 So can you hear me? I think so. OK. Now you have to turn it off. No, it's turned off. I can hear the, good. First thing, thank you for the opportunity to lecture here in a very nice school. So what I'm going to tell you about or try to tell you about is an interesting connection between black hole entropy and Schrodinger's theorem. So I think that by now, we all know that entropy of a black hole, or Beckinsen-Ockey entropy, has this formula here, area of the horizon over 4g with h bar equals to 1. And this is really dimensionless, OK? There's no Boltzmann constant. And on the other hand, Trim Simon's theory is a three-dimensional theory on compact manifold with Lagrangian as the following form. And I'll try to explain how we can use Trim Simon's theory to compute black hole entropy. Now the first thing you might be wondering is that, well, this is a very geometric quantity, right? It depends on the area on the metric. So how can we describe black hole entropy with Trim Simon's theory if there's no metric here? It's completely topological, right? It's topological theory. But I'll try to show that in these lectures. The Trim Simon's theory can be used to probe the black hole at the microscopic level. It will know about some discrete properties of this black hole entropy. So what do I mean by this? In the full quantum theory, as Samir explained, this S, which will be the logarithm of some number omega n, in a certain regime, will be described by the Backinstein-Ockin area formula. But this number will suffer corrections schematically of this form. So this is dimensionless. I can expand in this parameter. OK, you'll have things like this. So this is a very general. So this will be a series of perturbative corrections to the entropy. But in order for this to be a log of an integer number, you need that you also include non-perturbative corrections of this form, where these gammas, a and b, since these are constants. So it depends on the theory. So if you're a god and you want to compute the full black hole entropy, you'll find this type of corrections. What I will explain or try to describe is how this Trim Simon's theory can help you figure out all these numbers, including these non-perturbative corrections here. And what I'll try to show is that, once you do this Trim Simon's analysis, in principle, not completely, but you'll be able to recover the full integer, or the entropy of the log of an exact integer. Proturbative series. Yes? No, no, not a little bit. I was looking at the clock. That's right, it's a little bit. Yeah? Proturbative series. Is this border solved or not? I don't know. But I want to check. In this case, it will be very simple. And it will be, it's not even divergent, asymptotic. It depends on the limit you take. In one particular limit, I'll show that it's asymptotic, but there's a closed form expression for that. So it will be a border or some more. I'll show the formula. I'll show the formula. And you have the exact formula. OK, so. And then I will describe some connection with a similar story by reducing a five-dimensional theory to a four-dimensional theory. OK, so a very, very brief outline of what I'm going to do is that in the first part, I'll basically describe what you are asking for, the exact answer across the topics. And the second part, I will review some concepts of Schoen-Simon's theory. Theory, flat connections, topology, knots, all this story. Review very schematically. And the third part, I'll try to do the computation with Schoen-Simon's and compare with some exact answer. So I'll basically describe some SUGRA solutions and do Schoen-Simon's computation, OK? So before going to the microscopic, I just want to describe a bit of the context of the politicals I'll be considering that are important for this computation. So Samir talked about the AdS2 times S2 in an engaged program. So everything we'll say it's supersymmetric engaged supergravity, OK? But the interest of this computation is actually, so this is 4D, all is having in my mind five-dimensional. So there'll be an additional circle in the story. You can think of this as the anterior circle. Now it happens that for interiors with eight supercharges, solutions that preserve the full supersymmetry must be of this form with this circle actually fibered over the space, AdS2 times S2. So basically, I have a family of black hole solutions, on which case you have one extremum. One side of this family solution you have three sphere, this is S2 times S1. So a circle fibered on S3. And on the other hand, you have the AdS2 times S1, where S1 just fibered over AdS2, and it's two sphere. And this space here is locally AdS3. So it has a negative cosmological concept. And in between you have a complicated configuration which I will not be considering. So everything I'm also going to say is about this problem, this type of configuration. Because then you can just reduce on the circle and obtain the solution. And why I want to do that, because this is locally AdS3. And then I'll use the AdS3 safety correspondents to compute some path integral on this space. OK, I think some time. Yeah? How about a black string? Was there a black hole? There's always an AdS2, there's a black hole. Like a bit is a black hole, but it's rotating. Yeah. So