sc Example.}  Let $N=5,7$.  If $N$ is prime then the degree of the
covering is $\frac{(N^2-1)(N^2-N)}{N-1}$.  Therefore $N=5$ gives $d=60$
and $g=0$ (the Galois group of the covering in this case is $A_5$.)
Similarly, $N=7$ yields $g=3$ and $d=168$.  (Remark: For $p\geq5$ prime, the
group $SL(2,\Z/p\Z)/\pm1=L_2(p)$ is simple.)

{\sc Example.}  What is the genus of $X_0(N)$, for $N$ a prime?

It is an exercise to show that there are two cusps: 
$\left( \begin{array}{cc} 1 \\ 0 \end{array} \right)=\infty$
$\left( \begin{array}{cc} 0 \\ 1 \end{array} \right)=0$.
The covering $X_0(N)\to X(1)$ is easily seen to have degree $N+1$, since
an elliptic curve has $N+1$ cyclic subgroups of order $N$.  Here
$\infty$ is unramified and $0$ has ramification index $N$.
\[  2g-2 = N+1 -n_\infty-n_{1728}-n_0 \]
with $n_\infty=2$, $n_{1728}$ approximately $d/2$ and 
$n_0$ approximately $d/3$.  

To calculate $n_0$, we need the number of isomorphism classes $(E,C)$
with $E$ fixed with $End(E)=\Z[\mu_6]$.  $\mu_6$ acts on the
set of $C$'s.  Since $\pm1$ is acting trivially, we really have an
action of $\mu_3$ with $(E,C)\cong(E,\zeta C)\cong(E,\zeta^2C)$ with
$\zeta$ some third root of unity.

If we consider $E=\C/\O$ with $\O=\Z[(-1+i\sqrt3)/2]$, then 
$E[N]=\O/N\O$ as an $\O$-module.  In fact
\[ \O/N\O = 
  \left\{ \begin{array}{ll} 
     \F_N\oplus\F_N & \mbox{if N splits in $\O$} \\
     \F_{N^2} & \mbox{if N does not split in $\O$.}
The above covers all possibilities, because $N>3$ can't ramify in $\O$.
By Kummer's theorem, $N$ splits in $\O$ iff 
$\left(\frac{-3}{N}\right)=1$, and
there are exactly two $\O$-stable submodules of $\O/N\O$.  In the second
case, which happens iff $\left(\frac{-3}{N}\right)=-1$, $\O/N\O$ has no
$\O$-stable submodules.  Therefore
\[ n_0 = 
  \left\{ \begin{array}{ll} 
     \frac{N-1}{3} + 2  & 
           \mbox{if $N$ splits in $\O$, i.e., $N\equiv1\pmod3$.}\\
     \frac{N+1}{3} & 
           \mbox{if $N$ does not split in $\O$, i.e., $N\equiv2\pmod3$} 
In the first case, note that the two $\O$-stable submodules have to be
counted separately.

Similarly we obtain
\[ n_{1728} = 
  \left\{ \begin{array}{ll} 
     \frac{N-1}{2}+2 & \mbox{if $N\equiv1\pmod4$} \\
     \frac{N+1}{2} & \mbox{if $N\equiv3\pmod4$}
depending on whether$N$ splits in $\Q(i)$.

{\sc Examples.}  For $N=37$ the genus is $2$.  Using the formula, we get
$2g-2=36-(2+18)-14=2$ (so it works).

Using the formula we can also obtain $g(X_0(13))=0$ and $g(X_0(11))=1$.

It is therefore clear that the genus of $X_0(N)$ is approximately
$N/12$.  In the article by Serre in the Lecture Notes in Mathematics 349
(Antwerp), we find the following table.  Write $N=12a+b$ with $0\leq
b\leq11$.  Then
  $b$  &  $1$  &  $5$  &  $7$  &  $11$  \\ \hline
  $g$  & $a-1$ &  $a$  &  $a$  & $a+1$.  \\
  1 + g(X_0(p)) = {\rm dim}\,M_{p+1}(\modgp).

It was Serre's idea to think of ``modular forms mod $p$'', for some
congruence subgroup 
$\Gamma\ni\left(\begin{array}{cc}1&1\\0&1\end{array}\right)$, like 
$\Gamma_0(N)$ or $\Gamma_1(N)$.  We could use our moduli theoretic
interpretations, but instead we'll define
   M_k(\Gamma,\F_p) \subseteq \F_p[[q]].

By Shimura's cohomology trick, we know that $M_k(\Gamma,\Z)$ is a
lattice in $M_k(\Gamma,\C)$.  Hence we can set
  M_k(\Gamma,\F_p) = M_k(\Gamma,\Z) \otimes_{\Z} \F_p.
Then {\em Serre's equality} states that for a prime $p$,
  M_{p+1}(\modgp,\F_p) = M_2(\Gamma_0(p),\F_p)
in $\F_p[[q]]$.  The philosophy is {\em mod $p$ forms with $p$ in the
level can be taken to mod $p$ formswith no $p$ in the level, but of a
higher weight}.  So for example 
$M_k(\Gamma_1(p^\alpha N),\F_p)$ is a subset of
$M_?(\Gamma_1(N),\F_p)$ of forms of some higher level.

