 to this week's installment of our number three web seminar, just the usual kind of disclaimer. So please you're strongly encouraged to ask questions. I think the best way we've found to do that is to either raise your hand or ask something in the chat. And one of us will hopefully will check that out and interrupt Peter. But without further ado, I'm very happy to introduce Peter Sarnak from the Institute for Advanced Study and Princeton University. We'll speak to us today on integer points on affine cubic surfaces. Peter. Thank you. And it's a great pleasure to be part of this. I looked at some of the previous lecture titles and in fact, slides, looks like it's a fantastic success. And let me say that my talk will intersect two of your recent lectures by Browning and by Drew Sutherland. You will see as I go along, but I don't think that that's a negative. Maybe it means the subject has some interest. And yeah, so let me start. Let me also make clear that I welcome, I've given a few of these zooms now and it's nice to actually know there's somebody there on the other side. Sometimes I panic that maybe I don't know what I'm doing here and I'm just talking to myself. So feel free to interrupt, not just through chat, but just it's not interrupt, interact is the word. And so, all right, let me start. So this is all joint work with Amit Ghosh, the more recent work and some of the earlier work which has still been worked out is with Alex Gumbad and the Jean Bougain. Unfortunately, Jean Bougain died a few years ago, but Gumbad and I are still working on, we're writing up some of these details. All right, so I'm gonna talk about cubic forms, but before I do so, let me talk about, let me set up the general problem. We are interested in Diphantine equations of the integers. We'll be looking for integer solutions, not so much rational solutions. There's a difference when you're working over the rational things are a little better because you can divide and the theory is a little more complete every now and then you have groups where you wouldn't have if you just have integer solutions, but of course they're very closely related. So this is sort of the difference in the case that I'm interested in in this lecture, only one equation between a homogeneous polynomial and the more general polynomial in n variables integral. And I'll assume that throughout f and f minus any k because I'll be interested in an affine equation, f equals k, where k is also an integer. So I'll assume everywhere that f and f minus k are absolutely irreducible over the complex because if it factorizes the equation, of course simplifies, I have some device at least to point with, that's this little cursor. So if it reduces, especially if it factors over z, then you would have a product of two numbers equals k and you would be able to study each factor separately. So I'll assume that everything's irreducible. So it already takes away, for example, the case of the Riclet's theorem where you have a homogeneous form of degree and in n variables which factors into linear factors. That is a model case and I will refer to it occasionally but that's not what I'm interested in. So we look at the affine variety defined over whatever ring we're interested in and we look at the set of x such at f of x equals k, that's this hyper surface and we're interested in integer solutions. There's an obvious necessary condition, local congruence obstructions that allow us to study things locally and studying the equation locally is much simpler. I'll get back to that for the kind of equation I'm interested in in a second. So of course, if I have a solution of the integers, I have a solution modulo q for every q and checking the solutions modulo q, especially for the kind of equations which are absolutely irreducible than at least in three variables. I'll be looking at two big same three variables. The condition, the local condition of where they can solve modulo q by the Chinese remainder theorem simplifies tremendously and it just becomes a condition on arithmetic progression so that this equation f of x congruent to k modulo q, which of course counting the number of solutions would give you some of the extensions, field extensions, p, p squared and so on would give you the local zeta function. But whether their solutions or not is easy as I say. So if we in the situation, any situation that we'll encounter, we will know what the local obstructions are and we'll always impose them. And if we pass these local obstructions then we would be very happy if there's a global solution and if there's a global solution maybe many global solutions. So this is the local to global principle we're interested in. And we say that we have a hazard principle because this is based on our review of hazard theorem in a second. We have a hazard principle if the local obstructions these trivial local obstructions are the only ones. So if you have a linear equation let's dismiss that case immediately. So it's just a linear equation like that. Linear form equals K then of course if you're gonna be able to solve this over the integers you better have the GCD of the A's divide K and that's also sufficient and once you have a solution you have many solutions. So the linear equations easy but the quadratic equations already as difficult as pretty much most things can get in the subject. Although we do understand this case and I wanna explain a little bit why we understand this. So if it's a quadratic equation to be solved over the integers in a number field or over the rationals in the number field this was what Hilbert, this is a well-known Hilbert's 11th problem. He posed this after seeing putting together work of Smith and Mankowski. It's a difficult problem. He understood that class field theory which was not, it didn't exist at the time but which he foresaw at that point by looking at quadratic extensions would play a big role in this. And there has some Mankowski theorem solves has solved Hilbert's 11th problem in case A, the easy part of the problem which is to solve, do you have a local to global principle for a quadratic where you are asked to solve this for rational numbers in a number field. And of course the necessary condition is if you can solve it in rational numbers in the number field, you'd have to be able to solve it at every completion KV of K. And the converse is true and that's the beautiful Hassan and Kovsky theorem. The case of the integers like which numbers are some of three squares of a number field is much, much harder. Ziegel spent a tremendous effort to solve Hilbert's 11th problem over in rings of integers. The Ziegel mass formula was born out of this. He made tremendous progress on it. Of course he had to start off with the hazard principle to get started. But the answer is much more difficult in the case that you have quadratic equations, quadrics in two variables, binary quadratic forms. That's really class field theory that's understood through the theory of fields. But once you get to three variables, it's already very difficult. And Ziegel was able to do it in five variables. He actually developed a hardy little wood