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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

$$g \cdot \phi(x) = \phi(g^{-1} \cdot x) \cdot c(g) \quad \quad (g \in G, \phi \in \mathcal{O}_X(V), x \in V).$$

Define $\mathcal{F}(U)$ to be the invariants $(\mathcal{O}_X(V))^G$ of this action.

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.

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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 $\pi^{*}\mathcal{F}\cong \mathcal{O}_{X}$. You can use this to conclude...

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Without the twist, it's entirely trivial! Indeed, without the twist, $\mathcal{F}$ is just $\mathcal{O}_Y$... – R McOwen Apr 13 '11 at 5:32
Actually, I just looked there, and Proposition 3 of that section of Mumford has the result I wanted (twist and all). Thanks! – R McOwen Apr 13 '11 at 5:42
yes it is trivial without the twist, but i forgot to say that $\mathcal{F}$ can be any quasi-coherent $\mathcal{O}_{X}$ module... – Peter Toth Apr 13 '11 at 5:44
i mean $\mathcal{O}_{X}$ can be substituted by any coherent $\mathcal{O}_{X}$ module... – Peter Toth Apr 13 '11 at 5:48

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