Theoretical Physics Research Unit Meeting

0:00 This is part of the last lecture that I'm still trying to sort out, that we're trying to sort out. The sort, summary of the sort one uses in knot theory, the mathematics to describe the topology of the Penrose Tower. The mathematics to this seems to be using the same structure. The question that I've been looking at is why?

2:30 The one thing that is, just to show you, the thing that I've been concentrating on and trying to understand... is this Ratelli diagram of the temporary leader. 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 explaining this, bringing it into primitive implements, 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 clippet 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 out, another type of group. What's the diagram again? It's a Bertelli diagram. It tells you how these algebras are stacked. You really got... I remember... You remember I went back... We talked about this at the end of the video, didn't we? Yeah, you remember I... The essential algebra, if you have any bright ideas, go and 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 A2, A3, 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 spacetime, your continuous spacetime. That's one sort of direction we might be moving in. Sets of algebras are actually described very nicely by means of the tau algebras, but the tau algebras are specific types of these algebras, of which the temporary lead is one of these algebras, and it's the one that Jones deals with a lot, and that is the one that starts with a 1, 2... That's Jones's book to look at. I'm looking at my head at the moment. Then what do we do? We go two and... Yes, it's either in Jones, it's in several of his papers. You can't get away from it. Two, two and one is three, one, and then you go... If you don't draw this right you go skewed. So I've got 5, 4, 1, and then on, on, on, on, on, and that must be 5, 5 and 4 is 9, 4 and 4 is 5, 1. Okay? And these are the dimensionality of the algebras that are in this chain of the temporal loop. So, Tom, what do you mean by power algebra? Is that the same thing as von Neumann algebra or is it von Neumann algebra? In terms of von Neumann algebra. He sets this whole thing up in the context of von Neumann algebra.

7:30 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 it's not standard. Now then, if you take... He's got a series of three going to be there, four going to be there. At this point, what these 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... etc. That's Euclidian Algebra. The alert reader will recognise the Euclidian Algebra. The reason for that is, I've discovered, that for the odd dimensionalities, the Euclidian Algebra factorises into two. It's semi-simple and it factorises into two algebras. This one here. This one is an even algebra, so again, that does not matter. 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 rose with a simple explanation for it. It's to do with the order meaning of the orthogonal groups, one little dimension. And we know we've always got that problem, remember? But you have this problem with all of these numbers and dimensions.

10:00 Okay, so there you've got the clip of an algebra. Then, he says, one, one, one, two, one. Right, and now I'm at level three. That's at level two. I've got these numbers. 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. I've got one more coming here, that's 3, that's 2, and then I need to go bang, bang, bang, and then I get 5, 3, and then I get 5, 8, 9, and I draw one back on there, and I just stop there, that's 8, 3, and so on. And this, it says, the alert reader, oh, it doesn't say that, it just says, publicity numbers, of course, okay? Now, it turns out that the reason why this one has 12 to the minus 1 equal to 4 cos squared pi, and that's just equal to 2, which is where the 2 comes from. The total of minus one is just four cos squared five by five, and that's the important thing, and that's the big and large and so on, and it goes up to the next level and so on, whereas here, all the total of minus one are greater than or equal to four, and this is the trace, remember, the trace of the matrices, and this is where the tower algebras come in, they're characterized by this.

12:30 This is the coupling constant in the volume, or the trace. The trace on 4? The trace on 4? The, the, er, the trace. There's only one of those two, isn't there? So how can that be? This is not, this is, I don't know what that is. The trace on 4. That one you were doing. But this is not. But each one, you can see what happens as you take one out. t minus 1 equals 4 cos squared by 8 to the sixth. 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, actually, this index. It's the index of the alphabet. But the problem that I have is how do we get that out? Okay, so there's the background. We've got these towers and all. It's very tight. The whole thing should be less than or equal to four. Oh, wait a minute. Two over minus one. Although it's two, then. Okay, two. I don't know what you're... Sorry if I got it round the wrong way. It's... I'm kind of blind. I've got two over minus one this way. It might be the... It might be the other way around. Yes, Tor is greater than Tor. 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.

15:00 And that was the one you spoke of, L-A-N-D-E, wasn't it? This one is Landy. Key actually relates this structure to what is a pattern and what a young, young rapist, yeah, rapist, yeah, the pattern, yeah, the pattern, yeah, and there you had that algebra, what was the algebra, the rotor algebra, and you remember when I was doing my algebra of process, I essentially had a rotor algebra. And Luke Hasman, in generating the clippet 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 factoring practice motivated there? That is... Causally ordered, causal sex. That's right, it's causally ordered. Causal sex. 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 more of a... Take them yourself in one. And this was what moved Krautman, who was a sort of Feynman path integral, right? That's three and that's one. And then if you put a sum over two in here, you know, it was I. In general, he, he, uh, Rathis and De Patrim didn't have an I in here, they just had a single number.

