How to compute the Hecke operator on an Eisenstein series?

Question

In my current course on Modular Forms we are now discussing Hecke operators and we are asked the following: Prove that for any even integer $k \geq 4$ and prime $p$ we have $T_pG_k = \sigma_{k−1}(p)G_k$ for the Eisenstein series $G_k$ and the Hecke operator $T_p$ on $M_k(SL_2(\mathbb{Z}))$. First of all, I calculated that $\sigma_{k-1}(p)=1+p^{k-1}$. Furthermore, by definition, for $\alpha=\begin{pmatrix} 1 & 0 \\ 0 & p \end{pmatrix}$ we have, by definition of $T_p$, that $T_pG_k=\sum_{\gamma \in \Gamma' \setminus \Gamma}G_k|_k\alpha\gamma$, for $\Gamma=SL_2(\mathbb{Z})$ and $\Gamma'=\Gamma \cap \alpha^{-1} \Gamma \alpha$. We have, for $\gamma \in \Gamma$, $\gamma = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ \begin{align} &G_k|_k\alpha\gamma(z)\\ &=\frac{det(\alpha \gamma)^k}{(pcz+pd)^k}G_k(\alpha \gamma z)\\ &=\frac{p^k}{p^k(cz+d)^k}G_k(\alpha \gamma z) \\ &=\frac{1}{(cz+d)^k}G_k(\alpha \gamma z) \end{align} where , so now we have: \begin{align} T_pG_k=\sum_{\gamma \in \Gamma' \setminus \Gamma}\frac{1}{(cz+d)^k}G_k(\alpha \gamma z) \end{align} and we want this to be equal to $(1+p^{k-1})G_k$. I don't know how to proceed at this point, since we cannot just take $\alpha \gamma$ out out of $G_k(\alpha \gamma z)$ since $\alpha \gamma$ is not in $SL_2(\mathbb{Z})$.

Answer 1

The Hecke operators for $\Gamma=\mathrm{SL}_2(\mathbb{Z})$ can be written down explicitly: if $f\in M_k(\Gamma)$, then $$ T(n)f(z)=n^{k-1}\sum_{\substack{ad=n,a\geq 1\\0\leq b<d}}d^{-k}f\Big(\frac{az+b}{d}\Big) $$

In particular, if $p$ is prime and $$ G_k(z)=\sum_{m,n\in\mathbb{Z}}{}^{'}(mz+n)^{-k}$$ is the Eisenstein series, then $$ T_pG_k(z)=p^{k-1}\Big[G_k(pz)+\sum_{b=0}^{k-1}p^{-k}G_p\Big(\frac{z+b}{p}\Big)\Big] $$ $$ =p^{k-1}\Big[\sum_{m,n}{}^{'}(mpz+n)^{-k}+\sum_{b=0}^{p-1}\sum_{m,n}{}^{'}(mz+bm+pn)^{-k}\Big]$$

In the second term, if we separate out the terms with $p|m$, we obtain $$ \sum_{b=0}^{p-1}\sum_{m,n}{}^{'}(pmz+bpm+pn)^{-k}=p^{-k}\sum_{b=0}^{p-1}\sum_{m,n}{}^{'}(mz+bm+n)^{-k}=p^{1-k}G_k(z)$$ since for each $0\leq b\leq p-1$, the inner sum is just $G_k(z+b)=G_k(z)$.

Now consider the remaining terms. If $(u,v)$ is an element of $\mathbb{Z}^2$ with $p$ not dividing $u$, then there is a unique element $(m,n)\in \mathbb{Z}^{2}\setminus\{(0,0)\}$ and a unique $0\leq b\leq p-1$ such that $(u,v)=(m,bm+pn)$: take $m=u$, $b$ such that $u^{-1}v\equiv b$ mod $p$, and $n$ such that $pn=v-bu$.

If $(u,v)\neq (0,0)$ and $p|u$, then there is a unique element $(m,n)\in \mathbb{Z}^{2}\setminus\{(0,0)\}$ such that $(pm,n)=(u,v)$, namely $n=v$ and $m=\frac{u}{p}$. Therefore the remaining terms in the expansion for $T_pG_k(z)$ can be collected to yield $$\sum_{u,v}{}^{'}(uz+v)^{-k}=G_k(z) $$

Thus we have shown that $$ T_pG_k(z)=p^{k-1}[p^{1-k}G_k(z)+G_k(z)]=(1+p^{k-1})G_k(z)=\sigma_{k-1}(p)G_k(z) $$ as desired.

By the way, it turns out that $T(n)G_k(z)=\sigma_{k-1}(n)G_k(z)$ for all $n$. My guess is that for general $n$, it would be easier to prove this by looking at the Fourier expansions.