circle method. I'll return to that in a minute in order to do that. Knazer found a very beautiful treatment of four variables in the slightly weaker form. So you can't expect a local to global principle of a number field, especially for definite quadratic forms. That is that you can solve for every K if and only if there are no local integral obstructions. But you can demand that that be true if K is large outside some well-known exceptions. And that's sort of a stable local to global principle. And that's known now all the way down to three variables. And that's the work of Ziegel, Knazer, Duke Ivanich made a very important breakthrough in Kogdel, Piotetsky, Shapiro and myself were able to settle this. And the reason that three variables is the critical and hard case here. And the reason that one can solve that, so four variables or five variables, as I said, Ziegel already was doing with the circle method. Four variables can be handled by a method that I'll return to, close to months, improvement or development of the circle method, allowed you to deal with four variables even over a number field today. But three variables, the only way we know how to deal with it is to use the full force of modular forms. There is theory of theta functions which connects the solvability of F of X equals K by making modular forms. So the important thing about a quadric, which is F of X equals K where X is a quadratic form, is that it's a homogeneous space for orthogonal group. And it's that extra piece of algebra that it's got the symmetry of an orthogonal group underlying it that allows you to bring in modular forms through theta functions. And then the techniques that are required to prove this on highly non-trivial, but that's the feature that allows you to understand quadratic forms. So I worked on that many years ago and I've now been working on the case of cubics. And cubics are much, much more difficult, especially over the integers. We essentially know nothing in three variables, which is the case I wanna talk about, but let me review what we do know in many variables. So I'm going to stick to affine cubic, not homogeneous as I said. And the reason is I would love to stick to homogeneous only and deal with rational points on cubic forms, but I have nothing to say there in terms of any new methods. I'll review what is known. So I will have to allow because the case that we're going to study in great depth is not gonna be a homogeneous polynomial. It'll have lower order terms, but the leading part I always want to be cubic. I wanted to be non-degenerate, so it should involve all the variables. That's what I mean by non-degenerate here. And I'll assume as before that f and f minus k are absolutely irreducible. And we look at the level sets, this affine hyper surface, and we ask when k is admissible. And in this case, really the condition on local solvability is simply that k lies in a progression. We'll see this in every example that we look at. That's a general fact that follows from Langouay theorem very easily. And the question we're interested in is to produce points, not to show they aren't any points. And the production of integer solutions or rational points more generally for algebraic varieties is very limited. We don't have many tools and that's why the subject, even examples are very interesting. So I'm interested in the case that it's rich, that we may have a local to global principle, a hazard type principle. And if the number of solutions is non-empty, in other words, we do have an integer solution, then do we have infinitely many or there's a risky dense and are they rich enough to even satisfy some form of strong approximation? So strong approximation means that we have enough solutions. So we're looking at vk, f, z. We're looking at this affine variety. And we ask, can we actually solve, if you give me a local solution, so you give me a solution modulo q, does it come from a global solution? That's what strong approximation asks. So is it possible that the solutions are so rich that not only do we have a hazard principle, but we really have many, many solutions enough to cover all local solutions. All right, so we look at a cubic form and I get to talk about cubic forms in three variables. And that's because that's the critical threshold and by far the most difficult case for which we know almost nothing. If we look at a cubic form in two variables, so if let's say it's irreducible over q, so it doesn't fit into what I had before because in fact there's over c, but that's a base, it doesn't factor over q. So we're looking at a binary cubic form. Then there's a famous theorem of tour seagull that the number of solutions is finite. So in that case, we certainly not gonna have a rich number of solutions. And I'll return to heuristics in a minute, but it's clear why there are very few solutions and Schmidt even shows by extending to a seagull's method for this kind of equation. Doesn't have to be degree three, any degree. And not only is it finite, but for most case, so if we fix f and we vary k, which is what I'm interested in, for most case, there are actually no solutions. And for most case in a very serious sense, all but k in this case k to the two thirds. And if you look at the local obstructions in this case, which are a bit trickier than in the case of more variables because they can be local obstructions, it's easy to compute what they are. And the number of k's which are admissible is still, the number of k's up to capital K, which admissible is still capital K over log K roughly, while the number of cases for which it's empty is much, much larger. So you certainly don't have any hazard type principle in this case, and I'll call that the supercritical case. So the cubic forms in two variables, we understand they're very few solutions. So let's allow many variables. Perhaps the strongest theorem that's known today is the theorem of Browning and Heath Brown. I'll call this the subcritical, this is where we have many variables. And that is, let's allow these non-homogeneous cases. So if the f naught, the homogeneous part is non-singular and we only look of course at admissible k, then we will always have an integral solution and the integral solutions will be the risky dense and you'll have full force of strong approximation. The point is that in improving this, they're essentially proving it by the circle method. So the circle method of Hardy and Littlewood is a harmonic analysis method to solve these kind of equations. And if the number of solutions is extremely large, so let's try and understand how many solutions there are. We're looking at fx1, f is cubic. So if I look at fx1 up to xn, if I have n variables and I let the x's go up to b, say in size, I will have b to the n numbers. When f transforms any one of these numbers, which in each coordinate is at most b, since it's cubic, the number will be at most b cubed in order of magnitude. So that means that we are filling boxes, b cubed boxes by b to the n points, b to the n elements in the domain of this f. And so if the world were fair modulo, the local obstructions, each box would get roughly b to the n minus three. So if n is bigger than three, there should be a lot of solutions, at least on average. And maybe the world is nice and each guy that's supposed to be hit, except for the congruent obstruction, is hit. And when the circle method works, it proves that. And not only does it prove it, it proves that every kind of feature that you want of this nature is true. So that's in the subcritical case, but we need a lot of variables. And of course, a lot of the work in the subject is reducing the number of variables, especially if you're using something like the circle method. Hooley, before he passed away also maybe a year ago, I think this is the last paper that maybe that's published at this point, I'm sure more will be on the way. So the kind of, he explored the kind of question that I want to explore for Cubics in three variables. And that is, he got down to four variables, but not being able to deal with any f, but to deal with, not to be able to deal with every k, but to deal with almost every k. So the question is, is it true that almost every admissible k is actually represented by f? And you can apply a technique, which he developed many contexts, which is to compute the variance of the expected number of solutions from the number of solutions. Then you double the number of variables when you compute this variance, and then you into eight variables, which is getting closer to your 10 variables where the circle method works, and where he'd actually showed that if you assume the Riemann hypothesis for certain Hasse-Vael functions that intervene in his method, then you actually do get this almost all result. He, I mean, grounding, sorry, yeah. There's almost all here, me. Ah, yeah, yeah, yeah, yeah, yeah, yeah. Okay, this is very, you can see Bjorn who's both an algebraist and an analyst. Yeah, if I'm talking almost all, it's all, but density one, so it's in the sense of densities, that's a much weaker result, yes, of course. I should also add that when you produce a point by this method, by this averaging method, you produce a point, but you don't produce infinitely many points on any one of these fs, or any one of these ks. In the same, there's a recent work which actually was presented to the seminar by Browning, but he's a work with LeBoudeck and Sauen, where you average not over the values that an f takes, but you average, he's looking at the projective case and rational points, but he averages over the modular space of fs. That's a very big averaging, and again, proves that you have a Hesse principle in the case, which is similar to this in the sense that we expect many solutions. So this averaging method will maybe produce a point, but it never produces as a risky dense set of points, because you don't, as you increase your parameter space over which you're averaging, you produce points on the different, on different varieties and not points. Nobody knows how to produce them all in the same variety. So in this case, where I fix f and vary k, or in this Hooli case, you don't get for any f that you actually have more than some finite number of points, but you do get a Hesse principle for almost all level sets for f under this strong assumption, if you have four or more variables. So the cubic case is the case, which is about far the hardest. So two, we vary very few solutions. I explained heroistically why we would expect you if we look at balls of boxes of size b, that they're only going to be b to the zero number of solutions. So you're going to be, you're going to hit maybe finitely many times and this bodes poorly for analysis. So it's going to be kind of tough. You may be able to circumvent that difficulty in a way that I'll explain in a minute. But what I would love to say something about, I have nothing to say and people, there was a lecture on this also in the last month in the seminar, here's the sum of three cubes. That would be the ideal problem, but this is extremely difficult problem. By the way, we're allowing X to be negative. Yeah, if you put X to be positive, the problem that would restrict the values and it's very easy to see this, even the almost all result is not true. It's not true that almost all cases, some of the three positive cubes are just not enough points around to fill the boxes. It's a, requires a little bit of an argument, but it's not the real. Anyway, as I promised, that in any of these cubic problems, the condition that you admissible is very easy to determine. He had, so you shouldn't be four, five, nine. And if you're not four, five, nine, as far as the mankind knows, you probably, it's possible that you always do have a solution. We don't have any tool to produce solutions. We don't have any tool to produce almost all case. I'm afraid in this case. So one goes to the computer and I'll tell you what's known in a minute. But the talk of Sutherland explained very beautifully. They break through by Booker, who used the method of Elkies and then Sutherland, as far as I can see, explained much more detail of this method. It wasn't the first number that was not known to be some of three cubes, but which could have been a sum of three cubes was 33. And as you can see, the solutions are very sparse. So this is the, was the first solution that Booker found last year. And then he with Sutherland found many more solutions, including a solution for 42. You might ask, what are these numbers 33 and 42? They're the first that weren't seen, but there's a reason that they weren't seen, as I'll explain in a second. So these are found by a computer. They don't produce more than one solution. They may look for more and find more solutions. But it's a very difficult problem. There's no local obstruction. Nobody's found an algebraic obstruction. There's one case where we know something a little more. It's a very beautiful theorem of Lehmann developed further by Boykers, which you can show that for the equation equals one, sum of three cubes equals one, of which you can put two equal to zero and one, that the solutions are dense by a very clever use of Pelz equation. I don't want to go into it, but it's the one case where the level set is not only got a solution, but it's got a dense set of solutions for the sum of three cubes. Perhaps the best theoretical thing we know about sum of three cubes starts with an observation of castles using cubic reciprocity to show that the strong approximation for these sum of three cube problems must fail at least for some congruence class. In fact, I would suggest maybe there's only one congruence class or limited number of congruence classes for which it fails, but they always at least one congruence class. And this was observed first by castles for the right-hand side being three. So if you have a solution to x one cube plus x two cube plus x three cube equals three, then the solution has to have satisfy a congruence module of nine, which is not locally forced on you. So this is a failure of strong approximation and it shows that the solutions to the sum of three cubes equals three are gonna be rather sparse because they restricted to certain arithmetic progressions. This uses cubic reciprocity and Paul Joghter Leiden in Wittenberg realized this obstruction that castles found using cubic reciprocity by Broward groups for the surface. And that's a very important advance because when you just found obstructions as many people might by some trick here or there, you don't limit what you are doing. So when you understand that all certain families or country examples of a certain type, you also know the limit of what you can do and when you find something which doesn't fit in that category, you've learned something. So Colliot, Thelen and Wittenberg explain that strong approximation always fail for the sum of three cubes equals a number, admissible number in some congruence class using this Broward-Mannien obstruction. And they however never have found an example which would actually violate the hazard principle. So as far as we now repeat, every number which is not four or five modulo nine might well be a sum of three cubes and not only might it be a sum of three cubes but these high, these level sets might be a risky dense for all we know. And you can conjecture one way or the other I don't think anybody will prove it. This prove you will prove anything unless they have some really big new idea. So with that said, and having spoken for 25 minutes I now will turn to what I wanted to talk about. And that's some very special cubic surfaces that I've been working with the Bougain and Gumbard and Amit Gosha for the last eight years or so. And my one excuse for looking at special cases is we really don't have exotic equations, enough exotic equations, but in which we can say something non-privile and for which... Sorry, what? Is there a question? Can you hear me? Yes, we are here. Please continue. Okay. So I must carry on. Okay. I don't want everybody to be moved off into the waiting room. I'm sorry. All right. Yeah, so let me get back here. So I'm going to talk about Markov cubic-like surfaces. This is a family of surfaces, cubic surfaces, they're not homogeneous, for which there are many viewpoints and many different tools that can be used. And because of that, we can say something non-privileged. So let me introduce you to the basic case. So the cubic form is sum of three squares minus the product of three numbers. So it's something very special and we're going to see very many properties that allow us to look at it from many different points of view. Again, I'm looking at the level sets. Same question. I will single out two cases here. The Markov surface itself, he looked at the case where K was zero and there was an amazing number three. I'll return to that a bit later. And K equals four, which is the Cayley cubic, which is very important in the theory of cubic surfaces because this is the surface that Cayley used to prove the 27 lines on a cubic. But we have this one parameter, K, that's our level sets. And we want to look at the Diphantine solutions to it. So Markov looked at this first, he studied it in connection with what's called the Markov spectrum and approximations of quadratic serves, how well you can approximate them, very closely connected to the Lagrange spectrum. And he connected the two. There's a very beautiful paper by Frobenius explaining Markov's ideas. And if you look at Frobenius's paper, you will pretty much see the modern view of the subject. Harvey Cohen in the 60s observed the connection between the Markov solution to the Markov equation and simple closed geodesics on a congruent subgroup of the modular surface. So a connection between closed geodesics, hyperbolic geometry, and integer solutions of the Markov equation. And I will use that connection indirectly later. These Markov equation comes up in algebraic geometry in work of Rudakov and Gorinsteve. And even there's a paper on the archive just two days ago by Koti and Varchenko on the Markov equation for Laurent polynomials, a star Markov equation where they introduce a star operation and evolution, where you're trying to solve the Markov equation in polynomials, Laurent polynomials, and that's very much connected to this algebra geometric work of Rudakov. There are other cases. There's a Markov equation comes in left-shift vibrations as an obstruction, work of Dennis Arou. But the most important to me will be that the Markov equation comes up as the relative character variety. This is how we'll generalize it, and I'll explain at the end, of representations of the character varieties that is up to conjugation of representations of the once punctured torus into two by two matrices. I'll return to that. And that's a viewpoint that's extremely important. It brings in this hyperbolic closed geodesics is also very important because it brings other tools. And that's exactly what I like about this problem is that we can use the circle method, but we can also use other tools and even combinatorics to try to attack the problem of the richness of the integer solutions. The Markov equation arises in algebraic classification of monodromy groups of panel of A6. This I won't use other than to point out one of the theorems that we proved was already proved earlier by the Brovyn and Matsukawa in this connection when I gave. All right, so that's the Markov equation. What does the surface look like? So as I said, the case of K equal to zero is the Markov equation itself. So here are the real points drawn in a picture of the level set V zero for this K. So in V zero, there's a point, the origin is a solution. You remember the equation. And then there are four components at this, this is V zero. There are four components which are the same. And that's what the solution set looks like for V zero. As you change K, so if you increase K from zero to two, the zero, the component, the solution, which was just a point that the origin grows to a little ball here, little compact set. So they're four pieces like that with a compact set. Then as you increase in K is four, they all just touch exactly with the conic singularity. That's the very special case K equals four. Everything I say for K equals four is false. It's the reason that K, they like that case. It's a very simple case. It's almost linearizable, so to speak. And when K is bigger than four, then these all connect up into something like that. So that's what the real points of the Markov equation look like. And the reason one can study VKZ, so remember we were trying to look at the solutions to the Markov equation of the integers, is there's a group acting on the surface and it allows us to dissent. This group is already used by Markov and in any one of these interpretations of the Markov equation as a parameter space of something or as a modular space of something, this group comes up naturally in its own way. But the group can be described very easily. It's a group of automorphisms, polynomial maps of affine three space which preserve the surface. And that's very easy to see that if I take the easiest way to think of it is if I take the Markov equation M equals to K and if I fix two of the coordinates there, say I'm making K equal to zero, if I fix X2 and X3 and I look at X1, I happen to have an integer solution. Then the integer solution is a solution of a quadratic equation. You use high school algebra, the leading coefficient is one. So if you use a formula for quadratic equation, the conjugate solutions also an integer solution. So you get a map which takes you from an integer solution to an integer solution, just as go in pollution. And that map I call there are three, the one which, and you work out what it is. So it fixes X1 and X2, and it flips X3 into X1, X2 minus X3. That's a non-linear transformation and that looks harmless at the first step. But when I start to iterate R3 and R1 and start mixing them up, I get a very complicated non-Abelian group. The group generated by that and all the permutations, for example, is just PGL2Z abstractly as a group. And it acts on these level sets and it preserves integers. So it takes integer points to integer points. And that's the reduction that allows us to study the integer points on this Markov equation, which we can't do, for example, without any great new idea for some of three tubes. So as I said, when case four, everything I say is false. So ignore K equals four. It's a much simpler case, it's a singular case. And otherwise VKC consists of finitely many orbits. So it really goes back to Markov, Hurwitz-Mordell. They just observed that if you start off with any solution, you can make it smaller by using one of these three involutions and then you reduce at your descent and you get into some finite compact piece. And hence you have VKZ has got a finite number of orbits. And if nothing else, you also have a decision procedure to decide whether Markov equals K has an integer solution or not. And of course, if it has a solution, then you'll have many more solutions by iterating by using this very large, potentially large group. The group's large, but acting seriously on the variety. So we don't even have a procedure for sum of three cubes equals K, but we certainly have here, and this descent is what allows us to say something. So let me call HK the number of different solution, just something like orbit count or class number. I use the same notation because we'll see, it's motivating how I'm thinking here. So if HK is positive, is VKZ infinite? So is it's a risky dense? Very likely since we can start with a point and then move around unless we stuck in some little closed small orbit. If it's a risky dense, do we have strong approximation? Are there solutions to this Markov equation, which is a cubic, very rich once there is a solution? These are the questions that we've been trying to understand. Peter, may I please interrupt you for a moment? Yes. There is a question please from Sherry R. Sikander, would you please ask your question? Thank you. I was just wondering if one can think of this group gamma as the mapping class group of the surface 1, 1, sigma 1, 1. Absolutely. I'll return to that. Absolutely. OK, thank you. And just one more point. So maybe you'll also talk about this, but how does the parameter K enter in the relative character variety in the definition? It's the trace of the boundary component. It's a trace. So you're fixing the quantity of the class. I'll explain this in general, yeah. OK. Thank you. Yeah, good question. All right, so for all K, which again, it's given by progression. So if K is not 3 mod 4 or plus or minus 3 mod none, those are the Ks for which is an obstruction. Otherwise, there's no local obstruction. There are no obstructions periodically. And so we'll restrict to Ks, which are admissible, and only those. And for K small, you've got this reduction. You can work out what's going on. So I'll return to K equals 0, the Markov case. He found there was one orbit there only. That's his famous Markov tree. But let me first remove some trivial points that exist. So when some K, you have a solution where one of the coordinates is either 0, 1, or 2. And if you have such a solution, K is already determined because if you fix one of those coordinates and then you ask what can K be, then you're looking at a binary quadratic, a quadric in two variables. And that, of course, we have a complete theory for. And so I'll remove these special, easy to describe solutions without loss of understanding and we'll call the solutions generic. And in our work with Gauss, Gauss found something very, very beautiful. And if there's nothing else you take away from this lecture, you take away that there's the Gauss fundamental domain that everybody knows for the modular group. And there's the Amit Gauss fundamental domain for this. It's as beautiful. And this is the theorem that if K is positive, it's a little tricky when K is negative. A generic, you want to find all the solutions so you can do this reduction. And the set of solutions is very much like Gauss reduced. So we say it's Gauss reduced. The solution is minus x1, x2, x3 lying on vkz. If you can bring it into the form where x1 is less than or equal to x2 less than or equal to x3, greater or equal to 3. And of course, it satisfies the Markov equation. I've got a minus x1 there. And the observation of Gauss, which is very beautiful, is that you have one element inside this for every orbit. In other words, the class number that I called hk is exactly equal to the number of solutions in this region. And that's a finite region. And it's very easy then to numerically compute hk for any k, which he did. And we used it immediately, theoretically. So using that, one can prove that h of k, the number of orbits under this Markov or mapping class group, as it was pointed out, is at most k to the third. And the most important thing we observed, number two here, is that on average, so if I average either through positive k's or negative k's, the average value of the number of distinct orbits is growing like log k squared. And I'm going to return to this. This is the driving force of the main theorem. And that it means that for typical k, there are more and more solutions. And this allows us to at least be in the situation where these other averaging, where we want to maybe prove a local to global for almost all k, remember I said that if we take a box of size length b, we'll hit each person a bounded number of times. This is saying that these class numbers are growing, at least on average, slowly, log k squared. So there's a chance to do some kind of theorem which says that almost all k have a, has a principle is true. And once we have such a point, then we can start to use the group to becomes a risky dance, which something, so we use everything that's available. From the numerical experiments we found that the number of k's up to k for which there has a principle, we only was talking about admissible k's of course, for which it has a principle fails, is roughly k to the point nine. So they are a tremendous number, but still zero density failures from the numerics. And that's what we set out to prove. Here's an example of, there's the fundamental domain of gauche inside the triangle there and you count so that h of k is six, I think in this example. And here's our main theorem that we have proved and still busy refining in, as I'll explain at the end. So firstly, as the numerics indicated, there are infinitely many has