most recent 30 from http://mathoverflow.net 2013-06-19T23:36:36Z http://mathoverflow.net/feeds/question/108395 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/108395/invariant-lattice-of-algebraic-surface Pooya 2012-09-29T08:59:35Z 2012-10-06T09:34:45Z

<p>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 $S$. </p> <p>I would appreciate it if someone could tell me what condition we need if this is not true in general. </p>

http://mathoverflow.net/questions/108395/invariant-lattice-of-algebraic-surface/108958#108958 rita 2012-10-05T20:51:24Z 2012-10-06T09:34:45Z

<p>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 $2h$. </p> <p>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.)</p>