# Equivariant Coefficient ring action on singular cohomology

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 $$\mathcal{R}=\mathbb{C}$$)?

• 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. – user43326 Jan 3 at 19:30

$$EG$$ 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 $$H^\ast(X/G;{\mathbb Z}) = {\mathbb Z}[x]/x^{n+1}$$.
• 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)? – BrianT Jan 4 at 1:22
• I'm not sure what you’re asking. It’s the same as the action on $X_G$ under the identification you gave. – Steve Costenoble Jan 4 at 13:27
• @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. – Mike Miller Jan 4 at 18:01
• Let's take the case of $X=EG$. Then the action of $H^*(BG)$ on $H^*(X/G)$ is just the ring multiplication. – user43326 Jan 4 at 18:59
• 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? – BrianT Jan 5 at 7:43