 Okey, so Pepe told me that he's not going to introduce me because he thinks everybody knows me. I've given many courses on complex hyperbolic geometry for many, many years and looking around the audience I think probably at least half of the people have been to at least one of them and it's going to be the same. Weithu'n cael ei awdur yna, oherwydd mae wedi bod ymateb yn bwysig, ac mae'n dwi'n gweithio'n cael ei wneud losau, oedd mae'r wneud eich gweithio yn cael ei wneud. Mae'n werth sphereus oed yn ei fod yn gywir gyd. Ond mae'r gwirioneddau yn gwybod, mae'n gyfer y gallwn wahanol Fy hwnnw i'r gweithio am hwyl ac mae'n gweithio oed y mark. Mae rhan o'r booki sydd yn f sustained gwiyddio ar y penwyr cyfrodd o'r drwng gan osybbio maen nhw'n gynnig adeilad o ddweud, mae'r ffaldig gan llog, llog, llog dŵr. Dw i'n gyda'n gilio bydd yn fawr o'r epsalon yn sgrannu. Rhywb bod yma maen nhw'n iddo yn fawr. Ac rwy'n dweud am rydych chi i gy parenting dysgu ddaeth ffordd. Mae ydych chi'n ddatblygu ar y popeth, ychym ni'n mynd i chi mewn. A mi ddweud bod nhw wedi ddweud ond rwy'n ddweud mewn bydd na'w. Fel rwy'n ddweud ond oherwydd, rydyn ni'n ddweud beth yna mor gweithio ffrinddiol i ddechreu. Rydyn ni'n ddweud bod nhw'n ddweud iddyn ni'n ddweud a ddae. Rydyn ni'n ddweud, er mwyn hynny. I sort of got two aims. First I want to give you enough background in complex hyperbolic geometry so that the people who next week are giving talks in the subject, you will at least have some basic knowledge. But also this week is supposed to be a school and we have many topics and I want to try to link in what I'm talking about to a little bit about the very beginning, what Todd talked about. We're going to see connection to what Bertrand talked about, what Pepe is going to talk about and maybe even what Francois is going to talk about. But I'm going to try to at least give you sort of pointers as to how you can glue these little mini courses together so that you have a bigger knowledge. So most of the time I'm going to be thinking about H2C, which I write like this, which is complex hyperbolic two space. And I'm also going to be thinking about C21, which is a complex vector space, which is the complex version of the thing that Todd talked about. And I'm going to do both of those to keep everything definite, although these notes are exclusively about those cases. But just before we start, I thought I would just sort of try to paint a slightly bigger picture, one of which includes some of these other things that we've been talking about. And incidentally, because all of the mathematical details are in here, I'm going to be a little bit freer about trying to explain ideas and draw pictures. And if you want to know the proper mathematical proofs, you can follow them in here. So the main context, I mean I'm going to use some fancy words now. Well, some of us like to write grant proposals and if we put fancy words into our grant proposals, then they're more likely to get funded. Which reminds me of the story that Jonathan Hillman, who's a not theorist from Sydney, told me that they had a very big grant proposal. He worked in not theory, so the title of the proposal was something like not theory. And he said that it would be working on it, it was thick, lots of money, lots of post-docs, lots of grants, lots of conferences. And the day before he was due to submit it, somebody who got their ear open to something going on higher up with the Ministry of Education or whoever it was, said, they're not going to fund it because the title sounds too simple. So they just changed the title to, on co-dimension two embeddings, fantastic, fantastic, and they got funded. So sometimes you have to use complicated words to describe simple things in order to get funded. So here's some complicated words, not because of funding but just to give you a little bit of context. So the rank one symmetric space is a non-compact type. So symmetric space is a space where every point looks the same. Well, why is it that? What's the real definition? It means that it's a space where you have the tangent space at every point, so it's going to be sufficiently nice to have a tangent space. Then minus the identity, which you always have on every tangent space, that pushes down to an isometry. And because of that, then it has to look the same everywhere. So if I have two points, I can join them by a geodesic because I'm assuming it's sufficiently nice that I have geodesics. Then I take the midpoint and then that minus the identity switches the direction of that geodesic. So what it looked like there was what it looked like over there. So the symmetric space, that's roughly what that means. Rank one means that if I have a geodesic, then that is a locally isometric embedding of the real line. Rank one means that that is the highest dimension of any Euclidean space that I can locally isometrically embed into my space. So a sphere would be a case because even though with a sphere, when I go round a great circle, that's not an embedding of the real line, but locally it is an embedding of the real line. So that's why I'm saying locally