All right, thanks so much. It's nice to be here. So, everything I'm going to say today is joint work with Josh Lam.
All right, so in some sense my talk is a counterpart to Joe's. Joe told us about the arithmetic of discrete dynamical systems, and I want to tell you about the arithmetic of continuous dynamical systems, also known as differential equations. This is an area where there are a lot of conjectures, and I mostly want to tell you about those. But I hope by the end of the talk I'll give you some sense about recent progress.
Okay, so maybe let me start with a question. As far as I know, in this generality, this was first asked by Lazarus Fuchs in the 1870s, but special cases go back at least to the 1830s. And the question is: given an ordinary differential equation, so I don't know, the nth derivative of f equals some function of its first n minus one derivatives, when can you explicitly write down a solution? This is kind of the kind of question we like to ask in mathematics a lot of the time. This maybe you can abstractly prove the existence of a solution or uniqueness, but you still might want to actually write one down. And the way Fuchs characterized it explicitly was the following. He said, given an ODE, when does it have an algebraic solution?
So, what does algebraic mean? It just means in the algebraic closure of the field of rational functions. In other words, it satisfies a polynomial with polynomial coefficients.
Great. Okay, so now you're probably wondering why I'm talking about this question at a number theory conference. You all thought you were free and clear from differential equations a few years ago. In fact, I never took an ODE course, and I thought I would never have to think about them in my life, but then I got kind of obsessed with them. So, let me... I claim this is actually a question of number theory. And let me just do the simplest possible example to convince you that maybe that's true. So, I don't know, what's the simplest possible family of differential equations? I would say it's something like the following. So, you take d/dz minus a over z f of z equals zero, where here a is a number.
Okay. So, does anyone know what the solutions to this differential equation are? So, now I'm making you do calculus. Sorry. Zero is a solution. So, what does the general solution look like? Yeah, does anyone know what a general solution looks like? So, you're looking for a function where if you take its derivative, it's the same as dividing by z and multiplying by a.
Z to the... yeah, yeah. So, to include the solution we've already heard, you want a constant times z to the a, right? So, the solutions are a constant times z to the a. Here you can think of this as like e to the 2 pi i log a, if you would want to think about how to take z to a complex number.
Okay, so when are these solutions algebraic numbers? Algebraic functions, sorry. Yeah, so these are algebraic if and only if a is a rational number. Well, clearly if a is a rational number, you can take it to the power of the denominator and you get a polynomial, so it satisfies a polynomial with polynomial coefficients. If a is not a rational number, this function doesn't even have finitely many branches, so it can't be an algebraic function.
So, being a rational number, I would call the most basic condition in number theory. So, maybe in this example it tells you that this question is related to number theory.
Okay. So, at least for linear differential equations, there's now a very well-known conjectural answer to this question. So, let me tell you what I think is the main conjecture here, and in fact I think this is the most basic conjecture in the theory of algebraic differential equations. So, it's called the Grothendieck p-curvature conjecture.
So, a relatively general form of a linear ODE is the following. So, they conjecture that all solutions to the differential equation d/dz minus A f of z equals zero, where here A is a matrix of rational functions, are algebraic, meaning they're algebraic functions, if and only if the following holds. You take this linear operator d/dz minus A to the p-th power, and you ask that this is identically zero mod p for almost all p.
Okay. So, this is some weird arithmetic condition. Maybe let's just try to unpack it a little bit. So, A is a matrix of rational functions which have finitely many coefficients, and in those coefficients only finitely many denominators appear. So, for almost all primes, so all the primes which don't appear as denominators, it makes sense to reduce this expression mod p and evaluate on polynomials and see if it's identically zero.
Okay. So, this conjecture is very open, but let's prove it in this example. So, maybe let me call this differential equation star. So, proof for star. Okay, so you can do the following computation. What you see is that d/dz minus a over z to the p applied to the monomial z to the n is the following thing. It's n - a times n - a - 1 all the way up to n - a - p + 1 times z to the n - p. And when is this zero mod p?
