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.  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.  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.  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?  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.  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.  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.  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.


Welcome back to This is Alden Library. In this installment of the series, we will be giving you a crash course on what you need to know about the fifth floor. The Cornelius Ryan Room is dedicated to the work and legacy of World War II correspondent Cornelius Ryan. Currently on display is not only Cornelius Ryan's work, but pieces documenting the experiences of others during World War II, such as women and citizens of Athens at the time. The room is open Monday through Friday from 9am to 4pm. Located across from the Cornelius Ryan Room is the George V. Voinovich Seminar Room. Anytime it is not being used, the room can be used as a quiet study space. There are tons of great study spots across the entirety of the fifth floor of Alden Library. From elevated tables to individual booth style seats, there is space for anyone to come and work. Also located on the fifth floor is Ohio University Library's Government Documents Collection. The collection contains U.S. Government Documents, Ohio Documents, United Nations Documents, and Library of Congress classified maps. Anyone interested in using something from the collection should head to the fourth floor desk for assistance. If you've ever wondered what the thin, horizontal drawers are throughout the fifth floor, they contain Ohio University Library's map collection. Adjacent to a quiet study area and tucked away on the fifth floor is the United Meditation Room. The room's neutral colors and minimal furnishings encourage tranquility and peace for anyone who wants to use it. The Mons Center for Archives and Special Collections includes University Archives, Manuscript Archives, Documentary Photography Archives, and the Rare Book Collection. To access the archives, students should make an appointment via a form on the University Library's website or by emailing the curator in charge of each collection. I'm Dr. Miriam Intritor. I use she, her pronouns, and I'm Special Collections Librarian for Rare Books. I'm responsible for managing all aspects of the Rare Book Collection. We are here first and foremost to help students with their research, and that's what we like doing best. Thanks for checking out the fifth installment of This is All in Library. Follow Ohio University Libraries on social media to make sure you don't miss the next part of the series all about the sixth and seventh floors of Alden, which is coming soon.


Jason Kidd, by the way, which I find to be an interesting part of all this that will now be in charge of coaching Kyrie Irving and sounded pretty optimistic about it. Here he is. We have a relationship. He's all about basketball. He wants to win and he wants to be coached. And this is a great opportunity for me to have someone like this to help to split my time between those two and the rest of the team. It's a great challenge and it should be fun. Look, I mean, I was with the Nets when JK was a point guard there and he had some moments that were very dramatic and roller coaster like, although I don't know if they even compare to some of the stuff that we've seen lately. But Eddie, are you buying this? Are you buying that he is underneath all of this chill about what's about to happen? I think so. And I think he's looking at it as talent. They needed talent on that roster and he wanted a dynamic score next to Luca. This is the coach that, you know, for as much of a line as Jason Kittes as a coach in his Brooklyn tenure and then the way his Bucks tenure ended. He's the guy who said, yo, Yanis is a point guard. We need to put the ball in his hands. He's the guy who has been hard on his point guards, demanded a lot, but also got a lot from them. And he's one of the greatest point guards of all time. It'd be very interesting to see how he decides to mess the two, but he's familiar. And he's familiar with the years he spent coaching LeBron as well with kind of dealing with that heliocentric part of that offense and ways to milk their other guys on the side. I'm interested to see what he does with them. And, you know, I think there's only one Luca style, but they got to find a way to get the most out of Kyrie as well if they want to contend. And I think they're going to, I actually think it's a really good pairing between player, player and coaches will. God, people seem so sure about this. Maybe I'm just a negative person, Chandler, but I don't, I don't necessarily think this is a slam dunk a hundred percent. No, it's not a hundred percent. But, and again, let's not forget he, Kyrie Irving just played for another really good point guard and Steve Ness that just got fired. So listen, it's at the end of the day, I think it's worth the risk to go and get a talent like Kyrie Irving. He is a generational talent. He's 30 years old. There is still a lot of basketball to be played. And by the way, they have his bird rights so they can go and sign him for three to five years. If it works out, go get your bag, play well, make a run, mesh, lead these young guys, lead by example. He has, he has the perfect opportunity right now where he doesn't have to be the best player. He's got an X NBA player coach that's so relatable. He's got guys like Nico, who he knows from Nike. He's got Michael Finley. He's got Mark Cuban. He's got the coolest staff that you could ever assemble around him to be successful. So if this doesn't work out again, it's probably it for him. No team is going to take a flyer if he now kind of is a negative addition to another team. Everyone's looking at Kyrie Irving.  