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?



Rigidification and good moduli space (morphism) in the sense of Alper
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.


