A Mind for Madness

Musings on art, philosophy, mathematics, and physics


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Néron Models 1

Let’s start with the exciting news that today I broke 100,000 views!! Thanks to all my long-time and new subscribers. I’ve decided to talk about Néron models. The other idea was just too overwhelmingly big that I’d probably use that as an excuse not to post. Let’s start with the general idea of Néron models. Suppose you start with some abelian variety, {A_K}, over a number field {K}. You could view this as an abelian variety that is defined generically over the ring of integers of {K}, say {R}. From this you can “spread out” the variety and get a smooth scheme over an open subset of {S=Spec R}. You could define the spread out {A} over all of {Spec R}, but it won’t be smooth because it will have bad reduction at finitely many places.

The definition of Néron model will not require that {A_K} be an abelian variety. For this series of posts we’ll probably keep switching between {A_K} and {X_K} for the notation. It essentially won’t make any difference whether or not the scheme has a group structure. The reason for the confusion is that most applications right now have to do with abelian varieties.

The idea of a Néron model is to construct an honest smooth, separated, finite type scheme {X} over {Spec R} which will be canonical in some sense and also has the property that for any smooth {S}-scheme {Y} and any morphism {u_K: Y_K \rightarrow X_K} there is a unique {S}-morphism {u:Y\rightarrow X} extending {u_K} (the Néron mapping property).

For the rest of the post we’ll just describe properties we want the Néron model to have along with some related remarks. The generality that we will mostly work with is to let {S} be an arbitrary Dedekind scheme with field of fractions {K}. Given any scheme {X_K} over {Spec K} we call a scheme {Y} over {S} an {S}-model for {X_K} if {Y\otimes K\simeq X_K}. Without specifying more properties, we clearly won’t have a unique or canonical model since we can just change {Y} in sufficiently high codimension away from the generic fiber and get non-isomorphic {S}-models.

One property we will want our Néron model {X} to satisfy is that {X(R^{sh})\rightarrow X_K(K^{sh})} is surjective. One way to think about this property is that every rational point on our starting variety has some integral point on the model that specializes to it. Later we might relate this to a notion of boundedness.

We can say that Néron models are “minimal” in the sense that if you have any other {S}-model {Y}, then there is a unique morphism {Y\rightarrow X} restricting to the identity on the generic fiber. Unravelling this further we find that the Néron model is canonical in the sense that {X} is uniquely determined up to canonical isomorphism by {X_K}. One interesting question that you should be asking is if we are given some arbitrary {X/S}, can we tell if it is a Néron model of its generic fiber? A quick condition to check is that this happens if and if only for all closed points {s\in S}, the {\mathcal{O}_{S, s}}-scheme {X\times_S Spec \mathcal{O}_{S, s}} is a Néron model of its generic fiber.

Going back to the abelian variety examples. If we build a Néron model of {A_K}, the group scheme structure uniquely extends to the model. This is further reason to not worry about abelian varieties in the general theory. The previous paragraph simplifies greatly when we work with a group scheme. If {A/S} is an abelian scheme, then it is always a Néron model of its generic fiber. This just follows from the criterion given and Weil’s extension theorem.

The last thing to point out today is that Néron models are often not the obvious choice of model. For example, take {X_K=\mathbb{P}^1_K}. The variety {X=\mathbb{P}^1_S} is a smooth, separated, finite type {S}-model of {X_K}, so we might hope that it is the Néron model, but it isn’t. It even satisfies this extension property for étale points (a consequence of the Néron mapping property, but not equivalent to it). Next post we’ll have a technique to actually check this isn’t the Néron model, but for now just take it as a word of caution.


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Galois Deformations 5: Flach’s Theorem

I’ll begin today by taking a poll. Usually I’d do this at the end, but I assume no one will be reading at that point. Since my research has a bit of an arithmetic flavor, one thing I’d like to do is post about some standard algebraic number theory where about a 10 week graduate class would leave off with an eye towards class field theory. I should probably be more comfortable with these things. My other idea is to blog about Neron models which I feel like I should understand better as well. If you strongly want to see one of these over another, then speak up soon in the comments (I’ll probably do both eventually).

