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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.

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.

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.

So I am asking here what the standard terminology is.

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  • $\begingroup$ Why do you say that a proper morphism `as usual it should be representable'? There are proper morphisms that are not representable. For instance, the morphism from the classifying stack of a finite group to a point is proper. This is an example of a complete Deligne-Mumford stack (I also prefer to use 'complete' for spaces and 'proper' for morphisms). $\endgroup$
    – t3suji
    Jan 26, 2010 at 18:58
  • $\begingroup$ This is exactly what I'm asking. Can you argue a bit more on the standard terminology in an answer? The definition I have seen says that a morphism of stacks has property P iff it is representable and and every morphism between schemes obtained from it by base change has property P. $\endgroup$ Jan 26, 2010 at 19:58
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    $\begingroup$ If you look in Deligne--Mumford, you will find the definition, as well as a statement of the valuative criterion. $\endgroup$
    – Emerton
    Jan 26, 2010 at 20:03

2 Answers 2

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As requested, an answer on terminology My favorite reference on basics for DM stacks is Edidin's paper, which I find much easier to read than Laumon & Moret-Bailly (who of course deal with Artin stacks).

Short summary. Suppose $P$ is a property of morphisms $f:X\to Y$ in the category of schemes:

  • 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).

  • 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 ${\mathcal Y}$.

  • 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.

So proper morphism of stacks need not be representable.

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You can define properness for (not necessarily DM) stacks, this is in Olsson's book and in terms of t3suji's answer is a "special definition".

Definition: (Olsson "Algebraic spaces and stacks", p.210) A map of schemes $f:\mathcal{X}\to\mathcal{Y}$ is proper if it is separated, of finite type and universally closed.

  • $f:\mathcal{X}\to\mathcal{Y}$ is separated if the diagonal $\Delta: \mathcal{X}\to\mathcal{X}\times_\mathcal{Y}\mathcal{X}$ is proper (as $\Delta$ is always representable, so you can define proper as in point two of t3suji's answer: it means that the pullback of $\Delta$ along $Z\to \mathcal{X}\times_\mathcal{Y}\mathcal{X}$ (for $Z$ a scheme) is a proper map).
  • A map $f:\mathcal{X}\to Y$ to a scheme is closed if the image of every closed substack $\mathcal{Z}\subseteq\mathcal{X}$ is closed.
  • A map $f:\mathcal{X}\to\mathcal{Y}$ is universally closed if its pullback by any map $Y\to \mathcal{Y}$ (for $Y$ a scheme) is closed.

So for instance, is $BG$ proper? The diagonal map is $BG\to BG\times_{\text{pt}}BG = B(G^2)$, and taking a pullback \begin{array}{ccc} G&\xrightarrow{}& \text{pt}\\ \downarrow&&\downarrow\\ BG& \xrightarrow{} & BG\times BG \end{array} we see that $BG$ is not proper unless $G$ is proper. I think conversely that $BG$ should probably be proper if $G$ is.

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