 Better? Yeah, perfect. I put the microphone on which is one, but the next step is turning it on, fantastic. H2 is going to be the off diagonal like that. And this is going to lead to the ball model, and this is going to lead to the Ziegeldomain model, which you should think of as a generalization of the Poincaré disc, and this you should think of as a generalization of the Poincaré upper half space. And we're going to see what this looks like in detail later on. And then if I have A in U of H, or P U of H, S U of H, P S U of H, etc. Then this is going to be defined by it preserves the Hermitian form for all vectors Z and W. And so we saw what that meant that we had H was A star H A, or equivalently A inverse is H inverse A star H, where A star is the conjugate transpose or Hermitian transpose of A. So what I want my goal for today is to look more closely at the geometry of the space that we're talking about. I want to look at the geometry of the boundary. I want to look at some special subspaces, and I want to try to classify isometries. So those, that's where we're going today. Excuse me. So the first thing I want to do is I want to classify and look at the geometry of the boundary. And for that, we're going to use the Ziegel domain model because we've already, and the thing we want to generalize is this thing that we already know. We've seen many times that, so this is the motivation, that the boundary of H N plus 1 R is R N union infinity, which is SN, perhaps I ought to do, if I do N minus 1, that's probably going to be better. Which is a beautiful picture. We've seen this in many different contexts. We've seen it for limit sets. We've seen it for complex analysis. I mean, I guess I should say, in low dimensions, that would be C union infinity. And in other dimensions, you get H union infinity and even the octonians union infinity. So what would be the analog in this case? So let's think about the second emission form in the Ziegel domain. So I want to think about what is the boundary of the Ziegel domain. So remember that we can always, for both of these two models, we can choose the chart in CP2 where Z3 is equal to 1. So we think of C2 as the chart of CP2 with Z3 equals 1. So a little bit like Pepe's picture. I mean, I really get disturbed when Pepe draws this picture. I've lost my colors, have lost some colors on the front here. When he draws that picture, because I always want to think of that picture. Which is completely different because here's my Z3 equal, so that would be 1. And so here's Z1 and Z2. And so really that's a plane like that, right? So that picture that he's drawn several times for the complex numbers and the quaternions and so forth is exactly the same as the thing I was talking about where we're choosing Z3 to be equal 1. And in this case, if I have Z1, Z2 in C2, that standard lift, so that would be Z is equal to that. The standard lift would be Z1, Z2, 1 in C2, 1. And if I take the emission form of Z with itself, then I'm going to get twice the real part of Z1 plus 1 Z2 squared equals, equals, well, is equal to that, sorry. That kind of indicates to me that I should do something that makes my life a little bit simpler because I have Z2 as a variable. Once I have Z2 as a variable, I can immediately determine if I'm in the boundary, this is going to be 0, so it determines the real part of Z1. But then I still got one free variable, which would be the imaginary part of Z1. And so I'm going to think about C cross R and my variables, I'm going to be calling zeta and V. Now, that's how I draw my zetas. I had some grief from some people that they didn't think that looked like a zeta. So I asked one of my Greek friends and he said, that looks like a zeta to me. So when in, well, when in Rome doers, the Greeks do, no. So that's a zeta and we've got this, I don't like making halves because certainly if you're in latech it kind of messes up everything. So let's make Z2 to be the square root of 2 times zeta. And so then if Z is in the boundary of the Zegel domain, we would have that 0 was twice the real part of Z1 plus mod Z2 squared, which is twice the real part of Z1 plus 2 mod zeta squared. And so I can write Z1 as minus mod zeta squared plus i times V. So zeta is a complex number, V is a real number. And so then that works for all of the points that lie in this particular chart of CP2. There's only one point that does not lie in that chart of CP2, which is going to be the point at infinity. So if I have Z in the boundary of H2C not in this chart, then this is going to correspond to the point infinity, which is going to have the standard lift of 1, 0, 0. So the first attempt to answer my question, what is the boundary of the Zegel domain, is I've just told you that it's C cross R union infinity. That's perfectly fine. I mean, if you really wanted to, you could think of that as R2 cross R, otherwise known as R3, union infinity. So in fact, topologically, this is just going to be a 3-sphere. It's the boundary of a unit ball in C2, so it's going to be a boundary of a 3-sphere. But actually, what's the more