17:30 But you can do it in terms so that the intermediates sum over into the interspace and then you've got the sort of thing that Luke Hadley was suggesting, that you've got to sum over this string to the path integral. So you're collapsing the path integral into the structure. I can't remember. One of the recent, very recent ones. I remember him talking to Alessa and Pete about this. He came, yes, he put it out in the paper probably two years ago, one year or two years ago. 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 on 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 then that stops, but that's not as interesting as these hierarchies, and so you can't construct the terms from that. So you were saying that tarot and altruism are not metric countries? This is the end, yeah. I'm not answering yes or no, I'm saying... Because you said that the motivating example was Jones, Gauss, and the monominal. Yes, yes, which I'll make it so. The question is where the matrix is. But you see, we've got a problem, because in the templated ligandism, 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? Well, am I right in saying that any final answer is a legislative answer? 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 final answer.

20:00 I don't know the answer. Sorry, I should know the answer, but it's escaping me. I'm fairly sure you can. I was just trying to get my head around why you've got a problem with that. I can make it into, oh, Lee Lewis told us how to make it into a question, but 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 times. Thank you. No, you can't tell 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 not periodic. That's an interesting... I've not looked at it in terms of the Bertelli diagram. I've not looked at it in terms of the Bertelli diagram. Well, if it's equivalent to the Fenrose pilot, 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. Talking about where Penrose got his ideas from for the Penrose time. My admission was, and I wasn't sure, but I do know that when I came here in the first year of Chippendale's presence, then when I arrived, I was talking with John Penny, who had been working with J.D. Bernal, on liquid structures, and he was asking what is the difference between a solid and a liquid. Long range ordering. So what about a record? 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 he's going to build this one. 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 kind of football glider and sucking all the air out and making them, each ball bearing had a blue tacky stuff on it.

22:30 And then you make it into a heap. You see the difference between a heap, which is completely random, what looks like a big random structure, to pouring... I used to do a lot of experiments on 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 tore them apart and they looked at the coordinates of where they were touching and drew that out into a bond. And then they built this irregular structure. And then they analyzed water. These were the catonic solids that were in. And they found that the majority of them were pentagons. So if you want to heat, what dominates is the pentagon. Okay, so now that we've got two dimensions, it's much easier. We're talking about pentagons. And what Penrose was saying was, how can we cover the two-dimensional circuit? But then he started saying, well, what can we do? And to cut a long story short, he said very simply, let's cut. So that we've got two types of triangles in the pentagon. You've got this thing here, and you've got, now what he then did was say, all right, let's make this into a kite, and we make this one into a... I think I've probably drawn this. I might have got the manga the wrong way. Anyway, you get the idea.

25:00 So you've got these sides a unit, this side is, well, you can look at what it is, it's a square root of 2 or something like that, 1 plus 1. It's taut, I usually call it taut. And then these two sides are taut, whereas this is one. So you've got these triangles, and you can work out what that is because you know the angles. Oh Bob, can you imagine what the internal angle is of? 72, was it? 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. Okay, I don't mind. Because now what you can say is, when I can bisect these angles, which angle is bisected, or these angles are the same, maybe cut them into three, and what you're essentially dealing with is an angle of pi by pi, the basic angle of pi by pi, the various combinations of pi by pi, which immediately makes you sort of say, oh yeah, obviously there's going to be some connection between pi by pi. All of these things cover the surface with parts and parts and parts, 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. What you do is you've got your...

27:30 Penrose tiling, which you can see, these darts and kites are woven and from them you can see, you can combine them together to actually form a frequency that can be constructed. And the way you characterize each tiling is by taking a point in the tiling and then slowly taking the tiling to pieces by pulling arms out and so using 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. You find that you actually get 0, 1, okay? So you're just pulling things out and saying, what is the point now? So the total number of distinct timelines is going to be 0, 1, because that's the first point, because it can be a large instrument. 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 zero and one. What you find is you can never put two ones together because the way the thing forms a column. 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 to see that. So what you generate then, am I doing them all right?

30:00 And then we go 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0 1, 0, 0, 1, 0, 1, 0, 1, 0, 1, and that should be, he should be the, the, the, the, that's 3, that's 2, 3, that's 5, and this is 1, 2, 3, and then you do it again and you're going to get 8 here and you're going to get 5 down here and so on. Now, what the trick is to get it into this is to treat these as the dimensions of, treat these as a vector space, so this is a column of three objects, sorry, five objects, and then these algebras then decompose into an algebra which is m, d, n, plus m, d, n, plus m, d, n, plus m, d, n, plus m, d, n, plus m, d, n, plus m, d, n, plus m, d, n, plus m, d, n, plus m, d, n, plus m, d, n, Where this will be, if we start at the bottom, this will be a 2x2, let's start again. This will be a 2x2, this will be a 1-dimensional object, which will just be a 1. Then this will be a 3x3, this will be a 2x2, this will be a 5x5, and so on.

