Theoretical Physics Research Unit Meeting (contd.)

0:00 The mathematics to describe the topology of the Penrose Tower. The mathematics to this we're using the same structure.

2:30 I find this very interesting, the question that I've been looking at is why? The one thing that is, just to show you, the thing that I've been concentrating on and trying to understand is this Bratelli diagram of the temporary levers. Last week I thought I had it, but in fact I had the first three lines and then suddenly it started going wrong. When I pushed it through down here, so obviously what I was doing in terms of swaying the main speed into primitive components was not working, and then I discovered that I didn't have a complete set of numbers and I found that I didn't have people to follow and so on. But the surprising thing is that if I stop this diagram there, then I generate Clifford Algebra, so all that comes out. If I go one further down, I think I then get the Penrose timing. And if I go one further down, I get the symplectic structure coming, another type of group. What's the diagram again? It's a Bertelli diagram. It tells you how these algebras are stuck. You really got... I remember... You remember I went back... We talked about this at the end of December. Yeah, I remember I... The essential argument... Do you have any bright ideas, Graham? Yell. You see, what we're looking at is approximately finite algebras. Okay. And the idea there is that you start with an algebra A1, you embed it, you build the algebra in 2 and 3, and eventually you get up to A infinity.

5:00 Now one of the ideas that I had here was that you sort of slowly increase the complexity of the algebra and you get up with your space time, your continuous space time. That's one sort of direction we might be moving in. These sets of algebras are actually described very nicely by means of the tower algebra. One of these algorithms, it's the one that Jones deals with a lot, and that is the one that starts with a 1, 2, 3, 4, I'm looking at my head at the moment, then what do we do, we go 2, yes, it's either in Jones, it's in several of his papers, you can't get away from it, 2, 3, 4, I'm looking at my head at the moment, then what do we do, we go 2, 3, 4, I'm looking at my head at the moment, then what do we do, we go 2, 3, 4, I'm looking at my head at the moment, then what do we do, we go 2, 3, 4, I'm looking at my head at the moment, then what do we do, we go 2, 3, 4, I'm looking at my head at the moment, then what do we do, we go 2, 3, 4, If you don't draw this right, you go skewed. So I've got 5, 4, 1, and then om, om, om, om, om. And that must be 5 minus 5 and 4 is 9, 4 and 1 is 5, 1. Okay? And these are the dimensionality of the algebras that are in this chain in terms of least. So for least a tower algebra, is that the same thing as one-line algebra?

7:30 Jones treats the towers in terms of one-line algebra. He sets this whole thing up in the context of von Neumann algebra, but if you're only talking about algebra like the temperate leave, I don't see why you need the von Neumann algebra. That's essentially why I stand there. So then if you take, he's got a series of three goes in there, four goes in there, five etc. At this point what the leaves are saying is that if you stop at three You get one algebra, if you stop at two, you get another algebra, and so on. If you stop at, here, if you stop at three, which is the algebra one, one, one, two, then you don't put another one in there, two, two, four, and et cetera. That's a clip of algebra. The alert reader will recognise the clippet algebra. The reason for that is, I've discovered, that for the odd dimensionalities, a clippet algebra factorises into two. It's semi-simple and it factorises into two algebras. This one here does not. This one, it's an even algebra, but again, that does not factor. And then, and that's C1 plus 2. In fact, I looked back at my clip at algebra books, and there it was, and the alert reader, well, there's a simple explanation for it.

10:00 It's to do with the order of a number of groups, one little dimension, and we know we've always got that problem, remember? But you have this problem with order of even number of dimensions. Okay, so there you've got the clip on algebra. Then, he says, one, one, one, right, and now I'm at level three, that's at level two, I've got these numbers there. I come to this one, and now I don't put any more on there, so this is one, this is two, that's one, oh, I've got one more coming here, that's three, that's two. And then we're going to go bang, bang, bang, and then I get 5, 3, and then I get 5, 8, 5, and then I draw one back on there, and I just stop there and say 3, and so on. And this, it says, the alert reader, no, no, it doesn't say that, it just says, Now it turns out that the reason why... I hope he doesn't say that. I hope he says Fibonacci. What do you want me to say? Fibonacci? Fibonacci. Fibonacci. Fibonacci. Fibonacci. Fibonacci. Fibonacci. Fibonacci. Fibonacci. Fibonacci. Fibonacci. Fibonacci. Fibonacci. Fibonacci. Fibonacci. Fibonacci. And that's just equal to 2, which is where the 2 comes from, and this one, 4 and minus 1 is just 4 cos squared times 5, and that's the important thing, and that's the big and large event, it goes up over the next level, whereas here, all the 4 and minus 1 are greater than or equal to 4, and this is the trace, remember, the trace of the matrices.

12:30 And this is where the tower algebras come in, they're characterised by this, this is the coupling constant in the volume, or the trace. What's the question for? The question is for? The, the, er, the trace. So one of these two, so I can actually... This is not, this is, I don't know what that is. The question for. That one you were telling. But each one, you can see what happens as you take one out. t minus 1 equals 4 cos squared by 0.6, t minus 1 equals 4 cos squared by 0.7, and so on. Now, I'm not explaining where you get those from, Stephen, I'm just saying that in the analysis that we've been through previously, you connect these to this index, it's the index of the alphabet. But the problem that I'm having is how do we get that? Thank you for watching. Oh, yes, sorry, it's great. It might be the other way around.

15:00 Yes, Tor is great. Sorry, I thought I was conscious. You write it for me. But then there's also another paper that I was looking at by a guy called Landy, not the old Landy, the Italian Landy, L-A-N-D-I. And that was the one you spoke of, you know, the L-A-N-D-E. This one is Landy. He actually relates this structure to what is a patron and what a young, young rapist, yeah, rapist, yeah, the patron, yeah, the young rapist, and there you had that algebra, what was the algebra, the rotor algebra, and you remember when I was doing my algebra process, I essentially had a rotor algebra work, yeah. And Luke Hasman, in generating the Clifford algebra, was using a similar algebra to mine, the dual algebra. So lying at the back of this is exactly the same algebraic structure, and it's generating this rather rich structure. And my task has been to try to understand how it all fits together. And hopefully when I see the mathematics I can then try to understand it. Can you remind me, how's the patrion and raptus motivated there, is that this... Causally ordered, causal set, and what was it, it was something like, if I, I think it's almost quantum mechanical, wasn't it, but if I had this for one process, and then I had this for the process, and that can be equal to one, three, so it's a layer of... Take them yourself in one. And this was what Lou Krautman drew as a sort of a Feynman path integral, right? That's three and that's one.

