 that you got by email and on the big poster where I made an abstract, I wrote there would be either three lectures or four depending on audience's interest and on whether I succeeded in solving a problem related to the MiraQuintic by December 6th, which is today. So the audience's interest is, the number has started low and has shrunk, so it's not very great, but above all, I decided at 10.30 this morning to give up any hope of solving the problem by 2.30. So I have a partial solution which I had a month ago, that's why I announced this course of lectures that don't have a solution, but I haven't told you that problem anyway, so I will at the end I'll tell you how it connects up with what we've been doing, but I was unable to present the complete solution I hoped. So what I want to do instead for the first part is a continuation of what I tried to explain last time, which is for general culture, essentially every pure mathematician has heard people use the word motives. And most of them get very red in the face and angry, I had a friend who would get furious if he heard anyone use the word motive, he would say people had been talking about these things for 30 years, they don't even have a definition, they don't even have conjectures, it's just words, words. So a little like I feel about some of mathematical physics, no, I didn't say that. So it's of course not true, but I want to explain what the underlying idea is also in terms of the concrete exams we saw, I explained something last time, but I want to give you an analogy which is very useful. Consider the notion of a group, nobody complains that it's not well defined, but we don't know what it is that it's not useful. So we know a group is a set with a binary operation closed under multiplication, inversion and so on. But when groups were first discovered there was no axiomatic description. Groups were a list of groups, Abel found some groups which happened to be a billion, that's why they're called, by looking at points and elliptic curve and discovering you could add them, and Galois found that you could do something composing permutations of roots of an equation. They were isolated examples, but a lot of group theory was developed in the 19th century without a definition initially. And the idea was you know what a group is when you have it. So for instance, you have an object, like an icosheedron, I forgot to bring for an undismodel, you can rotate in a certain number of ways and those things, if you compose two rotations that preserve the symmetry, it still does. That's not an axiomatic definition, but you say now we have a group. And so if you wanted an actual definition, you could do it in the 19th century for a finite group. You could say, if I have some finite set and I have some symmetries of it or some permutations which are closed under product, I call that a group. And we know today that that's a complete and correct definition of finite groups because every finite group can be realized as permutations of a set, for instance of itself. So every group embeds, you can think of every group as a subset of SN. So all you have to do is explain about permutations and then say, if I have a subset of permutations that's closed under product, I call that a group. But that's a terrible definition because the same group, the group A5, of course it sits in S5, but it also sits in S6, at something else, in S20, at some of the 20 vertices of the dodecahedron and so on, it sits in many SNs. We want to think of an abstract group. So the idea, even if you didn't know the axiomatic way, is if you have two subsets of two symmetric groups, but there's a bijection between which preserves the operator, you say it's the same group. So at least a group now becomes an abstract thing which is sitting in many different places. And even if you don't have an intrinsic axiomatic description, you can work with that and you actually lose nothing except clarity of thinking. So motive is like that. If you have, as I started to explain last time, a variety, for us, let's say, smooth projective variety, so I don't want to get into any of the technicalities of mixed motives and so on, a smooth projective variety. So defined over Q. So that means it's given by equations which are polynomials. It's a lot of variables, a lot of equations. They're homogeneous equations because I'm in projective space and they have rational coefficients or integer coefficients if I multiply by the denominator. Then, as I explained last time, well, there are several things you can do. So one thing that I talked about last time is there's an L function called the Hasse-Wey, or the Hasse-Zehde function. Hasse-Wey, not vial. And I wrote the definition last time, I'll write it again. Z X of S is defined if the real part of S is large by the product, sorry, by the exponential of a sum which is of course the same as the product, the exponentials over all primes and all integers greater than or equal to one, so it's actually p to the n, it's a prime power. And then we take the number of solutions of the equation over the finite field of p to the n elements. Every finite field has p to the n elements for unique prime p and integer n. And you take the, so you just write down your equation and count the number of solutions where all the coordinates now are not rational numbers or complex numbers, but numbers multiple p. You count it, it's finite, because they're only finite in many variables. Each one is p to the n. You take that, you divide by n and you multiply by p to the minus ns. This converges for S large and that's the Hasse-Wey-Zehde function. And then as I explained, we know, these are very, very deep theorems, that if the dimension of X is d, that this thing will factor as the product of L series indexed L0 up to L2d with the even ones, even index ones occurring in the numerator and the odd index ones occurring in the denominator. And each Lj of S, these are now deep theorems that I talked about last time, so I'll just remind you very briefly. Each Lj of S has an order product of the form product, so it's equal to the product over all primes of some polynomial in p to the minus S. Of course, the polynomial depends on X, I might write it. And pp of X over t is a polynomial with integer coefficients, which starts with 1 and it ends with a coefficient, which is again an integer, which we always know, the top coefficient is easy. And it's got a fixed degree. The degree is the same for all primes. Well, for a finite number of primes, this question mark might be zero. The degree drops, but for almost all primes, it's the same degree. And bj, that's one of the deep insights going back to Angrévet, that bj is the jth-beddy number of the varieties. So if you take your variety and don't think that it's having rational coefficients, but simply complex coefficients, and you look at the complex variety it defines, I mentioned 2d as a real variety, so j indeed goes from 0 to 2d. And then you have the homology groups, like for elliptic curve, the bedding numbers that I mentioned are 1, 2, and 1 for curve genus G, 1, 2, G, and 1. So these bedding numbers are something we know very well. And one of the deep insights is that the geometry of the variety over the complex numbers, but then thought of as a real variety with ordinary homology, that somehow knows about the point counting. This is completely amazing. Now you might say, how do we know that it's the same? Because there could be cancellation. How do I know that, you know, so I could add something. I haven't told you what L2 is in L3. And so I could multiply L2 by something in L3, it could be another representation. And then that would change the degrees. But in fact, another statement which is known, but that was the Riemann hypothesis of Veyer, the Vey conjecture that Delinio got the Fields Medal for, is that if I factor this as the product of roots, new from 1 up to the Bedouin, so let's call the roots alpha, well they depend of course on P. So I mean, by Gauss we know that every polynomial factors into roots, then the Riemann hypothesis of the Vey conjecture, which is Delinio's theorem, says that all of these roots have the absolute value P to the minus J over 2. Oh sorry, P, this is J, well P also of course, and J, so this also had to J somewhere. And it of course depends on the variety X, it depends on everything, but here each of these roots has the same absolute value. So there can be no cancellation because no roots, so you put the roots where they should be. And then there are also other statements that are completely conjectural, LJ of XS has an analytic continuation. Well sometimes meromorphic, but let's say analytic. So in S, for all S and C, not just S very large, where this is provable, that's known for a few cases, but for instance as I said last time, Wiles's huge discovery that in particular included the