a failures and we are able to prove that by using reciprocity. So we found a obstruction which only produced k to the half of a log k has a failures. While I told you that it appears to be k to the point nine has a failure. So we certainly not explaining all of them. And that's a warning that algebra or these row of money in types of obstructions which I'll return to in a second, are not decisive, but they at least give something. And the main theorem that we prove is that almost all k's, h of k is positive. In fact, h of k goes to infinity, gets larger and larger for almost all k's. So that means that the has a principle is true for k admissible for almost all k's. That's the analog theorem that we've seen when you have many solutions. But yeah, we only growing like log when we set it up. And we still able to prove this in three variables. So we quite satisfied with that. And that's establishing what we saw numerically. All right, the has a principles we found all come from the couldn't be explained by the Cayley cubic modulo some prime but they use quadratic reciprocity. Koljot Thelen and coworker Desheng and Shu and also Lachron and Nankin independently have shown that the examples that we showing by our quadratic reciprocity that fail has a principle can all half of them I should say can be explained. This is the beauty of the brahmanin is it also limits what you can do. They explain all the examples that can be explained by brahmanin and they are essentially half of our examples. The other half combined that kind of argument with dissent which brahmanin does not build in dissent. So they can explain all our examples if you like by using dissent and brahmanin and that's I think very important that they explain conceptually not just that we produce these examples. But I repeat again that there are many more failures indicating that we should be very, very nervous about a sum of three cubes hitting every number. It may be that you can't explain the failure by some simple obstruction which involves reciprocity which the brahmanin and our examples are of that nature. Let me say another word about the brahmanin obstruction but our proof of the almost all because that's kind of subtle and I think maybe hasn't been looked at that much is we have, because if we restricted the variable say to be positive then we would hit every number bounded number of times and you could never prove almost all. We need this log growth or some kind of growth and that we do by taking certain tentacle regions where the number of points is growing using this gauche fundamental domain. The number of points is growing slowly like log and then we have to and because it's growing the number of diagonal solutions which when you're looking at cubic form in three variables is already significant number is just washed away so that we can get rid of it and it becomes technically very tricky especially to control the diagonal. So there's a calculation of the expected number of solutions and we take the difference, we square it we get to a cubic form in six variables but the cubic form is Mx minus My of course and we have the luxury in that situation of fix then this part doesn't use any of this Markov group. This part is just using geometry. If you like, it's not the circle method but it boils down at some point to the circle method meaning we fix two small variables to make a tentacle something that you should be doing with the sum of three cubes as well by the way and we are able to analyze when you fix two of the variables the situation that arises properly because we can deal with quadratic forms quadrics in four variables which is we have six variables two of them are fixed which we're going to make very small and then we have to see how many solutions there are in a region compared to what Hardy Little would predict and close to months methods allow you to do that especially in its modern incarnation to control the diagonal, the close demand method or any such theory is not strong enough and we actually have to get into divisor theory and count the number of devices and there's a very beautiful paper of Granville and Bloma which does the preliminary stuff that we need so it's quite delicate the main theorem here part two and it's a case where we don't have many solutions and we only averaging on one parameter but we're still able to do it because of the special structure of Mx minus My and it applies to such cubic forms which we've identified in the paper what that feature is. All right in the remaining time I want to talk about the richness which is the work that I've been doing much longer over a longer period with Bergen and Gumber this is where we started and suppose that we already know that VKZ is non-empty so we have one point and now we want to say a G I've got one point, I've got a tremendous number of points and here's where the mapping class group and all the geometry is going to come in. I'll stick to the Markov case the general case is quite a bit more complicated but can be understood completely and let me explain why. One thing that we're going to do is we're going to start with an integer solution on one of these level sets and then apply these Markov moves to make new integer solutions. I hope, Michael can you hear the lawnmowers? This is the trouble with Princeton in the morning is your neighbors always got lawnmowers going and it just didn't occur to me that this would be a problem. They are outside here, I suddenly hear them. Okay, as long as you don't hear them, that's fine. So getting back to these orbits so if you have a point on one of these VKZ and you start applying these Markov moves you have to make a new integer point so you would imagine you're going to be Zyrusky dancing. Your enemy is you might be stuck in a finite orbit. Maybe this is a finite orbit for this action on F1 and 3 space. So all of a sudden you need to know what are the finite orbits? In characteristic zero of a Q bar because anything that appears in characteristic zero will appear by the Schabotary field of a finite field of P for many P's. So we have to face that. And so we had to classify what the finite orbit of the action of this mapping class group bar on F1 and 3 space. And what we did and for that we used one of my favorite theorems, Lang's GM conjecture which was solved by, we only need the torsion part of that which is much simpler than the full conjecture which was solved by Michel Laurent in the 80s. We used that to make this class of applications on the 11th rule. But it was kind of difficult for us to me, I couldn't believe that nobody had looked at this because there are many people who work in Kleinian groups to study the action of the Markov mapping class group on the characteristic varieties of the circle. And they know it's very complicated. They looked at much more things, but somehow they had not identified the finite orbits which I was really critical. But the problem in Machokko should be, did actually determinists and they, to me, quite shocking. How did they do this? These are two mathematical physicists. How did they do this without using Lang's GM? Well, it turns out that