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](http://en.wikipedia.org/wiki/Hearing_the_shape_of_a_drum#Weyl.27s_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


$$ N(\lambda\_i) = \sum\_{j=0}^i n\_j = \sum\_{j=0}^i 1 = i + 1 \approx i = \lambda\_i. $$



Growth of Laplacian eigenvalues on a compact domain?
[Weyl's formula](http://en.wikipedia.org/wiki/Hearing_the_shape_of_a_drum#Weyl.27s_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.


