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Dec 4, 2017 at 13:54 comment added Steve Costenoble @DenisNardin: Sorry, I was thinking of Bredon cohomology. Yes, you're right.
Dec 4, 2017 at 13:54 comment added Denis Nardin I said, Borel cohomology, which is just the cohomology of the homotopy orbits (yes, that is also $RO(G)$-graded)
Dec 4, 2017 at 13:41 comment added Steve Costenoble Related, but actually rather different. $RO(\mathbb Z/2) \cong \mathbb Z \times \mathbb Z$ and the $RO(\mathbb Z/2)$-graded cohomology of a point is quite a bit more complicated than this calculation.
Dec 4, 2017 at 13:38 vote accept Steve Costenoble
Dec 4, 2017 at 12:54 answer added Matthias Wendt timeline score: 9
Dec 4, 2017 at 8:22 comment added Denis Nardin This looks like the $RO(G)$-grading on the Borel cohomology of the point
Dec 4, 2017 at 3:33 history asked Steve Costenoble CC BY-SA 3.0