isometric embedding. And non-compact type because I don't want to think about spheres or CPN or whatever. I want to think about the other ones. And so these are sometimes called the hyperbolic spaces and these are usually called HNR. People often forget the R there, HNC. So this is real hyperbolic space and people often will not use real. So this is complex hyperbolic space. This is quaternionic hyperbolic space. Notice this is why I use a bold face, an ordinary bold face H in my notes because as soon as I'm wanting to do things over the quaternions, I have a problem if I have a blackboard bold H sub blackboard bold H because I don't know whether it's hyperbolic quaternionic hyperbolic space or hyperbolic quaternion space. And I have H2O, which some people think is water, but in fact the two goes up and down. This is octonionic hyperbolic two space. And that's a complete classification. And well, the last one's a little bit of an exception, but the general framework that I'm going to talk about will work in all of those cases. And if you're prepared to be sufficiently broad minded, you can make it work in this case as well. So the things that Todd was talking about and what Bertrand was talking about all fit. Well, actually what Todd was talking about fits in here. What Bertrand was talking about fits in here, but with an equals one. And there's this low dimensional accident that real hyperbolic two space and complex hyperbolic one space are the same. And that is exactly the same as the low dimensional accident that Cp1, which is the compact analogue, is the same as S2. But as soon as I go to higher dimensions, Cpn is not the same as Sn for any other of these low dimensions. Okay, and we're going to see both of those pieces of that phenomenon coming in. Sorry, what? Okay, so we can also have one, but H1O would be the same as HR8. Because again, there's a low dimensional accident there. And you don't have any more because the problem is associativity. Okay? In the same way that the compact type symmetric spaces of rank one would be the real spheres, the Cpn's, the Hpn's, and whatever it is. OP2, the same thing happens there. Right, that's right. So H1C is H2O, H2R, normal, H1H is H4R and H1O is H8R. And I think if I've read the abstracts correctly, we're going to be seeing something about this connection later on probably next week. Okay? So this is talking about four dimensional real hyperbolic space viewed from the point of view of it being one dimensional quaternionic hyperbolic space, which is also a very interesting topic which I've thought about at some point. Incidentally, you can sort of continue this sort of line here. Well, really this sort of line. But if you're prepared to go up into Clifford algebras and you can describe real hyperbolic n space in terms of Clifford algebras, there's two by two matrices over Clifford algebras, that would be an entirely different mini course. And I've also got a set of notes about that, which you can find on my web page. But that also fits into this broad context that I'm talking about. So I'm going to give you some formulae and I'm going to mainly concentrate on this case, but I will tell you when they apply in this amount of generality. So let's start slowly because, well, it's the first day of the conference. It's after lunch, it's raining, all these things. When Todd was talking this morning, he took transposes of his vectors. Perhaps this is a good question for the problem session. So you end up with, the Clifford algebra is a problem because it has zero devices. And so you'll have to look inside the Clifford algebra, you get the Clifford vectors, and then you have the Clifford group, which is the multiples of Clifford vectors together with zero. And then you take matrices over those, but they act on this space of vectors. And you get, and the matrices, these Clifford matrices, there's certain compatibility conditions, which I guess were found by Barlin in 1902, and we discovered by Mars, and then popularized by Alphors, but those conditions are exactly the same as your matrices having to preserve a Clifford Hermitian form. And so you can rephrase the whole setup in terms of Hermitian forms, which is what I'm going to be talking about, and linear algebra. But as I say, that's something we can either talk about in the problem session, or you can look at my notes, or we can talk about privately. But it's going to take me too far away from what I wanted. So, right, let's kind of, let's try to get some of the, rather than talking about abstract things that's come down to earth. So I'm going to be thinking about what is C, let's just start with Cn1. N's going to be two very quickly, but C could be R or A. I want you to think about this as vectors like this, which would be, I always think of column vectors, because, well, my first postdoctoral advisor was the same as Todd's thesis advisor, and he likes to think in columns and his influence is the same way. And I always, when I want, we always like when we transpose, we like to, to a Hermitian transpose. So this is going to be, right? So whenever you see a star, that