The Continuum Hypothesis in Topos Theory and Algebraic Set Theory (contd.)

0:00 Before they got messed up. Well, you see, there was a post which you can get out of here. I don't know if you're going to get a feeling of telling things off. That would be very worth it, but I don't think it's going to be that much. I mean, you can sort of see it as part of the lecture. That's it. Yes, you see, originally... He did involve a split into right and left and right and left and right and left and right and left and right and left and right and left and right and left and right and left and right and left and right and left and right and left and left and right and left and right and left and left and right and left and left and right and left and left and right and left and left and right and left and left and right and left and left and right and left and left and right and left and left and right and left and left and left and right and left and left and right and left and left and left and right and left and left and right and left and left and right and left and left and left and right and left and left and right and left and left and left and right and left and left and left and left and right and left and left and left and right and left and left and right and left and left and left and right and left and left and left and right and left and left and left and right and left and left and left and left and right and left and left The extreme people who were making a big noise of saying they could work out an entropy of black holes and so on and so forth. And so the astrographic people said, well, don't lay out where the other things are really black holes. It's all a nonsense. Roughly speaking, it's all a nonsense. Let's do it properly. So then they looked at black holes with their actual horizons. And they come up with the rhinoceros, provided they produce this thing which is called the Euclidean space-time-space-quickness. You look at the spatial directions, and you put something into a tangential direction, and according to the literature, you say one plus i times the other, and then they say, well, what we find with our black holes is that would give us an i in expression. It doesn't matter if it's a right answer or a wrong answer, it's also a very complicated number. And this complicated number means that it's one plus one plus this funny parameter. And they say, well, okay, it's the same classical theory as you know, but that's about the reason I don't know.

2:30 Entropy is all different circumstances. They say, oh, maybe this is the real thing. This is, to me, going the wrong way. It has no geometrical basis. It shouldn't have been done on my side. And that's clear geometry. But it doesn't seem to give the right answer to all of our questions. And that is at one point the way that suddenly they discovered the language they were learning in each one of these assessments to put the computer on and we definitely need to project that. Heather, I don't know what the question is really. I don't know if she wanted to, but she may be interested. It was just as a general issue. If she wanted to find us, she would need to know the points. I think today she must have some knowledge of what she has decided to do, and so let's see if somebody knows either the coordinates or how to get here, she might be interested to see this in some way. Thank you for your attention. No, but that's a good way to put it, but yes, I'm pretty sure that the trouble is that I can see the mathematical benefits through the language of this seminar.

5:00 Thank you very much for your attention. I can't believe that you're leaving on Sunday. Yes, it's what happened. I don't know if it's your flight back today. Oh, no, no, no, no, no, no, no. Yeah, I turned it around the other day. Yeah, we had to expand on it. That's okay. Whatever carries the... No, I know you're leaving on Monday. Oh, okay, yes, yes. So the devil shall open my eyes and let me make a leaving. Yes, I've checked with them. I have talked to them in other matters and I'd say that I'd like you to in Sunday and Monday is absolutely no problem. But I think this weekend is the bound for more than normal. I'm not going to leave. Why don't you go somewhere else? I forget where I'm going. It was a problem I always had with my students. I left it as a challenge to find a very nice representative. And I could never think of one myself. And every now and again, the students would come and say, yes, I have a very nice one. Either I couldn't see it had anything to do with secondary biology, or I couldn't understand a word they were saying.

7:30 So I've never found a good description. I think that's where the EPR... I think that's where you might even believe that the EPR itself might serve. Thank you very much for your time. I think my best way of explaining it, if you are John Gooding and not a former semester student of the British Curriculum, is probably telling me about the impossible triangle, the way of introducing the first cohomology group, so I think that sounds really good. Stuart, do you really consider the teacher-students on cohomology? Oh, maybe it's not something I can handle at all. Now accessible, it's normal for people to get across on a, you know, by computer graphics to come and live with a second car. Maybe I'm thinking along the wrong lines, because the kind of thing I thought of is a room, a very rich room, you see, and there's a wall, sort of like a dome, a sphere, which you sit inside. And as you look around on the wall, everything looks the same. But the whole thing is invisible. But if you open a door somewhere, as soon as you open the door... It's consistent because the problem can disappear through the hole of the doors. Or it could be anywhere on the stairs. Now, the kind of thing that... Now, the kind of thing that... Now, the kind of thing that... Now, the kind of thing that... Now, the kind of thing that... Now, the kind of thing that... Now, the kind of thing that... Now, the kind of thing that... Now, the kind of thing that... Now, the kind of thing that... Now, the kind of thing that... Now, the kind of thing that... Now, the kind of thing that... Now, the kind of thing that... Now, the kind of thing that... Now, the kind of thing that... Now, the kind of thing that... Try to draw such a rule, but how would we have done it? Of course, it would probably be on me, but at least I have to speak with you. Yes, yes, yes. I hope you will stay with us. I hope you will stay with us. I hope you will stay with us. I hope you will stay with us. But maybe there is another way. But maybe there is another way. Yes, diatomites, you know, chemicals and diatomites, right?

10:00 Oh, yes, I know, it's been discussed as well. Thank you. I hope you enjoyed this video, if you did, please like, share, and subscribe to my channel. At the same time, we don't change our styles, it doesn't matter where I work, I'm modest, and somewhere else, then you see that we've got all different styles, and I mean, you're trying to, obviously, have nothing else to do with what you're doing. Thank you for your attention. So, I'm going to put the computer on this table. Your computer, I know, is connected to the projector, and you need to put the projector in the center. I don't know if it's possible to do it. Well, who can do it? We all can do it. But, um, of course I'll have to do it. Oh, well, let's do it. Let's do it. On television. I didn't understand. I didn't understand. Yeah. Allegedly. I didn't understand. But can you watch the video? Because I don't have enough time. Thank you for your attention.

12:30 I will just drop this home. Here's your, yeah, yeah, yeah, yeah, yeah, yeah, yeah, yeah, yeah, yeah, yeah, yeah, yeah, yeah, yeah, yeah, yeah, yeah, yeah, yeah, yeah, yeah, yeah, yeah, yeah, yeah, yeah, yeah, yeah, yeah, yeah, yeah, yeah, yeah, yeah, yeah, yeah, yeah, yeah, yeah, yeah, yeah, yeah, yeah, yeah, yeah, yeah, yeah, yeah, yeah, yeah, yeah, yeah, yeah, yeah, yeah, okay. Yes, you use all the beauty of the ingredients and the connections of the numbers. I need a sock. What, a piece of the office? This one? Oh, too far? If you have any questions, please feel free to ask them in the Q&A box below. Yes, it's working, I'm sorry. No, no, no, no, we've still got another hour to sit and be amused. Enjoy. Oh, yeah. I'm sorry, I don't... Thank you very much for your attention and I hope to see you again soon. I think they said they picked me up at 2 a.m. tomorrow.

15:00 2 a.m. tomorrow? I'm leaving on Monday morning. You're leaving on Monday morning? Yeah. They picked me up at 2 a.m. tomorrow. I think the same. Thank you for your attention. Thank you for your attention. That's all I have to say today. Thank you very much. But yes, the same question, all right, so let's go back for one minute. Yeah, it took us, it took us about, I don't know, maybe, uh, four or five minutes to finish it. Yeah, two, two, now, I don't know, it's four to two, but we are finished, that's about it. All right, so I want to try to finish rather early, earlier than that. Okay. If you want to finish up here, that's all right, let's end it here. No, no, no, that's great. Yes, I'm sorry, but it's all good, I guess. 6.5 minutes. Thank you. So you have to teach tomorrow. If there is an appointment, that would be nice. We'll have this, get ourselves organized, and then we'll get away from it. Thank you for your attention.

17:30 So this is another vignette, and then we'll do something longer, but I decided to talk about this for a little bit because it's maybe related to what Georg was talking about, about identicals and things. So you could make up variants of set theory by choosing some, or models, by choosing some situation where there's an asymmetrical relation. So let's decide that A belongs to B means that you have a line labeled A which goes underneath another line labeled B. And the line goes along like that. Then we could make diagrams of sets using the knot diagrams. And immediately there are some non-standard sets. That is, sets that don't obey the well-foundedness axiom because they have infinite descending chains of membership. For example, a knot diagram with a little twist in it is a member of itself. The A is labeling the entire line, okay? It doesn't stop at the breaks. So you have sets of the members of themselves. You have sets of the members of each other, a link, or chains of the like. So, for example, that little chain link on the left has A as B as a member of A. But A is also a member of B, and so is C, and so on, and you get a chain like that. And we could wonder to what extent this... The scheme is topological, so let's just spend a couple of minutes on it. Some diagrammatics of it are nice, like here's counting. What I've labeled zero is an empty set, so it doesn't have anybody going underneath it, and then what I've labeled one has the empty set going underneath it, so it's the set consisting of the empty set, and two has two members, the set consisting of the empty set and the empty set, and three has...

20:00 All the previously created sets as numbers and four and so on each set each integer is each grand natural number is the collection of all the previous ones which is the usual construction going back to von Arnhem and maybe farther back to Frege but in this case the notation writes it out without driving you crazy with brackets so that's a feature and then there doesn't There doesn't really seem to be a Russell paradox in the usual sense here because if we allow topologically to have membership or not, self-membership or not, by allowing the topological move, then you get the peculiar situation that A belongs to A is equivalent to A does not belong to A and it's not logically possible. On the other hand, these sets, if we want to make them topological, of course aren't multi-sets to begin with. B consists of two identical copies of A. That is, A occurs, or referentially occurs, in B twice. And there is no distinction between those As. They are twins, perhaps, in Bjork's terminology. And it's topologically equivalent to empty sets in this case. So you see that the occurrence of identicals should cancel in pairs if we want to make this theory topological. And it certainly would behoove us to make the theory topological since we use a knot diagram. And as far as the third Rheinemeister move is concerned, it just reorders things a little bit. It doesn't change anything about the corresponding sets. So that, in a way, is the end of the remark. You can make it topological. Linking remains. Certain linking patterns remain. So that some of the mutualities and so on, they're there and they're in the model. And you have an interesting small model of non-well-founded set theory in this form. Of course, it really perhaps shouldn't be called knot set theory, because the knots really are not known to it. If you take a knot, it's all labeled A, and it will either...

