A Mind for Madness

Musings on art, philosophy, mathematics, and physics


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The Stack of Pitch Class Sets

I know it’s been a while since I’ve talked about either of these topics, but I’ve always been meaning to point something funny out. I thought I might formally work it out and write it up to submit to a music theory journal, but no one would probably accept it anyway. So I’ll sketch the idea now. Back here I talked about stacks as a useful way to generalize what we mean by a “space.” Back here I talked about the math behind the idea of pitch class sets.

I know Mazzola wrote a whole book on using topos theory in music, but I’ve never dug into it very deeply. I fully admit this is probably just a special case of something from that book. But it’s always useful to work out special cases.

Recall that a pitch set (or chord) is just converting notes to numbers: 0 is C, 1 is C#, 2 is D, etc. A given collection of pitches can be expressed in a more useful notation when there isn’t a key we’re working in. For example, a C major chord is (047).

A pitch class set is then saying that there are collections of these we want to consider to be the same. For one, our choice of 0 is completely arbitrary. We could have set 0 is A, and we should get the same theory. This amounts to identifying all pitch sets that are the same after translation.

We also want to identify sets that are the same after inversion. In the previous post on this topic, I showed that if we label the vertices of a dodecagon, this amounts to a reflection symmetry. The reflections together with the translations generate the dihedral group {D_{12}}, so we are secretly letting {D_{12}} act on the set of all tuples of numbers 0 to 11, where each number only appears once and without loss of generality we can assume they are in increasing order.

Thus a pitch class set is just an equivalence class of a chord under this group action. It is not the direction I want this post to go, but given such a class, there is always a unique representative that is usually called the “prime form” (basically the most “compact” representative starting with 0).

Here’s where we get to the part I never really worked out. The set of all “chords” should have some sort of useful topology on it. For example, (0123) should be related to (0124), because they are the same chord except for one note. I don’t think doing something obvious like defining a distance based on the coordinates works. If you try to construct the lattice of open sets by hand based on your intuition, a definition might become more obvious. Call this space of chords {X}.

Now we have a space with a group action on it. One might want to merely form the quotient space {X \rightarrow X/G}. This will be 24 to 1 at most points, but it will also forget which chords were fixed by elements of the group. Part of the “theory” in music theory is to remember that information. This is why I propose making the quotient stack {[X/G]}. It seems like an overly complicated thing to do, but here’s what you gain.

You now have a “space” whose points are the pitch class sets. If that class contains 24 distinct chords, then the point is an “honest” point with no extra information. The fiber of the quotient map contains the 24 chords, and you get to each of them by acting by the elements of {D_{12}} (i.e. it is a torsor under {D_{12}}). Now consider something like the pitch class set [0,2,4,6,8,10]. The fiber of the quotient map only contains {2} elements: (02468T) and (13579E). The stack will tag these points with {D_6}, which is the subgroup of symmetries which send this chord to itself.

pitch stack

Now that I’ve drawn this, I can see that many of you will be skeptical about the simplicity. Think of it this way. The bottom thing is the space I’m describing. Each point in the space is tagged with the prime form representative together with the subgroup of symmetries that preserve the class. That’s pretty simple. Yet it remembers all of the complicated music theory of the top thing! If the topology was defined well, then studying this space may even lead to insights on how symmetries of classes are related to each other. Let me know if anyone has seen anything like this before.


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The Functor of Points Revisited

Mike Hopkins is giving the Milliman Lectures this week at the University of Washington and the first talk involved this idea that I’m extremely familiar with, but am also surprised at how unfamiliar most mathematicians are with it. I’ve made almost this exact post several other times, but it bears repeating. As I basked in the amazingness of this idea during the talk, I couldn’t help but notice how annoyed some people seemed to be at the level of abstractness and generality this notion forces on you.

Every branch of math has some crowning achievements and insights into how to actually think about something so that it works. The idea I’ll present in this post is a truly remarkable insight into geometry and topology. It is incredibly simple (despite the daunting language) which is what makes it so fascinating. Here is the idea. Suppose you care about some type of spaces (metric, topological, manifolds, varieties, …).

Let {X} be one of your spaces. In order to figure out what {X} is you could probe it by other spaces. What does this mean? It just means you look at maps {Y\rightarrow X}. If {X} is a topological space, then you can recover the points of {X} by considering all the maps from a singleton (i.e. point) {\{x\} \rightarrow X}. If you want to understand more about the topology, then you probe by some other spaces. Simple.

Even analysts use this idea all the time. A distribution {\phi} (on {\mathbb{R}}) is not a well-defined function, so you can’t just tell whether or not two distributions are the same by looking at values. Instead you probe it by test functions {\int \phi f dx}. If these probes give you the same thing for all test functions, then the distributions are the same. This is all we are doing with our spaces above, and this is all the Yoneda lemma is saying. It says that if the maps (test functions) to {X} and the maps to {Y} are the same, then {X} and {Y} are the same.

We can fancy up the language now. Considering maps to {X} is a functor {Hom(-,X): Spaces^{op} \rightarrow Set}. Such a functor is called a presheaf on the category of Spaces (recall, that for your particular situation this might be the category of smooth manifolds or metric spaces or algebraic varieties or …). Don’t be scared. This is literally the definition of presheaf, so if you were following to now, then introducing this term requires no new definitions.

The Yoneda lemma is saying something very simple in this fancy language. It says that there is a (fully faithful) embedding of Spaces into Pre(Spaces), the category of presheaves on Spaces. If we now work with this new category of functors, we just enlarge what we consider to be a space and this is of fundamental importance for many reasons. If {X} is one of our old spaces, then we can just naturally identify it with the presheaf {Hom(-,X)}. The reason Mike Hopkins is giving for why this is important is very different from the one I’ll give which just goes to show how incredibly useful this idea is.

In every single branch of math people care about some sort of classification problem. Classify all elliptic curves. What are the vector bundles on my manifold? If I fix a vector bundle, what are the connections on my vector bundle? What are the Borel measures on my metric space? The list goes on forever.

In general, classification is a hugely impossible task to grapple with. We know a ton of stuff about smooth manifolds, but how can we leverage that to make the seemingly unrelated problem of classifying vector bundles more manageable? Here our insight comes to the rescue, because there is a way to write down a functor that outputs vector bundles. There is subtlety in writing it down properly (and we should now land in Grpds instead of Set so that we can identify isomorphic ones), but once we do this we get a presheaf. In other words, we make a (generalized) space whose points are the objects we are classifying.

In many situations you then go on to prove that this moduli space of vector bundles is actually one of the original types of spaces (or not too far from one) we know a lot about. Now our impossible task of understanding what the vector bundles on my manifold are is reduced to the already studied problem of understanding the geometry of a manifold itself!

Here is my challenge to any analyst who knows about measures. Warning, this could be totally ridiculous and nonsense because it is based on reading Wikipedia for 5 minutes. Construct a presheaf of real-valued Radon measures on {\mathbb{R}}. Analyze this “space”. If it was done right, you should somehow recover that the space is the dual space to the convex space, {C_c(\mathbb{R})}, of compactly supported real-valued functions on {\mathbb{R}}. Congratulations, you’ve just started a new branch of math in which you classify measures on a space by analyzing the topology/geometry of the associated presheaf.


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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.


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Volumes of p-adic Schemes

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.


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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.


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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.


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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.

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