is going to mean conjugate transpose. And this is equipped with a Hermitian form. So this would be H, which is an N plus one by N plus one matrix. So again, H star is supposed to be H conjugate transpose, and being Hermitian means that that equals H. So this is just the complex analogue of being symmetric matrix. And so I can now form, have a form, which I'm going to always use angle brackets. And this is where, well, there's some sort of convention. And I don't know whether it's because I was, I learned linear algebra in England where they drive on the left, but I always like my Hermitian forms to be linear in the first variable and conjugate linear in the second variable. And I think that's because we drive on the left and that seems a natural way to do it. It's a little bit like I tell my second year, my second year complex analysis students that the reason why you go anti-clockwise around a contour when you're integrating is because Cauchy was French, and therefore when he went round roundabouts, he went round them on the right-hand side, whereas the English people would have gone the other way round. That way they remember which way to integrate round a contour. It's great. So I always have to switch here, so it's going to be W star H. So it's a nice linear algebra form so that you should just think about that that is the same as X dot Y is Y transpose X, which in that case would be X transpose Y. It's the same analog of that formula, but that's because we would be in the real case and of course we notice that simply because when I transpose, I reverse the order while taking transposes. So when I conjugate, I have to conjugate, reverse the order. Right, good. So now I need this, that would be a general thing because as we heard, some people don't like to call things inner products if they're indefinite, but perhaps we can just call it a product. And I want H to have signature N comma 1, which is the reason for this N1, i.e. N positive eigenvalues and one negative eigenvalue. Notice that that's used up all my dimension and that would have to be 0, 0 eigenvalue. So that's very much the same as the product that Todd was denoting by a dot, except that he was in the real world this morning, I'm in the complex world. And I could go into the quaternionic and if I was a bit more careful in the octonionic world. You see, in the octonionic world, I fail at this stage because I no longer have a vector space because scalar multiplication is no longer associative. And therefore I have to go through all sorts of gymnastics in order to avoid ever mentioning the word vector space, but once I go through those gymnastics, I can deal with matrices, I can deal with all sorts of stuff, but I just have to be very careful because I don't have a vector space. So there are going to be some standard examples. So I'm going to read some standard examples. I'm going to call H1, the matrix with a whole load of plus ones down the diagonal followed by a minus one. And so in this case, we would have ZW1. If I am independent of which form I'm choosing, I'm not going to put a subscript, otherwise I'll record it there just to help things. That would just be Z1 W1 bar plus ZWN bar minus. And this is the most basic one. We call this one the first Hermitian form. That notation from a paper of David Epstein. And H2 is going to be... I'm going to divide my matrix up into blocks. I'm going to have ones there. I'm going to have the identity matrix of size N minus one there. Zero is everywhere else. And so in particular in the case when N equals three, that's just going to be the standard off diagonal matrix, which we're going to call the second Hermitian form. And so that would mean that this would be Z1 WN plus one bar plus Z2 W2 bar minus one. No, ZN WN bar plus ZN plus one W1 bar. It's a small exercise to show that that's also got signature two one. And if you're bored at this stage, you can prove it for me. What I want to do is I want to, as far as possible, avoid getting bog bog down in particular normalisations and give you formulae at work in general, which you can then apply to your favourite situation and then use that. So that saves me, you'll see, saves me from having to kind of worry about the upper half plane versus the unit disk. I can just do everything at once and then say, OK, just specialise here, specialise there. Another Hermitian form which we might see next week when Martin's talking is if you have a triangle group, I'm now skipping way ahead, but we'll come back to this, but this is just sort of a little flag for you. So if I have a triangle group, so that would be the group generated by complex reflections in certain complex lines. Let's just stick to n equals two here. As Todd was talking about, you would have these light cone pictures and then any complex line sitting inside there has got a normal vector. And if I have a triangle sitting in there, I can't possibly draw three of these, then I would get three normal vectors which would be linearly independent. And so I could choose n1, n2, n3, which would be by definition going to be positive vectors that don't yet have a