 I actually was an official at a wedding. Actually, it wasn't a real wedding. They had been married before, but they were having their big ceremony. And I was the official at their wedding. And the girl who was, and I call her a girl because I remember her as a girl, she's a friend of John's daughter, Bill's. Lizzie, Lizzie. So when she was a little girl, her father, who's Bill Goldman, took her to school one day. And he took her to a class. He was giving a graduate class. Actually, John knows this. Were you there, John? And at the end of the class, he said, well, how's everything going? And Lizzie piped up, said, daddy, daddy, daddy, I want to understand more. Could you draw more pictures? And I thought, and Bill said, oh, that's a good idea. I should draw more pictures. And as I rushed through the last talk, I thought, I should draw more pictures, too. So I'll try endeavor to draw more pictures. Young Jo is in charge. You're supposed to say, you need to draw more pictures, Todd. That's your job today. All right, so where were we before? We were talking. I want to kind of go back and kind of revisit everything in a more general context than we did before. And start to use words that start to define some words that will make things a little bit easier to say as we go ahead. All right, so we're going to talk about representations of groups a little bit. Just a very little bit. The whole idea of representation theory sets my mind a Twitter. Well, no, that's wrong. It sets my nerves a Twitter. And so a representation of group is just this homomorphism from this abstract group into GLNM, if you want. You can do it more generally. And so you can ask it. Most of my representations are going to be faithful. They're going to be one-to-one. But I'm also going to be talking about an affine representation, which is going to go into the isometry group of this affine group, e to the nm, this different signature. And that's actually a subset. You can think of it as a subset of GLNM or a bigger, actually GLN plus 1, n plus 1, I forget, or plus 1, general linear group. And plus that, well, it's just, oh, shoot. It's n and m. OK, we're going back. N and m? That's just your signature of your group. You can put it into GLN. Can I do that? Yes, I can. Well, I don't have to do that. I'm doing isometries later. I can just put it into a group. I can just put it into a general linear group. Any representation? Yeah, n plus n. Yeah, oh, yeah, I should have n plus n. Oh, you're asking for money. That's what it is. She whizzed. I'm supposed to give money to Ravi? I don't think so. I don't think I'm giving any money to Ravi. That would ruin his final typos on me. Anyway, we like to think of these as these elements, which we wrote before. We wrote these things before. This is the S, O, N, M, an element of this looks like A, A, where this is an element of SN. And we have the standard multiplication. So you think of how this acts. This is the linear group. This is the linear action. This is the translation. And you can act on this. This goes AX plus A. And now if you want to write two of these together, OK, I'm sorry. So if we write two of these, OK, so this is VB, AX plus A. So this is BX plus A plus B. And so BAX plus BA plus B. So you can see that they're in the group. And you can see that you have this twisted product here where when you take BB times AA, this is just BA, the linear parts just multiply as usual. But the translational parts, the vectors that we're adding on, get this twisted part. So it goes like this plus B. So that's the group that we're talking about. And we already use this group. But I want to kind of write this out because we're going to think about these elements here. OK, just the projection from this group, OK, let me just erase this for a second. But the projection from this group to just picks out that element, which I actually have written up there. I don't know why I had to write it twice. But I'm writing more, slowing myself down. And G is usually we think of these G's as abstract groups. And the gammas are now sitting in this S, O, N, M. No plus there. Thank goodness, John. I'm not. This map, there's also a map, right? So the map goes from the element just to this part. See, the linear part goes just to the linear part. But the element goes to this other one. And that's the U map, right? There's the U map, right? It takes this group. It is actually from not the abstract group, but the element that you're going along, right? This gamma into A, right? Where that's the translation, OK? You can put it from the abstract group if you want. All right, so here's where I want to start really some drawings and people. So I have these representations, and I want to think about these things, all right? So one of the things that we have is we're going to take a group, and we're going to deform it. Well, what do I mean by deforming it, OK? So we have this group, G, this abstract group. And we take a representation phi. But we can also think of this. This is just living on some path of a bunch of representations. And so you can see what's happening is as you move along in this group, you're deforming it. You're changing the image of that group. So you can think of this as a deformation of gamma. And you want this to be continuous, right? GLN lives inside, you know, RN, right? S-O-N. So you can just think of these things as continuous group. Same old continuous crap that you learn when you were wee lads and lasses, as if you ever were. But everyone's here from Scotland. You have to be a lad or lad from Scotland. Well, there's another deformation that I want to talk about. This is a continuous deformation. This is the thing that we're going to talk about a little bit, OK? What? You can do that. It can change from this could be a faithful to a non-faithful representation. But in general, I want these all to be faithful to generate, one-to-one maps. But one of the things that you can think about this is that you can think about, well, let's say that you have, let's say that gamma is inside of here. And then, well, let's not do that. I'll do that in a second, OK? There's another representation, which is what I mean, deformation, which is an affine deformation. So it's not really the same thing, right? It's really just taking this group and you're mapping it into gamma. Well, you're mapping it into the isometry of what's really into this from gamma here. But you can think of gamma as isomet. Especially if you have a one-to-one map, you can think of G as isomorphic to gamma. And then you can map this into the affine group, the isometry of M