# Compact surface with arbitrarily large eigenvalue

Consider a compact surface $M$ with genus $\gamma \geq 2$ and fix a positive real number $V$. Is it known whether it is possible to produce a metric $g$ on the surface $M$ such that $(M. g)$ has volume $V$ and the first eigenvalue of the Laplacian $-\Delta_g$ is as high as one wants? If the answer is yes, what could one make the same conclusions if the metric were additionally required to be negatively curved (but of course not constant negative curvature)? This is mainly a reference request. Somehow, from general heuristics, I am very much willing to believe that the answer to the first question is yes, but I am not at all sure about the second one.

Update: Okay, I just found this paper. See in particular Corollary 1 on page 4, which gives very general bounds on the eigenvalues depending only on the conformal class on a compact manifold of dimension $n$. More explicitly, given a compact Riemannian manifold $M$ and a fixed conformal class $[g]$, it says that the $k$-th eigenvalue is bounded above by a constant $c^{(k)}_{[g]}$ depending on the conformal class, and the aforementioned corollary gives lower bounds for this constant $c^{(k)}_{[g]}$. However, it is not at all clear that this constant $c^{(k)}_{[g]}$ remains bounded as we vary the conformal class.

Furthermore, being not familiar with literature on this kind of investigation, there are a few more questions: (1) If we consider the above Corollary on a closed surface and in the particular case $k =1$, does there exist a smooth metric $g$ in every conformal class such that $\lambda_1(g) = c^{(1)}_{[g]}$? (2) What can we say about the curvature properties of such maximizing metrics?

I am hoping that someone sufficiently familiar with the literature can answer this. Thanks, and sorry if I the question is too long.

• Presumably one should at least consider only metrics with some bound on the total volume. Otherwise, by scaling the metric by a constant, one can scale the eigenvalues by the inverse of that constant, so one can adjust $\lambda_1$ at will. In particular, one can make $\lambda_1$ as large as desired if one is allowed to scale the metric by a constant. Whether the metric is negatively curved or not has no effect on the problem. Nov 17, 2015 at 0:24
• @RobertBryant Oops, you are right, fixed that! Nov 17, 2015 at 0:36
• As a side note, considering a normalized spectrum $V^{n/2}\lambda$ is equivalent to restricting to fixed volume $V$ and may in some circumstances lead to more elegant formulations.
– Neal
Sep 14, 2016 at 1:16

Yang and Yau proved that for a surface of genus $\gamma$, $\Sigma$ with a metric $g$, the first eigenvalue satisfies $$\lambda_1(g) Area(g) \leq 8\pi (1+\gamma).$$ So, the answer to your first question is "no" you cannot make it as large as you want. [On the other hand, I think that you can make $\lambda_1(g)$ very close to $0$ by choosing $g$ to have a very long, very skinny tube with volume close to the whole volume of the manifold.]
There has been quite a bit of work on maximizing $\lambda_1(g)Area(g)$ on surfaces. It turns out that you for a maximizing sequence of metrics, after bubbling off some $S^2$'s the sequence converges to a metric that admits a branced minimal immersion into some sphere $S^n$ by first eigenfunctions! So, there is a very unexpected connection to minimal surfaces. I think this is proven here, but there are several other references for this fact. See talk as well references to further work. There is a huge literature theses days pushing this question further.
For example, the sharp upper bounds for $\lambda_1Area,\lambda_2Area,\lambda_3Area$ are known for the sphere, and sharp upper bounds for $\lambda_1Area$ are known for the projective plane, torus, klein bottle, and genus two surface! I think that this paper contains references to all of these facts.