17:30 And then if you put a sum over two, it would make this i. In general, he...Raptis and de Bracken didn't have an i in here, they just had a single number. That's all they did. But you can do it in terms so that the intermediate sum over intermediate space and then you've got the sort of thing that Lee and Witten were suggesting, that you've got a sum over the history of the path integral, so you're collapsing the path integral into the structure. I can't remember. One of the recent, very recent ones in the last 30 years. I remember him talking to our lecturer and teacher about this in Cambridge. Yes, he put it out in the paper probably two years ago, one year or two. He didn't talk about it at the meeting, but he did talk about it in his film on the first day, the day before you came up. Well, you see, this again was something that started me a little in this because... He had those different idempotents P and Q and R, I think, and he was getting something like the beginning of the temporary algebra, but he was not getting the complete structure, and he was getting his idempotents, they were easy to see, they were 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0. And then that stops, but that's not as interesting as these hierarchies, and so you can't construct the towers from that. So you're saying the towered ultras are not matrix countries? This is the end, yeah. I'm not answering yes or no, I'm saying that... Because you said that the motivating example was the Jones data and the von Neumann. Yes, yes. Which are matrix countries. The question is where the matrix is. But you see, we've got a problem, because... What have we got? We've got generators which are like this, and the question is, how do you represent an object like that as a matrix?

20:00 Am I right in saying that any finite object is a matrix? You're not going to do well because you don't know the answer, but I think that's right. I'll let you know the answer. I don't know what? I think you can express any finite object. I don't know the answer. Sorry, I should know the answer, but it's escaping me. I'm fairly sure you can, but I need to check. I was just trying to get my head around why you've got a problem with that. I can make it into a question. Lewis told us how to make it into a question. The question is, it's not as straightforward as you think it is. I'll go through that in a minute. Okay, and then of course there's also the interesting... This is the better of time, non-periodically. Thank you. No, you can't take it from that. I'm just telling you that it so happens. But if you extended it far enough you'd see that it was non-periodic. That's an interesting... I've not looked at it in terms of the brutality done, of course. I've not looked at it in terms of the Penrose diagram. Well, if it's equivalent to the Penrose tile, then it must be. There's something aperiodic about it, but whether you can see it from the diagram or not, I'm not sure, because I haven't seen it. But, okay, no, I was talking about where Penrose got its ideas from for the Penrose tile. My admission was, and I wasn't sure, but I do know that when I came here in the first chapter of the Penrose Arise, I was talking with John Finney, who'd been working with J.D. Bernal on liquid structures, and he was asking what is the difference between a solid and a liquid. So a gas we know is completely random. A solid, long-range ordinary. So what about a liquid? And his idea was that, alright, let's have a random structure. So he built in his office, which you now see in... I don't think it's quite built this way.

22:30 He first had a full spoke model, which he just suddenly stuck things in, somebody would interrupt him, then he'd stick another one in, and he built this mess, this heap. And I think they also took ball bearings, and I remember somebody putting them in a football glider and sucking all the air out and making each ball bearing had a blue tacky stuff on it. And then you make it into a heap. You see, there's a difference between a heap Which is a completely random, what looks like a completely random structure, to pouring, and he used to do a lot of experiments on it, pouring marbles, if you pour it into a base which is fixed, it actually forms a tetrahedron, it's all good. But if you just let it, you know, it's all over the place. What they did was they took the ball bearings, and they pulled them apart, and they looked at the coordinates of where they were touching, and drew that out into a ball machine. And then they built this irregular structure, and then they analyzed what were the catonic solids that were in the structure, and they found that the majority of them were pentagons. So if you want a heap, what dominates is the pentagon. Okay, so now that we've got two dimensions, it's much easier. We're talking about pentagons. How can we cover the two-dimensional circuit? The answer is you can't. But then he started saying, well, what can we do? And to cut a long story short, he said very simply, let's cut these things here, triangles, so that we've got two types of triangles in the pentagon. We make this into a kite, and we make this one into a dart.

25:00 Sorry, that's supposed to be a dart. Am I doing it right, Brian? I think the face looks good. Like this. I think I've probably drawn this. I might have got them wrong the wrong way. You get the idea. You get the idea. So you've got... These sides are units. This side is, well, you can work out what it is. It's a square root of 2, or something like that, or 1 plus 1, or whatever you want to call it. It's taut. I usually call it taut. And then these two sides are taut, whereas this is 1. So you've got these triangles, and you can work out what that is because you know the angles. Can you remember what the internal angle of 72 was? And that is pi by, pi by 5 comes in here somewhere. Is 72 pi by 5? No, no, no. It's 2 pi by 5. It's 2 pi by 5. OK. I don't mind. Because now what you can say is when I can bisect these angles, which angle is bisected? Well, these angles are the same. And now what Penrose does is he tries to cover the surface with tiles, with coats of darts, and then you can do it. But the interesting structure that you get out of that is that they're aperiodic. And the Fibonacci numbers come in fairly easily.

27:30 What you do is you've got your... You've got your Penrose polygons, which you can see these darts and kites and whatever. And what the way you characterize each tiling is by taking the point in the tiling and then slowly taking the tiling to pieces by pulling arms out and saying this is a large, this is called a small, this triangle here is called a small triangle. And this triangle is called a large triangle. And as you slowly pull bones away, you put either a 1 if it's in a large triangle or 0 if it's in a large triangle. 1 in a... And what you get then, as you do this systematically, it's quite beautiful if you look at one of the books that does it, So you're just pulling things out and saying, what is the point now in? So the total number of distinct timings is going to be a 0, 1, because that's the first point. Then you do it again. If it's in a small you do that, or it could be in a large report you do that, or you can have 0, 1. What you find is you can never put two ones together, because the way the thing pulls apart, two smalls, if you find it in a small, the next line will never give you a small, okay, that's just, you have to look at the last, you know, to see if it's as destructive as you like it.

30:00 So what you generate then, am I doing them all right? You did them before, yeah. Okay, is that right? Yeah, that's pretty much it. And then we go zero, zero, zero, zero. 1-0-0-0, 0-1-0-0, 0-0-1-0, 1-0-1-0, and in goes 0-0-0-1, 1-0-0-1, 0-1-0-1, and that should be, we should be able to do it all right, that's three, that's two. Now, what the trick is to getting into this is to treat these as the dimensions, treat these as a vector space, so this is a column of three objects, oh sorry, five objects. And then these algebras then decompose into an algebra which is m, d, n, plus m, i, n, n, where this will be, if we start at the bottom, this will be a 2 by 2, let's start again, this will be a 2 by 2, this will be a 1-dimensional object, this will just be a 1, then this will be a...

32:30 This will be a 3x3, this will be a 2x2, this will be a 5x5, and so on. But you're now treating these descriptions of the Penrose tiling as a vector. I don't know what that means, because you have linear superposition and so on. And that makes me think that linear superposition is nothing but doing the interferences. That's the way we see it physically. But in order to describe these things, we have to treat these linear superpositions as vectors, and then we actually sum adders and subtracts of them. And this then allows us to classify all the Connors terms. Quite extraordinary. But you see here, what I've been drawing here is actually translated into this diagram. 3, 1, 2, 3, 5, 3. You're not going to know much in the way that... Okay, you say it's nothing intrusive to do with it, Bill, but it would recover that distinction, wouldn't it? This puts us into K-theory, because the way we analyze this is through topological K-theory, totally up to speed. But that K-theory comes out of treating these things as vectors. And then using the multiplication rule that I had down here, then we wrote the multiplication rule. So I've been reading a little bit about the history. I keep saying that, you know, what we're looking at is really the Gottenbeck matrix or something like that, and I haven't tied it up with that yet. So even that's here, Mike. So I somehow feel this, you see, this is something central to, to what? To the description, I think, to the description of the internet.

