Topological Entanglement & Quantum Entanglement

0:00 It is almost observed, not only the rotation, but the rows and columns in this table play the part of spatial axes. You couldn't simply say that, but there is that underlying feeling about it which they go somewhere later on as I should describe to you, you put more There is a kind of spatial aspect to it because it's also an angular momentum aspect as I was showing yesterday. Yes, I'm sorry, I know that we're going to have to get that. We haven't got it in some vaguely, and I expect it to have a regular membership yet. So I'm picking out little bits, suggestive bits, after all that. This is for all my humanity. I know I'm getting a little... Oh, sorry. So, for you, from this time on, you start putting physical interpretations, like anger and momentum, spin, and so on, and there we see a big parting of the ways, which I hope is a creative one, For example, if these are rotations, if they speak of spin at this stage, what kind of thing is the spin? They usually cast the language and they can't get away from the classical language. It's a set of operators that commute with each other like the angular momentum ones do. What? It's a set of operators which you can grow from various places, which commute with each other like the angular momentum ones do. Oh yes, yes. You could put all that stuff on it, but you've got to do that.

2:30 We are forgetting so many methods to do that without an awful lot of hard work. Yes, I agree about that. But that's the method of identification, that's what I mean. On a crude level, it fits me. You were mentioning the spinning thing, and yes, they think it is a rather, the concept of spin is quite classical for them, it means something rather like it. Then you say, well, how far, this is all rather tiny, isn't it? How far outside does it spin? Do the spatial axis spin, and they start looking at yours like a bit queer, and you say, well, I mean, we know the whole world doesn't go round and round, what is it that goes round and round, where's the boundary? No, the answer is nothing goes round and round. What you are adding, you're adding in to, in the case of fermions, a vector onto the classical angular momentum one, and the sum vector turns out to be a constant of a motion. Well, I'm prepared. And that means you can measure it. So the total angular momentum, which is the classical stuff, plus this funny thing, right, is something you can measure. And the funny thing turns out to have two possible values, right? And so, I'll bring you to the half H. And so you tend to describe it as spin because it has the rules of angular momentum. But one can't tell whether anything is spinning in a physical sense. You know, whether there's any extension or whether there's something really going round because nobody has observed any of that. I'm sure you keep your ears up, but I'm quite sure equally that there are a lot of cruderm who think you're much more cruderm. And there's a case for it, you see, you can't, there isn't a point in which the quantum canter labels from the, of the connectoral labels actually, so to speak, makes it. Frisch, Otto Frisch. I asked him, if you took that lead and you prepared a beam of particles, No, only from what I do, but the spinning part was part of all the spins, or whatever you like, of one state of spin.

5:00 And you fired it into this suspended upper leg, would it start to rotate? Good question. And he looked at me and he said, I don't know, but I can find ALT. And he did, and the odds that came from the camera, as I suspected, is of course it doesn't go around with that. The simplest idea that you mustn't speak as though the language of acting the momentum is to be taken literally and classically is not, and pardon me for not following up your sophisticated way of handling this question, because obviously you've thought deeply about it, but a lot of people haven't. I've tried to understand Dirac. They haven't thought deeply, I can't. See, I think what is very interesting about this tripartite discussion here is Tony has put forward a view which fits very nicely onto the fresh leg thing, but it's a view in which spin is actually based on the classical angular momentum business to start with. it's an extra bit so you can't have that extra bit without the original I don't I'm not sure about Peter but Ted I think if pressed by Peter to say that there was angular momentum here would have to say well, no sorry if there was spin here would have to say well maybe there is but we haven't got that classical part yet The big problem about the classical part is you can define it for the coordinates of one particle or another bloody universe of particles and you don't know where to stop. I'm going to have to cut in a later way and I will... Now, I'm pressing on.

7:30 This is an expression which is certainly used by RMC. Now, it's very important what they mean by it, there's an awful lot of trick and goodness here, because they're going to have to be a normal one here. Well, these are big computers, and this wouldn't be a problem. But, never mind, I can't think of all that. There is a very complex, a very subtle argument that RRC produced, which shows that the strong interaction should be treated as the first case, because one of them has to be treated, in the sense that you've got to preserve the symmetry on the opposite of each other and it's only possible to have one row with... what they say is that rotations are altering the unit charge as they say whether that puzzle might the whole time but this is undetectable if and only if their fails are problems and therefore force force forbid the rotations to affect E and W In other words, you're giving a uniqueness to the strong charge. And this enables them, it's from these spaces that they're able to construct their very complicated tables which represent the different possibilities that there are, and it falls to the next stage in going towards the conventional policy. In other words, it's the effective exchange of a single strong charge between three bound corks, which prevents the identification of any white cork by its colour. And that's a very important simulatory principle for them. They then go on, though I won't go on, to produce similar combinatorial arguments, specifying structure for bringing in new concepts, in the case that we need to have the E, this is E, and they deal with them separately one by one.

10:00 So, I refer to this as progressive child exclusion, and indeed they do use that term. Then we get on to, this is going to lead me to my real point, this isn't my expression, this is my expression. Now, using this charge exclusion by which you progressively demobilise one charge after another, you get to, they postulate that this is done mutually with charge and mass as a ratio. In other words, they're saying that, as I would, well, they don't say, I would put it, that this is fundamentally where quantization begins. And you quantize charge and mass and that things proceed together. And then, as they say... Which is why we have particles of certain masses with certain charges and not something else. That's right, yes, but they're not particles. Yes, but that's right. What they say is that... I get this... I'll read a piece I quoted from their paper on the Higgs mechanism. The Higgs mechanism was introduced earlier to account for the spectrum of particle masses. We've previously shown, as I quote, The brachial symmetry which produces mass is largely a result of the production of zero states of charge. Mass and charge acting as an invariant in the same manner as space and time. Well, I hear what they say about space and time. Anyway, it's mass and charge acting as an invariant.

12:30 And zero charges represent complete cutting to the ex-fuge. non-zero charges represent a reduction of the vacuum state to less vacuum. So the decoupling to a zero charge may be equivalent in energy to one unit of mE divided by alpha, the fine structure constant. Well, you will see that here we have a very strong tie-up with a hierarchy picture because they are saying is run by one until we get to the limit, and they presuppose the limit, which is defined by alpha. Well, correct. For us, well, we actually calculate alpha by having similar set of ideas. We, too, can't distinguish at that stage charge and mass, and would have Since they both appear in the fine structure constant, M appears through via Planck constant, we are also committed to treating M as a unit to be established once, and this principle, the mutual treatment charge of mass, ties up very nicely because we've got the, as we once said, whatever happens, we've got the Maxim Garden and they have not hit their belt. We've actually got this calculation, by they I don't mean RNC, I mean people in general Now, I seriously say to them that they are very seriously underselling their product because it's very proper here to say, well, this is the first step which explains quantization. Now, of course, those people brought up in quantization think it's the most obvious idea of the world. It's just a number. And we know all about the experiment of origins and so on. For some of us, it's a mystery. And we need to have explained now. It arises in the first place.

15:00 This is where it arises. I mean this is how old football we've seen for a long time that it was implicit in the combinatorial calculation of fine structure constants and the other constants but what we didn't have was a means of breaking it down and saying this is this is how you get back to the individual components of it, maybe charge the mass which are not yet separated of course in the hands of R and C they do get separated So this is a strong point. I'm almost at the end of your time, but I'll just make a few general comments. Firstly, these are general comments about the cross-modules of the two pictures. Firstly, the extra magnetic level provides a discrete standard, making sure to charge the uncorrected receptacle of the fine structure constant as its maximum, 1-3-7, that's what I've just been saying secondly both E and M appear in the dimensionless form of this constant, M and H and we're committed to a mutual generation of M and E, we are the guideline we don't get the one without the other, that's called the position of one C thirdly the correspondence between the two theories which is something I've had to chuck out of this paper for recent times It's more evident if we recall that in both cases the mechanism comes from the group transformations directly. It's a bit late in the dataset, but of course their double turn-in is to be compared with our construction, a level construction of one quadratic group and then the product, an outer product group operating upon it. It's not the same as truck construction, but it's got sort of eerie overtones of similarity. It should differ from the standard model. I think, for instance, if we regard this coupling of the constant, say, of interactions as a product, as a unity then the corresponding big group should not be a direct product when

17:30 should be something say something more if we want to treat them a like a unit you know the way you multiply say you bring a charge together yes the possibilities are missing. Yes. Of the direct product. Yes. You understand a direct product is like the trivial case, what we have in the standard model. Yeah. You say it is what we have in the standard model? Yes, yes. The big group say SU3 cross SU2 cross U1. That's very interesting. and I don't quite know how to respond to it. The algebraic situation is much more complex than that, but it's good to incorporate. I mean, once you've down two levels is a very subtle place in mathematics, really, in which you compose your situation by adding up the number of... subsets which have closed under the discrimination operation that's the basic thing and then you have to combine that with the thing which is identifiable with the direct product of two of the sets as a separate kind of thing and then you go on comparing the numbers of these right upscale until you get to the gravitational constant where it all stops so there's a lot to be said about that but thank you well I'll stop at this exciting point because I haven't gotten 30 really RNC is saying they have suggested ways of counting up the zero charges in the multiplex and I look forward to spending more time seeing how that goes you know I'll stop thank you The paper I did yesterday shows how that now fits in to the more conventional Higgs way of doing things. To the more what, sorry? To the more conventional way of Higgs doing things. Oh, yes, yes. Because the one I had yesterday has that final formula which we argued about whether it's an equation or not. Which is not an equation, it's just a formula. But the first term in that, which is the bulk of the mass term in those particles, is in fact due to this mechanism.

