The answer to the following question might be obvious but I haven’t found a full proof yet (neither by myself nor in the literature). So my apologies if it is trivial.

Let $X$ be a (for simplicity quasi-projective and non-singular) complex variety $X$ on which a finite group $G$ acts. Deligne has shown in Hodge III that the equivariant cohomology group $H_G^k(X,\mathbb{Q})$ for any $k$ carries a natural mixed Hodge structure (MHS). On the other hand, the Leray-Serre spectral sequence for the Serre fibration \begin{equation} X \to X\times_G EG \to BG \end{equation} degenerates (complex topology) over $\mathbb{Q}$ and hence yields an isomorphism

\begin{equation} H_G^k(X,\mathbb{Q})\cong H^k(X,\mathbb{Q})^G \end{equation}

(cf. this MO question: Equivariant cohomology of finite group actions and invariant cohomology classes). Clearly, $H^k(X,\mathbb{Q})^G$ inherits a MHS from the MHS on $H^k(X,\mathbb{Q})$. Now my question is:

Are these two MHS on $H_G^k(X,\mathbb{Q})$ naturally isomorphic? More precisely, is there a simplicial version of the above Serre fibration yielding an isomorphism $H_G^k(X,\mathbb{Q})\cong H^k(X,\mathbb{Q})^G$ of MHS?

A natural candidate for such a simplicial version is

\begin{equation} [X/G]_\bullet \to B_\bullet G, \end{equation}

where I (essentially) use Deligne's notation from Hodge III. Since I'm still learning simplicial methods, I was not sure if this is maybe too naive.

Any thoughts/references/comments are very welcome!

  • $\begingroup$ Yes, these mixed Hodge structures should be the same. I'll say more later if someone else doesn't first. $\endgroup$ May 2, 2017 at 15:53

1 Answer 1


As I said in my comment, the mixed Hodge structures are the same. Here is the outline. From [Hodge III,], $$[X/G]_n = (G^{n+1}\times X)/G$$ One has a descent spectral sequence $$E_1= H^q([X/G]_p,\mathbb{Q})\cong (G^{p+1}\times H^q(X,\mathbb{Q}))/G$$ abutting to $H^{p+q}([X/G]_\bullet, \mathbb{Q})=H^{p+q}_G(X,\mathbb{Q})$ [Hodge III, (], and this is compatible with MHS [Hodge III, (8.1.15)]. Now use the fact that the complex $E_1$ is the bar complex, which computes group cohomology; in this case it is trivial except in degree zero. So in conclusion $$H_G^*(X,\mathbb{Q})\cong H^*(X,\mathbb{Q})^G$$ as MHS.

  • $\begingroup$ Thank you, Donu, that was very helpful! I was trying to use a fibration-type argument because it seemed to be more appropriate for my situation that I'm actually working on. However, I'll check if your argument also works for my purposes. In any case, no reason to not accept your answer :) (I can’t help but have to add: Some of your papers, Donu, already helped me learning MHS beyond the definitions. So thanks for that as well :)) $\endgroup$
    – Florian
    May 3, 2017 at 11:29

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