Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I recently became interested in Maass cusp forms and heared people mentioning a "multiplicity one conjecture". As far as I understood it, it says that the dimension of the space of Maass cusp form for fixed eigenvalue should be at most one.

Since Maass cusp forms always are defined for a Fuchsian lattice, I wonder 1) for which lattices this conjecture had been conjectured? 2) what is the motivation for this conjecture? 3) to whom this conjecture is due? 4) is it published somewhere? 5) is it proven in some cases?

share|improve this question

4 Answers 4

up vote 3 down vote accepted

I rethought your question and have discovered a partial answer for 1) and 2). I add this as a disjoint answer, since my other answer adresses a totally different (negative) issue.

In Sarnak's article http://web.math.princeton.edu/sarnak/baltimore.pdf, he recalls one famous conjecture (Conjecture I, due to himself) that $\Gamma \backslash \mathbb{H}$ should have very few Maass forms for "most" Fuchsian lattices $\Gamma$.

Btw, he attributes the simplicity conjecture (Conjecture 3) for $SL_2(\mathbb{Z})$ to Cartier.

Wolpert (Theorem I) has shown that Sarnak's conjecture would follow if the simplicity conjecture holds for the congruence subgroup $\Gamma(2)$.

Also GH last conjecture that the multiplicity is uniformly bounded in $N$ would suffice for Sarnak's conjecture, but current knowledge is that the multiplicity of an eigenvalue of magnitude $T$ is at most $\ll_N \sqrt{T}/ \log T$ and not $\ll_N 1$.

In fact, Sarnak conjectures that there exist $\Gamma$ with only finitely many Maass wave forms, but this does not follow from the simplicity conjecture for $\Gamma(2)$.

share|improve this answer

This conjecture is usually stated for $\mathrm{SL}_2(\mathbb{Z})$, and it is widely open. I think it is folklore, and is stated in several papers, e.g. in Luo: Nonvanishing of $L$-values and the Weyl law (before (3)).

The motivation, I think, is similar as with the conjecture for the multiplicity of the Riemann zeta zeros. The belief is that there is no "accidental" algebraic independence among the eigenvalues of the Laplacian or the zeros of an automorphic $L$-function. For example, the Laplacian eigenvalue $1/4$ is expected to "come from" an even Galois representation, while the zero $1/2$ is expected to "come from" rational points of infinite order on an abelian variety. For $\mathrm{SL}_2(\mathbb{Z})$ or $\zeta(s)$ we don't know of any object that would "impose" any algebraic independence on the data, hence we believe that in those cases the data is entirely transcendental.

For congruence subgroups $\Gamma_0(N)$ the "multiplicity one conjecture" is false, because the eigenvalue $1/4$ is known to occur with multiplicity for some $N$'s. The known examples come from even Galois representations. I think it is safe to believe that the multiplicities are bounded for any $N$.

share|improve this answer
Is it also know for newforms of $\Gamma_0(N)$? –  Marc Palm Jul 19 '12 at 14:29
I think the eigenvalue 1/4 occurs with multiplicity even among newforms of some level. All we need is two even Galois representations with the same Artin conductor. –  GH from MO Jul 19 '12 at 19:16
Sorry for bumping this old question, but is the multiplicity bounded uniformly with respect to $N$? –  user31814 May 15 '14 at 18:07
@user31814: I think the multiplicity is bounded (conjecturally) for a fixed $N$, but it can get arbitrary large (provably) for varying $N$. –  GH from MO May 15 '14 at 18:11
@GHfromMO, thanks! Is there a reference for what you said? –  user31814 May 16 '14 at 1:36

Multiplicity one refers to something else, related but much weaker.

For the analogue question for lattices, there are trivial counter examples: Induction by steps for example suggests on the level of Lie groups $$ Ind_{\Gamma(N)} ^{PSL_2(\mathbb{R})} 1 \cong Ind_{PSL_2(\mathbb{Z})} ^{PSL_2(\mathbb{R})} Ind_{\Gamma(N)}^{PSL_2(\mathbb{Z})} 1$$ and e.g. by the Peter-Weyl theorem, we know that $$ Ind_{\Gamma(N)}^{PSL_2(\mathbb{Z})} 1$$ the multiplicity of an irreducible representation equals its dimension.

