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Here I read:

Another insight of Grothendieck and his school was, how important it is to represent functors in algebraic geometry - regardless of what you want at the end. [as Mazur reports, Hendrik Lenstra was once sure that he did want to solve Diophantine equations and did not want to represent functors - and later he was amused that he represented functors to solve Diophantine equations.]

My question is already in the title: Why does representing functors help solving Diophantine equations?

As a category theory enthusiast I am fascinated by the fact that sometimes category theory is helpful to prove concrete things (here: solving Diophantine equations). So I'd really want to know how that works, roughly.

Timothy Chow
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user481980
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4 Answers4

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E.g. let $f(x,y, z)=0$ be a smooth projective plane curve with $f$ a rational polynomial of degree $\ge 4$. Then Mordell conjectured, and Faltings proved, that this has only finitely many rational solutions. I hope you agree that this is a concrete statement about Diophantine equations. Faltings' original proof uses the moduli stack of principally polarized abelian varieties. This stack represents a certain functor in some sense (or in an exact sense, if you replace it by the moduli space with of ppav with level structures).

Added remarks I'm far from an expert, but I'll try to add a few words of explanation. Faltings managed to prove several conjectures; among them a conjecture of Shafarevich that given a finite set of primes $S$ in a number field $K$, the set of isomorphism classes of principally polarized abelian varieties of given dimension over $K$ with good reduction outside $S$ is finite. The relevance of the moduli stack should be clear for this. Faltings then used a trick of Parshin to show that a counterexample to Mordell's conjecture would lead to a counterexample to the Shafarevich conjecture QED.

Donu Arapura
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    Good answer. The original quote was apparently by Hendrik Lenstra; any idea what Lenstra may have been referring to? – Timothy Chow May 10 '22 at 13:22
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    Maybe he meant Mazur's universal Galois deformation ring. –  May 10 '22 at 13:40
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    I don't know what, if anything, Lenstra was referring to other than the fact that Grothendieck style algebraic geometry, which is useful for Diophantine equations, uses a lot of categorical ideas. Also it's funny. @Mastrem and anyone else: feel free to add answers if there's more to it. – Donu Arapura May 10 '22 at 14:30
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    @DonuArapura Thanks! Is there any non-technical high-level explanation of why representing functors is useful here to prove your statement? – user481980 May 11 '22 at 12:30
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I am fairly sure the reference is to Barry Mazur's paper "An introduction to the deformation theory of Galois Representations", which is based on lectures that Mazur gave at a 1995 conference on the proof of Fermat's Last Theorem. There Mazur writes

Hendrik Lenstra, in his lecture in the conference, recounted that twenty years ago he was firm in his conviction that he DID want to solve Diophantine equations, and that he DID NOT wish to represent functors - and now he is amused to discover himself representing functors in order to solve Diophantine equations!

The functors in question are indeed the one represented by Mazur's universal coefficient ring, and variants thereof.

Lenstra's lecture from the conference is available on Youtube here. It's at around 23:30. My transcription:

... the whole problem is, given your functor, can you represent it? And that was really a very hot issue in the late sixties. I remember, I think it was around 1970, I went to an algebraic geometry meeting in Oslo. Everyone was representing functors there, and they were carrying Schlessinger's paper under their arms and asking questions about it. I sort of looked at this and I decided that I wanted to do real mathematics and I started solving Diophantine equations. Oslo is a great city for solving Diophantine equations. And nowadays we know that, if you want to solve Diophantine equations, you have to start by representing a functor, so that's what I'm doing today.

Of course this isn't an answer to "why does representing functors help solving Diophantine equations?"; perhaps someone else would like to attempt that.

David E Speyer
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  • Ah, this answer is already here. I just spoke with Lenstra today, and he recounted exactly this, including the reference to the 1970's Oslo conference. I'm very happy to find a link to the video. – R. van Dobben de Bruyn May 11 '22 at 15:59
  • (Also, given his ability to produce memorable quotes, I find it very fitting to find an MO question dedicated to untangling a joke he made several decades ago.) – R. van Dobben de Bruyn May 11 '22 at 16:03
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Let me give an answer that pertains to the Diophantine equation that, according to David Speyer's answer, Lenstra was specifically talking about.

How does representing functors help solve the Diophantine equation $a^n + b^n=c^n$?

A solution to this Diophantine equation defines an elliptic curve $y^2 = x (x-a^n)(x-b^n)$ (Frey). The Galois group of the rational numbers acts on the $\overline{\mathbb Q}$-points of this curve, thus on the $\ell^m$-torsion points for each $m$. Taking an inverse limit as $m$ goes to $\infty$, we obtain a Galois action on a rank two free $\mathbb Z_\ell$-module (the Tate module).

We also obtain Galois actions on rank two free $\mathbb Z_\ell$-modules from modular forms, after Eichler-Shimura and Deligne. In fact, for many different rings $S$ we can obtain Galois actions on rank two free $S$-modules.

