Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $H_{*}(M/G;R)$ ?

share|improve this question
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). –  Chris Gerig Jun 13 '13 at 4:54
@Chris, Does Ken Brown's book deal also with Lie groups? –  Fernando Muro Jun 13 '13 at 7:50
Compact discrete Lie groups ;-) –  Chris Gerig Jun 13 '13 at 15:28
I was guessing. –  Fernando Muro Jun 13 '13 at 23:21

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.