 I kind of partially surveyed, partially at the end, a little more advanced lectures on the Ecosahedron and connection with multiple forms and many other topics. So today I want to finish what I started on Wednesday, which is to describe the Ecosahedron. I'll repeat a little bit how that worked. So you remember, even if you weren't taking notes. And then I want to give a very brief introduction to modular forms, which are the kind of key players in this story and in a lot of number theory, and explain how the two are related. And I think that's as far as I'll get this week. And then next week, we'll see that the story actually goes much further. So remember, we had the, oh, I forgot to bring the model. Where's Fernando? He's there. Fernando gave me a nice model of an Ecosahedron being built this morning out of plastic pieces. And I had it on my notes and forgot to take it down. I carefully moved it. So now you have to imagine there's a beautiful red plastic I call the Ecosahedron. What? You can go get it. Well, you can go get it. It's just on my desk. And I didn't, wait, unless I locked. Should I give you a key? Yeah, I mean, you, what? Well, I can give you a reminder. Here's an extra one I want. It might actually be hers because she didn't enter to me a couple of days. So she might not. So I drew an Ecosahedron, but now I don't have to draw it because we'll have a model very soon. But roughly, you had the North Pole in the unit sphere, the South Pole. And here, two cycles, P1, P2, and so on. And here, Q1, Q2. So we have the North Pole, the points Pj, the points Qj, and the South Pole, where j is an integer model of 5. And there were two completely different descriptions, each very pretty. I'll remind you very briefly, but I'll be using one of them and not the other. So one of the descriptions was that we take the points. Remember, I was always going to use the notation phi for the golden ratio. And maybe I'll also use, I think I used that already, e to the 2 pi i over 5 for the standard fifth root of unity. So this is a rotation around 72 degrees. And you have the relations zeta plus zeta inverse is 1 over phi. So the inverse golden ratio and zeta squared plus zeta to the minus 2 is minus phi. So the Pj's, in one of the descriptions, we took the three points, well, I don't know which one I want to take now. In one of the descriptions, we used the numbers, we used the description of S3. So that's the unit sphere, S2, I mean, in R3. R3, we thought of a C cross R with coordinates w and t. So here we have the set of points where w squared, absolute value squared plus t squared is 1. And in those coordinates, so here's the w and t coordinates. So the north pole is 0, 1. The south pole is 0, minus 1. So that's minus n. The point Pj is the point 2 over the square root of 5 times the jth power of our fifth root of unity. And this number is 1 over the square root of 5. And this is the negative. Oh, great. So here, courtesy, doubly courtesy, first for building it and then for getting it. Thanks. Fernanda is an actual courtesy even built this morning by hand from little special pieces that you can make any nice pictures. So this will serve as inspiration for you and for me. And we have to remember that Fernanda gets it back at the end of the hour. So if you took this, as I said, this is a very easy calculation, if you think of the north and south pole, it's clear that wherever you put P1, you have to rotate by 90 degrees. So it has to be some number times powers of z to j. And here, it has to be the negative. These have to be the antipodal points, because the whole thing is symmetric under rotations through the center of the two-sphere. And therefore, the only question is how big t is. Because if this is some positive number t, then this radius have to be squared of 1 minus t squared. And if you solve the equation, it's very easy. You find 1 over square root of 5. So that's one description. And I'll give a slightly different description later in terms of a new parameter, z, and maybe I'll do that right away, with stereographic projection. But the other description, which we start with the points 0 plus or minus 1 and plus or minus the golden ratio. So then you can see immediately that you have a golden rectangle, because the ratio of the sides of this rectangle is 0, 1, and 5. And then you take all cyclic permutations. So x, y, z goes to y, z, x, and x, y. And so that gives you 12 points. And those are not the same 12 points. It's rotated in three-space. That's another model. And so in this model, we see this golden rectangle. And I explained last time that there's another golden rectangle, the three golden rectangles, if you put the 0 in the three possible places, lie somehow orthogonally. And there are five different ways of doing that. There are five different ways of, so this tells us that the set of vertices, so I'm repeating from last time, v is the set of vertices, e is the set of edges, f is the set of faces, the cardinalities, there are 12 vertices, 30 edges, and 20 here. And you see Euler's relation, 20 plus 12 minus 30 is 2, as it should be. And so what we can do is we can divide v as the four vertices of a regular of a golden rectangle union, the four vertices of another golden rectangle union another. So we take three disjoint decompositions of the 12 vertices. I think it's, I'm not sure, it's this pair and a pair here. So you have three disjoint collector vertices of a golden rectangle, but this can be done in five ways. And that means that the group, which I call the i, the icosahedral group, which is the rotation of the sphere which send the icosahedral to itself. This is a group of order 60, and it acts. It permutes them by i. And therefore i, and it permutes, it's easy to see it's injective. So i is a subgroup of s5. And since s5 is order 120, and i is order 60, it's a subgroup of order 2. There is only one, so it has to be abstractly a5. But the actual realization is realized by this. It acts by even permutations of these. So that was one of the two descriptions. And the other description was with a, so I gave you two descriptions, one with 0, 1, 5, and the cyclic permutations. Here you see the golden ratios. And this is a different description. And the other way you can see it, I drew the picture last time, is you pick, and I won't be able to do it correctly, but you pick three, four faces. There's a way to, so the faces are all triangles, last time I drew it. You can find four faces that are completely disjoint. So once you pick one of them, the others are fixed. Then the central points of those four things form the vertices of a regular tetrahedron. And again, you can do that, actually that you can do in 10 different ways. But if you take these four vertices and you rotate around one vertex, then this particular tetrahedron, you rotate around vertex, gives you five tetrahedron. That's not all the possibilities there are 10. The other one, it's like black and white squares on a checkerboard. But you can choose these five tetrahedron, and they have disjoint vertices. So I can also write this as V is the union of the vertices of a regular tetrahedron, so there are four of them, and then five times, five of these, and these are also, excuse me, there aren't five of them, there are three of them, but the ways of doing that, again, it's the same story, they get permuted by A5. OK, I told that last time, I don't especially want to say that, no, but I just wanted to remind you the picture, but what I did want to do now is the more convenient coordinate, so the unit sphere in S3 maps stereographically, so SP, stereographic projection. I remind you how that looks, you have the plane, here's the complex plane, you put the point, so this point, the plane is at height minus one, so the south pole of the sphere sits on this plane at the origin, and now you project from the north pole, so any point on the unit sphere goes to a unique point in C. And so this gives you a map from S2 to C, except there's also the point at infinity, because when you project N, it goes off to infinity, so this is the projective line. OK, so we have, and this is an isomorphism, so we can think of this, and that's even more convenient, so originally we would have had three coordinates, X, Y, under U, V, and T, and then these would be some horrible numbers involving cosine, it's any complex coordinates, it's already better, but it'll be even better if we, so here we have three coordinates, all real, here we have two coordinates, one real, one complex, now we have only one real coordinate, P1 of C, and so you can write down what the numbers are, so the north pole goes to infinity, the south pole goes to zero, this goes to Z to plus Z to inverse, which I just told you, is one over phi times Z to the fifth, and these go to Z to minus Z to inverse, sorry, Z to squared plus Z to the minus two, which I wrote here as minus phi times Z, sorry, Z to the phi to the J, so that's even more elegant description of the 12 points, in P1C we have a point zero, infinity, these five points, and these five points. Now, I'll come back to that in a moment when I talk about invariance, first I want to talk a little more about the acosahedral group, so I said last time that if you take one vertex, here v0, take an edge that contains and take its middle point, and take a face that contains that edge, and take its middle point, so the middle point of a face, and then the v is the rotation by 72 degrees, so 2 pi over 5, around v0, that preserves the tetrahedron around v0, e is the rotation by 180 degrees, so by pi, around e0, so that flips this side, and therefore flips the whole acosahedral, and it's the reflection in the line between e0 and the antipodal point, and f is the rotation by 120 degrees, 2 pi over 3, around f0, then you see that v to the fifth, and e squared, and f cubed are all one, but you also have one more relation, I may get the order wrong, because, well, it's certainly correct if I don't tell you which way I rotate, then you have one relation that if you rotate around e, then around f, and then around v, you get the identity map. So this is a presentation of the acosahedral group, and if you think of this identification that i is isomorphic to the a5, then v will correspond simply to the cyclic permutation of five numbers, that's an even permutation, because five is an odd number, and it certainly has order five, then e, you cannot take an involution, that would be an odd permutation, but five is big enough to have two involutions, so you interchange one and two, interchange three and four and fix five, and then f, I won't get the order right, but I think if I do it like this, it's something like two, four, five, and one and three. Anyway, it's a cyclic permutation of three of the five numbers, and you fix the other two. So we can also make all of these isomorphisms completely explicit in that way. Now I want to say a little more about this, and maybe before I come back to the key thing, which is the polynomials that have this root, I want to say a little more about this, and I'd like to mention because it's one of the most beautiful stories, even though I don't directly need it for my story, I want to mention the connections between the icosahedron and the E8 lattice. So the E8 lattice, like the icosahedron, is one of those absolutely unique objects in mathematics. It's the only lattice which has properties, like the properties it has. It's one of those very, very special things. So this is a lattice, which is a lattice in eight dimensional space. If you look at lattices in n dimensional space, you can ask how densely packed they are. So you can ask, for instance, this famous problem, if you put, well, in the plane, I could put a unit circle. It does not be a lattice. There could be any collection of points, but let's say a lattice. Then you can put a unit circle around each point and try to get them to be as big as possible, or the lattice as tight as possible, such that they're all disjoint. So that's the problem of densest packing. And for two dimensional space, the B is already solved. It's the honeycomb lattice. Each point has one, two, three, four, five, six neighbors, which all touch each other. And that's the unique solution. In dimension three, Kepler claimed 300 years ago that the solution was a particular lattice, which was only proved a few years ago using the computer very heavily. But in eight dimensions, E8 was conjectured for a long time to be the densest packing of any point. So if you have a lot of unit spheres, unit balls, in eight dimensional space, and you try to pack them as densely as possible, you can't beat E8. And that was proved only two years ago by Marina Vyazovskaya, who actually had done her doctorate with me. But it had nothing to do with what was the interdoctoral thesis. So we have the E8 lattice, which is this densest thing. So I'll tell you briefly how to get it, but it doesn't really matter. Here's an actual model. It's the set of points x1 up to x8 in R8. Well, all of the points should be integers, or half integers, in one half z. And I demand that either, well, I demand that they're all congruent to each other, model of one. So that means it's either eight integers, or else it's eight half integers. And also that x1, that the sum of all eight of them, which by this will be, since each xi is an integer, this will be an integer, but it should be even. So that's a famous model. And if you want to see it even more concretely and see why it's called E8, if we take in the standard base of R8, I take the point 0, 0, 0, 1 at the i-th place, minus 1 at the i-plus-first place. And here i goes from 1 to 6. And then E7 is the point 0, 0, 0, 0, 0. But instead of 1 minus 1, I take 1, 1 and 0. And for E8, I take the point minus a half, I think. All eight coordinates are minus a half. So these are all in the lattice. That's a z basis. So this is the sum z, Ej, j from 1 to 8. And then if you look at the scalar products, you find that Ei squared is always 2. So all of these vectors have total length 2, like 1 squared plus 1 squared is 2, a quarter 8 times is also 2. So they all have length 2. And Ei times Ej is either 0 or minus 1, which means that the angle is 90 degrees. So two of these lattice points are either orthogonal to each other. That means that the scalar product is 0. Or they're at an angle 120 degrees, which means that if I take the third one, this is a rotation. And if I draw a picture, when this happens to the minus 1, then I get the famous E8. So if I put E1, E2, E3, E4, E5, then E6 is down here. And then E7 and E8. Then you see the E8 diagram. And of course, this has to do with the roots of a famous 248 dimensional Lie algebra called E8. So that's the lattice. And no, I claim that this is intimately related in a couple of ways. And because it's important and such a nice bit of general culture, I thought I would tell it. And it's important in general. It's not really important for me here. So let me tell that story a little more. So for that, I remind you what the quaternions are. H is the Hamiltonian quaternions. So this is generated by the numbers, well, 1, i, j, and k. So I can think of it as a four-dimensional space over c with the famous relations that i squared, j squared, and k squared are all minus 1. But ij is minus ji. And that's k. Well, there's some duplication. And similarly, cyclically permuted. So this is the famous discovery that Hamilton made crossing a bridge in Ireland that he realized that you could allow non-commutative elements and then get a nice algebra of dimension four. So this is the famous thing. You can also, of course, write that as c plus cj. So if I write it this way, a typical quaternion would be a plus bi plus cj plus dk. And then it has a norm. Well, the norm squared is just the usual nucleotide norm, a squared plus b squared plus c squared plus d squared. So in other words, if I identify this with this space with r4, the norm of the quaternion is the usual norm. But if I do it as cj, then I would write q as some alpha plus beta times j. And then the norm would be alpha squared plus beta squared. And explicitly, I can embed. If I embed this into the 2 by 2 matrices, sending 1 to the identity, I don't have much choice, I to I. But I don't want the i times the identity. It shouldn't be proportional to 1. So I take minus i, then it squares minus 1. j goes to, I never remember the sign. I'll try minus 1, but it might be the other one. And k goes to 0i i 0. Then you see that this satisfies these things. And so then I can think of h also as 2 by 2 matrices. So c plus cj in this space would be alpha beta. And here beta bar, alpha bar, where alpha and beta are in c. So we can also identify with c2, sorry, with the unitary group, or rather with unitary matrices. And so in particular, that tells you that if you take h of norm 1, so this norm is multiplicative. That's the key point that was discovered actually long before Hamilton, if you multiply two quaternions using these non-commutative rules, the norms multiply. And in particular, then things of norm 1 form a subgroup. And of course, as a manifold, this is just s3. So s3 becomes a group. And this is very unusual. It's not true for s2, or s4. In fact, it's not true for any other sphere, except maybe s6 where there's, well, I think that's also no. It's not true for any other sphere. That you have a group structure. And if I think of it this way, then this group is just the special unitary group, SU2, which is the set of 2 by 2 matrices. So SU2 is the set of such matrices where alpha squared plus beta squared is 1, in other words, determinant 1. So the three sphere has a group structure, and the three sphere itself. But that's in four space. We have the two sphere. But now if I use the stereographic projection, you have a connection from that to a, so let me say it like this. h contains the elements h of trace equal to 0. So the trace matrix means that if q, well, the trace of the matrix reduce trace, it just means that there's no component of 1. So it means it's this subspace spanned by i, j, and k. Well, then if q is in the unit quaternion, then I can conjugate. Actually, if q is any non-zero quaternion, I can conjugate by q. And that will preserve the trace. It preserves the property of not involving 1. And so that means that this group acts on R3 also, because this thing is a copy of R3. And it again preserves the absolute values. So it in particular also acts on the sphere. And so that means we get a map from SU2 to SU3, which is orthogonal transmission. But it's a 2 to 1 map, because if you take minus q, then obviously conjugating gives the same thing. So we have a 2 to 1 map. And that means that we have two groups. So we have SU2, which is equal to the Hamiltonians of norm equal to 1. And it says as a topological manifold, it's the three sphere. And this maps 2 to 1 onto SU3. And in SU3, we have our subgroup, which I called i. And here we have a group which is often called 2 times i. This is the icosahedral group. This is called the binary icosahedral group. So it's just you pull back these elements, but each one is 2 inverse m, which differ just by sign. So this is order 120. And one now also has, I don't know in what order to tell this. Maybe I'll come to the z5 in a second. I now want to say how this actually works. So now, since the group acts, and this group acts, we can now see the vertices in four-dimensional space instead. So this group 2i, let me call it g, so I don't have to keep writing 2i. So g is a group of order 120. And it is 120. If I take one point, this gives me the 120 vertices of another regular polytope. So the platonic solids are the regular polytopes in dimension 3. But this is in dimension 4. And this is called the 600 cell. Because just like the icosahedral and the dotechiedron, one has 20 vertices and 12 faces. The other is 12 faces and 20 vertices. This thing has 120 vertices, but 600 faces, which are regular tetrahedron. And there's a dual one with 120 and 60 reversed. By the way, here's something that maybe not everyone knows. I discovered it when I was a teenager. And when I was 16, and when I met Atia, which was two years later, I mentioned it to him. Of course, it was very well known, but I didn't know. And he was very happy. It's the only thing I ever told him until I was at least 40 that he ever liked. But when I told him I discovered the fact I'm about to tell you, he said, oh, when I was 17, I also discovered that fact. I was also very proud of it. But of course, it was already known then, too. And that's the fact, if you count here n, and here's the number of regular polytopes in n space. Well, if n is one, it's pretty obvious. I mean, up to scaling, so let's say in the unit sphere, there's only one. If n is two, there are infinitely many, because you can have a polygon of any number of sides. If n is three, there are five. That's exactly what Plato discovered. If n is four, there are exactly six of them. And if n is anything greater than or equal to five, there are exactly three. That's what I found. I think Coxeter found that first. I don't know whose theorem that is. It's an amusing exercise. If you've never thought about it, I recommend trying. Well, actually, what I showed is that there were at most six. There are at least three, because you always have the cube, which is just the unit cube multiplied by itself n times. You always have the tetrahedron, which is n plus 1 points all equidistant from each other. And you also have the octahedron. It's no longer got eight sides, but it's the generalized octahedron. How do you get the octahedron in three space? Three coordinate axes and intersect with the sphere. That gives you six points. So if you do it in n space, this is two n points. You intersect the n axis with the unit sphere. This is n plus 1 points. This is two to the n points. Those are distinct numbers, so they're different. And by an easy combinatorial argument, you showed that that's all there is. But here, you have these three, but what I showed is there are at most three more. And one of them, that was the 600. I couldn't prove it really existed. Of course, I didn't know about the quaternions itself. But it does exist, and here it is. Okay, so part of this story is more general culture and not for the specific thing I'm aiming at. So we have this six cell, but now we can realize that explicitly because the group is acting. So I can, if I identify h with r4 the way I said, and just take g and take one point, then I find that the vertices of g, so the elements of g now becomes a subset of s3, which is a subset of r4, which is the unit quaternions, but I'll just think of this r4. And it's the following points. First of all, they're not unit quaternions. Sorry, yes, unit quaternions. So first of all, you can take points with all plus or minus a half. So they're 16 like that because