## Serre-Tate Theory 2

I guess this will be the last post on this topic. I’ll explain a tiny bit about what goes into the proof of this theorem and then why anyone would care that such canonical lifts exist. On the first point, there are tons of details that go into the proof. For example, Nick Katz’s article, Serre-Tate Local Moduli, is 65 pages. It is quite good if you want to learn more about this. Also, Messing’s book The Crystals Associated to Barsotti-Tate Groups is essentially building the machinery for this proof which is then knocked off in an appendix. So this isn’t quick or easy by any means.

On the other hand, I think the idea of the proof is fairly straightforward. Let’s briefly recall last time. The situation is that we have an ordinary elliptic curve ${E_0/k}$ over an algebraically closed field of characteristic ${p>2}$. We want to understand ${Def_{E_0}}$, but in particular whether or not there is some distinguished lift to characteristic ${0}$ (this will be an element of ${Def_{E_0}(W(k))}$.

To make the problem more manageable we consider the ${p}$-divisible group ${E_0[p^\infty]}$ attached to ${E_0}$. In the ordinary case this is the enlarged formal Picard group. It is of height ${2}$ whose connected component is ${\widehat{Pic}_{E_0}\simeq\mu_{p^\infty}}$. There is a natural map ${Def_{E_0}\rightarrow Def_{E_0[p^\infty]}}$ just by mapping ${E/R \mapsto E[p^\infty]}$. Last time we said the main theorem was that this map is an isomorphism. To tie this back to the flat topology stuff, ${E_0[p^\infty]}$ is the group representing the functor ${A\mapsto H^1_{fl}(E_0\otimes A, \mu_{p^\infty})}$.

The first step in proving the main theorem is to note two things. In the (split) connected-etale sequence

$\displaystyle 0\rightarrow \mu_{p^\infty}\rightarrow E_0[p^\infty]\rightarrow \mathbb{Q}_p/\mathbb{Z}_p\rightarrow 0$

we have that ${\mu_{p^\infty}}$ is height one and hence rigid. We have that ${\mathbb{Q}_p/\mathbb{Z}_p}$ is etale and hence rigid. Thus given any deformation ${G/R}$ of ${E_0[p^\infty]}$ we can take the connected-etale sequence of this and see that ${G^0}$ is the unique deformation of ${\mu_{p^\infty}}$ over ${R}$ and ${G^{et}=\mathbb{Q}_p/\mathbb{Z}_p}$. Thus the deformation functor can be redescribed in terms of extension classes of two rigid groups ${R\mapsto Ext_R^1(\mathbb{Q}_p/\mathbb{Z}_p, \mu_{p^\infty})}$.

Now we see what the canonical lift is. Supposing our isomorphism of deformation functors, it is the lift that corresponds to the split and hence trivial extension class. So how do we actually check that this is an isomorphism? Like I said, it is kind of long and tedious. Roughly speaking you note that both deformation functors are prorepresentable by formally smooth objects of the same dimension. So we need to check that the differential is an isomorphism on tangent spaces.

Here’s where some cleverness happens. You rewrite the differential as a composition of a whole bunch of maps that you know are isomorphisms. In particular, it is the following string of maps: The Kodaira-Spencer map ${T\stackrel{\sim}{\rightarrow} H^1(E_0, \mathcal{T})}$ followed by Serre duality (recall the canonical is trivial on an elliptic curve) ${H^1(E_0, \mathcal{T})\stackrel{\sim}{\rightarrow} Hom_k(H^1(E_0, \Omega^1), H^1(E_0, \mathcal{O}_{E_0}))}$. The hardest one was briefly mentioned a few posts ago and is the dlog map which gives an isomorphism ${H^2_{fl}(E_0, \mu_{p^\infty})\stackrel{\sim}{\rightarrow} H^1(E_0, \Omega^1)}$.

Now noting that ${H^2_{fl}(E_0, \mu_{p^\infty})=\mathbb{Q}_p/\mathbb{Z}_p}$ and that ${T_0\mu_{p^\infty}\simeq H^1(E_0, \mathcal{O}_{E_0})}$ gives us enough compositions and isomorphisms that we get from the tangent space of the versal deformation of ${E_0}$ to the tangent space of the versal deformation of ${E_0[p^\infty]}$. As you might guess, it is a pain to actually check that this is the differential of the natural map (and in fact involves further decomposing those maps into yet other ones). It turns out to be the case and hence ${Def_{E_0}\rightarrow Def_{E_0[p^\infty]}}$ is an isomorphism and the canonical lift corresponds to the trivial extension.

But why should we care? It turns out the geometry of the canonical lift is very special. This may not be that impressive for elliptic curves, but this theory all goes through for any ordinary abelian variety or K3 surface where it is much more interesting. It turns out that you can choose a nice set of coordinates (“canonical coordinates”) on the base of the versal deformation and a basis of the de Rham cohomology of the family that is adapted to the Hodge filtration such that in these coordinates the Gauss-Manin connection has an explicit and nice form.

Also, the canonical lift admits a lift of the Frobenius which is also nice and compatible with how it acts on the above chosen basis on the de Rham cohomology. These coordinates are what give the base of the versal deformation the structure of a formal torus (product of ${\widehat{\mathbb{G}_m}}$‘s). One can then exploit all this nice structure to prove large open problems like the Tate conjecture in the special cases of the class of varieties that have these canonical lifts.

## Serre-Tate Theory 1

Today we’ll try to answer the question: What is Serre-Tate theory? It’s been a few years, but if you’re not comfortable with formal groups and ${p}$-divisible groups, I did a series of something like 10 posts on this topic back here: formal groups, p-divisible groups, and deforming p-divisible groups.

