Top cohomology of resolution of singularities - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T21:05:43Z http://mathoverflow.net/feeds/question/107881 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/107881/top-cohomology-of-resolution-of-singularities Top cohomology of resolution of singularities MathStud 2012-09-23T04:53:17Z 2012-09-24T21:31:37Z <p>Let $X$ be a projective variety over $\mathbb C$ of dimension $n$. Let $\tilde{X} \to X$ be a resolution of singularities. Suppose that $H^n(\tilde{X}, \mathcal O_{\tilde{X}}) = 0$. What can we say about $H^n(X, \mathcal O_X)$? When is it $0$? </p> http://mathoverflow.net/questions/107881/top-cohomology-of-resolution-of-singularities/107884#107884 Answer by Karl Schwede for Top cohomology of resolution of singularities Karl Schwede 2012-09-23T06:04:02Z 2012-09-24T14:02:50Z <p>First, let me point out that $H^i(\tilde{X}, O_{\tilde{X}}) \cong H^i(X, O_X)$ if $X$ has rational singularities for all $i > 0$.</p> <p>Indeed, if $X$ has rational singularities if and only if </p> <ol> <li>$R^j \pi_* O_{\tilde X} = 0$ for $j > 0$ and </li> <li>$\pi_* O_{\tilde X} = O_X$.</li> </ol> <p>It immediately follows from the Leray spectral sequence that $$H^i(\tilde{X}, O_{\tilde{X}}) \cong H^i(X, O_X)$$ for all $i \geq 0$. </p> <p>In fact, for any Cartier divisor $D$ on $X$, the same argument implies that $$H^i(\tilde{X}, O_{\tilde{X}}(\pi^* D) ) = H^i(X, O_X(D))$$ for any $i \geq 0$ since the projection formula can be applied in the cases of 1. and 2. above.</p> <hr> <p>Now, without rational singularities, you can run into trouble. For example, suppose that $X$ is a normal Cohen-Macaulay variety with an isolated singularity $x \in X$ that is not rational. Consider the exact triangle in the derived category: $$O_X \to R \pi_* O_{\tilde X} \to C \xrightarrow{+1}$$ Because $X$ is a normal Cohen-Macaulay, and has an isolated non-rational singularity, we know $C = M[-n+1]$ is a nonzero module supported at $x \in X$ (shifted over by $n-1$). See Lemma 3.3 in <em>Rational, Log Canonical, Du Bois Singularities: On the Conjectures of Kollár and Steenbrink</em> by Sándor Kovács. </p> <p>Then we have the following exact sequence by taking (hyper)cohomology $$0 \to H^{n-1}(X, O_X) \to {{H}}^{n-1}(\tilde{X}, O_{\tilde{X}}) \to {\mathbb{H}^{n-1}}(X, C) \to H^n(X, O_X) \to H^n(\tilde{X}, O_{\tilde{X}}) \to 0$$ where the two end points are zero since $C = M[-n+1]$ an Artinian module with a shift. On the other hand, $\mathbb{H}^{n-1}(X, C) = H^0(X, M) \neq 0$ for the same reason. </p> <p>Now, if $\tilde{X}$ is for example Fano and we are in characteristic zero, then $$H^i(\tilde{X}, O_{\tilde{X}}) = H^i(\tilde{X}, O_{\tilde{X}}(K_X-K_X)) = 0$$ by Kodaira vanishing for $i > 0$. But then $H^n(X, O_X) \neq 0$ from the exact sequence. </p> <p>Beyond the Fano case, you might luck out of course, but I don't see any reason why it would hold in general.</p> http://mathoverflow.net/questions/107881/top-cohomology-of-resolution-of-singularities/108008#108008 Answer by Piotr Achinger for Top cohomology of resolution of singularities Piotr Achinger 2012-09-24T21:31:37Z 2012-09-24T21:31:37Z <p>A very easy counterexample: let $X$ be a nodal cubic in $\mathbb{P}^3$. Then the resolution of singularities of $X$ is $\bar X = \mathbb{P}^1$, so $H^1(\bar X, \mathcal{O}_{\bar X}) = 0$. On the other hand, the exact sequence of sheaves on $\mathbb{P}^2$ $$0\to \mathcal{O}(-3)\to \mathcal{O} \to \mathcal{O}_X \to 0$$ shows that $H^1(X, \mathcal{O}_X)$ is one-dimensional.</p>