they found an old paper chronicle where he gives a proof of the classification by Schwartz of hypergeometrics with finite monotromy. This is a non-linear version. It's kind of a non-linear version of the land. And this proof of Schwartz's theorem by Kroneck uses roots of unity, not by, the proof is by, I mean, notice it was Gordon, Gordon, Gordon from Turkey. But it uses work of Kroneck. And if you look at that carefully, you'll see that that work actually has the roots. That probably is the first step towards the proof of Lang's GM long before anybody else. In the 60s, 70s, 80s, what it was looking at. So they found this proof and they found the finite orbits. And if you want to state whatever I'm about to state for arbitrary K, you must take into account those finite orbits. Let me say take the case K is zero and I'll put it three years because that's really exactly Markov's equation. Let me restrict to that. Then the moves are not what I had before but there's the three over there. And then Markov's big theorem is that the integer solutions to this equation, yeah, especially if all coordinates are positive, have one root, one orbit. So there's the zero, zero, zero, that's one orbit, that's special orbit. We of course remove it. I'll put the start to show the movement. And then all other integer solutions are gotten from this one solution. And the problem we're interested in is do we have strong approximation? So I look at all the integer solutions of P. I look at all the integer solutions period. I know that all these integer solutions must be gotten by using the mapping class group or this Markov group. So if I start with one, one, one, that's how I'm going to get every solution. If I work modulo p, I can count how many solutions there are. They're p squared. And if I reduce modulo p, this Markov formatting class group acts as a permutation group on this finite set of solutions. A strong approximation is completely equivalent to this group acting, this permutation group acting transitively. And that's something that we construct just through for every p and in the general K case, for every sufficiently large p, there are counterexamples for small p's, something that stabilizes when p gets to long points. Yeah, one can check for small p that this is true. This action is transitive. And this is the big problem, the big construction that we have that we first looked at numerically is that the Markov group is acting as richly as possible. Should be acting transitively. And as long as we can prove this conjecture, as long as p squared minus one is not very smooth, there's an enemy difficulty in proving it completely. So we have an approximate result. And this is the main theorem, two theorems. The first theorem is that if you look at the solutions y star p, there is an orbit, a giant orbit, whose complement is at the most size p to the epsilon. So the number of solutions is p squared. And we look at the connected components acting by this division group. And there's always one mass of giant component. That's our starting point, that giant component is at most off of the giant component at most p to the epsilon points. And any, and this uses the characteristic zero argument, any component must grow with p at least like a lot of p to the one. Improving this is very important because it might allow to completely resolve this strong approximation conjecture, which we only proved for all by very few p. And Konyagin and coworkers have improved this bound. They improved from p to the epsilon to exponential log and improved the exponent and the log bounded in a very recent paper that I just heard of a few weeks ago. They've improved this exponential log to log p to the a and the sub natural conjecture and that is very important because it starts to maybe allow you to sort of hold this whole wagon and maybe prove the strong approximation without any exceptions. But this all remains seriously open and I happy that I can talk on this public platform to popularize this problem. Anyway, but again, we're able to use this to prove that this strong approximation conjecture fails for very few p. The number of p's up to t for which it's not proved is at most t to the epsilon. That's already enough for many of the applications we had in mind, but the real truth is we want to know that this group really acts handsomely on these finite states. And in fact, the real problem is to act on the p addicts rather than p addict integers. So there's a nice theorem using finite group theory from last year. I mean here we included which say that if p is one of the p's for which we've proved that we are transitive on these solutions, more p. And you ask what permutation group are we actually looking at? So it's a transitive permutation group and now all theorems in finite group theory which ensure that transitive permutation groups which are triply, doubly, transitively, already big. And I use that to show that if p is one of four, then automatically this act, the group is the full symmetric group with alternating group and you can tell which one it is very easily. So not only is it acting transitively, but it's massive, but they have to assume p is one of four and unfortunately we don't have to do that. And they also show that the action is minimal on the p-addict integers. So when you look at this Markov permutation process group acting on the p-addict solutions which are integral, it's apparently very minimal action, meaning every open one wants to be dense and that's what you can prove by putting these things together at least in piece one, one, one, four. In fact, if I can go again, but then I can improve this without putting p one, one, four but this will be possible. Why do I say that this is very useful? There's this beautiful theorem of Gorsuch which is the notion of relatively prime group supposes that I want to get the strong approximation for products of integers rather than just for p's themselves. Supposes I know it for p's and I want to get it for the product. I come back again and I found a way to do the product by developing methods which I haven't explained at all in this context but once you know the group is one of the alternating or the symmetric group you can use the final group theory rather beautifully as follows. Supposes I want to prove I'm onto the product so what I know is I have a subgroup of the product of the groups and the question is if I have a subgroup of product of the groups which rejects onto each factor onto each factor is it true that subgroups is a whole product? Well, if it was Z mod mz and Z mod nz if m and n are relatively prime then this is automatic but how do you make relatively prime at the level of final groups? The answer is there is such a theory called Gossatz Lemma you have a subgroup of G cross H which predicts onto both G and H of final groups and the Jordan hold this series of G and H have nothing in common then automatically you're onto the product for free that's the kind of thing I like things that are for free like this and because the alternating group is simple and the symmetric group is essentially simple there are factors that are relevant