# Mathematical Reason for Uncertainty in Quantum Mechanics

Today I’d like to give a fairly simple account of why Uncertainty Principles exist in quantum mechanics. I thought I already did this post, but I can’t find it now. I often see in movies and sci-fi books (not to mention real-life discussions) a misunderstanding about what uncertainty means. Recall the classic form that says that we cannot know the exact momentum and position of a particle simultaneously.

First, I like this phrasing a bit better than a common alternative: we cannot measure perfectly the momentum and position simultaneously. Although, I guess this is technically true, it has a different flavor. This makes it sound like we don’t have good enough measuring equipment. Maybe in a hundred years our tools will get better, and we will be able to make more precise measurements to do both at once. This is wrong, and completely misunderstands the principle.

Even from a theoretical perspective, we cannot “know.” There are issues with that word as well. In some sense, the uncertainty principle should say that it makes no sense to ask for the momentum and position of a particle (although this again is misleading because we know the precise amount of uncertainty in attempting to do this).

It is like asking: Is blue hard or is blue soft? It doesn’t make sense to ask for the hardness property of a color. To drive the point home, it is even a mathematical impossibility, not just some physical one. You cannot ever write down an equation (a wavefunction for a particle) that has a precise momentum and position at the same time.

Here’s the formalism that lets this fall out easily. To each observable quantity (for example momentum and position) there exists a Hermition operator. If you haven’t seen this before, then don’t worry. The only fact we need about this is that “knowing” or “measuring” or “being in” a certain observable state corresponds to the wavefunction of the particle being an eigenfunction for this operator.

Suppose we have two operators ${A}$ and ${B}$ corresponding to observable quantities ${a}$ and ${b}$, and it makes sense to say that ${\Psi}$ we can simultaneously measure properties ${a}$ and ${b}$. This means there are two number ${\lambda_1}$ and ${\lambda_2}$ such that ${A\Psi = \lambda_1 \Psi}$ and ${B\Psi = \lambda_2 \Psi}$. That is the definition of being an eigenfunction.

This means that the commutator applied to ${\Psi}$ has the property

${[A,B] = AB\Psi - BA\Psi = A\lambda_2 \Psi - B \lambda_1 \Psi = \lambda_2\lambda_1 \Psi - \lambda_1\lambda_2 \Psi = 0}$.

Mathematically speaking, a particle that is in a state for which it makes sense to talk about having two definite observable quantities attached must be described by a wavefunction in the kernel of the commutator. Therefore, it never makes sense to ask for both if the commutator has no kernel. This is our proof. All we must do is compute the commutator of the momentum and position operator and see that it has no kernel (except for the 0 function which doesn’t correspond to a legitimate wavefunction).

You could check wikipedia or something, but the position operator is given by ${\widehat{x}f= xf}$ and the momentum is given by ${\widehat{p}f=-i\hbar f_x}$.

Thus,

$\displaystyle \begin{array}{rcl} [\widehat{x}, \widehat{p}]f & = & -ix\hbar f_x + i\hbar \frac{d}{dx}(xf) \\ & = & -i\hbar (xf_x - f - xf_x) \\ & = & i\hbar f \end{array}$

This shows that the commutator is a constant times the identity operator. It has no kernel, and therefore makes no sense to ask for a definite position and momentum of a particle simultaneously. There isn’t even some crazy, abstract purely theoretical construction that can have that property. This also shows that we can have all sorts of other uncertainty principles by checking other operators.

# How Hard is Adding Integers for a Computer?

In our modern world, we often use high level programming languages (Python, Ruby, etc) without much thought about what is happening. Even if we use a low level language like C, we still probably think of operations like ${1+1}$ yielding ${2}$ or ${3-2}$ yielding ${1}$ as extremely basic. We have no appreciation for how subtle and clever people had to be to first get those types of things to work.

I don’t want to go into detail about how those actually work at the machine level, because that would be a pretty boring post. I do want to do a thought experiment that should bring up some of the issues. Suppose you want to make a new programming language to see how one does such a thing. You think to yourself that it will at least be able to add and subtract integers. How hard could that be?

To play around a little you decide you will first make an integer take up 4 bits of memory. This means when you declare an integer ${x = 1}$, it gets put into a space of size ${4}$ bits: ${0001}$. Each bit can be either a ${0}$ or a ${1}$. Things seem to be going great, because you are comfortable with binary notation. You think that you’ll just take an integer and write its binary representation to memory.

