Jones Polynomial & Yang-Baxter Equations from the Ising Model

0:00 Thank you. Hello, Basil. Hello, Martin. Hi. You all right? Err, about what? Your lamb's not coming in this evening. Oh, very surprising. He has had the decency to telephone me and say something's cropped up. Yes, I could tell you what, because in fact... Can you speak to me? Oh, yes, yes, it's all come to a head. I don't really know. No, no, no. I just wanted to tell you that he's... No, he's blown me out completely and definitely absolutely not when... Don't, don't do that. Ah, I'm just, you know... Well, I wouldn't go home, so... I'm just going to make someone dance in front of your colleagues, but I've certainly got a few right words to say to him when I do their homework. Anyway, never mind. There we are. But that's why he's chickened out, you know? Yeah, certainly. That's certainly... You asked him why, didn't I? He's certainly not going to want to look me in the face for some time, he can't. He's wanted to see me privately, so I immediately put one and one together. Well, don't say anything. No, I'm not saying anything, I just wanted you to know. No, well, it's good of you to tell me. Oh yeah, let's have a... Oh yeah, thanks. Is it all right if I go through this icing model and the...

2:30 I'd like to see it, yeah, I know. What was it you were saying on the phone that you had something else that you've... Just a... Is that going to be enough? Yeah, that's for a... Diet Coke, I'm looking for you. Yeah. Can you manage a coffee? Or will you be okay with, you know, do you have a pair of hands? Can I have a pair of hands? I mean, with all this, you know, nonsense today, I never can't do what I wanted to do, which was to make a copy of Aditya's talk yesterday to give it to you, or at least to bring along the notes I made. It was an interesting talk. Well, just to flash something past you, just flash past you. Uh-huh. What I've been doing is going back to my old stomping ground. Right, yes. We've got you, bangs are yakster, Mr. How do you get the, remember those permutation relations of the primitive in-potence, and ui, ui plus one, ui equals ui, to see how they're coming out in the iso model. Yeah, yeah. Now, one thing, the POTS model, which I, you know, don't worry too much. No, you mentioned it last week. I remember asking you about it. Well, if you get the POTS model, it's very much like colouring graphs. Okay. And you can have the four colour problems set up on a finite network. And so you can begin to deal with the colouring problem. Very interesting. But there are some rather interesting things. If you get the chromatic polynomial that the guys in the colouring... I don't know any problem, do you want to come and draw, so look, I did. And what they find is, as they try to estimate the zeros of this polynomial, you do it by slowly increasing the number at the end, the number of terms in your polynomial. Yes, yes, it's kind of, you know. And you find that they cluster around what is called a... There are a number, which is given by this expression of two cos of a pi of n, four squared.

5:00 Have you seen that anywhere before? Wait, hang on. Actually, yes, I think I have. Can I move my hand and show you where you've seen it before? No, please do. Now it comes to von Neumann Algeist. I'll tell you where I think I've come across this, just by chance. I was looking up some of the Phil Warmore's papers in maths reviews about a week or two ago. And there was, on the same page, there was a review of a paper about the Yann Baxter equation, which obviously leapt out of my eye. And I read this, and this would be, oh, about 10 years back. Yeah, because this work I'm doing is about 82 years ago. Yeah, it was about this, well, it might even have been as far back as that. about 1989 1990 and I'm quite sure that it mentioned this connection with yeah because you know that the whole of the Jones work depends upon the fact what he calls a factor now remember last time I was embedding n in n yeah now this one now when you embed any n and you've got a type 2 to the infinity then that That factor is exactly the same as the one you get in this chromatic polynomial. Isn't it weird? Even what Ty was saying yesterday is absolutely weird, the number of directions in which this structure keeps cropping up. That's what I've been telling these two guys here while I've been waiting. I mean, he said that, and he actually mentioned almost in passing at the end, he gave him, obviously, a masterly talk, and it was very, very well-timed, you know, to finish just in a spot on the hour. But he said almost at the end that I could have had a bit more time. The thing I would like to have talked to you about is the thing which I don't understand myself at all, and which I've only just begun to learn about, which is how this all connects up with matrices. That's an extraordinary statement, isn't it? Yes. Because that's what I'm doing. I know. That's exactly what I almost felt. Then, of course, are there any questions, and I was Michael, because all these very, very bright imperial research students,

7:30 the young Turks who didn't, I kept quiet, and I was wondering whether I had the courage to put my hand up and said, could you expand a little bit on your last remark about matrices, because I know a person who's also working on the connection between the terms polynomial and matrices. So, well, I didn't, of course, but it is fascinating. Anyway, so go through this derivation. I can't. Oh, OK. But this is just a remark. That it's crops up from this extraordinary... Well, I know... One of the things I have to do is to understand this von Neumann structure. Yeah, yeah. I mean, he brought in the von Neumann structure quite a lot of what he was talking about yesterday. He was, of course, entirely in terms of these holonomies on these manifolds. the g2 manifolds which which which allow this classification of open and closed strings you know donald's modular spaces yeah he's talking about that do you know what is at the bottom of that group gromov's theorem yeah he mentioned gromov's theorem actually he was saying that because you can look at all of this as just a beautiful piece of mathematics it gives you two it gives you a beautiful deep duality theorem about the equivalence between the different ways of getting the jones polynomial um you know but but there obviously is all this physics in it too the physics the physics that he was talking about yesterday of course all connects up with string theory because he's been talking to witten and no no no but it well no it clearly does connect up with that because he he he produced this rabbit out of the hat yeah but you don't start there no no no no no no that that's where when you're fiddling about with these terrible path integrals, you suddenly find that something simplifies when you do this. But there is absolutely no reason at all why it simplifies, as far as I can see. But isn't this... It is extraordinary. It's remarkable. Absolutely remarkable. Thank you, Graeme. Yeah, thanks a lot. I didn't know if you wanted to shoot. No, no, I'm fine. Well, what I thought I'd do today, if you're interested, is to show how this comes out of the icing level. It's not going to be a complete story, because I'm still struggling with bits and pieces. I did keep a very cool set of notes of a tie yesterday. I've got pretty well everything, and I've also got a recording, so I will let you have that as soon as I get back from Paris. Another solid comment, you see, is that when Colin talks about the Penrose Lattice,

