Exotic spectrum of Laplace operator - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T05:38:43Z http://mathoverflow.net/feeds/question/91397 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/91397/exotic-spectrum-of-laplace-operator Exotic spectrum of Laplace operator Alex 2012-03-16T17:57:00Z 2012-03-19T13:03:47Z <p>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, $0,a_1,\cdots,a_k$.</p> <p>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 ?</p> <p>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...)</p> <p>[Edit : precisions]</p> http://mathoverflow.net/questions/91397/exotic-spectrum-of-laplace-operator/91416#91416 Answer by Renato G Bettiol for Exotic spectrum of Laplace operator Renato G Bettiol 2012-03-16T21:43:43Z 2012-03-18T15:37:40Z <p>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 <strong>Weyl's asymptotic formula</strong> (see <a href="http://www.amazon.com/Eigenvalues-Riemannian-Geometry-Applied-Mathematics/dp/0121706400" rel="nofollow">Chavel's book</a>, p.172): let $(M,g)$ be a compact manifold with $\dim M=n$ and $0=\lambda_0&lt;\lambda_1\leq \lambda_2\leq\dots$ 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</p> <p>$$N(\lambda)\sim vol(M,g)\frac{vol(B^n)}{(2\pi)^n}\lambda^{n/2}, \quad \mbox{as} \quad\lambda\to+\infty,$$</p> <p>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,</p> <p>$$(\lambda_k)^{n/2}\sim\frac{(2\pi)^n}{vol(B^n)}\frac{k}{vol(M,g)}, \quad \mbox{as}\quad k\to+\infty.$$</p> <blockquote> <p>Thus, the asymptotic behavior of the eigenvalues <strong>cannot be prescribed</strong> - it has to satisfy the above.</p> </blockquote> <hr> <p>Also, as far as I understand, Colin de Verdière's result is stronger than stated. Namely, given <strong>any</strong> 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 $0\leq a_1\leq\dots\leq a_k$.</p> http://mathoverflow.net/questions/91397/exotic-spectrum-of-laplace-operator/91566#91566 Answer by Liviu Nicolaescu for Exotic spectrum of Laplace operator Liviu Nicolaescu 2012-03-18T19:37:48Z 2012-03-18T19:37:48Z <p>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</p> <p>$$\lambda_1, \lambda_2,\cdots$$</p> <p>Using Weyl's asymptotic expansion we conclude that </p> <p>$$\lambda_n\sim const . n^{2/m}.$$</p> <p>Thus for any polynomial $P$ of degree $>1$ we have</p> <p>$$\lim_{n\to\infty} \lambda_n/P(n) = 0.$$</p>