 Okay, good morning, everyone. Welcome to day three of the school. So we start with Yifan's third lecture, right? About non-invertible symmetries. Fourth, oh, sorry, please. All right. Good morning. Good morning, everyone. So today will be the last lecture from me on non-invertible symmetries. In the previous lectures, we have been focusing on the big picture, the general structure of non-invertible symmetries, and especially we spelled out a lot of this general structure in detail in two-dimensional CFTs. And in this lecture, we'll apply the knowledge we have gained to identifying non-invertible symmetry in a very simple CFT, just to get our hands dirty. That is in the case of one of the standard first non-trivial CFTs that you probably encounter in any CFT class, the icing CFT in two-dimension. As I explained, this very simple CFT already gives an interesting instance of a non-invertible symmetry generated by a topological defect line. So let us start with a brief reminder about, well, a brief review of the CFT. So first of all, the CFT has a center charge equal to one half. So it's one of the simplest CFTs. And the operator spectrum, equivalently, the hubris based on S1, contain these primaries with subscript denoting the scaling dimensions, identity operator, the spring operator of dimension 1, 16, 1, 16, and the energy operator of dimension 1 half, 1 half. And together with, because we are talking about two-dimensional CFT, the operator spectrum organizing two-virazoral primaries and descendants, here I'm listing the primaries, and then the rest of the states in the hubris space are generated by the virazoral descendants. This CFT has a famous symmetry, a very simple symmetry, that is the Z2 splint-flip symmetry. The symmetry act on the CFT, this symmetry act on the CFT in a very simple way. All of the operators in this hubris, corresponding states in the hubris space are even under the symmetry, except for the sigma operator and this virazoral descendant. The sigma operator and their virazoral descendants are odd under the Z2 symmetry. And in terms of the notation that we have introduced, let's call this symmetry generated by a two-virazoral defect line eta. And in terms of the picture we have been drawing, this means that with eta acts on the local operator sigma, by enclosing it, whenever we encounter this kind of graph in your correlation function, we're free to shrink this graph and obtain just the sigma operator by the way it's opposite sign at the same location. As I've told you in the very first lecture, a hallmark of symmetry is that it gives rise to selection rules. In particular in this case because sigma operator is odd, this implies immediately any correlation function that involves odd number of sigmas, in particular the case with three sigmas in certain arbitrary locations, better be zero. But this CFT is very simple. It's so simple that you can actually compute all the correlation functions of these operators. I'll not explain the technology, but you have to trust me that you can compute all the correlation functions. And after you compute them, you will discover that in fact there's some other kind of hidden selection rule that says any correlation functions that involve an odd number of epsilon operators in certain arbitrary points is again zero. So if you try to extrapolate between from this picture down to here, you want to say that there may be some hidden selection rule that could be explained by some hidden symmetry. And that is the symmetry that will be identified. It turns out that will be a non-inverterable symmetry. Just a historical remark, so there is a non-inverteral symmetry, the kind of the signal there, this non-inverteral symmetry goes all the way back to Kramer's one year. For some already suggestion of such a non-inverteral symmetry in this simple model, since the work of Kramer's in the one year, I believe 1930s. Back then, people studied a lattice version of the CFD. It's the last icing lattice model. You can think about the statistical model in two space dimensions and it's described by a Hamiltonian normalized by the temperature in terms of here's neighbor coupling of the sphinx on the square lattice. Imagine you have some square lattice and on each side you have this sphinx know by si and they can take two values plus minus one and this notation means that you introduce this kind of nearest neighbor coupling for each pair of nearest neighbors and K is the dimensionless coupling constant. And as I said, I was briefly saying the same model, there's a very similar model related to this that's described by one-dimensional transfer sizing model. That's the quantum version of this statistical model. This model is so simple that you can actually study its entire phase diagram as a function of K and because we have observed the temperature over here, you see that this is equivalent to tune K is equivalent to tuning the temperature. So in particular, high temperature will correspond to the small K part of the phase diagram and low temperature will correspond to large K. And in this limits, it's very simple to solve the model. I will now solve it here, but you find that high temperature, you have some non-degenerative value. And on the other hand, you have a low temperature, you have a doubly degenerative value that corresponds to spontaneous breaking of the V2 splintlet. As a consequence, you expect there to be a phase transition between these two different phases. This is usually referred to as the disorder phase. This is related to referred to as order phase because the spontaneously symmetry breaking. There's some phase transition over here, which happens at a special value of K, which is denoted by Kc, that's a critical value of K. And over here, the description is given by there's a second order phase transition and the phase transition described by the ICIN CFT, which is the CFT we reviewed over here. In particular, the sigma operator, the spin operator is the CFT version of this individual spins that live on individual lattice sites. What is the statement of the Kramer's Wiener duality? The Kramer's Wiener found when they studied this ICIN model as a function of K, there's some mysterious duality between the high temperature and the low temperature phase of the theory. They find, what is the duality? They find the following equivalence relation, that under the identification between a dual coupling, which are called K dual, this mysterious relation, ICIN, and you already see that if one is small, the other is large, and the correspondingly, the duality relates to the low temperature and high temperature limit of this phase diagram. And in more detail, it was found out that the precise statement is that ICIN at the low temperature is equivalent or is dual to ICIN at high temperature in the infinite volume. But to be precise, you also need to introduce another D2 gauge field. The reason being that if you don't worry about the global issue, if you care about the global issue in particular, like the number of the ground states, obviously these two phases will not match. But introducing this additional D2 gauge field, which gauges the ICIN D2 symmetry over here, fixes the problem, so that it's actually an exact duality. And because this duality relates to the low temperature and high temperature IC model, and I should also say that in terms of the CFT, this phase diagram can be captured by the CFT, the ICIN CFT, couples to a relevant deformation, but deformed by a relevant deformation, figured by the precisely this energy operator, which essentially dual to the Hamiltonian over here, where m square is the coupling. And translated into this phase diagram, this point corresponds to m square equal to 0. So we are at the fixed point. And to the high temperature phase corresponds to m square bigger than 0. And the low temperature phase corresponds to m square smaller than 0, in my convention. So you may have different conventions for your definition of the action, and there could be an opposite. These two phases could be flipped. But this is my definition. In any case, in any case, this duality can be written more explicitly as a transformation on this coupling from m square going to minus m square, because it goes from low temperature to high temperature and vice versa. And correspondingly, because how it couples to the operator epsilon, your ICIN CFT, this is equivalent to sending epsilon to minus epsilon. So this is already some hint that the Kramers-Weyne duality may be related to this hidden symmetry that explains the selection rule at the fixed point. But the duality relation doesn't end here. There's also the famous duality relation that relates to the sigma operator to the mu operator. So we call sigma is the other operator. It corresponds to this spin degrees freedom on the lattice. When it takes a non-trivial expectation value, that corresponds to the spontaneous symmetry breaking phase of the ICIN model. On the other side, this is the desoldered spin operator. This is the operator that takes VEV on the desoldered phase or high temperature phase of the ICIN model. So when I draw this diagram, this is in terms of this other parameter, sigma. Then at the critical point, which in this case coincides with the self-dual point of this relation, meaning that pinch 2Kc is equal to 1, this duality first implies, sorry, this duality implies that the ICIN CFT is equivalent to this Z2 orbital. So this operation of coupling a quantum field theory to Z2 gauge field is essentially implementing the Z2 orbital. And as is common for dualities, duality typically relates to descriptions of the same system. And if there's some parameter you can tune, then typically what happens is that a special point at the parameter where the duality becomes a self-duality, the self-duality can be described by a symmetry. For example, you see this happens a lot on the modular space coupling manifold of two dimensional CFTs. One typical example is the two duality of a compact boson. Two duality maps the compact boson radius R to 2 over R, in certain units. That is not a symmetry in general, but at the self-dual radius, which is described as Z2 level 1, the two duality becomes a very special Z2 symmetry in the S2 level 1 WSW model. So we're expecting something similar here and we will discover that is indeed the case. But the difference is that in the case of S2 level 1, from the compact boson special radius, that symmetry is invertible. Here we'll discover the corresponding symmetry would be non-invertible, which is correlated with this very special feature. So this is a question we want to ask. So there's many ways to ping down what this symmetry is. Okay. So let me give you one argument that uses this very simple picture that we draw many times. That is the modular invariance. For different perspectives, two different perspectives you can adopt to view the torus planning function twisted by insertion of some putative toboggio defect line. Okay. So this is again the object we were considering is simply the torus planning function of this CFT twisted by this toboggio line. Okay. So right now we are not assuming what this toboggio line is. We're agnostic about the details of a line, but we'll deduce constraints from this very simple equation. Okay. And once again, this is the same object, but with the toboggio line oriented in a different way. Sorry, the other way around. Okay. And as we said before, this equality leads to the following relation between a planning function, between a planning function that's twisted in the time direction by the line defect insertion weighted by the Hamiltonian associated with the CFT, generated by L0 and L0 bar. And on the other hand, we have the trace with no other temporal insertion of the hubris space twisted by the insertion of the toboggio line. And the astral version of the torus modulus enter into this expression. Okay. So I'm writing this for general C, but here C is equal to one half in this specific set. And the fact that the toboggio line preserves the stress sensor means that both left-hand side and right-hand side will have decompositions into the basic building block of 2D CFT, representation theory building block of 2D CFT, namely characters of the soror algebra center chart equal to one half. That is very constraining because you need to representations of your sororal algebra as at center chart equal to one half only comes in three families. Okay. There's the there's one-to-one correspondence with the operators that we are talking about in the Ising model in the absence of any additional insertion. Okay. The chiral representation. So that means the left-hand side is let me call it is a linear