 So we're talking about n equal to 2 supergravity and half bps black holes in n equal to supergravity, which carries charges qi and pi. These are dionic. And the idea is to focus on the near horizon region. That's actually a full bps solution of the theory. And there's some a2 times ns2. And today we're going to use. So this is Euclidean a2. Something I didn't mention was the symmetry generators. There's a symmetry, u1 symmetry generator, which I'll call l0. That's just the rotation of the disk. And another one here called j0, which is just the rotation of the sphere. And in this geometry, we set ourselves the task to calculate the partition function of ads2, which is, we argued, it's only a function of the charges and it's given by a functional integral of something called a renormalized action of all the fluctuating fields of the gravitational field theory. And the fluctuating fields means you fix the boundary conditions according to the classical theory, namely according to the charges, fixed completely by the charges. And in the interior, you let everything fluctuate, and that's what you're integrating over. So that's our task. And this s renormalized of the gravitational fields is, so the input is some bulk, the Gaussian or action. So yesterday, two days ago, we wrote down, maybe yesterday, the action of n equal to 2 supergravity. It's s bulk of phi graph minus there's a Wilson line with respect to all the electric fields in the theory. That's at r equal to infinity. And then there's a boundary term, which is also local Gaussian variant function of all the fields. And that's chosen in such a way that this s renormalized is finite and also super symmetric. So that's a recap. Any questions about this so far? Either about the general formalism or philosophy or anything. Or what I'm going to do, if that helps you, perfect. Everything's very clear. OK. So today, I want to do a calculation. So I really want to calculate this for this example. And the strategy is to use localization. So in order to use localization, the first condition is that you need some super symmetry. And so that we're going to use this fact that this new horizon region is fully BPS. So in fact, there are eight supercharges. And I only need one in order to do localization. So we saw the formalism of localization. This was explained very nicely in more than one lecture. So we're going to take one supercharge, which is some kind of variation by a Killing-Spinner epsilon on this geometry. So there are eight of them. So suppose you write this geometry down, you can just ask, what are the super symmetries? There should be eight. You can just solve for the Killing-Spinner. And there's one combination, which squares this l0 minus j0. And that's the one we're going to take, all right? Yes? So you're trying to localize supergravity. But there might be many more field singles. If this is a twin theory, then you're already deciding that many, many other things are not connected. No, no, I didn't decide anything of the sort. So I said something which may have been a little cryptic. And today I'll repeat it more. So the question is, what do we mean by five graph? And what do we mean by the action? So five gravitational, I repeat, is a collection of all fields in the theory. Suppose you want to think of string theory. Then it's a graviton, dilaton, homo-romo field. And then higher spin, massless, massive fields. The whole tower of states, everything in the theory. Of course, this is an impossible task to actually write down. But what we can do very well is to write an effective action for the light modes by integrating out all the massive fields of string theory. This has been done since the beginning of string theory. So we can write some action. But now what is going to happen is that that action is no longer two derivatives. It's going to be an infinite derivative action. That's what I mean by this. So I've not lost any information. But your Q should be a Q for the fourth. It is. So the way I'm going to do it is think of this functional integral. I was going to talk about this later, but I can do it now. So there are many new issues in localization when we apply it to supergravity, both technical and conceptual. So this is one of the issues. So the strategy is the following. There's no proof of this. Strategy is the following. So a priori, your functional integral is not well defined. It's a gravitational functional integral. So there's an IR divergence, which we cured. But there's a UV divergence of gravity, which I've not cured. And that's not my goal in this. The idea is that because this observable is supersymmetric, we're just going to try to use the rules of localization for our noses, and just apply it, and see what you get. And if you're really sort of a purist, you should take that as a definition of this class of functional integrals of quantum gravity, just a second. So in practice, what I'm going to do is to write down an effective action for these fields. So the assumption is that I have a UV completion, which is consistent with supersymmetry, with this supersymmetry. So that I'm going to assume. And then I write an effective action for these low energy fields, including all kinds of derivative terms. It's supersymmetric, so I can use this. And the idea is not to integrate out all the fields until 0, because then you get these massless, these non-local things, which is the whole reason I started this problem. So I'm going to integrate