Modular Forms II: The Hecke Algebra

We continue where we left off. It turns out that the space of modular forms $M_k$ admits a lot more structure, as a module over a ring called the Hecke Algebra. We give an overview of the construction here.

Here's the starting point: how might we try to get some extra structure on $M_k$? We might try to see what happens if we let something else in $\mathrm{GL}_2(\mathbb{R})^+$ act on $f$. Unfortunately, in general, if $f$ is a modular form and $\gamma \in \mathrm{GL}_2(\mathbb{R})^+$, then $g = f \underset{k}{|} \gamma$ is not a modular form. Indeed, given a $\delta\in \Gamma(1)$, then it acts on $g$ by
\[ g\underset{k}{|} \delta = f\underset{k}{|} \gamma\delta = (f\underset{k}{|} \gamma \delta \gamma^{-1})\gamma\]
and usually $\gamma \delta \gamma^{-1} \not \in \Gamma(1)$. In fact the normalizer of $\Gamma(1)$ in $\mathrm{GL}_2(\mathbb{R})^+$ is generated by $\Gamma(1)$ and $aI$ for $a \in \mathbb{R}^*$.

It turns out we need to act in a smarter way.
Consider a group $G$, and $\Gamma \leq G$. The idea is to use the double cosets of $\Gamma$ defined by
\[  \Gamma g \Gamma = \{\gamma g \gamma' : \gamma, \gamma' \in \Gamma\}. \]
One alternative way to view this is to consider the right multiplication action of $G$, hence $\Gamma$ on the right cosets $\Gamma g$. Then the double coset $\Gamma g \Gamma$ is the union of the orbits of $\Gamma g$ under the action of $\Gamma$. We can write this as
\[  \Gamma g \Gamma = \coprod_{i \in I} \Gamma g_i \]
for some $g_i \in g\Gamma \subseteq G$ and index set $I$.

In our applications, we will want this disjoint union to be finite. By the orbit-stabilizer theorem, the size of this orbit is the index of the stabilizer of $\Gamma g$ in $\Gamma$. It is not hard to see that the stabilizer is given by $\Gamma \Cap g^{-1} \Gamma g$. Thus, we are led to consider the following hypothesis:
 

Hypothesis (H): For all $g \in G$, $(\Gamma: \Gamma \Cap g^{-1} \Gamma g) < \infty$.

Then $(G, \Gamma)$ satisfies (H) iff for any $g$, the double coset $\Gamma g \Gamma$ is the union of finitely many cosets.

The important example is the following:

TTheorem  Let $G = \mathrm{GL}_2(\mathbb{Q})$, and $\Gamma \subseteq \mathrm{SL}_2(\mathbb{Z})$ a subgroup of finite index. Then $(G, \Gamma)$ satisfies (H).


Before we delve into concreteness, we talk a bit more about double cosets. Recall that cosets partition the group into pieces of equal size. Is this true for double cosets as well? We can characterize double cosets as orbits of $\Gamma \times \Gamma$ acting on $G$ by
\[  (\gamma, \delta) \cdot g = \gamma g \delta^{-1}.\]
So $G$ is indeed the disjoint union of the double cosets of $\Gamma$.

However, it is not necessarily the case that all double cosets have the same size. For example $|\Gamma e \Gamma| = |\Gamma|$, but for a general $g$, $|\Gamma g \Gamma|$ can be the union of many cosets of $\Gamma$.

Our aim is to define a ring $\mathcal{H}(G, \Gamma)$ generated by double cosets called the Hecke algebra. As an abelian group, it is the free abelian group on symbols $[\Gamma g\Gamma]$ for each double coset $[\Gamma g \Gamma]$. It turns out instead of trying to define a multiplication for the Hecke algebra directly, we instead try to define an action of this on interesting objects, and then there is a unique way of giving $\mathcal{H}(G, \Gamma)$ a multiplicative structure such that this is a genuine action.

Given a group $G$, a \term{$G$-module} is an abelian group with a $\mathbb{Z}$-linear $G$-action. In other words, it is a module of the group ring $\mathbb{Z} G$. We will work with right modules, instead of the usual left modules.

Given such a module and a subgroup $\Gamma \leq G$, we will write
\[  M^\Gamma = \{m \in M : m \gamma = m \text{ for all } \gamma \in \Gamma\}.\] For $g \in G$ and $m \in M^\Gamma$, we let
  \[    m|[\Gamma g \Gamma] = \sum_{i = 1}^n m g_i,\tag{$*$}\]
  where
  \[    \Gamma g \Gamma = \coprod_{i = 1}^n \Gamma g_i.  \]
The following properties are immediate, but also crucial.

• $m | [\Gamma g \Gamma]$ depends only on $\Gamma g \Gamma$.
• $m|[\Gamma g \Gamma] \in M^\Gamma$.

