Suppose I have a certain (contravariant) moduli functor $M:Schemes \to Groupoids$ that is represented by a quotient stack $[X//G]$ where $X$ is a scheme and $G$ a linearly reductive group. Roughly speaking $[X//G]$ is the fine moduli space for $M$; hence it carries a universal family and there's a correspondence between families of objects of $M$ over a scheme $S$ and maps $S\to [X//G]$. Now, I claim that also $X$ carries a universal family, obtained via pull-back from $[X//G]$, and it is constant along the fibers of the quotient map $X \to [X//G]$.

If $S$ is once again a family of objects of $M$, what can one say of maps from $S\to X$? That the family over $S$ induces a map to $X$ that is unique up to $G$-action? That such a map is unique as long as $X\to [X//G]$ has a section?

Basically, by the existence of the universal family over $X$ a map $S\to X$ induces a family over $S$ but not in a unique way. So, in the other direction, given a family over $S$ this induces an orbit under $G$ of maps, but there's no canonical choice of a particular map in this orbit.


1 Answer 1


First a couple corrections. If $M$ is represented by quotient stack, it better be a contravariant functor, and moreover, it should probably take values in groupoids, not set. Anyway, here's what going on:

$X$ does not carry a universal family, but it carries a "locally" universal family $v \in M\left(X\right)$, in the sense that, if $S$ is a scheme, and $x \in M\left(S\right)$ then you can find a cover (say in the etale topology, if that's where your stack is) $\left(f_\alpha:S_\alpha \to S\right)$ such that each $f_\alpha^*\left(x\right) \in M\left(S_\alpha\right)$ arises, up to isomorphism as $g_\alpha^*\left(v\right)$ for some $g_\alpha:S_\alpha \to X.$

I haven't used the group $G$ yet. Lets do that. A morphism $S \to M$ corresponds to a choice of a $G$-torsor $P \to S$ and a $G$-equivariant map $P \to X.$ So the map $S \to M$ need not factor through $X,$ unless the torsor is trivial, but it always factors through etale-locally.

  • $\begingroup$ Ok let's say that I composed with a forgetful functor $f: Groupoids \to Sets$.... :) Just kidding, it is a good remark, I edited. That's more or less what I felt: that I can lift the map $S$-globally iff the torsor $X \to [X//G]$ is trivial over the image of $S$. $\endgroup$
    – IMeasy
    Apr 11, 2013 at 15:13
  • $\begingroup$ But I still don't understand one thing. By pulling back the universal family from $[X//G]$ to $X$, don't I get a "universal" $G$-invariant family on $X$? $\endgroup$
    – IMeasy
    Apr 11, 2013 at 15:18
  • $\begingroup$ What do you mean by "universal"? To me, universal means that every family is a pullback of it, which is not true here, but is locally- that's why I said locally universal. If instead, you mean canonical, then sure. $\endgroup$ Apr 11, 2013 at 15:57
  • $\begingroup$ By the way, in some more detail, there is a canonical map $$\left[X//G\right] \to \left[pt//G\right]=BG$$, which is faithful. This is how you get a $G$-torsor out of a map $$S \to \left[X//G\right]$$ (it is classified by the composition into $\left[pt//G\right].$) I should be a bit careful, since I am used to working with topological stacks, but I think everything should work. $\endgroup$ Apr 11, 2013 at 16:01
  • $\begingroup$ Yes, I should have said canonical. Universal has a categorical meaning which is not true here. Thank you! $\endgroup$
    – IMeasy
    Apr 11, 2013 at 16:16

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