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?
$\begingroup$
$\endgroup$
9
asked Jun 14, 2013 at 16:13
-
$\begingroup$ 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 $\endgroup$– Dan PetersenCommented Jun 14, 2013 at 16:22
-
$\begingroup$ 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? $\endgroup$– Peter CrooksCommented Jun 14, 2013 at 16:24
-
$\begingroup$ In particular, what can one say about freeness when $X$ is smooth and quasi-projective? $\endgroup$– Peter CrooksCommented Jun 14, 2013 at 16:38
-
1$\begingroup$ @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 $\endgroup$– Jim HumphreysCommented Jun 14, 2013 at 22:56
-
1$\begingroup$ @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. $\endgroup$– Reladenine VakalweCommented Jun 15, 2013 at 17:58
|
Show 4 more comments
2 Answers
$\begingroup$
$\endgroup$
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.
answered Jun 15, 2013 at 20:33
$\begingroup$
$\endgroup$
This is a helpful example for you. Consider Torus-equivariant cohomology of weighted projective spaces WPS (complex, projective, non-smooth in general, but with finite singular locus under some easy conditions on the weights). It has been computed by A.Bahri,M.Franz and N.Ray.(Unfortunatly, their computations are toric. I mean they are not in terms of the intrinsic geometry of WPS's (which do not need any fan to exist !). But however their toric calculations allowed them to answer a question on Chern classes (conjectured by Al-Amrani) in a particular case.).
