MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $M$ be a compact smooth manifold, let $g$ a riemannian metric and let $\Delta_{g}$ the Laplacian operator on functions induced by $g$. Is there a (topological?) bound on the dimension of $n$-th eigenspace of $\Delta_{g}$?

Does the answer change if $M$ is a compact complex manifold and $g$ is a kahler metric?

share|cite|improve this question

In dimension $n\geq 3$, there cannot be any sort of bound on the multiplicities which does not depend on some geometric input. This is because it is a theorem of Colin de Verdière (mathscinet and article (in French)) that:

If $M^n$ is a closed (smooth) manifold of dimension $n\geq3$, then any sequence $0=\lambda_1<\lambda_2\leq \lambda_3\leq\dots\leq \lambda_m$ is the first $m$ values of the spectrum of $\Delta_g$ for some metric $g$ (taken with multiplicity).

In fact, I'll remark that a theorem of Lohkamp has generalized this to the following result (mathscinet and article)

Given $M^n$ closed with $n\geq 3$ and $0=\lambda_1<\lambda_2\leq \lambda_3\leq\dots\leq \lambda_m$, as well as constants $V > 0$ and $K < 0$ there is a metric $g$ whose first $m$ elements of the spectrum agrees with the given sequence (with multiplicity) and also has $Vol(M,g) = V$ and $Ric_g \leq Kg$.

On the other hand, in dimension $n=2$, there is such a bound, depending only on the topological type of the surface, due to Nadirashvili (mathscinet and article).

share|cite|improve this answer
Thank you very much, for the kahler case is anything known? – student Jun 21 '13 at 17:25
You're welcome! Nothing is known to me, but this does not mean much... There do not seem to be any relevant papers on mathscinet that cite Colin de Verdière's article. Of course, the Nadirashvili result says that what you want is true for complex curves, so maybe thats a good place to start looking. – Otis Chodosh Jun 21 '13 at 17:51

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.