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.

Gerbes 1: The Definition

Here’s a nice short definitional post. If you think that defining stacks and now we have even more definitions is just completely absurd, abstract, solipsism bear with me for just one more post. In the next post we’ll see what the point of all of this is. It is not just pointless abstraction. Figuring out something is a gerbe actually gives you an amazingly powerful tool to work with.
A gerbe is just a special type of stack. Let’s go back to thinking topologically, since if I have non-AG readers, this probably feels most comfortable. Consider a stack ${\mathcal{S}}$ over ${\text{Top}(X)}$. For instance, line bundles on ${X}$.

So we get some stuff associated to every open set of ${X}$. Recall, we think of these as lying over these open sets. No part of the definition of stack guarantees that we must have things lying over open sets (i.e. the collection of things over a particular open set could be empty). The first condition for a stack to be a gerbe is that for any open set ${U}$, there is a covering ${\{V_i\subset U\}}$ such that ${\mathcal{S}(V_i)\neq \emptyset}$. In other words, we can always shrink our open set to get an object over it.

Let’s think to our line bundle example. Check. We at least always have the trivial bundle.

The other condition for a stack to be a gerbe is that everything is locally isomorphic in the following sense, whenever we have two objects ${\eta, \eta '}$ over some open set ${U}$, then there is some covering ${\{V_i \subset U\}}$ such that we get ${\eta|_{V_i} \stackrel{\sim}{\rightarrow} \eta '|_{V_i}}$.

Let’s think to our line bundle example. Check. By definition we have local trivializations, which are all isomorphic. So ${L(X)}$ is a gerbe. I didn’t do a good job at examples of non-stacks, but it might actually be useful to give examples of stacks that are not gerbes. The stack ${M_g}$ that was briefly mentioned last post is not a gerbe (in fact, I haven’t really told you what I mean by ${M_g}$, and it turns out if you define the moduli space with respect to the Zariski topology ${M_1}$ isn’t even a stack).

Again, most importantly for us the deformation stack (of a smooth scheme, ${Z}$) from last time is also a gerbe (mostly for the same reason as the bundle example, you have the trivial one and locally everything becomes the trivial one).

Stacks 2: An example

This will hopefully be a short, yet enlightening post in which the concept of a stack starts to make more sense than the abstract nonsense of the last few posts. Recall that we formed a category of line bundles on a manifold ${L(X)}$ and had a natural forgetful functor: ${L(X)\rightarrow \text{Top}(X)}$.

If one is not writing a blog and wants to be much more careful, one should probably check that ${L(X)}$ is indeed a category and the given functor is actually a functor. Most readers that have made it this far probably aren’t concerned with this part, though.

Is this fibered in groupoids? Well for checking what sorts of things lie over certain objects ${U\in \text{Top}(X)}$, the situation has been rigged so that the objects over ${U}$ are precisely the line bundles on ${U}$ as a manifold. The first type of “square” we have to be able to complete is as follows ${\begin{matrix} & & (V, L_V) \\ \\ U & \hookrightarrow & V \end{matrix}}$. Well, all we need to be able to do is find some ${(U, L_U) \rightarrow (V, L_V)}$. But by definition of our category this would consist of an iso ${L_V|_U \rightarrow L_U}$. These are line bundles, so we can always restrict to get another one, so just take ${(U, L_V|_U)}$ lying over ${U}$ to complete it.

What about the second diagram of being fibered in groupoids? The base is just ${U\hookrightarrow V \hookrightarrow W}$. Now suppose we have line bundles over these ${L_U, L_V,}$ and ${L_W}$ and isomorphisms ${L_W|_V\rightarrow L_V}$ and ${L_W|_U\rightarrow L_U}$. This certainly tells us that there is an iso ${L_V|_U\rightarrow L_U}$, and uniqueness is just from the fact that it has to be the one that makes the composition what we said it had to be. You could think of this coming from the fact that ${L_V|_U\rightarrow L_U}$ is unique up to automorphism of ${L_U}$, and we know which automorphism it from the other condition.

Now we check that Isom forms a sheaf. Let ${U}$ be some open set. Let ${L}$ and ${S}$ be two line bundles over ${U}$ (in this case, this literally just means line bundles on ${U}$ as a topological space). Now we want to check that the presheaf (of sets) ${\mathcal{F}(V)=\{L_V\stackrel{\sim}{\rightarrow} S_V\}}$ is actually a sheaf. This is a presheaf just because isomorphisms restrict. It is a sheaf because all the information is local. If you have isomorphisms defined on open subsets that agree on overlaps, then you can glue them to make an isomorphism on the union. These are just two basic properties of line bundles that most people have already seen. So Isom is a sheaf.

