# Equivariant cohomology vs. invariant cohomology vs. cohomology of quotient space

Given a space $X$ and an action of a group $G$ on $X$, the $G$-invariant cochains with coefficients in an Abelian group $A$ define a sub-cocomplex $\mathcal{C}^{\bullet}_G$ of the cocomplex $\mathcal{C}^{\bullet}$ of cochains with coefficients in $A$. I will call the cohomology $H^{\bullet}(\mathcal{C}_G)$ the "invariant cohomology" of $X$. What is the relationship between the follwing three objects: (i) the invariant cohomology, (ii) the usual equivariant cohomology $\mathcal{H}_G^{\bullet}(X, A)$, and (iii) the cohomology of the quotient space $H^{\bullet}(X/G,A)$? Note that I am most interested in the case where $A$ is a finite Abelian group.

I think that there are always homomorphisms $$H^n(X/G,A) \to H^n(\mathcal{C}_G) \to \mathcal{H}_G^n(X,A).$$ The first map comes from pulling back the projection $X \to X/G$, and the second can be seen from the interpretation of equivariant cohomology as the total cohomology of the double cocomplex of group cochains of $G$ valued in $\mathcal{C}^{\bullet}$. But how does one characterize the kernel and image of these maps? Is there a simple statement if $X$ is contractible, say?

[Edit: regarding the relationship between $H^n(X/G,A)$ and $H^n(\mathcal{C}_G)$, please see the comments on Mark Grant's answer. The upshot is that with simplicial cochains, $H^n(\mathcal{C}_G)$ seems to depend on the choice of triangulation. But, if one defines $\mathcal{C}_G$ in a suitably triangulation-independent way, i.e. either singular cochains, or simplicial cochains on the abstract simplicial complex containing all simplex embeddings into X, then $H^n(\mathcal{C}_G)$ and $H^n(X/G,A)$ are almost certainly not the same.]

In what follows I will assume that $G$ is discrete and that $X$ is a simplicial complex with regular $G$-action (see Bredon's "Introduction to compact transformation groups", Chapter III.1). The regularity condition can be ensured by passing to the second barycentric subdivision if necessary. Under these conditions (unless I have overlooked something), the invariant cochains on $X$ are the same thing as cochains on the orbit space $X/G$.

A $k$-cochain on $X/G$ is a function $\phi$ from the $k$-simplices of $X/G$ to $A$. Let $\pi:X\to X/G$ denote the orbit projection. Then $\pi^*\phi$, which assigns to a simplex $\sigma\subseteq X$ the value $\phi(\pi(\sigma))$, is an invariant $k$-cochain on $X$. (That $\pi(\sigma)$ is a simplex of $X/G$ is ensured by regularity.)

Conversely, given an invariant $k$-cochain $\tilde\phi$ on $X$, we obtain a cochain $\phi$ on $X/G$ whose value on a simplex $\sigma\subseteq X/G$ is given by $\tilde\phi(\tilde\sigma)$, where $\tilde\sigma\subseteq X$ is any lift of $\sigma$. Regularity ensures that these lifts all differ by an element of $G$, so that this is well-defined.

Hence under these (topologically quite mild) conditions your invariant cohomology $H^*(\mathcal{C}_G)$ is the orbit space cohomology $H^*(X/G)$. You can then interpret your homomorphism $H^*(\mathcal{C}_G)\to H^*_G(X;A)$ as being induced by the map $$X_G:=EG\times_G X\to X/G$$ which collapses $EG$ to a point. When the action is free, this is a (weak) homotopy equivalence, and so $H^*(\mathcal{C}_G)\to H^*_G(X;A)$ is an isomorphism. In general, this map can be analysed using the Leray spectral sequence (see Borel's "Seminar on transformation groups", for example).

If you are interested in $G=\mathbb{Z}/p$ or $G=S^1$, you should look up "Smith-Richardson theory".

• "the invariant cochains on $X$ are the same thing as cochains on the orbit space X/G". That was what I originally thought, but I was confused by the following example. Let $X=\mathbb{R}^2$, and let $G = C_2$ act on $X$ by $x \mapsto -x$. We will work with $\mathbb{Z}/2$ coefficients. Then $H^2(\mathcal{C}_G) = \mathbb{Z}/2$ (the non-trivial class is the Poincare dual of the 0-chain which assigns 1 to the origin), whereas $X/G$ is still homeomorphic to $\mathbb{R}^2$ so $H^2(X/G) = 0$. – Dominic Else Mar 1 '17 at 16:16
• Just to be clear, by "$x \mapsto -x$" I am referring to vector $x$, or in coordinate notation $(x,y) \mapsto (-x,-y)$ – Dominic Else Mar 1 '17 at 16:23
• @Dominic: I don't see a way to triangulate the plane to make your action a $G$-simplicial complex, so I think it's excluded from my setup. Even so, I'm not a 100% confident with my answer. I'll think about it some more. – Mark Grant Mar 1 '17 at 16:44
• Doesn't a simple triangular tiling, with (0,0) at a vertex, work? Of course, then the problem seems to be that you can't represent the non-trivial G-invariant cochain as a G-invariant simplicial cochain... en.m.wikipedia.org/wiki/Triangular_tiling – Dominic Else Mar 1 '17 at 18:26
• @Dominic: Looking back at your example, you seem to be assuming that Poincaré duality holds for invariant (co)homology, to get a non-trivial $H^2(\mathcal{C_G}$. This might need checking. Or maybe you can see an invariant $2$-cocycle in the triangulation? – Mark Grant Mar 2 '17 at 7:45

If $G$ is a finite group acting by simplicial automorphisms on a simplicial complex $X$, then

$$H^{\ast}(X/G;\mathbb{Q}) = (H^{\ast}(X;\mathbb{Q}))^{G} = H^{\ast}((C^{\ast}(X;\mathbb{Q}))^G),$$

where the superscripts indicate that we are taking the invariants with respect to $G$. For a proof, see Proposition 1.1 (and its proof) in my note "The action on homology of finite groups of automorphisms of surfaces and graphs", available on my webpage here. It is definitely necessary to take $\mathbb{Q}$ coefficients here (the referenced proof will make it clear why this is the case).

• Thanks. Unfortunately the case where the coefficients are a finite Abelian group is really the one I am interested in. – Dominic Else Mar 3 '17 at 17:02
• Note that Andy's Proposition 1.1 works with any coefficients, and asserts an equivalence of cochain complexes $\mathcal{C}^*(X/G)\cong(\mathcal{C}^*(X))^G$, as in my answer. So the outer terms are always equal. Only in Lemma 1.2 are rational coefficients used, to get equality with the middle term. This depends on having a suitable triangulation, of course (so that simplices of $X/G$ correspond to orbits of simplices in $X$). – Mark Grant Mar 3 '17 at 18:46