 Okay. Thanks a lot for the introduction. So, in fact, I debated when I knew I was coming here. I had to choose between two talks. And one of them Pierre advertised this morning. So I don't need to talk about it. So now I can talk about the other one. So the other one is related to this beautiful picture. And by the end of the talk, I hope I can tell you what they are and how we obtain them. But I'll go slowly because in fact these are supposed to describe what, how many, 12 groups, I think 12 groups, or the fundamental domains for 12 groups. And two of them, these are all fundamental domains for lattices. Two of them are arithmetic. So if you get bored, during my talk, just try to think which ones are arithmetic. So let me tell you a little bit of a general background. So X is going to be, for the beginning of my talk, and just for a few seconds, X is going to be a symmetric space of non-compact type. So non-compact type means non-positive coverage. It could have flats in a priori. But in a second, I'll get rid of flats. But Mg is going to be its isometric group. Well, okay, let me say up to find that index. Most of the time, that's what will happen. Every time then, you can even have something further from the isometric group than just up to find that index. But I think I'll try to stick to that convention. And as you probably know, X is going to be a portion of G by a maximal compact subgroup. Now, so the main interest of object in my talk, object of interest in my talk will be lattices. So I'll spend a second defining that. So subgroup Mg is called a lattice. This is not completely standard. In different parts of mass, this means different things. But here, it means gamma is discrete and the volume of the quotient. So you can take the quotient of anything you want. Either you take G mod gamma or G mod k mod gamma. This is the same thing. So I'll just write X mod gamma here. You want this volume to be finite. So one simple way, quote mark simple way of making this volume finite is to take something where the quotient is compact. But okay, it shifts the difficulty, but it's not so easy to come up with lattices. So probably you know some examples. You've seen a lot of them go by the last few days, I think. But so I'll just tell you a handful of examples. And then we can discuss some more about what those are. So the simplest example probably you can take is SLNZ inside SLNR. Now, this is the isometric group of a symmetric, of course, for n equals 2, you know this very well, right? And bigger n, okay, it's slightly more complicated, but this always gives you a lattice. It's not completely trivial as it is. And then I like hyperbolic geometry. So the next simplest example is you do the same thing, but you take integer matrices, but that preserves a certain Hermitian form. So you take your standard Lorentzian form and you take matrices with integer coefficients. This is a subgroup of SON1, and it gives you a lattice. Again, this is not completely trivial, but it's a fact. Now, of course, you can soup this up in many ways. One of the things is these give you non-co-compact lattices. And if you want to get, well, okay, I should say what SON1 is. So I said you take the standard Lorentzian form. So you take the subgroup of GLN, the determinant one that preserves this Hermitian, this quadratic form. Okay, and this turns out to give you a non-co-compact lattice, but there's a trick, a standard trick in this business that if you throw in some extra coefficient here, then you can soup this up into something co-compact. So if you look at SOQ with entries in the ring of integers in a certain number field, then you still get a lattice there. So the point is if the group is still the same, SOQ is the same as SON1. This root two has absolutely no incidence on your group, but, well, the real group. But then the integer points change a lot here. So, okay, is the ring of integers two extended root two. Now, again, it's not completely obvious that this is a lattice, but let me just say that the key, so I should say this is co-compact lattice. I don't want to give you a proof, but I just want to say that it's a lattice, essentially, because when you take SOQ, cross SOQ, sigma, where sigma, okay, I define this as being k. So you take the automorphism of k that's exchanging the root two and minus root two, okay? The roots of the polynomial x squared minus two there. And you take this product here. It's pretty standard to see that this group is defined over the rationals. And you should think of the integer points in that group to be given by matrices with entries in the ring of integers there. And the key is that when you look at, okay, I should say SOQ sigma is a compact group, that's the key. So one consequence is that when you take the projection onto SOQ, you have a projection with compact kernel. And it's easy to see that, well, okay, it's a big theorem actually, but when you take integer points in here, you get a discrete subgroup. And because of the kernel is compact here, it's a consequence that the image is discrete also. That gives you a proof of discreteness. You