Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

For congruence subgroups of $PSL(2,\mathbb{Z})$, the Weyl law for the eigenvalues of Maass cusp forms had been proven by Selberg. How is the status of such a Weyl law for eigenvalues of Maass cusp forms for arbitrary arithmetic nonuniform cofinite Fuchsian groups? Has it already been proven? Does one know at least that there are infinitely many linearly independent Maass cusp forms for these lattices?

share|cite|improve this question
I decided to ask the following follow-up question:… ... Perhaps you have an example at hand? –  Marc Palm May 8 '13 at 7:38

2 Answers 2

The Weyl law has been proven in great generality, e.g. for congruence subgroups of $PGL(n,\mathbb{R})$, and beyond. See the introduction in Lindenstrauss-Venkatesh: Existence and Weyl's law for spherical cusp forms, GAFA 17 (2007), 220-251 (manuscript available here). See also Marc Palm's recent thesis.

share|cite|improve this answer
The only arithmetic, nonuniform cofinite Fuchsian groups are (after a conjugation in $SL_2({\mathbb R})$) commensurable with $SL_2({\mathbb Z})$. So the issue is congruence subgroups vs. non-congruence subgroups of $SL_2({\mathbb Z})$. –  Venkataramana May 4 '13 at 23:42

Consider the normalizer $\Gamma$ of $\Gamma_0(N)$ in $SL_2(\mathbb{Q})$ for $N$ squarefree, which is not a subgroup of $SL_2(\mathbb{Z})$, then you obtain the same Weyl law as classical. For my taste, these groups should also be called congruence subgroups, although they are not contained in $SL_2(\mathbb{Z})$. They are important if you want classically distinguish between nonisomorphic supercuspidal reps contained as factors of automorphic reps associated, which are associated to ramified quadratic extensions of $\mathbb{Q}_p$ at $p |N$.

I don't know a single example what happens if $\Gamma$ does not contain a congruence subgroup, but is arithmetic.

Btw, your question makes perfect sense, because the modified Rolecke-Selberg conjecture states (see the introduction that apart from arithmetic lattices, there are at most finitely many discrete eigenvalues.

There is some computational evidence by Hejhal for this conjecture: Hejhal, Dennis A.(1-MN-SM) On eigenvalues of the Laplacian for Hecke triangle groups.

Copy&Paste from MathSciNet: This paper is part of a series of articles in which computer experiments are performed to numerically compute the eigenvalues of the Laplacian. In this paper the Hecke triangle groups generated by z→−1/z, z→z+2cos(π/N) are considered. The basic results are a computation of eigenvalues for the groups with N=3,4,6 (the congruence groups) with eigenvalue 14+R2 with R<25, and the conclusion that no even cusp forms exist when N=5, R<60 and N=7, R<40. The last result, which is similar to results of Winkler, gives evidence in support of the Phillips-Sarnak conjecture that one should have few if any cusp forms for nonarithmetic groups (except for the obvious ones caused by symmetries). The procedure used is essentially the collocation method. One must also be careful in evaluating the K-Bessel functions that appear in the Fourier expansion of the forms.

What is sure is that if $\Gamma$ does not contain a congruence subgroup, then the Maass cusps forms on $\Gamma \backslash \mathbb{H}$ can not be lifted to a vector of an automorphic adelic space $SL_2(\mathbb{A})$ in any obviuous way, and one cannot expect them to have a nice $L$-functions. Also the contribution of the continuous spectrum is most likely not expressable in terms of classical Dirichlet L-functions. Recall that this is the main point to be understood for Weyl laws of non-uniform lattices.

I know that the Belyi's theorem generates an isomorphisms between nonsingular algebraic curves over $\mathbb{C}$ and $\Gamma \backslash \mathbb{H}$ compactified at the cusps, where $\Gamma$ is an arithmetic subgroup of $SL_2(\mathbb{Z})$, but noncongruence in general. This might be a source of such lattices and at least modular functions on them should exists (by Riemann Roch?).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.