## Thoughts on Nicholson Baker’s Case Against Algebra II

The debate over standards in high school math has been going on for a very long time, but things seemed to come to a pretty nasty head last year when the New York Times ran the article Is Algebra Necessary? Bloggers and educators were outraged on both sides and started throwing mud. In the most recent issue of Harper’s (Sept 2013), Nicholson Baker wrote an essay basically reiterating the arguments from the NYT’s piece and responding to some of the criticisms.

I’ve been trying to stay out of this, because I honestly have no idea what high school is designed to do. The real argument here doesn’t seem to be whether or not algebra is “useful in the real world,” but rather about whether or not we should force students to learn things in high school that they are not interested in. Is the purpose of high school to teach students the basics in a broad range of topics so that they have some fundamental skills that will allow them to choose a career from there? Is the purpose to allow students to learn topics that are of interest to them? Something else?

I don’t know, and it is impossible to participate in this debate without clearly defining first what you think the purpose of making students go to high school is (of course, the arguments are muddied by the fact that no one actually defines this first).

Here is Baker’s main argument in a nutshell (he is a fantastic writer, so you should read the full thing yourself if this interests you). Algebra (II) is unnecessary for most people, i.e. the 70% of the population that do not go into a STEM field. It causes excessive stress and failure for basically no reason. Why not just have some survey course in ninth grade where some great ideas of math throughout history are presented and then have all future math courses be electives?

I assume for consistency this means that since English, foreign languages, history, and all other subjects taught in high school are also not directly applicable to most people’s daily lives that basically you’ll do ninth grade as a taste of what the subjects are about through survey courses, and then literally everything is an elective afterwards.

Honestly, I agree with Baker that this would probably make high school a lot more enjoyable and useful for everyone. A lot more learning would take place as well. It just boils back down to what you think the purpose of high school should be, and since I don’t know, I can’t say whether or not this is what should be done.

Here’s two thoughts I had that don’t seem to be raised in the main discussion.

1. How do you know whether or not taking algebra will be useful to you? Having core standards in some sense protects the high school student who isn’t equipped to make this type of decision from making a really bad decision. I’ll just give an anecdote about my own experience as someone who really loved all forms of learning and went into math and who still made a really bad decision when given a choice of electives.

When I was going into my senior year of high school, I knew I wanted to be a composer. I knew this so confidently that despite being advised against it, I decided to not take physics since it was an elective. My reasoning was that I would never, ever need it for my future career as a composer. Let’s ignore the fact that I didn’t realize that understanding the physics of sound is an extremely important skill for a composer to have and so made a poor decision for that reason. Let’s assume that physics really was useless for my intended career.

After my first year of undergrad I switched to a math major. I really regretted not taking physics at that point and ended up loving physics in college so much that I minored in it. Here’s the point. Almost no one in high school knows what they are going to do. So how in the world are the going to know if algebra is necessary for their career? Even if they know what they are going to do, they could still end up mistakenly thinking that it is unnecessary.

My guess is that if we switch to a system where practically everything is an elective, then when people get to college and their interests change they won’t have the basic skills to succeed. They’ll have to fill in this lacking knowledge on their own, because math departments definitely cannot offer more remedial classes. We have so many students and classes as it is we can barely find enough people to teach them all.

2. This seems much ado about nothing. What I’m about to say might seem harsh, but algebra II is not that hard. You don’t have to be good at it. You don’t have to like it. But it isn’t a good sign if you can’t at the very least pass it. Baker himself points out that Cardano in the 1500′s was able to do this stuff. Since then we’ve come up with much easier and better ways to think about it. The abstraction level is just not that high. We’re not talking about quantum mechanics or something. Students in other cultures don’t seem to struggle in the same way, and I don’t think we’re inherently dumber or anything.

Depending on where you look, 30%-50% of students fail algebra II. Let’s say it is closer to 30 because a large number of this statistic does not take into account that there are lazy/rebellious/apatethic/whatever students who can easily handle the abstraction, but just don’t put any work in and fail for that reason. I’d imagine the number of people who try really hard and still fail is pretty low (maybe 20% or less? I’m just making stuff up at this point, but probably way less if you count people who never pass it).

Is it too insensitive and politically incorrect for me to say that someone who can’t handle this level of abstraction probably isn’t cut out for college in any subject? Is college for everyone? I can’t remember what the proper response is to this anymore. What if the number who never pass is around 5%? Is saying this 5% isn’t cut out for college still too much? Sure, give them a high school diploma if they can’t do it, but college may not be the best fit. It seems a good litmus test.

What major won’t require abstraction at least at the level of algebra II? STEM is out. English? Definitely out, unless you somehow avoid all literary theory. Business? Most business degrees require some form of calculus. Music? I hope you can somehow get out of your post-tonal theory classes. History? There has been a recent surge of Bayesian methods in historical methods.

I guess the point is that if a high school diploma is meant to indicate some level of readiness for college, then algebra is probably a good indicator. This does not mean that you will use it, but will just point out that you have some ability to do some abstract things. I’m not saying it is the only way to test this, but it is probably a pretty good one.