need to work a little bit to see that it has finite volume also. I will not do that, but we just remember this fact here. That when you take a quadratic form, define over a certain number field and such that when you take its non-trivial Galois conjugate, you get something definite here. Definite orthogonal groups give you compact groups. Okay, that's the key. Now I should define what an arithmetic group is and I will not actually. I know I should, but unless really asked, I will not. So in general, I'm G, so I take any G as before. It's called arithmetic. We discussed a number of pronunciation issues. How do we feel that? Is it arithmetic? Arithmetic? So this stress is on the I. Okay, arithmetic. Okay. Okay. Probably if I ask an American, we get a different answer. In fact, in fact, probably there's a lot more Roman speaking. Oh, maybe not. I don't know. Roman language is speaking people. Probably would not mind of the way I say arithmetic anyway. So let me just give you a rough definition and it's given as let me say H of Z inside H of R for some group H defined over the rationales up to some standard operations or up to commensurability and up to taking this kind of homomorphisms with compact corner here. Taking subjective homomorphism of the ambient with compact corner. Okay, I could give you a more precise definition, but I think probably I'll make mistakes in writing it. I will not tell you the right conditions on G you need in order for this to make sense and so on. I think it's irrelevant. Just keep this in mind and I hope this is enough to follow the rest of my talk. If it's not, you tell me and we'll discuss this some more. The important theorem is that most lattices are this kind. So this is an important result of Margulis. Frank of X is at least two. And every lattice is actually arithmetic. Every lattice, I should say irreducible. So this is a consequence of a big theorem. This is a consequence of super rigidity, which I'm probably not going to discuss too much today. But when I first, this is a very problematic theorem. It influenced my whole career. So earlier on, it kind of got me away from rank, higher rank symmetric spaces. Because I thought, okay, so the story is done there. We know everything. In fact, there's lots of things we don't know about lattices and about discrete groups and so on. I mean, it's a very interesting topic. But so anyway, because I heard about this a long time ago, I looked mostly at rank one symmetric space. In fact, in a second, I'll tell you I'll be even more restrictive. So the rank one symmetric, one symmetric space is this way. They come into four families, which I'm not going to describe in detail. Probably all of you have heard of these before, even the students have, because they were here last week. And, okay, these are not completely separate lists, because there are some coincidences in low dimension. Most of the time, from what I'm going to say, I'm going to take n to be at least two, so that things are interesting. Otherwise, of course, H2r is extremely interesting, but I'm not going to say too many things about it. And there's a big theorem. So, of course, when you prove this, the techniques do not allow you to handle these things. Okay, but actually, super rigidity still holds in these two classes of groups. So lattices holds here. And in particular, you get the same result as Margoudi. So lattices, they are arithmetic also. Lattices are all arithmetic here. But, okay, so as you see, again, earlier in my career, I also heard about this. So I ended up thinking only about real and complex analytics. There's many, many interesting things to do there, too. Then, okay, then in fact, it's not true in HNR. NA is always going to stand from non-arithmetic. Non-arithmetic lattices in iso-HNR. For any end, she's a construction due to Gromov and Pioteski Shapiro. I'll say a little bit more in a second. And then, of course, when you see this, everyone thinks, okay, the same thing has to be true for HNC. It's not known for HNC. It's not only in low dimensions. Of course, I mean, if you hadn't seen this line, maybe you'd think, okay, if it's true for quaternions and octonions, it should be true for real and complex, too. But, okay, that's not the case. And in fact, there are, so it's not, again, the corresponding statement for HNC is not known. All that is known is that there are examples in iso-H2C and in iso-H3C. So I will need to, this is going to be, let me write holomorphic here. In the corresponding group, this is PU21 here and PU31, four dimension three. But in dimension four or higher, we have no example. So rather than telling you all the details about this construction, I just want to give you a rough idea how you construct lattices. Maybe I should say non-construction of lattices because, okay, of course, students here may find new ways of constructing them. I don't know. I'll tell you a little panorama of what we know, but there may be many other ways. It's left, it's up to your imagination. So, of course, the basic thing is you can do a arithmetic construction. This is, again, a very