Yang-Baxter Equation, Spinors, Bi-Algebras, Hopf Algebras & Quantum Groups

0:00 Thank you. That's right, that's what you said you had. So this is just what we did last time. Now, one of the things that one would like to be able to do is to get a matrix representation of these. What? For these multiplication rules. the TL algebra is in this one but you'll see what I mean now remember what I did towards the end of last room was tensors tensors for knots Did I do this? I didn't know. No, I don't think you did actually talk about the tensors last week. Yeah, last week you got finished about just the dreaming of the young bastard. Yeah, well, that's the same thing. That's the tensors. Yeah, you did sort of sketch that in very briefly in the last five minutes. You didn't actually mention tensors, although I do see the connection, but there might be an idea too. The idea was that we want to have some way, rather than do the diagrams,

2:30 okay, we've got rid of the diagrams here because we've got to temporarily leave generators, but I want to do this a little bit more generally, connecting it up with scaffolding. Remember, particle scaffolding. So what we do, what it's done, is that this cross here is denoted by, I'm going to call it the black cross even though it's a white cross and I'm going to have remember it was A going in B going in C coming out D coming out and the sort of convention is it's A, B, you read it, I'm sorry you read it English fashion And then you call this thing R, A, B, C, D. And it's like a scattering diagram, and because Lou is a mathematician, instead of doing it the way I would do it, time going up, it looks as if we've got time coming down. But it really doesn't, don't tie in with time at the moment, but it's a scattering. I've just got down here the arrows are just for keeping tabs of the labels and then this one here is just a right circle C, D again and that is the conjugate definition and then under the reader master moves I did this last time you did a bit about the reader master moves on the reader master moves here we get that if you want which is equal to reading master would give you this then you get that, right so now we do it in this diagram, I won't do them all you do this diagram this is

5:00 this one, so it's got to be a white hole and then this one is this one so it's going to be a black spot and then you get A, B, I, J, C, D and that must then equal to A, C, B, D And then you get that R, A, B, I, J, R, I, J, C, D, D, D, D, D, D, D, D, D, D, D, D, D, D, D, D. Okay, so you're getting a tensor notation for unscrambling lots of your eyes. Okay, well, there's another one, which is to do with the arrow change, and that's just everything. And you sum over all the intermediate states, if you're doing a scattering problem, you Now then, this is reader master 2 plus, and then there's a reader master 2 minus, but now I want to do the Baxter equation, which comes out of 3. And the three, remember, was this one One, two, three Oh, no Do you want to make a shift, Rob? This is really your chair, isn't it? No, but if you see me, I'll punch one that again. Well, I'll tell you that as a hint.

7:30 You take the thing that's underneath and pull it through so that this one goes above it. And these two are equivalent to each other. This is a break group equation. and then remember last time I showed this was sigma 1, sigma 2, sigma 1 and this is sigma 2, sigma 1, sigma 2 and that's where you get this great relation I don't want to generalise at the moment because there's too many things going on at the moment Now we will be following notes I mean, not followers Not followers, yeah Not followers and note followers I'm feeling thirsty, by the way I'm feeling thirsty, yeah Oh, but You're feeling good, you're feeling good It's like a ceremony A ceremony of the drinks I'm just realising what I did last time I'm just revising what I did last week. Really, Ria ought to be in charge of this. I'm chucking his tea ceremony. He ought to have Ria in charge of this. I'll have a pop of tea ceremony. OK, so we've got that. Now, can I, on top of this, let me know... Don't rub that bit off, though. I didn't remember it from last week. Yeah, let's give it last week. True, but you know, as you say, there's nothing like insistent repetition. You'll get more of the story... Why don't you rub off the bit on top? I just want to mess this diagram up. No, not quite. Hang on. That's because I can't see it through your back. I'm sorry, I'm very... It's alright, not at all. You're very non-transparent. I'm well aware of that. there's a new edition lots of mistakes in there many times

10:00 well we're used to I'm used to that Brazil changes in the middle of this middle of one equation doesn't matter You have to be consistent. Yeah, yeah, yeah. It should be consistent. But as I've told you before, my lectures are never consistent because... I think I've got... Okay. Now, can I mutilate this diagram? Yeah, mutilate away. Because what it means is just simply that that now is a... These... If I now want to turn this into an R language... these are all of this type ok they're all so that means that this is if I now put A B C on here I J K D E F A, B, C I, J, K D, E, F then another way of writing this another way of writing this R, A, B, I, J R, J, C, K, F, R, I, K, D, E, and that's equal to R, A, B, C, I, J, R, A, I, D, K, R, K, J, D, F. That's just consistently following through the rules of definition, and it's just tedious you should just sit there and make sure you've got the other thing to sit on it.

12:30 But this is the Yang Backstreet. Some other words are better. How goes black and white deeper? One goes over the top and the other one goes underneath. That's all there is in this game. As Graeme said, how the hell is just that little move? The third possibility, you know, in virtual knots Blue is introducing the third possibility, neither up, neither down, sort of like a tunnel. I don't know what he got out of it. And so simply you write the brain or art in a relationship in a different way, and you call it the Jan Baxter's brain. Absolutely. And this Jan Baxter obviously founded in whatever chemistry, organic chemistry. Not at all. Yang found it in one dimensional Schrodinger equation particles in a line with a repulsive core so they don't sit on the same point Baxter found it because of the star triangle transformation Potts model icing model Baxter was working on the icing model like I was only much more intelligent Okay, so it came from two different places, quantum mechanics, particles on a one-dimensional particle, and particles in one dimension. Do they have spins practically in one dimension? No, these are spin-less. Nothing to do with spin. Spin starts At the end of the next page This has got nothing to do with spin Nope, I haven't mentioned spin

15:00 There's no need for spin here Now what I want to do, excuse me, is to find a solution of that equation. But what does it mean, a solution? There is no variable, there is no solution. I mean, sorry, I couldn't... I want to find something, well in fact we know what it is. That is the solution of that equation. no I'm just going to have all this on this is going to totally that which is equal to R A B C D so this expression if I put in there it will satisfy that equation the question is how do I write that in tensor notation So if I call this A, B, C, D, then this is going to be A, C, B, D, this is going to AB, sorry I can't get it in there, can you see it? AB, C, D So this suggests I write R, A, B, C, D How to solve equations without solving equations Delta A, C Delta B, D plus A to minus 1 delta A to B, delta C to B, and if you feed that in there, it's the solution of the collision. But what about those other indices, like E, J? exercise for the troublesome student, show that this is a solution of that equation, you don't have to.

17:30 But you see, solution, because here you have three of them, three of them. Ah, so you, I understand. You identify, you know, put i, j, and then do the summation. OK, it's our pages page of the world Stack it into an equation And show it Yeah, sure But it's all so crazy, isn't it? Yeah, of course. It's, you see, it's mathematics of brackets and coefficients which is going mad, and you get all of that. and then if you take trace delta AC that's equal to delta AA which is equal to N where N is the dimension of the index set normally it's 1 and we take it 2 and that's why that delta is called delta and also you get N Changing notation again, I'm afraid. Minus A squared, minus A and minus 2. I call this delta on the previous transparency. That's fine.

