## 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.

## Mori’s Bend and Break

I noticed that recently people were clicking a lot of the links I had on my blogroll. Since many of the blogs were defunct, or I didn’t read them anymore I chopped a lot off. I also added a few that I found myself returning to frequently (including no longer active ones). So that has been updated for the first time in years.

The last little bit we’ll do that is related to moduli spaces and deformation theory is something called Mori’s bend and break argument. It says that if ${X}$ is a nonsingular projective variety of dimension ${n}$ over an algebraically closed field, ${k}$, of positive characteristic ${p}$ and if there is an irreducible curve ${C\subset X}$ with ${C. K_X<0}$, then ${X}$ contains a rational curve. In this context a rational curve is an integral curve whose normalization is ${\mathbb{P}^1}$. The condition on ${K_X}$ is sometimes stated as being not numerically effective (not nef).

Suppose ${C_0}$ is an integral curve such that ${C_0. K_X<0}$. If the normalization ${C_1\rightarrow C_0}$ has genus ${0}$, then we are done. Let ${g:=g(C_1)}$. Choose ${r}$ large enough so that ${-p^r(C_0. K_X)\geq ng+1}$. Define ${q=p^r}$. Let ${F:C\rightarrow C_1}$ be the ${q}$-th power, ${k}$-linear Frobenius map and denote ${f:C\rightarrow C_0}$ the composition. We only changed the structure sheaf and not the topological space, so ${C}$ still has genus ${g}$.

Fix some point ${P\in C}$. Let ${Hom_P(C,X)}$ be the quasi-projective scheme that represents the functor of families of maps from ${C}$ to ${X}$ that keep the image of ${P}$ fixed (it's a subscheme of the usual Hom scheme). Standard deformation theory tells us that the tangent space is ${H^0(C, f^*T_X(-P))}$ and the obstruction space is ${H^1(C, f^*T_X(-P))}$.

We didn't do this, but it is all part of the package that we've talked about. When the functor is representable we get the natural estimate that ${\dim Hom_P(C, X)\geq h^0(f^*T_X(-P))-h^1(f^*T_X(-P))}$. This just comes from the fact that if every possible obstruction is realized, then each one will cut the dimension down, but often despite an obstruction space being non-zero the obstruction itself might vanish. This will only make the dimension bigger.

Now Riemann-Roch gives us that ${\dim Hom_P(C, X)\geq -q(C_0. K_X)-n+n(1-g)\geq 1}$ by our choice of ${q}$. In particular, we can find a nonsingular curve ${D}$ and a morphism ${g:C\times D\rightarrow X}$, thought of as a nonconstant family of maps all sending ${P}$ to the same point ${P_0}$. You can argue here that ${D}$ cannot be complete, otherwise the family would have to be constant.

So let ${D\subset \overline{D}}$ be a completion where ${\overline{D}}$ is a nonsingular projective curve. Let ${G:C\times \overline{D} \dashrightarrow X}$ be the rational map. Blow up a finite number of points to resolve the undefined points to get ${Y\rightarrow C\times \overline{D}}$ whose composition given by ${\pi: Y\rightarrow X}$ is an honest morphism. Let ${E\subset Y}$ be the exceptional curve of the last blow up needed. Since it was actually needed, it can't be collapsed to a point, and hence ${\pi(E)}$ is our desired curve.

This is one of those interesting things where it is easier in positive characteristic than characteristic ${0}$ because you have the Frobenius at your disposal. It allowed us to jack up the tangent space without affecting the obstruction space to produce our curve. Mori actually does relate this back to varieties in characteristic ${0}$ to prove Hartshorne's conjecture which says that a nonsingular, projective variety with ample tange bundle is isomorphic to ${\mathbb{P}^n}$.

## Moduli of Vector Bundles on Elliptic Curves

We’ve been talking about moduli problems, and one notoriously hard type of moduli problem is to “classify” vector bundles on some variety. Even when you restrict yourself to some special case like a specific surface and try to classify only vector bundles of certain rank or Chern class you run into trouble. To this day, these types of problems are a very active area of research.