Again, if a high school diploma isn’t meant to indicate readiness for college, then who cares what you do?

*Cringes and waits for backlash*

## Movie Hidden Gems 1

I’ve decided I’m going to periodically do a “hidden gems” post. I realized this is one of my favorite uses of the internet as a reader. When I’m looking for a new good book, movie, music, or anything, I tend to hunt out people’s lists of hidden gems. The random lists I’ve found out there have been extremely helpful in learning about things I’d never have found otherwise.

I’ll start with a few movies. These may not be what I consider to be the greatest movies I’ve ever seen, but I think they are absolutely excellent and the main goal is to pick ones I think the least number of people will have been exposed to.

Quiet City (2007). Back when mumblecore was a thing (why did this genre die out again?), I was obsessed with it. I’ve seen pretty much everything that came out of those few years under that heading. Now that that phase is over I would never recommend most of what I saw, though there are certainly a few standouts.

Quiet City took the idea to a new level and created a devastatingly beautiful work of art. This is not for fans of fast pacing and strong story. The whole movie takes place in the course of a single 24 hour period, and follows the slow forming of a new relationship from a chance, accidental meeting in the subway.

It often feels like an urban reshooting of a Terrence Malick film. Instead of interspersing long nature shots, there are long beautiful shots of industrial NYC. The film brilliantly draws out the awkwardness of trying to think of things to do while bored with someone you’ve never met. From an impromptu running race in the park to randomly visiting a friend, there are always things going on even if they are unconventional events for a movie.

If you think of it as action oriented, then you’ve totally missed the point. This is a film whose purpose is to show a natural and realistic development of a relationship, and it succeeds like no film I’ve ever seen because mainstream movies tends to force contrived situations to keep the watcher interested. The film also tries to capture a demographic native to the mumblecore genre: the poor, jobless, artsy, bored city dweller. The attention to detail is amazing and readily manifests itself in such situations as serving wine from an oversized coffee cup.

This movie is definitely not for everyone, but it is a hidden gem if you give it a chance.

Ink (2009). Maybe this isn’t so hidden anymore, because you might have clicked on it when it became available on Netflix (this is how I stumbled upon it). This is a low budget epic fantasy. The concept is fascinating and done extremely well. I can’t say too much without giving away what’s going on, but essentially there is a sort of dream world and the real world and they aren’t completely separate.

I was far more captivated by this film than most high budget movies of the same genre, mostly because they did a great job of revealing what’s going on in just enough increments to keep you guessing. Then once you get the hang of it, you are far enough along that it is high suspense to a carefully orchestrated ending that ties everything together.

The character design is fantastic. You have everything from standard epic fantasy tropes to totally original concepts. The “bad guys” have these static screen faces which are really creepy. The fight scenes are brilliantly done, because they aren’t happening in the physical world even though the physical world is the setting, so as they break things in the fight the things snap back together since they haven’t really broken. The cinematography, artistic rendering of characters, and special effects are actually really impressive for such a low budget.

I can’t remember for sure, but there might be some sort of loop paradox that happens (i.e. an accidental plot hole). If this type of thing bothers you, then you might want to skip this movie. If you are just looking for something thoroughly entertaining and original in the fantasy genre, then this is the hidden gem for you.

Todo Sobre Mi Madre (1999). This is not hidden if you’re into foreign films, because Pedro Almodovar is one of the biggest name directors out there. It’s just that most Americans I’ve talked to haven’t watched any foreign language films at all. I’ve seen everything Almodovar has done many times. I love all them for vastly different reasons, but I’ve picked Todo Sobre Mi Madre, because I think it is less well-known than Volver, but still possible to find.

As with most Almodovar films, it is hard to say there is a “story.” Instead, it covers a huge number of issues with a vast number of characters weaving in and out of each other’s lives. The kernel story-line is essentially a retelling of the classic film All About Eve (hence the title of the movie: All About My Mother). I highly recommend being familiar with that first if you watch this.

The cinematography is stunning, and there is an air of unrealness to the hyper-real plot/subject matter just by the magic quality of the vibrant shots and the consistent overt symbolism strewn throughout. The movie tackles everything from sexuality, identity, and AIDS, to mother-son relationships, acting, and drug use.

One consistent thing with Almodovar is that all of the relationships between the characters are very fluid. People run into trouble when they try to peg labels on the characters, and it is fascinating because we don’t tend to think of how we relate to others in terms of some label. Almodovar really wants people to think about this issue and move past the cliche relationships you’d find in more mainstream movies.

This movie is extremely broad in scope, yet doesn’t spread itself too thin. It has its topics and examines them thoroughly. Despite being such demanding topics, the film always stays thoroughly entertaining. This is one of those movies that I could probably see 100 times and not get bored with it. Everything is done with such attention to detail. This is a must-see intro to Almodovar and a hidden gem for most people within the US.