So, this is zero mod p if, maybe let's assume a is in Q-bar, if a is in Fp mod p for almost all p.
Okay. And does anyone know why an algebraic number with this property is rational? Yeah. Yeah, exactly right. So, that's a consequence of the Chebotarev density theorem. So, now you apply Chebotarev. And I don't know a proof of this conjecture in this example which doesn't use Chebotarev. So, I think that should tell you the easiest possible case of this question is related to some relatively deep number theory.
Okay. All right. So, very little is known about this conjecture. Maybe let me tell you what the main evidence for it is. So, it's a theorem of Katz from 1972. So, I guess this one I think was from 1872, this is from 1972, which is that the conjecture is true for a class of differential equations called Picard-Fuchs equations. And if you don't know what those are, don't worry. I'll try to give you a sense soon.
All right. So, I'll come back to linear ODE, but I want to point out that this question of Fuchs is not about linear ODE. It was about ODE. We now have an answer for linear ODE. So, I would like to propose some... discuss some conjectures for non-linear ODE. So, some of them go back to maybe the '90s, and then some of them appear to be quite new. And then I'll talk about some progress towards those conjectures.
Okay. So, maybe the first question is what is a non-linear ODE? So, for me it will just be a vector field for this talk. So, what is an ODE? Let me just draw a picture. So, here's some space. At each point you have a little arrow. It's a vector field, and the solutions are given by flowing along the arrow. So, you would like to find a curve whose tangent vector at each point agrees with this vector field. That's what an ODE is.
Okay. So, I think the first conjecture here is about when all the solutions to this ODE are algebraic. And you will note that's what the Grothendieck-Katz conjecture is about, too. So, here's a conjecture. I think probably it was in some sense known to Grothendieck-Katz. It first appeared in an unpublished paper of Ekedahl, Shepherd-Barron, and Taylor from the '90s. And then I think it first appeared in print in work of Bosch in the early 2000s.
So, it's the following. So, all solutions to the ODE corresponding to a vector field V are algebraic. So, in other words, all these curves are algebraic curves, if and only if the following holds: V to the p is identically zero mod p and V for almost all primes p.
Okay, so let me say what I mean by this little subtlety here. So what does it mean to take the p-th power of a vector field? This is not something you can do in characteristic zero. I guess if you do it, you get some random differential operator, but it's definitely not a vector field, and it doesn't make sense to reduce it mod p. In characteristic p, if you apply a vector field to a function p times, you get a new vector field, you get a derivation. And the reason why is that, well, what does it mean to be a vector field? It means it satisfies the product rule, like differentiation satisfies the product rule. And then if you do that p times, you get a bunch of cross terms which have binomial coefficients in front of them. So, and those p binomial coefficients are zero mod p in characteristic p. So this is some new vector field in characteristic p, and I want it to be zero mod p and V for almost all p.
The main results about this are due to Bosch, but I would say it's fair to say the conjecture is totally open in this generality. And there's something also a little bit unsatisfying about it I want to point out, which is that this is a conjecture about when all solutions to an ODE are algebraic. In the linear case, you don't lose that much. For a linear ODE which is irreducible in some sense, if one nonzero solution is algebraic, they all are. For nonlinear ODE, that's definitely not the case. So nonlinear ODE like the Painlevé VI equation, which you heard about briefly in Joe's talk, there can be sporadic algebraic solutions while a general solution is wildly transcendental.
I also think this conjecture is a little hard to understand, so I want to propose a sort of more basic conjecture that is also stronger. I think this conjecture must have been considered in the past, but as far as I know, it first appeared in print in joint work with myself with Josh, where we kind of prove some things about it.
So, it's the following. So, a given solution to an ODE is algebraic — I guess when I say ODE, I always mean an algebraic ODE in this talk — is algebraic if and only if its Taylor expansion at a non-singular point of the ODE, so a point where this vector field is defined, lies in the following: lies in Z adjoin 1/n double brackets little Z for some n. So, in other words, saying this is that you should be able to test if a solution to an ODE is algebraic by looking at its Taylor expansion at any point and checking if only finitely many primes appear in the denominators of the Taylor expansion.
