We know that $H^2(\mathcal{O}_{\mathbb{P}^3})=0$. I am looking for blow-ups $$\pi:X \to \mathbb{P}^3$$ such that $X$ is non-singular and $H^2(\mathcal{O}_X)>0$. Of course, if we blow-up along smooth curves or points, then $H^2(\mathcal{O}_X)=0$. I will be grateful if somebody can suggest a method or a reference.
$\begingroup$
$\endgroup$
6
-
4$\begingroup$ If you are in char. 0, then $\dim H^2(\mathcal{O})$ is a birational invariant. $\endgroup$– Donu ArapuraCommented May 14 at 7:36
-
$\begingroup$ @DonuArapura Thank you! Could you suggest a proof or a reference. I was thinking of using the long exam sequence with support and using the result that $H^i_E(\mathcal{O}_X)=0$ for $i \ge 2$, when $E$ is the exceptional divisor associated to $\pi$. Is this in the right direction? $\endgroup$– user45397Commented May 14 at 8:04
-
2$\begingroup$ Any birational map can be factorised into a sequence of blow up and blow downs in smooth centres. Thus the argument you propose handles the general case. $\endgroup$– Daniel LoughranCommented May 14 at 8:57
-
$\begingroup$ @DanielLoughran Thank you! I understand. $\endgroup$– user45397Commented May 14 at 9:02
-
1$\begingroup$ It is also true in characteristic $p > 0$, by a result of Chatzistamatiou and Rülling. But in characteristic 0, it is a direct consequence of Hodge symmetry and the birational invariance of $H^0(X,\Omega_X^2)$ for smooth projective varieties. (This last result is Hartshorne, Chapter II, Exercise 8.8.) $\endgroup$– R. van Dobben de BruynCommented May 14 at 14:57
|
Show 1 more comment