## An Application of p-adic Volume to Minimal Models

Today I’ll sketch a proof of Ito that birational smooth minimal models have all of their Hodge numbers exactly the same. It uses the ${p}$-adic integration from last time plus one piece of heavy machinery.

First, the piece of heavy machinery: If ${X, Y}$ are finite type schemes over the ring of integers ${\mathcal{O}_K}$ of a number field whose generic fibers are smooth and proper, then if ${|X(\mathcal{O}_K/\mathfrak{p})|=|Y(\mathcal{O}_K/\mathfrak{p})|}$ for all but finitely many prime ideals, ${\mathfrak{p}}$, then the generic fibers ${X_\eta}$ and ${Y_\eta}$ have the same Hodge numbers.

If you’ve seen these types of hypotheses before, then there’s an obvious set of theorems that will probably be used to prove this (Chebotarev + Hodge-Tate decomposition + Weil conjectures). Let’s first restrict our attention to a single prime. Since we will be able to throw out bad primes, suppose we have ${X, Y}$ smooth, proper varieties over ${\mathbb{F}_q}$ of characteristic ${p}$.

Proposition: If ${|X(\mathbb{F}_{q^r})|=|Y(\mathbb{F}_{q^r})|}$ for all ${r}$, then ${X}$ and ${Y}$ have the same ${\ell}$-adic Betti numbers.

This is a basic exercise in using the Weil conjectures. First, ${X}$ and ${Y}$ clearly have the same Zeta functions, because the Zeta function is defined entirely by the number of points over ${\mathbb{F}_{q^r}}$. But the Zeta function decomposes

$\displaystyle Z(X,t)=\frac{P_1(t)\cdots P_{2n-1}(t)}{P_0(t)\cdots P_{2n}(t)}$

where ${P_i}$ is the characteristic polynomial of Frobenius acting on ${H^i(X_{\overline{\mathbb{F}_q}}, \mathbb{Q}_\ell)}$. The Weil conjectures tell us we can recover the ${P_i(t)}$ if we know the Zeta function. But now

$\displaystyle \dim H^i(X_{\overline{\mathbb{F}_q}}, \mathbb{Q}_\ell)=\deg P_i(t)=H^i(Y_{\overline{\mathbb{F}_q}}, \mathbb{Q}_\ell)$

and hence the Betti numbers are the same. Now let’s go back and notice the magic of ${\ell}$-adic cohomology. If ${X}$ and ${Y}$ are as before over the ring of integers of a number field. Our assumption about the number of points over finite fields being the same for all but finitely many primes implies that we can pick a prime of good reduction and get that the ${\ell}$-adic Betti numbers of the reductions are the same ${b_i(X_p)=b_i(Y_p)}$.

One of the main purposes of ${\ell}$-adic cohomology is that it is “topological.” By smooth, proper base change we get that the ${\ell}$-adic Betti numbers of the geometric generic fibers are the same

$\displaystyle b_i(X_{\overline{\eta}})=b_i(X_p)=b_i(Y_p)=b_i(Y_{\overline{\eta}}).$

By the standard characteristic ${0}$ comparison theorem we then get that the singular cohomology is the same when base changing to ${\mathbb{C}}$, i.e.

$\displaystyle \dim H^i(X_\eta\otimes \mathbb{C}, \mathbb{Q})=\dim H^i(Y_\eta \otimes \mathbb{C}, \mathbb{Q}).$

Now we use the Chebotarev density theorem. The Galois representations on each cohomology have the same traces of Frobenius for all but finitely many primes by assumption and hence the semisimplifications of these Galois representations are the same everywhere! Lastly, these Galois representations are coming from smooth, proper varieties and hence the representations are Hodge-Tate. You can now read the Hodge numbers off of the Hodge-Tate decomposition of the semisimplification and hence the two generic fibers have the same Hodge numbers.

Alright, in some sense that was the “uninteresting” part, because it just uses a bunch of machines and is a known fact (there’s also a lot of stuff to fill in to the above sketch to finish the argument). Here’s the application of ${p}$-adic integration.

Suppose ${X}$ and ${Y}$ are smooth birational minimal models over ${\mathbb{C}}$ (for simplicity we’ll assume they are Calabi-Yau, Ito shows how to get around not necessarily having a non-vanishing top form). I’ll just sketch this part as well, since there are some subtleties with making sure you don’t mess up too much in the process. We can “spread out” our varieties to get our setup in the beginning. Namely, there are proper models over some ${\mathcal{O}_K}$ (of course they aren’t smooth anymore), where the base change of the generic fibers are isomorphic to our original varieties.

By standard birational geometry arguments, there is some big open locus (the complement has codimension greater than ${2}$) where these are isomorphic and this descends to our model as well. Now we are almost there. We have an etale isomorphism ${U\rightarrow V}$ over all but finitely many primes. If we choose nowhere vanishing top forms on the models, then the restrictions to the fibers are ${p}$-adic volume forms.

