# Mixed Hodge Polynomial for Algebraic Stacks

Let $$X$$ be a complex algebraic variety. The numerical invariants associated with the Mixed Hodge Structure of $$X$$ can be encoded in a polynomial in three variables called the mixed Hodge polynomial $$H(X, x,y,t)$$. There is also a version for compactly supported cohomology $$H_c(X, x,y,t)$$. The polynomial $$E(X,x,y)=H_c(X,x,y,-1)$$ is called the $$E$$-polynomial of $$X$$. See, e.g., $$\S 2$$ of this paper for a review.

Question: Are there analogs of these polynomials for a complex algebraic stack $$Y$$?

As a first step, we can consider a quotient stack $$X/G$$. Thus, I'm asking whether the constructions of mixed Hodge structures and associated polynomials work for equivariant cohomology.

When $$G$$ is a finite group then the answer is yes, see, e.g., this post. I'm interested in the case that $$G$$ is not finite; e.g., $$G=\mathrm{GL}_n(\mathbb{C})$$. One hopes that in favorable situations $$E(Y)=E(X)/E(G)$$.

• Did you already look at Kai Behrend's PhD thesis? – Jason Starr Feb 5 '19 at 18:56
• The Borel construction of equivariant cohomology, as made algebraic by Totaro and Edidin-Graham, should allow you to define mixed Hodge structures. – naf Mar 3 at 4:01

The answer is yes, these constructions make sense also for stacks, although you typically obtain a power series rather than a polynomial. Funnily enough, the best reference I know is Deligne's Hodge III, even though it doesn't talk at all about algebraic stacks.

Before asking how to put a mixed Hodge structure on the Betti cohomology of a complex algebraic stack $$\mathfrak X$$, it makes sense to ask how to define the Betti cohomology in the first place. There are several possible approaches. The most direct is to choose an atlas $$U \to \mathfrak X$$. This choice gives rise to a simplicial scheme $$U_\bullet$$, with $$U_n = U \times_{\mathfrak X} U \times_{\mathfrak X} \ldots \times_{\mathfrak X} U$$ ($$n+1$$ factors). Taking levelwise analytification gives a simplicial topological space, which can be geometrically realized to a bona fide topological space. If I had chosen a different presentation I would have gotten a weakly equivalent (in fact homotopy equivalent) geometric realization, so the homotopy type of the realization makes sense as an invariant of $$\mathfrak X$$. We may then take its cohomology.

The point is, then, that Deligne in Hodge III tells us precisely how to put a mixed Hodge structure on the cohomology of a simplicial scheme (of finite type over $$\mathbf C$$). As you may know, Deligne needs to pass to simplicial schemes in order to define the mixed Hodge structure on a singular variety $$X$$ (via a "hyperresolution of singularities"), but he notes that he obtains a general definition of mixed Hodge structure on any simplicial scheme, and as an example application he computes the mixed Hodge structure on the classifying stack of a linear algebraic group.

Unfortunately I don't know off hand a reference that connects the dots between what's in Hodge III and generalities on algebraic stacks, though it is all certainly something "known to experts". Perhaps you can look into Noohi's papers on topological stacks.

Regarding your question whether $$E([X/G])=E(X)/E(G)$$, this is not true in full generality. Think about the case when $$X$$ is a point and $$G$$ is finite, in which case $$E([X/G])=E(X)=1$$, but $$E(G) = \vert G\vert$$. But if you assume that your group $$G$$ is connected then you're good. You can prove it by analysing the Leray-Serre spectral sequences for the two fibration sequences $$G \to \mathrm{pt} \to BG$$ and $$X \to [X/G] \to BG.$$ The first sequence implies $$E(G)E(BG)=1$$, and the second that $$E(X)E(BG)=E([X/G])$$. The key point is that if we assume $$G$$ connected then $$BG$$ is simply-connected, so the local systems in the spectral sequence trivialize.

• Hi Dan, Aji Dhillon's thesis contains a detailed treatment along these lines, but it's probably hard to obtain. His paper arXiv:math/0310299 contains a synopsis. There is also a paper by Behrend-Dhillon which has some related things. – Donu Arapura Mar 3 at 15:39
• Hi Dan. Thanks a lot. This makes sense. What about my second question? Is it true that E-polynomial of a quotient stack is the same as the quotient of the E-polynomials? – Dr. Evil Mar 4 at 18:26

It has been shown by Toën https://arxiv.org/abs/math/0509098 and Joyce https://arxiv.org/abs/math/0509722 that the Grothendieck group $$K_0$$ of Artin stacks of finite type with affine stabilizers (over $$\mathbb{C}$$) is isomorphic to the group obtained from $$K_0$$ of varieties after inverting the class $$\mathbb{L}$$ of the affine line, and the classes $$\mathbb{L}^i-1$$ for $$i\geq 1$$. In particular, all additive invariants of varieties which are invertible on these particular classes (such as the E-polynomial viewed as a rational function) extend to algebraic stacks with affine stabilizers.

• To be fair, we should mention Ekedahl's work as well: arxiv.org/abs/0903.3143 It is also useful for this MO question to point out the fact that Toen's paper considers the case of Hodge polynomials explcitly. – Denis-Charles Cisinski Mar 7 at 12:45