Just for a quick refresher, for 4 bits this means that your integer type can only encode the numbers from ${0}$ to ${15}$. Recall that you can go back to base ${10}$ by taking each digit and using it a coefficient on the appropriate power of ${2}$. Thus ${0101}$ would be ${0\cdot 2^3 + 1\cdot 2^2 + 0\cdot 2^1 + 1\cdot 2^0 = 5}$.

Things are going well. You cleverly come up with an adding function merely by manipulating bits with allowable machine level operations coming directly from the operating system. Then you test ${15 + 1}$. Woops. You overflow. This is the first problem, but it isn’t the interesting one I want to focus on. Even if you have a well defined integer type and a working addition function, this doesn’t mean that adding two integers will always result in an integer! There is an easy rule you think up to determine when it will happen and you just throw an error message for now.

Now you move on to subtraction. Oops. You then realize that you have no way of representing negative numbers with your integer type. If you haven’t seen this before, then you should really take a few moments to think about how you would do this. The “most obvious” solution takes some thought, and turns out to be terrible to use. The one that people actually use is quite clever.

The first thing you might try is to just reserve either the first or last bit as a way to indicate that you are positive or negative. Maybe you’ll take ${1xxx}$ to be negative and ${0xxx}$ to be postive. For example, ${0001}$ is ${1}$ and ${1001}$ is ${-1}$. First, notice that this cuts the number of postive integers you can represent in half, but there isn’t a way around this. Second, there is a positive and negative “0” because ${1000}$ is supposedly ${-0}$. This will almost certainly cause a bigger headache than it is solves.

Lastly, that adding function you wrote is meaningless now. Fortunately, people came up with a much better solution. It is called two’s complement notation. We just weight the most significant bit with a negative. This means that ${1010}$ would convert to ${-2^3 + 0\cdot 2^2 +2^1 + 0\cdot 2^0 = -6}$. This makes all the numbers that start with 1 negative like our earlier example, except there is only a single 0 now because ${1000}$ is ${-8}$ (our most negative integer we can represent).

Moreover ${3-2 = 3 + (-2) = 0011 + 1110 = 0001 = 1}$ (if we chop off overflow, yikes). So plain old addition works and gives us a subtraction. Except, sometimes it doesn’t. For example, take ${0111 + 0001 = 1000}$. This says that ${7+1= -8}$. This is basically the same overflow error from before, because ${8}$ is not an integer that can be represented by our 4 bit type. This just means we have to be careful about some edge cases. It is doable, and in fact, this is exactly what C does (but with 32 bit integers).

Just to wrap up, it seems that to make this hobbled together solution of merely representing and adding integers work we want to make sure that our language is strongly typed (i.e. we know exactly how big an integer is so that we know where to place that leading 1 indicating a negative and the type isn’t going to change on us).

Just consider if we tried to prevent overflow issues by making a “big integer” class that is ${8}$ bits instead of ${4}$. We try to do ${3-2}$ again, and upon overflow we switch the type to a big int. We would then get ${3-2= 0001 \ 0000}$ which is ${16}$. This means we have to be really careful when dealing with multiple types and recasting between them. It seems a minor miracle that in a language like Ruby you can throw around all sorts of different looking types (without declaring any of them) with plus signs between them and get the answer you would expect.

That brings us to the main point of this post. It is a really, really good thing we don’t have to worry about these technicalities when writing programs. The whole point of a good high level language is to not be aware of all the tedious machine level computations going on. But this also means that most people have no appreciation for just how complicated something as simple as adding two integers can be (of course, this is all standardized now, so you probably wouldn’t even worry about it if you were writing your own language from scratch).

# 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.

# 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.

# A Quick User’s Guide to Dieudonné Modules of p-Divisible Groups

Last time we saw that if we consider a ${p}$-divisible group ${G}$ over a perfect field of characteristic ${p>0}$, that there wasn’t a whole lot of information that went into determining it up to isomorphism. Today we’ll make this precise. It turns out that up to isomorphism we can translate ${G}$ into a small amount of (semi-)linear algebra.

I’ve actually discussed this before here. But let’s not get bogged down in the details of the construction. The important thing is to see how to use this information to milk out some interesting theorems fairly effortlessly. Let’s recall a few things. The category of ${p}$-divisible groups is (anti-)equivalent to the category of Dieudonné modules. We’ll denote this functor ${G\mapsto D(G)}$.