interesting subject. Okay, there's lots of things we don't know about arithmetic groups also, but I'm not going to talk about that today. Then, Ruth told you yesterday about another nice way to construct an arithmetic lattice. One of them is you just coxeted groups or coxeted diagrams. Okay, so I'll just show you an example. And if you get bored during my talk, same thing, try to prove what I'm going to write. So if you take, in fact, I never did it, I must say. I'm just quoting something that people quote all the time. I never checked this myself. So you take this diagram. So the dotted line with a C here means that the corresponding hyperplanes are at a parallel, and I'm going to give you the distance here. So you take C to be square root of 3 cosine 2 pi over... Okay, I have no idea where this formula comes from. I think probably I copied this correctly, but I don't even know. Okay, there's a formula like this. This is our example due to Makarov. And the claim is that for all but finitely many m, this is non-arithmetic. If you write any diagram like this, it's not clear that there exists a hyperbolic realization of this, right? But the claim is that at least for, I think for m bigger than 7, you get a hyperbolic group. And then this is non-arithmetic, finitely many values of m. I don't know what the values are. Again, m bigger than 7. What's dimension 3? In fact, I don't know of a similar construction in dimension 4. I wish I did. So maybe some experts know about this. I don't know. Maybe Ruth would know. Okay, and I should say something that I believe Ruth... I'm not sure whether Ruth mentioned this or not yesterday. Oh, actually I don't know if I believe what I told you anymore. No, that's okay. Now, what I want to say is that, okay, this group is generated by five reflections. And you remember what Ruth told you about what this diagram means, right? This means that the mirrors between the third and fourth generators there make an angle pi over 4. These make an angle pi over m. And these don't intersect. In fact, I gave you the distance between these hyperplanes. Okay? And the key is that you get a presentation for the group. It's generated by five reflections, such that, well, rk squared is always the identity for any k between 1 and 5. Then if you take r1, r2 to the m, you get the identity. If you take r2, r3 to the fourth, r3, r4 to the fourth, and I should also have a bunch of commutation relations because these nodes are not joined by anything. So you take rk, let's say rj commutator with j minus k, because you're going to recruit it. This actually gives you a presentation. This is a general fact. So all the relations you get in the group come from the reflection relations and the angle between the mirrors. In fact, the way you prove this is by showing that the corresponding polytope is a fundamental domain for a group. You use the Poincare polyhedron theorem. You know that this polyhedron is going to tile real hyperbolic space. And the tiling gives you this presentation for free with a little bit of work, but it's not term. Okay, now the next one is let me write gps for Gromov-Pioteski-Shipiro or variations. And the basic idea, I'm not going to go much into the details, but what you do is you glue pieces. Let me just say arithmetic pieces. It's not really important that they are arithmetic, but you glue pieces of hyperbolic manifolds along totally geodesic sub-manifolds. So let me draw a naive picture. In fact, it's not so... Well, let me just put it this way. You take one arithmetic manifold that contains a totally geodesic sub-manifold and another one. So it's said that you have an M1 and an M2 here. And you suppose that they are really different. You want these two not to be commensurable. Then one way to do this is you, in a way... Okay, I shouldn't... Maybe I should draw this as being shady or being gray, the other one being white, because they come from different manifolds. You take pieces of different manifolds and you assume that there's a common interface to glue and then you just glue them to each other. What you get is neither M1 nor M2. In fact, it's something complicated enough not to be arithmetic. So what I want to discuss briefly is, okay, what's left of this? These three types of constructions in SUM1. Now, as I mentioned, you can do the same trick with SUM1. You just take the... Here, you just replace this by Lorentz form. And you don't take just a real field because that's not enough. You need to take an imaginary quadratic extension. There's a variation on this. So arithmetic constructions still work. Yeah, so I should have mentioned... Okay, I had a title in my notes that I didn't copy on the board. So I should have mentioned that this was the case of SUM1. I should say that this concept of construction is kind of annoying because it works only until a certain dimension. You will not get examples in all dimensions by doing this. Arithmetic, you will. Romer-Fateski's chapter also. The description clearly shows that it's independent of the dimension. Okay, now what about SUM1? Of course, again, you can do arithmetic also. That's