9
$\begingroup$

Let $X$ be a smooth finite type separated connected Deligne-Mumford stack over $\mathbb C$.

Does there exist a finite etale morphism $Y\to X$ with $Y$ a scheme?

What if $X$ is an algebraic space (i.e., trivial stabilizers)?

Edit: I changed the old question to a different question which should be more clear. An answer to the new question would help a lot in answering the old question.

Improve this question
$\endgroup$
6
  • $\begingroup$ Take $X$ a stacky $\mathbb{P}^1$ with $1/2$-structure at $0$ and $1$, and $1/4$-structure at $\infty$. This is a specialization of non-hyperbolic stacks that have $1/2$-structure at each of $4$ points; these have finite etale covers by elliptic curves. But the specializations of those etale covers are not smooth curves. So I suspect that $X$ is hyperbolic. However, there is a finite (non-etale) cover that is an elliptic curve: namely, first double-cover by $\mathbb{P}^1$ branched over $0$, $\infty$, then cover by an elliptic curve branched over the preimages of $0$, $1$, $\infty$. $\endgroup$
    – Jason Starr
    Commented Oct 23, 2014 at 13:20
  • 1
    $\begingroup$ My example doesn't work: consider the homomorphism $\alpha:\pi_{1}(\mathbb{C}P^1\setminus\{0,1,\infty\}) \to \mathfrak{S}_8$ that sends the loop around $\infty$ to $(1234)(5678)$, sends the loop around $1$ to $(15)(28)(37)(46)$, and sends the loop around $0$ to $(18)(27)(36)(45)$, so that the product of the loops goes to the identity. The image is a transitive subgroup. By Riemann-Hurwitz, the finite etale cover of $X$ is $\mathbb{P}^1$. $\endgroup$
    – Jason Starr
    Commented Oct 23, 2014 at 13:31
  • $\begingroup$ @JasonStarr Are you giving a counterexample in your second comment? I don't quite follow. The assumption in the question is that any atlas (finite etale or just etale surjective) of $X$ is hyperbolic. In the case of orbifold curves this certainly implies that X is hyperbolic, as an orbifold curve with a hyperbolic finite etale atlas has universal covering $\mathbb H$. $\endgroup$
    – Ariyan Javanpeykar
    Commented Oct 23, 2014 at 13:52
  • $\begingroup$ My second comment shows that the example in my first comment is not a counterexample. I do not give any counterexample (I proposed an example, but it turns out not to be a counterexample). $\endgroup$
    – Jason Starr
    Commented Oct 23, 2014 at 13:56
  • $\begingroup$ Ok, sorry about that. I was just confused and worried for a second. :) $\endgroup$
    – Ariyan Javanpeykar
    Commented Oct 23, 2014 at 13:57

2 Answers 2

Reset to default
9
$\begingroup$

It seems to me that the answer is NO if $X$ is a DM stack. If I'm not mistaken, it suffices to give a smooth finite type separated connected Deligne-Mumford stack over $\mathbb{C}$ which is simply connected (since such a thing has no non-trivial finite etale covers, let alone finite etale covers by a scheme). But this paper of Behrend and Noohi shows that the weighted projective lines $\mathcal{P}(m, n)$ (constructed by taking the stack quotient of $\mathbb{A}^2\setminus\{0\}$ by the $\mathbb{G}_m$-action $\lambda\cdot(x,y):=(\lambda^m x, \lambda^n y)$) are simply connected. The proof is easy; one just uses the long exact sequence for homotopy groups associated to the fibration $$\mathbb{G}_m\to \mathbb{A}^2\setminus\{0\}\to \mathcal{P}(m, n).$$

Added later: The answer seems to be no for algebraic spaces as well. Example 5.7 here is simply connected if I'm not mistaken, and is not a scheme by Remark 3.4 in the same paper.

Improve this answer
$\endgroup$
0
4
$\begingroup$

To give a more simple example than Daniel's, you can just consider for X a projective line with a single orbifold point. By Riemann-Hurwitz X is simply connected and so there is no non-trivial finite étale morphism Y→X. This holds over an algebraically closed field of characteristic zero say (but would work in characteristic p as well by defining precisely X as a stack of roots in the sense of Vistoli - see Charles Cadman, Using stacks to impose tangency conditions on curves, for the precise definition).

Also, you may want to consider the following closely related notion, taken from

Fundamental Groups of Algebraic Stacks Behrang Noohi http://arxiv.org/abs/math/0201021

"An algebraic stack being uniformizable means that it has a finite étale representable cover by an algebraic space (roughly speaking, its “universal cover” is an algebraic space)."

The author proceeds to show that, roughly, a DM stack X is uniformizable iff all morphisms from the stabilizers to the fundamental group of X are injective.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Isn't this the special case of my answer, $\mathcal{P}(1, n)$ (at least in characteristic zero)? (Indeed, it must be, since Behrend-Noohi show that the only smooth proper DM curves are $\mathcal{P}(m, n)$ in characteristic zero.) $\endgroup$
    – Daniel Litt
    Commented Oct 31, 2014 at 15:41
  • $\begingroup$ Oh, and one must be careful in characteristic $p$; for example, $\mathbb{P}^1/(\mathbb{Z}/p\mathbb{Z})$ where $\mathbb{Z}/p\mathbb{Z}$ acts by translations is manifestly not simply connected, even though it has a single orbifold point at $\infty$. $\endgroup$
    – Daniel Litt
    Commented Nov 1, 2014 at 21:06
  • $\begingroup$ About your first comment: it seems better to me to give the simplest possible example than a class including pointless complications such as non-generic stabilizers. About your second comment : I was careful about what you mention. Your quotient stack is interesting but is definitely not a stack of roots. $\endgroup$
    – Niels
    Commented Nov 2, 2014 at 9:05
  • $\begingroup$ Fair enough; the root stacks need not be DM in char p, though, as I'm sure you know. $\endgroup$
    – Daniel Litt
    Commented Nov 2, 2014 at 14:24

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .