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Let $S$ be a smooth surface defined over a finite field $K$ of char. $p$. Let $G$ be a finite group of automorphisms of $S$. Let $Z\to S/G$ be the minimal resolution of the quotient of $S$ by $G$. Suppose that the fixed point set of every elements of $G$ is defined over $K$.

Let $\ell$ be a prime, $\ell\not=p$. Is it true that $$H^{i}(Z,\mathbb Q_\ell) \simeq H^{i}(S,\mathbb Q_\ell)^G$$ for $i=1,3$ and $$H^{2}(Z,\mathbb Q_\ell)\simeq H^{2}(S,\mathbb Q_\ell)^G+\mathbb Q_\ell C_1+\dots +\mathbb Q_\ell C_k$$where the $C_i$ are the exceptional curves of the resolution $Z\to S/G$ ?

This question is an echo of the question "Are there any known formulas about the Hodge-Deligne structure of quotients by action of groups ?" formulated in this forum 2 or 3 weeks ago.

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1 Answer 1

Xavier,

As per your request, I'll give a partial answer on mathoverflow assuming, for simplicity, that $G$ has isolated fixed points on $S$. Let $H^i(S)$ etc. denote the $\ell$-adic cohomology of $S\otimes\bar K$. I'll ignore Tate twists. Rather than taking the minimal resolution $Z$, let me take the minimal "log" resolution, so that the exceptional divisor $C\subset Z$ has normal crossings

Let [M] = Milne, Etale Cohomology. I'm referring to the book, but the online notes should work just as well.

Let $\iota:\Sigma\subset S$ be the set of fixed points, $U=S-\Sigma$ and $V=U/G$. Then by Hochschild-Serre [M, p105] $H^i(V)= H^i(U)^G$. By Poincare duality $H^i(V)= H_c^{4-i}(V)$. From the sequence $$\ldots H_c^i(U)\to H^i(S)\to H^i(\Sigma)\ldots$$ coming from $0\to j_!\mathbb{Q}_\ell\to \mathbb{Q}_\ell\to \iota_*\mathbb{Q}_\ell\to 0$ we deduce that

$H^i(V)= H^i(U)^G=H^i(S)^G$ for $i=1,2$.

If you combine this with the Gysin sequence $$0\to H^1(Z)\to H^1(V)\to \bigoplus H^0(C_k)\to H^2(Z)\to H^2(V)$$ you should be in good shape. To get the sequence above, apply the version of the Gysin sequence on [M, p 244] to the complement of the set of double points $Z'=Z-C_{sing}$. Use the isomorphism $H^2(Z)\cong H^2(Z')$ and do a diagram chase.

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Great, thank you very much. –  Xavier Roulleau Feb 17 '12 at 15:03

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