# Lecture 16: Structure of the Hecke algebra

$$\DeclareMathOperator{\Jac}{Jac} \newcommand{\SL}{\mathrm{SL}} \def\mat#1#2#3#4{\left( \begin{array}{cc} #1 & #2 \\ #3 & #4 \end{array} \right)} \newcommand{\bT}{\mathbf{T}} \newcommand{\bZ}{\mathbf{Z}} \newcommand{\bQ}{\mathbf{Q}} \let\wt\widetilde \DeclareMathOperator{\End}{End} \let\ol\overline \newcommand{\fh}{\mathfrak{h}} \newcommand{\bC}{\mathbf{C}} \newcommand{\bR}{\mathbf{R}} \newcommand{\id}{\mathrm{id}} \newcommand{\rH}{\mathrm{H}} \DeclareMathOperator{\Jac}{Jac}$$

In this lecture, we establish basic results on the structure of the Hecke algebra and some of its natural modules. In particular, we show that the Hecke algebra $\bT$ is a finite rank free $\bZ$-algebra and that $\bT \otimes \bQ$ is semi-simple. We also show that the space of weight 2 cusp forms (at prime level) is a free module of rank 1; this is the multiplicity one theorem.

## Review

Recall that for a modular form $f$ for $\Gamma(1)$ and weight $k$, we defined the action of the Hecke operator $T_n$ on $f$ by

$$(T_nf)(\Lambda) = n^{k-1} \sum_{[\Lambda:\Lambda']=n} f(\Lambda').$$

In fact, the same formula can be used to define $T_nf$ when $f$ is a modular form on $\Gamma_0(N)$ of weight $k$ if $n$ is prime to $N$. From now on, we only care about the $k=2$ case.

We proved the following results:

• The $T_n$ commute with each other.

• If $n$ and $m$ are coprime then $T_{nm}=T_nT_m$.

• We have the recurrence $T_{p^{n+1}}=T_pT_{p^n}-pT_{p^{n-1}}$.

• If $f=\sum_{n \ge 1} a_n q^n$ and $p$ is prime then

$$T_pf = \sum_{n \ge 1} (a_{pn}+p^{k-1} a_{n/p}) q^n$$

In particular, $a_1(T_pf)=a_p(f)$. In fact, $a_1(T_nf)=a_n(f)$ holds for all $n$ prime to $N$.

We define $\wt{\bT}$ to be the infinite polynomial ring $\bZ[T_p]$ with $p$ prime to $N$. We define $\bT$ to be the image of $\wt{\bT}$ in $\End(S_2(N))$. Our goal today is to understand this ring and the structure of $S_2(N)$ as a module over it.

Let $f,g \in S_2(N)$. Then $f(z) dz$ and $g(z) dz$ are 1-forms on the upper half-plane invariant under $\Gamma_0(N)$. It follows that $\ol{g(z) dz}$ is also invariant, and so $f(z) dz \wedge \ol{g(z) dz}=2i f(z) \ol{g(z)} dx dy$ is also invariant. We define

$$\langle f, g \rangle = \int_{\fh/\Gamma_0(N)} f(z) \ol{g(z)} dx dy.$$

This is called the Petersson inner product. The integral converges since $f$ and $g$ decay rapidly at the cusps (as they are cusp forms). It is clear that $\langle, \rangle$ is a positive definite Hermitian form on $S_2(N)$. Furthermore, we have the following result:

Proposition. The operators $T_p$ are self-adjoint. That is, $\langle T_pf, g \rangle=\langle f, T_p g \rangle$.

Corollary. The algebra $\bT \otimes \bC$ is semi-simple. The space $S_2(\Gamma)$ admits a basis consisting of Hecke eigenforms (i.e., forms $f$ which are eigenvectors of $T_p$ for all $p \nmid N$).

We call a Hecke eigenform $f$ normalized if $a_1(f)=1$. We note that for such forms, $a_n(f)$ is the eigenvalue of $T_n$ acting on $f$, for $n$ prime to $N$. If $f$ is an eigenform then we get a ring homomorphism $\alpha \colon \bT \to \bC$ by mapping $T_p$ to the eigenvalue of $f$ under $T_p$. One calls $\alpha$ a system of eigenvalues. The space of cusp forms decomposes as

$$S_2(N) = \bigoplus S_2(N)_{\alpha},$$

where $S_2(N)_{\alpha}$ is the space of forms $f$ with $T_pf=\alpha(T_p) f$ for all $p \nmid N$.