pretty clear. Now, I'm going to call them just like people call them, complex reflection groups. But it's a very bad name. It should really be called rotation groups and not complex reflection groups. So rotation groups is my own terminology. In the literature, everyone calls these complex reflection groups. And the point is you try to do something like what we did over there. You write some coaxial diagram. Now, you have a little bit more freedom than what you do in a real hyperbolic case because you have more freedom, less freedom at the same time. The important thing is that let me draw a schematic picture. I have no idea how to draw it any better. So if you take the complex hyperbolic plane, okay, there's two kinds of, let me write HNC for now. And look at total geodesics of manifolds. Well, you don't have so many. Of course, you have, I don't know if it's of course, but it's pretty natural, I think, to think that you will have copies of complex hyperbolic smaller dimension there. And then it's pretty natural, I think, also that you will have real hyperbolic space because you just take the real points. Think of this as the complex unit ball in CN with the appropriate metric. And then you just take the real points. That gives you a real ball that naturally carries the real hyperbolic metric here, which KC inside HNC. And still you are going to have a K less than or equal to N there. But the important thing being that nothing is going to bound in complex algorithm space because these are a very low dimension. This is co-dimension two in real sense. So none of them bound anything. So you don't have a good analog of walls of these complicated diagrams. Okay, people don't care about this so much. They take hyperplanes. They take complex hyperplanes. So I'm going to draw a picture for H2C. And I'll draw just the, I always draw the same picture because I cannot draw anything. I'm sorry. The other ones are drawn by the computer. So what I'm trying to draw here is three copies of H1C. Of course, okay, this is another thing I should say is that you, of course, it would be natural to take H2R also, H1C. Okay, it just goes completely in a different direction than what I'm going to talk about today. I want to focus more on H1C. And when you look at isometry, it's fixing a copy of H1C. Well, you still have a, since it's co-dimension two, I think it's quite natural to think that you have a one parameter family of isometry doing that corresponding to rotations. Indicated this way. You just take rotations around these copies of H1C. So I can, well, okay, I don't know. It's probably irrelevant that I'd write a formula. But in fact, this is what Mostow did, quite old paper in 1980. And he was, he was investigating groups generated by three complex reflections. Now, as I mentioned, there is an angle of rotation. So the P here indicates that you take a rotation of order P and take something that rotates by an angle of two pi over P. And then the node between these guys means that when you take, in fact, I can have any Q here on R2R3. And the node, the edge joining these two nodes here means that you want to take R1R2 to the Q over 2. You get the same thing as R2R1. So if Q here is an integer, not necessarily an even integer, I should say what it means when Q is odd. So if Q is odd, the Q over 2 means when you take a product of one or two or one, blah, blah, blah, blah. And you finish with R2. One, sorry. And here there are Q factors, but they're not the same, right? Some of them are ones, some of them are two. And Mostow was looking at a special case of this. He was taking Q equals three and some specific values of P. And the node is that in there there are some, there are some lattices. And some of them are non-arithmetic. So let me make a word of caution here. So to say that you have rotations on one or two or three that satisfy this relation does not determine a group. When you have a concentrated diagram like that, it may not, if you draw a random diagram, there will not exist a point of these angles. But if there exists one, it's unique. Well, okay, if you write random, actually, this is not quite true. There is one with finite volume. It's unique. Here it's not unique at all. If I said, think these relations, okay, there's still a parameter up to conjugation. There's still in fact a one parameter family of groups. So the coccited diagram, so the word of caution, so the coccited, the complex coccited diagram does not determine a group who P you to one conjugation. And in fact, there's, well, maybe it's not true for all PQ. I think it is. Anyways, okay. I'm not completely sure about that. But usually the typical situation is that you get a one parameter family. Typically, there exists a one, a real one parameter family of representation. Okay. And you have continuous families of representations. In fact, most of them will not be discrete. Most of the time, they're discrete. And sometimes they will be lattices, okay. But this is, so most of shows that are well chosen, the formation parameter, you get non arithmetic lattices. And in fact, I don't know if I