it at some low but finite energies, and then use this effective action to localize. And then you'll see that the answer looks good. And so then you can declare that this is the kind of thing to do. People do similar things in five dimensional gauge theories. It's a very similar idea. Kiri? So what's the mean theory? So maybe I should also. Good. So all these points I was planning to do, but let me do it right now. So here, one was like, what is effective action? What is the action? What is the meaning of the action? S. So at least I gave some idea of that. The other is, what are we computing? So maybe I should take some time now and do it here. Can I do it here? Yeah, I think so. So yesterday, one of the things we saw was that in this AdS2 times S2 geometry, if you look at a Maxwell field, there are two modes. And then the mode that dominates is the charge mode, charge carrying mode. And therefore, the thing should be in the micro canonical answer. That's why it's a function of only q and p. But now, let's put it together with another fact. So let's stick to four-dimensional n equal to two black holes in asymptotically flat space. So it's a fact that supersymmetric black holes in this context do not rotate in four dimensions. There are no supersymmetric solutions which rotate. Now, what that means is that, so rotation is discharge J, J0. So that's the J3 of the, so there's an isotope rotation symmetry, and that's the J3 of it. Now, the fact that the supergravity solution doesn't carry J0, it means that if you think of this as a quantum ensemble, in the quantum theory, the expectation value of J0 to 0. That's just a fact you can find out by doing supergravity. Combine this with the fact that there is a micro canonical ensemble. It means that every state in the theory must have J0 equal to 0. Because micro canonical means that the charge is fixed, and if the average charge is 0, that means every state is 0. So plus micro canonical, J0 equal to 0 on all states, on each state. That means that whatever this ensemble is, this quantum ensemble is, let me use a Hamiltonian picture just to make, it's just a shortcut. I can do it in the path integral language. But that means that e to the trace minus 1 over J0. J0, remember, was a rotation charge, so fermions have half-intercharge with that, 2 J0. So minus 1 to the 2 J0 is just minus 1 to the f. That's just equal to trace of 1. So that's an index of the supersymmetric index. It's just minus 1 to the f. It's just f. And that's an entropy. So it means that whatever quantum gravity thing I'm doing here, what I'm calculating is both an entropy and an index. And this is the reason why a lot of the older work actually carries through to be on the leading order. So what Baconstein-Hockeying want us to calculate is this. What Strowman-Jainwaffe actually calculate in their microscopic things is this. So the idea is that you start from near horizon, you have entropy, you generalize this to quantum entropy. That's equal to that. Then you use this argument to say it's an index. And now you can go to the microscopic regime. So continue this, and that's protected. So that's one comment I want to make yesterday. So thanks for bringing that up. So the slogan is index equals entropy. Adhesion, thank you. Exponential of there. If you want to know more details, this is the philosophy, and this is the basic argument. There's one paper by Schock-Sehn, I think 2012, something like this, and then another one by Sehn, Davulker, Gomesch, somewhere here, and myself. Let's sort of flesh this out. Any other questions of this philosophical type? Then let's move on. So I want to use localization. That's a supercharge. So let me recall the strategy of localization. So recall, what is the strategy? Strategy, so I'm going to be brief because this was done already in Francesco's lectures. So start with some z, which you want to compute. So first you have some q squared equal to h. Here, q squared equal to h. Can I erase this? It's on the whole board. So I have q squared equal to h. That's my starting point for this kind of localization. q is some supercharge. h is some compact u1 generator. And what's the idea? I start with some functional integral of this type, d phi e to the minus s, where s is supersymmetric. So q of s is 0. s doesn't necessarily mean action. I'm just saying this is the action plus operators, everything like here. It's the Wilson line plus operators. So the whole observable, I'm going to call e to the s. So it means actually the supersymmetric observable times e to the minus s. So what you do is you deform it. So z, you deform to z of lambda by something called qv, where v you choose. So you can have many choices. And I'm going to choose the one that Francesco explained. So just psi dagger cubed psi. Then you argue that dz d lambda, what that does is it brings down a qv. e to the minus s of phi minus lambda qv. And now you think of q as some, write it out in the field space. And it's some differential operator in field space. So you do some integration by parts. So this is qv of this whole thing. So that means it's q of that, which is a total derivative, assuming there are no boundary terms in field space, minus q times this, but q of s is 0. And q square of v is also 0. So this is 0 because q of s is 0. And q square, I've chosen v such that q square of v is 0. And also I've used the fact that q, the measure, is also invariant, is supersymmetric. So these are the conditions. And