Let ${W:=W(k)}$ be the ring of Witt vectors of ${k}$ and ${\sigma}$ be the natural Frobenius map on ${W}$. There are only a few important things that come out of the construction from which you can derive tons of facts. First, the data of a Dieudonné module is a free ${W}$-module, ${M}$, of finite rank with a Frobenius ${F: M\rightarrow M}$ which is ${\sigma}$-linear and a Verschiebung ${V: M\rightarrow M}$ which is ${\sigma^{-1}}$-linear satisfying ${FV=VF=p}$.

Fact 1: The rank of ${D(G)}$ is the height of ${G}$.

Fact 2: The dimension of ${G}$ is the dimension of ${D(G)/FD(G)}$ as a ${k}$-vector space (dually, the dimension of ${D(G)/VD(G)}$ is the dimension of ${G^D}$).

Fact 3: ${G}$ is connected if and only if ${F}$ is topologically nilpotent (i.e. ${F^nD(G)\subset pD(G)}$ for ${n>>0}$). Dually, ${G^D}$ is connected if and only if ${V}$ is topologically nilpotent.

Fact 4: ${G}$ is étale if and only if ${F}$ is bijective. Dually, ${G^D}$ is étale if and only if ${V}$ is bijective.

These facts alone allow us to really get our hands dirty with what these things look like and how to get facts back about ${G}$ using linear algebra. Let’s compute the Dieudonné modules of the two “standard” ${p}$-divisible groups: ${\mu_{p^\infty}}$ and ${\mathbb{Q}_p/\mathbb{Z}_p}$ over ${k=\mathbb{F}_p}$ (recall in this situation that ${W(k)=\mathbb{Z}_p}$).

Before starting, we know that the standard Frobenius ${F(a_0, a_1, \ldots, )=(a_0^p, a_1^p, \ldots)}$ and Verschiebung ${V(a_0, a_1, \ldots, )=(0, a_0, a_1, \ldots )}$ satisfy the relations to make a Dieudonné module (the relations are a little tricky to check because constant multiples ${c\cdot (a_0, a_1, \ldots )}$ for ${c\in W}$ involve Witt multiplication and should be done using universal properties).

In this case ${F}$ is bijective so the corresponding ${G}$ must be étale. Also, ${VW\subset pW}$ so ${V}$ is topologically nilpotent which means ${G^D}$ is connected. Thus we have a height one, étale ${p}$-divisible group with one-dimensional, connected dual which means that ${G=\mathbb{Q}_p/\mathbb{Z}_p}$.

Now we’ll do ${\mu_{p^\infty}}$. Fact 1 tells us that ${D(\mu_{p^\infty})\simeq \mathbb{Z}_p}$ because it has height ${1}$. We also know that ${F: \mathbb{Z}_p\rightarrow \mathbb{Z}_p}$ must have the property that ${\mathbb{Z}_p/F(\mathbb{Z}_p)=\mathbb{F}_p}$ since ${\mu_{p^\infty}}$ has dimension ${1}$. Thus ${F=p\sigma}$ and hence ${V=\sigma^{-1}}$.

The proof of the anti-equivalence proceeds by working at finite stages and taking limits. So it turns out that the theory encompasses a lot more at the finite stages because ${\alpha_{p^n}}$ are perfectly legitimate finite, ${p}$-power rank group schemes (note the system does not form a ${p}$-divisible group because multiplication by ${p}$ is the zero morphism). Of course taking the limit ${\alpha_{p^\infty}}$ is also a formal ${p}$-torsion group scheme. If we wanted to we could build the theory of Dieudonné modules to encompass these types of things, but in the limit process we would have finite ${W}$-module which are not necessarily free and we would get an extra “Fact 5” that ${D(G)}$ is free if and only if ${G}$ is ${p}$-divisible.

Let’s do two more things which are difficult to see without this machinery. For these two things we’ll assume ${k}$ is algebraically closed. There is a unique connected, ${1}$-dimensional ${p}$-divisible of height ${h}$ over ${k}$. I imagine without Dieudonné theory this would be quite difficult, but it just falls right out by playing with these facts.

