ordinary cohomology from equvariant cohomology

Question

Is it possible that the ordinary cohomology of a space can be obtained from its equivariant cohomology? action is algebraic torus action and space is nonsingular complete complex algebraic variety

If it is possible how can we do that? Are there any examples?

Answer 1

For a compact, connected Lie group $G$ acting on a manifold $M$, you have a spectral sequence starting at $\text{E}_1^{p,q}\cong {\mathfrak S}^p({\mathfrak g}^{\ast})^G\otimes_{\mathbb R} \text{H}^{q-p}(M;{\mathbb R})$ and converging to $\text{E}_\infty^\ast\cong\text{H}^{\ast}_G(M)$. The $G$-space $M$ is called equivariantly formal if this sequence collapses at the $\text{E}_1$-page, in which case you can deduce that there is a (non-canonical) isomorphism $\text{H}^\ast_G(M)\cong\text{H}^\ast(M;{\mathbb R})\otimes_{\mathbb R} {\mathfrak S}({\mathfrak g}^\ast)^G$ of ${\mathfrak S}({\mathfrak g}^\ast)^G$-modules. In particular, you can recover $\text{H}^{\ast}(M;{\mathbb R})$ from $\text{H}^\ast_G(M)$ up to isomorphism by applying $-\otimes_{{\mathfrak S}({\mathfrak g}^\ast)^G} {\mathbb R}$. You can read about these things in

Guillemin-Sternberg: Supersymmetry and Equivariant de Rham Theory, Chapter 6.