20:00 And the other two are the lesser minor adjustments. Well, one of them is the higher up, the three generations term. And the other one is the SU2 term, the up and down term. but the main one for the matters of composite particles is that first term and that shows how that now fits in to what we then hypothesised on that basis so I think you'd be interested to read that if I'm allowed to give this paper on the factor 2 you might be interested in what I have to say on spin then because it does intimately connect with this in my opinion it's a definition of discreetness, spin itself is a definition of discreetness itself because as in the Aronoff-Bohm effect you get changing the actual space that you are looking at when you have spin a half the change in the space is caused by being something discreet in the space that you doing it and you get a change of the topology of the whole so it's a discrete object within the space itself changing the topology of the space and that appears to be what spin the heart is all but i'd like to say more about that if i give that to them i'd like to just raise a rather naive question i mean i like the idea of the way in which come out of the corner and together. What's always puzzled me, though, about that idea is that the chart quantisation is so simple and clear you know, plus or minus E and so on, whereas the observed particle masses are only roughly multiples of something, indeed some of them not even that. Well, it's much better if you look at the whole picture given by that expression, which I produced yesterday, because really those original calculations are based on the first term of it, rather than all three terms.

22:30 The first term is the bulk of the term, but it's not all of the term. So I think you'll see that it corresponds more. In addition, when talking about using units for charges for quarks and so on, If you look at the standard Higgs mechanism, how it gets up and down, it can't possibly do something wrong with it. There's nothing wrong with the Higgs mechanism. What's wrong is the hypercharge values are completely wrong. You need two different hypercharge values to apply the Higgs mechanism to up and down quarks. But when you actually look at the actual hypercharges of up and down quarks, they're identical and they're also fractional. however when you look at the way the heat mechanism is done to up and down quarks it's 1 and minus 1 for the hypercharge which fits into integral charges it doesn't fit into fractional ones and they're also different which they ought to be if you use fractional charges they ought to be the same hypercharge so in other words you've got to completely invent something to apply heat mechanism to the way that particle physics is done at the moment that's a bit worrying isn't it well you don't in this I didn't realise this at first, I only realised this relatively recently. That this fits better with the Higgs mechanism than that does. If you look at the hypercharge values you use when you do the dual Higgs thing to get up and down, it's nothing like the fractional charge at all. I suggest we change R with a U and we transpose G and B and we call them the good, the bad and the ugly. I'm only talking for half an hour because I haven't got any results and I'm just wanting to tell you the problem which is in progress is handed over to who you might like to share it. And what I'm really doing is putting an appendix on the paper that's in the proceedings where I finished up with a vague look of horror at the fact that I've got a non-associative system and didn't really know much what to do with it. Let me, for those who haven't

25:00 read the proceedings yet and I haven't actually because I was distracted by the enjoyable historical beginning that Peter didn't and I went through it and tried to come and see what was wrong with it. I've got one point. In the long-standing exercise that Ted and have carried out to try and give sense to the Frederick Parker Road construction by means of the process theory we finished up with our new version of the hierarchy and that turned out to be almost exactly the same as Frederick so that was fine and in this new version at the lowest level, out comes, as one would hope, the quadratic group, CH, you see, and this in a notation which our letter pioneered and later dropped, we call G1,2, that is the group which is produced by two generators with a single signal for equality between them and usual discrimination And then in the Frederick construction, one says that you go up a level by looking at the DC subsets at this level and finding things which need them there, etc. But in our process procedure, you can't demand to go up a level, you have to let the system and do what it wants. Now a great thing is that there are no mathematicians peering over the shoulder to decide on what's going to happen. So it might not go up a level. And then a third element might come in and you start doing discriminations. And so then you're going to ask about G13, where there's still this single signal for equality, but there are now three generators and that's quite important in the Frederick construction because his construction relies on the fact that the next level when you've got two generators then has three and turns out to be the same as what you would have got if you had three So it's C2 cross C2 cross C2.

27:30 So there's no difficulty at C. Well then, two or three years ago, along came Aspect. Ted doesn't know why it's called that. Well, it's my stupidity that chose such a name for it. I called it Aspect because it was the realisation that you could look at things in two different ways, that is from two different aspects. I'd quite forgotten at the time, although I didn't know of the existence of physicists of the same name with a capital A, and so I'm awfully sorry that I didn't think of some other word for it. That's why I stumbled when I read the title, actually. Yes, because it's all in capital, so you don't know. I wish I'd called it something else, and no doubt in due course, well I'll change the name and confuse everybody once we introduce the fact that when we discriminate between A and B it's not exactly the same process as discriminating between B and A because if A was there already and then you bring B in and discriminate and if you get the answer that they're the same then what you've got to do is change all the B's you may have had up to there into A and not vice versa and if you do it the other way around then you've got to change all the A's into B. So they're not the same the proof that the operation was commutated in the book was a fudge well it was false actually I know where the mistake now is I shouldn't have made it I don't know. I take full responsibility because Ted will be the first to admit that the mathematics is not his doing. So at the first level, where one agrees that AB is not going to be the same as BA, one goes through the same construction which we lifted from Conway in the CH case, and we finish up with the group which now has two generators and it has two signals and that turns out to be quaternions and so when I initially found this out I made a number of rush assumptions about how when I

30:00 went up I was going to get all the clifford algebras and so on and in fact I think at the end of one of my anti-papers I stated that that's all false unfortunately because, well, for reasons which will become clear during the next half hour. Let's ask the question, though, and that's my interest today, let's ask the question, what if the system doesn't go up a level, but another element comes in, and so instead of this G13, which appeared in the original hierarchy, you now have G23, the two signals. well unfortunately there isn't any G23 if you have three generators and the required non-commuting thing, you're going to finish up with something which isn't associative so it's not fair to call it G23 because G stands for loop, so we call it L23 where L stands for loop because these sort of things have been looked at by mathematicians Now, the question then is about this algorithm three, which is what came up at the end of my last paper and produced a certain amount of disgust in the audience through its non-associative character. Can I just interrupt myself by saying, as a matter of fact, you might take the line, but it doesn't matter all that much. I refer back to a paper of two or three years ago that I gave a vampire where I looked into the question of even the original hierarchy of this G1-3. You see, if you have two elements, then in a very direct way, as Frederick's famous, you get this. When a third one comes in, you've got to test it against the two you've got already. You haven't gone up a level here, so you've just got to do it piecemeal. Test the new one against this, and then test it against the other, and so on. And since your access to the original two is random, it may be that it will take you quite a long time to determine whether you've really got a new one or not. In fact, each time you do a test, if it, I mean, if you,

32:30 the third element, say, let's call it C, and you test it against A, if C isn't the same as A, then the result of doing this test is not to give a signal, but to give a new element, and a discriminant of A and C. So now, instead of having three quantities in play, you've actually got four. And then when you do the next test, If, again, you have a failure in identifying it, you've got five. And so the number of things in play goes up all the time. And so it may take you quite a long time to find out whether you've really got a new one or not. And, in fact, if you work out the average time, you know, the sort of usual statistical argument, the average time turns out to be infinity. So it may very well be, it's very likely that you'll never actually manage to construct this in a finite time. On the other hand, you may be lucky. So G13 does come up, sometimes, and so it still interests me to ask questions about its structure. What would be a group if it were associated? Loop means several different things in the literature. One of them is a group without the associativity condition. Are the Octonians a loop? Let me make three quick remarks then. Firstly, it turns out that because of technical details I don't want to go into, this non-commutativity that one has with aspect actually exhibits itself as things anti-commuting we don't put the minus in, as it were it comes out of the structure I think that's an important point but it's there nonetheless so if the things are all anti-commuting and there are three of them it's very easy to see that we must have got non-associativity somewhere

35:00 For those familiar with Clifford Algebras, if you've got the Clifford Algebra with three generators, then the product of the three generators is in the centre, commutes with everything, and that's not allowed because everything's got to anti-commute. But that proof, of course, relies on associativity, therefore, conclusion that we must have a non-associative system. Well, given then that we've got to have non-sociativity, even when I get back to here, where I said G22 was isomorphic to Q, that resulted from choosing the two signals, which when it's turning blank in commuting will be 1 and minus 1, choosing the two signals in such a way that squares are minus one, otherwise of course you would have quaternions. But when one goes up a level, it rapidly becomes clear that that choice, although allowed, is a perverse one, and it would be much more natural to choose the signal for equality as one, and then even at this level you wouldn't have q but you'd have the structure which one gets from quaternion by just writing down the multiplication table and then altering the diagonal elements which are all minus one, altering them into plus one. So I call that q star. Moby could you repeat signal for what? The previous sentence? One has two signals here. One of them is a signal for equality. One does a discrimination operation between elements. If the two elements are the same, you get the signal that says they're equal. there's another one operating here because as you know in the aspect there is the second requirement that there must be a homomorphism of these systems back onto that one and the other equivalence relation which is shown by this other signal is two things are equivalent if they're mapped into the same one by that homomorphism And so you have one signal for equality and one signal for disequivalence. And I chose minus one for equality, so as to get quaternion, and plus one for the equivalence.