I have four signs. Then I can take a point which is plus or minus 1, 0, 0, 0, and permutations. So they're eight like that, right? Four positions for where it was. And then the last one is not so obvious. You have to divide by two. You take plus or minus the golden ratio, plus or minus one, plus or minus one over the golden ratio, and one. And now you take only even permutations. So when you take permutations of four things, there are 24. Here there were only four because it doesn't matter if you permute the zeroes. Here there would actually be 24, but even once it's only 12. But I also have three signs. That's eight. Eight times 12 is 96. And 96 plus eight plus 16 is indeed 120. So that's the list of the vertices. And now if we take what's called by Conway and Sloan, the Icosians, horrible words, you take the z span of all of these vectors in four space, all z linear combinations of them. Well, that's a, you can't even see what the dimension is, but you can embed this into R4 because it sits in R4. Sorry, this was in R4, nobody corrected me. But I can embed it in two different ways in R4. What's the other embedding? There are two ways to see it. One is to replace the word even by odd. There are exactly as many odd permutations as even, but that will be another copy of it, a different one, or equivalent like replace five by it's Galbakovsky, which is one minus the square root of five over two. I also get 120. So if I take both embeddings, then I get this. And the image, if you put them and make the metric right, is exactly the E8 lattice. That's kind of a beautiful connection. Okay, so that's a nice fact that you see the E8 lattice in R4, it's spanned by a bunch of points. In particular, I could write down here in some norm which are the E1 of E8 from before. And there's one more connection which I want to mention. Sorry. No, no, I didn't say that at all. It has a lattice of vectors. Of course, there's a module that splits in two. No, first of all, it doesn't split in two at all. This is not a direct sum of anything. This is a single lattice. It's not a direct sum of anything plus anything. I'm just embedding it by sending this point. When I have a point like this, I embed the first four coordinates are these and the next four are the conjugates. You place five by five prime. No, no, of course it doesn't split as a lattice. So I want to say one last thing about this. The E8 diagram of where it comes from. So this is especially in the work of Breeze-Corn and Arnold, but many later people, they're deep connects with the theory of singularities and we'll see that in a second. So I'll mention what I get to my equation of S2. I'll mention the connection with resolution of singularities. Okay, so I think now I've said everything I wanted to say about the icosidron except for one thing. And that one thing is the one that I care about the most. Sorry, I don't know which, I don't want to get into discussion of it now. This, none of this will generalize to any other situation. This is specific description for the icosidron and the unicrotonites. But I really don't want it to go into it now. It's not the subject of the lecture I just mentioned. We can talk afterwards just, I want to get. Sorry, I didn't use anything to describe E8. E8 is the unique densest lattice. It's a completely, no, this is the basis. Don't forget this, this is just to help because there are students in the room. E8 is the densest lattice in eighth space. It's unique. That's what it is. It happens to have this, I gave even two descriptions. Here's one basis. This is not a basis. I didn't choose a base here. Then in case people like a basis to see the E8 dashes, but that is nothing to do with it. It's unique lattice. I gave here three different descriptions. And of course you can write down the isomorphisms. But it's not relevant for my talk because I won't mention this again. I just, it was a digression. I don't want to spend more time, the time is short. So please, could we not discuss it now? But I gave three descriptions. Here was a description as a subspace of 1 1⁄2z to the 8th. So it's a sub lattice of this, but of some index. Then I gave a description with the basis. Then I gave another description with EI squared with E8 lattice. Then I gave a fourth description. And they're all the same by a certain amount of work, but it's not my subject. What I wanted to do is come back to this and use the description in complex space because this is the one that I really care about. So here, this set I just raised with V is a set. It is cardinality 12. Any collection of a finite number of complex numbers is the set of roots of a unique polynomial. But if it's in projective space, it should be since one of these roots is at infinity and one is 0, I could think that. So I will make a polynomial V of z, which will be z to the 11th. Now I have to get the sign right. So there are two choices of sign I can never remember which I'm using. So it's this polynomial. That's a polynomial of degree 12. But it has 11 visible roots, which are 0, these 5, and these 5, and these 5, that's 10, and one point at infinity. So it's really better to write in homogeneous coordinates. Then it becomes x to the 11th y. Well, I can factor it. x times y times x to the 10th plus 11 x to the 5th. y to the 5th minus y to the 5th. So that's the polynomial with the roots V. Now there should also be a polynomial. And Klein computed all of these in his book on the Aikos heat room. But there's a quick way to get them. I'll show you in a minute. But first, I'll just write them down. So remember that the cardinality of E, the set of edges, was 30. So this should be a polynomial degree 30. The cardinality of the set of faces was 20. So this one is x to the 30 plus 522. x to the 25 times y to the 5th minus 10,005. x to the 20, y to the 10. It goes in groups of 5 just like this. The next coefficient, x to the 15, y to the 15. It's antisymmetric. So it has to be 0. The next one is x to the 10th, y to the 20, up to the signs. It alternates. So these polynomials are invariant if you interchange x, y, and change one of the two signs. So that's a rather complicated one. And the last one, f, is x to the 20 minus 228 x to the 15. y to the 5th plus 494 x to the 10, y to the 10. Here, the middle one is not 0, because this paradigm is OK. And then it's, I'm not sure if I'm reading the sign, but I may have missed a sign. I should check it in my notes. I have a feeling it can't be right. One of these has to be a plus sign. So I should copy it from my notes, copy it by hand. And I think I screwed it up by, ah. Why don't I just put a minus sign, because I'm sure it's minus. I'll check the sign later. So there is a smooth way to get this. The first one is easy, because this polynomial is really easy, but the roots, as I told you, 5 plus 5 inverse is just the golden ratio, maybe up to sign. So when I take its 5th power, it's the 5th power of the golden ratio. And the 5th power of the golden ratio is 11 plus 5 squared to 5 over 2. That's a solution of the equation x squared minus 11x minus 1 is 0. Or it was actually minus the golden ratio. So it's x squared plus 11x. So the 5th powers of these solve the equation z to the 5th squared plus 11z to the 5th minus 1 is 0. So this, when you can do it practically in your head, all you have to do is compute to the 5th power of the golden ratio, which is very easy. The other two, if you compute just directly computing all the coordinates, of course, you can do it with a computer. It's even very easy. But in fact, there is a smooth way. And since it's also quite fun, I'll give it to you very briefly, but I have to get the coefficients right. So we already have v. I'll take the homogeneous version. So v is this polynomial, homogeneous. And now I take the Hessian. So you take the second derivative of v with respect to x twice, d2v by dx squared. The derivative v with respect to x and then y. The derivative v with y and then x, which is of course the same. And the derivative twice in y. And you take the determinant. This is what's called the Hessian. And then that is also the point is that under the icosahedral group, this polynomial has to go to itself. Because the icosahedral group preserves the set of vertices. And so it preserves the polynomial with those roots. But for the same reason, the other polynomials have to be invariant. So this is the beginning of invariant theory, one of the famous exams from the 19th century. How do you write down polynomials? So we have a very complicated group, which I'll tell you in a minute how to look. It is this icosahedral group. Non-commutative order 60, simple group. And I want polynomials that are invariant. Well, by definition, as I get, each of these is invariant. Because each of the sets, v, e, and f, is invariant under the icosahedral group, simply because it is the group of symmetries of the icosahedral to preserves the set of vertices, set of edges, set of faces. But this is a completely invariant way to make something. The group acts linearly. Then it preserves the Hessian determinant. And so up to a constant, actually, this has the right degree. It has to be correct. And the actual constant, that's the one I couldn't remember, is minus a 12. And similarly, once we already have v and x, we can take the Vronskin of those. So I now take vx and vy. But also now that I have fx as well, I take fx and fy. And this 2 by 2 determinant is again invariant under the group, automatically. And it is, in fact, the right thing. I've written minus 1 over, I think this 12 has to be 120. Now it's correct. My handwriting is illegical. But it's minus 1 over 11 squared. So it doesn't really matter. But just for fun, that's how you get these polynomials without computing any of these huge coefficients. You can compute them easily in one line on the computer with these formulas. I define them that way. So here, the icosahedron consists of these 12 points. But this was in the model c cross r. But under stereographic projection became a set of points in p1c. So these three sets, v, e, and f, are three sets all contained in p1c. They have cardinality, 12, 30, and 20. But any finite set of points in the complex plane, n points in the complex plane, is the set of roots of a unique polynomial of degree n minus 1. And if there's also a point at infinity and you make it homogeneous, there's one more. So v, e, and f, I defined originally as the polynomials whose roots. So if I made it homogeneous, x over y is z, then the roots of this polynomial would be v. You have to compute the coordinates I didn't give. But if you compute all of these points, look at which ones are adjacent. We can actually do it. The north pole is adjacent to these five. This one is adjacent to the one with j plus or minus 1. And here to the one with j plus or minus 2. So that gives all the j's and so I can write down all the midpoints. And then if I just do it, that's, yeah. Well, yeah. Maybe I forgot to say that. Yeah, sorry. E, thank you. I think I forgot to say it. I did say it when I drew the picture of v, e, z, and f arrow, but not for all the others. So the points in the sphere have to be, if it's an edge, an edge is not a point. You take the midpoint of the edge and the central, the face center of the triangular face. Those are points and they're premuted by the group. Sorry. And their cardinalities are 12, 20, and 30. And so you get three polynomials, which you can either