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A finite type Riemann surface is a finitely punctured compact Riemann surface.

It is easy to find lots of references proving that the de Rham cohomology of a compact manifold is finite dimensional; likewise, the Dolbeault cohomology of a compact complex manifold is also finite dimensional.

Qst 1) What is the easiest way to see that the Dolbeault cohomology of a finite type Riemann surface is finite dimensional? (at least I believe it is, if not, then what would be a counter-example?)

Qst 1 bis) What is the easiest way to compute the Dolbeault cohomology of a finite type Riemann surface?

Qst 2) If $X$ is a compact complex manifold and $S \subset X$ is a closed analytic hypersurface of $X$, is it true that $X \backslash S$ has finite Dolbeault cohomology?

EDIT

  • I now realize it is false that $H^{1,0}$ is finite dimensional in general: for example, $H^{1,0}(\mathbb C^*) \simeq \Omega(\mathbb C^*)$ contains the span of $\frac{dz}{z^n}$ for all $n \in \mathbb N$. This is coherent with the fact that in the absence of compacity, there is no Hodge decomposition and so $H^{1,0}$ can be infinite dimensional even though $H^{1}$ is not.

Now I wonder about $H^{0,1}$. For example, $H^{0,1}(\mathbb C)=0$ while $H^{1,0}(\mathbb C)$ is infinite dimensional.

So I guess my new question is:

0) Did I say anything incorrect above?

1) What is $h^{0,1}$ for a Riemann surface of genus $g$ minus $n$ points?

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All right, I found my answer so I will answer it for reference. This seems to be very classical, but I am reading these things by myself...

Apparently if $X$ is a Stein manifold, then for any $q \geq 1$ and $p \geq 0$ we have $h^{p,q}=0$. Also, any non-compact Riemann surface is Stein (hard theorem, though probably much easier for finite type Riemann surfaces).

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