Eigenvalues of Laplacian-Beltrami operator - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T20:31:03Z http://mathoverflow.net/feeds/question/64017 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/64017/eigenvalues-of-laplacian-beltrami-operator Eigenvalues of Laplacian-Beltrami operator Alberto 2011-05-05T15:53:17Z 2012-01-05T18:25:39Z <p>I am interested in the first non zero eigenvalue of the Laplace-Beltrami operator in a 2D compact manifold, and if there is a geometric characterization of its value.</p> <p>I am interested in the case when you fix the volume of the manifold to some value (say $Vol = 1$), and let the other modes of the metric fluctuate. The average curvature of the manifold is imposed by the Gauss-Bonet theorem, but can let the curvature to fluctuate from one point to another.</p> <p>My intuition says that the first non zero eigenvalue should approach 0 in the limit when the "fluctuations of the curvature" grow, but I can not give a precise meaning to this statement.</p> <p>so the question is: Is there any characterization of the first eigenvalue(s) of the Laplace-Beltrami operator in a 2D compact riemann manifold as functions of the curvature or its powers (i.g. $\int R^2 \sqrt{g} d^2x$).</p> <p>So let me be more specific. Imagine a manifold topologically equivalent to a Torus. The metric can be written as </p> <p>$ds^2=f(x_1,x_2)(dx^2_1+dx^2_2)$</p> <p>and the scalar curvature </p> <p>$R=−2\Delta \log(f)$</p> <p>For the case $f=cte$ we have a flat torus. Now expanding f as a fourier series we will have some regions of the torus with poritive curvature and some regions with negative curvature (this invalid some known theorems, that need wither the curvature to be always positive or always negative). Gauss bonnet says that: </p> <p>$\int R=0$ </p> <p>So the concrete question is: Is there a characterisation of the first eigenvalue of the Laplace operator in terms of $\int R^2$? </p> http://mathoverflow.net/questions/64017/eigenvalues-of-laplacian-beltrami-operator/64020#64020 Answer by Raphael for Eigenvalues of Laplacian-Beltrami operator Raphael 2011-05-05T16:43:04Z 2011-05-05T16:43:04Z <p>Yes, This is a classical result due to Lichnerowicz. If the Ricci tensor of a compact Riemannian manifold, is such that $Ric \geq kg$ for some k > 0, then $\lambda_1 \geq \frac{nk}{n - 1}$. In Your case, you just replace the Ricci curvature by $Kg$. You will find a proof in the book of Aubin <em>A Course in Differential Geometry</em> as consequence of the Bochner Formula.</p> http://mathoverflow.net/questions/64017/eigenvalues-of-laplacian-beltrami-operator/64021#64021 Answer by APG for Eigenvalues of Laplacian-Beltrami operator APG 2011-05-05T16:44:23Z 2011-05-06T10:45:30Z <p>The first eigenvalue of a compact surface can be made arbitrarily small (even for surfaces of fixed genus); see, for example, [1], [2], [3] (and references therefrom).</p> <p>However, as proved by Sarnak and Xue [4], there are arithmetic examples of (constant negative curvature) compact Riemann surfaces of arbitrarily high genus with the first eigenvalue bounded away from zero (see also [5] for a construction involving Selberg's $3/16$'' theorem).</p> <p>[1] B. Randol, Small eigenvalues of Laplace operator on compact Riemann surfaces, Bull. AMS 80, 1974 996-1008</p> <p>[2] R. Schoen, S. Wolpert, S. Yau, Geometric bounds on the low eigenvalues of a compact surface, Proc. Symp. Pure Math, vol. 36, AMS, 1980, 279-285.</p> <p>[3] P. Buser, On Cheeger’s inequality $λ_1 ≥ \frac{h^2}{4}$. Proc. Symp. Pure. Math. vol. 36, 29–77.</p> <p>[4] P. Sarnak and X. Xue, Bounds for multiplicites of automorphic representations, Duke Math J. 64, 1991, 207-227.</p> <p>[5] R. Brooks, E. Makover, Riemann surfaces with large first eigenvalue. J. Anal. Math. 83, 2001, 243–258.</p> http://mathoverflow.net/questions/64017/eigenvalues-of-laplacian-beltrami-operator/64022#64022 Answer by Alain Valette for Eigenvalues of Laplacian-Beltrami operator Alain Valette 2011-05-05T16:47:19Z 2011-05-05T16:47:19Z <p>Your intuition seems correct: assume that some part of your surface contains a long cylinder with a flat metric. Take a test-function on the surface, supported on that cylinder, equal to +1 roughly on one half of the cylinder and -1 on the other half, and of total integral 0. In this case, the $L^2$-norm of the gradient of your function will be very small compared to the $L^2$-norm of the function. By the variational characterization of the first eigenvalue of the Laplacian (= Rayleigh quotient), you can make the first eigenvalue very small.</p> http://mathoverflow.net/questions/64017/eigenvalues-of-laplacian-beltrami-operator/84983#84983 Answer by Liviu Nicolaescu for Eigenvalues of Laplacian-Beltrami operator Liviu Nicolaescu 2012-01-05T18:25:39Z 2012-01-05T18:25:39Z <p>This is a challenging problem. The precise meaning of randomness of the metric ought to play a role. I cannot think of a natural one in a general case but I suggest you consider first the case of the two-torus $T^2$ equipped with a metric of the form</p> <p>$$g = e^{2 u} d\theta_1^2+ e^{2v} d\theta_2^2$$</p> <p>where $u,v: T^2\to\mathbb{R}$ are independent random functions. Here you need to specify the nature of the randomness of $u$ and $v$. </p>