Spin Networks & Anyonic Topological Quantum Computing (MiniDisc)

0:00 In fact, you see it by forgetting, forgetting the fact that it's off in Sweden. Well, of course, sigma-i itself is a very good idea. So what do you think about representation? And one form of representation that's worth thinking about is assigning a vector space to each strand and then looking for some mapping from D tensor B to D tensor D, which generalizes permutation and thereby gets you a representation of a grade group that you put together in the right way. Now, why should we consider grades? Well, it's interesting to look for representations of a grade group which has so long operators. It's an interesting game. And the original idea behind that was that if you could represent everything in terms of grading, then maybe there would be models, physical models, which would have that property. on perturbation, just like grades are kind of stable on the pathological perturbation. And whether that idea is real or not is certainly up in the air in the 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 grading operators and think about it as a mathematical problem. So that's a theme. So, as I said, if you look for an R, like the R that I had on the previous slide, it B tensor B to B tensor B. Then on the one hand, you can use it to get representations

2:30 of the braid group, and if it's unitary, then you can try to think of it, you can try to see whether it might be useful as a plonk A. So, for that purpose, it's useful to have translated the R's into the braiding relation into the R's, and here's the relation that it is in the R's, so here's R, and then having the extra line, I need R tensor 1, and 1 tensor So this equation, which is called the Yang-Baxter equation a long time ago, is what I want braiding operator to satisfy. If I'm looking at it from this point of view, which is not the only point of view we're going to do, but I'm going to talk about this point of view for a moment. So the braiding operator operates on two strands? Yeah, so it's operating on... So R is taking V tensor V to V tensor V, and you could think of V as being one qubit space if you wanted to. And you can think of V as representing one single strand. so then I want to remind you about so quantum gig is just a unitary linear transformation and B could be taken to be C2 and for the purpose of this talk a quantum computer is just some cube like that and local unitary transformation in this language are mapped from that V to itself, which are elements of U or two. And then I remind you that it turns out that a single entangling two-cubit gate plus local unitary transformations is universal for quantum computation. So I'll say that a G taking V tensor V to itself is a universal gate, but this is so. 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 quasi-topological ways of representing quantum computing just exactly because they're entangling.

5:00 So the standard two-fubit gate, of course, is C0. You can use that. And that is indeed entangling. Maybe I should remind you of this. So the data 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 cubits that the state is entangled exactly when the determinant of the little 2 by 2 matrix coefficients is not zero. So it's quite easy to check whether a given 4x4 matrix is entangling or not. And so, for example, you can do the exercise to check that C0 is entangling. And if you're encountering something... So would it swap the entangling? Swap is not. Swap is not. But it is. But it is. Pardon me? It is. You could say it's completely entangling. But not entangling. There must be some problem of terminology here. When I say swat, I mean this guy. That's right. Yeah, that's right. Right. And the thing is, for example, on a four-cubit state, on the two middle-cubits, then creates a tangent. And all this does is permute the two middle ones, 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. If you have four qubits and you act on two of them, these two being already entangled, and these two being already entangled, and swap on the middle ones, then these two are linked to those type. Yeah, all right, I understand what you're saying, but what I'm saying is that I'm looking at, I'm just looking at two-cubit states and asking whether the four-by-four matrix can entangle that. In order to have this, we're talking about this criterion about computation in the universe out. So from the left point of view, this fellow here is not going to entangle any two-cubit

7:30 state that I'm doing. Yeah, but swap is not, as it happens, swap isn't with one cubit, it's not universal. That's the point. The thing is very close to the swap. If you change this one to a minus one, then that would be the first one. I wonder if we normally regard SWAP as entangling because it can create entanglement from subsets of parties that were not initially entangled. But the question is, can a gate that is not entangling, in your sense, be universal in the sense of public computation? So you say swap is entangling because you're thinking of two parties, but we are talking about computations, n parties, with n qubits, and you have to take care of the entanglement of all these n qubits, not just two. I think that's the reason. I've got more than one question, but I like Charlie's question, I haven't thought it through. It should be obvious that you can't... If you start the computation with a tensor product state, right, then you can never get anything but a tensor product state with a gate that's not entangling using this definition. Right, in other words, you couldn't make the entanglement that you fed into the... Although, maybe if you were just allowed to use once at the beginning, you'd get a lot of... We'd entangle them at the beginning, and we'd swap just to swap it around. But I mean, I guess it's at least logically possible from this definition that you could start with a very little bit of entanglement, and then you could create more somehow. but it seems unlikely that would actually happen okay but so i guess yes all right okay go ahead all right

