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Disclaimer: This is a very heuristic question and I will be satisfied with heuristic insights, if rigorous and precise answers are not possible.

All the examples of closed surfaces (or higher dimensional manifolds) whose spectrum I have seen evaluated explicitly have highly repeated eigenvalues. They also happen to be quite symmetric (non-trivial isometry group), which was instrumental in the calculation of the spectrum in the first place. I have heard many people quote this heuristic: the more symmetric a manifold is, the higher the chances of spectral multiplicity. I was wondering

  1. Is there a way to make this heuristic precise?

  2. What about the converse? If a manifold has high spectral multiplicity, does it need to have a nontrivial isometry group?

  3. If the answer to 2. is negative, and there are large classes of examples of surfaces/manifolds that are not symmetric at all, but have high spectral multiplicity, is there some other property shared by these examples which is causing the multiplicity? In other words, are there other criteria apart from symmetry that would make one suspect spectral multiplicity?

Edit: Okay, I just found High multiplicity eigenvalue implies symmetry? So, I know that the answer to 2. is negative. Also, Liviu Nicolaescu answers 1. below. I am still confused about 3. In other words, is there any criterion other than symmetry, that, if you know that a metric satisfies, you would suspect high spectral multiplicity?

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    $\begingroup$ The isometry group leaves the laplacian invariant and this implies that the different eigenspaces are representations of the isometry group. Of course, they could end up all being the trivial representation, but "heuristically" it is more likely than not that this is not case. $\endgroup$
    – José Figueroa-O'Farrill
    Commented Oct 30, 2015 at 14:08
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    $\begingroup$ There's well known conjecture of Sarnak (maybe Selberg?) saying the multiplicity of eigenvalues for say X(1) is $1$, and this is far from being resolved. Anyhow, X(1) is probably as symmetric as you may want. $\endgroup$
    – Asaf
    Commented Oct 30, 2015 at 14:29
  • $\begingroup$ You might find this useful. Apparently the higher the dimension the (possibility) bigger eigenspaces $\endgroup$
    – BigM
    Commented Nov 24, 2015 at 4:24
  • $\begingroup$ In that paper they consider manifolds without boundary and of genus zero though :-( $\endgroup$
    – BigM
    Commented Nov 24, 2015 at 4:27

1 Answer 1

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The group of isometries of a compact Riemann manifold is a compact Lie group. This group acts on the eigenspaces of the Laplacian. If the group of isometries is non-abelian, then it is natural to expect that that some eigenspaces will have large dimensions since most of the nontrivial irreducible representations of a compact non-abelian Lie group have dimensions $\geq 2$.

Looking for Laplacians with multiple eigenvalues is a bit like looking for a needle in a haystack since a result of K. Uhlenbeck states that for a generic Riemann metric on a given compact manifold the spectrum of the Laplacian will not have multiple eigenvalues.

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    $\begingroup$ do you mean will not have multiple eigenvalues? $\endgroup$
    – Fan Zheng
    Commented Oct 30, 2015 at 17:17
  • $\begingroup$ "for a generic Riemann metric on a given compact manifold the spectrum of the Laplacian will not have multiple eigenvalues." I don't understand how this is possible. The metrics on $S^1$ are classified, up to isomorphism, by the total length, and so each is isomorphic to $\mathbb{R}/x\mathbb{Z}$ for some real $x$, right? And for these manifolds the eigenspaces are all 2-dimensional for nonzero eigenvalues, generated by a sine and cosine function. What am I missing? $\endgroup$
    – Tom Price
    Commented Dec 16, 2016 at 22:33
  • $\begingroup$ The dimension has to be $\geq 2$. Here's Uhlenbeck's original paper jstor.org/stable/2374041 $\endgroup$
    – Liviu Nicolaescu
    Commented Dec 16, 2016 at 23:12
  • $\begingroup$ Ok, thank you very much, this is important for a paper I'm writing. Are you referring to theorem 8 in that paper? It strongly resembles your original claim but I can't be sure what it's saying since I can't find the definition of $\mathfrak{M}_k$ (the set of $C^k$ metrics perhaps?). Oddly I can't seem to find a condition on the dimension of the underlying manifold. $\endgroup$
    – Tom Price
    Commented Dec 17, 2016 at 0:20
  • $\begingroup$ She does state the dimension condition explicitly, but she uses $\dim>1$ in the proof and, towards the end of the paper, she mentions that the result cannot be true in dim 1. $\endgroup$
    – Liviu Nicolaescu
    Commented Dec 17, 2016 at 11:42

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