22:30 Yeah, because I allow the first right of meister move as well, so they cancelled in pairs, but also the last one could go away by the first right of meister move, which allows me to do that. So, there isn't any knot theory there at all. But on the other hand, there is linking, just no knots. And it gives you a picture of how non-well-founded sets can be thought about in terms of topology. Comments? That's just a very short... very nice. I'm a bit confused by the last one. You're a bit confused by the last one? Yeah, how do you... there's no cancellation. Well, remember, the A's cancel in pairs, and we're down to A equals the set consisting of A. But a self-reference of A inside itself is allowed to be removed as well. So... so that one will go away. By those, I didn't write down the moves, because of the, I mean the abstract moves are identical elements cancel in pairs and the self-reference is effectively removable, a direct self-reference. A very general comment, not I think particularly connected with the knock theory of what you've just said now, but you said part of the motivation for this is trying to see set theory in a topological. I was just going to say, exactly, that's exactly a major part of the motivation of the topos theory. And there, of course, you express conditions like the acts of extensionality, which we were talking about a little yesterday, are equivalent. To topological additions like relative uniform separability, for instance, so there is the connection there. But the question of identicals, the way that comes in, I thought was interesting in relation to where was that. Yes, because they really are identical, you know, I mean they're referentially identical. This is at about the same level as Axel's theory of nominal founded sets, where he uses graphs. Who do you say? Axel. Axel. Axel. Axel. Axel. Axel.

25:00 Axel. Axel. Axel. Axel. Axel. Axel. Axel. Axel. Axel. Axel. Axel. Axel. Axel. Axel. Axel. Axel. So just some basics from the beginning here. In this form of things... The quantum mechanics is abstract and more or less finite dimensional. Physical processes are modeled by unitary transformations. There is some measurement basis that's out there, assumed to be there, and when you measure you fall into one of the elements of that orthogonal basis or phenomenal basis with probability of the absolute square of the coefficient of the state vector. If we have multiple qubits, we're in a tensor product, and we could observe in one tensor line and be left with a state. And so, for example, when discussing entanglement in this formalism, you may talk about observing in the first tensor line, and then you're left with what's in the second tensor line, and somebody else could observe that. Our standard example of entanglement and I often like to use this as an example. Let's just look at it quickly just for warming up. Here I have the standard Mach-Zehnder interferometer. Two mirrors, one is an ordinary mirror and the other is a half-silvered mirror. The half-silvered mirror, if the state 0 comes into it, then you get a superposition of 1 and 0 coming out of it. And if a 1 comes into it, then you get a superposition of 0 and minus 1 coming out of it, making up a matrix which is unitary.

27:30 This is a finite-dimensional quantum process. You need that phase change in order to have H, the Hadamard matrix, be unitary, and the other mirror is just flipping, so it's certainly unitary. And then, if you think about building this device in the form up above, then you have all those different pathways that take you to end results. And you see that if you catalog the pathways, then you would get interference on the ones, and you would get constructive interference on the zeros, and you're adding up all the pathways when you look at it that way. If you multiply the matrices, which is what the process looks like if you think about it that way, then you get the same result. So this is finite dimensional path integral. And the reason why it's coming out, of course, is because when you multiply matrices, you're combinatorially summing over all the different pathways corresponding to the definition of matrix multiplication. So in this finite dimensional kind of quantum mechanics that happens in quantum computing, you have the path integral formulation all the time, in some sense, underlying just the matrix algebra. And there isn't a problem about the existence of functional intervals or anything of the sort. Then here's a nice example of a quantum algorithm that I will use at the very end of this bit. So I want to compute the trace of the unitary matrix. This symbol just means the trace of the matrix, those being the indices of the matrix on the left and on the right, and then I type them together, so that's UII summed on I. But that's not the quantum algorithm. How do you find the trace of a matrix with a quantum algorithm? Well, here's one way. This is called the Hadamard test. So there's H that we had before. It's on the screen again. And what you do is you take an extra tensor line, and you use that extra tensor line to form a control U, so that means that if the upper line has a zero on it, then U is shut off, but if the upper line has a one on it, then the U is operating, and you can augment the matrix into a larger one like that. So then you take, you know, what we're after is finding the value of a Psi U Psi, the Psi Psi entry, Phi U Phi, excuse me, the Phi Phi entry of U, and this works if you check the calculation, you take and apply H to zero on the first tensor line, so you have a superposition going into the control line, you put Phi into the lower line,

30:00 Then you apply H again on the output and you measure in the first tensor factor, in the top line tensor factor. And you measure for zero. You want to find out the expectation of zero. And you find, after you calculate it, that zero occurs with probability 1 half plus the real part of phi u phi over 2. So that means that if you run this enough times, then you can check from the frequency what the value of phi u phi is, the real part. And if you put a phase change in there and run the same experiment... Then you can get the imaginary part as well. So that's how to get the trace, that's how to get the diagonal entries of a matrix and thereby the trace with many repetitions. Plot's a good example of a quantum algorithm and one that we might be interested in using if we were going to calculate something like the Jones Polynomial. If you would actually do this, if you would actually do this, you would have to repeat the same thing many times. It gets done a very large 10 to the something number of times automatically with one experiment, but in other situations you might have to have a very repeatable experiment that you could set up and just keep doing over and over and over again. So you have to be able to set up the same, prepare it the same way a large number of times, either identically, essentially identically along a whole bunch of molecules, or set the machine again and again to the beginning. Is the worry that if you actually do the whole bunch of molecules that they might not be independent?