10:00 so a couple of examples This is entangling. That would be entangling in unitary if these are on the unitary circle. And here's the Bell basis makers, which is certainly very entangling. All three of these guys are solutions to the N-Bax request. So there are some natural examples which are topological. And even interesting from the point of view of a topologist. Sam and I just started on 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 the linking of the boronine rings, which doesn't have any linking numbers, for example. 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 grading operators. You still need arbitrary elements of U of 2 for the local unitary transformation. So these statements are some grading operator plus elements of U of 2 are generating. And that's the best you seem to be able to do. On the other hand, Friedman, Ketayak, and my collaborators suggested using what's called topological quantum field theory thought of as a theory of antigons, I'll explain what I mean, and make really deeper models for topological computation where everything would be represented by gradient. 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. 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 provortisms of manifolds so you have a manifold and another pardon, category of what? manifolds I'm going to explain so everybody knows you know what a manifold is it's like a service or a three dimensional space possibly with boundary possibly not here's a surface with boundary the boundary is a circle at this end

12:30 and two circles at the other end and one says that these these collections of circles co-board that is there's a co-boardism between these two circles and that one if you can fit a manifold in between them one higher dimension one higher dimension pardon Right. Yeah. And is category used in a technical sense or just... Yeah, it's a category in a technical sense where the objects in the category are the n-manifolds and the morphisms in the category are the n-plus-1-manifolds that move from one manifold to another when found in one. So in the spirit of category theory, a morphism doesn't have to be a set a theoretic map, it's just a relationship between two things that's of some use to somebody. So it's not a mapping from MN to M'N, 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, namely, you would think that here they're often mapping from v times of v to v since there's some fusion going on there. 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 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.

15:00 And then for three manifolds you can decompose your three manifold into two pieces. I wonder if I had a picture of that in the next one. You might decompose your three manifold into two pieces. which I'll just draw like this so that the union of them is the 3-manifold here's a surface in between and in this picture because this surface bounds something there's a vector in its vector space and because this surface bounds something there's a vector in its vector space and then you put them together and you hope that that's going to be the invariant of the 3-manifold that's going to be some topological invariant of the 3-manifold And there's a general, there are generalizations here of the ideas of putting bras and kets together. So a bra as, I mean, a bra or a ket as notation, in fact, is a cobordism of that sort. It's two points which together bound a little curve. And one can actually make use of that. How do I know which is the bra and which is the ket? Oh, I 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 I 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 let your feet in the left that's the one on the left is a bra this would be the bra and that's a cat it's another can to be real I wouldn't worry about it But it cannot be complex, it seems to matter which way through the picture. 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, you shouldn'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 the more common, the Fourier level. It's easy to work with. 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.

17:30 And Witten was the person who originally suggested that one should do this and that it should be related intervals that is he's going to integrate over the entire 3-manifold something and then the vector space that's associated with the surface has to be figured out it's not given 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 function and there's a whole story in 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 just draw them schematically like this but you see you can take a surface and you can decompose it into pairs of bands like that you can do it in various ways and 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 gradient 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 number. 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, the recoupling. The recoupling is written in various ways. 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 on another way of thinking about