32:30 I don't know what that means, because you have linear superposition and so on. And that makes me think that linear superposition has nothing to do with interferences. That's the way we see it physically. But in order to describe these things, we have linear superposition as vectors, and then we actually sum out a subtraction. And this then allows us to classify all the Penrose tiles. It's quite extraordinary. But you see here as well, what I've been drawing here is actually translated into this diagram, 3-1-2-3-5-3. It would be lovely if there was some way that the difference between constructive and destructive interference showed up in the diagram. Okay, you say nothing in truth could do with interference, but it would recover that distinction. This does, it's in the case theory, because the way we analyze this is through topological cases. 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 that I had in front of me. So I've been reading a little bit about the history of K-theory, and you talk about the history of what mathematics is about. But our friend Crottenbeek comes in here. Yeah, exactly. I keep saying that, you know, what we're looking at is really the gothic 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 what? I think to the description of the order. I was just saying why did Penrose go for pentagons and one idea was that they knew what they were doing.

35:00 Sorry? Well, I'm not sure they did. Stereochemistry. 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 coming in. 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, where I did the Bernoulli model and did a random walk. The number of n steps over the number of n minus one might be very, very, very, very, very, very, very, very, very, very, very, very, very, very, very, very, But now they were into protein folding. And they wanted physicists to understand protein folding. Well, one of the ideas is what is a protein? A protein is essentially a gas, you know, a liquid, a structure. 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. And then work out what the partition function of this thing was. Given forces between them, and also the three forces in the liquid, the van der Waals, that kind of thing, so it was designed for a firm.

37:30 But you see, this is something which the people trying to crack nanotechnology now worry about, how to do these problems. You see, so what I, the idea I have... I'll come back to that, sorry. If you plot these random walks against 1 over n, so that you can extrapolate to 1 equals infinity, okay, 1 over n equals 0, 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 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 at 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, 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 process going on. Now, the other bit of work I did, which was mentioned in Mandelbrot's book on fractals. How about that? How about that? And that was doing excluded volume walks, which came from the same point of the origin of Einstein, and they call them u-n, so if you plot a u-n against u-n minus one, or plus one, I don't remember, you actually get a buggering fact that then they settle, and you can't believe it, if you extrapolate that settling line, it meets the extrapolation point on the other side.

40:00 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 of the slope. And what I did with the Bernal model was to see, because these are regular tessellations, and 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, nowadays of course. All of this chemistry is telling me that we're up to thousands of molecules. It probably makes no difference at all what we've got with our 15 quarts just coming up. All we've done is confirm what we said it would be. Although, of course, it's taken all the errors off and made it much more exciting. No idea why this should happen. 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 in terms of, 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. Simply like this, a regular tessellation. And what happens if you have an irregular tessellation? Right now, in this year, John Finney had his motto, and I was sort of saying, well, what should we do with it? He said, well, let's break out a couple of random walks out. Exclude one random walk. Difficulty. When you've got a regular tessellation, every point is the same neighborhood. Random structure, this point is different from this point is different from this point is different from this point.

42:30 What you do is you have to do this quite differently. And then average. It doesn't matter how you average it, whether you do it directly to me 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 still knew it was in a three-dimensional. And the same thing for this. And what we didn't do was the Penrose tiling because Penrose tiling came in the 70s when I got to average it. But it would have been nice to do the Penrose time and then to see whether that was fitting one of the two-dimensionals. Does it make sense? I don't know, why can't you know? No, no, it's just so fascinating. It's such a... To me, this was the key... It's a lovely little... It seems to be due to the self-contained and... Absolutely. It's amazing that they've done it. Hamamesh, Hamamesh, no, Hemsley, the statistician at Cambridge, proved... Both of these things would go to the same level. That's sort of the rigorous mathematical theory. 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, does that relate to the... It's related to the critical point. You know, the ferromagnet has a spontaneous magnetization. That's the critical temperature when magnet... Where magnets magnetize as well. It gives you information about quantum mechanics, because that's where the susceptibility becomes weak. You know, I'm beginning to see all sorts of interesting connections here, because one of the motivations that Lou has for the subject of the instruction is to get out Feynman.

45:00 Pardon me? Pardon me, of course, Feynman. And Feynman did things in terms of this. Well, that's the Ising model, which is exactly the Ising model, which is also the thing which these people like to 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, but that's exactly what John Mabry's students in Lough has been doing with these, this sort of finite, he's trying to redo the whole of geometry based on this. 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 we're just trying to do it with arithmetic, but I think this is much more, I think this is much more fruitful. Well, it's richer, so you better tell me. No, no. I mean, I won't be against them doing it. But you see, the fact is that you already know with a 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 angle. And this whole idea of the spin level of the world, spin networks is essentially in there. I haven't seen a pin. The moon's got a lot on there. Spin networks come out from there. But this gives you a different level. But the interesting thing is you use the same algebraic. Now, if you're using group points, 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. Yes, exactly the same structure. If you look at the Penrose tiling in the classical way...