want to say much more about that. I want to mention something else that I think with interest at least a handful of people in the room, is that you can do some kind of uniformization construction. You take, let me just, again, some of you are dozing off because it's too, it's just too 30. It's the time for a nap, right? But let me give you a homework, a piece of homework so that you don't, so in case you're falling asleep, wake up now, and I give you homework. So you take this configuration of line and P2C, complex projective space and you take, okay, if you want, you can change the configuration also. But, okay, and the homework would be really hard if I didn't give you some hint. So the hint is you take some weight and in fact, what I really want to take is an orbit fold. I want to take P2C with weighted configuration of lines, okay. So at which line I'm going to assign a weight. These three have weight two, the other one have. What I claim is that this is, no, actually, okay, I need to go to say a few more things, but what I'm going to do is to blow up some of the triple intersections there at three points. You're going to have something slightly larger than complex projective space. There's still going to be these six, six, six, and then, okay, in fact, you don't have a choice for which weight you put on these guys. It's determined by the configuration, but I'm going to give you some hint also. I'll give you 12 here. So the claim is that this orbit fold is uniformizable by the ball. So this is, so this, what's the question? You would like to blow it up or? No, you need to blow this up because these are because there's no smooth branch covering with order six, six, and two of this configuration of lines here there is. Actually, I don't, I really have to blow it up for these three points, but for that one, I don't. If you blow, if you blew it up, I don't know, you'll get a blown up ball quotient. I don't know what, I don't have much to say about what I would, what good that would be. But this is a uniformizable. This is a ball quotient. This orbit fold is a ball quotient. I should say complex orbit fold. Okay, so your homework is not to show that because that's, okay, that would take you quite a bit of time unless you already know many, many things. But how do you show that? Why do you have to compute characteristic classes? The point is there's a, so you have to compute the orbit fold C1 squared of this configure of this orbit fold and the orbit fold C2. And in fact, this, this characterizes both portions. If you get that equality, you know that this is going to be a ball quotient. That's a big theory. And this relies on the solution of the Calabic conjecture by Oban Yaw. So this is why the homework without telling you what to do would be a bit hard, I think. But anyway, this is much more general. So people have explored this. In fact, the name that I should mention here is here's a book. Here's a book played with this quite a bit. I mean, he tried to list configuration that where you get this property. That's right. In fact, okay, this is the same relationship as what CP2 does. And H2C is the non-compact dual of CP2. So you have the book for personality. It tells you that you should have the same relationship for both portions. Now, the tricky part is there's this orbit fold here, but okay. So you have to know a little bit about orbit folds in order to be able to do this homework. But if you're bored and you know what orbit fold or other characteristic is, you compute that. This one is a little bit tricky. But it's not so bad. I should say that from this picture, it's not clear whether the corresponding ball quotient is arithmetic or not. It's a bit hard. Well, in fact, I chose it so that people who know about this stuff, actually, you can check very easily that it's not arithmetic by looking just at the total geodesic law sign in there. So if you look at these lines there, we need to work a little bit harder, but you will see that the corresponding group corresponds to a 4, 6, 12 triangle group. And this turns out not to be arithmetic. So this is giving you a sneaky way to show that it's not arithmetic. But it's in general very hard. I mean, this description will make it hard to check arithmeticity. Now, I should say what the champion construction is. Mahan, you have to speak louder. I'm deaf. I don't know. Some of them do because you use just the vibration. In fact, these clearly fibers over that P1. You just look at the vibration of P2 minus a point. There are fibers over P1. You look at the set of lines to a point. So it goes through the blow up. And sometimes you can get maps to remain surfaces. That way, I don't know if this one works. It takes a little bit of work to check it. I used to care a lot about these things. And now I have drifted toward it in other directions. So let me tell you what the champion construction is. And it's the former champion construction, I should say. But let me call it the Lin-Mostow and Thurston. And I don't want to say in detail what it is. But you just look at the modular space of metrics. I should say flat