therefore, what I originally wanted was z of 0, but that's equal to the limit lambda goes to infinity of z of lambda. And now this thing localizes. So what does that mean? It means that as lambda goes to infinity, this term dominates. And so you just look at critical points of, I'm sure all of you have heard this 100 times. So let me repeat it 100 first time, the critical points of this. The way we rig this up, the critical points of this qv, this is the perfect square. So it's just q psi equal to 0. So that just becomes integral. I'm going to call this the localization manifold, m log. That's defined as all fields such that q psi equal to 0. So all solutions of that. Then there's some induced measure. I'm going to call these the fields, the coordinates on this manifold. I'm going to call phi log localization. Then what I have to do is to evaluate the action on an arbitrary point on this manifold. And then again, because this is a perfect square, at q psi equal to 0, this itself is 0. So that doesn't contribute. But there's a one loop fluctuation determinant, which is lambda independent, which appears in the final answer. So it's one loop of qv. That's also a function of phi log. So that's your basic localization formula, which you already saw at least once, maybe twice in the last few days. So now I want to use this. I want to apply this to our problem. So what are the steps? The steps are the following. So steps. So the 0th step, I forgot to say something. So in this argument, there was some confusion about this. So I want to say it out explicitly. In this argument, at least in the most naive way of doing it, it's crucial that this algebra holds off shell. And the reason is that in going from here to here, sorry, in going from here to here to show that this is 0, I've used the algebra inside the function integral. I've used it for an arbitrary field configuration. So it's really crucial that I have it off shell. So step 0 is that I need an off shell formulation of n equal to 2 supergravity. I don't really need the full n equal to 2 supergravity. I just need this 1q that I want, or q squared equal to 1q. But that we already did. So Stefan explained to everyone how to do that. And I also reviewed that yesterday. So that's step 0. I mean, that's from the 80s. This is using the superconfirmed formulaism of Duvet and Fanolten and Franpryan. So these steps are outlined in this paper. 10, 12, 0, 2, 6, 5. So the second step is you want to find this localization manifold, but the first step. So the 0th step is a formalism. The first step is find the manifold. Then you have to evaluate S on phi log, an arbitrary point. And finally, calculate Z1 loop of qv. And then put it together and do the integral. And that's your answer. The point is that each of these steps is different. It's really qualitatively different from just rigid supersymmetric quantum filtering. I want to give you a sense of what is different and how we go about it. So not everything is a theorem unlike in quantum filtering. So two of the points were already made. One of them about the action. But I'll go through these things again. So I'll go through these things one by one. Any questions so far? Yes, I'm assuming all kinds of things like that which have been discussed already in the simple context of even free field theory or boring field theories. There was a lot of discussion about it. So I'm not going to repeat that. So there's some second level discussion of that type which. So there it's not so much different because it's gravity, but because it's non-compact. That brings about a distance. Let's postpone this very technical thing which let's postpone. So I don't think gravity brings about a new certainty about that point. Very good. So let me actually remind you of what the off-shell formulation is. So I never wrote it down. And neither did Stefan. So let me do it. So you have, remember, a vial multiplied and a vector multiplied. So we need to write down the supersymmetry variations. So this is the Gravitino, the two Gravitini. And there's some parameter epsilon. Like I showed, that's your killing spinner. This is some covariant derivative in gravity, minus 1 8th gamma a gamma b, t a b i j, gamma mu epsilon j, minus gamma mu eta i. Remember all these fields? So this was the Gravitino. This was this auxiliary in some sense t field, which was non-zero in the vial multiplied. That's the thing that's essentially became the Gravitin photon in the on-shell theory. It's this anti-symmetric, anti-self-dual tensor. And eta was the parameter. So epsilon is the parameter of q supersymmetry. And eta is the parameter of s supersymmetry. And this is some covariant derivative which I'll write in a second. And then there was lots of gauge multiplets, vector multiplets, and there's two gamma mu d mu of xi times epsilon i plus half epsilon ij fi mu nu minus gamma mu nu. I can't read my own handwriting. Epsilon j. No, there's no minus here. It's just that times gamma mu nu epsilon j plus y ij i epsilon j plus 2 xi eta j. So I just wanted to write it out once. Here it is. It's some very explicit supersymmetry transformation. It's slightly more complicated than the ones we've been seeing. There are these two is not about my handwriting. These two are actually different d's. They're both covariant derivatives. But it turns out that the Gravitino variation has a slightly different derivative here than this. It's just a fact. I could have written it out. I