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Nov 8, 2013 at 17:19 comment added abx Then the result depends on the dependence relations of the classes $[D_i]$ in $H^1(X,\Omega^1_X)$. If they are linearly independent, it is still the case that $H^0(X,\Omega^1_X(\log D))=H^0(X,\Omega^1_X)$.
Nov 8, 2013 at 13:43 comment added resid What happens if there are several $D_i$?
Nov 8, 2013 at 13:29 comment added abx In the case there is just one $D_i$, yes -- and therefore every such section comes from $H^0(X,\Omega ^1_X)$.
Nov 8, 2013 at 13:16 comment added resid Are you saying that every global section of $\Omega^1_X(\log D)$ has zero residues at all $D_i$?
Nov 8, 2013 at 12:51 comment added abx That doesn't change the problem : the map $\mathrm{Res}: H^0(X,\Omega ^1_X(\log D))\rightarrow H^0(D,\mathcal{O}_D)$ is still the zero map, for the same reason.
Nov 8, 2013 at 12:27 comment added resid Oh, thank you for this observation. What happens if we assume $H^0(X, \Omega^1_X) \neq 0$?
Nov 8, 2013 at 9:57 history answered abx CC BY-SA 3.0