37:30 But it turns out, as I say, that this is a perverse choice. It would be better to choose one. I mean, it's a free choice. Let me ask, what do you mean, tell us out? It turns out. It turns out to be perverse, you mean. Well, because when one goes up a level, one finds then that instead of having two signals, you have eight, in fact, C2 cross C2 cross C2. So you have one of them, which is the identity, and seven which aren't. And at that level, you've got seven equivalence relations and one equality relation. So a very strong indication then at that level that the equality, which is one out of eight, should have the signal one, because that's one out of eight. At the next level there isn't any minus one. There are a lot of non-ones. Seven non-ones. And there are seven non-equalities. And I think it's a free country as far as this is concerned because these things don't occur in the process and so it can be supplied in order to complete the structure. But I feel that when there are seven non-equality equivalences plus an equality, and then it turns out in this structure there are seven non-one signals and a one signal, I think it's natural to do it that way around but I don't want to that's my position I'll talk to you about it later if you like so we will even have here not G22 but L22 associativity has really in my opinion come to stay Excuse me, go to equals means equivalent.

40:00 No, isomorphic. Is that the same as equivalent? I have something here, well let's go after this one. I have a group here. I'm saying that the structure of that group is that of the quadratic group. I mean, maybe you'd rather write equals there, but it's one exhibition of the quadratic group. I've only mentioned that it doesn't mean that it's a popular word. Oh no, no. Or could you use the word representation, it's a representation. A representation of word. You still call it a representation of word. I try and avoid that. Well anyway, third remark then is general non-associativity, panic, how to pursue. Now during the year, a book landed on my desk which seemed to be a possible rescue. my point in talking to you today is to explain why it isn't but i thought it wasn't first and this book is by a a ungar um one of those clover series that's why landed on my desk because they send me one free each time because i'm on the list of editors inside i never see any of the manuscripts they like to have important names on the list this book's called Beyond the Einstein Edition Law and its Gyroscopic Thomas Precession not a very not a very inviting title I thought nonetheless I looked inside Agar had noticed something which as soon as I tell you it you will realise you all know it, and that is that if you look at the Einstein velocity addition relation, which we're always drawing our attention to over the question of light, and we're talking about ordinary velocities now, look at the Einstein velocity relation, the rule for relative velocities, let me put it that way, composition of velocities in special relativity, it's not commutative. As long as they're all going in one line, then it's u plus v over 1 plus u v over c squared, and that's complicated, all right. But of course, you've got the side bits, and the side bits get beta factors in.

42:30 And it's beta factors of one of the two velocities that come in. So if you do it the other way around, you get different beta factors. So the answer when you do it the other way around is different, it's not commutated. and indeed if you then take three velocities you can see at once that it won't be associated so if you look at an Einstein velocity set up we've got a non-associative how peculiar because all of this could be and always has been by chaps like me put in terms of the regs group and it's all as as splendid as it can be I was going to say, for instance, that depressed non-commutativity, I mean, when you add two velocities, at a group level, is reflected by the fact that boosts do not commute. Right. Absolutely. Yes. I mean, I'm going to point this out. This is the origin of the Thomas procession. Sure. but looking at the algebra it's a bit surprising that you should have this thing because if you then put it all into the Lorentz group everything's alright but of course these velocities are all got by a certain non-invariant projection so we take a certain time slice and then look at the three dimensional objects in breathing. However, that was his starting point. But what he did, I mean, it's a fat book, and what he did was very cleverly, I thought, he abstracted the properties of these velocities so as to get an abstract structure which is on this board here. He was able to show that what characterised the system of Einstein velocities was an algebraic system in which there was a left identity, so that that operating on x is x, and from that and the others you can prove there's a two-sided identity, that everything has a left reciprocal in which you can then prove it has two-sided reciprocal and instead of the

45:00 associative law you get u times vw is uv times something and that something is a certain function of w the function being determined by u and v so those three bits are the straightforward one And then he was able to show that in this special Einstein velocity case, these functions, phi u, v, satisfy this peculiar identity that if you replace the u there by u times v, and you get another element, those two phis are equal. and then I suppose this is what really got me going I mean first of all I flipped over the pages as one would and then I saw a table of what he called K16 that is he wants to show that he's got something a little bit more concrete than this vague Einstein totally continuous velocity business finite system with 16 elements which satisfies these conditions. So I thought oh 16 elements you see well what I was getting here or here rather were indeed 16 elements. People said why isn't it octonians? Well it was, I mean it's because octonians have 16 elements because you've got a count of plus and minus separately. Why wasn't it octonians? And here is a man who's getting 16 elements. Has he got what I've got? Well, it didn't take long to see how it actually, but then I started looking more closely, especially because by means of these four things, he cleverly shows how to embed the structures he has got into a group in a certain way. I mean, well, embed can't be quite the right word because you can't embed something and get rid of the non-associativity, but he's able to tie them up with groups in a way which allows him to lift all the theorems out of the group theory books and mostly translate them back into his system.

47:30 So I thought, well, there's something in it for me here. little work. But as I say, that system is not what I've got. It's easy to verify that from what I've got that those three are satisfied and it's fairly easy, unfortunately, to verify that this isn't. The fourth one, which plays an important part for him, is not satisfied for L2-3. You can see this without working the whole business out, otherwise I wouldn't be able to tell you about it at all, because L2-3 has various subgroups. Some of these subgroups are Q-star, some of them are another kind of quaternion-like non-associative structure, and that is the one you get absorbing a root minus 1 into 2 of the basis elements so that that swaps over some signs in the table and then those two basis elements will have their squares 1 which is your right and for the remaining basis element you impose that its square is 1 another sort of non-sociative paternities. They both arise as subgroups. And if you look at either of those subgroups by itself, the condition form isn't satisfied. So Ungar, in fact, hasn't provided us with the answer we were hoping. In some ways, in some sort of socias, I'm speaking sociology now, really, well, because when I read further through Angard's book, he seems to be a somewhat irascible person. He has three whole pages printed in italics, announcing someone who apparently has lifted his ideas without due acknowledgement. I mean, I read these pages rather carefully. without due acknowledgement it's not as so this person doesn't have mentioned on direct call

50:00 so I think oh yeah well maybe we'd all be better off by the fact that we aren't actually operating with this system nonetheless he has been very clever in getting these general properties which amongst other things allow this allows this kind of group and then gets lots of things he can prove. So I have to cross that one out because that doesn't hold to my system. So now I'm free of any charges of plagiarism. But what I do need, and that's the problem I'm proposing to you today, is that from the general structure which I've got at this stage, I'd like to know what properties other than the first three of Orngard, abstract properties there are which would then allow me to do some of the clever things that Ungar has done with his structure. I'll just show you one rather trivial result in order to conclude the talk to show you how little progress I've made, I have proved this, that the phi sub uv of u must always be equal to u. So that particular thing, it's a little bit, but it's nothing like as clever as it is. you can see this at once because you can see phi uv of u would arise when you try and alter the bracketing in that expression and because the things all anti-commute first assume v is not the same as u then when you flip those over you'll get u v with a minus in front and then if you flip that over with that you get another minus so that will turn out to be that u v that v u becomes u v and then you take the u over to the other side and so looking back here that shows that for this particular case phi u v of u is u itself so we've got one rather

52:30 general conclusion but what we really want is something much more clever you know more on the lines of Ungar's thing there and I haven't managed to find it anyone who wants to pursue it I'm very happy to give them a copy of the table the multiplication table for this thing for him to work on. Thank you. You said that the cliff reduction are generated by three elements and if you take the product it lies in the centre of course, so it must compute. Right. So that's not allowed. So that implies non-associativity. How about the Dirac? Oh, well, you've got four. Yeah, it's alright again. It's alright again then, but it's all the odd ones. The odd ones, for example. Yeah, I mean, this could be a way, if I want to avoid the non-associativity, You could perhaps have some kind of prohibition that you must never finish up with only odd numbers of generators. Somehow you're in an unstable state there. But I want a physical argument for this, of course. There is a somewhat weak argument that I hinted at during the lecture. get anywhere with three elements because it takes so many tests so there's kind of instability about this which isn't there but that wouldn't that would show not three because it would show not four even more and any greater none so that won't help me to choose the even one. It's a physical argument, I'll make a counter positive statement. For instance, a physical argument would go something like that. We have experiments where commutativity is violated. Can you think of any experiment that violates associativity?

55:00 I mean, in physics we use associative structure, we were very happy to abandon commutable structures. Yeah, I got the point, I got the point. I mean, it's going to be polarisation, photon polarisation, okay, biolation. There's three velocities. If you wanted to prove it experimentally, but if you tried to experimentally prove it, I mean, it would be difficult, I know, but supposing you said, all right, do we really believe this stuff, let's go out and try and compound the velocities in various orders and see what we get, if we do get agreement with the formal leave. I think your point is something I'd like to follow up. On the other hand, I'd say we haven't got a leveled claim, Philip, because any person who produces a non-associative system is told to go away and trade into something associative and, of course, what the experiment is... In useful, for instance, in strength theory, they have... Yeah, they're coming. But it does tend to, I mean, the experiments only have meaning as part of a theory, and the theory is very prejudiced towards associativity. I wonder whether we physicists don't cheat, actually, at times. Oh, sure! Everybody cheats. I mean, because if you take an algebra per se, that is not associated. Of course. Right. So what do we do? The new product, you mean? Yeah, the new product is more associative. What do we do? We actually go to the enveloping algebra and return back an associative property. We work in that enveloping algebra. I mean, and in fact, that is the kind of thing I'm trying to do. Well, I'm sorry, this really isn't a question about what you've put before us, but something which will obviously be obvious why I'm asking it. If you have these Dirac-type algebras where things are anti-commutative, and at least the squares are all the same, what happens if you continue them?

