Ariyan Javanpeykar said here in comments that,

$X\times_{\mathcal{X}}X$ being a scheme is equivalent to representability of $X\rightarrow \mathcal{X}$.

Context is as in this question.

Suppose $p:X\rightarrow \mathcal{X}$ is representable, then, for any scheme $M$ with a map of stacks $M\rightarrow \mathcal{X}$, the $2$-fiber product $X\times_{\mathcal{X}}M$ is a scheme. In particular, we can take $M=X$ then, $X\times_{\mathcal{X}}X$ is a scheme. Thus, $p:X\rightarrow \mathcal{X}$ is representable implies $X\times_{\mathcal{X}}X$ is a scheme.

Suppose it is given that $X\times_{\mathcal{X}}X$ is a scheme. Then, I want to see that $X\rightarrow \mathcal{X}$ is representable i.e., hen, for any scheme $M$ with a map of stacks $M\rightarrow \mathcal{X}$, the $2$-fiber product $X\times_{\mathcal{X}}M$ is a scheme. I wanted to write something like $X\times_{\mathcal{X}}M=(X\times_{\mathcal{X}}X)\times_X M$. But it does not make sense as there is no map $M\rightarrow X$.

any comments on how to see this are welcome.


2 Answers 2


This is not true even if $\mathcal X$ is an Artin stack. For example, let $G$ be a smooth group scheme over the base $T$, and let $\mathbf BG$ be its classifying stack (the category of $G$-torsors fibered over $Sch/T$). Then $T\times_{\mathbf BG}T=G$ is a scheme. However, the atlas $T\to \mathbf BG$ is not representable by schemes in general. An example where $G$ is an elliptic curve over a normal local scheme $T$ of dimension 2 can be found in the article Ineffective descent of genus one curves by Wouter Zomervrucht (arXiv:1501.04304).

What is true is that if $X$ is an algebraic space, $f:X\to \mathcal X$ is an effective epimorphism of fppf sheaves of groupoids on $Sch/T$, and $X\times_{\mathcal X}X$ is an algebraic space, then $f$ is representable by algebraic spaces. Indeed if $U$ is any algebraic space and $U\to\mathcal X$ any map, then $U\times_{\mathcal X}X\to U$ is representable by algebraic spaces fppf-locally on $U$ (namely after base change to $U\times_{\mathcal X}X$, where it becomes a pullback of $X\times_{\mathcal X}X\to X$), hence it is representable by algebraic spaces because algebraic spaces satisfy fppf descent (which is a somewhat nontrivial fact, see https://stacks.math.columbia.edu/tag/04SJ).

  • $\begingroup$ I thank you for your answer. I am only familiar with differential geometric version of stacks.. i am only recently started studying about algebraic stacks... I do not know much about algebraic spaces.. I can not ask you to explain more... Do you know any other easier ways to see in differential geometric set up of stacks? $\endgroup$ Dec 2, 2018 at 20:32
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    $\begingroup$ Things are even worse in the differential geometric case, since pullbacks fail to exist. $\endgroup$
    – David Roberts
    Dec 2, 2018 at 20:57
  • $\begingroup$ @DavidRoberts Ok. $\endgroup$ Dec 3, 2018 at 13:20
  • $\begingroup$ It seems that this is the best one can say about my question... I did not completely understand as I don’t know anything about algebraic spaces.. I have given bounty and a +1.. I hope I can understand once I read something about algebraic spaces.. Thank you.. I understand first paragraph in your answer. $\endgroup$ Dec 3, 2018 at 13:25
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    $\begingroup$ Many sources (including the original paper by Deligne and Mumford) only consider algebraic stacks with a separation condition that makes the diagonal representable by schemes. I guess this was implicit in your source. $\endgroup$ Dec 4, 2018 at 1:58

Following lemma is from Kai Behrend and Ping Xu's paper (page $8$, lemma $2.11$) Differentiable Stacks and Gerbes.

Let $f:\mathcal{X}\rightarrow \mathcal{Y}$ be a morphisms of stacks. Suppose given a manifold $U$ and a morphism $U\rightarrow \mathcal{Y}$ which is an epimorphism. If the fiber product $\mathcal{X}\times_{\mathcal{Y}}U$ is representable and $V\rightarrow U$ is a submersion, then $f$ is a representable submersion.

Take $\mathcal{X}=X=U$ in above lemma. Then above lemma turns to

Let $f:X\rightarrow \mathcal{Y}$ be a morphisms of stacks. Suppose that $f: X\rightarrow \mathcal{Y}$ is an epimorphism. If the fiber product $X\times_{\mathcal{Y}}X$ is representable and $X\times_{\mathcal{Y}}X\rightarrow X$ is a submersion, then $f$ is a representable submersion.

I guess relaxing the condition $X\times_{\mathcal{Y}}X\rightarrow X$ is a submersion conlcudes that $X\rightarrow \mathcal{Y}$ is representable.


Consider a morphism of stacks $f:X\rightarrow \mathcal{Y}$. Suppose that $X\times_\mathcal{Y}X$ is representable. If further $f:X\rightarrow \mathcal{Y}$ is an epimorphism then $X\rightarrow \mathcal{Y}$ is a representable submersion.

This is very much like Marc Hoyois's answer which says that adding the condition that the map is an epimorphism (along with algebraic spaces condition) concludes that the map is representable.


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