35:00 Oh no, I was just saying, why did Penrose go for pentagons, and what I knew was that they knew what they picked up. Well, I'm not sure they did. I was working with the chemists. Stereochemistry. But you see, I had a method of dealing with that, and that's what I was going to do with John Finney and Jeremy Bunnell. What I had was the... In fact, I was just talking to a chemist today. He said to me, I believe you were going to do something on random walks. Could you tell me the elements in the solution? Oh, this is going back a bit. I think that's random matrices. No, not random matrices yet. Just random walks. And playing about with this Penrose tiling again, it seemed to me that I could have done some random walks on the Penrose tiling and see if I got the same thing as I did when I did the random walks on the Bernoulli model. The number of n steps over the number of n minus 1 is much more than the ratio of the number of n steps. Was this something to do with attacking the protein? Yeah, I was in the background. I never got around to that. Yeah, but obviously that would be... No, this was at King's when DNA had, you know, Morris being in the corridor, DNA had been solved. But now they were into protein told them. And they wanted physicists to understand protein folding. Well, one of the ideas is what is a protein? A protein is essentially a gas, no, a liquid, a structure.

37:30 Think of it as a gas, only this and that molecule must always be close to this one and always close to that one, and yet allow it to wrap around and do what you like. ...forces between the... and also the three forces in the liquid. Right, yeah. The van der Waals. The van... that kind of... that kind of force. So it was designed for a firm. But you see, this is something which the people trying to, you know, crack a mountain out of technology now worry about, how to do these... What? How to do these problems? Yeah. Well, you see, so what I... the idea I have... I'll come back to that. If you plot these random walks against 1 over n, so that you can extrapolate to 1 over n equals infinity, you actually find that at first this ratio buggers about a bit, but it very quickly settles onto a straight line, and if you extrapolate that to a straight line, it intersects at a certain point. And you find that all three-dimensional lattices have exactly the same slope, you know, they're all base-centered cubic, body-centered cubic, or self-centered cubic, they all have the same slope, and it's some simple fraction like 7 over 3 or 5 over 2, I just can't remember the number. If you do it for two-dimensional lattices... There's a different slope, but again, they all form on exactly the same slope, so that slope is a function of dimensionality, and, you know, this is why I shoot that, but you can see where I got my original idea of trying to do all this in terms of knots, because in the two-dimensional surface, you can't go underneath, but in the three-dimensional, why you get a different slope is because you've got all this knotting force that's going on now. A little bit of work I did, which was mentioned in Mandelbrot's book on factors. How about that? How about that?

40:00 And that was doing excluded volume walks, which came to the same point as the origin of that stuff. And they call them u-n, so if you plot a u-n against u-n minus one or plus one, I can't remember, you actually get a staggering amount, but then they settle. I believe that if you extrapolate the Zeppelin line, it meets the extrapolation point on the other side. This is the open-ended, non-crossing, this is the closed, and these are fractals. It's again three dimensions, one slope, two dimensions, and the entire width is one slope. And what I did with the Bernal model was to see, because these are regular tessellations. And therefore you could argue that when you're only sampling, you're only sampling a small area, that these are about 15 walks, 15 steps, maybe down to 20, but nowadays, of course, we have all this chemistry of telling you they're up to thousands, but of course, it probably makes no difference at all what we've got with our 15 walks, which is kind of enough. All we've done is confirmed what we said it would be. Although, of course, it's taken all the errors off and made it much more exciting. No idea! Regular tessellation, all right, somehow you're picking up the regularity even in a 15-step environment. You know you're on a unit cell, you know the unit cell repeats itself. Therefore, I know where I'm going. See what I mean? Already by 15 walks, he says, oh, I'm on a space center. You know, I'm in three dimensions. He already knows after 15 steps he's in three dimensions. I mean, there's massive numbers here. Although it's only 15 steps, there are millions of them. There's a tremendous number of steps. So the question was, is this simply like this, a regular test solution? And what happens if you have an irregular test?

42:30 By now, we'll see that John Finney had his model. I'm sort of saying, well, what should we do with it? I say, well, let's work out what the random walks are. When you've got a regular tessellation, every point is the same narrative. Random structure, this point is different from this point is different from this point is different from this point. What you do is you count at different. And then average. It doesn't matter how you average, whether you do it directly or not. This three-dimensional doesn't suck in on this one. In other words, even though it was a random structure with no regular unit cell which repeats itself, it's still new. It was in three dimensions. And the same thing for... Now what we didn't do was the Penrose tiling, because Penrose tiling became... came in the 70s when I got to Harvard, but it would be... would it be nice to do the Penrose tiling to see whether that was fitting on the two-dimensional... Does it somehow... does it make sense? I don't know why I can't... No, no, it's just a fact that it's such a... To me, this was the key... It's a lovely little... It's a bit of a problem, but it seems to be due to the self-contained, and it's amazing that they've done it. Hamamesh, no, Hemsworth, a statistician in Cambridge, proved that both of these things would go to the same level. That's sort of a rigorous example. But, I mean, to me, the staggering thing was that simply by counting, you could tell whether you're in a three-dimensional space or a two-dimensional space. That value there, is that related to the critical point? It's related to the critical point. You know the ferromagnet has a spontaneous magnetization. That's the critical temperature when magnets magnetize as well.

45:00 It gives you information about the critical temperature, because that's where the susceptibility can be. But, you know, I'm thinking of all sorts of interesting connections here, because one of the motivations that Lou has... And of course, of course, Feynman. And Feynman, obviously, did things in terms of this. Well, that's the icing model. Which is exactly the icing model, which is also the thing which these people like experience. Because, you know, you couldn't make head or tail of what you were telling him a year ago when we were in Bristol. That's exactly what John Mabry's student in Lough has been doing, This sort of finite... he's trying to redo the whole of geometry based on this strictly finite arithmetic, and he's got this way of defining dimension which relies essentially on a random walk and random walk theory. This thing that I've got here is finite algebra. Yes, and they're just trying to do it with arithmetic, but I think this is much more proof. Well, it's richer, so I won't be against them doing that. But you see, the fact is that you already know with the very small section where you are. You know you're in a three-dimensional space. And that was the thing that fascinated me. And that's why I persevere with all this. You know, people say, why do you do all these finite things when you don't get these? But you do get them, you get them. You're getting dimensionality coming out of some sort of topological feature, isn't it? It's not free at bright angles. And this whole idea of the spin network, well, the spin network is essentially there, I haven't seen a pin, the moon's got a lot on there. The spin network's coming out from there. But this, this gives you a different network. But the interesting thing is you're using the same algebraic. You're using group, essentially behind all this, treating this as a vector space, is that you're using group points and the group point is exactly the multiplication rule that Heisenberg used with P and Q.

47:30 If you look at the Penrose tiling in a classical way, with Karte I gave a talk back in January in Paris about group lines, which unfortunately I missed, it was in French anyway, somebody recorded it, I was listening to it, French is very well, connecting rock with all this beautiful new stuff which I don't know about in symmetry, you weren't thinking about. Well, this is the sort of thing that I'd like to listen to, but it's... But he was also saying that, you know, there's new physics in there, which is very exciting, where the uncertainty relations might... There is, by the way, that the next one out of, like, seven or eight is the symplectic group and the metaplexic group, according to German, so that the, you know, and if you take Morris's story... I was going to say Morris would like this stuff. Well, I don't know, he seems a little bit... I don't know, maybe. It's a different technique. Yeah, it's much more algebraic. I like algebra. I was eight years old when I learned algebra. I was suddenly excited about it and it was 78 almost. Not quite. But it's still fascinating to me. Okay, so you see that the background there of the finite algebra is the fact that... And then you're trying to get... we're talking about priesthood, which is the quantum of it. And it's all these level hints that seem to suggest that this is the way you actually describe it. And the group point is essentially that mathematical and mathematical equation here. You have two symbols, A, A, B, and then you have A, B, C, the sum of the B's, and then A, A, C, which is just a matrix model.

50:00 Yeah, that's a good idea. That's because that came from the vile algebra. You see there's another algebra in BC. This is then the lecture I gave in Fougere. And the line that sees that, which I haven't put precisely in its place, but every time I turn over a page I suddenly see this y-algebra that I'm looking at. And remember that y-algebra is to say, I take an algebra defined by a, b, and cos, omega and ba, where omega is the nth root of unity. Now, nth root of unity, e to the i. Plus e to the minus i, cosine, there's a correction between these two, okay, it's not, this is a solidly, underlying the algebraic structure that we've got here, you've got this n through the units and stuff, and what I showed here was that this, in the limit, as omega goes, as n goes to infinity, becomes essentially the Heisenberg algebra, but the Heisenberg algebra with an idempotence. And the idempotence that I get out of here are simply the points on my space. So I've got a very primitive toy pre-space model in which I don't start with space and time, but I generate the phase space. But it's not a phase space of P and Q, P and X.

52:30 Now, if you remember my story about projecting out the quantum potential from the algebra, then here, you see, I've got to project that from this structure to a phase space, and there'll be a quantum potential, and I'm sure Graham is a quantum scientist. I'm sorry, this is the structure that I have for my toy prespace model. I'm creating the space-time coordinates out of the algebra itself. This is why I'm so interested in idempotence, Graham, because idempotence are going to be the points of my space. So if I know the structure of the idempotence, then I know the structure of my space. And now the difference is that these idempotences are central, whereas the idempotence I use here are left ideals. They're not central. So this is somehow classifying the central idempotence. Well, it's really what we should be looking at is the left-hand view. You remember what we had on here just now, the left-hand view. The amazing thing is that the dimensionality of the vector space determines what these things look like. It seems like magic all of us. We're struggling by taking the template, the algebra, and trying to get these matrices down. And yet the Penrose tiling comes straight away because of the way we've chosen to describe the order and the structure. So I still don't know why the groupoid is giving us these topological inferences. This is something I'd like to talk to Freddie about this evening. You haven't heard anything, have you? No, I haven't either. I checked my emails back two days ago. He must be in New Zealand. I think he must have come. Which gets him, hopefully, back again very soon.

55:00 You can link with what? Sorry, I've been rambling all over the place for a couple of hours. Um, there was no connection at all, actually. I don't know how the two things fit together. What's the link with the study that you mentioned, or did I mishear you? Didn't you say something about the spectrum of something that you studied? What do you have to fit in this field? I'm sorry, I misheared you. Do you know when you've been out at least for a little bit, it's been quite a bit, it's been quite a bit, it's been quite a bit. Yeah, they were, they were, it's whatever, it's whatever the five, five, make them symmetry, do whatever, whatever, yeah, I keep saying, um, I keep saying pentagons when I meant pentagonal, so it's whatever. By getting ball bearings. By getting ball bearings, sucking the air out, making the things stick together, and then some guy spent his whole life just picking one out, looking at the coordinates of it, mapping it onto a... And it was then put into the computer, and that's what John Finney did. He did a paper with him, didn't he? I did a paper with him where I discussed this. There were also mutilated body-centered cubic lattices. So basically, on the computer, if we just take a row of ball bearings, like a regular, I mean, is it a regular? Not with a heap, not with a heap. The heap has a characteristic feature about it. And what Bernhardt was really trying to do was trying to find out what characterizes the heap. Oh, then the liquid storm, you see. So how do you differentiate heat from the fact that it's mixed up with plasticine or something? Sorry, what? They're both able to mix up with plasticine. No, they're not mixed up with plasticine. They've got little plastic lacquer paint. Lacquer paint, which is wet. Then the air is... And then they're allowed to dry. And then as they're pulled apart, you know what happens, the paint comes on it. Where there's a point of contact. And that gives you a coordinate.

57:30 Ah, that's probably your cell phone. What are you going to say about liquids? Liquids. What about liquids, you're going to say? Oh, just that, um, you know, come down to this problem. See, what... DJ, I need some switch light on in a nutshell. What I did in my PhD thesis, I've marked them back, trivial stuff there. I also extended the Mayo-Costa theory to gases. Now, for a perfect gas you've just got boils in it. But now what happens when you get interaction between gas and gas? You've now got an imperfect gas, first approximation van der Waals, but what do you do in reality when it's not van der Waals, it's a real gas? And Meyer developed a method which was an expansion in terms of finite clusters so that your partition function... And then when I see Lewin is not writing down partition functions, I get rather excited. Your partition function is a sum over the stars, that's the irreducible finite clusters. Irreducible either polygons... 4p, 5a, it's got to be screwed in the way... It is amazing, isn't it, that Plato came back to life, and whoopee, I was right all along about the platonic solids being the key to understanding... And so on... So these, of course, the reason why I've got the S there is they're called stars, they're irreducible. As opposed to something like that, which is reducible, because all you do is cut it there and it falls apart. Whereas one of these, if you cut it anywhere, it doesn't fall apart, it's still... And then what you have on the end here is that end, and then you have some function of temperature. You have a partition function for the finite cluster that gives you what that contribution is going to be. Now my idea when I was going to use liquids and was going to be distracted by crazy only mechanics,

1:00:00 He really started there. I was going to take the stars that John Finney had evaluated, put some dynamical function in there for each star. What that was going to be was the real difficulty of what he meant. I mean, I had some ideas and you could do it as an approximation and just have a look and see what you did. Yeah, rather than finding the partition function of the Bernoulli. So I gave up. Now, the protein, some of you know this, what's protein folding? Oh, some of you. What that was, and that was part of my thesis. No, I didn't put it in my thesis, it was part of my thesis. All right, here we've got... You know, I'm at King's with Norris Wilkinson, shuffling down the corridor, proteins, DNA, GMO, how do you treat that as a mayotrophic structure? So what you've got here, you've got one additional difficulty, not only do you have the interactions between the particles like this, you also have this constraining feature, so that you've got, you can do a mayogas theory plus. So you've got an order along the chain. And then just apply, just apply the, the, the, the, the, the, the, the, the, the, the, the, the, the, the, the, the, the, the, the, the, the, the, the, the, the, All of these are very easy to do, so you could begin to get a closed expansion, which is different from a series expansion that anybody else would do.

1:02:30 So what kind of expansion did you get? I was getting closed form expansion. Closed form expansion. And I could actually sum up, in the solid state case, I could sum up all the contributions to all the poignants. The difficulty was these things which, you know, sort of stopped you in the track, they weren't quite in your course at some point. When you put it into this picture, they get even worse. And so I thought, no, it was something completely different. But I felt that this wasn't my idea, this was Michael Fisher's idea. He actually directed me and I just worked on analysis. Pages and pages, I think. But I did work out some of these individuals which require polygons, and the polygons were probably this contribution to the point that I've just described, and I thought that would contribute to the point that I've just described. I thought it was sort of taking a further stance. That's exactly what's come up for me. I wouldn't be very surprised if they hadn't. But when I added up these, it seemed that these were not very significant for the following problem, for the recording problem, I think. What we wanted, all the emphasis was in these. In other words, I have to do this kind of thing and this kind of thing. It's the ones that are really complicated, the stars, that seem to be much more important to the recording. It was on the way, but you can't do it without proteins and I don't see any structure here by speculating with the chemists as I speculated before, but if you think of this as a chemical compound with quantum mechanics playing a role and maybe the quantum potential to translate it into a whole language, it's an organizing...

1:05:00 This guy, his chemistry was very, very taken. He said, I want to make it. He didn't realize you could do it. That's something which, well, I don't know. Say, yes, let's do it, and then I sort of think, well, yeah, but this is at room temperature, and it's not really quantum coherence, and you just go to quantum coherence, and it's really that feature that, because it's so easy to lose the non-locality, and then, well, no, I hope, well, maybe there's some other idea that checks in. I don't think the problem is solved. No, absolutely not. I mean, I wouldn't kill myself to do this and think I'm going to find something else. They just did this kind of chucking more and more computer stuff. Yeah, but what we did was to actually put it on a lattice and do it by lattice counting, which makes everything look like you just count. The number of triangles, the number of two-thigh waves, and the measure of the mass. All this is done. It comes to the calculation, which I'm not going to do, with the quantum distribution. So you could certainly work this out, but I'm not convinced that that's going to tell you anything about the problem. Unless these forces, unless you characterise these forces in some way so that the particular... Coupling is a qualitatively different type of thing from this random thing we used.

1:07:30 And as I say, we put this on a lattice and it was a good framework in general chemical physics, where we actually showed how they've been coiling the temperature and the change of the solvent-solvent, the molecule-solvent interaction, and we showed it clearly now. A lot of people didn't like it because it was on the lattice, but then you see you've got the partition function, but then of course what Lu is doing, working out the partition function, is not. The method you're using there to get the partition is that you're using the expectation that you find it's a primitive input. This is a primitive input. It's actually these exceptions that I don't need to teach you that much, but this one's just something for the audience. Sorry, this is the temporary leave algebra. I thought you'd be suitably impressed that I've actually got the 42 elements of the temporary leave algebra. Not a memory. Oh, not a memory? No, you don't have a memory. Is that, where, where, is that in Jones's book, is it? That's a good source for that. Yes, sir.

1:10:00 That's a hell lot. In front of you on two pages like that. How much? How much? Oh, I don't know. A sort of telegraph, what do you ask? This is BJH. Oh, right. I've never seen, I've seen that. What am I copying, that? That's in Luke Gaffman's book. Yeah. But this I only got out Friday with the help of my granddaughter. Well, fantastic. There you are, you see. Very long time ago. Well, mathematicians sneer at me, but I couldn't find... I tried to do it with the algebra, and I found four missing, and I could not spot what the four was missing just by looking at the series of these chains of these things. As soon as I drew them out, it was immediately simple. Because everything's represented exactly once. There are mirror images, here's all the ones with... And so on. You've got them all classified like that. And that's all 42. So that's 2, 5, 14, and 42. How about that then, Mike? Pretty impressive, isn't it? It's just a question of sitting down. How old is your granddaughter? Eight. Yeah, well, to have that girl sit splashed at eight... No, she didn't work it out. We're just looking at it and saying which one was missing and so on. Well, I think it's pretty impressive she's into that. Well, there's a dog called Clifford that she watches on television and likes him. You know, he does articles, he does stupid articles. She's taking that on board. Oh, you mean your clip on algebra is ground down? No, we spotted this one. These were for Christmas, right? Yes, it was there twice. As you can see, there's a couple of them that have been put in there. Okay, sorry, one of my... Yeah, this. This is this. Now, remember that when... Sorry, I'm showing...

1:12:30 Remember when we had P, curly P, algebraic expression, curly P, expectation value, curly P. That was Markov's work when we were trying to understand this is the primitive hidden potent in the Schoenberg algebra. And whenever you're doing an algebra you always have a hidden potent and we always wondered how to get rid of this P, but of course you don't get rid of it. And this was Lute. I saw Lute had an equation like this and I said, how do you get it? And he said, the diagram is unchanged. And that's the diagram because these are the primitive components. And you close them off, right? So that one traces that that's completely closed. And if it's completely closed, you can pull it out. And when you get left, it's this, this, and this, and this. And that's why I'm interested in this lot, because the way you were looking at it last week, this is where you must come in, with the categories here as well. Well, which bit of is the characteristic? You're actually calculating the idempotence. You see, because, look, I mean, all of this algebra is about combining diagrams, and if you combine the diagrams just by drawing, you see, because what I'm doing is, I'm actually doing that rule of multiplication, which is just doing that. And whenever you've got a closed thing like that, it's just a trace. And the trace is just the expectation value. If you come to quantum mechanics, statistical mechanics, you trace with the density matrix and that gives you the mean value, and this is just a trace, but you get it not from what I was doing by actually enumerating stars, you actually do it from the algebra itself. So the algebra is doing the counting for you.

1:15:00 It's much better to have an algebra to do the counting for you than actually sitting down and going, one, two, three, oh god I missed one, one, two, three, and going back again. So you can do it like that. That's what amazed me. And all of this is part of that counting that's going on. So somehow the moment algebra is classifying the order. The structure that you're playing with, it's not to do with interference of waves interfering with each other. That's just, you know, which is a bit like these diagrams in a way, but the point is that from, so these algebras really are the, I mean if you want to be fancy about it, the algebras are the important part, and the realizations here like this kind of thing is just a little bit. Because right back at the beginning when you and David were first exposed to Schoenberg's work in the city, it was very interesting, you know, and he never would have gained any motivation, didn't he, for his quantum theory of mathematics? We don't get any motivation of where he got this curly P from, because remember, Marco and I used to go around saying, what's curly P? What's curly P? No, no, no, no, no, no, no. So he had another thing in the Clifford algebra, which is straight P, as he called it. But this one we have to bring in from outside, which always made us a little bit weary. What have you done? This is what I want to talk to you about, the Smush product. Yeah, I was going to say. Whether the Smush product really is what you were saying it was. And this is the category completion. And that paper that I gave you reference to actually talks about it. Which I'm really sorry I haven't got around to studying. I do intend, as soon as I get back to the show, to have a couple of nights of sleep on my list of the most important things to do. And that, you see, gets this diagram in a different way. Shall I just say a little bit of it?

1:17:30 Yeah. I'm trying to inspire people to actually take it seriously. Now, you see, when you're splitting these out, the whole point about the template-leave algebra is that your tower of algebras, sorry, of the Penrose timing, the tower of algebras is written as a matrix, it's isomorphic to a matrix algebra of dn, directly summed with a matrix algebra of dn, where these dimensions are actually different. Okay, they're close probably to these things. No, sorry, no, no, that's the long table. It's this table here that just goes up in two. Five and three, three and eight, and then you've got another one on the bottom there, and so on. This one goes up in two, and that's why you only have two. You always have two matrices, they just get bigger and bigger and bigger. Dimension of one of them goes up, three, eight, so then the other one follows up, but three, five, just one step behind it. You're just saying the dimensions in this thing are Fibonacci-like. Yeah, this is Fibonacci-like, but one of the reasons, one of the ways that one sees this is by essentially introducing two, in this case there'd be two, central components. See the amazing thing is that we've got this without actually talking about it. We've got it because we're taking information and thinking about it.

1:20:00 That's 0001. That tells me what, you know, without doing, without being clever, it tells me immediately what the size of this thing is. We've got to be able to construct these two. These two centuries of focus. We've even got a large model of it. And the same here in the temporary lead. This one's easier because there's only two of them. But we don't know anything about what goes in here, you see. Here at least we've got these algebras. Here we've got all the structure of the algebra, but we can't calculate things. Here we've got the structure of this, but we can't calculate things. We don't know what the algebra is. Or do we? I think we're missing something. I don't know what else to do. Let's go back to this one. You just sent me that thing, didn't you? The one that you were... I sent it to you as a reference and told you to download it from the... Yeah, which I... No, I think I did, I just haven't got around to looking at it. I'm pretty sure it was... Yeah, it was a link straight to the... It was straight to the... Yeah, yeah, I do. And I thought it would be easier for you to do that. I did, I did. Sorry, I just spent my quarter time in the room. But you see what not this guy does. Is this the paper you're talking about? I'm now talking about the paper that you should have read. No, no, no, that's not being mean. No, that's not being mean at all. I didn't mean that in a nasty way. Because you see what he claims to be. What's the guy's name again? Yamagami.

1:22:30 That's right. Segura, Seguru, Shiba, Shigeru, and I confess I'm not having a lot of having fun with them, I assure you. I'll tell you that weird thing that Keith sent me, the guy who submitted his book, Who's that guy? The guy from Newcastle. Yeah. I'm afraid you can't believe that. It's very serious. I'd like to speak to him. The paper is nonsense, I'm afraid so. Just to check, my impression was right. I sent it to the big man and he said that wasn't very rude, that wasn't his absolute self. No, you see, he says we can discuss the temporary legal algebra in terms of a category, a linear category. You've got a set of x1 plus x2. I was hoping that, you know, that this would tell me how the implotments come out of this. And then he has a lemma here. He says that essentially the far-right dimensional linear category is essentially semi-simple. P is equal to the sum of Pi in terms of, and these are minimal idempotents, and he doesn't say whether they're two-sided, central idempotents or just idempotents, and then he says the corollary to this is that it means that we can take our category and it becomes isomorphic to the direct sum of Mi copies of these Xi's. That's a good question. It's one of the objects in the category. It's just a linear category that he doesn't say, because he said it. And he's got two morphisms that you can take from this X.

1:25:00 This is like the identity, coming from the identity to the object, and then from the object back to the identity, such that beta i alpha j is equal to delta i j times the identity in the category. Beta 1 plus alpha 2, beta 2 plus etc. And this, Graeme, I think is just exactly what physicists use. Resolution of theory, beta. Yep. So it looks as if these, these homomorphisms that he's got here can be looked at. I've always thought that these, that a Brawler catcher would be thought of as a morphism, Mike. I didn't quite put it that way, I said it was process activity, but it's a, but what I wasn't, what it's doing is it's coming from, it's coming from the individual to the object, so that this thing here is a morphism which is taking me from x i, and the other one is, I mean you can almost see it, the arrow is almost self-suggested, isn't it, and that's from x, it's kind of a pullback.

1:27:30 Not, not, not, not the technical terms, but coming back the other way, so that, and it's nice that the arrow is pointing in the right direction, but I'm sure if you're actually the real answer. Mike, this is, this is what gives the conclusion. No, don't tell me anything, I must say, if I could only wish to God, I wouldn't say, you know, it might deflate me out, unfortunately. Given any linear category, its completion by idempotence is, by definition, The linear category is by definition the linear category where the objects of this consist of a pair with P as an endomorphism in X. Right, where X is a category. P is an idempotent, but you look at it as it's an element of the endomorphism. And om-sets, you know about them? Om-sets are set to be om-q-x equals om-y... What was that? You see, this is even published completeness. Yeah, he's saying this is a completion of category without follow-up.

1:30:00 I don't understand quite what he means by completion of a category without illustration. Well, it's a standard, it is actually a standard, what do you call it, can I just stretch a question, please? I did actually take it down as what completion, co-completion of category is, but I'm just wondering what the definition is. Completion, k-bar of cases, so it was L and L-bar, right? I should be able to answer that, but I'm just thinking I haven't learned. So the objects of L-bar are pairs. A is a different notation of A and E. So A is instead of X. L-bar is the completion of this linear category, once it's been completed by the idempotent. And the idempotent E is a mapping from, E is just a mapping from A to A, and it's idempotent. The morphisms of K at the L-bar from A E into V F, which is what I feel. There's nothing like as much as I do, I'm sure we've done. No, I'm a BF, you're a BK. Is amorphism alpha, okay, this is amorphism, alpha A to B, in K, in L. Ah, yeah, in L. In L, sorry, in L, alpha E equals X. In L, the original linear category, not in the completion, not in the completion, right. Such that, and then that's it. How do I think about that? You see, where's BF come from? What is F? Why is F suddenly on the other side? Now you can see why Freddie probably... I said I've got something which does, an idempotent which does, what was it, B times...