The subject of today is to discuss the obstruction space a little. Recall that whenever we’ve discussed the universal deformation ring in the previous posts we had the caveat “if the deformations are unobstructed, then … “. It would be nice to have some theorems that tell us when this actually happens, or if it doesn’t happen at least get some good control on what the obstructions are.

Let’s quickly recall how this works. Fix {S} a finite set of primes and some prime {\ell} (not in {S}). Fix a an absolutely irreducible residual representation {\overline{\rho}:G_S\rightarrow GL_2(\mathbb{F}_\ell)}. In this case, we know the deformation functor is representable by some universal deformation ring denoted {\mathcal{R}(\overline{\rho})}. The structure of this ring is immensely important since it completely determines what all the deformations are. We said that the obstruction space for deforming is {H^2(G_S, Ad(\overline{\rho}))}, so if we suppose this is {0}, then we know that {\mathcal{R}(\overline{\rho})\simeq \Lambda [[x_1, \ldots, x_d]]} where {d=\dim H^1(G_S, Ad(\overline{\rho}))}. This is why unobstructedness is so important. We can completely determine {\mathcal{R}(\overline{\rho})} up to isomorphism.

A theorem of Flach in 1992 tells us one case when we get unobstructedness. Let {E/\mathbb{Q}} be an elliptic curve having good reduction at {\ell} and {S} the primes of bad reduction together with {\ell} and {\infty}. Let {\rho=\rho_E:G_S\rightarrow GL_2(\mathbb{Z}_\ell)} and {\overline{\rho}} the residual representation. Suppose we have the following three conditions satisfied as well: {\rho} is surjective, for all {p\in S\setminus \{\infty\}} we have {E[\ell]\otimes E[\ell]} has no {G_{\mathbb{Q}_p}}-invariants, and {\ell} does not divide {L(Sym^2 T_\ell E, 0)/\Omega}, then the deformation problem is unobstructed for {\overline{\rho}}.

We definitely won’t worry about what everything in that last condition is. You might be worried that this never happens, but in some sense that can be made precise these conditions are almost always satisfied. Now to finish the post off we’ll just very generally give an outline of the proof. Since we are trying to make some {H^2} vanish, it isn’t surprising that we can reduce to checking that Ш{^1(G_s, E[\ell]\otimes E[\ell])\simeq}
Ш{^1(G_s, \mu_\ell)\oplus} Ш{^1(G_S, Sym^2 E[\ell])} vanishes. We can already start to see the appearence of some of the conditions.

Using standard techniques in Galois cohomology like Hilbert 90 it can be checked directly that the first factor vanishes, so we only need to check that second one. Now define {A=Sym^2 T_\ell E\otimes \mathbb{Q}_\ell/\mathbb{Z}_\ell}. It turns out that we get an injection
Ш{^1(G_S, Sym^2 E[\ell])\hookrightarrow H_f^1(\mathbb{Q}, A)} into the Selmer group, so we can reduce to proving this Selmer group is {0}. Using a very hard, deep theorem and the fact that {E} is modular we can conclude that {\deg \phi H^1_f(\mathbb{Q}, A^*)=0} where {\phi: X_0(N)\rightarrow E} is a modular parametrization. From this we can conclude the theorem of Flach.


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Galois Deformations 4: p-adic Deformations

Let’s start today by recalling that every elliptic curve over {\mathbb{Q}} is modular and every rigid Calabi-Yau threefold over {\mathbb{Q}} is modular as well. This just means that the Galois representation that arises from the etale cohomology is the same as the Galois representation of a modular form. It has also been found that many, many other types of varieties are modular. This leads us to a major theme of number theory which is to make precise and prove some variant on the question: Is every representation that comes from geometry modular?

