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Let $X$ be an Artin stack over the complex numbers. What can one say about the local structure of $X$, i.e. what is the simplest class of stacks by which were can always find a cover of $X$ by open substacks in this class? For example, does $X$ always have a cover by open substacks of the form $U/G$ where $G$ is an algebraic group acting on a scheme or complex analytic space $U$? I actually suspect that this is false, and that a counterexample is given by the moduli stack of nodal genus zero curves (I don't see how to construct such a neighborhood around $\mathbb CP^1\vee\mathbb CP^1$). If it is indeed false, what is the best one can say? I am very willing to pass to the analytification of $X$ if this helps get a stronger local structure result.

A related question: Stacks as local quotients or via atlases

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  • $\begingroup$ What do you mean by the "coarse space"? $\endgroup$
    – Angelo
    Commented Nov 23, 2019 at 14:25
  • $\begingroup$ @Angelo: Thank you, I've rephrased the question to avoid that notion. $\endgroup$
    – John Pardon
    Commented Nov 23, 2019 at 15:50
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    $\begingroup$ A DM stack has a coarse space, but it's still an algebraic space, and algebraic spaces can be pathological. $\endgroup$
    – Ben Wieland
    Commented Nov 23, 2019 at 20:36
  • $\begingroup$ Maybe I don't get the question, but if your stack is etale locally a quotient stack then surely it is also locally a quotient stack for the analytic topology? Since an etale cover can be refined to an analytic open cover. $\endgroup$
    – Dan Petersen
    Commented Nov 24, 2019 at 5:40
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    $\begingroup$ A stack which is the quotient by a group $G$ has inertia all subgroups of $G$. A finite dimensional algebraic group cannot contain infinitely many abelian varieties (up to isogeny?). Thus the moduli of (unmarked) genus 1 curves, which has inertia all elliptic curves is not locally a quotient stack. But I think that a finite dimesional family of unipotent groups can be embedded in a single linear group. $\endgroup$
    – Ben Wieland
    Commented Nov 25, 2019 at 2:19

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The stack of curves of genus 0 with at most one node is a quotient stack (see The integral Chow ring of the stack of at most 1-nodal rational curves, but Edidin and Fulghesu), so you are fine in this case.

On the other hand, Andrew Kresch proved that the stack of nodal curves of genus 0 with at most two nodes is not a quotient stack (Flattening stratification and the stack of partial stabilizations of prestable curves). Given that this stack has only three points, this should gives a counterexample.

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