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Let $X = [P/G]$ be a smooth finite type separated DM-stack over $\mathbb C$ given as the quotient of a smooth quasi-projective scheme $P$ by the action of a smooth (finite type separated) reductive group scheme $G$. Since $X$ has finite inertia, the coarse space $X^c$ of $X$ exists as an algebraic space.

At atlas of $X$ is an étale morphism $U\to X$ with $U$ an algebraic space.

I sm interested in the converse of the following statement:

Thm. If the coarse moduli space of $X$ is a scheme,then every atlas $U$ of $X$ is a scheme.

Proof. The map from an atlas $U$ to the coarse moduli space $X^c$ of $X$ is quasi-finite separated. Therefore $U$ is a scheme by Knutson Cor. II.6.16., p. 138 QED

So the converse would read as follows:

Q. Suppose that every atlas $U$ of $X$ is a scheme. Is the coarse moduli space of $X$ a scheme?

I expect the answer to be negative, but can't find a good example. Note that a counterexample can not be an algebraic space, as an algebraic space with the property that every atlas is a scheme is itself a scheme (the identity morphism being an atlas).

Edit: I have an application in mind of the above question, and in the context of my application the stack $X$ is even generically a scheme. Not sure if that helps...

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  • $\begingroup$ Do you have an example, without assuming your condition on the atlases, where $P$ is quasi-projective and $G$ is reductive, $[P/G]$ is separated DM, but the coarse space of $[P/G]$ is not a scheme? $\endgroup$ Commented Jan 15, 2022 at 18:28
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    $\begingroup$ @DoriBejleri I think (but I'd have to double-check) that Kollar's paper arxiv.org/pdf/math/0501294.pdf contains such examples. Do you agree? (BTW, this question is seven years old, and I somehow decided to revive it now, but I don't even remember the "application in mind" at the moment.) $\endgroup$ Commented Jan 15, 2022 at 18:51
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    $\begingroup$ Yes it looks like the example from Corollary 6 does the job! After taking a quick look at the proof, it seems like the condition that the coarse space is a scheme should be equivalent to something like the following: for each point $x \in P$, there exists a linearized ample line bundle $L$ such that $x$ is $L$-semistable. I'm not sure how this interacts with your atlas condition though. $\endgroup$ Commented Jan 15, 2022 at 22:49

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