20:00 Thank you. not so much what is the sound of one hand clapping that lamp was making a very irritating noise so he just got up and bashed it fine-tuned it fine-tuned Now where time comes in, or to help, not necessarily to help, is really to think of this as a scattering diagram with time going up there, then if you rotate this... Is that what the arrows are? No, not really, no, they're just, the arrows, if I was to redo this, I'd put A, B, C, D, and the arrows the other way. Because I didn't, I wanted to keep track of what... So that means you, well, that one is in the grid, that one, R... Hang on, I just want to know, I just want to find the relationship between R and R bar. That's all I want to do. Alright, alright. If you take this and rotate it through 90 degrees that way, you get that diagram. Take it 90 degrees and you get that. this has to go up to there this down here, this there but this will be flat time no, no, no, no, it's just a question of convention because if we keep time's arrow always in the same way then this is C A and therefore R bar C A

22:30 D D equals that and that tells me that that is equal to this which is R B D Now I've got that argument right. Oh, shit, I've got the labels on the room. No, I'm trying to do it too quickly. Okay, now then. R, Y, B, C, D, was equal to R, B, D, A, C. But is this moved by 90 degrees? Yeah, the moved by 90 degrees. in other words what I've got to do is the symbols, labels they actually rotate into each other but this is like a rotation so you move it, but you move it this way yeah, yeah, whichever way and therefore that enables you to show that R bar A B C D is equal to A to the minus 1 A, B, C, D plus A, A, B, C, D which is equal to A minus 1 delta A, C, Okay, so that then gives me the complete solution of the Young-Baxter's equations. I find the solution for and the solution for And you'll see it's just the opposite grade, which is what we should expect anyway. One sigma, one sigma to the minus one. But this shows that it's all consistent and the labels are all right.

25:00 But we have done nothing at all different from what we've been doing before. Well, you never do anything different in mathematics anyway. You think it's the same as another and you go on down there for A. OK, now I want to go on to something completely different. You ignore that. And Graeme knows what completely different means. It's the same subject. But by the way this going around 90 degrees is exactly the same as if you flip. However, when you flip you don't get the notation as you can. Do you ever flip? Question for investigation, what happens if you flip? This is not a trivia question, because this redemester move is that you flip. I don't know, I haven't looked at it. You haven't come across... I haven't looked at flipping whatsoever. I've been so trying to get this thing sorted out so there's a consistent story throughout that I haven't actually investigated the sidelines. Someone would say, of course, that they'd flip over there. Yeah, I think. I also wonder whether you already... But I love solving equations with diagrams. I'm sorry, Mike. No, that's the way to do it. Well, hey, I mean, you know... Yeah, I've been talking... Well, inspired by Bill. I mean, diagrams is what it's all about. I mean, these are not... Now then, do you remember that spinner paper I did with Fabio in the American Journal? Yes, I do. First thing of yours I ever read. Well, I thought this might. OK. This is going to be related to the spinner structure. Very interesting, yes, that's what I hear, what I hear. Particularly that lovely idea that you could get the, which of course is also there in Grassman, that you could get parts of a vector, parts of a bivector, and the other multivectors all inside the same structure. I don't know yet, but it's all lurking.

27:30 By the way, did you know that the Grassman algebra is not an exterior algebra? The Grasman's Grasman Algebra. It's not an exterior algebra. Yeah, I know. This is what I was learning from doing this. We gained in. We gained one by doing a certain restriction. Yeah, but it is, yeah. A hell of a lot of people just don't realize. Because, in fact, there are twice as many dimensions in the original Grasman Algebra. And it's got this difference operator, which is not there in the exterior. So, what was your first sentence, Grasman Algebra? It's not an exterior algebra, as all the textbooks tell you it is. Burbaki is not saying that. Oh, isn't he? Well, I haven't read by Burbaki as I should. Burbaki gets it right. I'm glad to hear that. But a lot of textbooks do say that, and it's not. It's something more subtle. It's a richer structure than that. It's not just an algebra or a vector space. It's got more structure in it than that. You are drawing a sphere or a circle? Sphere. It's my stereographic projection onto the plane through the middle of the sphere. So you are going into four dimensional knots now. Oh, no, no, no. Remember, I have a point P on my sphere, I have a north pole. And I can project, sorry, I'm going to miss things here. I have a point on my plane, my projected plane. SAS pole. Let me project this through the... Go to the diametrically opposite point, PD. project this and then we get Q Project North Pole Let me join the North Pole to PD Intersects through the plane at QNB

30:00 Well, it shows you where the flag picture is coming from I've got four projection points possible from the two poles. I've got a point on the sphere, and I project, stereographic projecting onto the plane. But I can take the mirror image or the diametrically opposite image and I can project that through the south pole to get another point. I can take this point and I can project it through the north pole. I can take this point and project it through the south pole. The purpose spinners. The spinner psi1 over psi2 and that is Penrose spinner ua upstairs where a1 and 2 and then I can write that as sigma n is equal to u1 over u2 I'm using homogeneous coordinates, and then what we do in physics is we write ua is represented by, perhaps what I ought to do is pi ua is equal to u1 u2, and that is why spin up and spin down. Do you write with false pens tonight? Do you write with one a bit and then another a bit? Er, no, I'm right-handed. I'm right-handed. No, but you've got two pens and she thinks you'd change over to cool it down. Something like that, yes. Some people, I mean, there are some ambidextrous people. More senses than one, but I...

32:30 The bar is complex conjugal in this story. Thank you. I was quite amazed when I discovered this. this is an SO2C Penrose had taught me that there were four types of spinners. There are four types of spinners. There are spinners with indices upstairs, one and two. There were spinners with indices downstairs, one and two. there was a complex conjugate spinner with the A upstairs, and you put a dot on it so you got a U-bar, a dot, and the corresponding Now when I was introduced to them, I thought to myself why have you got four spinners?

35:00 And the reason why you've got four spinners is because there are four ways of projecting points onto the projected plane. This is the Penrose Spinner, yeah. The Spinner of Quantum Theory is using that mathematical structure. But this has got nothing to do with quantum mechanics. this is to do with geometry the spin, we're told in physics or by some people, that spin only comes in in quantum mechanics yeah? rubbish! sorry thank you the spinner is already here without quantum mechanics Without relativity, it's in a stereographic projection. This was known to the ancient... Who was it in Coolemy? It's one of the... I think it's Coolemy. It's certainly one of the great Alexandrian geometers. One of the great Alexandrian geometers. I think it's a bit later than Coolemy. But he only got one of them. Why have you got four projects? I have no clue. I don't know why has everybody known. You're interested, let's look at it in terms of SU2. You're interested in rotations. Okay. Now normally when we're interested in rotations we rotate a vector. x, y and x and therefore you need three parameters in your rotation matrix to describe it but a rotation can be described by a point on a sphere but then there is any possible direction this is a problem why there are only four first of all you say How do I classify the spin with just two vectors?

37:30 The rotation, rather, with just a two-dimensional vector. And the way you do it is you say each one of these points corresponds to a rotation. If you try to cover this with a coordinate patch with just two dimensions, you can't cover the whole thing. You've got to do a double patch. you've got to put a coordinate patch on the top and a coordinate patch on the bottom that's why this is all to do with fibre bundles are you happy with me? you only need a two dimensional space to describe the rotation because every point on this sphere describes a rotation in what way? in what way? Because every rotation involves taking the vector v to the vector v prime, right? Which means it takes that point to that point. So two points, not one. Okay, but then we want the difference between these two points. All right, let's put two points. Because if two points, I understand, two points, because then it's obviously not in effect. All right, so you only need to, the point is you only need to affect the space of two dimensions. To describe the rotation, you need a two-dimensional vector space. Okay, good. So, but if you try to get this rotation on a two-dimensional space by using ordinary coordinates, you can't do it because you've got the problem of the singularity in the coordinate system. So you have to have two coordinate patches. And that means you've got to construct a fibre bundle. OK? I mean, that's technically only done with the way about that. But there's one way you can get away with one coordinate patch. And that is by projecting through the North Pole. But... Yeah? But adding the point at infinity. So it's the projection plane plus the point at infinity. And the way you do that is with ratios, because when psi 2 is equal to 0 you've got your point at infinity at least I have some direction of understanding

