# Serre-Tate Theory 2

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

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

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

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

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

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

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

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

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

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

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

# Serre-Tate Theory 1

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

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

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

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

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

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

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

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

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

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

# 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 ${\mathbb{P}^n}$.

# 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.

# Galois Deformations 3: Tangent and Obstruction

Recall last time that if we start with some finite field ${k}$ and some absolutely irreducible residual Galois representation ${\overline{\rho}: G_S\rightarrow GL_n(k)}$, the deformation functor is (pro)representable by some universal deformation ring which we will denote ${\mathcal{R}(\overline{\rho})}$ (again, this was known even in the original papers to be true in a much more general situation than this one). Today we’ll try to figure out some properties of this ring.

First we note something that better be true. We really only consider representations up to equivalence, so if ${\overline{\rho}}$ and ${\overline{\rho}'}$ are equivalent then there is a matrix that conjugates one to the other. By functoriality of everything involved we can consider the natural transformation of deformation functors induced by the group scheme homomorphism given by this conjugation action. By representability the natural transformation gives a homomorphism of the universal rings ${\mathcal{R}(\overline{\rho})\rightarrow \mathcal{R}(\overline{\rho}')}$. It can be checked that this is an isomorphism. Geometrically this means that the “local space” of deformations is the same for equivalent representations.

By similar arguments to the last time I talked about deformation theory, we can compute that the tangent space to the functor given by ${Def_{\overline{\rho}}(k[\epsilon])}$ is isomorphic to ${H^1(G_S, Ad(\overline{\rho}))}$ where ${Ad(\overline{\rho})}$ just means the adjoint representation. So as in the geometric situation, if ${\mathcal{R}}$ is smooth, we can compute the dimension of the space of deformations by computing the dimension of some first cohomology. Thus if the deformations are completely unobstructed, then we know that ${\mathcal{R}\simeq \Lambda[[x_1, \ldots, x_d]]}$ where ${d=\dim H^1(G_S, Ad(\overline{\rho}))}$.

The standard next question is to ask what is the obstruction space. Very standard arguments that can be made with almost any deformation functor show us the following. I’ll leave out the details, but tell you what to fill in. By an obstruction, we mean take a map in our deformation ring category ${R_1\rightarrow R_0}$ that has kernel ${I}$ with the property that ${I\cdot m_{R_1}=0}$ so that it is naturally a ${k}$-vector space. Take some representation ${\rho: G_S\rightarrow GL_n(R_0)}$. We want to know if there is some ${\rho': G_S\rightarrow GL_n(R_1)}$ so that upon composing ${GL_n(R_1)\rightarrow GL_n(R_0)}$ we get ${\rho}$ back. If you’ve seen this geometrically, then your intuition should be to show that “locally” you can do it, and then the ability to glue to get a global object is some class in ${H^2}$ causing the obstruction. This is again the idea here. You formally make the lift set-theoretically. The obstruction to being a homomorphism is a class ${Ob(\rho)\in H^2(G_S, Ad(\overline{\rho})\otimes_k I)}$. If the class is 0, then we can lift the representation, otherwise it is obstructed. We call ${H^2(G_S, Ad(\overline{\rho}))}$ the “obstruction space” to the deformation functor.

This gives us an expected dimension for the deformation space. In some sense, we’d expect that each dimension of the obstruction space would cut the dimension of the space by one. If ${d_1=\dim H^1(G_S, Ad(\overline{\rho}))}$ and ${d_2=\dim H^2(G_S, Ad(\overline{\rho}))}$, then we’d expect ${Kdim(\mathcal{R}/m_{\Lambda}\mathcal{R})=d_1-d_2}$, but this is still an open conjecture and is probably hard since it generalizes Leopoldt’s conjecture. What we do get is that ${Kdim(\mathcal{R}/m_{\Lambda}\mathcal{R})\geq d_1-d_2}$. The rough idea behind having an inequality is that each dimension of the obstruction space is introducing a relation, and so the relation at most cuts the dimension down by one. But maybe we get to the fifth relation and it is redundant, then the dimension won’t go down.