just don't want to fill up the blackboard. So this one starts something like. Both of them start like that. But this is d mu minus 1 fourth omega mu AB gamma AB. The usual thing. That's the spin connection and so on. And the so on contains all the fields under which psi mu is charged and similarly here. Just something. So if you have questions about this, you can ask. But if it's just a question of what is this or that, I'll just refer you to the book. But maybe it's important to ask me if something is not clear. So Epsilon, so remember there were q supersymmetry and there was s supersymmetry in this formulation. Any other questions? Yes, yes, yes. So there are other things I'm not writing down which. No, no, you have to write that down. It turns out that that is automatically satisfied for these things. But you have to, indeed, there's another spinner called chi. You have to write that out. All spinner variations should be 0. Other questions? Very good. So now I'm going to do step one, which is to find all solutions to this. So this is the kind of thing we saw a little bit in some quantum field theory example. But now even the philosophy is different, because if you stare at this, let's stare at one of these equations. Let's say this one. So here's the question. What are the independent fields? Suppose I'm a computer. Which fields am I supposed to solve for? This is supposed to be 0. So I put both of these to 0. Usually in quantum field theory, let's take this. This is some Gagino. You're given some background manifold which comes with the Killing spinner. And it comes with a metric. And then you're supposed to solve for these fields. All configurations of x and auxiliary fields. Over here, the problem is much more complicated. I didn't tell you what I sort of did, but I cheated. So here, now, so I said that the gravitational configuration, radius 2 times s2 configuration, has eight supersymmetries. And I'm going to take one of them. But that's when you really think of this radius 2 times s2 background. What I really want to do is a functional integral over all the supergravity fluctuations as well. When I think of that, then it means that the metric is also fluctuating. And so is the gravitino. And because the supersymmetry, you can think of it as generated by zero modes of gravity. Killing spinors can also fluctuate. The question is, what should I solve for? So this is a slightly, it's a hard philosophical problem. One has to sort of find rules. And I'll have a little bit more of formal discussion about this tomorrow. But let me tell you what the philosophy is. And if you think that that's not good enough, then let's have a discussion. So the philosophy is that because I'm doing a functional integral over gravity, I want to find all matrix by matrix, I mean all while-multiplet configurations. So metric and T and all the fields in the while-multiplet. That support some killing spinor epsilon. So that's what it means that it's supersymmetric. So I want to find all supersymmetric solutions such that as r goes to infinity, you get back the attractor-classical background. So that's AdS 2 times S2. And epsilon goes to this epsilon naught that I wrote down there. So epsilon naught is some fixed killing spinner of AdS 2 times S2. And what I'm giving myself to solve is to say I want to find all supersymmetric fluctuations of gravity, which means that it should support some killing spinner with the correct boundary conditions. So that sounds like a reasonable thing to do. And tomorrow I'll justify this even better. But any discussion about this? Yes, Francesca? How do you make singularity? Yeah, so here I'm going to choose. So today, I'll just assume that all fluctuations should be smooth. And at the end of the lecture, I'm going to make all solutions, all the localization guys should be smooth. And at the end of the lecture, I'm going to break it a little bit. Please ask me if I forget to mention this. Do ask. Because I want to talk about this today. OK, so that's the first step. And the second step is then for each such geometry or each such configuration, I'm just going to call it metric or geometry. But I really mean the full configuration. Find all solutions to the vector-multiplet, vector-multiplet solutions with configurations with killing spinner. So with delta omega delta epsilon, this epsilon equal to 0. So this is a problem we've seen already. But given some background, you just compute all possible. That background comes to the killing spinner. You just compute all possible supersimilar fluctuations. But the first step is the more important one. Very good. So that's the philosophy. And I said, I'll talk about this more tomorrow. But this was the thing that we followed in this paper. It sounds quite reasonable. So the next question is one of practice. So I said, this looks very nice. You have to find all possible solutions. But if you've ever tried such a problem that you have a Gravitino variation, you don't know anything about the right hand side except boundary conditions. And I'm asking you to find all metrics and all killing spinners that this admits. It's a very technically challenging problem. But so practice how to find all solutions. But thankfully, we've been helped by technically very good people like Witten. And then there was, I think, Todd. And there were many people. I think Witten was the first paper. And then what I'm going to say now is in this paper, 1208. Oh god, this thing is being recorded. 6, 2, 2, 1. OK, I should watch my tongue. And what is the idea? So I'm going to do a simpler example just to tell you the idea, because it's quite involved. So let me take a simpler example. So just take del dup psi mu i. And just take the following covariant derivative. So 2 delta mu minus 1 quarter omega mu a b gamma a b minus quarter times gamma a b t a b i j gamma mu epsilon. OK, so I've switched off all the other fields. There are many other fields in the vial multiplet. I've only kept the metric, the spin connection, and this field t a b. That's necessary to solve for supersymmetric configurations. Yes? Let me ignore all such things. Let me ignore all such things. So I just want supersymmetric solutions. So let's just take this after the class. So people who really know the super conformal algebra, we can have a small five minute discussion. I'll explain it. The answer is everything. I should keep both, of course. So what is the idea? So the idea is, so this idea was, go back to this, you form Fermi on bilinear. So that's, by the way, here there's another nice reference for the on-shell case. So all this was done in the on-shell case. And what was done in this paper was to use the same method for the off-shell case. So this probably the most recent and the paper I like the most is Gauntlet. Mania, please help me. Do you remember? Gdotsky, Hull, Parkes, and Real. Yes, good. And there's another one by Ortin and Patrick Mason, which is also nice, which is very relevant. OK, I should hurry up a little bit. So what is the idea? So form the following bilinear. So assume that there is some solution. So what we know? So assume that there is a supersymmetric solution. And the idea is that I'm going to use the BPS equation to find properties of the solution. And if I find enough properties of the solution, I'm going to reconstruct it. So from this thing, that's a scalar. So that's a bosonic thing. Form, that's F1. Probably with X. Probably with X. No, no, no. Psi, psi. Just a second, I'll explain this in a second. So firstly, I'm assuming that now I'm using some Dirac notation. It's just psi. So what I'm going to do is form bilinears of the gravitino, various bilinears, OK? And, oh, sorry. Yeah, I think you're right. So I meant epsilon. Yeah, so it's the gravitino. Yeah, sorry, sorry. You're absolutely right. So I don't know why I say that. So epsilon, well, just give me a second, because I think I've written it out in the language of psi. Yeah. Huh? Sorry? Again, Mania can help, maybe? That's right. Sorry, sorry. Yeah, there's an epsilon. Yeah. Yeah, so I think it was psi. No, the question is whether these bilinears are made of psi or epsilon. Yeah. Yeah, that's how I meant it. My memory is of the super-symmetry of that. Yeah. Yeah, OK, so let me call it psi. OK, thank you very much. Let me just call it psi. Yeah, so epsilon, maybe? OK, so this psi is epsilon. It's not the same as that. It's not the same as that. It's not this. OK, so let me just continue calling it psi, because I did that in my notes, and maybe even in the paper. That's why. OK, so the point is this. If I have some killing spinner, then I'm going to make bilinear sort of it. Let me make another one, and it will be very clear. Make this one. OK, so if this is a killing spinner, this is a vector, and it turns out that this is a killing vector. It's a bilinear on the killing spinner. It's a killing vector. And so this is some, yeah. OK, so let me continue. So that's a scalar. That's a pseudo-scaler. That's a vector. So thanks for the clarification. Sorry, psi is not equal to psi. There's an i, of course. OK, so then, and from f1 and f2, I'm going to make r e to the i theta. That's just the way of arranging it. All right, and now also r and theta are not related to anything else. So this killing spinner equation one, if you apply this to this bilinear, you get the following. The l mu of x, let me call this x, is 1 fourth T mu nu minus k nu, and T mu nu x bar is 1 fourth T mu nu plus k nu. OK, so just what you're doing is, you know that this should be equal to 0. OK, so you multiply this by epsilon bar or psi bar, and then you get some equations. And another equation you get is D mu k nu is minus 1 eighth T mu nu plus x minus 1 eighth T mu nu minus x bar. OK, so sorry for this confusion. I hope it's clear, right? You just, yeah. No, no, no. OK, sorry. Maybe I should have stuck to Leo's notation. This is just epsilon. So psi is not the... Sorry about that. That's right, that's right, but yeah. OK, now, so this side is manifestly anti-symmetric, and therefore if I symmetrize this, I get 0, so that means that k is a killing vector. So that's the kind of thing you get. So if you assume there's a killing spinner, it means that there's a killing vector in the geometry. I still haven't told you what the geometry is, but if the geometry supports a killing spinner, it must also support a killing vector. OK, then use some fierce identity. You get that k mu k mu is r square. OK, so it's all just algebra and using this one equation and spinorial algebra. So, sorry, minus r square. So that means that the norm of this, so k mu that means is either time-like or if r square is 0, it's light-like. There are two cases. So assume it's time-like, so k, then you can call just d dt. You can give, it means that the geometry must have a coordinate, which I can call t, which is time-like. OK, there's a time-like foliation. OK, so you get some information