40:00 ok, it's pure mathematics pure geometry the properties of not ordinary geometry it's projective point infinity it's projective geometry and so the coordinates of this space U1, U2. Okay. Now, the way these things transform is you make this one rotate, P rotated it, D rotates as well, and then you do all this up so that these things now all rotate in the correct fashion if you write them this way. But why Penrose... I mean, if you go to another point here to get P2, to get the rotation between P1 and P2, do all the projections, and it's equivalent to transforming these objects here. So this one always transforms... Is that the U... Yeah, this one always... This point always transforms as the spinner index upstairs. This point, which is Nd, this point always transforms as the complex conjugate upstairs, and so on. So the reason why those things transform the way they do is because of that figure. Hoff did this part. It is known as a Hoff projection, but it's also done by Haleigh Foscueur without realising that Hoff had done the same thing. It's a convention because U can be a complex object so you've got to take the complex conjugate of it and the A dot just says it transforms

42:30 as the dotted index. And if you go into it you'll see. So that's the geometry of it. Now what the hell has this got to do with? Remaster moves. right now then you're encoding the three-dimensional rotations on a two-dimensional net I don't know exactly what I'm doing when you're doing your shoe horses. So there's a much deeper relation even than I've thought about. Now there's one other thing that we've got to look at. Let's see if I understand this. This is why it's difficult to consider three-dimensional knots because you would have to be passing in four dimensions. Exactly. You also have to go up a dimension. Oh, God, I've got a step here, which I don't understand. I knew it this morning. Something's gone from me. I've got it in here. Since psi dS equals u2 over u1, that's all right, because all I've done is just take that. This one. and I say that is equal to minus psi n which is equal to minus u1 over u2 and therefore we have a relationship between the upstairs and downstairs index which we can write as ua equals psi log a b u b In a particular representation, that is equal to 0, 1, minus 1, 0. It's due to the symmetry of this figure.

45:00 These are the two points we'll consider. for some reason the giant escapes me at the moment you say those two points are identified in that way sorry man, it's just not coming there should be a good reason for it but it's a well known what is well known is this And you're not surprised because the two things aren't independent. This rotation and this rotation are obviously coupled, so you can either work with this object or you can work with that object. Yeah, because if, you know, it's all, you know, I think it's tied up tightly together. It's amazing that the Greeks got all of this, essentially got this, or gone with, without knowing any complex, without knowing about complex numbers. Yeah, well I don't think, they didn't get the whole story. No, but they got an awful lot of it. We got the idea. Okay, so now what have we got? We've got ua equals sin on a b, we've got sin on a b equals 0190 equals sin on a b. There's more work to do than I'm doing here, I'm just giving you a sketch of it. So this can all go off now, that was really put in there to show you where it all comes from. Okay, so that means if I use this, u1 is equal to epsilon 1 v uv which is equal to minus 2, and u2 is equal to epsilon 2v, and v which is equal to minus 1. Why not uv, but uv? Well, b is summed over. Summed over. over, sum, sum, sum, it's been summed over, repeated indices, I'm using the Einstein convention.

47:30 But the invariant we get out of this is uA, sin on AB, VB, which is equal to u1 to 2 minus u2 V1. why is it invariant because it's got no indices on it so it must be invariant so if i make t on u i get u prime where t is some rotation then that thing remains the same with the prime variables So this is invariant under SL2C. OK, so now let's use this notation. I have U A primed equals T A B U V and V primed A equals T A B V V where T A B is an element of SL2C. And therefore, u'a, sin on ab, e'b, is equal to 2a. I've got a funny notation here, which is all right in the end, but it's not the way you want to do it.

50:00 Therefore this is invariant provided epsilon is equal to T epsilon T transpose. That's the equation I really want to get. So for T to be an element of SL2C, you have to leave this invariant, Epsilon 0, 1, minus 1, 0. So what I want us to do is to concentrate on that, because that, leaving that invariant It tells me it's an SL2C. This is really hard. It's just index shuffling. Don't rub that last bit off, will you? Three index shuffling. Now then, what I want to have a look at, for reasons which will become apparent, is I want to look at what epsilon ab is equal to. It's equal to Epsilon 1 1 Epsilon 1 1 plus Epsilon 1 2 Epsilon 1 2 plus Epsilon 2 1 plus Epsilon 2 2 Epsilon 2 2. That's just what it means. Right, 1, 1 is 0, 2, 1 times 2, 1 is 1, 1, 2 times 1, minus 1, minus 1 is 1, that's 0, and therefore this thing is equal to 2. I just want to note here that epsilon a v, epsilon a v is equal to 2, and epsilon a v all squared is equal to minus 1.

52:30 But you mean it's matrices, because... Yeah, that's what I mean. What are you doing, maths? Because you are multiplying correctly. No, I'm saying what this means is this. Yes, and this is multiplication. This is zero and zero. Because it's the same. It's not matrix multiplication. It's a trace. No, it's not that either. What is the difference? If one of them is a matrix, this is what you have written there or not? Yeah, it's a matrix. None of them. So two of them is not multiplication but a number. It's a number. In what sense a number? In the sense that if I work this out, you agree with that? Yes, yes. So, I want to know what E 1 1 downstairs is. It's zero. I want to know what epsilon upstairs is. It's zero. So that's zero. I mean, this I don't understand. I mean... Well, look! If I take that, it's writing A1Y equals 0 equals... What? Yeah. So I'm missing something... Well, I'm not sure what you're missing because... Because I'm seeing... Please help me, someone else. It's just because you used the answer to me. What about if there was a sigma over the left-hand side? It would be clear. yeah oh please do because I'm using it all the way through wherever you see a downstairs and an upstairs index which is the same summer b upstairs or downstairs, that it equals this matrix state, I mean, this is what I'm worried about. This means that epsilon 1 1 is equal to zero.

55:00 Look, we write epsilon 1 a of b is equal to... I mean, maybe I should pop up for any mathematician, I don't know. The representation of this is epsilon 1 1, epsilon 1 2, epsilon 2 1, epsilon 2 2. Okay, so this is what I want to use. Okay, I've got you with me now. Okay, good. And everybody knows this because we use it in physics. Yeah, we use it all the time. I don't think anybody uses it. It's like you're writing sigma down. Yeah, yeah. I don't think I think the physicists will know what I'm doing I don't know why I'm doing it I am going to get exactly the same thing as we had on the previous transparency So it's very difficult to see when you're... Yeah, I'm sorry if you're mine. No, no, no, it's not your fault, it's just the way this room is configured means that when you're sitting here you can't see a bloody thing of what you're... Well, I'm trying to dodge backwards and forwards. No, no, it's alright, I'm not complaining, I'm just... As soon as I get in front here and I'm in front of someone else... We were not doing that, you see, when I was doing mathematics. Not a single time. I read about this convention only maybe a few months ago. Yeah, it's very queer when you... ...whether we can do it in Poland at all or what? Oh, you do Detrautman, who's Polish, very famous for Polish relativistic. Anybody who's doing relativistic physics knows. It's just laziness, because when you do Einstein's tensors, you've got all of these sigmas in there, there are too many of them, and all the other industries. This is how I came across it, like, what I understand, and then they were saying, Einstein Convention, what the hell is it? Well, that's what it is, it's nothing. Was it really Einstein who... Well, they call it the Einstein Convention. Was it him who did it? Okay, right. But he wasn't very good at the matter.