But our standard trick works again here. The isomorphism ${U\rightarrow V}$ pulls back the volume form on ${Y}$ to a volume form on ${X}$ over all but finitely primes and hence they differ by a function which has ${p}$-adic valuation ${1}$ everywhere. Thus the two models have the same volume over all but finitely many primes, and as was pointed out last time the two must have the same number of ${\mathbb{F}_{q^r}}$-valued points over these primes since we can read this off from knowing the volume.

The machinery says that we can now conclude the two smooth birational minimal models have the same Hodge numbers. I thought that was a pretty cool and unexpected application of this idea of ${p}$-adic volume. It is the only one I know of. I’d be interested if anyone knows of any other.

I came across this idea a long time ago, but I needed the result that uses it in its proof again, so I was curious about figuring out what in the world is going on. It turns out that you can make “${p}$-adic measures” to integrate against on algebraic varieties. This is a pretty cool idea that I never would have guessed possible. I mean, maybe complex varieties or something, but over ${p}$-adic fields?

Let’s start with a pretty standard setup in ${p}$-adic geometry. Let ${K/\mathbb{Q}_p}$ be a finite extension and ${R}$ the ring of integers of ${K}$. Let ${\mathbb{F}_q=R_K/\mathfrak{m}}$ be the residue field. If this scares you, then just take ${K=\mathbb{Q}_p}$ and ${R=\mathbb{Z}_p}$.

Now let ${X\rightarrow Spec(R)}$ be a smooth scheme of relative dimension ${n}$. The picture to have in mind here is some smooth ${n}$-dimensional variety over a finite field ${X_0}$ as the closed fiber and a smooth characteristic ${0}$ version of this variety, ${X_\eta}$, as the generic fiber. This scheme is just interpolating between the two.

Now suppose we have an ${n}$-form ${\omega\in H^0(X, \Omega_{X/R}^n)}$. We want to say what it means to integrate against this form. Let ${|\cdot |_p}$ be the normalized ${p}$-adic valuation on ${K}$. We want to consider the ${p}$-adic topology on the set of ${R}$-valued points ${X(R)}$. This can be a little weird if you haven’t done it before. It is a totally disconnected, compact space.

The idea for the definition is the exact naive way of converting the definition from a manifold to this setting. Consider some point ${s\in X(R)}$. Locally in the ${p}$-adic topology we can find a “disk” containing ${s}$. This means there is some open ${U}$ about ${s}$ together with a ${p}$-adic analytic isomorphism ${U\rightarrow V\subset R^n}$ to some open.

In the usual way, we now have a choice of local coordinates ${x=(x_i)}$. This means we can write ${\omega|_U=fdx_1\wedge\cdots \wedge dx_n}$ where ${f}$ is a ${p}$-adic analytic on ${V}$. Now we just define

$\displaystyle \int_U \omega= \int_V |f(x)|_p dx_1 \cdots dx_n.$

Now maybe it looks like we’ve converted this to another weird ${p}$-adic integration problem that we don’t know how to do, but we the right hand side makes sense because ${R^n}$ is a compact topological group so we integrate with respect to the normalized Haar measure. Now we’re done, because modulo standard arguments that everything patches together we can define ${\int_X \omega}$ in terms of these local patches (the reason for being able to patch without bump functions will be clear in a moment, but roughly on overlaps the form will differ by a unit with valuation ${1}$).

This allows us to define a “volume form” for smooth ${p}$-adic schemes. We will call an ${n}$-form a volume form if it is nowhere vanishing (i.e. it trivializes ${\Omega^n}$). You might be scared that the volume you get by integrating isn’t well-defined. After all, on a real manifold you can just scale a non-vanishing ${n}$-form to get another one, but the integral will be scaled by that constant.

We’re in luck here, because if ${\omega}$ and ${\omega'}$ are both volume forms, then there is some non-vanishing function such that ${\omega=f\omega'}$. Since ${f}$ is never ${0}$, it is invertible, and hence is a unit. This means ${|f(x)|_p=1}$, so since we can only get other volume forms by scaling by a function with ${p}$-adic valuation ${1}$ everywhere the volume is a well-defined notion under this definition! (A priori, there could be a bunch of “different” forms, though).

It turns out to actually be a really useful notion as well. If we want to compute the volume of ${X/R}$, then there is a natural way to do it with our set-up. Consider the reduction mod ${\mathfrak{m}}$ map ${\phi: X(R)\rightarrow X(\mathbb{F}_q)}$. The fiber over any point is a ${p}$-adic open set, and they partition ${X(R)}$ into a disjoint union of ${|X(\mathbb{F}_q)|}$ mutually isomorphic sets (recall the reduction map is surjective here by the relevant variant on Hensel’s lemma). Fix one point ${x_0\in X(\mathbb{F}_q)}$, and define ${U:=\phi^{-1}(x_0)}$. Then by the above analysis we get

$\displaystyle Vol(X)=\int_X \omega=|X(\mathbb{F}_q)|\int_{U}\omega$

All we have to do is compute this integral over one open now. By our smoothness hypothesis, we can find a regular system of parameters ${x_1, \ldots, x_n\in \mathcal{O}_{X, x_0}}$. This is a legitimate choice of coordinates because they define a ${p}$-adic analytic isomorphism with ${\mathfrak{m}^n\subset R^n}$.

