Gerbes 2: The Motivation


I’m going to make another definition, but I may as well get to the punchline first or else anyone reading this that doesn’t already know the punchline is going to skip reading it or tune out. If you have an abelian sheaf {\mathcal{A}} on {X}, then there is a notion of {\mathcal{S}} being not only a stack/gerbe over {X}, but an {\mathcal{A}}-gerbe. I’ll define this later.

Here’s the amazing part, Giraud did a lot of work for us and tells us that the global elements of the stack, {\mathcal{S}}, i.e. the objects lying over all of {X} are in bijection with elements of {H^1(X, \mathcal{A})}. Take the line bundle example, then {L(X)} is actually a {\mathcal{O}_X^\times}-gerbe and hence (iso classes of) line bundles on {X} are in correspondence with {H^1(X, \mathcal{O}_X^\times)}. Wait! We already knew that since {H^1(X, \mathcal{O}_X^\times)\simeq \text{Pic}(X)}.

We also found in our two examples of deformations before that this is true. We found that infinitesimal extensions by coherent sheaves are classified by {H^1(X, \mathcal{F}\otimes \mathcal{T})} here and here. It turns out this wasn’t a coincidence. Things are classified by {H^1} all over algebraic geometry and this is the underlying thread connecting them.

But it turns out Giraud didn’t stop there and we get even more. We actually get an obstruction theory as well. Giraud tells us that the obstruction to constructing a global object lies in {H^2(X, \mathcal{A})}. We may have not gone through it for our deformations with a tedious cocycle argument like the {H^1} exercises, but many books do go through this (see Hartshorne’s recent book on Deformation Theory). Good thing we didn’t go through it, we just get it from knowing that it is a gerbe.

I’ve tried to search for this, but can’t find it anywhere. This is why this theory is so cool and widespread. Think about that measure theory example from two times ago. If we knew that stack was an {\mathcal{A}}-gerbe for some {\mathcal{A}}, then we could use cohomology to determine whether certain measure theoretic constructions existed. I don’t think anyone has ever done this before. Knowing something is a gerbe is very powerful since it converts existence questions to cohomology computations.

Let’s get right to it now. Fix a sheaf of abelian (possibly not necessary) groups {\mathcal{A}} on {X}. Then an {\mathcal{A}}-gerbe is a gerbe {\mathcal{S}} on {X} such that for any open {U} on {X} we have a functorial isomorphism {\mathcal{A}(U)\stackrel{\sim}{\rightarrow} \text{Aut}(s)} for all {s\in \mathcal{S}(U)}.

Note that since {\mathcal{S}} is a stack, {\text{Aut}(s)} is a sheaf, so by isomorphism we mean an isomorphism as sheaves, and by functorial we mean given another object {t\in \mathcal{S}(U)}, the isomorphism commutes

\displaystyle \begin{matrix} \mathcal{A}(U) & \rightarrow & \text{Aut}(s) \\ Id \downarrow & & \downarrow \\ \mathcal{A}(U) & \rightarrow & \text{Aut}(t) \end{matrix}

In particular, we get that for any two objects {C, D\in \mathcal{S}(U)} we have that the sheaf {Isom(C,D)} is an {\mathcal{A}}-torsor. This gives that if there is some object over {U}, namely that {\mathcal{S}(U)\neq \emptyset}, then the set of isomorphism classes of obects in {\mathcal{S}(U)} is in natural bijection with {H^1(U, \mathcal{A}_U)}, as was pointed out in the motivation above.

One can form the {\mathit{classifying \ stack}} over {B} (the stacky version of a classifying space), {B\mathcal{A}} by taking {B\mathcal{A}(U)=\mathcal{T}ors(\mathcal{A}(U))}. So above an open set we get the category of {\mathcal{A}(U)}-torsors on {U}. A basic theorem about {\mathcal{A}}-gerbes is that an {\mathcal{A}}-gerbe, {\mathcal{S}}, is isomorphic to {B\mathcal{A}} if and only if {F(B)\neq \emptyset}. This says that {F} is isomorphic to the classifying stack if and only if it has a global object.

Advertisements

Leave a Reply

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

WordPress.com Logo

You are commenting using your WordPress.com 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 )

Google+ photo

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

Connecting to %s