A morphism of Artin stacks $f:X\to Y$ over $\mathbb Q$ is representable by algebraic spaces if and only if its geometric fibres are algebraic spaces. I would like to know if one can use this to prove the following statement.

Let $f:X\to Y$ be a morphism of finite type separated DM stacks over $\mathbb Q$. Suppose that, for any geometric point $x$ of $X$ with $y= f(x)$, the induced morphism on stabilizers $Stab(x)\to Stab(y)$ is injective. Then $f:X\to Y$ is representable by algebraic spaces.

  • $\begingroup$ Using pullback to an algebraic space over $Y$, your question is whether a DM stack $X$ (or more generally Artin stack) separated and finite type over an algebraic space $Y$ is itself an algebraic space when its geometric points have trivial Aut-schemes. This is true without char-0 hypotheses (so no need for Cartier!). See Theorem 2.2.5 in journals.cambridge.org/action/… (where Artin stacks are assumed to have diagonal separated and finite type); is this in the Stacks Project (maybe with weaker diagonal hypotheses)? $\endgroup$ – user74230 Mar 30 '15 at 1:40

This is http://stacks.math.columbia.edu/tag/04Y5 . I quote :



Let $S$ be a scheme contained in $Sch_{fppf}$. Let $f : \mathcal{X} \to \mathcal{Y}$ be a $1$-morphism of algebraic stacks over $S$. The following are equivalent:

1 for $U \in Ob((Sch/S)_{fppf})$ the functor $f : \mathcal{X}_U \to\mathcal{Y}_U$ is faithful,

2 the functor $f$ is faithful, and

3 $f$ is representable by algebraic spaces.


See also http://stacks.math.columbia.edu/tag/04YY for a fancy reformulation.

  • $\begingroup$ Thank you for your answer. Just to be sure I'd like to ask one question. In my situation, $S = Spec \mathbb Q$ and I only know that condition 1 holds for $U= Spec \mathbb C$ (or $U=$ spectrum of an alg closed field of char zero). Do I understand correctly that this is enough? I guess to prove this we just replace the target by a geometric point $y$ of $Y$, right? $\endgroup$ – user234 Mar 29 '15 at 14:31
  • 1
    $\begingroup$ Yes, it is enough to verify the property on geometric points : you need to check that the kernel say $K/U$ a is trivial, that is that the unit section $U\to K$ is an isomorphism. You can see these spaces as sheaves, and then use the fact that the geometric points form "a conservative family of points" see stacks.math.columbia.edu/tag/00YJ . $\endgroup$ – Niels Mar 29 '15 at 16:25
  • $\begingroup$ @Niels: Forming fibers of Aut-schemes at geometric points is that the OP wants to do, and that is very different from stalks as in the SP reference concerning "conservative family of points", so your suggestion for how to pass to considerations at geometric points seems a bit unclear. Are you sure that "see these spaces as sheaves" is sufficient for the purposes of the question posed? $\endgroup$ – user74230 Mar 30 '15 at 1:43
  • $\begingroup$ @user74230 : "see this spaces as sheaves" means to a space associate its functor of points ; this is sufficient since this Yoneda-like operation is fully faithful. Are you sure of "that is very different from stalks as in the SP reference" ? $\endgroup$ – Niels Mar 30 '15 at 12:36
  • $\begingroup$ @Niels: The difference is that base change on geometric objects along a map not "in the site" isn't generally computed by sheaf pullback for the representing functor. For example, if $X = Y \times \mathbf{A}^1$ over a scheme $Y$ then the scheme-theoretic pullback along geometric point $y:{\rm{Spec}}(k) \rightarrow Y$ is the scheme $\mathbf{A}^1_{k}$ whereas the fiber at $y$ of the functor represented by $X$ on the etale site is the constant sheaf over ${\rm{Spec}}(k)$ associated to the group of units in the strict henselization of $Y$ at $y$. So the SP reference seems not quite enough to me. $\endgroup$ – user74230 Mar 30 '15 at 14:59

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