Theorem  There is a product on $\mathcal{H}(G, \Gamma)$ making it into an associative ring, the \term{Hecke algebra} of $(G, \Gamma)$, with unit $[\Gamma e \Gamma] = [\Gamma]$, such that for every $G$-module $M$, we have $M^\Gamma$ is a right $\mathcal{H}(G, \Gamma)$-module by the operation $(*)$.

In the proof, and later on, we will use the following observation: Let $\mathbb{Z}[\Gamma\setminus G]$ be the free abelian group on cosets $[\Gamma g]$. This has an obvious right $G$-action by multiplication. We know a double coset is just an orbit of $\Gamma$ acting on a single coset. So there is an isomorphism between
\[  \Theta: \mathcal{H}(G, \Gamma) \to \mathbb{Z}[\Gamma \setminus G]^\Gamma.\]
given by
\[  [\Gamma g \Gamma] \mapsto \sum [\Gamma g_i],\]
where
\[  \Gamma g \Gamma = \coprod \Gamma g_i.\]

Proof  Take $M = \mathbb{Z}[\Gamma \setminus G]$, and let

\[ \Gamma g \Gamma = \coprod \Gamma g_i\]
  \[ \Gamma h \Gamma = \coprod \Gamma h_j. \]

  Then
  \[    \sum_i [\Gamma g_i] \in M^\Gamma,\]
  and we have
  \[    \sum_i [\Gamma g_i] | [\Gamma h \Gamma] = \sum_{i, j} [\Gamma g_i h_j] \in M^\Gamma,\]
  and this is well-defined. This gives us a well-defined product on $\mathcal{H}(G, \Gamma)$. Explicitly, we have
  \[    [\Gamma g \Gamma] \cdot [\Gamma h \Gamma] = \Theta^{-1}\left(\sum_{i, j} [\Gamma g_i h_j]\right).\]
  It should be clear that this is associative, as multiplication in $G$ is associative, and $[\Gamma] = [\Gamma e \Gamma]$ is a unit.

  Now if $M$ is any right $G$-module, and $m \in M^\Gamma$, we have
  \[    m|[\Gamma g \Gamma] | [\Gamma h \Gamma] = \left(\sum m g_i\right)|[\Gamma h \Gamma] = \sum m g_i h_j = m ([\Gamma g \Gamma] \cdot [\Gamma h \Gamma]). \]
  So $M^\Gamma$ is a right $\mathcal{H}(G, \Gamma)$-module. $\blacksquare$
Now in our construction of the product, we need to apply the map $\Theta^{-1}$. It would be nice to have an explicit formula for the product in terms of double cosets. To do so, we choose representatives $S \subseteq G$ such that
\[  G = \coprod_{g \in S} \Gamma g \Gamma.\]

Proposition  We write

\[    \Gamma g \Gamma = \coprod_{i = 1}^r \Gamma g_i\]
\[    \Gamma h \Gamma = \coprod_{j = 1}^s \Gamma h_j. \]

  Then
  \[    [\Gamma g \Gamma] \cdot [\Gamma h \Gamma] = \sum_{k \in S} \sigma(k) [\Gamma k \Gamma], \]
  where $\sigma(k)$ is the number of pairs $(i, j)$ such that $\Gamma g_i h_j =\Gamma k$.

The proof is counting and omitted. We are now done with group theory. From now on we take $G = \mathrm{GL}_2(\mathbb{Q})^+$ and $\Gamma(1) = \mathrm{SL}_2(\mathbb{Z})$. We are going to compute the Hecke algebra in this case.

The first thing to do is to identify what the single and double cosets are. Let's first look at the case where the representative lives in $\mathrm{GL}_2(\mathbb{Z})^+$. We let
\[ \gamma \in \mathrm{GL}_2(\mathbb{Z})^+\]
with
\[  \det \gamma = n > 0.\]
The rows of $\gamma$ generate a subgroup $\Lambda \subseteq \mathbb{Z}^2$. If the rows of $\gamma'$ also generate the same subgroup $\Lambda$, then there exists $\delta \in \mathrm{GL}_2(\mathbb{Z})$ with $\det \delta = \pm 1$ such that
\[
  \gamma' = \delta \gamma.
\]
So we have $\deg \gamma' = \pm n$, and if $\det \gamma' = +n$, then $\delta \in \mathrm{SL}_2(\mathbb{Z}) = \Gamma$. 