of E and M. OK? And we'll just consider them to be positive ones, where this is where you have this element where you're just taking phi of A is A plus this U map that we talked about before, right? This weird map that we talked about. It's not a homomorphism anymore, but it's this twist, the twisted part, OK? So here we have a deformation. We have the group and we're really deforming it. Here we're keeping the group the same in a sense, gamma, right? And we're changing it just by adding on this thing. Now what we're going to find out that in certain situations, especially in our situation, that we're interested, these are going to fit together very, very nicely. That there's going to be a nice way to think about one as in terms of the other. So an affine deformation is going to be related to a deformation of groups. All right, so one way to think about these things is that you have these groups, and we want to kind of parameterize this space. And that's the first part of my talk, is I'm going to talk about the parameter of these spaces, the parameterization of these spaces. And I want to, so these things are cobalt, these things are called co-cycles. E is just this affine space. It's affine space, this E right there? Oh, the identity map. So the linear part under that, it goes back to the identity. So it's just saying that you're just building on. So one is you're taking a group. You can think of this as deforming gamma, right? This map is now a deformation of gamma, but it's just adding on this linear part, OK? Yes, yes, yes, yeah. Because that's the tangent space. Remember, these are these maps, these U, these A's, right? The images of these are elements of your group gamma. These are vectors, right? So that's why they are, they're moving your point, right? It's A, X plus. I don't mean that's the same as this point. Yeah, yes, right. RN stands for the Tangents Voice of EN, right? And now, of course, one of the things with the flat group is that it's all the same, right? We can make the identification between ENM and RNM. OK, and so we can do these. These are called, this is called a co-cycle. This is the co-cycle, right, U? And then we have co-boundaries. Basically, co-boundaries are if you take a translation, if you conjugate by a translation, right? How much does U change, OK? So I don't want to worry about conjugating. I don't want to worry about where my origin is. I don't want to think about where the, so I want to throw that away, OK? And like we do in all things, co-homology, co-homology, you take co-cycles, mod, co-boundaries, like you did when you were a wee, well, not in kindergarten. Let's say that. I had a Swiss professor one time. You learned this in kindergarten. I'm like, I didn't learn that in kindergarten. You went to a different kindergarten. I guess Swiss kindergartens are really impressive, because he was trying to teach me general relativity, and I didn't get any of that in kindergarten. There you go. All right, so this space is this thing. So all this space is, it's a nice way of saying, I'm going to look at all these different types of affine deformations that I have, modulo the translational conjugate. Modulo changing the origin, in a sense, OK? So that's all that I'm interested in. All right, so now we're going to talk about ranked, I'm going to talk about three dimensions. I'm going to make myself easier, make my life easier. I'm going to talk about R21, like we talk about, and we're going to talk about ranked two spaces. So before I go into what we're going to do there, let me draw the ranked two spaces here. Free groups of ranked two, OK? I should say free groups of ranked two. And so the question is, how many different free groups of ranked two are there? And so this is an aside, OK? Well, if you have a free group of ranked two, how do you get a free group of ranked two? How would you get a manifold that has a free group of ranked two, or not even a manifold, but maybe a manifold with a point, some distinguished points? I take two circles, I put them, right? That has fundamental group pi z cross z, right? Not z cross z, z star z, OK? This has a free group of, and what you'll do is you'll fatten it up, right? You'll make it into a manifold, all right? And I'm going to fatten it up. You don't have to draw these on the back, OK? And so what I've done is I've taken this and I've fatten it up like that. There's my manifold right there. So let me draw this again without so many lines. It's basically I've taken, well, we don't have to draw like that, we can do it as a circle on the top, taking a circle, taking a circle like that, all right? And so how many different ways can we connect these two? What can we do to these two to connect them? Well, what we're going to do is we're going to fatten them up. We're going to connect the top with the bottom with just strips, all right? So how many different strips can you imagine, OK? Well, there's going to be four different ways to connect, OK? The first way is the obvious way. You just take the top one, you go like this, go to the bottom. These are called ribbon graphs because you're adding ribbons and I just do it like this. And there's my manifold, all right? Free group, the fundamental group is free. You see the figure eight right inside that, great. I'm happy, I'm happy as a lark. So this is the three-hold sphere. Well, what's the other way to do this, to make it a different manifold but still keep the same free group, all right? You can add a twist, all right? You can twist one of the ribbons, OK? So we'll do that again. Now, rank two is a little bit easier because you don't have to worry about all the different configurations that you have, but everything, all the higher ranks, with a little bit of tweaking, can be done in the same way. So what you're going to do is you're going to take a twist. It doesn't matter which way you twist. One twist, right? There's the twist. You've taken it and you just switched it, right? But these two, you keep the same. All right, so the manifold is this stuff right here. Here's the manifold, right? Here's the manifold, right? Just like this was the manifold. How many boundary components do you see? I see two. Do you see something different, John? Because sometimes I worry about you. I worry about John's classes. Now I see seven. Who knew? OK, but what do I do here? Because of this one twist, I've actually made this. If you go from here and you do the right, you know, put your vector pointing outwards, where does the vector pointing inwards work? When you go around the twist, it points back. But then you come back