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Let $[X/G]$ be a quotient stack such that $X$ is irreducible and $G$ acts trivially on $X$ (I am just adding automorphisms to every point). Under which hypothesis is $[X/G]$ irreducible as an Artin stack?

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I guess $[X/G]$ is always irreducible, as long as $X$ is (namely one doesn't even need to assume the $G$ action is trivial). If $[X/G]$ is the union of two proper closed substacks, or equivalently, there exist two nonempty open substacks that do not meet, then their inverse images in $X$ do not meet, too, contradicting the assumption. –  shenghao Jun 20 '11 at 21:37
    
Supplementing shenghao's comment, you can show that for any quotient stack $X/G$, the open and closed substacks of $X/G$ are in bijection with the open and closed subschemes of $X$ (i.e., they're all of the form $Y/G$ for Y open or closed in $X$). –  Mike Skirvin Jun 20 '11 at 23:24
    
Mike, I assume that you still mean that $G$ acts trivially on $X$, since $[Y/G]$ isn't defined unless $Y$ is $G$-invariant. –  Daniel Bergh Jun 21 '11 at 10:41

1 Answer 1

up vote 3 down vote accepted

If you have a presentation $s, t:R \to U$ of a stack $[U/R]$, the set of points of $[U/R]$ is just the equivalence classes of points in $|U|$ determined by the equivalence relation given by the image of the map $|R| \to |U|\times |U|$. In particular $|U| \to |[U/R]|$ is always surjective.

The topology on the underlying sets of points of stacks is characterised by the following two properties:

  • 1-morphisms of stacks give continuous maps
  • flat morphisms locally of finite presentation give open maps

These statements are can be found here and here in the Stacks Project.

The topological properties of algebraic stacks therefore behave as expected. It is a purely topological fact that if you have surjective continuous map $U \to V$ of topological spaces, then $V$ is irreducible if $U$ is. The corresponding statements hold for quasi-compactness and connectedness. As commented above, this applies in your situation with the stack quotient $X \to [X/G]$, regardless of the action of $G$ being trivial or not.

If the action, as in your case, is trivial, the equivalence relation on $|X|$ becomes trivial as well. Hence we see that the map $|X| \to |[X/G]|$ is a bijection. Assuming that $G$ is flat and locally of finite presentation (this is required if we want $[X/G]$ to be algebraic), we see that $|X| \to |[X/G]|$ is even a homeomorphism. This illustrates that stackiness is invisible to the Zariski topology. The stackiness may be explored pointwise by considering the residual gerbes.

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