1:32:30 Well there must be an arrow in this category which clearly can tell which is... Kearney p times q has an inverse of the component. Remember we had these conditions on my idempotence? Yeah. As soon as I put that down, he said, oh, it's the completion of the category of my idempotence. Right. There we go. Except that my p is identical. You see, he's got a different, what is this f? It's another idempotence in b. In l. No, it's a map in l of the original linear category. Now, of course, what I've got here is a particular case where alpha, you know, not alpha, but well, this is, you know, has a value of zero. I don't know how you put that, because here's an element of, it's not a mapping, you see, it's an element of, but we can think of p as a mapping, because momentum, remember, was a two-point function. In Heisenberg's picture, momentum is actually characterized by two points, because it's a matrix. And that's where the croupon came from. Could you not use the concrete data of the linear cataclysm? Well, this is why I was hoping that you might... Well, you know, a cat's vector space is the same as the vector space is. Yeah, the vector space is a linear cataclysm. Well, I was just looking at it. I can't see now that it's a special case. I mean, it's a special case, but yeah, a vector space is... Well, it's tantalizing. A vector space. Not in the sense that any vector space is a linear cataclysm. But if P and Q... But you see the body for years. Why are the alphas the same and the unimpotents the same, because what this is suggesting is that the unimpotent is the same and the mappings are different.

1:35:00 It's not like the left multiplication and right multiplication. It's supposed to be unimpotent, you see, whereas my idea, when I got this down, it pretty immediately garbled a message. ...completion of categories, which I don't know what's in the rubble at the moment. You could call it smashing progress. I don't have to have none. You know the way Freddie goes. Oh yeah, he was off on the horizon and there was me, sort of, blurry-eyed, trying to follow him over breakfast. Well, I will certainly get another chance to do that. But I'm just wondering why he... P is an element of the, is an endo. Well, no, no, look, here, here's an endo mapping of, of, of, in this, in this original linear category. Yeah, that's P. But, and then there you can think of that, if that is an endo. Yes, yes, but is, is F an endo? No. Well, unfortunately, they don't know. E is an endo, yes. But what, what's this arrow doing? So what is this arrow here? It's, it's taking a, it's taking an operational width. There's a morphism in K-bar, sorry L-bar. This is a morphism in L-bar from there to there. Yeah. I just want to understand what are the objects of the category it's taking, of which this morphism lives. This is the definition. It's the completion value of the original linear category, so these things are... It is also the left. It also acts as the left multiplication as well. At alpha. And then it's equal to alpha? Yes, but it's also the identity, which is... I don't understand that. Okay, well... Sorry, I've had a lot of fun. I'm sorry. I have to go away. I will have to go away. This is yellow yeah but I'm not well read the paper kit because you might know there might be some background here that I'm not familiar with yeah I know a little bit about and I'm not quite sure what this one after a concept yeah it's certainly not you see except that this is you know it's like left and right multiplication

1:37:30 That's, in fact, his own, you know, concept in the category of sex, I guess. Yes, very good. Unless, of course, alpha squared is equal to alpha. No, I don't know. It's a... Toyer used to say it's a mystery. Where did these things come from? You don't know, do you? Oh, no, I remember that song. You remember that song, do you? There's too many words in there. I don't know anything. I don't know anything else other than it's a mystery. Why does that go, why is that on the other side? Oh, I see.

1:40:00 Think of it in terms of, you know. Yeah, right. The guy who's got something here might be helping me. I'm taking credit, I'm working here. Sorry, Steve. You've got to make allowances from the other way. So you've got... E must be idempotent in category A, alpha is, and E must be idempotent in category B. Idempotent, has that property gone into itself? Is idempotent there used in exactly the same way as idempotent is used in the context of overalpha? Yeah, but if you apply it again... I suppose in one way, you can think of this, the fact that I've got the same, you see I've got the same in and potent in both categories, I'm not quite sure. Yeah, I'm sorry, I really am sorry, I think I'm useless to you. It's just that... So don't hide and I haven't looked at this paper but I want to know why was that good question I've seen in this network well if I do B followed by alpha I end up doing x followed by alpha and now can't be and it's x before alpha But how can I... It's a kind of speech. But Alper did that thing to make the speech.

1:42:30 No, you do... Excuse me, is it? No, you write Alper followed by F. He is doing that, and then it's followed by Alper. He's just that. And if I do Alper followed by F, then it's just that. Yeah, you've got to read Hindi. You don't read Hindi a lot. Okay, it's handy. I keep telling you to keep doing this. Yeah, there's always an argument right at the beginning of category theories. Which way you do it. Which way round you write your... Why do you think it wasn't an Indian who did it? There was some interesting story to hear. As to why they reversed the natural, it's to do with what, what I'm saying, it's to do with the order in which you perform operations. Um, so you always put the first operation second. Yeah, that's crazy, because... OK, that's helpful. I mean, I don't know if it really, I don't know. No, there was some reason why they adopted this strange. A lot of people resist. Do E first. OK, yeah. Yeah, and then do R. Yeah, and then do R. Well, that's what's normal, I think. Yeah, exactly. Why I say tending? Because normally you say do alpha first and then do E. Well, because you... Well, I can't... I take some credit for that. It could well be the case. It depends as to whether you're thinking of the verb or the object as being primary. It's whether you do... Do alpha and then E. Or do E to... Which way do I say it? The effect of alpha applied to E is. But that was the reason why there was this argy-bargy as to what order you ought to write. When you're writing computational functions or computational mathematics in general and category theory. And originally people did start off doing it the way it was traditional in linear algebra. You've definitely got to read it. You can't read it like algebra. You've got to read it like chemistry. I think they've got to read them as more meticulous through the diagrams.

1:45:00 That was the reason that, you know, it made people look at the diagrams. And you see them on the next page are the diagrams that we're all familiar with. So you go straight into... There's a technical definition of the way a character is complete. I don't know what it is. There's a technical definition of a complete, of a completion of a category. Which is different from a complete set of orthonormal projectors. Yeah, it is a technical notion of what a completion is. It's something to do with preservation of push-outs and something with preservation of the callbacks and copies, including the way that the category can be. Math, math, math, can we just, yeah, I presume that this means we've gone back to space and we've gone, yeah, it's the same, I think, good, good, good. They don't have to prove anything, you know, anything in the end of it, yeah. Why do they have to prove it? Yeah, well, that's, well, that's... All right, no, because it didn't probably complete, otherwise it would just be complete. It could be something to do with, I mean, it's just, as you say, that's true of any...

1:47:30 It's not you that's understood. You can't read it. Could you take that chair back? Thank you very much. This, from the inside out, he also introduced... What, this is the chaperon of this paper? Yeah, yeah, I'm just going on to get rid of the algebras now. Okay, I see, he's going to get rid of the algebras. Now we're going to talk about the algebra. Delta is one... And that, of course, is science. And if you're a physicist, you'd say that is like annihilation, as Lou does. Yeah, yeah. And that is creation. Creation operation, I know. Yeah, normally they have time going, Lou puts time going up in the standard form. And then, you see, I've changed all this by relying more on the sign. Because I don't like using Chinese, I'm not used to reading Chinese, that's why I put it that way, to say it's paper, but you could certainly, so really the diagrams that I'm being taught should be turned up. So you apply this, and then you apply this, and that's equal to your joint. Again, you join those two up, you get the circle, slip it out. That's the trace. And then you get left with this. So that's trace times, and that of course is just the reason why the interpotence in Jones algebra, and this causes a lot of problems, is E1 squared equals E, but in the 10th and D algebra, it's U1 squared equals E, a to the minus 1 microsecond.