Since ${D(G)/FD(G)\simeq k}$ we can choose a basis, ${D(G)=We_1\oplus \cdots \oplus We_h}$, so that ${F(e_j)=e_{j+1}}$ and ${F(e_h)=pe_1}$. Up to change of coordinates, this is the only way that eventually ${F^nD(G)\subset pD(G)}$ (in fact ${F^hD(G)\subset pD(G)}$ is the smallest ${n}$). This also determines ${V}$ (note these two things need to be justified, I’m just asserting it here). But all the phrase “up to change of coordinates” means is that any other such ${(D(G'),F',V')}$ will be isomorphic to this one and hence by the equivalence of categories ${G\simeq G'}$.

Suppose that ${E/k}$ is an elliptic curve. Now we can determine ${E[p^\infty]}$ up to isomorphism as a ${p}$-divisible group, a task that seemed out of reach last time. We know that ${E[p^\infty]}$ always has height ${2}$ and dimension ${1}$. In previous posts, we saw that for an ordinary ${E}$ we have ${E[p^\infty]^{et}\simeq \mathbb{Q}_p/\mathbb{Z}_p}$ (we calculated the reduced part by using flat cohomology, but I’ll point out why this step isn’t necessary in a second).

Thus for an ordinary ${E/k}$ we get that ${E[p^\infty]\simeq E[p^\infty]^0\oplus \mathbb{Q}_p/\mathbb{Z}_p}$ by the connected-étale decomposition. But height and dimension considerations tell us that ${E[p^\infty]^0}$ must be the unique height ${1}$, connected, ${1}$-dimensional ${p}$-divisible group, i.e. ${\mu_{p^\infty}}$. But of course we’ve been saying this all along: ${E[p^\infty]\simeq \mu_{p^\infty}\oplus \mathbb{Q}_p/\mathbb{Z}_p}$.

If ${E/k}$ is supersingular, then we’ve also calculated previously that ${E[p^\infty]^{et}=0}$. Thus by the connected-étale decomposition we get that ${E[p^\infty]\simeq E[p^\infty]^0}$ and hence must be the unique, connected, ${1}$-dimensional ${p}$-divisible group of height ${2}$. For reference, since ${ht(G)=\dim(G)+\dim(G^D)}$ we see that ${G^D}$ is also of dimension ${1}$ and height ${2}$. If it had an étale part, then it would have to be ${\mu_{p^\infty}\oplus \mathbb{Q}_p/\mathbb{Z}_p}$ again, so ${G^D}$ must be connected as well and hence is the unique such group, i.e. ${G\simeq G^D}$. It is connected with connected dual. This gives us our first non-obvious ${p}$-divisible group since it is not just some split extension of ${\mu_{p^\infty}}$‘s and ${\mathbb{Q}_p/\mathbb{Z}_p}$‘s.

If we hadn’t done these previous calculations, then we could still have gotten these results by a slightly more general argument. Given an abelian variety ${A/k}$ we have that ${A[p^\infty]}$ is a ${p}$-divisible group of height ${2g}$ where ${g=\dim A}$. Using Dieudonné theory we can abstractly argue that ${A[p^\infty]^{et}}$ must have height less than or equal to ${g}$. So in the case of an elliptic curve it is ${1}$ or ${0}$ corresponding to the ordinary or supersingular case respectively, and the proof would be completed because ${\mathbb{Q}_p/\mathbb{Z}_p}$ is the unique étale, height ${1}$, ${p}$-divisible group.

# p-Divisible Groups Revisited 1

I’ve posted about ${p}$-divisible groups all over the place over the past few years (see: here, here, and here). I’ll just do a quick recap here on the “classical setting” to remind you of what we know so far. This will kick-start a series on some more subtle aspects I’d like to discuss which are kind of scary at first.

Suppose ${G}$ is a ${p}$-divisible group over ${k}$, a perfect field of characteristic ${p>0}$. We can be extremely explicit in classifying all such objects. Recall that ${G}$ is just an injective limit of group schemes ${G=\varinjlim G_\nu}$ where we have an exact sequence ${0\rightarrow G_\nu \rightarrow G_{\nu+1}\stackrel{p^\nu}{\rightarrow} G_{\nu+1}}$ and there is a fixed integer ${h}$ such that group schemes ${G_{\nu}}$ are finite of rank ${p^{\nu h}}$.

As a corollary to the standard connected-étale sequence for group schemes we get a canonical decomposition called the connected-étale sequence:

$\displaystyle 0\rightarrow G^0 \rightarrow G \rightarrow G^{et} \rightarrow 0$

where ${G^0}$ is connected and ${G^{et}}$ is étale. Since ${k}$ was assumed to be perfect, this sequence actually splits. Thus ${G}$ is a semi-direct product of an étale ${p}$-divisible group and a connected ${p}$-divisible group. If you’ve seen the theory for finite, flat group schemes, then you’ll know that we usually decompose these two categories even further so that we get a piece that is connected with connected dual, connected with étale dual, étale with connected dual, and étale with étale dual.