47:30 Carter gave a talk back in January in Paris about group points. Unfortunately, I missed. It was in French anyway. Somebody recorded it for me. I was listening to it. French isn't very well. Connecting rock with all this beautiful new stuff that I don't know about in symmetry. But it literally gives a new way of thinking about symmetries in the problem. 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... 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 there'll be, you know, and there'll be taken viruses... 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. Yeah, I like algebra. I was eight years old when I learned algebra. There was something exciting about it. It was 78 or almost. Not quite. But it's still fascinating to me. Okay, so you see that the background there of the finite algebra is the practice. And then you're trying to get, we're talking about precepts, which is kind of the thing in quantum mechanics. And it's all these level hints that seem to suggest that this is the way you actually describe it. The group point is essentially that mathematical and mathematical function. Here you have two symbols. A, A, B. And then you have A, B, C. And the sum is B, B, A, A, C. Which is just a motorism. Yeah, that's a good idea. That's because that came from the vile algebra. You see there's another algebra. This is then the lecture I gave in Fougere.

50:00 One of the lying bits is that, which I haven't put precisely in its place, but every time I turn over a page I'll suddenly see this vile algebra that I'm trying to write. And remember that vile algebra is to say, I take an algebra defined by a, b, 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 algebraic structure that we've got here. You've got this and 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. Start the Heisenberg algebra with an idempotence. These 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, it's a phase space which is either x or, in other words, it's an x-brick in order. 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 has a quantum potential, I mean, that's not Jack, that hasn't he?

52:30 I'm sorry, this is the structure that I have for my choice of free space. I'm creating the space-time coordinates out of the algebra itself. And that's what I want to do here, you see. This is why I'm so interested in idempotence, Graham, because idempotence are going to be the points on my screen. 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 on the left are not. 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. You remember what we have on here, the amazing thing is that the dimensionality of the vector space determines all these things. It seems like magic all of us. We're struggling by taking the template of 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 function. But 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. Well, I checked my emails back two days ago. I must be in New Zealand. I think that's a problem. Which gets him, hopefully, back again very soon. There was no connection at all actually. I don't know how the two things fit together. Did you say something about the spectrum is something that you've got in you?

55:00 What do you have to fit in the spectrum? You know when you point out that these things are linked to each other, it's a bit of an extra textbook. Yeah, they were, they were, it's whatever, it's whatever they, five, five, make them symmetry, do whatever, whatever. Yeah, I keep saying, um, I keep saying pentagons, but I meant pentagonal. So, whatever. By getting ball bearings. By getting ball bearings, sucking the air out, making the, the, the things stick together, and then some guy spent his whole life just picking one out, looking at the coordinates, coordinates of them. And it was then put into the computer, and that's what John Finnegan did. He did a paper with him, didn't he? I wrote a paper with him where I discussed this. There were also mutilated body-centred cubic lattices. There were tall bombs out of it. If I sit 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 heat. Well, then the liquid storm, you see... How do you differentiate heat from the fact that it's mixed up with plasticine or something? Sorry, what? They're both mixed 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, and that gives you a coordinate. No, that's probably your cell phone. What about liquids? What about liquids? You're going to see. Oh, just that, you know, we've come down to this problem. See, what I did in my PhD thesis, I've marked those back.

57:30 I also extended the Mayor-Costa theory to gases. Now, for a perfect gas you've just got boils in it. But now what happens when you get interactions between gas and gas? You have an imperfect gas, first approximation down the valve. But what do you do in reality when, not down the valve, it's a real gas? I developed a method which was an expansion in terms of finite clusters, so that your partition function, and then when I see Louie 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. It is amazing, isn't it, that Plato came back alive, 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've got to do is cut it there and it falls apart. Whereas one of these, if you cut it anywhere, it doesn't fall apart. 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. Now, my idea when I was going to use liquids and got distracted by crazy only mechanics, We 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 a real difficult... I mean, I had some ideas and you could do it in approximation and just have a look and see what it did. But, as I say, I got, despite the point of view, it's maniacal.

1:00:00 Which is often a new review of the Schmier and the Nobel Prize winners, rather than finding the partition function of the Bernoulli. So I gave up. Now, the protein, somebody met all this before? 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, it was the same. All right, here we've got, you know, I'm at Keynes with Norris Wilkins, 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. You can do a Mayer gas theory plus the condition of A must be equal to B, C, D, so you've got an order along the chain. And then just apply the Mayer cluster expression. You see, one of the things that came out of the Avon method was that you could sum up... All of these were very easy to do, so you could begin to get a closed expansion, which is different from a series expansion that everybody else would do. 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 points. The difficulty was these things, which sort of stuck you in the track, and then they don't follow you, of course, at some point.

1:02:30 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 put in a word by word analysis. Pages and pages of it. But I did work out some of the individuals which required polygons, and the polygons were probably this, this, this, this, this, this, this, this, this, this, this, this, this, this, this, this, this, But when I added up these, it seemed that these were not very significant for the following point, for the recording of a mathematical problem. What we wanted, all the emphasis was in these ones. In other words, I have to do this kind of thing and this kind of thing. It's the ones that are really complicated to start that seem to be much more important to the recording. This was just putting a problem on the marker and solving it. It was on the way, but you can't do it without poking the same way. And I don't see any structure here in my theory that would actually do it by speculating with things that the chemists have speculated with before. But if you think of this as a chemical compound with quantum mechanics playing a role, and maybe the quantum potential is translated into the whole language, it's an organizing... You know, it's just a more big field. This guy, this chemist was very, very taken. He said, I want to make it. He didn't realize he could do it.

