Spin Networks & Anyonic Quantum Computing

0:00 Just in the bathroom, Lou. Yeah, well, it was Ian, but I didn't tell you anything about it before I could speak, so don't worry about that. Let's get everything set up here. I just want to... Ah, that's clever. There's no disk in it. Okay. Okay. OK, right. OK, folks, if you want to take your places, you are... It's all fair. It's all fair. It's all fair.

2:30 It's okay to me. Take your time. It's all right. Take your time. Well, welcome to the chair, everybody. Having finally got most of you here in one piece, we'll be joined late in the day, DP, by Morrison Gosselin from Sweden and his wife Charlie, and also by Fred Van Hoistel, who I think is known to some of you, from the University of Adler, who is your library, and then we'll have the whole program to complete. I've got a program here, which I'd just like to distribute to everyone, you can just take one of those, and you carry those around up there. And I'm afraid it's very, very amateurishly produced, because it had to be done last night. It says there that you have the program, a set of abstracts, some spiel about a little thing called the history and philosophy of the contemporary mathematics project, and some information about Fougere. Unfortunately, at this point, we only have the program at the meeting. By the time Lou was finished speaking this morning, I hope we will have the other things to be listed there. I think it doesn't have to be stuck on the brujere for us. Well, we may have to forget the information on brujere, but that's not the end of the world. But as I say, I will have a complete set of abstracts for everybody. And just very quickly previewing what's happening here. We're going to kick off this morning with Lou, giving the first of his talks, which, as you can see, goes on the subject of spin networks and anionic quantum computing. Did I say anionic right? Just before he starts, I'll just run through, just go through the manual very quickly. After coffee break about 11.15, Bantle will be giving the first of three talks, beginning with a general overview of his recent developments and his ideas about non-commodical algebraic structures in physics, particularly in the face space. Then we'll have lunch, which will be fairly sort of light and formal lunch, and I suggest that we take it in the little bar this four on the corner here. And then about 2.30, 2.45, depending on when people drift back from lunch, and how interesting

5:00 in the conversation about during lunch. Dmitry Pavlov, from Moscow, will be talking on the subject of numbers, our number, and the geometry of space-time. And in addition to his abstract, which will be in the set of abstracts I'll give out at the end of Lou's talk, there's also a copy of his paper, which I'll have copied and circulated for everybody. Because he will be speaking in Russian, and Sergei Zuparov behind you will be acting as his translator, We are going to allow rather more time for that talk, and whenever we wind up on that, we might have a general discussion of about 45 minutes or so before we break up to dinner. on the subject of timing of the talks every speaker has an hour and a half we would ask speakers to try and keep 15 to 20 minutes at the end of that spree for discussion and questions unless they've been particularly liberal in allowing questions during the talks in which case they can run a whole hour and a half if they want to we're fairly flexible now are there any questions before we kick off? do you have a cancer? Yes, they are. They are pens. Black and blue. And if you want me to move it over now, I will. No, I'm going to use the slide. Okay, well in that case, without more ado, I will introduce you to Lou Kauffman, who I think knows the very full introduction, and knows the most of the people here. Lou's work, it says well known to everybody, with an interest in the foundations of physics, and indeed the foundations of mathematics. He is often referred to as the only research mathematician who carries around a noose in his pocket. He proved last night. I thought it was odd. But I hope he's not getting into the deal as a practical demonstration on that. His work as a non-theorist is world-renowned, and he's interested in foundations of physics, and particularly those aspects of his research which bear on the quantum gravity research programme. also well known to most of you. And he's going to talk about some very exciting new ideas. His talk tomorrow, we held over his main talk well, we held over, as it were, the more general and perhaps the more philosophically exciting of his two talks until tomorrow so that we could ensure Morris and Freddie would get hearing. So this morning I'm going to ask him to kick off by talking to us about spin networks and analytic quantum computing. Thank you very much, Lewis.

7:30 That should be perfect in position. It's even focused. Except we can't, we try it, we don't get any back on the ball. Now that's fine. So, I'm going to talk about this type of braving spin. I'm going to begin by talking about braving. And then about gradient operators, which are linear maps, which represent gradient, and then how you might think about them in a quantum context. And then we'll get, finally, to looking at a model for any ontological quantum computation that comes out of this in a nice way. the idea of that isn't actually physically realized. One would like to have particles which when they move around one another will cause a change which is more than just plus or minus one and that would actually correspond to gradient but such things don't necessarily exist although they might exist in, for example, the fractional quantum all of them, things like that. It's so wrong I just want to remind you about braiding to begin with, but these are out-entry braids. A braid is an object like this where you have strands which start from some set of points and continue without backing up and end up with some other set of points and wind around on another. These aren't such interesting braids. There are many more interesting braids than that, for example. This is a long plate, which you can continue as far as you like. And braids are a group because if you have two braids, you can tap one onto the other from the top of the bottom and get a new one. And if you take any braid and separate it in a nice way, then you can see that it's a product of the generator. So, for example, this one is sigma-1, sigma-2 inverse, sigma-1, sigma-2 inverse, sigma-1, sigma-2 inverse.

10:00 And Arden, back in the beginning of the 20th century, by the 1920s, I think, heard it's defining the brain group. The basic relationship in the brain group is exactly this. Sigma-1, sigma-2, sigma-1 is the same as sigma-2, sigma-1, sigma-2, which you can see. I'm going slowly at the beginning just in case you haven't thought about this in a while. If you see, you can just move that strand across the top and push this one up a little bit. So you get that algebraic relation. The other relationship that's true in a break group is that these, if they're sufficiently far apart, commute with one another. So sigma one and sigma three commute with one another. And so that's an algebraic description of the break group, that the sigma i's satisfy this relationship, if i is next to i plus one, then you have a relationship of this kind, and sigma i, sigma j, will commute with one another if they're sufficiently far apart. So that's a very simple description algebraically. And one can look for representations of the break group. and one way to do that is to look for an operator from a vector space tensored with itself to a vector space tensored with itself so when you think of each one of these lines as corresponding to some vector space and this is a mapping from B tensor B to B tensor B which is more of that permutation and it should satisfy this equation if you turn this into an equation but what would it mean to turn that into an equation? Let's ignore the top of this slide for a moment, which is a comment I want to make afterwards. Look at the bottom part. If you have this, and there was a mapping from V tensor V to V tensor V, and each line corresponds to vector space V, then this is another V, and so this should be a mapping from V tensor V tensor V tensor V to V triple tensor V, and it does R on the first two tensor factors, and it does the identity on the last factor, so this can be encoded as R tensor I. The next one is I tensor R, and the next one is R tensor I.

