Question

Since $\Gamma(N)$ is normal in $\mathrm{SL}(2,\mathbb{Z})$, the quotient group $\mathrm{SL}(2,\mathbb{Z}/N)$ acts on the spaces of cusp forms $S_k(\Gamma(N))$. How do these spaces decompose into irreducible representations?

I can do the case $N=2$. I'm mostly interested in the case of $N$ a prime.

Answer 1

See Theorem 1.0.3 of Jared Weinstein's phd thesis (it uses equivariant Riemann Roch).

Answer 2

As usual, once I spot a question on here I have anything useful to say about, somebody has already answered it.

I can sum up that part of my thesis this way: let M be the induced representation of the character (-I) --> (-1)^k of the center up to all of SL(2,Z/NZ). Then S_k(Gamma(N)) is roughly k/12 copies of M, plus some error term which can be given precisely, with some effort.

When you ask instead about Hilbert modular forms over a totally real field K, the "1/12" becomes the absolute value of zeta_K(-1).

Answer 3

If you think about this question in terms of automorphic representations then it sort of becomes trivial. The space $Sk(\Gamma(N))$ can be re-interpreted as the direct sum of $\pi^{U(N)}$, where $\pi$ is running through the automorphic representations of $GL_2$ which are holomorphic of weight $k$. Each factor is $SL(2,Z/NZ)$-invariant and often irreducible but sometimes has small finite length. The representation of $SL(2,Z/NZ)$ that shows up on $\pi{^U(N)}$ is the "type" of $\pi$. For explicit $\pi$s one will be able to explicitly determine this representation.

Answer 4

You can almost do this with nothing more than Riemann-Hurwitz; in particular, by R-H you can compute the action of SL_2(Z/NZ) on H_1(X(N),C), which is just the sum of the representation you want with its dual.