 Okay, thank you very much. Recording progress. So for this lecture, I actually didn't get a chance to prepare a size in advance, so I'm going to write as I go. What I want to discuss today is how to do some computations with written diagrams. Just like in traditional ADSEDP. But in this top logical string context. What we're going to do is we'll build a chiral algebra by studying the top logical string on SLGC. And when you open them so far, we've related the dimensional symmetry algebras, the gauge theory side, and the holographic dual. We've seen that these are the same. And the main result we'll be able to do today. We'll use this fact in the proof that planar gauge theory chiral algebra is isomorphic to the holographic. So we'll be able to show that we can match all of the coefficients. And in particular all correlators. So the first thing we need to do to understand this. Is to understand how to build the holographic chiral. So we started discussing this a little bit yesterday. So we call yesterday that SL2C has a compactification boundary is CP1 times CP1. So I'm going to give an explicit description of the geometry near the boundary. It's on a path near the boundary. We have coordinates that in C. This will be the chiral algebra plane. So this, we thought it was the boundary of ADS3, W in CP1, and also N, which is normal to the boundary. So N really lives on a line bundle over CP1. So explain in a second, it's related to the hop-fibration, which means that N has a pull when W is infinity. In other words, N transforms like the W. These coordinates we're going to use are related to the coordinates on ADS3 times S3 as follows. If we write N equals OR e to the i theta, then Z, Z bar and OR are coordinates on ADS3. And the boundary is OR equals 0, W, W bar, and theta are coordinates on the trees here, which we're seeing as a hop-fibration CP1. The theta coordinate parameterizes the hop-fibration. So the goal is to build a chiral algebra living on the Z plane by studying a gravitational engage theory we have in the book. So to get there, we need a few more details of the geometry. Z is not a homomorphic coordinate. There is a Bultrami differential, which looks like mu is a factor of N. So we're going to use some factors of pi and squared. So this may look a little funny, but if you remember that N transforms like the square root of BW, so N squared transforms as W. This expression here includes a natural volume of CQ1. So it's a natural expression. But also, because there are bulk theory, we're going to be focusing on the open string set because it's easier. So the homomorphic transient is the homomorphic transient. So that involves a homomorphic volume from this geometry. So I need to write that down too. Oh my god, it has a whole order of three. I'm just going to look at my notes. The coordinates looks like this. I want to note that if we look over here, we can, there will be a piece of this, which looks like d theta, d times the volume form of the two sphere times N. So from this, you see that the integral over the three sphere of omega is equal to N, as we would expect. Okay, so far we've just set up the coordinates. Are there any questions? No question at this stage. Okay, I actually can't see you guys anymore. So recall, the homomorphic transient is a quarantine. We're going to have a gate field. There's no one form in this geometry, even SLA, and it looks like an integral trace A d bar A plus one of the sixth trace A, A, A. And this whole thing is hit with the homomorphic volume. So to describe the holographic capital algebra, we have to first describe boundary conditions at N equals zero is on the boundary of our geometry, and then states will violate the boundary condition at some value of Z. The boundary condition, which is a little time thinking about it. It's a little subtle. But we noticed that omega has a pole over three. It turns out that asking that A has a zero is a good boundary condition. So we ask that A goes like N near N equals zero is a good boundary condition. Wait, so now we can ask what is the state. And of course, we're going to modify the boundary behavior at some value of Z. For example, we could ask that a goes like delta function. That equals head zero. Plus, oh man, this is going to be the simplest lowest line state. So in a moment, I want to match these with the states in the current algebra. But for now, if one states ordinary ADCFT, he might say, well, what you really want is a solution to the equations of motion on the bulk geometry, which satisfies this boundary condition. And we can find such a thing. But there is a unique gate equivalence solution to the linearized equation of motion, which is just the D bar equation, which satisfies modified boundary condition. So this the explicit formula is a zero. Let's introduce a color index. Yeah. Plus. It's going to look like this. So the reason this satisfies the equations of motion. We should recall that there was a boundary differential so Z wasn't, wasn't homomorphic. So if I take a delta function, a fixed value of Z, that doesn't satisfy the D bar equation. But this adding on this correction from, from makes it so it does. I can draw a little picture of it by what this is. Yes, three. This is the boundary radius three. My state. Sorry Kevin, the gauging next in the second term is TA is multiplied by TA or what. Yes, yes, the second term is wrong. So what this looks like. I put the boundary. The boundary is an equal zero. As I put the boundary. It decays rapidly here with zero zero as I put the boundary. And if I move a little bit away from the boundary, it also decays. So this is actually said by the sense of apologies, decays as Z moves away from zero. So