1:05:00 He didn't realize he could do it. That's something which, well, I don't know, I keep... Say, yes, let's do it, and then I sort of think, well, yeah, but this is a good temperature, and it's not really quantum coherence, and what if we just go to quantum coherence? And it's really that feature, because it's so easy to lose the non-locality of the intelligence. I know I hope, well, maybe there's some other idea that checks in. I don't think the problem is solved. No, it has to be solved. 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. The number of triangles, the number of It's a qualitatively different type of thing from this random thing we used. And as I say, we put this on a lattice and it was a good way to look at general chemical physics, where we actually showed how they've been coiling the temperature, and the change of these solvents and the molecule-solvent interaction. And we showed it clearly now.

1:07:30 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 Loon 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 value times the primitive input. This is a primitive input. It's actually an exception. Sorry, this is the temporary need algebra. I thought you'd be suitably impressed that I've actually got the 42 elements of the temporary need algebra. Not a memory. Oh, not a memory. No, you don't have a memory. Is that, where, is that in Jones's book? That's a good source for that, just to have a lot in front of you on two pages like that. How much? How much? This is BJH. I've seen that. That's in Luke Atwood's book. But this I only got out. But I'd aim for the help of my granddaughter.

1:10:00 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 one line underneath and so on. You've got them all classified like that. And that's 4.2. So that's 2, 5, 14, and 42. How about that then, Mike? Pretty impressive, isn't it? Just sitting down. How old is your granddaughter? Eight. Yeah, well, to have that girl sit splashed at eight... No, she doesn't work it out. We're just looking at it and saying which one was missing. She thinks it's pretty impressive, she's into that. There's a dog called Clippet that she watches on children's television and likes everyone, you know, who does algebra. She says, Clippet, algebra. She's taking that on board. Oh, you mean you're Clippet Algebra's granddad? No, we spotted this one. It was their choice. 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. I remember that when, sorry, I'm shaking. 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 problem in the Schoenberg algebra. And whenever you're doing an algebra, you always have mid and potent, and I always wondered how to get rid of this P, but of course you don't get rid of it.

1:12:30 And this was Lou's. I saw Lou 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 primitives and the potents, 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, this, this, and this, and this, you can pull it out, 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 characters here as well. Well, where, which, which bit of, uh, is the, is the, is the, is the, is the, is the, is the, is the, is the, is the, is the, is the, is the, is the, is the, is the, is the, is the, is the, is the, is the, is the, And if you combine the diagrams just by drawing, 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... All of this is part of that counting that's going on.

1:15:00 The moment algebra is classifying the order of the structure that you're playing with. It's not to do with interference of waves interfering with each other. That's just the, you know, which is a bit like these diagrams in a way. But the point is that these algebras really are the, I mean if you want to be fancy about it, Because right back at the beginning when you and David were first kind of exposed to Schoenberg's work in the city, it was very interesting. He never would have gave it a major place, did he, because he wasn't quite so good at mathematics. We don't get any motivation of where he got this curly P from, because remember we, Mark and I used to go around saying, what's curly P? What's good? But this one we have to bring in from outside, which always made us a little bit weary. Have you, what have you done? This is what I want to talk to you about, the smash product. Yeah, I was going to say. Whether the smash product really is what you were saying it was. And this is the category completion. And that and 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, I've got 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 about that? 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 time, 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.

1:17:30 Okay, they're close, I'll be clear. No, sorry, that's the long table. It's this table here. It 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. The dimension of one of them goes up three, eight, so the other one follows up 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 that one, one of the ways that one sees IELTS, 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 about it is that you've got this without actually telling me. You've got it because you're taking information and things. That's zero, zero, zero, one. That tells me what the, you know, without doing, without being clever, it tells me immediately what the size of this thing is. But what if they add them and drop them? I guess it's really going to stop these two. These two centuries of problems. You've got a large model of it. And the same here, with the temporary lead, it just makes, this one's easier because there's only two of them.

1:20:00 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 it. Here we've got the structure of this, but we can't calculate it. We don't know what the algebra is. Or do we, you see? I think we're missing something. 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... Which you probably didn't. No, I think I did, I just haven't got around to looking at it yet. No, I certainly didn't. I'm pretty sure it was... yeah, it was a link straight to the... Well, it was straight to the paper. Yeah, yeah, I did. And I thought it would be easier for you to do that than I did. I did, I did, I did. Sorry, it's just so much water, so I'm going to bring it along. But you see what this guy does... Is this the paper you're talking about? I'm now talking about the paper which you should have read. No, no, no, that's not being mean at all. I didn't mean that in a nasty way. Because you see, what he claims is... What's the guy's name again? Uh, Yamagami. That's right, yeah. Segura. Seguru? Shigeru. Shigeru. No, no, no, I confess I'm not happy. I'm not happy about it. I'll tell you, I'll tell you that weird thing that Keith said to me about the guys who submitted for the full grant, that's what I thought it was relevant to the report, the two guys from Newcastle, yeah, I'm afraid we can't put it on because I don't think it's very serious, I'd like to keep it for the period.

