Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.


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.


share|improve this question
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

2 Answers 2

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 $\xi > 0$.

share|improve this answer
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 $\Vert(R(\xi)\Vert\geq r_0$.

share|improve this answer
$C_m$ 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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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