# Growth of Laplacian eigenvalues on a compact domain?

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

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

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