the recording. All right. So thanks for coming back. So we'll continue the journey on invertible symmetries. So this morning, I reviewed for you this notion of the general notion of symmetry as topological operators. And we saw briefly how this the ordinary symmetries of socialist continuous groups and discrete groups fit into this framework. And now we're going to generalize that using this based on this picture of symmetries as topological operators and dimension. So it's topological operators. And in the morning, we discussed the case when the symmetry associates a group. And this is defined by a topological operator on the co-dimension one manifold of a space time. We're ready to generalize the first generalization that retain the feature of being topological, as well as some structure of a group, is the notion of p-form symmetry, a high-form symmetry, generally a p-form symmetry. And this is realized by a higher co-dimension. In other words, p, co-dimension, actually my notation will be the p plus one co-dimension topological defect. So instead of defining a topological defect defined on the co-dimension one manifold, in general, you can have a topological defect defined on a higher co-dimension manifold. And they generate the p-form symmetry and the p enters in the co-dimension in this way. And the particular example is the one-form symmetry in four-dimensional gauge theories. And historically, they're known as center symmetry because the corresponding group coincides with the center of the gauge group. And in this case, p is equal to one. And the generators are two-dimensional topological operators. And they are generators, they're the charge, the unitary operators correspond to the one-form symmetry. And the operators in the theory that are charged under this one-form symmetry are both in lines. This is a kind of a characterizing feature of higher-form symmetries, is that the charged objects are no longer local operators or particles, but rather extended objects such as both in lines. And if you want to think about dynamical objects in your theory, think about say, strings. They'll be charged under this one-form symmetry. This is the generation number one. Another generation, as you probably already guessed from the title of the lecture, is to consider non-invertible fusion or non-invertible, which come from general fusion. So previously, the topological defects, as I said, the symmetry are characterized by the topological defects, their fusion, and how they act on local operators. How they act on local operators. In this case, it's generalized to the case of how they act on these extended operators. And the fusion previously always takes this form. They follow some group modification law. Okay. Instead of generalizing on the co-dimension, we can generalize this notion of a fusion product to something more general. So I'll denote two such topological defects by L, I, and L, J, okay, because I'm staying two-dimension. And the corresponding topological defects are one dimensional represented by these lines. We know these two topological lines, because they're topological, looked from far away, they look like a single topological object. And in general, it could be reducible. So it could be decomposed into a sum of individual topological lines, without assuming there's underlying group associated with this topological defects. And the point is, the point that on the right-hand side of this fusion product involves multiple topological defects rather than a single one is a signature of non-invertibility. So when you talk about group modification law, of course, you always only have a single term on the right-hand side. Okay. There's no meaningful decomposition. And as we'll see, sorry, can this be viewed as a generalization of the OPE expansion? That's a very good question. Indeed, I'll make a comment related to the OPE of more general defect in a moment. Yes. So this is a special case of OPE expansion involving extended objects here, line defects. Question? Why are there no signatures of the indices i and j on the right-hand side? Right. So what k appears here depends on i and j. So this step implicitly depends on the choice of i and j. But then being kind of a schematic here, later we'll write explicit equation. And in particular, that's a good point. So some of the lines here could be degenerate, meaning that I'm summing over, I'm just doing arbitrary decomposition here. So I'm not assuming these are each one appearing here is irreducible, for example. So if you write the right-hand side in terms of indecomposible lines, the coefficient may also carry the degeneracy of how many times they appear. We'll be precise about that later. Thank you. Here, just schematic. So the moment you have this non-trivial sum, having more than one term appear on the right-hand side, is already a signature of non-invertibility. Sorry. Question? Well, the definition of having more than one term needs the definition of some like building block. Some of lines is a line. That's right. So that's a very good question. So as we will get to that, from what I've told you here so far, there seems to be no kind of physical meaning of how I split the right-hand side. For example, we should say like when I have a single line as before, why can't I split the single unitary as a half? Here is obvious because if you take a half of a unitary, it's not a unitary anymore. Here, we'll also see that there's a constraint, precisely how to decompose this coming from the locality of the corner field there. So you can not arbitrarily scale your line defects. That's a constraint by locality. Okay. All right. So let me now come to this question that was just asked in relation to a notion of OPE between these extended operators in a more general context. So indeed, this can be thought of as a special case of OPE between defects in this case, I'm focusing on line defects saying D equal to 2. But this is a general notion you can consider in higher dimension. But in general, the OPE of the line defect is much more complicated. Okay. For example, just to make connection to existing knowledge about defects, let's consider an example of conformal defects. The large class of non-trivial defects are these conformal defects. The conformal defect lines can be equal to 2. Okay. So these defect lines are again kind of denoted by this in the same picture, but let me focus on this is a conformal defect in general. Okay. And it is defined such that the stress sensor, so in general it actually can be an interface. Okay. It's defined such that the stress sensor on this side satisfy this relation. No, let me think about this direction. So this is a z, this is a z plane. And for convenience, I'm calling this direction and imagine the z direction, which is a location of the interface. In general, there's a discontinuity in a stress sensor for a conformal defect. Okay. Even for in the same theory. Okay. The discontinuity in a stress sensor is captured by what's known as a displacement operator, which is the operator on the conformal interface. Let me call it I. The special case of topological defect correspond to the case when D is trivial. Okay. So the stress sensor is fully continuous across the defect. Now, for the general class of conformal defects, there's this notion of OP that you can discuss. Okay. So I take I1, I2, there are two conformal defects. Okay. But, and that's kind of the picture that you would want to have also for conformal defects, similar picture as that. But there's some, there's some important differences. For example, even if you take this two to be the, to be the same defect, okay, with opposite orientation, and you expect the identity defect to show up, the trivial line to show up here. Okay. But this coefficient would involve divergences in particular from the Casimir energy. So, so considering OP will be corresponding the limit where you take the two defects closer to another, and you will have this Casimir energy divergence that's proportional to a cosmological constant term integrated over the defect, okay, over the line. And moreover, you have subleading divergences due to relevant operators on the line. Okay. From this fusion product. So it's physically very distinct from what happens here. In the moment, you have this discontinuity. The discontinuity in the stress sensor is what give rise to this Casimir energy, which is essentially measuring like the some kind of imbalance between the stress sensor inside of this region and outside. Okay. And because in a non-tobodile case, you have this divergence piece. And they're so there's, there's not, there's the structure of this fusion, although you can write it down, but it's much messier. Okay. And in some, in some special cases, for example, if your system has supersymmetry, okay, in that case, because of the contribution from Bosonic, from Miannidur's freedom, and for super symmetric line, you can achieve the special case where this Casimir energy pieces of divergences is vanishes. Okay. And this is why in super symmetric theories, it's meaningful to talk about fusions of super symmetric rules online and so on. And this kind of fusion product is well defined. Okay. Sorry. And you may be able to derive similar relations, for example, from localization. There is no way to regulate this divergence then. There is no way to regulate the divergence. You can, but you know, like if we're for kind of a generic theory, there are many divergences over here for each of these relevant operators. If