Okay. So, that's the conjecture. Maybe let me remark that non-singular point here is really important, and you can see that, for example, by thinking of q-expansions of modular forms, which do satisfy some ODEs and definitely often have integral coefficients. Okay. And there are many other examples. Okay. So, are there any questions about the conjectures? Yeah.
Yeah, so one direction, so if a is rational, let's say a is 1/3, then you cube it and you get a polynomial, right? So, in other words, you can apply some polynomial to let it land in here. In other words, it satisfies a polynomial with polynomial coefficients. The other direction you need a little complex analysis. So, the observation is that algebraic functions have only finitely many branches. But you can check that if a is not rational, this function has infinitely many branches. That's an exercise I will leave it to you. Yeah?
At any one point, yeah. So then this is a condition at all... So this is a much stronger... Okay, so strictly speaking, this conjecture does not actually imply this one, but there's some minor... I can put an editor in that does. And maybe a small... For those who are experts, you can replace this with finitely generated Z-algebra, and it should still be true.
Yeah, maybe let me just briefly indicate what the connection is. So when you write down the Taylor expansion of a solution to a differential equation, you get terms like this to the p divided by p factorial. And so this being zero mod p says that division doesn't have a p in the denominator. That's somehow where this conjecture comes from. Okay. Great.
Maybe just one more word about the history here. So the forward direction of this conjecture is very easy. And I... well, I shouldn't say that. It's a theorem due to Eisenstein from 1852, and Eisenstein was a very smart guy, but maybe from a modern point of view, it's a little bit easier. It appears in the last paper Eisenstein ever wrote, right before he died of tuberculosis at the age of 29.
And in that paper, he says that as an application of the forward direction of this conjecture, he has all sorts of beautiful applications to solutions to certain differential equations whose power series expansions are well understood, but where one can't write down solutions explicitly. And then he died, so we don't know what those applications were. But I think it's just a sort of interesting historical question. I have a guess if you ask me after the talk.
Okay. So let me tell you a little bit about what we know. So, I first want to explain a little bit about this theorem of Katz. And then I want to explain a class of ODE where if you could prove this conjecture, or just this one class of ODE, you would already prove the p-curvature conjecture. And well, I don't know how to prove this conjecture for that ODE, but I know how to prove this one. And so, I want to talk... and we can also prove a little bit about this. I want to kind of explain that, too.
Okay. So, to start, I have to tell you what is a Picard-Fuchs equation. So, maybe let me start again by... This is section two. Where do ODE come from? I think the answer is often like physics or something, I don't know. But let me tell you another place where they come from. So, here's some kind of slogan. So, slogan: ODE come from interactions between algebraic geometry and topology.
Okay. So, let me give you an example. So, here's a kind of working-stiff definition of a Picard-Fuchs equation. So, it's an ODE satisfied by the integral over some cycle of an algebraic function, or family of algebraic functions here. I have f of s is this thing where little s is some algebraic function in many variables. Maybe I should put a dz1... dzn.
Okay, so this is a function because I have a family of algebraic functions here that depends on s, and so for each value of s I get a number. Let me give you an example of such a... this is called a period integral. Let me give you some example of such. So, example: f of t is the integral of dx over the square root of x times 1 - x times 1 - tx from 0 to 1.
So, this is what's called an elliptic integral. And you can check, well, essentially because it is an integral of an algebraic function, that it satisfies a differential equation, in this case of order 2. Okay. Are there any questions about this? So, it's a theorem, not obvious theorem, that these functions always satisfy algebraic ODEs. So, that fact in some sense was known to Picard and Fuchs, but maybe it was first written down carefully by Katz and Oda in the '60s. And this is the class of functions, or maybe a slight subset of the class of functions, for which Katz proved the p-curvature conjecture. Okay. Great.
I actually want to give you an idea of how the proof goes. So, to do that I need to explain a little bit more about what ODEs are. Maybe let me leave Katz's theorem up for now. Okay. So, I want to tell you just two facts.
