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Let $X$ be a smooth complex quasiprojective variety and let $G$ be a finite group acting on $X$. The quotient $X/G$ is then a variety. Let $\mathcal{F}$ be a $G$-equivariant coherent sheaf on $X$.

Question: Under what conditions will $\mathcal{F}$ descend to a coherent sheaf on $X/G$?

This will hold if $G$ acts freely (this can e.g. be found in Mumford's book on abelian varieties), but alas in the situation I am in this does not hold.

Dan Petersen's answer to this question says that this will hold if I am content to work in the category of stacks and use the stacky quotient, but I am not comfortable with stacks and would prefer to work in the world of varieties.

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    $\begingroup$ This is a well-known result of Kempf, see Drezet-Narasimhan, Groupe de Picard des variétés de modules de fibrés semi-stables sur les courbes algébriques, Invent. math. 97 (1989), no. 1, 53-94). The sheaf $\mathcal{F}$ descends if and only if for each point $x$ of $X$, the stabilizer of $x$ in $G$ acts trivially on the fiber $\mathcal{F}_x$. $\endgroup$
    – abx
    Aug 9, 2016 at 19:07
  • $\begingroup$ almost a duplicate : see mathoverflow.net/questions/234616/… . This version is much better formulated though ! $\endgroup$
    – Niels
    Aug 9, 2016 at 19:59
  • $\begingroup$ You're already working with stacks! Equivariant sheaves just are sheaves on the stacky quotient, so your question can be restated as a question about descent along the natural map from the stacky to the ordinary quotient. This map more or less has fibers the (stacky quotients of points by) stabilizers of the action, so it is natural to at least conjecture Kempf's result from here. $\endgroup$ Aug 9, 2016 at 23:00

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