Timeline for Spectrum of the Laplace-Beltrami operator on a domain of finite volume in the hyperbolic space $H^n$
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 26, 2016 at 12:55 | comment | added | Asaf | P.S. due to the Phillips-Sarnak result, it is widely believed that for "typical" B (where typical should be understood properly) there should be only finitely many points in the discrete spectrum. | |
| Jan 26, 2016 at 12:53 | comment | added | Asaf | 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 | |
| Jan 26, 2016 at 11:31 | history | edited | Vladimir | CC BY-SA 3.0 |
added 95 characters in body
|
| Jan 25, 2016 at 19:02 | comment | added | Alain Valette | @Vladimir: The link provided by Marcel contains a number of useful references. See also: en.wikipedia.org/wiki/Laplace–Beltrami_operator | |
| Jan 25, 2016 at 18:29 | comment | added | Vladimir | 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. | |
| Jan 25, 2016 at 17:50 | comment | added | Alain Valette | 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. | |
| Jan 25, 2016 at 17:40 | comment | added | Marcel | did you check this: mathoverflow.net/questions/68376/… | |
| Jan 25, 2016 at 14:13 | history | edited | Vladimir |
edited tags
|
|
| Jan 25, 2016 at 14:07 | history | asked | Vladimir | CC BY-SA 3.0 |