10:00 that has a trivial topology in that you can't treat it in a normal way yet if you use the non-commodative geometry line you suddenly get you get the golden number coming out I'm sorry all of this is a mystery but I mean it's such a tantalizing mystery it's a bloody tantalizing mystery certainly beats all these theologians worrying about the Trinity now I say this is not the complete story because I've been working like how I've got about a whole book of notes here which I've been trying to work out get stuck on hyperbolic cosines and hyperbolic tangents and hyperbolic 2k and but I'm now getting I feel, just a remark to Mike, there is a very, in the icing model, one of the things that always fascinated me in philosophy, there was a duality relation. But the way you get the critical points is by exploiting dual structures. Now these are topological duality. But this is precisely what the guy was talking about yesterday, the topological. Yeah, that's the main point of the talk. There's all these topological dualities that you get out. Not, of course, you wasn't connecting with the Eisenhower. No, but this Eisenhower is where it stands out. I mean, this is where Dave and I, we went to Alexandrov and then looked at the duality in Alexandrov, which is much more general. But we had in the, or at least I had in the back of my mind. And I was just saying, you see, if you get a... If you get a choratic lattice, you can then form a dual lattice, and the way you form the dual lattice is just to put points at the center of the inch, and then bisect the So you find that the quadratic lattice is self dual and from that you can predict the critical point of the paramagnet, thinking of this as spin up and spin down, you can actually predict the critical point without any physics at all.

12:30 Well, you put in the partition function, I mean, sorry, but it's a topological argument. Now, the Young-Baxter equation comes out of a star-triangle relation. OK, so we've got this tantalizing stuff, so we can ask you the next time we see it, it won't be long. Now, can I draw these any more? Not very well. No, I can't draw them any more. I used to think I'd draw these trident and a lattice is standing on my head. I'm sorry I can't draw them anymore. OK, we get the idea. No, you don't get the idea yet. I can't draw them. Let me try and make this one. I hate to say it, but it looks like a rabbit wire gone wrong. Yeah, well, I suppose we might have to be in those. I think maybe this will be it. That looks like a solitaire board. It might still be a board, isn't it? Oh, sorry to interrupt, but the other thing that Attire was tying this all in with yesterday,

15:00 these Jones polynomials, was the Churn-Simons? Yes, I don't know how that fits in with this lot. Well, that's one of the things he had, you know, a deal to say about. These are supposed to be triangles, and they're not very well drawn. He's also saying that there's a whole theory of time coming out of that. I'm sorry. We've had enough to like already have any crayons. No, I'm sorry, I didn't want to go off on a tangent. It's supposed to be a triangular lattice. Does it look like one, roughly? Yeah. Now, if you take the dual of the triangular lattice... ...you let the beehive... What's often why you question, why is that the jewel? By definition. You can ask me for an accuracy. That's what they always say when they're caught. Alexandrov introduced this idea that if you have any simplex at all, then you take the vice sector of the simplex, one simplex, this is a two simplex. You take the boundary of the simplex and you bisect the boundary. Right. And then where that meets, you put a point. Oh, OK. And that point, this is the dual of that triangle. Right, OK. And now you just generalize. So you think of these as triangular simplices, they're ordered, but you bisect each one, putting a point in the middle and drawing the vice sentence. And that's what Alexandrov meant by the dual, by the sympholism. This is all basic simplicial topology, right? Yeah, well, dressed up in IC model. Yeah, dressed up here in the physical guise in an IC model, which, as I say, connects with, well, it's a fairly generic model, isn't it? Except you were using it in connection with molecular bonds. I was looking at it as an explanation of ferromagnetism, as second order phase transition. So there's a general theory of symmetry breaking phase transitions. It's not specific to magnetism or something. My thesis was something about the theory of the system mechanics of co-operating systems.