The idea is the following. Suppose you have an elliptic curve ${E/k}$ where ${k}$ is a perfect field of characteristic ${p>2}$. In most first courses on elliptic curves you learn how to attach a formal group to ${E}$ (chapter IV of Silverman). It is suggestively notated ${\widehat{E}}$, because if you unwind what is going on you are just completing the elliptic curve (as a group scheme) at the identity.

Since an elliptic curve is isomorphic to it’s Jacobian ${Pic_E^0}$ there is a conflation that happens. In general, if you have a variety ${X/k}$ you can make the same formal group by completing this group scheme and it is called the formal Picard group of ${X}$. Although, in general you’ll want to do this with the Brauer group or higher analogues to guarantee existence and smoothness. Then you prove a remarkable fact that the elliptic curve is ordinary if and only if the formal group has height ${1}$. In particular, since the ${p}$-divisible group is connected and ${1}$-dimensional it must be isomorphic to ${\mu_{p^\infty}}$.

It might seem silly to think in these terms, but there is another “enlarged” ${p}$-divisible group attached to ${E}$ which always has height ${2}$. This is the ${p}$-divisible group you get by taking the inductive limit of the finite group schemes that are the kernel of multiplication by ${p^n}$. It is important to note that these are non-trivial group schemes even if they are “geometrically trivial” (and is the reason I didn’t just call it the “${p^n}$-torsion”). We’ll denote this in the usual way by ${E[p^\infty]}$.

I don’t really know anyone that studies elliptic curves that phrases it this way, but since this theory must be generalized in a certain way to work for other varieties like K3 surfaces I’ll point out why this should be thought of as an enlarged ${p}$-divisible group. It is another standard fact that ${E}$ is ordinary if and only if ${E[p^\infty]\simeq \mu_{p^\infty}\oplus \mathbb{Q}_p/\mathbb{Z}_p}$. In fact, you can just read off the connected-etale decomposition:

$\displaystyle 0\rightarrow \mu_{p^\infty}\rightarrow E[p^\infty] \rightarrow \mathbb{Q}_p/\mathbb{Z}_p\rightarrow 0$

We already noted that ${\widehat{E}\simeq \mu_{p^\infty}}$, so the ${p}$-divisible group ${E[p^\infty]}$ is a ${1}$-dimensional, height ${2}$ formal group whose connected component is the first one we talked about, i.e. ${E[p^\infty]}$ is an enlargement of ${\widehat{E}}$. For a general variety, this enlarged formal group can be defined, but it is a highly technical construction and would take a lot of work to check that it even exists and satisfies this property. Anyway, this enlarged group is the one we need to work with otherwise our deformation space will be too small to make the theory work.

Here’s what Serre-Tate theory is all about. If you take a deformation of your elliptic curve ${E}$ say to ${E'}$, then it turns out that ${E'[p^\infty]}$ is a deformation of the ${p}$-divisible group ${E[p^\infty]}$. Thus we have a natural map ${\gamma: Def_E \rightarrow Def_{E[p^\infty]}}$. The point of the theory is that it turns out that this map is an isomorphism (I’m still assuming ${E}$ is ordinary here). This is great news, because the deformation theory of ${p}$-divisible groups is well-understood. We know that the versal deformation of ${E[p^\infty]}$ is just ${Spf(W[[t]])}$. The deformation problem is unobstructed and everything lives in a ${1}$-dimensional family.

Of course, let’s not be silly. I’m pointing all this out because of the way in which it generalizes. We already knew this was true for elliptic curves because for any smooth, projective curve the deformations are unobstructed since the obstruction lives in ${H^2}$. Moreover, the dimension of the space of deformations is given by the dimension of ${H^1(E, \mathcal{T})}$. But for an elliptic curve ${\mathcal{T}\simeq \mathcal{O}_X}$, so by Serre duality this is one-dimensional.

On the other hand, we do get some actual information from the Serre-Tate theory isomorphism because ${Def_{E[p^\infty]}}$ carries a natural group structure. Thus an ordinary elliptic curve has a “canonical lift” to characteristic ${0}$ which comes from the deformation corresponding to the identity.

## 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!

## Derived Categories 5: Some Examples

Today we’ll just unravel some toy cases, because understanding these examples is really important, but they tend to be done in generality where the intuition goes away. The first fact is something that we’ll need to use to prove a structure theorem for objects in the derived category of a curve.

As usual, let’s fix ${X/k}$ a smooth, projective variety over a field and let ${D(X)}$ be the bounded derived category of coherent sheaves on ${X}$. In fact, our first fact will work for the derived category of any abelian category. It says that if we take an object (i.e. complex) ${A}$ whose cohomology vanishes for all ${n>i}$, then we can always put it into an exact triangle ${B\rightarrow A\rightarrow H^i(A)[-i]}$.

Let’s be extremely rough and vague to motivate this at first. Given any morphism ${B\stackrel{f}{\rightarrow}A}$ just from the axioms of a triangulated category we can always complete this to an exact triangle ${B\rightarrow A\rightarrow cone(f)}$. Roughly speaking the cone “behaves like a quotient.” I’ll remind you of the exact definition in a second. Our other motivating idea is that we always have a “truncation” process. I’ll chop the very last part of the complex off and then use the “inclusion” map. If the cone is like a quotient I’ll kill everything except that part that I chopped off, and I’ll just be left with the cohomology.