intrinsic structure that I can give to it that's actually going to make it not actually R3 union infinity, but it's going to make it something else instead? So in order to see that structure, I need to look a little bit more closely at what happens with the elements of Pu H2. So first of all, let's think about the things that stabilize infinity. Now this is, again, we want to have a little piece of thought. If I live in SL2C, PSL2C, or PSL2R, and I'm thinking of myself as a matrix ABCD, or Möbius transformation AZ plus B over CZ plus D, then stabilizing infinity is exactly the same as C being equal to 0. So my matrix being upper triangular, the same thing is going to be true here. So let's see. So I'm going to write my A in Pu H2. Well, it's going to correspond to a matrix, because I'm always going to write my transformations as matrices, because it's just cleaner than trying to do them as linear fractional transformations. So A, B, C, D, E, F. Now, if I was being purely alphabetical, I would want to put an I there. Unfortunately, I have lots of copies of the square root of minus 1, so I usually put a J there. Now that's fine until I start working with quaternions, and then I have problems. But for today, that's going to be my standard form for elements in here. And if I preserve this second Hermitian form, then I immediately, we had this formula for the inverse. So A inverse, which was H inverse A star H. It's a little exercise, which isn't so difficult, is that this is A bar, B bar, C bar, D bar, E bar, F bar, G bar, H bar, J bar. And if you look in the notes, there's all sorts of beautiful formulas relating these coefficients that you have. I can't choose the coefficients independently, because the group has a smaller dimension. But there's all sorts of formulae relating it. But that's what we want to just think about for a moment. And so if my matrix preserves infinity, not quite as sophisticated as Sullivan dictionary, there's a baby dictionary that goes between what happens to the geometry in my space and to what happens with matrices. We've already heard it referred to, is that a fixed point down on my space corresponds to an eigenvector for my matrix. And the eigenvalue is going to tell me something about the dynamics locally around that space. So if infinity is a fixed point of A acting on boundary of H2C, then 1, 0, 0 is an eigenvector. And it doesn't take too much effort to see what would happen if I multiply such a matrix by 1, 0, 0. I'm just going to get A dG. And for the inverse, I'm going to get J bar, H bar, G bar. And if this, that's my notion for projectively equal to 1, 0, 0, that's supposed to be projectively equal to 1, 0, 0, because it's an eigenvector. So there should be some eigenvalue sitting outside there. You can see that d, G, and H are all 0. So we've got something that we know and love from SL2C, which has got the exact analog in SU21. So what's the simplest sort of upper triangular matrix that I could think of? It would be one where I have 1s down the diagonal. So I just suppose that A has 1 on the diagonal and is upper triangle. So it's going to be of the form 1, 1, 1, 0s down here. And what have I got? I've got b, c, and f. Now what happens to remember that we had, I had my c cross r, which was zeta v, and that I had that z2 was root 2 zeta and z1 was minus mod zeta squared plus IV. So if I choose anything like that, so that's going to be minus mod zeta squared plus IV root 2 zeta 1, so that's exactly z1, z2, 1. Then I can find an element of this form taking the origin, which would be where I would have 0, 0, 1, because zeta and v were both 0 to this point. So this is going to immediately show that elements of this form are transitively on the boundary minus infinity. So if I have z to v is 0, 0, that's going to correspond to what I'm going to call the origin, O, which is 0, 0, 1. And I can find, I'm going to call it t, t as on the left-hand board, sending 0, 0 to z to v. So what's the shape of it going to be? Well, I'm going to have to have, I've got 1 there, so therefore I must have a root 2 zeta minus mod zeta squared plus IV. I don't know what that is, and I've got some 0s down below. And if I just play around a little bit with this formula for the inverse, I can immediately fill in what this missing entry is, and that's minus square root of 2 times the complex conjugate of zeta. And so this is going to be rather like a translation in SL2C, but it's my analog for SU21. So how does it act on points with those coordinates? I need to have a different letter if I see how it acts. So because it's translation, I like to remind myself that it's translation because it's so it's got letters that don't look like a t. So I'm going to have a big t, and I'm going to have tor and t for my parameters. This is in C cross R, which is going to be 1 minus root 2 tor bar 1, 0, 0, 1, root 2 tor minus mod tor squared plus i t. And how does it act on minus mod zeta squared plus IV root 2 zeta 1? Well, it's not so difficult to multiply this out. Last coordinate is going to be 1. That's great. Root 2 zeta plus tor. So