just compute on the computer and get this, or you can do it in this smoother way. And now, however, there's what's called the symmetry between them. So if I look at e squared, that has ordered a degree 60. If I look at f squared, that also is degree 60. And if I look at v, sorry, f cubed, and if I look at v to the fifth, it is degree 60, right? Because 12 times, well, you can just do it. 12 times 5, 30 times 2, and 20 times 3 are all 60. So these all belong in the set of points. But this is invariant theory. We're looking at c of x, y, but in fact, here of degree 60. And this thing, the dimension of this, you can compute from invariant theory. There's also, you can just do it directly on the computer. The dimension is 2. And so there has to be a relation. And the relation in the normalization that I've chosen, which is kind of the nicest with all 1s, is very simple. It's e squared minus f cubed is 1728 times v to the fifth. And in fact, now if I drop the condition of degree 60 and just look at all polynomials in x and y of any degree, but gamma invariance of the invariance, then there are all polynomials. There was all proved by Klein in this now. Very well. It's exactly the set of polynomials in three polynomials, e, f, and v, which are the ones that are written here, multilow one relation. And I want to end this part of the talk with that. That's the connection with singularity theory, because this thing is a singularity. It's actually called the e8 singularity. There's various names. This is a singularity in complex three space. So the set of points that satisfy this as a variety here is an algebraic surface, but it's got a very singular point. And in fact, because it's the invariance, you can see that all we're doing is we're taking c2. And remember, on c2 acts the group s22, and on that acts the al-Qaqsi-Hitru group, which I call g. When I divide by this group of order 60, then the invariance here are exactly this. This is a model of that. And so this singular surface is a quotient singularity of c2 by something. And when you have an isthe, singularity of a surface, you can blow it up. And it was shown by Patrick Duval in 1934, that when you blow it up, this bad point, I'm just doing this very quickly, consists of eight copies of p1. And they're joined in an E8 diagram. So there are eight copies of p1 and any two. Each one is self-intersection, I think, minus 2. And two of them intersect in one point if they're adjacent in the EI diagram. But nobody seems to know why to relate that point of view to the other. Nobody can actually go directly why this resolution is the same as that, or at least John Bay said he didn't know when. I certainly don't know. So so far, we now have these polynomials. So now I want to change gears entirely. We have this wonderful cisogy. And now for the rest of today, and actually most of the rest of the lectures, I want to talk about connections with multiple forms. So since I announced these lectures as completely elementary, I didn't want to assume anything about multiple forms. So I'm going to give the definition very briefly, but above all, examples. And the key thing about multiple forms there, too, one is that they're very hard to construct. There are actually only five ways that I know to construct multiple forms explicitly. So they will be function of a complex variable. There are many examples, but there are only five constructions that give at least arithmetic ones that I know about. Plus, once you have two multiple forms, you can add them, multiply them, take Hecchi operators, rank and cone brackets. You can make new ones out of the old ones. So you can generate everything from these five, but we don't know. So that's one important thing they're hard to find. On the other hand, they're easy to use. If you have several multiple forms, the multiple forms have invariants called the weight and the level. If you fix those invariants, it's a finite dimensional space, easily computed with an explicit basis if you want. So if you have a conjectural identity among multiple forms, it's easy. You just find what space they're in. If they're not in the same space, the identity is false. If they are, you write them in terms of basis. And if they have the same coordinates, they're the same. An infinite identity becomes something trivial to prove. So I want to give you all five known constructions of multiple forms. I think I'll remove all of this except the polynomial, which maybe I'll circle in red or something so I don't accidentally erase it because it's too much trouble to write those down. Again, not that I really need them. But just to remind you that we had these nice polynomials. So first of all, let me tell you the definition. So we have a group. Gamma is, for instance, the group I'll call the gamma 1, which is SL2Z. So I think everyone knows what that is. It's 2 by 2 matrices, integer coordinates. I'm not going to write it all down. A, B, C, and D are in Z, and the determinant is 1. It's in SL. But then, that might be EG. So I have a group gamma, which always will sit in SL2R. And other examples are, for instance, gamma 5, which is the set of A, B, C, and D in gamma 1, which are congruent to the identity of modulo 5. So and then another group that I'll sometimes use, it's called gamma 0 of n, is the group of A, B, C, D, and gamma 1. So this is gamma, and here gamma is congruent. Here C is congruent to 0 modulo n. And n is some integer. So soon I'll have an example where n is, for instance, 900. So it might be some strange integer. So we start with a group. And now a multiform should be. So the idea is that this group, and I'm sure, actually, all of you have seen this, this group acts. Actually, all of SL2R acts. On the upper half plane, the complex upper half plane, which I always write with the German capital H. So this is all complex numbers with positive imaginary parts. Here's fine middle. Here's 0 on the edge, but it doesn't contain the real line. So 0 is not in it. And the action, I'll use Taylor for the generic variable. The action is merbius transformations, or fractional linear transformations. So this is, for all of these groups, and for any subgroup of gamma 1 and find idindix, these are nice discrete subgroups. There's a fundamental domain, which for gamma 1, I'm sure you've all seen the picture. I won't talk about it, it plays no role. So we have these groups. And now a multidor function on gamma is a function, it's a holomorphic function, f from the upper half plane, though it takes complex values, I'm just saying scalar value, such that f of gamma tau is equal to f of tau for all gamma in the group, and of course, all tau in the upper half plane. So it's simply a holomorphic function on the quotient, h multidor gamma, which is a well-defined Riemann surface. But a multidor form has a weight, which is an integer. For us, it could be half integer, it could even be fractional. I'll come to that later. A multidor form of weight k. So k is maybe an integer for the moment of weight k. It's the same thing. If k is 0, it's exactly the same thing. It's a holomorphic function, a holomorphic f, again, from h to z, to c, such that. But now the difference is that instead of requiring invariance, you insert a factor. Now, if you've never seen this before or have only seen it vaguely and are not familiar, the most important thing about this definition is that you can forget it. It plays no role, sorry, and I left out the condition, plus a growth condition that I'll tell you in a minute when I get to it. You can actually forget the definition. The whole point of multidor forms is that this definition is so rigid that the space of multidor forms, as I said, of a given level, level means which group and a given weight. So if you fix k and gamma, the weight and the level, this space is finite dimensional. And it's computable. And so you don't actually care what the definition is. That's secret. That's the black box that tells you why these things work. But once you see it, you never actually use that properly, use the forms. But I do have to say a little bit to tell you the growth condition. So since my group contains two standard matrices, gamma 1, and it's generated by them, t is the translation. So it acts by tau goes to tau plus 1. And the other is always called s. And it acts by tau goes there for the minus 1 over tau. It's that Mobius transformation. So s squared is 1. And so OK, there's another relation that I'll come to later. So if you look at this condition for when tau is t, you see that in particular, even for a multidor form, not just for a multidor function with f of tau plus 1 is equal to f of tau. And therefore, f of tau has an expansion n in z, some coefficient, times e to the 2 pi i n tau. And the condition, the growth condition, is two conditions. First of all, there should be no negative powers. And secondly, that this is polynomial growth. So for some c, the coefficients only grow like a power of n. So if you impose that, that's just a technical thing, but I didn't want to lie, then you get a finite dimensional and computable space. And because it's finite dimensional, that's what I said, you have a free way to get non-trivial identities. If you can produce two multidor forms that count very different things, so they're coefficients, what we care about are these coefficients. Typically, we're interested in arithmetic. We'll be interested in the case when these are in q or maybe in a number field. So you have two multidor forms. You believe they're equal. So a n is b n. That's some theorem for infinitely many n. But once you know they're multidor forms since it's finite dimensional, you just have to check for a few values and that implies it. So I want to give you, as I say, not only several examples, but all five ways that I know of constructing. So there are five constructions, five explicit constructions. And I believe they're basically the only ones known. There might be some. So the first and the most familiar is called Eisenstein series. And the way I'll do it, I'll just do it for SL2C. Let me start with an example. E4 of tau, by the way, I will always write q to be e to the 2 pi i tau. It's a very convenient and very standard notation. So what was a Fourier expansion in tau becomes a power series expansion in q. So a multidor form always looks like this. So my first example is E4 of tau. It's going to start 1 plus 240q plus 2,160q squared. And the definition, the definition I want to use is the sum n from 1 to infinity sigma 3 of n times q to the n. Sigma 3 of n, sigma anything of n, is the sum of the divisors of n to the power 3. If I put sigma 5, it's the fifth powers. That is a multidor form of weight 4 on gamma 1. That's not at all obvious, but there's an easy other construction. And there's a construction, something which is obviously a multidor form and a one-page calculation of its Fourier expansion and you find this. But I want to take the definition already to be numerical. And then that's why I said this definition doesn't matter. You can't see that definition. You have to believe me that the things I write down are known to be multidor forms. But then they're just power series expansions. We're doing number theory now, not complex analysis. Although secretly, of course, it's also complex analysis. So this is E4. But there's also an E6. And actually, there's also an EK, which is, so this is an M6 of gamma 1. And this is an MK of gamma 1 for all positive even inches, except 2, where this is a slight problem. And each of these, so this one starts 1 minus 504 times the sum sigma 5 of Nq to the N. And again, it's not at all obvious that this has this invariance property, but it's a theorem. And EK tau for every k is there's some constant, which is well known. And then it's the same expansion. But now you take the k minus first powers of the divisors of N to get the coefficients. So all of these are multidor forms. So that's an example of the construction. But I promised you that they're nontrivial identities. So here I give an example of an identity. If I take E8, which will start 1 plus 480 times q plus sigma 7 of q squared, so the divisors 2 are 1 and 2. 