this is the kind of thing you want for a state right on the boundaries localize a particular point. As you move away from the boundary spreads out away from that point with some particular decay rate. The decay rate of course will be related to the two point function. Because this state is related to the current and the two point function of the current goes like one of us. So we're going to propose that. Hey, hey, said zero gets matched to the state. Ja, it's a zero in the car with this is the S away current. And that's where it was. And then there's some. Or an S one from one to eight and it's anti symmetric or S. So we'd like to also build to build the states correspond to, or to these, to these symmetrize places. And what we'll find is that the ways of modifying the boundary condition. At this point match precisely with the states in the CFT. That's the most general way we can modify the boundary condition. So we could say, a is like delta function. That said equals at zero. But there's nothing stopping me having more singular behavior in N. So I could have one over N disempower. But once I have that singular behavior in N, you'd also think, well, why can't I have a function of w two, and the remaining terms are going to be terms which are less singular and inclusion will be forced by solving the equations of motion. Never. What possible values do we have here for these indices. Well, I can have an arbitrary pole in N. But we know that N. As a pole, that w is infinity, which means one over N to the K. As the zero over K, having a pole in the W plane will be bad. Because it will prevent my field configuration in satisfying the equations of motion. So we can say that the index L ranges from zero to K. Well, this is perfect because these are exactly states with the white quantum numbers to match the states we see in the chiral logic that we propose that this corresponds to the polar factor. It corresponds to pi four x one L x two K minus I s. These states have the same quantum numbers. Any questions so far? Nope. I assume everything is crystal clear. All right, so now we've described the states on the homo apocalyptic dual side. We can ask, how can we compute their endpoint functions? Let's say the two and three functions to start. So the endpoint functions of these states computed in principle by Wittentagos. So let's start with the two functions. So, if I take my state, a z zero, let's go, and then square, I take two such states. Their two point function is obtained inserting into the kinetic term of homo apocalyptic dual side. It's going to be integral over all my coordinates A, A, zero, D bar, A, B, zero, and then when we contract indices, there's going to be a candidate. So, since these states correspond to the SOA current to my chiral algebra, these have two point function, which is like N, A, B, over C0 minus E1 squared. So this is homo apocalyptic dual side. So if I still remind you, the factor of homo apocalyptic dual side is even squared. You can see this because that's the standard central extension of the cascule, and it contains a cascule at level N just because that's the number of fields we have. These guys are like I, I, I, I, I know X is in the middle, and when I contract them, we do two with contractions, and there's a loop in the middle. So the loop in the middle contributes N and the two with contractions contribute over C0 minus E1 squared. The equation point introduced the volume formula. So let me scroll up a little bit so we can see the formulas each expression is pretty much the same. Here there'll be like a delta C and C1, and here there'll be the same kind of thing with the 1 over C minus C1 squared. So what I'd like to do is kind of quickly compute this integral explicitly, and we're going to be a little loose and I'll drop some terms, which are also important and just focus on the relevant terms. I can take the first one. I'll drop the color parameters too. This one to be a delta function. Here, I'm going to put d bar. So we have a 1 and the volume form looks like the N or N cubed, the WZ. This is the relevant term, the volume form. Because our geometry is non-compact, we're going to integrate over Z and W, or N is we're going to introduce a cutoff near the boundary of radius 3. Now, if we do integration by parts, this is a pure boundary integral. The boundary term, we're going to get a bit of the d bar, and we're going to find something like this, and N squared, dW bar, plus d squared, dN, or N cubed, dW, dZ. And I forgot a crucial term, which is 1 over Z minus C1 squared. And this term here really appeared originally over here. Okay, so all I've done is, you notice the integral is a total derivative, so you pick up the boundary term, and we just insert all the terms and we get something that looks like this. And then we see we get a really nice answer because this will correctly reproduce the chiral algebra 2.0. Why is that? The integral over W of dW, dW bar, 1 plus W squared squared. This is just a natural volume form in Cp1, which will be the study metric. So this is basically one, it's a multiple of pi. So that part goes away. The integral over Z, all because I have a delta function in the integrand, all that does is, produces 1 over C0 minus C1 squared. That looks good. And finally, we have an integral, contour integral over N, and we notice up here, we have N squared, there we have N cubed. So we get contour over N of dN over N, and that gives you node 2 by I, which is also important. So we do indeed find the holographic 2.0 function. Okay, any questions about this calculation? Yes, there is a question. Hi Kevin, I just got a little bit confused. Is the integral, if you go up just before you did the integral? Is