The standard examples to keep in mind for these four categories are ${\alpha_p}$, ${\mu_p}$, ${\mathbb{Z}/p}$, and ${\mathbb{Z}/\ell}$ for ${\ell\neq p}$ respectively. When we restrict ourselves to ${p}$-divisible groups the last category can’t appear in the decomposition of ${G_\nu}$ (since étale things are dimension 0, if something and its dual are both étale, then it would have to have height 0). I think it is not a priori clear, but the four category decomposition is a direct sum decomposition, and hence in this case we get that ${G\simeq G^0\oplus G^{et}}$ giving us a really clear idea of what these things look like.

As usual we can describe étale group schemes in a nice way because they are just constant after base change. Thus the functor ${G^{et}\mapsto G^{et}(\overline{k})}$ is an equivalence of categories between étale ${p}$-divisible groups and the category of inverse systems of ${Gal(\overline{k}/k)}$-sets of order ${p^{\nu h}}$. Thus, after sufficient base change, we get an abstract isomorphism with the constant group scheme ${\prod \mathbb{Q}_p/\mathbb{Z}_p}$ for some product (for the ${p}$-divisible group case it will be a finite direct sum).

All we have left now is to describe the possibilities for ${G^0}$, but this is a classical result as well. There is an equivalence of categories between the category of divisible, commutative, formal Lie groups and connected ${p}$-divisible groups given simply by taking the colimit of the ${p^n}$-torsion ${A\mapsto \varinjlim A[p^n]}$. The canonical example to keep in mind is ${\varinjlim \mathbb{G}_m[p^n]=\mu_{p^\infty}}$. This is connected only because in characteristic ${p}$ we have ${(x^p-1)=(x-1)^p}$, so ${\mu_{p^n}=Spec(k[x]/(x-1)^{p^n})}$. In any other characteristic this group scheme would be étale and totally disconnected.

This brings us to the first subtlety which can cause a lot of confusion because of the abuse of notation. A few times ago we talked about the fact that ${E[p]}$ for an elliptic curve was either ${\mathbb{Z}/p}$ or ${0}$ depending on whether or not it was ordinary or supersingular (respectively). It is dangerous to write this, because here we mean ${E}$ as a group (really ${E(\overline{k})}$) and ${E[p]}$ the ${p}$-torsion in this group.

When talking about the ${p}$-divisible group ${E[p^\infty]=\varinjlim E[p^n]}$ we are referring to ${E/k}$ as a group scheme and ${E[p^n]}$ as the (always!) non-trivial, finite, flat group scheme which is the kernel of the isogeny ${p^n: E\rightarrow E}$. The first way kills off the infinitesimal part so that we are just left with some nice reduced thing, and that’s why we can get ${0}$, because for a supersingular elliptic curve the group scheme ${E[p^n]}$ is purely infinitesimal, i.e. has trivial étale part.

Recall also that we pointed out that ${E[p]\simeq \mathbb{Z}/p}$ for an ordinary elliptic curve by using some flat cohomology trick. But this trick is only telling us that the reduced group is cyclic of order ${p}$, but it does not tell us the scheme structure. In fact, in this case ${E[p^n]\simeq \mu_{p^n}\oplus \mathbb{Z}/p^n}$ giving us ${E[p^\infty]\simeq \mu_{p^\infty}\oplus \mathbb{Q}_p/\mathbb{Z}_p}$. So this is a word of warning that when working these things out you need to be very careful that you understand whether or not you are figuring out the full group scheme structure or just reduced part. It can be hard to tell sometimes.

# Frobenius Semi-linear Algebra 2

Recall our setup. We have an algebraically closed field ${k}$ of characteristic ${p>0}$. We let ${V}$ be a finite dimensional ${k}$-vector space and ${\phi: V\rightarrow V}$ a ${p}$-linear map. Last time we left unfinished the Jordan decomposition that says that ${V=V_s\oplus V_n}$ where the two components are stable under ${\phi}$ and ${\phi}$ acts bijectively on ${V_s}$ and nilpotently on ${V_n}$.

