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Nov 13, 2011 at 17:35 comment added Will Sawin I meant to say $X - X^G$.
Nov 13, 2011 at 17:34 comment added Will Sawin I don't know about this specific cohomology theory, but does it have a long exact sequence property that could answer this? I would expect that this would happen when $X \ X^G$, or something like it, has trivial homology.
Nov 13, 2011 at 15:12 answer added Alex timeline score: 0
Nov 13, 2011 at 13:43 comment added Allen Knutson Incidentally, the definition of equivariant homology as the ordinary homology of the Borel mixing space is not a very nice one, e.g. the module structure over equivariant cohomology is locally nilpotent rather than free. Michel Brion has a (as always very nice) paper defining something that, for compact oriented manifolds, has a Poincar\'e duality to equivariant cohomology. projecteuclid.org/…
Nov 13, 2011 at 13:27 answer added Johannes Ebert timeline score: 2
Nov 13, 2011 at 13:19 history edited Johannes Ebert CC BY-SA 3.0
LaTeX fixed
Nov 13, 2011 at 12:48 history asked user2529 CC BY-SA 3.0