# Weyl law for Maass forms with nontrivial character

The classical Weyl law for $\Gamma = \mathrm{SL}_2(\mathbb{Z})$ counts the number of Maass cusp forms on $\Gamma \backslash \mathbb{H}$ with Laplace eigenvalue less than $T$. This is originally due to Selberg via the Selberg trace formula, and I believe the proof extends to any congruence subgroup $\Gamma$ of $\mathrm{SL}_2(\mathbb{Z})$.

What about Maass cusp forms with character? That is, let $N$ be a positive integer, $\chi$ a nontrivial Dirichlet character modulo $N$, and suppose that $\phi(\gamma z) = \chi(d) \phi(z)$ for all $\gamma = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in \Gamma_0(N)$. Does there exist a Weyl law for such Maass forms $\phi$ in the literature? Moreover, is there such a Weyl law with error term uniform in $N$ and $T$? I am sure at the very least that my first question is known, as the Selberg trace formula can be written down explicitly in this setting, but I can't seem to find a reference.

If we instead consider holomorphic cusp forms of weight $k \geq 2$, then the analogous question is determining dimension formulae for $\mathcal{S}_k(N,\chi)$, and these were discovered by Cohen and Osterle. Allegedly their proof uses the Eichler-Selberg formula, but it never appeared in print, and a proof was only published a couple of years ago by Jordi Quer using somewhat different methods. In the resulting formula, one finds that the main contribution as $k \to \infty$ doesn't depend on $\chi$, but the lower order terms do. For Maass forms, I'm not sure if one would be able to see this, as the associated lower order terms might be too small to distinguish from the existing error term for the Weyl law coming from the continuous spectrum, say.

• you may check recent work of Brumley and Milicevic. Nov 17 '14 at 18:16
• Unfortunately this seems to be unpublished and as-yet not available online... Nov 17 '14 at 18:38
• you can write them an email... Nov 18 '14 at 19:41
• I actually did a couple of months ago as I heard about their paper from Sarnak, but it wasn't yet complete at that stage. Nov 18 '14 at 19:54