# 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 $o(R^{(d-1)/2})$.) Jun 15, 2011 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? Jun 15, 2011 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. Jun 16, 2011 at 15:51