Lastly we need to check the stack condition. Maybe I should remark on the terminology here. A collection of objects and isos over a covering of an open set that satisfies the cocycle condition is called a descent datum. If the objects glue in the way of the stack condition, then that descent datum is said to be effective, so the stack condition is sometimes stated that every descent datum is effective.

Given a descent datum, the fact that you can glue to get an object over the whole open set is just a standard exercise or proven proposition in basically any text on manifolds. In fact all of the above things are true for any rank ${r}$ vector bundle. So we actually get the stack of rank ${r}$ vector bundles on ${X}$ for any ${r}$. Since I’m not sure we’ll return to this example, we’ll just temporarily notate it ${\text{Vect}^r(X)}$, and hence ${\text{Vect}^1(X)=L(X)}$.

If you’ve been following along, it should be pretty clear how to translate all of this over to a stack on the Zariski site rather than on ${\text{Top}(X)}$, but we’ll make that more explicit next time and get some more examples.

Stacks 1

Today we’ll actually get towards a what a stack is. Last we talked about what it meant for a category to be fibered in groupoids over another. This was a very general definition for any two categories. Today we’ll actually need to make use of more topological or geometric notions. Let ${X}$ be a topological space, then we can turn ${X}$ into a category ${\text{Top}(X)}$, which is just the standard topological site. The objects are the open sets, and ${Hom(U,V)=\left\{ \begin{array}{lr} \{U\hookrightarrow V\} & : U\subset V \\ \emptyset & : \text{else} \end{array} \right.}$

See this post for more information on what that category is and why it is a site. From these earlier posts we also have a notion of what a sheaf on this site is. Since this is just a topological space and hence the motivating example for the sheaf (on a site) definition, you can just think in terms of what a sheaf on a topological space is if you wish.

Consider some category ${\mathcal{C}}$ fibered in groupoids: ${F: \mathcal{C}\rightarrow \text{Top}(X)}$. Then we have a nice contravariant functor for any pair of objects lying over ${U}$, say ${\eta}$ and ${\eta '}$. We’ll suggestively call the functor ${\text{Isom}_{\eta, \eta '}: \text{Top}(U)\rightarrow \text{Sets}}$. The functor is just the set of isomorphisms on the open set you apply it to. So ${\text{Isom}_{\eta, \eta '}(V)=\{\eta |_V\stackrel{\sim}{\rightarrow} \eta '|_V\}}$. If ${\text{Isom}}$ is a sheaf for all pairs of objects and open sets, then sometimes we call ${F: \mathcal{C}\rightarrow \text{Top}(X)}$ a pre-stack. You may want to take a moment to absorb this. There are tons of things going on in these words.

A stack is just a pre-stack that also satisfies a cocycle condition: Given any covering ${\{U_i\}}$ of ${U\in \text{Top}(X)}$ and objects lying over ${U_i}$ say ${\eta_i}$ and isomorphisms ${\phi_{ij}: \eta_i|_{U_{ij}}\rightarrow \eta_j|_{U_{ij}}}$ satisfying the cocycle condition ${\phi_{j,k}|_{U_{ijk}}\circ \phi_{i,j}|_{U_{ijk}}=\phi_{i,k}|_{U_{ijk}}}$ then there is an object ${\eta}$ over ${U}$ with isos ${q_i: \eta|_{U_i}\rightarrow \eta_i}$ such that ${\phi_{i,j}\circ q_i|_{U_{ij}}\simeq q_j|_{U_{ij}}}$.

Don’t be scared off by this. If you’ve ever tried to do any gluing of objects in topology this should look familiar. It is just a purely formal way of saying that if you have locally defined things that are consistent on overlaps, then you can glue them together to get an object over the whole thing.

Another way to say these conditions is as follows. The first says that you can glue isos (Isom forms a sheaf, the non-trivial aspect of which is the gluing axiom). The second says that you can glue objects. If you want an exercise to see if you’ve parsed all this suppose ${X}$ is a manifold. Define ${L(X)}$ to be the category of line bundles on ${X}$. The objects of which are pairs ${(U, \mathcal{L}_U)}$ an open set and a line bundle on ${U}$. The morphisms are “restriction”, ${Hom((U, \mathcal{L}_U), (V, \mathcal{L}_V))}$ is empty if we don’t have an inclusion ${U\hookrightarrow V}$, and otherwise it consists of the inclusion itself along with the set of all isomorphisms ${\mathcal{L}_V|_U\rightarrow \mathcal{L}_U}$.

We have an obvious functor ${L(X)\rightarrow \text{Top}(X)}$ by just forgetting the line bundle. Show this functor fibers the category in groupoids over ${X}$, the pre-stack condition holds, and the stack condition holds. My next post will be me going through these details to help show what is going on in these definitions. Then we’ll generalize from a stack on a topological space to a stack on a general site (particularly the Zariski site) along with more examples. Then we’ll move on to what a gerbe is and how it has anything to do with deformation theory.