proof of Fermat's last theorem, the main thing you have to prove is that this was true for the simplest non-trivial case, which is the variety, which could of any dimension, is dimension 1, so it's a curve. Then the simplest case of the simplest case, that curve is dimension, has genus 1. Then it's an elliptic curve, the bedding numbers as I said are 1, 2 and 1, so already the case of L1, the first non-trivial case of an elliptic curve, is already Wiles's theorem. So then I put exclamation to it's true, but in general it's totally contractual, even for L1 of a curve of genus 2. And then if you want to get even wilder, we also conjecture and that has been checked, invariably, lots of case that these all satisfy the Riemann hypothesis, but as I explained last time, even in the trivial looking case when the variety is a point, just by, so the point counting you always get 1, this number is always 1, then you compute this immediately for a point, it's just the Riemann's data function, and as you all know, the Riemann hypothesis is not known, even for the original Riemann's data function. Okay, so that's one point of view. Now, I haven't yet said anything about motives, which is what I was trying to explain, but if the homology of X has some natural piece, so for instance, you might have, for instance, you take the Jacobian of a curve of degree 2, it's going to be on the surface, it's H1 is degree 4, but if that Jacobian is isogenous to the product of two elliptic curves, then the homology will split into two pieces, geometrically, and each of these pieces is natural, they're geometrically defined, so natural, let's say, means it comes from geometry, there are actually three definitions of natural and the huge medicinjectures that they all give the same answer, but for the moment, it's, for instance, the example I said, it takes an even a stupid example, let's say my variety is the product of two varieties, then the homology will split up, I couldn't have it into pieces, and those sub-pieces, so then each one will give, and this you can really do and show that it's true, there'll be an L series for each of these, and the whole one will be the product of the pieces, because the whole homology is the sum of these sub-pieces, and then the L series splits. Okay, so now the point is you do this, you count points on some variety, and you count points on some other variety, and then you find miracle, we have the same one, and since that's the main thing I'll do, let me give the key example right now, so example before I go on, there's this, the MiraQuintic, which was the beginning of the whole field of mirror symmetry, famous paper of Candelis and his three co-authors that sort of started the ball rolling, well introduced the idea of mirror symmetry, but I'll be very specific, so we start with a family which is called the Dwork Quintic Pencil, so psi is a variable, but for the moment think of it as a rational number, and then I have a variable, the associated Quintic, which is the set, it's going to be a three-fold, so it's going to sit in P4 over C originally, so it's given by, well in P4 you have five coordinates, and the equation is that the sum of the fifth powers is psi, this fixed rational number that we've picked, times the product of the Xi, and because it's the same five, it's homogeneous, so it makes sense, and this is a Kalabiyal three-fold, which is what the string theorists need for what they want to do, but it's a family because psi varies, so psi could also vary in C, and the statement of mirror symmetry is there's another family called the dual Quintic, which I won't write down, in this case dual family, there are two families, each one of them, like this one, would have an associated Picard-Fuchs differential equation, as you vary the parameter psi, and you look at the periods, so this is a Kalabiyal three-fold, that means that there exists a unique, up to a scalar, holomorphic three-form, and there's a nice cycle, and so you can integrate some cycle here, you can write it down, and you get a number, which depends on psi, and you can show here that it's given by a hypergeometric equation, I mean, well, it's given if psi's large enough, it's the famous... So this is a multinomial coefficient, 5n factorial over n factorial to the fifth, and then it's psi divided by 5 to the power minus 5n, if I remember correctly. So it's a hypergeometric function. But in general, mirror symmetry, you have two families of Kalabiyals, and one of them, it's called the A side and the B side in physics, one of them you look at the associated Picard-Fuchs differential equation, which tells you how the periods move as a function of the parameter, the periods are like this, integrals of a differential form. But the other side, you look at what's called the Gromov-Witton invariance, you look at things of rational curves or curves into the thing with given homology classes, and you use those numbers of embeddings to make a new generating function which also satisfies the differential equation, and after some change of variables, it's supposed to be the same differential equation. So it's a long, long story, I'm not going to talk about mirror symmetry, but this was the key example that started at the mirror family is known, I mean, the two families. But here, I want to show the... So if psi is generic, if psi is anything except 5, then, well, this thing is then non-singular, so it's singular if psi is 5, it's non-singular, and I'm interested in the middle homology of this thing, but its dimension, the third Betty number, is 204. That means that the polynomial I'm talking about, I'm not going to keep writing the variety, Px, so there are several pieces, but the Ps for L3 of Xs will therefore be the product, one over some polynomial of P to the minus s. This polynomial will start with 1, and it'll end, I think, so it'll be T to the power 204, and this is half of, I think, 306, if I remember correctly, because remember, the absolute values of the factors all should be P to the minus 3 halves. I should get 3 halves of 204, which is 306, so this doesn't look very promising, but actually it factors, and if I remember it correctly, it factors, so this was done also by Candela Set, I might get this wrong, it doesn't matter at all for what I'm saying, it factors, there are three factors, each of which has degree 4, but two of them have multiplicity 30, they come many times, and for the dual quint, it's even nicer, so for the dual quintic, for the dual family, you can also write the corresponding thing, and there, if I write the full L function, then you'll have in the denominator, this would correspond to j equals 0, 2, 4, and 6, so it's in the numerator and the L series, but in the denominator and the Euler factor, they would just be trivial factors correspond to Riemann's data functions, except that this one is to the power 1, and this to the power 101, this to the power 101, and this to the power 1, so the total is 204, that's the mirror symmetry, there's a duality, and here the numerator would only have one factor, R1 of t, where it's for the original one, the numerator would have several factors in the denominator in that case, in the original case, the numerator would be this, and the denominator would be 1 minus t, 1 minus pp to the first, the details don't matter, these are known theorems that the point counting, although the Betty number is huge, it's 204 dimensional, you pick out in particular one piece, which is the interesting one that occurs in both, which is a polynomial of degree t, and this R, so I can just call it R psi of t, it has the form 1 minus some integer, plus another integer, which if I remember correctly, is all this to be possible by p, minus I think p cubed a pt, plus p to the sixth, t to the fourth, again it has to be, this has to be 3 halves times that, because the absolute value, so a p and b pair are two numbers that you can compute on the computer, they are big tables, and of course they all depend on psi, everything depends on psi, but these are integers, and we also know some bound on their size, so you can make tables of them, and we conjecture that this thing also corresponds to a multidare object, but not a usual multidare form, but that's a very wild conjecture, because not a single value of psi has been found from any size, this has been calculated by several people, Candelis is one, but Fernando Viegas is another, right here, so we know lots of examples, but in no example have somebody found a Ziegel multidare form that could be calculated, and that for small primes gives the same factors. We believe it always exists, but it's very, very hard to compute Ziegel multiforms of degree 2 of high level, and so there are no examples, but if psi is 5, then this thing becomes singular, and now what happens is that this thing factors. First of all, the degree drops, it's