down here, it's the same way. So this is a non-oriented manifold. There's a name for this, which I'll say is not popular, but our old friend Bill Goldman is trying to make it popular. It's from John Conway's thing. He calls this a two-hold cross-surface. It's a RP2 with two punctures. Two-hold cross-surface, OK? All right, so what's the next one that we can do? Twist, twist. Doesn't really matter what the order is and straight. How many boundary components do you see now? Well, I only see one. I mean, John sings several, but we're not sure about him anymore. We thought we knew him, but this is also unorientable, right? Because if you go here, you take a vector that points out, points in, you go the whole way around, it's pointing the other way. So it's unorientable, one-hold, orientable. You think, what does it have to be? Well, it's going to be, I have to write it up here because it's down low. It's the one-hold climb low for other reasons, but, OK? The last one that we're going to deal with, ah, shoot, it's easy to do, twists all around. How many holes do you see again? How many boundary components do you see? I only see one. Again, John is seeing several, but again, we're worried about. This one actually is orientable. Still, it's back to being orientable because, let's say you take a vector here, you move around here, but the twist turns it to the back. But any way down also gives you a twist, so it moves it to the back. It's an orientable thing. Well, what's orientable? Free group, whose pi one is free group of rank two. It's the one-hold torus. So every, this is kind of a classification of all these, it's a topological construction of all the different surfaces that we can get. OK? All right. So, let's go, let's first, now we're done with the topology. These are the four objects that I'm going to draw pictures of. And then I'm going to get to the real heart of what I hope to do after that. Some of this is some calculations, which I'm not prepared to do today, but none of these calculations are hard. OK, so now let's talk about a rank two group, OK? So now it's a rank two group in, what's our favorite group of the week? Everyone's favorite group that PSL2R, what's just PSL2R? That's our favorite group, everyone's favorite group. Francois's favorite group, John doesn't have a favorite group because, well, he has weird favorite groups because they're much more complicated. He's much more, much more, I don't know why I took it out on you today. I know, I know, this is going to be really bad. Who's talking next? Oh, shoot, I shouldn't have done that. So you have two elements, right? A, B, C, D, right? A1, B1, C1, D1, and A2, B2, C2, D2. And again, OK, if I have any group, now it's not too hard to do. It takes a little bit of counting here. But you can tell, up to conjugation, again, up to conjugation, you don't worry about, here it's not an origin, but you just can conjugate. So I don't really care about conjugation. I can tell you what this group is by three numbers, OK? The absolute value of the trace of it. The generator, one generator, the absolute value of the trace of it. The other generator, and C. What's C? What's C? I don't know it. Oh, C is just A, B inverse, right? Because A, B, C is the identity, right? This is a different presentation of this group. The group that we're used to thinking about, we're used to thinking about a free group on two generators. The way I usually write it is like this, A, B, right? I just have A and B, right? But you can write that group another way. You can write it as a group on three generators with the relationship A, B, C is equal to the identity, right? C is just equal to A, B inverse, right? A, B, inverse is a thing. We like this, we like to write things like this because, well, because, well, first of all, you can tell what the group is by the three traces. Up to conjugation, you can tell exactly what the group is up to three traces, OK? And one of the things that I'm kind of being fast and loose here is also, I'm not worrying right now. I have to tell you where I'm worrying about it, where I want. I'm not really worrying about that I'm worried about PSL2R as opposed to SL2R. You really want to think about the trace of these three, these are actually three traces and you want them to be the sign to be in there. But let's just worry about the absolute value for the time OK? All right. You can take one of these, one of the ways you can do this, you could take two representations, two different representations of the group. You can take one, multiply one of the elements by negative one. Or you can change, and that'll change the other one to negative one. So there's this modification that you can do. You can take any two generators and look at their trace and say, oh, you can multiply two of them by negative one if you want. And that's the same group. It's the same, well, it's not the same group. It's actually a different group, but it acts the same on your space. No, it's just a counting argument. It's like you're solving a bunch of equations. Notice that because these determinants are one, you only have three variables. The trace takes out two other ones. The trace tells you some other elements. And then conjugation takes the three elements away, too. So that's why you only need three. OK, one of the things that we're going to do here is we're going to actually change the representations a little bit. The last talk of next week is by a young man called Sir Piao Tan. I call him young because he's my age. And he just loves things like this. He loves moving around these groups and changing the representations and following these paths as boat-ish kind of things. Oh, just get all giddy if Sir can get giddy, but he doesn't. OK, so that's the way we talk about the linear part. We have these three numbers that tell you what they are. Now, here's another thing. And again, this takes a little bit of time to do. I'm not going to do the calculations. It's a small calculation to do. But if you fix that linear part, if now you look at an affine deformation of this, we have the same sort of thing. We have three numbers that tell you what the affine deformation is up to conjugation by a linear part, by a translation. And those three numbers are, well, there's lots of three numbers you can pick. But the three numbers I pick are alpha of A, alpha of B, alpha of C. Now, the Margules invariance. And we're going to talk a little bit more about these Margules invariance. This is going to be a big player in our thing. OK, so that determines what your group is. So what you're saying, the translation is