We won’t prove this in general, but we’ll do it in the case of a very small complex to reacquaint ourselves with the cone construction and quasi-isomorphisms. This is one of those cases where this toy case is actually no less general than the general proof when you see what is going on.

Recall that a morphism of complexes ${f: A^\bullet \rightarrow B^\bullet}$ is a collection of morphisms ${f^i: A^i\rightarrow B^i}$ commuting with the differentials: ${d_B\circ f^i=f^{i+1}\circ d_A}$. The complex denoted ${cone(f)=C^\bullet}$ is given by ${C^i=A^{i+1}\bigoplus B^i}$. The differentials are a little tricky, but sort of the only obvious thing that actually makes it into a complex: ${d_C^i=\left(\begin{matrix} -d_A & 0 \\ f^{i+1} & d_B\end{matrix}\right).}$

For simplicity, let’s just assume that our complex has the following form ${0\rightarrow A^0\rightarrow A^1\rightarrow A^2 \rightarrow A^3 \rightarrow 0}$. We’ll assume that cohomology at the second spot is the highest non-zero cohomology. We’ll truncate there and then form the cone. The truncated complex just looks like ${0\rightarrow A^0\rightarrow A^1\rightarrow \ker (d^2)\rightarrow 0}$. Our morphism, ${f}$, of complexes in this easy example at the three different types of places is either the identity morphism, inclusion morphism, or ${0}$ morphism.

The cone complex of this morphism is (I’ll be redundant and put in the zeros for clarity):

$\displaystyle 0\rightarrow A^0\oplus 0 \rightarrow A^1\oplus A^0\rightarrow ker(d^2)\oplus A^1 \rightarrow 0 \oplus A^2\rightarrow 0 \oplus A^3 \rightarrow 0$

One important thing to notice is that at ${A^0\oplus 0}$ is sitting in degree ${-1}$ by definition. By construction the last few maps are again just the original ${d}$ maps. Thus ${H^3(cone(f))=H^3(A^\bullet)=0}$ by assumption. Also, ${H^2(cone(f))=H^2(A^\bullet)}$. If we can show that all other cohomology vanishes, then we will have shown that ${cone(f)}$ is quasi-isomorphic to the single term complex ${H^2(A^\bullet)[-2]}$. This will show the lemma.

This part just follows by “fun with the cone construction.” Note that the first differential is ${(a,0)\mapsto (-d(a), a)}$. If this element is ${(0,0)}$, then ${a=0}$. Thus the kernel is trivial and we get ${H^{-1}(A^\bullet)=0}$. The next one is even more fun. The map is ${(a,b)\mapsto (-d(a), a+d(b))}$. The image of the previous is in the kernel of the next by just doing the maps successively ${(a,0)\mapsto (-d(a), a)\mapsto (d^2(a), -d(a)+d(a))=(0,0)}$. Now we see why that minus sign was important to make the cone a complex.

To check that the image is equal to the kernel, now suppose ${(a,b)}$ is in the kernel. In particular, this means ${a=-d(b)}$ by looking at the second term. The claim is that ${(a,b)}$ is the image of ${(b,0)}$. Well, ${(b,0)\mapsto (-d(b), b)=(a,b)}$. Thus ${H^0(cone(f))=0}$. And now you get the point. There is only one more to check, but it follows the same way.

Of course you could do this with cohomology bounded below with the other truncation functor where you replace the lowest term with a cokernel. In general, this shows an incredibly useful proposition: Given any complex in the bounded derived category of an abelian category there is a finite filtration by truncation ${0\rightarrow A_k \rightarrow A_{k+1} \rightarrow \cdots \rightarrow A_j\rightarrow A^\bullet}$ where the “quotient” (i.e. cone) at each step is ${A_{n-1}\rightarrow A_{n}\rightarrow H^{n}(A^\bullet)[-n]}$. This notation is meant to reflect that ${H^n(A^\bullet)=0}$ if ${n}$ is not in the range ${[k,j]}$.

Here is a remarkable consequence of this proposition. Let ${C/k}$ be a smooth projective curve. Any object ${\mathcal{F}^\bullet\in D(C)}$ is isomorphic to the direct sum of its cohomology sheaves ${\bigoplus_i \mathcal{H}^i(\mathcal{F}^\bullet)[-i]}$. In particular, every object in the derived category is a direct sum of coherent sheaves.

Now we see the power of being able to filtrate by cohomology, because it will allow us to make an induction argument. We prove this by inducting on the length of the complex. The base case is by definition. Suppose the result is true for a length ${k-1}$ complex. Suppose ${\mathcal{F}^\bullet}$ has length ${k}$. By shifting, we suppose ${\mathcal{H}^i(\mathcal{F}^\bullet)\neq 0}$ only in the range ${1\leq i \leq k}$. Consider the first step of the filtration ${\mathcal{E}^\bullet \rightarrow \mathcal{F}^\bullet \rightarrow \mathcal{H}^{k}(\mathcal{F}^\bullet)[-k]\rightarrow \mathcal{E}^\bullet [1]}$.

