1
$\begingroup$

An algebraic stack or Artin stack is a stack in groupoids $\mathcal{X}$ over the étale site such that the diagonal map of $\mathcal{X}$ is representable and there exists a smooth surjection from (the stack associated to) a scheme to $\mathcal{X}$.

In Wikipedia's article on stacks I found in the excerpt a statement on local structure of algebraic stacks I do not understand:

[…] Given a quasi-separated algebraic stack $\mathcal{X}$ locally of finite type over an algebraically closed field $k$ whose stabilizers are affine, and $x \in \mathcal{X} (k)$ a smooth and closed point with linearly reductive stabilizer group $G_x$, there exists an etale cover of the GIT quotient […].

My question is what is here the stabilizer $G_x$ of $x$ at all? Recall we did not assume that $\mathcal{X}$ is a quotient stack, therefore it seems to me not to make any sense to speak about a "stabilizer group" of $x \in \mathcal{X} (k)= \operatorname{Hom}(\operatorname{Spec} k, \mathcal{X})$.

The point is that in order to talk about a stabilizer group $G_x$ of $x$ it is necessary to require the existence of a group $G$ acting on the set $\mathcal{X} (k)$ of $k$-valued points.

But for general algebraic stacks there is no reason that there is no reason that such group $G$ acting on $\mathcal{X} (k) $ such that $G_x \subset G$, right? Could somebody help me to resolve my confusion?

$\endgroup$
  • 1
    $\begingroup$ You still have a groupoid action you're killing as every algstack is a groupoid quotient, and the automorphisms at a point of a groupoid still form a group. The stabilizers are still subgroups of automorphism groups as well. $\endgroup$ – Harry Gindi Aug 9 at 15:24
  • $\begingroup$ I think I understand your argument. The (pre) stack $X$ is endowed with a fibred functor $F: X \to C$ to certain category $C$ and every fiber $F^{-1}(c)$ is a groupoid by definition. That means that for every $x \in F^{-1}(c)$ the set $Hom_{F^{-1}(c)}(x,x)$ equals $Aut(x)$. So for every $x \in X$ the group $Aut(x)$ acts on subset $\{f \in X(k) \ f(Spec(k)) =x \} \subset X(k)$ by composition $Spec(k) \to x \to x$. Is this the action you mean? $\endgroup$ – katalaveino Aug 9 at 15:41
  • 1
    $\begingroup$ Something like that. Loop spaces are always groups, and the stabilizer is the based loop space at a point. If you pull inertia back to a presentation, you literally get stabilizers wrt the smooth groupoid. $\endgroup$ – Harry Gindi Aug 9 at 15:45
  • $\begingroup$ @Harry Gindi: What do you mean by a "loop space" in this context? I looked up for a formal definition of a stabilizer $G_x$ of a $1$-morphism $x: Spec(k) \to X$. By definition it is defined as pullback (of stacks) of the morphisms $(x,x): Spec(k) \to X$ and the diagonal map $\Delta: X \to X \times_S X$. Also one can show that if $I_x$ is the inertia stack of $x: Spec(k) \to X$ then $G_x \cong I_x \times Spec(k)$. $\endgroup$ – katalaveino Aug 9 at 20:08
  • $\begingroup$ What I not completely understand is why and how this $G_x$ can be identified as subgroup of group $Hom_{F^{-1}(c)}(x,x)= Aut_{F^{-1}(c)}(x)$. Do you know literature where this identification is proved or give sketch of the construction? $\endgroup$ – katalaveino Aug 9 at 20:08
2
$\begingroup$

This was getting a little bit long for a comment, so I'll just write it here:

Let $X\simeq S//R$ be an algebraic stack presented by a smooth surjective map $S\to X$ with $S$ a scheme, then $R=S\times_X S$, and the pair of maps $R\rightrightarrows S$ has the canonical structure of a groupoid in algebraic spaces (with the additional structure coming from the higher simplices of the Cech nerve). Choosing a point $x$ in $X$ classified by some Zariski geometric point in $\operatorname{Spec}(k)\to S$, form the following big fibre square

$$ \begin{matrix} G_x & \to & S\times_X \operatorname{Spec}(k)&\to& \operatorname{Spec}(k)\\ \downarrow&&\downarrow&&\downarrow\\ \operatorname{Spec}(k)\times_X S&\to& R &\to & S\\ \downarrow &&\downarrow&&\downarrow\\ \operatorname{Spec}(k)&\to&S&\to&X \end{matrix} $$

In this case, the maps $G_x\to S\times_X \operatorname{Spec}(k)$ and $S\times_X \operatorname{Spec}(k)$ are injective, being pullbacks of injective maps, which gives an injective map $G_x\to R$, including as the literal stabilizer of the point $x\in S$ by the 'action' of $R$, it is including as the subgroupoid of automorphisms fixing $x\in S$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ One point isn't completely clear to me. In your comment you wrote: "every algstack is a groupoid quotient." Could your give a reference where an algebraic stack is introduced / constructed in that way? Or that this construction is equivalent to that one from Stacks project: stacks.math.columbia.edu/tag/026N $\endgroup$ – katalaveino Aug 10 at 13:17
  • $\begingroup$ As explaned in e.g. stacks.math.columbia.edu/tag/04TJ (here only for fppf) we can associate an alg stack to a smooth groupoid in algebraic spaces as groupoid quotient. But I not see why as you said every alg stack is obtained in that way. $\endgroup$ – katalaveino Aug 10 at 13:38
  • $\begingroup$ @katalaveino By choosing a smooth cover by a scheme or algebraic space, you can regenerate a morita-equivalent groupoid by taking the Cech complex. I don't know where this is in the stacks project, but it's easy to show that this always works. $\endgroup$ – Harry Gindi Aug 10 at 15:03
  • $\begingroup$ If you happen to find when you stumble upon a literature source that explains this construction in detail, could you name it? So far I have searched without successful in Olson's Alg Stacks and Laumon's & Moret-Bailly's Champs algebrques. $\endgroup$ – katalaveino Aug 10 at 16:13
  • 1
    $\begingroup$ @katalaveino It's the geometric realization of an internal Kan complex (in the category of (higher) fppf sheaves on Aff). The higher structure in the Cech nerve is not rich, it is coskeletal. You can see this from the fact that the diagonal of an algebraic stack is a relative algebraic space. I suggest looking through Jon Pridham's paper on simplicial schemes and algebraic stacks if you're still confused. $\endgroup$ – Harry Gindi Oct 1 at 4:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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