combination of the character associated with the identity operator. Okay. Okay. Let me call it alpha one. Okay. Alpha zero. Sorry. And the coefficient multiply the character associated with the 116, wait 116 representation. Okay. And the character associated with the, sorry, the h equal to one half representation. Okay. And they correspond to physically the three operators, identity, sigma and epsilon, as well as their descendants. And this coefficients, which are, which we are being agnostic about, determines how this operator acts on, this puritive line operators acts on those operators. Okay. The line operator acts on those local operators. Okay. For too many words about operators. Hopefully there's no confusion. And we can do the same thing for the right-hand side. Okay. But as we said before, the the right-hand side in some sense is more constrained. So even though we are, because we are agnostic about the detail of the line, we don't know much about this defect cube space. But just from the verisoral symmetry, we know whatever this defect cube space is, it must be built from the basically building blocks, which involve these current representations with weight zero, one half and one 16. Okay. So you have potential, you have a sum over two numbers. This, each of this number can take value in a set of three possible representations. Okay. And the summand involves this coefficient, which is a positive integer, non-negative integer that comes for the degeneracy of the representation in the given, just the hubris space. And they're multiplied by these characters. And this very simple relation. Okay. So you have this equality. This very simple relation with the constraint that this coefficient of positive integer is surprisingly strong. Okay. So I'll leave that as a homework to show that there are three independent solutions to this equation. Okay. Meaning that all the other solutions will be generated by positive integer combination of a solution I'll write down. Okay. The first solution is such that, let me write those three solutions. The first solution is when they are both equal to one. The second solution is one minus one and one. Okay. And the third solution is for two zero minus square root of two. Okay. The case when it's acting as one on any of the three operators, identity sigma and epsilon means that this is nothing but the trivial line, the identity line. The second one acts non-trivially, trivial on everything except for the sigma operator, as well as the sentence, identifies this operation, generally by this punitive line, as the z two flip symmetry, z two spin flip symmetry. Okay. So this corresponds to the identity line. This corresponds to this eta line. Okay. And this is the object that will give rise to this duality defect. Okay. So this is the only other possibility. And as you'll see, this is indeed a consistent solution, a consistent symmetry operation. Okay. In the sense that we'll give an alternative description, a deterioration of this result that does not rely on any assumption. Okay. But here it's like we assume the line exists. This is the most general form it can take. We'll give our turn argument such that this line must exist. Okay. Question. Yes. Is it okay that some coefficient is negative? So this coefficients appear on the left hand side. Okay. What's important, what's the equation, what's not true about the equation is that these coefficients are always non-negative. So it's your homework to see that having certain coefficients being negative is fine. But for example, you cannot have this coefficient being negative. But because in the left hand side, it's not a partition function. That's right. It's not a function. It's twisted in a temporal direction. Okay. So this left hand side keep track of how symmetry acts on the hyperspace with no twist. And sorry, what are the corresponding value of the coefficient n? You are keeping it arbitrary. Sorry, come again. The coefficient nij are arbitrary. Arbitrary. Totally arbitrary. So to solve this equation, all you need to know is that this coefficient has this property, but you leave it general. And you will find the solutions will be in general a positive integer linear combination. Sorry, non-negative integer combinations of these guys. Once you fix the alphas, the n going to be fixed. Yeah. So once you fix alpha, the n is obviously fixed. But to determine alpha, you don't need to make assumptions apart from what I have said. And sorry, just one other clarification. Why in the left hand side, you are assuming that it is diagonal? Right. So this is what we know from something we just erased. So we know that the way this Tawada defect line acts on the operators will be such that it cannot change its scaling dimension. In particular, it maps a virus or a multiplet to a virus or a multiplet. In the IC model, there is no degeneracy in the hyperspace without twist. So for a given virus or a multiplet, there is only one operator. So there is no other possibility than having just an overall number. And we can learn a bit more from just this very simple table. Just from this how it acts on the operators in the untrusted hyperspace, you can already infer the fusion rule. In particular, the fusion rule of obviously of eta squares identity, that's what we expect for z2 symmetry. Moreover, when you fuse the duality defect with eta in either way, you recover the duality defect. That is because this entry is zero. Furthermore, if you square the duality defect, you get one plus eta. So these rules are uniquely determined just by postulating the most general fusion rule with non-negative fusion coefficients and consistency with this table. This defines what's known as the icing fusion rule, name is not so surprising. And as I said before, once you have a set of duality defect lines and you have specified the fusion rules, there's the consistency condition just coming from locality of theory says that they should furnish to this whole fusion category structure, meaning that it should have consistency solutions to that symbol. So there's a consequence of locality and essentially the isotopy variance. In other words, the symbol associated with junctions formed by these topological defect lines would have to satisfy the Pentagon equation. And that has been solved. So here I'm listing the solutions. This was actually this problem was solved in generality that generalized this particular fusion rule in the case when the z2 is replaced by a zn by the work of Tambara and Yamagami. So in general, we call Tambara-Yamagami fusion category associated with the general abelian symmetry. They generalized the case with just eta. But here I'm listing the full data of the fusion category in the case of this particular icing fusion rule. So in particular, your simple objects are these three topological defect lines. They satisfy this fusion rule, which I showed before. And there's only one non-trivial junction where none of the three lines involve the identity. So whenever three of the external legs involve the identity that correspond to the trivial junction, as we said before, that comes from just bringing the identity operator in the bulk to the defect line. This is only non-trivial junction in this game. And then these non-trivial junctions enter into various non-trivial f-symbols associated with this fusion category, which I've listed here. So the important thing is about this sign, which is similar to what appears in the case of group-like symmetry case. The f-symbols are faces in general. But the characteristic of this non-inveritable symmetry case is that f-symbols are in general not just faces. It will involve these matrix elements, which are not faces. And we'll use this question. There's a question? Sorry, but we know that Tambara Yamagami's fusion categories are not uniquely fixed by diffusion rules. Very good. So here I jumped over one small subtlety, is that if you just use this particular fusion rule for the Z2 case, there are two solutions up to gauge freedom to the f-symbols. Here what I'm writing down is the f-symbol that will be relevant for the icing CFT. And there is some physical meaning to the choice of the resource, some physical meaning for the choice of this one of the two solutions. It's because the CFT actually realized just one of them. I think CFT realized one of them. The other is realized by the SU2W11 model at level two. And there is a way to see that. That's right. But I'm afraid I don't have time to explain that detail here. Well, thanks for the question. And just from how this solution looks like, and it tells you how this duality defect acts onto the hubris space, we have these diagrams, which we have been drawing before. So acting on identity, which is equivalent to inserting nothing, this gives you square root of two. We're just copying down the table over there into the diagram we have been drawing. This means that whenever you have this duality defect in circling nothing, you can just shrink it. And the cost is to include additional factor of square root two. And if you have instead the epsilon operator inserted in the interior of this loop generated by the duality defect, you have the cost of introducing an extra minus sign. And similarly, instead if you have the sigma operator inserted inside the loop, you will just have zero. It's completely annihilated. And this is another signature of this duality defect line being non-invertible. So it annihilates certain operators. But there will be in a sense where this operator will be recovered. But it will be recovered in a way that will be consistent with this duality relation. And this is what we will see next. So instead of, as we said before, a feature of this non-invertible defect is that it doesn't just give maps between the hubris space to itself. It also gives maps between defect hubris space twisted by a given line to defect hubris space twisted by a different line. In particular, it gives maps from the defect hubris space without twist to defect hubris space with twist. And such linear operations are represented by, again, this kind of diagram. So this is a diagram that we call lasso diagram when we discuss the generality, but with potentially with a non-trivial junction. So I'll be using the red to represent eta and using the, okay, this is not very red. The red line to represent this eta symmetry defect and the white circle to represent the duality effect. Oh, I should also have said that here the defects are involved in the icing fusion category. They're all self-dual. So I do not need to keep track of the arrow, okay? Just in case you worry about it. Question? Again, about the F symbols, it's not true that you always have the solution with trivial symbols. Come again? Is it not true that you always have a solution with trivial symbols? No. No, okay. So as I said before, F symbol, what F symbol is doing is giving you the change of basis between different representations of the junctions associated with four external legs. Because of the change of basis, it will always have to be invertible. So it cannot be trivial. Yeah, I don't mean zero. I mean, essentially you have no anomaly. Yeah, so, well, trivial, so trivial could be zero, trivial could be one identity matrix. Yeah, one, let's say one. So that would not be consistent because you have multiple fusion channels, okay? That'll be okay for, you know, that's the case for non-anonymous group like Symmetry. But because you have these multiple channels, you cannot put one here and here, okay? To be consistent. So for some fusion rules, you necessarily have anomalies. Come again? What you're saying is that for given the fusion rules, there are fusion rules that implies automatically that you have topped anomalies. I'm not sure. I get your question. Maybe it's better to ask that again in the discussion. Yeah. So what I was saying here is just that there's this additional actions, linear actions of the symbolic defects, the duality defect represented by these white circles, acting operators, okay? And sending operators not to the hubris space without twist, but to a hubris-waste twist. In this particular case, twisted specifically by the eta line, okay? So once again, because every junction is topological, so you can shrink this diagram. So this produced for you an operator that's attached to this eta line. And this is precisely the operator mu that's relevant here. How do we see that? Okay? So once again, because the theory is simple enough, you can actually specify the entire hubris space, okay? So every state that lives in here is a state in the hubris space twisted by eta, okay? So similarly to the to the IC model without twist, here the hubris space