your cohomology of this is three dimensional. It's defined by these three vectors, these three numbers, alpha of A, alpha of B, alpha of C. You know what alpha of A, alpha of B, alpha of C? You're understanding that you have to fix that linear part. You're not changing the linear part anymore. Fixing the linear part, you say, what kind of affine deformations do I have for that fixed linear part? And those depend on three numbers. So now we're back to where we were talking about last time. We're talking about proper deformations. Proper affine deformations. I should say that there should be affine up there. You owe me 50 cents because I found the typo before you did. I don't know why John suddenly owes me money. OK, I'm going to take away the traces, too. I'm not going to worry about that. The traces, too, is a bad, bad place. We're just going to throw, I should, in all of this, you're just really, you're looking for that cash. Are you feeling poor these days? No, the trace of two is a problem because parabolic elements have trace two, which we did not talk about too much. I haven't seen that too much today. But parabolic elements have trace two, and also the identity has trace two. And that's a problem. That's a big, big problem. But we're not going to deal with that. I'm just going to avoid that. I apologize. I should have said that. So what you want to think of, OK, I'm going to get rid of these. So now I'm fixing my linear part. And I want to look at the space of all deformations, OK? So I'm fixing gamma. And I want to look at H1 of gamma. This is in SO. Let's keep it to the identity component because that's a bad thing. SO1, R21, OK? It's a three-dimensional one. And I want to look at proper deformations, proper affine defs, OK? And what was the big theorem, the big lemma that I had before was the opposite sine lemma. The opposite sine lemma stated the following. If any one of the three, if you have two elements which have opposite sine, OK, then the group does not act properly, OK? So this space is going to be isomorphic to R3. So I can draw it. I can draw three dimensions. I can. Honest. You don't believe me. But I can draw three dimensions, all right? One of the axes is going to be alpha of A, the other alpha of B, the other alpha of C. The proper deformations, I think I'm going in the wrong order, but the proper deformations the first octant, or I don't know how to number my octants, the negative, negative, negative octant, right? The proper after iterations have to deal with both of these. There's also a simple calculation you can do, which I actually did because it's really simple. You can look at this and you can say, if I multiply all my linear, my translational parts by the same number k for two elements, it just multiplies the next one. I changed the order from A to AB just to annoy you, OK? But it just multiplies. So really, not only can you think of this three-dimensional picture and you can say, well, if I'm only interested in proper deformations, I'm only interested in what happens in the plus, plus, plus quadrant and the minus, minus, minus quadrant, I really don't even have to look at the whole plus, plus quadrant because this is going to be projective. If the aphiodine deformation up here, if this is 1, 1, 1, 1 is proper, then 3, 3, 3 is also proper because it's just multiplying everything by 3. It's the same thing. The calculations are going to work the same. And the same thing for minus 1, minus 1, minus 1 goes on. Yeah, it's the same thing. OK, so what we're going to do is we're going to often not draw a three-dimensional picture. It's not going to be really helpful. What we're going to do is we're going to take a plane, some plane. Hopefully it turns out that some of our pictures are not exactly the equilateral triangle that you'd like, but we're going to try to take a plane that intersects it. And we're going to look at proper deformations of just that part because that will tell us the three-dimensional picture. So the pictures you'll see are all related to this triangle. That triangle is really sitting inside a three-space because there's three dimensions to this. But it's projected because you can multiply by negative 1. And that'll change the aphids, too. OK, so let's start on our tour. Let's see what happens. OK. All right, so what was our first manifold that we drew? It was the three-hold sphere. How do you get a three-hold sphere? Well, there's the topological way to do it, which we showed. But the other way to do it is to take two elements, A and B, whose fixed axes, right? We talked about axes before, our disjoint. And this is, for some reason, I like to change my model. I write some things in Poincare and some things in Klein, but today was Klein, right? So this map takes A to there, right? That's the motion of A and B to there, right? You can see the three-holes, right? Where are the three-holes here? There's one of the holes, right? Those two points are connected. There's the other connection, right? There's the other hole. And here's the other hole, right there. It goes down here, around, right? So that's a three-hold sphere, okay? Okay. Now, what you can do is you look at alpha, A, B, and C, and you see which ones are... Where do you get all the three alphas being positive? But in fact, what you can do, you can do a computer experiment. You can ask your favorite idiot. It is my favorite idiot. It does exactly what I tell it. Well, not really. I don't really know how to tell the idiot how to do things. But it's an idiot, right? It just does what you tell it to do, right? And what you'll do is you'll march through the group. Well, you really only have to worry about the primitive elements, but I don't want to talk about what the primitive elements are. But you march through the group and you calculate where you say what kind of... You take the points here. Well, actually, we're not going to do... If you want to know whether an action is proper, you just march through the group and you say, what's alpha of that? What's alpha of that? What's alpha of that? And you make sure they're all positive. Well, that's not enough, but at least we're going to show you what's happening and the intricacies that can happen here. And you do this picture and you say... And you notice that when you do it, right there is your picture... Right here... Sorry. Now I have to go back, because I have to see it too. Right here is the... Those are the three axes right over there, right? That's that three... Right there. And when you go through it, you find that alpha... And what we're drawing here is the alpha of