here. Then similarly, there are one forms called, let me call them phi a, phi alpha, alpha is 1, 2, 3, which are closed. OK, so similar analysis will give you that. So that means that at least locally, you can write phi alpha as dy alpha. OK, so then you can check that k mu, so phi alpha is phi alpha mu is mu. You can check that this is 0, k mu, phi alpha mu is 0 and phi alpha mu, phi beta mu is r square times delta alpha beta. OK, so you get all these kind of properties of the one form. So what this means is that there are three sort of coordinates y, which are, so time foliates the manifold and then there are three coordinates y inside the spatial slice, which look like that. So you start to form a metric. So this means that the metric must look like that. It must look like minus r square dt plus some one form v square plus one over r square times dy alpha, dy alpha, alpha equal to 1, 2, 3. OK, so you start getting a metric. OK, just by these properties. All right, and then there's a long story. You keep going like this, which I'm not going to tell you. So the main things I'm going to use are like here are the BPS equations, smoothness, so that's for Francesco and the boundary conditions, which is Adius 2 times S2. OK, and a lot of patience and good collaborators who don't make mistakes with notation. And so with these three conditions, it turns out that you actually can completely nail down the problem in this case, almost completely. So what you get is ds square is e to the phi of x times r square minus 1. I'm writing the Euclidean theory directly, although I started in some Lorentzian theory. OK, so you get something pretty amazing. You get that essentially up to a conformal factor. It's a conformal factor. You get Adius 2 times S2. So this conformal factor is something like this. So if you think of this as Adius 2, the metric overall can fluctuate. That's square root of g. And these are the only, there's just a one function worth parameter, one function worth of solutions. So it's quite a strong statement. And then in fact remember that there was a dilutation gauge which we still haven't said, and we can use that to set this to 1. That just makes life easy. And in that case, you just get Adius 2 times S2, OK, attract. So this means that you're actually integrating all the metrics, but in this case you don't have to. You can, the only solution, you don't have to in the sense that the only solution that matters is the classical solution. Do you also have information about both having to have the same radius? Yeah, yeah, yeah. So this is really the attractive geometry. You get exactly this in this case because the boundary condition, essentially the boundary conditions in this case and smoothness in the Euclidean problem is very constraining. Because the first order equation, and you won't know singularity, that's the thing which makes it very constraining. OK, then you do the same thing for the Gagino. So rather all the Gagini, so this is equal to zero. And again, I'm not going to write too much. So this is D slash of xi times epsilon plus yi ij epsilon j plus gamma mu f mu nu i epsilon ij epsilon j equal to zero, OK. So given this, this has some killing spinner epsilon, OK. So use that to solve the vector-multiplied equations. And of course, one solution is the attractor equation itself, where yi j, attractor solution, where yi j is zero, that's the auxiliary field, x becomes x star, and f mu nu becomes f mu nu star. Remember, the attractor near-reason geometry was already BPS. It turns out that all the solutions, the whole solutions, all solutions are the following. So a mu actually doesn't change at all from the attractor geometry. Star means attractor like yesterday. Xi is xi star plus ci over r, where ci is an arbitrary real number, OK. Remember, here r goes from 1 to infinity, OK. So that goes off-shell. The on-shell value is x star. That goes off-shell in the sense it doesn't solve the equation of motion, but it does solve the supersymmetric equation because it's supported by yi j. Y12 is minus, is plus 2 ci over r square. So all the solutions are as follows. OK, this took a little more time than I wanted, but that's OK. So all solutions, the whole solution manifold, so m log is g mu nu, is just the attractor geometry g mu nu star, and for each xi I have some excitation of this type, OK. Now that thing is the real part of xi, OK. Near infinity, everyone gets their attractor values as they should, and there's one parameter family. That parameter can be just labeled by the height at the origin, which is r equal to 1, which is just ci, OK. So ci plus xi star, excuse me. So that thing is just I'll call it phi i, which is ci plus xi star, OK. So it's a one real parameter. So for each vector multiplied, there's one real parameter, OK. So it's an enormous reduction. We started with an infinite-dimensional manifold. We reduced to an nv plus one-dimensional manifold, OK. So this is for each vector multiplied 0 to nv, OK. So m log is just equal to phi i i equal to 0 to nv. So that concludes the first thing, namely how to actually calculate the localization manifold, OK. Let me not pause for questions here because I'm running out of time a little bit, we can look after the lecture unless there are very urgent questions. OK, so then let me move on to the action, OK. So the next point is the action. And again, there are many issues which are different compared to the quantum field theory, one of which was