1:22:30 No, you see, he says we can we can we can discuss the temporary legal algebra in terms of a category, a linear category, okay, and now what he has here is that You've got a set of x1 plus x2. I was hoping that, you know, that this would tell me how the importance come out of this. And then he has a lemma here. He says that essentially the five-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 idempotents, 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 could be. No, no, that's irrational. And he's got two morphisms that you can take from this X. Big family of objects.

1:25:00 This is like the identity, but coming from the identity to the objects, and then from the objects back to the objects, they might be called a category. Such that beta i alpha j is equal to delta i j times the identity in the category. And then, he says we've got 1 of x is equal to alpha 1. Beta 1 plus alpha 2, beta 2, etc. And this, Graeme, I think is just exactly what physicists use. Resolution of theory, beta 2, etc. Yeah. So it looks as if these, these 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 what it's doing is it's coming from the individual to the object, so that this thing here is a morphism which is taking me from XI. 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. Not, not, not, not the technical terms. No, but I see what you mean. Coming back the other way, and it's nice that the arrow is pointing in the right direction, but I'm sure you actually didn't realise that. Mike, this is, this is what gives the conclusion.

1:27:30 No, not coming into me, I must say, in fact I only wished it had gone, I wouldn't say, you know, it might have deflated 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 of X. And homsets, you know about them? Homsets are set to be homq equals q. I don't understand what you mean by completion of a category by 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 to completion of a category by illustration, but I'm just wondering what the definition is. Completion, k-bar of cases, because it was L and L-bar, right? I should be able to answer this, but I'm just thinking I haven't learned.

1:30:00 So the objects of L-bar are pairs. A is a different notation, right? Yeah. 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 just a mapping from A to A, and it's idempotent. The morphisms of K at the L-bar, A, E, into V, F, which is what I feel, Nothing like as much as I do, I'm sure we've done. No, I'm a BF, you're a BK. Is a morphism, okay, this is a morphism, alpha A to B, in K, in L. Ah, yeah, in L, sorry. In L, sorry, in L, alpha E equals X. In L, the original linear category, not in the complete, not in the completion, right, okay. 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... Well there must be an arrow in this category which clearly can tell which is... Curly P times Q has a... Remember we have these conditions on my idempotent. Yeah. As soon as I put that down, he said, oh, it's the completion of the category by idempotent. Right. Except that my P is, you see, he's got a different, what is this F? It's another idempotent in B, in L.

1:32:30 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 group all came from. Could you not use the concrete start of a linear cataclysm? Well, I mean, a cat's vector space is the same as a vector space is. It's not what you're looking at. I can't see now. It's a special case. 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 P thing is... Well, the vector space is... Sorry, that was badly put. Vector space is the object. Why are the alphas the same and the unimpotents the same? Because what this is suggesting is the unimpotent is the same and the mappings are different. Yeah, what you want to know is what the F mapping is. It's an infinite improvement from the original in the category. It's not like the left multiplication and right multiplication. But it's supposed to be unimpotent, you see, whereas my idea, when I put this down, pretty immediately garbled a message to do with...

1:35:00 ...completion of categories, which I don't think was in the rubble that time, I thought it was in the rubble. I don't have the whole money, you know, the way Freddie goes. He was off on the horizon with me, sort of, blurry-eyed, trying to follow him over breakfast. Well, we'll certainly get another chance to do that. But I'm just wondering why he... P is an element of the endo. Here's an endo map in this original linear category. That's P. And right there you can think of that. That is an endo map. Yes, but is F an endo? No. Well, unfortunately, they don't know. E is an endo. So what's this arrow doing? So what's this arrow doing? It's taking a... There's a morphism in K bar, sorry L bar, this is a morphism in L bar from there to there. I just want to understand what of the objects and of the category it's taking for which this morphism lives. This is the definition? It's the completed. The completion value of the presence of the original linear category, so these things are, I mean, the right is F, isn't it, is also the left, is also the X, the left multiplication as well, the Y, you know, that's the Y, and then it's equal to Y, yes, but it's also the identity, which is, I don't understand that, I don't understand that very well. Okay, well, thanks, Michael. Sorry, I'm out of time, sorry, I'm sorry. I have to go away. I really have to go away and read this paper. This is Yammer. Yeah, but I'm not... Well, read the paper, Kev, because you might know... There might be some background here that I'm not familiar with. Yeah, I know a little bit about linear factors. And I'm not quite sure what this one is about, but it's about concepts. Yeah, it's certainly not... You see, except that this is... You know, it's like left and right multiplication. There are a lot of examples of the concept of the category of sex.

