I'm trying to understand the definition of the quotient stack $[X/G]$ as defined in Frank Neumann's Algebraic Stacks and Moduli of Vector Bundles.

Explicitly, let $G$ be an affine smooth group $S$-scheme with right action $\rho:X\times G\to X$ on a noetherian $S$-scheme $X$. The quotient stack $[X/G]$ is the pseudofunctor $$ [X/G]:(\mathsf{Sch}/S)^{\operatorname{op}}\to\mathsf{Grpds} $$ defined as follows.

For an $S$-scheme $U$ let $[X/G](U)$ be the category whose objects are diagrams $$ U\xleftarrow{\pi}E\xrightarrow{\alpha}X $$ where $\pi$ is a principal $G$-bundle and $\alpha$ is a $G$-equivariant morphism. The morphisms in $[X/G](U)$ are the isomorphisms of principal $G$-bundles commuting with the $G$-equivariant morphisms.

For a morphism of $S$-schemes $f:U^\prime\to U$ let $$ [X/G](f):[X/G](U)\to[X/G](U^\prime) $$ be the functor induced by pullbacks of principal $G$-bundles.

It is not clear to me how to determine when $[X/G]$ is representable in general. My questions are:

Question 1. Is there a sufficient condition we can impose on $\rho$ to ensure that $[X/G]$ is representable? The Wikipedia article on quotient stacks says something about when the categorical quotient $X/G$ exists the canonical map $[X/G]\to\operatorname{Hom}(-,X/G)$ need not necessarily be an isomorhpism. This is somewhat opaque to me as I'm not even sure how $[X/G]\to\operatorname{Hom}(-,X/G)$ is defined.

Question 2. What is an example where $[X/G]$ is not representable?

If these questions are too broad, I'd be very grateful if someone could point me in the direction of a good reference.

  • $\begingroup$ Functors represented by schemes land in the category of sets. But as you can see, your functor has values in groupoids and not sets, except for "trivial" cases already mentioned in the answers. $[X/G]$ is still an algebraic stack, though. One doesn't need that $X$ is noetherian. $\endgroup$ – Martin Brandenburg Oct 12 '14 at 20:55

The functor $[X/G]$ is not representable whenever there is non-trivial isotropy of the action of $G$ on $X$.

Let us consider the most extreme case: when $X = \bullet$ is a point (the terminal object) and $G$ is any non-trivial group. In such a case, $Hom(-,\bullet/G)$ is a singleton, as $\bullet / G = \bullet$, which is still terminal.

However, $[\bullet / G]$ is much more interesting than that! If you trace through the definition provided above (which is worth doing at least once in your life), we find that $[\bullet / G](U)$ is the collection of principal $G$-bundles over $U$, which is non-trivial most of the time. Thus it follows that, as a stack, $[\bullet / G]$ is the classifying space of $G$.

Particular examples are $[\bullet / \mathbb{G}_m]$ being the collection of line bundles, etc.

As for the question about how the morphism $[X/G] \to Hom(-,X/G)$ is defined:

In your diagram, the morphism $\alpha : E \to X$ is equivariant. You can thus complete the diagram to

$$ \begin{array}{c c c} E & \to & X \\ \downarrow & & \downarrow\\ U & \to & X/G \end{array} $$

which yields the desired map.

  • $\begingroup$ Thanks for the example. Is it true that $[X/G]$ is representable if there is only trivial isotropy of the action of $G$ on $X$? $\endgroup$ – Brian Fitzpatrick Mar 3 '14 at 22:07
  • $\begingroup$ That is my understanding, yes. I've always held it near and dear that $[X/G]$ is representable as a scheme/topological space/whatever if and only if the action of $G$ on $X$ has no isotropy. In such a case, it is represented by the coarse quotient, which is defined about how you would expect it to be. $\endgroup$ – Simon Rose Mar 4 '14 at 3:45
  • $\begingroup$ I see. Do you know of any good references along these lines? $\endgroup$ – Brian Fitzpatrick Mar 4 '14 at 3:46
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    $\begingroup$ Hironaka constructed a 3 dimensional variety $X$ with a free action of $G = \mathbb{Z}/2\mathbb{Z}$, such that $[X/G]$ is not a scheme. The problem is that not all $G$-orbits are contained in affines. See section 4.4.2 at the end of Vistoli's notes arxiv.org/abs/math/0412512 $\endgroup$ – S. Carnahan Mar 4 '14 at 13:25
  • $\begingroup$ @SimonRose The action of $\mathbb G_m$ on $\mathbb A^1$ has non-trivial isotropy so $[\mathbb A^1/\mathbb G_m]$ is not representable. How would one describe this stack? $\endgroup$ – Brian Fitzpatrick Mar 4 '14 at 22:45

It follows from the definitions that $[X/G]$ is representable by a scheme $S$ if and only if $X$ is a $G$-torsor over $S$, i.e. the natural map $G\times X\rightarrow X\times _SX$ given by $(g,x)\mapsto (x,gx)$ is an isomorphism. Example 0.4 (Chapter 0, §3, page 11 of 3rd edition) in Mumford's GIT is a variety $X$ with an action of $G=SL(2)$ with trivial stabilizers and geometric quotient $\mathbb{A}^1$, but such that $X$ is not a $G$-torsor over $\mathbb{A}^1$, so $[X/G]$ is not representable.

  • $\begingroup$ Hi @abx! In the torsor condition that you quote, you really mean isomorphism of schemes, or is it enough to have a bijection on closed points? (let us say $S=\textrm{Spec }\mathbb C$ for simplicity) Thanks in advance! $\endgroup$ – Brenin Aug 26 '15 at 13:54
  • $\begingroup$ In the case $S=\mathrm{Spec}\mathbb{C}$ I think bijectivity is enough: since $G$ acts transitively $X$ is normal, and a bijective map of normal varieties is an isomorphism (in char. $0$). $\endgroup$ – abx Aug 26 '15 at 16:25

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