And this is his chance to kind of put all the drama behind him finally, which we always say, but go there and play basketball and win games. And I think he has a lot of basketball to play left. Y'all are so cute. I just love the optimism and the idea that people change. This is a Dorbsie's part of this show. Shoms final package in the trade included Mark Heath Morris, which I feel like he's not getting enough attention. We haven't even mentioned him. Where do we stand on how people are receiving that part of the trade? I mean, he was part of the trade to make it a two for two. So he's part of it in that way. But listen, Mark Heath Morris, great veteran player, him and Kyrie Irving toward the facility yesterday in Dallas. I'm told it was a great meeting for Kyrie Irving with Mark Cuban with Nico Harrison. I think they all hit it off. I think they all feel very, very optimistic about this partnership and really Kyrie Irving's entry into a new organization, a new culture. It's a totally different dynamic with Mark Cuban as Chandler knows well, just a really hands on owner who treats all of his players, I think first and foremost as human beings and trying to relate with them on a daily basis. They're not always going to agree on everything, I'm sure, as Mark Cuban and Chandler Parsons, I don't think agreed on everything. But I think at the end of the day, try to see eye to eye on every single factor. And I think the biggest thing has been, how does Luca Dancic feel about the trade of Kyrie Irving? From what I'm told, it's been nothing but optimistic feelings. And I think just positive vibes from Luca Dancic about the arrival of Kyrie Irving. I think Luca Dancic views Kyrie Irving as a guy that can shoulder a lot of the load, can decrease his minutes at times, decreases his pressure and burden to score and create a lot of the similar to the way that Jaylen Brunson did last year for him. But like Jay Kid said last night, no knock to Jaylen Brunson. We're talking about Kyrie Irving. So I think overall there's a lot of optimistic vibes right now in Dallas. I mean, worst case, you get two good months, right? There you go. There's the silver lining. You've just seen a couple really great months in a row and make it work.


together. Well, a new study is showing the U.S. could drastically lower its number of coronavirus cases by simply wearing a mask. According to the data from UC Berkeley, if 80% of us wore a facial covering in public, our COVID-19 infection rate would drop to about one-twelfth the number of cases. So why is it that only 14 states require it? Joining us now is someone who's really striving to change that co-founder of the Global Masks for All movement, Jeremy Howard. Jeremy, thank you so much for being with us. Bring us up to speed on your efforts. Thanks so much for having me. So I'm one of the over 100 signatories to a letter that was released yesterday. It does talk about that research and other research. And the letter asks that all governors require masks in public. This is a really big step you mentioned earlier in the segment about the kind of relationship between roughness and debate. But let me tell you what the scientists on this letter include a Nobel Prize-winning virologist, a Nobel Prize-winning economist, the editor-in-chief of Nature, the world's top science journal, the editor-in-chief of the Lancet, the world's top medical journal. This is something which scientists consider so important that they are coming out publicly and saying, the science says we have to do this. If we're going to see these lockdowns end, avoid a second wave, avoid the death count getting bigger and get the economy back up and running again. Jeremy, can I ask where do you live? So I'm in Houston, San Francisco. I'm currently with my family in Houston, Texas. Okay. Do you see a difference between those two places as far as how people react to wearing the mask? Oh, yes, absolutely. So both of those states do not require masks. Only 14 states require masks, but my hometown of San Francisco does, as does Houston. So there's a kind of a difference in between cities, between states. What I've seen is there's a lot of kind of confusion, which is quite reasonable. Let me ask you, is it a confusion or is it culture? Because I've lived in California and I've lived in Texas. And I think in Texas, you're going to have a hard time requiring the whole state to wear masks. It could be. I mean, we'll have to see. I think what we've seen happening, though, is when the death count goes up really high, like happened in New York, that it wasn't until that people started saying, okay, we'll take this pill. So if winter comes and the death count shoots up in Texas, I think people might say, well, given a choice between my family dying and wearing a mask, maybe I should just wear the mask. Is that what it's going to take? Because we're, we in the United States are kind of hard headed. We can't be. Yes. We can be. So I can understand how all the data you have is probably absolutely true. And yet Americans, they don't like being told what to do. I get it. I really do. But look, surely it's better than being locked up inside. Don't we want to get back to work? Right. Here's the thing. Sometimes we accept some little infringements. For example, I don't want to go out and sit on a restaurant seat where the previous customer was not wearing pants. So we do have laws in this country that say, put on some pants before you go out. You have a great sense of humor, Jeremy.   So please wear one. You think sometimes, though, Jeremy, it takes almost a tragedy. It almost takes you being personally affected by it. Say, for example, my mother-in-law, unfortunately, I lost her to COVID-19. So for me, you know, there are times when I take a mask, I try to be responsible. But I think that I would have a little bit more of a feeling like, oh, this is something that I should do for myself and for others. But you know what? That mask isn't really protecting you that much, just a little bit. The main thing it's doing is it protecting the people around you. So, you know, I have people in my family who have serious health risks. In fact, 80% of Americans, I think, are considered to have comorbidities, reasons that they'll be more at risk. And the thing is, they're not safe unless all the people around them are wearing masks. So if we want the economy to really open up, getting rid of the lockdown won't be enough. We actually need people to feel safe. And that means people need to wear masks. Jeremy, thank you very much for joining us and having such an interesting conversation. And good luck in your