here and you get for free that we are onto the product and now we really start to have a serious form of approximation for Markov equation because we can do products and we can do prime powers and that's we modular the starting point which we haven't completed we only know that but there are very few exceptions there should be no exceptions that Markov that's very pleasing and this immediately solves long standing problems of vanities for example if you make Markov numbers Markov numbers a number which has a coordinate which is a solution to the Markov equation the very Markov equation that I've been talking about in the last five minutes these are numbers that people have speculated and looked at for years and years and very hard to say much about them other than the growth rate but for vanities immediately observe the congruences that maybe must be supplied for example if p is 3 Markov 4 and p is not 3 then any Markov number cannot be 0 plus minus 2 3rd as Markov p the question is is that the only abstraction and the answer is how strong approximation the conjecture certainly implies that means we prove this conjecture for very many primes but maybe this is the way of doing this without proving the full strong approximation but it's a beautiful problem and the answer we can use this to prove something about Markov numbers I'll skip that except to say it does use some work of Mirza Khan where the interpretation of the Markov troubles is in terms of time in terms of closed geodesics and modular surface in order to do the accounting so the number of Markov numbers up to t is about t squared it's very lackenry and you can ask questions about the infinitely many Markov primes that seems way too hard but we can show that almost all Markov numbers are composite all right I have I think three or four but I need to say something about where this is going I was going to say something and if people ask questions I'll come back and say something about the proof because the proof at least of this began and gambit paper result that we have uses a finite field and uses something beyond the Riemann but what is the general setting up promise that I must finish with that so really this is a very special this cubic equations very special it's very special cubic with many points of view even as a modular space and that's what allows us to study and as I said I like the idea of getting exotic examples and this fits into a completely general theory that if you look at the fundamental group of genus G with n-pages look at the representations to an area divided by conjugation that's called the character variety turns out that variety is always defined as Z like the case of the once punctured torus which gives exactly the mark of equation where the K this was asked earlier where the K is the trace of the boundary component of the punctured guy because the element the two by two matrix its trace for the puncture is the number K and the mapping class becomes a Markov group and takes integers to integers and all of this is true in general so you could ask is the action of the mapping class group on the integer points does that produce to financially many orbits and Jay Wang according Peter Wang in his Princeton thesis from a couple years ago made two big steps at least two big steps in direction firstly before you start to say that your variety is a threshold case like the case of three variables as I explained in some terms the right way to understand whether the variety is in the threshold is log club meaning it's just in the situation where integer points could go either way it's not finite by conjectures avoid but it's not rich yet because the number of variables and that's connected with taking the variety the affine variety it's compactification and looking at some properties which are log club we are in terms of the divisor at infinity and economical class at infinity and he proves that these relative varieties of log club which shows that we're in the same position and more striking theorem beyond that shows that you can do a full descent on the integer points I could fund any of your essentially shows that and unlike the case of model in code you can't do the reduction by at least we don't know how to do the reduction by some people starting with a point and say I'll find a simple point he uses differential geometry in non-linear harmonic maps to find using non-linear differential geometry which I've never seen so these more character varieties higher character varieties of a set of integral affine in varieties for which integral points are very rich and which are threshold and need to be understood there's much to be done to put it in the same category as understanding of the Markov all right the real question here is the mapping class group act on these varieties and the real points that work of Goldman on the complex point and Thurston and Goldman and on the periodic points I'm telling you this seems to be a very rich theory simple the other integral points seems to act at least the genus one n equals one case is true concerning finite orbits some work run out of time I just want to end with the following this is very important in dive and team geometry when you work over the integers you if there's something to be stable it should be true that whatever you do also works as integers or integers of the number and while some of what we do is to emphasize I want to emphasize immediately the decision and final of orbits fails and that's because when I increase the ring I'm looking at Markov equation z one over five over zero two I increase the ring I increase the group the group is a group of symmetries and so all of a sudden it becomes deficient compared to what the ring size is and so this feature is lacking and it's the thing we're working on hardest right now and unfortunately we haven't quite understood everything but that is needed before I would say that we have a theory of Markov equations over s integers I'll end with one problem that when you increase the s integers something fantastic happens and then we end with a question which is very similar to some of the three cubes and I've just challenged it out there and our techniques relate this in two by two cases and three by three cases so here's a simple question which I don't even know if it's true or false just like some of the three cubes you can say one way or the other way or argue you can change your mind daily three by three integer matrices it's known that to commutate the subgroup of s integers 3z is everything the group generated by commutators is everything so the first homology group is true but you could ask is every element a one commutator is it true that you can solve the following equation in three by three integer matrices x you give me b is it true that b can be written as x y x inverse y inverse there's no local obstruction that's a beautiful theorem there's no local obstruction in any finite quotient and every finite quotient is congruent so there's no local obstruction so is this true or false thank you and sorry for getting over time I hate people who would like to unmute and clap it's a great time