does the finite dimensionlity of the first cohomology group ($ H^1 $) of the sheaf of meromorphic sections of a holomorphic line bundle on a compact riemann surface follow easily from the finite dimensionality of the cohomologies of the sheaf of holomorphic functions on the surface?

1$\begingroup$ Out of interest, are you asking out of curiosity, or because you need this in some research you're doing? $\endgroup$ – Yemon Choi Nov 24 '10 at 9:05

1$\begingroup$ simply to trying to prune down the analysisintensive parts of Gunnings RiemannRoch treatment to the bare minimum..turns out he had the same motivation ! $\endgroup$ – faquarl Nov 26 '10 at 7:09
The derivation is indeed "easy", in the sense that no additional results from analysis are needed. Consider the sheaf map $\mathcal{O}(L) \to \mathcal{M}(L)$, it is injective and the cokernel is the sheaf $\mathcal{H}(L)$ of principal parts of meromorphic sections of $L$. The sheaf $\mathcal{H}$ is fine. Therefore you get from the long exact sequence associated with
$$0\to \mathcal{O}(L) \to \mathcal{M}(L) \to \mathcal{H}(L) \to 0$$
a short exact sequence
$$ H^1 (X;\mathcal{O}(L)) \to H^1 (X;\mathcal{M}(L)) \to H^1 (X;\mathcal{H}(L))=0 $$
and the finitedimensionality of $H^1 (X;\mathcal{O}(L))$ implies the result.
Addendum: it is known that any line bundle admits a nonzero meromorphic section $f$. Multiplication by $f$ induces an isomorphism $\mathcal{M}\cong \mathcal{M}(L)$, so $dim (H^1 (X;\mathcal{M}(L)))$ does not depend on $L$. Since there is a line bundle $L$ with $H^1 (X;\mathcal{O}(L))=0$ (this happens if the degree of $L$ is a sufficiently large positive number). This show that $H^1 (X;\mathcal{M}(L))=0$ for ANY line bundle.
The proof that $H^1 (X;\mathcal{O})$ is finitedimensional (which requires quite a bit of analysis) generalizes to general line bundles. Or one can use the result for trivial line bundles plus the existence of meromorphic sections, see the answer by Francesco.
The existence of meromorphic sections is not easy to establish; it follows from RiemannRoch. There is also a more direct argument, using the analysis involved in the proof of RiemannRoch.

$\begingroup$ Maybe I'm missing something, but how do you know that $H^1(X, \mathcal{O}(L))$ is finite dimensional, just knowing that $H^1(X, \mathcal{O})$ is so? It seems to me that you must use the argument I gave in my answer. (Ok, I just proved finitedimensionality of the sheaf of holomorphic sections, since I slightly misread the question). $\endgroup$ – Francesco Polizzi Nov 24 '10 at 14:08

$\begingroup$ The proofs of the finitedimensionality of $H^1 (X;\mathcal{O})$ that I know apply with only notational changes to general line (even vector)bundles as well. Your answer gives a derivation, using the finitedimensionality of $H^1 (X;\mathcal{O})$ as a black box. Also note that your argument assumes that you know that $L$ has a nontrivial meromorphic section. $\endgroup$ – Johannes Ebert Nov 24 '10 at 14:50

$\begingroup$ which is of course true, but not trivial. $\endgroup$ – Johannes Ebert Nov 24 '10 at 14:51

$\begingroup$ Yes, it is probably the same proof I know. And of course this is also true for any coherent analytic sheaf (Grauert or Cartan, I think). I used the finitedimensionality of the $H^1$ of holomorphic functions as a black box just because the question required this... $\endgroup$ – Francesco Polizzi Nov 24 '10 at 15:16

$\begingroup$ a) +1 for the elegant presentation b) Finite dimensionality of cohomology of coherent sheaves on compact manifolds is due to CartanSerre, CRAS 237, 1953, 128130. c) You want the degree of $L$ to be a positive large number, not a negative one, to ensure vanishing of $H^1(X, \mathcal O (L))$. $\endgroup$ – Georges Elencwajg Nov 24 '10 at 21:00
The answer is "more or less yes", depending on your definition of "easily". The following is a possible approach. We use the language of divisors, and we assume the standard facts that the $H^0$ of a skyscreaper sheaf is finitedimensional and that its $H^1$ is zero.
Assume first that $\mathcal{O}(D)$ is effective. From the short exact sequence
$0 \to \mathcal{O} \to \mathcal{O}(D) \to \mathcal{O}_D \to 0$
and from the vanishing of $H^1(\mathcal{O}_D)$ one obtains a surjective homomorphism
$H^1(\mathcal{O}) \twoheadrightarrow H^1(\mathcal{O}(D))$,
hence the finitedimensionality of the first group implies the finitedimensionality of the second.
Now assume that $D$ is any divisor, and write $D=E F$, with $E$, $F$ effective.
Then by
$0 \to \mathcal{O}(D) \to \mathcal{O}(E) \to \mathcal{O}_F \to 0$
we deduce as before
$H^0(\mathcal{O}_F) \to H^1(\mathcal{O}(D)) \to H^1(\mathcal{O}(E)) \to 0$.
Since $H^0(\mathcal{O}_F)$ and $H^1(\mathcal{O}(E))$ are both finitedimensional, the claim follows.
EDIT. This argument proves the finitedimensionality of the sheaf of holomorphic sections. The finitedimensionality of the sheaf of meromorphic sections $\mathcal{M}(D)$ follows from the fact that the cokernel of the natural injection $\mathcal{O}(D) \to \mathcal{M}(D)$ has trivial $H^1$, see Johannes Ebert's answer.