here's the non-rotating case. And then you have rotating black holes. So this here is twisted. So it's not really. And because the circle is fibered over S2 and AdS2, there's a rotation and a charge, a charge. This case is particularly useful, as I'm going to mention. So you will have this metric here. I'll explain this in more detail perhaps tomorrow or on Saturday, but just to give you an idea. So this is the black string. So you have a metric of this form. And you have the circle, which is fiber over AdS2. So this close the client gauge field has support on this space here. And the solution in 5D is very simple. So you have this metric. And all the gauge fields are just flat connections and carry some phi i's here, some was aligns. And then you have also magnetic fluxes on the sphere. So I'll put here A0 such that the A0 equals the volume form on the sphere. So it's the rack gauge field. What is nice about this solution is that, also the dimensional reduction, for example, you just put the phi i's and then di's plus A. And then you have to subtract the phi i, A. Then you have P i times this cosine of theta phi. So this is the gauge field that Samir is working with. This will be the four-dimensional gauge field. Then it becomes, sorry, this is, yeah. So this close the client gauge field, hey, you can see this is a phi 0 minus 1 dt. This is the coordinates that Samir was using this metric. Then you get exactly the gauge fields that Samir was considering. This is just the dimensional reduction of that. But the key point is that, why I'm saying this is because in phi d, what you'll have, the solution is basically this locally ADS-3 solution. And then all the gauge fields have this flat connection part and the magnetic part. And all the scalars in phi d are just constant. And then you can show that the metric and the gauge fields are solutions of 3-D gravity. So these are solutions of 3-D gravity with a constant plus there will be some ability gauge fields and also will be some non-ability gauge fields. So sorry for being a bit sketchy. So for the moment, it's just motivation. The, I'll be more precise next, bigger, OK. What here? So it's like the minimal, the minimal theory can construct such that this is a solution, OK. That's when the Schrodinger-Simons theory will come in. OK, but let's get to something a bit more precise. So this ADS-2 times S1 is somehow can see as a states in ADS-3. So it's a configuration in ADS-3. So you'll be looking at states on the two dimensional CFT. So what I'm going to argue is that the number of black hole states will be equal to the number of dps states in a 2D super conformal field theory. Now it was mentioned before that to count dps states, well, it was mentioned in 3-D. But I'll do the same, I use the same idea, which is to use what is called alytic genus partition function. So you take a trace with minus 1f for the f from your number. Then you have your on the cylinder, you have the Hamiltonian. L0 is the reservoir generator. You have the Q bar part. And then you do this chemical potential for the RC motor generator. So this F will be some combination of L and Jr. So I'll be using a notation from 2 comma 2 super conformal materials. So these are the RC motor generators. Why do I want this? Because in many cases, you can compute thislytic genus when the theory is free or with a couple. So you just have to solve the Hamiltonian, count the states, dps states, and so on. But because it's a protected quantity, can I extrapolate this partition function to a regime where the theory is strongly coupled, which would be the black hole? It's the usual story. Now in my ask, if thislytic genus really computes the entropy of a black hole, I'm not sure if Sameer is going to explain that. I will not explain that too. So I'll just assume that you can. Yeah, there can be a difference. Central charges, sorry, left and central charge. So just a small comment. This is completely different from computing. That's a statistical partition function where you do not put the minus 1f, you just keep the 1 here, d4, q bar, l bar, and so on. So this is not protected. But that's what you should compute, actually. It can be compared to a black hole entropy. Now because you have supersymmetry, roughly speaking, the input is supersymmetry. So in this case, it'll be like on the two sides. This operator is roughly speaking q closed. This is the supercharged. This is the q square. And so when l not bar, c r to d4. So you have a state for which this is different from 0. As you know, you can always act with a q because it commutes with that. And you find a pair of states with different fermion numbers. So when this is different from 0, you find a grading by the supercharge. So they carry different fermion numbers. And the q bar, the contribution will cancel because of this minus 1f. So the usual with an index argument. So this object ends up to being allomorphic, depending only on q. So this chi tau z then becomes this trace jl, where I just put l not minus c r to d4 over d2 d4 equals 0. So it is allomorphic. One important comment, I will actually be super interested in 0.4 sub-comform of filterias. So just a comment here. I'll be interested in 0.4 sub-comform of filterias as I relate it to the MSW CFT that