=\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
\[ \left( \begin{array}{cc}
a & 1 \\
-\epsilon(p)p^{k-1} & 0
\end{array}
\right)
\]
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
different.
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
$U_p$.
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
\[ \left( \begin{array}{cc}
0 & 1 \\
0 & 0
\end{array}
\right)
\]
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 $U_p$).
\bigskip
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 }
\bigskip
In last lecture, the following theorem was stated:
\begin{thm} Suppose $\rho : G \rightarrow \GL(2,\F_{l^\nu})$ satisfies the
usual hypotheses, that is, $l > 2$, irreducible, modular, $\det \rho = \chi$,
and $\rho$ is semistable.
Suppose $A$ be a complete local Noetherian ring with residue field
$\F_{l^\nu}$;
$0 \rightarrow \M \rightarrow A \rightarrow \F_{l^\nu} \rightarrow 0$.
Then, $\tilde{\rho} : G \rightarrow \GL(2,A)$ is modular if it satisfies the
following:
(1) $\tilde{\rho}$ lifts $\rho$,
(2) $\det \tilde{\rho} = \tilde{\chi}$ (the $\l$-adic cyclotomic character),
(3) $\tilde{\rho}$ is ramified at only a finite number of primes,
(4) $\tilde{\rho}$ satisfies the following condition (*) which is defined
according to two cases:
Case 1. $\rho$ is finite flat at $l$. In this case, $k(\rho) = 2$ and
$\rho | I_l$ is given by two fundamental characters $I_l \onto \F_{l^2}^*$ of
level 2. In this case, (*) requires that $\tilde{\rho}$ is also finite flat
at $l$ which means that for every $n \geq 1$,
$\tilde{\rho} \mod \M^n : G \rightarrow \GL(2,A/\M^n)$ comes from a finite flat
group scheme over $\Z_l$ provided with an action of $A/\M^n$.
Case 2. $\rho | D_l \sim
{\left(\begin{array}{cc}\alpha&*\\0&\beta\end{array}\right)}
$, $\beta$ is unramified, and
$\alpha | I_l = \chi$. In this case, (*) requires that
$\tilde{\rho} | D_l \sim
{\left(\begin{array}{cc}\tilde{\alpha}&*\\0&\tilde{\beta}\end{array}\right)}
$ and
$\tilde{\beta}$ is unramified ($\tilde{\beta}$ is a lift of $\beta$).
\end{thm}
\begin{remark}
In case 2, it is possible that $\rho | D_l$ can be finite flat without
$\tilde{\rho}$ being finite flat.
\end{remark}
\begin{dfn}
Let $\Sigma$ be a finite set of prime numbers. A class of liftings of
$\rho$ of type $\Sigma$ consist of liftings $\tilde{\rho}$ with
$\det \tilde{\rho} = \tilde{\chi}$
which satisfy the following properties (which make $\tilde{\rho}$ like
$\rho$):
(1) $p \neq l$.
\ \ (1-1) $p \in \Sigma$. No condition.
\ \ (1-2) $p \notin \Sigma$.
\ \ \ \ (a) If $\rho$ is unramified at $p$, then $\tilde{\rho}$ is unramified
at $p$.
\ \ \ \ (b) If $\rho | I_p \sim
{\left(\begin{array}{cc}1&*\\0&1\end{array}\right)}
$ is ramified but unipotent, then
$\tilde{\rho} | I_p \sim
{\left(\begin{array}{cc}1&*\\0&1\end{array}\right)}
$.
(2) $p = l$.
\ \ (2-1) $l \in \Sigma$. No condition.
\ \ (2-2) $l \notin \Sigma$. If $\rho$ is finite flat, then $\tilde{\rho}$ is
finite flat.
\end{dfn}
\begin{eg}
Let $\tilde{\rho} = \rho_{f,\lambda}$ ($\lambda | l$). Then
$\ord_p N(f) = \ord_p N(\rho) + \dim(\rho)^{I_p} - \dim(\tilde{\rho})^{I_p}$.