57:30 No, he was very bad. Well, there's a good... He's not enough, though. I dispute that. That was the story which Hilbert put around. I wish I was as bad a matter of... He made this famous cutting remark about, you know, when he'd just been beaten by a matter of two or three weeks you know, they'd been beaten by a man who knew no more mathematics than a schoolboy. But in fact, Einstein was, I think, not just because he had amazing physical intuition, I think he actually had remarkable mathematical powers. Remember, I made that point to you, he independently discovered the Duran formula without realizing that's what he'd done. Well, in terms of certain domains of integration. the whole transformations without knowing what I was doing, just by looking at the structure of the thing. Yes, exactly. Well, he discovered, you know, large bits of homology and homography theory, but in terms of behavioural cyclic domains of integration, which he published as a paper about the Bohr-Sommerfeld quantization conditions, the orbitals, without any of the physicists who looked at it realising that he'd actually discovered the fundamental theorem of modern topology in that paper. No, I mean, I think he was much underrated as a mathematician, Einstein. of them is I've always discovered things which were discovered 200 years before. Because he clearly was so great as a physicist, but I think he was a lot under a seriously underrated. If he'd lived in the 19th century, I think people might have appreciated... Well, I mean, if he'd lived around the same time as Maxwell or Wierstra, if he'd lived in the Soviet Union, if he'd been a contemporary of people like Hamilton and Grassman, I think he's just mathematical. Because of the direction that maths had taken by then, and Hilbert and this kind of structuralist, I think. I'm glad you can say that I've been trying to say it all weekend Yeah, I saw that perhaps you And he was a school teacher Yeah, he chose to be Yeah, he did He was a school teacher in Berlin also, briefly He actually did that after he gave up

1:00:00 He gave up because nobody paid any attention to his There's just been an English translation, incidentally, did you know of Grassman's... Lewis, which is reviewed, which, I don't know if you met here the other day, was just reviewed by Historia Mathematica. That's good. Which we will hope to be in heaven. I see, you are coming. I think perhaps Kaufman was working backwards from all the stuff and this is how he got all the stuff. That's a very interesting idea. When he starts to make noises like that, I know something really exciting is happening. It's not quite right though. Ah, because we don't know anymore. Yeah. Right, we'll just stay there for long enough for me to look at it. The only problem is that if you had a and a minus 1, you can't get plus 1 and minus 1.

1:02:30 Because a is sort of like a, a minus one cannot be made of like a minus one. Hang on. Is this a catastrophe? Don't rub off the bottom bit. No, I'm not going to rub off the bottom. I've only seen it yet. Don't spell it. No, no, it's not. I read it in your paper. No, it's just grey. But it's coming from spinners. So it is almost like a commutator and anti-commutator. No, I don't know. We can get it. This is not topologically invariant. That's the problem. So if you want to use it in knot theory, this is not topologically invariant. But I remember exactly the same sort of sentence in Lou's paper right on the train, that he was saying, oh, but it is not invariant. I've been reading Lou, so all I'm doing is repeating what Lou says. And he was saying, no, no, other way around. And he was saying that this is great, but I don't understand why. And I don't know where he's talking about this, but this is not topologically invariant. But we can very easily change it if we write side on AB Z equals to 0 I minus I0 Change their numbers And that's the Pauli's sigma Y matrix if you remember It is sigma Y I mean plus sigma Y This is lovely stuff, but I can't, to be honest, see how this connects with the way that you and Fabio Foscuro went about it. No, I'm sorry, I don't. I'm just saying that here's the epsilon, which is in the spinner, and it seemed to be fitting in with the structure.

1:05:00 That's all I'm saying. Yeah, okay. And then I'm saying if we do the i, then what you get is that, so if I call this prime just for a minute, sine on prime ab, sine on prime ab, then that's equal to minus 2. And sine on prime ab, all squared is equal to 1. change and these are pen roses negative dimensional spin this thing here this D This is the D that comes out in the... OK, when you do this... Can I rub this off now, Mike? Yeah, that bit, yeah, sure, sure. I'm now going to... Please notice... I'm now going to drop the... Oh, no, I won't for a minute. Can you just... No, it's not. OK, so what I've got here, then, is epsilon prime AB prime is equal to minus delta AB delta AC. is T plus delta AD delta DC. And that is A equals 1. The solution is the Yang-Baxter equation. Which, which, which, if you... Did not really. And remind us again of the two roots by which Yang and Baxter respectively came to the equation. One dimensional, three particle with n particles on a line.

1:07:30 They're not allowed, no two particles are allowed to stay on the same point. Right, yeah. And the Baxter equation is the POTS model, icing model. Right, which you talked about the other day. Which I talked about, which has got the braid in it. That's not surprising. What is surprising is it's in quantum mechanics. And that's what we haven't connected. It's trying very much to connect it to quantum mechanics. And all we're noticing here is that if you use Penrose Spinner and you use the I, so notice we brought in the I here, so it's not quite the projection that we put on according to the complex numbers to do it. Yeah, this is what I was about to say. You were having to bring in the complex number 2s here. You have to bring in the complex numbers. You can't do this with the stereographic. I can't do this in the purely classical. Is it you who has derived it? No, this is Lou, this is Lou. I just put it into the context that I came to the spinners via projection because I wanted to show you that it wasn't just not theory, it actually comes from the geometry of projection, the projective geometry in the rotation group. But here you can't get this simply out of the projective geometry You've got to bring in the complex numbers to do it this way. What else is a neutral metric, if you replace the... Well, that's the question that we've always been worrying about. So maybe it makes this project neutral metric into some complex project. That sort of stymies your idea of knots on two surfaces and so on, so it means you've got to bring in the complex... See, that's the route that Rogers Penrose has largely gone down. But the remarkable thing is that you, well, as you've shown me and I think you've talked about it to the group, you can actually even get the twister out of that holographic, eventually out of the holographic, well, not out of the stereographic projection. You can get even the twister out of that without bringing in complex numbers at all. It's interesting to look at that because, again, you do that without bringing in the, essentially bringing in the complex numbers. If I have a few minutes to spare, I'll try and think about that.

1:10:00 I've been so busy getting it all in a year-end story. I think telling that story about the twister again sometime will be very interesting. Not tonight, maybe. OK, so if we've got this, then this thing is just equivalent to what we've been talking about. I'll put it in smalls. so this is essentially that a to the 1 sorry? what is a to the 1 oh because if you this, if I draw it in diagram form is just that but if we compare it with a to the minus 1 and a or a to the minus 1 whatever, it's just for putting a equal to 1 And notice, here we do not distinguish between over and under. It's just a fact. Now whether there's any significance to it or not, I'm not sure. It's just notice that it does not distinguish. because there is no difference between a and a to minus 1 so if you would like to write the other equation it would be exactly the same I'm going to do something dramatic in a couple of seconds are you going really? I've done all the work for you. What? I've done all the work for you guys. Oh, no, no. You've seen talk about the brave groups. You weren't here when I talked about brave groups. It was a paper which was sent the same, at the same time actually, it was impossible. It cost me a lot of effort and if I wipe out I'm not able to come. Okay, I'm sorry, I'm just saying that I had sympathised with you but I am saying that I actually... I wish we could have seminar during the day but... So do I but 90%... Is anybody working here during the day?

