# What’s up with the fppf site?

I’ve been thinking a lot about something called Serre-Tate theory lately. I want to do some posts on the “classical” case of elliptic curves. Before starting though we’ll go through some preliminaries on why one would ever want to use the fppf site and how to compute with it. It seems that today’s post is extremely well known, but not really spelled out anywhere.

Let’s say you’ve been reading stuff having to do with arithmetic geometry for awhile. Then without a doubt you’ve encountered étale cohomology. In fact, I’ve used it tons on this blog already. Here’s a standard way in which it comes up. Suppose you have some (smooth, projective) variety ${X/k}$. You want to understand the ${\ell^n}$-torsion in the Picard group or the (cohomological) Brauer group where ${\ell}$ is a prime not equal to the characteristic of the field.

What you do is take the Kummer sequence:

$\displaystyle 0\rightarrow \mu_{\ell^n}\rightarrow \mathbb{G}_m\stackrel{\ell^n}{\rightarrow} \mathbb{G}_m\rightarrow 0.$

This is an exact sequence of sheaves in the étale topology. Thus it gives you a long exact sequence of cohomology. But since ${H^1_{et}(X, \mathbb{G}_m)=Pic(X)}$ and ${H^2_{et}(X, \mathbb{G}_m)=Br(X)}$. Just writing down the long exact sequence you get that the image of ${H^1_{et}(X, \mu_{\ell^n})\rightarrow Pic(X)}$ is exactly ${Pic(X)[\ell^n]}$, and similarly with the Brauer group. In fact, people usually work with the truncated short exact sequence:

$\displaystyle 0\rightarrow Pic(X)/\ell^n Pic(X) \rightarrow H^2_{et}(X, \mu_{\ell^n})\rightarrow Br(X)[\ell^n]\rightarrow 0$

Fiddling around with other related things can help you figure out what is happening with the ${\ell^n}$-torsion. That isn’t the point of this post though. The point is what do you do when you want to figure out the ${p^n}$-torsion where ${p}$ is the characteristic of the ground field? It looks like you’re in big trouble, because the above Kummer sequence is not exact in the étale topology.

It turns out that you can switch to a finer topology called the fppf topology (or site). This is similar to the étale site, except instead of making your covering families using étale maps you make them with faithfully flat and locally of finite presentation maps (i.e. fppf for short when translated to french). When using this finer topology the sequence of sheaves actually becomes exact again.

A proof is here, and a quick read through will show you exactly why you can’t use the étale site. You need to extract ${p}$-th roots for the ${p}$-th power map to be surjective which will give you some sort of infinitesimal cover (for example if ${X=Spec(k)}$) that looks like ${Spec(k[t]/(t-a)^p)\rightarrow Spec(k)}$.

Thus you can try to figure out the ${p^n}$-torsion again now using “flat cohomology” which will be denoted ${H^i_{fl}(X, -)}$. We get the same long exact sequences to try to fiddle with:

$\displaystyle 0\rightarrow Pic(X)/p^n Pic(X) \rightarrow H^2_{fl}(X, \mu_{p^n})\rightarrow Br(X)[p^n]\rightarrow 0$

But what the heck is ${H^2_{fl}(X, \mu_{p^n})}$? I mean, how do you compute this? We have tons of books and things to compute with the étale topology. But this fppf thing is weird. So secretly we really want to translate this flat cohomology back to some étale cohomology. I saw the following claimed in several places without really explaining it, so we’ll prove it here:

$\displaystyle H^2_{fl}(X, \mu_p)=H^1_{et}(X, \mathbb{G}_m/\mathbb{G}_m^p).$

Actually, let’s just prove something much more general. We actually get that

$\displaystyle H^i_{fl}(X, \mu_p)=H^{i-1}_{et}(X, \mathbb{G}_m/\mathbb{G}_m^p).$

The proof is really just a silly “trick” once you see it. Since the Kummer sequence is exact on the fppf site, by definition this just means that the complex ${\mu_p}$ thought of as concentrated in degree ${0}$ is quasi-isomorphic to the complex ${\mathbb{G}_m\stackrel{p}{\rightarrow} \mathbb{G}_m}$. It looks like this is a useless and more complicated thing to say, but this means that the hypercohomology (still fppf) is isomorphic:

$\displaystyle \mathbf{H}^i_{fl}(X, \mu_p)=\mathbf{H}^i_{fl}(X, \mathbb{G}_m\stackrel{p}{\rightarrow} \mathbb{G}_m).$

Now here’s the trick. The left side is the group we want to compute. The right hand side only involves smooth group schemes, so a theorem of Grothendieck tells us that we can compute this hypercohomology using fpqc, fppf, étale, Zariski … it doesn’t matter. We’ll get the same answer. Thus we can switch to the étale site. But of course, just by definition we now extend the ${p}$-th power map (injective on the etale site) to an exact sequence

$\displaystyle 0\rightarrow \mathbb{G}_m \rightarrow \mathbb{G}_m\rightarrow \mathbb{G}_m/\mathbb{G}_m^p\rightarrow 0.$

Thus we get another quasi-isomorphism of complexes. This time to ${\mathbb{G}_m/\mathbb{G}_m^p[-1]}$. This is a complex concentrated in a single degree, so the hypercohomology is just the etale cohomology. The shift by ${-1}$ decreases the cohomology by one and we get the desired isomorphism ${H^i_{fl}(X, \mu_p)=H^{i-1}_{et}(X, \mathbb{G}_m/\mathbb{G}_m^p)}$. In particular, we were curious about ${H^2_{fl}(X, \mu_p)}$, so we want to figure out ${H^1_{et}(X, \mathbb{G}_m/\mathbb{G}_m^p)}$.