gamma for a particular deformation defined by the three alphas, alpha of A, alpha of B, alpha of C. And then these... All these different lines to give you a different element in the group, right? And you set it equal to zero and you draw the line. A plane over here, of course, but a line there. All right? So this... So for instance, I don't know what this is, which one is which, but this might be... This is given the... Given the point in three space, right? A, B, and C defined by the alphas of A, B, and C. This will be alpha of B, A, B squared C. Something like that. Right? And you'll draw that line. And you'll see that all of those lines, those alpha of B equal to zero lines are outside the original triangle. Okay? And it might suggest to you that that's... That if you know alpha of A, alpha of B, and alpha of C, that all your alphas are positive. It could suggest to you it's not a proof by any stretch. Okay? But although it's not a proof, I will tell you that it is true. That it... That it happens like this. Okay? Trying to look... When did I start again? Now I'm going too slow. Okay. It takes a while to prove this. You can prove it several different ways. You can look at the stuff that I'm going to talk... That I'm going to indicate what happens later on. You can prove it by crooked planes, too. Okay? Let me indicate the proof of this. Let me indicate... Just indicate what we do with the... By the crooked planes. How you prove this. It's not... For this particular example, it's true. Yes. If all the three alphas are positive, then all the alphas are positive. Oh. That's also not true. I mean, we're proving something a little bit deeper here, but that's something that I'll show... I'll indicate what happens later in a different situation. It's true, again, in this case, that if all the alphas are positive, that these are... That it's a proper deformation, but it's really only true in this case. And I'm going to indicate it's going to happen later. There's going to be some other thing that's going to happen later. Okay. So where are we? So, here we have... Do I go down? Nope. I go up and down B. Now, this is the shocky groups. These are really the shocky groups that Pepe was talking about. They're shocky groups in two dimensions. They're not circles. Well, they are circles, if you want to think about it, but it's really the shocky groups. The Klein model is making everything straight where you might think of as circles. But you can screw this up model. There's a little... You know, shocky groups are a little flabby. Unlike some people who are speaking... I'm a fit young man. You can change... You can change these choices of these domains or a little bit. If I moved it over here just a little bit, well, that would move this in here. Oh, go like that. You can change it. It's not fixed. There's not one set of intervals that you pick. The intervals on the circle that you pick. So, what happens is that you can change it so much and you can show this. Again, I'm saying a lot of things. You can show. Okay. You can show this so that what happens is... I think it's... Yeah. It's like this, something like that, where this is B, this is A. Okay. Let me draw it again here. So, draw it straight. B. It's the same group. It's the same group except now it cut all these intervals. They all touch, right? They all touch here. Okay. Now, one of the things that you do when you start off with a group like this and you start... This is the fundamental domain, which I really haven't defined, but the fundamental domain is a picture of the group. You just act on... The group acts by the fundamental domain, moves the fundamental domain to another iterate of the fundamental domain. And those iterates don't overlap except on the boundary, right? You get three. And it fills... And as you build it, it's a tiling, right? You take this domain and you tile it. You get bigger and bigger and bigger, right? And this covers... When you build up this whole thing, it covers the entire domain. Okay. Generically, when you have three elements which are hyperbolic, but let's just say it, we're... Generically, whenever you... You can always draw this picture with these four, but you don't cover the whole space. You cover the Nielsen region, right? The space that contains all of its closed geodesics in a way. And so you get a bunch of intervals, a bunch of spaces which you don't get to cover. You build these out, keep building them out, keep building them out, and you never get to cover everything. So it's a fundamental domain, but it's not a fundamental domain of anything I'm used to. Why would I care about the Nielsen region? I mean, it's an interesting thing, but I like these things, okay? But it turns out that we take this... This picture, and what did I tell... What were we talking about yesterday? We were trying to make a fundamental domain in affine space. Okay. And we'll take this picture, and we take four crooked planes, seated at the same place, right? And all of the crooked planes share a wing with the neighboring crooked plane. Okay? And what we'll do is we'll move, and we're going to move in a direction that's kind of this direction, right? And if we move... All of these are moved, translated by that direction, they can all separate. You think, well, then I can separate, then I can find a fundamental domain of the affine space, of the entire affine space. So we start with this fundamental domain of the Nielsen region, not of the entire hyperbolic plane, and somehow we get a fundamental domain of the entire affine space, which is just weird. And believe me, it bothered me for several trips up and down the Northeast Corridor, as I curled and involved, no, it can't be true. And Virginie says, I'm just quoting your theorem. It was the meanest thing she ever said to me. There's a whole direction. So let me draw another fundamental domain. When I say I go in this direction, there's a bunch of directions I can go in. So all I want to do is, what type of directions can I pick so that one fundamental domain, as I move it, stays inside that fundamental domain. So let me draw that picture, too, over here. Just one fundamental domain here, now. Let's do it to the side. Okay, and let's say this is the interior. This is the inside. This is the crooked half space that we talked about yesterday. The crooked half space that we're in. What are the directions that I can go in? Well, I can pick a lot of directions, but the directions I'm going to pick are always in this plane, and they're in this quarter of the plane. I think I even called it that. It's the stem quadrant. It's