What is a proper stack? - MathOverflow most recent 30 from http://mathoverflow.net2013-05-23T02:12:31Zhttp://mathoverflow.net/feeds/question/13058http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/13058/what-is-a-proper-stackWhat is a proper stack?Andrea Ferretti2010-01-26T18:11:40Z2010-01-26T22:26:23Z
<p>I have seen the use of the word "proper Deligne-Mumford stack". Now, it is clear to me what it means for a morphism f of stacks to be proper: as usual it should be representable, and every morphism between schemes obtained from f by base change should be proper.</p>
<p>Now, first I guess that "proper" here actually means "complete". A scheme over a field is complete when the structural morphism to a point is proper. But it does not make sense for a stack to ask that the morphism to the point is proper. Indeed it would be in particular representable, and since a point is a scheme this would imply that the stack itself is a scheme.</p>
<p>Another possibility is that the sentence means "a stack with a proper atlas", so that one cannot speak of proper stacks, but only of proper Deligne-Mumford stacks.</p>
<p>So I am asking here what the standard terminology is.</p>
http://mathoverflow.net/questions/13058/what-is-a-proper-stack/13076#13076Answer by t3suji for What is a proper stack?t3suji2010-01-26T22:26:23Z2010-01-26T22:26:23Z<p>As requested, an answer on terminology
My favorite reference on basics for DM stacks is <a href="http://arxiv.org/abs/math/9805101" rel="nofollow">Edidin's paper</a>, which I find much easier to read than Laumon & Moret-Bailly (who of course deal with Artin stacks).</p>
<p>Short summary. Suppose $P$ is a property of morphisms $f:X\to Y$ in the category of schemes:</p>
<ul>
<li><p>If $P$ is local on both $X$ and $Y$ (`local' in appropriate topology, e.g., etale for DM stacks), it makes sense for morphisms of stacks (pass to compatible presentation of both stacks).</p></li>
<li><p>If $P$ is local on $Y$ only, it is easy to define for representable morphisms $F:{\mathcal X}\to{\mathcal Y}$ by changing base to a presentation of ${\mathcal Y}$.</p></li>
<li><p>If $P$ is local on $Y$, but you want to make sense of it for all morphisms, you have to make a special definition --- there is no general approach that works for all properties.
This is what happens with definitions of separated/proper morphism of stacks.</p></li>
</ul>
<p>So proper morphism of stacks need not be representable.</p>