If this triangle splits we are done, because then the middle term will be a direct sum of the outer terms and by the inductive hypothesis ${\mathcal{E}^\bullet}$ splits as a direct sum of its cohomology. We may as well write it this way

$\displaystyle \mathcal{E}^\bullet\simeq \bigoplus_{i=1}^{k-1} \mathcal{H}^i(\mathcal{E}^\bullet)[-i].$

A sufficient condition to show the triangle splits is to check that ${Hom_{D(C)}(\mathcal{H}^{k}(\mathcal{F}^\bullet)[-k], \mathcal{E}^\bullet [1])=0}$. But by what we just wrote this is the same as

$\displaystyle Hom_{D(C)}(\mathcal{H}^k(\mathcal{F}^\bullet), \bigoplus_{i=1}^{k-1}\mathcal{H}^i(\mathcal{E}^\bullet)[k+1-i]) \simeq \bigoplus_{i=1}^{k-1}Ext_C^{k+1-i}(\mathcal{H}^k(\mathcal{F}^\bullet), \mathcal{H}^i(\mathcal{E}^\bullet))$

Since smooth curves have homological dimension ${1}$ and the exponent is always strictly larger than ${1}$ the Ext groups all vanish. This shows the triangle splits and hence any object in the derived category of a smooth curve splits as a sum of its cohomology sheaves. This exact same proof also shows that for any abelian category with homological dimension less than or equal to ${1}$ the objects of the derived category split as a sum of their cohomology.

## The Derived Category 4: A Nice Spanning Class

Recall that we are assuming that ${X/k}$ is a smooth projective variety. Let’s also say it is of dimension ${n}$. We’re going to be lazy (i.e. sane) and all functors will be derived when talking about the derived category even though the ${\mathbf{L}}$ and ${\mathbf{R}}$ will be omitted. Our derived category always has some (auto-) functors. For example, we definitely have the shift functor that comes with any triangulated category ${[k]}$ just by shifting where the sheaves occur in the complex.

Also, given any coherent sheaf we have the functor ${\mathcal{F}\otimes - : D(X)\rightarrow D(X)}$. In particular, we could tensor with the canonical bundle and shift by ${n}$. This functor is so useful it has a name and notation ${S_X(\mathcal{E})=\mathcal{E}\otimes \omega_X [n]}$ (again, ${\mathcal{E}}$ is any object of ${D(X)}$ and hence a complex even though I didn’t write ${\mathcal{E}^\bullet}$). We call this the Serre functor.

This name just comes from the fact that the generalized form of Serre duality for the derived category says that there is a functorial isomorphism

$\displaystyle \eta: Hom_{D(X)}(\mathcal{E}, \mathcal{F})\stackrel{\sim}{\rightarrow} Hom_{D(X)}(\mathcal{F}, S_X(\mathcal{E}))^*.$

Notice that if ${\mathcal{E}}$ and ${\mathcal{F}}$ are honest sheaves sitting in degree ${0}$ we can use that ${Ext^i(\mathcal{E}, \mathcal{F})=Hom_{D(X)}(\mathcal{E}, \mathcal{F}[i])}$ to derive the special case ${Ext^i(\mathcal{E}, \mathcal{F})\simeq Ext^{n-i}(\mathcal{F}, \mathcal{E}\otimes \omega_X)^*}$ which is the standard form of Serre duality given in classic texts like Hartshorne.

For the rest of today let’s look at a very important concept from triangulated categories. One might wonder how much we can know about certain triangulated categories just from knowing certain special classes of objects. A collection of objects ${\Omega}$ is called a spanning class for a triangulated category ${\mathcal{D}}$ if the following hold:

If ${Hom(A, B[i])=0}$ for all ${A\in \Omega}$ and ${i\in \mathbf{Z}}$, then ${B\simeq 0}$.

If ${Hom(B[i], A)=0}$ for all ${A\in \Omega}$ and ${i\in \mathbf{Z}}$, then ${B\simeq 0}$.

It is not in general true that these two conditions are equivalent, but it is easy to check that Serre duality for ${D(X)}$ will allow us to only have to check one of the conditions. The idea of spanning classes (which may not come up for awhile) is that you can check certain properties just on these objects to get properties on the whole category. For example, one can use this idea to prove necessary and sufficient conditions for a Fourier-Mukai transform to be fully-faithful.

Since our triangulated category ${D(X)}$ is somehow built out of ${X}$, to any (closed) point of ${X}$ we have a natural object associated to it that we’ll call ${k(x)}$. This is just the skyscraper sheaf at the point ${x}$. One hope would be that the set of objects of this form is a spanning class. This intuitively makes sense, because checking a property on this class in the derived category is sort of like checking a property on “points” of variety. It is indeed the case that this forms a spanning class.

Suppose ${\mathcal{F}}$ is a non-trivial object of ${D(X)}$. We’ll check the second condition. This says that we must produce some closed point ${x}$ and some integer ${i}$ so that ${Hom(\mathcal{F}, k(x)[i])\neq 0}$ (well, almost, we used Serre duality again to flip the i over to the other side). We will use the standard local-to-global spectral sequence

$\displaystyle E_2^{p,q}=Ext^p(\mathcal{H}^{-q}(\mathcal{E}), \mathcal{G})\Rightarrow Ext^{p+q}(\mathcal{E}, \mathcal{G}).$

If we plug in ${\mathcal{E}=\mathcal{F}}$ and ${\mathcal{G}=k(x)}$ we get

$\displaystyle E_2^{p,q}=Hom(\mathcal{H}^{-q}(\mathcal{F}), k(x)[p])\Rightarrow Hom(\mathcal{F}, k(x)[p+q]).$

Let ${m}$ be the maximal ${m}$ such that ${\mathcal{H}^m(\mathcal{F})\neq 0}$. The sheaf itself is assumed non-trivial, so there exists ${m}$ with that sheaf non-zero, but ${X}$ is regular so there are only finitely many non-zero and hence such an ${m}$ exists. We will now argue that ${E_2^{0, -m}=E_{\infty}^{0, -m}}$ by showing that all differentials with source and target ${E_r^{0,-m}}$ for any ${r}$ must be trivial.