We have $\ord_p N(\rho) + \dim(\rho)^{I_p} = 2$, and $\dim(\rho)^{I_p} -
\dim(\tilde{\rho})^{I_p} \geq 0$. So, $\ord_p N(f) \leq 2$. If $p \notin
\Sigma$, then $\ord_p N(f) = \ord_p N(\rho)$. If $\rho$ is ramified at $p$,
$\ord_p N(\rho) = 1$ and hence $\ord_p N(f) = 1$.
\end{eg}
\begin{dfn}
$$N_\Sigma = \prod_{p \in \Sigma - \{l\}} p^2 \cdot
\prod_{p \notin \Sigma \cup \{l\}} p^{\ord_p N(\rho)} \cdot l^\delta$$
where $\delta = 0$ or $1$, and $\delta = 1$ if and only if (a) $k(\rho) = l+1$
or (b) $l \in \Sigma$ and $\rho | D_l \sim
{\left(\begin{array}{cc}\alpha&*\\0&\beta\end{array}\right)}
$.
\end{dfn}
\noindent{\bf Guess.}
Suppose $\tilde{\rho} = \rho_{f,\lambda}$ belongs to the class defined by
$\Sigma$. Then $N(f) | N_\Sigma$.
\begin{exercise}
Justify, a priori, the definition of $\delta$. Suppose, for example, that
$\rho_{f,\lambda}$ satisfies (*). Prove $l^2 \not| N(f)$.
\end{exercise}
Our goal is to prove
\begin{thm}
Every $\tilde{\rho}$ in the class defined by $\Sigma$ comes from
$S_2(\Gamma_0(N_\Sigma))$.
\end{thm}
\begin{remark}
(1) $\rho$ comes by reduction from a $\rho_{f,\lambda}$ with $f$ in
$S_2(\Gamma_0(N_\Sigma))$ because $N(\rho) | N_\Sigma$ and $k(\rho) = 2$.
%or $l+1$(if $k(\rho) = l + 1$, then $\delta = 1$ and so $l | N_\Sigma$.
(2) Let $\T = \Z[\cdots T_n \cdots] \subseteq \endo(S_2(\Gamma_0(N_\Sigma)))$
where $(n,lN_\Sigma) = 1$. Then $\exists \M \subseteq \T$ such that
\ \ (a) $\exists \tilde{\rho} : G \rightarrow \GL(2,\T_\M)$ such that
$\tr \tilde{\rho}(\frob_r) = T_r$.
\ \ (b) $\tilde{\rho}$ is universal for liftings of type $\Sigma$.
\end{remark}
We have $\rho : G \rightarrow \GL(2,\F_{l^\nu})$, $l > 2$, irreducible,
modular, $\det\rho = \chi$, semistable.
Let $\T = \Z[\cdots T_n \cdots]$ where $(n,l N_\Sigma) = 1$, and
let $R = \Z[\cdots T_n \cdots] = \T[T_l;\cdots U_p\cdots]$ where $p|N_\Sigma$.
(Note that $\rank_\Z \T = \sum_{M|N_\Sigma} \dim S_2(\Gamma_0(N_\Sigma))$ and
$\rank_\Z R = \dim S_2(\Gamma_0(N_\Sigma))$.)
Since $N(\rho) | N_\Sigma$, there is a new form $f$ of level $M | N_\Sigma$
whose coefficients are in $\O_\lambda$ and there is a map
\begin{diagram}[tight,height=2em]
\T &\rTo(2,0) &\O_\lambda \\
&\rdTo(2,2) &\dTo(0,2) \\
& &\F_{l^\nu}
\end{diagram}
such that
$$T_r \mapsto \tr\rho(\frob_r) \in \F_{l^\nu}.$$
Let $\M = \ker(\T \rightarrow \overline{\F})$.
We need to show that $\T_\M$ is Gorenstein.
We want to find $\M_R \subseteq R$ such that $\T_\M = R_{\M_R}$.
To do this, find an eigenform $h$ of level $N_\Sigma$ satisfying
$h|U_l = ({\rm unit})h, {\rm unit\ } \in \O_\lambda$,
$h|U_p = 0$ for$p \notin \Sigma - \{l\}$,
$h|T_p = a_p(f)h$ for $p \in \Sigma$.