12:30 So I can write this gadget here as R tensor I, I tensor R, R tensor I. This one is R tensor R, R tensor I, I tensor I. And so if you wanted your matrix that you would set up from B tensor B to B tensor B to satisfy the gradient relation, then it should satisfy this relation among the tensor problems of those matrices. It's called the Ann Baxter equation because it arose in physics originally and then was taken up by topologists. And in physics it arose for Baxter because he was using the R as the vertex weights for a statistical mechanics model. He wasn't thinking of it as necessarily having to do with topological braiding. But this relationship turned out to be very useful for Baxter for thinking about computing the eigenvalues of the model. On the other hand, Yang was thinking about particles. And I don't want to be thinking about particles, so I want to say a word about how Yang was thinking. Yang was thinking, oh well, I was going down, but I guess I'll go up because I'd like tying people up in my graph. So you just imagine there are three particles here, and they're interacting with one another. and then this one interacted with that one, and then these two interacted and ended up up here. And the interactions could be conceivably indicated by them going around one another if you wanted a two-dimensional model where the base is two-dimensional and time is one-dimensional, and then you imagine the particles as moving around one another and something happens. So you can imagine that this actually is a picture of some particles interacting. But in Yang's original model, it was one dimension this way and one dimension that way. He wasn't actually thinking about braiding. But what he wanted was that in this one plus one dimensional physical theory, it shouldn't matter what the probability is for getting from here to here, and what the amplitude is, except depending on the permutation of the lines. from the probability amplitude where these three particles interact in this pattern and end up here, and then also the probability amplitude here, the amplitude should be the same, depending only on the permutation, that one ends up at 3, 2, 2, and 3 at 1.

15:00 And not on the specific way in which things bump into one another. And so we worked on that equation in this simplified 1 plus 1 dimensional model and thought about it. of the same equation. But these metaphors of thinking of the braid as something like particles interacting are quite useful. And so that's this equation that came up in physics. On the one hand, it is, of course, obviously as some use of topologists comes to the idea of finding solutions to this equation because you can make representations of the braid group out of it and you can also make non-invariants out. So there are two contexts here. One is that braids and representations of the braids, and then the other context, which is really, in a way, a more recent context, is to just say, well, I would like the R to be unitary as well, because if R is unitary, then I could just think of it as a quantum transformation, because in quantum theory, a quantum process is a unitary transformation. And at the most general level, all that happens in quantum theory is that some vector is being rotated by a unitary transformation in that measure. Again, again, again. So it would be interesting to know of some unitary R's and see whether you can use them. And then I use the term quantum gate because I'm thinking about quantum computing where this would be a gate in a quantum computer, which also would be just a unitary transformation. So let's say a couple more words about that. So a quantum gate is just a unitary transformation. Usually from B to the other tensor power to itself, where B is a two-dimensional complex space. That's for the context of quantum information theory because one thinks of generalizing a bit in a computer to a qubit, which is an element in a two-dimensional vector space, a linear combination of a bit called zero and another vector called one. So when you measure in V, you end up with either zero or one with probability, the absolute square coefficient of either zero or one. Vectors, by the way, are represented by unit vectors, of course, so that the sum of the

17:30 probabilities, the amplitude is equal to 1, the absolute value of squares, the probabilities themselves add up to 1. When I write vectors, sometimes in this talk, I will usually not bother to normalize them, so they aren't written as length 1 or as close to 12. I was taking in length. And a quantum computer as such, as an imaginary entity, is actually just a unitary transformation from B to the end of tensor power to itself, which is equipped with preparation and measurements. So it's assumed that certain states can be given to that transformation and that measurements can be made with respect to some basis, perhaps the standard tensor basis of B to the end of tensor power. And remember what measurement means. that one element in the basis occurs after the measurement and the probability is the square of the absolute square of the coefficient of that element after you have spent lives. So you can start your computer in the state 1011, apply you, and then try to measure 0110. But trying to measure 0110 is a long process because you have to keep measuring and you don't necessarily get 0, 1, 1, 0, but sometimes you get 0, 1, 1, 0, and then you check the frequency with which you get 0, 1, 1, 0, and that's going to be the absolute square of this inner product. You knew you, but you don't know you, so you're finding out about you. So all that a computer actually does is it gives you an experimental way to find out about the entries of this U. But of course you're trying to design the U to do something for you. And the problem on the mathematical side is to design the U so that they do something interesting. And the problem on the physical side, which is much harder, is to make a device which actually will do this for lack. So that's the quantum computer. And so when I speak the quantum computer I'm actually just talking about the structure of the unitary transformation And then there's a term which is important, namely a local unitary transformation is just one of those maps of rotations of the qubits themselves, elements of U of two. And then it turns out, and this is actually a result about the structure of unitary transformations, it turns out that a single entangling, and I'll say what I mean by that, two qubit gate,

20:00 that is a mapping of B tensor B to B tensor B. Plus, local unitary transformations is universal for quantum computation. That is to say, any unitary transformation can be approximated by compositions of such things. So we say that a given gate of G from B tensor B to itself is universal, if that's so. So one way of thinking about how to make a quantum computer in the abstract is to look for an element of mapping from V tensor V to itself, which has this nice property of V entanglement, which I'm about to tell you what it is. And an example, entanglement occurs at the bottom of this slide, but an example of a universal gate is the gate C0, which is a very nice analogy to 0 in ordinary computing. is, it leaves the second two qubits alone if the first one is zero. And if the first one is one, it actually does not change in zero to one, one, zero. So it's a control of non-gate. You could make it as a digital gate, but here we're working with the linear combinations of those states. So it's a unitary transformation. And it's an example of an entangling gate. We'll talk about that in a moment. So let's talk about entangling. A gate is said to be entangling if when you apply it to some vector, it gives you an indecomposable element in v tensor v, somebody which is not in the form of something tensor something else. And of course you want to start with somebody that's decomposable. So we'll take somebody that's decomposed and entangling. And the results due to the Brzevinskys, is that if the gate is entangled, then it's universal for quantum computation. So if you're looking around at 4x4 matrices and you want to know that you can add that to elements of U of 2 to generate all of U of N, or U of 2 to the N, then you can just check whether it's entangled and that'll do it. So let's see what entangling means. Well, entangling, as I said, means indecomposable. And here's the basic fact, which is easy to verify. That if you have a two-cubit state, which of course I can represent as a little two-by-two matrix

22:30 by thinking of the coefficients as a00, a01, a10, and a11, then the determinant of this little matrix should be non-zero. That's what it means for that to be entangled. It's quite easy to see that it's entangled exactly like that happens. You just multiply some things together and see what happens. But then what does C0 do? If C0 was applied to a state with coefficients A, B, C, D, you end up with this state, A, B on 0, 1, D on 1, 0, and C on 1, 1 because of the flip. So instead of being the determinant A, B minus B, C, which would be the determinant of that, the determinant you get here is A, C minus B, that'll make it split around a little bit. how C0 will manage to entangle somebody. Here's an example. Let's take 0 plus 1, 0 minus 1. So that multiplying it out is 0, 0 minus minus 0, 1 plus 1, 0 and minus 1, 1. So if you apply C0 to that you have 0, 0 minus 0, 1 then 1, 1 And then the determinant is the product of the coefficients of 0, 0, 1, 1, which is 1, minus, oh, I made a mistake. Sorry. No, I did. No, wait. Yeah, I'm sorry. I made a mistake. So that's not an example. I'm sorry. Excuse me. I was just getting up this morning. So I'll give you an exercise to find an example of a state which is in town. It looks wrong to me in my heart city. It looks wrong to me in my heart city. No, because the determinant is this times this, which is one, minus this times this. One times one is ten, minus one. So I leave it as an exercise because you can write down a state which does in town. So, the digression here is about entanglement. Here's a simpler example of an entangled state. And Einstein would also get Rosen worried about mentioning the properties of entangled states because of nonlocality. That is, the left and the right cubits of this state could be in different places.