completely fixed. It consists of the operator that corresponds to the remnant of the fermion before you bolonize the size of dimension one half on either side, okay? And just one other operator of dimension one 16, one 16, okay? And for the same reason as before, this topological operation of shrinking this diagram encircling this operator does not change the skinning dimension. And because there's a single operator of dimension one 16, one 16, it has to be proportional to this guy. And the coefficient can be fixed by asking this operator to have normalized 2.5, okay? And what you find is that the coefficient with that particular choice of normalization is square two. And similarly, okay, just to be consistent with such a duality operation, another thing that you may wonder if is this kind of diagram, this again a lasso diagram, but encircling instead the mu operator, okay? So like this. So for this to be consistent, this better give you something similar to this, okay? So it should be zero. Why is this zero, okay? This is why I draw it over here, okay? This minus sign is crucial. So look at this region, apply that as smooth because this operator scalar, you can freely rotate this red spoke all the way around. And because in the process of moving all the way around, it crosses this point once at which you use this operation, it gave up a minus sign. It was asked if that would mean that this diagram is equivalent to minus times itself, that implies this diagram vanishes. And furthermore, a similar exercise to what we did over here tell you that if you have this diagram, okay? Again, shrink it. So what this diagram means is that it's a map induced by the Toba defect line on the hubris space twisted by eta to into the hubris space without twist, okay? So because when you shrink it, there's no dangling, there's no dangling line defect, okay? So this better produce the state in the hubris space without twist. And what you find is that it's indeed you recover this operator, okay? The spring operator, okay? So what we see from this diagram is the following. If we just focus on the how the duality defect acts restricted to the hubris space without twist, it looks invertible. So it looks non-invertible because the annihilates the operator. Somehow this non-invertibility is not, can be, this invertibility can be recovered once you realize there are other non-trivial maps that max one defect hubris space to another, in particular the hubris space without twist to the hubris space with twist. So there's a sense in this particular non-invertible symmetry that once you include, once you're taking into account how the duality defect acts on all the defect hubris spaces, it is really invertible, okay? So you are not losing information. And that is why it makes sense to talk about it as a, you know, as a remnant of the dual, remnant duality at a self-do point. So there's another, the picture that is helpful to represent the way the duality defect acts on local operators is to instead imagine that you have the duality defect over here and some local operator inserted around it, okay? This is the diagram we draw when we discuss the topological feature of these defect lines. And as we know, as we have drawn this many times, okay? You can deform the duality defect line as a consequence of topology invariance, isotopic invariance, these diagrams produce the same observable when you insert the incorporation functions. But because, because of this diagram, okay, this implies, okay, and you apply the fusion rule over here. So this is the fusion rule here is important. So let me, so you apply the fusion rule for these two corners, okay? Using, sorry, you apply the f-mool for that two corners using that the rule. What you find is one over a square root of two times this diagram, okay? Plus attached to the eta line and then further attached to the duality line, okay? And then you use what we have already discussed over here, okay? So there's one diagram I didn't draw, which is the case when you have this line attached to the epsilon. I'll leave that exercise to convince yourself that this is zero from a similar argument. And this implies combining with this fact that this is equivalent to flipping the sign of epsilon, okay? So the cost to move across this energy operator to move the duality defect across the epsilon operator is to flip this sign. You can do a similar analysis. And this is, okay, and this explains this selection rule, okay? So we have achieved in explaining the selection rule using this symmetry. And the way you do that is to put the duality defect line at one end and move it all the way. You pick out all the faces, but you can also annihilate to the vacuum. So, and you get a contradiction if this is down zero, and that's how you argue this selection rule. A similar exercise in view of time. Let me again leave as a homework by doing the same procedure. Again, using this, the f-symbols we wrote over here over there, what you will find is precisely the expected relation that says that when you move the duality defect across the sigma operator, which is charged on the z2, you will get instead the dissolved operator, which lives in the hubris space twisted by the d2 line, okay? This explains this duality mapping between sigma and i mu, okay? And you can do it also inversely, move the mu across, and you bring the sigma. Okay, so combining this picture, this explains the Kramers-Weyne duality at a self-dual point with this non-trivial transformation rule that sends this basic operator in i-sync CFT, these operators, which naturally are local operators in the i-sync CFT after the dual operator, okay? So the mu operator is not a good operator, it's not a good local operator before you do the operator, because it is attached to a non-local line, okay? But after you do the operator, this becomes a local operator, and this operation explains in what sense this is the precise symmetry, okay? So let's now discuss, so that's the kind of explicit example of how non-invertible symmetry works in motion, okay? Let's now, given this example, let's now deduce some consequences, okay? So the main point here is that this symmetry, just like a z2 symmetry, okay? We discussed it in the very simple CFT, like i-sync CFT, but the structure of the symmetry is quite general, okay? It shows up in various interesting systems, and it can be used to deduce non-trivial consequences, for example, on RG flows, okay? So this is what we'll discuss now, the dynamical consequence for RG flows, okay? So what is the general picture for RG flow, okay? So you have some, we focus on the case when you have some quantum field theory for, which we'll call TUV, which is the UV description for some potential RG flow, okay? You can trigger some RG flow, for example, by some relevant deformation or by gauging, okay? This will end up with a natural IR phase in general, okay, if the theory is strongly coupled, and there are several possibilities for what the IR phase may look like. So possibility number one is trivially gapped, meaning that there's a unique vacuum that's gapped, okay? Only massive excitations about the vacuum. Possibility number two is vacuum degeneracy, or in other words, a well, gapped with multiple vacua, okay, but discrete, okay, because it's gapped. This correspond to the case, for example, when you have some discrete symmetry that is spontaneously broken, okay, in two-dimension context. And the third possibility is gapless, in other words, described by a non-trivial CFT, okay? So these are the three general possibilities for an RG flow, okay? So let's focus on the two-dimension case from now on. And for example, consider RG flow triggered by a CFT in two-dimension, perturbed by a relevant operator, okay? That's the meaning of being relevant, okay, and also scalar, so that we preserve the Lorentz symmetry. Okay, so symmetry that's preserved by RG flow generally leads to constraints on what the IR phase diagram may look like, okay? And here, we want to consider this non-inverterable symmetry. So we're first going to introduce, in what sense is a non-inverterable symmetry preserved by RG flow of this form? A non-inverterable symmetry that's generated by a two-dimension defect line called L is conserved. It is transparent to the deformation operator. We can move it across the operator without introducing any phase factor. That means it's transparent, just like how stress sensor is always transparent to topological defects. This ensures that the topological defect remains topological under this deformation, okay? And then there's a very simple theorem, okay, once we have defined what it means for a symmetry generated by a non-inverterable topological defect to be preserved on RG, there's a very simple theorem we can state. The theorem, the very simple theorem, states the following. If a quantum field theory admits a topological defect line, L, okay, such that it's VEV, okay, it's quantum dimension, it's VEV on the cylinder, it's not integer, positive integer, okay, but preserved on RG, then IR theory, the IR phase, has to be trivially gapped. So the only possibility will be a non-trivial TQFT, which is described in multiple vacuums, okay, or a gapless CFT. And this is a very simple theorem, there are refined statements that constrain how precisely the symmetry is broken. Let me just quickly give you the argument for this theorem, which is very simple to prove. So the proof proceeds by contradiction. Essentially, the same steps that you will go through in solving that homework problem will lead to the proof of this statement, okay, and the proof actually only uses something even simpler, okay. So assume, because we're proving my contradiction, we assume the theory close to a TQFT or a gapped phase with one vacuum state, meaning that there's only one operator of dimension zero comma zero in this TQFT, okay, and while it preserves a topological defect L. So we can simply consider this again, this torus primary function, twisted by the topological defect in the time direction, in temporal direction, and which is related by a modular S transformation to the configuration with topological defect line inserted in the twisting the spatial direction, okay. But in this TQFT, we can compute the primary function very simply. On the left hand side, we have a trace over the hubris base S1, okay, weighted by twisted by this insertion, okay, the same L had before, because of the TQFT, the Hamiltonian trivial, you literally just get a trace, there's no non-trivial Q dependence, okay, and because there's only one state in this hubris base, okay, there's only one state in the D by R, this just give you the number and that's nothing but the VEV of this line operator. But using the right hand side of this representation of this relation, on the right hand side, instead you have a trace over the defect hubris base that's punctured through by this defect line, okay, and there's nothing inserted because there's no more twist and it just corresponds to one, okay, and just because you have a hubris base, this has to be a non-negative integer, okay, so we arrive at the contradiction to the assumption in the theorem, thus we have proved the theorem. So if the number of states in the twisted sector is bigger than one, no, would you still call it a trivially gapped phase or? So it could be that in the twisted sector there are more than one state, as long as there's only a trivial vacuum, a single state in the untreated sector I would still call it a trivially gapped phase, okay, so according to this definition, okay, that's where, that's the circumstance where the theorem applies, but it can be refined, okay, so does this also apply to higher dimensions then or? So the very same argument applies in higher dimension and indeed that's how the higher dimensional version of this dynamical constraint is taking this, taking this form, having derived, okay, thanks. Of course in higher dimension you have more choices for what you call a torus depending on the choice of the space e minus one dimensional slice, okay, so let me now use the other side, the other table that I wrote before to apply this dynamical constraint to another less trivial example, okay, beyond icing, what is the simplest and less simplest next non-trivial CFT is the tri-cruelizing CFT, okay, so let me just write over here, for example, of an application of this theorem, okay, so here I've written down the the operator content of the tri-cruelizing CFT, this is the next minimum model of 7 over 16, it has six verisoral primaries and because the theory is so simple you can actually solve the entire, you can actually identify all the topology defect line is theory and there