TITLE: Dimension of eigenspaces of Laplacian on a compact Riemannian manifold
QUESTION [9 upvotes]: 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?

REPLY [13 votes]: 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).