So, first of all, it turns out that these Picard-Fuchs equations are very well-behaved. So, all Picard-Fuchs equations have the form d/dz times a vector of functions can be written in the following way. So, you have some matrices over z minus lambda i, f of z, where here the a i are just r by r matrices, and the sum of the a i is zero.
So, here the lambda i's are the singularities of this function. And so, what you should observe here is that these have poles of order one. So, this says these things have regular singularities for the experts in the room. So, that's a fact that will come out shortly.
And the other fact is that these are in fact purely topological objects. So, I want to explain that briefly. So, well, it's a fact from, I guess, your first analysis course that locally any ODE with analytic coefficients has a basis of analytic solutions. So, here's a maneuver you can do. You can take a basis of analytic solutions, and you can analytically continue them in a loop. You get a new basis of analytic solutions. Right? And so, what that means is I've associated to each loop in my parameter space a matrix, a change of basis matrix.
So, what we've said is that analytic continuation of solutions yields the following: yields a representation from pi one of C minus these points lambda one through lambda n to the set of invertible matrices.
Okay, so let me just say that again. So you choose a local basis of solutions, you analytically continue them in a loop, you get a new basis, and you send the loop to that change of basis matrix. Implicitly here I'm claiming that that matrix only depends on the homotopy type of the loop, or the based homotopy type of the loop. And that you actually can do this analytic continuation. Okay. Great.
Okay, and then finally the remaining fact is that for these special kinds of differential equations where you have a pole of order one, this is actually an equivalence. You can go back. So that's called the Riemann-Hilbert correspondence.
Okay. So using this we can at least sketch the proof of Katz's theorem. I mean I should warn you it's like a 100-page paper, so I'm not going to give you the meat of it, but I want to give you the scaffolding. So, proof of Katz's theorem. So when I say Katz's theorem, I mean Katz has proved many theorems. I mean this specific theorem of Katz.
Right, so it's enough to show, it turns out, enough to show that this representation rho has finite image. So the reasoning there is that in fact the orbits of rho acting on C to the r here are the branches of the solutions to this function. So showing that this representation rho has finite image says the solutions are finitely branching, and then this regular singularities property tells you that that's actually enough to conclude algebraicity. So it's some version of the Riemann existence theorem.
Okay. So, the fact is that in fact, up to conjugacy, rho, well, first of all, factors through GLrZ. And this comes from the theory of singular cohomology. It's not at all obvious that there's any relationship to singular cohomology. My hint is that integration is about the comparison between de Rham cohomology and singular cohomology. So, that's how singular cohomology is appearing.
And the other fact is that rho preserves a Hermitian inner Hermitian form. So, a quadratic form, which is... here, sorry. Well, a form given by a matrix whose conjugate transpose is its inverse.
Okay. And what Katz observed is that in fact if this p-curvature vanishing condition happens, so if d/dz minus A to the p is zero mod p for almost all p, this form is positive definite. So, it's a unitary Hermitian form. So, here this comes from Hodge theory. And this comes from what I would call the beginnings of maybe the beginnings of p-adic Hodge theory, some kind of mod p Hodge theory.
Okay. So, given this, can anyone prove the theorem? So what we just shown is that rho factors through the unitary group intersected with the general linear group of invertible matrices with integer entries. And this group is compact. This group is discrete. So can anyone characterize the discrete compact topological spaces? They're finite, yeah. So discrete plus compact implies finite. Okay. Great.
So what I would like to explain for the rest of the talk is that there exists a fam- Oh, yeah, is there a question? So what is the representation? Yeah, so in this case it's the representation of SL2Z on like the universal rank two local system coming from the first homology of the universal elliptic curve. So it's like it's the topological representation of SL2Z on a two-dimensional vector space.
Yeah, but this does not satisfy the hypotheses of the... This is a very interesting ODE which does not satisfy the hypotheses of the theorem. I mean, in fact, one way to prove that is to observe that this is not an algebraic function and then apply Katz's theorem. Yeah, so this is a hypergeometric function. It's like F 1/2 1/2 1. Yeah. Any other questions? Yeah?
Not on the board right now. I can explain it to you after the talk. But yeah, this is called Katz's formula. So what it does is... maybe what the one-word summary is that some component of this, meaning this is graded with the set of the conjugate filtration, is up to the Frobenius twist the same as the Kodaira-Spencer map. So that's the content.
All right. So, I want to introduce another class of differential equations where kind of understanding their algebraic solutions would in fact imply the classical p-curvature conjecture. Right. So, where do ODEs come from? They come from interactions between algebraic geometry and topology. And well, here I see some topology of an algebraic variety. All right. So, fundamental group of an algebraic variety, maybe that should also give me a differential equation.
So, let me describe for you a new kind of differential equation. So, here's another differential equation. So, this is called the Schlesinger system. So, here A i are functions from C to the N modulo the diagonal consisting of pairs of numbers which are equal to each other, to R by R matrices over C. So, i is 1 to N.
There are as many functions as variables. I'm just going to write down a differential equation. It's the following. So, dA i d lambda j, here the coordinates of the lambda is the commutator of A i with A j over lambda i minus lambda j if i is not equal to j. And then for the rest of this differential equation, I'll say the sum over i of dA i d lambda j is zero for all j.
Okay. So, this is an ODE. Let me tell you what it means. So, the solutions are, well, they're tuples of functions. And well, the A i and the lambda i are here. So, the question we're asking is if we move the lambda i, how should we change the A i so this representation doesn't change? Right, so if you move the lambda i, this group is not changing. You're just moving some points around in the plane that doesn't change its fundamental group.
And the solutions to this system are exactly the families of differential equations whose monodromy representation, where this representation doesn't change. So, solutions are families d/dz minus sum of A i lambda over C minus lambda i such that rho is independent of lambda.
And in fact, if N is 4 and R is 2, and you look at matrices with trace zero, this is the Painlevé VI equation that showed up in Joe's talk. Yeah. Yeah, it's no longer algebraic. That's correct. Yeah, so in general, if you integrate an algebraic function, there's no reason for it to be algebraic. Think about like, what's the derivative of arctan? It's an algebraic function, but arctan is definitely not algebraic. Other questions? Okay.
Right, so this is a special case of something called an isomonodromy ODE, which I won't give you the complete definition of unless you ask me after the talk. But you should think it's some kind of non-linear version of a Picard-Fuchs equation.
Okay, so let me tell you the main theorem of the talk. Well, maybe theorem A is that proving that solutions actually just to the Schlesinger system in Z join 1 over N double bracket C, so solutions whose Taylor coefficients have only finitely many primes in their denominator, are algebraic. So, this would imply the p-curvature conjecture.
Okay, so this one ODE is some kind of master ODE that controls the arithmetic of linear ODE. So, this theorem is actually joint work with Brian Lawrence from a few years ago. And theorem B, which is a bit more recent, so this is a combination of work from 2025 and 2026, joint with Josh Lam.
We unfortunately cannot prove this statement that the Taylor expansions of solutions to the Schlesinger system characterize their algebraicity. Well, you would have heard about it by now because we would have proven the Grothendieck-Katz conjecture. But what we can do is we can prove the other conjecture I stated about linear ODE, and we can sometimes prove this. So, there's two parts.
So, let V be the vector field corresponding to a nice isomonodromy ODE. So, for example, the Schlesinger system. Then all solutions are algebraic if and only if this condition we saw in the beginning holds. So, V to the p is equal to zero mod p and V for almost all primes p.
Okay. So, you should think of this as a non-linear version of this theorem. And then two: solutions to the Schlesinger system in Z adjoin one over N double bracket C are algebraic if you're taking the Taylor expansion around some special points.
So, if d/dz minus sum of A i... Here, remember, a solution to a Schlesinger system is a tuple of matrices. So, I can use that to form a differential equation. And we can prove that the Schlesinger system... So, I guess I should put it like a zero, the point I'm expanding around, is a Picard-Fuchs equation.