Categories Fibered in Groupoids

Sorry about the delay, I’ve been really busy with other things. Most (probably all) people have completely forgotten what I was talking about. Luckily you don’t need to in order to follow this post!

Today we’ll look at what it means for a category to be fibered in groupoids over another one. Suppose we have a (covariant) functor ${F:\mathcal{C}\rightarrow\mathcal{D}}$. I’ll refer to ${\mathcal{D}}$ as the “base” category and ${\mathcal{C}}$ as lying over ${\mathcal{D}}$. We’ll say ${\mathcal{C}}$ is fibered in groupoids over ${\mathcal{D}}$ (by ${F}$) if the functor satisfies two conditions.

First, suppose we have objects in the base with an arrow between them ${A\rightarrow B}$ and an object lying over ${B}$ (i.e. some object ${Y\in \mathcal{C}}$ such that ${F(Y)=B}$). The condition is that whenever we have this situation we can “complete the square”. This means that we can find an object and an arrow ${X\rightarrow Y}$ such that the arrow maps to the base arrow. So ${F(X)=A}$ and ${F(X\rightarrow Y)= A\rightarrow B}$.

It might be good to visualize this in the following way: Given ${\begin{matrix} & & Y \\ & & \\ A & \rightarrow & B \end{matrix}}$ you can always complete to ${\begin{matrix} X & \rightarrow & Y \\ & & \\ A & \rightarrow & B\end{matrix}}$

The second condition is that whenever you have objects and arrows ${A\rightarrow B\rightarrow C}$ in the base category and you have ${X, Y, Z}$ lying over ${A, B, C}$ respectively with the two arrows ${Y\rightarrow Z}$ lying over ${B\rightarrow C}$ and ${X\rightarrow Z}$ lying over the composite ${A\rightarrow C}$, there is a unique arrow ${X\rightarrow Y}$ so that everying lies over ${A\rightarrow B\rightarrow C}$.

There is a nice way to visualize this as well, but I am still awful at making nice things in wordpress, so we’ll do it as follows, you have the following two pieces of information

${\begin{matrix} Y & \rightarrow & Z \\ & & \\ B & \rightarrow & C \end{matrix}}$ and ${\begin{matrix} X & \rightarrow & \rightarrow & \rightarrow & Z \\ \\ A & \rightarrow & B & \rightarrow & C \end{matrix}}$

the whole thing can be completed ${\mathit{uniquely}}$ to ${\begin{matrix} X & \rightarrow & Y & \rightarrow & Z \\ \\ A & \rightarrow & B & \rightarrow & C \end{matrix}}$.

To get a feel for this, let’s look at a motivating example for the terminology. Consider any functor ${F: \mathcal{C}\rightarrow \mathcal{D}}$, then if we pick an object ${D}$ in the base we can form the ${\mathit{fiber \ category}}$ over ${D}$, which we’ll denote ${\mathcal{C}_D}$. This is just a category whose objects lie over ${D}$ i.e. all ${X}$ so that ${F(X)=D}$, and the morphisms are the ones the functor maps to the identity morphsim ${id_D: D\rightarrow D}$.

Claim: Any fiber category that is fibered in groupoids (via the functor that we are taking the fiber of) is a groupoid. By groupoid here we just mean that every morphism is an isomorphism.

Suppose there is a map between ${Y}$ and ${Z}$ in ${\mathcal{C}_D}$, say ${Y\stackrel{f}{\rightarrow} Z}$. Then by definition of the category ${F(f)=id_D}$. So we can build the situation of the second condition. The base is just ${D\rightarrow D \rightarrow D}$ all the identity and hence the composition is the identity. We have ${Y\rightarrow Z}$ lying over the second identity map, and we have the identity ${Z\rightarrow Z}$ lying over the composition, so we get a unique map ${Z\rightarrow Y}$ such that the composite ${Z\rightarrow Y\rightarrow Z}$ is the identity. We can now repeat this process with this unique map ${Z\rightarrow Y}$ to get ${Y \rightarrow Z \rightarrow Y}$ which is the identity on ${Y}$ and by uniqueness the completed map must be the original ${f}$ and hence ${f}$ is an isomorphism. Thus the fiber category is a groupoid since any morphism is an isomorphism.

Towards Stacks 1

Let’s start working towards what a stack is. I don’t usually like to skip a lot of material, but I know of at least two other blogs that have done some of the preliminary work I need. So today will be very sketchy. I’ll just blurt out a whole bunch of stuff without explaining it, but I’ll give references to other blogs.

