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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 the absolute value of the first eigenvalue of the Laplacian 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 ChristianRemlings comments.

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.

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 spectrum of the Laplacian is non-positive 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 the absolute value of the first eigenvalue of the Laplacian 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 ChristianRemlings comments.

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 the absolute value of the first eigenvalue of the Laplacian 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 ChristianRemlings comments.