## Towards Stacks 2

In all honesty, I’m not going to blog all the gory details of all the definitions of things that go into the definition of a stack. This leaves me with a problem I haven’t quite figured out how to solve: What do I show? The task seems so massive right now, but there must be a non-offensive way to sort of give the idea of a stack by being more precise than the general vague way people try to explain it, but without doing more than say 5 more posts on it (I also would like to avoid cheating, i.e. go through descent rather than just saying “a stack is a sheaf on a site”).

So today we’ll jump back a little and try to figure out why one would care about stacks. Recall the Yoneda Lemma. Colloquially this says that if we have a (small) category ${\mathcal{C}}$, then the functor of points, which is ${h_X: \mathcal{C}\rightarrow Set}$ by ${h_X(Y)=Hom(Y, X)}$ (note for us it’s contravariant) completely determines the object ${X}$. So what is happening is that we are “testing” what the object looks like, and if we test it against everything, then we completely know what the object looks like.

If we’re thinking about differential geometry, then by functor of points, we really mean “points”. Let ${M}$ be a smooth manifold in the category of smooth manifolds. Then we’ll test what the point-valued points look like. Well, apply the functor of points to a point, ${h_M(pt)=Hom(pt, M)\simeq M}$. Each map is completely determined by which point of ${M}$ the point goes to. It sounds silly, but this is just saying “the points of ${M}$ are the point-valued points”, which we already knew.

Let’s think about this from a very classical algebraic geometry standpoint. Let’s work in the category of affine schemes. We call ${h_X(T)}$ the ${T}$-valued points of ${X}$. This because if we have something like ${X=Spec\left(\mathbb{Z}[x, y, z]/(x^3+y^3-z^3) \right)}$, then by ${Spec(R)}$-valued points we really mean ${h_X(Spec(R))=Hom_{Aff}(Y, X)\simeq Hom_{Ring}(\mathbb{Z}[x,y,z]/(x^3+y^3-z^3), R)}$ what are the solutions to ${x^3+y^3=z^3}$ when ${x,y,z}$ can take values in ${R}$. So if we think of ${h_X(\mathbb{C})}$ it is the complex curve ${x^3+y^3=z^3}$, or ${h_X(\mathbb{Z})}$ are the integer solutions (only trivial ones by Fermat’s Last Theorem). So scheme theoretically this tool is really cool. It groups all of this information into one package.

In reality, we actually usually think of Yoneda in terms of the embedding it gives us. By identifying ${h_X}$ with the object ${X}$ (i.e. the functor represents ${X}$ in the usual sense of a representable functor) we get an embedding ${h: \mathcal{C}\rightarrow Func(\mathcal{C}^{op}, Set)}$ by the obvious thing ${h(X)=h_X(-)}$. So our category ${\mathcal{C}}$ is actually sitting inside the category of functors from ${\mathcal{C}^{op}}$ to ${Set}$.

To quickly recap in the language of schemes, given a scheme ${X}$, we get a functor. But the other way around is not always true, i.e. not every functor is represented by a scheme. Here is where the notion of a stack (or algebraic space) might be nice. Sometimes it is really easy to write down what the functor of points should be for something. Now the question becomes, is this a scheme? Or maybe a better question should be does it matter if it is or is not a scheme? Exactly how loose do we allow the definition to get before we shouldn’t consider it as some sort of “geometric space”?

Just a last quick example to show this really does happen. Suppose you want to figure out what the space of curves of genus ${g}$ looks like. Call this space ${M}$. Then you can almost immediately write down that the functor of points for this space needs to be something like ${h_M(T)=\{C\rightarrow T\}/\simeq}$ where ${C\rightarrow T}$ is proper, smooth, and the fibers are curves of genus ${g}$ and we mod out by appropriate isomorphisms (maybe if we think about moduli later we’ll be more careful with that definition). This turns out to NOT correspond to a scheme. But it is still quite nice, and is in fact a stack. Moral of the story: If we think of schemes as a certain class of functors, it is very natural to come to a more general version of schemes.

