# closed substack of quotient stack

The question concerns quotient stacks. I am not very comfortable with stacks, so feel free to edit the question if I am saying nonsense. It is also the reason why I may be spelling in too much detail some obvious facts...

Fix a base scheme $S$ for everything that follows. Say we have a quotient stack $[X/G]$, where $X$ is a scheme and $G$ acts on $X$ (e.g. $G=GL_n$) and consider $F$ a substack of $[X/G]$, such that $F\hookrightarrow [X/G]$ is representable and a closed immersion.

(If it helps, think of $[X/G]$ as some moduli space of some objects and $F$ a moduli space classifying those objects with an extra property $P$).

If I understand correctly, we have a map (of stacks) $X\to[X/G]$ corresponding to the trivial torsor $G\times X$: \begin{array}{c c c} G\times X & \to & X \\ \downarrow & & \\ X & & \end{array} where the vertical map is projection and the horizontal is the action.

Q1: is this map universal in any sense? (It seems so to me, but I can't make it a precise statement). In the moduli space setting, it seems to correspond to ''the most general object'' you can write, assuming the objects can be cut out by equations). Does this make sense?

Edit/addendum to Q1: In the moduli space setting, the "distinguished" element $X\to[X/G]$ corresponds to some object $A$ over $X$. What can be said about $A$?

Continuing, take the fiber product \begin{array}{c c c} Z & \to & F \\ \downarrow & & \downarrow\\ X & \to & [X/G] \end{array} gives a closed subscheme $Z$ of $X$ (by the assumption on $F$).

Q2: Under what conditions on $F$ (or is it always true), do we have $F=[Z/G]$? A priori, it's not even clear that $Z$ has a $G$-action, but I believe I checked that and I think I can write a map $[Z/G]\to F$.

And the last question: Q3: Again, is there any "universality" to this? It seems to me that it says something along the lines of: take a scheme $U$ over $S$. Then a map $U\to[X/G]$ corresponds to an object $A$ classified by our moduli problem and the assumption on $F$ shows that the fiber product $V$ is closed in $U$: \begin{array}{c c c} V & \to & F \\ \downarrow & & \downarrow\\ U & \to & [X/G] \end{array} Then $A$ (over $U$) has property $P$ if and only if it factors through $V$? Or maybe (if $P$ can be checked on points), $V$ is the ''locus'' where the object $A$ has property $P$. Does this make sense?

Thank you.

• I wouldn't have thought that the map $X\to [X/G]$ satisfies a universal property the category of stacks (without explicitly referring to $X$). For example, you can write a $pt = [G/G]$ for any group $G$... more generally, a given stack may be expressed as a quotient stack in multiple ways, so $X$ is not uniquely determined. Apr 13, 2018 at 2:56
• I believe the answer to Q2 is "always": $Z\to F$ is the pullback of the $G$-torsor $X\to [X/G]$, so in particular is a $G$-torsor itself. Apr 13, 2018 at 3:00
• Here's a silly remark. Consider the circle $Y = S^1$. Let $X_1 = R^2 - 0$, so that $Y = X_1/G = [X_1/G]$, where $G = R - 0$ (as the action is free, the brackets [ ] don't do anything). On the other hand, let $X_2 = R$. We could view $Y = X_2/Z$. The only thing intrinsic about Y is Y itself, and $X_1, X_2$ are giving you two presentations of $Y$ as a quotient. I think this is partly why Q1 and Q3 lack a good answer. On the other hand, the answer to Q2 is 'always true', as SG said. That's bcause the pullback of a G-bundle is a G-bundle, and the map $X \to [X/G]$ is a G-bundle (aka torsor). Apr 13, 2018 at 4:26
• Thanks @SamGunningham and Yosemite Sam (can't tag both for some reason). What if in Q1 and Q3 we restrict to the category of $X$-schemes? (and hence objects to be classified over schemes over $X$). Also, any thoughts on the last part of Q3 (about objects having property $P$)? (it doesn't depend on a specific choice of $X$..) Apr 13, 2018 at 5:51
• I modified Q1 a bit. I think it's less vague now. Apr 13, 2018 at 9:12