32:30 Well, it can be done so that they're essentially the same. Okay. Oh, this is an abstract comment about tenderly devolved algebra. Maybe it's a little bit out of place here, but... You usually think of temporary, I mean, I tend to think of temporary algebra in terms of tensor lines so that they look topological and they get pulled apart, but if you just take two projectors, and I've made an abstract formalism for them here so that the Dirac ket bra is, I'm calling it an extainer, and the bra ket I'm calling a container, just to give it a name in this abstract form. So P is a projector and Q is another projector. It could be, if you want it to be real algebra, then you can put a ket bra with a given vector direction and another ket bra with another vector direction. But then notice that you see P squared is a multiple of P, scalar multiple of P. Q squared is a scalar multiple of Q. You get a scalar in the middle, and the cat bra that's left is p, so you get a multiple of p. And if you form qpq, then you get the same scalar and a multiple of q. So this is a small temporary lead algebra, up to multiples, if in your temporary lead algebra you want pqp to be equal to p. So they arise naturally in an algebra of projectors. And it turns out actually to be easier to make unitary representations of the break group by using this kind of projector algebra than it is using tensor algebra. But in any case, temporarily leave the algebra. It's just part of having projectors and it's good to understand that. It of course occurs in this way over here with the tensor lines. So this slide is just a summary of what that looks like. If you think of cobordisms of zero manifolds, right, that's what you're looking at is some cobordisms of zero manifolds. And this is a temporary V category. And this slide was designed to fit into the kind of language that Bob was using. So you see over here

35:00 The same kind of categorical zigzag situation that could be pulled out and written as a product of some matrices and thereby figuring out teleportation by using it if one wants to. So I'm not going to dwell on this. I just wanted to bring it up. So now let's talk about topology and quantum computing. How many gates do you need in order to build a quantum computer? Well, you need something like CNOT. This is controlled not. This is just like the control view. If the first bit, in this case, is zero, then it doesn't do anything. It's the identity. If the first bit is a one, then it flips. So that's controlled knot. And what you need in order to build a quantum computer is very few gates, actually. It's almost like saying that to build an ordinary computer, you need a knot gate and maybe an AND and an OR, very little. Here, you need a two-qubit gate. And CNOT will do. And then you need local unitaries because you need those little rotations in two dimensions. And Hadamard and a couple of others are sufficient to generate what you need. So it's fairly cheap on the mathematical side to create a quantum computer. You just need a few matrices and then you can generate everything else. But you might wonder what you could substitute in here instead of C-naught, and there's a very nice theorem due to the Grelinskys, which says that you can, given that you can get the local unitaries that you need, then if you choose another gate, another two-cubic gate, that is entangling, then you're good. So, by an entangling two-qubit gate, I mean a gate that has the property that it takes an unentangled state, a tensor product of two single-qubit states, and it makes it entangled after you apply it. And if you have a two-qubit gate that does entangle some state, then it's universal if you add it to...

37:30 So you can replace CNOT by any other date you like as long as it's entangling. So for example, and then if you care to do a little algebra, it's easy to check that a two qubit state The determinant is entangled exactly when the determinant of its obvious corresponding matrix is non-zero. See, it has 0, 0, 0, 1, 1, 0, and 1, 1 entries, so it looks like a 2x2 matrix. If that determinant is non-zero, then it isn't a tensor product. So that's what entanglement means. Like, if you take the 0, 1, and 1, 0 state, with b and c equal to 1 and a and d equal to 0, then of course the determinant is non-zero. That's the standard one. There's a famous matrix transform, which I've called R here, which takes the basis 0, 0, 0, 1, 1, 0, and 1, 1 over each one into an entangled state, and this is usually thought of for purposes of, it's called the Bell basis change matrix, and it's often thought of for the teleportation procedure as the right basis for measuring in. On the other hand, this is certainly a very entangling two-cubic gate. So you could use the Bell basis matrix as a universal gate. You have the Bell basis matrix plus everything else. But the interesting thing about that is that it's topological. That is, it satisfies the Yang-Baxter equation. The R that I just showed you. All of this is actually an R that satisfies the Yang-Baxter equation, which is the braving equation. It's better to look at it as a diagram. So here's the basic braving move for the braver, right? Sigma 1, Sigma 2, Sigma 1 equals Sigma 2, Sigma 1, Sigma 2. But if you wrote it in this term, then you have a mapping on a two-fold tensor product, and then you augment it with the identity, and that gives you R tensor I, or you have the identity augmented by R, and that's I tensor R, and then the braiding equation reads R tensor I, I tensor R, R tensor I, should be the same as I tensor R, R tensor I, I tensor O.

40:00 Somebody wrote a symbol backwards somewhere. No. No, that's right. Okay. So that's the Yang-Baxter equation. And if you have an R which satisfies the Yang-Baxter equation, then you get a representation of the Brae group. As long as the R is invertible so you can have somebody for the reversed crossing. These are the same as having certain kinds of representations of the break group and so what I'm saying is that that Bell basis change matrix which is written up there at the top, that's a solution to the Yang-Baxter equation and it's also interesting on the quantum computing side and you can use it as a universal gate. So we can imagine quantum computers that use that gate, and then they would be almost topological, but they wouldn't be quite topological because you would have the following funny situation then. If you wanted to draw a picture of the unitary transformations that were doing your quantum computer, they might look something like this, but then on these individual tensor lines there would be some local unitary transformations sprinkled around that were needed in order to make the unitary transformation that you were interested in that does this quantum computer. So you would have an extension of the break group by the elements of this group on the line. And then, that's sort of quasi-topological. If these local unitaries are out of the way, then you can do the braving, and it does look topological. But these local unitaries don't necessarily commute with the R, and so it gets kind of complicated to think about it. So it's a primitive picture of... of how you might begin to bring some topology into the quantum computing. And Sam and I were playing with this because we were curious about another question, namely, what's the relationship between topological entanglement and quantum entanglement? And instead of asking it in the arrogant way, we were thinking, well, here's an R, which is the solution to the Yang-Baxter equation, so you can think about it in relation to topology.