Alright. You’re now probably wondering what in the world to I do with the étale cohomology of ${\mathbb{G}_m/\mathbb{G}_m^p}$? It might be on the étale site, but it is a weird sheaf. Ah. But here’s something great, and not used all that much to my knowledge. There is something called the multiplicative de Rham complex. On the étale site we actually have an exact sequence of sheaves via the “dlog” map:

$\displaystyle 0\rightarrow \mathbb{G}_m/\mathbb{G}_m^p\stackrel{d\log}{\rightarrow} Z^1\stackrel{C-i}{\rightarrow} \Omega^1\rightarrow 0.$

This now gives us something nice because if we understand the Cartier operator (which is Serre dual to the Frobenius!) and know things how many global ${1}$-forms are on the variety (maybe none?) we have a hope of computing our original flat cohomology!

# Heights of Varieties

Now that we’ve defined the height of a ${p}$-divisible group we’ll define the height of a variety in positive characteristic. There are a few ways we can motivate this definition, but really it just works and turns out to be a very useful concept. We’ll mostly follow the paper of Artin and Mazur.

We could do this in more generality, but to keep things as simple as possible we’ll assume that we have a proper variety ${X}$ over a perfect field ${k}$ of characteristic ${p}$. The first motivation is that we can think about ${\mathrm{Pic}(X)}$. One way to get information about this group is to use deformation theory and look at the formal completion ${\widehat{\mathrm{Pic}}(X)}$.

The way to define this is to define the ${S}$-valued points (${S}$ an Artin local ${k}$-algebra with residue field ${k}$) to be the group fitting into the sequence ${0\rightarrow \widehat{\mathrm{Pic}}(X)(S)\rightarrow H^1(X\times S, \mathbb{G}_m)\rightarrow H^1(X, \mathbb{G}_m)}$.

So ${\widehat{\mathrm{Pic}}(X)}$ is a functor which by Schlessinger’s criterion is prorepresentable by a formal group over ${k}$. Notice that ${\widehat{\mathrm{Pic}}(X)(S)=\mathrm{ker}(\mathrm{Pic}(X\times S)\rightarrow \mathrm{Pic}(X))}$, so there is a pretty concrete way to think about what is going on. We take our scheme and consider some nilpotent thickening. The line bundles on this thickening that are just extensions from the trivial line bundle are what is in this formal Picard group.

There is no reason to stop with just ${H^1}$. We could define ${\Phi^r: Art_k\rightarrow Ab}$ by ${\Phi^r(S)}$ is the kernel of the restriction map ${H^r(X\times S, \mathbb{G}_m)\rightarrow H^r(X, \mathbb{G}_m)}$. In the cases we care about, modulo some technical details, we can apply Schlessinger type arguments to this to get that if the dimension of ${X}$ is ${n}$, then ${\Phi^n}$ is not only pro-representable, but by formal Lie group of dimension ${1}$. We’ll call this ${\Phi_X}$.

When ${n=2}$ this is just the well-known Brauer group, and so for instance the height of a K3 surface is the height of the Brauer group. We also have that if ${\Phi_X}$ is not ${\widehat{\mathbb{G}}_a}$ then it is a ${p}$-divisible group and amazingly the Dieudonne module of ${\Phi_X}$ is related to the Witt sheaf cohomology via ${D(\Phi_X)=H^n(X, \mathcal{W})}$. Recall that ${D(\Phi_X)}$ is a free ${W(k)}$-module of rank the height of ${\Phi_X}$, so in particular ${H^n(X, \mathcal{W})}$ is a finite ${W(k)}$-module!

Remember that we computed an example where that wasn’t finitely generated. So non-finite generatedness of ${H^n(X, \mathcal{W})}$ actually is related to the height in that if the variety is of finite height then ${H^n(X, \mathcal{W})}$ is finitely generated. Since we call a variety of infinite height supersingular, we can rephrase this as saying that ${H^n(X, \mathcal{W})}$ is not finitely generated if and only if ${X}$ is supersingular.

Just as an example of what heights can be, an elliptic curve must have height ${1}$ or ${2}$ and a K3 surface can have height between ${1}$ and ${10}$ (inclusive). As of right now it seems that the higher dimensional analogue of if the finite height range of a Calabi-Yau threefold is bounded is still open. People have proved certain bounds in terms of hodge numbers. For instance ${h(\Phi_X)\leq h^{1, 2}+1}$. For a general CY ${n}$-fold we have ${h\leq h^{1, n-1}+1}$.

This is pretty fascinating because my interpretation of this (which could be completely wrong) is that since for K3 surfaces the moduli space is ${20}$ dimensional, we get that (for non-supersingular) ${h^{1,1}=20}$ since this is just the dimension of the tangent space of the deformations, which for a smooth moduli should match the dimension of the moduli space. Thus we get a uniform bound (not the one I mentioned earlier).

But for CY threefolds the moduli space is much less uniform. They aren’t all deformation equivalent. They lie on different components that have different dimensions (this is a guess, I haven’t actually seen this written anywhere). So this doesn’t allow us to say ${h^{1,2}}$ is some number. It depends on the dimension of the component of the moduli that it is on (since ${h^{1,2}=\dim H^2(X, \Omega)=\dim H^1(X, \mathcal{T})}$ using the CY conditions and Serre duality). So I think it is still an open problem for how big that can be. If it can get unreasonably large, then maybe we can arbitrarily large heights of CY threefolds.

Next time maybe we’ll prove some equivalent ways of computing heights for CY varieties and talk about how height has been used by Van der Geer and Katsura and others in a useful way for K3 surfaces.

# 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.