things are looking good. That looks like a translation to me. But then things get slightly a little bit more interesting in this top entry. So I'm going to have minus mod zeta squared plus IV minus 2 times tor, well, let's call it zeta tor bar minus mod tor squared plus i t. And if I have now running out of space, I don't want the wet one, I don't want the dry one. I can simplify this. Look at the real part of this. Well, I see that that's nothing other than zeta plus tor squared. And then what do I get for the imaginary part? I have IV plus i t minus zeta tor bar plus tor zeta bar. And so what's happened is that I've got something where I have a nice translation here. I have a nice translation here. But then something gets twisted in this imaginary part. So in fact, what happens is this is giving c cross r the structure of the Heisenberg group. So I'm going to translate on the left by tor t with zeta v, which is tor plus zeta t plus v plus, well, maybe let's do minus twice the imaginary part of zeta tor bar. So that's the addition rule for the Heisenberg group. Now, there are too many topics with an h. We've already seen quaternions and hyperbolic. So I always like to call the Heisenberg group n because it's a nilpotent group. So what have we seen? We've now seen that I can endow the boundary of complex hyperbolic space, so the boundary of h to c. I can think of as the one point compactification of the Heisenberg group. The same argument is going to work for all of my other hyperbolic spaces. So let's just do an aside. But the boundary, and let's just call this, I'm going to call this 2 plus 1 because I've got a complex and a real. So the boundary of h, n, c, 1 plus 1, sorry, is the boundary of h, h, no, I'm going to do 2 plus 1, sorry. That's good. Yeah. The boundary of h, n, c is going to be the Heisenberg group with whatever it would have to be. I'm getting myself confused here. Something plus 1, unit infinity. So it's supposed to have total dimension 2n minus 1, so it's 2n minus 1 plus 1. Good. The boundary of octonionic hyperbolic space is the Heisenberg group for n minus 1 plus 3, unit infinity. Here I have the n minus 1 copies of the quaternions. And the other part is the imaginary quaternions. And the boundary of h2o is n, 8 plus 7, unit infinity. So it's another Heisenberg group. But now I have the octonions and the imaginary octonions. So how does real hyperbolic space fit into this? Well, it would be the Heisenberg group where I would have the real part and then the imaginary part. But there isn't an imaginary part for the real numbers, so it just becomes here. And we see in the first coordinate, I just act as I would do in rn. And so really, it's a sort of a degenerate case of a Heisenberg group otherwise known as the real, rn. Right, I want to draw some pictures. I always like drawing pictures because it helps us to see what's going on. It's inside the stabilizer of a point. We've also got some other things. So it's like if I were to do the Riemann sphere, the stabilizer of infinity contains translations. It also contains other things, like rotations and dilations. Yeah, so this is like a translation subgroup. There's going to be dilations and rotations as well. So when I draw pictures, there is my copy of C with my parameter zeta. And here is my parameter r with v. The reason why v was chosen, this is actually a notation I think Bill Goldman chose, v stands for vertical. Yeah, it's an easy way to remember what's going on. So what happens if I do this translation? What happens to a point here? Well, it's going to get translated along by some value tau. And maybe it's going to get moved upwards by a t and then there's going to be something funny that happens. And it's always a little tricky to draw this, but I will see if I can manage. So I'm going to get lines that are preserved and they're going to, as I move out on the perpendicular to tau, they're going to move around because this last coordinate, this twisting. So what happens is that as I move outwards, then there's a shear that goes on in the vertical direction. So it's a translation, but with a shear. So there's not a rotation, but it's shears, okay? So there's an up and down movement. So there's a little exercise that we could do. And I'll let you do it as an exercise by multiplying things out. But if I have a pair of translations in different directions, so I could, so let's suppose that t tau t and t sigma s are translations and that tau over sigma is not a real number. So they're really genuinely in different directions. Then piece of chalk. And if I were to do translate along, let's just start at the origin, why not? We translate along by tau tt and then we're going to go somewhere by tau sigma s. Now let's come back to tau tt inverse. And I come back, I'm going to end up above where I started from by, this is t sigma s inverse. Because if I just looked in the plane, then I'm going to just always come back to a square because I'm just in the first coordinate, it's just straightforward translations. And I know that commutators there just take me