42:30 And on the other hand, it's a universal gate, so you can think of it in terms of quantum computing. And it's entangling, so it's a universal generator for quantum computation. Does it do anything interesting topologically? And we find in the examples, but it's not really a general theorem, we find in the examples that when it's entangling so that it's a universal gate, it also usually can do something for you as a topologist. So there's some hint of the relationship there. In the case of this guy, it can detect the Borromean rings. That is, there's a Lincoln variant associated with it, and that Lincoln variant can see the Borromean rings. So somehow the fact that it's entangling is related to the fact that it can see the Borromean rings. It's interesting to think that. There aren't that many 4x4 solutions to the Yang-Baxter equation that are unitary. These are the basic ones. I'll do various basis changes and you can find a paper by Heather Jai that details that on the web. And the same kind of phenomenon happens elsewhere about the entanglement. But not always interesting topologically. For example, R0 down there, which is just a swap gate with an extra phase, that's entangling and you can use that instead of C0. And you might find that useful since it's really simple. Maybe it's just as good as having C0 for your purposes as quantum computing. But from the point of view of topology, it doesn't really do anything much. The other ones can detect some winking numbers. So the most interesting one was the Bell-Reyes exchange matrix, which is very entangling and can detect the Borrelman-Burns. Anyway, that's the story, but we're not really doing... Topological quantum computing, if we do it this way, we're doing partial. And maybe that partial, if topology ends up being useful for quantum computers and people are really making them, maybe the partial thing will be of some use.

45:00 But I want to talk about how to make them so that the local unitary transformations are also part of the representation of the grading, so the whole thing is together. The other day, I went through a very quick picture of that and I thought it would be a good idea to do it a little more slowly so you can see how this bit of machinery works. So, part of the motivation of this is the quantum Hall effect. I have very little understanding of, I'd like to learn more about the quantum law effect, but it's a remarkable effect that happens if you have a very cooled metal plate in a magnetic field, and you have a current going through it, and you measure the transverse resistance, the transverse resistance becomes quantized in a subtle way, and people try to explain this, and the paper that Connected with topology directly is this paper back in 1991 by Moore and Reed where they suggest using the braiding representations in conformal field theory to see the structure of collective excitations of the electrons. In this effect, and this goes back to Laughlin's work on wave functions that would describe collectivities of electrons, where the collectivity corresponds to these points zi in the complex plane here in this wave function, and these guys suggest that the The conformal field theory related to the Chern-Simons theory is what you want, and the braiding representations that come out of that are what you want. So what we're going to be looking at here is not that, but the bare bones formalism of those kind of braiding representations from the point of view of the topologists. But what we are looking at is essentially what they're talking about, but without the analysis. And here we are again where we were the other day and remember I said that we were going to make these process spaces out of some recoupling theory and then we would have neighboring particles that could be measured directly the braiding of them and that would be some phases and if they aren't neighbors then we would be using basis changes in the recoupling to get at the braiding.

47:30 In the case of three-strand braids, it would be like this. We would have a local braiding for sigma 1, but a non-local braiding for sigma 2, and that one would be given by F inverse RF, where F is the basis change. And the process spaces can be larger than, much larger than the small one that I just said. But, and then remember we were talking about this, where we have a particle that interacts with itself to produce itself, or a neutral particle. And then you've got the single qubit space like this, with an action of the three strand break group on it. So far so good. Let's see what this will look like in a little more detail. I'm using the Q-deformed spin networks as before and maybe all you need to remember about this for the moment is that it's based on using the bracket polynomial model for the Jones polynomial expansion, that it has this interesting structural characteristic that it's an exact generalization of the original. Penrose-type spin networks with the binary identities at the bottom, but there are these projectors which are songs over all the different symmetrizations, all the different permutations, and the projectors can be calculated in temporary leave algebra, and when you want to form three vertices, you just need to add up the lines in the right way, and then we can get a model which has this abstract particle in it, like this. And to see that you need that third root of unity, well I'm going to talk about that in a moment, I mean that fifth root of unity, well it would be a fifth root of unity in the usual Jones variable. It's a tenth root of unity here, right? Anyway, let's think about the process space for a moment. So the black dots are the p's, and the star is the neutral black point. So this is an example here of one possible process. Two peas interact to produce a pea. Two peas then interact to produce a neutral particle. The neutral particle interacts with a pea to produce a pea, and so on down the line.

