2
$\begingroup$

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.

Improve this question
$\endgroup$
6
  • $\begingroup$ Stable coherent sheaves are simple and so have a $\mathbb{G}_{m}$ of automorphisms. $\endgroup$
    – dhagbert
    Commented Apr 29, 2013 at 22:00
  • 11
    $\begingroup$ 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. $\endgroup$
    – Jeremy Pecharich
    Commented Apr 29, 2013 at 22:09
  • 1
    $\begingroup$ 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". $\endgroup$
    – André Henriques
    Commented Apr 29, 2013 at 22:50
  • 1
    $\begingroup$ 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. $\endgroup$
    – David Roberts ♦
    Commented Apr 30, 2013 at 3:44
  • 8
    $\begingroup$ Why voting to close? The question has a simple answer: 'no'. $\endgroup$
    – Matthieu Romagny
    Commented Apr 30, 2013 at 8:14

2 Answers 2

Reset to default
8
$\begingroup$

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.

Improve this answer
$\endgroup$
4
  • $\begingroup$ 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
    Commented Apr 30, 2013 at 23:24
  • 4
    $\begingroup$ 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. $\endgroup$
    – Jason Starr
    Commented May 1, 2013 at 1:40
  • 2
    $\begingroup$ @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
    Commented May 1, 2013 at 2:33
  • 2
    $\begingroup$ 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. $\endgroup$
    – Jason Starr
    Commented May 1, 2013 at 10:20
7
$\begingroup$

No (per the many examples in the comments).

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ 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
    Commented May 1, 2013 at 3:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.