no longer 4, it's only 3, but it factors as a linear factor, which I can even tell you, it's the same Legendre symbol that we already had early in this series of lectures, 1 minus p over 5 times pt, times a quadratic factor, and when I talked about these things, yes, two days ago, I told you that whenever there's a quadratic factor, that's part of the general conjectures, it should be the coefficients coming from a multidare form, and here that's actually a theorem, so this is a true factor, this when psi is 5, for the singular phi, but the degree drops from 4 to 3, because of the so-called vanishing cycling homologies, so the effective homologies, 3-dimension instead of 4-dimensional, but this piece drops its dimension, and this AP, you have a completely explicit multidare form, I'll call it f25 of tau, because the important thing about it is that it's a multidare form of weight 4 on the group called gamma 0 of 25, but just so that you actually see it, I'll actually write down what the function is, what this multidare form is, that I don't remember by heart, it's 8 of 5 tau to the 10th, I won't repeat 8, many times in these lectures, anyway it's the simple product expansion, so 8 of 5 tau to the 10th over 8 of 25 tau, plus 5 times 8 of tau squared, times 8 of 5 tau to the 4th, plus 8 of 25 tau squared, so explicitly this starts q plus q squared, plus 7q cubed, minus 7q to the 4th and so on. Okay, and this is actually a theorem, this was discovered first experimentally by Chad Shudden, but as I mentioned last time, it's a theorem due to Delinia, that for every multidare form that is a Hecke form, like this one is, so that means these coefficients are multiplicative, so without even computing, I know that the coefficient of q to the 6 will be 7, because it has to be the coefficient of q squared times q cubed, Delinia showed that any multidare form with that multiplicative property is geometric, its L function will show up in the commodity of a certain variety called the Cougar-Salter variety that you can write down, and so we know that this is geometric, and this one is already geometric, it comes from the commodity of the Quintic, and that means in the fancy language both of them associated Galois representations, and you check on the computer that AP agrees for the first 200 primes, and then there's a wonderful, very deep theorem of Feltings, that if two Galois representations are the same weight if enough coefficients traces of Frobenius they're called agree, some fixed finite number that's effective, then they actually agree, and that number was made actually effective not by Feltings, but by Sarah, and other people improved it, and so Chad Schoen knew that if you check this up to P equals 100, that actually proved it was true for all P, and you can check much further, you can check hundreds of values, so this is actually a theorem, so that's like what I said about groups, you don't yet know what the motive is because I haven't really told you, but we know that in this Cougar-Sauder variety, where Delinio showed that there's a piece of the homology, whose L-Series will be the L-Series of this cusp form, that's a motive because it's a piece of homology, but in this 204-dimensional Betty, a homology group which drops to 203 because we're at the singular point, there's also a two-dimensional piece, but that is the same piece, in the sense that the L-Series is the same, and that's what a motive is, so that motive is an intrinsic motive, you can think of it as a piece of the homology of the Quendic, or you can think of it as the motive associated to multi-reform, but tomorrow it may show up in some other variety, it's itself, it's like an abstract group, it can be embedded in lots of different things, and that's what a motive is, it's something that occurs, so it wouldn't be interesting if the same motive didn't occur in different homology groups, then we wouldn't have the notion, but the whole point is actually now a lot of progress, there are now formal definitions, subclasses like mixed-tape motives are completely rigorously defined, several others, there are two or three competing definitions for what a motive should be, and they're all legitimate definitions, and there are some partial proofs that they agree, but they're not yet known to have all of the properties that we want of the category of motives, and so people haven't yet officially decided, that's what a motive is, because to say it's the right definition, it's a question of, it's a sociological question, do we decide that's the right one, only when we can prove the main structural theorems that we want, we say now we know it's the right definition, so it's not that there are no definitions, but the feeling, it's more than a feeling, if you find pieces of two varieties, now, so you see that this, so here I was talking about this, so this is at the level of, well, L functions, but then another way is at the level of topology, so the homology groups, but then you don't want just, and that's what you see, but what you actually want is the actual topology, or algebraic geometry of these things, in other words the conjecture, well first not a conjecture, let's have some variety X, and I have another variety Y, and that's what it's called a correspondence between them, so I have some other variety that maps to both of them, then, if I have a piece of the conmolegy here, and if I have a piece of the conmolegy here, some subspace of H3 of X, and some subspace, I can pull them back, and that might be the same piece, geometrically, they might coincide, and when I say that it's not just at the topological level, there's also something called a laticur et al conmolegy, and on that the Galois group acts and has its eigenvalues of Frobenius, so if you have a piece here, and a piece here, and they pull back to the same thing, then since it's the same here, the traces of Frobenius, which are these AP, have to agree, and now we know why that piece agreed, so if you have this geometric map, then automatically, by using the theorem, the theory, which is a huge theory, you get the coincidence of the L functions, but the conjecture, so if I make a map from motifs I'll start the same picture, but I'll do it again, with a map to L functions, but the conjecture is that this is essentially injective. In other words, if you have twice two different things, and you find the same L series, they must actually be geometrically the same object, which means you will find such a correspondence. The corresponding piece of the conmolegy here and here are not just random piece that when you calculate the traces, they give you the same L function, but they really are the same piece, it's not the same variety, so there's a series of maps up and down, but you can always have just one big map. So you can just think there's something which maps to both and explains, and so the injectivity is completely not known, but the map is certainly known. So if you have the geometric thing, you get L functions, and they have all of these properties, some of which are conjectural, but the analytic continuation and the Riemann hypothesis, but many of them are known. But we have a completely different way, which you go on to the complex numbers, that was going to be this thing, so I go to C. This is what's called the period map. And so the idea is very easy, so I wrote a long-exposed story article with Konsevich 20 years ago called Periods, where these things are explained with lots of examples, and a period, by definition, is simply the integral of some kind of a cycle in some variety of some differential form. This is a differential form, but on the variety, so it's defined in the differential form, defined over Q or over Q bar. So that makes a countable set, because there are only countably many algebraic numbers and rational functions. So for instance, well, every algebraic number is a period, but for instance Pi is a period, because you can write Pi as the integral, minus infinity to infinity is a closed path in P1 of R, or P1 of C, and then you take dx over 1 plus x squared, and so that's all it means. A period is simply a number that can be given by an integral, but of course I have to say with rational coefficients, because otherwise, if I have a number that we think is not a period, I can obviously write it as, you know, if I just integrate the constant function E from 0 to 1, of course I get E, so I need that the integral has rational coefficients, otherwise I'm beginning every number. So to get a countable set, these are called periods, and there's a metaconjecture that we actually make there is an actual conjecture, although I usually don't use the word conjecture in a paper unless I'm willing to put on my right hand and you cut it off if it's wrong, I should be really sure, but we do say some words, so I'll say conjecture one is preceded by