We then considered a strange consequence of what happens on the part on which it acts bijectively. If ${\phi}$ is bijective, then there always exists a full basis ${v_1, \ldots, v_n}$ that are fixed by ${\phi}$, i.e. ${\phi(v_i)=v_i}$. This is strange indeed, because in linear algebra this would force our operator to be the identity.

There is one more slightly more disturbing consequence of this. If ${\phi}$ is bijective, then ${\phi-Id}$ is always surjective. This is a trivial consequence of having a fixed basis. Let ${w\in V}$. We want to find some ${z}$ such that ${\phi(z)=w}$. Well, we just construct the coefficients in the fixed basis by hand. We know ${w=\sum c_i v_i}$ for some ${c_i\in k}$. If ${z=\sum a_i v_i}$ really satisfies ${\phi(z)-z=w}$, then by comparing coefficients such an element exists if and only if we can solve ${a_i^p-a_i=c_i}$. These are just polynomial equations, so we can solve this over our algebraically closed field to get our coefficients.

Strangely enough we really require algebraically closed and not merely perfect again, but the papers I’ve been reading explicitly require these facts over finite fields. Since they don’t give any references at all and just call these things “standard facts about ${p}$-linear algebra,” I’m not sure if there is a less stupid way to prove these things which work for arbitrary perfect fields. This is why you should give citations for things you don’t prove!!

Why do I call this disturbing? Well, these maps really do appear when doing long exact sequences in cohomology. Last time we saw that we could prove that ${E[p]\simeq \mathbb{Z}/p}$ for an ordinary elliptic curve from computing the kernel of ${C-Id}$ where ${C}$ was the Cartier operator. But we have to be really, really careful to avoid linear algebra tricks when these maps come up, because in this situation we have ${\phi -Id}$ is always a surjective map between finite dimensional vector spaces of the same dimension, but also always has a non-trivial kernel isomorphic to ${\mathbb{Z}/p\oplus \cdots \oplus \mathbb{Z}/p}$ where the number of factors is the dimension of ${V}$. Even though we have a surjective map in the long exact sequence between vector spaces of the same dimension, we cannot conclude that it is bijective!

Since everything we keep considering as real-life examples of semi-linear algebra has automatically been bijective (i.e. no nilpotent part), I haven’t actually been too concerned with the Jordan decomposition. But we may as well discuss it to round out the theory since people who work with ${p}$-Lie algebras care … I think?

The idea of the proof is simple and related to what we did last time. We look at iterates ${\phi^j}$ of our map. We get a descending chain ${\phi^j(V)\supset \phi^{j+1}(V)}$ and hence it stabilizes somewhere, since even though ${\phi}$ is not a linear map, the image is still a vector subspace of ${V}$. Let ${r}$ be the smallest integer such that ${\phi^r(V)=\phi^{r+1}(V)}$. This means that ${r}$ is also the smallest integer such that ${\ker\phi^r=\ker \phi^{r+1}}$.

Now we just take as our definition ${V_s=\phi^r(V)}$ and ${V_n=\ker \phi^r}$. Now by definition we get everything we want. It is just the kernel/image decomposition and hence a direct sum. By the choice of ${r}$ we certainly get that ${\phi}$ maps ${V_s}$ to ${V_s}$ and ${V_n}$ to ${V_n}$. Also, ${\phi|_{V_s}}$ is bijective by construction. Lastly, if ${v\in V_n}$, then ${\phi^j(v)=0}$ for some ${0\leq j\leq r}$ and hence ${\phi}$ is nilpotent on ${V_n}$. This is what we wanted to show.

Here’s how this comes up for ${p}$-Lie algebras. Suppose you have some Lie group ${G/k}$ with Lie algebra ${\mathfrak{g}}$. You have the standard ${p}$-power map which is ${p}$-linear on ${\mathfrak{g}}$. By the structure theorem above ${\mathfrak{g}\simeq \mathfrak{h}\oplus \mathfrak{f}}$. The Lie subalgebra ${\mathfrak{h}}$ is the part the ${p}$-power map acts bijectively on and is called the core of the Lie algebra.

Let ${X_1, \ldots, X_d}$ be a fixed basis of the core. We get a nice combinatorial classification of the Lie subalgebras of ${\mathfrak{h}}$. Let ${V=Span_{\mathbb{F}_p}\langle X_1, \ldots, X_d\rangle}$. The Lie subalgebras of ${\mathfrak{h}}$ are in bijective correspondence with the vector subspaces of ${V}$. In particular, the number of Lie subalgebras is finite and each occurs as a direct summand. The proof of this fact is to just repeat the argument of the Jordan decomposition for a Lie subalgebra and look at coefficients of the fixed basis.

