How is the equivariant cohomology of a space related to the cohomology of the corresponding associated bundle

Question

Let $X$ be a manifold with a left $G$-action, and let $\Sigma$ be a Riemann surface. How is the equivariant cohomology $H^*_G(X)$ of $X$ related to the de Rham cohomology of the associated bundle $H^*(P\times_GX)$, where $P$ is a $G$-bundle over $\Sigma$?

The following is my attempt at a solution. We know that $H^*_G(X)=H^*(EG\times_GX)$, where $EG\rightarrow BG$ is the universal or platonic $G$-bundle, with the base space $BG$ (the classifying space).

Also, any $G$-bundle is a pullback of $EG$, e.g., given a map $f:\Sigma \rightarrow BG$, we have \begin{equation} P= f^*EG \end{equation} for a $G$-bundle $P\rightarrow\Sigma$.

Therefore, we find that $H^*(P\times_GX)= H^*(f^*EG\times_GX)$. If $H^*(f^*EG\times_GX) = f^*H^*(EG\times_GX)$, then we have \begin{equation} H^*(P\times_GX)= f^*H^*_G(X). \end{equation} But is it true that $H^*(f^*EG\times_GX) = f^*H^*(EG\times_GX)$? References would be appreciated.

Answer 1

I'm not even sure what you mean by $f^*H^*(EG \times_G X)$. The image of this cohomology under the pullback map? Before you make any other guesses, think very carefully about the case where the $G$ action on $X$ is trivial: $H_G(X)=H(BG)\otimes H(X)$, but $P\times_G X\cong \Sigma \times X$. The pullback $f^*$ is obviously not surjective in this case.

Instead, there's a natural fiber bundle $P\times_G X\to X/G$ (don't worry about what the latter means; it's an Artin stack, but that fact is a distraction) with fiber $P$; you have to apply the Leray-Serre spectral sequence to this bundle.

Answer 2

The de Rham cohomology of the associated bundle $H^∗(P×_GX)$ is indeed the pullback of the equivariant cohomology $H^∗_G(X)$. See the commutative diagram (2.10) in 'Equivariant Cohomology and Gauged Bosonic $\sigma$-models' by Figueora-O'Farrill and Stanciu (https://arxiv.org/pdf/hep-th/9407149.pdf).