25:00 And when you measure, well, one person could measure the right qubit and the other person could measure the left qubit. So if somebody in London measures the left qubit, the state goes into a definite, telling you definitely that it's either going to be a one or a zero in the right qubit, which is often in the other location. So one can worry about that. They worry about that. If you want to slow down and stop and solve the exercise, we could, but I think I'll just do it. Here are some more examples of universal gates that are also solutions to the Ann-Baxter equation and are unitary, okay? So that was my crossword puzzle, remember? My crossword puzzle was I would like some solutions to the Ann-Baxter equation which solutions to the index equation. Also, I would like them to be entangling so that they could be universal gates. Here are some simple examples. This one is very easy to see that if you apply it to say 0, 0 plus 0, 1 plus 1, 0 plus 1, 1, it will be entangled. Because what will happen here is that the last one will switch its sign and then the determinant won't be zero more. And this is a nice gate. This is switching, just switching lines, permuting, and then one little bit of phase change. Our thoughtful solution to the antiraptic question, here's another family where A and B are elements of the complex numbers unit circle and A squared is not equal to B squared. And that will be entanglement. That's the condition of the determinant, the non-zero. And that one turns out to be interesting from a topologist's point of view, because you can use it to calculate linking numbers of links. And you will find, if you do that, that you calculate linking numbers of links non-tribulately with this one, exactly when it's entangled. So entanglement, in this case, entanglement entanglement on the topological side in the sense of linking are related to one another. This leads to a metaphorical question which is quite an intriguing question. What's the relationship between

27:30 bottom entanglement and topological entanglement if there is any? And of course the reason it's an intriguing question is because when you talk about entanglement in the sense of Einstein but also in Rosen, you're putting in space in some way to a formalism that doesn't know anything about space to begin with. And so putting in space is of course putting in topology. So already topology and non-locality are certainly interrelated. But whether it's related to things like linking and that is another question. And so this is one way of playing with that question, is looking at certain operators which are both, like this one, which are both topological on the one hand and quantum on the other. how those questions relate on the level of those operators. Here's another example. This is called the Bell Basis Change Matrix. If you think about what this one does, it's quite clear that it's making a lot of entangled states. Because what does it do to zero? It takes zero to zero, I mean zero, zero. It takes zero, zero to zero, zero minus one, one. And that's certainly entangled. It takes 0, 1 to 0, 1 plus 1, 0. And in each case it takes something to something that's entangled and that new basis of entangled states turns out to be very useful for various things. And it's called Bell basis. And this is a solution to the Ann-Baxter equation as well. And Sam Ramonicko and I this talk is trying to work with Sam Ramonicko We discovered that this one can detect linking when you find out what kind of invariance of knots and links it can do. It can even detect linking of things as subtle as Boromian rings, which, if you're familiar with them or not, are pictured here, and they're a very nice example of a link where linking numbers is not enough. You see, if you remove the red component, the green and blue components fly apart. If you remove any component, the other components fly apart. And yet they're linked. And a person without thinking without, who didn't necessarily know any topology, that's why they're linked, would probably end up describing how they're linked in the following way, I think. Red lies over green. Green lies over blue. The blue lies over red. So it comes

30:00 around in the cycle and ends up linking. But that's a higher order kind of linking, that triplicate relationship and in order to prove that their NPR could do some work one way or another. And this matrix actually sees them. You can find out more about that in this paper. So where are we at this stage? Braining operators as universal gates, there are lots of them, I gave you a few, but they still need arbitrary elements of U of 2 for the local unitary transformations, so making your quantum computer topological by putting in one of my R's is nice because the R is very specific, there it is, and it's topological. On the other hand, the elements in U of 2, well, I haven't made that topological. You could say, well, they're geometrical. They're these rotations. But the practical question that's in back of this, nobody really knows. But there's a speculation that maybe if you made your design for the quantum computer completely topological, it would be more stable under perturbation. Because after all, topological things are stable under perturbation. thought experiment or question about the structure of unitary transformations to wonder whether you could make all the unitary transformations you needed out of one topological scheme and then somehow find the realization of that scheme. I don't know about the realization but I want to tell you about how single topological schemes can actually do this. But they do it in a much more complicated way. And that's one reason why I started with simple operating operators. And so this suggestion was carried out by Friedman, Gutierrez, Carson, Wang, collaborators, who suggested using typological quantum field theory, followed as a theory of anion, followed as a theory of particles, perhaps moving around in a planar situation to make deeper models. So what I'm going to do now is I'm going to show you a a way of making one of their models that I discovered over the last month or so that can be done using basically very elementary means having to do with using the Jones Polygonium. And the ideas coming from Penrose's spin networks.

32:30 If you look at their papers, you'll find that they are basically using this, look closely at their papers, they're using the same kind of background, but the papers look a lot more technical. This starts from the beginning. And so I think it's interesting to think about it this way. So what I want to do for the next few minutes is just talk about what topological quantum field theory is and then we'll see how these models work. So TQIP began in the hands of Witt and Matthea as a way to conceptualize invariance that arose from Gaite's theoretic functional interval, three manifolds and the idea is that you're going to be looking at cohortisms of manifolds. You're going to be looking at a manifold, say some dimension, I haven't specified it yet, and another manifold W whose boundary is a couple of these manifolds. So the boundary of W is n prime union n. And then the idea is that associated to a cobordism of manacles, there should be a mapping of a vector space associated with one boundary part to the vector space associated with another boundary part. And if there was no boundary part, that's another possibility. Suppose you had something like this. M, W, and the boundary of W is equal to m. Then what you expect is that from the scalars there will be a mapping from the scalars to the vector space associated with m. So that's the case where one boundary is actually. Of course that's all consistent, right? If you have v of m going to v of m prime, then that's the same as saying that there's an element from the tensor product of v of m and the dual v to the complex numbers. So either the complex numbers to, or from this to the complex numbers, if things are equivalent to their duals, then you get all those things at once. If I have a map from the complex numbers to the vector space, then dualizing I have a map from the vector space to the complex numbers. So this is a generalization, if you like, of a capital.