# Frobenius Semi-linear Algebra: 1

Today I want to explain some “well-known” facts in semilinear algebra. Here’s the setup. For safety we’ll assume ${k}$ is algebraically closed of characteristic ${p>0}$ (but merely being perfect should suffice for the main point later). Let ${V}$ be a finite dimensional vector space over ${k}$. Consider some ${p}$-semilinear operator on ${V}$ say ${\phi: V\rightarrow V}$. The fact that we are working with ${p}$ instead of ${p^{-1}}$ is mostly to not scare people. I think ${p^{-1}}$ actually appears more often in the literature and the theory is equivalent by “dualizing.”

All this means is that it is a linear operator satisfying the usual properties ${\phi(v+w)=\phi(v)+\phi(w)}$, etc, except for the scalar rule in which we scale by a factor of ${p}$, so ${\phi(av)=a^p\phi(v)}$. This situation comes up surprisingly often in positive characteristic geometry, because often you want to analyze some long exact sequence in cohomology associated to a short exact sequence which involves the Frobenius map or the Cartier operator. The former will induce a ${p}$-linear map of vector spaces and the latter induces a ${p^{-1}}$-linear map.

The facts we’re going to look at I’ve found in three or so papers just saying “from a well-known fact about ${p^{-1}}$-linear operators…” I wish there was a book out there that developed this theory like a standard linear algebra text so that people could actually give references. The proof today is a modification of that given in Dieudonne’s Lie Groups and Lie Hyperalgebras over a Field of Characteristic ${p>0}$ II (section 10).

Let’s start with an example. In the one-dimensional case we have the following ${\phi: k\rightarrow k}$. If the map is non-trivial, then it is bijective. More importantly we can just write down every one of these because if ${\phi(1)=a}$, then

$\displaystyle \begin{array}{rcl} \phi(x) & = & \phi(x\cdot 1) \\ & = & x^p\phi(1) \\ & = & ax^p \end{array}$

In fact, we can always find some non-zero fixed element, because this amounts to solving ${ax^p-x=x(ax^{p-1}-1)=0}$, i.e. finding a solution to ${ax^{p-1}-1=0}$ which we can do by being algebraically closed. This element ${b}$ obviously serves as a basis for ${k}$, but to set up an analogy we also see that ${Span_{\mathbb{F}_p}(b)}$ are all of the fixed points of ${\phi}$. In general ${V}$ will breakup into parts. The part that ${\phi}$ acts bijectively on will always have a basis of fixed elements whose ${\mathbb{F}_p}$-span consists of exactly the fixed points of ${\phi}$. Of course, this could never happen in linear algebra because finding a fixed basis implies the operator is the identity.

Let’s start by proving this statement. Suppose ${\phi: V\rightarrow V}$ is a ${p}$-semilinear automorphism. We want to find a basis of fixed elements. We essentially mimic what we did before in a more complicated way. We induct on the dimension of ${V}$. If we can find a single ${v_1}$ fixed by ${\phi}$, then we would be done for the following reason. We kill off the span of ${v_1}$, then by the inductive hypothesis we can find ${v_2, \ldots, v_n}$ a fixed basis for the quotient. Together these make a fixed basis for all of ${V}$.

Now we need to find a single fixed ${v_1}$ by brute force. Consider any non-zero ${w\in V}$. We start taking iterates of ${w}$ under ${\phi}$. Eventually they will become linearly dependent, so we consider ${w, \phi(w), \ldots, \phi^k(w)}$ for the minimal ${k}$ such that this is a linearly dependent set. This means we can find some coefficients that are not all ${0}$ for which ${\sum a_j \phi^j(w)=0}$.

Let’s just see what must be true of some fictional ${v_1}$ in the span of these elements such that ${\phi(v_1)=v_1}$. Well, ${v_1=\sum b_j \phi^j(w)}$ must satisfy ${v_1=\phi(v_1)=\sum b_j^p \phi^{j+1}(w)}$.

To make this easier to parse, let’s specialize to the case that ${k=3}$. This means that ${a_0 w+a_1\phi(w)+a_2\phi^2(w)=0}$ and by assumption the coefficient on this top power can’t be zero, so we rewrite the top power ${\phi^2(w)=-(a_0/a_2)w - (a_1/a_2)\phi(w)}$.