1:12:30 OK, can I now go on? One thing I'll let her say, though, which is perfectly vague, you did actually promise us about the end of the week before last, and again last week, that you were going to talk about the hot falgebra. Oh, come on! Oh, I'm patient! That's why I'm saying I'm going to do something staggering now. Infinitely patient. I'm sorry about it. Look, unless we build this up slowly, slowly, catch the ruddy monkey. Yeah, I was going to say, if you try and do it all in one bite, then. OK. Because it is the... I'm going to put I's in here. Minus, plus... All right. There we go. Actually, if we had the camera, Mike, you should... You know, I was exactly about ten minutes ago when he was doing the stereographic projection for the spin-off. I was thinking that. I was actually thinking that. I might do that next week. I guess that wasn't costed out. I haven't worked out what it is with the icing. OK, so I've now moved from the one and minus one to the eye. So I have sometimes done that at conferences and meetings. The speaker said it's OK. It's a good way. Yes, I might do that. Are we ready? Epsilon AV. I'm now going back to small notation because I'm going to not theory now. Those little disposable ones with flash are fine to get some in the border. So maybe I should just put these... In other words, when you see A and B, A and B, I'm using the I, whereas when I'm using the capitals, I'm using the... I don't know why I've tried it. Small A and B means I've got the I in there. A and B is the pen, actually. Hang on, Basil's trying to let us know something. We're getting sidetracked as usual. What I'm saying is, instead of using 1 and minus 1, I'm using I and minus I. And I'm changing the sign onto A and B so I don't have to carry little a and little b because I don't have to carry a slash. Okay, what you've got down before is a song with a slash on it.

1:15:00 What happens if a is not equal to 1? a is not equal to 1. Consider. And those a, do they hold some positive? Yeah, what kind of? I think it's just a number. This capital A has all the properties that Lou wants to attribute to it. In other words, it's just a number. And D is equal to minus A squared minus A to minus 1, minus 2. Let's look at M-A-B M-A-B and that's equal M-1-1 Do I need to put this down? I have to put it down because I don't know how to do this thing other than writing it all out and I'm very glad it's only indices a and b are one of two. And therefore we see that's equal to d. And that's very nice, because if we look at M, A, B, and we look at M, A, B, in other words, if we think of this one downstairs, I'm not sure if I got the, yeah, that stands, downstairs is this AB and then we multiply it by this and this is now

1:17:30 Japanese AB so that forms a loop and therefore we get D which is N in the previous same with n-dimensional space and if we look at what m a real squared is it's one and therefore we have a b c d equals r a b and it's the bar one Lou's got this wrong again in his book, he changes the notation. A, M, A, B, N, C, D, plus A, the one is Y, delta, A, C, delta. So we have another solution of the Young-Baxter equation, where we have actually got the off-diagonal element A. So you can see this is a deformed structure. So now I begin to see where Hopfizer has come from. Sorry, that's all right. Is it the same A? This is the same A. I want the same A, yes. Okay. Because if you speak, I would still wonder... No, no, okay. Only a right to check. No, this is all the same A. This is all the same A. The A that I put equal to 1 up here. So this is a deformation. So I'm deforming the epsilon AB into MAB. So this is another solution, yeah? This is another solution. Okay. And then you might say, well, so what? But now I can connect it with the day groups. And the day groups are governing the deity of parafermials?

1:20:00 No, no, no, no, forget about that. It's nothing to do with it. I know, but I read it yesterday. Oh, well, I mean, forget about parafermial. I'm not doing physics at all yet. This is geometry. This is Euclidean geometry. And you can get the same thing for the R, OK? Without the bar on it. It's just the opposite. Now when you go to this M-A-B, you now distinguish between over and under. Right. when you deform it into the a's then you take the difference between over and under and you can see why you have no difference between over and under in the case of a equals 1 because if you turn a and a minus 1 round when they're 1, you turn round 1 and therefore it doesn't that's why it doesn't distinguish between over and over but I just told you that I'm stupid No, no, you said, oh yes I just wanted to emphasize it because I find this quite staggering Okay, so what I've got then I've got R, A, B I'll just write this out is equal to A delta A, C delta BD plus A to the minus 1 I've got the bar now this is the Pucker one now remember sigma was the braid operator corresponding to that which is equal to A.

1:22:30 Now this U, I'm sorry, is the U of the Leigh, Tempoli-Li algebra. This U is the Tempoli-Li algebra. Or I should read first. So, this then corresponds to this, and this, and this. So is this algebra having this matrix representation? So this is the matrix representation of u So that u equals m, a, b, u m, c, d Right, right So now if you want to look at what that is M, A, B, M, C, D is equal to M tensor product M and then you've got a matrix which is 0, 0, 0, 0 0 minus A squared 1, 0 minus A to the minus 2, 0, 0, 0, 0 I just drew that the same way as I did this one. I don't know what Bob's put in the Coca-Cola. OK, so what I've got then in general notation is a 1 cross 1 plus a minus 1 m cross m. So if I'm going to write this in... It's beautiful now without any indexing. Oh, this is it. Let me have a little confidence. So that means in order to solve this problem you have to double the dimensions of your representation space. Yeah. Sorry? Absolutely, absolutely, absolutely.

1:25:00 Just as advertised and promised. to you people who are not but you have to know the rules for this because not any old you have to realise the indices are important even though we've thrown them away We've only thrown them away because they bring out the structure of the thing. But that direct product is not the same as the direct product that I might use. Because it's defined in this way. You have to define these direct products in the way you're defining them. OK, now then, the next thing is quantum groups, i.e. that function is good. Now, for some reason, I have to regress something like that. That is the operator U in the temporary lead algebra, which is to do with. But also the tensile product of it. Yes, yes. So this whole thing. Remember the symbol. whenever I have a symbol like that it's a tensor product yeah, the two strokes are just 1-1 so that's going to be M cross M cross 1 cross 1 I'd rather draw the diagrams and write all that crap out that's what we're doing

1:27:30 but so it was before yeah sure I haven't done anything wrong that's why I say Graham, Monty Python now for something completely different now for something much the same but it seems to be because the structure is richer now mathematically it's not probably I mean the same Well, I mean, you could certainly try and make something flexible. Okay. Now, I want to ask the following question. What leaves... Underneath. Asylum bar equal to... I should point it out. Zero A minus A to the minus one zero invariant. The war in Iran. Pardon? The war in Iran. War in Iran? Has it started in Iran? No, no, no. This is the answer to your question. What leaves the infallion? The war. Well, that's true. It's not a very exciting thing to do. And this is not what you want. This is not what I want. OK, so what I have to do... You notice I haven't got any eyes in there. It's not a mystery. I haven't got any eyes. No, I noticed that. In fact, it doesn't matter, does it? Because I can put an I in there on both sides and I'll cast that. What I want to do is I want to find out the T. So it doesn't matter whether I put an I in there or not. This invariant is independent of where I's there or not. I want to find what transformations T, T, in other words T, is an element of what? Well, we don't know yet. No, it's not. Remember when we were doing the spinners, T was the rotation.

1:30:00 The element of the SL2C. So what is it in this case? It's not SL2C, because I haven't got 1 minus 1 down there. So let's find out what it is. Let T equal A, B, C, D. T is equal to just that matrix line. Okay? And then that's T transpose. Okay? And now I want that to equal 0A. And so the question is what properties... Okay, I want... So what you find is that if A squared is equal to Q, because you're going to get some A squareds in there, no you're not, you do get some A squareds in there, You find BA equals QAB, DA is equal to 1 minus QCB, BC equals Q into AD minus 1, and you get DC is equal to QCB. Let's just multiply matrices together. Ok, the first comment is that the elements in the matrix are non-committed objects. it's not non-combat but non-commutative

1:32:30 they only become commutative if Q equals 1 which is SL2C so once again it's a deformation of SL2C we're talking about deformed by a parameter Q now you also need to do the following that also holds so you find another set of relations from there relations are CA equals Q and C C B equals Q A D minus 1 D A equals Q B C plus 1 and D B equals Q Therefore our A B C and D are determined by these two sets of required chips No, no, no, no, no, no, no, no. You actually find that BC commute with each other. Those two commute, the rest don't. And you can work out all that. It's not very exciting. Now, the problem is, the problem is, okay, so that's just, and this, this object is called a quantum group. And it's called the special linear group Q of T2C. But it's not a group. No, it's not a group. And you have it. Okay? I'll rub. Can I rub? Sure. but it is not a group so the question is first of all, why isn't it a group?