already discussed. And I'll come to this again. So you have S bulk of phi. You want to compute this on the localization manifold. OK, that turns out so it's equal to the Lagrangian of n equal to supergravity that I showed you on the localization manifold. So this is a relatively simple step. I've given you the solutions. I've given you the Lagrangian. You just have to go and evaluate, OK. So it turns out that the answer is very pretty. So evaluate. What you get is that this bulk of Lagrangian, phi log i, let me just, from now just let me call it phi i. Phi i, that's phi i, OK. That's the localization coordinate. And it turns out that has a very simple form. 2i times, it's actually a total derivative, r d dr of r times f of xi minus f bar, OK. So that's an exercise which you can do. You write this big Lagrangian that I showed you. And there was one term which was just f, the n times t squared or something. That turns out to be the only one that essentially contributes. So that means that the, sorry, I heard some mumbling. Is there a question? That means that s bulk equals 8 pi squared. You do the integrals of the four dimensions. Integral from 1 to r naught dr. And that's minus 2 pi r naught imaginary of f at x star plus ci over r naught. That's the boundary term, oops, l. OK, so there are two terms. This is a total derivative, OK. So the action just gets the value of this at infinity, which is r naught, minus the value at 1. So that's the value of infinity. Plus 2 pi imaginary f of phi i plus i pi. OK, because at r naught equal to 1, x just becomes phi. OK, sorry, at r equal to 1, x becomes phi. r equal to r naught is equal to that. All right. So now to this, you have to add. So that's all. So the bulk action is very pretty. It's just a sum of two pieces. And now you begin to see things fall into place. This term is divergent. There's an r naught here. And this is going to be canceled. But let me spend two minutes on that. This x star plus something over r naught. So in fact, when you expand this at large r naught, you'll have two pieces. One is just f of x star. Completely divergent and will be renormalized away. But there's a finite piece coming from the first derivative of this. The first derivative of f is called fi. So what you're left with is minus 2 pi r naught into mf of x star minus mf of x fi. The first derivative times this, which is ci, and the r naught will cancel, plus 2 pi mf of phi. So that's the divergent part. And those two are the finite pieces. Very good. Then remember you have to add this thing. And this you can just evaluate. It's just 2 pi. That's the Wilson line. 2 pi qi ei star times r naught minus 1. Remember the gauge field went as r naught minus 1. And now you want to add. So there's also a divergent piece here. And there was a finite piece here. And remember that the real part of xi star is just ei star. That's in the attractor geometry. And also here, this imaginary fi at xi star. So that's just the attractor. So xi star means attractor. So you're asking what is the imaginary part of fi at the attractor point. And that was the attractor equation, which was just qi. That was the second attractor equation. So put all this together. You'll see that there's some divergent part and a finite part. And you have to add a s boundary, which is minus 2 pi r naught times qi times the real part of xi star plus imaginary of f of xi star. And then you can easily check that it's finite. I've been fairly careful with the concentration. You should be able to check it. But you can also kind of see that it's supersymmetric. Because you see the bulk action already is supersymmetric by definition. Up to this boundary piece. The boundary piece is being cancelled. There's only that thing remaining. So the thing that you have to check supersymmetry of is the Wilson line itself. So the Wilson line, as written, is not supersymmetric. But this is the same fix as Diego made, Maldesena made. Maybe Su-Jong. I don't know who did this first. It's that you just add, essentially, delta Su-J of a plus x is 0. This a varies one way, x varies the other way in the same multiplet. So that's what this thing is doing. So at the end of the day, what you get is s renormalized is minus pi qi pi i plus 2 pi imaginary f pi i. So that's s renormalized. Let me box it in a second. So that means that z ads 2 is now integral product i equals 0 to nv d pi i. So here there might be another measure there, which I'm not talking about explicitly, times z1 loop at. So that's the main equation. So you've essentially solved the problem. You've reduced the problem to an n plus 1 dimensional, nv plus 1 dimensional integral. It's an ordinary integral now. And s renormalized is given by this. There's some z1 loop which I haven't done yet. This is discussion for tomorrow. But what has happened is the ci and the ei. Remember this was an ei that becomes phi i. So it's an ei here. Sorry, what? On what? It does. So sorry, why does or why it doesn't? No, no, this is s. You do an integral. Lagrangian doesn't. But this is the integral of the local counter term. It's the integrated action I'm talking about. The induced metric is just the length of the boundary. That's all I'm saying. The Lagrangian is some local counter term that I'm adding. The length was just this. The length was that. So z1 loop I'll talk about tomorrow. But notice what happens already. That the classical problem was also written like this. In terms of exactly this. The classical