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This is basically the last step of problem 5.3.7 in Huybrechts' Complex Geometry.

Let $X$ be a complex manifold, $x \in X$, $E$ a holomorphic vector bundle on $X$ and $L$ a positive line bundle. Denote by $ \hat{X} = \operatorname{Bl}_{\{x\}} X$ the blow-up of $X$ at $x$ and let $\sigma: \hat{X} \to X$ be the projection map and $F = \sigma^{-1}(x)$ the exceptional divisor. I'm trying to show that $$ H^1\left(\hat{X}, \sigma^* E \otimes \mathcal{O}_{\hat{X}} \left(-F\right) \otimes \sigma^*L^k\right) = 0 $$ for $k$ large enough. Any ideas?

This is of course almost Serre's vanishing theorem: $\sigma^*L$ is positive for every tangent vector except those of $TF$ and one can show that $\mathcal{O}_{\hat{X}} \left(-F\right) \otimes \sigma^*L^k$ is positive for $k$ large enough, but the problem is that we can't take powers of $\mathcal{O}_{\hat{X}} \left(-F\right)$...

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  • $\begingroup$ This question was migrated from math.SE $\endgroup$
    – Carlos Esparza
    Commented Aug 20, 2020 at 16:29
  • $\begingroup$ 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. $\endgroup$
    – Devlin Mallory
    Commented Aug 21, 2020 at 4:12
  • $\begingroup$ 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 $R^1\sigma_* \mathcal O_{\hat X}=0$. $\endgroup$
    – Devlin Mallory
    Commented Aug 21, 2020 at 4:14

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