It is fairly well-known (due to Grothendieck, but a problem in Hartshorne as well) that any finite rank vector bundle over ${\mathbb{P}^1}$ is just a finite direct sum ${\oplus_i\mathcal{O}(n_i)}$. The next interesting case would be to move up to genus ${1}$ curves. It turns out that Atiyah in the 1957 paper, Vector Bundles over an Elliptic Curve, worked out a classification. I just learned about this a month or two ago, and it is pretty cool so I’d like to briefly describe the idea.

Fix an algebraically closed field ${k}$ of characteristic ${0}$, and let ${E/k}$ be an elliptic curve. Define ${V(r,d)}$ to be the set of indecomposable vector bundles (up to isomorphism) on ${E}$ of rank ${r}$ and degree ${d}$, where degree just means the degree of the determinant. It is well known that ${V(1,0)}$ can be identified with ${E}$, because ${V(1,0)=Pic^0(E)}$ is the set of degree ${0}$ divisors. In particular, this says that the moduli space of degree ${0}$ divisors (line bundles) on ${E}$ is fine and representable by ${E}$.

Let’s continue with this idea of classifying degree ${0}$ vector bundles. It turns out there is a unique vector bundle ${T_r\in V(r,0)}$ with the property that ${\Gamma (E, T_r)\neq 0}$, i.e. there are non-trivial global sections. Now, recall how ${V(1,0)}$ works. Let ${0\in E}$ be the origin. Then our isomorphism ${E\rightarrow V(1,0)}$ is given by ${P\mapsto \mathcal{O}(-P)\otimes \mathcal{O}([0])}$. Our ${T_r}$ is going to play the role of ${\mathcal{O}([0])}$ here. In complete analogy we get a bijection ${V(1,0)\rightarrow V(r,0)}$ by ${L\mapsto L\otimes T_r}$.

This finishes off the case of degree ${0}$ vector bundles, because we get that ${E\simeq V(1,0)\simeq V(r,0)}$ (roughly speaking, of course there’s a lot more structure we need to know about to say something about the moduli spaces).

In more general situations we can use the same sort of trick. Note that we always have a bijection ${V(r,d)\rightarrow V(r, d+nr)}$ given by ${V\mapsto V\otimes \mathcal{O}(n[0])}$. Thus one reduction Atiyah makes right away is that (by the Euclidean algorithm) we only need to consider ${V(r,d)}$ for ${0\leq d < r}$.

Now suppose the rank and degree are non-zero, and ${n=\text{gcd}(r,d)}$. We can establish a bijection ${E\rightarrow V(r,d)}$, so that the composed map ${E\rightarrow V(r,d)\stackrel{det}{\rightarrow} V(1, d)\rightarrow E}$ is the map multiplication by ${n}$. The idea again is that one can find a certain vector bundle ${T_{r,d}\in V(r,d)}$ that is unique up to isomorphism from which all the others can be produced. Most people call this the Atiyah bundle. As a corollary, we see that if ${(r,d)=1}$, then the moduli space of indecomposable rank ${r}$ and degree ${d}$ vector bundles on ${E}$ is fine and again representable by ${E}$.

In modern language, we could work out that stable or semi-stable sheaves are the appropriate things to classify if we want a hope for the moduli space to be fine. It turns out that this condition of certain numerical invariants being coprime often implies that the sheaves are stable. For example, on an elliptic curve, a K3 surface, or even when dealing with relative moduli of sheaves on a K3 fibration. We may get to this another day.

## Basic Properties of Moduli Spaces

I realized I left off in really strange place last time. Sorry about that. There should be a burning question in everyone’s mind. To recap, we’ve seen that sometimes coarse moduli spaces don’t exist. When they do, sometimes there is no universal family. When there is, sometimes it is not a fine moduli space. Despite all these examples I’ve shown, I forgot to show you an example where a fine moduli space exists! So the question should be: is fine too strong a condition to ever exist?

