Timeline for Growth of Laplacian eigenvalues on a compact domain?
Current License: CC BY-SA 3.0
18 events
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Jun 17, 2011 at 17:26 | vote | accept | TerronaBell | ||
Jun 17, 2011 at 17:26 | comment | added | GH from MO | @fuzzytron: For the torus $\mathbb{R}^2/\mathbb{Z}^2$ the multiplicities are very small (far from linear), but I am sure you know this. I am no expert in these matters, but I would think that on a generic surface (e.g. with no symmetry, variable curvature etc.) all multiplicities are equal to one. | |
Jun 17, 2011 at 12:57 | history | rollback | Anton Petrunin |
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Jun 17, 2011 at 8:53 | comment | added | TerronaBell | @GH: Thanks -- yeah, I'm not terribly surprised to hear that. Are there known examples of 2-manifolds where the multiplicities are anything other than linear (apart from the conjecture you mention)? Really I'm interested in surfaces that represent "everyday objects," e.g., the boundaries of (reasonably smooth) objects you might have sitting around your house. [I wish there were a concise term for such surfaces -- something like smooth, compact, connected, orientable 2-manifolds with small genus.] | |
Jun 16, 2011 at 15:54 | history | edited | GH from MO | CC BY-SA 3.0 |
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Jun 16, 2011 at 15:51 | comment | added | GH from MO | @fuzzytyron: In general it is a very difficult question how eigenvalues are distributed, what are their multiplicities, how many are there in short intervals etc. For example, for the modular surface (upper half-plane modulo $\mathrm{SL}(2,\mathbb{Z})$) it is conjectured that each eigenvalue has multiplicity one, but we only have very weak results in that direction. | |
Jun 16, 2011 at 14:53 | comment | added | Helge | fuzzytron: Both scenarios you propose happen. Symmetric examples are highly degenerate. Break the symmetry and you lose degeneracy ... Of course a small perturbation will still have "almost high multiplicities". | |
Jun 16, 2011 at 11:29 | comment | added | Anton Petrunin | @GH, hope it is correct now. | |
Jun 16, 2011 at 11:27 | history | edited | Anton Petrunin | CC BY-SA 3.0 |
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Jun 16, 2011 at 11:19 | history | edited | Anton Petrunin | CC BY-SA 3.0 |
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Jun 16, 2011 at 11:12 | history | edited | Anton Petrunin | CC BY-SA 3.0 |
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Jun 16, 2011 at 8:29 | comment | added | TerronaBell | Very interesting -- thanks! Unfortunately I don't think relationships like Weyl's formula answer my original question. For instance, suppose (for the sake of argument) that $N(R) \approx R$. Then the growth of eigenvalues is still ambiguous. For instance, $\lambda_i$ could be quadratic in $i$ and have multiplicity linear in $i$, but alternatively $\lambda_i$ could be linear in $i$ and have constant multiplicity. Either scenario satisfies $N(R) \approx R$. | |
Jun 15, 2011 at 23:15 | comment | added | GH from MO | The constant above is incorrect. For the correct formula and some modern developments see math.uni-bonn.de/people/mueller/papers/weyllaw.pdf | |
Jun 15, 2011 at 22:32 | comment | added | TerronaBell | I must be interpreting one of your constants incorrectly -- for the unit sphere in $\mathbb{R}^3$ the distinct eigenvalues are $\lambda_i = i(i+1)$ appearing with multiplicity $2i+1$. So then the number of eigenvalues with value no greater than $\lambda_i$ is $\sum_{j=0}^i 2j + 1 = i^2 + 2i + 1$. In other words, we have $R(i) = i^2 + i$ and $N(R(i)) = i^2 + 2i + 1$, hence $N(R) \approx R$. But for $d=2$ and $V=4\pi$, Weyl's formula says $N(R) \approx R/4\pi$. Where does the factor $1/4\pi$ come from? | |
Jun 15, 2011 at 21:58 | comment | added | TerronaBell | Great -- thanks for the clarification. (Also, looks like it should be $o(R^{(d-1)/2})$.) | |
Jun 15, 2011 at 21:33 | history | edited | Anton Petrunin | CC BY-SA 3.0 |
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Jun 15, 2011 at 21:11 | comment | added | TerronaBell | Not sure I understand the connection (also: the link is broken; had to Google it). I guess I should also mention that I'm not interested with surfaces with boundary (or Dirichlet boundary conditions). | |
Jun 15, 2011 at 21:04 | history | answered | Anton Petrunin | CC BY-SA 3.0 |