It turns out that the Galois representations arising from curves of the form $y^2 = x (x-a^n)(x-b^n)$ have very special properties on congruence mod $n$, properties which the Galois representations arising from modular forms cannot have (Ribet). So if we can show that every Galois representation arising from an elliptic curve also arises from a modular form, we can obtain a contradiction from any solution to the Diophantine equation.

Now here's where we introduce the functors to be represented. We consider the category of complete local rings $S$ of residue characteristic $\ell$, and the functor that sends each such ring to the set of isomorphism classes of rank two free $S$-modules with a Galois action that is congruent mod $\ell$ to some fixed Galois representation, satisfying some conditions known to hold for elliptic curves. We can consider another functor that sends the ring $S$ to only the set of isomorphism classes that arise from modular forms.

We'd like to prove the natural transformation from one of these functors to the other is an isomorphism, which is equivalent to the statement about every Galois representation arising from a modular form. In general, this is a hard problem.

But if we can represent these functors by complete local rings of residue characteristic $\ell$, then it becomes equivalent to proving that a map of local rings $R \to T$ is an isomorphism. That's a question which is much easier to tackle, because we can use all the commutative algebra theory of local rings. In particular, we can hope to prove a criterion for the map to be an isomorphism that depends on concrete properties of these rings which can be expressed in terms of the original functors. (The simplest case of this is that a map of smooth local rings is an isomorphism if and only if it's an isomorphism on the residue field and tangent space, which both have a functorial interpretation in terms of very simple rings.) We then reduce the problem to checking some new properties are satisfied by the functors. These properties turn out to be much more tractable than proving the isomorphism directly. In part, this is because they connect to areas where there is a pre-existing theory (related to $L$-functions, the class number formula, Iwasawa theory) that can be applied (though of course deep new ideas were needed as well).

Will Sawin
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    What I don't understand about this procedure: the representing object and the functor determine eachother, so are "equivalent data". In this sense the representation business is nothing more than a dictionary, that translates data into a different but equivalent form. So how can a question be fundamentally easier to solve after a translation? Could be not translate every step in the proof and obtain an equivalent proof without the need for translation? Moreover, shouldn't every step of the translated proof be equally "difficult", or "natural" or "obvious" as before translation? – user2520938 May 12 '22 at 16:36
  • @user2520938 For one thing, not every functor is representable. So we can't translate a proof into a proof about only the functor if the theorem to be proven is not true for functors that aren't representable. Maybe another thing to say is that, when we prove a result about a functor from a category, we are free to consider any object in the category at any step in the proof. When we prove a functor is representable, that suggests a particular object it may be helpful to consider, i.e., the one representing it. – Will Sawin May 12 '22 at 16:57
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    @user2520938 Suppose you studiously step-by-step translated the proof to avoid representing the functor, and found that the proof repeatedly mentioned the same particular object of the category, without ever mentioning that it's the object representing the functor. What would doing this have bought you? Just made the proof harder to read. The existence of some object satisfying such-and-such conditions, to which various arguments can be applied (considering the elements, subobjects, etc.) would still be the crux of the proof. – Will Sawin May 12 '22 at 17:01
  • Maybe, but I'm still not convinced. I still don't see how it can be true that representing the functor helps solve the equations, when no new information is produced in the process, it is just formulated in a different language. I do agree though that it is more intuitive to reason about object than about functor satisfying some abstract representability criteria. – user2520938 May 12 '22 at 20:34
  • @user2520938 One could say for every theorem proving that two properties are equivalent that it doesn't produce any new information and just formulates one property in different language. Indeed all of our theorems are tautologies so arguably none of them produce any information. – Will Sawin May 12 '22 at 20:58
  • Yes you are right, in fact I'm skeptical about the value of much of modern mathematics, but that's a different discussion. – user2520938 May 12 '22 at 21:33
  • @user2520938 It's very frequently the case that the simplest example of the application of some technique does not convincingly demonstrate that the technique is indispensable. Because it's a simple example, you can see ways to solve the problem without the technique. But if one were to start with a more complicated example, it would be difficult for a beginner to understand. Simple word problems in algebra can be solved without the machinery of middle-school algebra. The area under a parabola can be found without calculus. This doesn't mean that algebra and calculus have no value. – Timothy Chow May 16 '22 at 13:29
  • @TimothyChow that’s not the same thing though, calculus is more than a dictionary, it’s a toolbox. A dictionary that translates info into equivalent info formulated in a different language does not in an essential way add anything. – user2520938 May 16 '22 at 14:58
  • @TimothyChow e.g. anything in calculus that can be solved using polar coordinates can also be solved without first rewriting to polar coordinates, because changing coordinates does not in any way provide essentially new information – user2520938 May 16 '22 at 15:00
  • @user2520938 I don't think it's different. The categorical point of view opens up a big toolbox. That may not be evident from the simplest example, but it is evident in more complicated examples. By the way, even in your example of polar coordinates, surely you know that the Gaussian integral is more easily evaluated using polar coordinates. – Timothy Chow May 16 '22 at 15:17
  • @TimothyChow easily, sure, and the same thing goes for the categorical pov. It makes some things easier or more obvious, but anything can also be done without it, again because it just provides a new perspective on the same objects, it doesn’t introduce anything new. – user2520938 May 16 '22 at 15:24
  • @user2520938 Wondering about the necessity of the categorical point of view is very different from wondering about the necessity of representing functors! Indeed they are somewhat apposed because representing a functor replaces a more-categorical object by a less-categorical one. If one asked to what extent the categorical point of view was really necessary in Wiles's argument, one would probably point to a different step of the proof, for example the use of étale cohomology theory. – Will Sawin May 16 '22 at 16:04
  • Whatever you want to call it, that’s just semantics. But anyway I don’t think we are going to agree on this. – user2520938 May 16 '22 at 16:08
  • @user2520938 Any mature field of mathematics is a toolbox, since it consists of a large number of results potentially useful for a given problem. Certainly ring theory is a toolbox! In Wiles' argument, representing the functor allows ring theory to be used. The point of translating information into a different language is often to allow an existing toolbox to be used. In any kind of pragmatic terms, this is not necessarily any less useful than creating a new toolbox. – Will Sawin May 16 '22 at 16:08
  • @user2520938 You're right that this comment thread is not going to convince you, but let me give a few other historical examples where people complained "this is nothing new; what's the point?" but which have eventually come to be recognized as being of practical value: Random variables versus probability distributions; adeles and ideles versus ideal theory; nonstandard analysis; umbral calculus. It's always a fair question to ask what an abstraction actually buys you, but at the same time, history has shown that simply arguing that the abstraction is eliminable is not a cogent argument. – Timothy Chow May 16 '22 at 16:35
  • @TimothyChow I'm not arguing against abstract though, I'm arguing against the idea that a dictionary can provide fundamentally new insights. If I change the basis of a vector space to make a matrix upper triangular, it becomes easier to determine whether or not the matrix is invertible, that's true. However, I could've also just computed the determinant without doing a basis transformation to determine this. This representing bussiness is the same thing; some things might be easier from one perspective than the other, but nothing is essentially different between them. – user2520938 May 16 '22 at 16:41
  • @user2520938 If the argument is just about the definition of "fundamentally new," then there's nothing to argue about. However, a dictionary can lead to the solution of a difficult problem that resisted many efforts. An example that I learned about early in my career: Richard Stanley (and others) established a dictionary between combinatorial questions about polytopes and simplicial complexes, and questions in commutative algebra. This led to the solution of longstanding problems about polytopes, because the algebraists and geometers had already proved the appropriate theorems. – Timothy Chow May 16 '22 at 16:50
  • If you wish to argue that the dictionary in this case did not provide any "fundamentally new insight," then I will concede the point. But if we can solve major problems and make major mathematical advances without fundamentally new insights---I'll take it! – Timothy Chow May 16 '22 at 16:50
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Solving a Diophantine equation is the same thing as showing that the functor defined by a certain scheme $S$ of finite type over $\bf Z$ gives a non-empty set when evaluated at ${\rm Spec}({\bf Z})$. So it is about a functor in the first place.

