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What is known about the ($L^2$) spectrum of the minus Laplace-Beltrami operator ($- \Delta$) with zero boundary conditions on $B =H^n/\Gamma$, where $H^n$ is $n$-dimensional hyperbolic space ($n>1$), $\Gamma$ is a discrete subgroup of the isometry group of $H^n$? Is it possible for the spectrum to be discrete for some $\Gamma$, when $B$ has finite volume but is not (sub)compact? If so, it could be generalized for a large class of (generic) $B$ (with smooth boundary) of finite volume. (When the closure of $B$ is compact the spectrum seems to be discrete.) Are there universal bounds (from below) on the spectrum of ($- \Delta$) on $B$ of finite volume (with smooth boundary)? The case of special interest is $n=2$ and $\Gamma = PSL(2,{\mathbb Z})$ - well-known modular group.

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  • $\begingroup$ did you check this: mathoverflow.net/questions/68376/… $\endgroup$
    – Marcel
    Commented Jan 25, 2016 at 17:40
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    $\begingroup$ You should check Chapters XII, XIII and XIV of Serge Lang's book ``$SL_2(\mathbb{R})$'' (where SL doesn't stand for Serge Lang): you will see that, for $\Gamma=SL_2(\mathbb{Z})$, the spectrum of the Laplace operator has both a continuous and a discrete part. Everything is very explicit. On the other hand, it is a classical fact that, on a compact Riemannian manifold, the Laplace operator has compact resolvent, hence has discrete spectrum. $\endgroup$
    – Alain Valette
    Commented Jan 25, 2016 at 17:50
  • $\begingroup$ Many thanks! I apologize that I asked the question without any preliminary studying of the subject (I was reading in physical papers about possible presence of continuous spectrum in some cases but have doubts about generality of that). It looks that the same situation may take place for general case, when $B$ is non-compact but of finite volume and arbitrary $n$. It will be good to have also a standard citation about the classical fact you have mentioned for the compact case. $\endgroup$
    – Vladimir
    Commented Jan 25, 2016 at 18:29
  • $\begingroup$ @Vladimir: The link provided by Marcel contains a number of useful references. See also: en.wikipedia.org/wiki/Laplace–Beltrami_operator $\endgroup$
    – Alain Valette
    Commented Jan 25, 2016 at 19:02
  • $\begingroup$ The spectrum will never be discrete, due to the presence of Eisenstein series in non-uniform lattices (morally, by the Kazhdan-Margulis theorem - non-uniform lattices implies existence of unipotents...). Nevertheless, by the Gelfand-Graev theorem, the cuspidal spectrum is always discrete! (hence the Laplacian is essentially compact). Randol has shown how to construct quotients of $PSL_{2}$ with arbitrarily small spectral gap, and therefore any meaningful bound towards Ramanujan will involve number-theoretical considerations/constructions $\endgroup$
    – Asaf
    Commented Jan 26, 2016 at 12:53

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