Exotic spectrum of Laplace operator - MathOverflow

Exotic spectrum of Laplace operator

2012-03-16T17:57:00Z

Given a closed Riemannian manifold and a generalized Laplace $\Delta$ operator, it is well known that $\Delta$ has discrete spectrum $(\lambda_n)_n$ (arranged in a increasing way, not counting multiplicities). By a (consequence of a) result of Colin de Verdière, given any finite strictly increasing sequence $a_1,\cdots,a_k$ of strictly positive numbers, there exists a Riemannian manifold and a Laplace operator on it such that the first k+1 eigenvalues are exactly, $0,a_1,\cdots,a_k$.

My question is about what's happening at infinity. More precisely, since usually the spectrum of a Laplace operator is a quadratic polynomial (in the sense that {$\lambda_n : {n\geq 0}$} is of the form {$P(n) : n\geq 0$} where $P$ is a quadratic polynomial), is there a Laplace operator (on a closed manifold) such that there is no $n_0$ such that {$\lambda_n : {n\geq n_0}$} is of the form {$P(n) : n\geq 0$} where $P$ is a quadratic polynomial ?

My question could be reformulated : do you know example where the explicit (exact) eigenvalues (not asymptotics of them) are (after a certain rank) not given by a formula of the form $a n^2 + bn +c$ where $a,b,c$ are constants (for example some kind of fraction P(n)/Q(n) where the degree of P is 3 and the degree of Q is one...)

[Edit : precisions]

Answer by Renato G Bettiol for Exotic spectrum of Laplace operator

2012-03-16T21:43:43Z

Regarding the asymptotic behavior of the spectrum of the Laplacian (or, as the OP puts it, the behavior at infinity), the most basic result is Weyl's asymptotic formula (see Chavel's book, p.172): let $(M,g)$ be a compact manifold with $\dim M=n$ and $0=\lambda_0<\lambda_1\leq \lambda_2\leq\dots$ be the eigenvalues of the Laplacian, each distinct eigenvalue repeated according to its multiplicity. Denote by $N(\lambda)=\sum_{\lambda_j\leq\lambda} 1$ the number of eigenvalues (counted with multiplicity) that are $\leq\lambda$. Then

$$N(\lambda)\sim vol(M,g)\frac{vol(B^n)}{(2\pi)^n}\lambda^{n/2}, \quad \mbox{as} \quad\lambda\to+\infty,$$

where $vol(B^n)=\frac{\pi^{n/2}}{\Gamma(n/2+1)}$ is the volume of the unit ball of $\mathbb R^n$. In particular,

$$(\lambda_k)^{n/2}\sim\frac{(2\pi)^n}{vol(B^n)}\frac{k}{vol(M,g)}, \quad \mbox{as}\quad k\to+\infty.$$

Thus, the asymptotic behavior of the eigenvalues cannot be prescribed - it has to satisfy the above.

Also, as far as I understand, Colin de Verdière's result is stronger than stated. Namely, given any compact connected manifold M, with $\dim M\geq 3$, and any finite sequence $0\leq a_1\leq\dots\leq a_k$, there exists a Riemannian metric on $M$ such that the first eigenvalues in the spectrum of its Laplacian are $0\leq a_1\leq\dots\leq a_k$.

Answer by Liviu Nicolaescu for Exotic spectrum of Laplace operator

2012-03-18T19:37:48Z

For a generic metric on an $m$-dimensional the manifold the eigenvalues of the Laplacian are all simple. Fix such a metric and denote the coresponding eigenvalues by

$$ \lambda_1, \lambda_2,\cdots $$

Using Weyl's asymptotic expansion we conclude that

$$\lambda_n\sim const . n^{2/m}. $$

Thus for any polynomial $P$ of degree $>1$ we have

$$ \lim_{n\to\infty} \lambda_n/P(n) = 0. $$