metrics on S2 with pre-scribed. Okay, flat metrics on S2 don't exist. You have to do something. It's going to be flat except for some isolated points with pre-striped con angle singularities. Okay. And it's the Lin-Mostow using a very different description of this space showed that this carries as an HNC structure. Whatever, so what you do is you fix the angle, you fix the number of con angles, and you fix the angle there. And if you do it under some assumptions on the angles, you get a complex hyperbolic structure. And sometimes the metric completion is a complex hyperbolic obi-fold for a lattice gamma. And it's fairly easy if you work through the construction to determine when these guys are arithmetic. So this gave the champion. So when you work this out, you'll get nine non arithmetic commensurability classes in PU21. And one non arithmetic commensurability class in PU31. That's the limitation of this construction. Okay. So let me state what we did with, I could state it as a theorem, okay. I don't know if it's, I like to write statements, okay. But I'll just tell you what the statements are. If I write everything out in detail, it would be very long. So let me just state it like this. So it's a recent result. Maybe, I don't know if I can even say 2015. It's online anyway. It's published online. I don't know what the official date of the publication is. But we produce nine, doesn't look right? Yeah, I think that's great. Okay. Nine more non arithmetic lattices. It's 12 minus two, if I, okay. Ten more non arithmetic lattices. Okay. Ten non arithmetic lattices. And at least, leaving at least nine more commensurability classes. So, okay. The ten non arithmetic lattices we produced, none of them was in the, during most of the construction that we proved. And then almost all of them are different. There's just two of them where I don't know. Okay. So, of course, you notice that two of them look very much alike, you know. Okay. I will tell you what I mean by looking alike in a minute. But the point is the construction we did, sees very, sees nothing a priori about the arithmeticity. Okay. Arithmeticity is a completely different check. When you're, you want lattices, of course, you will not see the difference. This is the point of this picture. We all look the same. And the commensurability classes are hard to spot, of course. This is, there's some work there. I will tell you a little bit about how to check these things. So, let me tell you a word about how we obtained these lattices. It's a long story. As I was saying before the talk actually started, it's hard to tell when this started, but it's a decade project, I think. You can say that. It started around 2005, fair to say. And the origin is, well, we played a lot with triangle groups. So, but I'm going to say it's slightly differently. So, the origin, there are a lot of ways it's triangle groups. Triangle groups are everywhere in this story. So, what I'm going to do is to take R1, a complex reflection. Again, I should call it a rotation, right? But no one wants to call these a rotation. So, I should take this opportunity to rotate. You guys, you know, these young people in the room should call these rotations. And J is going to be a regular elliptic order 3. Like Pierre was mentioning in his talk this morning. And then, I'm going to take R2, R3, which are obtained just the same way as these triangles there. It's going to be R1, R4, R3, except they are 2, R3 come by conjugation by J. So, you just take R2. R3 is going to be JR2. What's your middle name, Robert? This is John Robert Denizer to be a P or Parker. So, let me tell you who P is. P is going to be R1J. And if you're bored, again, do this exercise. Actually, PQ is R1, R2, R3. That follows from the definitions there. Now, if you want to parametrize isometric classes of pairs of rotation and regular elliptic like this, I'm going to assume it's not degenerate probably. I want this complex line not to go through the fixed point of J here. But you can parametrize. Buy or trace. Let me write tau for the trace of R1J, trace of P, trace of John, Parker. Okay. And now, okay. It's just like for coxata diagrams. There's not always exist configuration of these regular elliptics and rotation there with given P and tau. In general, it doesn't exist. There's a condition that's kind of not so complicated to state, but I don't really care about. So, there are some conditions on P and tau for group to exist. But if it exists, then it's unique up to conjugation. Okay. So, now, there's going to be a number of steps in what I'm going to tell you next. Guess which choices. I'll give a discrete group. Okay. So, these took us four years maybe. And then step two is you're going to check your guess by constructing a fundamental domain. Now, these took us another four or five years. I don't know. I mean, okay. We had candidates fundamental domain. It took us a long time to prove. Then step three, what's that? Yes. Okay. That's going to be step nine. So, you're going to check arithmeticity. Step four, we need to check commensurability. So, my co-authors are very, they