17:30 And they also use it in molecular sciences, don't they? Oh, yeah, you use it for spin glasses, spin networks, you know, a whole lot of spin networks are much more general. We even apply it to the metaprotein-folding problem now, doesn't it? Well, my thesis was about the end-to-end distance between polymers, which of course is... ...related to the polymers, because that's what I was originally interested in, but found it too difficult. Well, considering that sort of 40 years on, it's still pretty intractable, you've made the right decision to bail out of that one. doing your PhD you don't do the most difficult thing you always keep the difficult picture in the background but then try and focus on some simple thing within that picture okay so now let's have a look it's not going to look like there's supposed to be a honeycomb lattice come out of this I'm sorry I'm just can you see a honeycomb lattice coming out of it triangle now the problem here is you see that here's self-duality and the problem came out very easy even in my day this is the triangular is dual to the honeycomb and the honeycomb is dual to the triangular so you don't have self-duality in this case but nevertheless you can actually solve the problem this is where the end of actual equation comes in because you've got If you take one of those, and you take a triangle, so that's a triangle, and you're looking at, it's supposed to, right, this bisects that, this bisects that, that bisects that, and there's the center. So that is where it is upside down, if you see what I mean. Then you have things like this is K1, this is K2, this is K3, and this is L1, this is L2, and this is L3.

20:00 And therefore there is a relationship between the partition function based on the K, on the triangular lattice, and on a bit of the honeycomb. and this is known as the triangle transformation transformation and that is where the Yang-Baxter equation comes from just looking at the duality between these two I don't expect you to see that that's just a marker that's the way we're heading and then from this duality you can actually get the critical point of the honeycomb and the triangle lattice out exactly as well. Okay, so these are, the duality is very powerful in getting critical points where the symmetry breaking takes place. I don't know what that would mean topologically. Okay, so this is the game I played when I was a kiddie. Not being clever, just doing counting. Graham doesn't believe that. I'm still sure your theory was a little more than counting, but wasn't at all, was it? Well, I have read most of it now, actually. It actually was substantially more than counting, yes, I have to say. He's being a bit of, you know, slightly false modesty there. One-dimensional icing. So that means I've got just a, you know me, I love starting with, this doesn't have a critical point in it by the way, it doesn't have a base point in it, it doesn't have a base point in it, it doesn't have a base point in it, it doesn't have a base point in it, and all I've got is that it's either spin up or it's spin down, and I'll take sigma, so that's equal to plus or minus, so if you see pluses and minus coming in, plus it's spin up, minus it's spin down. Okay, so I've got a set of spins, which is I, and so what I do, instead of drawing arrows, I actually say a particular configuration would be something like that. And that's one configuration. And what I've got to do is to work out the energy of that configuration, and then the partition function, because I sum over all these energies to get the partition function.

22:30 So I'm using the partition function is equal to exponential, I think it's minus e over k2 over n, and I sum over n. Okay? So now my partition function in this case, I simply write the sum over sigma, that's sum over n. I'll put that in the middle. Excellent. K sigma j equals 1 to n sigma j sigma j plus 1. So this is the energy of nearest neighbors. So if they're plus, plus, it's one sign, and if they're plus, minus, it's another sign. so that gives me the two energy states so i'm interested in the energy state between plus plus and minus minus they have the same energy and then i have the plus minus minus plus so this would be e1 and this would be e2 okay so that would be my two energies but i have to do it over all the bonds all the sorry all the sides and then there's one other term which comes in here which is not very important as far as the topology is concerned and that is if it's in a magnetic field then there is an interaction energy between the magnetic field and the spin-out because you either get plus or minus so if that wasn't there it would be the paramagnet Ah, okay? If that wasn't there it would be paramount. And if that's, well, I mean, I will mainly put this equal to zero because this would not be interesting. But you need that in order to calculate the susceptibility. Okay, the susceptibility is the magnetization by the H. The ease of magnetizing it, so you need to have the H in there for that. And then put H equal to zero. So, Mike, this is where the... the susceptibility is where the random walks,

25:00 extremely random walks come in. Right, right. It's already on the spirit camping there. Right, yes, and that's... And that was telling me where the critical points were. Right, exactly. Which is like the bra-ha-ha or whatever. Yes, and the bra-ha-ha is what connects up with the face, with the second order... It's got something to do with it, but I'm saying I'm just... But it seems to be this weird structure, the same that you have in the, well, as you say, in the Jones polynomial, but also in this, I hadn't realised the connection with graph theory in the four-colon problem, because I've never heard of a chromatic polynomial, but it seems quite astonishing, it seems astonishing the way this thing keeps cropping up in so many different directions. OK, now what we'll do now is we want to write Zn as equal to sum over sigma v sigma 1 sigma 2, v sigma 2 sigma 3, v sigma 3 sigma 4. In other words, we want these v's as 2 by 2 matrices. OK, now what's V? Where V, sigma sigma prime, is just equal to the exponential of K. Oh, I haven't put K, K by the way, is equal to J over KT. Plus KT, just in case you want and where the temperature is going. OK, I don't know. So those k's are functions of temperature. Sigma sigma prime plus h over 2 sigma plus sigma prime. And the only tricky thing here is the mean value of those spins that you use here. I'm saying, don't worry about that. The main is to keep your eye on this. Okay, because this sum means that they're actually a product of exponential sum. And therefore, you bring that product out by calling these. And then we roll it V sigma. This needs thinking about a little bit.

