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Much has been said about bounds on Laplacian eigenvalues, and the literature can be tough to sort through! I am specifically interested in the case where the domain is a closed compact surface, and am seeking bounds on the difference between the two smallest (in magnitude) distinct nonzero eigenvalues. Any good pointers are greatly appreciated!

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The smallest non-zero eigenvalue can have multiplicity $>1$, so in this case is the gap zero? Or are you counting without multiplicity? – Ian Agol May 2 '13 at 5:27
Probably one should skip the "nonzero" requirement as in this case, the gap is always positive. – Matthias Ludewig May 2 '13 at 9:15
Are you interested in upper bounds, lower bounds or both? – Rbega May 2 '13 at 10:21
@Algol: without multiplicity. Otherwise you are very right that the answer is not very interesting. :-) – TerronaBell May 2 '13 at 13:29
@Kofi: I say nonzero because I don't care about the zero eigenvalue corresponding to the constant function. – TerronaBell May 2 '13 at 13:32

The question is a bit vague, but I'm assuming you're asking for an upper bound on the gap between the smallest non-zero eigenvalue and the second smallest non-zero eigenvalue (ignoring multiplicity), let's call this number $E$, taking the supremum over all closed hyperbolic surfaces. If $\lambda_j$ is the $j$th eigenvalue, counted with multiplicity, then Otal has shown that $\lambda_{2g-2}>\frac14$. Moreover, Buser has shown that this estimate is sharp: one can find for any $\epsilon >0$ a surface with $\lambda_{2g-3}<\epsilon$. Now, I believe that in these Buser examples, one could actually arrange that $\lambda_1=\lambda_2=\cdots=\lambda_{2g-3}$, by adjusting the moduli (recall moduli space has dimension $3g-3$). If so, this would show that $E\geq \frac14$.

On the other hand, a result of Besson implies the maximal multiplicity of $\lambda_1$ is $4g+3$, so we know that $E\leq \lambda_{4g+4}$. There is some universal bound on $\lambda_{4g+4}$, since by the Margulis Lemma, there exists $R>0$ so that on a Riemann surface of genus $g$, one can find $4g+5$ disjointly embedded disks of radius $R$ (there is such an $R$ for any linear function of $g$). By Corollary 4.65 of Gallot-Hulin-Lafontaine, one has $\lambda_{4g+4} \leq \lambda_1^D(D_{R})$, where $\lambda_1^D(D_R)$ is the first Dirichlet eigenvalue on a hyperbolic disk of radius $R$. Thus, $E\leq \lambda_1^D(D_R)$. One ought to be able to get explicit bounds on $R$ and $\lambda_1^D(D_R)$ using computations of Margulis constants to get an estimate of $E$.

It is conjectured by Colbois and Colin de Verdiere that the maximal multiplicity of $\lambda_1$ equals the chromatic number of a genus $g$ surface, which grows like $\sqrt{g}$. Given Huber's result that $\sup \lambda_1 \to \frac14$ as $g\to \infty$, this would imply that $E_g$ is asymptotic to $\frac14$, where $E_g$ is the maximal difference between these eigenvalues for genus $g$ hyperbolic surfaces.

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I think that in general $vol(X)$ is the best possible upper bound for the multiplicity of the first non-trivial eigenvalue. This can be proved rigorously for compact Riemann surfaces with the Selberg trace formula, but I would guess this holds more general via an analysis related to Weyl laws.

For certain arithmetic compact Riemannian surfaces (associated to division algebras), the first non-trivial eigenvalue is assumed to be larger or equal $1/4$. Some better upper bounds in this particular case are due to Michel and Venkatesh (the Jacquet-Langlands correspondence has to be applied).

The Langlands correspondence also suggests that the multiplicity can be arbitrary large.

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Thanks Marc. I wasn't clear about this in my original question, but my real interest is in the difference between the first two distinct nonzero eigenvalues. So I don't need to determine anything about multiplicity. Sorry for the confusion, and thanks for the help! – TerronaBell May 2 '13 at 13:38
Sorry, but I suspect that you can't distinguish between close distinct eigenvalues and eigenvalues with higher mutiplicity than one. – Marc Palm May 2 '13 at 16:39

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