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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!

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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.

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