27:30 is equal to the matrix v plus plus v plus minus v minus plus v minus minus this is because you're summing over all signals so you've got a sum over plus and minus And that then is equal to e to the k plus h, e to the minus k, t to the minus k, e to the k minus h. And of course you can regard this as matrix multiplication because you've got a common signal between the two. It's just matrix multiplication. to multiply maple seeds like this. And then, because you're summing over sigma, your partition fraction could be traced b to the n. V to the end because you've got elephants. And V, by the way, is called the transfer matrix if you want to, if I ever use the word. V is the transfer. Y is called the transfer matrix, today I don't know. What's the origin of that? No, I don't know. I never used it. What's getting transferred? I never used it in my bed. No, but I just think that's not the origin of the... I think it's because you're adding. You can think of it as building up by adding something... Yeah. No, I don't know. Well, I was going to say... I mean, why is that more the case than we're hitting on? Okay, remember, once you've got a partition function, Then there are a standard set of differentiations to get the free energy, the Helmholtz free energy, to give free energy, the entropy. That's all the standard in one of the textbooks like Terrell Williams.

30:00 Okay, so that's what we've got then. And now what we would like to do is to find the argon values, you don't know. Okay, so we want to find v, we want to find a second xj. these are two, xj are two by two column, sorry not two by two, column two two column, because v is two by two so we want to find the eigenvalues and eigenfunctions of v so what we do is we say okay let's write v as equal to p to the minus one sorry p lambda one zero zero lambda two p to the minus one the representation of which the v is diagonal. And then trace vn is equal to trace, because you can bring the trace of the p and the p minus 1 out so you get rid of that, so that's just trace lambda 1, 0, 0, lambda 2 to the nth, which is just equal lambda 1 to the nth plus lambda 2 to the nth. So your eigen, it is very, very easy to do with your eigenvalues. And then if you want the free energy, and you take the largest eigenvalue, And we'll take lambda 1 to be the largest element with lambda 1 greater than lambda 2. Now, why do you want the eigenvalues? Well, the free energy is equal to minus k-teen. The limit of n goes to infinity of 1 over n log. That should be long, so now. Remember the free energy is equal to minus kT times 1 sec. But because we're going to... n goes to infinity to get the limit in the infinite lattice.

32:30 Which is what we have to do in this game. So we do it that way, and therefore n minus 1 set n to log n, log lambda 1 plus n the minus 1, log 1 plus lambda 2, lambda 1 to the n. And then clearly as n goes to infinity, we get three energies, just minus kt9 and pi. And then all you've got to do is to find lambda from that little matrix. You can then find every... Okay, so that's how you find all the thermodynamic quantities. So the secret there is to find the eigenvalues, but I mean, we never solved the problem this way, so that's just to show you how. So this is the most trivial example, no phase transition here. I should be able to tell you why there's no phase transition here. I think there's no similarity in the solution for that equation. So this is the most uninteresting equation. But it just gives you, I hope, the feel of what's going on. But now we've got to go with two dimensions. One dimensions. Okay, can I...? Yeah, sure. That'll be all level. No, I've just got to have a little bit of notation, because the notation was what killed me, gave me that headache three weeks ago. So the notation is that V, the vector V, is equal to V sigma sigma prime plus sigma

35:00 4, and then I wrote that as V plus plus V plus minus V minus, I'm just repeating, I'm Sorry, what I wrote before. Okay? Now, then, the matrix S equals 1, 0, 0, minus 1, which I think we would recognize as sigma Z, is written as S sigma sigma prime equals sigma delta sigma sigma prime. What does that mean? means that I do the following, plus, delta plus, plus, plus delta plus minus, minus delta minus minus, minus delta, oh, hang on, plus, minus, minus. I thought that was the one comfy chair in the place, that's right. and so now you see delta plus plus is just one so that's plus one delta one minus zero delta minus one is zero and that's minus one okay this notation just needs a bit of getting used to yeah it's a bit it's a bit clumsy i'm sure there's a simpler i think there probably is but Mike, on the subject of notation, I will say one thing I agree with Keith about, the notation in that Con Rebelli article is really a horrible area. I think there are two people writing the paper and they're changed. Yeah, so they couldn't get their act together, I think that's a problem for them. I'm doing the icing model, Bob, because there's lots of vain Yaxter equations coming up.

37:30 Something has cropped up. Okay. Now what I want to do is to roll with a two-dimensional model. Okay, this is just notation. So you can come back and... And this notation really did give me a terrible headache. Well, I don't know, I mean, I've looked at Jones's notation and I haven't created it. But this is Baxter's notation. Okay, we want to build up the lattice. so we start with the one dimensional and then what I want to do is one two four and I want to put this is supposed to be symmetrical we just started on this yeah I've already done the one-dimensional. I've just done this one. I'm just building it up now. I think you'll pick it up without it. As you say, the one-dimensional case is virtually trivial, isn't it? It's trivial. There's no placement. Now let's add another point. I've added another point. So this adds a point, I'm sorry we should all be on the same line, and then add an H. age you see what I'm doing is adding another point on this second row okay so that's the idea then we have to grow so we're going two ways we grow along and we grow up someone was a little you know we're actually a chief growing up we just grow along now I want to introduce I'm sorry this you just have to sit back in suspense why

40:00 the hell is he doing this I don't know but you'll come you'll see in a minute I want to introduce I want to introduce a matrix, which is 2n by 2n. I've got n points, and each point has plus and minus. So this is a gigantic matrix. And I write it, and he writes it, sigma i, delta sigma 1, sigma 1 prime, delta sigma 2, sigma 2 prime, right up to delta sigma n, sigma n prime. And this is equivalent to 1 cross 1 cross cross sigma z cross where this is in the i-th position Now, to see that, delta sigma, sigma prime, remember, is equal to the matrix delta plus plus, delta plus minus, delta minus plus, delta minus minus. sorry, this is an aside delta plus plus is equal to 1 delta plus minus is equal to 0 delta minus plus is equal to 0 and delta minus minus is equal to 4