Now we use the same silly trick as before. Suppose ${\omega=fdx_1\wedge \cdots \wedge dx_n}$, then since ${\omega}$ is a volume form, ${f}$ can’t vanish and hence ${|f(x)|_p=1}$ on ${U}$. Thus

$\displaystyle \int_{U}\omega=\int_{\mathfrak{m}^n}dx_1\cdots dx_n=\frac{1}{q^n}$

This tells us that no matter what ${X/R}$ is, if there is a volume form (which often there isn’t), then the volume

$\displaystyle Vol(X)=\frac{|X(\mathbb{F}_q)|}{q^n}$

just suitably multiplies the number of ${\mathbb{F}_q}$-rational points there are by a factor dependent on the size of the residue field and the dimension of ${X}$. Next time we’ll talk about the one place I know of that this has been a really useful idea.

## BSD for a Large Class of Elliptic Curves

I’m giving up on the p-divisible group posts for awhile. I would have to be too technical and tedious to write anything interesting about enlarging the base. It is pretty fascinating stuff, but not blog material at the moment.

I’ve been playing around with counting fibration structures on K3 surfaces, and I just noticed something I probably should have been aware of for a long time. This is totally well-known, but I’ll give a slightly anachronistic presentation so that we can use results from 2013 to prove the Birch and Swinnerton-Dyer conjecture!! … Well, only in a case that has been known since 1973 when it was published by Artin and Swinnerton-Dyer.

Let’s recall the Tate conjecture for surfaces. Let ${k}$ be a finite field and ${X/k}$ a smooth, projective surface. We’ve written this down many times now, but the long exact sequence associate to the Kummer sequence

$\displaystyle 0\rightarrow \mu_{\ell}\rightarrow \mathbb{G}_m\rightarrow \mathbb{G}_m\rightarrow 0$

(for ${\ell\neq \text{char}(k)}$) gives us a cycle class map

$\displaystyle c_1: Pic(X_{\overline{k}})\otimes \mathbb{Q}_{\ell}\rightarrow H^2_{et}(X_{\overline{k}}, \mathbb{Q}_\ell(1))$

In fact, we could take Galois invariants to get our standard

$\displaystyle 0\rightarrow Pic(X)\otimes \mathbb{Q}_{\ell}\rightarrow H^2_{et}(X_{\overline{k}}, \mathbb{Q}_\ell(1))^G\rightarrow Br(X)[\ell^\infty]\rightarrow 0$

The Tate conjecture is in some sense the positive characteristic version of the Hodge conjecture. It conjectures that the first map is surjective. In other words, whenever an ${\ell}$-adic class “looks like” it could come from an honest geometric thing, then it does. But if the Tate conjecture is true, then this implies the ${\ell}$-primary part of ${Br(X)}$ is finite. We could spend some time worrying about independence of ${\ell}$, but it works, and hence the Tate conjecture is actually equivalent to finiteness of ${Br(X)}$.

Suppose now that ${X}$ is an elliptic K3 surface. This just means that there is a flat map ${X\rightarrow \mathbb{P}^1}$ where the fibers are elliptic curves (there are some degenerate fibers, but after some heavy machinery we could always put this into some nice form, we’re sketching an argument here so we won’t worry about the technical details of what we want “fibration” to mean). The generic fiber ${X_\eta}$ is a genus ${1}$ curve that does not necessarily have a rational point and hence is not necessarily an elliptic curve.

But we can just use a relative version of the Jacobian construction to produce a new fibration ${J\rightarrow \mathbb{P}^1}$ where ${J}$ is a K3 surface fiberwise isomorphic to ${X}$, but now ${J_\eta=Jac(X_\eta)}$ and hence is an elliptic curve. Suppose we want to classify elliptic fibrations that have ${J}$ as the relative Jacobian. We have two natural ideas to do this.

The first is that etale locally such a fibration is trivial, so you could consider all glueing data to piece such a thing together. The obstruction will be some Cech class that actually lives in ${H^2(X, \mathbb{G}_m)=Br(X)}$. In fancy language, you make these things as ${\mathbb{G}_m}$-gerbes which are just twisted relative moduli of sheaves. The class in ${Br(X)}$ is giving you the obstruction the existence of a universal sheaf.

A more number theoretic way to think about this is that rather than think about surfaces over ${k}$, we work with the generic fiber ${X_\eta/k(t)}$. It is well-known that the Weil-Chatelet group: ${H^1(Gal(k(t)^{sep}/k(t), J_\eta)}$ gives you the possible genus ${1}$ curves that could occur as generic fibers of such fibrations. This group is way too big though, because we only want ones that are locally trivial everywhere (otherwise it won’t be a fibration).

