The fibred category $\mathcal A$ of algebraic spaces over a scheme $S$ is a stack (over the category of affine schemes with the etale topology). This is proved in Laumon and Moret-Bailly's book (see (1.6.4) and (3.4.6)).

Let $\mathcal G$ be the fibred category of group algebraic spaces over a scheme $S$. Is $\mathcal G$ a stack?

My guess is that the forgetful functor $\mathcal G\to \mathcal A$ is "representable" in some sense (as the stabilizers of $\mathcal G$ are smaller than those of $\mathcal A$). But I can't see how to make this rigorous.

Sidenote. Note that $\mathcal A$ is not an algebraic stack (see Claim 3.1 in http://arxiv.org/pdf/math/0602646v1.pdf).

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    $\begingroup$ My guess is yes. Consider the more general case of the category of group objects in a category that is a stack. Effective descent is (or should be) easy. The hom-presheaf of the group-object category is a sub-presheaf of that of the original stack given by a limit, so should be a sheaf. $\endgroup$ Jun 24, 2015 at 14:14
  • $\begingroup$ Dear @DavidRoberts , can you explain how the sub-presheaf you mention at the end is given by a limit? Also, effective descent should be easy indeed, but it requires showing that "group structures" can be glued. $\endgroup$ Jun 24, 2015 at 14:27
  • $\begingroup$ It seems that this question got downvoted. Is it inappropriate for this site, or unclear what I'm asking? $\endgroup$ Jun 24, 2015 at 15:16
  • $\begingroup$ @DavidRoberts I'm a bit skeptical about descent (but I'd be happy to be wrong). Work in the category Top and take S to be a point. Let G1, G2 be two groups and let Gij be an open subgroup of Gi. We are given an isomorphism G12 $\to$ G21 and we wish to glue G1 and G2 along it. Now, there is a candidate topological space G, defined with this datum. We need to show it's a group. Certainly it's pointed (the neutral elements glue). But how do you define the multiplication? I mean, if you take g in G1 - G12 and h in G2 - G21, what is g*h in G? $\endgroup$
    – pro
    Jun 24, 2015 at 16:16
  • $\begingroup$ on the other hand pushouts of groups do exist, so maybe there is descent, but it won't coincide with descent for spaces, which seems very weird. no? $\endgroup$
    – pro
    Jun 24, 2015 at 16:17

1 Answer 1


Here's a sketch proof. First, recall that inside the category of presheaves, sheaves are closed under taking limits. Second, inside the set of maps between the underlying algebraic spaces, the group homomorphisms are constructed by taking an equaliser: $$ Hom_{Grp}(G,H) \to Hom_{Sp}(G,H) \rightrightarrows Hom_{Sp}(G\times G,H) \times Hom_{Sp}(\ast,H). $$ Thus the condition that the hom-presheaf is a sheaf holds for the fibred category of group algebraic spaces, and so at the very least it is a prestack.

I claim that one can use the existence part of the universal property of descent (consider the map of descent data that is the multiplication map on the underlying groups) to show that given descent data for group algebraic spaces, the descended underlying algebraic space inherits a multiplication, and again by universality (uniqueness, this time), this makes the multiplication that of a group object. I can fill in details if needed, just not right now.

  • $\begingroup$ @Stackystudent is it sufficient? That is, can you see how to get effective descent from what I wrote? $\endgroup$ Jun 25, 2015 at 6:56
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    $\begingroup$ I think so. Let $\{U_i\to U\}$ be an etale covering of a scheme $U$, and let $\{(X_i\to U_i, m_i, 0_i, inv_i)\}$ be group algebraic spaces over $U_i$. (Here $m_i$ is multiplication on $X_i$, etc.) Assume the usual compatibility. Then the $X_i$ glue to an algebraic space $X\to U$ as the fibred category of algebraic spaces is a stack. But (and this is what I got from your answer) the same applies to the morphisms $m_i :X_i \times_{U_i} X_i \to X_i$, i.e., these also glue to a morphism $m:X\times_U X\to X$. Similar reasoning applies to $0_i$ and $inv_i$. $\endgroup$ Jun 25, 2015 at 7:34
  • $\begingroup$ Does that sound about right? $\endgroup$ Jun 25, 2015 at 7:35
  • $\begingroup$ @Stackystudent yep, and then the group axioms hold due to uniqueness of glueing morphisms (one has to look at the two products of three elements etc) $\endgroup$ Jun 25, 2015 at 7:37
  • $\begingroup$ Yes, thank you very much again for this answer. $\endgroup$ Jun 25, 2015 at 7:38

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