form. I could just choose those to be the standard basis and then I would immediately pick up a matrix whose entries would be the different Hermitian products and that's sometimes called a Gram matrix. So I could have h, it's a Gram matrix, which would be something like the matrix n, i, n, j. So the i, jth entry would be whatever it was. So I need to find this, I can find this somehow by geometry what these numbers should be and then I just plug it in and providing this has got signature 2, 1, then that should also be a good example. And the cases that I've been interested in when dealing with Marta and where there's a lot of symmetry so you would have something like 1a, a bar, 1, 1. You would have something like this where a is some complex number and you would then have to convince yourself that this has got signature, the right signature which is some condition on the a. So if I just were to take a Hermitian product of a vector with itself then that has to be equal, I can swap the two factors around. If I swap the two factors around I complex conjugate and I get the same answer because they're both the same. I'm sorry it must be real. And there's a great thing about real numbers. They're either positive, negative or zero. So I'm going to have v plus is the set of all z in c21 such that zz is positive, v0. I don't like the origin because I'm going to, very soon I'm going to do complex projectivisation to get down to cp2. I don't like the origin so I'm going to get rid of that. So all the people who are equal to 0 and v minus and I'm going to draw pictures that look like a light cone. Out here is v plus here is v0 and in there is v minus and of course I'm removing the origin so there's a little hole there. This is very schematic because I don't have to worry about future and past because future and past are all one to me because future is plus one past is minus one but I have a whole circle worth that goes between them and so really that picture is a little bit misleading but that would be the picture if I intersected with the real numbers for the first emission form because I would have my z3 axis here and my z1 axis and my z2 axis and that really would be the usual light cone that you would see. On the other hand if I were to do the second emission form it's not so difficult to discover that if I have my z3 my z1 my z2 maybe it's good to have a slightly different colour at this stage and let me try to get this right that my cone should sit it has both the z3 axis and the z1 axis on it and it sort of sits like that. So it is an illustration of what I was saying that we're going to have different formulae involved in different cases but I can just sort of move things around by changing my emission form and everything looks the same. Bless you. The chalk is all very short so you'll notice that speakers keep going down there with the chocolates in there. Right. So I want to say I want to take you from c21 I'll do a canonical projection onto cp2 perhaps I should say minus the origin and that's the usual canonical projection. I just multiply identify all complex scalar multiples of all of vectors notice that I've gone down to 2 already and have it and then I'm going to call this p and then complex hyperbolic 2 space the projective model is going to just be p of v minus and its boundary is going to be p of v0 Oh, look at this. No expense spared. You can tell the U end behind this. Okay. You watch me break it into tiny fragments in 30 seconds. I really try not to. Okay, so those so this is my space it's a projective model just like we saw the projective model this morning but maybe we don't like to actually think too hard about a projective model because that can be a little bit tricky so let's try to think of something more concrete models so I'm going to think first of all about the first emission form and so this would be the set of all points with z1 well let's just stick to that being negative notice that z3 cannot be 0 in this case so I can divide through by z3 so this means z3 is non-zero and so if I just change this the name of this into a new inhomogeneous coordinate then I just get the unit ball in C2 and that would be the same as intersecting with the plane where z3 was equal to 1 plane complex complex 2 space and so here that would be the plane where z3 is 1 and you can see that I intersect my light cone in a nice well what looks like a disk there but really if I'm doing even the correct dimensions should be a copy of the unit ball in C2 similarly if I do the same thing for H2 again I'm just going to let z3 be equal to 1 then I would get that twice the real part of z1 plus mod z2 squared has to be less than 0 and this is some sort of paraboloid that you can see and if I do that over here then that's supposed to be z3 is 1 and over here I'm going to get the paraboloid and this is called the Zegal domain see, it happened already so let's just stop for a moment and think about what would happen if I decided that 2 was a bigger number for me and that I had just taken 1 instead if I had taken just 1 instead in this case it's quite easy to see that I would be