1
$\begingroup$

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.