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Let $X$ be a manifold acted on by a Lie group $G$. The $G$-equivariant cohomology of $X$ with coefficients in a ring $\mathcal{R}$ is defined as the cohomology ring $$ H_G^*(X; \mathcal{R}) := H^*(X_G; \mathcal{R}), $$ where $X_G := (X \times EG) / G$ is the homotopy quotient, $EG \to BG = EG / G$ being the universal principal bundle of the group $G$.

The natural projection $$ X_G \to BG $$ obtained by collapsing the elements of $X$ gives rise to a ring homomorphism $$ H^*(BG; \mathcal{R}) \to H_G^*(X; \mathcal{R}), $$ or equivalently to an action of the ring $H^*(BG; \mathcal{R})$ on the equivariant cohomology $H_G^*(X; \mathcal{R})$ of $X$.

Suppose now that $G$ acts freely on $X$. In this case, the cohomology groups $H_G^*(X; \mathcal{R})$ and the singular cohomology $H^*(X/G; \mathcal{R})$ of $X/G$ agree.

My question is the following: how does $H^*(BG; \mathcal{R})$ act on the singular cohomology of $X/G$ (if it can make things easier, one might take $\mathcal{R}=\mathbb{C}$)?

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    $\begingroup$ Since there is already an answer, this comment may be superfluous, but you can combine your first statement and your second statement to see how one gets the action. $\endgroup$ – user43326 Jan 3 at 19:30
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$EG$ is also the universal free $G$-space, meaning that, if $X$ is a free $G$-space (let's assume of the $G$-homotopy type of a $G$-CW complex), there is, up to $G$-homotopy, a unique $G$-map $X\to EG$. Taking quotients, you get a map $X/G \to BG$ which induces the action of $H^\ast(BG)$ on $H^\ast(X)$ with any coefficients.

Edited to add: I really should have mentioned that $X/G \to BG$ is the classifying map of the bundle $X\to X/G$. I think this makes it a little less mysterious where the map comes from.

Second edit to add (simultaneously with Mike Miller's comment): To give an example where the action is nontrivial, let $G = {\mathbb Z}/2$ and $X = S^n$ with $G$ acting as $-1$. Then $X/G = {\mathbb R}P^n$, $BG = {\mathbb R}P^\infty$, and $X/G \to BG$ is the inclusion. $H^\ast(BG;{\mathbb Z}) = {\mathbb Z}[x]$ then acts in the obvious, nontrivial way on $H^\ast(X/G;{\mathbb Z}) = {\mathbb Z}[x]/x^{n+1}$.

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  • $\begingroup$ Thank you for your answer. I’m not sure I understand. What exactly is the action? My guess is that $H^*(BG)$ acts by $0$ on $H^*(X/G)$. Is it true (or when is it true)? $\endgroup$ – BrianT Jan 4 at 1:22
  • $\begingroup$ I'm not sure what you’re asking. It’s the same as the action on $X_G$ under the identification you gave. $\endgroup$ – Steve Costenoble Jan 4 at 13:27
  • $\begingroup$ @BrianT That is certainly not true. Try an example, like $X = S^3$ with the circle action coming from the Hopf fibration. That will be true if $X \cong G \times (X/G)$ as a $G$-bundle so that the map $X/G \to BG$ is null-homotopic. There may be some bizarre non-trivial cases where all positive degree elements of the cohomology ring act as 0, but I don't know them and expect this to be very very rare. $\endgroup$ – Mike Miller Jan 4 at 18:01
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    $\begingroup$ Let's take the case of $X=EG$. Then the action of $H^*(BG)$ on $H^*(X/G)$ is just the ring multiplication. $\endgroup$ – user43326 Jan 4 at 18:59
  • $\begingroup$ Thank you for your answers. In an article of A. Givental math.berkeley.edu/~giventh/papers/tor.pdf in the proof proposition $6.3$ (last lines of p.46) it is stated that the coefficient algebra $H_{T^k}^*(pt)$ acts trivially (by $0$ if I understand well the equation below the statement) on singular cohomology, where $T^k$ is a torus of dimension $k$ acting on certain sublevel sets of $\mathbb{C}^n$. Does someone understand this statement? $\endgroup$ – BrianT Jan 5 at 7:43

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