Simple Equivariant homology [no borel-Moore] - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T05:37:49Z http://mathoverflow.net/feeds/question/93958 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/93958/simple-equivariant-homology-no-borel-moore Simple Equivariant homology [no borel-Moore] laerne 2012-04-13T13:39:42Z 2012-04-13T15:47:45Z <p>Hey. I'm working with Bredon's equivariant cohomology. At some point I need to compute the $4$th equivariant cohomology group of $S^1 \x D^3$ relatively to its boundary for the antipodal action of $\mathbb{Z}_2$.</p> <p>I found a paper that just use "equiviriant poincaré duality" to bring the problem to compute the $0$th equivariant homology which is said to be $\mathbb{Z}_2$. This sounds trivial, but Bredon does not define what equivariant homology is, just cohomology, hence he does not talk of Poincare duality either (At least not in the 30 first pages of his paper "Equivariant Cohomology Theories" I use.).</p> <p>I'm trying to search for a nice definition of equivariant homology, but every paper I find use concept I don't master well or at all (Vector bundles, groupoids, Borel-Moore Homology). I also tried to design my own definition of equivariant homology, simply letting $H_0$ be the quotient of equivariant chains $H_0 = C_0^{\mathbb Z_2}/\partial(C_1^{\mathbb Z_2})$ with $C_i^{\mathbb Z_2} = \lbrace c \in C_i| (1+\mathbb Z) \curvearrowright c = c \rbrace$. But using the relative exact sequence of chain, one finds $C_0^{\mathbb Z_2}(S^1\times D^3, S^1 \times S^2)$ is a subgroup of $C_0(S^1\times D^3, S^1 \times S^2)$, which is $0$, so $H_0^{\mathbb Z_2}(S^1\times D^3, S^1 \times S^2) \approx 0$, not $\mathbb Z_2$ !</p> <p>Many thanks for any help !</p> http://mathoverflow.net/questions/93958/simple-equivariant-homology-no-borel-moore/93968#93968 Answer by Peter May for Simple Equivariant homology [no borel-Moore] Peter May 2012-04-13T15:47:45Z 2012-04-13T15:47:45Z <p>I'm afraid this is not an easy subject to get into. There is no problem defining Bredon homology. Maybe first in print in a 1975 memoir of Soren Illman. A more recent summary is in my Equivariant homotopy and cohomology theory''. However, just that won't help you. Tautologically, Poincar\'e duality is about duality. A manifold M embedded in Euclidean space is (after adding a disjoint basepoint) Spanier-Whitehead dual to the Thom complex of its normal bundle. Equivariantly, you must start with an embedding of M into a representation, and then to understand Poincar\'e duality you must use $RO(G)$-graded Bredon homology and cohomology, which is only available for those coefficient systems that extend to Mackey functors. Even then, equivariant orientation theory is very subtle. There is a long paper by Costenoble, Waner, and myself that explains what is going on conceptually and geometrically. Unfortunately, unlike nonequivariantly, unless one soups up cohomology to deal with equivariant fundamental groupoids, orientations are not (as far as is known) definable in purely homological terms. This is a fascinating area, still in its infancy, but not an easy one. </p>