Now when trying to solve that (extremely hard) problem, one will be led to study different and more tractable functors on the category of schemes of finite type over ${\rm Spec}({\bf Z})$. In order to say something about these new functors, one will wonder whether they too have geometric meaning, which roughly translates to asking whether they are representable in the same category.

As explained in Donu Arapura's answer, this is at work in Faltings's proof of the Mordell conjecture but an easier example is Buium's proof of the Manin-Mumford conjecture for curves. In this situation, one starts with a curve $C$ over a number field $K$ and one takes a prime $\mathfrak p$ of $K$, which is unramified over $\bf Z$ and where $C$ has good reduction. A property of a prime to $\mathfrak p$ torsion point of the Jacobian of $C$, which lies on $C$, is that it provides a point of the curve $C_{{\mathcal O}_K/{\mathfrak p}}$ (the curve over a finite field obtained by reduction), which lifts to an element of $C_{{\mathcal O}_K/{\mathfrak p}^2}({\mathcal O}_K/{\mathfrak p}^2)$, which has the property that it is divisible in the Jacobian by the prime number $p$ lying under $\mathfrak p$. Buium's proof shows that there can only be finitely many such points and to do this, he uses that fact that the set $C_{{\mathcal O}_K/{\mathfrak p}^2}({\mathcal O}_K/{\mathfrak p}^2)$ can be seen as the set of ${\mathcal O}_K/{\mathfrak p}$-points of a certain variety over ${\mathcal O}_K/{\mathfrak p}$, which represents a functor, which appears as an adjoint to a naturally representable functor. If one could not represent this adjoint, one couldn't understand the set $C_{{\mathcal O}_K/{\mathfrak p}^2}({\mathcal O}_K/{\mathfrak p}^2)$ geometrically and it would not be clear how to approach the statement proven by Buium.

So I would say that one solves Diophantine equations by representing functors, because a Diophantine equation is a question about a functor, which can be studied by replacing it by simpler ones. If you can show that the new ones are also representable then you will be led to another Diophantine problem and you can start all over again. This is exactly what happens in the example given above.

Damian Rössler
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