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Setup: Let's work in the category of schemes over $\mathbb C$. Let $G$ be a finite group. Let $BG=[pt/G]$ be the classifying stack of principal $G$ bundles. This is a fiberd category over the big etale site of complex schemes. The groupoid of sections of $BG$ over $U$ is the category of principal $G$ bundles over $U$. I will also appreciate it if someone can answer the question in the topological category.

Question: What are the automorphisms of $BG$, as a stack over $\mathbb C$?

What to Expect: Since automorphisms of $BG$ are functors, $Aut(BG)$ should be a category. It does not quite make sense to say that two automorphisms are the same. We can only say that two automorphisms are isomorphic. An automorphism of $BG$ might have nontrivial (2-)automorphisms.

What I've Tried: I trived the case when $G$ is abelian and got lost even for this simpler case. Imaging $BG$ as a point with automorphisms $G$, I expect that the objects of $Aut(BG)$ are $Aut(G)$, since a functor sends the point to itself and sends automorphisms to automorphisms. But when I look at the groupoid $BG(U)$, I have some difficulty. Suppose $a,b\in BG(U)$ are different objects and $\phi\in G$, there is no canonical way that $\phi$ acts on $Hom(a,b)$. However, although this is a big trouble, it's not hopeless. It seems that we can pick an distinguished element in $Hom(a,b)$ so that we can identify it with $G$. We can pick an isomorphism of a groupoid with its skeleton.

My Guess: Let me have a guess. I guess that $Aut(BG)=[Aut(G)/G]$. This is the quotient groupoid given by the conjugation map $G\to Aut(G)$.

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    $\begingroup$ Note that your $[Aut(G)/G]$ is none other than the group stack associated to the crossed module $Ad\colon G\to Aut(G)$ of group schemes over $\mathbb{C}$. This is sometimes called $AUT(G)$. This goes back (though not in so many words) to Giraud's work classifying $G$-gerbes. $\endgroup$ – David Roberts Apr 7 '15 at 5:58
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Yes that's correct. An element of $Aut(BG)$ gives, for each scheme $X$, a functor from the category of $G$-bundles on $X$ to itself.

First look at how the functor acts on the category of $G$-bundles on a point. This is a functor from the category with one object and automorphisms $G$ to itself. These functors are given by automorphisms of $G$, and isomorphisms between two functors are ways of writing the ratio of the two automorphisms as conjugation by an element of $G$. So this is exactly the groupoid you describe.

Now, given how the functor acts on $G$-bundles on a point, we can determine how it acts anywhere else. We can see how it acts on trivial $G$-bundles by pullback. Then every $G$-bundle can be made by gluing trivial $G$-bundles on an etale cover, and this determines how your automorphism acts on it.

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  • $\begingroup$ Thanks very much for the answer! Although there are still details to work out, the confirmation of my guess is very helpful! $\endgroup$ – YZhou Apr 7 '15 at 8:22

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