20:00 the patterns of this is in terms of algebra because if I have x y and z and I think that this is x times y and then this is x times y times z then you see recouple, I'm now looking at the pattern of x times y times c. So recoupling is reassociating. 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 pants, 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 have these then need to satisfy various conditions they need to satisfy conditions of being of braiding being there and then some other conditions which i'll show you pentagon naturality and hexagon yeah this equation with the sum is that are you working in some free algebra with these diagrams 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, you'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 the 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 can end up with W because X and Y can interact to give me something, and then these can interact to give me something. And I look at all of the different intermediate possibilities X, Y, Z to W, I have a little vector space there with a basis. And then this formula would mean that a basis of this vector space with these starts, and would say these three starts and that end, would be rewritten in terms of another vector space

22:30 with basis these three starts and that end. So those are the two main ways of thinking about it. There's directly a vector space associated with it, or else it's evaluations. Is that it? Speaking of catcher. 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. I know that means I can close. 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 them 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, bases of 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 that. These are the simplest things that could happen, that you would have identities in both of these cases. 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 in pairs of hands. 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 basis of angular momentum recoupling theory a little bit so that it has some topological variable and that's why we want to talk about Penrose's spin matters so so let me tell you what Penrose did Penrose a long time ago was

25:00 making up pictorial tensor calculus to handle various things that he was interested in and this one arose from thinking about SU2 or equivalently at the we algebra level SL2C. SL2C can be thought of as the two-edged matrices that under conjugation, what do you call it when you multiply by the transpose. A epsilon A transpose equals epsilon in any case, where epsilon is the little antisymmetric guy like that. In general, if you took a two by two matrix and you did that, you'd get the determinant of A multiplied by epsilon. 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 one. 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 identity. 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 and 1 and 2 these are chronic or deltas a little line with that so it has a label on top and a label on the bottom is equal to 1 if the labels are the same and 0 otherwise by definition diagrammatic tensors or diagrammatic matrices so 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 sine 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 two 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

27:30 also changes sign. And of course there's the identity. And if you tie two epsilons together top to bottom like this, adding, you sum over the indices that are tied together, just like you do in tensors. So this means that you summed over the common indices that are not three n's. You'll 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 calculus. So he redefines the epsilons by multiplying them by the square root of minus one, and the crossing by putting in an intrinsic minus sign. So when you see this crossing, it means 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 Convention Well, this choice, which is meant to simplify the diagrammatic calculus, has some very nice consequences for him. He discovers that not only does it help with the science, but it makes the calculus later 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, minus that one smooth that way. So I could 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, the sense that if two lines are crossing through one another and holding them apart. 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

30:00 movements back in the 60s. Of course, from present vantage point, it's very obvious that one should try to generalize this because there's lots of interest in knot theory, but probably in his context, he wasn't thinking about knot theory at all, so he didn't try to generalize this to include over and under crossings, 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. 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. And this is the underlying bit of algebra that corresponds to the irreducible representations of SU2 in this theory because you do a symmetrization process of that type to make the representations. So these are gadgets 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 crossings, if you want. And so, for example, here, I've taken the two-symmetrizer, which is just 1 over 2 factorial times identity minus switch, and then I expanded this using the biner 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 these symmetrizers. So, for example, I said that if you tied it together at the bottom, it would be zero. You see that here in these terms, because you see when you tie something on the bottom

32:30 of this, you get that loop, which is minus two, which cancels out. And when you multiply them together and you expand one of them, you see you get this 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 3-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 formulas like this one. Penrose. This is actually due to Wenzel and Jones later in a slightly more general context. But there are recursion formulas like this. So for example, the three can be recursed in terms of the twos, and then I can continue to write that out. So there's this technology of the projectors. And then, with the projectors in place, one gets to the angular momentum recoupling theory in a completely diagrammatic way. In this way, you need three vertices for your angular momentum recoupling. 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 congruent to C minus two. 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 if you, 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, and a C particle comes out, and you can try to figure out what kind of relationships there are. So then in calculating with these networks, there are some basic facts, like this one corresponds to Schur's lemma.

