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Jan 24 at 11:14 comment added Matthias I see, thank you for pointing this out. So is it true that $Ext^1(\Omega_C, \mathcal{O}_C)$ is isomorphic to $H^1(\omega_C^\vee)$ then?
Jan 23 at 21:49 comment added Jason Starr The group $\text{Ext}^1_{\mathcal{O}_C}(\Omega_C,\mathcal{O}_C)$ does not equal $H^1(C,\textit{Hom}_{\mathcal{O}_C}(\Omega_C,\mathcal{O}_C)).$
Jan 23 at 11:31 history edited YCor CC BY-SA 4.0
removed capitals from title
Jan 23 at 11:30 history asked Matthias CC BY-SA 4.0