35:00 When you have a vector, what's a vector? complex numbers to the vector space, right? Because if you have a vector in your vector space and you take one in the complex numbers, then one goes to some specific vector v, right? And that determines that mapping from the complex numbers because if you multiply by some complex number, then it goes to z times v. So a vector is just a mapping from the scalars to the vector space. And a co-vector is a mapping from the vector space to the So what one is thinking about is having a puncture from cobordisms to vector spaces like this in such a way that things are topologically invariant. In particular, the dimension in which people were thinking about this, and I'll explain in a moment with the next slide, was three-manifolds, where a three-manifold, like the three-dimensional sphere, can be thought of as a union of two three-manifolds of boundary along some surface, cover along a surface. And then that surface is the thing that had the cobordism. You see there's somebody, this three-manacole has boundary this surface, and so there should be a vector in a vector space associated with this surface from the left. Call it this. And there should be a vector in the vector space corresponding to the right. And you take their inner product, or you turn one into a co-vector and develop it. And that's supposed to be the invariant. Where does that come from? I said from functional integrals. So this slide got written earlier and the vector space turns into H here otherwise it's the same slide, right? What I call B on the other slide is called H on this slide, alright? So it came from this. You see, Witten said that you ought to write down a certain integral related to fields on the 3-manifold and that that would be the invariant of the 3-manifold. and then if you restricted his integral to half then it becomes a function you see, you choose this T-maniple and then you're able to plug in the other one and evaluate and then he said many things in this fundamental paper of his back in the late 80s among which he told you how to write down a vector space associated with the surface having to do with functional field theory, this is all things a way to associate a vector space to the surface

37:30 and to reformulate this theory about this functional integral in the kind of framework that I was talking about. And he and Atiyah pioneered the way to formulate this in terms of cobordisms in this way. So that idea and that way of thinking about things is called topological quantum field theory. Now, what does that have to do with anions or little particles moving around? Well, surfaces can be decomposed into pairs of pants. That's a pair of pants. And here's a surface decomposed into pairs of pants. By which I mean that if you were a couple on all these circles, it would fall into three pairs of pants, right? Now, a pair of pants, schematically, is just a triumphant vertex. And so underlying the structure, and then it turns out that underlying the structure of the vector space that you're going to associate to this surface is a theory of little vector spaces that are associated to these pairs of pants, to these vertices. I'm just being sketchy about this. But the vector space, if you have, so it turns out that if you have an appropriate theory of particle interactions with fusion and creation vertices like this, and appropriate properties, then you can make the vector space corresponding to the surface. All right? And so that's how some kind of imaginary theory of particle interactions turns out to be related to what we're calling topological prompt field theory in the sense of its beginnings. So that's how it's done in three dimensions. But that's the only really successful example of a topological quantum field, I should say. People have tried to do the same thing in dimension four, and it's very hard. No one has anything that's really working very well. The algebra becomes more complicated. Here, the cognitoris finally falls down into something about knots and links and particle interactions like this. It's concrete enough and simple enough to actually have. And what kind of relationships do you need?

40:00 Well, the sort of relationships that you need are that you need some compatibility conditions. There should be braiding because we were really starting with a topological situation and actually the original invariant, in the original invariant you have the possibility of points moving around on the surface forming braids or links in the manifold. So there were already knots and links already implicit in the original situation. So you want these vertices to grade. And there should be some grading operator like this. And that should satisfy something like the end factor equation. So that's one of the conditions we need. Then another thing that you need in order to make things work is change of basis in the vector space. You see, if I... Maybe I should draw a picture here so that you see what I'm talking about. If you're looking at the diagram on the left, then you're thinking of the circle having been drawn here. If you're looking at the diagram on the right, then you're thinking of the circle having been drawn here. two ways of decomposing that basic four-fold sphere into trinians into, excuse me, into pairs of pandas. Now, that means that you're going to walk up to this surface and decompose it into pairs of pandas, and then out of that you will construct, by writing down these graphs, you will construct this vector space. But suppose you did it a different way, then you want the vector spaces to be isomorphic to one another. That means that you're going to re-expression of the basis, of a basis of things that's written this way in terms of these. And so you need a transformation like this. A recoupling formula, so-called, that takes you from here to here. So you need a recoupling formula, and then you need some other relations about these formulas. Now all of this formalism actually goes back to the theory of angular momentum. So for people who are, if you think back to your experience if you have some with learning about the angular momentum recoupling kind of theory then you know that there is exactly this sort of thing

42:30 if you have particles with spin this would be spin a halves if it was by angular momentum a halves, b halves it can produce some spin c halves and then there are relationships like this 6j coefficients and tabulations of things like this diagrammatic theories of this angular momentum recoupling. Well, angular momentum recoupling theories indeed have some extra relations, and those are the kind of relations that are needed here. So I'll show you what the relations are. Of course, we're doing some topologies, so there are some topological relations as well. But you're not putting in a rotational group here, are you? Well, eventually I'm going to be talking about something that's related to this too. Now, you thought that would be the But here I'm talking very generally, I'm talking very general, so there could be 6J symbols There could be 6J symbols for something else. Okay, sure. So you need braiding, you need the R to satisfy a master type of relationship. You need naturality, that's what I wrote. Naturality meaning that the braiding in relation to the fusing or creating should be natural. so if Yang had continued to re-generalize his theory he would have said this if in his theory he had had particle interactions like that then he would have said this too then there's the pentagon the pentagon is a fundamental bit of combinatorics if you have trees and you're recoupling them I was making dots for those vertices so yeah so you see From here to here, what you've done is you've done one recoupling along the black edge. It takes some getting used to, actually you should draw, but in order to be able to look at this and see, oh yeah, I did a recoupling here, but that's what happened. In general, this recoupling that I'm doing, let's go back to it for a moment, just so you get used to thinking about it. if you have something like this and you fix on one of these vertices and you decide I think I'll move it as though I can pick it up and move it and I'll stick it over there instead