So it shouldn’t be surprising that the classification of such things is given by the Tate-Shafarevich group:

Ш $\displaystyle (J_\eta /k(t))=ker ( H^1(G, J_\eta)\rightarrow \prod H^1(G_v, (J_\eta)_v))$

Very roughly, I’ve now given a heuristic argument (namely that they both classify the same set of things) that ${Br(X)\simeq}$ Ш ${(J_\eta)}$, and it turns out that Grothendieck proved the natural map that comes form the Leray spectral sequence ${Br(X)\rightarrow}$ Ш${(J_\eta)}$ is an isomorphism (this rigorous argument might actually have been easier than the heuristic one because we’ve computed everything involved in previous posts, but it doesn’t give you any idea why one might think they are the same).

Theorem: If ${E/\mathbb{F}_q(t)}$ is an elliptic curve of height ${2}$ (occuring as the generic fiber of an elliptic K3 surface), then ${E}$ satisfies the Birch and Swinnerton-Dyer conjecture.

Idea: Using the machinery alluded to before, we spread out ${E}$ to an elliptic K3 surface ${X\rightarrow \mathbb{P}^1}$ over a finite field. As of this year, it seems the Tate conjecture is true for K3 surfaces (the proofs are all there, I’m not sure if they have been double checked and published yet). Thus ${Br(X)}$ is finite. Thus Ш${ (E)}$ is finite. But now it is well-known that if Ш${ (E)}$ being finite is equivalent to the Birch and Swinnerton-Dyer conjecture.

## Newton Polygons of p-Divisible Groups

I really wanted to move on from this topic, because the theory gets much more interesting when we move to ${p}$-divisible groups over some larger rings than just algebraically closed fields. Unfortunately, while looking over how Demazure builds the theory in Lectures on ${p}$-divisible Groups, I realized that it would be a crime to bring you this far and not concretely show you the power of thinking in terms of Newton polygons.

As usual, let’s fix an algebraically closed field of positive characteristic to work over. I was vague last time about the anti-equivalence of categories between ${p}$-divisible groups and ${F}$-crystals mostly because I was just going off of memory. When I looked it up, I found out I was slightly wrong. Let’s compute some examples of some slopes.

Recall that ${D(\mu_{p^\infty})\simeq W(k)}$ and ${F=p\sigma}$. In particular, ${F(1)=p\cdot 1}$, so in our ${F}$-crystal theory we get that the normalized ${p}$-adic valuation of the eigenvalue ${p}$ of ${F}$ is ${1}$. Recall that we called this the slope (it will become clear why in a moment).

Our other main example was ${D(\mathbb{Q}_p/\mathbb{Z}_p)\simeq W(k)}$ with ${F=\sigma}$. In this case we have ${1}$ is “the” eigenvalue which has ${p}$-adic valuation ${0}$. These slopes totally determine the ${F}$-crystal up to isomorphism, and the category of ${F}$-crystals (with slopes in the range ${0}$ to ${1}$) is anti-equivalent to the category of ${p}$-divisible groups.

The Dieudonné-Manin decomposition says that we can always decompose ${H=D(G)\otimes_W K}$ as a direct sum of vector spaces indexed by these slopes. For example, if I had a height three ${p}$-divisible group, ${H}$ would be three dimensional. If it decomposed as ${H_0\oplus H_1}$ where ${H_0}$ was ${2}$-dimensional (there is a repeated ${F}$-eigenvalue of slope ${0}$), then ${H_1}$ would be ${1}$-dimensional, and I could just read off that my ${p}$-divisible group must be isogenous to ${G\simeq \mu_{p^\infty}\oplus (\mathbb{Q}_p/\mathbb{Z}_p)^2}$.

In general, since we have a decomposition ${H=H_0\oplus H' \oplus H_1}$ where ${H'}$ is the part with slopes strictly in ${(0,1)}$ we get a decomposition ${G\simeq (\mu_{p^\infty})^{r_1}\oplus G' \oplus (\mathbb{Q}_p/\mathbb{Z}_p)^{r_0}}$ where ${r_j}$ is the dimension of ${H_j}$ and ${G'}$ does not have any factors of those forms.

This is where the Newton polygon comes in. We can visually arrange this information as follows. Put the slopes of ${F}$ in increasing order ${\lambda_1, \ldots, \lambda_r}$. Make a polygon in the first quadrant by plotting the points ${P_0=(0,0)}$, ${P_1=(\dim H_{\lambda_1}, \lambda_1 \dim H_{\lambda_1})}$, … , ${\displaystyle P_j=\left(\sum_{l=1}^j\dim H_{\lambda_l}, \sum_{l=1}^j \lambda_l\dim H_{\lambda_l}\right)}$.

This might look confusing, but all it says is to get from ${P_{j}}$ to ${P_{j+1}}$ make a line segment of slope ${\lambda_j}$ and make the segment go to the right for ${\dim H_{\lambda_j}}$. This way you visually encode the slope with the actual slope of the segment, and the longer the segment is the bigger the multiplicity of that eigenvalue.

