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It is well known that on any compact Riemannian symmetric space $X$, the eigenvalues of the Laplacian have very high multiplicity (comparable with the Weyl bound), and the resulting actions $\operatorname{Isom}(X)\to \operatorname{SO}(W_\lambda)$ for eigenspaces $W_\lambda$ give many representations of the Lie group $\operatorname{Isom}(X)$.

Suppose one has an unknown compact Riemannian manifold $X$ ($n=\dim X$), but where the eigenspaces of the Laplacian have large dimension (I don't have a precise definition of "large" here; the weakest definition would probably be something like $\dim W_\lambda>1$ for infinitely many $\lambda$. I'd be happy even with a much stronger assumption, say $\dim W_\lambda$ is at least $\epsilon$ times the Weyl bound $\operatorname{const}\cdot\lambda^{(n-1)/2}$ infinitely often). Can one conclude that $X$ is a symmetric space, or close to one in some sense?

EDIT: I would be interested in any result which takes as a hypothesis some assumption of large multiplicity in the Laplace spectrum, and whose conclusion is some sort of symmetry of the underlying manifold.

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Wouldn't even a nontrivial $S^1$-action on $X$ imply an infinite number of multiple eigenvalues? –  Tom Goodwillie May 23 '11 at 4:09
    
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. –  Helge May 23 '11 at 5:47
    
@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. –  John Pardon May 23 '11 at 14:53
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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. –  Robert Bryant May 25 '11 at 14:11
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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). –  Terry Tao Mar 17 '12 at 21:30
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4 Answers

One situation where people have thought hard about this issue is the multiplicity of the spectrum of the Laplacian on the modular surface. This is a notoriously difficult problem.

Conjecturally, the (discrete) spectrum is simple, but as far as I know, the best known bounds on the multiplicity of a Maas cusp form of eigenvalue $\lambda$ are somewhere in the neighborhood of $O(\lambda^{1/2}/\log \lambda).$

(Weyl's law in dimension $2$ says that the number of eigenvalues of size at most $N$ is $O(N)$).

See e.g. section 4 of the following survey of Peter Sarnak: http://www.math.princeton.edu/sarnak/baltimore.pdf

So I would say that some of the stronger versions of your conjecture are almost certainly out of reach.

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Well, the funny thing is that a result of this sort DOES hold for the length spectrum: Takeuchi (I believe) showed that if a surface has linear number of traces smaller than $N,$ if and only if it is arithmetic, which is morally a closely related result. –  Igor Rivin Mar 18 '12 at 2:54
    
@Igor: The most natural way to estimate the multiplicity of the eienvalues is to use the trace formula, which relates it to a sum over the closed geodesics (i.e. the length spectrum). But the problem is that you get a sum over exponentially many geodesics, and you end up trying to add up e^\lambda terms of size O(1), while the actual answer is also O(1). This is why it is so hard. –  Alex Eskin Mar 18 '12 at 3:32
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I might be totally confused, but in this paper of Hubert Goldschmidt (Infinitesimal isospectral deformations of symmetric spaces), he seems to construct the very deformations of the title. It is not obvious that these can be integrated, but if they can be, it gives you a family of manifolds isospectral to the symmetric spaces and not symmetric, which would provide a negative answer to the OP (note that if you read Goldschmidt's paper, you will see that various spaces are not NOT to be so deformable).

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My guess is No. You do not need a one parameter Lie group of symmetry to have infinitely many double eigenvalues. Just one involution suffices. And one involution is not enough to make a symmetric space. For instance, every complex curve whose equation is real has this property; then the involution is the complex conjugation. For most such curve, there is no other non-trivial involution.

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My answer is rather a question. Suppose that $(M,g)$ is a compact Riemann manifold and for any positive integer $N$ there exists an eigenvalue of the Laplacian that has multiplicity $>N$. Can one conclude that the group of isometries of $(M,g)$ has positive dimension?

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The Poincaré homology sphere $P$ is a counterexample, right? It's the quotient of $S^3$ by a finite group $G=2.A_5$ acting freely by isometries. Any automorphism of $P$ lifts to its universal cover $S^3$,and is thus a coset of $2.A_5$ in the normalizer of $2.A_5$ in ${\rm Aut}(S^3) = O_4({\bf R})$; but this normalizer is $2.A_5$ itself, so ${\rm Aut(P)}$ is trivial. But for even $d$ the space $H_d$ of harmonic polynomials of degree $d$ has a $2.A_5$ invariant subspace that is an eigenspace for the Laplacian on $P$ of dimension asymptotic to $\dim(H_d) / 60 \rightarrow \infty$ with $d$. –  Noam D. Elkies Mar 17 '12 at 22:50
    
The Poincare sphere is also a Seifert manifold. I have to admit that I am a bit confused about how the Seifert structure fits in the above picture. –  Liviu Nicolaescu Mar 18 '12 at 0:32
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