1:35:00 why isn't it a group? because if you have T epsilon T T equals N because there are only two groups of all the parts well it could be but you can actually prove that it's not going to be right then if you want then it should be if it was a group you'd have this relationship if st was an element into the group, then you must have that. This doesn't satisfy this relation because TST is not equal to STTT. Because it's non-computative. Okay, so the set of objects, S, T, T, S, etc., are not elements of a group. So Ts do not satisfy group axioms Do not satisfy group axioms but we've already got a sort of a hint because we've got m cross n so maybe we have to make Make a bigger structure in order to make this work. Introduce Co-Product

1:37:30 In other words, introduce a product which is... I'm going to replace Epsilon bar by capital E now. Okay? Just for the notation. That's a mapping. where e is some invariant, I don't have to specify it. And then, with the definition, and this is important, I'll put the sum for... because it's ambiguous in this case, as it's a strange thing you've got to do. So that is the nature of this tensor product, it's very specific, if you don't do that you won't get the thing coming out correctly. And then we want TET transpose equal to T. We want to find T such that. So then E would be this A, A the minus, minus A the minus one down the diagonal. Yeah, this is looking very, very sort of functorial because curve product is always a function. Okay, you'll have to tell me where it comes in. I'm going to have to try and see how to do this. So knot theory can be done by category theory? Well, I think it can. Anything in maths can be done by category theory. There are all these functions lurking there in the background somewhere. OK, so what we've got to do... Certainly anything that involves algebra or geometry.

1:40:00 Now we've got to show... Why do we want to show this? Because we want this to be a map from an algebra, a map of algebras from this algebra on to this algebra so we want to make sure that delta T a i e i j delta T b j equals e a b times i where i is equal to 1 cross 1 shall I go through that because it's I think I'll start you off because it takes a bit of getting a physicist it takes a bit of getting used to I don't know about mathematicians the 12th is obvious this is true well otherwise you wouldn't put it on the black that's not that's hardly a reason it's also it's a calculation you get it make a mistake that's right yeah but I mean the problem is Because you're doing funny things, the legitimacy of things you do... Oh, wait a minute. There are certain things you look at and you say, well, can I do this? And I suppose you would say, well, if it gives the right answer, obviously you can do it. So this is the left hand side here expanded, here's the comprise sum over K, sum over L. Einstein convention, right? Now, because of the nature of this, you can actually pull the Eij out to one side. So that this is equal to TAK multiplied by TBL, OK?

1:42:30 This one with my boy, this one with my boy, this one with my boy. So that's equal to T-A-K-T-B-L, I can't do anything with that, direct point. Let's make sure we pull this in between here, T-K-I-E-J-T-L-J. I've got a summation in there and I know what that's equal to because I must have that from the condition I've been using before Now then, remember that this is essentially, implicitly on this side, 1 cross 1. 1 cross 1 is in there, implicitly. So that then is equal to T-A-K-E-K-L-T-B-L crossword. If you don't put this one in, it looks a mystery where this comes in, but this is floating around in the background. I don't know whether to put it. Lou doesn't put it in.

1:45:00 It assumes you're skilled enough in tensor products to know it's there. So that's equal to EAB cross 1, which of course is equal to EAB1 cross 1, and that by definition is just EAB times pi. And therefore this is a map. from one algebra on to another algebra now then, if you do the product T star S remember we couldn't get the group properties right because of T and S, and define this by sum over K, TAK tensor product, S, K, V, then you can show that t star s t star s transpose in other words you've got that s star s to minus 1 is not equal to 1 So it's still not a group, even with that new definition, because there is no inverse. so now what does this do to our structure that we've got previously we're not only

1:47:30 So we go back to our algebra that we've just done with the epsilon rather than the e and this. And then not only do we have the relations between the a's and b's, the non-comitative relations, we also have the following. that the co-product of A is equal to the co-product of T11 because A is in the T11 position and that's equal to T11 direct product T11 plus T12 direct product T61 and that's equal to A cross A OK. I'm just coming to my co-associativity now. You're going to be here next week, aren't you? Well, the thing is I wanted to ask you if you could very kindly countersign that LMS application. Do you remember the one I showed? Yeah. That's great. So, I'll... Just remind me a bit again. I really want to try and get it sorted out before I move. Thanks a lot. 461679. Is that an aye? Okay, so that defines the co-products on the individual elements in there.

1:50:00 Oh, because those are the things that go into the previous proof. You could put these in, because remember I had delta T-A-I. E I J delta T V J OK, all I've done is tell me what those deltas are in terms of my original A's and V's Sorry, but that's just showing the way I've already used these results I'm in, so I can't rub everything off. I've been talking about bi-algebras without having defined what I mean by bi-algebras. Yeah, sorry, excuse me, it's been a bit slow tonight. Sorry. Can I rub this all along? Yeah, as far as I understand. And the first condition is it's got to be co-associative.

1:52:30 And what do I mean by co-associative? Oh, Mike's in his element. Something that looks... Exactly. But this is co-associativity. Yeah, I realise that. Wait a minute, there's something wrong here. No, it's all right, it's all right. It's all right. Oh, have I got the arrow in the right direction? No, it's all right. Oh, there's no left one. The opposite one. Yeah, she'll be going in the other direction, yeah. Sorry, that looks like. We've got to be co-associative. A has to be associated I'm not even in the right order A, the algebra A is associated with a unit it's got a

1:55:00 inner product well it's just got a product A cross A onto A, that's a usual form of product. And it must also add a co-unit. which is sign on A goes to C, C-complex field, whatever, some field. OK, and then you've got for the collage to die, where the co-units satisfies the following. Thank you. and if your algebra's got all those properties then it is a bi-algebra and then what I the question is have we got a bi-algebra and what we've got in that

1:57:30 I'm not sure if he can keep an eye on it I think it's in there anyway so you can see where I've got all my stuff All the details are in there somewhere, but they're not necessarily in the order that I give them to you. I think that's the difference, if it's ordered in a certain way, I think it's more... I find the story I tell a little bit less confusing than the story that Lou tells, because he goes up in different directions, there's a lot of unnecessary work in there. Surely one of those arrows needs to be reversed, doesn't it, on the right, on the left? I don't know, I haven't really checked them, I mean... Because it's a definition you've seen. What's the definition of a fire? Yeah, this is a definition, so I can't really... This is a co... No, this is just the condition on the co-unit. Yeah, but how is the k-unit if it's on the, I'm sorry, the projection, the downward arrow on the right-hand side of the right diagram, surely it should be an upward arrow, shouldn't Oh, I'm completely missing something. let me look at this one this one's wrong no I think that one's right but surely the one on the right should be up I think it should be up well ok we'll check that out later

2:00:00 I mean the way you sections and fibres of maps if it's a co-unit. I don't know. I haven't checked. I've just checked this one. Okay. Yeah, that one's fine. I haven't checked that one. I think that should be an upward arrow. It's not. There's mistakes in that. Yeah, well, I think that... When I say mistakes, typos. Yeah, typos. I think that is a tie. I think that should be an upward arrow. I can't remember. I don't want to see how it's a co-unit, but no. But I've got the gist of the conditions anyway. Do you want me to just do the co-associativity, or is it obvious? Yeah, I'm just thinking in terms of that, I'm being selfish, because people may have longer to stay. Are you off? I can do it, it's just about four, to show the co-associativity. Oh, yeah, we'll just talk in five minutes. Yeah, as long as I'll be away before night. No, because that's finished now. Alright? Right. Perhaps I ought to put the co-associative diagram... OK, I'm going to rub this off. Now, what was the co... You don't have to rub the diagram, actually. A cross A. A cross A cross A. A cross A. So we've got to show that one cross delta that's equal to 1 cross delta aj cross that's equal to cross delta t j pi and that's equal to