42:30 treating as a delta function so if x is not equal to x prime it's equal to 0 funny notation But this is the notation that the big boys use, and Jones even obscures that notation. Okay, so that's... Can I rub this off now? Can you just keep that on for a second? You can rub your diagram up. No, I just want to come underneath there. No, I can't... Now I want to introduce... Keep the rest of it on there. Yeah. Now I want to introduce... So that's an important... This is an important thing. result. I just put the i under there to show you it's in the i-th position. You can actually read it from here. Maybe you could do that. You see the i's there. It means put the sigma z in the i-th position. Then I want to introduce a ci sigma sigma prime which is equal to delta sigma 1, sigma 1 prime, delta sigma i minus 1, sigma i minus 1 prime, delta sigma i minus sigma i prime, delta sigma i plus 1, sigma i plus 1, etc. So this is a spin flip. And can be written as 1 cross sigma x cross y.

45:00 And then this is in the odd position. So we've got ci, sigma, sigma, that gigantic matrix. Hmm? Sorry, I didn't catch that. C is asking about the sigma. No, C, C is where, S is where that sigma Z, and C is sigma X, otherwise everything else is the same. Definition. Is this on the second line up? There's sigma in that line. Oh, well, OK. Spin flip. And how do you spin flip? You spin flip with the... So you're taking the sigma x to... See, sigma x is equal to 0, 1, 1, 0, right? So let's have a look. Sigma X on 1, 0 equals 0, 1, 1, 0, 1, 0 equals 0, 1. So at that spin up, Sigma X flips it to spin down. That's all it would say. That's why I called it the spin blink. This is where, why Onsaga used clippinality books to solve them, I've seen one before, because these are essentially confined clippinality. And now you can see where the clippinality book comes from. I'm pulling this one out. Sorry. The thing to do is just to concentrate. S has got the sigma Z in the i-th place, C has got the sigma X in the i-th place.

47:30 Forget about that. because we're going to build up and then what you can show is that Si squared equals Si squared equals 1 and remember Remember when you're A cross B times C cross D is equal to A C cross D, yep, so wherever you see the 1, this 1, 1, 1, the only thing you do is let sigma Z squared, that's equal to 1, and here the only thing is sigma X squared, and sigma X squared is equal to 1. So that's almost trivial. No, it is trivial. And then what you find also is S i C i plus C i S i equals 1. and the reason for that is because sigma z, sigma x sigma x, sigma z and you add them together equals so all you're doing is you're taking the Clifford algebra defining I mean it's a trivial defined by sigma z, sigma x and you're transferring it to the hole. The hole, gamut, gamut, gamut, gamut. And you also got that SJSI commutator S I C J commutator equals C I C J commutator and that's quite obvious because once these things are not looking at each other in the same column you just move this one up here

50:00 for the J and this one down here and it's obviously not really trivial Now this at the moment has got absolutely nothing to do with the partition function because we've got no exponentials. But remember if we want to look at the partition function we've got exponential k sigma i sigma i plus 1. So now we want to construct pi k sigma sigma prime is equal to exponential k sigma i sigma i plus 1 delta sigma-1, sigma-1 prime, delta, sigma-n, sigma-n. Remember we product the exponentials together. So this is just one term in that V, V, V, V, V, and we're still only in the row here, we haven't got to the columns yet. it. Now, e to the k sigma i sigma i plus 1 is simply equal to cosh k times 1 plus and they've got sine k into sigma i sigma i plus 1.

52:30 All you want to do here is you just expand this exponential and whenever you see this square or the 4, it's unity. And therefore you pull out the cosine term and the hyperbolic science term just has left. Because this to the cube is the same as this. It's the same as you do the cosine sign. Now this is essentially the gigantic one cross one cross one cross one cross one so that pk in that notation is equal to cosh k times the identity matrix which is one cross one cross one cross one cross one at the end, plus now this is the one that's rather interesting, it's simply one cross cross one cross sigma z in the ith cross sigma z in the i-th plus one, cross one. In other words, all you do here is you put sigma z in the i-th position in the i-th plus one position. But if we just look at that, we'll see it's xz cross, I'm sorry, sigma z cross sigma z. You actually look at it as 1 minus 1 cross 1 minus 1,

55:00 which is equal to 1 minus 1, minus 1, 1. So it's a 4 by 4 matrix. 1 minus 1, minus 1, 1 down the triangle. And you can work that out from here. This sort of thing takes a bit of sorting out because the notation is a bit... OK, have everybody got... I'm sorry this is tedious, but I don't know how else to do it other than to... Because these things are ultimately going to be your Joneses polynomial. Just to keep them... Okay, so what we can do then is we can actually write b i k is equal to exponential k s i s i plus y. Now you see why I have put the s i in, because I now express this. And I don't think I'll go through the details of it. You just expand that out exponentially, and then collect terms together. And you find it turns out once again, when you get square terms of number of terms, it's the same thing. So that operator is a key operator in working out the partition function. Now what we want to do is to put in the partition function which generates a column.

