Timeline for Eigenvalues of the Laplace-Beltrami operator on a compact Riemannnian manifold
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| May 18, 2019 at 21:21 | answer | added | Alexandre Eremenko | timeline score: 13 | |
| May 18, 2019 at 17:16 | history | became hot network question | |||
| May 18, 2019 at 16:48 | answer | added | user25309 | timeline score: 10 | |
| May 18, 2019 at 15:12 | comment | added | user80744 | @Nate Eldredge: By elliptic regularity I mean: If $A$ is an elliptic operator of order $m$, then $Au\in H^s(M)$ implies $u\in H^{s+m}(M)$. Sobolev embedding is involved only when subsequently passing from Sobolev to $C^k$ regularity. I can see no danger of circularity here. | |
| May 18, 2019 at 15:11 | comment | added | Nate Eldredge | You get some "weights" from the derivatives of the coordinate functions, but they are all bounded. | |
| May 18, 2019 at 15:09 | comment | added | Nate Eldredge | @MaxSchattman: I think you just use a partition of unity. When $M$ is compact, can find a finite number of smooth functions $\psi_k$, each compactly supported in a coordinate chart, whose sum is 1. Now if I have a sequence of functions $f_n$ in my Sobolev space, the functions $f_n \psi_1$ are in the corresponding Sobolev space of the coordinate chart. Apply the Euclidean version of Rellich to pass to an $L^2$ convergent subsequence, and repeat iteratively for $\psi_2, \psi_3, \dots, \psi_k$. Adding them up, you have a subsequence of the original $f_n$ converging in $L^2$. | |
| May 18, 2019 at 14:57 | comment | added | Nate Eldredge | @SönkeHansen: I was thinking about that, but I seem to recall that the proof of elliptic regularity goes through the Sobolev embedding argument itself, so I wasn't sure off the top of my head whether that is circular. | |
| May 18, 2019 at 14:30 | comment | added | Max Schattman | @Nate: So I guess this changes the question into a request for an abstract/formal approach to the Rellich embedding theorem :) I've never actually seen a proof for a general Riemannian manifold . . . might an easy proof exist for spaces with enough symmetries. For example, what does Rellich look like for a compact homogeneous space? | |
| May 18, 2019 at 14:29 | comment | added | user80744 | To Nate Eldredge's route, which is a standard one, I would add: $\Delta_g$ is self-adjoint and, by elliptic regularity, its domain is the Sobolev space $H^2$. So the resolvent factors through $H^2$, hence is compact by Rellich's Theorem. | |
| May 18, 2019 at 14:04 | comment | added | Nate Eldredge | I guess the route I think of is like this. You show that the resolvent (or the semigroup) of $\Delta_g$ maps $L^2$ into an appropriate Sobolev space. The latter is compactly embedded in $L^2$ by Rellich's theorem, so the resolvent is a compact operator. So the eigenvalues of the resolvent (with multiplicity) converge to zero, and hence the eigenvalues of the Laplacian converge to infinity. | |
| May 18, 2019 at 13:23 | history | edited | Max Schattman | CC BY-SA 4.0 |
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| May 18, 2019 at 11:37 | history | asked | Max Schattman | CC BY-SA 4.0 |