Of course not. We can sort of cheat to make an example by working backwards. Our moduli functor ${\mathcal{F}}$ is fine if it is naturally isomorphic to ${h_M}$ for some scheme ${M/k}$. Thus take any scheme ${M/k}$ and make the moduli functor ${h_M}$. This is the moduli space of points on ${M}$, so it is not surprising that ${M}$ represents this functor. Still, this is a good example to understand as sort of the easiest of all possible moduli problems.

Lots of other schemes are defined to be the scheme whose functor of points is some moduli problem. The Hom scheme, the Quot scheme, the Hilbert scheme, and the Picard scheme are all examples of schemes (when they are schemes) that are fine moduli spaces for less trivial moduli problems.

Let’s look at some standard properties of moduli spaces. The moduli functor is called bounded if there is some finite type scheme ${S/k}$ and a family ${X\in \mathcal{F}(S)}$ such that for any object ${Z\in \mathcal{F}(k)}$ there is some fiber ${X_s}$ such that ${Z\simeq X_s}$. This is just saying that the objects you are trying to make into a moduli space fit into some finite type scheme.

The moduli functor is called separated if for any nonsingular curve ${S/k}$ and a fixed ${s_0\in S}$ if ${X, X'\in \mathcal{F}(S)}$ with all fibers ${X_s, X_s'}$ isomorphic for all ${s\neq s_0}$ then ${X_{s_0}\simeq X_{s_0}'}$. This is essentially the functor of points translation of the valuative criterion for separatedness, so the term makes sense. Intuitively this is just saying that if you have a family of objects over a punctured curve, there is at most one way to fill in an object over that point to make a family over the whole curve.

The moduli functor is called complete if you can always fill in a family over a punctured curve. If there is a fine moduli space ${M}$ associated to ${\mathcal{F}}$, then these properties translate exactly to the corresponding properties for the scheme ${M}$. Namely, if ${M/k}$ is of finite type, then ${\mathcal{F}}$ is bounded. Also, ${M}$ is separated if and only if ${\mathcal{F}}$ is separated, and ${M}$ is proper if and only if ${\mathcal{F}}$ is complete. The proofs are that these are exactly the respective valuative criteria.

Fix a positive integer ${N}$ and a Hilbert polynomial ${p}$. Let’s do a different type of example for today. The Hilbert functor assigns to ${S\in Sch_k}$ the set of subschemes ${Y\subset \mathbb{P}^N_S}$ flat over ${S}$ whose fibers all have Hilbert polynomial ${p}$. We won’t prove that there is a fine moduli space for this problem, since this is a fairly long proof due to Grothendieck.

One way to prove that the functor is bounded is to convert the moduli problem to one that involves the ideal sheaves of the closed subsets. Once this is done, bounded becomes equivalent to finding a single ${m_0}$ such that all coherent sheaves in the problem are ${m_0}$-regular. This can be found using Castelnuovo-Mumford regularity. Thus the Hilbert scheme ${Hilb_p(n)}$ is of finite type.

The other two conditions we talked about hold for more trivial reasons. A family over a punctured curve ${S_0}$ is a closed subscheme of ${Y\subset\mathbb{P}^N_{S_0}}$, flat over ${S_0}$. Thus we can always fill in the punctured point by taking the scheme theoretic closure of ${S}$ in ${\mathbb{P}^N_S}$ where ${S}$ is the filled in curve. This is the unique way of doing it. This shows that ${Hilb_p(n)}$ is proper. Actually, it is projective, but this takes more work.

## Fine Moduli Spaces

To make things easy, recall we are assuming we have a fixed algebraically closed field ${k}$. Last time we talked about what it means for a moduli functor ${\mathcal{F}: Sch_k\rightarrow Set}$ to have a coarse moduli space associated to it. Even though this notion is fairly weak (recall that it essentially only requires the closed points to match up with the objects ${\mathcal{F}(k)}$ we are classifying) it can still sometimes fail to exist.