But this way of encoding the information gives us something even better, because it turns out that all these ${P_i}$ must have integer coordinates (a highly non-obvious fact proved in the book by Demazure listed above). This greatly restricts our possibilities for Dieudonné ${F}$-crystals. Consider the height ${2}$ case. We have ${H}$ is two dimensional, so we have ${2}$ slopes (possibly the same). The maximal ${y}$ coordinate you could ever reach is if both slopes were maximal which is ${1}$. In that case you just get the line segment from ${(0,0)}$ to ${(2,2)}$. The lowest you could get is if the slopes were both ${0}$ in which case you get a line segment ${(0,0)}$ to ${(2,0)}$.

Every other possibility must be a polygon between these two with integer breaking points and increasing order of slopes. Draw it (or if you want to cheat look below). You will see that there are obviously only two other possibilities. The one that goes ${(0,0)}$ to ${(1,0)}$ to ${(2,1)}$ which is a slope ${0}$ and slope ${1}$ and corresponds to ${\mu_{p^\infty}\oplus \mathbb{Q}_p/\mathbb{Z}_p}$ and the one that goes ${(0,0)}$ to ${(2,1)}$. This corresponds to a slope ${1/2}$ with multiplicity ${2}$. This corresponds to the ${E[p^\infty]}$ for supersingular elliptic curves. That recovers our list from last time.

We now just have a bit of a game to determine all height ${3}$ ${p}$-divisible groups up to isogeny (and it turns out in this small height case that determines them up to isomorphism). You can just draw all the possibilities for Newton polygons as in the height ${2}$ case to see that the only ${6}$ possibilities are ${(\mu_{p^\infty})^3}$, ${(\mu_{p^\infty})^2\oplus \mathbb{Q}_p/\mathbb{Z}_p}$, ${\mu_{p^\infty}\oplus (\mathbb{Q}_p/\mathbb{Z}_p)^2}$, ${(\mathbb{Q}_p/\mathbb{Z}_p)^3}$, and then two others: ${G_{1/3}}$ which corresponds to the thing with a triple eigenvalue of slope ${1/3}$ and ${G_{2/3}}$ which corresponds to the thing with a triple eigenvalue of slope ${2/3}$.

To finish this post (and hopefully topic!) let’s bring this back to elliptic curves one more time. It turns out that ${D(E[p^\infty])\simeq H^1_{crys}(E/W)}$. Without reminding you of the technical mumbo-jumbo of crystalline cohomology, let’s think why this might be reasonable. We know ${E[p^\infty]}$ is always height ${2}$, so ${D(E[p^\infty])}$ is rank ${2}$. But if we consider that crystalline cohomology should be some sort of ${p}$-adic cohomology theory that “remembers topological information” (whatever that means), then we would guess that some topological ${H^1}$ of a “torus” should be rank ${2}$ as well.

Moreover, the crystalline cohomology comes with a natural Frobenius action. But if we believe there is some sort of Weil conjecture magic that also applies to crystalline cohomology (I mean, it is a Weil cohomology theory), then we would have to believe that the product of the eigenvalues of this Frobenius equals ${p}$. Recall in the “classical case” that the characteristic polynomial has the form ${x^2-a_px+p}$. So there are actually only two possibilities in this case, both slope ${1/2}$ or one of slope ${1}$ and the other of slope ${0}$. As we’ve noted, these are the two that occur.

In fact, this is a more general phenomenon. When thinking about ${p}$-divisible groups arising from algebraic varieties, because of these Weil conjecture type considerations, the Newton polygons must actually fit into much narrower regions and sometimes this totally forces the whole thing. For example, the enlarged formal Brauer group of an ordinary K3 surface has height ${22}$, but the whole Newton polygon is fully determined by having to fit into a certain region and knowing its connected component.

## More Classification of p-Divisible Groups

Today we’ll look a little more closely at ${A[p^\infty]}$ for abelian varieties and finish up a different sort of classification that I’ve found more useful than the one presented earlier as triples ${(M,F,V)}$. For safety we’ll assume ${k}$ is algebraically closed of characteristic ${p>0}$ for the remainder of this post.

First, let’s note that we can explicitly describe all ${p}$-divisible groups over ${k}$ up to isomorphism (of any dimension!) up to height ${2}$ now. This is basically because height puts a pretty tight constraint on dimension: ${ht(G)=\dim(G)+\dim(G^D)}$. If we want to make this convention, we’ll say ${ht(G)=0}$ if and only if ${G=0}$, but I’m not sure it is useful anywhere.

For ${ht(G)=1}$ we have two cases: If ${\dim(G)=0}$, then it’s dual must be the unique connected ${p}$-divisible group of height ${1}$, namely ${\mu_{p^\infty}}$ and hence ${G=\mathbb{Q}_p/\mathbb{Z}_p}$. The other case we just said was ${\mu_{p^\infty}}$.