45:00 and then you look at the common torts of the resulting thing you see there it's exactly what that other one is on the right. So, recoupling combinatorially in terms of the graphs means that you want to plug the vertex and then you slide it along the graph and then plug it in after you move it across some other vertex. That way, that's my way of looking at it when I'm trying to see what's a recoupling when it's in some funny angle. So, understanding that, let's go back to this picture. then you unplug this one and slid it over here and then I did it again and I end up here on the other hand I can go around the other way I can do a recoupling here and now I can unplug this one and slide it over there and now I can unplug this one and slide it and I get back to the same place two this way and three that way now that means that if we have a formula get a formula involving 6j symbols like this one, then that would mean that I would apply the formula, I would apply it again, or I would apply it three times in the other direction, and I should get the same result. That's what the identity means. Now, of course, there's no reason why you should get the same result in general for some transformation of this kind, but it turns out that in angular momentum recoupling, you do, and that's called the pentagon relation. And so it turns out that that's what you want to happen in order for combattability to work out in this vector space theory. So that's one of the things you need. You need the recoupling formulas and the theory should satisfy the pentagon relation. And then I also need to satisfy the hexagon relation, which is an interrelationship between naturality and recoupling of the brain. It's just one of the simplest things that could go on. So you start here, and you recouple over to here, and then you braid, because braiding occurs at a vertex, you see. The braiding transformation occurs at a vertex. And then I recouple again, and then I can do a braiding, and then I can do a recoupling, and then I can do abrading. And when I get all the way over to here, I'm back to topologically being the same as the other one by naturality. So that's the hexagonal relation. And now, of course,

47:30 this is a whole bunch of different things you could like to have happen. You'd like to understand why they would occur, right? If I gave you a whole list of axioms like this and then claimed that something satisfied it, you would have a whole lot of verification. So the question is, are there some natural models for something like this? And there about. And the way I get to a natural model, the one that I like a lot, and that something Ms. Lins and I wrote a book about, is by starting with ideas that come from Penrose's spin networks and then adding the topology to it. And then you get things that are very naturally modeled So that's what I'm going to talk about. Next. So I'm going to tell you about Henrose's spin networks. Now, I don't remember when I started and when I'm supposed to finish. So we should have got about 15 minutes to go out from my first step to one hour. So we'll We'll do a half an hour or more, I think that'll be just fine, and then we'll open it up to 12. So this is how Hanrow started. He's thinking about tensors for SL2C. And one way to define SL2C is through the epsilon. The epsilon is this little bit. the epsilon is this little matrix here which is 1 for epsilon 0 1 minus 1 for epsilon 1 0 and 0 otherwise and if you take any matrix A 2 by 2 matrix and you form an A epsilon A transpose you'll get the determinant of Um, I'm sorry, don't write the slides and it just went up in the morning. Um, right. 32 are two matrices. If you form A epsilon A transpose, you get the determinant of A times the identity matrix. But this is one way of thinking about the nature of the epsilon.

50:00 It's somebody which does it. Can you still do it? Right? It's times epsilon. The epsilon is invariant, the complication by A. That defines epsilon, too. And so you can think of SL2C as being basically the epsilon in some sense. And he was making these networks in relation to the algebra of the epsilons. And he wanted diagrams for the epsilons, like this. But the epsilon satisfies a very nice identity, which is this, where these are chronic regalas. values must be equal to that. So the identity of the epsilons is this, in indices, right? These are other indices and these are lower indices. And he's making tensor diagrams out of this, and I'll show you how he builds the rest of it. I've just been sure how the networks are built. But the networks represent the representation theory for SL2C or for SU2, depending on which way you want to think about it. He set it up initially so that his diagrams would be topologically invariant in the plane. And diagrams made out of these are not topologically invariant in the plane because if you were, for example, to cross these two lines, it would change sign. So if you think of this as a little curl when you do that, it changes sign. So you have to keep track of things like this, that this is equal to minus this. So he wanted to eliminate that kind of behavior. So he adjusted his numbers a little bit in order to do so. And made up a very elegant-looking calculus, basically by re-normalizing the epsilons like this, both by each one by root minus one from a cup and a cup, and changing the sign on the cross into minus. So then the epsilon identity becomes minus, that's root minus one squared, equals this plus this because this has the minus embedded in it. And you could write it this way that cup cap plus parallel lines plus crossing equals zero. And the value of the loop, the value of the closed loop,

52:30 that is the contraction of the epsilons top and bottom. See, square minus one squared. That's the value of the closed loop. Right? where you contract, where that means you contract these epsilons with each other if you sum over A and B. And you contract. That's a value of close loop. But of course this is 2, so you get minus 2. And he wrote a very nice paper about all this kind of comment where it's called On Applications of Negative Dimensional Tensors. This is a precursor to bomb groups and things like that where lots of strange evaluations of dimension occur. These are negative dimensional in that sense. And this is the bottom of his Biner calculus. And his Biner calculus has the following property, that if you have a little curl in it, it'll be not seen. If you have two things that cross over one another like that, it's as though you could pull them apart. We're at a flat right, and that's just three-move works. So it's topologically in a plane. didn't do, this was way back in the late 1960s, or early 1960s maybe I think, when he originally did it. He has planar topological invariance of his diagrammatic tensors. He didn't try to make non-theorality. It just wasn't in his context to try to make non-theorality. It would be interesting if he had encountered the notion of making non-theorality that he could have invented some things that didn't get invented until later. But that theory, I want to describe how that theory then ends up giving you vertices and everything you need before generalizing it. So the next thing that happens is this, that you define an anti-symmetrizer or a projector in the theory, and this is just the sum of all permutations written diagrammatically, so like that's a diagrammatic switch, right? And two lines this way and one that way would be another Just read the permutations as diagrams, and you sum over the signs of the permutations that I lay in factorial. Then you get a gadget, which if you multiply it by itself, gives itself back as a projector. And for example, if you tie two lines together, that's if you make these indices the same, then it's zero.

55:00 So this is a nice bit of algebra that is important if you're doing this kind of tensor calculus. and he's writing it diagrammatically, and remember he has the binary identity, so that means he can do tricks like this. Here's the two symmetrizer. 1 over 2 factorial times the identity minus the switch, but the switch net minus the switch is cup cap plus parallel lines, right? And so adding this up, you have two parallel lines so the dividing of 2 factorial, that's just one identity, plus one half of the cup cap. So then you Suppose that I closed it at the bottom. Then you get a loop. The loop has value minus two. Multiplied by one half is minus one. And then it comes out zero. So the anti-symmetrization property is now coming out of the binary calculus in this situation. And he makes larger networks out of this. by the following beautiful definition of freedom, which is the key. So you see, there are an A-symmetrizer and a B-symmetrizer that are there to interact with a C-symmetrizer. Or a particle of spin A-amps interacting with a particle of spin B-amps interacting with a particle of spin C-amps. And how can that happen? Well, combinatorially, the lines coming in A have to divide to I and J, and B has to divide it into J and K, and C has to divide it into I and K, so it all matches up. I plus J is equal to A, and so on. And then you can make this. That's by definition. So the three vertex is, by definition, this rather complicated thing, which can be expanded out. And it only exists, and it exists uniquely when it does, when you can solve these equations integrally. And if you think about that a little bit, you see it means that a plus b minus c is greater than or equal to zero, that a plus b is compared to c mod 2. Just add the muffin, you'll see that has to work because you get 2j, 2i, 2k, right? A plus b plus c. And all cyclic permutations of those equations. Now, those are the well-known conditions for angular momentum recombination. You might probably remember that. And this is a combinatorial model for it. So what we're getting out of this is a combinatorial model for these little particle interactions at this level.