On the one hand, ${E_2^{p,q}}$ is the ${p}$-th Ext group between coherent sheaves, so when ${p<0}$ it always vanishes. This means that any differential with target ${E_r^{0,-m}}$ must be trivial. On the other hand, our choice of ${m}$ maximal implies that any differential with source ${E_r^{0, -m}}$ is trivial.

Now ${\mathcal{H}^m(\mathcal{F})}$ is non-trivial, so in particular it has non-trivial support which is a closed set and hence contains some closed point ${x}$. This tells us that ${E_{\infty}^{0, -m}=E_2^{0, -m}=Hom(\mathcal{H}^m(\mathcal{F}), k(x))\neq 0}$. But this says that ${Hom(\mathcal{F}, k(x)[-m])\neq 0}$ which is what we set out to prove and hence the collection of skyscraper sheaves of closed points do form a spanning set.

## The Derived Category 3: Derived Functors

Let’s get back to some math. Today I’ll restart a series that I started forever ago on the derived category of a variety. I briefly described what the derived category of a variety is in order to give a very sketchy outline of what Homological Mirror Symmetry is. If you’ve forgotten the construction you can go read about it. I’ll do a one paragraph recap here, so if you don’t care about the details then that should suffice.

For us, in this series we will assume our varieties are all smooth, projective (and irreducible) over a field ${k}$. The notation ${D(X)}$ will mean the bounded derived category of coherent sheaves on ${X}$. It is no longer abelian, but it is a ${k}$-linear, triangulated category. One way to construct ${D(X)}$ is to form the category of complexes of coherent sheaves, then consider morphisms of complexes up to homotopy equivalence, then invert all quasi-isomorphisms. Thus a morphism ${\mathcal{A}^\bullet \rightarrow \mathcal{B}^\bullet}$ is a “hat.” There is a complex ${\mathcal{C}^\bullet}$ quasi-isomorphic to ${\mathcal{A}^\bullet}$ and a morphism in the homotopy category ${\mathcal{C}^\bullet\rightarrow \mathcal{B}^\bullet}$.

Recall how we form derived functors in classical-land. The most beloved derived functor in algebraic geometry is probably the global section functor, because the ${i}$-th right derived functor is just sheaf cohomology: ${R^i\Gamma (X, \mathcal{F})=H^i(X, \mathcal{F})}$. What do we do? We take our sheaf and replace it with an injective resolution (an exact sequence) ${0\rightarrow \mathcal{F}\rightarrow \mathcal{I}^\bullet}$. Then we take global sections of each term (and chop off the first guy) to get a complex which is possibly no longer exact. The ${i}$-th derived functor is now just cohomology at the ${i}$-th spot.

One of the beautiful things about the derived category is that we can keep track of all of this information all at once using a “total” derived functor. Secretly what is going on is that we started with an additive functor ${\Gamma: Coh(X)\rightarrow Vec(k)}$ to vector spaces over ${k}$ (in general, between any two abelian categories). When we first started talking about this we noted that the homotopy category of the full subcategory of injectives is isomorphic to the derived category: ${\mathcal{K}^+(\mathcal{I})\simeq D^+(X)}$. So step one of finding an injective resolution just amounts to replacing the complex ${\cdots \rightarrow 0 \rightarrow \mathcal{F}\rightarrow 0 \rightarrow \cdots}$ with a quasi-isomorphic complex of injectives (i.e. use this equivalence!).

In the homotopy category it is perfectly fine to take global sections of everything and get another complex. Then we just go back to the derived category by the universal quotient functor (of inverting quasi-isomorphisms). What did this do? Well, we may as well generalize. If I have a left exact functor between abelian categories (and I have enough injectives) ${F: \mathcal{A}\rightarrow \mathcal{B}}$, then I can make a total derived functor ${\mathbf{R}F: D^+(\mathcal{A})\rightarrow D^+(\mathcal{B})}$ by replacing a complex by a quasi-isomorphic complex of injectives and applying ${F}$ to everything (and strictly speaking passing back to the derived category).

It takes a complex to a complex, but it is keeping track of all the information of a classical derived functor, because it is literally the exact same process but we just omitted that last step of taking cohomology. So what we end up with is a complex with the property that ${\mathcal{H}^i(\mathbf{R}\Gamma(\mathcal{F}))=R^i\Gamma(\mathcal{F})=H^i(X, \mathcal{F})}$. Since we’re in our nice variety situation all higher cohomology vanishes so we can actually stay in the bounded derived categories.

Note that this isn’t merely a way to keep track of all the ${R^iF}$ at once. It wouldn’t be that useful if this was the only thing it was doing. If we have any of our common left exact functors ${F:Coh(X)\rightarrow Coh(Y)}$, we get a functor ${\mathbf{R}F: D(X)\rightarrow D(Y)}$. So we apply the derived functor to complexes and not just objects of ${Coh(X)}$! This is a vast generalization. A word of warning here. Strictly speaking we have to keep making use of two facts (presented previously).

First, the natural functor ${D(X)\rightarrow D^b(QCoh(X))}$ induces an equivalence between ${D(X)}$ and the full subcategory of complexes of quasi-coherent sheaves with coherent cohomology. This allows us to actually have enough injectives to form the resolutions. Second, we have enough conditions on ${X}$ so that we can do things in either the bounded below or bounded above versions of the derived category, but keep landing inside the bounded derived category since our resolutions may not necessarily be finite a priori.