One example is to make the moduli problem of vector bundles of rank ${2}$ and degree ${0}$ over ${\mathbb{P}^1}$. You can form a family over ${Spec(k[t])}$ that is ${\mathcal{O}(-1)\oplus \mathcal{O}(1)}$ over ${\mathbb{A}^1\setminus \{0\}}$ and becomes ${\mathcal{O}\oplus \mathcal{O}}$ over ${0}$. The existence of such a family that jumps at this closed isolated point rules out the possibility of a coarse moduli space. This is why stable vector bundles turn up in the theory of moduli of vector bundles. Maybe later we’ll elaborate on this example, but it isn’t worth it right now.

Today we’ll discuss a stronger notion than a coarse moduli space. If our functor ${\mathcal{F}}$ is naturally isomorphic to ${h_M}$ for some ${M\in Sch_k}$, then ${M}$ represents the moduli functor and we call ${M}$ a fine moduli space for ${\mathcal{F}}$.

The reason this is a strictly stronger condition is that we automatically have the universal property since they are isomorphic, and we have a bijection ${\mathcal{F}(S)=h_M(S)}$ for all ${S}$. In particular, if ${S=Spec(k)}$ we have a bijection. This shows that every fine moduli space is a coarse one. Moreover, even coarse moduli spaces may not have a universal family, but a fine moduli space always does since we can just take the object corresponding to ${1_M\in h_M(M)=Hom(M,M)}$.

Let ${X\in \mathcal{F}(S)}$ be a family of objects. We say the family is trivial if it is the pullback of some object over ${Spec(k)}$, i.e. ${X\rightarrow S}$ is obtained by the base change:

$\displaystyle \begin{matrix} X & \rightarrow & P \\ \downarrow & & \downarrow \\ S & \rightarrow & Spec(k) \end{matrix}$

A family ${X/S}$ is called fiberwise trivial if all fibers ${X_s}$ are isomorphic. One very important property of fine moduli spaces is that any fiberwise trivial family must be trivial. Suppose ${M/T}$ is the universal family. By definition, every family must be obtained by some map ${S\rightarrow T}$. Thus our fiberwise trivial family is the pullback

$\displaystyle \begin{matrix} X & \rightarrow & M \\ \downarrow & & \downarrow \\ S & \rightarrow & T \end{matrix}$

But every fiber ${X_s}$ is isomorphic to all others, so the image of ${S\rightarrow T}$ is a single point. Thus the family ${X/S}$ is the pullback over a point and hence trivial.

Let’s now use this fact to show that the example we did last time that everyone probably thought was way too simple and easy to illustrate anything actually has no fine moduli space.

Let ${\mathcal{F}}$ be the functor that assigns to ${S}$ a smooth family of genus ${0}$ curves over ${S}$. Since ${\mathcal{F}(k)=\mathbb{P}^1}$ we have a nice coarse moduli space ${M=Spec(k)}$ for the functor. It even has a universal family ${\mathbb{P}^1\rightarrow Spec(k)}$. From the chapter on surfaces in Hartshorne it is a standard fact that if you have a ruled surface ${X\rightarrow C}$ with a section, then there is a rank ${2}$ vector bundle ${\mathcal{E}}$ on ${C}$ such that ${X\simeq \mathbb{P}(\mathcal{E})}$. But every fiber is isomorphic to ${\mathbb{P}^1}$ and hence we can make fiberwise trivial families that are not trivial (${X}$ is not just a product ${C\times\mathbb{P}^1}$).

A high-level explanation of this is that if ${\mathcal{F}}$ were represented by some scheme, then the functor would have to be a sheaf, but sheaves are determined completely by local data. Thus if you are locally trivial, then you must be globally trivial.

## The Coarse Moduli Space

On this blog we’ve extensively looked at lots of things from deformation theory over the past few years. Deformation theory is in some sense a local examination of a more global object called a moduli space. Today we’ll start a brief series on moduli spaces.

Roughly speaking a moduli space is a “space” whose “points” are objects you are trying to classify. You could make the moduli space of elliptic curves in which the points are elliptic curves. You could make the moduli space of rank ${3}$ vector bundles over some ${X}$. Each point of this space would correspond to a vector bundle of rank ${3}$ on ${X}$ (up to isomorphism). You could make a moduli space of morphisms between ${X}$ and ${Y}$.