back. But what happens is I'm going to end up above where I started from. And this height is going to be four times the Euclidean area, signed Euclidean area of the parallelogram spanned and sigma in C. So if I move around in a circle somehow, and this is square, then when I come back, I'm vertically above where I was and the height that I'm vertically above depends on the area that I've swept out in the projection. So if I were to go somehow the other way, I would end up below where I was, it's a signed area. There's something you might notice about this. There's an intrinsic orientation on the Heisenberg group. It is oriented and everything that preserves it must be orientable. Maybe there's a chirality if you want to use fancy words. Now, let's just think back a moment. We talked about what the isometries of complex hyperbolic space are. And we said that they were elements of U21 or SU21 or PU21. Those are holomorphic, that's great. They preserve orientation always. And the other sort was complex conjugation or something coming from complex conjugation. Now, in two complex variables, or even number of complex variables, complex conjugation preserves orientation. So in fact, for complex hyperbolic two space, all isometries are orientation preserving. So we've had several talks where people have been carefully put in a little plus on their isometries and saying, well, we only want to think about the ones that preserve orientation. In this case, what I really want to say when I put my plus is holomorphic isometries. Because if I'm on the Poincare disk, being holomorphic and preserving orientation is the same thing, whereas for an isometry, whereas being anti-holomorphic and reversing orientation is the same thing. So there's a little trap and I have seen papers in the literature where people have talked about orientation preserving isometries of complex hyperbolic two space. And if I'm refereeing such papers, then I always say, no, it's not orientation preserving, it's holomorphic, right? Because all isometries preserve orientation. That's a little trap. So hopefully that gives me less work to do when I'm refereeing future papers of people in the audience. See, this is all labor-saving devices. Right, where was I? Yeah, good. So this is a picture of what's going on in the boundary. One of the things we often do when we're doing real hyperbolic geometry is we actually, and we've seen it when we've been talking about inversions, is we actually intrinsically use the Euclidean metric on RN, or RN minus one or whatever it is for the boundary and we do things there and then when we're dealing with things. So particular examples might be, you would look at translation lengths of something fixing infinity or if you were doing dilations, how does it expand distances? So what is the metric that we should be using here on the Heisenberg group that actually plays the role of the Euclidean metric? Well, there are several choices and those choices depend on your taste. I'm gonna indicate to you two of them. These two metrics are equivalent in the sense that distances in one are bounded above and below for distances of the other. So they're equivalent metrics. So the first one which we might think about is called the Carnot metric or sometimes the Carnot-Karathaeodori metric. So this is a sub-Riemannian metric and it would be more or less defined as follows that if I was at the origin in the Heisenberg group, then what does the tangent space at the origin look like? Well, it looks like I've got C cross R and now if I start to move away from the origin, I'm gonna get other tangent spaces. I'm gonna get, they're also gonna split as a C cross R, but it's no longer a parallel C and they're going to be, these planes are going to get tilted. If I move on a way away from the origin, they're gonna tilt by this shear. They're never gonna rotate, but they're gonna tilt and they get further and further until if I went to infinity, they would be approaching the vertical plane. If I went in a different radial direction, they would start to tilt in that direction and eventually they would tilt up towards. If I go the opposite way, they tilt in the opposite direction. And so at each point here, this is going to be called the contact plane. And I don't want to actually give the proper definition of the Carnot-Carothea Doreometric, but it's essentially formed by integrating along curves whose tangent vector always lies in the contact plane. These are called CR horizontal curves. So CR horizontal curves, they're rectifiable curves whose tangent vector always lies in the contact plane. And then the Carnot-Metric is formed by integrating along these and these would form, Carnot, they'd form geodesics if I chose my path sufficiently well. Yeah, I can't remember exactly what the formula is, but there's a formula in this. It's going to be invariant by these translations. It's going to be invariant by rotations. It's not going to be invariant by dilations, but that's what you would expect. Because if you think of the Euclidean metric on the boundary of real hyperbolic