The common functors to which I referred above are the pushforward of a map ${f: X\rightarrow Y}$. If I input a sheaf and take cohomology, then we recover the higher direct images ${\mathcal{H}^i(\mathbf{R}f_*(\mathcal{F}))=R^if_*(\mathcal{F})}$. We have ${\mathcal{H}om(\mathcal{F}, -): QCoh(X)\rightarrow QCoh(X)}$. Again, smoothness saves us and we get a functor on the bounded derived category with the property that ${\mathcal{H}^i(\mathbf{R}\mathcal{H}om(\mathcal{F}, \mathcal{G}))=\mathcal{E}xt^i(\mathcal{F}, \mathcal{G})}$. We could keep going, but the only other major one I foresee coming up in the near future is the derived tensor product. Of course this will be left derived and so we have to do the whole process above but for right exact functors.

The subtlety about reversing everything is that when you unravel the definitions you’ll find that ${\mathcal{H}^{-i}(\mathcal{F}\otimes^{\mathbf{L}}\mathcal{G})=\mathcal{T}or_i(\mathcal{F}, \mathcal{G})}$. A note to the detail oriented reader. I may forget to put bullets in the superscripts, but everything from here on out should be read as a complex of sheaves and not just a sheaf. I also may get lazy and leave off the R and L for right and left derived, but functors between derived categories will ALWAYS be derived.

Overall, this was just an annoying technical post I had to do. Next time I want to get to some actual geometry of the derived category!

## This Week’s Finds in Arithmetic Geometry

It has been going around the math blogosphere that in honor of John Baez’s 20 year anniversary of doing This Week’s Finds we all do one in our area. So here’s a brief This Week’s Finds in Arithmetic Geometry. Hopefully this will raise awareness of the blog that essentially pioneered math/physics blogging (and if you’re into arithmetic geometry some papers you might not have caught yet).

Since the content and style of Baez’s This Week’s Finds vary so much, I’ll just copy what Jordan Ellenberg did here and give some papers posted the last week that caught my attention.

The fields of definition of branched Galois covers of the projective line by Hilaf Hasson caught my attention because I just met him and saw him speak on this exact topic a few weeks ago at the Joint Meetings. Although the results are certainly interesting in themselves, the part that a blog audience might appreciate is the set of corollaries to the results.

Recall that a major open problem in number theory is the “Inverse Galois Problem” which asks which groups arise as Galois groups. I even posted an elementary proof that if you don’t care what your fields are, then any finite group arises as ${Gal(L/K)}$. In general (for example if you force ${K=\mathbb{Q}}$), then the problem is extremely hard and wide open.

If you haven’t seen this type of thing before, then it might be surprising, but you can actually use geometry to study this question. This is exactly the type of result that Hilaf gets.

Next is New derived autoequivalences of Hilbert schemes and generalised Kummer varieties by Andreas Krug. This topic is near and dear to me because I study derived categories in the arithmetic setting. I haven’t taken a look at this paper in any depth, but I’ll just point out why these types of things are important.

In the classification of varieties one often tries to study the problem up to some type of birational equivalence otherwise it would be too difficult. Often times birational varieties are derived equivalent, but not the other way around. So one could think of studying varieties up to derived equivalence as a slightly looser classification.

When trying to figure out what two varieties that are derived equivalent have in common, a typical sticking point is that you need to know certain automorphisms of the derived category (i.e. autoequivalence) exist to get nice cohomological properties or something. When papers constructing new autoequivalences come out it always catches my attention because I want to know if the method used transfers to situations I work in.

Lastly, we’ll do La conjecture de Tate entière pour les cubiques de dimension quatre sur un corps fini by François Charles and Alena Pirutka. If you don’t know what the Tate conjecture is, then lots of people refer to it as the Hodge conjecture in positive characteristic.

If you’ve ever seen any cohomology theory, then you should be at least passingly familiar with the idea that certain sub-objects (subvarieties or submanifolds etc) can be realized as classes in the cohomology. Sometimes this is due to construction and sometimes it is a major theorem.

The particular case of the Tate conjecture says the following. Consider the relatively easy to prove fact. If you take a cycle on your variety ${X/k}$, then the cohomology class it maps to (in ${\ell}$-adic cohomology) will be invariant under the natural Galois action ${Gal(\overline{k}/k)}$ (because it is defined over ${k}$!). The Tate conjecture is that any Galois invariant cohomology class actually comes from one of these cycles.

The fact that mathematicians can have honest arguments over whether or not the Tate conjecture or the Hodge conjecture (a million dollar problem!) is harder just gives credence to the fact that it is darned hard. If you weren’t convinced, then just consider that this paper is proving the Tate conjecture in the particular case of smooth hypersurfaces of degree ${3}$ in ${\mathbb{P}^5}$ just for the cohomology classes of degree ${4}$. People consider this progress, and they should.

## Brauer-Manin Obstruction

We will continue with the Brauer group after this post, but today let’s answer the question: Why should we care about the Brauer group? We gave one good reason back when talking about rational surfaces. It provided something that might be computable to detect whether or not our variety had good reduction.

Today let’s consider what is probably the most well-known use of the Brauer group. Suppose we have ${X/K}$ a (smooth, proper) variety over a number field. We want to know whether or not there are any ${K}$-rational points. Of course, classically if you hand me this variety using an equation, then the problem reduces to a Diophantine equation. So this problem is very old (3rd century … and one could argue we still don’t have a great handle on it).