First, recall that a site is a category equipped with a Grothendieck topology. You can read about these at Rigorous Trivialites or at Climbing Mount Bourbaki. This is just a way to extend the notion of a topology to a general category.

Some of the standard examples in AG are the Zariski site ${X_{Za}}$, which is just the category of open immersions to ${X}$ with obvious morphisms (the ones that respect the immersion), and the coverings are open immersions ${\{U_\alpha\stackrel{\phi_\alpha}{\rightarrow} V\}}$ such that ${\cup \phi_\alpha(U_\alpha)=V}$. Notice this is just a more formal way of saying that the coverings are Zariski open sets that actually cover the set. Likewise we can define the étale site or fppf site by requiring our maps to be étale or “faithfully flat and locally of finite presentation”.

Sometimes we may want to distinguish between “big” and “small” sites (we’ll see why later). The difference will be that in the big site we allow all scheme maps to be the objects in the category. The small site will be that in the category we only allow maps of the type specified by the site (which is the one I technically wrote above).

If a category, ${\mathcal{C}}$, has two Grothendieck topologies ${\mathcal{T}}$ and ${\mathcal{T}'}$, then there is a notion of the two topologies being equivalent. An easy way to define this is that each of the topologies are refinements of eachother. Another way to define it is if there is a continuous map between the two sites ${F: (\mathcal{C}, \mathcal{T})\rightarrow (\mathcal{C}, \mathcal{T}')}$ that satisfies three conditions:

1) ${F^{-1}}$ is fully faithful.

2) Every open set in ${U}$ in ${\mathcal{T}}$ has a covering of the form ${\{f^{-1}(V_\alpha)\rightarrow U\}}$ where ${V_\alpha}$ are open in ${\mathcal{T}'}$.

3) A collection ${\{V_\alpha\rightarrow V}$ in ${\mathcal{T}'}$ is a covering if ${\{f^{-1}(V_\alpha)\rightarrow f^{-1}(V)\}}$ is a covering in ${\mathcal{T}}$.

Note that an equivalence of topologies in not the same thing as the two sites being “isomorphic”. Equivalence is a weaker notion.

Now that we have a generalized notion of a topological space (on a category), we can try to generalize sheaves on sites. Again, this has been done in two other places (here and here), so we’ll hit the highlights.

Recall that a sheaf on a standard topological space, ${X}$, is just a contravariant functor from the category of open subsets of ${X}$ plus some stuff that makes it “local”. Since all of these things were just stated categorically, it extends in a natural way to any site. Thus we get a category of sheaves on a site denoted ${Sh(\mathcal{T})}$.

It turns out that if you have a category with two equivalent topologies, then the pushforward induces an equivalence of categories ${F_*: Sh(\mathcal{T})\rightarrow Sh(\mathcal{T}')}$, and hence the natural map of cohomology is an iso ${H^i(\mathcal{T}', F_*\mathcal{F})\rightarrow H^i(\mathcal{T}, \mathcal{F})}$.

So for instance you could define the site ${X_C}$ to be the category with objects holomorphic maps from analytic sets ${U\rightarrow X(\mathbb{C})}$ to the ${\mathbb{C}}$-valued points that are local homeomorphisms and coverings to be if the union of the image actually covers ${X}$. Then we have a continuous map ${F: X_C\rightarrow X_{et}}$ since given an \'{e}tale map ${U\rightarrow X}$ if we look at the underlying analytic sets ${U(\mathbb{C})\rightarrow X(\mathbb{C})}$ this is a local homeo. One can check that this is actually an equivalence of topologies. Thus we get that computing complex analytic cohomology or étale cohomology will give the same answer.

Here is why the big site is important. We can only compare Grothendieck topologies on a category if, well, the underlying category is the same. Taking the category as all scheme maps into ${X}$, and then designating certain ones as “special” by the topology allows us to compare the topologies. Notice the underlying categories in the small sites are not the same. The category of open immersions ${U\rightarrow X}$ is not the same as the category of étale maps ${U\rightarrow X}$. I’ve never seen this reasoning for the big site pointed out, and it confused me for awhile, so that’s why I’m making a big deal out of it.

(New edit:) It seems that the above point I just made isn’t universal in the literature for the following reason. There are obviously continuous maps between two sites where the underlying categories aren’t the same. For instance, any of the big sites have a continuous maps to the (same) small sites just by sending the map to itself. It is possible for a continuous map between two sites with different underlying categories to be an equivalence. The comment above was mostly based on Vistoli’s stack notes, in which he only defines equivalence of Grothendieck topologies on the same category.

That seems enough for today. Just to reiterate, the ultimate goal is to figure out what a gerbe is, but in order to do that we need to know what a stack is.