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For some work I'm doing, I need a version of the Hodge to de Rham spectral sequence for stacks. I am not at all an expert on stacks, so please excuse me if I make minor technical mistakes in stating it.

I only need to deal with quotient stacks. Let $X$ be a smooth quasiprojective variety over $\mathbb{C}$ and let $G$ be a finite group acting on $X$. I am trying to understand the cohomology of the quotient stack $X/G$. Sheaves on $X/G$ should be the same thing as $G$-equivariant sheaves on $X$ and sheaf cohomology should be the derived functors of the $G$-equivariant global sections functor. What I think should be true is that there should be a Hodge to de Rham spectral sequence converging to $H^{\ast}(X/G;\mathbb{C})$ with

$$E_1^{pq} = H^p(X/G;\Omega^q).$$

Here $\Omega^q$ is the sheaf of $G$-equivariant holomorphic $q$-forms on $X$.

This should be able to be proved following the usual proof of the Hodge to de Rham spectral sequence: one first proves that the constant sheaf $\mathbb{C}$ on $X/G$ is quasi-isomorphic to the de Rham complex $\Omega^{\ast}$, and the one looks at the hypercohomology spectral sequence. The only place where I can see any issue is proving the exactness of the de Rham complex, which would require an equivariant version of the Poincare lemma. However, this should be able to be derived from the usual Poincare lemma by averaging.

Question: Am I correct that this spectral sequence exists, and if so can anyone give me a reference for it?

I can find plenty of papers that investigate situations where a version of the Hodge to de Rham spectral sequence for stacks degenerates, but all of them are working in much more generality than I am and it is hard for me to verify that the spectral sequence they deal with is the same one I (tried) to state above. I don't need the spectral sequence to degenerate, though it presumably does in the situation I am working in.

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Short answer: yes.

As I recall, Teleman constructs such a spectral sequence for fairly general stacks. You can look at his paper The quantization conjecture revisited Annals 2000. But this may fall into the category of "more generality than you need".

The case you want can probably done by hand. The point is if $G$ is a finite group acting on a complex quasi projective variety $X$, then the quotient variety $X/G$ exists (see Sarah's comment below for an explanation). This is not the same as the quotient stack $[X/G]$ for a nonfree action, but there is relationship. The cohomology of $[X/G]$ with coefficients in a sheaf $\mathcal{F}$ would be equivariant cohomology $H^*_G(X,\mathcal{F})$. This can be taken as the derived functor of $G$-invariants composed with global sections as you said. This yields a spectral sequence $$H^a(G, H^b(X,\mathcal{F}))\Rightarrow H^{a+b}([X/G],\mathcal{F}) $$ Since $G$ is finite, this will collapse to an isomorphism when $\mathcal{F}$ is a sheaf of $\mathbb{C}$-vector spaces by Maschke's theorem. This remark applies to $\mathcal{F}=\mathbb{C}$ or $\Omega_X^q$. In this case, we can further identify $$H^i(X,\Omega_X^q)^G= H^i(X/G,\tilde\Omega_{X/G}^q)$$ where the sheaf on the right is defined in Steenbrink, Mixed Hodge structure on vanishing cohomology. The upshot is that you may as well work on $X/G$, where the details about how to construct Hodge to de Rham can be found in Steenbrink's paper.

Added explanation of second isomorphism. By definition $\tilde \Omega^q(U) = \Omega^q(\pi^{-1}U)^G$, where $\pi:X\to X/G$ is the projection. This implies that, we have an isomorphism of Cech complexes $$C(\{U_i\}, \tilde \Omega^q)=C(\{\pi^{-1}U_i\}, \Omega)^G$$ and therefore on cohomology (you need to use exactness of $(-)^G$ here as well).

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  • $\begingroup$ Thanks! Can you say a little more about the isomorphism in your second math display? I see how to prove it for $i=0$ (where it is essentially proved in Steenbrink's paper), but I'm not sure how to do it when $i>0$. $\endgroup$ – Sarah Aug 11 '16 at 21:45
  • $\begingroup$ " if G is a finite group acting on a complex quasi projective variety X, then the quotient variety X/G exists" - Am I getting confused or there was a counterexample given by a smooth projective variety with an action of Z/2 such that the quotient exists only as an algebraic space? $\endgroup$ – Qfwfq Aug 11 '16 at 23:45
  • $\begingroup$ Yes, I'm aware of Hironaka's example and of what I said. The key word in what you quoted is quasiprojective. See Knutson, Algebraic Spaces, page 180, prop 1.5 $\endgroup$ – Donu Arapura Aug 12 '16 at 2:09
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    $\begingroup$ @Qfwfq: The general condition is that if $G$ is a finite group acting on a variety $X$, then $X/G$ exists as a variety if and only if for all $x \in X$, the orbit $G \cdot x$ is contained in an open affine. The basic idea is that this condition allows you to cover $X$ by $G$-invariant open affines. It is easy to construct quotients of affine varieties (their coordinate rings are just the rings of invariants), and you then glue them together. It is an easy exercise to see that this condition is satisfied for quasiprojective varieties. $\endgroup$ – Sarah Aug 12 '16 at 3:29
  • $\begingroup$ Ok, Hironaka's X (for the purpose of this example) was only an abstract non-quasiprojective variety (so, in particular, X would be a scheme but X/G would not be a scheme). I misremembered the example: I thought even something worse could happen, i.e. X projective and X/G not a scheme, but this is ruled out by Knutson. Thanks for the comments. $\endgroup$ – Qfwfq Aug 12 '16 at 19:51

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