57:30 For example, if you have three of those, I can construct for you the Pauli algebra, the Pauli operators, and that's fine. If I have more, because each Pauli thing is paired with one of the elements, I get lots more Pauli things. And if I split them off in threes, the original thing, I get little separate clumps of powerly operators. But if I don't split them off, like this, I have shared objects which have peculiar properties. One needs to answer this question, I think, because it has to do with what happens if you have that linear Hamiltonian form and you say, I don't know how many particles are there. I don't think it's just a question of the standard business about, well it's I think, yes, I mean, I see what you're saying there, not a question, so I'm not going to answer it. For your, I guess it's the L23, that's your non-associative group with 16 elements? Sorry to interrupt. It's rather like Octonians, but some of the signs are wrong. OK. So it is not the Octonian group, but it's related to the Octonians. Can I just like a comment back on, if I've understood what you're saying, it sounds very much like a paper that Weill wrote entitled Hengist Spinos. I think it's relevant. I don't think it covers everything, but I think it must be relevant. It depends on how you generate those powers and where you get them. Also, you didn't cheat, but I think unconsciously, when you absorb the square root of minus one to some quaternion units,

1:00:00 that's what you get in the powers. I mean, it is a higher power. If you take it's not the units, take them in all three units. Yes, of course. It's a hybrid. That's what I said. So you have two columns and one quaternium. That's right. And that's not associated. Yes, yes. Well, it is if I put ones down the diagonal. Yes, that's right. It's a hybrid stuff. Basically, you've got this kind of shifted quaternium curve, but from that you're generating a shifted octagon. Right. you're going to ask a question I think well I think that it's very late you ask me over coffee either one short comment first of all it's perhaps difficult to go to a higher level with non-associativity even if you manage to find some theory because they are it's possible to be non-associated in different ways some of those ways are more easy and the theories will cover this is one other thing I was supposed to talk this year but I decided not to and in this talk which I'm not giving I wanted to show that which I found last year that it's that to come from system to associative system demands that we introduce a set of equivalences in non-associative systems. And then we get associative systems. So maybe there is a similar way, like you lump certain elements together and treat them as one element, and you come to associative system which is incredible. It is, yeah. In the same way that you can restrict the octomials to sub-sectors that they are commutative. But not even, I'm saying not even in that in view, not that you want really to get this call and you do it. You get it anyway. I got it. But the associative system you come to is then smaller. Sounds to me as if it will then be

1:02:30 And actually it seems that it's quite a general principle that in order to get non-associating system from associating, like this equivalence relationship seems to be something universal, which connects associativity and non-associativity. You see, this has arisen, the L2-3, the whole construction which gives it, has as part of it the requirement that at every stage there's a homomorphism back onto there. That homomorphism determines equivalences. And those equivalence classes then become associated with it. It's the same kind of construction. Yeah, and even commutative, but in between. There's still a lot of structures which are very well behaving, I don't know why you insist on non-associativity, but I don't know why you insist on non-associativity. No, I don't insist on anything. I think, let me do it. I mean, I'll just conclude by this kind of proclamation that I'm trying, despite all the pressures of my past life, not to insist on anything but only to look at what the physics is telling us. I'm trying not to be the mathematician who puts a particular framework on it. Anyway, let's have coffee. Okay, so this is the main theme of this talk, topological entanglement and quantum entanglement. You could make an analogy between a topological entanglement, which is certainly non-local in its way, and a mechanical entanglement, like this guy, which is not a tensor product, and is in a state which has the property that if you observe it, it goes to one thing or another. Could you stand to be off? Stand here. So there's an analogy to not being a tensor product with state and being linked in the sense of linking, topological linking. Is it an analogy? Is it a metaphor? Or what? Right? So that's a nice question. And I want to try to consider it a bit. And this is pretty preliminary.

1:05:00 There's a paper by a fellow named Erevind. And he points out some things. He points out that you could consider this state, for example, which is three spins, three particles with a spin, and that's just alpha 3. It got illegible there. But here's a nice entangled state. But if you were to observe it, it becomes disentangled. If you observe, then you go to one of these, and it's a tensor product. So it's rather analogous to the Borromean rings, where you have this beautiful link here, which has the property that if you remove any one of its components, then it falls apart. So in some sense, this is a Borromean state. But Aravind also points out that if you change basis, observe in the x direction rather than the z direction, rewrite the state, then if you observe in the x-direction, the first one, then you still get an entangled state. So if you were trying to make a link diagram for it, and it was supposed to hold up for all of the different changes of basis that you might have, then you couldn't do it. If you change basis, you get a different linking pattern, like this, that will describe this. What happens if you write your font and place in a way which is independent of places? Well, but that's the point, it just isn't independent. You were talking about how do you observe the state. If you change the way you observe the state, then you can get a different result. Yes, quite fair. Now maybe you should, I'm not sure what the moral is here. I'm only presenting this part as a problem because I don't know how to deal with it exactly yet. But one way would be to take a given state like this state here and classify all of the different linking patterns that you can get out of it. And that doesn't have to do with links particularly. I think it's just worth doing. I haven't done it even in this case. But you could look at all the different kinds of patterns that come out of it. And, of course, you could presumably draw a link for each one, pretty, but it looks like it's a many-to-one-on-both-sides kind of relationship, and not obvious.

1:07:30 On the other hand, there certainly are interesting links to draw. For example, you can draw what are called Brunian links, which are links of many components, and they still have the same property. If you remove any one component, the whole thing falls apart, and yet it is linked. corresponds to this state in this particular basis. So the question is to investigate the collection of linking patterns that describe the entanglement of the different state. So there is something to be done here, but it's more in the line of the question than anything else. Is the idea that you measure just the spin of one of the particles and see what happens? You don't do any more... Right, but of course when you change basis you may be measuring not a lot. are you changing the tensor product decomposition or just measuring a different component of the spin of the same particle? I think I would rather just measure the spin of one of the particles. Yeah, so you're not actually... In a different direction. In a different direction. It doesn't seem fair to measure some linear combinations. So you're holding fixed decomposition of the space as a tensor product. So I think there's a nice problem there. And that's all I know about it. Can I just really go out to that picture? Is that an Escher type thing or is there closure there? Pardon me? Sorry? What's your question? Is it an Escher type thing or is there closure you see? I'm just trying to figure it out. You're talking about the mic? No, one of these Escher types. They go around and around. The Escher type is just... Could you actually make it? Dutch. Can I make this? Look what happens if you take say this one away, then this one falls out of that one and then this one falls out of that one and this one falls out of that one and this one falls out and so on and the whole thing falls apart. Now, if you do investigate the possible linkages given a certain number of these groups, are you suggesting that perhaps this might tell you the possible ways of disentangling the

1:10:00 Yes, but one is a linear algebra problem, which should be investigated even for these simple states like that. How many different ways are there of having this and seeing this entanglement if you're allowed to change bases? And on the other hand, there's another thing here. I mean, a link is a macroscopic thing, but it has a nonlocality involved with it. This is a quantum thing. This is a linear algebra thing. This is a topological thing. So to begin to actually get these to speak to one another, I think we should do something else. And that's what I'm going to talk about in a little bit. So I want to take an operator viewpoint here. And let me give you an example. Here's the controlled knot of the quantum computing people. All right? This has the property that if the first bit is 0, then it doesn't change. But if the first bit is 1, then it changes the second bit. That's controlled knot. And if you apply controlled knot to this state, tensor 0, then you see you get this flip here to 1. And this is entangled. Could you move the slide at first one? Pardon? Oh, yeah, sure. Yeah, let's try again. So if you apply control-naught to this little tensor product here, it becomes a tank. So you can, this is part of the standard ideas that people use in quantum computing. You could entangle a state by applying certain unitary operators, and you have some lexicon of unitary operators that you like for building your quantum computer. And so this isn't topological, but it could be. Let's think about it. The operator could be topological. That is, if I went back over to the braiding situation, thinking about braids, then I could imagine an operator which takes an untangled braid and entangles it. And in fact, such an operator could be a braid itself, which is a good point to make. typology. I can let the operator be this little braid, and when I attach it to whatever is

1:12:30 on the left, then it becomes entangled. And I can make a lexicon of braiding operators, and in fact, that's what generates the braids. Any braid that you see is actually a product of certain simple braiding operators applied to the identity braid on the left. So the So the R, the operator here, could be an elementary braid. And so then the question becomes, how would you associate a unitary operator to an elementary braid? So let's look at this as though we've never seen it before. Now, of course, you may know how this story goes. but we certainly can look at it as though we were trying to reinvent it for these purposes. Here's an elementary braid. Here's an elementary three-stranded braid, and here's another elementary three-stranded braid. And if I had associated an operator R to this two-stranded, then I could get a new one on the three-stranded by taking it to be the identity here, taking the tensor product with the identity matrix, or on the other side. So what I really need are operators for two-stranded braids, up-level them to get operators for n-stranded brains like that. I have the identity for a while, and I have the r, and I have the identity for a while. And then, of course, the r will have to be invertible. Well, that's part of it being unitary. So the other one, if I've associated this one to r, then the other one will have to be associated to r prime, who is the conjugate transpose of the other one. And then it would be unitary. And then we need also invariance under the basic topological transformation on braids, which is this, that you have a line which is going down here like that. And if you pushed it downward and slid the other one up a little bit, then you end up in this configuration here. So that's the basic topological transformation that generates braiding. And reading this off, in terms of these operators, I have R tensor 1 and 1 tensor R and R tensor 1. That's on one side. And I have 1 tensor R, R tensor 1, 1 tensor R on the other side.