the following sentence, a widely held belief based on the judicious combination of experience, analogy, and wishful thinking is the following. In other words, nobody has any idea if it's true, but it might be true, and that conjecture will say, so this is the conjecture on that this map is also essentially injective in the same way. So the statement is this, we have rules of calculus, so if I have such an such an integral, then I can transform it in another integral, so let me just do it in one variable, remind you this is first year undergraduate, the integral of f of x dx from a to b plus the integral from b to c is equal to the integral from a to c, and that generalizes to higher dimension, if you split up the domain of integration into two disjoint pieces, then it's additive. Secondly, you'll see they're all trivial, if I integrate from a to b f of x plus g of x dx, or also lambda f of x plus g of x, then of course it's the sum. So it's additive in the integral, completely trivial. Then if I integrate from g of a to g of b, some function f of y dy, then if I write y as g of x, this will become the integral from a to b f of g of x g prime of x dx, so that's called change of variables in first year calculus, and the last integral is that if you integrate a derivative, so this is the fundamental theorem of calculus, so the Newton-Leibnitz formulae, you get f of b minus f of a. All of these generalize to higher dimension, so if I integrate an n form over an n cycle, then if I add two forms, I add the integrals, if I add two cycles, or we can, not cycles, you know cycles, the chains, so they can have boundaries, they add, this says that you have pulled back, pulled back formulae, and this is the Stokes formulae, that the integral of a derivative over a manifold is the same as the integral of the original form of the boundary. And the conjecture is that those are the only rules, so the wishful thinking conjecture, written explicitly in our papers, conjecture one, but as I say, we say we don't really, any idea if it's true is if, so if two periods are equal, then always for a trivial reason. A trivial, but that sounds rude, so I'll call it geometric or trivial reason. In other words, if, so you, I gave you one example here for pi, let me give you a slightly more non-trivial example, z of three famous number is the triple integral zero less than x less than y less than z less than one of dx dy dz, it's very easy to show this, but the point is it's true and therefore shows you explicitly that z of three is a period, there it is. Now let's say that you're very smart and you go home and you think of some completely different, maybe a 17-dimensional integral, and you prove it's also z of three, or you don't prove it, you calculate it to 100 digits on the computer, you're sure it's z of three. Then the conjecture says if that integral is really equal to this, then the reason is you can get from this integral to that one at the level of the integrals, not just to the numbers, by just applying these four rules, just by the rules of calculus. Now nobody knows if that's true, there are no counter examples, obviously if it's a conjecture, but there are examples connected with the Mahler measure, Fernando is one of the world experts on this, where we have explicit conjecture that two numbers should be equal, each one is a period, that they're each a Mahler measure or a NEL series of things, but we don't know, but we believe that we will be able to prove and many such theorems have been proved. So what that tells you in a fancy language is that the map from a motive to periods, where again we take the homology groups. So here you represent a homology class by some sub-manifold, let's say, or some cycle, sigma, and then you have point-ray duality that no, sorry, not duality, you have just duality. Comology acts on homology, and by Durham's theorem every homology class over C can be represented by an algebraic form, and integrating that form over cycle is the pairing. And then the period is the integral of omega over sigma, but we're now assuming that omega is not rational in the sense of being a rational homology class. So this is not a rational number, it's rational in the sense of being an algebraic form with rational coefficients. So this number is completely transcendent, like presumably Z of 3. But the idea is that this map should also be injective, that the only way, so this meta-conjecture, that if two periods agree, the actual integrals can be made to match up. That says roughly that you aren't losing anything when you go from homology and homology to periods. It's written very explicitly in the last section of the paper with Konciiewicz. So you have two, by the way, this type of conjecture is called Tate conjecture, so that's a special case, and this type is called Hodge conjecture. That's also a special case. So the meta-conjectures say that if you have two motives, meaning you have two things that you've found in real life and you compute, for instance, a period of one of them to 100 decibels and you compute a period for the other to 100 decibels and they're equal, then you're sure that there is actually a geometric correspondence between your two varieties, called them X and Y. There is a Z which maps to X and Y. See here again, if I have a sigma here and an omega here and an omega prime here, then I can integrate the pullback of pi star of omega over sigma and this, forget it, you can push up, push forward and pull back and push forward. If you have maps, an actual correspondence, then of course the integral by the change of variables that are minded can be pulled back to an integral in Z and the conjecture that if two periods are equal, they can be shown to be equal by enough changes of variables means exactly that in some Z they simply become identified. So it tells you that you can recognize motives in two completely different ways. All of this is conjecturally general. If you find the same L function twice, as Chad Schoen found for the singular quintic and this particular cup's form that he had to find, of course, then you believe that there's a physical correspondence and if there is a physical correspondence then the periods will have to agree and you can check that on the computer. So in this case, since Chad Schoen found this, we know that the L functions do agree for this cup's form, so if I take my example I did the mirror quintic for 5 and this cup's form, this modular form of weight 4 and level 25, we know that the L functions agree, because as I said you check that for the first 200 primes and then by Falking and Ser you know that that's enough, it's really true. But then if the Tate type conjecture, that means that these are both actual motives this anyway because it's given by an equation and this by DeLinio, so there are pieces of true variety, there's nothing conjectural about them but we believe it's the same motive and that we don't know, all we know is that they had the same L function, but if this meta conjecture, generalized Tate conjecture is true, that all the experts believe then it is the same motive but that means there must be an actual correspondence, something that goes between the coca soda variety that I didn't really tell you about and this Q5 but if that's true, then every integral of one becomes the integral of the other and therefore the periods, here there are also periods, just by integrating this one form this is the period I'm talking about so I'll be talking, well it's F and the derivatives of F at 5, at Psi plus 5 the periods should be equal to the periods of the mother form so mother form has periods actually at HECU form has two periods and actually it has four, two periods that have been known for 60 years and something called quasi-periods that was invented independently in 2015 by Francis Brown and Oxford and by me it can actually be something I was doing with Vasily Goldschiff in Bodin, but then it turned out that actually it was first we followed this from the general theory, abstractly but it had been more or less explicitly done by either 40 or 50 years ago and forgotten but anyway there's a thing called quasi-periods so just believe me if you have a HECU form there are four complex numbers associated, two periods and two quasi-periods and so the conjecture the prediction was that since the L-functs agreed there should be a geometric map we don't know that geometric map but if it exists then the periods have to coincide but here the periods you can compute on the computer and also on the other side you can compute what you actually do is you look at this hyper-geometric equation the hyper-geometric equation has order four so there are four solutions this is just one, there are three other independent ones and you can do those at the original point where I'm expanding psi equals infinity or t equals zero where t is five