## Multiplicity one

Theorem. Suppose $N$ is prime and $f,g \in S_2(N)$ are two normalized Hecke eigenforms with the same eigenvalues for all $p \ne N$. Then $f=g$.

Proof. We first introduce a piece of notation. For $\gamma \in \SL_2(\bR)$ and a function $f$ on the upper half-plane, define $(f | [\gamma] )(z) = (cz+d)^{-2} f(\gamma)(z)$. Then one can show that this defines an action of $\SL_2(\bR)$. Furthermore, $f$ belongs to $S_2(N)$ if and only if $f \vert [\gamma] = f$ for all $\gamma \in f$, and $f$ satisfies the appropriate holomorphicity conditions.

Now, for $p \ne N$ we have $a_p(f)=a_p(g)$ since the $T_p$ eigenvalues of $f$ and $g$ are assumed equal. It follows that $a_n(f)=a_n(g)$ for all $n$ prime to $N$ by the multiplicativity and recurrence properties of the $T_n$. Let $h=f-g$. Then $a_n(h)=0$ unless $N \mid n$. It follows that $h(z+1/N)=h(z)$. Thus $h | [\gamma] = h$ if $\gamma$ belongs to $\Gamma_0(N)$ or if

$$\gamma = \mat{1}{1/N}{}{1}.$$

The group $\Gamma_0(N)$ and the above matrix together generate the group $\sigma^{-1} \Gamma(1) \sigma$, where

$$\sigma = \mat{N}{}{}{1}.$$

Thus $h | [\sigma]$ is invariant under $\Gamma(1)$, and therefore belongs to $S_2(1)$. But $S_2(1)=0$, so $h=0$ and $f=g$.

Corollary. For any system of eigenvalues $\alpha$, the space $S_2(N)_{\alpha}$ is one-dimensional.

Corollary. There is a bijection between homomorphisms $\bT \to \bC$ and normalized cusp forms.

Corollary. $S_2(N)$ is free of rank 1 as a $\bT \otimes \bC$ module.

Remark. There is also a stronger version of this theorem: if $f$ and $g$ are two normalized Hecke eigenforms that have the same eigenvalues under $T_p$ for all but finitely many $p$ (or even all $p$ in a density 1 set of primes) then $f=g$.

Remark. The above theorem is false for $N$ composite, but for a somewhat silly reason. If $p \mid N$ and $f \in S_2(N/p)$ then both $f(z)$ and $f(pz)$ belong to $S_2(N)$, and they have the same $T_{\ell}$ eigenvalues for $\ell \nmid N$. This is the only problem. More precisely, we define the old subspace to be the subspace of $S_2(N)$ spanned by $f(z)$ and $f(pz)$ where $p|N$ and $f \in S_2(N/p)$. We define the new subspace to be the orthogonal complement of the old subspace. Then multiplicity one holds on the new subspace.

## Hecke correspondences

In terms of lattices, $(T_pf)(\Lambda)$ is defined by summing the values of $f$ over the index $p$ sublattices of $\Lambda$. If $\Lambda$ corresponds to the elliptic curve $E$, then the sublattices correspond to isogenies $E \to E'$ of degree $p$. The set of such isogenies is exactly the fiber of $X_0(pN) \to X_0(N)$ above $E$. Summing over the $E'$ the corresponds to pushing forward along the map taking $E \to E'$ to $E'$.

Let us be more precise. We previously defined a $\Gamma_0(p)$ structure on $E$ to be a cyclic subgroup $G$ of order $p$. But it is equivalent to define a $\Gamma_0(p)$ structure to be an isogeny $E \to E'$ of degree $p$ with cyclic kernel. For the moment, it will be more convenient to think of $X_0(p)$ as the space of such isogenies. We think of $X_0(Np)$ as parametrizing data $(E \to E', G)$ where $E \to E'$ is an isogeny of degree $p$ and $G \subset E$ is a cyclic subgroup of size $N$.

Let $p$ be prime to $N$. There are two natural maps $X_0(Np) \to X_0(N)$, namely, take $(f:E \to E', G)$ to $(E, G)$ or $(E', f(G))$. Call these maps $p_1$ and $p_2$. The diagram

$$\xymatrix{ & X_0(pN) \ar[rd]^{p_2} \ar[ld]_{p_1} \\ X_0(N) && X_0(N) }$$

is called the Hecke correspondence.