are six of them, okay, the details are important, what is important there are two distinguished sub-categories, this is one called the Fibonacci category, it's also called the Young category, it is generated by the single topology defect line W and it has this very interesting fusion rule, it squirts to not quite itself but with an additional identity line also appears, okay, and this line is also same as its dual, okay, and the icing category appears also in this tri-cruelizing CFT and this is correlated with the fact that similar to the icing CFT, the tri-cruelizing CFT also have the cramers when you're typing it's self-dual under the two gauging, okay, but what's important about this particular sub-categories, generated by the topology defect line W and N, is that this W and N, they have the common feature that their webs are non-integral, in particular the quantum dimension of this duality defect line is square root of two, okay, on the other hand the quantum dimension associated with this W line because they have to solve the same polynomial equation given by the fusion rule you can find out is a golden ratio, okay, and so if you have any RG flow that preserve the symmetries and from the general theorem we just discussed, then it's guaranteed to land on a non-trivial IR phase, so we just have to go through the table to find out which operators, if there are, preserving these lines, okay, and as we said preserving means that you have this line over here and you can move it across without introducing any factor, okay, so this is equivalent to to that this diagram is equal to the the VEV of the topology defect line multiplying this operator, okay, so if you in terms of this diagram this will be the equivalence relation, so we just have to go through this table and look for operators phi when L is either the W line or the duality line that this equation is satisfied and it's easy to spot that for the for the duality line the relevant operator, so you're looking for h because h is equal to h bar you're looking for operators which smaller than one, the relevant operator will be this one, okay, so this is the operator that preserve the duality, preserve the duality defect, okay, and if you look for operator preserve this this W line, okay, then the only option is the sigma prime operator that is relevant, so what this means is that the tracheonol we have the immediate prediction that tracheonol ising be formed by sigma prime, okay, and be formed by epsilon prime, okay, in this case we preserve the Fibonacci category, okay, in this case we preserve the ising category, okay, so they cannot be trivially gapped, so if gapped, if gapped there has to be non-trivial vacuum degeneracy, okay, and you can you can actually derive a stronger result which I did not explain, so you can actually show the minimal number of degeneracy is two, and here the minimal degeneracy from a similar argument is equal to three, okay, but note that there's no other symmetry for this deformation there's no other symmetry if we don't know about this non-invertebral symmetry having some degeneracy typically you want to interpret as something being spontaneously broken that leads to this degeneracy, okay, just like the V2 degeneracy in the ising case in the low temperature phase, right here there's no if you don't have this non-invertebral symmetry there's no other symmetry that's responsible for these degeneracies, okay, instead these degeneracies are enforced by non-invertebral symmetries, in particular, this is a prediction and one can go ahead and check if this is actually the case given the tracheotidal ising model, okay, this rg flow is integrable and you can check this statement, okay, so if it's gapped indeed which corresponds to one particular sign of deformation indeed you have three degenerative vacuum, okay, this side is not integrable but you can check numerically and you will find out that indeed has a two degenerative vacuum, okay, it turns out that for this deformation there's another possibility as I said the general theorem rules out the possibility of having a single vacuum, a single gap vacuum, there's another possibility of being non-trivial CFD saturating this symmetry and that is nothing but the ising CFD with the opposite sign of this deformation, so okay, so this is just a very simple application of the general theorem we derived and this is similar, the theorem can be applied in higher dimension to deduce similar statements, I think I'm running out of time so let me just summarize these lectures by listing some other aspects which I did not discuss but you are free to ask me afterwards, so let me just summarize, so in these lectures we spend a lot of the basics of non-verbal symmetries and hopefully through these lectures I'll convince you that they are as good as the usual symmetries, in particular they give rise to selection rules and leads to constraints on RG flows, okay or IR phase diagram of some UV description, what I didn't discuss is that in these lectures I focus on the bosonic theories and the water defects in the bosonic theories, there are extension to the fermionic case just like the extension of bosonic group-like symmetry, the fermionic case including the fermion parity and that story can be tied together with what I discussed using the boson relation duality, okay in two-dimension and higher and as I just discussed this non-verbal symmetry, similarly to usual symmetries, can undergo symmetry breaks, boundary symmetry breaking, okay, and it can be used to explain the degeneracy of the vacuum and something I also didn't discuss is that this non-verbal symmetry if they're not anomalous they can be gauged, okay, so there's a precise way to gauge non-anomalous, in some sense non-anomalous non-verbal symmetries and this is the way to produce other CFTs, okay, starting from CFTs with non-verbal symmetries, okay, and the list goes on, essentially the list contains all the nice things we like about the euro symmetries and it's a more general, it's kind of a more general richer framework, okay, and lastly let me just try to connect to the other lectures, the very nice lectures by the other lecturers at the school, okay, just posing some questions for you to think about, okay, obvious question in relation to Laura's lectures is to look for this non-verbal symmetries, okay, in the context of a celestial CFP, okay, and the potential applications of this non-verbal symmetries on constraining the asymmetrics of massless particles and in relation to Matias and Kevin's lectures, a puzzle, current puzzle is to understand the non-verbal symmetries in the context of ABS CFP and in relation to the notion of no global symmetry in quantum gravity, okay, so in particular, in the context of the non-verbal symmetry in the context of the ABS3 tensionless strength, okay, which is a very nice explicit playground to discuss the non-verbal symmetry in the quantum gravity, in that case describe some explicit string theory, which we will hear more from Matias, and similarly in higher dimensions, perhaps in the twisted horography, higher dimension, the string theory in the bulk is much more complicated, perhaps one can make some progress using the twisted horography, okay, so that is the end of my lectures, but hopefully this will not be the end of your journey in the world of non-verbal symmetries, so sorry for going over time, I'll take questions. Okay, let's stand for three, maybe let's wait one more minute since people are still coming back, okay, I think we can start now, so Matias is there, is the microphone okay, okay, welcome back, so we now have the lecture by Matias about strings in ABS3, please. Okay, well thank you very much, so let me remind you where we got to last time, so we are describing the strings on ADS3 cross S3 cross T4, and so far we've been trying to understand how you describe strings on ADS3, and we've concentrated so far on bosonic strings on ADS3, and the idea is that ADS3 is geometrically a group manifold corresponding to SL2R, except we have to be careful about the periodicity in T, which we will have to undo, and I'll come back to that, so ADS3 is really the universal covering group of SL2R, and that factor we can describe in terms of an SL2R vasomino-viton model, if we are in the situation where we have pure Nervyshvods flux, so it's a vasomino-viton model based on the Li-algebra SL2R, and as I explained to you, once you add the vasomino term, you get conserved currents, these conserved currents are defined by this expression, and there's a corresponding formula for the left-moving version of those, and they give rise to an affine-cut-smoody algebra, which is this infinite-dimensional Li-algebra that I wrote down, and that's characterized by level K. And then the idea was that instead of trying to describe all classical solutions, given the fact that this theory has this enormous symmetry, we can use this as a way of organizing the space of states of this theory, and we argued that it should be given the Hilbert space, but Hilbert space is the wrong word, because it's definitely not a Hilbert space, it has a non, I mean it's not positive definite, it's the infinite-dimensional vector space of physical excitations, and it has to organize itself in terms of representations of this affine-cut-smoody algebra, the left-moving version and the right-moving version, and therefore schematically it should be of the form that you have some representation on the left and some representation on the right, and initially you would naively expect that these representations will be highest rate representations, and what I mean by a highest rate representation is that its fox space is generated by terms of the form negative modes acting on some ground states, and the ground states are annihilated by the positive modes, so these are conventional highest rate representations, so they're killed by the positive modes, and the negative modes freely generate some fox space, and normally you would expect your spectrum just to be described by highest rate representations on the left and highest rate representations on the right, and then the question is what should this sum over J run over, and what the sum over J labels are these highest rate states, and what I mean by that is that we have an action of the zero modes of SL2R on this highest rate state, and specifically I choose the conventions that the plus mode just shifts, I mean you have obviously seen bracket JM before, if you know the representations there of SU2, so J will stand for the spin, and M will stand for the magnetic quantum number, so J30 on JM will be just M, and J plus will move you up one step, and then J minus will move you down one step, and in these conventions obviously there's a, I mean sometimes you write this funny square root factor here, I've decided to rescale my states, so as to absorb this square root factor, and then it bites you here, there'll be a factor here, and the factor that appears here then is M into M minus 1 minus the casimir applied to JM, and what is the casimir, the casimir operator is the generator J A0, J A0, and if you write it out in terms of this plus minus the generators, it'll be of the form a half times J plus zero, J minus zero, plus J minus zero, J plus zero, minus a half times J30, J30, so just like for SU2 what you show is that this combination of generators commutes with all the zero modes, so it's what's called the casimir operator, therefore it takes a definite value in each irreducible representation, and what you take the value of this to be in the conventions, and so all of these things look like SU2 except there is a random number of minus signs scattered throughout, so for SU2 there would be a plus sign here, for SL2R there's a minus sign here, for SU2 this would be J into J plus 1, and for SL2R it's minus J into J minus 1, so there's always a 50% chance that a plus sign turns into a minus sign, well actually judging by that is bigger than 50% because there are more minus signs than plus signs here, but that's I mean okay you have to work a little bit to fix these signs, but it's basically SU2 with some small bells and whistles giving rise to signs, so then the question is what is the, so what are the values of M that run, so I mean when I specify J what I mean by that I specify the representation, the representation will be given by specifying what value J takes, and therefore what value the Casimir takes, that specifies the action of the zero modes then uniquely, and then I have to specify which values of M mod integer will appear because the generator has moved the values of M up and down by integers, so I was mumbling words like Peter Waal