over psi or you can do it at the point psi equals five which is called the conifold point, the singular point and then there's a transition matrix when you have a differential equation you have four solutions at that point four here, if you follow a path you just take an interval let's say you go from one to the other then these four become linear combinations of those four you get a four by four matrix that means 16 complex numbers you can compute them all and you can make them real by choosing your base as well 16 real numbers you can compute to a hundred digits actually we computed them to a thousand digits the first row and the first column are known so seven of the 16 are known that leaves nine unknown ones but then the conjecture the prediction was that the four periods the two periods, the two quality periods of this specific model of form with weight four and level 25 should show up in that matrix and with this is joint work that we've talked about here in ICT last year with Albrecht Klemm and Emmanuel Scheidecker and we've checked that to very high precision for the mirror quintic and actually the mirror quintic is one of only 14 families that were identified by no I think the families were known but there are 14 families where you have a family of Calabi al-3 volts as here and where the associated Picard-Futz differential equation is hypergeometric and Chad Schoen found for the mirror quintic this model of form of weight four and level 25 but there should be one for each of the 14 families I think one or two were known that Fernando found them all so we have a list of 14 families of Calabi al-3 volts 14 explicit cuss form multiple forms all of weight four smallest level is eight then after a while you have the 25 and the biggest one is 864 and in this joint work we've succeeded in finding the two periods and the two quasi periods for all of them the 864 we only got six weeks ago it's tremendous computations because you have spaces of dimensions many thousands it's quite tricky so in other words to repeat the philosophy we have two different ways of recognizing a motive how do you even know that you think that you have a motive in common you have two completely different varieties I gave one example last time it was Ron Livner's example I can remind you what it was in projective nine space you have ten variables you have the equation sum xi is zero and the sum xi cubed is zero so that's a cubic hyper-surface in P8 so it's a seven-dimensional variety and he counted the points and they correspond to some other form of weight instead of ten so when you do that then you know that the same motive sits in both you don't see it by looking at the varieties you don't see a connection between this and some other multiple form called f10 because it is level 10 instead of level 25 and the only reason you know that you expect them to be the same is you compute the L series you count something, count points and you get the same numbers but then you expect a physical correspondence so here it should mean multiple forms, fi of tau of some weight which satisfies this identity that the sum of the equations is zero, the sum of the cubes is zero and when you compute the integral in multiple forms you will find the right integral but as far as I know in this example it's never been done but the thing has a very strong predictive power because if the L functions agree you believe there's a physical correspondence if there were then the periods would agree and the periods you can compute on the computer to under digits and they do agree and so you're sure that everything is true now if it were the other way if we could prove that the periods agree which we can't but we couldn't prove that the L functions agree we could make the prediction the other way because the periods can be shown to be agree or even if you can't show but on the computer they agree to 200 digits then we're sure that they do agree then by this we're sure that there is something then we would have known we should look at the L function and we will see the same L function occurring here it didn't happen historically but if you assume both conjectures you can go from equality of periods conjectually to equality of or geometric correspondence between the varieties motives and then that implies that the L functions agree so you use the Hodges conjecture to go up and then you go down or if you have the L function occurring at two like here then the type conjectures tells you the motives should be geometrically related and the periods should agree and indeed they do okay so that's the end of my mini course on motives but I hope you'll remember some of that so when you hear the words you say okay it may not be a finished theory but it's not nonsense it's really saying something it makes very concrete predictions that you can often check numerically and in some cases prove now in particular however in this case as I said Chad Shun that's like 1980 or something I could look up the date but it's at least 30 years ago maybe 40 so it's been known for a long time I've known this example for ages all the experts in the field know it and so the L functions agree and so we assume that there's a geometric correspondence and then it was pointed out actually first by Vasily Golachev thinking of the usual periods since we expect that you should find the numerical periods of the cuss form this modern form weight 25 in that transition matrix and we found them and then a couple of months later I remembered that I'd found these quasi periods I didn't know yet that Brown also had them and I said well then the quasi period should also be there and we managed to compute them and there they were but now that leaves the question can one in this very concrete example so I'm now talking about the singular to work quintic so it's the same equation in p4c but now with 5 this particular one should be related to this cuss form we know morally we're sure it's related because the L series agree that's proved it was first numerical the first 200 coefficient then because it felt sincere the whole L series agrees and the periods agree that's not proved but it's true to 200 digits so we're pretty sure that they're really the same numbers all four of them occurred so there must be a map and so the question is can we make a map but it's not necessarily a map directly from multiple forms to here but it means something like a multiple parameterization of this just like you do for an elliptic curve where you have an elliptic curve y squared is x cubed plus 7x plus 5 and then the term the arm of a conjecture which is now Wiles's theorem says you can actually parameterize by multiple functions or multiple forms that satisfy that equation identically so you can realize the geometric thing by actual multiple objects so number of theoretical objects so here we can ask the same so now this is the idea that I had six weeks ago how to do that I'd often discussed with my co-authors Simon Schoidecker whether we could do it in particular of the MiraQuintic example because it's the most important it's kind of the central example of these 14 and it looked fairly hopeless but then so here was my thinking process and then I thought now this gives me an attack I can actually try to construct some multiple forms that satisfy this equation so otherwise you can just wish for them but how do you go about finding you can't just look at tables of multiple forms and then pick out five different multiple forms that satisfy something like this so my idea was this equation has one very striking property which is that the group S5 acts on it right you can permute the i's and the sum of the fifth powers and the product doesn't change and actually there's another group which is you can multiply each x i by an arbitrary fifth a root of unity so you can take mu five which is a group four to five the fifth root by arbitrary you don't change that but here on the product you will change by fifth root of unity so it's only mu five to the fifth where the product is equal to one so that's a group of order five to the fourth but actually because it's projective it's up to a common root of unity that would be the same point but it's still order 125 and that's five factorial so 120 times 125 who could do that form it's 15,000 so here we have a symmetry group of order 15,000 that should help us a lot so I thought where could I find a symmetry group well here you have S5 and I knew like most people know that S5 well not quite S5 but A5 is the same as PSL2 or sorry SL2 I've talked about this in earlier courses which is the same as the group SL2z modulo gamma five is this group of order 60 so I thought since I'm looking for multiple forms why don't I look for multiple forms of on gamma five not just any old multiple forms but of gamma five five of them but which are permuted by S5 exactly the way that