Using some hard algebraic number theory we can actually convert ${d_1-d_2}$ to something only involving ${H^0}$, but it is beyond the scope of these posts. We’ll end here for today. Next time we’ll specialize to Galois representations that are modular, since that was our motivation and see what more information we can get in that case (this means if you are following along with the Gouvea article, we will be skipping lectures 5 and 6).

# Galois Deformations 2: Representability

Today we’ll describe what is meant by a deformation of a Galois representation. Since our motivation was Taniyama-Shimura we’ll quickly recall the type of Galois representations that came up there. There we technically had what are called ${\ell}$-adic representations, because we considered ${\rho_X: \text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})\rightarrow GL_n(\mathbb{Q}_\ell)}$. Our only caution was that we could have problems at places of bad reduction or ramification, so we will build that into the representations we consider.

Let ${S}$ be a finite set of primes. Define ${G_S=\text{Gal}(\overline{\mathbb{Q}}_S/\mathbb{Q})}$ to be the Galois group of the maximal algebraic extension of ${\mathbb{Q}}$ unramified outside ${S}$. Note that ${G_S}$ will always have the profinite topology. The term “Galois representation” from now on will mean a continuous representation ${\rho: G_S\rightarrow GL_n(A)}$ where ${A}$ is a topological ring. Maybe this is a little loose because when we write ${GL_n(A)}$ we really mean ${Aut(A)}$, but we’ve chosen a basis to get actually matrices. Thus we really only want to consider two representations as different if they can’t be conjugated to one another. This is standard in representation theory, so we won’t dwell on it.

The idea of deformations of Galois representations is roughly to extrapolate information when ${A=\mathbb{Q}_p}$ or better $\mathbb{Z}_p$ by using information about representations when ${A=\mathbb{F}_p}$. If we think to the last post we can almost see how the deformation functor formalism can play a role here. We will set ${k=\mathbb{F}_p}$ in which case ${\Lambda=\mathbb{Z}_p}$ is a local Noetherian ${\Lambda}$-algebra with augmentation to ${k}$. In fact, suppose ${A\in \ _\Lambda Noeth_k}$, then the augmentation gives a natural map ${GL_n(A)\rightarrow GL_n(k)}$, so if we fix some ${\overline{\rho}: G_S\rightarrow GL_n(k)}$ called a residual representation a deformation of ${\rho}$ should be a continuous representation ${\rho: G_S\rightarrow GL_n(A)}$ which is (up to equivalence) ${\overline{\rho}}$ when composed with this map. The functor ${Def_{\overline{\rho}}}$ is now defined in the same way to be the set of all deformations of ${\overline{\rho}}$.

Now as was pointed out last time, in order to define our functor there is ambiguity about whether to define it on the completed category or on the full subcategory of Artin rings. It turns out that since our functor is continuous, for the purposes of representability we can check it on this full subcategory. In order to prevent a large amount of tedium and space there is a lot being brushed over here. I highly recommend GouvÃªa’s great article on Galois Deformations in the book Arithmetic Algebraic Geometry for a more precise discussion of these points.

The punchline is that ${Def_{\overline{\rho}}}$ is actually a deformation functor. Moreover if ${\overline{\rho}}$ is absolutely irreducible, then Mazur showed that the functor satisfies Schlessinger’s criterion and hence is prorepresentable (a far more general case was actually considered). Let’s unravel why this is important. What this says is that there exists a universal deformation ring (in the completed category), ${R}$, so that given any ${A\in Art_k}$ we have ${Def_{\overline{\rho}}(A)=Hom(R, A)}$. Even stronger we know there is a universal deformation ${\psi: G_S\rightarrow GL_n(R)}$ so that the correspondence ${Def_{\overline{\rho}}(A)=Hom(R,A)}$ is actually given by ${\phi:R\rightarrow A}$ goes to the deformation given by composing ${G_S\stackrel{\psi}{\rightarrow} GL_n(R)\rightarrow GL_n(A)}$. This is wonderful. If we can somehow get our hands on this universal ring and universal deformation it will completely control all deformations.

