Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $X$ be a quasi-projective variety over the complex numbers, equipped with an action of a linear algebraic group $G$.

Is there a natural mixed Hodge structure on its equivariant cohomology?

Is it pure if $X$ is smooth projective?

What if we ask the analogous question for $l$-adic equivariant cohomology for varieties over finite fields?

More generally is there a formalism of "mixed equivariant derived categories" such that the six functors have the expected effects on weights?

share|improve this question

2 Answers 2

The answer to the first question is yes. Although the words "equivariant cohomology" don't appear there, this goes back to Deligne's Hodge III, since he defines the mixed Hodge structure on a simplicial variety and you can define the Borel construction simplicially in the usual way (via the resolution with $X$, $G \times X$, $G \times G \times X$, and so on). In particular he proves that $H^\bullet(BG)$ is pure of Tate type. If $G$ is connected, then there is a spectral sequence $$H^p(BG) \otimes H^q(X) \implies H^{p+q}_G(X).$$ If $X$ is smooth and projective this must therefore degenerate immediately for weight reasons and so the equivariant cohomology of smooth projective varieties is very simple: $H^\bullet_G(X) \cong H^\bullet(X) \otimes H^\bullet(BG)$. In particular the answer to your second question is also yes.

I would imagine that there is a suitable theory of "six functors" for simplicial varieties as well, but I don't know the literature well enough to say anything meaningful or give a reference.

share|improve this answer

In the $\ell$-adic setting a reference would be Laszlo-Olsson's papers on six operations for Artin stacks:

http://arxiv.org/abs/math/0512097

http://arxiv.org/abs/math/0603680

http://arxiv.org/abs/math/0606175

Also see:

http://arxiv.org/abs/1211.5948

Regarding mixed equivariant derived categories in the Hodge sense, as far as I know there isn't any canonical reference in the literature for this. However, I claim that the Bernstein-Lunts approach (for $X/G$, $G$ linear algebraic) works just fine for mixed Hodge modules (or even $\ell$-adic coefficients, but I am more comfortable with the mixed Hodge setting so not quite as sure about $\ell$-adic). The point is that Bernstein-Lunts mainly just use the six operations in their approach. So everything goes through formally in the same way for mixed Hodge modules. The only place where you might worry is something funny happening with weights. However, there is no problem if all your approximation spaces are algebraic (hence the linear algebraic requirement) and all pushforwards are along proper (algebraic) maps and pullbacks are along smooth (in the algebraic sense) maps.

I remember going through things carefully for $B\backslash G/B$ a couple of years ago. Many of these checks are also done in O. Schnurer's thesis. A condensed/article version of the latter can be found here:

http://arxiv.org/abs/0809.4785

I would also suggest just asking Wolfgang.

I do not know what the state of the art in defining mixed motivic sheaves at the moment is.

Some comments not directly related to the question (but I am reading between the lines and assuming this is where you are coming from Jan): As far as graded representation theory type applications go, the desire for a mixed equivariant derived category usually manifests itself in trying to prove splitting of some sequences, purity/formality type results, and/or to get a grading on Ext-spaces. In each of these situations one can avoid having to invoke a high powered theory of mixed equivariant categories by just working on an approximation space and the mixed (non-equivariant) derived category on this space (we are just unwinding the Bernstein-Lunts approach/Borel construction). I would claim that often even working with a mixed (non-equivariant) derived category can be avoided, since usually the only way one has of getting a handle on Ext-spaces is by interpreting them as cohomology of some space, or a convolution algebra formalism, or using a suitable fibre functor (I am thinking along the lines of Soergel bimodules, Geometric Satake, etc.). So as long as you know there is a functorial mixed structure on cohomology groups you can get away without explicitly invoking mixed categories. Having said that, there is one caveat: coming from the graded representation theory perspective, mixed sheaves offer the opportunity to impose gradings in a functorial way (note: Ext-groups in the mixed derived category are ordinary vector spaces, it is Ext-groups in the non-mixed derived category that inherit mixed structures via realization). I don't know how one can (in general) cheat to get these mixed strucures on Ext-spaces that behave functorially. In special situations you can get by via devices like Soergel bimoules, working with $\mathbb{C}^*$-coherent sheaves on cotangent bundles, etc. But these are replacements that require a lot of extra work (often worth it for characteristic $p$ applications). In characteristic $0$ the conceptual approach still would be to go through mixed categories. Of course, getting gradings this way is not at all a triviality: a grading corresponds to having split Tate structures which can be quite difficult to prove (eg. I don't know how see directly that Ext between Vermas, in just the ordinary $G/B$ case is split Tate). In representation theory type settings, there is the (somewhat mysterious) phenomenon that intersection cohomology has a basis given by algebraic cycles (I am thinking Geometric Satake, fibres of the Bott-Samelson resolution etc.). As Wolfgang would (probably) say, if we had a fully functional theory of motivic sheaves all the pain would go away!

A comment on Dan Peterson's answer: I am not sure about a 6 functor formalism for simplicial varieties. There is a brief discussion in Bernstein-Lunts about some of the difficulties in defining six functors and proving their properties in this setting (I don't remember the precise section).

share|improve this answer
    
Thanks for your comprehensive answer! One problem when working with approximations is, that you need to construct them. For example I would not know how to generalize O. Schnurer's thesis to the $P\G/Q$ case since I don't see how to construct the relevant approximations. Also approximations involve choices and choices usually lead to trouble in the long run. –  Jan Weidner Jun 17 '13 at 8:54
    
I have looked into the other references you suggest and one thing I am missing (though it could be hidden somewhere, I don't understand them very well...) would be the formula $f_!=f_∗$ for proper $f$. Without it, I can't even deduce that $H_G(pt)$ is pure. Do you know if $f_!=f_∗$ is in there or follows easily? –  Jan Weidner Jun 17 '13 at 8:57
1  
Regarding choice of approximation: say $X$ was the $G$-variety; if $V_k'$ and $V_k''$ are both approximations for (the classifying space) for $G$ then so is $V_k = V_k' \times V_k''$. Pulling back along $G\backslash (X\times V_k) \to G\backslash (X\times V_k')$ and similarly for $V_k''$ should give an equivalence showing that your category doesn't depend on the approximation. (I think I am just remembering the argument from Bernstein-Lunts, sorry am traveling right now so can't be more comprehensive). –  Reladenine Vakalwe Jun 17 '13 at 15:06
    
Regarding pushforward: so we are given a proper equivariant map $f\colon X\to Y$, then the pushforwards $f_!$ and $f_*$ are basically going to be a family of pushforwards (on the approximations) given by base change. Each of these will be proper, so $f_* = f_!$. Am I missing/misunderstanding something? –  Reladenine Vakalwe Jun 17 '13 at 15:13
    
I should add that I am assuming above that the approximations being used above (this is relevant for the argument showing that choice of approximation is irrelevant and nothing messes with weights) is smooth. –  Reladenine Vakalwe Jun 17 '13 at 15:15

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.