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Let $(M,g)$ be a Riemannian manifold and $-\Delta$ the Laplacian on M (acting on smooth functions). Then the resolvent $R(\xi)$ of $-\Delta$ is a compact operator.

Is it possible to find for every $\epsilon>0$, a point in the resolvent set $\xi$, s.t. $\Vert R(\xi) \Vert\leq \epsilon$?

Maybe it is very easy to prove, but I'm not so familiar with spectral theory. I hope you can help me.


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The resolvent of the Laplacian is compact if $(M,g)$ is compact. In general, the spectrum of the Laplacian will not be discrete. Consider the classical example $-y^2( \partial^2_x + \partial^2_y)$ on $SL_2(\mathbb{Z}) \backslash \mathbb{H}$ for example. The spectrum has a continuous part. – Marc Palm Apr 4 '13 at 7:01
Assuming a self-adjoint realization of the Laplacian: Yes. The resolvent $R$ of a self-adjoint operator always satisfies the estimate $\|R(z)\|\leq |\mathrm{Im} z|^{-1}$ if $z$ is not real. – Sönke Hansen Apr 4 '13 at 7:02
Just to be sure that I am not confusing the OP. I was only saying that the resolvent will not always be compact. Nevertheless, you get your estimate as a chep consequence of functional calculus as Nik Weaver explains. @Soenke Hansen: I would believe that your estimate needs at least some constant, or not? – Marc Palm Apr 4 '13 at 7:12
@Marc: I think Sönke's estimate is right. As I mentioned, $\|R(z)\| = 1/{\rm dist}(z, {\rm spec}(-\Delta))$. The distance from $z$ to ${\rm spec}(-\Delta)$ is always at least $|{\rm Im}\, z|$ since ${\rm spec}(-\Delta) \subseteq {\bf R}$. – Nik Weaver Apr 4 '13 at 7:44
However, I thought it was simpler to take $z < 0$ since actually ${\rm spec}(-\Delta) \subseteq [0,\infty)$. – Nik Weaver Apr 4 '13 at 7:45
up vote 10 down vote accepted

Sure, since $-\Delta$ is a positive operator, by the spectral theorem it can be realized as multiplication by a positive function $f(x)$ on some $L^2(X)$ space. Then $R(\xi)$ is multiplication by $\frac{1}{\xi - f(x)}$ and its operator norm is the sup norm of this function. If $\xi < 0$ then the function $\frac{1}{\xi - f(x)}$ is bounded by $\frac{1}{|\xi|}$ in absolute value, so as $\xi \to -\infty$ we have $\|R(\xi)\| \to 0$.

The point is that $\|R(\xi)\|$ equals one over the distance from $\xi$ to ${\rm spec}(-\Delta)$.

If you restrict $\xi$ to be positive the question is more interesting. On the unit circle ${\bf T}^1$ the eigenvalues of $-\Delta$ are the square integers, and for any $\epsilon > 0$ we can find $\xi > 0$ such that $|\xi - n^2| > \frac{1}{\epsilon}$ for every $n \in {\bf Z}$, so we can still ensure that $\|R(\xi)\| \to 0$. In other words, the eigenvalues are spaced farther and farther apart so $\xi > 0$ can be chosen arbitrarily far from the spectrum of $-\Delta$. But on ${\bf T}^4$ the eigenvalues of $-\Delta$ are of the form $a^2 + b^2 + c^2 + d^2$ for $a, b, c, d \in {\bf Z}$, which means that every positive integer is an eigenvalue. Thus any $\xi > 0$ is at most $\frac{1}{2}$ units away from an eigenvalue, and therefore $\|R(\xi)\| \geq 2$ for every $\xi > 0$.

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Nik, I think you meant $T^4$ instead of $S^4$. – Liviu Nicolaescu Apr 4 '13 at 9:17
@Liviu: you're right, I'll correct it. – Nik Weaver Apr 4 '13 at 16:52

This is mostly enhancing Nik Weaver's comments. Suppose that $M$ is compact of dimension $m$. If $m\geq 2$, then for a generic metric $g$ on $M$ the eigenvalues $\lambda_k$ of the Laplacian $\Delta_g$ are simple. In general, for any $m$, Weyl's spectral estimates imply that

$$\lambda_k \sim C_m \left(\frac{k}{{\rm vol}_g(M)}\right)^{\frac{2}{m}}\;\;\mbox{as $k\to\infty$}, $$

where $C_m$ is an explicit universal constant that depends only on $m$. (Hat-tip to Marc Palm!) In particular this shows that for $m\geq 2$ and a generic metric we have

$$0<\lambda_{k+1}-\lambda_k =O(1). $$

Now Nik Weaver's argument shows that there exists $r_0>0$ such for any $\xi\in [0,\infty)\setminus {\rm spec}\;(\Delta)$ we have $\Vert(R(\xi)\Vert\geq r_0$.

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$C_m$ also depends on the volume of $M$? – Marc Palm Apr 4 '13 at 9:51
@Marc Palm: Yes, the constant is proportional to the volume. The basic principle of Weyl asymptotics is that the number of eigenvalues $<\lambda$ behaves, asymptotically as $\lambda\to+\infty$, as the symplectic volume of $\{(x,\xi)\in T^*M; |\xi|^2<\lambda\}$. – Sönke Hansen Apr 4 '13 at 10:11
@ Marc Palm I've edited my answer to incoporate the dependence on volume. – Liviu Nicolaescu Apr 4 '13 at 11:52
The eigenvalues of the laplacian on the circle of radius $R$ are $\lambda_k=(k/R)^2=(k/R)^{2/m}$, $m=\dim S^1=1$. Note that the volume of this circle is $2\pi R$. – Liviu Nicolaescu Apr 4 '13 at 17:41
You missed the words "generic metric". As I mentioned in the first paragraph of my answer for a generic metric the eigenvalues are simple. This is an old result of Karen Uhlenbeck, Amer. J. Math. vol.98(1976), p. 1059-1078. – Liviu Nicolaescu Apr 5 '13 at 12:04

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