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Let $\mathcal{X}$ be an Artin stack. In "Abramovich, Graber, Vistoli - Twisted bundles and admissible covers", the authors describe a procedure to rigidify $\mathcal{X}$ by a central subgroup $H$ of the generic stabilizer and obtain an Artin stack $\mathcal{X}^H$. Roughly speaking, the objects of $\mathcal{X}^H$ are the same of $\mathcal{X}$ and the automorphism groups of each object in $\mathcal{X}^H$ is the quotient of the automorphism groups of the corresponding object in $\mathcal{X}$ by $H$. One of the properties of the rigidification is that it preserves the coarse moduli space, i.e., if $\mathcal{X}$ has coarse moduli space $X$, also $\mathcal{X}^H$ has coarse moduli space $X$.

In the following, I shall ask if the rigidification preserves also the good moduli space (morphism) in the sense of Alper (Alper - Good moduli spaces for Artin stacks).

Definition. Let $\mathcal{X}$ and $\mathcal{Y}$ be Artin stacks. Assume that $\mathcal{Y}$ has quasi-affine diagonal. Let $\phi\colon \mathcal{X}\to\mathcal{Y}$ be a quasi-compact mophism. We say that $\phi$ is a good moduli space morphism if the following properties are satisfied:

  • the functor $\phi_*\colon QCoh(\mathcal{X})\to QCoh(\mathcal{Y})$ is exact,
  • the natural morphism $O_{\mathcal{Y}}\to\phi_* (O_{\mathcal{X}})$ is an isomorphism.

If $\mathcal{Y}=Y$ is an algebraic space, we say that $\phi$ is a good moduli space.

Question. Let us consider an Artin stack $\mathcal{X}$ which admits a good moduli space $\pi\colon \mathcal{X}\to X$. Is $X$ also the good moduli space of the rigidification $\mathcal{X}^H$? Is the natural morphism $\mathcal{X}\to\mathcal{X}^H$ a good moduli space morphism?

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won't in general the rigidification morphism have higher pushforwards? (so that $\phi_*$ fails to be exact) I guess in characteristic zero this won't happen if you're killing off a finite group. –  Jacob Bell May 15 '13 at 17:29

1 Answer 1

up vote 9 down vote accepted

It is certainly not true that $\mathcal X \to \mathcal X^H$ is a good moduli morphism, unless $H$ is linearly reductive, because when you push forward the cohomology of $H$ will come into play.

On the other hand $\mathcal X^H \to X$ is a good moduli space, because the pushforward $QCoh(\mathcal X^H) \to QCoh(X)$ can be factored as the pullback $QCoh(\mathcal X^H) \to QCoh(\mathcal X)$ followed by the pushforward $QCoh(\mathcal X) \to QCoh(X)$, and both are exact.

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