corresponds to basically M-fibrane wrapping divisor in Calabi-L. But in that case, sorry, it's a bit more complicated. It will take me a lot of time to explain. Yeah? OK, so this is 0.4 CFT, conformal filterias. This is the MSW, M-fibrane wrapping divisor in Calabi-L. But in that case, the story is a bit more complicated to analyze. It will take me a lot of time. So I will restrict to 4.4 or 2.2, where these objects are much simpler to analyze and tell you what I'm looking for. Now these objects, because they're computing on a conformal filteria, super conformal filteria, it will be a reparameterization variant, including, so this object chi tau z is a modular object. So what is a modular object? Let me just give a simple example. So here you have two complex variables. But, for example, take just one variable. This is a function. So a modular function is a function that takes tau in the upper-off plane to the complex numbers. So it's just the fact that the imaginary part of tau is bigger than 0. And it has the following property, that phi of h tau plus b tau plus d omega phi tau, where this a, b, c, d is an acetyl-z matrix. So everything is integer and determinant 1 or 4. Sorry? I think in the mathematics section, strictly speaking, this would be called a modular form. And a modular function is when w equals 0. Yeah, so that's true. So thank you. And because if you think, if you put a function on a CFT, you expect it to be a modular variant, reparameterization variant. So what Sameer is saying is that this weight tells it as some sort of anomaly. So you should call it as a form. So there's some sort of volume dependence that will scale with this weight such that everything is invariant. But OK, one nice example is a modular form that you know is the dead kind function of tau. Discriminate function is the following form. This q and 24 is a modular form. It has weight 12. But these things, I suppose you have seen before. Now, when you have one more complex coordinate, you have additional transformations. Oh, no, no, this is on the cylinder. And then, OK, you can see the torus by the time. And then it has a complex structure. Or because on a CFT, the boundary of space will be a torus. So we will be putting the conformal filter on the torus. But what I'll be interested is on Jacobi forms. So it's just a generalization of what I told you when you have a complex parameter. Yeah, no, no, no. You can put many boundary, up to a conformal factor, you can put a human surface. So a Jacobi form is now an object which has some weight and some what is called index, which I'll explain. This object transforms in the following way. So you have, again, the weight. But now, because of this z dependence, you get this strange phase factor. Again, this A, B, C, D is always a set of z. But besides this transformation, you have what is called elliptic translation. So we have what is called elliptic. Now, if you fix the tau, then you shift z by lambda, tau, and mu, where lambda and mu are just integer numbers. That's what it called translation. But along the elliptic variable tau. So this transforms as, so now we can use these properties to show that these functions phi omega m have a Fourier expansion. I can just leave it as their size. I mean, it's obvious, but it's a Fourier expansion in tau. It's a variant of translations. It also has a Fourier expansion in z. So people might be complaining, which are experts. m is always an integer. Could be alpha integer. You'll have some minus sign. But I'll keep it just an integer. So you have these Fourier expansions, these Fourier expansions, which means that you can expand phi omega m in the following way. And for simplicity, I'll restrict this n to be bigger or equal to 0. In some cases, you can have negative here. It'll be called nearly allomorphic. But for what I want to show, I'll restrict to this n to be equal to 0. And one of these conditions is called a weak Jacobi form. So everything what I'm trying to do is give you definitions. And then I will show you what I want to extract. I mean, I'll use all these definitions to extract these Fourier coefficients, which count the black hole states. So for that, I need to define a polarity. Polarity is just this combination n minus l square over 4m. So when you have this condition n bigger or equal to 0, you can have states for which delta is negative. But they are bounded by minus l over 4 besides all the other states, which delta bigger or equal to 0. That's basically the weak condition means that you have delta where that can be positive and negative. And these states with delta negative are called polar terms, which will play a crucial role in this story. Maybe I'm going too fast. I don't know. No questions, OK? Yeah? You said it's an integer, but it's always going to be positive? Yeah. So if the speed is OK, we're going to speed up. It happens that because the analytic translations, you can write the Jacobi form the following way as h mu tau tau z. So and this is some of our mu. I'll just explain what is that. Actually, this mu is just the following. It's a cos-satore representative of z mod 2 m z. It's going to be, for example, mu from m to minus m, OK? So choice. This theta, sorry, to m and mu, this theta and mu are the data functions. And