coming down to Cp1 which is the reman sphere and in here I would be having mod z2 where z2 is less than 1 I would be having the unit disk in the reman sphere so let's just record this fact that if 1 we get Cp1 which is the reman sphere and the h1 is going to lead to the unit disk and in the second case I don't have a z2 anymore I'm going to be having the real part of z is less than 0 this is a left half plane which I don't really like the left half plane because it's far more conventional to use the upper half plane and so what I could do oh dear put this on top of there didn't you was that a mind no I told when you were timing how long it took me to reduce all the chalk oh gosh right the head took something like a permission form let's just call it h2 the tilde is in one dimension 0 minus ii 0 then this is going to give me the upper half plane so this is in some sense really complicated language but what I've derived so far are these very very familiar objects of the unit disk and the upper half plane the spaces at the moment we haven't yet got a metric we haven't yet got isometries but they're coming very quickly and I'm actually told you how to generalise it well to two complex dimensions or actually to n complex dimensions actually to n real dimensions to n quaternionic dimensions doing a similar thing you can do stuff with Clifford algebras and with a little bit of gymnastics you can get the octomians so somehow what we saw as a little seed of this stuff that we know you can by putting some fertilizer on it it grows into all sorts of things and it's a great fun game to then take results that you know for the unit disk and the upper half plane and to see what's going to be true if I generalise them to some of these contexts it's such fun that I've been doing this for rather more years than I'm still not bored with it right for arbitrary real n I'm just going to get the standard SON SON1 model yes so in that case what you would end up with is you can have two different emission forms you can have the emission form like that yeah and that's going to give you the unit ball model now we're working over the Clifford algebras right emission form over the Clifford algebras and we have to be a little bit careful about stuff but we will be that careful if you want or I could have 0-ik ik and that's going to give me an upper half plane model and in fact it's this one that you can find in algebras paper but he doesn't have it he never mentions the word emission form but actually perhaps I want a k plus 1 or n plus 1 maybe n is a better number there maybe it's n or n plus 1 I don't know depends on where you start from 0 it needs a little bit of work because the fun thing you're dealing with there is you're dealing with Clifford vectors which when you add them together that gather give vectors multiply them don't give vectors and you're multiplying but also working with a Clifford group which when you multiply them together give you elements of the Clifford group but when you add them doesn't so you have to be very careful about separating out your adding and your multiplying and making sure that when you add you only ever add vectors and when you multiply you only have a multiply things in the group but it works so that was here's a formula for the metric now I don't have a good way to motivate this perhaps your motivation could be like my motivation was I was reading Mosto's book on symmetric spaces and it came from there I don't know how Mosto found it but here's a formula I'm going to put a minus 4 there to determine the curvature but if you want to have different curvatures you put a different constant there that's a formula and seems to have got a round bracket and a closed bracket let's and that formula is going to give you the metric the real hyperbolic metric if you're in the real spaces it's going to go with the decline daltrally metric or it's going to give you the the complex hyperbolic metric the Bergman metric or in one dimension the Poincaré metric or it's going to give you the quaternionic hyperbolic metric and you have to, as I say, do a little bit more gymnastics because there's an extra term to deal with non-associativity for the octonians but it's a single formula that does for everything and that's always how I remember what the formula is for the metric in any particular case so if you want we could well perhaps we don't want but if you want to have a little bit of something to do this evening because it's still be raining then you could use the Hermitian form given by 0 minus i i 0 and check that you pick up exactly the metric that we saw from Bertrand his factor of 4 is exactly the same as my factor of 4 so if you want to try to do it over the reels and to try to pick up the metric that Todd didn't actually mention this morning but was implicit in what he what he said then you would find you probably don't have a factor of 4 there because in fact this is going to give you in that case you won't have something with curvature minus 1 and this is going to give you curvature minus a quarter if you use that formula in that case so providing you don't