57:30 Now, to generalize, we'll put K for the strength of the force horizontally, and L for the force vertically. And then what we need is a qil sigma sigma prime, and that's going to be written as delta sigma 1 sigma 1 delta sigma i minus 1 sigma prime i minus 1. So in this place we put exponential L sigma i, sigma i prime, delta sigma i plus 1, sigma i plus 1, delta to the n. So all we've done is in the i-th place put this exponential, and that gives us then the energy when we're going up, when we're going up. Now the reason why we've got the same i there is because this is i and this is i prime, and we're doing, the other one was ij, so I probably didn't make that clear. The other one had I and J in it, it's going along that way, this one just has II because we're going up that way. So on the same column, there's a bit of notation there in fitting these two things together. OK, now let's have a look at this thing here. E, the exponential of L sigma i sigma i prime, is just equal to E to the L, E to the minus L, e to the minus l, e to the plus l. That's using the definition of the functions v. And that's equal to e to the l times 1 plus e to the minus l times sigma x.

1:00:00 Obviously, sigma x has not done my orders. And therefore, qi sigma sigma prime equals e to the l times the identity plus e to the minus l times times 1 cross 1 cross sigma x cross 1 cross 1. And so that's equal to e to the l times identity plus e to the minus l times c i. Matrices from which we can construct a partition function for the whole lattice. Okay, so what I've done here is just rearrange the partition function so that it's built up by adding a site in the horizontal direction and adding bonds in the vertical direction so this is just when we put all the other junk in it as well this is just oh that's all that sorry these are just the two components the matrix and just multiply them together and we get the

1:02:30 I'm sorry it's so messy, which is the reason why many of us didn't see what was going on. Now what I want to do, do you remember when we were doing the Tampoli-Lieb algebra, when we were doing the Jones polynomials, we were introducing matrices? And cetera, yeah? Well, I'm nearly there. Now the next step, and I don't think I'll do all the details, is to introduce... E to the 2L star equals... Now remember when I was talking about the dual lattice? I could let... I didn't do it very thoroughly, but I think I can get away with it. Let me call the dual lattice L-star, the force, the interaction between the two spins on the dual lattice L-star, whereas the force on the ordinary lattice is L. Now this is just introduced, really by hindsight, okay, and I cannot motivate it, I just have

1:05:00 to accept it, maybe there is a very easy way of seeing why there is that duality relation. But if you use that duality relation, then after about a page of working, which I'm not sure you're really interested in, is you show that QI, sorry, this should be L, that's the second time I've got K in there, sorry, that's L. I didn't notice that you had L. Yeah, you had L at the bottom now. OK, sorry, I don't know where I've got that. down as well. I think I must have cracked it. Just, just, just. There's enough mystery in this one. No, I had actually written it down as L without even looking at it. More by luck than the other one, I think. Well, no. You actually know what to do. I'm not quite sure. I'm not sure I do. Then you can write this as equal to 2 sin H L all to the That is just fiddling about with cosines, sorenches, tanches, it doesn't look as if it's coming out, but it does come out. Okay, so now what we're going to do is to call this, now can I just, have you got all this down anywhere? Uh, not, oh, hang on, I think that last line. Okay.

1:07:30 So what I've done is just write this in a way which just makes it a little bit neater, a little bit more, so you've got exponential symbols, and the L star is just a conversion thing at the moment, you don't have to tie it up with a duality, but it's just a conversion factor. And then we define simply p j k equals exponential k s i s i plus 1 if i equals to If i is equal to 2j, then we put it equal to 2 sinh 2l, which should be 2 in there, otherwise you're going to say where the hell I am, I just haven't copied it, but there's a 2 in there. 2L minus 1 half QJL equal to exponential L star CI if I is equal to 2J minus 1. so we've really got u is equal to that even this is equal to that i that's all it is and that's where the even and odd comes in, the exponential, that K is... Is that? That's L-C-I. No, but on the upper line, the exponential... Oh, these are S's, sorry. Yeah, no, those are S's, but is that K? K, that's K, that's K. I thought it was, but it's ambiguous, but it looks... No, no, no.

1:10:00 And this is where the U's and the Tendley-Lieb are coming from. and the R's in the young Baxter. Right, yeah, well, of course, they have the R's in the L's in the Templar League, the U's rather than the Templar League. Yes, OK. Can you remind me, in the Templar League, how he introduces the U's and what they are actually meant to be in the Templar League? I don't, I haven't got that. This is, this is pure icing, let's not worry about that. I mean, well, no, it's always interesting to see... It's obviously going to be related to... Yeah, well... I'm sorry, I haven't got all the answers. I mean, it's obviously abstract, you know, structural homology. It's not, um... Since Kemperley-Liebe has obviously nothing to do with the icing model, at least not directly, it's, uh... Well, it is, I'm sure. Yeah. Kemperley-Liebe. Did it? Oh, I'm sorry, I completely misunderstood the story on that. I didn't realise that. Kemperley was an icing model. What? I didn't know that. I didn't know that, I thought it was Peter Schoen, Peter Nath that they were playing around with, because he was on a combinatorial polychaun. Oh, sorry, I don't need to pick that. So the Template Leave was, was, work on, was, you know, derived from the Ising Mall, so that was, that was long before Jones did his work on the Colin Hill. Oh, yeah, yeah, long before. I'd have been lying around for years. Yeah, definitely. He was very active when I was young. Yeah. He was a Peterhouse man, wasn't he? Yeah, he was a good... Yeah, but he was a Peterhouse man, wasn't he? He went to all of us. I don't know when... I think he was a fellow Peterhouse, I'm sure he was. He was a Cambridge man. Yes, he was certainly a Cambridge man. I'm pretty sure he was in my college, pretty sure he was a Peterhouse. I mean, in those days... I think he was a sort of emeritus professor or something. Oxbridge man. Yeah. Well, they're not all rubbish, I mean, one or two reasonably good bright guys. But he went to all of us, he couldn't go down to electric. He was struggling, yes, a lot of very bright people are. I used to have, as a matter of fact, there was a magician at Peter House called Matt Adrian Mathias, who was absolutely open. Yeah, that's exactly how Adrian Matthias was. Well, I have great respect for the man.

