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Let M be your favorite moduli stack over the field of complex numbers.

Is it reasonable to expect M to be a Deligne-Mumford stack?

I know this is true for the moduli space of curves of genus g, ppav's and K3 surfaces. I'm just wondering what I should expect when considering other moduli stacks.

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  • $\begingroup$ Stable coherent sheaves are simple and so have a $\mathbb{G}_{m}$ of automorphisms. $\endgroup$
    – dhagbert
    Apr 29, 2013 at 22:00
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    $\begingroup$ My favorite is $Vect_{n,d}(C)$ of vector bundles on a curve of genus $g\geq 2$ of fixed rank and degree up to isomorphism. The automorphism group has positive dimension, hence not a Deligne-Mumford stack. Maybe what you should expect is to replace Deligne-Mumford by Artin. $\endgroup$ Apr 29, 2013 at 22:09
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    $\begingroup$ I'm voting to close this question as it has no real answer. Or maybe the answer is "it depends on your taste", and on the meaning of "favorite". $\endgroup$ Apr 29, 2013 at 22:50
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    $\begingroup$ The problem is, one can define a moduli stack (which is Artin) for any groupoid in the category of complex manifolds. Generically these won't be étale, hence the stack won't be Deligne-Mumford. $\endgroup$ Apr 30, 2013 at 3:44
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    $\begingroup$ Why voting to close? The question has a simple answer: 'no'. $\endgroup$ Apr 30, 2013 at 8:14

2 Answers 2

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If the objects under consideration all have finite automorphism groups, you should expect your stack to be Deligne-Mumford. Otherwise, it isn't Deligne-Mumford; but that is no cause for alarm.

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  • $\begingroup$ It would be a bit more accurate to say "finite etale automorphism schemes" (though since offered just as an "expectation", perhaps one cannot insist on too much precision). $\endgroup$
    – user28172
    Apr 30, 2013 at 23:24
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    $\begingroup$ As nosr points out, this answer is incorrect. For a very natural example, the stacks of Kontsevich stable maps are not (usually) Deligne-Mumford in positive characteristic. $\endgroup$ May 1, 2013 at 1:40
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    $\begingroup$ @Jason: For another example, $X_0(N)$ (appropriately defined as a proper flat Artin stack over $\mathbf{Z}$) is not Deligne-Mumford in characteristic $p$ when $p^2|N$. $\endgroup$
    – user28172
    May 1, 2013 at 2:33
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    $\begingroup$ Correction: As userN points out, I missed the OP's hypothesis that the stack is defined over $\text{Spec}(\mathbb{C})$. For stacks over $\text{Spec}(\mathbb{C})$, I agree with Ravi: every algebraic stack with finite diagonal is Deligne-Mumford. This follows from Artin's theorems. $\endgroup$ May 1, 2013 at 10:20
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No (per the many examples in the comments).

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    $\begingroup$ The question specifies 'over the complex numbers', which rules out all of the examples listed so far. I think Ravi's right that we should expect DM in the case of complex algebraic objects with finite automorphism groups. $\endgroup$
    – user1504
    May 1, 2013 at 3:05

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