there's only a single divergence coming from the, you know, the cosmological constant term, then that's more kind of a, there are ways to deal with, but we're facing the entire tower of supplicating divergences from relevant, relevant operator means dimensions smaller than one in this context. Okay. On the line. There's no obvious way to achieve a meaningful answer. Okay. Question. Sorry. This might be a knife question, but is this divergence related to both the UV and IS structure of the theory? Because in one case, you're taking delta to zero, which is the short distance structure, but you're also integrating over some non-compact direction, presumably. Right. So here you can make it compact, but the divergence stays is because of the limit. So the divergence is mainly UV divergence. That's right. That's right. Thank you. That's the Casper image. Okay. So that's just a quick comment. All right. Well, let's fulfill this promise to provide some toy example before we jump into general structure, some toy example for non-invertible symmetries. Okay. The toy example is not going to be, the toy example of this university would take place in a very simple theory, which is the discrete gauge theory with gauge group, a discrete group G. Okay. Someone of you already asked me about this during the discussion, while informal discussion. So in this case, there are interesting topological operators, which are extended objects. These are Wilson lines. And these are the Wilson lines that you would have. They're just the analog of Wilson lines you would have for a continuous gauge theory. Okay. So they are labeled by representations of the corresponding group. In particular, the indecomposable Wilson lines are labeled by d-reps of the group. Okay. So explicitly, you can write them in terms of the exponential of your gauge field over a curve. Okay. This follows. Now, in general, continuous gauge theory, like in Yamio's theory, this Wilson line would not be a topological object. Okay. That's because the connection in general is not flat. But here it's topological because discrete gauge fields are always flat. Okay. It doesn't support. Discrete gauge field cannot support any curvature. As a consequence, this provides possible realizations of the topological defect operators that we may interpret as symmetries. Okay. And they claim they're non-invertible. They give us non-invertible symmetries. If G is non-obedient. Okay. So what is the simplest question? Yes. So what happens when the connection is not flat? Ah. So when the connection is not flat, then this Wilson loop is not going to be a topological defect. But they will fall into this realm of conhomal defects. And as I said before, their fusion in general is much more complicated. There are these divergent resist, and from these divergent resist, there is the regularization and degree of peace. But in special cases, like with supersymmetry, you can still have this well-defined fusion products when all the lines preserve the same superchargers. And how can you assure that in this case you have a flat connection? Oh, it's essentially by construction. If you're talking about the discrete gauge field, there's just no, there's no curvature. Okay. The discrete gauge field backgrounds are classified just by how long that is. Sorry. Just by the transition functions. Thank you. Sorry. What are the connections for a non-abelian discrete group? So this is the same thing that I drew before. The, to specify a gauge field background equivalent to, you know, specifying the gauge connection for a discrete gauge field is equivalent to decorating your operating function with a network of symmetry defects. Each of them will be labeled by, you know, this UG, okay, where G is element of the group. Okay. And they're joined by this topology functions, which you introduced in the last lecture. So this specify piling of your space time by networks like this, specify a particular gauge connection background. And the discrete gauge theory means that you are summing over all these configurations. Right. And if you, if you don't like this notation, well, okay. So this is, this is, I think this is the best I can do for general non-abelian groups. For discrete, discrete abelian group, sometimes you can introduce some kind of Lagrangian multiplier field and write everything in terms of U1 gauge field. But this is the, this is the, this will be the, the, what, what do we need to do for, for this general non-abelian discrete gauge groups. So another question. As you also draw in this, in this picture and as you presented this morning, there are maybe two different point of view on, on fusion, both as picking defect and stacking on the same manifold or as junction as you draw here. Is it obvious that this is exactly the same thing? And I mean, maybe in, in four lines, it is almost obvious because you can put lines only on three-dimensional manifolds. But in, for higher dimensional defects, you, you might put your, your similarities on different two-dimensional manifold, for example. And I don't, I don't know how to see that. Yes. Yes.  That was correct. So when I draw this picture, this represents a discrete gaging in two-dimension. There's a similar picture, but it's harder to draw to, for the three-dimensional gaging. In this case, in that case, it will not, the basic junction will not be of this shape, but it will, it's similar basic building blocks and similar in higher dimension. Okay. So, but the, the, the two point of view are always the same, both as, as, as stacking defects and, and with junction are always the same. So, so the, so, so as I said before, you have the, you have the topology defects and they satisfy some fusion rule. And on top of that, there are these junctions, the existence of junctions can be inferred just from the topological property of the lines. Okay. But there are some extra data, which is the, you know, the specifying each junction. Okay. And that goes into defining this gauge theory background. The, the key point is that the freedom that you have in defining these junctions would not affect physical quantities like anomalies or surface, the corresponding symmetry. So in this case here, for example, I'm talking about the discrete gauge theory, in which case I have dynamically gauge the group. So I'm implicit already assuming there's no anomaly associated with the symmetry in this setup. All right. So I'll get back to this, this claim. So the simplest, perhaps the simplest example of a non-impedient group is the case when it's the permutation group of three elements. Okay. In this case, we have three irreducible representations, the trivial representation, right as one, for the moment, for the moment it will become clear why I call it one. And there's additional one-dimensional non-trivial representation as the sine representation. Okay. And then there's a vector representation, which is the, which is the two-dimensional representation for S3. Let me call it V. Okay. And as is true for Boston loops in general, okay, you can decompose at the level of the representation data associated with the gauge group. We can decompose the fusion product between two Wilson loops into Wilson loops labeled by the representation that appear in the tensor product. Okay. So in particular, the Wilson loop associated with the sine representation, when you fuse it with itself, you just get the identity. That's because the tensor product of sine representation with itself is the identity representation. Similarly, identity, when you fuse with the sine representation, it does not change the representation, and the same is true for the vector representation. Okay. So far I'm writing fusion rules where only one term appear on the right-hand side, and that's the signature of so far we're just uncovering some invertible symmetries in a theory. Now, non-invertible symmetry will be generated by the Boston line, the Boston loop associated with the vector representation. Now, the key is that the tensor product of two vector representation of S3 decompose into three representations. All these three representations appear in the tensor product. As a consequence, on the right-hand side, you have this combination of Wilson loops appearing. Okay. And this is the non-invertible symmetry that we promised in this very simple toy theory. And if you are still confused about something that I said before, let me just focus on the case for D equal 2, and then this is literally the picture that it means for, you know, this is literally kind of the network you sum over to define this discrete gauge theory. Okay. But this construction, what I was trying to emphasize before is that this construction works in general dimensions, and you always, for any discrete gauge theory in any dimension, you will have this non-invertible symmetries generated by these Wilson lines. Okay. As long as the gauge group is non-Abedian. When it's Abedian, all the reduced representations are one-dimensional. So, and you will now have this non-trivial tensor product that involves more than one term in the decomposition on the right-hand side. Okay. And this is, this is a very, very simple model that I realized this symmetry. And it turns out that this symmetry is usually denoted by S3, wrap S3. Okay. Meaning that the individual generators of the symmetries are one-to-one correspondence with representation, you reduce representation of S3. Okay. And later we'll see this kind of trivial symmetry will show up in the interesting two-dimensional