were very antsy when they discovered this. So, in fact, we did step three first. Because, of course, you knew we knew this would take a long time. So, if you're going to spend a long time, you want to be sure it's worth it. Okay. So, Julian actually spent a few, he started by computing these guys and said, okay, if we find lattices, they're going to be in the arithmetic. And the difficulty was to prove that they were lattices. Okay. That's the really difficult part. Arithmeticity, you have to know how to do it. Commensurability, I think we were kind of lucky. Okay. So, there's a nice necessary condition. I should say commensurability invariant. You take the field of traces in the adjunct representation. And, well, we were lucky because this adjunct trace field, this is not a complete invariant. I'm not saying that if you have the same trace field, you're commensurable. But at least if you're commensurable, you have the same trace field. And we were just lucky. This happens. There's just two groups there where you get the same trace field. And that's why I wrote nine and not ten there in this list. Okay. We don't know, two of them, we don't know if they're commensurable. So, again, we were lucky about this. Let me just write check here. Arithmeticity, okay, there's a number of computations. How do we do this? Well, you have to justify that this is correct and this is not so obvious. But actually what you do, you write these groups as preserving, I could write Hermitian forms for you to give you homework. But, okay, so if mu is e to the two pi i over three p, you're going to have a Hermitian form. You can write these groups explicitly. So, and then you make this Hermitian. Alpha is going to be two minus mu cubed minus mu bar cubed. Beta is going to be okay. And that I know now off the top of my head. You have alpha mu bar squared minus mu tau. Okay. And R1 is very simple. R1 is just given by that form. And what you do is you compute, okay, so these actually live in a number field. In fact, you can make it a psychotomy field if you want. And then you just apply psychotomy, cotomorphism to the Hermitian form. You compute the signature of Galois conjugates. And the group is, you have to check that this is true. But the group is arithmetic if and only if the non-trivial Galois conjugates are, if there's at least one that's indefinite. Remember in the Arithmeticity example I gave earlier, in the very beginning, Arithmeticity was obtained by having only definite conjugates. Here, if there's one that's not, that's not definite then you're done. What I said is not true as such. You have to be careful of which number field you use. And the number field you use is this adjoint trace field. You have to apply Galois automorphisms of, well, actually a quadratic extension of the adjoint trace field. But that's, there's a criterion for this. So let me say check here. Let me tell you a few words about how we guessed. Because I think that's a nice story. And then I'll be out of time probably. Well, let me give you the baby version of, of step one. Well, as I said, you're looking at a group generated by rotations. So just think of your simplest situations where you have two rotations. I'm thinking of this actually. I should say elliptic elements in each one. And you just join them by genetic. Now, if you want a fundamental domain for the action of B, you just take a sector like that. Fundamental domain for the action of A, you get a sector like that. And in fact, to know whether this group is discrete or not, I'm assuming that this guy has, has angle two power over P than one two power over Q. Well, okay. If the, you just look at the product AB. If the AB, if AB is loxodromic, then this gives you a group of infinite covolume. If AB is parabolic, you get the lattice, but that, that's not co-compact. And then when AB is elliptic, well, you need to check whether it has finite order or not, whether it rotates by, so let me just say if AB works so parabolic, I told you what happens. If it's elliptic, so if AB rotates elliptic and rotates by, by two pi over R, then you have a co-compact lattice. It's not quite a necessary and sufficient condition. If you have an elliptic element of finite order that does not rotate by two pi over R, it could still be discrete by accident in a way. But, so anyway, I'm going to summarize this by saying that AB controls the work, the group controls this greatness. It's a rough way of saying it, but now you do the obvious analog in our case. Okay, so you just, so in our case, what we guessed is, okay, product should control the word, should control this greatness R1J and R1R2, control this greatness. Okay, this is not true, but it's a rough idea, this greatness. You hope that this could give you something, okay? It's neither necessary nor sufficient, okay? It's just a way to guess what's going on, okay? But this is what we did. I mean, this is you hope, you hope or you try. You try this. I mean, this is a silly idea, but actually it gives you something. So what happens is, you just write down, okay, so tau is the trace of R1J. Now, if you