Eigenvalues of Laplacian - MathOverflow most recent 30 from http://mathoverflow.net2013-05-25T23:34:27Zhttp://mathoverflow.net/feeds/question/316http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/316/eigenvalues-of-laplacianEigenvalues of LaplacianIlya Nikokoshev2009-10-11T21:42:03Z2009-11-16T23:34:52Z
<p>What's the most natural way to establish the asymptotics of $\Delta$ on a compact Riemannian manifold M of dimension N? The asymptotics should be
$$\#\{v < A^2\} = \mathrm{const}*\mathrm{vol}(M)*A^n + o(\mathrm{something})$$
(Perhaps one could consider first a case of Kahler manifold? The Laplacian is partiularly simple there).</p>
http://mathoverflow.net/questions/316/eigenvalues-of-laplacian/1315#1315Answer by Lior Silberman for Eigenvalues of LaplacianLior Silberman2009-10-20T00:06:01Z2009-10-20T00:06:01Z<p>The most natural way is to study the short-time asymptotics of the heat or wave kernel on M. For example, you can use the heat kernel $p_t(x,y) = \sum_i e^{-lambda_i t} f_i(x) \overline{f_i(y)}$ where $f_i$ are the eigenfunctions with eigenvalues $\lambda_i$. This is a fundamental solution to the heat equation.</p>
<p>When $t$ is small then you can construct a good approximation to $p_t$ near any particular $x$ by hand, using Fourier analysis in local co-ordinates. The end result is that that $p_t(x,x) \approx C t^{-n/2}$. Now integrate this estimate $dx$, noting that $\int_M p_t(x,x)dx$ basically counts eigenvalues with $\lambda_i \leq 1/t$. </p>