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Given a homomorphism f:G→H between smooth algebraic groups, we get an induced homomorphism of algebraic stacks Bf:BG→BH, given by sending a G-torsor P over a scheme X to the H-torsor Px^GH, whose (scheme-theoric) points are {(p,h)|p∈P,h∈H}/∼, where (pg,h)∼(p,f(g)h).

Is every morphism of algebraic stacks BG→BH of the form Bf? If not, what is an example of a morphism not of this form?

flag	asked Oct 22 2009 at 4:34 Anton Geraschenko♦♦ 13.7k●3●48●108

Bhargav · Accepted Answer · 2009-10-22 05:11:55Z UTC

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Depends on the base scheme and the topology being used. For example if you're working over a field k in the etale or the flat topology, and take the group G to be trivial, you're asking if H^1(k,H) is trivial, which is obviously false in general. This is, in a sense, the only obstruction: for any base scheme S, giving a map from BG to any stack Y (in stacks/S) is the same as specifying a point y of Y(S), and a homomorphism G -> Aut_S(y). In particular, if BH(S) is connected (i.e., if H^1(S,H) = *) then the answer to your question is positive.

answered Oct 22 2009 at 5:11

Bhargav
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	Is it true under the conditions of your last sentence that the groupoid of maps from BG to BH is equivalent to the category of functors from BG to BH as ordinary groupoids, i.e., the groupoid whose objects are conjugacy classes of maps and where the group of automorphisms of a map is its centralizer? – Reid Barton Oct 22 2009 at 7:20
	I think yes, though the triviality of what I'm thinking makes me think I'm missing something. Map(BG,Y) is equivalent to the groupoid of pairs (y in Y(k),f:G-> Aut(y) alg. homomorphism) with obvious groupoid structure. This obvious groupoid structure comes just from Y. Now assume Y(k) is connected. If we pick a point of Y(k) with k-automorphism group scheme H, then the resulting description is exactly what you suggest. – Bhargav Oct 22 2009 at 8:03

Anton Geraschenko · Answer 2 · 2009-10-28 05:09:38Z UTC

Taking Bhargav's answer to its logical conclusion, we get the following result.

If G, H, and K are smooth groups over a base scheme S, then isomorphism classes of morphisms BG→BH are given by

Hom(BG,BH) = H¹(S,H) × Hom_gp(G,H)

with composition Hom(BH,BK) × Hom(BG,BH) → Hom(BG,BH) given by

(Q,h) ^_o (P,f) = (Q + h_∗P, h ^_o f).

To see this, note that a morphism from BG to any stack X consists of a point P ∈ X(S) and a group homomorphism G→Aut_X(P). In the case of X=BH, this amounts to a choice of H-torsor P over S (i.e. an element of H¹(S,H)), which is where you send the trivial G-torsor over S, and a group homomorphism f:G→Aut_X(P)=H.

S. Carnahan · Answer 3 · 2009-10-22 06:22:37Z UTC

Bhargav said this first in different words, but (by analogy with the homotopy picture) you need your map to be basepoint-preserving. In particular, the point corresponding to the trivial G-torsor should be taken under composition to the point corresponding to the trivial H-torsor. Once that is satisfied, then the homomorphism G -> Aut_S(basepoint of BH) is the homomorphism to H.

Mike Skirvin · Answer 4 · 2009-10-22 04:46:54Z UTC

I am not an expert on this topic, so someone please correct me if I'm wrong, but I believe the answer to this question is yes.

The stack BG (resp. BH) is represented by the simplicial scheme also usually denoted BG (resp. BH) which is obtained by covering BG (resp. BH) by a point and then taking the nerve of this covering. Then a map from BG \to BH should just be given by a map of the corresponding simplicial schemes, which in particular includes a map G \to H (these are the 1-simplices). However, I think that this map completely determines the map BG \to BH (this should have something to do with the fact that BG and BH have no nontrivial homotopy groups beyond \pi_1, so we really only need to work with groupoids).

Does every morphism BG-->BH come from a homomorphism G-->H?

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