57:30 If you want to interact two of these particles to get a third, then they have to match up combinatorially this way. And the underlying effort evaluations tell you everything about the underlying mathematics that you want to calculate, the covenant pool issues, and so on. So that's how the Penrose-Spin-Everd Theory works if you wanted to do just standard angular momentum recombination. What did he do with it? Well, his idea was that this was a way of starting from pure combinatorics without really invoking SU2 because it's just the combinatorics of these networks and it's very natural. And then once you think of abstract networks, abstract graphs made out of just these trigonal vertices labeled, think of a large graph of this sort so it's kind of like a huge graph a lot of little extra n's coming out of it and wonder what kind of physics does that graph have and the graph has some physics in the sense that you can take two edges in the graph and interact them by putting in two new vertices, let an interaction happen and measure an amplitude for that going on, and the amplitude is measured by evaluating the networks in a certain way and he finds that when he does this and the network is large and repeatable basically repeatable. It needs to be large in order to be repeatable. Repeatable means that the calculations you get for an interaction between these two spines sticking out of it are the same after you do it again. Well then you get angles between these that come out of the amplitudes that you calculate that are compatible with angles in three dimensional space. So a large repeatable network looks like a collection of directions in three dimensional space. So he's hoping that he could reconstruct space and time by using the networks, by just starting with abstract networks which don't have anything to do with space and time except having to derive from certain rules having to do with rotations, of course, and what he ended up getting was directions but not distances, and in order to get distances, you have to to do something more complex. And people these days work on generalizations of spin networks of various sorts, but they tend to more or less ad hoc put the distances in one way or another. There isn't a perfect generalization of this

1:00:00 that's doing the method part. So it doesn't end up giving you quantum general relativity in some instantaneous way. It just doesn't, but one could still hope. In any case, it's a very elegant theory, and it does have a very nice generalization, which fits directly into things topological, and that's what I'm going to use. And that generalization goes like this. There's a way of writing down a invariant of nons and links, which I call the bracket state summation model, the bracket model. And the bracket model works like this. Here's a knot, K, and here's one crossing in K, and here's two associated diagrams obtained by removing the crossing and smoothing it this way or smoothing it that way. In the interest of time, I won't go into this in any more detail than just at the level of showing you the formula, but you can imagine a knot. In fact, maybe it would be good to imagine a knot and think about what I just said, and I won't do more. But here's a crossing. And you see, if I eliminate it one way, I get this length. And if I eliminate it the other way, I get this unknotted curve with some curls in it. So this is the one that corresponds to A inverse in this formula. And this is the one that corresponds to the A formula for that crossing. Okay? And so in terms of talking about knots and lengths, value of the polynomial that's associated with this is the sum of A times the polynomial associated with that plus A inverse times the polynomial associated with this. And then if there's an extra loop, you multiply by delta, where delta is this. And this procedure assigns a round polynomial to an R of a big K, that is the polynomial in A and A inverse, which is invariant under the topological second-Granmeister move and third-Granmeister move. This is a combinatorial version of being invariant under the grading groups and so on. And it's normalized by some factors under the first Rademeister group, a little twist. We can normalize it to get something that's invariant under all three Rademeister groups by using an appropriate grading factor here, which I won't bother you with.

1:02:30 And it turns out to be a model for the Jones polynomial which was discovered in 1984 by a somewhat different graph. The thing to notice here about this non-invariant is that this looks just like a binary identity with a different value of coefficient. Because you see, if you took a equal to minus 1, it would be the binary identity, right? And you're probably deep to, if you put a equal to minus 1 here, you get the minus 2. So this polynomial invariant could be regarded as a generalization of Penrose's binary calculus if you look at it this way. And there's more to say about that. What about A equals minus 1? A equals minus 1, yeah. so therefore, we go back and say, okay, let's try to repeat what Pembroke's did and make a topological recoupling theory. And that recoupling theory It turns out, once we set it up, we'll satisfy all those relations that I told you, pentagon, hexagon, and so on. Because you see, there's a gradient in it, and so the hexagon could be a photograph and the others. So you start here, and you define the symmetrizer this way. Now, this is the tricky one, but it really does have a cute definition that's exactly in parallel to Pembrox's. You need a different coefficient here rather than minus one. You take the number of switches in the permutation that will take you back to the identity, and you have to represent the permutation by a least version. And then you lift the permutation and turn it into a braid. I have a picture on the next slide. You divide by a factorial, but a factorial is a generalization of a factorial, a deformed factorial. If you summed one, if you summed on ones in there you would end up getting a factorial course if you put a to the minus one sum then you would just be getting a one that many times so this is a deformed factorial and this is a deformed topologically in here and algebraically in there

1:05:00 a deformed version of the projector and it is a projector in exactly the sense that we want. So what happens is that you use a radius like this. Then the radius gets expanded by the bracket polynomial calculus instead of the minor calculus. So here's how it looks in our example. Here's our little two-strand particle. The factorial is 1 plus A to the minus 4. Here we get parallel line plus permutation multiplied by A to the minus 3. This gets expanded, and remember this gets multiplied by A. and A inverse, and you collect all terms, you have 1 plus A to minus 4, that just comes out the identity, as it should. Here you get minus 1 over delta, delta being minus A squared minus A to minus 2. And the delta was minus 2 before, remember? I mean, right. And minus 2 times minus 1 was 1 half, so that's how it's generalized. It looks just like the other one, but delta is replacing minus 2. If you close this little bit, you'll get a delta, and that will cause the cancellation just like before. So this has the same property that if you close two parts, it will go away. And if you multiply two of these together, you get one back. It's a projector. So this is a projector in the algebra of all lines connected to one another like this, which is called temporary lead algebra. and then the three vertices are defined in the same way as before exactly the same as before so I won't repeat but then now we can start to think about what a little particle reaction theory might look like and here's the one that I'm going to talk about I want to have one particle called two in this case because it's got two lines my particle is this guy there are two things that can happen with two two can interact with two and produce a two or two can interact with two and produce nothing that's it I want to make the simplest possible so I do not even want two plus two equal to four which is conceivable right 2 and 2 interact to produce 4. Which means that the fourfold symmetrizer should be equal to 0 in this theory.