## 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.

## A Non-projective Proper Surface

Before moving on to some more exotic topics let’s use these past couple of posts to construct an important example: A proper surface over an algebraically closed field of characteristic 0 that is not projective. This is important because every such curve is projective.

Using the tools we’ve developed, here is the construction. Let ${X'}$ be a non-trivial infinitesimal extension of ${X=\mathbb{P}^2_k}$ by ${\omega :=\omega_{\mathbb{P}^2}}$ (the canonical sheaf). How easy was that? Such a non-trivial extension exists because we’ve already shown that iso classes of extensions are in bijective correspondence with ${H^1(\mathbb{P}^2, \omega\otimes \mathcal{T})\simeq H^1(\mathbb{P}^2, \Omega^1_{\mathbb{P}^2})\neq 0}$.

Here’s how we check it is not projective. We have an exact sequence by the definition of an infinitesimal extension ${0\rightarrow \omega \rightarrow \mathcal{O}_{X'}\rightarrow \mathcal{O}_X\rightarrow 0}$. This induces a sequence ${0\rightarrow \omega\rightarrow \mathcal{O}_{X'}^*\rightarrow \mathcal{O}_X^*\rightarrow 0}$ (just by the first map being ${x\mapsto 1+tx}$ where ${t}$ is the first map of the previous sequence).

This gives a long exact sequence of cohomology: ${\rightarrow H^1(X, \omega)\rightarrow Pic X' \rightarrow Pic X \stackrel{\delta}{\rightarrow} H^2(X, \omega) \rightarrow }$. Now we know the canonical sheaf is really just ${\mathcal{O}(-2)}$, so by knowledge of cohomology of projective space ${H^1(X, \omega)=0}$ and ${H^2(X, \omega)=k}$. Also, the Picard group of projective space is ${\mathbb{Z}}$ generated by ${\mathcal{O}(1)}$.

So if we can show ${\delta}$ is injective, then ${Pic X'=0}$, and hence can’t be projective. Since ${char k=0}$, this just amounts to showing that ${\delta (\mathcal{O}(1))\neq 0}$. We’ll derive a contradiction if it does map to zero. If it does map to zero, then this is the zero map and hence ${Pic X'\rightarrow Pic X}$ is surjective. i.e. there is some invertible sheaf ${\mathcal{L}}$ on ${X'}$ such that ${\mathcal{L}\simeq \mathcal{O}(1)/\omega}$, which is what it means for a sheaf to map to ${\mathcal{O}(1)}$.

So we examine the exact sequence ${0\rightarrow \mathcal{O}(-2)\rightarrow \mathcal{L}\rightarrow \mathcal{O}(1)\rightarrow 0}$. It doesn’t matter what ${X'}$ is to know ${H^1(X', \mathcal{O}(-2))=0}$, so we get that ${H^0(X', \mathcal{L})\rightarrow H^0(X, \mathcal{O}(1))}$ is surjective and hence we get a map ${\mathbb{P}^2\rightarrow X'\rightarrow \mathbb{P}^2}$ which is the identity on ${\mathbb{P}^2}$. This gives us a splitting of ${0\rightarrow \omega \rightarrow \mathcal{O}_{X'}\rightarrow \mathcal{O}_X\rightarrow 0}$. Thus ${\mathcal{O}_X'\simeq \mathcal{O}_X\oplus \omega}$. Ack! This means it was the trivial extension, which we assumed it wasn’t. Thus ${\delta(\mathcal{O}(1))\neq 0}$, which means ${X'}$ is not projective.

Of course this isn’t the most concrete example of one of these things, but I don’t know of any other examples that are this easy to construct and prove. If you wanted something a little bit more concrete, you could actually pick a non-zero class in ${H^1(\mathbb{P}^2, \Omega^1)}$ and piece together your infinitesimal extension.

I’m not making any promises, but next time I might start trying to figure out what gerbes are for a few posts, and then maybe try to tie them back in with how they relate to deformation theory.