Gerbes 3: Another Example and Some Caution

This might be my last post on gerbes (explicitly for gerbe’s sake), so as in my last ‘stacks for stack’s sake’ post I’ll try to clarify some things with more examples and then give some cautions. Last time I mentioned the classifying stack {BA}. Let’s first actually construct it better than the quick idea I gave.

Let {B} be a topological space, and {A} a sheaf of abelian groups on {B} (note that I’ll use {A} instead of {\mathcal{A}} to avoid typing the script, but it is a {\mathit{sheaf}} and not just a group, otherwise we’ll just recover the classifying space).

Define a functor {BA: \text{Top}(B)\rightarrow \text{Grpds}}, where {\text{Grpds}} is the category of groupoids, by {BA(U)=} groupoid of {A_U}-torsors over {U}. This is a sheaf, say {T}, on {U} with an action {A_U\times T\rightarrow T} such that if {T(V)\neq \emptyset}, then {A_U(V)} acts simply transitively on {T(V)}.

Again, this is just fancy language for something that is probably familiar to you. Since we have a sheaf of groups, just think an open set at a time. {A_U(V)} is a group, call it {G}. Then {G\times T\rightarrow T} is really just an honest group action, and “acting simply transitively” means that if we pick out some {t\in T}, then we have a way to identify {G} with {T}, namely {G\stackrel{\sim}{\rightarrow} T} (as sets) via the action {g\mapsto g\cdot t}.

You could also think of this as a “relative” principal bundle. The group that it is a principal bundle of gets to change locally, but if it is a constant sheaf and hence not changing, then we really do just get the classifying space.

I told you {BA} as a functor to Grpds and not as a functor to {\text{Top}(B)} which is how stacks were defined, but recall if we have a sheaf on {B}, then we can convert it to that form by taking our category to have objects the pairs {\{(s, U)\}} where {s\in BA(U)}, and the maps in the category are inclusions and restricting to the right thing. If we were doing all the details we’d have to check all of this and then check it is actually a gerbe and that it is actually an {A}-gerbe, etc, but we’d be stuck here forever and these are all straightforward enough that it would make a great exercise if you don’t see it right away.

Recall last time that an {A}-gerbe, {G}, is isomorphic to {BA} if and only if it has a global object. Recall that {\text{Vect}^1}, the stack of rank one vector bundles, was a {\mathbb{G}_m}-gerbe, and it has the trivial bundle as a global object, so {B\mathbb{G}_m\simeq \text{Vect}^1}.

Let’s actually prove it now. If {G\simeq BA}, then {A(B) \in BA(B)} and hence {G(B)\neq \emptyset}. For the reverse direction, suppose there is some {s\in G(B)}, then we get a map {G\rightarrow BA} which we’ll denote {t\mapsto \text{Isom}(t,s)}. One can check that this induces the isomorphism. In fact, one can check that whenever you have a map {G_1\rightarrow G_2} in the category of {A}-gerbes, it will be an isomorphism.

This is why I wanted to bring this example up. Here are some of the cautions that jump to my mind. Something might feel fishy to you right now. That’s because I haven’t really told you the proper way to think about these things. When I say “isomorphism” what does that mean? Well, it really means as {2}-categories.

Also, suppose you are an algebraic geometer and you say you have a gerbe on the étale site of {X}. This isn’t precise enough, since funny differences can happen whether or not you’re on the big or small site. I guess because of all that I’ve left out in an attempt to bring the concept out, my main caution is to consult the literature and not any of these blog posts if you want to know if something is true.


Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s