S5 acts on five things okay and then I can hope that those five will satisfy this equation so then I started looking at these articles that I spoke of last week by John Bayes and Oliver Nash that explained in detail how you construct the acosahedron remember I had lots of pictures and we had a model so we constructed it and you could identify the two-sphere in which this sits with p1 of c and then the vertices and the edges and the faces of which there are 12 and 30 and 20 were given by explicit polynomials that I wrote up before so the first one is x to the 11th y minus plus it doesn't that's a question of convention x to the sixth y to the sixth minus but this one has to be opposite of that one that's not convention and the other two I won't write again so there were three explicit polynomials homogenous polynomials of degree 12, 30 and 20 which if you think of the roots of the polynomials lines in two-spaces that's points in p1 then under spherical rapid projection those are the vertices the centers of the edges and the midpoints of the faces of the acosahedron so but they also remember to satisfy wonderful equation which is that e cubed which is degree 60 because e sorry e squared minus v cubed I might get a sign wrong sorry minus f cubed so f is degree 20 so f cubed is the same degree 60 and then it was 1728 v to the fifth where this also is degree 60 because v is degree 12 right so we had this wonderful equation going back to Klein that says that but that reminds one very strongly of the fact which I also explained in the second lecture that if you take ordinary multiple forms well now I got a sign wrong so I have to if I'm honest then I have to do it correctly so this I should have changed the sign of something to make the analogy nicer if you take the cube of the Heisenstein series that I explained this is the one that starts 1 plus 240 times the sum n cubed q to the n or for 1 minus q to the n and this is very similar with 1 minus 104 times something similar with 3 replaced by 5 then you get 1728 times delta of tau these are modern forms e4 is weight 4 so this is weight 12 e6 is weight 6 so this is weight 12 and this is weight 12 all by itself and you have this identity so if you compare those two things you feel that the geometry of the icosahedron is right and remember we had the two Rob's Ramanutian functions which I called I think g1 and g2 these were modular functions so weight 0 on gamma 5 so that meant they're invariant under any tau goes to a tau plus b over c tau plus t where the matrix is congruent to the identity of modulo 5 so I won't write down the things again but this was also some theta series that I called theta 10 1 divided by the dedicated eta function and this is the same theta series with 3 instead of 10 so each of these the theta series of weight half, eta's weight half you had these and the action g1 and g2 jointly are jointly acted on by sl2z in other words if you take abcd in the full modulo group like minus 1 over tau or tau plus 1 then g1 of tau is not or g1 of gamma tau is not a multiple of g1 of tau it's not a modulo form on that group it's only on the subgroup gamma 5 the g1 of gamma tau is a combination of g1 and g2 so together we have sl2z individually with the subgroup gamma 5 and you can write down the action of the elements of sl2z on this but because gamma 5 acts trivially that action factors through the finite group which is a5 and so we have an action when you look at the formulas you find it's exactly the same action as the action here of psl2z on x and y so that means that if you take the polynomial v of either g1 and g2 it's nicer to work with the holomorphic one so if I let me just call this the theta1 and theta2 because I'm too lazy to keep writing it if I take those my three polynomials for the vertices the edges and the faces and I evaluate them at either the mass-reminution functions or the corresponding theta functions it doesn't matter because there's a common factor all these polynomials are homogenous that's just a power of 80 you can forget it but then if I do this well remember v was at degree 12 is a polynomial the theta is a degree a half so this is a multiform weight 6 this has weight 30 over 2 so it's weight 15 and this one is 20 over 2 so it's weight 10 well if you compare this and this we kind of expect unfortunately it's reversed we expect e to correspond to e6 of tau that is weight 6 15 minus 6 is 9 so maybe you could have 8 at the end of the 18th and indeed when you check that on the computer that's true and similarly v well f should correspond to e4 of tau but this is weight 10 this is weight 4 the difference 10 minus 4 is 6 8 is weight a half so it should be 12 which is in the ratio 2 to 3 just as this is that's good everything should be homogenous and then this thing has weight 6 and it should be simply 8 of tau to the 12 and it is and so if you take these 3 equations you find the Klein's equation the relation between these 3 and variance at the accuracy of the group f5 e squared minus f cubed is 1728 p to the 5th that's an identity if you remember these were huge polynomials with gigantic coefficients they satisfied that identity but here when you evaluate them at the Rotis Ramanutrient Functions you get this so what was my idea how to construct this map very simple we want to realize the group we want to realize the group that's actually the Icazzi-Hitkel group not as SL2 of f5 which I already have by this identification but as a5 so I want to see the Icazzi-Hitkel group physically as permuting 5 things by only even permutations and I explained that in the first lecture there were actually 2 ways but one of them of the 20 faces you picked 4 faces and I drew a picture on the board and you take their midpoints so these are 4 of the 20 faces and they're disjoint and you can do it in several ways you pick 4 disjoint triangles of your Icazzi-Hitren ok and then you take their midpoints and it's completely irregular they form a regular tetrahedron but I've only used 4 out of 20 faces so now I take the next face the adjacent one which is also adjacent to that but they don't have anything in common and I find another regular tetrahedron and I find in the end 5 tetrahedron by dividing up the 20 faces into 5 disjoint 4 tuples so I label each face with one of 5 colors and one of the colors is for instance red then they're exactly 4 red things and their vertices form a regular tetrahedron but then there are 5 blue ones and they form regular so I get 5 tetrahedron and of course the Icazzi-Hitren group permutes them but there are 2 different ways of doing that that was what I read in the books so what I did then is I just took this equation I labeled everything I worked out which ones form Icazzi-Hitren and I multiplied so I get a polynomial p1 of tau p1 of x and y which corresponds to the 4 tetrahedron in one of these things so I choose out of my 20 roots so I take this polynomial f which we had for the faces I factored on the computer in its algebraic numbers and then I take the 4 individual factors that I have but now I don't see where I wrote it, I write it and I found that if I just take those 4 so if I take all 20 faces and multiply I get this huge polynomial that I wrote up before and won't write again but now if I just take the 4 I get a polynomial degree only 14 only 4 and it's this polynomial so it's x to the fourth minus 1 minus the square root of minus 15 over 2 x cubed y then 3 minus the square root of minus 15 then over 2 x squared y squared plus 1 plus the square root of minus 15 over 2 x y cubed and then plus y to the fourth and of course then there's a second one as any number of theorists will see if you have one you have to have the other where square root of minus 15 goes to its negative the Galois conjugate and now indeed if you check that if you multiply j mod 5 and you take the first polynomial or the second and you get the same answer if you multiply then since this polynomial has 4 roots those 4 roots are the centers of my 4 vertices of the tetrahedron but if I multiply by 5th roots of unity it's the other choices so I get them all when I multiply them all I have to get the face polynomial indeed I checked that on the computer this probably had all been done before but I didn't know so this polynomial degree 20 factors over q of squared of minus 15 sorry it doesn't factor into 5 polynomials over that field only one polynomial factors and then the rest has degree 20 but it factors over the 5th roots of unity as well as squared of minus 15 in this way and of course the same is true for p2 so I can also make the polynomial if I multiply both of them p1 times p2 so that when now it's integer coefficient so I might as well write it out although it doesn't make any difference to anything I say you certainly shouldn't care so multilay typing transcription errors that's the polynomial and