CFT, namely the three-state POS model. So here I discussed the kind of trivial example, a trivial theory, where some non-invertible symmetry shows up. But the general lesson is that given non-invertible symmetry can show up in many theories, in particular we'll see that show up in more interesting theory, the three-state POS. Okay. Once we have settled this very simple term example, let's start with some general generality of non-invertible symmetries. We'll spell out the general structures in more detail. Okay. And once again, I'll be focusing on dimension two. Okay. And as I've alluded to before, the mathematical structure here, we'll see the mathematical structure here generalize the groups to what's known as the fusion category. Okay. And we'll unpack the ingredients that go into this definition step by step. So the basic starting point of defining a non-invertible symmetry in two-dimension in general quantum field theory is to first specify the set of topological defect lines, which will abbreviate as TdL. Okay. And it will denote by Li visually. Okay. And analogous to the fusion rule that we wrote before, and we generalize to the case when there are more term appearing on the right-hand side, this topological defect lines generally satisfy some fusion algebra or fusion ring, which tells you how to decompose products of this generators in the fusion product. And when I write the equation like this, it should always be interpreted in this way. Okay. So you are taking two, you are taking this topological defect lines to lie on manifolds, which are, which are homologous. Okay. And you're taking a limit where they go to, they approach one another and decompose this configuration into the individual topological defect lines. And in particular, among this set is this distinguished element, which we'll call one, that correspond to the identity line, or correspond to the case there's no insertion at all. Okay. And you're free to apply this to any, fuse this with any of the line over here, and you return, it will return to you the line that you started with. Another notion is something that's already asked in the context of discrete symmetry that generalized beyond the fusion ring is the notion of topological junctions. Okay. The existence of topological junctions can be inferred from the topological property of these lines, but to specify these junctions individually requires additional data. Okay. So this, this topological junctions are defined as follows for any line, I call it L3, that appear in the fusion product between L1 and L2. So whenever I write a product of a line, it's interpreted as in the sense of a fusion product. Okay. For any line that appear in the fusion product, you have a topological junction of this type, generalizing the topological junction that we saw when these are lines that generate group-like symmetries. Okay. The junction relies on a specific, specification of this, this point. Okay. So we can be thought of a point operator, which are denoted by V. Okay. In general, when you specify the external topological defect line, there may be multiple choice of V. And this V in general is a vector in the vector space, which we'll call the junction vector space, topological junction vector space to be precise. Okay. And this topological junction vector space, because these are point-like operators, you can equivalently think about this as kind of a TDL or topological defect line changing topological operators. Okay. In particular, in a special case, when L2 is identity, V will be the topological changing operator between the defect line L1 and L3. And in general, V will be the topological defect changing operator between the fusion product L1, L2, and L3. Okay. And later, we'll give a more precise kind of a picture for what this topological junction vector space is when we discuss this topological defect line in two-dimensional CFD. Okay. The last ingredient to fully specify this notion of fusion category is the notion of F symbols. Okay. And this captures possible factors you get from a topological change, a topology change in a network of topological defect lines. Similar to the phase that we encountered when we changed a network of part of the background describing a part of the background for a discrete symmetry. Okay. So this topological change is commonly referred to as an F move, which is a basic move to change the topology of this kind of diagram describing a two-dimensional background for the discrete symmetry and also generalize to this background of TPL network. Okay. So I'm just redrawing the diagram we draw before, but in a different way. That's conventional to, in convention to the literature. So one, two, three, four denotes four topological defect lines. Okay. And A labels this internal topological defect line. Okay. And all these junctions are topological junctions. So they're specific specified implicitly by some specific vector in this vector space. Okay. The F move corresponds to a topology change in this network by reconnecting the second topological line to another leg. Okay. And in general, some other intermediate topological defect line may appear over here. And this change of basis is captured by some numerical coefficient known as the F symbol. Okay. So here this thing can be thought of as a matrix. The one, two, three, four are external labels that specify the external lines that enter into this diagram. And this index is specified the change of basis between these two configurations. Okay. So why do I think, why do I say this is a change of basis? What I mean precisely is that this is a change of basis in the junction vector space between four external lines. L one, L two, L three, and L four. Okay. So you think about this pair of trivalent trivalent vertices as a basis for the topological junction between four external lines. Okay. Because there are two ways to resolve this four point junction. There are, there are junctions. So there are two, that give you two preferred basis for this space. And this F matrix captures the change of basis. This question. Can you comment a bit about operational change of this arrow? Because in the first picture, you draw the two in go, like in the vertex, you have two in going and one out going. But in the other diagram, you have two out going and one in going. Sorry, which picture? For example, for this picture, like picture. Yeah. How important is the two in going and one out going? So it's just because I want to make sense of this notation. So, so I can draw, I can draw the in going and then it will be the L three bar. I'll come back to that. You will define what you mean by bar. Okay. Yes. Thanks for the question. All right. I think I can use the question. Yes. Can you explain again why there's a whole vector space of junctions? Right. So, so, okay. So it's not, I think that notion is not completely clear at the moment. But it's, you can think about this way. So the point is that this is like a point like operator. Okay. So having one junction, you can, you can, you can multiply this point like operator by arbitrary complex number. That's still topological. It's just a different way to define this junction. So if you have two independent such junctions, then you have, you know, the two dimensional complex vector space. And that's one way to think about it. Later, we'll give a physical interpretation of how this junction vector space comes about in a two dimensional CFT. It will be related to the hubris space of the two dimensional CFT on a circle with defect points. And this will be, they will be characterized by specific states in a hubris space. It's a subspace. We'll make that precise. But in order to define this f symbol, why don't you need to specify the junction? Very good. Very good. So I cheated a bit. So I said, I said that but I didn't write it because it just makes the equation more complicated. So on the left hand side, there's a choice of v1 and v2 over here. Okay. And on the right hand side is also the choice of junction vectors here, v3 and v4. And this matrix will involve this additional indices, v1, v2, and v3, v4. And you're summing over all these choices. And for simplicity, I'm just drawing the case when the junction vector space is one dimensional. So that the information is completely captured by this coefficient. Otherwise, there's a further further indices. Is there some interesting case in which the junction are always one dimensional? Yes, of course. So welcome to that. For example, the duality defect, which shows up in two dimensional IC model as well as in this three dimensional and four dimensional gauge theories, they all fall into this category that having one dimensional junction vector space. As we'll explain, the dimensionality of this junction vector space will be tied to the coefficient in the fusion ring when you write to the right hand side in terms of indecomposable objects. And whenever the fusion product such that only coefficient one shows up in the decomposition, as we'll explain, that corresponds to the case when this is one dimensional. Thanks for your questions. But I was slightly jumping ahead. So if you don't understand that last comment, don't worry. We'll come to that. All right. So these symbols are not arbitrary. So satisfy some consistency condition. This is something that you can already infer from the fact that when you specialize the case when the defects are invertible, this symbol is nothing but the phase that we introduced before that captures the anomaly associated with