this an integral over Cp1 times Cp1, or is it over S2 times S3? Right, up here, this was an integral over SLTC, but I removed the neighborhood of the boundary. So here, when I go down to the boundary integral, this is really an integral over an S1 bundle over Cp1 times Cp1. Yeah, so where's the fifth coordinate then? I just got confused. It's N, you see, because N, the absolute value of N is fixed. So we're left with the angular value of N, that's the fifth coordinate. And that appears when we, you know, it's an integral, I can just remove, I can just compute it in a dense open set, as long as it converges. So I don't worry too much about the fact that it's a non-trivial sort of bundle. So in which case the integral becomes an integral over C times C times S1, and the integral over the S1 just gives us this common integral. Does that make sense? Yeah, thanks. The three-point function. Then we will play the same game. If I take three states, I'm going to get like, F, A, B, C integral with a six dimensional geometry. Freezing it like this though, it turns out to be a little tricky. And I personally prefer, well, in any holographic computation, I prefer to use the OBE rather than the three-point function. So how do we find the OBE? Well, the OBE, which when inserted into the two-point function gives, with AC, gives the three-point function. So what is that? This is just the definition of the OBE in general. But there's a nice formula for it. The OBE is given by D bar inverse. This expression is, and you can imagine that as follows. There's my first state. And then here D bar inverse is the propagator coming off of that. So why is this the OBE? If I insert AC, B2 times, times the D bar, D bar of that, well, clearly the D bar is cancel, is the three-point function. So the OBE is something that's actually easier to compute than the two-point function. So let me quickly explain the technique for doing that. So the technique for computing the OBE is, well, A, AC0, select this. And it turns out that this is D bar, TB times D bar of one over C minus C1. So I can remove this state by a gate transformation. However, in the presence of the other state, this gate transformation has a non-trivial effect. It sends this other state, C0 to one over C0 minus C1 F, A, B, C, T, C, delta, C equals C0. That's the other track. Now here we see the correct cascading OBE. And of course we can play this game with all the other states. So it's a little more complicated. There is a question, Kevin. In the last equation, how does the B index get soaked up? Because it remains free on the right-hand side, but on the left-hand side, it's just A, A. Right, so the B index is present in the gate transformation. The gate transformation is like... Oh, I see. ...one over C minus C1, and I'm looking at the effect of this gate transformation. Thanks. Okay, so in this way, one can, in principle, are just by working really hard and thinking about standard written diagrams, try to compute all of the correlation functions of the tiring algebra. However, that's really hard. And there's a shortcut. The shortcut is that we know, we've already checked, that on both sides, there's this really big symmetry algebra. And that's going to constrain everything really tightly. And that will really help us to cut down the problem to a much, much simpler problem. So let me very briefly say, why, how... It's already kind of obvious that this symmetry is present on the holographic side, because it's just the gate transformations. But it can be helpful to think about the mode algebra from the holographic point of view. So just like a local operator is obtained by modifying the boundary conditions at a point, so we get a field that looks like a delta function at a point in the boundary, where the mode will be a field configuration. For example, some power of z, a delta function that's where z is on the circle, maybe there might be some negative powers of n, plus some other stuff, the negative powers of n are localized on the circle. So we need to see, you know, the commutators of modes can be seen, can be computed using the same gate transformation trick we just mentioned. So I want to briefly explain why does functions on SL2C, these are homomorphic functions, tensor SOH, live inside the mode algebra. The answer is very simple. If I take f, which is a homomorphic function on SL2C, f times ta times delta function at z equals one, this expression satisfies the field equation, plus a, even though z is not homomorphic. So geometrically, what we're doing is we're just taking our homomorphic function and calling it by a delta function on a five-manifold, which add infinity to this circle, this circle where z is one like that. And the reason it satisfies the field equation is just, this is closed, you know, the delta function, any five cycles close one form, and drf is zero, because it's a homomorphic function. So in this way we see that this global symmetry algebra really lives inside the modes of our chiral algebra. And how does that help us? We know that the commutators of modes of our states are given by the algebra function on SL2C, SOH. If I take in any chiral algebra, the commutator of modes is some expression, is an expression in terms of the ecofissions. So this relationship, it turns out to constrain OBEs and so correlation functions uniquely except the two-point function of the stress energy tensor, which goes like this. So in other words, once we know this result, now we have this huge symmetry algebra, and once we can identify the quantum numbers of the states, everything else is determined by a single grammar and the graphic algebra is isomorphic, the planar. So I think I'll stop there. Thank you.