now if I multiply this one of degree 8 there are 5 I get 40 of course I'll get f squared so f squared honestly factors into 5 factors over q f doesn't but f squared does sorry doesn't factor over q because again only one of the factors has rational coefficients the others you multiply by 15 so this is not z to j it's 8 to 5 to the power j I just rotate my icosetron around one vertex so now that tells me exactly what to do now I take p of the two theta series that I had since p this p has weighed 8 this is weight 4 but then it turns out you're still allowed to divide by 8 of tau to the 4th no squared wait a second p is degree 8 8 over 2 is 4 I won't wait 2 so this should be 2 so I should divide by 8 of tau to the 4th this function is still holomorphic and so you just do it on the computer and it's you get something and it starts like this q plus q to the 7th sorry it actually doesn't because there are fractional powers it starts q to the 7 over 30 plus so let's call this f of tau over 30 because it's too much trouble to write all these numbers then f of tau has integer coefficient I mean integer exponents and that one starts q plus q to the 7th plus 7 q to the 13th plus 7 q to the 19th plus 11 q to the 39th and so on so it's got a power series so now if I haven't made any mistakes then automatically sorry I can't hear what no p is what I wrote but I unfortunately erased p is the product I had 2 polynomials I just erased p1 and p2 had coefficients in q of squared of minus 15 and one is conjugate to the other but I want rational coefficients so I multiplied p1 by p2 so this is this polynomial there is rational coefficients and it is the property that if I rotate by 5th you know by 72 degrees multiples to 72 degrees it doesn't matter if you do it for x or y if you multiply over j from 1 to 5 then this will be the phase polynomial squared which means that when I insert the Rob's Ramanujan functions if I put theta 1 and theta 2 we already saw that f of theta 1 and theta 2 is e4 so that probably will become e4 of tau so here you get this function so now I'm going to go through the rest because it's already 3.30 I'll just say maybe 5 minutes to finish we started I think 8 minutes late so this thing has to be a multiple form of weight 2 of course on something and ok that's 0.1 and 0.2 is it must satisfy that if I define fj of tau to be f of tau plus j over 5 but this thing is periodic because it's a power series in Q so if I set tau to tau replace tau by tau plus 1 it doesn't change so this only depends on j multiple of 5 with this definition and if I define f to be this thing then and I do this then it must satisfy my equation and indeed it does so I've solved partly my problem the problem was to get a geometric correspondence between the mirror-quintic for q5 which is this and a multidar object well I have it this is the multidar so I was very happy that was 5 weeks ago but I'm not done because my multidar object here is only a multidar curve sitting in here I mean the homogeneous thing since all the fj's have the same weight in projective space it's just a point on the multidar curve so it's on the curve x of 5 but that's only a curve and I cannot realize a third homology group in h3 in the homology curve I need a 3-fold or even a bigger variety but delinear represented realized the cuss form the L series of a mother form of weight k in a variety dimension k minus 1 here it's weight 4 it should be a 3-fold then my correspondence should be more or less a map like a rational map so I want a 3-fold and this is only a 1-fold but I felt I know exactly the delinear's theorem there are things called Jacobi forms which I've co-invented so I know them well and delinear's theorem tells you if you have a form of weight k then you should realize that it's called the cougassade variety but in terms of multidar forms it means Jacobi forms with one multidar variable I'm not going to define them and k minus 2 complex variables so here we need to lift this multidar form by inserting two further variables which are elliptic variables somehow elliptic in C1 and C2 and I thought it must be very easy and I already so in other words I want functions like this such that if I do the same thing phi j is phi of tau plus j but it should again have period 5 then the sum phi j to the 5th should be 5 times the product but now there are 3 variables and then I would have a 3-dimensional multidar object called it is called the cougassade variety X of 5 and the fiber is the square of the elliptic curve over that point and so if I could find such a multidar function with this identity then I would have my map then of course I'd still have to check that that period pulls back but that would just be a routine verification it's called the work so I'm sure I nearly had it and I bravely announced this series of lectures but I've been trying for 5 weeks I cannot lift this thing from a multidar function to a Jacobi things I found a lot of interesting things I found a function with 2 variables tau and another variable u which does satisfy the identity so now I have a 2-dimensional variety which is automorphic but this variety is still multidar but this is not at all an elliptic variable this is not a Jacobi form but it's called the Picard multidar form in this case for the group s u of 2 1 which corresponds to the 3-ball but I want to lift to a Picard even if it were a Picard I would need s u of 3 1 the 3-ball to have many variables and I can't do it I tried several things and all of them I could prove fail so every idea I've had and I had to believe me I had a whole bunch and then I found a way to vary this by adding a quasi-multidar form to it and I found 2 lines through each of my points so on the quintic the quintic has the property that certain points on it there are many many lines on it that's true for any kubesai and they actually form 2 families on any kubesai this is a discovery made by Turkish mathematician Mustatah and then improved and made a beautiful version by Kandeles van Rehmann van Straten and also Kandeles's wife and then I heard a lecture Kandeles gave and found an even better presentation relating it to the modular space of 5 points on the genus zero curve but there's a nice description that there are 2 families of lines on this and each one is indexed the base of the family is a curve this is really hallucinating a curve of genus and now I've actually forgotten as well as I think 626 but it's very close to that I knew it this morning and I've already forgotten so you have 2 different families of lines and I found lines through this so I have as well as my FJFTL I have a line on this and I do have a modular parameterization by quasi-multiforms but unfortunately the lines on this a family of lines parameterized by surface is too small it's only 2 dimensional and we're on a 3 fold so not every point is on the line you don't get the whole 3 fold that way and actually I realized this morning you're getting a rather boring sub 3 fold of this so the final answer is I do not have a geometric map yet that explains the 30 or 40 year old coincidence of the L function of the singular merequintics of the conifold point and the L function of the chat shown in multiform or the equally conjectural but verified by us numerically identity of the periods each of those separately are known one is proved one is proved to high accuracy each would be explained completely if you wrote down a geometric map and then there's nothing to prove once you find it you just check that it's correct that's the one I hoped to find and as I said I have a partial success but I want to end with one last thing because this function even though so this function has one nice property already told you that the sum of the f of tau plus j j mod 5 the 5th is 5 times the product of the same thing so that was the connection with the quintic but the other property is this any number of theorists in the audience will look at this say aha this is a multiform weight 2 it starts with q it is multiplicative coefficient believe me I checked the first thousand and therefore by the taniamov a conjecture which is a theorem it must be the elliptic curve the L function associated to the elliptic curve now it's by counting and so if that's true there must be a symmetry under tau goes up to sign under tau goes to minus 1 or for n tau for some n so I looked on the computer for every n until it worked I knew it had to work for 900 so therefore I knew that this had to be the L series for the elliptic curve of level 900 you look in paris you can do an elliptic curve search elliptic search for 900 there are 20 elliptic curves but only 8 that are non-isogen so there are 8 L series I computed them all with paris paris computed them all one of them was this and so and then again once you know the L series you know it's true so E is the following elliptic curve