MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $\mathcal{M}$ be a compact Riemannian manifold and let $\Delta$ be the (scalar) Laplace-Beltrami operator on $\\mathcal{M}$. Then $\Delta$ has a discrete spectrum and if we order its distinct eigenvalues $\lambda_i$ by magnitude then some very simple examples suggest that the magnitude of $\lambda_i$ might be roughly quadratic in $i$. For instance, on the circle $S^1$ eigenfunctions have the form $\cos(nx)$ or $\sin(nx)$ for $n \in \mathbb{N}_0$; hitting these functions with $-\frac{\partial}{\partial x^2}$ yields $n^2\cos(nx)$ and $n^2\sin(nx)$, respectively. Similar analysis can be done for the geometrically flat torus $T^2$. On the 2-sphere, we have $\lambda_i=i(i+1)$. This rough idea of "differentiating twice leads to a square" makes me suspect that a similar relationship might hold for other domains -- what can be said in general? I'm particularly interested in smooth surfaces embedded in $\mathbb{R}^3$.

Update: Weyl's formula provides some valuable information about the Laplace spectrum, but does not determine the asymptotic growth of $\lambda_i$. For instance, suppose we have a manifold such that $N(R) \approx R$, i.e., the number of eigenvalues with value no greater than $R$ is roughly equal to $R$ itself. Letting $n_i$ be the multiplicity of $\lambda_i$, this relationship holds for, say, $\lambda_i = i(i+1)$ and $n_i = 2i+1$ (which is the situation on the sphere), since

$$ N(\lambda_i) = \sum_{j=0}^i n_j = \sum_{j=0}^i 2i+1 = i^2 + 2i + 1 \approx i(i+1) = \lambda_i. $$

But it also holds for $\lambda_i = i$ and $n_i = 1$ since then

$$ N(\lambda_i) = \sum_{j=0}^i n_j = \sum_{j=0}^i 1 = i + 1 \approx i = \lambda_i. $$

share|cite|improve this question
up vote 11 down vote accepted

Weyl's formula:

$$N(R)=\frac{1}{(4{\cdot}\pi)^{d/2}{\cdot}\Gamma\left(\frac d2+1\right)}{\cdot}V{\cdot}R^{d/2}+o(R^{d/2}).$$ where $d$ --- dimension, $V$ --- volume, $N(R)$ --- number of eigenvalues $\le R$. It works for any compact Riemannian manifold.

share|cite|improve this answer
Not sure I understand the connection (also: the link is broken; had to Google it). I guess I should also mention that I'm not interested with surfaces with boundary (or Dirichlet boundary conditions). – TerronaBell Jun 15 '11 at 21:11
Great -- thanks for the clarification. (Also, looks like it should be $o(R^{(d-1)/2})$.) – TerronaBell Jun 15 '11 at 21:58
I must be interpreting one of your constants incorrectly -- for the unit sphere in $\mathbb{R}^3$ the distinct eigenvalues are $\lambda_i = i(i+1)$ appearing with multiplicity $2i+1$. So then the number of eigenvalues with value no greater than $\lambda_i$ is $\sum_{j=0}^i 2j + 1 = i^2 + 2i + 1$. In other words, we have $R(i) = i^2 + i$ and $N(R(i)) = i^2 + 2i + 1$, hence $N(R) \approx R$. But for $d=2$ and $V=4\pi$, Weyl's formula says $N(R) \approx R/4\pi$. Where does the factor $1/4\pi$ come from? – TerronaBell Jun 15 '11 at 22:32
The constant above is incorrect. For the correct formula and some modern developments see – GH from MO Jun 15 '11 at 23:15
@fuzzytyron: In general it is a very difficult question how eigenvalues are distributed, what are their multiplicities, how many are there in short intervals etc. For example, for the modular surface (upper half-plane modulo $\mathrm{SL}(2,\mathbb{Z})$) it is conjectured that each eigenvalue has multiplicity one, but we only have very weak results in that direction. – GH from MO Jun 16 '11 at 15:51

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.