1:07:30 Now, it turns out that in order for the fourfold symmetrizer to be equal to 0, you can't use any A. You have to use an A of the form, maybe I pi k of 10. Something like that. You see. But if you choose this very specific value of A, then you have this extremely simple theory which has only one particle in it itself, either produce itself or nothing, okay? And the recombin theory for this one turns out to have enough structure in it to give us a topological quantum theory that can represent quantum computing. Now, that is to say there are enough unitary transformations in this thing. Now, I ran out of time in the slides. Can I just ask one question a little about those? AIDS, they often turn out to be roots of unity. Is this obvious why they should be? I mean, I'm just linking up with the Joneses' work on the House of Algebras, and yeah, there are various amazing agencies, but there are various instances where the roots of unity are nice. Why are they nice, or when they come out, that's a question I'm really asking. Well, first of all, in Britain's integral, you need the corresponding things, the things that it corresponds to are things with roots of unity because of some integrality and desire for something to be integral in the exponential. So then on the other hand in the recoupling theory, in order to control the number of things that you get when you start recoupling, roots of unity tell you that things will cut off. This is a typical example of something cutting off. You see, this theory cuts off right there. It will be a cutoff as soon as you use a root of unity so that you don't end up having infinite sums in the formulas that you write. If you're trying to write down corresponding formulas that you would get for the invariants, you end up with infinite sums when you're away from roots of unity. And then you can worry about whether they might converge. Not only I'm playing with a model of roots of unity,

1:10:00 So there might be some corresponding reasons that there's a connection between the two. But here you can see exactly, it's just I want this cut on, and now I end up having to be at a vertical unit. So, as I said, I ran out of time making slides, so let me just show you. You're not allowed to tell a little something about it. So let me compress this, but show you a little bit, and then you can discuss. So we have our particle here, and then the interactions look like this. Two particles can interact to produce a particle, or two particles can interact to produce nothing, like that. And I want recoupling formula here, like this, that there should be a change of basis like this. So this is the simplest conceivable recoupling theory you could write down just to, you know, just that many vectors that could be non-trivial at all. And there's a matrix here, f, which is going to be a, b, and a prime, b prime. So we could say that f applied to this is equal to, I'm sorry, f applied to the vertical guys is equal to the horizontal guys. wiggly line, because you can't see my dotted line. The wiggly line is nothing at all. Okay, it's the nothing part. So this is what f does. That transforms like that on a basis. And so, of course, I get the same formulas if I rotate all these glitz by 90 degrees. So that implies that f squared is equal to the identity. That's called the orthogonality relation for the recovery coefficient of generalization. Now, that doesn't mean that F is equal to its own transpose or unitary. It doesn't. But it's coming close, right?

1:12:30 At least the square is equal to the identity. What is it that I want here? What I want is that F should... You see, I'm also going to have a gradient operator. Right? which in this case is just going to multiply by something, depending on the coefficients here. These are all the same coefficients. There will be another grading operator for the case where there's no particle there. So this and this. So this is going to be actually equal to this times something, and this is going to be equal to this times something. I'll write them down later. They'll just be phases. So the local interactions of the particles are just phase changes. that's all you need in order to have this but then if this were unitary that F is unitary right then I have I have the R matrix which is 2 by 2 R is just going to be equal to phase, phase okay, that's what's going to be I'll write the phases later and the F is unitary and then we're going to be looking at explain what we're going to be looking at. We're going to be looking at things like this. Suppose that I, let me see how many I need in order to get something interesting. Well, it doesn't matter. Suppose that I started with a situation like this, and I start with 2s. I'm calling my non-trivial particle 2, okay? And then 2 and 2 could interact to give you 0, and 2 and 0 could interact to give you 2, and 2 and 2 could interact to give you 2, and 2 and 2 could interact to give you two, two, two, two, two, two, zero, all right, that is, I'm thinking of all the different interactions that could happen, this is an element in here, that could start with twos and end up with zero, okay, that's a nice vector space to think about, right, I start with particles and I end up with nothing, and I want to know how many different ways I can do it, well, it's a Fibonacci number of different ways you can do it, which is That's why this is called Fibonacci-9. And you just think about the different possibilities in here. And this is one vector in that space. Now the braiding operates on this space. You see, let's take an example like this. And we ask, well, what will happen here?

1:15:00 Right? What will happen here? What is this? assuming that the braiding here is you only know the braiding in a vertex and then it just multiplies by a face these guys are a little bit separated but you can recall then you can form F then you can excuse me then you will end up with then you can do the braiding I'm sorry that's all just do the braiding and then you can do R inverse and you end up back here alright so what does that mean, that means that this is equal to something times this and then an R and in the end you get a little matrix which takes you from the vector space back to itself in terms of the basis right, so so what you have is R and then you also have things another matrix which is FRF because F squared is equal to the identity which I could call this one R. R is equal to say R naught and FRF is equal to R1. So I have two mappings of this vector space of this one to itself R9 and R1 and then if you think about what we were saying about compatibilities and so on that satisfies the braiding relation. So you get it. And if you use more of these, you get representations of all the different grade groups, you see, this way. And so you get lots of radian representation going on here. And furthermore, let's go back to, well, I just want to show you size and vector space. Let me see. That's the one I want. I want to think about B2, 2, 2, 2, and 0, and think about the elements in this space. So let's just think about it. 2 and 2 can interact to give you a 2, and then I want 0 to come out. So if 0 is to come out, then it can't be 0 and 2 because 0 and 2 interact to give you 2. So this has to be 2. So that's one element in this vector space. And let's look for another. 2, 2, 2, 2

1:17:30 and 0 well we could interact with you 0 and then again this has to be 2 in order to drop to 0 and that's legitimate, 0 and 2 interact with you 2 so we might call this one 0 by definition and this one 1 and this vector space here is dimension 2 and those are the cubits but now I have a cubit And that means that I can do grading on the cubit. And so it turns, and those gradings are the matrices that I call R, which is some phases, E to the I, something or other, which I could write down to you. And the F, which in the model turns out to be taw, minus taw, root taw, and root taw. And I'm going to skip the calculations. For the calculations that get us over to this matrix, I was going to show you, but we're really a little out of time, and it's a little bit technical. You just have to go through all the formulas for the two-fold recouplers and find out what the recoupling matrices are and find out that it ends up being this, where Fibonacci number is tau squared plus tau is equal to 1. This is the inverse of the problem ratio. Unitary. unitary so there's a little action of U of 2 going on in there on the level of the single cubits as well as braiding in the large and then it follows from approximation theorems of Kutaiya and Soloway that you get in U of 2 that's the first part Like this gives you a dense subset of UO2 when you iterate those with each other. And similarly for the other vector spaces. So you get enough unitary transformations to do the job. And that's about, although they have some of our slightly more complicated models than this one, this is the simplest model. And it's very simple if you're looking at it from this point of view, which is what was exciting that you see, because everything is completely natural. you just look at this theory in the simplest possible case and it lands you on that model. And these calculations about the recoupling of course are quite easy to do except they have come out wrong about five times before they come out right.