2:02:30 t a j cross t cross okay so that takes us from here onto here now I've got to make sure that that is equal to this So I need to have a look at delta cross 1 on delta t. And that's equal to delta cross 1 of t k a c t k i. Which is equal to delta t a k c t k i. That means you can go either that way to there, or you can go that way to there, in this case the arrow is up so I did have it wrong the first time. OK, so that's the game you play. now the Hoth algebra just puts in an antipode rather than an inverse but I won't do that now I'll do it

2:05:00 Ok, so what have we done? We've seen the Tempoli-Liebe Algebra, and we've seen how it's related to the Spinner and if you want to generate that to something bigger than the spinner structure then we use the deformation parameter which in this case is the A normally it's called Q but this is the A when we do the deformation process process, we then find that we have to find out what the transformation is on the spinners, because there are spinners in that structure. We cannot find that transformation except through non-commutative elements of the matrix. And the problem is that because of that non-commutativity we do not get anything which looks like a group because we don't have S and if S and T are both elements of the group TS is not an element of the group so we've got to bring a bigger structure in which to embed this thing and that bigger structure is the bi-algebra and then ultimately the ultra-algebra but the bigger structure involves the co-product, so you're essentially doubling the algebra you've got to double the algebra and that gives you the co-product now what I just to finish off this story is to see how that relates to the more abstract notation that they use for the Yang Baxter equations that's where books like Majid actually start so when you're reading Majid

2:07:30 you think where the hell is he getting all this from all of this background has to be there before you begin to understand what he's talking about. Which is... Because he's just slapping down formulae and you sort of say, well, why is he doing this? Why is he... I don't understand. Well, at least here, you've got a reason why they're doing it. Yeah, yeah. You can see that instead of the ordinary spinners, you've got to now go to something more complicated. And the complicated stuff can only be handled in this bi-algebra. Now the interesting thing is that this bi-algebra is already there in the R, A, B, C, D things. So really, to handle the Yang-Baxter equations, you've got to double up anyway. So this is just showing how the group theories double up in that. And therefore all this topology, all this invariant knot theory, when it's done in algebraic terms it's done in terms of these bi-optimals and Hoth algebra is ultimately if there's an antipode present now exactly where that leaves the original story I'm not quite sure but that's I think I really can't just stop and take stop now and say okay we were looking at. And the original problem was the idempotence in the von Neumann altruples. Yeah. Does it take spin the period further than doing it generalizing? Yeah, it takes a general, whether there's anything in quantum mechanics that uses this, well we've already seen there is in that Yang problem, the pollution of particles, permutation because that's where the original Yang-Baxter equation came from. So it is in quantum mechanics. I wonder if there's any connection with this stuff. And is it in von Neumann, more general? Yang did a very special case. Particles on a one-dimensional line, making sure that no two particles stay at the same point. which is not a very thrilling

2:10:00 system to deal with but then if you read people like the quantum groups comes in with the oh my memory's going what are those not quantum dots I can't remember where they have circles in two dimensions Can you remember what that was? Where you get fractional quantum numbers coming out. What the hell is that subject called? You don't mean, what, fractional? You don't mean parastatistic? No, no, no. This is in solid state physics, where you've got electromagnetic fields present. And you get quantised orbits coming in. No, not spin resonance. It's got a special name. I don't know that bit of Con's work at all. I haven't read Con as I should have done. Oh, the Hall effect. Oh, no, no, it's not the Hall effect. It's related to that. The fractional Hall effect. The fractional Hall effect, that's it, the fractional Hall effect. But I think the interesting thing for me is whether this lot ties up with the von Neumann algebra, because that will then give you a more general setting of the Yang-Baxter equations, in quantum mechanics in general, and therefore to see how this relates to the hidden potent stuff that I was dealing with. algebras are all classified by their hidden potens. And there are different classes of hidden potens. I'm starting to be learning von Neumann out of different classes of hidden potens. And the ones that I've always been dealing with are the discrete because it's easier. The most difficult ones are the... The tricky ones are when you go up to one infinity, two infinity, and three. Those are the type 7, isn't it? I don't know what one. It's all to do with the topology. It depends which topology you're putting on your... structure, on the operator space. I've now got another couple of very interesting articles on this which I'm hoping I can actually get to the bottom of it at long last.

2:12:30 I was wondering whether any of this ties up with that stuff of faked alphas about getting the kind of fermion and boson statistics out of these topological geons. could well be the thing I haven't done is I haven't done spin networks and that's the next chapter at the end of the book because he does this sort of thing twice this is earlier on in the book where he sticks down these hot articles you're talking about Lou now I'm talking about Lou's book it's all come out of Lou's book in one form or another been trying to do is just to see what it is that you know what the key feature is my key feature is this I mean it's magic isn't it that's all I'm discussing that is all I am discussing you get the icing model the pots model the Schringer in one dimension. Isn't it only one is up to one you get the plus thing? Yeah, I'm sorry, I've drawn this carelessly. I should really do whatever it is. I can't remember which way away. Yeah, over. Or just symbolically. Well, there was the old joke about, you know, if we ever get the kind of finesse secret of the theory of everything, you've got to be able to put in a T-shirt. You can certainly get that on a T-shirt. Yeah, I mean, I think that's I don't know why we don't buy Lou a T-shirt with that on there. Actually, wouldn't I? Has he got a birthday coming up or something? Is Lou coming in this April? I've no idea. He's got this meeting in the States, in Buffalo, the one which Bill is refusing to go to because he thinks it's promoted by the National Security Agency. But that's not until June or July. I thought, he certainly thought they did,

2:15:00 because he was chattering away 19 to the dozen in Bristol about the imaginary... What was that for? Oh, yes, that's right, thanks very much. Did you have forgotten? No, I was being very polite. Because he suddenly, when we were talking to John Mabry in Bristol, about completely different stuff, about this business of discreet and co-discreet spaces and set theory. How he sees that set theory is fitting into this corner of this geometric Galois theory and these strange properties of the left and right hand joints. How it fits inside a tower of structure where you capture these different degrees of cohesion. This is all that stuff he was talking about to Michael. I was very disappointed he thought the continuum was real. Oh yeah, no, that's absolutely key to Bill's thinking. He thinks the continuum is absolutely basic. Anyway, Bill has got all this rich structure inside. It seems much, much richer than the arithmetic continuum. The arithmetic continuum is trivial. But not the smooth wheels, he's got something bigger than the smooth wheels. Yeah, yeah, but he's explaining all that at some length with John Mabry, who's of course a fellow mathematician and therefore he kind of opens up a bit more. You see there's this additional right adjoint which gives you these fractional exponents so you can go beyond exponentiation. But no, he suddenly started talking about the real and imaginary parts of the Schrodering equation. And it sounded exactly like what you've been saying about the real and imaginary parts of the Schrodering equation in and out of season, you know, for the last... Well, since I've known you, I've known you for a lot longer than that, you know. so it's very strange how he kind of arrives at these things by the back door well they're probably sort of stored up there I only hope you thought you got something out of it I hope you didn't think he was too bearish he's not the easiest man in the world no he's fine no problem with him as a person well some people do find him a bit bearish not going to be worried about that No, I think that stuff he was coming out with. Oh, he promised to send me this paper of Steve Shannwell's, you know, about these upper triangular matrices getting non-cognitivity out of it. You know, that really is interesting. You see, he explains how, given his conviction that functoriality and,