2 to the 7 plus 1 is 129. So it starts like this. And now, the identity is that this is the square of E4. And the proof is very easy. One knows that the space of multidor forms on gamma 1 is a dimension form, and the dimension is 1. So therefore, E4 squared and E8, since they're both in there, have to be proportional. Since they both start with 1, you're already finished. So we didn't even need to check that 480 was twice 240, just the 1 agreeing. And now we get a completely cheap proof of a formula. I won't write it out. If you just compare coefficients, you can write the sum of the seven powers of the divisor of any number in terms of the sums of the cubes of the divisor of A and B for all A and B which add up to N. There's no elementary proof of that. So this is a typical example that we write down simple expressions that have some interesting arithmetic meaning here it's the sum of divisors function, which is a standard function in number theory. And then we get a very non-trivial identity simply because we know that they're multidor forms. But you have to show that there's a multidor form that takes a little bit of work. OK, so that was my first example. The second example I should get the order right is the A to function. I'm not giving a course. I'm giving a survey. I didn't prove it. It's a half hour proof. Of course I didn't give it. I said that it's not very hard that it's a one-page calculation. So there's another definition which makes the multidarity clear, but I don't care. That's the point. I'm not doing analysis. I'm doing number theory. I want the definitions used in the coefficients. And then you have to believe me if anyone wants to know I can do it separately, but I don't want to do it now to prove that these things are true. But once you know that these are multiforms, that's what I told you. It's hard to know something is multidor. But once you know it, then you get identities like this for free. So here's the identity. So the second construction is the Dedekind A to function and the Ramanujan delta function, which is almost the same. So A to tau is q. Remember, q is e to the 2 pi tau. And it's the infinite product I'd already written this last time. And delta of tau is the 24th power of that. And then what's not clear is that this is a multiform of way to half on sl2z. But we saw multiplier. So explicitly, if you shift tau by 1, then it's kind of obvious that you just multiply A to tau by the 24th root of unity. So I always write z to n, just like I did for n equals 5, is e to the 2 pi over n. If you change tau by 1, you don't change q. So you don't change this part, but you change this by 24th root of unity. That's obvious. But what's not obvious, and there are many, many proofs, is that A to the minus 1 over tau is the square root of tau up to an A through the unity times A to the tau. It's square root of tau over i. That has to be proved. I'm not proving it. But because of that, you see that it's way to half, because I have c tau plus d to the 1 half. And when I take the 24th power, this one is actually a multiform of way 12 on the full multiform. But now, again, we get a free identity, because one knows that this space is two-dimensional. I mean, for weight 8, I told it was one-dimensional. Therefore, E8 and E4 squared had to be proportional. This space is two-dimensional, but we already have three functions of it in it. E4 cubed, because E4 has weight 4, so its cube is weight 12. E6 that I just wrote down there has weight 6, so its square has weight 12. Delta, I just told you, is weight 12. And the identity with the one line proof, once you know its multiter, the proof of this identity is you have three vectors in a two-dimensional space. There has to be some identity. If people can't see it, maybe I'll wrap it up here. There has to be some identity. You just look at the first two coefficients, and that's enough. And then if you're not convinced, you look at the first 100,000 coefficients on your computer, and they're all true. But you've already proved it. That's the beauty of multiforms, to prove any identity becomes a triviality. To discover multiforms is not a triviality. So now, this should ring a bell. If it doesn't ring a bell, you're probably not this to get all, except I removed the equation. But here, we have the E squared minus F cubed, with 1728 times V to the fifth. And you can hardly believe that this thing and that thing have nothing to do with each other. And of course, they have everything to do with each other. I'll come to that very shortly. So that's where the connection between the two fields that I'm talking about, multiforms and the icosahedron and the bell-destroy-moodledron, will come in. But first, I want to finish with the next. So the third general way is theta series. So I'll give several identities here. So here, EG, simple example, theta of tau is just the sum q to the n squared, sum over all integers. So since every square is a square, but except for 0, it's a square of two numbers, plus or minus its square root, the coefficients of this will all be 0 or 2. So it starts like this. Now, identity. Well, let's take an example. So another example would be theta 4 of tau. Well, that will have an expansion, theta of tau to the fourth power. So this one's weight, 1 half. The proof there is relatively easy. Any theta series, the theta series in general, is q to some quadratic form. In possibly many variables, and sometimes there's a polynomial in front. There are variants, but roughly it's a quadratic form. So if I take theta of tau to the fourth, then you see immediately that it is numbers which are very interesting. Namely, r4 of n is the number of four tuples of integers, such that the sum of the four squares is n. Right because if I multiply, theta is q to the a squared, the second theta is q to the b squared, q to the c squared, q to the d squared. When I multiply the fourth power and take the nth coefficient, I've cried n is the sum of four squares. Now, a very, very famous theorem of now elementary number theory, not elementary then, is Lagrange's theorem that every integer, every positive integer, is a sum of four squares. So we know that this r4 of n is never positive. But no, an identity of multiple forms tells you this immediately because you find theta of tau to the fourth. We take the function that I told you before, e tau, that are replaced tau by four tau. And I multiply by four thirds. I mean, again, you can't see in your head why this is true. But if you know a little about the theory of multiple forms, you know that this thing is a multiple form. e2 isn't quite, but this combination is a multiple form. It's of level 2. It's on the group gamma 0 of 2, gamma 0 of 4, excuse me. And it is weight 2. But this also is of that level. And the space is one-dimensional. So these are proportional. And you just compute a single coefficient to get this. And so that immediately tells you the following. I think this was me to solve this first. I don't know anymore. The number of representations of any positive integer as a sum of 4 squared is 8. Remember, e2 had a coefficient. There was a c2, which is minus 24, as it happens. It's sigma 1 of n, the sum of the divisors. So here, I have to sum the divisors. That would be e2. But actually, because of this combination, I'm taking the divisors only that are not multiples of 4. And so this is a theorem. And in particular, it's positive because one divisor of every number is 1. And that's certainly not visible by 4. So we have at least one, actually at least eight representations. That's a beautiful example of multiple forms. That you don't need to do anything. You just have to believe somebody who tells you that this is a multiple form. That's a multiple form. They both have level 2. They both have level 4 and weight 2. It's a one-dimensional space. Even for a 100-dimensional space, you would compute 101 coefficients. So any such identity is totally trivial to prove. You don't have to prove it. You just check it numerically. So now, just for fun, let me check this. So if I take r4 of 2, if I want to write 2 as the sum of two squares, then I need 2 plus or minus 1s and 2 zeros. I can choose the positions of the plus or minus 1 in six ways. The sign's in four ways. So therefore, this is 24. And that's indeed equal to 8 times 1 plus 2, which is what the theorem says. And if I take r4 of 3, the only way to get 3 as the sum of three squares is to take 3 plus or minus 1s and 0. There are four positions here, eight possibilities. So this is 32. And that is indeed 8 times 1 plus 3. And you can check as many as you want. So here, you get a deep theorem of number theory. I don't actually know any elementary proof. The proof of Lagrange, it's given in Hardin Wright's book. The proof, even that every number is a sum of four squares, is a hard theorem. But to know that the exact number of representations is 8 times the sum of the divisors that are not visible by 4, that's not an obvious thing. But with multidiforms, you don't have to do any work. It comes. And as you see, we never use the invariance. We use the invariance secretly to construct these functions. We have to show that they're multidorm. And for that, we, of course, need this definition that I wrote. But for the purpose of the user, which is us today and maybe you anyway, you don't have to know what a multidorm is. It's just a black box. There's spaces of multidorms. All you need to know is that they have a weight and a group. And that it's a finite dimensional space since you get identities. So this was the first three. Actually, I want to give two more examples. So here are two more examples of the hell with that. I don't think I don't really care what these gophes you are. You remember there were three polynomials, and they satisfied the relation. I'll leave the relation there. So I want to give you two more identities using a theta series. So identity two, still for application number or construction number three, which is theta series. Identity two, this is two identities. One is due to Euler. One is due to Jacobi. So if you take the function 8 of tau, you remember it was q to the 24th times 1 minus q times 1 minus q squared and so on. So Euler discovered this function, even though it's called the Dedekind function. He multiplied this out to, I think, 50 coefficients. And he discovered something. Well, he didn't do it quite like this, but he should have. He discovered that you only get perfect squares. And so the actual statement is here n is congruent to 1 multiplied by 6 minus 1 to the n minus 1 over 6, q to the n squared over 24. So you see that's exactly a theta series. I told you a theta series is q to a quadratic form, and that there could