positive linear combinations of these two vectors. I can show it to you. I've got my models. Last time I'm going to use the models, I promise. So you take it like this, and you can move it straight like that. It's inside. The originals, that's the crooked half space that we're talking about. And we're moving it inside. But there's a whole, see the quadrant that you get. It's along this direction. It's the positive linear combinations of that one and that one. This is what we call our stem quadrant. Okay, so the translations are always picked in the stem quadrant. What do I mean by stem? Remember, the stem was that these, it's the same plane as those two triangular regions, but they go on forever. Okay, all right. So what we do, and it takes a little bit, again, a calculation. You can just calculate these things. This is a work by myself, Charrette and Goldman. You can calculate these things, and you can represent any one of these alphas that you pick, any one of these alphas here in this triangle, right, any one of these alphas in the triangle by one of these translations which separate the planes. Many of them, actually, but at least one. Okay, so I'm going to go a little bit fast through the next ones because I want to just show you the implications of what's going on. But you have to go through it. So this is really interesting. By the way, well, anyway, I can't tell you the end result of this, but I will tell you in a little bit. Okay, where is that little sucker? There he is. That's our first manifold. That's our three-hills sphere. One of the things I haven't mentioned here is that you can allow for parabolic elements. Everything works through. Your alphas a little bit different, but it's okay. We can deal with it. You can actually, and in fact, this could be, it can all work. It all works very nicely. Okay? This is now our two-hold cross-surface. Right? That's how you build a two-hold cross-surface. I think it's the way you build a two-hold. Yes, it is. Yes. Right? Here's one hole right there. I like to go up the board. There's one hole right there because this point is identified with that. There's one hole. The rest of it joins up together to get the other hole. Okay? Orientation reversing. And at the end of the lecture, I told you orientation reversing. Surface doesn't mean orientation reversing. Manifold. Three-manifold, right? Because it takes two, it keeps one dimension the same and it flips the other one. So it has this nice, you can do it. Okay? All right. Do the calculations. There's a whole, lots of things about the one-hold, the two-hold cross-surface, which were fun and exciting. Talk a lot, there's a lot of discussion about all the different, instead of the three, the numbers you can get, the different ways you can represent this group. But what you get here is this is the first indication that weird things are going to happen. Okay? And here's our triangle. Right there is our triangle. Right there, oh, there's, uh-oh, it's cut off. Right? We don't have, you can have elements which have alpha of A, alpha of B, alpha of C positive, which live outside that line. Those points out there. Right, right. There, it goes the whole way out there. There's our triangle. That's the, it should have been an equilateral. I have no idea. I'll blame the person who built it. No, I can't because I didn't draw the pictures. Right? And so you have elements out here. What are these elements saying? There are elements out here. This is where alpha of some element is zero. So on the other side of this line, alpha of that element is negative. It's not proper. I know we have, okay? And then we have to do the same sort of proof. We have to build up a fundamental domain of this. And again, we do the same Nielsen region like before. Okay? So this is a four, we got a four-sided domain for this. The proper actions have a four-sided domain. Okay? I'm going to go out of order from what I did in the board. But, so that's exciting. I like that. The triangle is the triangle, this triangle, the original triangle that I did, the positive, positive, positive triangle is right there. It goes the whole way out. It's cut off a little bit by this picture. It goes there and up. I can do it better with my finger. It goes the whole way out. That's positive. These points are positive for the three, right, a point in here would be positive for the three generators that you took. But it would be negative for some other element. Okay? By the way, these are, and all of these are done by three generators plus another generator. You can rewrite it into A, B, X, Y. Oh, did I, oh, okay, let's do this one. Let's do this. I'm going to not do this one. I'm just going to say it's inside. But let's do that. I actually did it in the right order. Okay, this is the one-hole torus. Right? There's your one-hole torus. How do you make a one-hole torus out of in hyperbolic space? You take two hyperbolic elements, cross their axes. This transformation takes B to B, right? The left hand to the right hand up to that. That's how you make a one-hole torus. Look at it. Build it together. Take some paper. Paste the sides together, right? That's what you're identifying these points. Okay? All right. So let me tell you about a weird thing that happened because Bertrand had to ask. I can't believe you had to ask these hard questions, okay? Okay, so this is a little bit confusing because there's a little bit, there's too many lines on this picture. Okay, these, let's not worry about these black lines in the middle. Let's not worry about the black lines in the middle. Okay, let's worry about only the colored lines. Those are alpha of an element is equal to zero. On one side, it's positive. On the other side, it's negative. Okay? When you start to draw this picture, you start to find there are an infinite number of sides. There's an infinite number of sides to this. You always get a bunch of sides between any two sides. So like a, you might have a side where, like right here, where alpha of some word is zero. And over to the right, you have some other word zero. They never meet. There's always a side in between them. Okay? And these limit points are exactly what Bertrand was asking. These limit points are where these particular examples where alphas of all the elements are positive but they're accumulating to zero and so you don't have this. So there's these limit points, limit points where it's not proper but it is all the alphas are positive but the alphas are not growing fast enough. And I'm not, we're not, that's not what this whole, what I want to talk about today but I