Suppose ${Spec \ K\rightarrow X}$ is a rational point. Then we can use the embedding ${K\hookrightarrow K_v}$ into various completions of ${K}$ to get ${Spec \ K_v\rightarrow X}$ points over all the local fields. This isn’t saying much. It says that if we have a global point, then we have points locally everywhere. The interesting question is whether having points locally everywhere “glue” to give a global point. When a class of varieties always allows you to do this, then we say the varieties satisfy the Hasse principle.

This brings us to the content of today’s post. We will construct the Brauer-Manin obstruction to the Hasse principle. We’ve already considered the following pairing in a previous post. For any prime ${v}$, we get a pairing ${Br(X_{K_v})\times X(K_v)\rightarrow \mathbb{Q}/\mathbb{Z}}$ which is just given by pullback on étale cohomology using the map defining the point ${Spec \ K_v\rightarrow X_{K_v}}$ followed by the canonical identification ${Br(K_v)\stackrel{\sim}{\rightarrow}\mathbb{Q}/\mathbb{Z}}$. We could write ${(\alpha, x_v)=x_v^*(\alpha)}$ or more typically ${\alpha(x_v)}$.

We can package this into something global as follows. By restriction we have ${\displaystyle Br(X)\hookrightarrow \prod_v Br(X_{K_v})}$ which we will write ${\alpha\mapsto (\alpha_v)}$. Now we just sum to get ${\displaystyle Br(X)\times \prod_v X(K_v)\rightarrow \mathbb{Q}/\mathbb{Z}}$, so our global pairing is ${\displaystyle (\alpha, (x_v))\mapsto \sum_v \alpha_v(x_v)}$.

Recall that we have an exact sequence from the post on Brauer groups of fields ${0\rightarrow Br(K)\rightarrow \oplus_v Br(K_v)\rightarrow \mathbb{Q}/\mathbb{Z}\rightarrow 0}$ by summing the local invariants. If our local points glue to give a global point ${x\mapsto (x_v)}$, then the pairing factors through this sequence and hence ${(\alpha, (x_v))=0}$ for any choice of ${\alpha}$.

This is our obstruction. The above argument says that if ${(\alpha, (x_v))\neq 0}$, then ${(x_v)}$ cannot possibly glue to give a rational point. We will give this set a name:

$\displaystyle X^{Br}=\{(x_v)\in \prod_v X(K_v) : (\alpha, x_v)=0 \ \text{for all} \ \alpha\in Br(K)\}$

We will call this the Brauer set of ${X}$ (earlier we called it being “Brauer equivalent to ${0}$” since we saw this type of condition was an equivalence relation on Chow groups). It is now immediate that ${X(K)\subset X^{Br}}$. We should think of this as the collection of local points that have some chance of coming from a global point. Now we have two obstructions to the existence of rational points. The first is that ${\prod_v X(K_v)\neq \emptyset}$ which is just the trivial condition that there are local points everywhere. The second is the Brauer-Manin obstruction which says that ${X^{Br}\neq \emptyset}$.

If the local points condition is necessary and sufficient for the existence of rational points, then of course that is exactly the Hasse principle, so we say the Hasse principle holds. If the Brauer-Manin condition (plus the local condition) is necessary and sufficient, then we say that the Brauer-Manin obstruction is the only obstruction to the existence of rational points. It would be fantastic if we could somehow figure out which varieties had this property.

Caution: It is not an open problem to determine whether or not the Brauer-Manin obstruction is the only obstruction. There are known examples where there is no B-M obstruction and yet there are still no rational points. As far as I can tell, it is conjectured, but still open that if the Tate-Shafarevich group of the Jacobian of a curve is finite, then the B-M obstruction is the only obstruction to the existence of rational points on curves over number fields. So even in such a special low-dimensional situation this is a very hard question.

## More Complicated Brauer Computations

Let’s wrap up some of our Brauer group loose ends today. We can push through the calculation of the Brauer groups of curves over some other fields using the same methods as the last post, but just a little more effort.

First, note that with absolutely no extra effort we can run the same argument as yesterday in the following situation. Suppose ${X}$ is a regular, integral, quasi-compact scheme of dimension ${1}$ with the property that all closed points ${v\in X}$ have perfect residue fields ${k(v)}$. Let ${g: \text{Spec} K \hookrightarrow X}$ be the inclusion of the generic point.

Running the Leray spectral sequence a little further than last time still gives us an inclusion, but we will usually want more information because ${Br(K)}$ may not be ${0}$. The low degree terms (plus the argument from last time) gives us a sequence:

$\displaystyle 0\rightarrow Br'(X)\rightarrow Br(K)\rightarrow \bigoplus_v Hom_{cont}(G_{k(v)}, \mathbb{Q}/\mathbb{Z})\rightarrow H^3(X, \mathbb{G}_m)\rightarrow \cdots$

This allows us to recover a result we already proved. In the special case that ${X=\text{Spec} A}$ where ${A}$ is a Henselian DVR with perfect residue field ${k}$, then the uniformizing parameter defines a splitting to get a split exact sequence

$\displaystyle 0\rightarrow Br(A)\rightarrow Br(K)\rightarrow Hom_{cont}(G_k, \mathbb{Q}/\mathbb{Z})\rightarrow 0$

Thus when ${A}$ is a strict local ring (e.g. ${\mathbb{Z}_p}$) we get an isomorphism ${Br(K)\rightarrow \mathbb{Q}/\mathbb{Z}}$ since ${Br(A)\simeq Br(k)=0}$ (since ${k}$ is ${C_1}$). In fact, going back to Brauer groups of fields, we had a lot of trouble trying to figure anything out about number fields. Now we may have a tool (although without class field theory it isn’t very useful, so we’ll skip this for now).