problem was written in terms of function of this type. The entropy which I talked about yesterday was exactly of this type. It was the Lagrange transform of the imaginary part of the pre-potential at the attractive value. The quantum problem becomes an integral of that on this phi i. But now this phi i, this was actually a guess made by, based on work by Bernoulli-Witt and Cardoso and Mohopt and Capelli. Oguri-Stromingenwaffe made such a guess. And this is kind of a derivation of that formula. I want to say one more thing before stopping, which is it won't take too much time. Maybe two minutes about the action again. So far we took to a two derivative action. I just wrote down the action yesterday and I said think of that action. But as we know and as we already discussed. So this discussion is 10, 10, 21, 50 and 13, 0, 6, 3, 7, 9, 6. So in strict theory or something like that, or if you have seen examples, this pre-potential will look like this. There's a cubic divided by x0. If you've worked with this, you'll immediately recognize this. And that's the thing that gives you the two derivative action, this cubic potential. But the fact is that the formalism actually allows for arbitrary higher derivative actions of this type. As long as you have a certain type of f. So yesterday I said f must be homogeneous of degree 2, like this one. But in fact, you can add by a formalism, slight extension of the formalism, which I'm not going to talk about. You can add terms of this type. These are linear. This is homogeneous of degree 0. And the way the homogeneity is absorbed is by introducing another field here called w, which has weight 2, if I'm correct. I'm being a little bit schematic. So please forgive me if the details are not correct or correct me if there are experts. So that has degree 2. And that has to do with, actually let me call it w squared. That has to do with four derivative terms in the action. So this one had to do with two derivative terms in the action. That's four derivative terms. And in fact, you can just put an infinite sequence. You can put whatever you want. So there's something with six derivatives and there's an infinite sequence. As long as the whole thing is homogeneous, it all works out. So the formalism of n equal to, of shell formulation of n equal to supergravity easily allows for inclusion of this kind of higher derivative terms. You just change f. And in fact, in string theory, these things have very direct meaning. So this one is some intersection matrix of the Calabiow. This one is some second chance class of the Calabiow and so on. There's some function. So the f terms are of this type. So in super space, it looks like some d4 theta times f of some super field x. So this is in super space. That's how this formalism is made. So as long as f has the correct properties, this is guaranteed to be super symmetric. So all you have to do is to write down what f is. It's an infinite series. In string theory, this is given by some topological string and so on. So that's the homomorphic pre-potential. But the fact is that in everything, my whole derivation, I only used of shell q. I never used what the action actually was, although I said keep the two derivative thing in mind. The whole thing goes through with this f. It doesn't matter. So this equation still holds. So this is a way of taking into account the whole class of higher derivative terms. And now, this is not the only type of terms you can add to the action. Like in super symmetry, there are d-type terms. Let me go here. So there are d-terms, which are d-terms in super space. So these are holomorphic functions of super space. d-terms have this type of, it's like the scalar potential n equal to 1. It's a non-holomorphic function. And something like this, you can have xi, xi bar, for instance. It's a non-holomorphic function. And there is no known classification of all possible d-terms, but there's a large class of d-terms that people have written down. So this is this paper by DeWitt and Kathmanders, I think. Yeah. And one can check. So that's what it's done in this paper. That's the paper with Val. So again, because of the localization, I never need to know what the action is. This formula holds. I just have to evaluate the renormalized action, whatever action is given to me. So I can just take this, take the same formula and just evaluate this. Instead of the f-term action, I can also add this. And what you get is that s renormalized of the d-term just vanishes. So I'm essentially done. So that means that this formula is actually really true. So it started with all kinds of assumptions. So let me just tell you the, sort of, turn it around. So start with some string compactification, if you want. So at low energies, it has some two derivative action given by an f-term which I wrote yesterday. Then at higher and higher energies, it has all kinds of corrections. So there's f-term corrections and the d-term correction. You include all that and I want to do a path integral with all of that. Turns out the d-terms actually don't matter. And the f-terms do matter and that's captured by this formula. So this formula is not some two derivative formula. It's a non-linear formula. So that's getting towards the exact entropy. Tomorrow I'll finish this program. Tomorrow I'll discuss what z1 loop is. And maybe if you ask me in the question session just now about the singular metric, then I will answer this. All right, thank you very much.