theorem, so if you look at the situation where K is large you can think about this geometrically, and then you can ask what is the L2 space of SL2R, and it's described in terms of tensor products of representations of the finite dimensional, of the finite dimension group as SL2R, and then the logic is that in the quantum theory, in the in the string theory you should sum over the same set of representations, so what are these representations for the case of SL2R, what you should, there are two classes of representations that are so-called discrete representations, and they're characterized by J being a real number, and then M minus J must be an integer, in fact M minus J must be a positive, or the discrete representations of this kind, there's another family of discrete representations, but I'll just look at those, and in fact in order for this theory to satisfy the Nogos theorem that is one to impose the physical state condition, you end up with a positive definite space of states, you have to restrict this J parameter to a finite range, namely it has to satisfy that it's bigger than a half, but bigger than a half is sort of, that goes for free, because this is a quadratic relation, so for a given value of Casimir you can always choose, if it's real, you can always choose J to be bigger than a half, so it's bigger than a half, but then the Nogos theorem tells you that it also has to be less than K plus 1 over 2, but you see in the K goes to infinity limit you don't care, and this is the analog of what you know for SU2, for SU2 the spin runs from a half to K over 2, so this is just the sort of SL2 analog of that, so these are the discrete representations, so here I have to sum over all of them, but you see this is now an integral, because J is real, J is not quantized, because SL2R is non-compact, and therefore you integrate over all the J's that run in this range, so this is really some sort of integral sum, and then the second class of representations are the continuous representations, and these will be the real heroes in our story, so we will treat this a little bit, we will sweep them under the carpet, and I'll give you an explanation for why we're allowed to do this later on, today we'll just pretend that it doesn't exist, but we'll come to that later, and the continuous representations are characterized by the spin formally, well the spin taking the value a half plus IP, and then if you plug this into the cosimere, what this tells you is that the cosimere is a quarter plus p squared, where p is a real number, and then you see, I mean this condition guarantees, if you stare at this formula, that J minus zero on the state JJ is equal to zero, so what these discrete representations look like is they have a state here where m is equal to J, and then J plus acts freely, and J minus stops here, so it basically looks like an infinite line, and you have J plus zero going to the right, right, and you have J minus going to the left, so you increase the eigenvalue of m with J plus, and you decrease it with J minus, but there is a place where it stops, namely when m is equal to J, this prefactor goes equal to zero, and you see this is sort of like the SU2 representations you are familiar with, except for the SU2 representation you would look at this range, and here you look at this range, so it's basically, that's what it is, and for SU2 you would choose J to be half integer, whereas here we choose J to be any number, and we just look at what happens to the right, whereas the continuous representations, so they are sort of semi-infinite, right, they run to the left at infinitum, they never stop, but they stop to the right, there is the smallest value of m, but there's not the largest value of m, and then the continuous representations, they are unbounded in both directions, they run all the way, because you see once c is of the form of a quarter plus p squared, or rather when j is equal to a half plus i p, since m is equal to real, this prefactor can never become zero, so therefore this sort of a row of J minus operators and J plus operators will never stop, you will go as far to the left as you can go to the right, so this extends infinitely in both directions, and then because it never stops, there aren't any preferred values for m, so then m can really take any value, so the way you say is that m is in the set z plus alpha, so these continuous representations are actually labeled by two parameters, they're labeled by the spin, which in this language corresponds to the real parameter p, and they're labeled by this parameter alpha, which tells you the quantization condition of m, a more integer, and remember I told you that the folklore is that strings on ADS-3 can't, at the pure level Schwarz-Röhr-Sperz background, which is what I'm describing here, can't be due to the symmetric orbit fold, and the reason I gave was that they have this continuum coming from the long strings out at infinity, and this continuum is exactly discontinued, so this p parameter is basically the momentum out at infinity, and the fact that this is continuous means there is really a continuous spectrum here, even after you impose the physical state condition, and therefore this looks totally different than your symmetric orbit fold theory, which has a discrete spectrum, so that's the folklore, and obviously we will have to break the folklore in order to identify the world sheet theory that's exactly due to the symmetric orbit fold, because the symmetric orbit fold, I mean on the face of it, it has no chance to match any of these theories. Okay, so this is, this is what you would naively expect, but that's not quite the right answer, and it's not quite the right answer for reasons that has to do with the fact that we're really interested in the universal cover of SL2R, rather than the group manifold SL2R itself. Namely, what this part of the spectrum describes, if you think about it in terms of solutions, I mean, so you can ask what are the equations of motions that are, I mean there's a sort of two dictionaries here, you can either write down the vessel me know written model and you can look at the classical equations of motion on G, and what you find is the solutions for G are functions of the form sigma and tau, that you can write as functions of G plus of X plus times G minus of X minus. This is the, this is the family of, of solutions of the vessel me know written model equations of motion. Right, so, so you could try to start with this and then quantize the face space produced by all of these solutions. Now here we've sort of gone down a different route. We've said, okay, we know that this theory has this symmetry, and therefore we know that this space must organize itself in terms of representations of the alpha and cut smoothie algebra. So you can ask, how does this language translate into writing down this fox space of physical states. And the logic is that the states that are described by this fox space correspond to those maps that have the property that G plus of X plus plus two pi. Remember, X plus and X minus are tau, tau plus or minus sigma. So these are the world sheet light current coordinate. And it obviously has to be, the world sheet has to be periodic in sigma goes to sigma to two pi. So therefore, this function has to be periodic in sigma, when sigma goes to sigma plus two pi. And how do you arrange for that? Well, you arrange for that by saying that G plus of X plus goes to under rotation by two pi goes to G plus of X plus times times fixed matrix and G minus of X minus minus two pi goes to M to the minus one to G minus of X minus. Then obviously the product will be periodic because you see this M will kill this M to the minus one when I multiply them together. So this will describe periodic solutions. And there is this sort of mental dictionary that the matrix here is fixed up to an element up to conjugation. So it lives in the carton torus and the different values of J. You can effectively think of describing the different matrices here. So roughly speaking, I don't want to explain this in detail, but the way you should think about it, the different values of J describe the different monotremies this solution describes, but they all describe strictly periodic solutions for SL2R. Because this is like the analog of what you do for SU2 and for SU2, there is nothing else. But for SL2, there is something else. So this doesn't account for all the interesting solutions because the T parameter here is not periodic in two pi. I mean, when you write it as a matrix in SL2R, it appears to be periodic in two pi, but it shouldn't be periodic because we want to look at this covering space and therefore we have to include additional solutions. And the additional solutions we have to include, they are of the form that you... So suppose you have a solution of this kind, I'm going to engineer a new solution that will account for the fact that T is not periodic. So what I do is I define G left of WR of X plus to be given of the form of E to the I times WR over 2. You'll see in a second why this is a smart thing to do. X plus times sigma 2 times what I call G0 left of X plus. So this is a solution of this kind. So let's call them G0 here. These are... And they satisfy this sort of periodicity where M is a fixed matrix in SL2R. And now the new solutions are taken, old solution of this kind, and I multiply G left WR by this factor and G right of WR... Sorry, this should probably... This is actually an R and then this is an L of X minus. What you do there is you multiply it from the other side. So I should have called this right. So this is left X minus and then I multiply it E to the IWL over 2 X minus times sigma 2. So okay, so what I'm claiming is, suppose you have a solution of that kind, I propose let's look at the solution that comes from this function for G plus and this function for G minus. What's the difference? Well, the difference is not as... A monkey can see the difference is obviously the factors of E to the I times something over there. But now remember, and that's the reason why I wrote out these equations over here, how we parameterized the group manifold to start with. So you see on the left we have E to the U times sigma 2 and the right we have E to the V times sigma 2. And the way I'm modifying these solutions is by multiplying them on the left by... I must admit this is the world's most stupid convention to call the thing that stands on the left right and the thing that stands on the right left. And as you see, I've confused myself, but so this is acting on the left and this is acting on the right despite appearances. So what this means is I'm taking a solution and relative to the solution I have before what I do is because of this factor, you see this will effectively shift U as a function of X plus. So U will go to U goes to U plus W R over 2. You see it's sigma 2. So it's W R over 2 times tau plus sigma. And it will shift V, which is the factor standing on the right here. You see this is the V. So V is what stands on the right and U is what stands on the left. So U gets shifted by this factor and V gets shifted by the corresponding factor W L over 2 times tau minus. And now why is this a smart thing to do? Well, remember that U and V are the light cone coordinates in target space. So if I translate them back into T and phi, T is the sum and phi is the difference. So what this means is that T gets shifted by V plus the sum. So there will be a term proportional to tau, which will go like a half times W right plus W left. And there will be a term proportional to sigma times a half times W right minus W left. And then phi gets shifted by phi plus tau times a half. And now I have to take the difference. So here there will be W right minus W left. And here there will be a half times W right plus W left. Okay, so suppose I have a solution that's described by this G zero. And once I modify it in this way, I get a solution whose T and phi dependence, I mean previously it was periodic. And now it picks up these additional pieces. And now why is this interesting? You see, we want T not to be periodic. So we have to make sure that this factor is equal to zero. Right, because when sigma goes to sigma plus two pi, that describes must describe the same point. And in order to avoid T being periodically identified, I have to make sure that this factor is equal to zero. So this tells me I have to choose W R is equal to W L. So when I do this, and this term goes, and this term goes. And what I see here, this just becomes W R equals to W left equals to W. So what I find is T goes to T plus W times tau who cast tau is not periodic T is not periodic. And and phi goes to phi plus