the corresponding discrete group like symmetry. And as we said before, the anomalous phase is subject to this consistency condition that essentially tells you that physical cycle. And that's where the group cohomology classification of the anomalies comes in. And here we can be more explicit and recall where the consistency condition comes from. The consistency will come from looking at a five-fold junction. So it's a topological junction involving five external legs. Okay. So for simplicity, I'll now keep track of the arrow. And then there are two internal legs labeled by A and B. So this specified, if you once we specify the junction vector over here, this specified a particular junction in the space of between five external lines, one, two, three, and four. One, two, three, four, and five. Okay. Now there's a conservative change of basis you can do. Okay. I think I was too ambitious with my space management. What I'm going to draw will not fit in the region. So let me move over here. I'm going to start with this diagram once again. One, two, three, four, and five. Okay. A and B. And as I said, this specifies a particular junction vector in this junction vector space with five external pvls specified. One, two, three, four, five. Okay. And of course, there's various ways to resolve. So if I think about this as specifying the, you know, one, two, three, four, the choice of junctions living here. Okay. So this is what the, so this is the general general junction, topical junction with five external legs. Okay. And this, for any vector here is going to be element inside this big vector space. Now there are different ways to represent the vectors in this space using the topology, using the topological nature of this pvls. So in other words, you can try to resolve this very singular looking junction into like a triplet of a trivalent junction in this picture. Okay. And each different resolution will give you a different basis for this junction vector space by considering arbitrary choice of internal legs over here and arbitrary choice of junction vectors at this trival junctions. Okay. And there's a consistency coming from comparing this to the following. Okay. So let me draw it over here. So this is the famous Pentagon equation. We can perform this change of basis. Okay. Given by this f move from this basis to this basis, by using this f matrix. Okay. And we can do it multiple times. Now do the f move focusing on this part. Okay. And you get to this picture. And we can do it again, focusing on this part and apply the same f move and we get to this picture. Of course, there will be different, different co-efficient or different matrices that enter into each step correspond to a different labeling of the external labels as well as the choice of internal lines that appear in that channel. Okay. But then the point is that going through this round, we establish the change of basis between this basis for the five full junction vector space and this basis for the five full junction vector space. Okay. But there's a different way to reach the same change of basis matrix is to do first to this picture. Okay. Which corresponds to doing f move on this block. Okay. So you reconnect three with this other side of this diagram and then do f move on this region and you get to this basis again. And in this process, another different, the same f matrix with again with different external indices shows up. And by consistency, because you're just discussing change of basis between two special basis in two different ways, you must get the same result that schematically look like a product of three f matrices with certain indices that's been contracted is equal to the product of two f matrices going this way. Okay. And this is the famous Pentagon equation because this looks like a Pentagon, which is the generalization, the co-cycle condition for g symmetry. Okay. For invertible symmetry where g is discrete. And in which case the Pentagon equation is the same as the co-cycle condition that defines the g analysis. So now we have introduced all the basic ingredients that define what is known as the fusion category. But it's not something mysterious. It's just some objects that generalize groups and knows about the anomaly associated with groups to the case where the building to body of defects are non-invertible. Oops. I'll pick it up later. Together, so I was saying the topology of defect lines, their fusion junctions. So there's our individual building blocks for this underlying structure and f symbols. They lead to the fusion category, the structure of the fusion category. Okay. So this is just some references in case that you have heard about the category before and you want to make connection to the notion of category in this language. In this case, the underlying objects for the fusion category are the topology of defect lines. The fusion will describe the tensor product between these objects in this fusion category. The fusion category has in particular a structure of a tensor product. And this junctions leads to morphisms between objects in the fusion category, in particular between objects and their tensor products. Okay. And this is why the trivalent junction defines the morphism. And these f symbols are known as associators, which specifies isomorphisms between the tensor product of three objects. Okay. Okay. It actually has six external labels. That's why it's also called a 6j symbol. And this fusion category structure, as we already saw from various parts of this structure, it generalizes discrete group symmetry. Okay. In particular, in a special case, when all the topology of defect lines are invertible in corresponding some group, the fusion category is nothing but the usual discrete group symmetry, but at the same time, taking track of the anomaly through the f symbol. So it's in a sense more kind of contain more data than the group. But the data it contains is precisely what we care about in physics. There are anomaly associated with discrete symmetry. And the second comment is that the structure of fusion category is extremely constraining. Of course, if you insist, if you don't introduce this non-year-old defects, then it's just as constraining as it is for constraining anomalies associated with a given group symmetry. Okay. But here, the constraints manifest in the following way. Let me just give you an example. If you start from a set of two water defect lines, which you postulate that describes some fusion category symmetry, and you postulate the fusion rule, okay. For example, if you just postulate a single topology defect, and you postulate the fusion rule of the form L squared is equal to one, which is a trivial defect plus N, which is some degeneracy times L itself. Okay. This is only if a fusion category or N equal to zero or one. Okay. So, this is very surprising because why can't we have a single non-evertible defect that satisfies such a fusion rule? Okay. And the constraint is coming from the S symbols. And in particular, the Pentagon equation that constraint F symbol. And the punchline is that when the N is, so the story is that when N is large enough, there's just no solution to the Pentagon equations. Okay. And as well later, we'll see all the structure of fusion category symmetry wouldn't necessarily comes out from a locality of a quantum field theory. So, that tells you that this symmetry is just simply not possible in the quantum field theory. Okay. Sorry about the question. Yes. So, does this constraint come from a unitarity or it's independent of unitarity? This statement is independent of unitarity. It's just, well, okay. So, let me start. It depends on how unitary you want the theory to be. So, to be absolutely sure, I think we're implicitly assuming some notion of unitary because if we forget about unitary, then for example, the fusion products may have funny coefficients and that would ruin this story. But here, the fact that we're talking about the fusion category actually relies on some notion of unitary. That's a very good question. Otherwise, there's this general notion of some pseudo fusion category. Right. Very good question. Sorry. When you look at the solution of the Pentagon equation, in which range do you look for F? Should be a phase that it should be that is some condition. Very good question. So, very good question. Very good question. So, as I mentioned briefly over here, for each junction vector, there's ambiguity, right? Even if the junction vector space is one-dimensional, you have an ambiguity coming from the complex number. And if you keep track of that ambiguity in this equation, you see that depending on your choice of normalization of these vertex here and here, that will change the F matrix. So, in general, this F symbol is not completely unambiguous. But a statement here is that there's just absolutely no solution, whatever choice of normalization you take for the vertices. But in general, you will have family of solution for the F symbols, but we only study them up to this equivalence relation. And that is the analog of the exact cosycles in the context of group cohomology. So, the phase coming from the F-move for the group symmetry case, that's subject to ambiguities coming from the exact cosycles. And that is the physical origin of that is coming from the redefinition of this point-like operators. Another comment I want to make over here is that, just like in the case of groups, there are ways to classify groups, and that is known. So, you may ask if there's a