compute the trace of R1R2, you get something that you just compute. Now, if those are going to control your group, you need this to be the trace of elliptic elements of finite order. Okay, so you need to write this as something like e to the i lambda plus e to the i mu plus e to the minus i lambda plus mu. For some lambda and mu, there are multiple, rational multiples of pi. So these are going to be roots of unity and you do the same thing here. You're going to want, in fact, this is going to be real. So you want it to be 1 plus 2 cosine 2 theta, first theta, rational multiple of pi. Now, take these two equations and solve them for tau. This is okay. I would never think of doing this, but John does somehow. I guess he's used to doing that. What's that? Okay, so I'm lying a little bit. But anyway, okay, you asked John what you need to do. Ah, okay, p equals, let me, okay, I'm doing the p equals 2 case and then, okay, turns out that the answer is the same for p higher than 2, but it's for slightly mysterious reasons I never understand. But, and you solve it, you get something that, you get a relationship between these guys. So you need theta, lambda, and mu cannot be arbitrary. That's the point. You get a very strong restriction on lambda, on lambda mu and theta. So you get an equation like this and you need to correct the preprint on your web page, I think. But anyway, you get a sum of, remember, you want lambda mu theta to be rational multiples of 2 pi. And you have a relationship between cosines there that has to be actually a very simple cosine. This gives you a rational relationship between cosines of rational multiples of pi. Well, okay, if there's relatively few cosines, this is very restrictive. So you just use this beautiful paper of Conway and Jones, classify these sums, the possible sums as above, mu theta in q pi. And okay, there's an infinite list. Okay, probably you can guess that because if you take supplementary angles, they have opposite cosines, you can get silly relations there. So there are infinite families, but on top of that, there's discrete isolated family situation. This is what Julien and John had done when I joined the project. So you get that if this, so it's quite tricky, I must say, to use the Conway-Jones paper to get the list. Like I was mentioning John the other day, I tried to do this a while ago and I just started filling papers of computations. After two pages, I gave up and I thought, oh, this is what John had done. I don't know how many pages you need to do it, but it's pretty tricky. But so the point is that if this happens, then you get the solutions are of, let me say, two types. Either you get subgroups of master groups, which I don't care about because these are not going to give us anything new, or tau is in a finite list, tau is in a list of 18 values. Aha, this means I'm out of time. Here are the values. They worked out this list. Again, it's not so easy to do. The ones in small letters means I don't find any lattice there. I don't really care about them. The ones in bold there actually contain lattices, but still you have to choose p. So you cannot take p arbitrary there. Well, you should remember there are three relevant values of tau there, and there are three families there. Actually, there are three families on the left, three families on the right. These are the same pictures projected differently. The point is the parker isometry there is regular elliptic. It has two invariant complex lines, and I'm doing orthogonal projection on either one of these axes. So that's why there are two pictures, the one on the left correspond to one axis, the other one. So these are the same pictures, really. At the bottom, it doesn't look like they're the same, right? The point is the sum. So if you take p to the five in that case, you get a complex reflection. So there's a bunch of things that line up in one of the projections. It's a little bit difficult to understand, as I say, but it's actually quite natural. These are not quite fundamental domains actually. They're invariant under p. And p has somewhat large order. In the bottom groups, p has order 30. So this is why the picture is very complicated. It's invariant under an isometry of order 30, but actually the combinatorics are not so bad. It's just you have quite simple faces. Let me show you what the faces look like. I thought I was going to tell you what the numbers are on these edges of the polytop. Let me just tell you in one word. So when I started playing with this, John was writing the pictures. He was drawing this picture by hand and drawing all these things by hand and so on. I never understood how he did it. So I spent, I don't know, months asking him, but what are you doing, John? Okay. And now we finally understand this is why the computer can tell you the picture. The computer can draw everything because, okay, I think we have clarified it. We know exactly what John was doing. And we have a proof that these guys are lattices, that they are non-arithmetic, that at least nine of them are new. And I think I'm going to stop here because I'm out of time.