1:50:00 And he calls this pairing and this co-pairing, and then the co-product. Remember in the Hopped Algebra we talk about the co-product? Well that's why I'm saying you're really dealing with Hopped Algebra. Well you're beginning, I haven't got the Hopped Algebra, because the Hopped Algebra has the... Oh, that, that. The line, what is, sorry, close it a second. Can you just point us together the distance between the temporary leave and the Jones? Oh, it's because of the, it's because of the temporary leave always shoves the D in like this. And then it has E, I, E, I, plus or minus 1, E, I, E equals 2. This one is equal to just Ui, whereas this is Ui, Ui, that's the Jones, the first one. This is Jones. That's Jones, right. And then the Jones. And then this one is equal to just Ui. Okay, right. All you've done is you've just slipped the normal, it's like a normalising thing. Right. You can't get rid of it. And I'm not sure what the relation between the tor and the d is, it's either in, this tor is this tor that I've been putting in. Just to connect, you know, it's not something now completely different, but actually this trace, the tool that comes out of here, and it's a trace.

1:52:30 And I think this might be, well, this is a trace in the diagram sense. Because that, you can see why you need it in the diagram sense, because, so you really, these are called, should really be called views. So you don't get mixed up halfway through saying what E's am I playing with. Are these E's that I'm using temporal E's, Jones E's, or are they count loose E's? Loose E's. And then he talks about a provenius transfer. Provenius transfer. Provenius transfer. Provenius record problems. Yes. No. He draws a provenius. Is that what you're going to ask about? Well, one of the things I'm just going through. You were just going to ask me a question when I interrupt you. Frobenius! I remember Bill giving an excellent talk about it in Bangor. That was interesting to me. Yeah, good. Is there a Frobenius transformation in categories? This is a Frobenius. So this is your object. So can you say anything? Is there any reason why we should take this one and point it out? What did Frobenius have to do about it? We thought it was a key to a very general construction which showed up all over category theory, including in the way that you keep quantifiers and categories in logic. There's a rule about the transformation of quantifiers and the cycles and the restrictions. And the whole thing drops very naturally out of this construction, which is also, Fabrini's reciprocity, also thought that you can get the canonical connotation relations out of it. Is that Fabrini's theory? Yes, yes, that's Fabrini's theory. He's a human mathematician, he's called Fabrini. Yes, it's a very old time, Fabrini's been around. But this reciprocity is something to do with the way that, you know, you compute integrals and things like that.

1:55:00 And you're beginning to do the trace because you can see if you put on the top of the ear. But what I do remember is that it was always people connected with the origins of quantum, you know, the quantum uncertainty. But I say this was 15 years ago. So you're going from, you're somehow reducing the number of objects you've got in your, you're going from, well not really, you're going from one object, you then create a pair, you then have this to annihilate it, and this is the, and he actually distinguishes between these two things, he calls this, this is where we ended up last time, where he calls this... So he's got these two dual transformations. Yeah, one of them is bent round and goes up with the time flow, sort of time reversal. And this one is time reversal here, so the process is time reversal. I mean, I look at it as just your clues and things like that. Well, you're not quite, you see, because this is already there and you're just closing this one here. You're actually, you're constructing something to close it. And this is where they're done. Now, these x's, Mike, are they just, do we think of these as the x's? Are these the objects? Because I'm taking these two objects and I'm putting it into the identity because I'm in love with it.

1:57:30 Yeah, yeah. That's what I mean. Thank you very much for your time, and I look forward to seeing you again soon. Now you see, what Graham and I were talking about earlier on, if you take A2, that's just the algebra with E1 for the identity. And then you can decompose this into complex numbers times E1. And this, these are really... Oh, see, I should really be using these. Let's not worry about it like that, it's a general structure. This C should be it. Here's my semi-simplicity. My A2 is becoming too simple. And then he says that this is equal to C plus C, only the complex number is not going to be equal to 1. And this corresponds to the first level in the diagram, 1, 1, because 1 means just the complex number. It's a one-dimensional matrix. But now he says something which I don't like, I don't understand. The linear category, the linear subcategory generated by 1, x, and x2, is semi-simple, and the simple objects are given by 1, x, and x sub 2, where x is equal to some even potent f2.

2:00:00 So, really, am I right? This should be the identity, this should be the line, and this should be the loop. This is the x. Well, you're sure the first is the identity. I'm not sure about the identification of the other two. It sounds very plausible. x is one vector space and x is two vector spaces. I'm not including them. But, it's because this thing is like this that we, it's not x squared. The x squared would just be that. Yeah. And therefore we must close it with this. We must close it. Yes, and that's what this idempotent does. And that's what this idempotent does. But this idempotent is actually... Yes, because you would make the point it's this business of a closed form. But then you see I have a problem because he writes F2. And he actually relents all this contemptuously in his paper, doesn't he? Yes, he gets the tower out. I'm sorry, I've not seen the paper. He gets the tower out. And I'm trying to understand how the tower comes up, and if someone's got a different way of looking at it, who's going to show some light on it? Allegedly. Now, what does x squared, what does that equal in diagram? This is the formula, isn't it?

2:02:30 Yeah, here's paper. I don't know what it means. E1x squared is equal to the isomorphic to the isomorphic, and x squared is isomorphic to one tensor product x squared two. So if you say two lines for a loop of a column, then the other x squared will be the loop of a column. In other words, it would be like this Frobenius, this diagram of Frobenius with a box or something like that. X2 is looped. Well, if X2 is at the top of the loop, then X2 is at the top of the loop. If this is a loop here, then this one would be a loop. Well, we don't know. This, when it's written like that, is not a loop. It's just two objects. Well, how do you differentiate between the one on the right and the one on the left? This, this and this. I don't know, that's what I'm puzzling about. That's what I'm supposed to be able to answer. But this looks as if it's something... this has got the loop in it. That's my... Yeah, x, lower k. Yeah, x with a lower 2. x with a lower 2 is the one with the loop. But why, you know, he adopts that convention. What's the argument? That I just spoke, I'm sorry, I really can't help you. Okay, well this is something to look out for when in the paper, it's on page 11. I am able to assist you, Lord, in that matter, if you may, sir. What minus is that? It's just, it's just the two dots, we got, we sorted that out last week. It's just the, it's just, it's just that minus that. And you just treat it as a minus in the multiplication. Just treat it as associative. X minus Y. So when you're multiplying that all squared, you just do this, multiply it all squared, which means that you put that.