1:37:30 Sorry, where the hell does the Y come from? It's like A and B in there. So what are you doing? This is... Well, you see, if this was the case, then I would probably understand. Then they would be the same. Yes, then it would just... Unless, of course, alpha squared is equal to alpha. No, I don't know, it's... Toya used to say it's a mystery. Where do these things come from, you don't know, Craig? 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. And that's what a mystery is. Why does that go, why is that on the other side? Why does, think of it in terms of, you know, the arrows. You've got to have got something in here that might be helpful. 43. 43. I'll take the credit, I'm sorry. You've got to make allowances from the other way.

1:40:00 So you've got E plus E to the category A. Yeah, yeah. You know, alpha is, and then you've got B, and then you've got C, and then you've got A, and then you've got B. I mean, is it impotent that that property's going into itself? Is it impotent there used in exactly the same way as it impotent used in the context of all of algebra? Yeah. But if you apply it again, it's the same method? Yeah. I suppose in one way... You can think of this, the fact that I got the same, you see I got the same in both categories, but I'm not quite sure. Yeah, I'm sorry, I really am sorry, I'm being so useless to you. It's just that, uh, SODOC hired, and, um, I haven't looked at this paper, but, uh... No, that's okay. Oh, what? So why was that? Well, that's a good question. That's a good question. I'll speak and you'll see. Let's work. Well. We're looking at if I do D followed by alpha, I end up doing F followed by alpha, and alpha after E, and it's F before alpha, so how do I do that? But alpha doesn't have to be related to B. No, you do... It's Hindi, is it? No, you write alpha followed by F. If E is doing that, and then followed by alpha, it's just that. And if I do alpha followed by F, it's just that. Yeah, you've got to read Hindi. If you don't read Hindi, you're not... No, that's all right. It's Hindi. Keep telling people it's Hindi. Yeah, there's always an argument right at the beginning of kind of category theory, which way you do it, which way round you write your, you write quite a bit of arrows.

1:42:30 Why did they, wasn't it an Indian who did it? Uh, there was some, actually there's an interesting story to hear, uh, 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... Okay, that's helpful. I mean, I don't know if it's really a thing. No, no, no. There was some reason why they adopted this strange. A lot of people resisted. Do E first. Okay, yes. And then do alpha. Yeah, and do alpha. Well, there's no other way of doing it, is there? Yeah, that's right. Well, why is there a tendency? Because normally we say do alpha first and then do E. Well, because you work hard on the metaphor, yeah. It depends as it were whether you're thinking of the verb or the object as being primaries whether you know do you know do do do do do do alpha and then e or do e to Whichever way I should say it, the effect of alpha apply 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 competition of functions or competition of the mappings in general and capital K, and originally people did start off doing it the way it was traditional in linear algebra and then... You've definitely got to read them. You can't read it like algebra. You've got to read the function. I think they've got to read them as more physical through the diagram. Yeah. That was the reason that, you know, it made people look at the diagrams. And you see on the next page are the diagrams that we're all familiar with. So you go straight into... There's still nothing to say about what Newton poses as completeness, but with E, he doesn't pose it as completeness either. No, I don't know if I'm claiming you're right. I don't know. Why is it completion?

1:45:00 Yeah, that's what I'm... well, that's what I actually asked. I said I'm not quite certain what the notion is. It's the word completion that is bugging me. There's a technical definition of it where a character is complete or completely. There's a technical definition of a complete, of a completion of a category. Which is different from a complete set of orthonormal projections. Yeah, it is a technical notion of what a completion is. It's something to do with preservation of push outs, something with preservation of the pull backs, I presume that this means we've gone back to space. All right, no, because it's even probably complete, otherwise it would just be complete. It could be something to do with, yeah. Okay, it just doesn't say that picture of any... It's not you, Ben. I'm just doing it for the system. You're going with it. Could you take that chair back with you? They're coming round.

1:47:30 This... He also introduced... Well, this is the chaperon that we just played from. Yeah, yeah, I'm just going on to temporarily leave algebra. OK, I see. He's going to get temporarily leave algebra. Now we're going to temporarily leave algebra, OK? Delta is 1, and that, of course, is 0. And if you're a physicist, you'd say that is like annihilation, as Lu does. Yeah, yeah. And that is creation. Creation operation, yeah. So, yeah, if you, normally they have time going, Lu puts time going up in this kind of form. And then, you see, I've changed all this by relying more on the sign, because I don't like using, I'm not used to reading Chinese. That's why I put it that way, so it's paper, which you can certainly, so really I should, the diagrams that I'm being taught should be turned up for a minute or two at a time, so you apply this, and then you apply this, and that's equal to, you join, sorry, you join those two up to get the circle slivered out, that's the trace. And then you get less and less and less and less. Trace times... And that, of course, is just the reason why the inner potency of Jones' algebra, and this causes a lot of problems, is e1 squared equal z, but in the tension z algebra, it's u1 squared equal v1. And v is just a to the minus 1 microsecond. And he calls this pairing and this co-pairing.