classification of this fusion category symmetries. And the answer is that mathematicians are still working on that. And there's no classification as of yet, except for low-rank cases. Rank means the number of indecomposable objects, which we'll come to shortly. But on the other hand, from physics, as we'll see, we'll be able to produce tons of tons of examples for the fusion categories. So, it's really down to the mathematicians to actually pursue a classification that would include all the examples we've come up. And lastly, it's related to the question. There's some structure associated with the Pentagon equation that is up to this gauge freedom that we have, associated with redefining each point-like operator that lives at the junction. There's topological point-like operators. The Pentagon equations have only discrete solutions. Okay? So, again, this is up to the gauge freedom that we talked about, redefining these junctions. Okay? And this is very nice. Mathematically, this is known as the Oknienu rigidity. This is something that the mathematician has proved. So, it says that the Pentagon equation does not allow continuous family-all solutions. When you remove the potential continuities, ambiguities coming from these redefinitions. And this is very nice because this is the similar feature that we have for anomalies. Anomalies are quantized and they cannot change under continuous deformations. And because of this feature, they are bound to make non-trivial constraints on quantum field theories, in particular for RG flows, which is the continuous process. So, that is the mathematical framework underlies this non-invertible symmetry in two dimensions. We have not really put in any physics. As we will see, studying these symmetries in the context of two-dimensional quantum field theories would make many of the structures more transparent and it will also lead to additional physical information, which is not obvious at all from these basic building blocks. Yifan, just in principle in two minutes, we should move to the discussion, but maybe you got a lot of questions so you can take a little bit more if you want. Oh, how much? This is already, okay. Yeah, I think if I can have, it can have 10 minutes. Okay, I think that's good. So, there's some extra structures. Let me explain. Coming from studying the symmetry in two-dimensional QFT with this non-invertible symmetry. So, to be explicit, we'll restrict to two DCFTs. We'll discuss the symmetry in the context of two DCFTs. And this is not really a loss of generality because, as we alluded to over here, the structure of the student category is rigid. So, to the QFTs that's obtained from CFTs by RG flows, it's naturally captured by the statement we'll be making here, okay. But the CFT or TQFT, which is a special case of a CFT, okay. And, but the notion of two DCFTs will help me to make some explicit statements using, so this is convenient because we'll be able to use radial quantization. And as we'll see shortly, they will help me to specify what these operators are in radial quantization, okay. Because in this case, in the CFT, I have radial quantization on top of that. I have what defined eigenvalues for the L0 and L0 bar generators for the conformal symmetry and their H bar and H, okay. This is really where I'm using the CFT structure. Okay, all right. And the key, as I already mentioned, coming, that leads to this actual structure in CFT for this field and category symmetry is the locality, okay. And this is something true for general quantum field theory, which can be thought of as consistency in cutting and gluing CFT or quantum field theory pattern function with observables, okay. So you can cut and glue, cut your, your observable function with observables in different ways and glue them back together. It will lead to the same results, okay. That's the, that's one notion of locality, okay. One way to phrase the notion of locality. And as we'll see, this locality will lead to this fusion category structure starting from the basic objects, the topological defects, okay. All the others will just follow. First of all, so we start from the topological defects, which we postulate as a symmetry in the theory, okay. The topological defects acts on local operators, okay. And we can be quite precise in the context of two dimensional CFT. So imagine you have a local operator over here, okay. As I said before, the way a unitary operator acts on the local operator is by enclosing it, okay. And because this operator is topological, it means that in particular it commutes with stress sensor, okay. So it will preserve the conformal way that social resistance local operator. And it will produce another operator, which I'll denote by L hat acting on operator phi, okay. It's an operator with the same, same h, h bar as the regional operator before you act by, by the topological defects, okay. And moreover, this also implies that the topological defects will map berocellular primaries in the two dimensional CFT to berocellular primaries. And they will map the entire multiplets to multiplets, okay. And equivalently from the radial quantization, you can think about this picture as meaning on the cylinder, okay. Where on the bottom of cylinder insert to the corresponding state correspond to this local operator. So it's a primary operator, okay. And it's a state inside the hubris space on S1. And this is the line insertion, okay, that acts on this hubris space that correspond to, that comes from this, this radial mapping, okay. The coordinate change that we use in the radial quantization. And this is the, this is the equivalent meaning that tells you how, you know, this, this is why we use the L hat, because L hat can be also thought of as the operation of this symmetry defect acting on the states, okay. So this is the, if you wish, this is the characterization of what Wickner wrote for unitary, but now for this more general operation linear operation L hat. So I just introduced the last, another ingredient and questions, okay. The other important ingredient coming from the quantum field theory in the presence of this topogen defects is the defect hubris space. And this is, this is the consequence of the locality of this topogen defects, okay. And this is the generalized notion of the twisted sector of a two-dimensional CFT that we associate with discrete symmetries, okay. The idea is that you study again the theory on the cylinder, but you can choose to align the topogen defect now in the time direction, okay. In which case the hubris space gets deformed and the general state inside the hubris space, I'll denote it by psi, hubris space gets deformed, okay, into this twisted hubris space. In a special case, when it's L correspond to a group-like symmetry defect, this is nothing but the twisted, the hubris space with the twisted boundary condition imposed on the spatial circle. But in the case of a non-year-old defect, because of locality, you can still define this defect hubris space. And by the same kaffoba map that relates the brittle concentration to the cylinder, this is equivalent to specifying this operator is equivalent, sorry, specifying this state is equivalent to having a local operator that sits at the end of topological defect line, okay. And once again, because of the topological nature of these lines, this again fall into representations of the two copies of your solar symmetry. And in particular, we'll assume the following condition, which is a physical assumption, let me call it the type of condition, that is only the identity line can end topologically. And this is equivalent to having a faithful representation of the symmetry, in this case, the non-invertible symmetry generated by the supportive defect on the hubris space on S1. The reason is that if other topological defects can end topologically on operators, so in this case, psi in general is operator with h and h bar, that's not equal to zero comma zero, okay. So it's not, so in the case of a topological junction will be the special case where h bar equal to zero, okay. But in general, this is not possible. And indeed, for CFTs, we want to impose the condition that this is always not possible. So having h and h bar equal to zero for psi, it's always impossible when the topological defect line is not an energy. And that's equivalent to requiring the symmetry that you are studying to be represented physically on this hubris space. I think I'm running out of time, right. So maybe let me stop here and answer questions and we'll catch up next lecture. Okay, so let's thank the speaker. And before we move to the discussion, I think there is a group picture to be taken. I don't know if there are some instructions. Just the bottom, the bit. We cannot see the bottom part, but is it scrolling down continuously? I see, okay, that's good. Can you hear me okay? Yes, yes, we can hear you very well. So we will start in about five minutes. Okay. Because it's continuous. If anybody is missing this was found on a desk, there was a watch that was found on a desk. If somebody is missing it, please come to me. Otherwise, I will leave it. Kevin, as you probably know, it's like one hour of lecture and then there will be 15 minutes for informal discussion. Just let me know what I should start. Yes, now people are still entering the room, the volume of Zoom. I guess the time has come. So we are very happy to have Kevin Costello doing us remotely from Perimeter Institute and he's gonna tell us about twisted holography. So I will just start