Does every morphism BG-->BH come from a homomorphism G-->H? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T01:32:46Z http://mathoverflow.net/feeds/question/1809 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/1809/does-every-morphism-bg-bh-come-from-a-homomorphism-g-h Does every morphism BG-->BH come from a homomorphism G-->H? Anton Geraschenko 2009-10-22T04:34:02Z 2009-10-28T05:09:38Z <p>Given a homomorphism f:G&rarr;H between smooth algebraic groups, we get an induced homomorphism of algebraic stacks Bf:BG&rarr;BH, given by sending a G-torsor P over a scheme X to the H-torsor Px<sup>G</sup>H, whose (scheme-theoric) points are {(p,h)|p&isin;P,h&isin;H}/&sim;, where (pg,h)&sim;(p,f(g)h).</p> <p>Is every morphism of algebraic stacks BG&rarr;BH of the form Bf? If not, what is an example of a morphism not of this form?</p> http://mathoverflow.net/questions/1809/does-every-morphism-bg-bh-come-from-a-homomorphism-g-h/1813#1813 Answer by Mike Skirvin for Does every morphism BG-->BH come from a homomorphism G-->H? Mike Skirvin 2009-10-22T04:46:54Z 2009-10-22T04:46:54Z <p>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.</p> <p>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).</p> http://mathoverflow.net/questions/1809/does-every-morphism-bg-bh-come-from-a-homomorphism-g-h/1819#1819 Answer by Bhargav for Does every morphism BG-->BH come from a homomorphism G-->H? Bhargav 2009-10-22T05:11:55Z 2009-10-22T05:11:55Z <p>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.</p> http://mathoverflow.net/questions/1809/does-every-morphism-bg-bh-come-from-a-homomorphism-g-h/1826#1826 Answer by S. Carnahan for Does every morphism BG-->BH come from a homomorphism G-->H? S. Carnahan 2009-10-22T06:22:37Z 2009-10-22T06:22:37Z <p>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<sub>S</sub>(basepoint of BH) is the homomorphism to H.</p> http://mathoverflow.net/questions/1809/does-every-morphism-bg-bh-come-from-a-homomorphism-g-h/2999#2999 Answer by Anton Geraschenko for Does every morphism BG-->BH come from a homomorphism G-->H? Anton Geraschenko 2009-10-28T05:09:38Z 2009-10-28T05:09:38Z <p>Taking Bhargav's answer to its logical conclusion, we get the following result.</p> <blockquote> <p>If G, H, and K are smooth groups over a base scheme S, then <em>isomorphism classes</em> of morphisms BG&rarr;BH are given by </p> <blockquote> <p>Hom(BG,BH) = H<sup>1</sup>(S,H) &times; Hom<sub>gp</sub>(G,H)</p> </blockquote> <p>with composition Hom(BH,BK) &times; Hom(BG,BH) &rarr; Hom(BG,BH) given by</p> <blockquote> <p>(Q,h) <sup><sub>o</sub></sup> (P,f) = (Q + h<sub>&#8727;</sub>P, h <sup><sub>o</sub></sup> f).</p> </blockquote> </blockquote> <p>To see this, note that a morphism from BG to any stack X consists of a point P &isin; X(S) and a group homomorphism G&rarr;Aut<sub>X</sub>(P). In the case of X=BH, this amounts to a choice of H-torsor P over S (i.e. an element of H<sup>1</sup>(S,H)), which is where you send the trivial G-torsor over S, and a group homomorphism f:G&rarr;Aut<sub>X</sub>(P)=H.</p>