space, it's invariant under translations and rotations, but not by dilations. And so the Carnot is formed by integrating, and I can't remember what you integrate by, along these horizontal, CR horizontal curves. This is not the metric that I usually use. So an example would be, if I were to take a pair of points on this vertical axis, if you were being naive, you would say that it's Carnot-geodesic should be going straight upwards. That looks like the shortest distance. However, I can't do that because I'm certainly, I'm now not in my, at least at the very beginning, I'm not in the contact plane there. So in fact, the way you should think about it, well, we've got a staircase out here. How will I go up to the terrace? Well, the shortest distance where we go straight up, but I can't manage that. I have to go round on that staircase. And so that's exactly what we would do. We would take a path that lay on a cylinder, it's a spiral path, and the area of that circle is exactly going to be four times, so the distance up here is four times the area of that circle. So that tells me which circle I should be on exactly. So if I'm looking from above, then I would just see, this point would be the whole axis, and I would just see a circle there. I have many geodesics, so exactly. I have another geodesic there, and so forth. So it's not uniquely geodesic. If you also think about it, what would happen if I go around twice around a smaller circle? Well, that's also going to be. So there's also, there's all sorts of interesting things that would happen. That's the sort of metric that you should use if you are the sort of person that really likes to differentiate and integrate. When I moved to Durham, which was like 21 years ago or 22, where 21 years ago, the group that I was becoming part of was called the Differential Geometry Group. So I said, well, I'm due geometry, but I don't like to differentiate. And so I made a campaign, and eventually now we're called the Geometry Group, because they realize that there's more sorts of geometry than people who like to differentiate. I'm a geometry who doesn't like to differentiate. So if you don't like to differentiate, there's another metric which has some better properties and some worse properties. And that's the Sagan metric, so sometimes called the Heisenberg metric. And from my point of view, this is a better adapted metric to the sorts of things that I want to do. So we have N, which is the collection of Z to V in C cross R. And so I first of all want to define a norm on the Heisenberg group, and then the metric is going to be the naturally induced metric associated to that. And so I'm gonna have the norm of the Sagan metric. Well, to start by thinking it might possibly be looked something like that. Don't write this down because it's gonna get erased in a moment. Because that's what we would imagine the norm should be because that's what we like to think about. Then you notice there's a problem. When I do this translation, I had in the imaginary part, I had a quadratic term in terms of the Greek variables. So that's gonna be bad. And so what I want to do is I actually I want to put that and then take the one quarter power, or if you like. So that's like them taking the norm of the first coordinate, which is another way. Thank you. Which would also be the same as taking the Hermitian form of this minus mod zeta squared plus IV root two zeta one, one zero zero zero one to the one half. So it's nicely adapted to the Hermitian form. And then you can sort of guess what to do here. You have to be slightly careful because the Heisenberg group is not commutative. So the reason it's called the Heisenberg metric, so I'm told is if you look at the associated Lie algebra, that's connected to some of the things that Heisenberg did when he was doing quantum mechanics. This metric, no, this metric, there's a whole family of things you can do with different powers there. And Karani was studying those. And this particular one was proved to be a metric that I'm gonna tell you about in a moment by Sagan. So Sagan is a Polish mathematician and this is sometime in the 70s that this was proved to be a metric. I don't know of any earlier references than that. Well, so Sagan and Karani when he was doing, they were doing like complex analysis, harmonic analysis on this. So there was all sorts of analytical things. And well, my motivation for using it is it's well adapted to my Hermitian form so I can calculate it easily. And many of the properties that I'm interested in to generalize, my way of thinking is I take SL2C and things that I know there and I say I want to generalize this to the complex, the higher complex dimensions. And I want something that actually generalizes nicely. And this metric, it has some bad properties but it generalizes nicely. And I'm going way, way, way too slowly. Nevermind, perhaps nevermind. We'll do classification of isometries tomorrow. Right, so if I have, so that the Sagan distance between Z to one, V one, Z to two, V two is going to be the Sagan norm of Z to one, V one inverse composed