For ${ht(G)=2}$ we finally get something a little more interesting, but not too much more. From the height ${1}$ case we know that we can make three such examples: ${(\mu_{p^\infty})^{\oplus 2}}$, ${\mu_{p^\infty}\oplus \mathbb{Q}_p/\mathbb{Z}_p}$, and ${(\mathbb{Q}_p/\mathbb{Z}_p)^{\oplus 2}}$. These are dimensions ${2}$, ${1}$, and ${0}$ respectively. The first and last are dual to each other and the middle one is self-dual. Last time we said there was at least one more: ${E[p^\infty]}$ for a supersingular elliptic curve. This was self-dual as well and the unique one-dimensional connected height ${2}$ ${p}$-divisible group. Now just playing around with the connected-étale decomposition, duals, and numerical constraints we get that this is the full list!

If we could get a bit better feel for the weird supersingular ${E[p^\infty]}$ case, then we would have a really good understanding of all ${p}$-divisible groups up through height ${2}$ (at least over algebraically closed fields).

There is an invariant called the ${a}$-number for abelian varieties defined by ${a(A)=\dim Hom(\alpha_p, A[p])}$. This essentially counts the number of copies of ${\alpha_p}$ sitting inside the truncated ${p}$-divisible group. Let’s consider the elliptic curve case again. If ${E/k}$ is ordinary, then we know ${E[p]}$ explicitly and hence can argue that ${a(E)=0}$. For the supersingular case we have that ${E[p]}$ is actually a non-split semi-direct product of ${\alpha_p}$ by itself and we get that ${a(E)=1}$. This shows that the ${a}$-number is an invariant that is equivalent to knowing ordinary/supersingular.

This is a phenomenon that generalizes. For an abelian variety ${A/k}$ we get that ${A}$ is ordinary if and only if ${a(A)=0}$ in which case the ${p}$-divisible group is a bunch of copies of ${E[p^\infty]}$ for an ordinary elliptic curve, i.e. ${A[p^\infty]\simeq E[p^\infty]^g}$. On the other hand, ${A}$ is supersingular if and only if ${A[p^\infty]\simeq E[p^\infty]^g}$ for ${E/k}$ supersingular (these two facts are pretty easy if you use the ${p}$-rank as the definition of ordinary and supersingular because it tells you the étale part and you mess around with duals and numerics again).

Now that we’ve beaten that dead horse beyond recognition, I’ll point out one more type of classification which is the one that comes up most often for me. In general, there is not redundant information in the triple ${(M, F, V)}$, but for special classes of ${p}$-divisible groups (for example the ones I always work with explained here) all you need to remember is the ${(M, F)}$ to recover ${G}$ up to isomorphism.

A pair ${(M,F)}$ of a free, finite rank ${W}$-module equipped with a ${\phi}$-linear endomorphism ${F}$ is sometimes called a Cartier module or ${F}$-crystal. Every Dieudonné module of a ${p}$-divisible group is an example of one of these. We could also consider ${H=M\otimes_W K}$ where ${K=Frac(W)}$ to get a finite dimensional vector space in characteristic ${0}$ with a ${\phi}$-linear endomorphism preserving the ${W}$-lattice ${M\subset H}$.

Passing to this vector space we would expect to lose some information and this is usually called the associated ${F}$-isocrystal. But doing this gives us a beautiful classification theorem which was originally proved by Diedonné and Manin. We have that ${H}$ is naturally an ${A}$-module where ${A=K[T]}$ is the noncommutative polynomial ring ${T\cdot a=\phi(a)\cdot T}$. The classification is to break up ${H\simeq \oplus H_\alpha}$ into a slope decomposition.

These ${\alpha}$ are just rational numbers corresponding to the slopes of the ${F}$ operator. The eigenvalues ${\lambda_1, \ldots, \lambda_n}$ of ${F}$ are not necessarily well-defined, but if we pick the normalized valuation ${ord(p)=1}$, then the valuations of the eigenvalues are well-defined. Knowing the slopes and multiplicities completely determines ${H}$ up to isomorphism, so we can completely capture the information of ${H}$ in a simple Newton polygon. Note that when ${H}$ is the ${F}$-isocrystal of some Dieudonné module, then the relation ${FV=VF=p}$ forces all slopes to be between 0 and 1.

Unfortunately, knowing ${H}$ up to isomorphism only determines ${M}$ up to equivalence. This equivalence is easily seen to be the same as an injective map ${M\rightarrow M'}$ whose cokernel is a torsion ${W}$-module (that way it becomes an isomorphism when tensoring with ${K}$). But then by the anti-equivalence of categories two ${p}$-divisible groups (in the special subcategory that allows us to drop the ${V}$) ${G}$ and ${G'}$ have equivalent Dieudonné modules if and only if there is a surjective map ${G' \rightarrow G}$ whose kernel is finite, i.e. ${G}$ and ${G'}$ are isogenous as ${p}$-divisible groups.

Despite the annoying subtlety in fully determining ${G}$ up to isomorphism, this is still really good. It says that just knowing the valuation of some eigenvalues of an operator on a finite dimensional characteristic ${0}$ vector space allows us to recover ${G}$ up to isogeny.