1:12:30 In fact, he wasn't even, you know, he just couldn't lecture. Couldn't have a heard from him. Adrian Matthias, through whom I learnt, you know, what little set theory logic over learnt, was like that. And he was even worse, because he not only would stand at the ball, you know, but he would kind of mumble into his shoes. You couldn't even hear what he was saying, let alone, you know. Well, that's exactly what happened to Adrian. Well, they wouldn't give him a job in the department, so he ended up just with a college fellowship. In fact, in here, Michael Redhead found him a job at the History and Crosswood Science Department. Then, of course, he got kicked out of Peep House because he had a big row with the Master, with Dacre, over various things. Yeah, well, I shouldn't pick flags with the master, I should go when she says go to yoga, I should go to yoga. Yeah, yeah, yeah, well. Anyway. Okay, so what we've got, really, is really two simple expressions for the U. That's all I've done. Okay, now, you know, this can be scrapped as well. So there is the u's from which everything follows. So, really, what we've done is simplify by using the s's and the c's. I've just got simple expressions from which we can build up the partition function. but they give us directly the due things and then the start which I'm not going to prove this the start triangle relation gives i plus 1 L2 K2 Ui plus 1 K3 L3 equals Ui K3 L3 Ui plus 1 K2 L2 K2 Ui K3 L3 Ui plus 1 K2 L2 K2 Ui K3 L3

1:15:00 And there is the Banyakster Equation. okay so what to get this you've obviously got to go through and use remember we have the You had it this way, didn't you? You had a star triangle? Which was this was K1, K2. I hope I've got the same notation, L1, L2, L4. So it's the duality between the grinder and the halter. That does want to be a question that one should be, uh, but for me it was rather interesting that simply by building up the lattice in this way to build up the lattice you've got to have the Yang-Bachstrom equation satisfied otherwise you won't get you won't get anything coming out consistently and this is with nearest neighbors so of course i always ask the question is this something particular just the nearest neighbor interactions or what happens if we have two next nearest neighbor interactions i'm sorry i don't have

1:17:30 to keep going through my mind particularly remember the counting was just so now we have the next nearest labour interaction which is what i did i apparently got my ph too with jeff joyce i think it was we did two next nearest neighbours and then talked about long range forces so long range forces actually i was very pleased this guy in Brazil came up to me and said, thank you for writing your paper for your choice. A long way to do your action. The Salis is entropy. Does that connect that with very magnetism? Yeah, it's all the same. It's a very interesting question. All of this is New York Native forces. What happens when the forces start to change the whole range? which could not be interesting, so... I don't think it's like the whole... own... game... that you've got. OK, now the question is... I can see from the triangle... Is that useful? Yeah, that is useful. The most useful thing for me is the little diagram, because seeing that I can see how it connects up relation, you know, connects up with these ideas in, uh, in, in, uh, graph theory, and with the, uh, well, I can get a feel of the chromatic, the chromatic polynomial, how it connects with that, because you can see how it is, it's simplistic, the kind of weightings you assign to these faces and edges in the graph. I would suspect that there are quite a lot of theorems in the simplicity of apology that connects up with too that's what i was going through i was just too excited about the interpretation of quantum mechanics to worry about the real stuff real is absurd yeah no but this is all this always fascinated me these relations that kept coming cropping up in you know to what extent is it something merely icing model It seems to be something much more... And is it merely something which is topological? And then, you know, the topology comes into it because we're doing...

1:20:00 we're interested in configurations and lattices and so on, so we're really dealing with topology. And clearly there are all these deep dualities which connect all of them, and indeed, you know, connect with other stuff as well. Well, shall I try and do why duality... how duality works. I think I've tried this. Are you guys ready for it? Ready for it, yeah. I'm probably going to push up a bit early tonight because of, you know, events day by events, as they say. Half an hour? Oh yeah, yeah, you can get in half an hour. Well, I mean, I don't want to miss anything exciting. I'm not sure whether it's exciting or not. I'm just satisfying my own curiosity. OK, so we now know that, and later on I'll connect it up with the more formal mathematical with Hopf algebra and so on. I thought, I wanted to know where it was coming from, from what I had been used to dealing with, so that we could see exactly what it is. Well, that's always the best way to tackle any news. You're not going to go to the hot fountains with us tonight, are you? No, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, it's just that I would hate to leave just when you were about to. If I, I'm sorry, Bob, if I give you a, um, pound, can I, can you get us a coffee, is that possible? Oh, I'm sorry, I should have started to lose now. It's a bugger, I really wanted to prepare something to talk to you about, about this thing of the tyres yesterday. Yeah, I mean, you know... It was just that today has been a suck-up bloody nightmare. Yeah, modular spaces and Gromov, it's all... Yeah, well he was talking all about modular spaces, of course. He didn't actually mention Gromov very much in passing. Well, that's all Wittgen does in his paper, so obviously they've influenced it. Well, in fact, he had I mean very typical attire. He just said it lost away that some well I need about two years ago. Anyway, this was the state of the subject There were these beautiful realities that nobody really understood I mean, I never really understood how it was going to affect them here and all this