mind about changing your constant in the different contexts then that's a formula and it's going to give you a metric in all of these different cases in case you don't like the metric you can derive a distance function and this is essentially going to be a little bit like the Cauchy Schwartz formula so this is going to be so we're going to have a distance function so this should be in the context I'm dealing with this is the Bergman metric where it's generalised and then we have a distance formula so I'm going to have row of points I know what I've forgotten to tell you but never mind I'll write down the formula and I'll tell you what I've forgotten I had it in my notes but if I looked at my notes I wouldn't care for you so I grew up where the hyperbolic metric was always called row my advisor was Alan Bearden that's what he used in his book that's what I learnt from being however high I was when I was a PhD student and I can't possibly ever imagine any other name for the hyperbolic metric so some people from other cultural backgrounds find using row for a metric is a little bit odd but I'm afraid this is what it is so this is the distance function so that's what it is row is a distance function and it's given by Cauch squared of the distance between z and w over 2 is z w w z z z w and what I forgot to tell you was how I'm relating a point here without an underline to a vector here with an underline and that's to do with exactly these pictures that I'm about to erase so if I have z 1 z 2 in c 2 that's going to correspond to z it's going to correspond to the vector z z 1 z 2 1 and I'm just going to call c 2 1 and I'm going to call this a standard lift what's my motivation for that well when we're thinking about cp 1 we know that cp 1 is c union infinity and we do this by taking z 1 z 2 and identifying it to z 1 over z 2 1 which we could think of as z union 1 0 and so because I kind of grew up in this complex analysis world I always think of it as the Riemann sphere with c union infinity and there's my variable c when I want to lift it up to a vector then it becomes something like that and so that's what my motivation for is over here so this point corresponds to infinity when I'm dealing with the unit ball I can extend this this can go the unit ball sits inside here but the Ziegledomain as you can imagine from this picture that I've just missed off there that goes off to infinity and so I'm going to have a point infinity it's going to have its standard lift as 1 0 0 which is just the natural generalisation of this standard lift of infinity here and that's exactly why that's how I think about it so when I started doing this game I had all this knowledge about the Riemann sphere and hyperbolic space and discrete groups of merbious transformations and I just wanted to see well how is this going to generalise in this world and so the things that I started then and they're now so ingrained in me I couldn't possibly change they all have this sort of motivation good so it may be that I've taken something that you know and love and I've written it in a really bizarre way because and told you that it's going to generalise probably true but now I'm going to tell you why what I've just done is actually going to make life a lot easier for us because I've just done a lot of hard work and if you do hard work you want to have a payoff so the payoff is I'm going to be able to write down isometries of all of these metrics really really easily okay so where should I go let's just keep the formula for the metric at the moment so I'm going to say that if H is going to be a Hermitian form U of H is the unitary group with respect to this form so this is going to be all matrices A such that A, Z, A, W equals Z, W for all Z, W I've deliberately not quite told you where my spaces are because that's going to work for any dimensions well unitary is implying that there's complex here if I had done this over the reals this would have been O of I guess a symmetric matrix over the reals and it would have been called SP standing for symplectic over the quaternions and so the same thing so let's deconstruct what this really means just in matrix terms so let's first of all write out the right hand side the right hand side was that we had Z, W was W star H Z now this is supposed to be equal to A, Z A, W so let's use the same formula here so I'm going to get A, W star H A, Z but just like with ordinary transposes when you take a transpose of a product a permission transpose of a product it's the permission transpose of the elements but the opposite way round and so this is equal to W star A star H A Z and that's supposed to work for all W and Z so that means that I can just get rid of those get rid of those and so I would have a formula that H is equal to A star H A and that is the generalisation of the standard formula that you know for when H is the identity matrix and you're dealing with over the reals that if you have an orthogonal matrix then its inverse is its transpose because if I was over the reals this star would become a transpose and if I had the identity matrix I'd just have A transpose