with Z to two, V two, SIG, right? I have to take my inverse on my, so inverse just would mean that would be the same as minus Z to one, minus V one. You can check that's the inverse exercise for you. But I want this to be left invariant by translations and so if I put in a translation here, acting on the left, then I want to have the inverse here and so that this would be invariant. If I chose this inverse on the other side, it would no longer be invariant under those translations. Let's just do a little, oh, in the notes you will find a proof that this is a metric. So it's clearly going to be, well, the distance to a point to, it's not so difficult to show the three properties of a metric, right? First one, it should be, the distance is zero if and only if the point is the same. It's not so difficult to show. Second thing is that it should be symmetric and it's fairly clearly symmetric. The third one you have to satisfy is the metric in the sense of a metric space metric, not in the sense of a Romanian metric. And the third thing you need to satisfy is the triangle inequality. And when you're satisfied trying to prove the triangle inequality, you'll discover that something very interesting happens. No, no, no, it's not a path metric. That's what I'm about to tell you, yeah? It's not a path metric, right? And so I can take a pair of points here, zeta one, v one, zeta two, v two, and let's just suppose that I could find a path between them whose length is this metric, is this distance. Then if I chose any point on the path, then the distance from this point to that point plus the distance from this point to that point should be the distance of the whole path. So in other words, for any pair of points, I must be able to find a path where I get equality in the triangle inequality. But it's not so difficult to exhibit pairs of points where for every other point, I get strict inequality in the triangle inequality. So this does not happen. So it's not a path metric or a geodesic metric. So that's a bit bad, but you have to sort of live with some bad things and some good things. And that's the bad thing that I choose to live with. So quite often I'm trying to build fundamental polyhedra, maybe using things like this and some spheres. And you would want to say, well, two spheres are disjoint if the distance between their center is bigger than the sum of the radii. Well, if the distance between the centers is bigger than the sum of the radii, then that's gonna be fine because of the triangle inequality, but it's not going to be going the other way around. So there's all sorts of issues that come in there. Right, I've been going a little too slowly, but what I want to do is, I want to talk about, switch gears slightly, talk about some totally geodesic subspaces of complex hyperbolic space. And I'm gonna try to draw some pictures of what happens in the boundary so that you get a bit of a feeling for them, which is why I wanted to draw these pictures and we'll then kind of gently come to a close. All right, the first thing we need to say is one of the biggest nuisances or challenges or things that makes it very, very interesting, depending on your point of view, whether you're a glass half empty or glass half full person, is there are no totally geodesic real hypersurfaces in complex hyperbolic space. Gen C, let's just even say that. This is for N, obviously that's false, N equals one. Well, let's just, let's do, I mean, this is no totally geodesic. So this would be two N minus one real dimensional. I mean, something of real co-dimension one. Real co-dimension one, yeah? Obviously I can have complex hypersurface, which would be complex co-dimension one or real co-dimension two. Let's just do a little thought exercise. Suppose that there were something that was a totally geodesic real hypersurface and that should be fixed by an isometry which interchanges the two half spaces. That would naturally be orientation reversing and so that's a very soft proof that that couldn't happen in complex, in even complex dimension. So I'll now tell you what the totally geodesic real hypersurfaces are. So the totally geodesic subspaces are the, the, well, let's just stick it again to the case when N equals two and then I'll tell you what happens in general. So the complex line, which is, if I think about maybe I could switch to the ball model, would be the intersection of a complex line in C2 with the ball. So that's going to intersect it in a disc and so these are going to be isometric to the Poincare disc with curvature minus one with the normalization that I've chosen. The second case would be totally real or Lagrangian subspaces, e.g. R2 sitting inside C2 and these are going to be isometric to the Klein-Beltrami model. So I noticed that we're having our history of hyperbolic geometry yesterday. We're in Italy, but Beltrami didn't get a mention. Now, as far as I'm aware, Beltrami is the first person to have constructed a concrete model of something of curvature, constant curvature minus one in about 1860. So about 20 years before Poincare and Klein