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May 23, 2013 at 22:18 comment added Piotr Achinger I mean $E$, not $E(t)$, because if we prove vanishing of $H^1$ for all rank two vector bundles, then in particular we have it for $E(t)$.
May 23, 2013 at 15:42 comment added Fei YE I guess that you mean that $H^1(\mathbb{P}^n, E(t))=0$ for any $t\in \mathbb{Z}$.
Aug 13, 2012 at 22:45 history answered Piotr Achinger CC BY-SA 3.0