1:20:00 And so then you're left with the following questions, you see, which I don't know the answer to. I think there's probably quite a bit of number theory involved in actually figuring out what you're getting here. I mean, this is a very slim little subset of U of 2, but it's going to generate all of U of 2, and suppose I wanted something, some obvious matrix in U of 2, and how do I actually get the good approximations, what's the algorithm? I'd like to know a little more about that, and I'm not sure they know either. And then the same thing at the next level, I can write down a nice gradient matrix like I did at the beginning to hand you a picture of the gate, the basic gate, but I don't know in a simple way here, but I think it's going to be easy to construct entangling gates because there's lots of breathing going on, so there's some interesting calculations to do with the model, and this method will allow us to calculate some things. But you see, it's not. If you were hoping that, on the one hand, the physics would hand you a wonderful theory of anions where there would be these phases happening just the right way and satisfying the the axioms for topological quantum field theory, nobody knows. And even if it did, it still wouldn't be easy to construct the computer because it would only be giving you approximations from way out there and things like that to the transformations that you wanted. So there's still lots more work to be done in this notion of doing topological quantum computing to make anything really practical. But from a mathematical side, I think it's very intriguing that you can generate U of 2 to the end, essentially, by doing some very simple things like that that are fundamentally topological and coherently topological in that way. So it certainly means a lot of smart thinking. So we probably have about five minutes for discussion. What other people wanted to... Another question? I'd say I would be the chairman. That's fine. Yeah. But why don't we... I talked a little bit too much, so let's just wait a minute and see. What happened to Mike? At least he's doing administrative work.

1:22:30 So, any questions? Any of you guys who asked for something? It sounds like, I don't know, how many other moments in 4-mere contrasts are more than in 3-mere, but what with more high-mere contrasts, 5-6, and so on? It's talking about the 4-mere contrasts, which is obviously more difficult to learn than the 4-mere contrasts. And what about the high-mere contrasts for this brain therapy? You can use my order of unity and do a similar... The question actually is that it's known that four-dimensional space is somehow specified for it's the most complicated and not difficult space to work in. Oh, four dimensions. Four dimensions. Oh, you mean four-dimensional manifolds? Yeah. And you tell just in a roundabout way that it's a more difficult problem to deal with all this theory in the four dimensions. But what about the higher dimensions? Because it is known that for five or six dimensions, all the biological problems are much easier than for the four dimensions. So what can be said about this theory? It's not a logical problem. Yes, it's not a logical problem. Well, again, if you try to do it in this direct way, it's still not easy to figure out what the algebra should be. It's not just obviously possible to say what is the difference and what are the problems separately in the three-dimensional, four-dimensional and high-dimensional cases. You know the regular way to start this theory for high dimensions, but can you say about of their character problems in this case? Well, the four dimensional case is the most interesting one because we know

1:25:00 that there's very many trivial things happening which should be measured by this kind of theory if one had it. In higher dimensions one understands the classification of metaphors better and so it doesn't make it doesn't make us much of a difference. On the other hand, when you're thinking about the relationship between the differential category and the pathological or piecewise linear categories, then in our dimensions there may be something. You see... It's not that they're something, but maybe they're similar. Yeah, you see, one doesn't have a completely combinatorial description of what it means to have a differential structure. c-infinity differential structure and yet in studying classification of differential structures one has topological or homotopy theoretical theories for that and so classifying say exotic structures on the seventh sphere turns into a combinatorial problem, turns into a problem of combinatorial topology, but by very indirect of characteristic classes of bundles and other things. And if it was being measured by topological quantum field theory, then it would be being measured directly in terms of the way the space is fitted together out of local pieces. But there are gaps in understanding how you might conceivably do that because one doesn't have a direct hold on what it means to be differential structure in a combinatorial way. So it's interesting to think about that. As soon as you get up above three, the tensors that you're dealing with are not like ordinary tensors. Tensors in three come down to tensors that are modeled by little lines that are separate and they look just like ordinary tensors, T, upper, A, B, C, lower, C, D, E. But as soon as you go up a dimension, then the transformations have as their base geometry, like curves or higher dimensional things. So, categorically speaking, it means that the morphisms are morphisms of morphysics, and you're in a higher category, and it's hard to do higher categorical algebra. So there are a lot of people out there who are trying to do higher categorical algebra, like John Bison's students and other people with Lewis Kreen and so on,

1:27:30 because they figured that if they could understand higher categorical algebra in the right way, then they could make these theories work. There are lots of problems. Thank you. Thank you. I'm sure I don't know quite clearly, why is that sort of like stabilization, that you are just rising up dimensions. That's all the point also, that you should just, wouldn't change anything. That's what you quote. Yeah, right. In higher dimensions there will be some stabilization, but when you try to go from 3 to 4 apart, you get a stabilization. And when you go from four up, again, it won't take a specialization. And then maybe from there on things start to look more uniform. So what does the fact about the pattern for three dimensions is really specific for three-dimensional paint? Well, it's not a guess for three dimensions. But for four dimensions, of course, one would like it very much. Because you would like to have a variance of four-dimensional manifolds in this form. Or, if you're thinking about quantum gravity, then you would like to think, well, an invariant of this sort involving co-ordisms would be something like there's a metric here and there's a metric here and there's a co-ordism between them. A co-ordism is something like an amplitude for going from one metric to the other. And so it would look like some phase of quantum gravity. And also, I want to ask a question. Where is the place for, I don't know, I'm going to answer for the audience. Okay. At some moment I felt that they were just driving, they just had them, but then it was a picture. I was using the Jones file. The bracket is the Jones file. Right. Um. Yeah, I was saying the phone only was not appearing in this discussion.

1:30:00 but not but if you don't just directly to not then they will appear oh yes of course right and there are models of the alexander polynomial which have the same form um in terms of solutions to the actual equation that sort of thing but uh the uh the sanctity momentum type representation It's curious, the Alexander polynomial doesn't have an easy expression in terms of functional interval either. It does have state systems. any further questions well I think can I thank Lou for a fantastic opening performance thank you very much indeed Practicalities, can I just, if anybody wants bickies, they are here. And there is now a complete set of abstracts, which you can pick up from here. Unfortunately, I haven't yet had time to copy your actual paper, Dimitri. Can you just let Dimitri know, I've got the abstracts, but I haven't yet had a chance to copy his paper. I'll have it by the start of the month. So, one other practical thing, sorry, coffee. Guys, sorry, practicalities. Because I've been taking delivery of vegs and getting all these abstracts finished, I haven't had time to make the coffee. So what I suggest we do, since we're not starting to get at 11.30, there is a very nice little bar on the corner called the Barley Sports, which will do us excellent coffee. or indeed if you want something more substantial meantime, help yourself to the abstracts the abstracts are here, Luke, sorry that's Dimitri's that is yours well, yeah, not quite because I didn't have enough folders but that's yours yeah, that's yours

1:32:30 why don't I put them on the table over there yeah, right guys, I'm going to put the abstracts on this table It's about five o'clock left.