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Suppose $X$ is a smooth variety and $G$ is a finite group acting on $X$. $X/G$ is not locally factorial.

Let $h: X\rightarrow X/G$ be the quotient morphism. Suppose there is a coherent sheaf $F'$ on $X/G$ which pullsback to $F$ on $X$. Since $X$ is smooth, one can define the determinant of $F$.

Can one define $det(F') \in Pic(X/G)$, such that $det(h^*F)=h^*(det F')$ ?

I looked into Mumford-Knudsen" preliminaries on Div and det", but could not understand if this is possible.

thanks

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2 Answers 2

No, usually not. The quotient of the "standard action" of $\{\pm 1\}$ on $\mathbb{P}^2$ is a singular quadric cone $Q$ in $\mathbb{P}^3$. The ideal sheaf $F'$ of a line in this cone is a coherent sheaf. The pullback of this coherent sheaf to $\mathbb{P}^2$ has determinant $\mathcal{O}(-1)$. But there is no invertible sheaf on $Q$ whose pullback to $\mathbb{P}^2$ equals $\mathcal{O}(-1)$.

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thanks very much. –  john Apr 11 '13 at 20:22

Unless you know that $F'$ has a finite locally free resolution, and this will tend to fail when $F'$ is the skyscraper sheaf of a singular point.

But maybe you just want to work with $G$-equivariant coherent sheaves on $X$.

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