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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 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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6 Answers 6

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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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  • $\begingroup$ 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? $\endgroup$ Oct 22, 2009 at 7:20
  • $\begingroup$ 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. $\endgroup$
    – Bhargav
    Oct 22, 2009 at 8:03
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There is some result in the case of Lie groupoids and I believe this is related.

Given Lie groupoids $\mathcal{G},\mathcal{H}$ a morphism of stacks $B\mathcal{G}\rightarrow B\mathcal{H}$ comes from what is called a $\mathcal{G}-\mathcal{H}$ bibundle $P$. This bibundle comes from a morphism of Lie groupoids $\mathcal{G}\rightarrow\mathcal{H}$ if and only if the anchor map $a:P\rightarrow \mathcal{G}_0$ has a global section.

This can be found in proposition $3.36$ of Orbifold as stacks. I think similar result in case of Algebraic geometry can be said.

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As a complement to the answers above : it is kind of well-known (at least I thought it was) that the natural morphism

$$\operatorname{\mathbf{Hom}}_{gr} (G,H) \to \operatorname{\mathbf{Hom}}(BG,BH)$$

is an $H$-torsor (where $H$ operates on $\operatorname{\mathbf{Hom}}_{gr} (G,H)$ by conjugacy). In other words, the internal Hom stack $\operatorname{\mathbf{Hom}}(BG,BH)$ identifies with the quotient stack

$$\left[\operatorname{\mathbf{Hom}}_{gr} (G,H) \right/H ] \simeq \operatorname{\mathbf{Hom}}(BG,BH)$$

of the Hom sheaf $\operatorname{\mathbf{Hom}}_{gr} (G,H)$ by the conjugacy action of $H$.

This of course answers the original question completely (globally, no; locally, yes).

The proof is straightforward enough. For instance here is a sketch of the local surjectivity: let $\alpha : BG\to BH$. If $s_G$ is the canonical section of $BG$, we get a morphism $G=\operatorname{\mathbf{Aut}}(s_G)\to \operatorname{\mathbf{Aut}}(\alpha(s_G))$. Since $\alpha(s_G)$ is isomorphic to $s_H$ locally, this defines locally a morphism $G\to H$, up to conjugacy by $H=\operatorname{\mathbf{Aut}}(s_H)$ of course.

I have learnt this from Angelo Vistoli, and since this is nice and quite useful, I am happy to share it back with you. For the context: this isomorphism is a frequent guest in non-abelian cohomology, and in tannaka duality.

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  • $\begingroup$ I did not see this kind of things before... By $H$-torsor you mean principal $H$ bundle? So, you mean $\text{Hom}_{gr}(G,H)\rightarrow \text{Hom}(BG,BH)$ is a principal $H$-bundle. What is topology/smooth structure on $\text{Hom}_{gr}(G,H),\text{Hom}_{gr}(BG,BH)$? Just like any principal $G$ bundle $P\rightarrow M$ with $P/G=M$, you have here $\text{Hom}(G,H)/H=\text{Hom}(BG,BH)$.. You want to take quotient stack $[\text{Hom}(G,H)/H]$ (I am only recently reading about Quotient stack mathoverflow.net/questions/319038/… so I dont know much)... $\endgroup$ Jan 2, 2019 at 14:48
  • $\begingroup$ I did not follow anything after "This of course answers the original question completely (globally, no ; locally, yes)". Please consider explaining last but one paragraph. $\endgroup$ Jan 2, 2019 at 14:49
  • $\begingroup$ @Praphulla Koushik : for your first question, this is mainly a sheaf/stack theoretic statement, there are no considerations about algebraicity (or representability). I don't understand your second question, please be more specific. $\endgroup$
    – Niels
    Jan 3, 2019 at 8:34
  • $\begingroup$ Sorry for not being specific (I thought I was, I will give one more try)... Can you suggest some reference where $\operatorname{\mathbf{Hom}}_{gr} (G,H) \to \operatorname{\mathbf{Hom}}(BG,BH)$ is seen as a principal $H$-bundle... I am not able to understand what you said in your last paragraph... $\endgroup$ Jan 3, 2019 at 9:07
  • $\begingroup$ @Praphulla Koushik : it is very likely you can find this in Giraud's book on non-abelian-cohomology, but honestly the best is to work it out by yourself. What don't you undertstand in the last (but one) paragraph: what I am trying to do, or the way it is done ? $\endgroup$
    – Niels
    Jan 3, 2019 at 9:34
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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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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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    $\begingroup$ 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... $\endgroup$
    – Bhargav
    Oct 28, 2009 at 7:12
  • $\begingroup$ 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). $\endgroup$ Oct 28, 2009 at 16:03
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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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