# Quasi-projective orbifolds and algebraic line bundles

The notion of quasi-projective orbifold is generally accepted to contain at least the following: let $X$ be a (simply-connected) complex manifold, $G$ a group acting on $X$ by biholomorphisms, and containing a finite-index subgroup $G' \subset G$ acting freely on $X$ so that $X/G'$ is a quasi-projective variety. Then the orbifold $X/\!/G$ is quasi-projective.

If $G$ does not act effectively on $X$ (with $H \subset G$ the finite normal subgroup of elements which act trivially everywhere) it is "unlikely" that one can find a finite-index $G'$ acting freely. Suppose that $X/\!/(G/H)$ is a quasi-projective variety in the above sense: should $X/\!/G$ be considered quasi-projective?

Question 1 : If $G$ does not act effectively on $X$, can $X/\!/G$ be ever considered a quasi-projective orbifold in any sense?

In the above case, an algebraic line bundle on $X/\!/G$ can be taken to be a $G$-equivariant holomorphic line bundle on $X$ such that induced line bundle on $X/G'$ is algebraic.

Question 2 : What should one take as definition of algebraic line bundle if $X/\!/G$ is not effective?

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Does $X//G$ denote the GIT quotient? – Mike Skirvin Jul 28 '10 at 13:37
@Mike: It denotes the orbifold / stack quotient. – Oscar Randal-Williams Jul 28 '10 at 13:48
In that case, there is a quite general definition of line bundle on an algebraic stack. If $X$ is a stack, then a line bundle on $X$ is the assignment of a line bundle $L_S$ to every scheme $S \to X$ such that if also $T \to X$ factors through $S$ via some map $f,$ then there is an isomorphism $f^{\ast}L_S \simeq L_T.$ Finally, these must also satisfy a cocycle condition if we add a third scheme above. In fact, we could have replaced line bundle with quasi-coherent sheaf above to define quasi-coherent sheaves on stacks. For quotient stacks, this is equivalent to usual equivariant sheaves. – Mike Skirvin Jul 28 '10 at 14:15