# Is there a non-quotient stack with affine stabilizers whose good moduli space is a geometric point?

Definitions: One says that a map $\pi\colon\mathcal X\to X$ from an algebraic stack to an algebraic space is a good moduli space if $\pi$ is cohomologically affine and universal for maps to schemes. (A good (but not equivalent) way to think about a stack whose good moduli space is a single point is that for any two points, the closure of their orbits intersect.)

One says that a stack $\mathcal X$ is a quotient stack if it is the quotient of an algebraic space by a subgroup of GL_n (see for instance EHKV, which also gives a criterion for a stack to be a quotient stack in terms of vector bundles).

One says that a stack has the resolution property if every coherent sheaf is a quotient of some vector bundle; Totaro's paper The resolution property for schemes and stacks relates the property of being a quotient stack to the resolution property.

Question: Let $\mathcal X$ be a stack with a good moduli space $\mathcal X \to X$ such that X is a geometric point (i.e., X = Spec k, where k is a separably closed field). Suppose further that the stabilizers are affine linearly reductive groups. Is $\mathcal X$ a quotient stack?

(See this answer for the definition of stabilizer of a point of a stack that isn't a quotient stack.)

Remarks:

1. The condition on stabilizers excludes things like BE with E an elliptic curve.

2. The condition that k is separably closed excludes non-trivial gerbes.

3. I'd be just as happy with an answer to "Does the resolution property hold for $\mathcal X$ ?".

• This is an interesting question. Here is a specific instance: let $n$ be a positive integer, and let $\mathcal X_n$ be the stack over $\mathbb C$ of nodal proper curves of genus 0 with at most $n$ nodes. One shows that it satisfies the conditions above. It is a quotient stack for $n \leq 1$, but I don't know the answer for any $n \geq 2$. – Angelo Feb 28 '11 at 7:03
• The stack $\mathcal{X}_n$ is not a global quotient stack for $n\geq 2$ (Kresch: Flattening stratification and the stack of partial stabilisations of prestable curves). But, as noted by David Zureick-Brown in arxiv:1208.2882, $\mathcal{X}_n\to \mathrm{Spec}(\mathbb{C})$ is not a good moduli space so this does not answer the question. – David Rydh Feb 5 '13 at 19:10

This is answered in general by Theorem 13.1 in our paper The étale local structure of algebraic stacks (arXiv:1912.06162). Let $$\mathcal{X}$$ be a stack with a good moduli space $$X$$ such that
1. $$\mathcal{X}$$ has affine stabilizers,
2. $$\mathcal{X}$$ has separated diagonal, and
3. $$\mathcal{X}$$ is of finite presentation over an algebraic space.
Then $$\mathcal{X}$$ has the resolution property étale-locally (and even Nisnevich-locally) on $$X$$. In particular, if $$X$$ is the spectrum of a field or a henselian local ring, then $$\mathcal{X}$$ has the resolution property.
Conditions 1 and 2 are necessary. Note that having the resolution property locally on $$X$$ implies that $$\mathcal{X}$$ has affine diagonal.
In general $$\mathcal{X}$$ does not have the resolution property Zariski-locally on $$X$$. There is an example in SGA 3, Exp X, §1.6 (cf. Remark 2.5 in our paper), of a 2-dimensional torus $$G$$ over the nodal cubic curve $$C$$ such that $$G$$ is not locally isotrivial. This means that $$G$$ cannot be embedded in $$\mathrm{GL}_N$$ for any $$N$$, not even Zariski-locally on $$C$$. It follows that $$\mathcal{X}=BG$$ does not have the resolution property Zariski-locally on its good moduli space $$X=C$$.