Hey there chameleons, it's Camille and today I'm going to show you guys how to properly bleach your roots and fix banding So the first step we're gonna put on our gloves as usual protect your fingers And I'm starting with some coconut oil already in my hair that I put on last night to help protect it Then I'm gonna take two scoops of my favorite bleach Which is L'Oreal quick blue and I know I've answered this question before if I had a favorite bleach And I didn't until recently so there you go Now I'm gonna use a 40 volume developer I'm using the cream developer because it's the easiest to maneuver just the salon perfect brand and some Olaplex and the only reason I'm using 40 volume is because the Olaplex brings it down to a 30 volume and You don't want to fry out your hair and the reason I'm using Olaplex is because it protects your hair from some of the damage That bleaching normally does to it So now I'm gonna section my hair into four sections starting down the middle and then doing a crosswise section So kind of like if you were playing four square just like that Just do like four squares Then you want to use some pins to just collect your hair into the four quadrants You can leave one section out because obviously you have to work on one section first So it wouldn't really make sense for you to pin all of them up I'm also pulling one side to the other side so that it's really out of the way So when I'm doing the bleaching it will just be pulled to the other side and make it a lot easier Anyways for this I have really overgrown roots And I'm taking the more professional approach for this and I'm actually using a brush this time And I'm only applying the bleach to the sections that aren't Directly next to my scalp So I'm leaving about a quarter of an inch free and applying bleach on the rest of that Then I'm using my gloved fingers and just kind of squishing the hair that has the bleach together Just to make sure that the bleach is very saturated on all the sections of the hair I'm pulling very thin sections of each quadrant and just coating it and bleach leaving that quarter inch of my roots For later because that quarter inch processes faster than the rest of your hair So if you want to have more of an even bleaching you have to do it this way Which I never really did before but I'm really just trying to fix any banding and just do this as properly as possible So then I can have a more even base for my color this bleach and developer combo doesn't really swell So you can just put it exactly where you want it to be So now you're gonna take out your next section that you're gonna work with and do the same thing You're going to section each section lots of section and going on here into very thin strips And just use the brush to paint on the bleach on both the top and the bottom sides of each of the section to sections You don't really have to squish it together if you're doing it properly or professionally But I like to since I'm doing on my own and here's a close-up you can see that I'm leaving a little bit of the roots untouched as I mentioned earlier and That also does help with the bleach feeling so it's not going to be touching your scalp So it's not going to burn at least initially so that's something that I enjoy not burning And then once it's all covered and you look like a freak It's time to cover that all up with a plastic bag Now you want to leave this on for as long as your bleach should be left on which is a little bit different for any bleach I'd recommend about 20 to 30 minutes But you want to just look at your hair and make sure that it's lightening And then you're gonna go back in with your bleach and just coat all of your roots now while those process You want to also add a little bit of extra bleach to any parts that have banding So the overgrown section of your roots that aren't directly on your head need more processing time And then the roots and the banding need to process less So you do those at the same time and then leave everything on for an additional however long it needs to be on I'd say about 20 to 30 minutes and then once you're in set out I'm going in with step two of Olaplex to help repair those bonds that I just broke Because once you break the sin ting you need to fix it. So that's one way to help restore health to your hair I recommend doing any bleaching at least a day apart. This is a different day Where I've just done a round two and then I put bleach on the bottom banding also Leah was there when I was doing this So hey girl, look at that cute face Anyways, so now I've rinsed out the second round of bleaching where I did exactly the same thing I'm putting in Olaplex step two throughout the bleach parts and then just running it through my hair And now I'm doing a very simple bleach bath, which is just a scoop of bleach Some shampoo some people add developer I didn't this time and then I'm putting that on the green sections of my hair and This is just going to be kind of a gentler way like compared to bleaching of getting out the remaining dye in your hair So once I rinse that out another day later. I am toning my hair So this is the blonde brilliance toner and developer I've used platinum before but I decided to do lavender this time You want to use a full tube of that and then half of a can of developer for that and I just use 10 volume just so it will Just work instead of lifting my hair