## algebraic group G vs. algebraic stack BG

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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.

In particular, Scott Carnahan mentioned here 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 only deformations?

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## 4 Answers

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.

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What about setting $pt \rightarrow G$ to be the unit $e$ of $G$? – Qfwfq Apr 21 2010 at 6:19
Thanks for catching the typo -- it should read $pt \to BG$, i.e. a choice of universal bundle $EG$ (unique up to equivalence).. – David Ben-Zvi Apr 21 2010 at 14:21

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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.

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.

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That sounds very reasonable, but why couldn't BG deform to a gerbe, which is some kind of a twist of a BG that only has a point locally? The answer is probably, "go learn more about gerbes." – Anton Geraschenko♦ Oct 16 2009 at 16:30
Of course, the other answer might be that Scott and I are a bit off.  Though I tend to think of gerbes as "the land where deformations that shouldn't exist live" as in obstructed defomrations really exist if you're willing to think of them as objects over the gerbe corresponding to their obstruction class.  Though, maybe that's wrong. – Ben Webster♦ Oct 16 2009 at 16:53

Hello Ben,

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...

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.

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You should add this as a comment to Ben's answer rather than as a separate answer. Like this: – Anton Geraschenko♦ Oct 16 2009 at 18:05
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. 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. – Anton Geraschenko♦ Oct 16 2009 at 18:06
If you actually want to comment on other answers, you can just leave a comment, rather than writing a new answer.  I think you have a good point about keeping the section.  You might want to try to edit this into an actual answer to the question, since mine may well not be right.  For the pullback question, you may have just not understood what I was doing in that section; I was talking through (in part for my own benefit) how to reconstruct G by looking at the stack BG and taking the automorphisms of the trivial bundle on all k-schemes. – Ben Webster♦ Oct 16 2009 at 18:07
i just learned how to use edit/comment. – shenghao Oct 16 2009 at 20:05

what i would expect is that the group G is basically the same thing as the pointed stack BG, where you point it by the trivial G-bundle.

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But there's only one k-point (where k is your base field), so what on earth does pointed get you? – Ben Webster♦ Oct 17 2009 at 4:21

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there isn't only one point -- there's a connected groupoid of points, with automorphisms G.. if BG were a scheme you'd be right, but it's a stack.. or topologically it has nontrivial pi_1.. – David Ben-Zvi Oct 20 2009 at 1:30
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