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Apr 8, 2016 at 4:41 review Close votes
Apr 16, 2016 at 3:01
Apr 8, 2016 at 4:26 comment added abx Assuming you mean $\mathcal{O}_{\mathbb{D}}(D)$: any divisor on $\mathbb{D}$ is principal, so $\mathcal{O}_{\mathbb{D}}(D)$ is isomorphic to $\mathcal{O}_{\mathbb{D}}$, and of course its higher cohomology is trivial.
Apr 8, 2016 at 3:54 comment added Will Sawin You mean $H^1(\mathbb D, \mathcal O_{\mathbb D}(D))$, right? Or what do you mean?
Apr 7, 2016 at 23:16 history asked user76513 CC BY-SA 3.0