Timeline for High multiplicity eigenvalue implies symmetry?
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13 events
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| Mar 17, 2012 at 23:04 | answer | added | Alex Eskin | timeline score: 10 | |
| Mar 17, 2012 at 21:30 | comment | added | Terry Tao | This is not exactly the same thing, but there is a paper of Sogge and Zelditch at ams.org/mathscinet-getitem?mr=1924569 which links maximal eigenfunction growth (in the sense that the sup norm of L^2-normalised eigenfunctions is as large as possible given the energy level) with being a Zoll manifold (or more generally having a positive measure set of closed geodesics of a given length). | |
| Mar 16, 2012 at 23:14 | answer | added | Igor Rivin | timeline score: 6 | |
| Mar 16, 2012 at 22:49 | answer | added | Liviu Nicolaescu | timeline score: 4 | |
| May 25, 2011 at 14:11 | comment | added | Robert Bryant | The multiplicities, even for a compact symmetric space, need not be all that high. For example, consider the symmetric space $X = \mathbb(R^n)/\Lambda$, where $\Lambda\subset\mathbb{R}^n$ is a generic lattice. For nearly all lattices, the multiplicities of the eigenvalues of the Laplacian will be at most $2$ since the only other vector in $\Lambda$ that has the same length as any given $v\in\Lambda$ will be $-v$. I think you probably meant to consider only Riemannian symmetric spaces of compact type. | |
| May 23, 2011 at 15:59 | comment | added | Helge | You need a bound of the form $\lambda_n = O(n^{1/d})$. See arxiv.org/pdf/0810.2088 . But it would be quite some work checking these conditions (one has too untangle the definitions involved and be tricky). | |
| May 23, 2011 at 15:05 | history | edited | John Pardon | CC BY-SA 3.0 |
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| May 23, 2011 at 15:03 | comment | added | John Pardon | @Helge, For any Riemannian manifold $X$, the eigenvalues of $-\Delta$ satisfy the property that $\zeta_M(s)=\sum_\lambda\lambda^{-s}$ has analytic continuation to the entire plane (en.wikipedia.org/wiki/…). So your statement that "any sequence of eigenvalues corresponds to a space" probably needs some additional qualifications (if by space you mean a Riemannian manifold). | |
| May 23, 2011 at 14:53 | comment | added | John Pardon | @Tom Goodwillie, Yes, I guess the multiplicity has to be on the large end for anything like this to be true. I would be interested in any result which takes some multiplicity assumptions as hypotheses and whose conclusion is some sort of symmetry of the manifold. | |
| May 23, 2011 at 8:01 | answer | added | Denis Serre | timeline score: 6 | |
| May 23, 2011 at 5:47 | comment | added | Helge | My guess is that: No. There are inverse spectral results, so that any sequence of eigenvalues corresponds to a space. Just figure out a sequence that does not arise as the eigenvalues of a symmetric space and you are done. | |
| May 23, 2011 at 4:09 | comment | added | Tom Goodwillie | Wouldn't even a nontrivial $S^1$-action on $X$ imply an infinite number of multiple eigenvalues? | |
| May 22, 2011 at 20:57 | history | asked | John Pardon | CC BY-SA 3.0 |