Let $M$ be a smooth $G-$manifold, where $G$ is a compact Lie group. From the result of S.Illman that $M$ admits an equivariant triangulation. Moreover, we can construct the equivariant singular homology $H^{G}_{*}(M;k)$ where $k$ is a covariant coefficient system for $G$ ovr a ring $R$. For every standart equivariant $n-$simplex $\triangle_{n}(K_{0},...,K_{n})$ and $G-$map $$T:\triangle_{n}(K_{0},...,K_{n})\longrightarrow M$$ the induced map by the $G-$action is $$u:\triangle_{n}\longrightarrow M/G.$$ What is the relation between the equivariant singular homology group $H^{G}_{*}(M;k)$ and the homology group $H_{*}(M/G;R)$ ?
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1$\begingroup$ I want to guess that this is an equivalent construction to what is ordinary referred to as equivariant homology, $H_\ast^G(M)$ being the homology of the Borel construction. Then when $G$ acts freely, they are the same; otherwise, there is a spectral sequence relating them (check out Ken Brown's Cohomology of Groups for this construction). $\endgroup$– Chris GerigCommented Jun 13, 2013 at 4:54
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$\begingroup$ @Chris, Does Ken Brown's book deal also with Lie groups? $\endgroup$– Fernando MuroCommented Jun 13, 2013 at 7:50
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$\begingroup$ Compact discrete Lie groups ;-) $\endgroup$– Chris GerigCommented Jun 13, 2013 at 15:28
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$\begingroup$ I was guessing. $\endgroup$– Fernando MuroCommented Jun 13, 2013 at 23:21
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