9
$\begingroup$

The Laplacian on a compact Riemannian manifold has a discrete spectrum. For example on a circle of perimeter $L$ the $n$-th eigenvalue starting at $0$ is $-\lambda_n = -(2\pi/L)^2 n^2$.

On the other hand the Laplacian of a non-compact manifold may be continuous. For example on $\mathbb{R}$ the spectrum of the second derivative operator is $(-\infty,0]$ (I use the convention that the Laplacian is negative semi-definite).

I was wondering if it is always the case that when a sequence of pointed Riemannian manifold $(M_n,o_n,g_n)$ converges smoothly to a limit Riemannian manifold $(M,o,g)$ then the spectrum of the corresponding Laplace operators also converges. I don't think this is true but a semi-continuity property should hold (and I'm guessing is well known, hence the reference-request tag).

Given a Riemannian manifold $M$ let $-\lambda_1(M)$ be the maximum of the spectrum (i.e. the closest point to $0$). Then $\lambda_1$ satisfies the following "domain monotonicity": If $\Omega$ is an open submanifold of $M$ with boundary then $\lambda_1(\Omega) \ge \lambda_1(M)$ (e.g. $\lambda_1(\mathbb{R})=0$ but $\lambda_1([0,2\pi]) = 1$). If one knows that $\lambda_1(M)$ is the limit of $\lambda_1(\Omega_n)$ over an increasing sequence of bounded submanifolds then this is enough to prove upper semi-continuity of the absolute value of $\lambda_1$ under smooth convergence.

To start I'd like to know if the above reasoning is correct and if there is a good reference. Concretely I'd love to have a good reference for the answers to the following questions:

  1. Is $\lambda_1$ upper semi-continuous with respect to smooth convergence?
  2. Is there a simple counterexample for continuity?

The question: What about the rest of the spectrum? isn't definite enough for this site so I'll try to make it more concrete.

By the spectral theorem the (unique self-adjoint extension of the) Laplacian on the limit manifold is unitarily conjugate to multiplication by some (non-positive) function $\varphi$ on $L^2(\mu)$ for some $\sigma$-finite measure $\mu$ on a Polish space $X$. Let $\varphi_n,\mu_n$ and $X_n$ be defined similary for the sequence.

Is it true that $\mu(\varphi^{-1}((-a,0])) \ge \limsup \mu_n(\varphi_n^{-1}((-a,0]))$ for all $a > 0$?

Edit: (Hopefully) Fixed some issues (positive vs negative definite Laplacian, first eigenvalue vs maximum of the spectrum) in response to comments.

$\endgroup$
  • $\begingroup$ Also in question 1, I assume by "first eigenvalue" you mean "top of the spectrum"? Because as you note, when $M$ is non-compact the Laplacian may have no eigenvalues. $\endgroup$ – Nate Eldredge Aug 23 '14 at 3:28
  • $\begingroup$ For your last question, since $(X,\mu,\varphi)$ is not unique, are you sure that $\mu(\varphi^{-1}((-a,0]))$ is well defined? (If you drop the requirement that $X$ be Polish, it definitely is not.) $\endgroup$ – Nate Eldredge Aug 23 '14 at 3:31
  • $\begingroup$ Hmm, you're right that $(X,\mu,\varphi)$ is non-unique (e.g. multiply $\mu$ by a constant). However I'm hoping there's some way to normalize correctly. When the spectrum is discrete the right thing to do is clearly to replace $\mu(\varphi^{-1}((-a,0]))$ by the number of eigenvalues in $(-a,0]$. I'm not sure what makes sense in general. Maybe $-\int\varphi 1_{\lbrace \varphi > -a\rbrace} \mathrm{d}\mu$ (i.e. minus the integral of $\varphi$ on the set where it is larger than $-a$)?. $\endgroup$ – Pablo Lessa Aug 25 '14 at 21:54
6
$\begingroup$

This is only a partial answer since it concerns only compact manifolds. There is a result due to T. Shioya http://projecteuclid.org/download/pdf_1/euclid.jmsj/1213023969 a special case of which says that if a sequence of compact $n$-dimensional Riemannian manifolds $M_i$ converges to another compact n-dimensional Riemannian manifold M in the Gromov-Hausdorff sense and such that the sectional curvature is uniformly bounded from below, then for any $k$ the $k$-th eigenvalue of the Laplacian of $M_i$ converges to the $k$-th eigenvalue of the Laplacian of $M$. Notice that smooth convergence implies the above type of convergence.

$\endgroup$
1
$\begingroup$

By the theorem in

  • Andreas Kriegl, Peter W. Michor, Armin Rainer: Many parameter Hölder perturbation of unbounded operators. Math. Ann. 353, 2 (2012), 519-522. (pdf)

each eigenvalue is a Lipschitz function on the space of all Riemannian metrics of a fixed compact manifold. See also the following paper for related results.

  • Andreas Kriegl, Peter W. Michor, Armin Rainer: Denjoy-Carleman differentiable perturbation of polynomials and unbounded operators. Integral Equations and Operator Theory 71,3 (2011), 407-416. (pdf)
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.