be congruence conditions. So here it depends on n multiplied by 12. If it's not 1, not 6, you get 0. Otherwise, you get plus or minus 1. And similarly, Jacobi, if you take 8 of 2 cubed, which is q to the 1 eighth times 1 minus 3 q and so on to the next term would be 3 q to the 9 eighths, then you find you only get squares over 8. It's the sum n congruent to 1 multiplied by 4 minus 1 to the n minus 1 over 4, q to the n squared over 8. And again, the proof of these in each of these is that you can prove those two in elementary ways, but it's quite tricky with the Jacobi triple product identity, but you don't need to. This thing has way to half. This is way to half. If a couple of coefficients are equal, you're finished. This is way 3 half. This is, sorry, I forgot the most important. You see the 3. It's n times q to the n squared over 8. I told you there is a variant where you can allow certain polynomials. So again, the proofs of each of these, once you know everything is a model of form, is easy. And the next one, the third identity for the Theta series, I take the Theta series for the E8 lattice. I could do this for any lattice. So I take x in the E8 lattice that I spoke of before. And each norm is always an even number. So I take the norm of x, the scalar product of x itself divided by 2. Well, this one starts, 1, of course, because the only vector with x squared equals to 0 is, of course, x is the 0 vector. But it's a general theorem. This lattice is even and unimulted or whatever that means, and such a thing, the Theta series, is a model of form. The weight is half the number of variables, so it's 4. But because it's even and multidare, it's on the full multidare group, sl2z. And I already told you that this is spanned by E4. But this thing, obviously, starts with E4. So the next term has to be 240. And the next term has to be 2,160. And so you get immediately that the number of roots, which are, by definition, the x in the lattice such that the length is 2, the number there is 240. And the length where x dot x is 4 is 2,160. And the length where it's 2n is 240 times the sum of the divisors of n. You get that with no work at all. Because we know with the theory of Theta series that this thing is always a multidare form of some level and of weight half the number of variables. So here, weight 4. Because this lattice is even and unimulted, we know it's on sl2z. And as a result, here it's one dimensional. So you get the counting for the number of points in the E8 lattice of any length with absolutely no work. It takes one minute. So now I want to give the last two. The last two have a completely different nature. The first three are systematic. They're elementary. I mean, you have to do some work with the Poisson summation formula and so on. But in principle, they're elementary. Now we have something that you need luck. So let's take these are the q-hybrid geometrics functions. So I talked about this last time. So first of all, I had the Reynolds-Raminutin functions which are one of the two themes in the title of the series. g of q was this, and h of q was the sum 0 q to the n square plus n. And the Reynolds-Raminutin identities were that this thing is the product over all n which are congruent to either 1 or 4 mod 5 of 1 over 1 minus q to the n. And this one is all n which are congruent to either 2 or 3 mod 5. So those are the identities. But now you can use these r. So now I define g1 of tau to be q to the minus. I wrote this actually last time, but I'll write it again. g1 of tau is this, and g2 of tau, q, is q to the 11 over 6. Remember, q is e to the 2 pi i tau times h of q. Then these are modular functions. These are, but again, that's not at all obvious. These are modular functions on the group that I just erased that I called gamma 5. And the group gamma 5 is the subgroup of sl2z of elements of congruent to the identity, modular 5. So that's a fact. And in fact, if I make the vector-valued form, so g1 of tau, g2 of tau as a vector, then as a vector, it's invariant under all of sl2z. Remember, sl2z is generated by tau plus 1 and minus 1 over tau. So if I shift by 1, it's obvious. Because if I shift tau to tau plus 1, I don't change q. So I don't change g of q and h of q. But here I get the 60th root of unity to the power minus 1. And here I get the 60th root of unity to the power l. So this is 11. So this is obvious, even if I took some other stupid power series. But what's not at all obvious is that if I take the other generator of sl2z, then you get an explicit matrix, which I'll check in a minute. But I'm pretty sure I have it right. It's sine 2 over pi, sine pi over 5, sine pi over 5. I think it's right. If it isn't, then I've got a sine wrong somewhere. So although the functions individually, g1 and g2, only are invariant under the subgroup of index 60. In fact, in 60, in sl2z, but as a pair, g1 and g2 jointly are invariant under all of sl2z. You have these transformation laws. But this is completely miraculous. And there's no general way. I mentioned this last time at the end in response to some questions, so I won't repeat it. But for instance, if I put other quadratic forms here, so if I put, well, let me write immediately g1 and g2 but therefore have a minus the 60th and a plus 11 over 60, then there's no, if I change these three coefficients, 1, 1, and 1160 or 1, 0, minus 6, do three arbitrary rational numbers. There are only seven, that's a theorem, only seven cases. And you have to really prove it, it's several page proof, that those are the only ones that are multiple functions. And no one can produce them a priori. So although there are constructions, for instance, from vertex-upgrade relative, that sometimes, if you're lucky, produce a specific q-hypergeometric series, which means the sum of products of this general form that are modular, we have no idea how to classify them. There's a lot of interest in that question. So this is miraculous, there's no systematic way. But in this particular case, it follows from the quadratic identity, so now I also give an identity and at the same time the proof of modularity. Namely, this series with the q to the minus 160 is the ratio of a theta series by the eta function. And this is the ratio of another theta series. By the eta function, I don't even have to tell you, because if I take these two identities and write them, if I write this as theta, 6 comma 1, and I write this as theta 4 comma 1, then you can see the theta of n comma a, well, m comma a is the sum n congruent to a multiple of m minus 1 to the n minus 1. It's the same definition. You do the same instead of 6 and 1 or 4 and 1 with 10 and 1 and 3 and 1, and you get this. And so that implies that it's multiter. If you knew that it was multiter, again, the proof of this would be a triviality. You just multiply through by 8 of tau, and then you check to find that number of coefficients. But there is no easy proof that these are multiter. In fact, the usual proofs, you first prove the broadest ruminational identities, and then the so-called Jacobi triple product identity, which is elementary and 200 years old, tells you that this product, that is multiter. Any such product is multiter. But so just to say, that's not a systematic way. And then there's the last way, which is from algebraic geometry. And this is not very systematic. It's completely a miracle after miracle. And when it works, the proofs can be hundreds of pages or thousands of pages. So this is arithmetic algebraic geometry. You have x is a variety defined by equations with coefficients in z, let's say. And then you can look at x over the finite field of fp, or actually over all finite fields. And that's a finite set, and you can count it. And so the collection of these cardinalities for all p prime, you have to do a little more, gives you something called the L series, or it's a much longer story. But it sometimes gives you a connection with multireforms, sometimes. And even when it is not a lot clearer, and that will be this I'll talk about more next week. But the famous case is this. You let e be an elliptic curve. And since I want the coefficients of my finite equation to be in z, I'll take it to be an elliptic curve with coefficients in z, or at least in q. It's good enough, because it's easily normalized my example. Then you take the number of x and y, modulo p. So the theorem. But this is the theorem that was a conjecture for 50 years, and then became the theory of, so it was originally, I'll just put t w, tan the arm of a, and it's now Taylor Wiles, still t w, it's very convenient. The theorem is that e is modular. And what that means is the following. There exists an integer, some positive integer. And there exists an f, which is a modular form, in the sense that I told you, on the group that I also defined, gamma 0 of n. So let's write f, as f of tau you can remember, always write. It's a Fourier expansion here. There's no constant term, so it's an q to the n. And the theorem is, let me, sorry, I'm writing backwards because I didn't want to erase, such that the number of solutions of this equation, for all primes, at least not dividing n, the number of solutions of this equation, maybe not dividing 6n, the number of solutions multiplied by p of this congruence, for every prime, is always p minus ap. So you get infinitely many, just by counting points, you get infinitely many coefficients, a, p, and then there's a standard way to get the other an. So you get the modular form. But the proof that it's a modular form, the proof is probably 10,000 pages. If you take Wiles's proof, it's already very long. But all the things he uses, which is essentially all of modern, Golden League, Algebraic, Geometry, SGR, and AGR, and lots of things about deformation theory and Maser and Son, it's certainly at least several thousand pages, maybe 2,000 or 3,000, maybe 10,000. So this is incredibly difficult. It's completely different from the other statements where there's an elementary proof of modularity. But it's conjectured in a wide class of examples. And the reason I'm giving this lecture series is because there's a beautiful example that will, therefore, be explained on the next week on Friday. A conjectural example, I can even tell you what the variety is because many of you have seen it. Certainly every physicist has seen it. This is, you take the mirror quintic, which is very famous in conformal field theory and in all the physics and in all of mathematics. And not that I space theory, but it's the beginning of the theory of mirror symmetry. You put here a constant, but I'm going to want the singular fibers. So the constant will be 5. You take this variety. So there are five coordinates, but it's homogeneous. So it's in p4. It's a three-dimensional variety. So this is a famous Kalabiyaw three-fold. And now we know in this case that there is a model of form which is related in the same way you count the points of this. That model of form has weighed 4. And it's on gammas there of 25. And that I'll talk about. But here I just wanted to give an example because I've been giving examples of everything. And this example become relevant because it will play a role for this. I just discovered it last week. I mean, the example was known, but the connection. I take that's why I wanted rational numbers. I want this very simple equation. y squared equals x cubed plus l to the fourth. So that's my e. Then the f sub e. So to each such elliptic curve, there's this canonical f sub e. And it's got a q expansion. It always starts with q. Here it continues q minus. Maybe it's plus. Let me give you just the beginning of the q expansion. q plus q to the seventh. All of the coefficients are 1 mod 6. So the next is q to the 13th. But it's 7. Sorry. 