50:00 And if you think about these processes for a moment, you see that there's no way you could get two stars in a row. It's a detail, but it's an important detail. Once you have a star, it interacts with a p, the way this is set up, and so the next one will have to be a p. No way to get two stars in a row. So you're looking, if you're wondering what the dimensions of these spaces are, then you have to think about how many sequences of stars and p's are there, such that you have no two stars in a row. You can have as many p's in a row as you like. And this matches directly into An old picture that everybody knows that you generate a Fibonacci tree by having a young one, which is the star, which grows old, but the old one being mature can give birth to a young one and it continues on. So you have a two-fold generation from an old one and a single-fold generation from a young one. And then you form, and then you start with, say, a young one, and you build this tree, and at every level you get a Fibonacci number, two and then three, then five, and so on. And then if you go out to a branch and walk back along the unique path to the root, you see that it's a path which has no repetitions of start. And if you think of each one of the elements of the branch as a Fibonacci number, then you see that there's a one-to-one correspondence between Fibonacci numbers and paths of that type. And so that shows you that these spaces are going to have dimensions that are Fibonacci numbers. So I think that's key. And that's why this is called the Fibonacci model. So the dimensions of the spaces that you get are going up exponentially as you make more and more trees, and there are Fibonacci dimensions, and we can start with dimension two if we want to, as I did. Then what does it look like to figure out what the coefficients and parameters are for this model? Well, you're using a two-strand projector, so you need to do some exercises, and if you don't want to worry about the formulas in the general recoupling theory, you can just work them out.

52:30 You've got a two-strand projector, you can figure out what the three vertex is, it looks like that, and expand it. You're going to need some things like what happens when you tie a projector back into itself. That one I've written out. You see, you tie a projector back into itself and expand the projector and you get two parallel lines and a turnaround. So the two parallel lines is a delta squared, delta being the value of the loop, and the other one is a delta times one over delta. So you get delta squared minus one. You get down to an evaluation of it. You need theta networks. You need tetrahedral networks. So you do a few diagrammatic calculations and you have those, and then you have some undetermined coefficients for the recoupling, the dotted line being the neutral particle and the solid line being the p, and play around with that and figure out what the coefficients are. I'll come to the general form of how you play around, but when you close things, you can... You can close things on both sides of the equation and kill off some things and actually figure out the coefficients. So then you can figure out what the recoupling transformation is and you calculate it and you find out to your surprise that it isn't symmetric. Not in this formulation. And then you have to say to yourself, oh, oops, I wanted unitary. It's not even symmetric. It's real. It's not, what am I going to do? And you realize that this theory has been desymmetrized in some way, although it's quite natural to write it down in this form. And you should multiply all the vertices by an appropriate constant. And once you do that, it becomes symmetric. And the F, the transformation, turns out to be tau root tau root tau minus tau, where tau is 1 over the golden ratio. So the golden ratio comes in as well. And the phases... Come in and you get this as the end result. So you don't need to know the whole recoupling theory in order to build a model this way. You can just do a few base calculations and arrive at the phases and and the F. And then the remarkable thing about this model is that it is quantum universal.

55:00 The braiding representation that you get at three-strand braids is dense in the unitary group there, group two, and there's enough entangling operations in here so that you can use it for quantum computing. So this is the simplest way that I know to get to that point which was originally made by Michael Friedman by using a little more machinery of topological quantum field theory and then it was remarked by Katya that you could do this kind of Fibonacci model and then we I took a look at it from the point of view of the temporary lead recoupling theory and saw how you can see it really simply this way. So in principle, you can build quantum computers using this braiding representation. Then, what would it look like if you wanted to generalize this? Well, I just thought I'd show you that. So here's a little more about the recoupling theory. The general change, the general recoupling transformation is going to take things that look like the little guy on the left, a horizontal double y, and turn it into a vertical double y, and there should be some coefficients, which we'll call our 6j coefficients for the time being, but they will have to be fixed a little bit, just like I told you, and we need these delta a's, which are the sort of... Quantum traces for the projectors. And we're going to need some theas. And there's a very important kind of orthogonality formula about this little bubble over here. You see that bubble on the upper right? It has a top line A and a bottom line B. And you will find, if you do the calculations, that that little bubble is zero unless A is equal to B. What its value should be, that it will be a multiple of the identity A, I mean the identity line A, which really has a projector on it. And almost all the formulas here can be figured out by using that and just working in an exterior way.

57:30 So for example, suppose I don't know this coefficient, but I want to know what it is, and I do know that A is equal to B. Then I close both sides and I get a delta A on this side. And what happens when I close this side is that I get a theta ACD. So I get theta ACD is equal to x times delta A. And so x must be theta ACD divided by delta A. So that's how you can figure out the coefficients. You justify that on the basis of what's under the hood there, which are all these sums. Formalism, you can figure out almost everything by closing things up that way. I think the next slide shows what happens with that. Yeah, so if you want to figure out a 6j coefficient, you close top and bottom on the network in the same way, and on the left you get a tetrahedron, you see, and on the right you get bubbles. And those bubbles don't work unless j is equal to k, so it shuts it down except for one term, and then you just solve for the 6j coefficient. And you find that the six state coefficient is a tetrahedral evaluation times a delta divided by two theta's. So that pinpoints everything you see. You end up getting out of a formalism exactly what the basis change transformations have to be. And then the local breaking is a calculation. We did expand in terms of the bracket polynomial. We're not flat and you keep track of everything and you find out that you get a coefficient like that. That's the coefficient formula for A and B and C. Then, in order to symmetrize things, it turns out that this works. It's kind of amazing that it works in as much as you get reality in the cases that we're interested in. You need a double square root of the product of delta is divided by a square root of theta and use it as a symmetrization coefficient and redefine all the vertices that way so just multiply the whole theory by that that thing and you get the same theory but now it's been shifted a little bit and when you do that then the matrices of the transformation end up being symmetric