just want to show you the complications that can go. This is a much more interesting picture and it's more interesting of what's going on, right? The way you prove it is you do the same picture. It's the same sort of thing, right? Over and over again. But you only get like a piece of the interior. You don't get the interior by looking at one sort of generator. You have to look at a bunch of generators and I'm not going to go into that because that would take me far afield although that's fun and exciting. We could ask Sir to talk about it. This is all related to that three elements and you take the change of generators. This is to go from one triangle so you find out this is proper, this triangle is proper and then you flip, this is a change of generators that we talked about before. Everything feels a lot of Nielsen. There's a tree, right? You can see a trivalent tree in here. Take this, oh, here's a neighbor, here's a neighbor, here's a neighbor. Every triangle that you have has three neighbors. This is just to indicate and this is the first three examples, the first two examples were kind of, they're not really the cool ones, right? This is the cool example. This is the one that's robust. This is the one that happens all the time. You're more than likely to get something with an infinite number of sides than you are to get three sides, okay? When you go up in higher dimensions you will get things with lots and lots of sides. Okay, that's, now I got way too slow. I went too fast. The other day I went too slow there. Okay, now we get to do, I can't believe it. I don't believe anybody said these words before this week. I can't believe I'm the one to say Lie algebra. Okay, we have a, what's a Lie group? A Lie group has, it's a manifold, it's also a group, right? SL2R, I like that Lie group. You can call it PSL2R if you want. We know that it's isomorphic to SO21, that's a lie. Let's just say it's close enough, plus or minus a little bit. Yeah, you take PSL and you, anyway, we're okay. We can deal with it. It's a lie, but it's okay. That's a Lie group. And we want to think it's Lie algebra. What is a Lie algebra? In the basic sense of a Lie algebra, what is Lie algebra? It's the tangent space at the identity of the Lie group. Just like we had accidents that John was talking about, accidents upon accidents, here we have another beautiful accident. The accident is the Lie algebra of SL2R, 2 by 2 matrices whose determinant is 1, are 2 by 2 matrices whose trace is 0. There is a natural inner product on those matrices, 2 by 2 matrices whose trace, and that matrix is the multiple of the killing form, which we'll write as B, okay? And what you do is you take, we'll take a half of the trace of the product, okay? Should we do a little calculation? I like to do a little calculation here. So a Lie algebra is a linear subspace. It's a linear subspace, right? You can add, and it has this inner product on it, too, which is invariant under conjugation by SL2R, but I'm not going to go that far down the rabbit hole, okay? So let's take, for instance, so I claim that E1, E2, and E3 are a basis for SL2R. Every traceless matrix, 2 by 2 matrix, can be written as a linear combination of these three elements, right? E1, E2, E3. You believe that? You believe anything right now. You're just thinking, I'm hungry for lunch. That's what I am, but that's it. Now, but I have an inner product on it. Let's do the inner product. Let's do a couple inner products, okay? Let's do E1, E1. It's half the trace of one, half the trace of one, which one, I put negative one down there. I don't know if I, okay? And I multiply them, and I get the identity. I take its trace. Its trace is two, and I take half of that. That's one, right? That's how we do the calculation. It's not a hard calculation to do. E2, E2. It's also one, right? You get, when you take E2, E2, that's, its product is the same thing, right? That's one. How about the next one? E3, E3 minus one. I claim that for I not equal to J, that's equal to zero. What's the magic that happens? The Lie algebra is the affine space. Well, it's the tangent space. It's R21. That's the accident that's happening. That doesn't happen in many other spaces, right? This is really weird. The Lie group is, the Lie algebra of the Lie group is the space that the Lie group works on. I mean, in some sense that's true, but it's the one that you'd think it would work on. Well, it's really because SO21 is isomorphic to SL2R. That's the amazing thing, and then this comes for free. What a shocking thing. I'm shocked. I'll take your word for it. I'm not thinking right now. Okay, so I'm going to do this a little bit quick, but I'm going to do a calculation a little bit quick, but the calculation is up here. Okay, I'm going to look at the Margulis invariant in a different way. I'm going to look at the Margulis invariant as what happens to... So SL2R sits naturally inside of GL2R, two-by-two matrices of any trace, right? And we do this amazing thing. I don't know where this trick... That's a great trick, right? You just take A and you subtract the half the trace of the identity, half the trace of the identity off of it. Well, what's the trace of P of A, Pi of A? What's the trace of A? It's one. It's anything. It's the trace of A, right? But then you're subtracting off half of trace, half of trace. What's your trace now? You've got a traceless matrix out there. So you took a matrix, a general matrix in SL2R and you did this stupid thing to it. You just subtracted off a thing and it's a traceless matrix. So we got from an element of SL2R we got a natural element of the Lie algebra SL2R. You can see what these do and it turns out through some of this calculation that what this matrix is, up to a multiple. Up to a multiple. It's not exactly the right thing exactly because all of this crap, I'm not going to go over this, but all of this crap right here, that's all just trying to make the length of that element one. But what you have, this element, is in the direction of A0. Remember what A0 was? It was the fixed direction. And you can see kind of why it works. All right. So alpha of this is going to be, you're going to take, yeah. Okay, so you're going to take, what was alpha of an element? Okay. You take A of, remember, one of the things that I told you right from the beginning, when I introduced the Margulis invariant, originally the meaning of the Margulis invariant was when I picked my point on the closed, on the closed, the unique fixed axes. But I don't have to take it. I can take it anywhere I want. So I'm going to look at where A of the