7 q to the 13th. The next is again 7, but with the minus sign minus 7 q to the 19th. Well, I've lost more coefficients. And this thing is a model of form with weight 2 on the group that I actually mentioned when I defined gamma 0 of n. I said n might be 900. That's why I took 900. And this particular one, you don't need the theory of tan yamove. This is a so-called curve with complex multiplication. It's much simpler. And for those curves, it had been known for 50 years. And there's an explicit way to construct this. And I'll talk about that next week. So now I want to, ah, the time is, I'm still OK on time. I hurried on this last part a little because I wanted to say the key thing that connects all of our themes together. But let me, I know I went fast, but it's not meant to be a course. It's meant to be a bird's-eye view of a big subject. What I've told you about model of forms is that there is a definition, which is elementary with an infinite symmetry group. But what we care about is not the definition. It's the fact that every model of form has a Fourier expansion, some a and q to the n. And typically, these numbers a and n are interesting numbers. They might be the number of ways to write n as the sum of four squares. They might be the number of vectors in the E8 lattice of length 2m. They might be the number of divisors of the sum of the divisor n to some power. They might be many other things. They might be partitions. They might be combinatorial things that count dimensions of a vector space of a vertex operator algebra in various weights and so on. So the a n's is what we care about. And if you have a sequence, an interesting sequence of, let's say, integers, any sequence, you can always make the generating function. But if, by some miracle, you happen to get a model of form, then you're in good shape, because we know all model forms, you get a formula. And so I gave the five basic examples that I know. Eisenstein's series with this coefficient is the sum of divisors. The a to function, where I gave the formula for a to the a to the a to the series, by the way, you can also take one over a to the tau. That's a meromorphic motor form of weight minus 1. That's the one that Euler actually originally discovered. That's the generating function of the partitions, but it's another famous function. So the whole point is we care about the numbers. And each of these constructions gave us something where these coefficients were interesting numbers, like partitions or sums of divisors, numbers of vectors in a lattice of a certain length. Here are some combinatorial things in the hypergeometric example. And here's the number of points multiplied by p on some variety. OK, but now I want to come to the thing that ties up the two. And I've already, of course, given the game away before, because I told you here that we also have the relation E4 cubed minus E6 squared. That's why it didn't work, because this is E cubed, but E4 cubed minus E6 squared is 1728 delta. So I would have put in a minus sign. I'll put it in a second. So these are these three big polynomials that I wrote down before. v started x, y, x to the 11, y. So it had degree 12. E started x to the 30. And f started x to the 20. So there were homogenous polynomials of degree 12, 30, and 20, and they satisfied that identity. Now, I already mentioned that the two numbers, g1 and g2, are multiple of, sorry, I said that they're individually multiple functions on gamma of 5. And therefore, they're, but they have the same, well, they both are way 0, but I actually want to take the quotient. But if I look at this form, I can see that this is equally 10 3 of tau divided by theta 10 1 of tau. So since 10 3 starts with the q to the 11 over 6 to this minus 1, 11 minus minus 1 over 6 is 12 over 6 is a 5th. This starts q to the 1 5th times 1 minus q and so on. And that's the famous continued fraction that I wrote down last time that Ramanujan found. And now this function is invariant, of course, under gamma 5, because I told you that actually each of these functions, which is already invariant to their quotient. But this is a very good function because if I take the upper half plane, multilow gamma 5 and add some points called cusp. So it's, unless I add some points or if I subtract some points here, it's not quite true. This is an isomorphism. It turns out that for this group, which is index 60 in SL2z, that the quotient, which is a Riemann surface, which might have had a large genus, actually this gen is 0. And this R function is an identification. But actually the way I've done it, there are three. So I subtract three points. So I might have to take a multiple of R. You subtract three points, 0, 1, and infinity, which are the images of the so-called cusps. And then you get this. But we can see that in terms of what we already did. Because remember I had the icosahedron. The icosahedron gave me 12 points, z1 up to z12, which were the north pole, the south pole, the pj, and the qj, and then the stereographic projections. It gave me 12 points, but those 12 points lie in a two-dimensional space. So remember the z, each cj was really the ratio x over y. I should really think not of a point in p1, but a point in p1 is a line in c2. So I should think of xy. Then the icosahedral group acts. Well, actually, so g, which was twice the icosahedron binary. And that's exactly the group that we had, SU of 2. Remember the icosahedral group sat in SU of 2, which is contained in particular in 2 by 2 complex matrices. So it acts on these things. And I gave the identifications. So therefore, we have an explicit action of the group. But on the other hand, we also know, and that's very easy, that the icosahedral group, which remember is isomorphic to a5, it's also isomorphic to PSL2 of the field of five elements. So that's 2 by 2 matrices over the field of five elements, determinant 1, but up to sign. Since I wrote that, let me give you the remaining isomorphs. But I'll check my notes so I don't get one of them wrong. Now, of course, I can't find my notes. So maybe I'll just do it by memory. Yeah, here. So we have PSL2 of f5, which is 2 by 2 matrices determinant 1. So it's simply the same as if I took SL2 of z divided by gamma 5. So that would be SL2 actually. So it's PSL2 z divided by P gamma 5. OK? So on the other hand, we also have the binary icosahedral group G, which was 2i. And that goes, this is a 2 to 1 map. And here we have a 2 to 1 map, 2 by 2 matrices over f5 of determinant 1. And this is order 60. This is order 120. Sorry, this is order 120. And that's very easy to see. If you just count elements in SL2, it's very easy to count that they're 125. So we have this. And by the way, you might also ask what happens to s5, which also is order 120. But it's not at all this. This a5 is a quotient of this group. This group of order 120 contains it as a subgroup. And then under this isomorphic, this would be actually Pgl2 of f5. So you take the determinant could be arbitrary, but up to skaters, then there are only two possibilities for the determinant, up to skaters, which are 1 and 2. So this is index 2. This is index 2. So with this exception, isomorphisms. But now if we think of the way the Icosahedral group acts, then our three generators that I called v, e, and f. Remember, e squared was f cubed was v to the 5th is equal to v, e, f, which is equal to 1. That's in i. And if I drop this, then in g, it's just the same, but you don't put that it's equal to 1. It's a central element of order 2. So we have this presentation. But if you know sl2z, we have the two elements, 0 and minus 1, t, which was the translation, and u, which is the element st. And it's very well known that s squared and u cubed are minus 1. That means there are plus 1 in psl2 at z, so up to sign. And in fact, they're generators, and this is the standard presentation of the infinite group psl2z. But t has infinite order. On the other hand, if I reduce multiple of 5, this makes perfectly good sense multiple of 5, so does this. But over f5 rather than over z, t to the 5th is the identity, because t to the 5th is the matrix 1501 over the field of 5 elements. That's the identity. And so now you see the Icosahedral group staring in the face with three elements of order 2, 3, and 5, whose product is 1. Now if I make that identification, it means that my group acts on everything. But now if you look at the thing that I wrote, luckily it's still here. Well, I removed the left-hand side, so I'll put it back. So if I applied t, which was tau plus 1, and here s, which was minus 1 over tau, to my matrix, then when I applied t to the two rod-as-ramonutin functions, I got this particular matrix, and here I got another particular matrix. Up to scalars, that has to be the same representation as the one we had before. It's simply isomorphic. So what's happening now is extremely nice, namely g1 and g2, or theta 1 and theta 2. Remember, it doesn't matter if I work projectively, because g1 and g2 were these two theta series, theta 10, 1, and theta 10, 3, divided by a common factor, 8 of tau. So the point to project the space is the same point. And the group is really acting the Icosahedral group is acting on projective space. So now the action of, so this is, if I think that up to scalars, it becomes a point of p1 of c. And on that, the Icosahedral group is acting of order 60. And what I showed you, I mean, these identities tell you that the usual s of order 2, t of order 5, and s times t of order 3 act exactly the way they act, and this is all compatible. So in particular, if I now take my two polynomials, v of theta 10, 1, and theta 10, 3, or I could equally up to, no, I do want to do it. So if I take theta 10, 1, and theta 10, 3, there might be a minus sign, because I kept changing my mind. If I take these particular polynomials, but I substitute for x and y, theta 10, 1, and theta 10, 3, they're equivalent to the up to a factor of eta that would be the same as g and h, or little g1, little g2. Well, remember, v is degree 12. This is degree 30. This is degree 20. The thetas are 1 over 8 at times the original revolution functions. If I preferred, I could call this 8 of the minus 12, 8 of the minus 30, and 8 of the minus 20 times v of the relative to the revolution functions themselves, g1 and g2, which were g and h up to the powers of q. So that's just a choice. Since it's homogeneous, I can take either one. But now, since this is polynomial degree 12, and the theta series of weight I have, this is weight 6. This is weight 15. And this is weight 10. Well, what can it be? This one is going to be 8 of tau to the 16, 0 to the 18, times e6 of tau. So 8 is weight 1 half, 8 of the 18 is weight 9, 9 plus 6 is 15. That's good. Similarly, the next one will be 8 of tau to the 12. So again, that's a scalar thread. Since everything's homogeneous, these factors you could forget. So this would be e4 of tau. And the last one, whether it's a minus sign because we changed the sign, it's minus 8 of tau to the 12th. And you can check on the computer that all of these identities are true. But actually, they have to be true because these things are invariant