Finally, consider the map from the right hand side to the left hand
side in Serre's equality.  Recall that
\[ G_k = \frac{-B_k}{2k} + \sum^\infty_{n=1}\sigma_k(n) q^n
    \in M_k(\modgp). \]
By Kummer, ${\rm ord}_p(B_{p-1}) = -1$, so
  E_{p-1}=1+\frac{-2(p-1)}{B_{p-1}}\sum\sigma_{p-1}(n)q^n \equiv 1 \pmod{p}.
Hence we got the map from the right to the left: 
multiply by $E_{p-1}$ to get to $M_{p+1}(\Gamma_0(p),\F_p)$.  Then take
the trace to get to $M_{p+1}(\modgp,\F_p)$.  The trace map is dual to
the inclusion and is expressed by
  \tr(f) = \sum_{i=1}^{p+1} f | \gamma_i
\section*{February 14 and 16, 1996}
\noindent{Scribe: Jessica Polito, \tt <polito@math>}

Our goal, for these two days, is to define the modular curves $\X$ over
$\Q(\mu_N)$, and  
$X_1(N)$, and $X_0(N)$ over $\Q$.
These notes will spell out the construction of $\X$, with some 
discussion of the construction of the other two types of curves.
The idea comes from Shimura.

Let $\Q(t)$ be the function field of $\P^1/\Q$, and pick an elliptic 
curve $\E/\Q(t)$ with $j$-invariant $t$:
$$\E: y^2 = 4x^3 - \frac{27t}{t-1728}x - \frac{27t}{t-1728}$$
(Note that the general formula for the $j$-invariant of a curve $y^2 = 
4x^3 - g_2x -g_3$ is $j = \frac{1728g_2^3}{g_2^3 - 27g_3^2}$, and 
that $j(E)$ determines the isomorphism class of the given curve $E$ 
over the algebraic closure of the field of definition.)  

By substituting in a given value $j$ for $t$, we would get a formula 
for an elliptic curve over $\Q(j)$ with $j$-invariant $j$; for $j= 0 $ 
or 1728, we could pick a diferent formula for $\E$ which would give an 
isomorphic curve over $\Q(t)$, for which that substitution would make 

Notice that, in general, if we have an elliptic curve $E/K$, with $K$ 
some field of charictaristic prime to $N$, then we can consider 
$E[N](\K)$, the set of all $N$-torsion point of $E$ defined over $\K$, 
which is isomorphic to $(\Z/N\Z)^2$.  Then we let $K(E[N])$ be the 
smallest extension of $K$ over which all the points of $E[N]$ are 
defined.  Notice also that $E/K$ and $N$ together define a 
representation of the Galois groups $\gal(\K/K)$ into the 
automorphisms of the $N$-torsion points of $E$, as they are defined by 
polymial equations with coefficients in $E$.  We get
$$\rho_{E,N}: \gal(\K/K) \into \aut(E[N]) \isom \GL_2(\Z/N\Z)$$
with $\gal(\K/K(E[N])) = \ker(\rho_{E,N})$.  Unsurprisingly, we will 
frequently leave out the $N$, writing simply $\rho_E$.

Now, to return to our construction, where $K=\Q(t)$.  We will show 
that $\Q(t)(E[N])$ is the function field of a curve defined over 
$\Q(\mu_N)$ (so $\Q(t)(E[N]) \cap \overline{\Q} = \Q(\mu_N)$), 
and that this curve corresponds to $X(N)$.

We will actually do this, not by looking at $\rho_E$, but instead at 
$\rh$, where $\rh$ is given by reducing the image of $\rho_E$ mod 
$\pm 1$:
\gal(\K/K) & \rTo^{\rho_E} &\GL_2(\Z/N\Z) \\
           & \rdTo_{\rh}   & \dTo\\
           &		   &\GL_2(\Z/N\Z)/\{\pm 1\} 

Note: $\rho_E$ is surjective iff $\rh$ is. 

\pf If $\rh$ is surjective, then 
$$\pm \left(\begin{array}{cc}
0 & -1 \\
1 & 0
\end{array} \right) \in \im(\rho_E).$$
By squaring, we see that 
-1 & 0 \\
0 & -1 \end{array}\right) \in \im(\rho_E),$$
and thus $-1 \in \im(\rho_E)$, so $\rho_E$ is surjective.  

if $\rho_E$ is surjective, so is $\rh$.)

Why will we use $\rh$ instead of $\rho_E$?  The kernal of 
$\rho_E$ is the field generated over $K$ 
by the $X$ and $Y$ coordinates of the non-zero points of $\E[N]$, 
whereas the kernal of $\rh$ is the field generated by just the $X$ 
coordinates of those points. This follows because $X(P) = X(-P)$, 
while $Y(P) = -Y(-P)$, as then the $X$  but not the $Y$ coordinates 
are fixed by $\pm 1$.  (We are assuming that we have a Weierstrass 
model for $\E$ here.)

\begin{remark} Let $E_1$ and $E_2$ be elliptic curves over $K$ with equal
$j$ invariants.  (In other words, $E_1$ and $E_2$ are isomorphic over
$\overline{K}$.)  Then $\rho_{E_1}$ is surjective iff $\rho_{E_2}$
is. (We'll assume that either $N  \neq 2, 3$ or $E$ does not have complex multiplication.)