A is the identity so that gives me a clue well maybe I could actually write the inverse of my matrix A and so H is invertible it had no zero eigenvalues and so this means that I can immediately write that A inverse is H inverse A star H so another exercise no no another exercise would be if I took this zero minus I I zero and I put complex matrices in and I said that the inverse had to satisfy this formula let's just suppose that the determinant was equal to one then what would drop out would be that entries had to be real and so I would immediately be getting SL2R ah SL2R those were the isometries of my metric well it's clear because of this formula that if I apply A both of these formulae only involve the commission form and so it's automatic that everything that preserves the commission form is an isometry because they preserve these formulae so that's a sort of a very convoluted way of saying that elements of SL2R act on the hyperbolic plane or half plane as isometries right was what we heard in the last lecture but I've probably changed it into such a bizarre and complicated statement that it's a little bit difficult but it's my motivation it's the thing I should keep in the back of my mind that this is what I'm just generalising and a similar formula is going to work okay so that's some of the isometries can we think of any other isometries? yes we can well if I were to send if I just complex conjugate my vector Z what happens to my commission form well it just complex conjugates and so I immediately see that that I get complex conjugation here so let's record this A in U H isometry and complex conjugation then I need to show that those are all of the isometries and perhaps we could mimic this sort of argument we heard earlier but a much more low tech way of doing it and you can find that as a proof in the notes which is somewhere let's see 10 in the notes that these are all the isometries or rather they generate all the isometries because I can compose an element of U H with a complex conjugation and the way I prove this was I mimicked the analogous proof in Bearden's book namely I take any isometry I like because this well I didn't actually tell you that A acts transitively on my space it does there's a little proof in the notes then I can always assume that it fixes the origin in the ball model now I can compose with a rotation so a point somewhere on the real axis that point always goes to somewhere on the positive real axis I think I chose a half that always goes to a point on the positive real axis again by applying elements of U of H and then once I have that data it's easy to show that that actually it fixes the half then it has to fix all points on that real axis and then it has then you keep on going and it has to fix points on the imaginary axis and eventually you discover that that map has to either be identity or complex conjugation and so that shows you that any element any isometry is either in U of H or it's U of H followed by complex conjugation very very low tech argument which I quite like I like low tech now how does this act how does this act on my space so well A of Z well the first thing I have to do is I have to take my standard lift of Z and then I multiply by my matrix and then I projectivise and that's exactly the argument that we got earlier on when we were seeing how elements of SL2 are acted on the upper half plane because what happened was I took my so that was the same so think Z goes to Z1 goes to A, B, C, D Z1 which is A, Z plus D, C, Z plus D that's just multiplication goes to by projectivisation A, Z plus B over C, Z plus D it's the same argument as we had before and it works here so what I could in fact do is I could write these transformations as complex projective maps in two variables where the denominator in both cases is the same I've never found a really useful way of using that formulation but it's true and so just like we had before we have to see what would happen what happens with there's a kernel here so the identity clearly acts as the identity but there are other maps that would act as the identity and it's not so difficult to see that all scalar multiples of the identity are going to act trivially and so what I'm going to go from is U of H that was what I was dealing with but then scalar multiples of the identity so let's just say lambda I acts trivially because over there I'm going to get lambdas at the thing and then I projectivise it goes away and so I'm going to define P U of H U of H quotiented out by lambda I by the way because it's unitary it's determinant it's going to have to be one in absolute value so that's just a circle I'm getting rid of I can also think of S U of H by making sure my determinant is equal to plus one and then P U of H when N equals two is going to be perhaps let's not do it for any old N is going to be S U of H divided by well N plus first roots of unity times the identity so when in particular when N equals one then this is just S U divided by plus or minus one so because of the size of my matrix I don't have to have more so in the case I'm interested in I'm going to