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?



The vanishing of the 2nd plurigenus  of a sextic threefold
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. 


