Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $(M,g)$ be a conformally compact surface.

An example situation is a hyperbolic surface of infinite area like the quotient $\Gamma\backslash \mathbb{H}$, where $\mathbb{H}$ is the hyperbolic plane and $\Gamma = \langle T \rangle \subset PSL(2, \mathbb{R})$ a Fuchsian group that is a cyclic group generated by a hyperbolic transformation $T$.

Let $\Delta_g$ be its associated Laplace-Beltrami-operator on functions. I read in some math papers that the orthogonal projection to $ \ker \Delta_g$ vanishes here - in contrast to the case of a closed Riemannian manifold. How to show this this fact? Are there "nice" characterizations of the nullspace of the Laplacian?

Thanks for your help!

share|cite|improve this question
up vote 0 down vote accepted

I received a deciding hint now:

Since the area of these surfaces isn't finite, the constant functions are not $L^2$-functions any longer. In particular they are not 0-eigenfunctions as in the case of closed surfaces and therefore $\lambda=0$ is not in the discrete spectrum of $\Delta_g$.

It's well known that the continous spectrum of $\Delta_{\mathbb{H}}$ is $[1/4, \infty)$, so zero is not in the continous spectrum either. That's why $\ker \Delta_g = \{ 0 \}$ and therefore the projection onto it vanishes.

share|cite|improve this answer

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.