Let $A$ be a sheaf such that $$A(U) = \{ f \in \mathbb M(U): f \in \mathbb{O}(U \backslash\{p_1,\ldots, p_n\}) \ \mbox{with at worst a simple pole at}\ p_i \} $$ where $\mathbb M(U)$ means the set of meromorphic functions on $U$ and $\mathbb O(U)$ means the set of holomorphic functions on $U$. What is the cohomology of $A$ on the Riemann sphere?
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$\begingroup$ In this context, $\mathcal{O}$ is more common notation than $\mathbb{O}$. $\endgroup$– Michael AlbaneseCommented Mar 18, 2018 at 22:33
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The cohomology of these sheaves are always 0. This follows for example from the case $\{p_i\} = \emptyset$ by induction and using the sequences $A_{p_1,..,p_{n-1}} \to A_{p_1,...,p_n} \to \mathbb{C}_{p_n}$. In general, the isomorphism class of this sheaf depends only on the number of points, and a representative of the class for $n$ points is denoted $\mathcal{O}(n)$. The first cohomology of these sheaves is $0$ for $n > -2$ and of dimension $-n - 1$ for $n \le -2$ (for negative $n$ this is given by imposing zeros instead of allowing poles). Finally, there are no higher cohomologies since $\mathbb{P}^1$ is one dimensional.
answered Mar 18, 2018 at 22:09
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$\begingroup$ Any book on Riemann surfaces will give you the answer. Your sheaf is usually denoted $\mathcal{O}_{\mathbb{P}^1}(p_1+\ldots +p_n)$. As S. carmeli says, it is isomorphic to $\mathcal{O}_{\mathbb{P}^1}(n)$, and has no higher cohomology. This is completely standard, and not appropriate to MO. $\endgroup$– abxCommented Mar 19, 2018 at 2:41
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$\begingroup$ As mensioned above, this is very standard computation. What I said is that you have a sub-sheaf of functions that are holomorphic at $p_n$ and that the quotient is a skyscraper sheaf at $p_n$, i.e. gives $\mathbb{C}$ to open sets containing $p_n$ and $0$ otherwise. Then you can write down a long exact sequence of cohomologies and compute by induction. It requires some analysis of the maps on the $H^0$-s, but it is not hard to see that $H^0(\mathbb{P}^1\mathcal{O}(p_1 + ... + p_n) \to H^0(\mathbb{P}^1,\mathbb{C}_p)$ is onto. $\endgroup$ Commented Mar 19, 2018 at 6:42