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.

  • $\begingroup$ 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
  • 4
    $\begingroup$ 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


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.


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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