# The stack of group algebraic spaces

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).

• 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. Jun 24, 2015 at 14:14
• 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. Jun 24, 2015 at 14:27
• It seems that this question got downvoted. Is it inappropriate for this site, or unclear what I'm asking? Jun 24, 2015 at 15:16
• @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?
– pro
Jun 24, 2015 at 16:16
• 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?
– pro
Jun 24, 2015 at 16:17

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 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 $inv_i$. Jun 25, 2015 at 7:34