In theory, any type of mathematical thing you think up you could try to make a space of them. Algebraic geometers do this a lot, but I see no reason why you couldn’t try to study Borel measures on some metric space by trying to make a space whose points correspond to Borel measures (maybe up to mutual absolute continuity or something).

We’ve talked about the deformation functor of an object. Roughly speaking you should expect something along the following lines. Fix an object ${P}$. This corresponds to a point on your moduli space ${M}$. The tangent space of ${M}$ at ${P}$ should correlated to the first order infinitesimal deformation of ${P}$. Nearby points to ${P}$ on ${M}$ should correspond to more similar objects (whatever that means) and far away objects should correspond to quite different objects.

The reason I want to keep this series brief is that the subject turns incredibly technical quickly because there are lots of conditions that people impose on their objects to get the moduli space to be small enough and nice enough to study. We’ll restrict ourselves to some fairly straightforward examples.

The general idea behind constructing a moduli space is that by specifying what the type of object you want to make a space out of, you’ve told me what the functor of points of the space is. In order for it to be some sort of “space” all that we need to do is figure out what space represents this functor.

Let’s start with making some of this more precise. Unfortunately, even the easiest cases have some strange technical points that can’t be avoided. Fix an algebraically closed field (unnecessary in general), ${k}$. The moduli functor will be a functor ${\mathcal{F}: Sch_k\rightarrow Set}$ from schemes over ${k}$ to sets. The set ${\mathcal{F}(S)}$ is the set of equivalence classes of our objects “over” ${S}$ (which will have a meaning depending on the type of object).

The coarse moduli space (if it exists) for this functor is a scheme ${M}$ (a highly restrictive condition we’ll remove if this series goes very far) over ${k}$ with the property that there is a natural transformation ${\mathcal{F}\rightarrow h_M}$ such that ${\mathcal{F}(k)\rightarrow h_M(k)}$ is bijective and satisfies a universal property: given any other natural transformation ${\mathcal{F}\rightarrow h_N}$ where ${N\in Sch_k}$ there is a unique map of schemes ${M\rightarrow N}$ so that the original map factors ${\mathcal{F}\rightarrow h_N\rightarrow h_M}$.

If our moduli functor has a coarse moduli space ${M}$, then we define a universal family for the moduli problem to be a family of objects ${X}$ over ${M}$ (i.e. an element of ${\mathcal{F}(M)}$) with the property that for each closed point ${m\in h_M(k)}$, the object ${X_m}$ over ${M}$ is the one corresponding to ${m}$ under the bijection ${\mathcal{F}(k)\rightarrow h_M(k)}$.

How should we think of this? Well, our coarse moduli space is just a scheme ${M}$ whose closed points are the objects we are considering. This was our motivating definition. (Un)fortunately, schemes have a ton more structure than their closed points. This is what is meant by “coarse”. Other than being the universal space in some sense and actually having as its points the objects we want the rest of the scheme structure is basically irrelevant for a coarse moduli space. The universal family is essentially geometrically designating to each point of ${M(k)}$ the object that it corresponds to.

We’ll end on an extremely simple example that will become somewhat annoying next time when we want a better notion of moduli space. Let ${\mathcal{F}}$ be the functor that classifies smooth, projective, genus ${0}$ curves over ${k}$ up to isomorphism. We need to make precise our notion of a relative curve over some ${S\in Sch_k}$ if we want a well-defined functor on all of ${Sch_k}$.

Define ${\mathcal{F}(S)}$ to be the set of ${X\rightarrow S}$ which are smooth and projective with geometric fibers curves of genus ${0}$. This is exactly what one would expect a family of genus ${0}$ curves to be. The functor’s value on ${Spec(k)}$ is just a single point ${\mathcal{F}(k)=\{\mathbb{P}^1\}}$ because all genus ${0}$ curves (over an algebraically closed field) are isomorphic. This easily shows that ${M=Spec(k)}$ is the coarse moduli space and there is a universal family which is to just put the one object over ${M}$, i.e. ${\mathbb{P}^1\rightarrow Spec(k)}$ is the universal family.

