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A friend in physics asked this question, and I didn't know the answer.

Are there lower bounds on the first eigenvalue of the Laplacian of a Riemann surface equipped with a metric of constant negative curvature?

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2 Answers 2

up vote 3 down vote accepted

I guess you mean constant curvature $=-1$; otherwise you get an example by rescaling.

On a sphere with two handles there are metrics with curvature $\equiv-1$ which look roughly as a long neck attached to two tori. Such metric has as small eigenvalue as you want.

On the other hand upper diameter + lower curvature bound (in particular curvature $\equiv-1$) imply a lower bound on the eigenvalue.

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What sort of bound? Can you give a reference? –  userN Mar 16 '11 at 0:57
    
I am sorry, I do not know a ref, but I am sure it is written somewhere. Hopefully someone will help you... I know it since it follows from Gromov--Levy isoperimerical inequality. You could look in Gromov's appendix in a Milman's book. Formally the proof is written only for positive curvature bound but rough estimates still work if bouds are negative. –  Anton Petrunin Mar 16 '11 at 1:10
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Naturally, to a physicists anyway, the curvature and also the eigenvalues are unit-full numbers. So I'm sure the question is looking for bounds on the unitless ratio, which is the same as asking for bounds when the mean constant curvature is fixed at $-1$. –  Theo Johnson-Freyd Mar 16 '11 at 1:11

All hyperbolic Riemann surfaces are quotients of the upper half-plane ${\mathbb H}$ by a Fuchsian group $\Gamma$. When $\Gamma$ is a congruence subgroup, Selberg's eigenvalue conjecture says that the smallest eigenvalue is at least $1/4$, and, in fact, he proved it was at least $3/16$ (for $SL_2({\mathbb Z})$, a better lower bound is known: $3\pi^2/2$). This lower bound has been improved somewhat since then (I think the current record is something like 0.228 due to Kim and Shahidi, via functoriality for the symmetric cube).

For general hyperbolic Riemann surfaces, the lower bound is zero. See this paper. Apparently, this, too, but I don't have access to it, so I can't confirm.

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This is a really nice answer. Thanks for the reference! –  userN Mar 16 '11 at 18:07
    
You are welcome. –  B R Mar 16 '11 at 20:21

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