1:00:00 So I've denoted it by Bosch, ABCD, and IK. And that's the real 6J coefficient that one wants to use. It now gives you a unitary transformation when it's real because it's a symmetric matrix. So you get... and in that way you can produce unitary... These are all representations of the big group for more general roots of unity by exactly the same procedure that I was showing you for the Fibonacci. So you get them for roots of unity like this in terms of A. And those could be used for quantum computing as well. Although why you might want to do that would depend on an application more than anything else since in principle the simpler one is good enough. On the other hand, an application may be exactly what you're interested in. So for example, here I'm curious about whether I could use this machinery to calculate a knot polynomial. So the knot polynomial I want to calculate is the eighth Jones polynomial, or the eighth colored Jones polynomial, and that means the following sort of thing. It means that I would take my knot... And instead of thinking of it as a single line, I would think of it as eight parallel lines with a projector on it, like that, okay? And then I would calculate the bracket polynomial of that, and we could call this the eighth bracket polynomial of the knot. So suddenly you have infinitely many invariants of knots associated to the one polynomial, one for each irreducible representation of the group, if you're thinking of it in terms of the group. So there are lots and lots of them, and these are very interesting things to work with because of the fact that it turns out that if you sum over A belonging to the appropriate set, which I'll just write as admissible, Bracket polynomial of K in the eighth one. Then this is equal to the Witten, Reshetikhin, Tarayev invariant for the 3-manifold obtained by surgery on K.

1:02:30 That's one of the reasons why they're interesting. That is, Witten defined a 3-manifold invariant originally by functional interval, and then Reshetikhin and Tarayev showed that you could make it by using non-invariants. And what you need in order to make a 3-manifold invariant is that you think of the 3-manifold as obtained by surgery on a knot or a link. You cut the link out and you put it back with a twist and a reversal and you can produce any 3-manifold this way. But then you need some extra conditions about invariants. You need weird stuff like this. If you have a link nearby another component, then you would need that you could exchange it for this. What did I do? I doubled this link into two parallel components, this knot into two parallel components, and I connected the other component with it. That's a big change from the point of view of knot theory, but the 3-manifold that you get by surgery on this is exactly the same as the 3-manifold that you get by surgery on that. I'm just digressing a little bit about the topology of this invariant. So then you need to manufacture, if you want to do things this way, you need to manufacture a Lincoln variant that's loose enough so that it doesn't see the difference between these two. And it turns out that if you average over the representations in this way, that happens. Those are things I don't want to do. I deal with them in detail, but the point is that the WRT invariant then gets expressed in terms of sums of colored Jones polynomials, if you like. So, if we could write down a quantum invariant for, a quantum algorithm for computing colored Jones polynomials, then that would be interesting because you could then, at least in principle, say that you could compute Witten's three-manifold invariant as a quantum algorithm as well. So, what would a commutation look like here? Well, here's my link. It's the plat closure of this brain, so I've closed the top and the bottom of the brain. It's not the usual closure, but it's the other usual closure.

1:05:00 So I cut the top off, and now I think it looks something like a braid being acting on a network except there isn't any tree. So let's, now these are labeled A here, and they look like the upper branches of a tree, so let's stick a tree on the bottom of it, and label it zero to start, okay? And now it looks like a braid acting on a tree, where the tree is this. This little tree here, it has output zero and A and A and A and A and these two are the variables for the processes, right? So our braid is acting on that tree and we know that we have a unitary representation of the braid group for this parameter, which is the root of unity. We get a sum over coefficients, over all the different possible things that can happen in there, and there's some combinatorics to figure out what they are, of course. If I want to evaluate the polynomial, I can just close the top and work out the evaluation, because remember, when I do close things, that evaluates it. So I close the top over here on the algebra, and I'm left with these lollipops. But then there's another recoupling fact about the lollipop. The lollipop will be zero unless the leg is equal to zero. Okay, you could have fun doing an exercise with that, with the case two and see how it works. It's because in the combinatorics you get some turnarounds in there if p is not equal to zero. Turnarounds into a projector. Remember, if you have a turnaround into a projector it will shut off. Well, there's all these little interesting exercises, but anyway, it's true. So that means that when you close it over there on the expansion formula, The only thing that survived was the zeros. Everybody else went away. And you're left essentially with the coefficient B zero zero. So the evaluation of the plot closure of this link at A is B zero zero for our braiding representation at A multiplied by delta A squared.

1:07:30 So that means that what you need in order to compute it is one diagonal entry of this big unitary matrix. So now, at least in principle, we can say we have a quantum algorithm for computing the eight-colored Jones polynomial, you form this big unitary matrix from the Brady representation, and then you use the quantum computer that you have on your desk to compute it by the Hadamard test, and that computes the polynomial, okay? It computes the value of the polynomial, of course. It didn't compute the polynomial. So these are statements in principle, of course, but it's interesting to realize that you can actually do that and therefore you can compute the WRT invariant as well. That leads to interesting speculations that I don't know what to do with, like the WRT is originally defined by Whitman as a functional interval, which is also a quantum mechanical entity, but a more difficult one to handle, so at least in principle it's telling you that a quantum computer could, in principle, find a value of a functional interval. Of course, very indirectly. We don't have time to stop. Of course there are lots of questions here. I think it's really quite interesting that a very simple representation of a grade group like the Fibonacci representation, really simple, just generated by two by two matrices that are configured in such a way that you can build essentially coherent representations of lots of dimensions of grade groups. Because of the recoupling theory is enough to generate all the unitary transformations that you would need for finite dimensional quantum mechanics, right? The fact that it may be really hard to use them directly out of the quantum Hall effect is not stopping the mathematical interest in the fact that great groups are rich enough.