The other equation is

$\displaystyle \begin{array}{rcl} b_0w+b_1\phi(w) & = & b_0^p\phi(w)+b_1^p\phi^2(w)\\ & = & -(a_0/a_2)b_1^pw +(b_0^p-(a_1/a_2)b_1^p)\phi(w) \end{array}$

Comparing coefficients ${b_0=-(a_0/a_2)b_1^p}$ and then forward substituting ${b_1=-(a_0/a_2)^pb_1^{p^2}-(a_1/a_2)b_1^p}$. Ah, but we know the ${a_j}$ and this only involves the unknown ${b_1}$. So since ${k}$ is algebraically closed we can solve to find such a ${b_1}$. Then since we wrote all our other coefficients in terms of ${b_1}$ we actually can produce a fixed ${v_1}$ by brute force determining the coefficients of the vector in terms of our linear dependence coefficients.

There was nothing special about ${k=3}$ here. In general, this trick will work because it only involves the fact that applying ${\phi}$ cycled the vectors forward by one which allows us to keep forward substituting all the equations from the comparison of coefficients to get everything in terms of the highest one including the highest one which transformed the problem into solving a single polynomial equation over our algebraically closed field.

This completes the proof that if ${\phi}$ is bijective, then there is a basis of fixed vectors. The fact that ${V^\phi=Span_{\mathbb{F}_p}(v_1, \ldots, v_n)}$ is pretty easy after that. Of course, the ${\mathbb{F}_p}$-span is contained in the fixed points because by definition the prime subfield of ${k}$ is exactly the fixed elements of ${x\mapsto x^p}$. On the other hand, if ${c=\sum a_jv_j}$ is fixed, then ${c=\phi(c)=\sum a_j^p \phi(v_j)=\sum a_j^p v_j}$ shows that all the coefficients must be fixed by Frobenius and hence in ${\mathbb{F}_p}$.

Here’s how this is useful. Recall the post on the fppf site. We said that if we wanted to understand the ${p}$-torsion of certain cohomology with coefficients in ${\mathbb{G}_m}$ (Picard group, Brauer group, etc), then we should look at the flat cohomology with coefficients in ${\mu_p}$. If we specialize to the case of curves we get an isomorphism ${H^1_{fl}(X, \mu_p)\simeq Pic(X)[p]}$.

Recall the exact sequence at the end of that post. It told us that via the ${d\log}$ map ${H^1_{fl}(X, \mu_p)=ker(C-I)=H^0(X, \Omega^1)^C}$. Now we have a ridiculously complicated way to prove the following well-known fact. If ${E}$ is an ordinary elliptic curve over an algebraically closed field of characteristic ${p>0}$, then ${E[p]\simeq \mathbb{Z}/p}$. In fact, we can prove something slightly more general.

By definition, a curve is of genus ${g}$ if ${H^0(X, \Omega^1)}$ is ${g}$-dimensional. We’ll say ${X}$ is ordinary if the Cartier operator ${C}$ is a ${p^{-1}}$-linear automorphism (I’m already sweeping something under the rug, because to even think of the Cartier operator acting on this cohomology group we need a hypothesis like ordinary to naturally identify some cohomology groups).

By the results in this post we know that the structure of ${H^0(X, \Omega^1)^C}$ as an abelian group is ${\mathbb{Z}/p\oplus \cdots \oplus \mathbb{Z}/p}$ where there are ${g}$ copies. Thus in more generality this tells us that ${Jac(X)[p]\simeq Pic(X)[p]\simeq H^0(X, \Omega^1)^C\simeq \mathbb{Z}/p\oplus \cdots \oplus \mathbb{Z}/p}$. In particular, since for an elliptic curve (genus 1) we have ${Jac(E)=E}$, this statement is exactly ${E[p]\simeq \mathbb{Z}/p}$.

This point is a little silly, because Silverman seems to just use this as the definition of an ordinary elliptic curve. Hartshorne uses the Hasse invariant in which case it is quite easy to derive that the Cartier operator is an automorphism (proof: it is Serre dual to the Frobenius which by the Hasse invariant definition is an automorphism). Using this definition, I’m actually not sure I’ve ever seen a derivation that ${E[p]\simeq \mathbb{Z}/p}$. I’d be interested if there is a lower level way of seeing it than going through this flat cohomology argument (Silverman cites a paper of Duering, but it’s in German).