6
$\begingroup$

Consider a compact, connected $n$ dimensional Riemmanian manifold $\mathcal{N}$ and its $m$ dimensional closed submanifold $\mathcal{M}$ (with the metric coming from from the one defined on $\mathcal{N}$).

Question: Is there any way to give a lower bond for the first nonzero eigenvalue of Laplace- Beltrami operator ($\Delta f = - div(grad f)$) defined on $\mathcal{M}$? In particular, is such lower bound related to the first nonzero eigenvalue of the Laplacian on $\mathcal{N}$?

I am especially interested in in the case when $\mathcal{M}$ is of "small" codimension and is given by the intersection of level sets of some smooth functions defined on $\mathcal{N}$. The original motivation for this question comes from the problem of "incheritance" of measure concentration by $\mathcal{M}$, when it is known that it occours for $\mathcal{N}$.

$\endgroup$

1 Answer 1

10
$\begingroup$

In the general case (especially for high codimension) there will not be such a relation: Every compact Riemannian manifold can be isometrically embedded into the Euclidean space $R^N$. As the image of the submanifold is compact, you can change the metric "near infinity" such that it defines a metric on the sphere $S^N.$

But in special cases, there might be a relation, see for example Yau's conjecture: The first non-zero eigenvalue of a compact embedded minimal hypersurface in (the round) $S^n$ is n. You should take a look into the paper of S. Montiel & A. Ros, Minimal immersions of surfaces by the first eigenfunctions and conformal area, Invent Math.

$\endgroup$
2
  • 1
    $\begingroup$ You forgot a "minimal" in the statement of Yau's conjecture, right? $\endgroup$ Commented Jun 13, 2012 at 8:51
  • $\begingroup$ Thank you for your answer. As I mentioned I am interested mostly in the case of low codimension ($\frac{n-m}{n}\ll1$ and in the case when $\mathcal{M}$ is a level set of a function on $\mathcal{N}$. Maybe more is known in this case? $\endgroup$ Commented Jun 13, 2012 at 10:51

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .