Spin Networks & Anyonic Topological Quantum Computing

0:00 Oh, thank you. So when you think about representations of the grade group, one form of representation that we're thinking about is assigning a better space to a strand and then looking for some of the on-mapping, which generalizes the reputation, and there might be a representation of the grade group that we put together. Now, why should we consider grades? Well, it's interesting to look for representations of the braid group which are also quantum operators. It's certainly an interesting game. And the original idea behind that was that if you could represent everything in terms of braiding, then maybe there would be models, physical models, which would have that property that would be stable under perturbation, just like braids are kind of stable under pathological footprint. And whether that idea is real or not is certainly up in the air at present time. And on the other hand, one can just look at this as a question of to what extent can you represent unitary operators by rating operators, and think about it as a mathematical problem.

2:30 There's another thing behind me, there's something behind me that's making a topic for you. So, as I said, if you look for an R, like the R that I had on the previous slide, it takes B tensor B to B tensor B, then on the one hand, you can use it to get representations of the brain group, and if it's unitary, then you can try to see whether it might be useful as a common gate. So for that purpose, it's useful to have translated the R's into the Brady relation into the R's. And here's the relation that it is to the R's. So here's R, and then having the extra line, you need R tensor 1. Then one tensor R and R tensor 1. So this equation, which is probably getting back to the equation in the rows a long time ago, is what I want Brady-operated to satisfy. If I'm looking at it from that point of view, it's not the only point of view we're going to do, but I'm going to talk about this point of view from one of these. So the radio operator operates on two strands? Yeah, so it's higher. It's operating on, um, so, hold on. So R is taking V tensor V to V tensor V, and you can think of V as being one qubit space if you wanted to. And you can think of V as representing one single strand if you want. So then I want to remind you about what I've put in the quantum gate. So, quantum gate is just a unitary and v could be taken to be c2 so that it's a space of cubits. And for the purpose of this talk a quantum computer is just on u like that. And local unitary transformations in this language are matched on that v to itself, which are elements of u. And then I remind you that it turns out that a single entangling of local unitary transformations is universal compliant computation. So I'll say that a G taking B tensor B to itself

5:00 is a universal gate. So all you need for a two-cubit gate in order for it to be universal in the presence of local unitaries is that it be entangled. With that, there are lots of examples that are partly topological ways of representing quantum computing, just exactly because because they're entangled. So the standard two-pubic date, of course, is C0. You can use that. And that is indeed entangled. Maybe I should remind you of this. So the date is entangling if some decomposed tensor product of things when G is applied to it is not decomposed. So, it's easy to see, in the case of two qubits, that the state is entangled exactly when the determinant of the little 2x2 matrix of coefficients is non-zero. So it's quite easy to check whether a given 4x4 matrix is entangled or not. And so, for example, it would be the exercise to check that c-not is entangled. And if you're encountering some... Would it, would it, would it, would it, would it be entangling? Swap is not. Swap is not. Swap is not. But it is. It is. Pardon me? It is. You could say it's completely entangling. Yeah, but not entangling. It must be some problem of terminology. When I say swap, I mean... And the thing is, it becomes a four-cubic state, two middle-cubics. And all this does is pernive the two middle ones, and it will turn it around and make it so nice to set. That's right. So it doesn't entangle any pure state inputs, but if you give it halves of entangled states, then it entangles across the... If you have four qubits, and you act on four qubits, these to be already intact, and these to be already intact, when you act swap on the middle ones, then these two are like... Alright, I understand what you're saying, but what I'm saying is that I'm just looking at two qubit states, and asking whether the four-by-four matrix can entangle that.

7:30 In order to have this, we're talking about this criterion about uh, complication in the cell. So the last part of you, this is a Well, this fellow matter is not going to be entangling any two-qubit state. As it happens, slope isn't one-qubit is one-qubit is one-qubit. That's the count. If you change this one to a minus scale, I wonder if we normally regard SWAP as entangling because it can create entanglement from some sets of parties that were not initially entangled. But the question is, can a gate that is not entangling, in your sense, be universal in What doesn't really create a pattern if you're trying to do a way to get across positions, right? So you say Swamp is entangling because you're thinking it's a two-party. But we're talking about computations, in-cubits, in-cubits. And we have to take care of the maintenance of all these in-cubits. They're looking at two. Right? I think that's the reason. I don't know. I like Charlie's question. I have it. It should be obvious that you can't. If you start the computation with a cancer product state, right, then you can never get anything but a cancer product state in the case that's not a entanglement. Right. In other words, you couldn't make the entanglement that you fed into these. Although, maybe if you were just allowed to use it once at the beginning. So, you get a lot of it when you find it back, I guess. When you find them at the beginning, you'd swap JavaScript and swap it out. But, I mean, I guess it's at least logically possible for your definition that you could start with a very little bit entirely, and then you just create more. Yeah, I doubt that. Okay, but so I guess... Yes, alright. Okay, go ahead. Alright.

10:00 So a couple of examples. This is entangling. And here's the Bell basis, which is such a burden. All three of these guys are solutions to the dandrex proportions. So there are some natural examples which are topological. And even interesting from the point of view of topology, Sam and I have to start analyzing this one, that on the topology side, which I won't go into here, This can detect non-trivial linking, so it can detect a linking of a below-name branch which doesn't have no linking. So there are sometimes non-trivial invariants associated with that. But the problem with thinking about this is that you aren't getting a complete representation of quantum operators for quantum computing in terms of trading operators. you still need arbitrary elements of U of 2 for the local unitary transformations. So these statements are, some grading operator plus elements of U of 2 are generated, and that's the best you seem to be able to do. On the other hand, Friedman, Kupai, Gerson, and my collaborators, suggested using what's called topological quantum field theory, thought of as a theory of anion, so I'll explain what I mean, and make really deeper models for topological quantum complication, where everything would be represented by grading. So what I want to do is show you a simple example of how such a model comes about. Thinking, yet again, not theoretically, but I'm going to talk for a few minutes about the background of what one means by TQFT. And so on. So, there are some nice ideas here. Not all of which I have to use in the end. One idea is that you think about the category of cobordisms of manifolds. So you have a manifold and another category of what? Manifolds. Select surfaces. I'm going to explain. So everybody knows what a manifold is. It's like a surface or a three-dimensional space. Possibly with boundary, possibly not. Here's a surface with boundary. The boundary is a circle at this end and two circles at the other end. and one says that these collections of circles co-board.

12:30 That is, there's a co-boardism between these two circles and that one. If you can fit a manifold in between them, it won't hide it much. What kind of co-board is a circle? Pardon? Yeah. Is category used in a technical sense? Yeah, it's a category in a technical sense, where the objects in the category are the end manifolds, And the morphisms in the category are the n plus 1-malfolds that move from one-malfold to another, and I found it. So, in the spirit of category theory, the morphism doesn't have to be a set theoretic map, it's just a relationship between humans, that's of some use to some of it. So it's not a mapping from m-end to m-prime-end, it's a relationship between them. And then what you want for the idea here is that if there is a relationship of this kind then there should be a mapping from here to here of some vector spaces that are associated with these. So if this were a lecture on string theory then these could be strings and there might be some vector space associated with these strings as particles which would be going to that. Or if I went back to point pictures, then this fits with the sort of thing that we were just talking about. We would think that here they're often mapping from V tensor V to V since there's some fusion going on here. So this is a language for talking about this sort of thing. And sometimes one uses higher dimensional pictures rather than just pictures like this. Sometimes one takes a higher dimensional picture and turns it into a lower dimensional picture. if a manifold just bounds something like this then one thinks of that as corresponding to a mapping from the scalars to the vector space that's associated with this end so one is looking for a functor in this sense, and then for three manifolds you can decompose your three manifold into two pieces I wonder if I have a picture of that you might decompose with three manifolds into two pieces

15:00 which are just rusting like this so that the union of M is the 3-manifold here's a surface in between and in this picture because this surface found something there's a vector in inspector's space and because this surface found something there's a vector in inspector's space and then you put them together and you hope that that's going to be the invariant of the 3-manifold, it's going to be some topological invariant of the 3-manifold generalizations here are the ideas of putting bras and kets together. So a bra or a ket as a notation, in fact, is a cohortism of that sort. It's two points which together bound will occur. And one can actually make a piece of that. How do you know which is a bra and which is a ket? Oh, I'd call this a bra and that a ket. So in the other picture, the one on the left is the bra and the one on the right is the cat? That's a bra, this is a bra and that's a cat. The one where you're like sewing two socks together, so you can't put your feet into it. That's the one on the left is the bra. This would be the bra and that's the cat. But I mean, that doesn't, if you know what you have to be real, I wouldn't worry about it. But it cannot be complex. It seems to matter this way. Yeah, yeah, it does matter. So I'll be more careful. I sometimes am a little bit cavalier about who's who. But kets for me are vectors. And you'll notice that this is a vector. That is, a mapping from the scalars into a vector space is just a vector. So this is arranged correctly. Don't worry. because I'm not actually going to work at this level, I wanted to show you how this level ends up connecting with them like, on the Fourier level. But in this level, one is talking about invariance of manifolds by somehow associating a vector space with the surface and making this sort of thing. And Witten was the person who originally suggested that one should do this and that it should be related to physics by means of functional intervals. entire three-manifold someday, or the H-fields over the three-manifold in some way, and then

17:30 the vector space that's associated with the surface has to be figured out. It's not hidden, but in his famous paper, he starts with this functional interval, and then he explains how, heuristically, one should associate a certain vector space associated with the surface and that that should be the same as doing this functional loop and there's a whole story in the back of that but some of the patterns of that are worth understanding. So in particular one of the patterns that comes out of that is thinking about decomposing surfaces into little pieces that look like particles fusing or particles being created if you take the bits of surface the cobordisms and just draw them schematically like this. But, you see, you can take a surface and you can decompose it into pairs of hands like that. You can do it in various ways. And the underlying structure of this business involves fusions and creation vertices like this. And it's that combinatorics that I'm interested in talking about. So, you're going to have kind of an abstract particle theory, where you have some particles that interact in these ways, and it's going to be topological because of the context that we're talking about. It's going to have some braving in it, and when you twist near a vertex, there's going to be some operator that will be applied. It will usually, in these cases, just be multiplying this by some complex limit. And there will also be rewrites of these patterns of fusion and creation. So you could have this pattern of fusion and creation getting rewritten in terms of this, so recoupling. The recoupling is written in various ways. I guess I wrote one of the other ways of thinking about it over here. You can think about recoupling the pattern, the graphical pattern of it, as unplugging this vertex and then just moving it along and then plugging it back in. Another way of thinking about the patterns of this is in terms of algebra. Because if I have x, y, and z, and I think of this as x times y, and then this is x times y times z, then you see that if I recouple, I'm now looking at the pattern of x times y times z.

20:00 So recoupling is reassociated. And because of the fact that in these theories one wanted the vector spaces that were associated with the surfaces to be independent of the way you decompose them into pairs of hands, you need some kind of change of basis formula that takes you from one way of decomposing it to the other way of decomposing it, which is exactly this pattern. And these then need to satisfy various conditions. They need to grading being there, and then some other conditions, which I'll show you, pentagon, naturality, and hexagon. Oh, this is a question with the stamp. Are you working in stamp-free algebra with this diagram? Oh, this is up into some interpretation. It will depend on what I have in mind. I'm going to show you some models later where what I have in mind is certain non-invariants that I'm going to evaluate, and these represent bits of the non-invariant, and I'm going to evaluate them, and this formula then says that this is equal to that, where this being equal to that means that when you plug this into some larger network and evaluate, we'll get the same result. That's one way. Another way could be that I'm thinking about different possibilities of interactions from the labels of the particles, maybe a set of labels, and I may have a vector space where I start with x, y, and z, and I'm supposed to end up with w, and there are various ways in which I could end up with w, because x and y can interact to give me something, and then these can interact to give me something. I look at all of the different intermediate possibilities from x, y, z to w. vector space there with a basis. And then this formula would mean that a basis of this vector space with these starts and let's say these three starts and that end would be rewritten in terms of another vector space with basis these three starts and that end. So those are the two main ways of thinking about it. There's a directly vector space associated with it or else it's in valuations. Is that the picture? So I have a tendency to just draw it as abstract formalism, keeping in mind that I might interpret it one way or the other.

22:30 So what do these funny conditions look like? They all come out of certain models, and I'll show you that in a moment. But some of them are obvious that you would want them. Like you would certainly want something like this if the topology is to interact with it well. On the other hand, other things come from fundamental aspects of the combinatorics, like this one. If you take an associated product and start rewriting it, you'll find that you can, starting from here, A times left associated, you can do two things and get over to here, or you can do three things and get over to the same place. Now, there are transformations going on of bases and vector spaces here, or formulas that say that something should be equal to something else. It's not obvious that if you go through my formula, the one that I wrote there before with two transformations going this way, or three that way, that you get the same result. But you can ask to get the same result. And similarly here, topologically you should be able to move this line up. And then you can go through recouplings and bits of grading and find yourself getting to the same place by going all the way around there. These are the simplest things that could happen, that you would have identities in both of these cases. and that turns out to be a sufficient condition for the vector spaces associated with those surfaces to be well defined. It doesn't matter how you decompose them as it can. So that's one of the places where this comes from. But it also comes from angular momentum recoupling theory and that's actually where I'm going to start. So I'm going to build for you a theory of this type by starting with the basics of angular momentum forming it a little bit so that it has some topological variable. And that's why we wanted to talk about Penrose's thing now. so let me tell you what Penrose did. Penrose, a long time ago, was making pictorial tensor calculus to handle various things that he was interested in. And this one arose from thinking about SU2 or equivalently at the algebra level SO2C. SO2C can be called as the 2x2 matrices that under conjugation, under A conjugation, what do you call it when you multiply by the transpose? A epsilon A

25:00 transpose is epsilon in any case. For epsilon is the little in general if you took a 2 by 2 matrix and you did that you get the determinant of the name of the 1 by xl so if you think of sl2c this way then the epsilon tensor is the underlying bit that's important for thinking about it and he was looking at tensor diagrams involving sl2c so he made diagrams for the epsilon in this form and there's a basic identity about the epsilon that I need to use, which takes epsilons, two of them, and turns them into Kronecker deltas. This is just a little switching, I think. You see, if this is a 1 and this is a 2, and this is a 1 and that's a 2, then you have epsilon 1, 2 times epsilon 1, 2, which is 1, right? And here you have 1 and 2, 1 and 2. These are Kronecker deltas, a little line with epsilon, which has a label on top and a label on the bottom, if the labels are the same and zero otherwise by definition of diagrammatic tensors or diagrammatic indices. So this one will give you a 1, but if you switch the order of one of these pairs of indices, 2 to 1, then this is the one that turns on and it's got a minus sign and so the sign switches correctly. So this little diagram represents this identity, and he wants to make various diagrams of tensors of this sort. I'll show you what kinds of diagrams. and he wants to have his diagrams calculate for him so he wants the diagrams to be as simple as possible and he runs into the following problem about the epsilons if you switch to legs of an epsilon it changes sign that's its definition of course if you run two epsilons into one another like this which looks like a little topological wiggle it also changes sign and of course there's the identity and if you tie two epsilons together top to bottom like this sum over the indices that are tied together, just like you're doing tensors. So this means that you sum it over the common indices that are not three n's. You get two, of course. So this is the story if you calculate with epsilons. Here's a chronicle delta. Here's the epsilon. Well, Penrose regarded these two in particular as being quite awkward and would rather not have that sign change going on in his depth. So he redefines the epsilons by multiplying And the crossing, by putting in an intrinsic minus sign.

27:30 So when you see this crossing, it means minus the Kroniker deltas of the crossed lines. And then you get a different identity. The epsilon identity turns into this one because of the switches of signs. The sum of the cup and the cap plus the sum of the parallel lines plus the crossed lines is equal to zero. And the value of a loop is minus two. resulting in a paper that he wrote entitled Applications of Negative Dimensional Tensors. Well, this choice, which is meant to simplify the di-dramatic calculus, has some very nice consequences for him. He discovers that not only does it help with the signs, but it makes the calculus planar topologically invariant. I've indicated it here just for fun. So, for example, if there's a little loop here, then you see this formula says that you can expand a crossing into minus this one smooth that way, that one smooth that way. So I can smooth it this way and a little loop will appear. Or I can smooth it this way and it just becomes a little cap. The loop has value minus 2, so this is 2 cap minus cap, which is cap. And so the little loop can be thrown away. If you have two lines that are passing across one another like this, and you do the same exercise, expand, this now doesn't matter, and this one expands, and these cancel, and you see it's topologically invariant in the plane, in the sense that two lines are crossing through one another and similarly, three lines crossing through one another can be shifted. So he found this basic calculus for his tensors, which is invariant under flat topological That exists. Now, of course, from present vantage point, it's very obvious that once you try to generalize this, because there's lots of interest in non-theory, but probably in this context, it wasn't thinking about non-theory at all, so you didn't try to generalize this to include over-and-undercrussings, but that's what I'm going to tell you about in a moment. But I want to tell you about the rest of the technology of Penrose's theory, which is really about angular momentum recoupling done in these diagrams. So this is some of the technology.

30:00 one of the pieces of technology is to take the sum over all possible permutations of n strands minus the sine of the permutation and divide by n factorial, this gives you projectors this is the underlying bit of algebra that corresponds to the irreducible representations of SO2 because you do a symmetrization process of that type you make the representations so these are in this network theory, and they're projectors. If you multiply two of them together, stick them together, you get the same thing back. If you have any turnaround in it, it'll be equal to zero. And because of the binary identity that underlies everything, all of these get expanded in terms of diagrams that have no process, if you want. And so, for example, here, I've taken the two-symmetrizer, which is just one over two factorial times identity minus switch, and then I expanded this using the binary identity and collected terms. There's an identity and an identity. You get the identity plus one half of this guy here. So this is part of the technology of this, all of these symmetrizes. and let's just take a look at so for example I said that if you tied it together at the bottom it would be zero you see that here in these turns because you see when you're tying something on the bottom of this you get that loop which is minus 2 which cancels out and when you multiply them together and you expand one of them you see you get a turnaround and that cancels out. So you can see how the calculus in these projectors works at the level of these sums of diagrams without any crossings at all. If you go to more complicated things of course it's complicated. For example, the three symmetrizer is this sum, which I haven't bothered to write out in terms of things that don't have any crossings, but it will sum over all 14 things that are there. There are recursion This is actually due to Wenzel and Jones in the general context. But there are recursion formulas like this. So for example the three can be recursed in terms of the twos and I can continue to write that out. So there's this technology of the projectors. And then with the projectors in place

32:30 one gets to the angular momentum recovery theory in a completely diagrammatic way. In this way, you need You need A half spins, B half spins, being projected to C half spins. And the possibilities for doing that are, you may be familiar, if you've looked at angular momentum recoupling theory in a book on quantum mechanics, you see these conditions. That A plus B minus C is greater than or equal to zero, and A plus B is conjugate to C minus two. These are twice the spins. Here, you just connect things up. you take a symmetrizer and you connect it up in such a way that i of its lines go down here and j over here you solve for that and you find out that you need these conditions in order to stick things together in that way so those are the vertices in the theory and then you then have a kind of an elementary particle theory where you think of the little symmetrizer itself as a particle and then these are the particle interactions you take an a particle and a b particle and you stick them together in this way particle comes out and you can try to figure out what kind of relationships that are. So then in calculating with these networks there are some basic facts like this one corresponds to Schurzmann. If you have a little loop in here and you have a here and a prime here, then the only way this can be non-zero is if a is equal to out the coefficient. If you won't worry me about why A has to be equal to A prime, we can figure out the coefficient very easily by imagining that there should be a coefficient, and closing this and closing that. And here you get the value on the right hand side you get lambda times the value of whatever this is when it's closed. And on the left hand side you get a theta graph labeled A, B, and C. So the lambda is the ratio of the value delta A, which is the value of this. So calculations look like that in this theory. And there are recoupling formulas. And by the time you get to this level of the structure, everything satisfies all those identities

35:00 that you could have at this point. Pentagon. Pentagon is what's called the Elliott-Biedenheim identity for recoupling coefficients. and we don't yet have any grading but we'll get to grading soon enough. So we've got transformations like this and again this principle of let's close this up and put a line here if you close this up and put a line here then because of the because of the Schur's Lemmon kind of identity this edge here would have to be equal to the edge out here and so all the terms go away except the one you're looking for to put in J and you get coefficients in terms of little network evaluations, figuring out this tetrahedral network, figuring out this little loop, and some fatas. So that one can go ahead and actually get all the calculations here if you want. So at this stage, what are the values that A, B and C take? So, yeah, so A and B and C are taking values, integral values, and you're summing over things by the admissibility conditions for, remember, a plus, like a plus b minus j is squared equal to zero, I'm sorry. H, A, B, C, O, zero, equal to zero. And how does that make this spin? Oh, just about two. If you have two lines and it's spin, it's spin. Okay, so that's what this looks like. You see it's almost a topological quantum field theory in the sense of learning all those identities. And what did Penrose do with it? Well, Penrose originally wanted to put in these networks as an abstract substitute for space, as a kind of process background for getting space to space-time. So he thinks of abstract spin networks like this, basically arbitrary trigly-owned graphs. And he proved a theorem that has the following form, that spin network evaluations can be used to define angles between three M's in such a network, and that for appropriately experimentally repeatable large-ish spin nets,

37:30 the angles satisfy restrictions of directions in a free space. so you can find out what the angle is between two of these by imagining that they exchange some spin and doing a calculation about it and then if you assume that the measurement we can make by doing such a thing is repeatable then the angles come forward to angles in a true space of course you started with information about the rotation group to begin with and then you left it there, but it's all combinatorics at this point. So there's a hope that you might be able to use this as a substitute for space in some discrete sense. But in fact, all you get is angles, and you don't get distances. So more work needs to be done, and people are still working on what you might do to spin that steering of the work for you in that sense. Next. Transition to anthropology. Well, it turns out that Penrose's ideas fit directly into a generalization that gets you tautological invariants. What you can do is you can take the binary identity. This isn't the way this got discovered, but it fits, so I can say it this way. You can just generalize the binary identity. You see, if I took A equal to minus 1, this would be the binary identity. But now it's in a different context. It's in the context of making a non-invariant. And there aren't any matrices in the background. It says that if I have a knot with a crossing like that, then I can think of smoothing the crossing in two different ways. And get related pictures. And then this becomes a recursion relation that I can work with to calculate something about the knot. And it turns out that if you choose the coefficients this way, A and A inverse, and the value of the loop, minus A squared minus A to the minus 2, then the resulting thing is invariant under topological moves, which are generalizations of those flat planar moves that we were looking at before. In fact, these are the braiding moves, right? So if we were restricting ourselves to grades, this would be invariance under braiding. And then when you upload some little curls in the diagram, you get some extra multiples like this. This calculation can be normalized.

40:00 It's well-defined if you think about it. And you can normalize it by multiplying by an appropriate factor that has to do with this behavior on curls so that it's invariant under curls as well. And then it's basically the Jones polynomial. So the Jones polynomial is sitting here from this point of view as a very close relative to Penrose's basic final calculus. I think I've got one more illustration. Sorry, I didn't understand what is more general now. So everything is not expected to be greater already still, but you have something more that it's nice to do? Well, remember what minor calculus was. Minor calculus was a flat calculus, and it was actually describing a certain test in Here, this looks like minor calculus, and it's normally identical to it if you take A equal to minus 1. Then you have minor calculus, let's draw it. Minor calculus, which is the underpinning of the spinence, says flat crossing equal minus smoothing minus other smoothing, value of root equals minus 2. That was the underpinning of the technology that I was just describing to you a moment ago, right? And it was planar flat invariant. So what happens is that by some serendipity, the whole thing generalizes very nicely to actual invariants of nons and links in three-dimensional space by deforming minus 1 into an A. And you get an extra algebraic parameter happening. and then braiding is happening. And in fact, and I'll sketch for you how it works, everything generalizes, the recoupling theory is there, but now it's got braiding in it, and so we're actually about to look at an example of TQFT in the sense that I was talking about earlier, something that satisfies all of those kinds of relationships. Does that answer that question? Maybe it's too much, but I just thought I'd show you what that looked like. You see, if you started with an arbitrary A and a B, and you had that idea that you would like it to be invariant topologically, then you're not working on matrices, you're just looking at the properties, the formal properties of expanding. And then if you take this situation that you would like to be equal to, this, and expand it, using those rules, and collect terms,

42:30 you find out that this part here ought to be zero if you were going to get it to do what you wanted it to do, and that this should be one. And so that leads to the choice of B equal to A inverse of D equal to whatever it is, and then you have the topological So it's a very elementary step. So there's a book about this that we wrote, Croftman, I wrote, and with Sostanus-Livings, if you want to see this whole recoupling theory detail. But I'm about to sketch how it works. So you generalize the binary relation to this. You define the symmetrisers. Now, it's amusing that you actually can't define the symmetrisers exactly in analogy to the old symmetrisers, but you take the braids and you, you take the permutations and you lift them to braids. I'll draw a picture on the next slide. And The sign of the permutation to some variable raised to the number of transpositions that minimally take the braid back to the identity. And you have to generalize the factorial to a 2D form factorial. It works. The lift of the braid is a standard lift of the braid like this. here's the, I'm sorry, the loop to the permutation is to do it so that the crossings are all the same. So here's an example. Here's our two symmetrizer again, and this was the factorial. I know you don't remember that, but no matter, that was the factorial, and this was the coefficient. And when you collect up the terms and expand, just like we did in the Biner calculus, you end up with this gadget. This is the gadget you can look at. It's a projector just like the other one was a projector, but now instead of being minus one-half, or plus one-half, it's minus one over delta, where delta is that generalization of minus a squared minus a to minus two. So all these things deform in that way, and now we have a larger range of particles to interact. In particular, the recombling theory, just like before, goes through, and now there's braiding of vertices, and the braiding looks like this. It's a multiplication by some power of A. depending on who's there. And this thing now satisfies all of those identities hexagon. So this is a natural model of all those things that's available.

45:00 And one can try to see what one can do with it. Well, you can make three manifold invariants out of it and all of that sort of thing. So let's talk about this for the matching antions. Well, I can take about 15 more minutes with the outside. Sorry. I'm pushing the end of this. I didn't mean to. But finally we got to the thing I wanted to show. So here's our, we want to make a particle period. I'm going to use two symmetrisers. and you see with two symmetrisers you could have two of these and they fit together and go to one or you can have one of them and it will fit together and go to two or you can have two of them and they will connect up like this and go to nothing or you can have nothing giving rise to two so I have particles which I'm calling two because of this reverse labeling and two plus two should be equal to zero or two but I don't want four you could have four right two plus two could be equal to four that's a possibility but I have to get rid of it theory that I could. I want a theory with one particle and none. Or one particle with a charge and one particle with no charge. And this particle, the one with the charge, can interact with itself to either produce the particle with no charge or the particle itself. That's what I'm after. Just the simplest example that I could possibly make. Right. So for the recoupling, we want very simple formulas like this. We want that the 0 will resolve into 0 and 1 and 2 with some coefficients and similarly here. Now if you think about the recoupling for a moment, you see you have a matrix. It's a little 2 by 2 matrix that describes it. And then you could think of it this way symbolically, that these two basis elements are equal to this matrix times these two. So if you rotated everybody by 90 degrees, then you see that F squared will be equal to the identical. That's called the orthogonality relation. It's a generalization of it. You can look down a little thing. But that doesn't mean that F is unitary. And what I have in mind is that F should be unitary, so that I would have a little braiding operator,

47:30 and I would also have a unitary operator F. which could be used to describe other gradients, as we'll see. So we have to keep an eye on F and see whether we might be able to get it to be unitary. What are the calculations going to look like? Remember this. I'm just illustrating it here in the case of two lines. If there's nothing coming out, that's the case where this is not equal to that, and you expand it, it ends up being equal to zero. I'm just doing the exercises here. Otherwise, you get theta over delta times a line, where you have twos here. figure out the theta, which is all 2s, and the delta, which is the loop with 2. The delta, which is the loop with 2, if you just expand out, is little delta squared minus 1, where little delta hasn't been figured out yet. In order to make this work so that 2 plus 2 is never equal to 4, we're going to have to specialize the a. And I'm leaving it open for the time being to see what I have to specialize the A to. To what I have to specialize. So now some amusing calculations, which I'll just indicate to you. You need to figure out this theta. And you do some expanding and calculating and you find out that it's equal to this. And you need the tetrahedron. And you do some expanding and calculating and you find out that it's equal to this. Now, you needed the tetrahedron things and figure out some coefficients. So, for example, if I was interested in B prime as a coefficient, then I close this top and bottom like this, where the solid line is the particle 2 on 2, and the dot is empty. And I find that this one is zero, and that one is whatever it is, and then I get B prime by calculating like that. So we don't want seem to actually be a calculation of the type of result. So, here's the result. You do a little calculating of this sort, and you find that B prime is tetrahedron, prime is delta divided by theta squared, and similarly, calculations for A prime and A and B, and there's the matrix, the recoupling matrix. And then, to see what we needed, I'm going to square it and ask, when is it the identity? You see, if 4 is equal to 0, then that will be consistent with F squared equal to the identity. can do it another way and just check if the final result is going to make 40 equal to 0. This is the quickest way to find out what restrictions you're getting. So here's f, and if there aren't any other terms in the recoupling, then m squared will

50:00 be equal to the identity squared, and you find that 1 over delta squared plus 1 over delta had better be equal to 1. And there's the reason why this is Fibonacci, because this is what's satisfied by the reciprocal of the former issue. So then, if little delta squared is little delta plus 1, so that's actually the problem being little delta, let little delta be the problem being, that's the loop value, single loop value, then capital delta is actually equal to little delta, which makes life easy. and you need that one is equal to this and you find that capital delta is delta you can calculate the theta and the tetrahedron and you find that indeed F squared is equal to the identity and that F is this which isn't symmetric and I would like F to be symmetric and then if you've played around with these a little bit you realize that it's kind of amusing The theta could be adjusted any way you like, and this would still satisfy the recovery point. And in fact, you can adjust the model and make this matrix symmetric very easy. Because, you see, what I have in mind, sorry, I should have explained a little more carefully earlier on. I was going to explain it later. We know what the braiding is going to be like here. what I want to find out about is what the braving will be like when I'm over here because I have in mind to make great group representation by letting the bravings act on these guys and return them to things at the same time now these guys can be thought of as little process vector spaces where you start with some spins and you end with a certain spin And then you go through, like a billiard ball machine, various possibilities in between each firm basis. And then if you put a braiding operator on the top of it, it will reconfigure this as some sum of things of that type. So the question is, how do I find out, knowing what this braiding is, how do I find out what this braiding is? And the answer is, you should recouple.

52:30 Do your braiding. If you do that, if you start here, you recouple it, then you know what this braiding does, do it, and then come back, you get the braiding operator here, between other things that are just adjacent. and those are given by in this case by conjugating by F. So if F is unitary and R is unitary then the new grading operators that I'm generating in this way will also be unitary and I'm in good shape. So that's why I want F to be unitary because I'm trying to make unitary grade. I should have said that earlier but but F is very adjustable It's easy to adjust. All you need to do is multiply each vertex by some constant, and then take a look at the factors that you get, and you see, sorry that's a little messy, but if you just multiply each vertex by an arbitrary constant, everything will still work out right, and it will have the effect of doing this to the matrix. So I can just choose alpha so that the matrix is symmetric. And after you do that, you find out that you're looking at this matrix. Remember, tau was the solution to tau squared plus tau equals 1, and tau is 1 for delta. Here you have the square root of tau. And this may look familiar if you heard a talk earlier in the semester by John Pressfield talking about this, so you saw his notes, words, and so on. So there's the little f. and then the braiding well we wanted delta to be equal to the golden ratio and if you take a to be equal to e to the i 3 pi over 5 then you see that delta is the golden ratio so that's the a and then the braiding formula that I showed you long ago tells you then that for the particle The result of doing braiding is this, and for just a bear situation, the braiding is this, so that the braiding matrix is this, and the F is that. And that's exactly what you'll find, for example, in Presco's notes.

55:00 r is a little 2 by 2 grating matrix. Now, what have we got here? What do we have? We have the following situation. Perhaps I should have started with this side. You have this little theory of particles called 2 and 0 and you can consider these vector spaces of processes where you start with 2's and you end up with 0. For example, suppose you started with four twos and you asked about the diameter of zero. Well, you see that two and two can give you a two, and then those can give you two and zero, or two and two can give you zero. But what's going to give you a zero here from a two has to be a two, because zero and two combine with two. So there are only two possibilities. This vector space with four starts and one end has exactly two elements in it, and that's the cubic space. But that's got internal structure of braiding operating on it. So you don't have to do this slightly complicated business of calling your qubits certain combinations of these little anemones. And then you get internal braiding going on, generated by F and R, or really R and FRF. And then you think about R and FRF in U2, and you see that it's not generating any finite subgroup, and you can argue that it's against the products that are going against in U2. Now, I think it's an interesting puzzle, and I don't know who knows what about this part, but it's an interesting puzzle to see whether you can get familiar matrices in U2, familiar local humanitarian transformations, by some simple formulas involving emerald. you really have to take limits to get your favorite matrices. It's not obvious to me. I was playing with just a low level on the computer. But maybe somebody knows a lot about that. In any case, that's the local unitary transformations. And then by using larger vector spaces of this type, you get the global transformations. So that's the way this model works. And it's nice to see that it's possible. that you can generate things entirely by using one coherent system of three. And are there really any anyons in the brain that can do this?

57:30 So let's see, I was trying to make up questions. You could make up a few more. Do there exist physical realizations for these kinds of representations? How efficiently can standard gates be made or approximated from R and F and the recoupling apparatus? So that's a mathematical problem. We'll get clearer as time goes on. How do questions about topological entanglement versus quantum entanglement resurrect in this context? since we have lots of large operators. SpinNATs and Q-deformed spinNATs have been used as substitutes for space-time and measurement operators in quantum gravity. There's lots of literature on this square. A lot of the quantum gravity theories done by Lee Smolin and his collaborators is written in terms of these kinds of networks. So you could re-examine quantum gravity from this point of view, i.e. unitary transformations in the quantum theory are generated by the spinNAT as well. And what do quantum algorithms look like at this time? I could mention that there are some knot polynomials that are associated with the two-strand gadget, but I think I'll leave that for discussion afterwards. There is a way to express, if you look at the knot invariants that are made by just taking your knot and replacing it by two strands and a symmetroxer, then that non-invariant can be described in various ways that are prevalent in reality. That's probably not a lot of the topic here. It's all stuff. Question? I wonder whether if you make something that approximates a cubit in some piece of physical material, and then you make another one, another piece of material, can I get them entangled? In other words, do I get, do I recover a simple tensor product structure ultimately if I have to go through some complication to get it of the raw materials on this material?

1:00:00 I'm not quite sure I'm getting a question from the different people. That was the slide, right? Okay. The zero and one are living in here, right? And the other, another one would be living in here. And you could put them together into some larger space. So then I could have zero tensor one sitting there. And then I could do some braiding, cross braiding among these. So that's the sort of picture I was thinking of. It's not natural. Well, I just, I'm not sure the matter does it. There was always a hope of getting some complicated, perhaps one of all effect type of actual physical material that would harbor these anyons and use certain operations on it to do quantification. But I was never quite sure that you would get a, whether you would get an over space that would grow exponentially in the amount of this material. As you do with ion traffic. The question is, if you have the anions, or these collections of amulets that correspond with the humans in this material, then how would you be getting them to interact? How would you actually make them do with it? Yeah, and would you get a familiar, at least to other approaches, familiar pincer product structure out? How fast does the, how fast does your, does the dimension of your space grow as you, as you attach it?

1:02:30 These spaces are growing exponentially. If you keep on attached, I mean, if you compare it. In the case of the model, that's the other reason why it's quite too notchy, Because if I take spaces like this, back to the end of the field. Right. I mean, so the space, like in the top one here, doesn't have an obvious tensor product decomposition. That's right. But in a big space, I think you can embed the tensor product structure. Only it's part of the space. And on that, then you can do the two-qubit operations and so on. The two qubit operations, do they work in small bits of the space, or do you have no access to the whole space to do two operations? No, they work in small bits, as they say, by doing graining in between the anions from one of those qubits to another one. So you tend to go outside this tensor product space in crust as we get back, maybe not exactly until we're very close to it. These drawings, they look much like the drawings one does in operas, they look like things involving, related to operands, you're saying? Some structures pull over. Right. And in that kind of context, you also have the same situation like with the associativity and re-sociativity. And you may not be putting a condition like the Pentagon condition on that things come back on themselves considering higher order relations. I'm wondering, what kind of work has any place here? Is it the open up kind of work? There's any role in your work? Not directly here, but that same... All those questions come up as you try to generalize the invariants, try to get according to the invariants and so on. I don't know how it's related to the . All right.

1:05:00 Thank you.