2:17:30 what the hell do you call it, you know, this function, he wants everything to live inside a structure where there are nice function-space properties. Right, right. And so therefore, of course, this is why he can't stand the idea of non-commutative geometry. It's quite clear after talking to him a bit more that it's not non-commutativity that he has a problem with, it's just people talking about non-commutative geometry. Non-commutativity is there, all right, in fact, you've got to have it, Well, he thinks it's kind of, and, of course, this idea that it might connect with this idea of getting H-bar out of Maxwell's, there is this, of course, very speculative, you know, except that could be completely wrong, but that you've got to get it inside some structure where you preserve these nice function space properties. And, therefore, that something like this Steve Chaniel way of treating the derivative, which shows that all functions are really analytic, could be the way to go. It's an absolutely fascinating paper, and I can't wait to study it. Steve Shannon was a very, very fine mathematician. I mean, just as a piece of pure math, it would be interesting. I mean, whether Bill's speculation has anything to do with the origins of quantum theory. I'm not going to be in London next Tuesday. Ah, okay. Right, okay. Yeah, actually, Marlette already said that. That's the week. That's Easter week. Easter Tuesday. Which is what day is that? 22nd. Yeah, I'll be around. I'll be going completely mad packing. Next week. Next week, no. I'm leaving. Where are you off to? Down to the south coast. Oh, just to have a break. Good. So, Tuesday 22nd of April. On Monday the 14th, I'm going down to the South Park. I'll do my damnedest. Well, I'm going to be out of the country. I'm going to take a group over. I'm going to go back to the 19th. 19th? 18th. 18th, you're going back. And then are you coming back again? Yeah. Well, don't ask me. Probably beginning of June. I'm going to be away from the 12th and the 15th in France, just for three days.

2:20:00 It's a lot of Tuesday. Tuesday. Tuesday. I don't know if I'll be able to get there that Tuesday, but I'll do my best, but that's the week I'm going to be moving most of my books and furniture on the 27th, and I suspect that I shall just be absolutely tearing my hair out in a sea of packing cases, and the whole of that week will just be a total nightmare. Can we book the farms? A crate? A crate? A crate? I had just taken delivery, I had just taken delivery of 80, of 80 crates. I mean, one of those things that put on lois... Oh, the containers, yeah. Well, I was thinking of having it containerised, but in fact I'm going to try and move it all over in... I'm hiring a removals van for a week. It's been about £1500. What? No, it's a bit less. What, the removals van? No, no, the containers. Yeah, well, it's going to be... I think it's going to work out a bit cheaper doing it well. I'm harming the van for a week, and it's 650 quid all in, with the fuel on top of that is probably going to come to at least 100 quid, although I'm taking a short C with all. I really have to do it. The House in France is the one thing that is completely sorted out. That's the easy bit. That's the one bit that's all sorted out. It's getting all my stuff over there. I'm setting aside a week to do it, but I'm going to need a week before that at least just to do the boxing up and you know I'm afraid so but it doesn't mean I'll sort of simply disappear I'm just keeping a little base over here so I'll be able to get in I hope fairly frequently but probably not as often as I have been I guess but then you can come over to Fougere and stay there we have some more talks with Bill which he certainly would like to do Yeah, I'd still like to learn a little bit more about how to do it. Yeah. Yeah. Exactly. And, um... Now, it's a hell of a lot out of the two days in Bristol that we had with Michael Driver, who, of course, is an old friend of all of us both of them, who didn't get much under the micro-redeat of... Well, that was just a social education, wasn't it?

2:22:30 It's, um, yeah, you're on the papers. I mean, I'm doing a first thing, but I wasn't expecting this to you. I'm very proud of what I can do. Michael's a stranger to the rest of it. Michael's a liar, but I think he's grown, he's very young. He's almost 30 years. I've known him for about 30 years, actually. I'm nearly getting on. I'm like Jenny, I think she's a lovely lady. Is it Jenny? Yes, yes, she is. No, I've known him since I was at the club, and just starting an end film on Bedford. Right. I came from Cambridge, so I've known him for a... Yeah, I've known him. He's looking a bit crazy. Well, he was very ill a couple of years ago. Because he's had this awful back pain. Yeah, yeah, that drains him. Which is why he has to water that screen. John Mayberry, who of course is an acquisition of rather physicists, although he has a great interest in physics. He's got these fascinating ideas about the natural numbers, that there is no single-isle-isle-mortem type of natural numbers, but that there are actually number-sistered, well, what did he think, or in a boring phrase, and it can be called simply infinite system of different length. And this fits in very nicely with how Bill's idea is about bending up. Because of this additional right-hand joint, here's the instruction. And of course, you have a galop there, which I think is the geomics of the agent. There's no such thing as a space with all the patient's sequences, rather than a real problem. OK, see you earlier. Have a good trip. Take care.

2:25:00 I don't know if there's no reason to inform that the government gets to one point from another. The government's well-centered attention to the topological characters of this place. People who can grow up as an emotion. So I think it's like that. Although, you know, the quicksing that he uses to think about this is what he's very unfamiliar and it seems at first to me absolutely to avoid stand-up. And when you actually see how it comes to the thing when you start with mathematics, or when it comes to a parallel to the kind of TV's role as a claim of, I think, literature, actually, you know, sticks, because of the old numbers, very dodging, and publishing analysis, but then when it comes to, you know, you need to keep on to something that you need. That's no question whenever using that prediction of the norm, I mean, that's very clear. When you actually see other mathematicians, you know, interacting, I think that they, when they get, you're talking about very smart people to be, you know, the absolute problem, whether any of you can exit visits or not, because it doesn't act all together. And he has got a very certain share of prejudices, which are coming from this, you know, and that it's, you know, I didn't see much of it. There's a lot more open and flexible suggestion, but he gets very hooked up sometimes. For instance, that paper of Carthage, that had a bit of work. Oh, yeah. We got chatting about that with John, and he hates it. He hates it because, you know, the idea of putting it in the court is really great. Oh, yeah, he doesn't like it. really is on the same footing as Gropenbeek, who often is in regard as gone. And he thinks that Korn's work during, you know, the idea that Korn took Gropenbeek's work

2:27:30 and fulfilled it and showed what was all meaning absolutely feels like a spitting fury. So, all known only, completely misunderstood, and Gropenbeek's work in quite a long direction. The way it should have gone is, and there's quite a little bit about what he was saying, very briefly to you that first afternoon about how he thinks that this idea of working into this kind of extraordinary structure in which inverse limits and direct limits indicate how everything's going to be filled in and that all this was about looking at the spectrum and said that is the basic structure and trying to build the space on top that's completely a long way to go. Our conceptuals are very interesting and of course I mean who knows in a hundred years' time, the mathematics of the late 21st century, who knows which of them will be proved right. I certainly wouldn't presume to sort of argue with somebody of the power of calm, but it's interesting to see that there is this different viewpoint. I think the answer is that one of them essentially sees geometry as fitting in, one of them sees algebra as fitting inside what is still essentially a vector structure, of the geometries as well as being reconstructed entirely on alternate places. Anyway, listen, have a very nice time at the South Coast, and enjoy your vacation. Are we going to see you again? Yeah, well, I hope I'll get in that week, the 22nd, but if I don't, I'll give you a ring. I may not, because I'll probably be busy boxing up. Once I've moved, I'll be back around from side to side. I'll give you a ring and obviously send you an invitation to book one. Cheerio, Bob. Take care. Thank you.