35:00 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 a prime. And then you can figure 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 on the left-hand side you get a theta graph labeled a B and C so the lambda is the ratio of the value of the theta graph divided by delta A which is this so calculations look like that in theory and there are recoupling formulas and by the time you get to this level of the structure everything satisfies all those identities all the identities that you could have at this point Pentagon Pentagon is what's called the Elliott being our identity for recoupling coefficients and and we don't yet have any braiding but we'll get to braiding soon enough so so got transformations 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 Lemma 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 if you put in a J and you get formulas for the 6 J coefficients in terms of the little network evaluations figuring out this tetrahedral network, figuring out this little loop and some things. So then one can go ahead and actually get all the calculations here as well. So at this stage, what values does A, B and C take? So, yeah, so A and B and C are taking values, integral values, and and you're summing over the things that satisfy the admissibility conditions for remember a plus like a plus b minus j is greater than equal to zero so they're just thinking of just an entity between right

37:30 positive energies or zero and how does that relate to the spin half oh it just just about by two. If you have two lines, then it's spin. It's just useful here to count the number of lines, so I saw that's what this looks like. You see it's almost a topological field theory in the sense of having all those identities. do with it? We're almost ready to make the foundation. Well, Penrose originally wanted to think of these networks as an abstract substitute for space, as a kind of process background for getting space or space-time. So he thinks of abstract spin networks like this, basically arbitrary trivalent 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 for appropriately experimentally repeatable large-ish spin nets, the angles satisfy the restrictions of directions in a 3-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 involving exactly this calculus that I showed you. and then and then if you assume that the angle headed that the measurement that you make by doing such a thing is repeatable then the angles conform to angles of course you started with with information about the rotation group to begin with and you left it there but it's all common towards at this point so so there's a hope that you might be able to use this as a substitute for space in some discrete sense but not not 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's to really make it work for you in that sense next transition to the ecology well it turns out that Penrose's ideas fit directly into a generalization that gets you topological 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 just generalize the binary identity you

40:00 see if I took a equal to minus one this would be the binary identity but now it's in a different context it's in the context of making a non-inherent and and there aren't any matrices in the background it's it's a recursion relation that says that if I have a knot with a crossing like that then I can and 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 two, then the resulting thing is invariant under topological loops, which are generalizations of those flat planar loops that you were looking at before. In fact, these are the braiding loops, right? So if we were restricting ourselves to braids, this would be invariants 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. 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 as a very close relative to Penrose's basic final calculus okay I think I got one more So everything is nice with respect to the grading operator still, but you have something more that is nice to? Well, remember what Biner Calculus was. Biner Calculus was a flat calculus, and it was actually describing a certain tensor in the plane, right? Here, this looks like Biner Calculus, and is formally identical to it if you take A equal to minus 1, right? Because then you have Biner Calculus, let's draw Biner Calculus. which is the underpinning of the spin heads, it said flat crossing equals minus smoothing minus other smoothing,

42:30 value of root equals minus two. 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 invariance of knots and links in three-dimensional space by deforming minus one into an egg and you get an extra algebraic parameter happening here 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 those kinds of relationships. So maybe it's too much, but I just thought I'd show you what that looked like. If 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 matrices are 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, 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 and D equal to whatever it is. And So, it's a very elementary step, once it's made. So, there's a book about this that we wrote, Croftman, I wrote, and with Sustenous Lins, 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 symmetrizers. Now, it's amusing that you actually can define the symmetrizes exactly in analogy to the old symmetrizes, but you take the braids and you take the permutations and you lift them to braids. I'll draw a picture on that slide. And you have to generalize the sign of the permutation to just some variable raised to the number of transpositions

45:00 that minimally take the braid back to the identity. and you have to generalize the factorial to a Q-deformed factorial. It works. The lift of the braid is a standard lift of the braid like this. Here's the, I'm sorry, the lift of the permutation is to do it so that the crossings are all the same, like that. So here's an example. Here's our two symmetrizer again, and this was the factorial. that no matter that that was the factorial and this was the coefficient and when you collect up the terms and expand just like we did in the minor calculus you end up with this gadget this is the gadget you could 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, the minus a squared minus a squared minus two. So all these things deform in that way, and now we have a larger range of particles to interact. In particular, the recoupling 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 those identities, hence the non. So this is a natural model of all those things that's available, 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, and I think I should try to talk about that. But let's talk about this Fibonacci Anteons. I'll try to, well I can take about 15 more minutes at the outside, sorry, I'm pushing the end of this, I didn't mean to, but here's, but finally we got to the thing I wanted to show you, so here's our, we want to make a particle theory and 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'll fit together and go to two, and they will connect up like this and go to nothing.

