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Let X be complex curve (complex manifold and $\dim X=1$).

For $x_1,x_2\in X$ we define the sheaf $\mathcal{K}_{x_1,x_2}$(in complex topology) of meromorphic functions vanish at the points $x_1$ and $x_2$.

Is it true, that $H^1(X,\mathcal{K}_{x_1,x_2})=0$?

In general, what are sufficient conditions for the $\mathcal{F}$ to $H^1(X,\mathcal{F})=0$ if X is curve?

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up vote 2 down vote accepted

The answer is yes for a non-compact Riemann surface $H^1(X, \mathcal K_{x_1,x_2})=0$ .

The key is the exact sequence of sheaves on $X$:$$0\to \mathcal K_{x_1,x_2} \to \mathcal K \xrightarrow {truncate } \mathcal Q_1\oplus \mathcal Q_2\to 0$$ where $\mathcal Q_i$ is the sky-scraper sheaf at $x_i$ with fiber the Laurent tails (locally of the form $\sum_{j=0}^Na_jz^{-j}$).
Taking cohomology we get a long exact sequence $$\cdots \mathcal K(X) \xrightarrow {\text {truncate}} \mathcal Q_1(X) \oplus \mathcal Q_2(X)\to H^1(X, \mathcal K_{x_1,x_2})\to H^1(X, \mathcal K) \to \cdots $$
The vanishing of the cohomology group $H^1(X, \mathcal K_{x_1,x_2})$ then follows from the two facts:
1) $H^1(X, \mathcal K)=0$
2) The morphism $ \mathcal K(X) \xrightarrow {\text {truncate}} \mathcal Q_1(X) \oplus \mathcal Q_2(X)$ is surjective because of the solvability of the Mittag-Leffler problem on a non-compact Riemann surface.

For a compact Riemann surface of genus $\geq1$ the relevant Mittag-Leffler problem is not always solvable, so that we have $H^1(X, \mathcal K_{x_1,x_2})\neq 0$ (however for the Riemann sphere $H^1(\mathbb P^1, \mathcal K_{x_1,x_2})=0$)

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