algebraic group G vs. algebraic stack BG - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T07:44:07Z http://mathoverflow.net/feeds/question/727 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/727/algebraic-group-g-vs-algebraic-stack-bg algebraic group G vs. algebraic stack BG Anton Geraschenko 2009-10-16T08:08:15Z 2009-10-20T01:18:50Z <p>I've gathered that it's "common knowledge" (at least among people who think about such things) that studying a (smooth) algebraic group G, as an algebraic group, is in some sense the same as studying BG as an algebraic stack. Can somebody explain why this is true (and to what extent it is true)? I can get as far as seeing that quasi-coherent sheaves on BG are the same as representations of G, but it feels like there's more to it.</p> <p>In particular, Scott Carnahan mentioned <a href="http://mathoverflow.net/questions/570/deformation-theory-of-representations-of-an-algebraic-group/690#690" rel="nofollow">here</a> that deformations of BG as an algebraic stack should correspond exactly to deformations of G as an algebraic group. I assume this means that any deformation of BG must be of the form BG', where G' is a deformation of G (as a group). It's clear to me that such a BG' is a deformation, but why should these be the <em>only</em> deformations?</p> http://mathoverflow.net/questions/727/algebraic-group-g-vs-algebraic-stack-bg/746#746 Answer by Ben Webster for algebraic group G vs. algebraic stack BG Ben Webster 2009-10-16T15:25:07Z 2009-10-16T18:08:57Z <p>If G is a group scheme over k (algebraic closed), then me talk through how to get G back by looking at the stack BG. The k points of BG (which is a groupoid) is one point whose automorphisms are the k points of G. The pullback of this point to Spec A for any k-algebra A has automorphisms given by the A points of G. If you think of BG points as principal bundles, I'm saying the automorphsims of the trivial bundle on Spec A are the A points of the group.</p> <p>So what happens if you deform BG? You still have this one point, you can't deform that to anything, so you can only change its morphisms. That's your G' (you get an algebraic group since you can pullback to all the Spec A's). How you see it's BG' is a little trickier, so maybe I should leave it to a real algebraic geometer, but I think that the idea is that BG is distinguished by being the sheafication of the trivial bundles in the smooth/fppf topology, and this won't change when you deform.</p> http://mathoverflow.net/questions/727/algebraic-group-g-vs-algebraic-stack-bg/766#766 Answer by shenghao for algebraic group G vs. algebraic stack BG shenghao 2009-10-16T17:16:05Z 2009-10-16T17:16:05Z <p>Hello Ben,</p> <p>a little comment: when you say "G is a group scheme over k", you mean k is a separably closed field, right? Because otherwise the groupoid BG(k) may not have only one isomorphism class of object; the set of isom classes is the Galois cohomology H^1(k,G). Also I got confused by "the pullback of this point". I think one should deform BG along the nilpotent embedding Spec k --> Spec A, rather than considering Spec A --> Spec k...</p> <p>The structural map BG --> Spec k has a section Spec k --> BG. So maybe one can deform BG --> Spec k together with this section, so that any gerbe becomes trivial.</p> http://mathoverflow.net/questions/727/algebraic-group-g-vs-algebraic-stack-bg/811#811 Answer by blah for algebraic group G vs. algebraic stack BG blah 2009-10-16T23:29:02Z 2009-10-16T23:29:02Z <p>what i would expect is that the group G is basically the same thing as the <em>pointed</em> stack BG, where you point it by the trivial G-bundle.</p> http://mathoverflow.net/questions/727/algebraic-group-g-vs-algebraic-stack-bg/1332#1332 Answer by David Ben-Zvi for algebraic group G vs. algebraic stack BG David Ben-Zvi 2009-10-20T01:18:50Z 2009-10-20T01:18:50Z <p>The stack BG only recovers G up to inner automorphisms, not canonically (as suggested by blah) - this can lead to serious issues in families or equivalently over a nonalgebraically closed field, as Shenghao's comment points out. One way to say this is the following : the loops in BG are G/G, the adjoint quotient of G. On the other hand, if you give a map pt --> G then the based loop space (fiber product of pt with itself over BG) is G, so you recover the group canonically.</p>