\pf Assume $\rho_{E_1}$ is surjective.  Then $E_1$ does not have complex
multiplication over $\overline{K}$.  (If $E$ has CM, then the image of
$\rho_E$ is abelian for some quadratic extension of $K$, but
$\GL_2(\Z/N\Z)$ has no abelian subgroup of index 2, as $N>3$.)

Then $\aut(E_1) = \{\pm 1\}$  Pick an isomorphism $\phi:E_1 \isom E_2$ over
$\overline{K}$.  Then, for any $\sigma \in \gal(\overline{K}/K)$, we have
${}^\sigma\phi:{}^\sigma E_1 \isom {}^\sigma E_2$, and ${}^\sigma E_i =
E_i$.  Thus ${}^\sigma\phi = \pm \phi$ for all $\sigma \in
\gal(\overline{K}/K)$.  So $\phi: E_1[N] \isom E_2[N]$ does not quite give
us an isomorphism of representations $\rho_{E_1}$ and $\rho_{E_2}$, but it
does give us an isomorphism $\overline{\rho_{E_1}}\isom
We then have that $\overline{\rho_{E_2}}$, and thus $\rho_{E_2}$, is

From here on, let $K=\C(t)$, $\E$ and $\rho_E$, $\rh$ as above.  We have
$$\rho_E: \gal(\K/K) \into \GL_2(\Z/N\Z),$$
so the determinant of $\rho_E$ is a cyclotomic character.  As the 
$N$th roots of unity are all already in $K$, this means that the 
determinant of $\rho_E$ is trivial, so
$$\rho_E: \gal(\K/K) \into \SL_2(\Z/N\Z).$$
Our goal is to prove that $\rho_E$ is surjective onto 

(For any field $F$ of characteristic not dividing $p$, there is an 
algebraic proof by Igusa that a ``generic elliptic curve'' (such as 
$\E$ here), has the maximal possible Galois action on its division 

We will now introduce another way of looking at $K$.  It is equal 
to $\C(j)=\ff_1$, the field of modular functions of level 1.  (A 
modular function of level $N$ is a $\Gamma(N)$-invariant function 
on $\H$ which is meromorphic on $\H$ and at the cusps.)  The set of 
all modular functions 
for $\SL_2(\Z)$ are generated over $\C$ by a single function $j$.  
Similarly, $\ff_N$ is the field of modular functions of level 
We then have the tower of field extensions
\K& = &\overline{\ff_1} \\
&&| \\
&&\ff_N \\
&&| \\

$\SL_2(\Z)$ acts on $\ff_N$, via the map $f(\tau) \to 
f\left(\frac{a\tau +b}{c\tau + d}\right)$, and fixes $\ff_1$.  By reviewing 
definitions, we see that $\SL_2(\Z)/\Gamma(N)=\SL_2(\Z/N\Z)$ acts on $\ff_N$, 
and that $-1$ acts trivially.  We will show that 
$\SL_2(\Z/N\Z)/\{\pm 1\}$ is that Galois group of $\ff_N/\ff_1$, by 
exhibiting a set of functions on whose permutations this group acts faithfully.

So, the entire picture so far is:
				&&	& \overline{\ff_N} \\
				&&\ldLine& \dLine           \\
K(\E[N])/\{\pm 1\}	&&	& \ff_N		   \\
				&&\rdLine&\dLine		\\
				&&	&\ff_1=K

(Recall that $K(\E[N])/\{\pm 1\}$ is the fiexed field of the kernal of $\rh$.)
We will show that $K(E[N]) = \ff_N$, at the same time that we show 
that the Galois group is what we want it to be.  To do this, we'll 
use the Weierstrass $\wp$-function attached to a given lattice $L$ of 
$\C$ (and thus to the corresponding elliptic curve $E/\C$).  So we 
will pick a specific $\tau \in \H$. (Then sending $t$ to $j(\tau)$ 
gives us a map from $\E_t$ to $E_\tau$, the elliptic curve over 
$\C$ with $j$-invariant $j(\tau)$, which is also given by the lattice 
$\Z+\Z\tau$.)  To the lattice $L=\Z+\Z\tau$ 
we attach $\wp{z}$, which is a modular form of weight 2 satisying the 
differential equation
$$\wp'(z)^2 = \wp(z)^2 - 4g_2(\tau)\wp(z) g_3(\tau),$$
which, replacing $\wp$ by $X$ and $\wp'$ by $Y$, is a Weierstrass equation for 
$E_\tau$.  Thus we have 
$\wp(\frac{r\tau + s}{N}, \Z+\Z\tau)$, with $r,s \in \Z$ and not both 
divisible by N,  running through the 
$X$-coordinates of  points in  $E_\tau[N]$.  This form is of weight 
2; we need to find modular functions, so let
$$f_{(r,s)}(\tau):\tau \mapsto 
\wp\left(\frac{r\tau+s}{N}, \Z+\Z\tau\right).$$
(The $f$ are certainly meromorphic on $\H$ and at the cusps, as $g_2, 
g_3$ and $\wp$ all are.) 
$\SL_2(\Z)$ acts on these functions $f$ by 
$\alpha:f_{(r,s)}(\alpha \tau) = f_{(r,s)\alpha}(\tau)$, where 
simply multiplies the row vector $(r,s)$ on the right.  Then we can 
see that $\{f_{(r,s)}\}$ is invariant under the action of $\Gamma[N]$, 
so they are permuted by $\SL_2(\Z/N\Z)$.  In fact, as $\wp$ is an 
even function ($X(P) = X(-P)$ 
for any point on the eliptic curve), $f_{(r,s)}$ is also fixed by $\pm 
1$, so they are permuted by $\SL_2(\Z/N\Z)/\{\pm1\}$.

