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A scheme is separated if the diagonal inclusion $X \to X \times X$ is a closed immersion. I what to know if there is a good generalization of `separated' for algebraic stacks?

My usual stack reference, Anton Gerashchenko's stack notes, doesn't seem to provide an answer.

In a previous MO question several related notions came up. The most similar is quasi-separated where you require the diagonal to be quasi-compact. You can check wikipedia for some relevant algebraic geometry terminology. How does this compare to separatedness?

The main obstacle that I can see in defining separated for stacks is that the property of a map of schemes $X \to Y$ being separated does not appear to be local in the target. Since maps between affines are separated, it seems that every map of schemes is locally separated. This means that we shouldn't expect the usual trick of replacing an algebraic stack by a scheme which covers it to work very well.

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Look at Def. 4.7 of Deligne--Mumford for the definition when $X$ is DM: they define $X$ to be separated if $X \to X \times X$ is proper (or equivalently, finite).

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    $\begingroup$ Does this reproduce the usual notion of separated when X is a scheme? It seems stronger. If it doesn't give back the usual notion, then it is not a good generalization. Also there are many many stacks which are not DM. I'd like to see a notion which works for general algebraic stacks. $\endgroup$ Jan 29, 2010 at 18:08
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    $\begingroup$ For a scheme, the diagonal is always an immersion, and so is proper if and only if it is finite if and only if its a closed immersion. I don't know the non-DM case well enough to be sure, but it doesn't seem unreasonable. $\endgroup$
    – Emerton
    Jan 29, 2010 at 18:26
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One can first define a 'proper algebraic space' $X,$ using its 'underlying space' $|X|,$ and then define a morphism of algebraic spaces $f: X \to Y$ to be proper if for any affine (or just quasi-compact) scheme mapping into $Y,$ the fiber product gives a proper algebraic space. Finally define an Artin stack $X$ to be separated if the diagonal (which is representable) is proper, i.e. for any algebraic space mapping into $X \times X,$ the fiber product....

One can also define a 'proper Artin stack' similarly. See Laumon and Moret-Bailly.

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    $\begingroup$ Precise reference: Laumon, Moret-Bailly. Champs Algebriques Lemme 7.7. $\endgroup$
    – S. Carnahan
    Nov 8, 2010 at 14:46
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Maybe I'm missing something, but I'm pretty sure "separated" for morphisms of schemes IS local on the target:

http://stacks.math.columbia.edu/tag/02KU

The problem with your reasoning is that if $U\hookrightarrow Y$ is the inclusion of an affine open and $f:X\rightarrow Y$ is your morphism, then the pullback of $f$ by $U$ may not be affine, and hence you can't conclude that $f|_U$ is separated.

Actually, it seems to me that the real obstacle to defining separated algebraic stacks by piggy-backing on the scheme version of separatedness is that for an algebraic stack $\mathcal{X}$ over a scheme $S$, the structure morphism $\mathcal{X}\rightarrow S$ is never representable unless $\mathcal{X}$ was itself already a scheme.

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  • $\begingroup$ If your stack has automorphisms then there is no chance of the diagonal being a monomorphism! (Indeed, just pull it back along a morphism classifying an object with automorphisms). $\endgroup$
    – Thiago
    Feb 4, 2022 at 19:55
  • $\begingroup$ Since closed imersions are proper monos (and diagonal morphisms of schemes are automatically monomorphisms), this motivates the Deligne-Mumford definition in the accepted answer $\endgroup$
    – Thiago
    Feb 4, 2022 at 19:56
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The work of Street "Two-dimensional sheaf theory" (JPAA 23 1982 251-270) is relevant I believe.

He has notions of 1-separated and 2-separated, and a 2-analog of the $\_^+$ functor which he calls $L$, with natural transformation $l:\text{identity}\to L$, such that

  • $P$ is 1-separated iff $l_P:P\to LP$ is mono;
  • for every $P$, $LP$ is 1-separated;
  • $P$ is 2-separated iff $l_P:P\to LP$ is $\textit{chronic}$;
  • for every 1-separated $P$, $LP$ is 2-separated;
  • $P$ is a stack iff $l_P:P\to LP$ is an isomorphism;
  • for every 2-separated $P$, $LP$ is a stack.

Here a morphism $f:P\to Q$ is called chronic when the functor $\hom(X,f):\hom(X,P)\to\hom(X,Q)$ is full, faithful and injective on objects for every $X$.

Comparing this with the properties of $\_^+$ seemingly leads to the conclusion that separatedness naturally splits into two notions, something like "$2/3$-separatedness" and "$4/3$-separatedness".

See also the (later) Question regarding 2-mathematics: Can you stackify a 2-functor without prestackifying it first? here on MO.

Update

After comments by David Roberts I must confess I no longer believe it is relevant. However I still don't understand why it is irrelevant, so let me leave it for a while. Sorry.

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  • $\begingroup$ I thought the question meant something else, namely geometric separatedness: it deals with stacks that arise from the stackification of a groupoid in schemes or algebraic spaces. $\endgroup$
    – David Roberts
    Aug 2, 2016 at 6:31
  • $\begingroup$ @DavidRoberts I am not acquainted well enough with that, but does not Street's setup subsume it? He deals with fairly general 2-sites. $\endgroup$ Aug 2, 2016 at 8:23
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    $\begingroup$ No, his notion of separateness is more analogous to a presheaf being separated for a Lawvere-Tierney topology. That is, if you like, orthogonal to the concerns about algebraic stacks above. $\endgroup$
    – David Roberts
    Aug 2, 2016 at 9:29
  • $\begingroup$ @DavidRoberts I still think there is something to it. For example, if you take a site of spaces over a fixed space, some of them are separated and some not. Switching to the étale space built from the sheaf of sections makes any of them separated. This more or less corresponds to switching from a space to the underlying set with discrete topology I believe. I am not sure I understand what I am typing here but I think there is some content to it which links to separatedness of morphisms of sheaves/stacks. At any rate your "is more analogous to" sounds as imprecise as my current comment, no? :D $\endgroup$ Aug 2, 2016 at 13:47
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    $\begingroup$ In any case, it is interesting that there is a clash of terminology, coming from different generalisations of what 'topology' means. $\endgroup$
    – David Roberts
    Aug 2, 2016 at 21:44

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