1:15:00 So that equation has to be satisfied by our grading operator, along with unitarian. And that's called the Yang-Baxter equation. And it has been studied for a long time in various contexts. It came up in statistical mechanics and what kind of thing. So can we see an example of this? Well, it's funny. At first you think that the requirement of unitarity makes it hard to find examples. And it does if you're thinking like a knot theorist, because it just has to do with the history of the subject. People didn't worry about unitarity in making invariants of knots and links to solutions to the anti-Baxter equation for many years. And I think that people doing statistical mechanics either, for one reason or another. So lots of examples are available which aren't unitary. And on the other hand, if you search through your lexicon, say, of four-by-four matrices that you happen to know, you might find this one. And this is a nice example, because it's so simple, and we can think about it a little bit. If I let C and C be the same here on the off-diagonal, and then I have A and B, and I let these be unit complex numbers, then that's certainly a unitary matrix. And it's a kind of a generalization of the controlled knot, as you see, where I'm switching in the middle instead of at the end, so I have to rewrite my basis. But it's a kind of generalized controlled knot, and I've written out what it actually is doing here. It's unitary, satisfies the Yang-Baxter equation, and actually it's kind of fun to see that it satisfies the Yang-Baxter equation. So, let me show you that. So what I'm supposed to worry about is if I had assigned entrance states here, I'm thinking of going down like that, entrance states say 0, 1, and 0, then what could happen from the matrix? Well, when things are different, then the only thing that can happen is the term that corresponds to permuting that, right? Remember how the matrix works. and so now 1 is here and 0 is here and there's a 0 here so these are still different so the next time that it comes up that we have to apply the matrix they have to again be permuted and on the other hand these are the same and so they remain the same

1:17:30 so that if you started with indices 0, 1 and 0 you'd have to come out with indices 0, 1 and 0 you'd pick up weights C and C and A On the other hand, if you look at the other side of the gradient equation over here, ignoring this, because I'm just thinking about one state, then now you're looking at the 1 and the 0, and they get permuted. And then the 0 is over here, and there's no permutation, and then there's a permutation. And you see you pick up C and C again in A, so the weight on this is AC squared in both cases. And that takes care of one case of the many cases that have to be checked. I've checked the other case over here. You'll notice that 0, 1 is happening here, and 1, 0 is happening there, and that's why I needed it to be symmetric, C and C. So once it's symmetric, then it ends up satisfying the Yang-Baxter equation. The other's a similar example. So that's why it's topological, this operator, and you can fool around with it. For example, if you wanted to see what it's doing on a bit of actual linking, then you could square it. And then you see that it's up phases, but it is picking something up and detecting some linking in that process. And it also does... Luke just said, would you ask for a second, that actually is a question, it may be a really dumb question, but no, the one before that, don't have a knot on the bottom, they have some The one I just used. The one you just have it. Yeah. Most of the pictures you have, when you cross the rope across, it goes on top of both of the ones that are there or underneath both of the ones that are there. But the one down below, that's a real genuine twist. Yeah, this is a really genuine twist, right? My question is, how do you get the over versus under? All you're doing is to keep track of all the way it was. But that's just it. This is not equivalent to, I'm sorry. This is not equivalent to this, right, which would have been an identity matrix. Right. Right. That's the whole point. You get the over and under because, I mean, it's not enough for the ones to switch. It seems to me to get the over and under. I'm missing something else. Now, maybe I should emphasize what I'm saying here.

1:20:00 If I draw out some braid like this, which is non-trivial, although I guess there's a lot of overpassing in there, but certainly it's a non-trivial braid, or more simply this one, and I do some topological transformation of this, each one of these is going to some matrix. This goes to some matrix M, this goes to some matrix N, yeah. And if a braid was actually equivalent to, say, the identity, then the matrix that this goes to, called N prime, will be equal to the identity. All right? So if two braids are equivalent, then they get the same operator. If they're topologically equivalent, they get the same operator. And at the level of the braids, you're seeing how the operator is detecting the linking by the fact that it's not the identity. I think he's saying. I'm just wondering. It seems to me you're numbering the braids 1 and 0. Oh, those are the states. Yeah, I can't see that just labeling 1 and 0 distinguishes whether. I think you may have missed the point about what I was saying. I've got time here because I think I've got time to digress a little bit because I've already told you the main point I wanted to make. And I should finish by 1.15, right? Yeah. Just a check. So let me go back to what I was saying here. Suppose I'm looking at this. And I'm saying that this is a matrix, all right? And that means that if I give it some input states like what I did over here, 0, 1, and 0, and some output states, which in this case are also 0, 1, and 0, then this will give me a matrix entry. So 0, 1, 0, 0, 1, 0, right? And there's going to be some complex number here, because it's a matrix. And I'm asking, well, what is the actual entry?

1:22:30 What do I get when I compute that? And then what I'm supposed to find out is that if I were to change the thing to a topologically equivalent one, like this, with the same entries, 0, 1, 0, then I should get the same number I should get exactly the same number from one as from the other that's what the Yang-Baxter equation is telling me I should get the same number I think the question is how you distinguish diagrammatically R from R inverse diagrammatically how do you distinguish R from R star dagger. It's the crossing rule going down. Yeah, yeah, yeah. This corresponds to R, and this corresponds to R star, and this distinction is being made with respect to a vertical line in the plane so that I don't have to orient the strings. Sometimes people like to orient the strings, and that's fine. So you put an orientation on each string, and you say, rule, and if I switch it, then it's a negative crossing by the right-hand rule, right? But I'm just noting that I have a right-to-left going up. So if I do an R followed by an R star, then I get a twist. If you do an R followed by an R star, then you get a knot. Like this. Right. So if I do R followed by R, then I get a twist. If you do R followed by R, you get twists. Oh, that's what I do. Okay. So the twist comes from the iterations of the same operator. Yeah, if you keep iterating R over and over again, or you do R, and then you do R in a different place, and you do it back, then you can get moving. Sorry. You just got more and more twists. Yeah, you can add lots more twists and different twists. I thought from the way you were writing these ones that if you R twice in a row, then undo it. But neither of these are undone. These are both non-trivial braidings because the permutation is non-trivial, right? This line ends up in the third place, so it's never going to be the identity. They have the same underlying permutation, but they're actually isotopic from one another. You can slide one together.

1:25:00 So they should give you the same matrix entry. And in terms of my simple example of a solution, the only thing that's needed in order to make sure that they do 0, 1, and 1, 0. The c's are the same. Would you be just as happy with 1 and minus 1? Sorry? Instead of 1 and 0, would you be just as happy with 1 and minus 1? Oh, yeah, I'd be perfectly happy with 1 and minus 1. I had some choices here, and usually when you're doing qubits, you use 0 and 1. And so I chose 0 and 1. Actually, I like minus 1 and 1 a lot for other reasons. Yeah? In this permutation structure that you've got, where does just a group of three parallel lines fit? I hope that's not a daft question. You mean like this? Yeah, just straight. Well, that's the identity permutation. It's the identity braid. And it gets the identity matrix. It gets the tensor product of a small identity matrix three times. So the matrix that it gets is 1, 1, 1. all right because it doesn't change anything if you use this as an operator sorry because it doesn't change anything it doesn't change anything if you remember that there there is this um i alluded to it but maybe we ought to look at it again i mean if i take if i take a couple of braids say this one and and this one, then I can multiply them together topologically by simply connecting the bottom parts of one to the top parts of the other. And that gives me a product. So the braids form an algebra with a product structure. And you can think of this product structure as corresponding to the multiplication of the corresponding matrices. This is indeed, that's the intent here. And it's a group, the braids form a group. Because if you take any braid, you can undo it by taking its mirror image. Like if I wanted to undo this braid, then I will draw exactly the same thing backwards, like this. I can't draw backwards very well. It's a mirror image as though it was standing over a pond and it was looking at its own

1:27:30 mirror image. That's what you can see down here. This is its mirror image in a little pond. And if you connect it up, I've set it up so that these cancel. And then once that happens, these cancel and the others cancel so that this is all gone. So that's the inverse, where the braids form a group. Nice discovery of Arden back a long time ago. So that means that topological entities can be regarded as algebraic entities when you do this. It's an interesting idea, actually. You start with a topological space, but you turn it into an algebraic operator. And that's what I'm doing here in this case. The linking of the entanglement as an operator. You only get a finite number of entanglements. Sorry? There can only be a finite number of entanglements? I'm assuming that. Yeah, I'm assuming that I'm dealing with, you know, the kind of knot or a braid that you can make out of rope so you don't have infinite amounts of winding. It's interesting mathematically to think about braids or knots with infinite amounts of winding on them. But that involves thinking of what it means. Like a DNA molecule, sir. Or, well, molecules, but still. DNA, sir. If you tap me, I'm probably a fine artist. Okay. So what about entanglement from this operator that we were playing with? Just to show that it really is strong enough to entangle something. Let's find out what it needs to do to entangle. operator, I'm going to apply it to, as an example, I'll apply it to 0 plus 1, tensor 0 plus 1, okay? So that means you get A times 0, 0, and B times 1, 1, and C times 1, 0, and C times 0, 1. That's our matrix expressed again. And is this an entangled state? So let's do a little calculation just for fun. so if it's a tensor product then it would be x0 plus y1 tensored with x prime 0 plus y prime 1 so that says a is x times x prime and c must be x prime y but also xy prime and b must be yy prime so let's see if my algebra is correct here

1:30:00 since I just did it to write this slide could be wrong a is xx prime, c is x prime y and I've solved for y prime so that y prime is c over x and I put it in the last equation, b is y times c over x and then I'm going to solve the first equation for x prime so that's x prime is a over x and put it in the second equation so I get c is a over x times y and xb is yc and xc is a y So that says that A over C is X over Y, and B over C, oh, I've got it backwards. A over C is X over Y, and C over B is X over Y. Now that's right. So A over C should be C over B, or AB should be equal to C squared, which is a special case. So if AB is C squared, it's not entangled, but otherwise it's entangled. And so in general, within the leeway of what I have for available numbers for A, B, and C, I get lots of entanglement here. So that means that my braiding operator maps to entanglement all right. And so you could continue to investigate this, and I intend to, but I don't know very much about thinking, again, about the correspondence. I mean, I can map it to entanglements, but you could ask lots of questions. So here are some questions. So we have that you can make an entanglement. You could ask to classify the quantum entanglements in terms of braids or braiding operators that could produce them. So that's another way of trying to classify entanglements, right? Is there a braiding operator that produces it? Which braiding operators produce it? Can all quantum entanglements be lifted to braiding? And then going a little further in, how do protocols for quantum computing look like from this point of view? So I could take the quantum teleportation algorithm and try to write everything in terms of grading operators for the entanglements that are needed. I intend to do that, but it's not clear to me that it won't just make it more complicated. But in any case, there is, in this way of thinking, a way of making the language of entanglements and bravings speak to the language of crime entanglements, and so I think it's interesting and I'm intending to explore it further.

1:32:30 Let's see, there are some other questions. Are the rules in this that you just have one R, you have to do everything with one R, or are you going to introduce more than one R? If I choose one R, then I can uniformly map braids into various entanglements. But I might find that I need to use more than one R if I'm to express all possible entanglements, I suspect so. What you might want to add is a sort of delay. I mean, these are sort of analogs of interferometry. I mean, it's one way of realizing this. You might want to add an analog of a phase delay, which would just mean putting a pure complex number of unit modulus on the line without any crossing or something like that. Another basic. Yeah, right. So you mean I might want to use a phase game, just phases. Yeah, yeah. There's one. Along those lines, there's another remark that I was making to myself that I think is interesting. I guess there's room here. For certain purposes that I use, it's useful to add another crossing to the theory of the knots and lengths, which isn't really there, this one. It's not really there. That is, it's just a detour of lines. It isn't over and it isn't under. It's just virtual. So I could think of situations in which I might want to do that. I want to come back out from here to here, but I don't want to go over or under. Maybe I want to leap through a higher dimensional space, but I'll draw a picture of it by drawing it like this and putting a little circle around it. So then you're going to extend knots, links, and braids to what I call virtual knots and links and braids, where you have the extra crossing. Well, it's natural to associate the extra crossing to a matrix which is just permuting a couple of things like that. Because that's all that's happening. Nothing's happening. We're just sending this line over here and that one there. So in other words, you can associate this to a swap gate directly. And then you'll notice the matrix that we were working with, A, C, C, and B, is a swap gate composed with a phase matrix. So if you allow these in to brew, then you can think of generating certain phase matrices directly out of the swaps and the solutions to the Yang-Baxter equation.

1:35:00 So you can get pure phase matrices out from the topology that way. In other words, you would start with this Yang-Baxter solution and then compose it with a swap, and it ends up being just phase. So that may be useful in terms of thinking about this language. But then as we were saying, I may need to use more than one R matrix, different parameters, which means that only locally will it be topologically invariant in some portions of the braiding, and then other portions won't be. I expect I need that in order to get everything. Okay. Okay, does the association of the braiding operators shed light on quantum computing algorithms? Okay, and what about computing algorithms for non-invariants or statistical mechanics models? Now, the remark here is the following, that you might take a knot or a link like this and set it up with respect to a vertical direction and go and compute a non-invariant. And what you do when you do that is naturally divided up into a... I guess I'll use this. is naturally divided up into a braiding portion and another portion which involves these creations and annihilations from the point of view of the diagram. Now, they're not unitary because of the structure that they have, but the operators do have to be put in in order to make the non-invariant work, usually. So you could regard this as a preparation phase, and this is a detection phase, and this as a computing phase. And as much as if the matrices associated to the braiding are unitary, then this is a little quantum computer which takes you from this state to that state. So then you can ask whether you could understand how to make the quantum computer compute the not-invariant by somehow getting enough information out of it

1:37:30 to take care of the rest of the stuff. I find when I try to do that is that I lose phase information and I can only get partial information about the non-invariant usually. So it's not so easy to say, OK, here's this Jones polynomial or whatever. I'll turn it into a quantum computer. But the Jones polynomial looks like a quantum computation. It's an interference of a whole bunch of different things. But in fact, you lose phase information in trying to translate it into a quantum computer. So it was a problem. In a sense, what you're hypothesizing the existence of is quantum computing made out of classical ropes. Quantum computing? Made out of classical ropes. Quantum computing related to classical ropes. Made out of. Made out of. He's making himself a little bit. Yeah, yeah, made out of, in some sense, made out of. I'm going to take some classical ropes situation, and it will be the design. This is the same argument. It's a representation of it, but it's not it. Yeah, I mean, that's what I'm saying. I'm saying that I could take some pattern of entangled ropes and this would be the design, this would be a picture of the design of a certain quantum computer. The ropes themselves aren't doing quantum computing. This is just a design picture. a certain graph that shows you the design of the quantum computer if you use a topological entity. If you did various transformations, you would be doing various quantum steps, right? If it came apart or not, which ones came apart, that could be the answer to the computation. Right. It's not just a graph. Certain aspects of the structure of the rope as a topological entity are telling you something about the structure of the corresponding computer. So there's a language going, there's a movement of language back and forth that I hope... I think you're trying to weasel off on my question. I want a straight answer here. If I had a bunch of ropes, and I entangled them in a particular way, then it's a semantic question.

1:40:00 are all unitary right and the global entangling conditions are the analog of the quantum entanglement So I am producing, as it were, an analog, in both senses of the word analog, quantum computation. So I can make quantum computations out of classical macroscopic robots. Yes, yes, but it's not making a macroscopic model, right? In order to build a quantum computer, you have to have quantum states that are being detected after a unitary transmission. The purpose of doing a quantum computer is to do a quantum calculation. You cannot just do quantum computations on a possible computer. You cannot just do quantum computations on a possible computer. You can do it, it is possible. Yeah, right, it's just a question of the time involved, but otherwise all the effects are the same. Well, I mean, of course, any time I specify a quantum computer, you can model it with a classical computer if you have enough space. No problem. Here are the matrices. See, you're not using exponential space to do this shit, huh? Yes, he is. Well, no. Nor are you. I mean, you're just multiplying some matrices together. Well, no. I'm talking about the ropes. I don't care what that means. I'm talking about the ropes. No problem. He is. Well, I don't see no. No, because he's got all these various rope configurations that are effectively equivalent, so he doesn't have to hook you to explicitly. If you're wrong, you will get it, right? They're being, all these various superpositions of states, they are being modeled by the distributed entanglement of the ropes. So you don't actually explicitly have an exponential number of the possibilities going on in the ropes. You have all these possibilities that we're being quoted. Well, let me put it this way. It may be, it may be, that we could illustrate this idea by saying, here's a certain quantum computer and I want to know how it behaves, and I make a rope model, and then by examining the topology of the rope model, I tell you about how it behaves in some ways, and maybe perspicuously, I tell you the answer to the problem you've had, and so you don't have to actually run the quantum computer or anything, you just think about the ropes. That's right. That's possible. That's the reason for setting up a language which goes back and forth like this.

1:42:30 It's just speculation entirely true. Can I ask you a question entirely? Sure. You know the length of the rope? The length of the rope, yeah. The length of the rope is a significant variable, yeah. It's very significant for following sense. Particularly if we want to make them like Mike wants to actually make them out of rope. I tell you one of my favorite problems about which I know nothing about. What's the least amount of rope needed to make a given knot? Okay, here's the truffle knot. You see, I need a certain amount of rope, given that I fixed the diameter, I need a certain length of rope to make this knot. It's a differential geometry problem of a very difficult sort to think about what the least amount of rope is that you need to make. Thickness, or given length. Symmetrical, right? So yeah, length of the rope is important, even though the topologist wants to work. How would that come out of the computer? No, what the length has to do with the properties of my computer. I have a feeling that it doesn't really model the quantum computer because it doesn't seem to be any analog of the massive parallelism. I mean, I know you can model all the functions of a whole new computer and it goes on. Well, but then I'll be a little more antiseptic, right? I mean, you can model a quantum computer the input lines and the gates, right? And I'm saying that I'm going to take that and I'm going to replace, in certain cases, the individual pieces of that by bravings so that the graphical picture then becomes a braid or a partial, right? And it is showing you the parallelism in the sense that you have multiple lines going through. But the internal mechanism of the computer, so far, is in those matrices. I would claim that the, I mean, if Lou's speculation is entirely correct, that the multiplicity of the paths is encoded in the global relationships between the streams. They're all really there.