We have defined $f_{r,s}$ for $r,s \in \Z,$, not both divisible by N.  In
fact, if $(r,s) = (r', s') \pmod N$, then $f_{(r,s)} = f_{(r', s')}$, so we
only need to consider $(r,s) \in \Z/N\Z^2/\{\pm 1\} \backslash (0,0)$
As $f_{(r,s)}(\tau)$ runs over all of the $X$-coordinates of points 
in $E_\tau[N]$ which are not zero for these $(r,s)$, 
we must have these $f_{(r,s)}$ all distinct.  
(There are $N^2 - 1$ non-zero points on $E[N]$, and $X(P) = X(Q)$ 
iff $P=-Q$, so there are exactly as many pairs $(r,s)$ as there are $X$
coordinates of points in $E[N]$.)

Thus we have shown that the non-zero points of $\E[N]$ have 
$X$-coordinates (in $\overline{\ff_N}$) given by the $f_{(r,s)} \in 
\ff_N$, and thus $\ff_N = K(\E[N])$.

(As a corollary, we have shown that the Galois group of 
$\ff_N$ over $\ff_1$ is $\SL_2(\Z/N\Z)/\{\pm1\}$, as this group 
acts faithfullly on the $f_{(r,s)}$.)

Thus, finally, we have shown that for any $E/\C(t)$ with 
$j$-invariant $t$, the associated representation $\rho_E$ has image 
We know that, if  we let $K=\Q(\mu_N)(t)$, $E$ an elliptic curve with 
$j$-invariant $t$, then $\rho_E$ has image contained in $SL_2(\Z/N\Z)$.  (This 
doesn't work for $K=\Q(t)$, as we need to have $\mu_n \subset K$ in 
order to have the image of $\rho_E$ contained in $SL_2(\Z/N\Z)$.)
Furthermore, if we replace $\C(t)$ by $\Q(\mu_N)(t)$, 
the image of $\rho_E$ can only get larger.  In the following diagram

	&		&\C(t)(E[N])	&		& 	\\
	&\ldLine^{G'}	&		&\rdLine	&	\\
\C(t)	&		&	 	&	&\Q(\mu_N)(t)(E[N]) \\
	&\rdLine	&		&\ldLine^{G}	&	\\
	&		&\Q(\mu_N)(t)	&		&	
$G$ and $G'$ are the Galois groups (and also the images of $\rho_E$ 
over $\Q(\mu_N)(t)$ and $\C(t)$ respectively).  Basic Galois theory 
tells us that $G' \subset G$, but we also know that $G' = 
\SL_2(\Z/N\Z)$ and $G \subset \SL_2(\Z/N\Z)$, so $G=G'=\SL_2(\Z/N\Z)$.

Next, let us consider $E/\Q(t)$, as always with $j$-invariant $t$.  
Then $\rho_E:\gal(\Q(t)(E[N]) \into \GL_2(\Z/N\Z)$.  The 
determinant of $\rho_E$ is again a cyclotomic character, but is not 
trivial this time; indeed, it is surjective and thus 
contains a set of coset representatives 
for $\SL_2(\Z/N\Z)$ in $\GL_2(\Z/N\Z)$.  We see that $\rho_E$ is again 
surjective, although onto $\GL_2(\Z/N\Z)$ this time.  Notice that, in 
this case, $\Q(t)(E[N])$ contains $\mu_N$, so we now have the diagram
\overline{\Q}	&	&	&	& \Q(t)(E[N]) \\
\dLine		&	&	&	&\dLine	 \\
\Q(\mu_N) 	&	&	&	&\Q(\mu_N)(t) \\
	   &\rdLine(2,4)&	&	&\dLine  \\
		&	&	&	&\Q(t)	\\
		&	&	&\ldLine&	\\
		&	&\Q	&	&	
%\overline{\Q} & & && && \Q(t)(E[N]) \\
%    |       & & && &&     \quad|       \\
% \Q(\mu_N)    & & && && \Q(\mu_N)(t) \\
%   &  \sel     & && &&      \quad|    \\
%    &        &\sel&&& &\Q(t)            \\
%    && &\sel  &&\swl &                \\
%    &&&&\Q&&                           
% \end{array}
% $$       
Basic Galois 
theory now tells us that $\overline{\Q} \cap \Q(t)(E[N]) = 