Schlessinger unfortunately only tells us it exists, but subsequent work does tell us these things. That will be the subject of the next post.

# Galois Deformations 1: Schlessinger

Today we’ll begin a short series on deformations of Galois representations. I know very little about this topic in general, but since I brought up Taniyama-Shimura and this is one of the main tools used in proving it I thought it would be interesting to take a look at what these are. It will also be a nice example of why abstraction is often better. Today we’ll review standard deformation theory from a geometric viewpoint, but we’ll frame it in very abstract terms (the way Schlessinger did in his fantastic paper). This abstraction will allow us to apply certain results that come from geometric intuition to any functor that satisfies certain criteria.

I’ve posted a little about deformation theory around here. I won’t recall it because our framework will be slightly different. Instead of thinking about individual deformations, let’s do something more Grothendieckian and see what happens when we consider all deformations at the same time. In other words, fix some smooth varitey ${X/k}$, then given some “deformation ring” ${A}$ (to be precise later), we consider the set of all deformations of our variety ${\tilde{X}/A}$. Here is where our formalism comes in. Our set changes with ${A}$ nicely enough that ${Def_X: _\Lambda Art_k \rightarrow Set}$ actually forms a functor.

Let’s talk about the category of deformation rings now. Let ${\Lambda}$ be a Noetherian ring. The category labelled ${_\Lambda Art_k}$ means the category of local Artinian ${\Lambda}$-algebras together with a choice of augmentation map to ${k}$. This last part is important, because the morphisms between two of these rings must be both ${\Lambda}$-algebra maps and maps that commute with the augmentation to ${k}$. The reason for this is that when I pick a deformation ${\tilde{X}/A}$ I’ve done more than just specified an abstract deformation. I also am saying that the pullback square gives me a choice of isomorphism of the special fiber:

${\begin{matrix} X & \hookrightarrow & \tilde{X} \\ \downarrow & & \downarrow \\ Spec k & \rightarrow & Spec A \end{matrix}}$

where that bottom arrow is the one on spectra induced by the augmentation ${A\rightarrow k}$. Based on this geometric picture, we abstract as little as possible and call any functor ${F: _\Lambda Art_k\rightarrow Set}$ a deformation functor if it satisfies certain properties that formally correspond to having a single deformation over ${k}$ (exercise for the uninitiated: prove that if X has a non-trivial automorphism, then the functor will not satisfy this condition if the seemingly strange augmentation condition is dropped. I’ve never seen this exercise written down, but it seems important to me), being able to glue when you ought to be able to, and this gluing being unique over first order infinitesimal neighborhoods. The exact conditions can be found here (I don’t feel bad linking to that nLab page since I wrote it).

A key theorem due to Schlessinger that will come up next time is that under mild conditions to check on this functor we actually get that it is prorepresentable. This just means if we “complete” the category formally by making a new functor ${\widehat{F}: _\Lambda Noeth_k\rightarrow Set}$ by ${\widehat{F}(R)=\lim F(R/m^n )}$ the functor is actually representable here. We won’t dwell on this because when thinking about Galois representations we will have stronger things going on by continuity of our maps which will make this point less important. All this will be made more precise next time, but it is worth pointing out how amazing the forsight of Schlessinger was to formulate this in terms of functors so that it would apply all over math. His criterion says if we check certain conditions based on geometric intuition the functor will be representable. Or more intuitively, we will have a “space” that universally parametrizes everything. You can read more about this at that nLab page.

Next time we’ll bring this back from this abstraction and talk about what this means for being able to “deform” Galois representations.