47:30 Or you can have nothing giving rise to 2. So I have particles which I'm calling 2 because of this perverse labeling. And 2 plus 2 should be equal to 0 or 2, but I don't want 4. You could have 4, right? 2 plus 2 could be equal to 4. That's a possibility, but I have to get rid of it because I'm looking to make the simplest possible theory that I could. I want a theory with one particle and nothing, or one particle with a charge and one particle without 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. 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 would see that F squared would be equal to the identity. That's called the orthogonality relation, and there's a generalization of it if you wrote down the whole 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, and I would also have a unitary operator F, which could be used to describe other variables, as we'll see. So we have to keep an eye on it and see whether we might be able to get it to be unitary. What are the calculations going to look like? Well, 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. And then we would need to figure out the theta, which is all twos, and the delta, which is the loop with two. The delta, which is the loop with two, if you just expand out, is little delta squared minus one, where little delta hasn't been figured out yet.

50:00 In order to make this work so that two plus two is never equal to four, 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 the A to. So now it's 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 tetrahedra, and you do some expanding and calculating, and you find out that it's equal to this. Now, you needed the tetrahedron because you wanted to close 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, the two line, two, 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, you know, I've seen it actually in the calculation, I'll tell you this off. So here's the result. You do a little calculating of this sort, and you find that B prime is tetrahedron times 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. check that the final result is going to make 4 equal to 0, but this is the quickest way to find out what restrictions you're getting. So here's f, and if there isn't any, if there aren't any other terms in the recoupling, then f squared will 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 of an auction, because this is what's satisfied by the reciprocal of the golden ratio. So then, if little delta squared is little delta plus one, so that's actually the golden mean, little delta. Let little delta be the golden mean. That's the loop value, the single loop value, then capital delta is actually equal to little delta, which makes life easy.

52:30 And you indeed get one is equal to this. And you find that capital delta is delta. You can calculate the beta 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 that theta can be adjusted any way you like and this would still satisfy the recoupling formula and in fact you can adjust the model and make this matrix symmetric very easy why it would be symmetric? oh, because you see what I have in mind sorry, I should have explained a little more carefully earlier on I'm just 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 braiding will be like when I'm over here, because I have in mind to make braid group representation by letting the braiding's act on these guys and see what they, and return them to things of the same type. Okay? 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 and various possibilities in between from the 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, what this braiding is. And the answer is, you should recouple do your breathing come back So if you do that, if you start here, you recouple it, then you know what this

55:00 braiding does, do it, and then come back. You get the braiding operator here, between other things than just adjacent points. And those are given by, in this case, by conjugating by F. So if F is unitary and R is unitary, then the new braiding 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 greater. Should have said that earlier, but here it is. So, 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 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, tau is 1 over tau tau. Here you have the square root of tau. And this may look familiar if you heard it talked earlier this semester by John Preston talking about this. I already saw his notes where it's detailed nicely in terms of this. 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 EDI 3 pi over 5, then you see that delta is the golden ratio. 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 Bayer 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 Cresco's notes. R is a little 2 by 2 Brady matrix. Now, what have we got here? What do we have? We have the following situation.

57:30 Perhaps I should have started with this slide. You have this little theory of particles called 2 and 0, and you can consider these vector spaces of processes, you end up with a zero. For example, suppose you started with four twos and you asked how the diameter is zero. Well, you see that two and two can give you a two and then those can give you two and a zero, or two and two can give you a zero. But what's going to give you a zero here from a two has to be a two, because zero and two combine to be 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 qubit of space. But that's got internal structure of braiding operating on it. So you have to do this slightly complicated business of calling your qubits certain combinations of these little anions, 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 and you see that it's not generating any finite subgroup and you can argue that it's dense 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 them or whether you really have to take limits to get your favorite matrices. I was playing with it just a low level on a 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. It says that you can generate things entirely coherent system of freight. And are there really any aliens in the world that can do this?

