Constructing invertible sheaves out of actions of finite groups - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T19:38:37Z http://mathoverflow.net/feeds/question/61505 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/61505/constructing-invertible-sheaves-out-of-actions-of-finite-groups Constructing invertible sheaves out of actions of finite groups R McOwen 2011-04-13T05:06:00Z 2011-04-13T05:30:29Z <p>Let $X$ be a quasiprojective algebraic variety over $\mathbb{C}$ with structure sheaf $\mathcal{O}_X$ and let $G$ be a finite group which acts freely on $X$. The quotient $Y = X/G$ is thus a quasiprojective variety in a canonical way; let $\mathcal{O}_Y$ be its structure sheaf. Assume that $c : G \rightarrow \mathbb{C}^{\ast}$ is a homomorphism. Define a sheaf $\mathcal{F}$ on $Y$ as follows. Let $\pi : X \rightarrow Y$ be the projection map. Consider an open set $U \subset Y$. Let $V = \pi^{-1}(U)$. The group $G$ acts in a natural way on $V$, and we get an action of $G$ on $\mathcal{O}_X(V)$ via the formula</p> <p>$$g \cdot \phi(x) = \phi(g^{-1} \cdot x) \cdot c(g) \quad \quad (g \in G, \phi \in \mathcal{O}_X(V), x \in V).$$</p> <p>Define $\mathcal{F}(U)$ to be the invariants $(\mathcal{O}_X(V))^G$ of this action.</p> <p>I'm pretty sure that $\mathcal{F}$ is an invertible sheaf (ie a line bundle). If we were working analytically rather than algebraically, then this would basically be obvious; however, since things are only quasiprojective we don't have GAGA available, and I can't figure out how to deal with Zariski open sets.</p> http://mathoverflow.net/questions/61505/constructing-invertible-sheaves-out-of-actions-of-finite-groups/61506#61506 Answer by Peter Toth for Constructing invertible sheaves out of actions of finite groups Peter Toth 2011-04-13T05:30:29Z 2011-04-13T05:30:29Z <p>This is classical, can be found in Mumford: Abelian varieties, II.7 (i think) group actions on varieties (however without the twist, you mention, but i think it is no poblem), and the main reason (again without the twist) is that <code>$\pi^{*}\mathcal{F}\cong \mathcal{O}_{X}$</code>. You can use this to conclude...</p>