Equivariant Cohomology of a Complex Projective Variety

Suppose that I have a complex projective variety $X$ endowed with an algebraic action of a complex torus $T$. Suppose also that the set $X^T$ of fixed points is finite. I would like to relate the equivariant cohomologies $H_{T}(X;\mathbb{C})$ and $H_{T}(X^T;\mathbb{C})$ using the existing localization theory. Yet, I am finding it difficult to verify some of the hypotheses conventionally imposed in the main theorems (ex. that $H_T(X;\mathbb{C})$ is a free module over $H_{T}(*;\mathbb{C})$) for the examples I am considering. What are some of the most general conclusions that follow in the situation described above, and what sorts of things should I investigate in order to relate the equivariant cohomologies?

-
If $X$ is moreover smooth, then $H_T(X) \cong H(X) \otimes H_T(\ast)$, so the freeness hypothesis is satisfied. See my answer to mathoverflow.net/questions/133406 – Dan Petersen Jun 14 '13 at 16:22
My variety is not smooth, but I could consider its smooth locus. Unfortunately, this space is no longer projective. What can be said in this case? – Peter Crooks Jun 14 '13 at 16:24
In particular, what can one say about freeness when $X$ is smooth and quasi-projective? – Peter Crooks Jun 14 '13 at 16:38
@PDC: This is not at all my specialty, but I'm aware of a lot of work by Michel Brion and others on equivariant cohomology, especially relative to "rational smoothness" in settings like Schubert varieties, with natural torus actions having finitely many fixed points. There's also a survey by Jim Carrell: "Torus actions and cohomology" in Algebraic quotients. Torus actions and cohomology. The adjoint representation and the adjoint action, 83–158, Encyclopaedia Math. Sci., 131, Springer, Berlin, 2002 – Jim Humphreys Jun 14 '13 at 22:56
@PDC: The only reference I can add on top of Jim's is one you are probably familiar with: Goresky-Kottwitz-Macpherson `Equivariant cohomology, Koszul duality and the Localization theorem'. The reason for suggesting intersection cohomology (for complete but not necessarily smooth spaces) is essentially because of what Dan Peterson says: as far as I know all methods that deduce freeness boil down to a weights/purity argument. – Reladenine Vakalwe Jun 15 '13 at 17:58

Usually one uses a Białynicki-Birula‎ decomposition derived from a general circle in $T$ to find cycles that give a basis of this free module. But this decomposition is less useful when the space is singular.
For a basic example, let $X$ be the triangle $\{ [x,y,z] : xyz = 0 \} \subset {\mathbb P}^2$. This space has $H^1$ and $H^1_T$, but its fixed points do not.