which I never seen before I have no idea if it's come up anywhere else in mathematics it has complex multiplication by q squared minus 3 so it's this curve y squared is x2 plus 4 fifths so the concrete theorem to just to end with something take the elliptic curve over q y squared is x2 plus 4 fifths then you count the number of points remember ap if p isn't different from 2 and 3 would be minus the sum x mod p of x cubed plus 4 fifths of the Legendre symbol so you count the number of points but you can do it completely for undergraduate mathematics it's the sum of the Legendre symbols then this will be the same as the ap defined by this very expansion that is provable and so that is this multiform so this elliptic curve and it shifts by integers but in this language you would have to shift by tau plus j over 5 because I multiplied it up by 30 actually 6j over 5 so if this function and shifts do satisfy the meruquintic so at least I have a multidar interpretation of part of the singular meruquintic but it's only a one-dimensional thing I have a two-dimensional thing using these bizarre Picard multidar forms and I'm still hoping to find a three-dimensional one using Jacobi forms but it hasn't yet worked so I can only report failure on that okay so thank you for those of you who are still here for still being here thank you and we can of course it's SU of 2 1 I wrote it I think I wrote it somebody maybe I removed here SU of 2 1 but the important thing is the symmetric space the symmetric space is the complex ball so SU of 1 1 happens to be the same as SL 2 it's one of these accidental isomorphisms so SL 2 acts on the upper half space also you can think of the upper half space as the unit disk but that generalized to the unit ball so the unit 2 ball is two complex numbers Z1 and Z2 with absolute value Z1 squared plus absolute value Z2 squared less than 1 then you forget that's the group that acts on that up to some renaming is called SU of 2 1 and so that group contains discrete subgroups which is essentially you take SU 2 1 which are matrices of a certain size of size 3 so you identify some identity but then you take SU of 2 1 not over C but over the ring of integers of Z of squared of minus 3 so you have to pick an imaginary quadratic field then this group will act on the 2 ball and on the 3 ball if I do SU of 3 1 and so what I have is a Picard one I don't want to write down the definition but I can actually write down the function because it's cute so one more thing about F I gave you two definitions of F one of them was as a polynomial this complicated polynomial the two broadest ramanutian functions divided by power of eta that's how I found F but then after rescaling it with inter-exponents I said to her this has multiple coefficients it comes from a elliptic curve it's this elliptic curve but since this elliptic curve is complex multiplication you can multiply X by cube root of unity it's a so-called grosson character L series which I talked about in my course in CSA this year and I'll write it down explicitly you sum over all alpha in the root of you just joined the cube root of unity to 1 that's the roots of the integers of Q of squared of minus 3 you take all alphas which are congruent to 1, multiple of 2 is squared of minus 3 every alpha that's prime to 12 by multiply by root of unity is you can make congruent to 1 an explicit character which depends you go from the ring of inches model of 5 which is cyclic well it's cyclic of order 24 subjectively to the 6th roots of unity and so you write down an explicit character and then you put alpha and here you put Q to the norm of alpha so this is kind of standard that's complex multiplication so you can do that but then to get the Picard form if I take a new function f of tau let's call it x and here I multiply by so I still have the Q to the norm of alpha Q is still in the e to the 2 pi tau but I multiply by a certain function which is a virus trust sigma function and here it's alpha times x so the virus trust sigma function sigma z is the so called virus trust sigma function now the certain elliptic curve which here is rational coefficients I forget exactly it's maybe 27 it's also called complex multiplication by Q squared of minus 3 but it's a curve I knew extremely well the function because in work I did again with Fernando I've mentioned him many times when we first met each other many years ago we did a lot of work on complex multiplication I talked about it in my course here and in CISA this year and we found wonderful things L functions to something and so we found that certain L functions had values and those values had a certain series of numbers and when I found these numbers now in the computer it took me a day to recognize them but then when I did it in the right coordinates I said I've seen those numbers before so the numbers start 1 1 minus 6 the next one is minus 552 so they're quite characteristic they're you know non-trivial enough that if you see them more than once you're sure it's not a an accident the next one is plus 18600 I mean I have the first as many as you want of course but these are they have many many interpretations in our paper we gave several different interpretations several different expanses of different multi-analyptic functions in which these same coefficients 1 1 minus 6 minus 52 18600 appeared and we showed that that sequence of coefficients is quasi-recursive so there's an elementary way to get those coefficients but anyway this is a well defined function sigma z and to my pleasure that was so what I tried here since I needed higher multi-reforms then if I expand in power series in the other variable I'll get linear combinations of alpha you know combinations of powers of alpha and maybe also alpha bar so I should get monomials in alpha to the i and alpha bar to the j if I look at those terms individually they're so called quasi-multi-reforms but I can look at a whole bunch of them and see I'm lucky and they satisfy my equation and I found that the only case that worked was just with polynomials in alpha and then if I started with alpha I could put alpha to the seventh times any common number but then the next one once I got to alpha to the 13th all the other coefficients had to be 0 this had to be a specific multiple of lambda squared which was very hard to recognize until I realized I should put in some factorials and then it was minus 6 and the next was minus 552 so just completely experiment that I discovered that if you ask that a power series in alpha x you just take a power series with unknown coefficients and now you ask that this new function has the same property that the sum when you shift tau by integers over 5 to the fifth that it's 5 times the product f of tau and x it's not very likely to happen but when you do it you find that you can make it happen even only if that power series has 25 coefficients which I recognized and then you can check but this kind of a function that's a mixture of a usual theta series q to a quadratic form and a virus stress sigma function I recognize that from my more recent work that I talked about here two months ago at the special EGAP lecture in honor of Boris Dubrovnik I gave a lecture and I had Picard multiple forms and it was the exact same function it was a slightly different chi but it was the same q to the n of alpha and then there were two functions just what I wanted now but it doesn't work here that was in connection with some non-integral nonlinear systems of differential equations integral systems very complicated but the function you had to take there was also a character but a different one it was the same form q to the n of alpha the same sigma but here there were two variables x and y just what I need here but here when I put in the second variable it does not satisfy the equation it just doesn't work because it has to match up with the linear theorem which realizes the motive in the world of Jacobi forms so it's all still a mystery but at least you see things are happening and that this motivic way of thinking leads you to expect certain miracles to happen and when you look miracles do happen not as many miracles as I needed but still I've written about five here that this thing is the comes from a elliptic curve that it comes from this that it generalized to Picard these are all kind of miraculous things there's no reason to just write them down and think this will happen but here it somehow had to happen because of this known connection between the group gamma 0 of 25 and the quintic and gamma 0 of 25 if you just change tau to 5 is the same as gamma 5 essentially so this is gamma 5 gamma 5 is the acosahedron the acosahedron vibrates the Ramanush and gives you multiple reforms and those multiple reforms solve our problem and it's only one third solved so I'm sorry maybe next year I can tell you the rest of the story