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Let $\Sigma$ be a compact smooth surface with boundary. Define

$$\Lambda(\Sigma) := \sup \{ \lambda_1(\Sigma,g) \operatorname{Area}(\Sigma,g) : g \text{ is a smooth Riemannian metric on $\Sigma$} \}$$

where $\lambda_1(\Sigma,g)$ denotes the first eigenvalue of the Laplacian associated to the metric $g$ with Dirichlet ($=0$) boundary condition.

In another post, we concluded that $\Lambda(\Sigma)$ is infinite when $\Sigma$ is a cylinder. Is it true that this quantity is infinite for every $\Sigma$? If this is not the case, what are the topological types of surfaces for which this supremum is finite?

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    $\begingroup$ some remarks: $\lambda_1$ is continuous at the $C^0$-topology in the space of metrics for $\Sigma$ fixed. Now since the classification of surfaces with boundary only depends on removing open discs from closed surfaces, I suppose that $\lambda_1$ can be parametrized by unions of circles, precisely, to understand $\lambda_1$ it suffices to understand the number $k$ of connected components on the boundary and the eigenvalue of each circle in the boundary. So I think that such a quantity is infinite for any surface with $k\geq 2$-boundary connected components. $\endgroup$ Commented Jan 5, 2023 at 0:05
  • $\begingroup$ @L.F.Cavenaghi I don’t see how to relate the eigenvalues of circles and surfaces. I’m assuming Dirichlet boundary condition… $\endgroup$ Commented Jan 5, 2023 at 0:23
  • $\begingroup$ Intuition behind the cylinder counterexample in the linked question is: if "most" of the volume is "close to" the boundary, then $\lambda_1$ goes like $(\mbox{distance between boundary components})^{-2}$. So I suspect the idea can be adapted for any topological surface with boundary to produce a class of metrics with constant area and $\lambda_1$ growing without bound. $\endgroup$
    – Neal
    Commented Jan 5, 2023 at 17:40

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