most recent 30 from http://mathoverflow.net 2013-05-22T07:49:40Z http://mathoverflow.net/feeds/question/103598 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/103598/if-x-is-a-smooth-and-proper-stack-does-it-admit-a-smooth-and-proper-atlas Jacob Bell 2012-07-31T11:50:30Z 2012-07-31T13:50:59Z

<p>Fix a ground scheme $S$ (a field say). By atlas for an algebraic stack I mean a smooth and surjective morphism $Y \to X$ from a scheme (or algebraic space or affine scheme) $Y$. If the stack $X$ is smooth then all atlases are smooth schemes over $S$.</p> <p>What about when $X$ is smooth and proper? Can I find a proper atlas?</p> <p>If in general the answer is no, what about for Deligne-Mumford stacks?</p>

http://mathoverflow.net/questions/103598/if-x-is-a-smooth-and-proper-stack-does-it-admit-a-smooth-and-proper-atlas/103599#103599 Jason Starr 2012-07-31T12:07:48Z 2012-07-31T13:50:59Z

<p>The answer is "no". Consider a smooth, proper $1$-dimensional Deligne-Mumford stack over $\mathbb{C}$ which has coarse moduli space $\mathbb{P}^1$ and which has a single "stacky" point (with any nontrivial stabilizer group at that point). </p> <p><B>Edit.</B> A more precise definition of the stack is as follows. Let $m$ be an integer, $m>1$. Let $A$ be <code>$\mathbb{A}^2 \setminus \{(0,0)\}$</code> with coordinates $x$ and $y$. Let $\mathbb{G}_m$ act on $A$ by $t\ast (x,y) = (tx,t^m y)$. The stack $Q_m$ is the quotient stack $[A/\mathbb{G}_m]$. The quotient morphism $A\to Q_m$ is a smooth atlas, implying that $Q_m$ is smooth (since $A$ is smooth). In fact, this morphism is naturally a $\mathbb{G}_m$-torsor; denote the associated invertible sheaf on $Q_m$ by $\mathcal{L}$. It is not hard to see that this is an $m$-torsion invertible sheaf; the $\text{m}^{\text{th}}$ tensor power of $A$ as a $\mathbb{G}_m$-torsor over $Q_m$ is the quotient stack $[(\mathbb{G}_m\times A)/\mathbb{G}_m]$ where the $\mathbb{G}_m$-action is $t\cdot (u,(x,y)) = (t^mu, (tx,t^my))$. This admits a "trivializing" morphism to $\mathbb{G}_m$, $(u,(x,y)) \mapsto u/x^m = u/y$, each defined on the appropriate open $\mathbb{G}_m\times D(X)$ or $\mathbb{G}_m\times D(y)$. The only nontrivial stabilizer group is $\mathbf{\mu}_m$ acting on the $\mathbb{G}_m$-orbit <code>$\{0\}\times \mathbb{G}_m$</code> inside $A$. So $Q_m$ is a Deligne-Mumford stack. Finally, $Q_m$ is finite over its coarse moduli space $\mathbb{P}^1$, where the $\mathbb{G}_m$-invariant morphism $A\to \mathbb{P}^1$ is just $(x,y) \mapsto [x^m,y]$. </p>