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?
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$\begingroup$ Here's the argument I was thinking of which uses Lefschetz, which I should seen immediately: By Lefschetz $H^{p,q}(\mathbb P^5)$ is isomorphic to $H^{p,q}(C)$ for $p+q<4$, where $C$ is the cubic hypersurface. Applying this again by cutting with the quadric $Q$ to get our $X$, we see that $H^{p,q}(X)$ is isomorphic to $H^{p,q}(C)$ for $p+q<3$. Since the $h^{p,q}(\mathbb P^n)=0$ for $p\neq q$, this gives our result. I suppose since one proves Lefschetz with Kodaira vanishing, it's all the same really, but this is what I had in mind originally. $\endgroup$– HNuerFeb 21, 2011 at 20:36
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$\begingroup$ Of course this is another possibility. The point is that this is a more general property of Fano manifolds... Of course all this needs $X$ to be smooth... $\endgroup$– diveriettiFeb 21, 2011 at 21:55
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2 Answers
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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?
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.
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.
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$\begingroup$ Do you need any explanation about how to use adjunction formula to find the expression of the canonical bundle in term of the tautological bundle of the projective space? $\endgroup$ Feb 21, 2011 at 19:31
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$\begingroup$ No, I understand why $K_X=\mathcal O_X(-1)$. $\endgroup$– HNuerFeb 21, 2011 at 20:27
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$h^{q,0}(X) = h^{0,q}(X) = 0$ 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.
answered Feb 21, 2011 at 18:09