\pf Let $F$ be the intersection of these two fields.  
It clearly contains $\Q(\mu_N)$.  If $F$ is bigger than $\Q(\mu_N$), 
then $F(t)$ is bigger than $\Q(\mu_N)(t)$, and thus 
$$\gal(\Q(t)(E[N])/F(t)) \subsetneq \SL_2(\Z/N\Z) .$$
  But we know (as above) that 
$$\SL_2(\Z/N\Z) = \gal(\C(t)(E[N]/\C(t)) \subset 
so we must have equality above, and thus $F = \Q(\mu_N)$. 

Recall that $\C(t)(E[N])/\C(t)$ is the function field for $X(N)$. 
We have shown that $\Q(t)(E[N])$ is a function field for the corresponding 
curve defined over 
$\Q(\mu_N)$, so we have defined $X(N)$ over $\Q(\mu_N)$.

We can similarly define $X_1(N)$ and $X_0(N)$ over $\Q$. Let
$L=K(E[N])$ (where $\Q(t) = K$).  Then $\gal(L/K)=\GL_2(\Z/N\Z)$.  Consider
the subgroup $H$ of this Galois group consisting of all matrices of the
form $\left(\begin{array}{cc}
		*&* \\
Then $L^H \cap \overline{\Q} = \Q(\mu_N)^H = \Q$.  As $L^H$ is an extension
of $\Q$ of transcendence degree 1, it is the function field of a curve
defined over $\Q$.  It turns out to be the curve $X_0(N)$. 
 To get $X_1(N)$,
we would use the subgroup $H= \left(\begin{array}{cc}
		*&* \\

\section*{21 February, 1996}
\noindent{Scribe: David M Jones, \tt <dmjones@math>}

First of all, we start with a correction from last time. Let $L/K$ be
a Galois extension. Let $E_{/L}$ be an elliptic curve. Suppose that $j(E)
\in K$ and for all $g\in\gal(L/K)$, ${}^g\!E\isom_{/L} E$.

Last time it was stated that we could conclude that there exists ${E_0}_{/K}$
such that $E\isom_{/L} E_0$. This is not correct. We may only conclude that
$E\isom_{/{\bar L}} E_0$. The point is that $L = {\bar K}$.


Let $E_{/K}$ be an elliptic curve. Let $g\in\gal(L/K)$ for some
extension $L/K$. Then let $\la_g: {}^g\!E \buildrel {\la_g} \over
\isomap E$ be a family of isomorphisms. Now, $\la_{gh}$ and
$\la_g \circ {}^g\!\la_h$ are both isomorphisms from ${}^{gh}\!E$ to $E$
so they differ by an element of Aut$(E) = \{\pm 1\}$.

Let $g,h\in\gal(L/K)$. We define $c(g,h)\in$ Aut$(E)$ by the
relation $c(g,h) \la_{gh} = \la_g \circ {}^g\!\la_h$.

$c(g,h)$ is a 2-cocyle.
First, let's rewrite $c(g,h) = \la_g\circ {}^g\!\la_h\circ {\la_{gh}}^{-1}$.
Then in order to prove that $c(g,h)$ is a 2-cocycle, we must show the
following (see Serre's ``Local Fields", p.113):
$${}^g\!c(g',g'')\cdot c(g,g'g'') = c(g,g')\cdot c(gg',g'')$$
Then $c(g,g')\cdot c(gg',g'') = \la_g\circ {}^g\!\la_{g'}\circ
{\la_{gg'}}^{-1}\circ\la_{gg'}\circ {}^{gg'}\!\la_{g''}\circ
{\la_{gg'g''}}^{-1} = \la_g\circ {}^g\!\la_{g'}\circ {}^{gg'}\!\la_{g''}
\circ {\la_{gg'g''}}^{-1} = \la_g\circ {}^g\!(\la_{g'}\circ {}^{g'}\!
\la_{g''})\circ{\la_{gg'g''}}^{-1} = \la_g\circ {}^g\!(c(g',g'')\cdot
\la_{g'g''})\circ{\la_{gg'g''}}^{-1} = {}^g\!c(g',g'')\cdot\la_g
\circ {}^g\!\la_{g'g''}\circ{\la_{gg'g''}}^{-1} = {}^g\!c(g',g'')\cdot

We want this 2-cocyle $c(g,h)$ to be trivial.

