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Given an algebraic surface $S$ with action of a finite group $G$. Is it true that the invariant lattice $H^2(X,\mathbb{Z})^G$ is generated by elements pulled back from the $H^2(X/G,\mathbb{Z})$ (or $H^2(\widetilde{X/G},\mathbb{Z})$)? This seems true for some examples such as an Enriques surface, but I don't know this works for any algebraic surface $S$.

I would appreciate it if someone could tell me what condition we need if this is not true in general.

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1 Answer 1

I think in general the answer is no. Take the map $\mathbb P^2\to \mathbb P^2$ given by $[x_0,x_1,x_2]\mapsto [x_0^2, x_1^2, x_2^2]$. This is the quotient map for the action of $G=\mathbb Z_2^2$ given by $[x_0,x_1,x_2]\mapsto [x_0,\pm x_1, \pm x_2]$. Since $H^2(\mathbb P^2, \mathbb Z)=\mathbb Zh$, where $h$ is the class of a line, $G$ acts trivially on $H^2(\mathbb P^2, \mathbb Z)$. On the other hand, the pullback of $H^2(\mathbb P^2, \mathbb Z)$ is the subgroup generated by $2h$.

ADDED: this example shows that the answer is no whenever $h^2(X)=h^2(X/G)$, because the intersection form on $H^2(X,\mathbb Z)$ is unimodular, while the intersection number of two classes coming form $X/G$ is divisible by $|G|$. (At least when $X/G$ is a smooth surface, I'm not sure what happens when $X/G$ is singular.)

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On the other hand, over $\mathbb Q$, since the category of $\mathbb Q(G)$-modules is semisimple, this cannot happen. –  Will Sawin Oct 5 '12 at 21:10

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