W times sigma. And sigma is two pi periodic and phi is two pi periodic. So therefore I've included now solutions that are not periodic in T, even if I go around the sigma by two pi. So I've made solutions that were periodic in T into solutions that are not periodic in T. And what I see is it's not obvious that this accounts for all the possible things, but you at least see that you are producing arbitrary winding in the phi direction. And but you're you've undone the fact that this would have identified also the T direction. Okay, so that's the proposal of Maldesine on Agoury was that what you have to do is you have to look at the solutions corresponding to this. But now I've translated it in terms of classical solutions. So in terms of classical solution, I also have to include these new solutions. But because I want to work in that language, I now have to explain, what does this modification do on the level of the representations I've written over there? Sorry, the winding in phi does have to be integer, so like W is an integer or it can be... Oh yeah, sorry, W has to be an integer. Sorry, I forgot to say it says absolutely. So W has to be an integer, otherwise I would... Yeah, yeah, yeah. Thank you. So here I've explained to you on the level of the classical solutions, what I have to do. But I'm an algebraic CFT type person. I want to work in that language. So now I have to translate this description into this language. So what I have to do is I have to calculate what happens to the currents when I do this. Remember, the currents are defined by the formula that I also wrote down here. You have to take the trace of the Li-algebra generator, and that's the convention I'm picking, times d plus gg to the minus one, and similar for the other currents. So let's work out what happens when I modify my solution in this manner. Okay, so what we do is we take a solution that was of the type g0, a periodic solution, rather fixed by this fixed monotony, and now we have to include this factor. So now what we want to calculate is that the j a of r is now k times the trace of ta. And now I'm writing out a d plus of g my of d of g. And so, okay, so this stuff doesn't, okay, so I'll write it as, let me write it. What happens? So I have to apply this to e to the i times w right over two x plus times sigma two times my old g right to zero of x plus. And I have g left of x minus. And because I'm interested in the plus derivative, I don't have to write it out. And then I have here g inverse, which is gl x minus to the minus one g right zero x plus to the minus one. And then I have to write the inverse of that e to the more blackboard space here e to the minus pi w over two x plus sigma two. And I should write now w because w right is equal to w left is equal to w isn't integer. Okay, so now I have to evaluate that. Now, obviously you see the dx plus I can stop the dx plus the action here because this doesn't depend on x plus. So this term obviously cancels against this term that makes it already less to write. And now there are obviously two derivatives that the d plus can hit here. So what am I going to get I'm going to get the term that looks like k trace ta. And now I'm going to get two term if the d plus hits here, I get an i times w over two times sigma two g zero. And then the g zero r will cancel against this g zero r. So that's all there is. And the other term is plus if the derivative hits here. And I get e to the i times w over two x plus times sigma two d plus of g are zero x plus times e to the minus I sorry time. So I'm so g zero are x plus to the minus one. So this is this time times e to the minus i w over two x plus sigma two. Right. And then so that's what I have to calculate. Now, if you think about it you see this is basically the old current right. This is the original current. And it's now conjugated by that. But the country I can either think of this conjugating or I can think of the conjugation, conjugating the ta so I can write this as k times trace of ta times i w two over sigma two us k times trace of e to the minus i w over two x plus sigma two times ta times e to the i times w over two x plus sigma two times. I'll just write it gr gr to the minus one the old thing the one I had before. And now now it depends a little bit on which component I'm looking at. So if you so in the convention so the conventions are I mean somebody should have worked a little bit better on these conventions because they're the sigmas and then the T's so the T three to confuse absolutely everybody and this has confused me for a long time I misread the paper many times is proportional to sigma two. There you go. And then t plus and minus is proportional to sigma three plus and minus sigma one. But that's what it is. The sigmas are the Pauli matrices so everything is totally explicit. So now we can just work this out. So let's work this out for J three of R. So when you pick ta to be T three, you see T three is just equal to by this funny convention minus i over two sigma two. Then you have a minus i over two times an i over two gives you plus one over four sigma two squared has trace equal to two. So this term just tells you this will be equal to K times W over two plus. So this comes from this term and then from this term if sigma two since T three is also proportional to sigma two obviously sigma two commutes with itself. So this time goes away and you just keep the old current. So what you read off from that is that the three component of the current just gets shifted by a constant term. And what happens to the plus minus component of the current. What you have to do for that is you have to ask what is this for this being equal to plus minus. And that's a little calculation which you probably better do in the privacy of your room than seeing me struggle on the blackboard. I mean it's not the rocket science you I mean each of the sigma two sigma two is an explicit matrix so you can write each of the i sigma two as an explicit matrix. It's basically cosine sine sine cosine and then you just multiply this rule and what you find at the end of the day and it's a very elementary calculation but I'm not going to do it for you here is that this is equal to each of the minus plus i times W times X plus times T plus minus. So the T plus minuses go back to themselves but they pick up a phase e to the i e to the minus plus i W X plus. So what this tells you is that j plus minus R you see for them this time is zero because the trace of sigma two with sigma plus minus is zero because the traces of the of the different sigmas are zero. So this time is absent and this term you just pick up this phase and then you have to again the old current so what you learn from that is that this is equal to e to the minus plus i W X plus times j plus minus R of X plus. That's what you calculate so that's the that's the sort of algebraic description of having introduced these additional solutions right I'm taking these additional solutions and I'm translating it what it means from the point of view the affine-cut smoothies symmetry and that's what it means. Yes there is a bit of confusing point here for me because you introduced these extra solutions because you want to describe universal cover but even if you want to describe just SL2R you should include them but WR is not equal to WL or not. No but I mean if I'm looking at SL2R then I have to be then it has to be periodic in T right so you're saying it's also periodic in T? Yes you just need that WR minus WL is a multiple of four pi. Well sorry in this convention oh yeah so you would have to use something like that yeah yeah maybe maybe you have to do this for SL2R as well yeah. So it seems to me that going from SL2R to the universal cover is not adding new solution but it's tricking. Well I mean but but then that's not compatible with this right are you saying WR minus WL has to be equal to a multiple of pi or two pi? Four pi maybe. Four pi whatever yeah so maybe you're right maybe you're right maybe there's also additional solutions for I think you're right there's probably also additional solutions if I don't go to the universal cover but if I go to the universal cover I have this constraint on this class of additional solutions and in fact this is I mean you see for SU2 you may ask why don't I have to do the same for SU2 and for SU2 if you were to do the same thing you would actually land on the same spectrum you're not introducing new degrees of freedom because I mean this will turn out to be spectral flow and the spectral flow maps the SU2 representations again back to standard highest rate representations so normally you're probably right one has to include the sectors as well but for SL2R they generate new representations as I'm about to explain and you're right the universal cover is not responsible for adding new solutions they're there anyway it's more responsible for removing some okay good thank you yeah you're right okay so let me explain to you what this means in terms of the affine cut smoothie algebra so remember we have this melt expansion j a r of x plus I said you can write as a sum over j a n and I think I forgot a minus sign here yesterday there should be e to the minus i and x plus otherwise I'm going to run into trouble now so that's the consistent solution with everything else I'm doing so when you translate this you see what this means is that j3 uh so it's j3n the new j3n is the old j3n and then the constant term means that the zero mode term gets shifted so plus k times w over two delta n comma zero and what you find for j plus minus n is that this is the old j zero plus minus but now n gets shifted up and down in terms of w because you see this factor basically just shifts the mode number because I mean this just multiplies to the exponent and thereby you shift the mode number and when you when you find this that this is the set of transformations you get and what this means is the new solutions are described you can think of them as being described on the fox space of the old solutions except that the sl2r now acts in a modified way so so so you have the old fox space on which the j zeros act like they acted before but now on the same fox space I define a new action where the new modes are defined in terms of the old modes plus corrections and in fact what you can check is that this is an automorphism of the affine cut smoothie algebra associated to sl2r I these guys satisfy the same commutation relations as the original guys if you redefine them in that manner so this is an automorphism and whenever you have an automorphism and you have a representation then the automorphism produces for you a new representation now there's no guarantee that it's a genuine new representation it may be the old representation in disguise but at least potentially it's a new representation normally you would think it's a new representation if the automorphism is outer then generically you get a new representation if the automorphism is inner then you won't then you're just relaping the states now you can check that this is uh that this isn't in general an auto automorphism and the reason for that is you see that it shifts the mode number in this funny way so if we start with the highest rate representation and so on the highest rate representation you remember we had j zero plus m on the state j m is equal to zero for n bigger than zero that was it what it meant to be a highest rate representation but now if I think about it in terms of the action of j plus n j plus n acts on these states by saying that this is the same as j zero plus n minus w acting on these states so this is only zero if n is bigger than w so what this means is that there are some negative j plus modes that will not annihilate the highest rate state and as a consequence it's not a highest rate representation it's a representation where there are certain negative modes that you can apply as many many times as you want and they will always be non-trivial and so so so these are generically not highest rate representations you can also see it in terms of the a virus or algebra so if you look at it in terms of the if you ask what does the virus error generators do under this modification now there is sort of an an abstract way of doing this because you know the commutation relations of the of the virus error generator with the currents I mean you know that the l n generators as to with the j a m generators have to satisfy this commutation relation and then compatibility with this will tell you how l transforms and what you find is that l m transforms as l m zero minus w times j zero three m minus k over four times w squared delta m comma zero this term you only see if you insist that they still satisfy a virus or algebra and this term you get by simply demanding that this is also an automorphism under this transformation and you see because of this term and the fact that the j three zero spectrum is always unbounded to the positive line if i take w to be positive then the zero mode of l zero will be unbounded so so so these representations are not included in what i had before because before i had a bounded l zero spectrum because i had the highest rate state and the positive modes killed it and l zero move you up and now once i've included a spectrally flowed sectors i've genuinely produced new representation with an unbounded l zero spectrum but this hinges on the fact that these representations are infinitely extended to the right and therefore that this eigenvalue can become as negative as you want in particular if you think about doing the same thing for s u two you see for s u two you have finite dimensional high state representations and what you find is that after you have applied this spectral flow to an s u two representation you just get another s u two representation and for s u two what you find i mean for those people who are sort of familiar with this sigma will map the j's representation to the representation associated to k minus two over j so it'll just flip around the so it's order two so sigma squared will be trivial will map a representation to itself sigma two is sigma squared is inner sigma is outer and in fact this has to do with the fact that s u two is a double cover of s o three and what this is really implementing is the quotient going down to s o three but this is just as a side comment but what's important here is that we genuinely get new representations and therefore this spectrum that i wrote down before was really too small because it didn't include any of these resolutions so i have to include my spectrum and now the correct ansatz for what's my my fox space of this world sheet theory should be is that there should be a sum over these winding sectors and i'll restrict myself secretly to w bigger than zero then i have this integral over j or sum over j and then i will have and i write this like like such so by this i mean the the representation induced by this automorphism sigma to the w and i apply this as simultaneously to left and right movers because w right times w left is equal to double so this is basically the maldicina or gurus spectrum and as i explained to you the spectrally flowed sectors come from the fact that you have to have this periodic solutions that are periodic and fine but not periodic and t well i'm sorry i mean that you have these additional solutions because s l two r has these additional solutions and then the fact that we are in the universal cover means that w right is equal to w left rather than the difference is something funny okay so this is this is the spectrum we have to work with and now the idea is that now we've identified this world sheet spectrum now the aim of the game is that we are going to work out the the physical states that satisfy the physical state condition but before we do that there is one more thing we have to do now i'm losing track of my notes oh yeah so so far we have done everything bosonic and bosonic is fine but it captures the essence of the spectral flow but in order to really get the symmetric orbit fold we have to deal with the super string so what's the super string version you see so far i've really only concentrated on the ads three fact done only concentrated on the bosonic spectrum now i have to look at the super string version so what's the super string version well so the the super string version is actually relatively easy to describe so what happens is you see we had this s l two r bosonic algebra level k and if i'm looking there's a natural supersymmetric generalization of it and the natural supersymmetric generalization consists of writing an upper index and writing a one that means it's n equals to one super conformal and what does this algebra consist of well this algebra consists of the generators j a m that satisfy an s l two r level k f i and cut smoothie algebra with the commutators i wrote down before but then in addition you have fermions that also transform in the adjoint representation of s l two r so there are three fermions psi three and sub psi plus minus and so the j's with themselves give you an s l two r and the j's with the with the uh with the step size give you just the they sit in the adjoint representation so they just transform as somebody who sits in the adjoint representation so i hope this is so these are the structure constants of s l two r so so three with plus gives you plus times plus three with minus gives you minus times minus and so on so it's uh it's just that they transform in the adjoint representation and then the size by themselves just satisfy their free fermions so the non-trivial anticommutators they satisfy of the form psi and again there's a funny minus sign because we are in s l two r so there's a minus i'm sorry there's a k over there's a k there's a factor of k here and there's a factor of minus k over two here so so going to the super string is basically means you replace the bosonic algebra s l two r by the sort of supersymmetrized version and the way you think about it this is like NSR normally functions i mean these are the bosonic degrees for freedom of your target space and for each boson dx mu you add the fermion psi mu so the fermion has the same labels it's also sits in the adjoint representation it transforms under the bosons in the adjoint representation you can't really do anything but and then it's a free fermion just because there's nothing else you can do okay so that's basically the suzi version for ads three so that's what you have to do for ads three and then you have to do the analogous thing for su two for su two you also enhance the su two level k prime affine cut smoothie algebra by adding in fermions in the adjoint representation of su two and they satisfy the analogous commutation relations that's that and then the t four is the t four now now we want to so the first thing i have to explain to you is why the level of the s l two r is quantized i was asked this earlier and the reason for that is that it must be equal to the level of the su two so let me explain how this comes about now when you look at this super conformal super affine algebras you have to you can ask yourself what's the central charge and the central charge there's a clever way of calculating the central charge and that's sometimes useful um you see this is a coupled system so i mean how do you calculate central charge you try to separate it into blocks and then you just add that pair wise commute with one another and then you add the central charges of the blocks now here you can't do that because the stupid fermions transform under the bosons so you can't just say it's the bosons plus the fermions because they're not you have to disentangle them now in fact you can disentangle them and that's what goes under this uh decoupled current so this is meant to be a different letter than this j this is a more curly version of this data j and what you do is you define new generators new currents which are the old currents and you correct them by terms that are bilinear in the fermions and they look explicitly like that so for j plus minus you just do that and then for j three you you take these j threes and then the way you decouple is that you add to it the term j minus j plus i mean these are to be understood as normal order products of this size so what you do is you take your currents and you add to them suitable fermionic bilinear terms and then what you can prove and that's a little exercise is that these currents then commute with the fermions you basically remove the fermionic piece of the currents so that these guys commute with the fermions and they still satisfy an s l 2 r affrancats moody algebra but their level has been shifted so they now satisfy s l 2 r not at level k but at level k plus two i mean for people who know what's secretly happening here is that the free fermions themselves build an s l 2 r algebra at level two and you are basically sort of taking the coset they're sort of taking them out and then a rather minus two taking them out and then you get this term plus the decoupled fermions so so but now it's very easy to calculate the central charge that comes from this factor so when i calculate the central charge of ads 3 so you see for ads 3 i have this s l 2 r supersymmetrized level k then i have s 3 then i have s u 2 supersymmetrized level k prime and then i have a t fall so let's calculate the central charges well so i have to calculate the central charge of this decoupled s l 2 r now the central charge of a of an of a bosonic s l 2 r at level k is three k into k minus two so now when i shift this what i get is i get a three k plus two into over k this comes from the from the decoupled s l 2 r currents because they're at level k plus two then i from the fermions i get three halves because three free fermions give me three halves each fermion contributes a half to the central charge then i do the same thing for s u 2 so for s u 2 what happens is for s u 2 level one level k is the same as s u 2 bosonic at level k minus two plus three fermions so here for s u 2 level k the central charge is three k minus so this is a k prime so this will give me so this is a k prime so this will give me three times k prime minus two over k prime plus three halves and then the towers will give me six because the towers is four bosons which gives me four and four fermions each fermions gives a half that's another two so all together six and this has to be equal to 15 that's the critical dimension of super strength theory right i mean the no ghost theorem tells me that i have to live in the critical dimension so that has to be c equals to 15 which corresponds to a super conformal theory in 10 dimensions 10 bosons and 10 times five fermions 10 times fermions 10 times a half fermions give you 15 and if you stare at this you see you get a three from here plus six over k plus three halves plus three minus six over k prime plus three halves plus six and now you see three plus three halves plus three plus three halves plus six is equal to 15 so therefore what you learn is that six over k minus six over k prime has to be equal to zero and therefore you learn that k has to be equal to k prime i mean this is also familiar from a supergravity perspective that's just saying that the radius of the ads space has to be the same as the radius of the three sphere that's the supersymmetric supergravity solution but as the c of t the version of it is that in order to get a critical strength theory the level of the su2 and the level of the sl2r have to be equal have to be equal to one another and therefore because the su2 level is quantized the sl2 level has now also turned out to be quantized because they have to be equal to one another there's a question sorry i have a question on the shifting on of the level maybe i'm confusing myself but with the different conventions but also in the bosonic in the bosonic w zero model you have a shift of the level by the local excited number well that only appears in the central charge formula of s of g level k is k times the dimension of g divided by k plus the dual cox the number so that's for example well for su so for su2 level k this gives you three times k because the dimension of su2 is three and it gives you k plus two because the dual cox the number of s is two yes and if i'm not wrong for sl2r is there is a minus yeah so for the sl2r exactly for sl2r i basically you can think of this the dual cox the number being equal to minus two so is it true that by adding fermions i have an opposite shift absolutely so this is actually a very important point so you notice here you get a factor of one over k and one over k prime here right in the bosonic case you would have had a factor of one over k plus two and one over k prime minus two or whatever now why is that important remember k is the radius so if you think about it like you weren't smart and you took this vasomino written model you turned it into Feynman diagrams and you would do a Feynman calculation then the powers of one over k is basically the higher loop corrections as you would calculate it from a Feynman diagram and then in the bosonic case you would have something like one over k plus the dual cox the number and the way you think about it is that this is one over k times one my one plus h dual over k to the minus one so this is one over k times the sum from n is equal to zero to infinity