Skip to main content
6 events
when toggle format what by license comment
May 2, 2011 at 16:12 answer added Sándor Kovács timeline score: 4
May 2, 2011 at 16:07 answer added Karl Schwede timeline score: 4
May 2, 2011 at 14:37 comment added Torsten Ekedahl You get a map $H^0(X,K_X)\to H^0(\hat X,K_{\hat X})$ by pulling back forms. A map in the other direction is by restriction $H^0(\hat X,K_{\hat X})\to H^0(\hat X\setminus E,K_{\hat X\setminus E})$, the isomorphism $H^0(\hat X\setminus E,K_{\hat X\setminus E})=H^0(X\setminus\{x\},K_{X\setminus\{x\}}$ and then the fact that (if $n\ge 2$, $n=1$ is trivial) $H^0(X\setminus\{x\},K_{X\setminus\{x\}}=H^0(X,K_X)$.
May 2, 2011 at 13:41 answer added Francesco Polizzi timeline score: 4
May 2, 2011 at 13:31 answer added Sasha timeline score: 4
May 2, 2011 at 13:09 history asked Lucas Kaufmann CC BY-SA 3.0