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Sep 23, 2019 at 12:20 vote accept Arnaud Maret
Sep 13, 2019 at 4:54 answer added Qiaochu Yuan timeline score: 3
Aug 7, 2019 at 20:32 comment added Ben Wieland There is such a thing as group homology. There is a perfect pairing $H^2(G;\mathbb R)\otimes H_2(G;\mathbb R)\to \mathbb R$. Thus integration is an element of group homology, just as it is an element of the homology of a surface. See also Poincaré duality groups and Bieri-Eckmann duality groups.
Aug 6, 2019 at 18:32 comment added Bombyx mori It is not an answer, just some vague idea. If I know a proof, I would have written down an answer already. The question is a good one. One thing I want to point out is that by taking the "integral" technically you are evaluating the residue. And you do not need de Rham cohomology to define the residue. Since formally hochschild homology is an analog of group cohomology, there is a similar residue formula in literature. But this is all I know.
Aug 6, 2019 at 14:03 comment added Arnaud Maret Thanks for your answer! Can you maybe elaborate a little bit (or share a link if you have a reference to suggest). Cheers ;)
Aug 5, 2019 at 21:02 comment added Bombyx mori I think you would have to use cyclic homology/hochschild homology, and the map you are talking about is essentially the residue map.
S Aug 5, 2019 at 11:55 history suggested Ali Taghavi
I add a tag.
Aug 5, 2019 at 11:55 review Suggested edits
S Aug 5, 2019 at 11:55
Aug 5, 2019 at 10:21 history asked Arnaud Maret CC BY-SA 4.0