Does the sphere have maximal first Laplace-Beltrami eigenvalue over all smooth surfaces enclosing the same volume?

Question

I think this from the eigenfunction of Laplacian on a rectangle $$ u(x,y)=\sin\left(\frac{n\pi x}{L}\right)\sin\left(\frac{m\pi y}{H}\right). $$ Seemly, there is a way to yield an eigenfunction on a sphere by gluing some eigenfunction on a rectangle. And the eigenvalue on a rectangle respect to eigenfunction is $$ \lambda_{nm}= \left(\frac{n\pi}{L}\right)^2 + \left(\frac{m\pi}{H}\right)^2. $$ So, when the surface enclosed same volume, $\lambda_{nn}$ is maximal just when $L=H$. (In fact, I don't know how to describe what is $L$ and $H$ on surface. For sphere, it's like meridian and weft.) So, I think the sphere has maximal first eigenvalue of Laplacian for all smooth surfaces enclosing the same volume. I want to try to prove it by volume preserving mean curvature flow. But I am not sure it is right, so I asked here. Thanks for any help.

Answer 1

The volume is not the right thing to be comparing against: the reason is that the Laplace-Beltrami operator is intrinsic and so cannot "see" the volume that is being enclosed by a particular embedding of the surface. You can create very highly-corrugated surfaces, with large surface area but small volume, that have large first Laplace-Beltrami eigenvalue.

You can ask whether the sphere maximizes the first Laplace-Beltrami eigenvalue for a given surface area, however. In the case of genus zero surfaces the answer is yes, a result due to Hersch. Larger eigenvalues are possible if you allow larger genus. I'm not aware of any tight bounds, though I believe Berger proved some bounds for the torus.