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Mar 1, 2016 at 17:11 vote accept Ron
Feb 29, 2016 at 22:48 answer added Sándor Kovács timeline score: 4
Feb 28, 2016 at 18:05 comment added Ray Hoobler But that is not enough to answer the question since $H^0(X-S,R^1 j_*\mathcal{O}_{X-S}) = Ker[H^0(X-S,R^2j_*\mathcal{O}_{X-S}) \rightarrow H^2(X,\mathcal{O}_X)]$.
Feb 28, 2016 at 7:46 comment added abx $R^1j_*\mathcal{O}_{X\smallsetminus S}$ is a sheaf supported on $S$, with fiber at $s\in S$ the group $H^2_{\{s\} }(\mathcal{O}_X)$, which is nonzero. See SGA2, §I and II.
Feb 28, 2016 at 7:33 history edited Ron CC BY-SA 3.0
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Feb 28, 2016 at 6:11 history edited Ron CC BY-SA 3.0
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Feb 28, 2016 at 5:37 comment added Ron @abx I am a bit confused. I know this fact. I do not understand, whether this proves or disproves the question. Could you please elaborate a bit more?
Feb 28, 2016 at 3:56 comment added abx $j_*\mathcal{O}_{X\smallsetminus S}$ is isomorphic to $\mathcal{O}_X$.
Feb 28, 2016 at 2:29 comment added Ron @dhy I have edited the question.
Feb 28, 2016 at 2:29 history edited Ron CC BY-SA 3.0
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Feb 28, 2016 at 2:13 comment added dhy Is there any reason to believe something like this should be true? I don't see why there should be any natural map between the two and it's false for $S$ empty...
Feb 28, 2016 at 1:41 history asked Ron CC BY-SA 3.0