the recording and please, Kevin, take it away. Okay, thank you very much. And I'd like to thank the organizers for the opportunity to give these lectures. I'd also like to apologize that I wasn't able to make it in person. I'd appreciate it if I'm not entirely sure the level and background of participants. So I really appreciate any feedback about how fast I'm going and the level of the material and the introductions are very welcome. So what I want to talk about is the title is twisted holography. So what I'll talk about is based on work by many people. And it's a way similar to in Mathias' lectures, twisted holography is a way of getting really sharp exact results from a supersymmetric subsector of ordinary holography. Mathias' lectures, he's not considering a supersymmetric subsector, but there are also exact results. So let me remind you a little bit about the background. So the original ADS-C of key correspondence, you know, we've all heard a great deal about it, but it's important to recognize that the original correspondence is still very much conjectural. So it's conjectural duality between n equals 4 super young mills and type 2b super gravity on ADS-5 to MS-5. So the reason why I want you to see this as conjectural is that both sides of the duality are incredibly difficult to understand. So on the gauge theory side, n equals 4 super young mills, we're supposed to be studying strongly coupled gauge theory. You know, strongly coupled constructing rigorously or in any sense a strongly coupled gauge theory is probably the most difficult problem in mathematical physics. It's one of the play millenium-prize problems. So this is really out of reach of any mathematical techniques. And on the other side of the duality, when studying type 2b strain theory on ADS-5 to MS-5, this is difficult for two reasons. If one wants to take the supergravity approximation, then we're studying quantum gravity. Quantum gravity is of course hard, because it's not normalized and so ill-defined at the quantum level. On the other hand, we might try to think of it as a strain theory. Strain theories are a way of defining quantum gravity, but it turns out that strain theory on ADS-5 to MS-5 is also very challenging. The reason that I don't want to get into you, but it involves the fact that this is a Riemann-Riemann background. So of course, people have done lots of checks, and there's like very heroic, extremely difficult computations of both sides. So it's very plausible that this works. But what I'm going to talk about today is a subsector of all of this, where both sides are really easy to understand, and you can really get your hands on things and check things completely. Precisely on both sides, we're going to select certain states which are preserved by supersymmetry. Now on the gauge theory side, why is that easier? You'll find it has no coupling constant. The gauge theory is basically free. It's very, very simple. But on the gravity side, we find that the gravity side is a topological strain, which is also much easier to understand than the physical spring. So in this context, in these subsectors, we can perform exact computations to try to match both sides. And as I mentioned, the ADS-3, a lot of the in Gabriel's talk is a different, not unrelated context, where exact results can be obtained. So there might be some parallels with the Metaeuses talks. Let me be able to see you guys can see a little bit more of the text. Is this legible or is it too small? No, it's good. Can I zoom out a little more? I think maybe it's the good size. Maybe if you make it smaller, it could be a little bit too small. All right. I didn't want to just kind of be scrolling too fast. So the starting point of all of this analysis of twisted supergravity is the concept of a twist of a supersymmetric gauge theory. This was introduced by Whitton in the late 1980s, pretty famous work, in which he related n equals 2 gauge theory to the Donaldson variance of a format. So I'm going to start by reminding you of how one builds the twist of a supersymmetric gauge theory. So the example will start with will be a four-dimensional quantum field theory of n equals 2 supersymmetric. So let me remind you what it means to have n equals 2 supersymmetry. The group spin 4 is, well, in Euclidean signatures, SU2 times SU2. So each copy of SU2 has its fundamental representation, and these are the two spin representations of spin 4. They're complex dimension 2. I'm going to use Penrose's notation for spinners. Spinners in S plus will have an index, a Greek index alpha like this, and in S minus, you'll have a dotted index like that. When we have an n equals 2 theory with n equals 2 supersymmetry, we're going to have two pairs of supercharges, q alpha i, where i equals from 1 to 2, and q alpha dot i, or also i equals from 1 to 2. So we should recall that the vector representation, complexly vector representation, is a tensor product of these two spin representations. So in Penrose's notation for spinners and gamma matrices, instead of having an index for vectors, we write a vector as having a pair of indices, one dotted, one un dotted. So the x alpha alpha dot indicates a vector. Now the commutation relations with the supersymmetry algebra is simply q q bar is delta ij times translation in the direction of this vector. So what does it mean to twist? If I have a theory with n equals 2 supersymmetry, well I have two pairs of supercharges with this x extra index i. And we want, so the super algebra has a symmetry rotating that index. SU2 symmetry, it's called the or symmetry. So we're going to assume that our field theory also requires the symmetry. So the twisting procedure says that we're going to change the spin of our fields by using the SU2 or. Now recall that spin4 is a product of two copies of SU2. So I'm going to change the spin of the fields by instead of making SU2 acting SU2 plus active and long way, I'm going to make it act by the diagonal map between the original SU2 plus and the or symmetry. So what does that mean? That just means that whereas before I had an index i in my supersymmetry algebra, I'm now going to make that index transform under the rents group. So it transforms as the index alpha. So whatever supercharges that become, qi alpha becomes q alpha beta, or again alpha beta and spinar indices. And q bar alpha dot i becomes q bar alpha dot alpha. We'll leave that in a minute. But you notice that q bar now transforms as a vector. Because it has two spinar indices of opposite elicits. So what this means is that because the i index now transforms as a spinar index, we can build a supercharge, which I'll just call q, which is a scalar. And it is automatic that in the supersymmetry algebra, this is a nil put in supercharge. The reason is, well, I take q squared, the square of any supersymmetry is a vector, but this is also a scalar, so it must be zero. So the key point of the twisting procedure is that given a supercharge q of square zero, we change the theory by replacing the original Hilbert space by the q-colonology. So the twisted Hilbert space is defined to be those states which are in the kernel of q, modulo those which are in the imitative q. And we do this for everything else. We do this for local operators. Of course, if it's a CFT, that's the Hilbert space on the three-sphere. It's a bit more challenging, but it is also possible to formulate this definition for extended operators. If one reads the older literature on twisting, the part which is really emphasized is the first step, this part here, changing the spin of the fields. But for me, I don't think that does very much. In fact, it does absolutely nothing. All that's happened is step one, where I've changed the spin of the fields, especially if I'm on class space, is I'm just calling things by a different name.  I still have the same number of fields. My Hilbert space is still the same. My Hamiltonian is the same. I have all of the complexities are there. However, step two, where I've replaced my Hilbert space by the q-colonology, is a really radical simplification. And this is what allows you to do computations in the twisted theory. Let's go. What I've described is the Donaldson-Witton twist of n equals two supersymmetrications. And it turns out to be a topological twist. So let me try and explain what that means. Although maybe unless anybody has any questions about the material so far. Sorry if that was awkward. Hello. Can you hear me? Can you hear me? Yes. So you have an index i for the su2r and an index alpha for the su2 plus. Why not to call them the same letter and then treat them as if they are in the same space? Because you take a trace at some point in defining q, right? That's right. So what I've done is I've changed the way su2 plus acts on everything. So the charge of su2 plus is the original charge it had before twisting, plus how it transforms under su2r. So let's scroll up a little bit. So I'm taking this step. So my new action of su2 plus is the original action of su2 plus, but then I embed it diagonally in the product of these two groups. Yeah, I see, I see, thank you. Okay, so let me move to, unless there are further questions, let me move to explaining what it means to be topological. So the hallmark of the topological theory is that the energy momentum tensor vanishes. So energy is the action of translation in time on the Hilbert space. And momentum is the action of translation in space. So let's explain why these operators vanish. When I twist my q bar supercharges, the