1:22:30 Relationship from the open and closed screens and whether there was any feet I had a conversation with Witten about three years ago in New York about all this, and I'm very, very interested in Shabner. I just threw in a couple of Charles' remarks. Anyway, about a year later, he phoned me back and suddenly was publishing his paper. Would I like to put my name on it? Well, I only contributed to a very modest remark. Well, of course, the modest remark that the tire contributed was, of course, the key, which has led to complete generalisation of this duality, to tide it up with the classification of all these, you know, invariants in Q2 manifolds, and allowed them to get, it is claimed on the physics side, you know, completely consistent M theory, which allegedly bridges the gap between the closed and the open strings. But there's no question at all that the string theorists think that it's a really, you know, just about the deepest result mathematical that they've got in the whole subject, the thing which really doesn't make M-theory rather mathematically respectable, and the topologists think that it's an incredibly beautiful and deep piece of algebraic topology, in other words, it's the only thing you'd expect the type to come up with. that he's still coming up with in the 70s is, you know, impressive, but then we all knew it was, you know, Alphaman. Yeah, yeah. But that's what this talk was about, and I mean, I don't know enough of the algebraic topology to be able to follow the fine detail of the classification of these manifolds, but it's the stock that started with Donaldson's theory, you know, and tying up QFT with these deep results in topology, but it's gone much further and deeper, of course, when it tries to get into new levels of abstraction. I'd love to get... That's where I got, you know, the primitive impotence from in the first one. Of course, the one thing that a tire's not on to at all is the algebraic dimension, the connection with the impotence, because he does think very, very much as an algebraic geometer. I think it's all about the classification of these beautiful structures on menopause, The manifold structures have got to be there to carry all those structures. I was not thinking in terms of the structures in the algebra that we need in focus and their significance. Well, no, he does. Well, you and just something to say, Morris, and other people have been pursuing that line.

1:25:00 That's amazing. No, I don't think so. I think you are on to the really deep, you know, in the really deep waters. which is why I want to get Bill to come and talk to you, which I now saw, it hadn't been called the same vacuum, it clicked while we were out of the room. I could really have murdered, and after I stepped to you on the debate, I finally managed to get home, and we had promised, we had sworn, when I was with him face-to-face last night, he was going to call me about 10 o'clock in the morning without fail. I finally got home with him at about 3 o'clock in the afternoon, Uh, it was quite an hour ago after I spent a year, and, uh, well, when are you guys like, oh, I'm in the... You'll have to start a little later, probably a little later. Thanks a lot, Mark. What's the latest number? Crikey, that's enough to give anybody a bloody high-quick low stream there, have you? I, I feel, Michael, I have, I think there's more in these two relations than I thought out at the moment. Well, I'm sure there is, because of the number of, you know, places, mathematically deep places that they're coming from. I mean, the icing model is fascinating, because obviously this is where you cut your teeth, to see that it comes out of something which is really very concrete. Well, I think of it as very, very kind of concrete physics. The sort of things that people do when they're calculating, let's say, molecular bonds are trying to understand polymers. You'd expect graph theory to be relevant to polymers and protein polymers, but, yeah. I must spend more time just thinking about these two relations here. We're just looking here, S, S actually. This is the one that's got the signal Z. and this is the one that's got the Sigma X because you have to really understand this we've got to step one because I haven't proved this for you and in fact it's a terrible thing it's just a long, weird slide which is why I'm going through it what one was saying there? it's Baxter's block oh it's Baxter's block Well, I thought I had to get it because everybody refers to him.

1:27:30 Exactly cells, models, and statistical mechanics, and that's the kind of great Sam's work, isn't it? This book was conceived as a slim monograph, but I'd hate to see the bubbly, but he thought it was conceived as a big, fat, multi-volume work. There's a lot of detail in there. Unfortunately, with my experience in the model, quickly step past it yeah so I go to place 334 was my first reason so I don't recommend people will actually go through them. So these, some, one of these is a spin flip and the other one is just an argument value in this. Hang on, let's see what this is, this is um, and this is where I, I can't see how Lou gets his diagram down because that's really what I wanted to do but I'm afraid I had to mix that I play with. Okay, shall I? Yeah. I just wanted to... I haven't shown how the graphs come out yet, have I? I haven't shown where the pictures are. All I've done is a horrible lot of... Oh, sorry. In fact, all I've done is I've just shown how sigma z and sigma x fits into a multi-dimensional lattice. And you're not surprised to find sigma z and sigma x in each point. In a what, right? You're not surprised to find sigma x and sigma z in this problem? No, no, no, certainly not. But it's the building up of the thing that's important. And then this relation is really telling you what happens when you're in that.

1:30:00 And now you can see why I said this was a duality relation, because it deals with this dual lattice. Okay, now, let's go back to the partition functions. And the partition function that we have here is Zn is the sum over sigma. Thank you.