This gives a bijection between cosets $\Gamma \gamma$ such that $\gamma \in \mathrm{Mat}_2(\mathbb{Z})$, $\det \gamma = n$ and subgroups $\Lambda \subseteq \mathbb{Z}^2$ of index $n$}

What we next want to do is to pick representatives of these subgroups< hence the cosets. Consider an arbitrary subgroup $\Lambda \subseteq \mathbb{Z}^2 = \mathbb{Z} e_1 \oplus \mathbb{Z} e_2$. We let
\[  \Lambda \Cap \mathbb{Z} e_2 = \mathbb{Z} \cdot d e_2\]
for some $d \geq 1$. Then we have
\[ \Lambda = <ae_1 + be_2, d e_2 >\]
for some $a \geq 1$, $b \in \mathbb{Z}$ such that $0 \leq b < d$. Under these restrictions, $a$ and $b$ are unique. Moreover, $ad = n$. So we can define
\[  \Pi_n = \{ \begin{pmatrix}   a & b\\     0 & d    \end{pmatrix} \in \mathrm{Mat}_2(\mathbb{Z}): a, d \geq 1, ad = n, 0 \leq b < d  \}.\]
Then
\[  \Big\{\gamma \in \mathrm{Mat}_2(\mathbb{Z}): \det \gamma = n\Big\} = \coprod_{\gamma \in \Pi_n}\Gamma \gamma.\]
These are the single cosets. How about the double cosets? The left hand side above is invariant under $\Gamma$ on the left and right, and is so a union of double cosets.
Lemma   

1. Let $\gamma \in \mathrm{Mat}_2(\mathbb{Z})$ and $\det \gamma = n \geq 1$. Then $\Gamma \gamma \Gamma = \Gamma \begin{pmatrix}n_1 & 0\\0 & n_2 \end{pmatrix} \Gamma $ for unique $n_1, n_2 \geq 1$ and $n_2 | n_1$, $n_1 n_2 = n$.
2. \[ \Big\{\gamma \in \mathrm{Mat}_2(\mathbb{Z}) : \det \gamma = n\Big\} = \coprod \Gamma \begin{pmatrix} n_1 & 0\\0 & n_2 \end{pmatrix} \Gamma,\] where we sum over all $1 \leq n_2 | n_1$ such that $n = n_1 n_2$. 
3. Let $\gamma, n_1, n_2$ be as above. if $d \geq 1$, then  \[ \Gamma (d^{-1}\gamma) \Gamma = \Gamma \begin{pmatrix}  n_1/d & 0\\ 0 & n_2/d \end{pmatrix} \Gamma  \]

The proof is apply the Smith normal form theorem, or, alternatively, the fact that we can row and column reduce.


Corollary  The set
$\{\Gamma \begin{pmatrix} r_1 & 0\\  0 & r_2 \end{pmatrix} \Gamma : r_1, r_2 \in \mathbb{Q}_{>0}, \frac{r_1}{r_2} \in \mathbb{Z} \}$  is a basis for $\mathcal{H}(G, \Gamma)$ over $\mathbb{Z}$.

So we have found a basis. The next goal is to find a generating set. To do so, we define the following matrices:

For $1 \leq n_2 | n_1$, we define $T(n_1, n_2)$
\[  T(n_1, n_2) = \left[\Gamma \begin{pmatrix} n_1 & 0 \\ 0& n_2  \end{pmatrix} \Gamma \right]\]
For $n \geq 1$, we write $R(n)$
\[  R(n) = \left[\Gamma \begin{pmatrix} n & 0 \\ 0& n \end{pmatrix}\Gamma \right] = \left[\Gamma \begin{pmatrix}  n & 0 \\ 0& n \end{pmatrix} \right] =T(n, n) \]
Finally, we define $T(n)$
\[  T(n) = \sum_{\substack{1 \leq n_2 | n_1 \\ n_1 n_2 = n}} T(n_1, n_2)\]
In particular, we have
\[  T(1, 1) = R(1) = 1 = T(1), \]
and if $n$ is square-free, then
\[  T(n) = T(n, 1). \]
Theorem:  

1. $R(mn) = R(m) R(n)$ and $R(m) T(n) = T(n) R(m)$ for all $m, n \geq 1$.
2. $T(m)T(n) = T(mn)$ whenever $(m, n) = 1$.
3. $T(p)T(p^r) = T(p^{r + 1}) + p R(p) T(p^{r - 1})$ of $r \geq 1$.

The proof is a counting argument based on the formula and omitted. What's underlying the proof of this is really just the structure theorem of finitely generated abelian groups. We can replace $\mathrm{GL}_2$ with $\mathrm{GL}_N$, and we can prove some analogous formulae, only much uglier. We can also do this with $\mathbb{Z}$ replaced with any principal ideal domain.e see how it helps us find a nice generating set for the Hecke algebra.


Corollary  $\mathcal{H}(G, \Gamma)$ is commutative, and is generated by $\{T(p), R(p), R(p)^{-1} : p \text{ prime}\}$.

One proves that $T(n)$ is in the subalgebra generated by the given set by induction on $n$. This result is rather surprising, because the group we started with was very non-commutative.