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Aug 21, 2020 at 4:14 comment added Devlin Mallory To see $R^1 \sigma_*\mathcal O_{\hat X}(-F)=0$: take $0\to \mathcal O_{\hat X}(-F)\to \mathcal O_{\widehat X}\to \mathcal O_F\to 0$, push it forward, and use that $\sigma_* \mathcal O_{\widehat X}=\mathcal O_X\to \sigma_* \mathcal O_F=\mathcal O_x$ is surjective and $R^1\sigma_* \mathcal O_{\hat X}=0$.
Aug 21, 2020 at 4:12 comment added Devlin Mallory I think the following works: I think $R^1 \sigma_*\mathcal O_{\widehat X}(-F)=0$, so by the Leray--Serre spectral sequence and projection formula we have that the thing you're trying to show is 0 is equal to $H^1(E\otimes \sigma_*\mathcal O_{\widehat X}(-F)\otimes L^k)$; now using Serre vanishing on $X$ this is 0 for $k$ large.
Aug 20, 2020 at 18:20 history edited YCor CC BY-SA 4.0
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Aug 20, 2020 at 16:29 comment added Carlos Esparza This question was migrated from math.SE
Aug 20, 2020 at 16:29 history asked Carlos Esparza CC BY-SA 4.0