algebraic group G vs. algebraic stack BG 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?
 A: 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.
A: If G is a group scheme over k (algebraic closed), then let me talk through how to get G back by looking at the stack BG.  The k points of BG (which is a groupoid) consist of 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.
A: 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.
A: 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
$pt \to BG$ then the based loop space (fiber product of $pt$ with itself over $BG$) is $G$,
so you recover the group canonically.