1:10:00 So what will come of that? It would be really interesting if more could come of that by understanding how to use it in other physical contexts than the very technical quantum hall effect. I'd like to know what would come of that. Comments? Any comments? Are we all ready for that? Would it be more efficient to compute this polynomial with a cloud computer than ordinarily? Oh, yeah, it is more efficient. It is. I haven't double-checked what happens in this instance, but in a very similar kind of setup, but using a different kind of gradient representation. Aronoff, Jones, and Landau worked out the efficiency of an algorithm to compute the Jones polynomial time for computing the specific values of the Jones polynomial. But at specific values, it's an NP problem in the usual situation and not understood to be polynomial time, so it is inefficient. Yeah, well, I lost something there when you were introducing this, mentioning the quanticized whole effects, and you showed a paper. Oh, the paper way back up the pipe here. Yeah, but the technique, the technique in that paper, how is that connected to your... Oh, yeah, of course, that's a missing link, because I just said it without telling anyone. First of all, Laughlin conjectured some time earlier wave functions for the quantum hall effect, which are of this form that you see here, where that's the collectivity of electrons, and you think of them as written in the complex plane, and this turned out to be a really good guess, and it's not so complicated.

1:12:30 Conformal field theory on a surface, which is a more complex kind of theory, which is related to the Chern-Simons theory, particularly by a witness paper about the Jones problem, and one has representations of the braid group that happen in that theory, so then particles moving around one another get braided. And they make a guess about how the conformal field theories should be related to the quantum model. We can try reading them in Hilbert. But it turns out that the sort of gradient representation that I was describing to you is essentially the same kind of thing. It's just been abstracted from physics. I sort of like this, I mean, the goal which you had over there, you had this sort of mind-break with the points on that lecture, but you said this was not practical. Well, it's not fully topological in the sense that... You see, the fully topological picture of a unitary transformation written as a representation of the braid group waves nicely in the breeze, right? There are these braiding generators, and if you move something topologically, they move, right? Here they're obstructed from moving because of the fact that there are these local unitaries which aren't particularly topological. And the idea is that if you could build all the transformations in the quantum computer and each one is a topological transformation, then maybe the whole computer is... A little bit impervious to perturbation because of the fact that the topological transformations will be the same if the pathways get moved a little bit.

1:15:00 So yeah, that was the dream of Friedman and people there. But could you think of these little dogs in a break, or could you sort of replace them by other breaks, and then you get something? You want to replace them by little breaks? Well, that's what I showed you how to do in a coherent way, actually. Because, you see, I said, we'll just change our tune. And we aren't actually working in attempts of representation anymore, but we're going to change our tune. Now we're acting. We have three-strand braids acting on these lines, but the space is two-dimensional, so we actually have unitary transformations being generated that are two-by-two matrices, so the whole construction starts at two-by-two matrices. Everything is braiding. Everything is built out of braiding. As I understand it, the kind of, you know, that final line, everything is built out of braiding, you really do intend that to be taken in a literal sense, and this is really being offered as a suggested fundamental ontology. Are we not to think of this? Yeah, I don't want to take it too seriously. I'm noting mathematically that I can generate all the unitaries I want by backing up and just mapping braids in, right? Yes, exactly. And then whether you should take that as an ontology is another question. It's another question, but I think if I was to really press you... Well, you know what I don't like, but it doesn't mean that it's the way the world works. But it's rather kind of close to the circuit of your original... I'm actually just interested in how it might work, right? If it worked that way, that would be nice. And if you could show me exactly, if we could see why it was different when you did it. That isn't a possible ontology, but it would be interesting. Yes, I know. But it is very interesting, because there seems to be a suggestive idea here of a possible kind of topo-dynamical ontology under that, which goes back, as you said, with your first slide from the other day, to Galvin, into the vortex atom.

1:17:30 Well, I think it's fun to play with ontologies, but you've got to underline the word. Yes, I completely agree. The other thing I wanted to talk about, but we can talk about it privately, is there has been this amazing development, and part of it coming directly off the bracket model with Jones polynomial, and part of it coming off other bits of knot theory, where people have figured out how to associate homology theories to the states. So you have something that looks like a partition function and you're usually just evaluating it, but instead, each state becomes a holder for part of a chain complex and you build a chain complex out of the states, and then the chain complex's homology ends up generalizing the original invariant, in this case, say, the Jones polynomial, you get a homology theory, Khovanov homology, which generalizes the Jones polynomial. Now, I don't know what this means physically. I would like to know what it means physically. That could be a conversation over dinner. The way you build it is quite easy to understand, actually, because in the abstract, if you want to build homology, what you need are some notion of adjacency and some notion of order. Something should be of higher dimension than the other, but it doesn't have to be a dimension. It just has to be some kind of order. And the states are ordered in terms of various things related to the way you set it up. This could happen in a partition function in real physics just as well as with knots. So the question is what would it mean to associate homology to partition functions in physics? But in the case of topology, there are various motivations that make a lot of sense for this construction, but the construction is simple on the one hand and very unobvious on the other, and it would be another lecture to talk about it, so we stop here.