index i becomes an index alpha. So it looks like this, q bar alpha dot alpha. So it transforms as a vector. And if one looks at the original commutation relations, we take our supercharge q called q is like, looks like that. When I can commute it with q alpha dot alpha, well, the only thing it can really become, if you think that one thinks about the symmetry, is the operator of translation by the space time vector x alpha dot alpha. So what this means is that on the q-core homology, this translation vanishes. So I didn't write up the equation that does that. So if I take a state psi, which is q closed, then it's derivative. If I hit it with a translation in some direction, it looks like this. Because here, the term where I move the q past the q bar is zero because q psi is zero. And this is equal to zero in commology. So that's why we're actually with this for local operators. The same thing holds. If I have a q closed local operator, then if I try to move it around, it's q exact. So what this means, if I'm studying the correlation functions of q closed operators, they're independent of where I put them. So let's see this explicitly. I'm assuming all my operators here, I'm assuming these guys are q closed. And I differentiate with respect to the position of the first one. Now I move the derivative inside the bracket and I replace it by q-q bar. Get q-q bar of the first one there, and then the other ones. And because q is a symmetry, I can distribute q over the other operators instead. But they're all q closed, so that's zero. So this is a summary of Witton's construction of the late 80s, that by performing this topological twist procedure, we find special operators have correlation functions which are independent of position. This is a very unusual feature, and it's a homework about having a TFP. Soon after Witton's construction, people thought about twists which were not topological. That's going to be the focus of my lectures. So in the mid 90s, there was work by Johansson, Necrosal and others, where one studies the holomorphic twist of a theory that equals one supersymmetry in dimension before. So this particular example is not going to really be the focus of these lectures, but I'm just going to spell it out to demonstrate how twists can be holomorphic theories. Now, if one only has n equals one supersymmetry, then we no longer have the index i in the supercharges. We just have, the supercharges are just consists of two spinners, one of each other's. So here, whatever we do when we twist, you can see it's not going to be possible for the twist to have the full spin-force symmetry. We started whatever supercharges select will break that. So we're going to break the supersymmetry to SU2 minus, where SU2 minus rotates the q bar alpha dot and fixes the q alpha. So I'll take my supercharge, q, to be q1. So just one of the two components of q. And then I'm going to write holomorphic coordinates on spacetime by z alpha dot is x alpha dot 2 and z bar alpha dot is x alpha dot 1. So what's going on is that once I've broken the Lorentz symmetry group to SU2, that SU2 preserves a complex structure in spacetime. And here, you can think about this should be a bar. The key thing we needed earlier was that q, q bar would give me all of the translations.  So now, what does this tell us? The same argument tells us that correlation functions are independent of z bar. So d by d z bar, say, for one, I differentiate the z bar direction first coordinate. By the same argument, I bring this inside, I replace it by q, q bar and I'm going to put it around. That's going to be zero. So the correlation functions are holomorphic functions of the position. So this equation here is, this is the Cauchy Moon equation. So once I spend some time, you know, it's interesting to study the behavior of these holomorphic twists. There's lots of fun things one can say about them. In general, the most general behavior would be something that's partially holomorphic and partially topological. But let's not get into that. This is mostly an example. Now, I want to move to the kind of twists we're really going to be focusing on to do twisted laundry. It's legible. So I want to focus on a construction, which seems a little odd at first, whereby when it starts with a four-dimensional theory, an n equals two theory, which you assume is super conformal, and then you turn it into a two-dimensional chiral theory. So this is a little bit funny. It's a little bit of a different thing than we did so far. There are two approaches to this. One is, which is perhaps the most widely studied, instead of taking q to be a supersymmetry, you take it to be in the larger algebra of the super conformal symmetries. And the second approach, which gives you the same answer, is to use a version of Necrosov's omega background construction. Actually, I'm going to spend a bit more time on this one. So this, while Necrosov did really kind of escalow version of this, this particular omega background construction is due to g1o and geniagi. This construction is a bit more technical than what I've described so far. So I'm not going to give all the details. So mostly what I want to do is explain the answer. You give me an n equals two theory, we will be able to figure out what the two-dimensional chiral theory is, and then we can do computations. But let's take an n equals two theory in dimension four. Actually, maybe I should. Paul, are there any questions from you? Hi. I have some questions from the beginning. I got a little bit lost. So by the twisting, do you mean working on the cumulogy of the Hilbert space? Yes. We're in the cumulogy of the Hilbert space and the space of local algorithms. Okay. And I think that after that, you mentioned that after performing this twisting, we preserve a complex structure, right? What does it mean physically? I think it depends. So this was the case with n equals one supersymmetry, and the complex structure arises by noting that if I take correlation functions of q-closed operators, they only depend on half of the coordinates. So in Euclidean signature, they depend on certain complex combinations of the coordinates which are called z, and they don't depend on z-bar. So the complex structure arises just from that equation. Let's see. Is that complete? Yes, more or less. So the fact that our correlation function is holomorphic is that a consequence of the twisting and a consequence of this preservation of the complex structure or are they related at all? They're related. The preservation of the complex structure, the fact that spin 4 is broken to SU2, maybe we should put that aside for now. And then the most important thing is that the correlation functions are holomorphic functions. And that statement is just a formal consequence of the supersymmetry object. It's just that d by d z-bar is the collider of q with something. So that's all. Okay, thank you. So in the n equal one example, can you say in a few words what is the set of operators that you are left with? Because I guess it's larger than the Karar ring of the theory. Yeah. Let's say we start with, let me find some black paper, like an n equals one Karar multibet. And we're left with one bosonic field and all its derivatives in z1 and z2. And one fermionic field and all of its derivatives. So if you look at the index of this, these are exactly the states that contribute to the super conformal index. Like for this one gives you kind of one over one minus q l, one minus p l, so on and so forth. And this kind of thing. But by looking at words in this field and its derivatives, if you take its character with respect to the group u2, which is this symmetry because it's conformal, you'll find it's the standard index of an n equals one super conformal theory. I see. Thanks. So maybe this is a comment for the experts. I think this is really interesting because suppose you wanted to give a character a proof of cyber duality, what you really should be doing is looking at the states of the twisted theory and identifying all the correlation functions. That's kind of accessible, but also really not trivial at all. That's the kind of thing I mean, Davide is looking at this kind of question. Okay. Are there any more questions? So let me try to explain that this construction of Yankee and L. So here I wrote down n equals two super symmetry algebra again. And I'm going to do something a little funny. We are going to take q, which does not square to zero squares through a particular transformation. So I'm taking epsilon to be a small problem. And I write my q as a linear combination of like the like this. And if you see just by contracting the tensors, and maybe the indices, my notation is not super clear possibly, but if you can just contract the tensors when you can be q squared, you find a q squared is epsilon times e translation. But if you look at the norm of this factor, you'll see it's epsilon times the space like transmission. So what we'll do is make this direction periodic. So really, we're working on this cylinder just coordinates x 11 x 2 x 11 dot x 2 2 dot plus another plane, which has coordinates x 12 dot x 21 dot. And it'll be important in a moment that here d by the x 12 dot is q exact. And this this other plane will be where the chiral algebra lives. I'm going to call x 12 dot z. And it's a homomorphic coordinate on this plate. Okay, so the result is, I'm going to take my cylinder here. And I'm going to fill it in. So that it becomes a cigar. So at the tip of my, you know, my, my