Welcome to MathOverflow

Question

3

1

Let $\mathbb{Z}/2\mathbb{Z}$ act on $\mathbb{A}^1$ as $x \mapsto -x$, and let $\mathscr{X}$ be the quotient stack. It has coarse moduli space $\mathbb{A}^1$ and a residual gerbe $B\mathbb{Z}/2\mathbb{Z}$ at 0. There is then a surjective map from cartesian powers $\mathscr{X}^n$ to $\mathbb{A}^n$. What are the closed substacks of $\mathscr{X}^n$? Will there be anything other than pullbacks of closed subschemes of $\mathbb{A}^n$?

flag	asked Nov 18 2010 at 17:17 Dima Sustretov 1,287●4●12

Jim Bryan · Accepted Answer · 2010-11-18 17:40:15Z UTC

5

A substack of $\mathcal{X}$ is given by a $Z/2Z$ invariant subscheme of $\mathbb{A}^1$ and hence given by a $Z/2Z$ invariant ideal $I\subset k[x]$. So for example, the ideals $(x^n)$ give substacks, but they only pullback back from subschemes of the coarse space for $n$ even: the coarse space is given by $Spec(k[x^2])$. The substack given by $(x)$ here is just a copy of $B(Z/2Z)$ embedded at the origin.

answered Nov 18 2010 at 17:40

Jim Bryan
2,659●1●7●22

Thanks, Jim! I have a quick followup question, which is a generalisation of the initial one. So, on schemes there is a 1-to-1 correspondence between closed subschemes and quasi-coherent ideal sheaves. Is there a similar correspondence for stacks? From your answer I would suppose that only those ideal sheaves on the coarse moduli space that satisfy some invariance condition define a closed substack. Is there such an invariance condition for a general stack (i.e. not necessary a group quotient)? – Dima Sustretov Nov 18 2010 at 18:23

1

When $G$ acts on $X$, then substacks of $[X/G]$ are the same as $G$-invariant subschemes of $X$, which are given by $G$-invariant ideal sheaves, which are the same as ideal sheaves on $[X/G]$. Sheaves on the coarse moduli space $X/G$ don't come into the equation, except for the fact that you can pull them back to $[X/G]$ and obtain some of the ideal sheaves on $[X/G]$. I use $X$ and $X/G$ because in your example, unfortunately both $X$ and $X/G$ are isomorphic to $A^1$, which adds some confusion. – ABayer Nov 19 2010 at 17:35

closed substacks of cartesian powers of a stack

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

1 Answer

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.