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](http://arxiv.org/abs/0709.3615))


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?



Equivariant singular cohomology
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).