more Anyways now I'm putting this toner onto my roots all of the yellowy sections Because this is going to get out the brassy tones all that yellow that you don't need ruining your life And it's just going to help bring your hair to more of a blonde and use its ashy cool toned colors to counteract the warm I'm putting that on all of the roots and then just bringing it down a little bit so it does fade into the rest of the hair So it's not just a stark transition, but you want to focus it on the roots of your hair Once that's in you don't have to wait long and it will already start working its magic You can see that the lavender has shown up on the whitest parts of my hair Which is to be expected, but that also just kind of washes out But anyways, this is what it looks like after I've bleached everything in the way that I described I've gotten rid of some of the banding and my hair is ready. It's light enough It's a blank canvas and it is ready for some coloring girl I hope you guys found this video helpful Don't forget to subscribe to my channel if you did and if you want to stay tuned so you can see what the next hair Color is I have links to all the products that use down below as well as my social media and my vlog channel So thank you Camille and so much for watching and I'll see you in the next video kisses


What's up everybody once again it's Brandon Shawn and we got to talk about this clip right here. What do you think for a new artist is a way that they can help like attract these brands?  Get a nigga that is a business manager or CPA something in that vibe and project yourself as a business like as a brand yo in this year I'm gonna make this much and in 2025 I'm gonna make this much make that shit like like like legible so you can see it right and in that same breath at that at that point in time if you can prove that numerically you can you can get that money right you can really get that money shoot an email to somebody like there's a lot of niggas with money that are trying to figure out a barrier a lot of niggas not paying taxes trying to figure out how to offset it we don't know who that is that is justice one of the founders of love renaissance love renaissance as in black love renaissance as in drum Rory group from Atlanta in bed with in a scope all that good stuff and what's important is what he was talking about actually the fact that artists need to be more professional in how they handle their business and that's ironically what we start to talk about late last year right that was the reason that I started to talk about packaging how do you bring yourself together not only from the fan base side of things but also the business side of things because those are two different types of branding particularly when we talk about packaging what looks good and resonates for a fan is going to be different than what resonates for a business because they need to connect with you and see that you can connect with the people that they're trying to reach but at the end of the day is going to come down to the bottom line and how you affect them as a business not whether they like your music or listen to you in the car again I said this in a video recently there's a lot of artists that do not have as many sponsorships or professional deals as artists with far smaller fan bases than them why it's because how they're branded how they're packaged and their teams understanding of how to navigate the business side of things not only do you need to be able to have your aesthetic look that brand understanding what that brand is but you also need to be able to codify this stuff in a true balanced accounting format you people need to understand how much money you're making how much money you're bringing in even if it's not a lot just the fact that you're tracking your money you're showing that you're tracking your money and maybe have projections that alone shows that you're handling your business in a certain fashion because there's three types of primary things you will encounter when it comes to these businesses some people wouldn't want to partner with you because they want some cloud right they got a good business going but they need to get into culture they need to connect with people but then you have some people that partner with you because they want some money so the cloud is good and whatever but they needed to affect their bottom line financially of course there's people who want both but then there's some people who still they might not necessarily care about what's working with you to affect their bottom line but they still will not take you seriously if they do not see that you candle yourself like a real business they don't see that you're profitable or working on being profitable having a plan to get there I understand for many it is a different language and that's one of the things that I'm working on this year specifically in terms of certain types of content but keep that in mind you need to carry yourself as a business as a business as a business I cannot say it enough because you are a business you cannot consider yourself an entrepreneur I know some people so many people say I'm a boss I'm an entrepreneur you know all that kind of stuff but if you want to do business with people you need to be perceived as a business that's a part of branding right we always talk about marketing and branding it's cool if you have a business technically on paper you're registered with the government all that good stuff but it doesn't matter if when you're in the mode of doing business with people or trying to do business with people if they don't perceive you as a legitimate business yourself you're trying to market and brand to them just like you're also trying to market and brand to the fans it's just different languages and you got to learn how to switch and that's it if you liked this video go hit the like button if you like your minds will share it and if you're not subscribed you know what to do hit that subscribe