What can be said about the spectrum of the Laplace-Beltrami operator on a non-compact, complete Riemannian manifold of finite volume? For example, is the point spectrum non-empty?
What would be a good reference for this?
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As far as I know, this is a very delicate question. That is, already in two dimensions, there can be only finite discrete spectrum. (See Phillips-Sarnak and Wolpert.)
Even on the modular curve $SL(2,\mathbb Z)\backslash \mathfrak H$, it was highly non-trivial to prove existence of infinitely-many $L^2$ eigenvalues, apparently requiring Selberg's invention and application of his "trace formula" in the 1950's.
Similarly, it has only been after 2000 or so that various people (S. Miller, Lapid-Finis, and others) have proven "Weyl's Law" for higher-rank quotients of semi-simple Lie groups, that is, that the asymptotics for discrete (even cuspidal) spectrum are what one would expect for a compact Riemannian manifold.
That is, based on this perspective, I'd be amazed if much could be said in general... maybe some negative results for small deformations, analogous to Phillips-Sarnak and Wolpert?
answered Dec 5, 2013 at 23:42
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$\begingroup$ May I ask what is the subgroup $\mathfrak H$? I have never seen it before. $\endgroup$ Dec 6, 2013 at 3:13
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$\begingroup$ $SL(2,\mathbb{Z})\backslash \mathfrak{H}$ is the upper half-plane in $\mathbb{C}$ modulo the action of the group $SL(2,\mathbb{Z})$. $\endgroup$ Dec 6, 2013 at 3:53
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$\begingroup$ Thanks Paul. I was indeed interested in whether one can read the volume in the spectrum of the Laplacian also in this non-compact case. $\endgroup$ Dec 7, 2013 at 8:26