## Deformations of p-divisible Groups

I’ve made the official decision to not do a proof of anything with the deformation theory of ${p}$-divisible groups, but now that I’ve motivated it I’ll still state the results. The proof is incredibly long and tedious. It should be interesting to look at what this functor is, since it probably isn’t what you think it is. We’ll construct the moduli functor, but it isn’t just ${p}$-divisible groups up to isomorphism or isogeny, but involves the notion of a quasi-isogeny.

A few definitions are needed. We’ll work in great generality, so let ${S}$ be a base scheme with your favorite properties. Our ${p}$-divisible groups will only be required to be fppf sheaves on ${S}$ (with the ${p}$-divisible group property). An isogeny ${f:G\rightarrow G'}$ of ${p}$-divisible groups is a surjection of sheaves with finite locally free kernel. An example is multiplication by ${p}$ since the kernel is ${G_1}$ which is by definition of a ${p}$-divisible group a finite locally free group scheme (of order ${p^h}$).

The group ${\mathrm{Hom}_S(G,G')}$ is a torsion-free ${\mathbb{Z}_p}$-module. We can make the sheaf version by taking the Zariski sheaf of germs of functions ${\mathcal{H}om_S(G,G')}$. A quasi-isogeny ${G\rightarrow G'}$ is a section ${\rho\in\mathcal{H}om_S(G,G')\otimes_\mathbb{Z} \mathbb{Q}}$ with the property that ${p^n\rho}$ is an isogeny for some integer ${n}$.

Now we have enough to write down the moduli functor that we want. We have everything over an algebraically closed (I think a descent argument allows us to do this all over a perfect field) field, ${k}$, of characteristic ${p>0}$ and ${W}$ its ring of Witt vectors. Consider the category ${\mathrm{Nilp}_W}$ of locally Noetherian schemes, ${S}$, over ${W}$ such that ${p\mathcal{O}_S}$ is locally nilpotent.

Fix a ${p}$-divisible group ${G}$ over ${\mathrm{Spec}(k)}$. Our moduli functor ${\mathcal{M}}$ is a contravariant functor from ${\mathrm{Nilp}_W}$ to the category of of pairs ${(H, \rho)_S}$, where ${H}$ is a ${p}$-divisible group over ${S}$ and ${\rho}$ is a quasi-isogeny ${G_{\overline{S}}\rightarrow H_{\overline{S}}}$. Then we mod out by isomorphism where an isomorphism ${(H_1, \rho_1)\rightarrow (H_2, \rho_2)}$ in this category is given by a lift to an isomorphism of ${\rho_1\circ \rho_2^{-1}}$, i.e. an isomorphism that commutes with the quasi-isogenies.

The theorem is that ${\mathcal{M}}$ is representable by a formal scheme formally locally of finite type over ${\mathrm{Spf}(W)}$. The way to prove representability is the usual way of finding particularly nice open and closed subfunctors. In our case, part of the breakdown is already in this post. Since the definition of quasi-isogeny involves an integer ${n}$ for which ${p^n\rho}$ is an isogeny, you can break ${\mathcal{M}}$ up using this integer. See the book Period Spaces for p-divisible Groups by Rapoport and Zink for more details.

It comes out in the proof that for the nice cases we were considering in the past two motivational posts we get that the functor is representable by ${\mathrm{Spf}(W[[t_1, \ldots, t_d]])}$ where ${d=\dim G \dim G^t}$. One really interesting consequence of this is that if ${G}$ has height ${1}$, then since ${\mathrm{ht}(G)=\dim G + \dim G^t}$ we have that one of ${\dim G}$ or ${\dim G^t}$ is ${1}$ and the other is ${0}$, so in either case the product is ${0}$ and we get that the deformation functor of ${G}$ is representable by ${\mathrm{Spf}(W)}$. In general, it is always smooth and unobstructed.

I’m not sure if I should continue on with ${p}$-divisible groups now that I’ve done this. Maybe I’ll go back to crystalline stuff or move on to something else altogether.