Let $G = \gal(L/K)$.
If $L = {\bar K}$, then $H^2(G,\{\pm 1\})\to H^2(G,L^*)$ is injective.
Consider the exact sequence
$$\0 \rightarrow \{\pm 1\} \rightarrow {\bar K}^* \buildrel
{\bullet^2} \over \rightarrow {\bar K}^* \rightarrow \0.$$
Looking at a piece of the long exact cohomology sequence we get
$$...\rightarrow H^1(G,{\bar K}^*) \rightarrow H^2(G,\{\pm 1\})
\rightarrow H^2(G,{\bar K}^*)[2] \rightarrow \0.$$
Now $H^1(G,{\bar K}^*) = 0$ by Hilbert Theorem 90 so $H^2(G,\{\pm 1\})$
is isomorphic to the 2-torsion in $H^2(G,{\bar K}^*)$. So
$H^2(G,\{\pm 1\})$ certainly injects into $H^2(G,{\bar K}^*)$. Since
$L = {\bar K}$, the result follows.


In order to show that the class of $c(g,h)$ in $H^2(G,\{\pm 1\})$
is trivial, by the above lemma, it is enough to show that the class
of $c(g,h)$ in $H^2(G,{\bar K}^*)$ is trivial. To do this we will
\vskip .5cm
Differentials: Pick a non-zero differential $\om \neq 0$ on $E$. That is,
$\om\in H^0(E,\Om^1)$. Let $g\in\gal(L/K)$. Then $\la_g: {}^g\!E \isomap
E$ so the pullback $\la_g^*: H^0(E,\Om^1) \isomap H^0({}^g\!E,\Om^1)$. Thus,
$\la_g^*\om\in H^0({}^g\!E,\Om^1)$, which has basis ${}^g\!\om$ so
we must have $\la_g^*\om = a_g\cdot {}^g\!\om$ for some $a_g\in L^*$.

What is ${}^g\!\om$? Let $E_{/L}$ be an elliptic curve and let
$L\hookrightarrow M$ be an injection of fields. Then $H^0(E,\Om^1)
\tensor_L M = H^0(E_{/M},\Om^1)$. In our case, $M = L$ and the
map $L\hookrightarrow M$ is given by the action of $g\in\gal(L/K)$.
So we get $H^0(E,\Om^1)\tensor_L L = H^0({}^g\!E, \Om^1)$. The
element of $H^0({}^g\!E,\Om^1)$ which corresponds to $\om\tensor 1$
under this map is ${}^g\!\om$.

Show that $c(g,h) = ({}^g\!a_h\cdot a_g)/a_{gh}$.
First, consider the following:

$a_g\cdot {}^g\!a_h\cdot {a_{gh}}^{-1}\om = a_g\cdot {}^g\!a_h\cdot
((\la_{gh}^*)^{-1} {}^{gh}\!\om) = a_g\cdot (\la_{gh}^*)^{-1}\circ
{}^g\!(a_h\cdot {}^h\!\om) = a_g\cdot (\la_{gh}^*)^{-1}\circ {}^g\!
(\la_h^* \om) = (\la_{gh}^*)^{-1}\circ {}^g\!(\la_h^*)\circ (a_g
{}^g\!\om) = (\la_{gh}^*)^{-1}\circ {}^g\!(\la_h^*) \circ \la_g^* \om
= ((\la_{gh})^{-1}\circ {}^g\!(\la_h)\circ\la_g)^* \om$.

But $c(g,h) = \la_g\circ {}^g\!\la_h\circ{\la_{gh}}^{-1}$ is multiplication
by a constant ($\pm 1$) so the dual is multiplication by the same
constant. Thus, we have $a_g\cdot {}^g\!a_h\cdot {a_{gh}}^{-1}\om =
c(g,h)\om$ and as $\om\neq 0$, we must have $c(g,h) =
(a_g\cdot {}^g\!a_h)/a_{gh}$.

This shows that $c(g,h)$ is a coboundary, and thus its class is
trivial in $H^2(G,{\bar K}^*)$.


Let $N>3$ be an integer. Recall that we wrote down an elliptic
curve $E_{/\Q(j)}$ whose $j$-invariant was $j(E) = j$.

Let $\ff_N$ be the field of modular functions of level $N$. That is,
$\ff_N = \Q(j)(E[N]/\{\pm1\})$. This field lies over $\ff_1 = \Q(j)$.
We have $\gal(\ff_N/\ff_1) = \GL_2 (\Z/N\Z)/\{\pm 1\}$.

Now define $\ff = \cup_N \ff_N$. This field, which is
the compositum of all the of the $\ff_N$'s, corresponds to a
projective system of modular curves.

The finite adele ring of $\Q$, denoted $\A_f$, is
$$\{ (x_p)\in\prod_p \Q_p :x_p\in\Z_p \mbox{{\rm { }for almost all }} p\}$$

We may also describe $\A_f$ as ${\hat \Q} = {\hat \Z}\tensor\Q$.

Notice that $\GL_2 ({\hat \Z}) \subset \GL_2 (\A_f)$.

The group $\GL_2 (\A_f)$ acts on $\ff$.

