Two mixed Hodge structures on equivariant cohomology for actions by finite groups

Question

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!

Answer 1

As I said in my comment, the mixed Hodge structures are the same. Here is the outline. From [Hodge III, 6.1.2.1], $$[X/G]_n = (G^{n+1}\times X)/G$$ 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, (5.2.1.1)], 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.