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Zhaoting Wei
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How does the kernel of the map $\Omega^{\bullet}(X)\rightarrow \Omega^{\bullet}(G\times X)$ relate to equivariant cohomology?

This question may be trivial for experts. Consider a (compact, connected) smooth manifold $X$ and a (compact connected) Lie group $G$ act on $X$. Then we have the action map $$ \mu: G\times X\rightarrow X $$ and the projection map $$ p: G\times X\rightarrow X. $$

We can consider the differential forms $\Omega^{\bullet}(X)$ as well as $\Omega^{\bullet}(G\times X)$. The maps $\mu$ and $p$ give the pull-back map $$ \mu^*: \Omega^{\bullet}(X)\rightarrow \Omega^{\bullet}(G\times X) $$ and $$ p^*: \Omega^{\bullet}(X)\rightarrow \Omega^{\bullet}(G\times X). $$

Now let's consider the map $\delta$ defined to be the difference of $\mu^*$ and $p^*$, i.e. $$ \delta=\mu^*-p^*: \Omega^{\bullet}(X)\rightarrow \Omega^{\bullet}(G\times X). $$ It is easy to easy that $\ker \delta\in (\Omega^{\bullet}(X))^G$. Moreover, for any vector field $\Theta$ on $X$ generated by the $G$-action, we have $\ker \delta\subset \ker\iota_{\Theta}$. Hence in the case that $G$ acts on $X$ freely, we know $$ \ker\delta=\Omega^{\bullet}(G\X). $$

$\textbf{My first question}$ is: if the $G$-action is not free, could we describe the equivariant cohomology $H^{\bullet}_G(X)$ in terms of $\ker\delta$?

My second question may be a little bit vague: for the action $G$ on X we can construct a simplicial manifold $$ \ldots G\times G\times X\Rrightarrow G\times X\rightrightarrows X $$ by actions, multiplications and projections and hence we get a sequence of differential forms $$ \Omega^{\bullet}(X)\rightarrow \Omega^{\bullet}(G\times X)\rightarrow \Omega^{\bullet}(G\times G\times X)\ldots $$

$\textbf{My second question}$ is: is there any reference for the study of the above sequence?

Zhaoting Wei
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