First, recall that $\ff = \Q(f_{(r,s)} : (r,s)\in(\Z/N\Z)^2\backslash
(0,0), N\geq 1)$.
$\GL_2 (\Z/N\Z)$ acts on the set of $f_{(r,s)}$ via
$$\abcd : f_{(r,s)}\mapsto f_{(ar + cs,br + ds)}$$


Consider the object $\plim X(N)$. It is not a variety but does
exist in some appropriate category. We have the following two
notions of ``points'' on $\plim X(N)$, which are contra-dual to
each other:

1. Pairs $(E,\iota)$ where $E$ is an elliptic curve and $\iota :
(\Q/\Z)^2 \isomap E_{\rm tors}$

2. Pairs $(E,\a)$ where $E$ is an elliptic curve and $\a : {\hat\Z}^2
\isomap Ta(E)$, where $Ta(E) = \plim E[N] = \prod_p Ta_p(E)$


We will now do a warm-up to show that $\GL_2(\A_f)$ acts on $\ff$.

Let $g = \abcd\in\GL_2^+ (\Q)$.
Look at pairs $(E,\a)$ with $E_{/\C}$ and $\a : {\Z}^2 \isomap
H_1(E(\C),\Z)$. Note that we have
{\Z}^2    & \isomap & H_1(E(\C),\Z) &                          \\
\bigcap | &         & \bigcap |     &                          \\
{\Q}^2    & \isomap & H_1(E,\Q)     & = H_1(E(\C),\Z)\tensor\Q \\
We denote this bottom map $\a : {\Q}^2\isomap H_1(E,\Q)$ as well.

Then we find
\a\circ g: & {\Q}^2  & \isomap & H_1(E,\Q) \\
           & \bigcup &         & \bigcup   \\
           & {\Z}^2  & \isomap & L'        \\
where $L'$ is defined as the image of ${\Z}^2$ under $\a\circ g$.

As we have a new lattice $L'$, it determines a new elliptic curve
$E'$. We also get a map $\la : L'\isomap H_1(E',\Z)$.

So $g : (E,\a)\mapsto (E',\a')$ where $\a'$ is the composed map
$\la\circ\a\circ g : {\Z}^2 \isomap L' \isomap H_1(E',\Z)$.


Let $\t\in\H, E = E_{\t}, L_{\t} = \Z + \Z\t
\isom {\Z}^2$. Let us denote the inverse to this latter isomorphism
by $\a_{\t} : {\Z}^2 \isomap \Z + \Z\t$.
Under $\a_{\t}$, $(1,0) \mapsto \t$ and
$(0,1) \mapsto 1$.

Check that $g : (E_{\t},\a_{\t})\mapsto (E_{\t'},\a_{\t'})$ where
$\t' = {{a\t + b}\over{c\t + d}} = g\t$.
First of all, $\a_{\t'}$ takes $(1,0)$ to $\t' = {{a\t + b}\over{c\t +d}}$
and takes $(0,1)$ to $1$.

Now we look at $\la\circ\a\circ g$. $\GL_2^+(\Q)$ acts on the right on
$\Z^2$ so $g$ sends $(1,0)$ to
$$(1,0)\cdot\abcd = (a,b)$$
and sends
$(0,1)$ to $(c,d)$. Next $\a$ sends $(a,b)$ to $a\t + b = (c\t + d)\t'$
and sends
$(c,d)$ to $c\t + d$. Now $\la$ just scales the result so that it is of
the form $\a_{\mbox{\rm something}}$. That is, $(0,1)$
gets sent to $1$. This is dividing by $c\t + d$ and we see that under
$\la\circ\a\circ g$, $(0,1)$ is sent to $1$ and $(1,0)$ is sent to

In the end we want to show that $\GL_2(\A_f)$ acts on the left on
functions and on the left on curves.

\section*{23 February, 1996}
\noindent{Scribe: David M Jones, \tt <dmjones@math>}


Today we start of with some philosophy about where we are headed
in this course and why we are headed there.

We defined modular forms over $\C$ early on in the course. We
also discussed a trick by which we could define modular forms
over any ring $R$. We also defined the Hecke operators and
algebras for these modular forms over $\C$.

Recently, we defined, following Shimura, modular curves and
modular forms over more ``arithmetic" rings (such as $\Q$).
The construction of the modular curves over these rings was
done on 2/14 and 2/16 and then the modular form were realized
at differentials on the modular curves; elements of $H^0(X,\Om^1)$
where $X$ is the modular curve.

Now we are setting out to define the Hecke operators on these
modular forms -- to define the Hecke operators and Hecke algebras
in a more algebraic (or ``arithmetic") way.


Recall: For $N'|N$, we had a tower:
X(N)       & & \ff_N    \\
\downarrow & & \uparrow \\
X(N')      & & \ff_{N'} \\
\downarrow & & \uparrow \\
And we also have $\ff = \dlim \ff_N$. That is, $\ff$ is the compositum
of the fields $\ff_N$.

Intuitively, $\ff$, which is a field, corresponds to the
``pro-curve" $\plim X(N)$, which is not an algebraic curve.


We were discussing the operation of $\GL_2 (\A_f)$, where $\A_f$ is
the ring of finite adeles of $\Q$. $\A_f = {\hat \Z}\tensor\Q$ and
$\A_f$ contains ${\hat \Z}$ as a subring.