1:45:00 Well, you know, there are, I mean, it's worth bringing up in this context that there are other ways to compute the same thing. So I'm saying take this non-invariant, which is basically a little quantum computer with some preparations and detections attached to it, right? But that isn't the only way to compute that information. There's another way. For example, you can put the whole thing, you set the whole thing inside a gauge field so that at every point there's some gauge field, right? And you do some integration process according to WIP, right? And you will pick up the same information. So you use a quantum field theory approach. And then you're supposed to integrate over all the gauge fields. But you can imagine some physical box that you put in like a microwave oven, and you put the knot into this microwave oven and turn it on and wait. It'd be some other kind of quantum computer, completely different from the kind I mentioned, using the gauge fields, but I don't have to do that, but there is, again, there is a lot of different language translations available to you, so maybe. What I'm after is sort of a frontal assault on the physicist's assumption that the only place where quantum things can happen is that at some microscopic level. That's a claim. There's nothing more. Well, it's fun to say that. Here we have a microscope or ballistic. They go simply ballistic. Well, we might as well then bring up another question, which is what about the knotting? How far down do you go before they're not going to lose the knotting? Now, personally, I don't think you, I think you could probably continue to get some kind of knotting phenomenon all the way down so that the knots can lie underneath the quantum level. But I don't know how that's going to work either. So, you see, the quantum computer could be made of knots. Quantum knots. I think it ought to be called the distaff computer. The what? The distaff computer. The distaff computer. The distaff. Distaff? What? Well, it's two minutes. Oh, the fact that it's a feeling of weaving. Spinning. Spinning. The distaff side is always the feminine side because they use these, they use the on the spinning wheel. Oh, right. The weavers were always made. Getting back to the knots, I mean, another thing that I always found a little strange

1:47:30 about the way algebra is related to knots and braids is you can represent knots and braids into algebra. But somehow the algebra ought to be coming out of them. It does, in some sense. And let me go on a little bit about that. So now I'm going to shift to remind you about Dirac's notation. And then I'm going to abstract it immediately. First, you set up a projector, not necessarily of norm one, right? I call that a projector. So if I square it, it's a multiple of itself. That's in Dirac notation. I'll abstract it. So I just have a little guy like this. Epistemologically, this is like taking two iterants and putting them back to back. I mean, this is a distinction. And maybe this is a distinction between A and B. be like Basil was talking about, or maybe it itself is an AB. But I'm just thinking of it as an exemplar of a distinction, and then another one mirror-imaged. And if you let them interact, if you let it interact with itself, then it does the Dirac thing and produces the other kind of whole form, which is an enclosure which we'll regard as movable, a kind of generalized scalar, and it comes out. OK, that's available. And then the very next simplest thing would be to take two of them and let them interact with each other. Now, I guess this probably falls in the category of incidence algebra, but I haven't been studying what people are calling incidence algebra. In any case, it could have been in Dirac, but I don't think Dirac did this. Here's two of them. I've just written them in abstract form. So P squared is diamond times P, and Q squared is squared times Q. And if you take P and Q and P, then you get a couple of diamonds times P. And if you take Q and P and Q, you get a couple of diamonds times Q. So there's a little algebraic structure in here that's very nice. It was discovered in a different context by Temperley and Lieb around 1970. So I don't think it was indirect. But it's just fundamental, because if you take two projectors and that's the way they behave with respect to one another. And this picture of Q and P and Q and P and Q and P is very reminiscent of what happened with the braids, sigma one, sigma two, sigma one.

1:50:00 I didn't use that notation before. First one, second one, first one, second one, first one, second one. Except that they're equal, and these are not equal, but they're controlled. And it leads to the fact that you can try to make representations of the braid group by taking some combination of the identity in P and some combination of the identity in Q to get that to factor off. This was discovered in terms of these algebras, which were discovered much later and not in this simple form, by Ron Jones, who found representations of the braid group in this way. And so in this sense, if you go below the knots and the links to the distinctions, then the algebras come up. And in another level, the knots and the links come up out of the distinctions. So beneath them both, you get the algebra and the braids. And then you can go representing one on the other. But there is something that's sitting down below both of them. And so maybe the answer to your question of unifying the quantum computation is somewhere below both. But I don't know how to point to it except in a kind of symbolic way. But in any case, it's interesting. And that's the, that in a way is the, I have one more remark to make, but that's the end of what I was saying, what I intended to say about braiding and quantum computing. Is there a comment on that, without going below both of them? Yeah, I think that I tried to do that once and it didn't seem to get any soon. It seemed to get worse, actually. A couple of years ago, the last time I was here, I think I talked about the ENCOA plane, what I thought meeting Parker Woods was getting with the ENCOA plane. And that was a world where you just had tuples, and it could detect, you could react to the lengths of the tuples, whether they were the same or not. And it turned out the number of operations and stuff you had to put on that was actually more than you were just doing state's distinction. And, you know, it also goes and it flies in the face of the Buddha's point that there are no essences. There are no essences. There are no essences, so that means that if you keep looking for a level underneath at every point, then you are going in the wrong direction.

1:52:30 Well, you have infinite regress. Other people say that. Well, mathematically, this is no regress. This is pointing out that there's a place from which these are actually being constructed. Well, maybe that there is some fundamental symmetry like zero equals zero down below. But what happens here mathematically is that you can go from a certain place from which and construct something. And then they end up being surprisingly far apart and hard to bring together, even though they came from the same place and are tightly related. Well, it's sort of like the ladder structure, right? I mean, you have plus on one side, and it turns out that's the same as multiplying the other side, and it all starts from zero. that also now my remark lies at the higher level of braids and their algebraic representation and vice versa and if the previous question a couple of slides back do all entanglements correspond to bravings. If there is an assertion that yes, one can show that all rectangles correspond to bravings, then this will give a blow to, since the algebraic breakability of braiding is the braided groups, right? Yeah. So which is associative, that will give a blow to it. And if arguably entitlement is the intriguing feature of, say, quantum phenomenon, then that will give a blow to experimenting with non-associative structures. Oh, perhaps, but if it comes in sectors with different braiding matrices working in the sectors, Look, I'm assuming we have to draw this to a closer look. If you want to continue it afterwards. Well, let me make the last comment quickly and finish. The last comment has to do with that functional integration thing. So it's just a very fast comment. You know, if the integral from minus infinity to infinity of some function is equal to another one, it would be equal if their difference was the derivative.

1:55:00 So suppose that I don't know how to integrate. Maybe I don't even know how to integrate from minus infinity to infinity. But certainly at the next level, I don't know how to integrate. I don't know how to integrate overall gauge fields. But suppose I said that two functions of gauge fields, or whatever, or paths, are equivalent if their difference is a derivative. Well, that means a functional derivative with respect to path, or with respect to gauge fields if it's path. Well, then I defined the integral to be the equivalence class of the functions, and all this is well-defined mathematically. And then if you think about it, you realize that 99% of the manipulations that physicists use in dealing with functional integrals work perfectly well with this definition, because that's all you ever did was integrate relative to something like that. So that's the remark. And I was going to talk about how that illuminates, for me, some things about the non-imvariance. But it's an interesting thing to think about. That an integral, in fact, I want to turn it around. I want to say an integral is just a particular way of evaluating this equivalence class, which is what you really ought to be thinking about. And that when you're integrating over all paths, which seems like an entirely unreal thing physically that you got used to, you're not. Now, you're just looking at the form of a certain functional on paths, and it happened that you could evaluate it by integrating overall paths, but you're really looking at a certain structural properties of it, which are encoded in an equivalence class like this. I'm going to stop you mid-sentence. The reason for this is actually because Clive and Ted need to read a need to read everything. We can continue. Well, you didn't stop me in mid-sentence. That was the end. Oh, that was good. I think we ought to have another session and make people's second thoughts come up on this. Cosmosy. They're a pile. Actually, I mean, London is a little bit. That's a bad one. By permanent helmets and series. Which part of London? Which part of London? That's hard to answer, but I'm staying near Russell Square and visiting different places like Imperial College, Queen Mary College, LSE.

1:57:30 That made me perfect a little bit. Many of those three. So, let me start by putting the title up, which I don't usually do. In this case, I put it up because these crucial red bits were left off of the published title. To me, they're crucial for a reason that may become clear later. But in particular, it's because I don't believe that there's a problem with time in the sense that people talk about it. So that's why I wanted to put it in quotes. There is, however, a kind of issue that comes up very analogous in some ways to the problem of time, and that's actually what I will be talking about. But I will be talking about this within the context of a particular theory of quantum gravity, or let's say an approach to a theory of quantum gravity, which is that based on the causal set. I mean, in the title, well, I could put it there, but in the title there was a discrete cosmology, but this is not just a discrete cosmology, it's a particular form of discrete cosmology based on a particular model for the deep structure of space-time, which is a causal set, which I'll define for you in a little bit, and based on a particular family of dynamical laws for the evolution of that causal set, which we call the dynamics of sequential growth, which I'll try, which is a kind of stochastic process, and I'll try to indicate for you how that goes, although there won't be time to define it fully. So having done that, or before, having said that, I should say a little bit about causal sets in general, what the program is and where it stands, just to give you a flavor of what this ongoing work is about. I guess there's no pointer here. Is that one? Oh, this one. Okay, yes. So, I've actually put the word phenomenology out here, which is in some ways an oxymoron in relation to quantum gravity, but maybe not, because I think that it's not entirely sensible to start doing that. So, this is partly as a result of recent developments in the theory itself, which in particular is the formulation of this family of growth dynamics that I've referred to, and it's in

