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Why does the first cohomology group $H^1(X, K)$ of the sheaf of meromorphic functions on a non-compact Riemann surface $X$ vanish?

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Is this homework? If so, please tag it. Have you tried anything and can share that with the community? Regards. –  Amzoti Jan 5 '13 at 23:23
    
It's a problem (26.6) from Forster's book (Lectures on Riemann surfaces). The first part of the exercise ($H^1(X,\mathcal{O}_D)=0$ for any divisor $D$) would follow from this fact and solvability of Mittag-Leffler problem on non-compact Riemann surfaces (plus some sheaf exact sequence), but I have no idea how to prove the statement above. –  mathdonk Jan 6 '13 at 11:23

1 Answer 1

Since for $ H^{1} $, derived-functor cohomology coincides with Čech cohomology, I'll use the latter and show that every $ \xi \in {\check{H}^{1}}(X,K) $ is zero.

The cohomology class $ \xi $ is represented in some locally finite open covering $ \mathcal U =(U_{i}) $ by some cocycle $ (\xi_{ij}) $ with $ \xi_{ij} \in K(U_{ij}) $, where $ U_{ij} = U_{i} \cap U_{j} $. Letting $ D $ be the divisor defined by the poles of all these $ \xi_{ij} $’s, we see that actually $ \xi_{ij} \in {\mathcal{O}_{X,D}}(U_{ij}) $.

However, $ {\check{H}^{1}}(X,\mathcal{O}_{X,D}) = 0 $ because on a non-compact Riemann surface, the positive-dimensional cohomology groups of coherent sheaves are zero (open Riemann surfaces are Stein!).

Hence, we can write $ (\xi_{ij}) $ as the coboundary $ \xi_{ij} = \xi_{j}|_{U_{ij}} - \xi_{i}|_{U_{ij}} $ of some $ 1 $-cochain $ (\xi_{i}) $ with $ \xi_{i} \in {\mathcal{O}_{X,D}}(U_{i}) $. We can interpret a fortiori $ (\xi_{i}) $ as a $ 1 $-cochain with $ \xi_{i} \in K(U_{i}) $ (because $ \mathcal{O}_{X,D} \subseteq K $), thus proving that $ \xi $ is the coboundary of a cochain of $ K $ and hence that its class is zero in $ {\check{H}^{1}}(X,K) $.

Conclusion: $ {\check{H}^{1}}(X,K) = 0 $.

A GAGA-type remark

If $ X_{\text{alg}} $ is the unique algebraic variety underlying $ X $ and $ K_{\text{rat}} $ its sheaf of rational functions, we also have $ {H^{1}}(X_{\text{alg}},K_{\text{rat}}) = 0 $, but this time, the nullity is trivial because $ K_{\text{rat}} $ is a flabby sheaf.

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