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