1:50:00 And then the coproduct. Remember in the Hopped Algebra we talked about the coproduct? Well, that's why I'm saying you're really dealing with Hopped Algebras. Well, you're beginning. I haven't got the Hopped Algebras. The Hopped Algebras has the, uh, I don't know, that, that. It's so unusual. But, Mike, what, what is... Sorry. Close it for a second. Can you just, can you just point us together? It's between the, essentially, leaves and the germs. Oh, it's because of the, it's because of the... Technically, Li always shoves the D in like this, and then it has E, I, E, I, plus or minus 1, E, I, equals 2, I, where this is U, I, U, I, plus or minus 1. That's the Jones, the first one you've done. This is Jones. That's Jones, right, and then the Jones. And this one is equal to just U, I. Okay, right. All you've done is you've just... 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, it's actually this trace that comes out of here, and it's a trace. And I think this might be, well this is a trace in the diagram, so, because that, you can see why you need it in the diagram. So these should really be called use. So you don't get mixed up halfway through and say, what ease am I playing with? Are these eases that I'm using temporary Jones leases, or are they loose leases? Loose eases, I'm sure. And then he talks about a Frobenius transfer.

1:52:30 A Frobenius transfer. A Frobenius transfer. A Frobenius record copy. Now, he draws a Frobenius. Is that what you're going to ask about? Well, one of the things I'm just going through is... You just wanted to ask me a question when I interrupted you. Prevenius traps. I remember Bill giving an interesting talk about it in Bangor. That was interesting, didn't you? The Prevenius transformation in categories. This is a Prevenius. So this is your object? It's a recording of it. It's about 15 years ago. 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 the category theory, including in the way that you could quantify the category theory in their logic. There was a rule about the transformation of quantifiers and the cycles about restrictions. And the whole thing drops very naturally out of this construction, which is also, Fabini's reciprocity, also thought that you can get the canonical computation relations out of it. I think Fabini sees something. Well, there's Fabini's theory. Yes, there's Fabini's theory. He's a human mathematician who's called Fabini. Yes, it's very old-timey. Fabini's been around. But this reciprocity is something to do with the way that, you know, you conclude in the cool. You're beginning to do the trace because you can see it from the top of the ear. But what I do remember is that this is all deeply connected with the origins of quantum, you know, the quantum uncertainty. 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, this is the, it actually distinguishes between these two things, he calls this, this is where we ended up now, where he calls this phi, he calls this phi.

1:55:00 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 you're closing things like this. Well, you're not quite, you see, because this is already there and you're just closing them. This one here, you're actually, you're constructing something to close it. And this is where they're at. Now, these x's, Mike, are they ju- can- 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 them. Yeah, yeah. It's just a temporary leave. Algebra and the, you know, this diagram rotation. I mean, the not diagram, but the rotation vector. But, yeah, I think the answer is yes, they are the answer, I think. I can't see anything wrong with that, because I've got a... Certainly the operations you're performing on will give you the... Because now I have a bit of the F's that we can answer.

1:57:30 Now, you see, Graham and I were talking about earlier, but if you take A2, that's just the algebra with E1 for the identity. And then you can decompose this into complex numbers times one, and this, these are really, see, I should really be using, let's not worry about it, this C should be it. Here's my semi-simplicity, my A2 becoming too simple, and then he says that this is equal to C plus C, I mean the complex number's not going to be there. He now responds to the first level in the diagram, 1, 1, because 1 means just a complex number, it's quite brutal, it's a one-dimensional matrix. But now he says something which I don't like, I don't understand. He now says, in other words, 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, which would mean any... Well, only that, you know, where you've got 1, those are brackets, those are generators.

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 loop. Is that what, is that, am I writing that correctly? 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 so confident. 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. This is making sense now. We must close it. Yes, and that's what this idempotent does. And that's what this idempotent does, because 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. Look, I mean, he actually relates all this to simply leaving it in his paper, doesn't he? Yes, he gets the... Okay, right, I'm sorry, remember I've not seen the paper. He gets the tower out. Okay, 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. What? X squared. What does that equal in diagrams? This is a formula, isn't it? Yeah, it's paper. I don't know what it means. So, x squared is equal to the isomorphic of the object, and x squared is the isomorphic to one tensor product x sub 2. So if it takes two lines for a lift of a solid, then the other x2 will be this, or what?

2:02:30 In other words, it will be like this Frobenius, this diagram of Frobenius with a box or something like that. The Frobenius construction. Yeah, right. Well, yeah. x2 is looped. Well, if x2 is the top of the loop, then x2 is the top of the loop. If this is the loop here, then this one would be the loop. Well, I'm going to get some new shit. 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 one on the right and one on the left? This, this and this. I don't know, that's what I'm puzzling about. That's what I was basically 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. Yeah. But why, you know, he adopts that convention. What's the argument? That I just don't, 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'm able to assist you, Lord, in that matter. 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. It's just like... Minus y. So when you're multiplying that all squared, you just do this, multiply it all squared, which means that you put...