Equivariant cohomology via equivariant sheaves

Question

Ordinary cohomology of topological space $X$ are known to be the cohomology of constant sheaf.

Question Is there analogous description for equivariant cohomology?

More precisely. Consider category of $G$ equivariant sheaves $\mathcal{Sh}_G (X)$. Denote by $\Gamma_G := Inv \circ \Gamma$ composition of two functors. I.e. we are taking global section of an equivariant sheaf $\mathscr{F}$ and get $G$-module $\Gamma ( \mathscr{F} )$, then we take invariants in this module.

Conjecture Equivariant cohomology are derived functors of $\Gamma_G$.

Answer 1

This question does not actually makes sense.

Let me explain all this in the case of point. If $G$ is discrete, then equivariant sheaves are just $G$ representations. If $G$ is a topological group, then one should consider topological vector spaces. The problem is that category of topological vector spaces is not an abelian category. Then you can not just refer to classical homological algebra and define derived functor. But a modification of classical homological algebra was done by Hochschild and Mostow.

Note, that finite dimensional representations of compact groups do not have higher cohomology. But

$$H^* (BG, \mathbb{R}) = S^* ( \mathfrak{g}^* )^G$$

In particular, there are higher cohomology. So, the answer on original question is "no".

Although the answer is no, those cohomologies are related. Here is a reference.

Answer 2

The answer is no. There is an counterexample. $G = X = U(1)$. More precisely $X$ is principal homogeneous space for $U(1)$.

If group acts freely then equivariant cohomology are equal to ordinary homology of quotient space. In this case the quotient space is just a point. Then $H_G^1 (X)= 0$.

On the other hand, consider exact sequence. $$0 \rightarrow \underline{\mathbb{R}} \rightarrow \Omega^0 \rightarrow \Omega^1 \rightarrow 0$$

It is just usual de Rham resolution. Note, that I do not claim that this is indeed an acyclic resolution in category of equivariant sheaves (I do not know whether it is true; if you know please write me). I just have a short exact sequence and then write down long exact sequence of cohomology (to be more precise $R \Gamma_G^i$).

$$0 \rightarrow \Gamma_G( \underline{\mathbb{R}} ) \rightarrow \Gamma_G ( \Omega^0) \rightarrow \Gamma_G ( \Omega^1) \rightarrow R^1 \Gamma_G( \underline{\mathbb{R}} )$$

It is easy to see that $\Gamma_G ( \Omega^0) \simeq \mathbb{R}$ and $\Gamma_G ( \Omega^1) \simeq \mathbb{R}$. From exactness, arrow $\Gamma_G( \underline{\mathbb{R}} ) \rightarrow \Gamma_G ( \Omega^0)$ is isomorphism. Again from exactness, arrow $\Gamma_G ( \Omega^1) \rightarrow R^1 \Gamma_G( \underline{\mathbb{R}} )$ is injective. So $R^1 \Gamma_G( \underline{\mathbb{R}} ) \neq 0$