The last computation we’ll do today is to consider a smooth (projective) curve over a finite field ${C/k}$. Fix a separable closure ${k^s}$ and ${K}$ the function field. First, we could attempt to use Leray on the generic point, since we can use that ${H^3(K, \mathbb{G}_m)=0}$ to get some more information. Unfortunately without something else this isn’t enough to recover ${Br(C)}$ up to isomorphism.

Instead, consider the base change map ${f: C^s=C\otimes_k k^s\rightarrow C}$. We use the Hochschild-Serre spectral sequence ${H^p(G_k, H^q(C^s, \mathbb{G}_m))\Rightarrow H^{p+q}(C, \mathbb{G}_m)}$. The low degree terms give us

$\displaystyle 0\rightarrow Br(k)\rightarrow \ker (Br(C)\rightarrow Br(C^s))\rightarrow H^1(G_k, Pic(C^s))\rightarrow \cdots$

First, ${\ker( Br(C)\rightarrow Br(C^s))=Br(C)}$ by the last post. Next ${H^1(G_k, Pic^0(C^s))=0}$ by Lang’s theorem as stated in Mumford’s Abelian Varieties, so ${H^1(G_k, Pic(C^s))=0}$ as well. That tells us that ${Br(C)\simeq Br(k)=0}$ since ${k}$ is ${C_1}$. So even over finite fields (finite was really used and not just ${C_1}$ for Lang’s theorem) we get that smooth, projective curves have trivial Brauer group.

## Brauer Groups of Curves

Let ${C/k}$ be a smooth projective curve over an algebraically closed field. The main goal of today is to show that ${Br(C)=0}$. Both smooth and being over an algebraically closed field are crucial for this computation. The computation will run very similarly to the last post with basically one extra step.

We haven’t actually talked about the Brauer group for varieties, but there are again two definitions. One has to do with Azumaya algebras over ${\mathcal{O}_C}$ modulo Morita equivalence. The other is the cohomological Brauer group, ${Br'(C):=H^2(C, \mathbb{G}_m)}$. As already stated, it is a big open problem to determine when these are the same. We’ll continue to only consider situations where they are known to be the same and hence won’t cause any problems (or even require us to define rigorously the Azumaya algebra version).

First, note that if we look at the Leray spectral sequence with the inclusion of the generic point ${g:Spec(K)\hookrightarrow C}$ we get that ${R^1g_*\mathbb{G}_m=0}$ by Hilbert 90 again which tells us that ${0\rightarrow H^2(C, g_*\mathbb{G}_m)\hookrightarrow Br(K)}$. Now ${K}$ has transcendence degree ${1}$ over an algebraically closed field, so by Tsen’s theorem this is ${C_1}$. Thus the last post tells us that ${H^2(C, g_*\mathbb{G}_m)=0}$.

The new step is that we need to relate ${H^2(C, g_*\mathbb{G}_m)}$ to ${Br(C)}$. On the étale site of ${C}$ we have an exact sequence of sheaves

$\displaystyle 0\rightarrow \mathbb{G}_m\rightarrow g_*\mathbb{G}_m\rightarrow Div_C\rightarrow 0$

where ${\displaystyle Div_C=\bigoplus_{v \ \text{closed}}(i_v)_*\mathbb{Z}}$.
Taking the long exact sequence on cohomology we get

$\displaystyle \cdots \rightarrow H^1(C, Div_C)\rightarrow Br(C)\rightarrow H^2(C, g_*\mathbb{G}_m)\rightarrow \cdots .$

Thus it will complete the proof to show that ${H^1(C, Div_C)=0}$, since then ${Br(C)}$ will inject into ${0}$. Writing ${\displaystyle Div_C=\bigoplus_{v \ \text{closed}}(i_v)_*\mathbb{Z}}$ and using that cohomology commutes with direct sums we need only show that for some fixed closed point ${(i_v): Spec(k(v))\hookrightarrow C}$ that ${H^1(C, (i_v)_*\mathbb{Z})=0}$.

We use Leray again, but this time on ${i_v}$. For notational convenience, we’ll abuse notation and call both the map and the point ${v\in C}$. The low degree terms give us ${H^1(C, v_*\mathbb{Z})\hookrightarrow H^1(v, \mathbb{Z})}$. Using the Galois cohomology interpretation of étale cohomology of a point ${H^1(v,\mathbb{Z})\simeq Hom_{cont}(G_{k(v)}, \mathbb{Z})}$ (the homomorphisms are not twisted since the Galois action is trivial). Since ${G_{k(v)}}$ is profinite, the continuous image is compact and hence a finite subgroup of ${\mathbb{Z}}$. Thus ${H^1(C, v_*\mathbb{Z})=0}$ which implies ${H^1(C, Div_C)=0}$ which gives the result that ${Br(C)=0}$.

So again we see that even for a full curve being over an algebraically closed field is just too strong a condition to give anything interesting. This suggests that the Brauer group really is measuring some arithmetic properties of the curve. For example, we could ask whether or not good/bad reduction of the curve is related to the Brauer group, but this would require us to move into Brauer groups of surfaces (since the model will be a relative curve over a one-dimensional base).

Already for local fields or ${C_1}$ fields the question of determining ${Br(C)}$ is really interesting. The above argument merely tells us that ${Br(C)\hookrightarrow Br(K)}$ where ${K}$ is the function field, but this is true of all smooth, proper varieties and often doesn’t help much if the group is non-zero.