under the acosahedral group. The initial functions are multiple forms for gamma 5. The acosahedral group exactly is the relation between s of 2, gamma 5. So these things, up to powers of 8, which I've written here, are multiple forms on s of 2z. There is only 1. Weight 4, there's only 1 in weight 6. And here, that's because 4, there's only 1 in weight 12. See, there's actually no choice. And therefore, these two equations, which look so similar, are really the same equation. The equation defined in the acosahedral and the cisogy and the singularity, the e8 singularity, is exactly the same as the relation between e4, e6, and delta. So that's the identity. First, it shows how two different random mathematics meet. But this is the identity that I'll use now to continue. So next time, I'll repeat everything of this that I need, which is a small part. This was an overview. And then, in a little more detail, how you go from the rodents-reminugin functions, g1 and g2. Remember, this was just the original hypergeometric series up to a fractional power of h. So it's the original capital G. Or you can multiply by 8 and get these Jacobi-theta series. These relations will be the key, we hope, to understanding, for instance, a model of parameterization of this thing. So that's where I plan to go. I'm sorry if today was a little chaotic. I didn't sleep enough, and I can't concentrate well. But I tried. So now I'm finished. And I hope somebody will have some questions. Now you can ask anything. I know that if I use a technical conformal field theory and say the characters of conformal field theory, I can construct millions of modular functions. First of all, that's so what? I mean, that's like saying I can construct millions of integers. I mean, there are infinitely many, and it's a very, very special construction. And it's not reversible. I agree it's a big list, millions. I don't know if it's really millions maybe. It's infinitely, actually many. Already the PQ examples, but there aren't so many. If you look at the PQ minimal models, all of the characters are given by uniform form. It's a quotient of theta series. But I agree you construct lots of examples. But so what? If you take any branch of pure mathematics and look for a few days or a few minutes, depending on which branch, you will find that there are examples from that field which are multiple forms. But they have nothing to do with each other, so I agree conformal field theory is one of the many places that leads to examples of functions that we know are multiple forms, and that therefore we can apply this magic, prove identities easily, find differential equations. But counting vectors in a positive definite lattice also gives multiple forms. Counting powers to advisers give multiple forms. Counting, you know, sum to squares. Well, that's the same. Counting partitions. There are many explicit constructs in mathematics, as I said, not so many, basically different. But each of those, there are many, many parameters. What I mean is, why don't you add a sixth method to compute modular functions? It isn't a sixth method. It's a sixth, I mean, there's, well, yes and no. I don't know if there's a direct proof that those are multiple, it could be that you're right. So then, if that's what you meant, I just gave a course on it last semester. It was called vector operator, vertex operator algebras and multiple forms. I love this subject, but I didn't actually give the proof, Dong's proof of the modularity. The ones I know, like minimal models, you don't need a general theory. You write down the character explicitly as a theta series divided by eta. And I think all of the examples deep down are like that. These characters have a denominator. It's like the denominator form. There's a denominator which always, it's not necessarily just eta, but it's a product of Jacobi, triple products. So it's got modular problems. And the numerator is some kind of a theta series. And so the modularity follows from these constructions. So that's more to say that if you know these two constructions, which is theta series and eta products, then many other functions that occur in mathematics, and in particular there, come. But I may be wrong. I think that Dong's proof actually uses Verlinder forms and said that there's an intrinsic proof that it's modular without writing it as a quotient of a theta series. So you might actually write, I never thought of it, that the theory of the characters of multi-server vertex operator algorithms is an honestly independent way of constructing multi-forms. Of course, each multi-form you get will be also one of these, because they have to repeat. That's clear. So yeah, that's a very good remark. Please. I saw two E8, the first one for the Eisenstein series, I think. And in the beginning, when we were talking about this group I. Yeah, sorry. I meant to apologize. Actually I was planning to call the E8 lattice, to call it lambda 8, because the E8 lattice has a theta series called theta E8, or theta lambda 8, which is the sum, as I told you, q to the power 1 half, the length of the vector. But this is not E8. It's E4, which is the square root of E8. So I'm using the same letter for two very different things. If that's confusing you, I meant to say it, and I forgot. Thank you. So E8 is actually not the name of this lattice. It's the name of an algebra with the Dinken diagram E8. It's a huge lattice. It's a 248-dimensional algebra. It's Carton sub-algebra is 8-dimensional. In that, there's a whole story. And then this thing is called the E8 lattice, because it comes from the E8 diagram. The E8 diagram, this is my great discovery about the E's. The reason it's called E is because you're going to write it like this. So that's E8. So it's that diagram, and that's why it's called E8. The other E8 is named after Eisenstein. They have nothing to do with each other. This is exceptional ABCDE series, and that's Eisenstein. So indeed, it's very confusing. The theta series of E8 is not E8. It's E4. There are actually two lattice, only two unimodal lattice in dimension 16. One is E8 plus E8. And the other is called usually lambda 16, or the wall's barn lattice. There are only two up to isomorphism. Any lattice, when you take the theta series, there's only one vector of length 0. So it always starts with 1. Since these both are 16-dimensional lattices, they have weight 8 multiple forms. And there is only one, as I told you before. So in fact, those lattice, which are completely different from each other, have the same number of vectors of any length, and that vector is E8. So I didn't mention that, but I apologize for that confusion. So it tells me that the number of vectors in E8 plus E8 of length x squared or x dot x, whatever you prefer, x squared equals 2n. Or in the other totally different lattice, which is lambda 8, the number of vectors of the same length, every vector is even length, it's an even lattice. In both cases, they're the same as each other. And by this theorem that it's E8, it's also 480 times the sum of the devices of n to the power 7. So for instance, if n is a prime, it's p to the 7 plus 1 times 480, a crazy formula. And so that's the confusion. These are the cofists of E8, but this is not the E8 lattice, but two copies of it. But this, by the way, has a nice application. Maybe you've heard the famous question of Mark Katz. Can one hear the shape of a drum? So the original question, you have a drum. So a classical drum. It has some boundary, a membrane in the plane, and this is a uniform plane. And you bang on it with your drumsticks, and you listen to all the tones you can produce to the whole spectrum. Question, can you find two different shapes in R2, which have the same spectrum? It was eventually found that you could, actually, by three people, one of whom was a good friend of mine, Scott Wohlpert. There are examples in two-dimensional plane. But the first example of two different manifolds, which have the same spectrum, but are not the same manifold, were found by Milner. And they're absolutely trivial if you know this example. You take E8 plus E8. So that's a 16-dimensional closed manifold. It's not a very ordinary drum, but it's a torus. And the other one would just top logically the same but metrically completely different is R16 over lambda 8. They're metrically different, because these are non-isomorphic lattices. But their spectrum exactly has to do with the number of geodesics of each length. And the number of geodesics is just this, and it's the same number. And so this fact, from multiple forms, there is no elementary proof as far as I know. This fact tells you that if you listen with a 16-dimensional ear to what happens when you bang with your drumsticks on these two tori, you cannot distinguish them. And the reason is because of this amazing fact about lattices. So this is the fact that M8 of the full-multi-group is one-dimensional spanned by E8. But the one that I use there is that M4 of gamma is one-dimension spanned by E4. And then the confusion was that the theta series for the lattice E8 happens to belong to this, so it has to be E4 and not with that E8. I should have said it and forgot to, so thank you. Sure. So E8, Marvelous, fine, I'm fine with that. But what is the role of the dinking diagram? Is there any role? In this story, num, they don't have, well, yes, you can take any dinking diagram, you can take any lattice. It doesn't have to be root lattices. Those are special ones, which are very nice root lattices. But you can take any lattice at all, any lattice, let's say with the length function, so x squared is, let's say, an even number, or an integer, it doesn't matter, you can divide by two. And then you can make the theta series with the sum q to the x squared. And this is always a multiform, it's such a huge number of applications of multiforms to the study of any lattices. There's a famous book by Conway and Sloan on lattice and cryptography and coding theory and packing and so on, and multiforms are used all the time. So that generalizes not just to those lattices. However, it's only if the lattice is even in unimulted or that you get multiple forms of the full multirgroup. In general, you'll have multiple groups. I mean, I'll give examples in the other two lectures. We'll have other theta series where the group is not SL2C. It might be, for instance, gamma zero of 900. That will be one of my examples, just the one right over there. So all lattices give you theta series and therefore multiple forms and therefore applications of this theory. And this is nothing, it's sort of a coincidence that there's this particular intersection between even unimulted or lattices and Dinkind diagrams and they're the only intersection is E8. By the way, the proof that it's the only intersection is very short. The dimension of an even multir lattice has to be a multiple of eight. So the first possibility is E8 and it's unique. The next possibility is 16 and they're exactly two. The next is 24, they're exactly 24. Neemire in general, they're a lot. And that fact also is a one line proof using multir forms. But the Dinkind, of course, the An, the dimension is n and so on, I mean, it's the wrong dimension even. So this is, again, this particular thing is a very special example, but of course each of these things is part of a huge theory, but those theories don't necessarily overlap as cleanly. Maybe it's time to go home. No more questions.