have cube roots of unity so it's going to be a triple cover but in general it's going to be you know whatever size and this is where there's a difference in those different schemes that I told you this is where you're going to be slightly careful when I'm when I'm dealing over the reels I get the same formula except there I've only got I've only possibly got two roots of unity there plus or minus the identity and sometimes I don't even get minus the identity because because I can't have a unit determinant minus identity doesn't have determinant one so for some cases so this is where we get to all these different four different components that Todd was talking about with plus one and minus one and forward pointing and backwards pointing here I've got a connected group and so everything's okay over the reels I have to be slightly more complicated over the quaternions scale of multiples of the identity do not act trivially so the only projectivisation I do when I'm doing over the quaternions is via plus or minus one dividing I've got the right dimension so there's some this is a significant difference and that's because of the fact that the quaternions don't commute whereas I do get because I'm always if I'm doing this carefully in the general context I always have the matrix acting on one side and the projectivisation happening on the other side and because I'm English I like to drive on the left and so my matrices act on the left and my projectivisation happens on the right and there are algebraists who like to do things the other way around but I don't know what side of the road they drive on but and so what would happen there would be another multiple of the identity would be acting on the left I'm doing it on my right your left very clever that and then the projectivisation would happen on the right my left and so these things don't interact and so I don't get a scale of them so I have to be a little bit more careful in that case and in the other cases as well the Clifford and so forth the same things are true but in this case I just get a and so in my case 2 equals 2 I'm just going to have a triple cover I've really been going there in 5 minutes goodness me that's fine I'm actually not so far away from where I wanted to be which is amazing so I'm going to finish by talking about how I would go switch between Hermitian forms and and so sort of link in so this would give a concrete analog of this going between the unit disk and the upper half plane and so what I can do is I've got two Hermitian matrices of the same signature so the same number of positive eigenvalues and negative eigenvalues then I take a basis of eigenvectors here and all I need is a linear map that takes one from one to the other and then I need to scale things a little bit but we know one of these theorems from linear algebra which when I was taught and therefore when I teach linear algebra is always called Sylvester's law of inertia and that is that given two Hermitian forms or quadratic forms in the real case of the same signature forms or matrices so let's call them H and H prime there exists C such that C star H C is H prime so in other words this is a change of basis matrix and I'm putting a star on there so that I'm still getting a matrix out of it and I call it C because this is usually called Cayley transform and the names of as many British mathematicians I can because they're usually all French or German and so if I have this actually this means that if I have A in P U of H then C inverse which way round if I've done this C A C inverse is in U I don't care about P is in U of H prime and so these groups are all just conjugate to each other and so quite often I don't care about which form I'm going to do I'm just going to have U signature so quite often I'm not going to actually bother telling you what my Hermitian form is the only thing that's important to me is the signature it's a little bit like doing vector spaces without giving a choice of basis and so this is all notation that's going to come up and again I could do a similar thing with ON1 and SPN1 the analogous group for the Antonians is one of the exceptional groups and it's called f4-20 which doesn't sort of fit into this nice language. Okay I think I'm going to just stop a couple of minutes early maybe I'll tell you in those couple of minutes tomorrow I want to go through I want to describe in more detail the geometry of complex hyperbolic space I want to describe the geometry of its boundary and I also want to perhaps this would be the third lecture would be to start classifying isometries and tell you several different methods of classifying them depending on what sort of mathematics you like so this is where we're going it's all in the notes and I hope that the notes are relatively error free there's a few places where I would like to improve the exposition but if you find any mistakes as you're reading please let me know and I will I don't promise to give you chocolate or I'll give you money or whatever for it but maybe if you find sufficiently many and I remember then you might get a place in the acknowledgments okay