## Generalities on correspondences

Let $C$ be a smooth projective curve. A correspondence $C \dashrightarrow C$ is a pair of maps $p_1, p_2 \colon C' \to C$ with $p_1$ and $p_2$ finite maps; we assume $C'$ is a smooth projective curve as well. We can think of a non-constant function $f \colon C \to C$ as a correspondence by taking $C'=C$, $p_1=\id$, and $p_2=f$. In general, one thinks of correspondences as multi-valued functions, where $x \in C$ is mapped to the set $p_2(p_1^{-1}(x))$.

Correspondences act on the singular cohomology $\rH^1(C, \bZ)$ by the formula $(p_2)_* p_1^*$. Here $p_1^*$ is the usual pull-back operation $\rH^1(C, \bZ) \to \rH^1(C', \bZ)$, and $(p_2)_*$ is the adjoint to $p_2^*$ under the cup product pairing.

Correspondences also act on differential forms, via the same formula. Over the complex numbers, the isomorphism

$$\rH^1(C, \bZ) \otimes \bC = \rH^0(C, \Omega^1) \oplus \ol{\rH^0(C, \Omega^1)}$$

from Hodge theory is compatible with the action of correspondences.

Correspondences also act on divisors, by the same formula. This action preserves principal divisors, and thus induces a map on the Jacobian. If $f$ is the correspondence $(p_1,p_2)$ then the induced endomorphism of $\Jac(C)$ is $(p_2^*)^{\vee} p_1^*$, where $p_i^* \colon \Jac(C) \to \Jac(C')$ is the natural map, and $(-)^{\vee}$ is the dual isogeny.

## Back to Hecke operators

By the above discussion, the big Hecke algebra $\wt{\bT}$ acts on $\rH^1(X_0(N), \bZ)$, and the isomorphism

$$\rH^1(X_0(N), \bZ) \otimes \bC = S_2(N) \oplus \ol{S_2(N)}$$

is compatible with the actions on each side. (Here we have identified $S_2(N)$ with $\rH^0(X_0(N), \Omega^1)$.) If an element of $\wt{\bT}$ acts by zero on $S_2(N)$, then it does so on $\ol{S_2(N)}$ as well, and therefore on $\rH^1(X_0(N), \bZ)$. We therefore see that the image of $\wt{\bT}$ in $\End(\rH^1(X_0(N), \bZ))$ is just $\bT$. Since this endomorphism ring is $M_{2g}(\bZ)$, we see that:

Proposition. The Hecke algebra $\bT$ is a free $\bZ$-module of finite rank.

Corollary. The Hecke eigenvalues of a normalized cusp form are algebraic integers.

Corollary. $\bT \otimes \bQ$ is a product of finitely many number fields.

We have shown that $S_2(N)$ is free of rank 1 as a module over $\bT \otimes \bC$. The same is clearly true for $\ol{S_2(N)}$. Thus $\rH^1(X_0(N), \bZ) \otimes \bC$ is free of rank 2 over $\bT \otimes \bC$. Now, $\rH^1(X_0(N), \bQ)$ is a module over the semi-simple ring $\bT \otimes \bQ$ which becomes free of rank 2 when tensored up to $\bC$; it is therefore necessarily free of rank 2 itself. We've proved:

Proposition. $\rH^1(X_0(N), \bQ)$ is free of rank 2 as a $\bT \otimes \bQ$ module.

## The Atkin--Lehner involution

The space $X_0(N)$ admits a natural involution $w$ defined as follows. If we think of the points of $X_0(N)$ as cyclic isogenies of degree $N$, then $w$ takes $f \colon E \to E'$ to the dual isogeny $f^{\vee} \colon E' \to E$. If we think of the points of $X_0(N)$ as elliptic curves with a cyclic subgroup of order $N$ then $w$ takes $(E,G)$ to $(E/G, E[N]/G)$.

Thinking of weight 2 cusp forms as 1-forms on $X_0(N)$, we get an action of $w$ by pull-back. In terms of functions on the upper half-plane, $(wf)(z)=f(-1/Nz)$. One verifies that this action of $w$ on $S_2(N)$ commutes with the Hecke operators $T_p$ for $p \nmid N$. Assuming now that $N$ is prime (or $N$ is arbitrary and we use the new subspace). Since $w$ commutes with $\bT$, it preserves the $\bT$ eigenspaces, and these are one dimensional. It follows that if $f$ is an eigenform then $wf=\pm f$.