Consider a pair $(E,\a)$, where $E_{/\Q}$ is an elliptic curve and
$$\a : {\hat \Z}^2 \isomap \prod_p Ta_p(E)$$
We refer to $\prod_p Ta_p(E)$ as $Ta(E)$, the Tate module of $E$.

Now, $Ta(E)$ is free of rank $2$ over ${\hat \Z}^2$. Let $V(E) =
Ta(E)\tensor_{\Z}\Q = \prod_p V_p(E)$ where $V_p(E) = Ta_P(E)

Since $\a$ maps ${\hat \Z}^2$ to $Ta(E)$, we may extend by $\Q$ to
get a map (also called $\a$) ${\hat \Q}^2 \isomap V(E)$.

If $g\in\GL_2(\A_f)$, then $g:{\hat \Q}^2\to {\hat \Q}^2$. So by
composing $\a$ and $g$, we get a new map ${\hat \Q}^2 \isomap V(E)$.

Look at the subring ${\hat \Z}^2$ in ${\hat \Q}^2$, and take its
image under $\a\circ g$, and we get a lattice $T'$ in $V(E)$
which is different from $Ta(E)$. The following picture illustrates
the idea:
            &                                   & Ta(E)     \\
            &                                   & \bigcap | \\
{\hat \Q}^2 & \buildrel {\a\circ g}\over\isomap & V(E)      \\
\bigcup |   &                                   & \bigcup | \\
{\hat \Z}^2 & \isomap                           & T'        \\
Our goal, as stated earlier, is to make the Hecke operators
$T_n$ and $<d>$ into objects defined over $\Q$.

The key idea is that $X_1(N)$ and $X_0(N)$ have Hecke operators
acing as correspondences.

A correspondence between two curves $X$ and $Y$ is a third curve
$C$ and two maps $\a:C\to X$ and $\b:C\to Y$.
Given a correspondence $(C,\a,\b)$ of curves $X$ and $Y$, we can
form the maps
$$\a^* : H^0(X,\Om^1)\to H^0(C,\Om^1)$$
$$\b_* : H^0(C,\Om^1)\to H^0(Y,\Om^1),$$
this latter map $\b_*$
being a trace map of sorts.

$\a^*$ is the standard pull-back map. $\b_*$ can be defined using
Serre duality. Recall that $H^0(C,\Om^1)^* = H^1(C,\O_C)$ and
$H^0(Y,\Om^1)^* = H^1(Y,\O_Y)$. So the dual of $\b_*$ (which may
be described as $(\b_*)^*$ if one likes) may be
defined via the pull-back
$$\b^* : H^1(Y,\O_Y)\to H^1(C,\O_C).$$


We will make this explicit for $T_p$, $p$ not dividing $N$.

In order to define $T_p$ (for $p$ not
dividing $N$) as a correspondence from $X_0(N)$ to
itself, we will say what $C$ is and what $\a$ and $\b$ are.

Let $C = X_0(pN)$. View
$$X_0(N) = \{ \mbox{\rm pairs } (E,C) | C\isom {\Z/N\Z}\}$$
$$X_0(pN) = \{ \mbox{\rm triples } (E,C,D) | C\isom {\Z/N\Z},
D\isom {\Z/p\Z}\}$$
Then we can consider the
maps $\a$ and $\b$ as given by:
$$\a : (E,C,D)\mapsto (E,C)$$
$$\b : (E,C,D)\mapsto (E/D,(C+D)/D)$$
One must also describe what happens at the cusps but we will gloss
over that.

We may also describe this complex analitically are follows:

$\G_0(pN)\subseteq \G_0(N)$ so $\G_0(pN)\backslash\H \rightarrow
\G_0(N)\backslash\H$, which corresponds to $\a: X_0(pN)\to X_0(N)$.
Also, conjugation by $\left(\begin{array}{cc} p&0\\0&1\\ \end{array}
\right)$ maps
$\G_0(pN)$ to $\G_0(N)$ since
$$\left(\begin{array}{cc} p&0\\ 0&1\\ \end{array}\right)
\left(\begin{array}{cc} a&b\\ pNc&d\\ \end{array}\right)
\left(\begin{array}{cc} p&0\\ 0&1\\ \end{array}\right)^{-1} =
\left(\begin{array}{cc} a&pb\\ Nc&d\\ \end{array}\right)$$
So we get $\G_0(pN)\backslash \H \to
\left(\begin{array}{cc} p&0\\ 0&1\\ \end{array}\right)\G_0(pN)
\left(\begin{array}{cc} p&0\\ 0&1\\ \end{array}\right)^{-1}\backslash
\H\to \G_0(N)\backslash\H$ and this is
$$\b : X_0(pN)\to X_0(N).$$

Now we consider pulling back differentials via these maps. As always,
recall that
$$H^0(X_0(N),\Om^1) = S_2(\G_0(pN))$$
so as we have
$\a,\b : X_0(pN)\to X_0(N)$,
we get
$\a^*,\b^* : H^0(X_0(N),\Om^1)
\to H^0