were working on this, that's my knowledge of the history I may be wrong. So I think as we're in Italy, we ought to at least say the name Beltrami once. I've said it's three times now. So that's used up my quota. So the Klein-Beltrami disc. So this has got curvature minus one quarter. So here I have two copies of the hyperbolic plane isometrically embedded in two distinct ways in complex hyperbolic space. That's really, really interesting. And what I will hopefully be able to do tomorrow is talk about what the analog of quasi-Fuxian groups are for those two cases and how you think about representing surface groups to tie in with some of the other courses we've had. And of course, we can also have real geodesics which are just their lines. So for example, around about a line, a real geodesic, that sits in a unique complex line, but it sits in a whole S1 worth of Lagrangian planes. So how do these look in the boundary? And I'm supposed to finish, we started two minutes late, so I think can I have, right? So I'm just gonna draw a few copies of the, I want to draw the boundaries of different. So first of all, what is the boundary? This is the boundary of complex hyperbolic space. And I want to draw the boundary of these totally geodesic subspaces. First of all, let's do the complex lines because they're the easiest. If I have a complex line passing through infinity, well, it's going to just be a vertical line in the Heisenberg group. So this is the boundary of a complex line through infinity when you can think of it as being zeta nought v for vr. So it's all points with constant zeta coordinates, just a vertical line. So there's the whole pile of them, right? You can draw as many as you like. Now, what about ones that don't pass through infinity? Well, I'm gonna start off by drawing the ones that would be set that they, well, they're centered at the origin. So you could think of a whole bunch of circles that sit in this zeta nought plane. But what I could do is I can move this around. So I've got a contact plane somewhere else. So they're going to be, they're gonna live inside the contact plane at that point. They're going to be, well, they're not quite circles centered at that point. They're going to be ellipses, but which live on a circular cylinder. So they're going to be, so these are the boundary of a complex line, not through infinity. They're going to be ellipses in contact plane at the center which lie on a circular cylinder. So there's, all right. So if I were to take a circular cylinder, it's not a very circular, circular cylinder. It's not very cylindrical. And I would imagine cutting it with an oblique plane. I'm going to get an ellipse. And it's exactly those ellipses providing that the center of this water bottle it passes through the point about which the contact plane is. And so when I'm at the origin, my contact plane is horizontal. I'm going to get genuine circles, but the further out away from the origin, I get the steeper it gets. And so a lot, these ellipses are going to get very, very long and thin. Now what about Lagrangian planes? Lagrangian planes, well, through the origin, they're going to be lines that lie in the contact plane. So these are going to be the boundaries of a Lagrangian plane through infinity. And similarly, if I have here, I can put a whole bunch of lines and they lie inside that contact plane at the center. So they're lines in contact plane through the center. So you can, with a little bit of thought, you can realize that any line, am I right? No, well, I won't say that because I think I'm wrong. Okay, so you get lots of lines like this. And in particular, you get something that looks very much like polar coordinates on each contact plane. Now, the complicated one, which I've left to last, is the boundaries of Lagrangian planes that don't pass through infinity. Well, these are space curves, which we sort of draw like this. They're so topologically a circle. And if I look down from infinity, what do I see? I'm going to see a Lemniscate of Bernoulli. So it's a curve like that, which crosses at right angles. So these are, this is the boundary of a Lagrangian plane, not through infinity. And the vertical projection, which I haven't defined, but it's the obvious thing. You just project onto the Zeta coordinate is a Lemniscate. If you want to, technically, it's a Lemniscate of Bernoulli, a particular sort of Lemniscate. And that's quite good because that encloses zero area because it has a plus area here and a minus area there. And so that's one of the things you would expect for something that was Lagrangian. It should actually enclose zero area, okay? And so you can start to build all sorts of fun, playing with these as building blocks. And I guess that's what I ought to stop. So tomorrow I'm going to do what I was supposed to have done the last half of today. I'm going to classify isometries. I'm also going to try to talk about representations of surface groups and talk about something called the Toledo invariant, which is the exact analog of the volumes of representations that we saw yesterday.