most recent 30 from http://mathoverflow.net 2013-05-22T05:01:00Z http://mathoverflow.net/feeds/question/56186 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/56186/the-vanishing-of-the-2nd-plurigenus-of-a-sextic-threefold HNuer 2011-02-21T16:30:50Z 2011-02-21T18:30:02Z

<p>I'm reading a recent preprint by Beauville on the nonrationality of a specific sextic threefold $X$ which is a complete intersection of a quadric and a cubic in $\mathbb P^5$. At some point he uses that $H^0(X,\Omega^2)=0$, and I was having trouble figuring out why that was. It occurred to me that it might use the Hodge decomposition, its symmetry, and two applications of the Lefshetz hyperplane theorem, but I couldn't get it to work. Anyone know a quick answer?</p>

http://mathoverflow.net/questions/56186/the-vanishing-of-the-2nd-plurigenus-of-a-sextic-threefold/56194#56194 Johannes Nordström 2011-02-21T18:09:20Z 2011-02-21T18:09:20Z

<p>$h^{q,0}(X) = h^{0,q}(X) = 0$ for any Fano $X$ and $q > 0$, because $-K_X$ ample implies $$ H^q(X, \Omega^0) = H^q(X, K_X \otimes (-K_X)) = 0 $$ by Kodaira vanishing.</p>

http://mathoverflow.net/questions/56186/the-vanishing-of-the-2nd-plurigenus-of-a-sextic-threefold/56196#56196 diverietti 2011-02-21T18:30:02Z 2011-02-21T18:30:02Z

<p>What you are asking for is not the second plurigenus: the second plurigenus is $h^0(X,2K_X)$. So do you need the vanishing of the second plurigenus or of the space of global holomorphic two forms?</p> <p>If you need just the second plurigenus, then this is very easy since by adjunction $K_X\simeq\mathcal O_X(−1)$ and so $h^0(X,2K_X)=h^0(X,\mathcal O_X(−2))=0$ since $\mathcal O_X(−2)$ is negative.</p> <p>On the other hand, if you need the vanishing of global holomorphic two-forms, this is quite easy, too. Just observe that since $-K_X\simeq \mathcal O_X(1)$, then $-K_X$ is positive. So, you get by Kodaira's vanishing $$ H^q(X,K_X-K_X)=H^q(X,\mathcal O_X)=0,\quad q\ge 1. $$ But now, by Dolbeault's isomorphism, $H^q(X,\mathcal O_X)\simeq H^{0,q}(X,\mathbb C)$. By the Hodge symmetry $h^{0,2}(X,\mathbb C)=h^{2,0}(X,\mathbb C)=h^0(X,\Omega_X^2)$, where the last equality is again thanks to the Dolbeault isomorphism, and you are done. </p>