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Aug 24, 2015 at 15:26 comment added Karl Schwede True. I guess it depends on exactly what statement you are looking for.
Aug 24, 2015 at 13:18 vote accept CommunityBot
Aug 24, 2015 at 4:02 comment added grghxy Strictly speaking, one has to know how the Grothendieck duality isomorphism is defined in order to affirm that the concrete isomorphism ${\rm{Ext}}^i(F,\omega) \simeq {\rm{H}}^{d-i}(X,F)^{\vee}$ that you obtain in the end really coincides (up to a universal sign depending only on $i$ and $d$) with the map built from Ext-pairings and a (suitable) trace map ${\rm{H}}^d(X,\omega) \rightarrow k$. So it is necessary to "look under the hood" a bit.
Aug 24, 2015 at 3:41 history answered Karl Schwede CC BY-SA 3.0