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Recall that an etale topological stack is a stack $\mathscr{X}$ over the category of topological spaces (and open covers) which admits a representable local homeomorphism $X \to \mathscr{X}$ from a topological space. Equivalently, it is a topological stack arising from an etale topological groupoid. It is well known that a differentiable stack is etale if and only if all of its automorphism groups are discrete, but the proof involves foliation theory. It seems this proof cannot be extended to the topological setting. However, clearly every etale topological stack has discrete isotropy groups. This begs the question:

If a topological stack has all of its isotropy groups discrete, is it necessarily etale?

EDIT: By a topological stack, I mean a stack $\mathscr{X}$ over the category of topological spaces (and open covers) which admits a representable epimorphism $X \to \mathscr{X}$ (not necessarily a local homeomorphism). This is equivalent to saying $\mathscr{X}$ is the stack of torsors for a topological groupoid.

Remark: This question is equivalent to asking if a topological groupoid all of whose isotropy groups are discrete must be Morita equivalent to an etale topological groupoid.

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  • $\begingroup$ What do you mean by a topological stack? Is it in the sense of Noohi? $\endgroup$
    – Angelo
    Commented May 8, 2011 at 4:21
  • $\begingroup$ Yes, depending on which article of his. I mean what he refers to as a "pretopological stack" in "Foundations of Topological Stacks I". I'll add this to the question. $\endgroup$
    – David Carchedi
    Commented May 8, 2011 at 10:43
  • $\begingroup$ Surely you want the representable map $X \to \mathcal{X}$ to have some additional property like being surjective? $\endgroup$
    – Chris Schommer-Pries
    Commented May 8, 2011 at 14:46
  • $\begingroup$ @Chris: Yes, I meant to say representable epimorphism. $\endgroup$
    – David Carchedi
    Commented May 8, 2011 at 21:31
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    $\begingroup$ This is not particularly important, but it might be good to change "begs the question" to "suggests the question", since the former is often used to mean something else. $\endgroup$
    – S. Carnahan ♦
    Commented May 10, 2011 at 13:14

2 Answers 2

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I don't think this is true. Let $X$ be the quotient of the action of $\mathbb Q$ on $\mathbb R$ by translation. This is a sheaf, and its automorphism groups are trivial. Suppose that there exist a local homeomorphism $U \to X$, where $U$ is non-empty a topological space. Let $V \to U$ be the pullback to $U$ of the $\mathbb Q$-torsor $\mathbb R \to X$; then $V\to \mathbb R$ is a local homeomorphism. By restricting $U$ we may assume that $V = U \times \mathbb Q$; but then $V$ can't be locally connected, and this is a contradiction.

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  • $\begingroup$ Angelo: Your stack certainly arises from an etale topological groupoid. So I don't think that it's a counterexample to what Dave is asking. $\endgroup$
    – André Henriques
    Commented May 8, 2011 at 12:24
  • $\begingroup$ Dear André, I am probably confused, since you do understand these matters much better than I do; but I still think that if I give $\mathbb Q$ the topology coming from the euclidean topology, the stack does not come from an étale groupoid (this is what the argument should show). $\endgroup$
    – Angelo
    Commented May 8, 2011 at 13:00
  • $\begingroup$ To clarify further, I am thinking of the groupoid $\mathbb Q \times \mathbb R \to \mathbb R$, where the maps come from the action by translation. Its associated stack is a non-representable sheaf, and it is what I had in mind. $\endgroup$
    – Angelo
    Commented May 8, 2011 at 13:02
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    $\begingroup$ Ahh. So there are two versions of this groupoid, corresponding to two different sheaves. Both are presented by action groupoids of $\mathbb{Q}$ acting on $\mathbb{R}$. In one $\mathbb{Q}$ has the discrete topology, and the result is an etale groupoid. In the other $\mathbb{Q}$ has the usual topology and the result is not an etale groupoid. Both stacks (sheaves) are covered by $\mathbb{R}$. It seems like the later is indeed a counter-example. $\endgroup$
    – Chris Schommer-Pries
    Commented May 8, 2011 at 14:38
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    $\begingroup$ To Chris: yes, that's what I was trying to say. $\endgroup$
    – Angelo
    Commented May 8, 2011 at 14:46
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Here's a counterexample: the stack associated to the relative pair groupoid of the map $$ ([0,1]\times\{0\}) \cup (\{1\}\times[0,1]) \cup ([1,2]\times\{1\})\;\;\to\; [0,2] \qquad\qquad\qquad\qquad\qquad $$

$$\qquad\qquad\qquad(x,y)\;\;\mapsto\;\;\; x$$

Equivalently, this stack can be described as the pushout in the 2-category of stacks of the diagram $[0,1]\leftarrow \{1\} \rightarrow [1,2]$ (where we identify a space with the stack it represents).

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  • $\begingroup$ As a pushout, this stack admits a map to the space [0,2]. How exactly does it differ from this space? $\endgroup$
    – Chris Schommer-Pries
    Commented May 8, 2011 at 14:43
  • $\begingroup$ @Chris Schommer-Pries: Let $X$ be the above pushout stack. As you noted, there is a map from $X$ to $[0,2]$, but that map is not an isomorphism. For any topological space $T$, the induced map $\hom(T,X)\to \hom(T,[0,2])$ is injective, and the subset $\hom(T,X)\subset \hom(T,[0,2])$ can be characterized. A map $f:T\to [0,2]$ comes from a map $T\to X$ iff $T$ has an open cover $T=T_1\cup T_2$ such that $f(T_1)\subset [0,1]$ and $f(T_2)\subset [1,2]$. $\endgroup$
    – André Henriques
    Commented May 8, 2011 at 19:23
  • $\begingroup$ Maybe I'm being dense. Doesn't any map f satisfy that property? Take $T_1$ to be the inverse image of the complement of [1,2], and likewise $T_2$ to be the inverse image of the complement of [0,1]? $\endgroup$
    – Chris Schommer-Pries
    Commented May 8, 2011 at 20:09
  • $\begingroup$ The two sets that you describe do not form a cover of $T$. You're missing the inverse image of {1}. $\endgroup$
    – André Henriques
    Commented May 8, 2011 at 20:20
  • $\begingroup$ Ooops! You're right. Okay, this also looks like a counter example to me. $\endgroup$
    – Chris Schommer-Pries
    Commented May 8, 2011 at 23:00

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