On the coarse moduli space of a stack - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T10:36:07Z http://mathoverflow.net/feeds/question/120567 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/120567/on-the-coarse-moduli-space-of-a-stack On the coarse moduli space of a stack Nullstellensatz 2013-02-02T01:08:15Z 2013-02-04T08:04:14Z <p>Consider a stack $\mathcal{X}$ over $\mathbb{C}$ as a category fibred in groupoids over the category of schemes. Let $\mathcal{X}^s$ be the $\pi_0$ of this category, i.e. objects of $\mathcal{X}^s$ are the objects in $\mathcal{X}$ and morphisms of $\mathcal{X}^s$ are the morphisms in $\mathcal{X}$ modulo automorphisms of objects. It "kills" the groupoid structure, so I think it is possible to consider $\mathcal{X}^s$ as a category fibred in sets over the category of schemes. Assume $\mathcal{X}^s$ is represented by a scheme. Should it be the coarse moduli space for $\mathcal{X}$?</p> http://mathoverflow.net/questions/120567/on-the-coarse-moduli-space-of-a-stack/120603#120603 Answer by Dan Petersen for On the coarse moduli space of a stack Dan Petersen 2013-02-02T16:49:26Z 2013-02-04T08:04:14Z <p>Yes, this would imply that $\newcommand{\X}{\mathcal X}\X^s$ is the coarse moduli space, but I don't think this is the "right" question to ask -- I believe that $\X^s$ will not even form a sheaf unless $\X$ happens to be a scheme/algebraic space to begin with. </p> <p>Anyway, any morphism from a groupoid to a set factors through $\pi_0$ of the groupoid. This implies in particular that any morphism from $\X$ to an algebraic space factors through the presheaf $\X^s$. And the map $\X \to \X^s$ is a bijection on geometric points because it's in fact a bijection on $S$-points for any scheme $S$. So if $\X^s$ is a scheme/algebraic space then it is the coarse moduli space.</p> <hr> <p>Addendum. I think you are confused about some basic issues. Let us see why $BG^s$ is not the coarse moduli space of $BG$. Let $G$ be a nontrivial finite group, say.</p> <p>Consider for simplicity the topological setting, so we have a topological space $X$ and an open cover <code>$\{U_i\}$</code>. If we have a $G$-torsor on $X$ then we can restrict to a $G$-torsor on each $U_i$, and on each overlap $U_i \cap U_j$ we have isomorphisms between the restrictions from $U_i$ and from $U_j$. These isomorphisms satisfy cocycle relation. Conversely, if we have $G$-torsors on each $U_i$ and isomorphisms satisfying the cocycle relation, we can reconstruct a $G$-torsor on the whole of $X$, unique up to canonical isomorphism. What this paragraph says is exactly that the functor $BG$ which sends a space to the groupoid of $G$-torsors over it is a <em>sheaf of groupoids</em>, that is, a <em>stack</em>. (In the usual Grothendieck topology on the category of topological spaces, where open covers are, well, open covers. And when I call $BG$ a "functor" I should say "pseudofunctor" or "fibered category".)</p> <p>On the other hand we can consider $BG^s$, which is now a priori just a presheaf of sets, mapping a space to the <em>set</em> of <em>isomorphism classes</em> of $G$-torsors over it. If we have an isomorphism class of $G$-torsor on $X$ then we get well defined isomorphism classes of $G$-torsors on each $U_i$ with compatible restrictions to each $U_i \cap U_j$. But it is NOT true that if we have an isomorphism class of $G$-torsor on each $U_i$ which agree on double overlaps, then we can reconstruct a unique isomorphism class on all of $X$: consider the case when $G$ is nontrivial on $X$ and <code>$\{U_i\}$</code> is a trivializing cover! What this says is that $BG^s$ is in fact only a presheaf - it is NEVER a sheaf of sets. Put simply, <em>one can not glue together isomorphism classes</em>.</p> <p>What this shows is in fact that if we sheafify $BG^s$, then we get a point. If we only remember isomorphism classes of torsors then every $G$-torsor becomes equivalent to the trivial torsor on some open covering of your space, which means that these torsors are identified under sheafification.</p> <p>The same arguments work verbatim in algebraic geometry, since every $G$-torsor is locally trivial in the étale topology.</p> <p>In any case, this is why I said above that this is not the "right" question to ask: it is not natural to expect $\X^s$ to be a sheaf in the first place. I would suggest reading Heinloth or Fantechi's notes on stacks (they are somewhere online) and thinking over just what question it is you want to ask.</p>