So, in other words, when you solve a differential equation, you choose an initial condition and you Taylor expand around it. If that initial condition is a Picard-Fuchs equation, then this conjecture is true. Okay. Are there any questions about that?
Okay, so I want to give you in the last... I don't know, five... Oh, no, I actually have some time. Okay. I want to give you in the last 10 minutes some idea of how you prove this part one here. Because it's actually... Well, you'll see. It's not too hard. In some sense.
So, proof of this theorem B part one. Okay. So, there's a very simple idea: just to copy Katz.
Okay. So, there's some challenges in executing that strategy. The main one is that this is a linear ODE, and so, you know, associated to it is a linear representation of a group. And, you know, well, for example, statements like factors through the general linear group over Z or preserves a Hermitian form or whatever else don't make sense if you don't have a representation floating around. And in fact, I do not know a way of associating a representation to an isomonodromic ODE. So, let me tell you what is associated to it.
Okay. So, what you get is a nonlinear... So, a discretized dynamical system in fact is associated to this. Okay, so here's the geometry. So, the geometric setup in which this all works is the following. One has a smooth proper family of algebraic varieties. Okay, maybe with some extra decoration like a divisor. Don't worry about it. And to this one can associate the following data.
The fundamental group of the base acts on these sorts of objects, representations. So, on representations of pi 1 of the fiber into, say, GLrC up to conjugation.
Okay. Why is this? Well, if you have a smooth proper morphism, so it's in particular by Ehresmann's theorem a proper holomorphic submersion, it's a fiber bundle. So, all the fibers of this map are topologically the same. If you have a representation of the fundamental group of a fiber, you can just drag it in a family around a circle in the base, and you get a new representation.
Okay. So, these are the dynamical systems associated to isomonodromy differential equations, just as these are the representations associated to ODE. So, fact: these sort of non-linear objects are the representations, and whatever they are, they're some kind of discrete dynamical systems associated to isomonodromy ODE.
And just like in Katz, it's enough to show that this action factors through a finite group. So, enough to show: so, given our hypothesis about something vanishing mod p, so, given that V to the p is zero mod p and V for almost all p, that this action factors through a finite group.
Maybe just to orient you, the example of a family you take for this system, so what I wrote over there, is you take S to be M0N, the moduli space of genus zero marked points, and X to be the compactification of the universal family with its marked divisor. Okay. So now what we have this action to copy Katz, we have to show, well, it's unitary. In other words, it preserves a Hermitian form in some kind of non-commutative sense, and we have to show it factors through GLrZ. Neither of those two words make sense.
Okay. Right, so how do we make sense of them? So maybe here's theorem C, which is also joint work with Josh. So in this setting, if V to the p is zero mod V and p for infinitely many p, so our arithmetic condition is satisfied, then the orbits of pi1 SS on this space of representations have compact closure in the Euclidean topology.
All right, so this is part of what being unitary means. If you have a unitary representation, the orbit of... you know, you have a group acting on a vector space, you take the orbit of any point, its closure in the Euclidean topology on that vector space is compact. Because it's in a level set for a positive definite inner product.
Okay. I will say very little about the proof of this theorem except to say a little bit about what it uses. So it uses... so Katz here used Hodge theory in characteristic zero and some kind of mod p version of Hodge theory. The proof of this uses something called non-abelian Hodge theory, both over C which is due to Simpson and others, and also in positive characteristic where it's due to Ogus and Vologodsky.
So if you ever are hoping to go to number theory because you don't want to learn any analysis, I have bad news for you. I had to learn a huge amount about the theory of harmonic maps to prove this theorem. And so yeah, there's... it's good to learn analysis is all I'm trying to say. Okay.
Now once you have this, so we've copied the part of Katz's theorem where we said we have a representation which preserves a positive definite Hermitian inner product. What's left over is the theory of singular cohomology, the fact that we had a representation into GLrZ. So for that what you have to do is produce a lot of representations of the fundamental group of a variety into, well, GLrZ-bar.
And that I do not know how to do in general, but for the affine line minus a bunch of points, it's very easy. And so maybe let me stop there and open the floor to questions. Are there any questions?