Note that GH's example is less trivial, since $$ Ind_{\Gamma_0(N)}^{PSL_2(\mathbb{Z})} 1$$ decomposes with multiplicity one.

share|improve this answer
Could you elaborate to what it does refer, please? –  Ruedi Meier Aug 15 '12 at 17:49
For every prime, you get also an eigenvalue. Multiplicity one theorem states that their exists only one eigenfunction with all the same eigenvalues. More rigorously put, you find this here: en.wikipedia.org/wiki/Multiplicity-one_theorem –  Marc Palm Sep 4 '12 at 9:33

At the risk of blowing my own trumpet, I feel like I ought to mention a recent preprint of mine that addresses this question.

Marc Palm answered your question 2, 3, and 4 reasonably well. (For question 4, the reference for where it is published is Cartier's paper.) For question 1, I know that Bolte and Johansson certainly expect the conjecture to be true for $\Gamma_0(q) \backslash \mathbb{H}$ with $q$ squarefree, provided one first removes all the Maaß oldforms (those coming from lower level), as they will always give rise to spectral multiplicity at least $2$. So for $q > 1$, the conjecture should be modified to only be about the eigenvalues of Maaß newforms.

Bolte and Johansson (and later Strömbergsson) describe a spectral correspondence between the eigenvalues of Maaß newforms on $\Gamma_0(q) \backslash \mathbb{H}$, $q$ squarefree, and eigenvalues of the automorphic Laplacian for the group of units of norm one in a maximal order in an indefinite quaternion division algebra over $\mathbb{Q}$. Bolte and Johansson conjecture that the spectrum of this automorphic Laplacian is simple (see the Hypothesis on p.61), and hence that the eigenvalues of Maaß newforms on $\Gamma_0(q) \backslash \mathbb{H}$ are simple when $q$ is squarefree. Here we are looking at the congruence subgroups \[\Gamma_0(q) = \left\{\gamma \in \mathrm{SL}_2(\mathbb{Z}) : \gamma \equiv \begin{pmatrix} * & * \\\ 0 & * \end{pmatrix} \pmod{q}\right\},\] \[\Gamma_1(q) = \left\{\gamma \in \mathrm{SL}_2(\mathbb{Z}) : \gamma \equiv \begin{pmatrix} 1 & * \\\ 0 & 1 \end{pmatrix} \pmod{q}\right\},\] \[\Gamma(q) = \left\{\gamma \in \mathrm{SL}_2(\mathbb{Z}) : \gamma \equiv \begin{pmatrix} 1 & 0 \\\ 0 & 1 \end{pmatrix} \pmod{q}\right\}.\]

On the other hand, I show in my preprint that if $q$ is odd but not squarefree, then the new part of the spectrum of the Laplacian on $\Gamma_0(q) \backslash \mathbb{H}$ is never simple; there is always a positive proportion of eigenvalues $\lambda \leq T$ for which the corresponding eigenspaces of Maaß newforms are at least two-dimensional. For $\Gamma_1(q) \backslash \mathbb{H}$, the situation is even worse: there is spectral multiplicity even if $q$ is squarefree (provided $q \neq \{1,2,3,6\}$), and the dimension of an eigenspace can be proven to grow with $m$ if $q = p^m$ for an odd prime $p$. The proof doesn't use anything about the eigenvalue $1/4$, as in GH from MO's answer, but rather looks at twists of newforms that have the same level $q$ after twisting; this gives rise to spectral multiplicity.

There is one remaining case where I know that spectral multiplicity must occur; that of $\Gamma(p) \backslash \mathbb{H}$, with $p$ an odd prime; Randol shows that there must be infinitely many eigenvalues with multiplicity at least $\frac{1}{2} \left(p + (-1)^{(p - 1)/2}\right)$. I don't think anything is known for noncongruence subgroups.

I still think that for any congruence subgroup $\Gamma$, the multiplicity of an eigenvalue of the Laplacian on $\Gamma \backslash \mathbb{H}$ ought to be uniformly bounded as $\lambda \to \infty$, with the bound depending on $\Gamma$. Unfortunately our only tool for attacking this question (finding an upper bound on the multiplicity) seems to be via getting better error terms in the Weyl law, and we are very far off making that approach be useful.

share|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.