# Equivariant singular cohomology

One can define the $G$-equivariant cohomology of a space $X$ as being the ordinary singular cohomology of $X \times_G EG$ --- I think this is due to Borel? (See e.g. section 2 of these notes)

Alternatively if $X$ is a manifold, we also have $G$-equivariant de Rham cohomology, defined in terms of $G$-equivariant differential forms --- I think this is due to Cartan? (See e.g. section 3 of loc. cit.)

I suspect this is extremely standard or obvious, but if it is, I don't know where it's written down: Is it possible to define equivariant cohomology of a topological space in terms of some notion of "equivariant singular cochains", that is, without using the Borel construction?

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Isn't Cartan's equivariant cohomology the hypercohomology of the de Rham complex? –  Mariano Suárez-Alvarez Dec 23 '09 at 2:35
Do you really mean just the de Rham complex? There's no $G$−stuff in the de Rham complex, and shouldn't equivariant stuff involve $G$−stuff? –  Kevin H. Lin Dec 23 '09 at 2:47
If $G$ acts on $M$, then $G$ acts on $M$'s de Rham complex $\Omega^\bullet(M)$. Now take $\mathbb{H}^\bullet(G,\Omega^\bullet(M))$. This gives you an equivariant theory. You can do it replacing $\Omega^\*(M)$ by $S^\*(M)$, the singular complex, of course. Since hypercohomology sees only the quasi-isomorphism type of its argument, you get isomorphisms between what you get from de Rham and what you get from $S^\*(M)$, &c. –  Mariano Suárez-Alvarez Dec 23 '09 at 2:52
Ah, ok, so this is, umm, hyper-group-cohomology? –  Kevin H. Lin Dec 23 '09 at 3:15
Those adjunctions can only exist if G is discrete. It's not clear from the question if we are assuming this or if G was allowed to be, say, a compact Lie group. –  Chris Schommer-Pries Dec 24 '09 at 13:36

Here's an answer which I learned from Goresky-Kottwitz-MacPherson's paper on equivariant cohomology and Koszul duality: they use some notion of geometric chain which is probably something like subanalytic chains, but anyway, the idea is as follows.

Suppose $G$ is a compact Lie group of dimension d. An abstract equivariant $k$-chain $c$ is a $(k+d)$-dimensional chain in some $\mathbb R^n$ (or perhaps it's better to say $\mathbb R^\infty$) equipped with a free action of $G$. Then if $X$ is a $G$-space, an equivariant chain in $X$ is a $K$-equivariant map from an abstract chain to $X$. You can obviously form a chain complex out of these things, and the result gives you the equivariant cohomology of $X$ (in the Borel construction say).

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Yeah, I was guessing that it might be something along those lines. But then how do you put a free action of $G$ on $\mathbb{R}^n$? And then you still have to show that the result is independent of this choice? –  Kevin H. Lin Dec 23 '09 at 3:09
An abstract chain just has to have a $G$ action, not the ambient space. The idea I guess is that a chain in the Borel construction can be pulled back to $X\times EG$ and then I can approximate $EG$ by a smooth finite-dimensional manifold $E$ containing the pulled-back chain. Embed $X\times E$ in $\mathbb R^\infty$ and you have geometric chain (or rather its graph). Anyway the paper is nicely written if I remember, so probably it's better to read what they said! –  Kevin McGerty Dec 23 '09 at 3:17
Oops, yes, I meant the chain. Ok, I'll try to take a peek at the paper. –  Kevin H. Lin Dec 23 '09 at 3:23

In 1965 or so, Glen Bredon defined ordinary equivariant cohomology, ordinary meaning that it satisfies the dimension axiom: For each coefficient system $M$ (contravariant functor from the orbit category of G to the category of Abelian groups), there is a unique cohomology theory $H^*_G(-;M)$ such that, when restricted to the orbit category, it spits out the functor $M$. Just as in the nonequivariant world, it can be defined using either singular or cellular cochains, the latter defined using $G$-CW complexes. This works as stated for any topological group $G$.

For an abelian group $A$, Borel cohomology with coefficients in $A$, $H^*(EG\times_G X;A)$ is the extremely special case in which one takes $M$ to be the constant coefficient system $\underline{A}$ at the group $A$ and replaces $X$ by $EG\times X$. That is,

$$H^*_G(EG\times X; \underline{A}) = H^*(EG\times_G X;A)$$

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You can think of it as the cohomology of the simplicial manifold $X\leftleftarrows X\times G \cdots$ where the $n$-simplices are $X\times G^n$ and the face maps either act on $X$ or multiply two consecutive entries.

Of course, some people will tell you that that is really the same as the Borel construction, but if you're willing to interpret things that liberally, you'll never get away from the Borel construction.

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How does one define cohomology of a simplicial manifold? –  Kevin H. Lin Dec 23 '09 at 17:12
There is an abelian category of "simplicial sheaves over a simplicial manifold" and there will be enough injectives. So you just take any injective resolution. The standard reference (in the scheme case) is "Etale Homotopy of Simplicial Schemes" by E. Friedlander. I think there was also some work by Dupont on exactly this equivariant case. Anyway, in the case you are looking at you can just use the usual de Rham resolution of Z in each level of the simplicial set. You will get a double complex (one direction from the de Rham differential, one simplicial). You just take the total cohomology. –  Chris Schommer-Pries Dec 24 '09 at 13:34
Thanks, Chris! That was very helpful. –  Kevin H. Lin Dec 24 '09 at 13:50
of course, "... each level of the simplicial set" should be read as "... each level of the simplicial space". –  Chris Schommer-Pries Dec 24 '09 at 15:59

This isn't singular, but in the same spirit as what you want (i.e. not the Borel construction):

Take the cellular chain complex $C(X)$ of a $G$-complex $X$. You can define cohomology groups with coefficients in a chain complex, and $H^*(G,C(X))$ is defined to be the equivariant cohomology of $X$ (for $X$ finite-dimensional and $G$ finite).

This is explained in Ken Brown's $\textit{Cohomology of Groups}$ (chapter VII).

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$G$ induces an action on the singular (co)chain complex $S(X)$ and you can use $H^*(G;S(X))$ in the same way as described by Chris. –  Ralph Feb 24 '11 at 23:46