cylinder goes along in the radius shrinks. I got this cigar geometry. And then the result is that that your n equals two theory extends across the cigar in a way which still preserves the super chart for a reason. So if we do this, we can do the following trick. So above we saw that q squared was d by d p that's rotation on the cigar. If I take an operator with that the tip of the cigar, it makes sense to ask that that operator is rotation invariant on charge under rotation. And on such operators q squared is zero because q squared is rotation. So then we can consider the q chronology of the s one invariant operators at the tip. And this is going to be the chiral algebra. So what's important for this construction is that it doesn't make sense to move my operators away from the tip of the cigar because away from the tip of the cigar q squared is rotation and there are no rotation invariant operators. So this is why we're effectively localized to a two-dimensional theory. On the plane which lives at the tip of the cigar, d by d z bar is q exact. So for the same reason we've had before, if I take this, I'm taking things which are at the tip of the cigar and have other points d i in the plane, then the z bar derivative of the correlation functions vanishes as long as my operators I'm inserting are q closed in rotation invariant. So in this way we've identified in our 40 n equals 2 theory, it must be super conformal, something very familiar, a two-dimensional chiral theory just like one might be familiar with from elementary studies of the bosonic spring or something like that. Sorry. Can you consider also extended operators on this cigar like a line that wraps the theta cycle and so on? Did you give something interesting in that today? Yeah, that's a really good question. So the answer is yes. So what happens is if instead of thinking about it as a 40 theory on a cigar, we'll compactify to a 3D theory along that circle, the effective 3D theory becomes topological in the bulk with the chiral non-topological boundary condition. But there are extended operators and these become the bulk operators in the 3D n equals 4 theory. From the 3D n equals 4 perspective, they're the ones which parameterize the cool mortgage. So they're kind of subtle and you realize that their 40 uplift must be the two glands rocking those that cigar. Okay, thanks. So let me very briefly mention a different perspective. Perhaps it might have been better to explain this one, but for my purposes, I kind of didn't want to get into that much detail about either way of doing it. I mostly wanted to say the answer. So the different perspective developed by Dean Lemos-Fiando, Peres, Mastelli, and Fanlis, they say that instead of taking an ordinary supercharge like we've been doing, they take a superconformal charge which lives in the superconformal algebra, which in this case is PSC2, 2, 2, 2. Now what they find, things in the superconformal algebra include rotations. So in their story, rotations in one of the planes will be exact. So that implies, if you think about it, that the operators that live in the cuec homology only exist on a plane inside of a 4. Just like we saw above, the operators only lived at the tip of the cigar. Then they take the cuec homology and it's the same chiral algebra. This perspective was historically the first one on building chiral algebra for my new PS2 theory. And there's a great deal of literature about it, so I'm sure people who are interested in it can find good things to read. But as I said, I mostly want to say, given N equals 2 theory, what is the algebra? So let me start on this. If we have a look wrong in theory, then we can just write down the algebra complicated as a table. So we'll start with the simplest thing, which is an N equals 2 free hyperbolic algebra. One one has N equals 2 supersymmetry. The smallest number of scalar fields, what has is 4. So I'll write it like this, 5i, 5r, and this notation is kind of selecting, is breaking the symmetry in some way. And the log of N for the free theory is the simple form like this. Okay, so what do we get when we take the chiral algebra and the way it's discussed, what is the sketch? These four scalar fields, the N equals 2 theory, become two scalar fields in the chiral algebra. Beta 1 and beta 2. Beta and the beta i's comes from the phi i's and the phi bar's don't contribute. These are spin half. So the OBE between the beta i's is beta i is 0, beta j of z, beta j of z is epsilon ij times 1 over z. So it's a little funny because the beta i's are bosonic, but the OBE is like that of free fermions. So this chiral algebra is called symplectic bosons because they're bosonic fields. What happens for gauge theory? The N equals 2 vector multiplied contributes some fields for the chiral algebra. And these are something that should be quite familiar to people who studied some theory. They're BC ghosts. Although in contrast to the BC ghosts of string theory, these are ghosts that implement gauge symmetry. They're not ghosts that implement the diffeomorphism invariance theorem. So the BC ghosts consist of BN to the spin 1 field and CA would have spin 0. And these are both fermionic. They have the simple OBE. The B times C is delta AB times 1 over z. Same will be one finds in free fermions, except there are spins 1 and 0. As in any ghost system, one has to add on the BRST charge. And the physical open space or space of local operators will be the BRST chronology. So here is the formula of the BRST charge. And again, it might remind you of something you've seen in the string theory literature, except it's a little simpler. And the final thing one wants to do is consider how to couple the matter content to the gauge symmetry. So all we do is we add the fields from the matter to the ghost from the gauge symmetry and build a BRST charge, which forces gauge invariance. So for the matter, while if I started with a single hyper multiplet, I would have two fields, beta 1 and beta 2. So the most general picture is I should take my beta I, my matter fields, to be in a representation, which is symplectic. So that means that I have some invariant tensor omega ij, which is anti-symmetric. So I'll give you some examples later. Using the fact that G-ax on the representation, we can build a current here. This is the current which generates the action of my gauge fields on the matter. And then we say the BRST charge is just that for the gauge theory, plus the C goes to times the current. So we're running out of time. We'll have time for questions. What we find is this construction works. The BRST charge squares to zero. If and only if, there's some trace identity, which is twice the trace in the joint representation of two of the outer elements is equal to the trace in the matter representation. So again, you might be familiar with such a computation from, you know, there's something like this in Poltinski. In the bosonic string, the BRST charge squares to zero, if and only if, or the critical dimension, this computation is much easier. I'll give you the computation in just a sec. But it's the useful fact is that this is the same trace identity one needs to guarantee that a 40 equals two theory is super conformal. Let me move down to do the computation. So from the gauge sector, we have this vertex that I want to consider the OBE of this bit itself. So there's going to be a contribution from tree level involving one width contraction. This vanishes the Jacobian. The one width contribution looks like this. I'm going to have C fields on the outside. We call the legs correspond to a width contraction. They connect to C and a B, because the OBE of C and B is one over C. So let's compute this. I have a contraction like that, and a contraction like that. Because there's two contractions, again, one over Z squared, A zero, C, B, Z. And then that's into the flavoring of this carefully. There's going to be twice times the trace, the joint representation, K, A, T, B. The factor of two is because B and C run around the loop. So let's take this expression. We notice C is fermionic, C A zero, C B zero, trace, T A, B is equal to zero, because this expression here is symmetric. So if we take the OBE and expand in series, we find the first non-zero term, looks like C A zero, derivative, C B zero, two, trace, T, T A, B. Now for the matter sector, it's going to be the same kind of calculation. The C goes, it's going to couple to two matter fields. I'm going to do a bit contraction where all the matter fields are contracted, and we find one over Z, C A, del, C B, that's trace and the matter representation, K, A, T, B. And to be careful, this one came with a minus sign because they were fermions and the other one comes with a plus sign. I take my BRST operator, this is as no O, if this condition holds. And finally I'm going to ask, you might have, I should have said this first, why does a pole in the BRST operator and the OBE of the BRST operator with itself have to do with the BRST operator and fined in no equivalent? Well if I compute the commutator of its interval, when I move the contrast past each other, we'll pick up a term where one of them is integrated around in a little circle around the other. That said, it all goes around like that. And if we look on the right-hand side of this expression, a first order pole in the OBE of the BRST operator with itself will contribute a term in the square of the BRST charge. This is not equal to zero unless the trace Okay, so I will stop there. Okay, let's thank Kevin.