1:00:00 So Alexi, 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 that 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 topological operators. spinnets and q-deformed spinnets have been used as substitutes for space-time and measurement operators in quantum gravity. There's lots of literature on this where 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 spin network as well. And what do quantum Okay, I could mention that there are some non-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 non-invariants that are made by just taking your knot and replacing it by two strands and a symmetrizer, then that non-invariant can be described in various ways that are relatively elegant. It's fun, but it's a little off topic here. So I'll stop. Whenever I hear about this stuff, or I hear from Tyab, so for example, I wonder whether if you make something that approximates a qubit in some piece of physical material, and then you make another one in 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,

1:02:30 if I have to go through some, some, uh, complication to get it, uh, from the raw materials? I'm not quite sure if I've got any questions, if you want to get my preview. That was the slide, right? The anions are living, they're living in here. The zero and the 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 in there. And then I could do some braiding, cross braiding among these, and they will end up entangled if you do that. So that's sort of the picture I was thinking about. Now, does that match what we're asking about? Well, I just, I'm not sure whether it does. There was always a hope of getting some complicated, perhaps, a quantum hall effect type of actual physical material that would that would harbor these right these anyons and use certain operations on it to do one rotation but i was never quite sure that you would get of whether you would get a overt space that would grow exponentially in the amount of this material. As you do with what you have, let's say, I am trying to do that. So your question is, if you have the anions or these collections of anions that correspond How would you actually make them do work? Yeah, and would you get a familiar, at least, other approaches to communication,

1:05:00 a familiar tensor product structure out of it? Maybe somebody else? How fast does the dimension of your space grow as you attach particles? Yeah, these spaces are growing exponentially. In the case of the model, that's the other reason why it's going through the notch. In the past, I take spaces like this. Then the dimensions are fibonacci numbers, so they're going around that's very much. But there's also a matter of the operations, and it's not enough that the space grows to the expansion, but somehow simple, two-cubit operations being conversal is sort of important to enable to use it. You don't want to have to access the whole thing. Yeah, right. I mean, so the space, like the top one you drew, it's not, it's not, it doesn't have an obvious tensor product decomposition, but in a big space like that, you can embed the tensor product structure only as 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 somehow small bits of the space or do you have to access the whole space to do two qubit operations? They work on small bits. As you said, by doing grading between the anions from one of those pieces, one of those qubits, another one. So you tend to go outside this tensor product space in the process of this, but you get back, maybe not exactly into what they're very close to. These drawings that you do, they look very much like the drawings one does in operas. That has anything to do with this work. They look like things involving, related to operas, you're saying? Yes. Some structures call operas. Right. Right, and in that kind of context, you also have the same situation like with the associativity and reassociativity. Yes, yes, as in the future, associativity and action on the side.

1:07:30 And you may not be putting a condition like the Pentagon condition on, the things come back on themselves, you may be considering higher order relations. I'm just wondering if that kind of work has any place here? Is the operas kind of work, place any roles in your work? Not directly here, but that same, all those questions come up if you try to generalize the invariance, for example, try to get a four-dimensional invariance and so on. I don't know how it's related to the quantum computing aspect. I didn't know you were going to be here, so I didn't say anything. That's okay. Yeah, well, I'm sorry. You're talking at 2 o'clock, right? In the computer. Where is that? The computer lab. Sir, down the road. Oh, it's okay. It's actually here on the street. Yeah. What did you say you didn't hear? What? What did you say you didn't hear? Oh, you recorded. Oh, that's right. Of course you did. Oh, did you record that thing? Oh, thanks. That's what you did. To be honest, I almost forgot about that. That was the John Coleman thing. Of course, since then, we've actually had our meeting in the chair. We're here somewhere. Yeah, we're on the club. It's actually on the Maddinger Road. It's all right. You go off to lunch. Okay, we'll just come and get it.