2:00:00 part because of developments in experiment or in astronomical observation. One of the oldest sort of forecasts, I would say call it maybe a heuristic prediction from causal sets, about 10 or 12 years old now, was based on some of the basic ideas of it, that the cosmological constant would not actually be zero, but it would be fluctuating. It would not have any definite value, but it would be fluctuating about that definite value, whatever it may be. If we take it to be zero, the fluctuations are predicted to be of the order of magnitude of 1 over the square root of n, n being the number of elements of the causal set in existence up till now, and if we assume a typical scale for the causal set discreteness, as anyone would, of the Planck scale in any discrete theory and plugging the values for the observable universe, you get a magnitude of about 10 to the minus 120. And of course, this is entirely compatible with what's been seen or probably been seen within the last two or three years. One can go further and we're trying to do that and make some slightly ad hoc models that would give you more information than just the order of magnitude, something about the time correlation of the fluctuation and remains to be seen whether or not that will remain compatible with the observations. Can I just interrupt a minute about this big N, which you said is the number of elements of the causal sector. When you first wrote it down, I had taken the big N to be the number of particles. No, it has nothing to do with particles. These are atoms of space-time, if you like, but not particles. So that's why it can be so much bigger than I expected it to be. Yes. This is one particle, this is one element per plant volume in that space-time So that's 10 to the minus 32 centimeters to the fourth is a typical volume. If you take that, divide it into the Hubble radius and take that to the fourth power, then you get about this number of 10 to the 240. That's where it comes from. Another thing that has recently been done, mainly by work of Jamal Du,

2:02:30 kind of kinematical evaluation of black hole entropy. I'm not sure that black hole entropy counts as phenomenology, but for the purchases of quantum gravity, I think it's fair enough to count it that way. And certainly one of the more established signposts that we have for our samples. So, I should stress kinematical, that means it's not based yet on a theory in which the entropy is given meaning in such a way that one could derive the second law of thermodynamics for it, but it's based on foreseeing that the entropy will end up counting something like the number of discrete units making up the horizon of a black hole, much in the way that the entropy of, say, a box of gas, of dilute gas, is basically up to a logarithmic factor counting the number of molecules in the box of gas. And this is a question whether one can identify something like these molecules and count them. If you do, you get an answer. Sorry for this little bit washed out slide, but if you do that, you get an answer which is indeed proportional to the area of the black hole horizon, which I think is not that remarkable because it had to more or less come out either infinite or zero or proportional to the area. but what is a little non-trivial is that this has been done both for an equilibrium black hole and a very dynamical one just in the process of formation where the horizon is expanding rapidly so in that way it's the only calculation I know where something like that has been done for a dynamical black hole and the coincidence of the coefficients of the area in the two cases trivial results another thing I'll just touch on is that we've studied in a certain way how to put a scalar field on a background causal set so this is much like doing quantum field theory in third space-time you just take the structure of the causal set fixed and you imagine a scalar field on it and you ask whether you can define something like an action from which some analog of the continuum action from which you could start to do a dynamics.

2:05:00 And one thing that you hit, one problem, serious problem that you hit right away, which I think is more generic than just causal sets as such, is that there's a kind of contradiction almost between discreteness, Lorentz invariance, and locality. and you can just maybe squeeze your way through that and we seem to know how to do it in two dimensions anyway and possibly in four dimensions excuse me, are you assuming that we all know what you mean by causal sex? no, I'm not so you will explain that the other thing I mentioned here is that you get a kind of blurring an interesting conceptual insight from this perhaps is that you get a kind of blurring of ultraviolet and infrared effects not in the sense of, say, the dualities in string theory which in some sense equate ultraviolet and infrared but it kind of, for example, you find that there's a because the universe is finite for cosmological reasons you find that there's locally an upper bound on the magnitude of a boost that can make sense I mean, it's a ridiculously large upper bound because the universe is so large but it's an upper bound on some local something you might regard as a local or ultraviolet thing coming, in fact, from an infrared effect this classical sequential growth dynamics I'll be saying more about let me mention that it's derived from a particular idea of how the causal set can grow or develop in time which we call sequential growth then if you impose on that general idea the two principles of general covariance and actually Bell causality, a certain form of causality, you're led uniquely to a family, a well-defined family of stochastic processes. So it's that family of stochastic, it's within the context of that family of stochastic processes that I want to discuss something more about general covariance, not in its role in leading but in its role in interpreting those theories, analogous, for example, to the, if you know, the sort of so-called whole argument or whole question that Einstein worried about, came up, which really in modern terms is coming to, is coming to the realization that it's not a metric, concretely, that has physical meaning, but a diffeomorphism,

2:07:30 equivalence class of metrics. The question is what's the analogous thing in causal sets and what does that allow us to say that has physical meaning about a causal set or about the dynamics of the causal set. I'll just run briefly. Because we have this, we can ask certain dynamical questions and it turns out that there's some interesting suggestive effects. I call them suggested because I should stress now and will mention again that these are still classical stochastic processes. This is not yet quantum gravity, this is a warm-up or a step on the way towards quantum gravity, but one may hope that there's enough similarity that things found here will have meaning for the quantum theory as well. And there's a kind of cosmic renormalization that goes on, that as the universe goes through despite a cycle of expansion, contraction, and re-expansion, the effects of coupling constants that define the dynamics are re-normalized in some well-defined way. You get another dynamics of the same form, but with different parameters. In this sense, the universe sort of chooses its dynamics as it goes along in a sort of historical way or evolutionary way. And it turns out it does so in such a way that if we could take this seriously as the dynamics of the actual universe, might be solving for us the so-called flatness problem of cosmology in a way that has nothing to do with the post-Planckian inflation. Let me stop there. There's a lot of computer . Oh, this part? Yeah, so there's a large library, which is on my webpage, the second release will come out soon, of LISP code for working with PoSETs. It's doing a large number if you have to take the transitive closure of a relation, all kinds of, it'll actually run, it's actually written in such a way that it'll run both in CommonList and Elist, so if you have Emacs, you can run this library. So if you're thinking of getting it, please do so, but also check in a few months or something

2:10:00 when I have an improved version of it, it will come out. Okay, so to answer Aletta's question, what is this causal set that I'm talking about? And then the next slide will be, what is this sequential growth dynamics? What looks like the top here is not actually the top. So is it causette or corset? Well, causal set was the original word, and this is our sort of nickname or abbreviation. And then I'm not quite sure how to pronounce this. Some say corset, some say corset. Corset. I thought of, in a sort of Slavic way, Anyway, it's a causal set. What it is mathematically is something very easy to describe. It's part of the allure of it, it's simplicity. It's a partial order. What is a partial order? Partial order is a set, a set of elements. Here, I call the partial order, I mean, the set itself C, and the elements we can call X, Y, and Z, and so on, provided with an order of relation, this thing indicated by the word cruelly, plus that sign. It's also represented graphically, usually, by means of what's called the Hasse diagram, which is just like a discrete space-time diagram. In such a diagram, these open circles here represent elements, and these purple lines represent relations, The convention being that upward relations show, you know, that rising lines show the direction of the relationship. So this element precedes this element, and if you can find a path of rising lines between any two elements, then there's a relation where this element also precedes this element, although I haven't tried any relation explicitly. Oh, I wanted to say also that you can view this as well as a kind of graph, you can think of it as a kind of family tree that actually this represents a descent pattern.

2:12:30 So you can think of this as being sort of the great grandparents of all of them. This one, these three are its immediate children, and it has some later generation descendants. These two, so to speak, got married and had a child. this one had a child without getting married I mean, without sex or anything like that whatever you want to call this so in your room, the bottom one is the first element so this is the first one this is not to show the order this is just this one is a minimal element in fact, it's the unique minimal element to this case why isn't this a causal set? this is a causal so why do we call it a causal nothing more, if the causal set is finite, then a causal set is nothing more than a post-set regarded as a kind of model for the deep structure of space-time. It's a post-set used in the context of quantum gravity. However, we would have occasion, at least in some idealization of letting the growth process run to completion, we have occasion to consider infinite post-sets. In that case, there's an extra axiom that you want to impose which represents the discreteness. So the real idea is that a causal set is a discrete postset. So after all, Minkowski space itself is a postset. It has an order relation, but it's certainly not a discrete model of space-time. So that causal set indicates also that, that idea of discreteness. So here are specifically the axioms for a causal set. And the first two are the axioms for a post-z or a partial order. The first one is called transitivity, and it states that if x precedes y and y precedes z, then x precedes z. Or in the genealogical language, if x is an ancestor of y and y is an ancestor of z, then x is also an ancestor of z. you also want to rule out cycles what you might call closed time-like curves if you were thinking in space-time language and that's done by demanding that no element precede itself which in the presence of the first axiom rules out the cycles because if you had x precedes y and y precedes x which would be a cycle then by transitivity x would precede itself so the real function of this is to rule out the cycles

2:15:00 it also contains within it a kind of convention in which we say that no element precedes itself some people use the opposite convention in which every element precedes itself then they need an extra axiom to rule out the cycles then the finiteness or discreteness comes in this axiom that's what's called locally finite which is the statement that all intervals are finite interval is the set of all elements causally between two fixed elements. So if you think in space-time, it's like a region, it's like a double light from the region between two elements to time-like related elements if it's not an empty interval. So this is the statement that between any two, there should be no more than a finite number. So in particular, for example, you couldn't have a whole, if you had an element x and element y, you couldn't interpolate an infinite number of ones, each of which was an ancestor of the next one. May a calcet have an infinite number of elements? Yes, it may have an infinite number of elements. In fact, I'll show some examples later, but if you want to think of one, for example, think of just a chain, that's one of the easiest ones to visualize. Go like this. But going on to infinity in the future direction, for example. That would be a very trivial kind of universe that you could grow from one of these sequential rows dynamics. For one choice of the parameters, you would get only this universe. Because it can be infinite, you have to have that extra accent. If for what? Because it can be infinite, we have this extra accent. Is it in case that any two other ones, since Tavanese, are taking it? So that there can be... This is not a lattice, this is only a positive. No, I appreciate that for that reason. Well, so that the interpretation of the ordinary relation as time means that there can be two things rather than time. So, in the cosmological context, you may want to strengthen this streak to say that...