Serre’s conjecture (now proven) tells us that if we start with an absolutely irreducible odd two-dimensional residual representation {\overline{\rho} :G_S\rightarrow GL_2(k)} then there is a modular form {f} such that {\overline{\rho} = \rho_f}. So one approach to studying the above question would be to figure out which deformations of residual representations are modular.

Let’s put ourselves in this situation. We could get massively bogged down in the details at this point, so we’ll just set things up by analogy to the earlier posts on modular forms. Fix {\mathcal{O}} a DVR that is a finite extension of {\mathbb{Z}_p}. Now we want to consider {f\in S_k(\Gamma_1(Np^\nu), \mathcal{O})}. We haven’t discussed {\Gamma_1}, but it is similar to {\Gamma_0} in that it is just matrices that are upper-triangular mod {Np^\nu} with the added property that the diagonal entries must be {1 \mod Np^\nu}. Roughly, you can think of {S_k(\Gamma_1(Np^\nu), \mathcal{O})} as modular forms where the coefficients of the {q}-expansion are allowed to be in {\mathcal{O}}. Our other way works as well where we consider a moduli space of elliptic curves with level {Np^\nu}-structure and take the {\mathcal{O}}-valued points. As before, we actually only want to consider certain eigenforms so that the theory works out.

To any such {f} we can attach (very non-obvious step here!!) a Galois representation {\rho_f: G_S\rightarrow GL_2(\mathcal{O})} where {S=\{\lambda : \ \lambda | N\}\cup \{p, \infty\}}. Suppose {\frak{m}} is the maximal ideal of {\mathcal{O}}. To get to our residual representation we can just take the reduction mod {\frak{m}} followed by the semisimplification {\overline{\rho}_f: G_S\rightarrow GL_2(\mathcal{O}/\frak{m})}. Now we can finally say something interesting. Suppose {f} and {g} live in completely different places, for example by altering {k} and {\nu}. As long as the coefficients of the {q}-expansion are the same mod {\frak{m}} it is easy to see that {\overline{\rho}_f=\overline{\rho}_g}. Thus we can produce lots of different deformations over {\mathcal{O}} of a given residual representation.

Let’s just reiterate a few things from last time. Take our {f} as above and suppose that the attached residual representation {\overline{\rho}} is absolutely irreducible. We know in this situation that the deformation functor is representable and by a Galois cohomology calculation (supposing unobstructedness) there is a universal deformation ring {\mathcal{R}(\overline{\rho})\simeq \Lambda [[x_1, x_2, x_3]]}. In order for the theory to be flexible enough for such amazing conjectures as given in the first paragraph we need to expand the class of modular forms we consider. We’ll just say that we want to be able to define modular forms over {p}-adically complete separated rings. It turns out that the classical modular forms are dense in this enlarged space and so bootstrapping off that we can make a universal modular deformation {\mathbf{\rho}: G_S\rightarrow GL_2(\mathcal{R}_m(\overline{f}))} even when our original {f} was not a classical modular form.

We won’t worry what this ring {\mathcal{R}_m(\overline{f})} is. The take away from this is that even though we allowed ourselves to move away from classical modular forms, the universal modular form cuts out a subspace of the deformation space consisting of the {p}-adic modular forms. Or in other words it is the Zariski closure of the set of classical modular forms. Since in many cases we can explicitly determine what this subspace is, we haven’t lost much by allowing ourselves this larger class of modular forms. In fact, by imposing extra deformation conditions it can sometimes restrict us to exactly classical modular forms, and this is exactly a key technique used to proved the Taniyama-Shimura conjecture.

Although all this is really interesting and fun, I’ve already skipped so many technicalities and complication that it would be pointless to drag this on much further. I think I’ll probably do one more post on this stuff since the main idea of what gets used in Taniyama-Shimura is in this post.


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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).


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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 Gouvêa’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.


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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.

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