of h bar over k minus this to the power n right so what you would see is you get infinitely many corrections in perturbation theory if you treated this naively in perturbation here now what happens in the sousy theory you see the sousy theory is sort of renormal not renormalized and that's manifesting itself that the central charge involves just the unshifted level and therefore it's one loop exact you don't get any higher order correction so that's the reason why this is exactly opposite and that's a hint that sousy is doing something good for you okay thank you okay so so this is so this is basically so now we know that we have to live at the same level and now we just have to put everything together and and analyze the physical spectrum now I'm out of my many many right so so now what we have to do is we have to so now so let's write down the spectrum and actually so here there's some small subtlety you see I explained to you spectral flow on the level of the sl2r by sumina written model but now you're going to ask yourself so should I spectrally flow the j's or should I spectrally flow the curly j's right I mean they're both sl2r by sumina written models and I told you how to flow them so which one should I flow and the answer is it doesn't matter so why does it not matter well because you see spectrally flowing the j's and spectrally flowing the curly j's differs by whether you spectrally flow the fermions or not there's an analogous way in which you can spectrally flow the fermions but for the fermions if you spectrally flow them you're just rearranging the states you're not generating any new states in your in your fox space just like with su2 level k where the spectral flow just maps the jth representation in the k over 2 minus j representation of the fermions it just re-assembles the states so whether you spectrally flow the fermions or not is up to you you're not describing a different spectrum you're just describing them slightly differently and it's more convenient to actually spectrally flow the coupled sl2r generators because the coupled sl2r generators are the geometric ones because geometrically the fermions will also feel the mobius symmetry so they should also transform under say the zero mode so decoupling the fermions is just a trick for the purpose of calculating the central charge but the real sl2r is the coupled one because the fermions I mean these are the tangent vectors they live in the tension space to the manifold they also feel the rotation of sl2r and to remove them would be artificial so what we are going to do is we're going to spectrally flow also the fermions i.e. we spectrally float the coupled generators rather than the decoupled generators and what this means is that the k that will appear here is really the k i'm talking about here rather than k minus 2 or k plus 2 for the case of sl2r su2 so i'm flowing the full generators and then on this resulting fox space which will look so there's indeed so there's the sl2 factor so again this will look like a sum over w then i will have the sum over j and then sigma to the w hj but now these are the super conformal representations but that we are happy with and then i have a similar factor for su2 this is like the covariant description of string theory right this is this involves the time direction ads3 contains the time direction so in order to describe the physical string states i have to impose the physical state condition so this will be the condition that gr of phi is equal to zero for r bigger than zero and i'm thinking of here working in the nervous ward sector there's always a nervous ward sector and a roman sector but a nervous ward sector is the interesting one and then you have ln on phi minus a half times delta n comma zero on phi is equal to zero for n greater equal to zero so these are the the physical state condition this is just like if you're open green schwarz mitten you do a flat super string that's what you have to impose in covariant quantization right it's the it's the usual n equals to one super conformal symmetry you have to impose and in the nervous ward sector the mass shell condition l zero has to be equal to a half that's the mass shell condition in order to get the no ghost theorem going so now what we have to do is we have to evaluate we have to look for the states that satisfy these conditions we have to enumerate all of them and then we want to find out what their space time charges are and that will describe for us the space time spectrum of this specific world sheet theory so what does so the interesting condition here is that l zero has to be equal to a half so what does this mean well remember we have to work with the real l zero and in the spectrally float representation l zero gets shifted in that way but what i'm going to do is i'm going to spectrally flow um yeah so this is i'm going to use exactly this formula so the l zero condition will amount to the following it will amount to the h zero of s l to r so for my for my level for my susie theory at level k plus the h zero of the rest so the rest will be s u two so the rest is s u two plus the torus so this will be the ground state conformal dimension then there will be some excitation number this is the total excitation number coming from everybody and that has to be equal to a half right that's the mass shell condition that i will have to impose on this spectrum right you have a you have the s l two factor and the rest the rest is s u two and t four this is the conformal dimension of the ground state of s l two that's the conformal dimension of the ground state of s u two that's the excitation number and i'm i'm thinking of looking at states that are in the in the vacuum of the t four i could also include that doesn't really matter for the analysis just to be simple and this factor for this factor i now have to use the fact that this is spectrally float so this will be the ground state energy before spectral flow and the ground state energy before spectral flow is the before spectral flow is the cousin here so this will be minus j into j minus one divided by k right because in the susie theory i get this shift it's the the the chukavara construction will have a level k down here for the same reason that we discussed before and then it'll be minus w times m so m i will call the eigenvalue of j zero three before i apply the spectral flow so this is the eigenvalue of this guy and then i have to subtract minus k over four times w squared so that is what this term is and then i have plus h zero rest plus n is equal to a half so this is the mass shell condition i have to solve and now i'm going to solve this mass shell condition for you and explain to you that we get something interesting so let's let's do this this is a maybe i can do this yeah perhaps uh you can take another five minutes yeah it'll be five minutes it's very it's a very simple i mean it's uh very very elementary except uh it's quite key so i would quite like to explain it today and then we can discuss its consequences so what i'm going to do and at this stage i'm just declaring this so i'm going to set k is equal to one that i've told you before that you are used to and now i'm going to say that j is equal to a half times i times zero so i'm going to look at the continuous only condition its representations i'm only going to look at the one for which p is equal to zero now the justification at this moment is non-existent but i'll promise you next time i'll begin to explain to you that this is what you have to do if you look at it from the hybrid formalism in this language it's not obvious why these are the right representations to look at but grant me just look at these representations and see what happens okay so what's the casimir well the casimir is a quarter so what does this equation become like this becomes a quarter minus w times m minus a quarter times the w squared is plus h zero and i stop writing this rest stuff now plus n is equal to a half okay so now this equation i have to solve and remember in the continuous representation the eigenvalues of j three zero are up to me because the continuous representations remember they are labeled by cj alpha and m has to be of the form alpha plus a dead but i can simply take this equation to solve for m and then declare that that means i'm looking at the representation with the corresponding value of alpha and because all values of alpha appear inside this direct sum so if i'm looking at a continuous representations i would sum over all values of j and alpha there will be one term in the sum where this mass shell condition would be satisfied so i have in mind i take a specific dissentant and i ask for which term in this integral does it satisfy the mass shell condition and i'm saying okay i'll just take this equation i solve it for m then m tells me which alpha and that picks out the term in this integral where it matches the mass shell condition okay so solve this for m okay that's not difficult i just bring w m to the other side so what is this w m is then equal to minus a quarter a half minus a half a quarter minus a half minus a quarter w squared matias sorry just to say that the equation should be equal to zero right so say again a question equal to zero so the the quality the question means is equal to zero i'm here it's equal to a half i'm in this is in a never short sector of a amount never short sector string it was a minus and the equal because i can't see from here i'm sorry sorry no this is this is equal to a half and this is this a half this is the usual a half ground state energy in the never short yeah sorry okay so so this is the equation okay but now remember that we are not interested in m m is the j three zero eigenvalue before spectral flow and we are interested in the j three zero eigenvalue after spectral flow because that's the real j three zero right the other thing is just a way of describing these representations so the real j three zero eigenvalue remember that j three zero is equal to j zero three zero plus k w over two and this is this eigenvalue if denoted by m remember that was the term that appeared in the shift formula for the conformal dimension so if i'm interested in this and remember this is the scaling dimension of the dual cft so this i can also call h and think of it as the conformal dimension as looked at from the point of view of the spacetime theorem so for that i should not look at m i should add to m k over w but first of all i should divide by w but that's not hard and then in order to find out the state the spacetime conformal dimension of the corresponding state what i have to do is i have to add k w over two which is just w over two but look there's w over two and this time is minus w over four so what i'm going to get is w over four minus one over four w which comes from here plus and let's set h zero two zero for simplicity plus n over w now now it depends whether you've seen the symmetric orbital before or not if you have seen the symmetric orbital before you should say aha if you haven't you should not say anything and from your reaction i assume you haven't seen it before so that's what i'll explain to you next time but this is exactly the conformal dimension spectrum of a symmetric orbital where this is the casimir energy of the ground state and this is what you expect for the symmetric orbital of t four and the excitation numbers in the spectrally flow in the w-cycle twist detector and i'll explain this next time are not integer moded they're fractionally one over w moded so this exactly reproduces the spectrum of the symmetric orbital so obviously i have so what i'll explain to you next time is the other way of getting at this formula from the symmetric orbital and then i have to answer all the nasty little questions is why was i allowed to do that and what happens to all the other degrees of freedom and all the rest of it and then the proper answer to that i will i'll give you some hand waving answer and then the proper answer will be the hybrid description where there's a very clean answer where this comes out of the representation theory of the super league algebra and where all the degrees of freedom do exactly the right thing to match then exactly the spectrum of this but here you see the first sign that you are on the right track did this looks to somebody you see the symmetric orbital spectrum before like the symmetric orbital spectrum so now you just have to dot the i's and cross the t's to make sure everything works out and it does work out but that requires a little bit more work but since i'm over time i'll stop here yes thank you okay what time for questions let me also encourage questions from zoom