I have stumbled upon some literature on Deligne-Mumford stacks, and it seems to me, at least superficially, that there is a strong link between DM-stacks which have "trivial generic stabilizers" and "effective etale stacks" in the sense of:

Intrinsic Characterization of when an orbifold (or more general stack) is effective?

However, the references I have found do not define what it means to have trivial generic stabilizers, but rather assume the reader is familiar with the terminology. Would someone mind telling me the precise definition?

Also, is there much currently known (or in the literature) about the precise link between DM-stacks with this properties, and effective etale stacks?

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    $\begingroup$ For any quasi-separated Artin stack, the locus of points with trivial automorphism scheme is Zariski-open and (by a theorem of Artin) is an algebraic space. One can ask that this open subspace be Zariski-dense, and it is equivalent to this open subspace containing all generic points. That is, the automorphism schemes at generic points are trivial if and only if there is a dense open substack that is an algebraic space. $\endgroup$
    – user76758
    Nov 28, 2013 at 2:12
  • $\begingroup$ @user76758: Thanks for your response. So, if I am understanding you correctly, in particular, is the following equivalent to $\mathscr{X}$ having "trivial generic stabilizers"? : There exists an \'etale atlas from a scheme $X \to \mathscr{X}$ such that for each generic point $p$ of $X,$ the stabilizer of $p$ in $\mathscr{X}$ is trivial? $\endgroup$ Nov 28, 2013 at 2:33
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    $\begingroup$ Yes, and it would be true for any etale atlas from a scheme. But isn't it more "geometric" to say that the given stack contains a dense open substack that is an algebraic space? $\endgroup$
    – user76758
    Nov 28, 2013 at 4:02
  • $\begingroup$ @user76758: It is more geometric yes, but I was just checking that I understood the definition correctly, as there was a lot of terminology in your comment. Do you perhaps know the answer to the question I posted as the comment to the answer below btw? $\endgroup$ Nov 28, 2013 at 4:17
  • $\begingroup$ I don't understand why one would expect a condition at generic points should be equivalent to a condition at "each point $p$" (which $p$?), nor what "acting faithfully" is suppose to mean (faithfully on what, and in what sense)? I also don't know anything about the topological context, so I'll bow out here. $\endgroup$
    – user76758
    Nov 28, 2013 at 5:03

1 Answer 1


The comment by user76758 hits the nail pretty much perfectly on the head. That said, it might be good to see an example of something that does not satisfy this condition:

Let $X$ be any scheme, and let $G$ be any non-trivial finite group. Then there is a trivial action of $G$ on $X$:

$$ G \times X \to X \qquad (g, x) \mapsto x $$

The stack-quotient of this, $[G/X]$ has $G$ as a stabilizer for every point; this is the simplest example of a DM stack with non-trivial generic stabilizer.

  • $\begingroup$ Thanks, this is the kind of example I was certainly expecting for something with non-trivial generic stabilizers. Can I make precise the following idea? : $\mathscr{X}$ has trivial generic stabilizers if and only if "each point $p$ has $Aut(p)$ acting faithfully"? This is the situation that occurs in the topological context. $\endgroup$ Nov 28, 2013 at 2:45
  • $\begingroup$ @DavidCarchedi It seems to me that if we look at the quotient of the affine line by reflection through the origin, then the generic stabilizer is trivial, but the action at the origin is not faithful. (I may be wrong; I am not quite fluent with stacks.) $\endgroup$ Dec 20, 2013 at 3:50

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