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

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 PxGH, 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?

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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.

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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 '09 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 '09 at 8:03

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) × Homgp(G,H)

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

(Q,h) o (P,f) = (Q + hP, 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→AutX(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→AutX(P)=H.

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I believe your description of Hom(BG,BH) if H is commutative, but not generally. As Reid and I discussed above, even in the usual topological case (i.e., S contractible), pi_0(Map(BG,BH)) = GroupHom(G,H)/conjugation... –  Bhargav Oct 28 '09 at 7:12
I missed that in the comments to your answer. It seems like taking isomorphism classes in Hom(BG,BH) is destroying more information than I'd like. I'll think about this some more later today and see if I can edit this answer into a description of the category Hom(BG,BH). –  Anton Geraschenko Oct 28 '09 at 16:03

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 -> AutS(basepoint of BH) is the homomorphism to H.

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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).

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