# Behavior of the spectrum of the Laplacian under pointed smooth convergence

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.

• 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. – Nate Eldredge Aug 23 '14 at 3:28
• 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.) – Nate Eldredge Aug 23 '14 at 3:31
• 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 $-a$)?. – Pablo Lessa Aug 25 '14 at 21:54

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.