origin, minus the origin, which is just A of the origin, right? Okay, thought of as a vector, dotted with inner product with A0. And I get this calculation right here. Okay. Now, wait a second. This is from that. That's the definition of alpha. That's what I know alpha is. Well, I see there, the trace of alpha and alpha, and the sigma is just, by the way, the sigma is just to worry about PSL2R as opposed to SL2R. You want to always pick, in some sense you want to always pick the ones with the traces positive, where the elements are positive. What about the trace, that's an identity, that's a traceless matrix. Oh, this other part, the u of a times this part, that doesn't have any trace. Because this is just a number, or just a multiple, right? This doesn't have any trace. So all of this part, this part over here, you don't see it up here because it doesn't exist, right? You're taking the trace and it's zero. So it's a nice clever little thing. Okay. All right. So that takes a little bit of time. I'm running a little bit of late, because I went too fast the other day. I had to go really slow today. Okay. And here's the thing. We're now going to think about a deformation, right? We're going to take a group element and we're going to deform it. Oh, shoot. That's not what we're going to do. We're going to take a group element and we're going to deform it. We're going to look at a path. And the path is going to start at A, right? A times the identity right there. Sometimes my identity is just a bold face. Sometimes it's a blackboard face. I do this just to annoy certain people in the audience. Okay. And I'm going to add, what do I want to add? I want to add something in the tangent space, right? It's your Taylor expansion, right? All right. So I add something in the tangent space. And then I have other crap. You know, oh, something of order t squared. And I take the derivative of this. And that derivative turns out to be this element. Trace of A times U of A. But I saw that before. Trace of U of A. This is defining the Margulis invariant. And this is now looking at, now I missed the magic words. I'm talking of the trace of the elements of this deformation. I'm looking at a bunch of A's, right? Here I have a line of A's, right? Here's A. And I'm deforming it along the direction of U of A. And I look at those traces of those elements. And what we haven't talked about yet today, but I'm sure somebody will, is trace is related to the length of the geodesic on the H2. So what is alpha really measuring? Alpha is measuring alpha. Well, first of all, something in R21, it's saying, I'm changing my trace. I'm deforming my group, right? So really, instead of thinking about the affine deformation, the affine deformation isn't really the affine deformation. It's the tangent of a real deformation, right? It's just the tangent directions of the real deformation. Because those things that I'm adding to it, U, right? That's just something in R21, which is in SL2R, the Lie algebra. And what all of this calculation is saying is that alpha is positive, meaning I'm expanding. I'm changing my manifold by making it longer. My closed geodesic becomes longer. So here, if I want to just think of H2 mod the group generated by A, here I'm getting a nice cylinder like that. And over here, positive alpha. Here's my transformation, translation, right? Positive alpha means I get a nice cylinder and it's a little wider. That's what's happening. That's what alpha positive means. I'm lengthening my underlying geodesic. So in some sense, all of this affine crap is just crap. What it's really saying is, what's the tangent space of my deformations, of my original surface? What am I doing to my original surface? I'm moving in a direction so that I'm expanding. What does negative alpha mean? It means that I'm contracting, right? So you can expand or contract. And alpha says, what's alpha saying? What are proper deformations? What are affine proper deformations? All my alphas are positive. What does that mean? It means that all of my geodesics are expanding. So in our original case, let's go back. We're not going to get the stripped deformations. I'm sorry, but it's the cool part. But we're not going to do that today. What was this picture? This picture said affine deformations, right? This picture said that if I know alpha of A, alpha of B and alpha of C, which is just AB inverse, are positive, then I know all my alphas are positive. There's a theorem that said that. It wasn't that thing, but all my alphas are positive. So what does that mean? I have a pair of pants. That's what a three-hole sphere is. I've got one-hole, two-hole, three-hole. I've got a hyperbolic pair of pants because I need some hyperbolic stuff helping me. And it's stretchable, right? And I stretch my belt, and I stretch that, and I stretch that. Every other closed geodesic is also stretched. Every other closed geodesic is also stretched. There's infinitesimal. There's also a more general theorem by Thurston, which says this in a not infinitesimal way. But it's this stretching, right? So it's another way of looking at this. Bertrand asked me a question earlier today. He said, well, can you have an underlying closed surface, right? A closed manifold, right? This is a little bit further. Well, who's affine deformation acts properly? Well, what would you need for affine deformation to act properly? Well, I'd need all the alphas to be positive. Well, what would that mean? I would need all of my closed geodesics to be expanding, and I'm just going to tell you that I'm ruining the surprise. I know it's going to ruin your day. But I'm going to tell you, I'm going to spoil it a little bit. You can't do that. You can't take a closed hyperbolic surface whose curvature is negative one and expand every closed geodesic. You just can't do it. Okay? This was a... And so a lot of this stuff, the calculations of every stuff, were originally done by Golden and Margulis and to prove this, basically, you can't have a closed surface as your linear part. Okay? All right. So I'm at my time. There are some other things that I may or may not talk about. I think tomorrow, the next day, Thursday, that I see you, you're thinking, why can't it be Saturday when we're not here? I'm going to change everything completely, not really talk about alpha too much, but change thinking about Lorentzian space and thinking about what can I do to a Lorentzian space? What can I do? How can I change it? How can I add points to it? How can I think about maybe adding a little curvature to it? And what does that mean? Okay? So that's a little bit...