\noindent Then $h$ gives us the following diagram:
\begin{diagram}[tight,height=2em]
\M_R &\subseteq&R & & & \\
& & &\rdTo(2,2)\rdTo(4,2)& & \\
\bigcup& &\bigcup& &\O_\lambda&\rTo(0.5,0)&\overline{\F}\\
& & &\ruTo(2,2) & &\ruTo(4,2) \\
\M &\subseteq&\T & & &
\end{diagram}
where $\M_R = \ker(R \rightarrow \overline{\F})$.
\noindent {\bf Construction of $h$.}
To construct an $h$, we make the following two changes to $f$. (The order the
changes are made in doesn't matter.)
\noindent
(1) Let $f = \sum a_n q^n$. Then change $f$ to $g = \sum a_n q^n$ where the
sum is over $n$ prime to each $p \notin \Sigma$. Then $g$ is an eigenform
for $T_p$ for $p \notin \Sigma$. We can do this because $g = (f \tensor
\epsilon) \tensor \epsilon$ where $\epsilon$ is a Dirichlet character ramified
at $\Sigma$. In terms of L-functions, roughly if
$L(f,s) = \prod_p (1- a_p p^{-s} + p^{1 - 2s})^{-1}$, then
$L(g,s) = \prod_{p \notin \Sigma} (1- a_p p^{-s} + p^{1 - 2s})^{-1}$.
\noindent
(2) Suppose $l | N_\Sigma$ but $l \not| N(f)$. This happens if $l \in \Sigma$
and $\rho$ is good ordinary at $l$. We have $f |T_l = a_l(f) f$, but $f|U_l =
{\rm \ junk}$. In this case, change $f$ to $g = f + (*) f(q^l)$. If we choose
(*) correctly, then $g | U_l = Cg$ where $C$ is a root of $X^2 - a_l X + l = 0$
where $a_l = a_l(f)$. $X^2 - a_l X + l = 0$ has exactly one unit root in
$\O_\lambda$ because $a_l \notin \lambda$. So take this unit root and then
$g|U_l = ({\rm unit})g$.
\begin{remark}
When $p || N_\Sigma$, then $p \notin \Sigma$ and $p || N(\rho)$. We have
$N(f) = N(\rho)$ or $N(\rho)l$. In this case, we have $f|U_p = a_p(f)f$. So
nothing has to be done.
\end{remark}
The following lemma will be used to prove $\T_\M = R_\M$.
\begin{lem} $R = \T[\cdots U_p\cdots]$ if $l \not| N_\Sigma$.
\end{lem}
\noindent{\bf Proof.\ } It is sufficient to show that $(R : \T[\cdots
U_p\cdots])$ is prime to $l$ (since then we cannot tell the difference if we
tensor both sides with $\Q_l$). We show that
$\T[\cdots U_p\cdots] \rightarrow R/lR$ is surjective.
Let $A$ be the image of this map, that is, $A = \F_l[\cdots T_n \cdots]$
where $(n,l) = 1$. We have the duality:
$$R/lR \times S_2(\Gamma_0(N_\Sigma);\F_l) \rightarrow \F_l.$$
To show $R/lR = A$ is equivalent to show that $A^\perp = 0$ in this paring.
Suppose $f \in A^\perp$ and $f \neq 0$. $f \in A^\perp$ means that
$a_n(f) = 0$ for all $n$ such that $(n,l) = 1$. That is, $f$ can be written as
$f = \sum a_{nl} q^{nl}$. Then $\theta(f) = 0$. This is a contradiction because
$w(f) = 2$ and $w(\theta f) = w(f) + l + 1$ if $l \not| w(f)$.
\begin{eg} Suppose $l = 2$ and consider $S_2(\Gamma_0(23))$. Then
$R = \Z[\frac{1+\sqrt{5}}{2}]$ and $\Z[\cdots T_n\cdots] = \Z[\sqrt{5}]$
where $(n,l) = 1$. This shows that the lemma does not work for $l = 2$.
\end{eg}
\end{document}