they have the following expression, just a sum over this lattice, q mu plus 2. Basically, it gives you mu, and then just shifts on this lattice, OK, by 2 m, 2 m n, 2 m n. And this h mu tau, which are allomorphic in tau, are called vector value modular because there's only tau. And forms because they have some weights and there are multiple formations. This h mu tau can be expanded in a free, there's a free expansion for h mu tau, but there will be a slight and very important difference from the rest. So it depends on mu. And r is an integer. They are equal to 0. But this a may not be integer, so it can be fractional, OK? It's an important difference. I'll just explain what is that. I'll just explain just the next few lines. So what I want to compute is the c and l, right? So I have to do two Fourier inverse transforms on d tau and d z just by m z, OK? That's the definition. These are just for your integrals. But now this is h mu tau theta mu z as a sum over mu. And now it's easy because all the z dependence is on theta. Then you can do the integral from here, this y. So the integral over z projects on sector mu, right? When this integral tells you what is mu and is mu, it's nothing more than l mod 2m. The remaining integral on tau, it tells you that n means that n must be equal to what? a plus r plus r. And then there's a q coming from theta, this one here, which is the l square. Oh, sorry, there's some 4m here missing. Plus l squared over 4m. So then you find that a plus r equals minus l square 4m. And you can show, I believe, this exercise that, so a plus r is just a polarity, just a polarity of the state. So these functions h mu only carry information about the polarities of the states, about this combination of n, l, and m. This seems very abstract, but I will give examples. I really have to speed it up. So to come back to your question, this theta and mu transform under the following way, under model transformations. So they're not really modular. They come back to themselves, but there's a matrix mu and u, which relates they rotate the data functions in that way. And this theta and mu equals 1z. So these get these phases here. And basically, it's this phase that will cancel this a here that I told you can be fractional so that you have allomorphicity. In the most general case, so if you pick an arbitrary model transformation, there's some matrix here that depends on a, b, c, d, mu, and u, and theta of m, mu, tau, z. So this matrix here is called a multiplier matrix. And now it's very easy to see that if you want phi, if you want the combination, so if you want the phi omega m, so this combination, h, mu, tau, theta, m, mu, tau, z. To have the properties that I mentioned before, model transformations implies that this h, mu must transform with an inverse matrix so they cancel. So this h, mu are what's called vector because of that property. So when you do a model transformation, it was d minus 1, so there's a minus 1, because the functions have weight 1 off. And then they have to transform with the inverse matrix. Is it mu, mu, h? Sorry, mu, tau. That's why they're called vector. They're following model of forms. OK, so it depends on the case. Then you can compute them. It's hard, but you can compute them. So I mentioned the MSW and the MSW just 10 seconds. In the MSW, you have an h, mu, tau, but the data functions are not allomorphic. You have some tau and tau bar, z. That's why I didn't want to explain detail that. And that's because in the MSW, you can have BPS states. So this is for experts. You can have BPS states due to nonlinear supersymmetry. So you can just excite the right side. You can have tau bar, but that's BPS states. So there's a theta, which depends on tau and tau bar. But what you want is this h, mu, which is allomorphic. And that contains information about the Fourier coefficients. So when you do the integrals, the Fourier integrals, I raised that part, that there were the two integrals. So the integral over z projects on the L sector. So you get a theta function. But all the coefficients of the theta function are something like this. So all the coefficients are 1. And all the information of c and l can be recovered from the h, mu. So roughly speaking, the h, mu, tau is enough to compute c and nl. You can just do it as an exercise. It's very easy. And then you also learned that in conformful theory lessons that something is called a cardi formula. So we can extract the asymptotic growth of the degeneracy of statistical partition function using modularity. And you can use the same ideas here to extract c, mu of r using the model transformations. Actually, there is an exact formula for these coefficients. c, mu of r. I'll put just n here. That formula is called the Hadamaker expansion. So basically, you can write the exact Fourier coefficient of h, mu, tau in a very complicated way, but very useful, too. So there's a sum over the c, mu, such that n bar minus n squared, small than 0. So it's the first sum over the terms which are polar. That is, you have the h, mu, tau. There are some terms here which have negative powers. And then you have the positive. So we just pick the four coefficients of these negative powers of q and plug it here. And then there's a sum from c equals 1 to infinity. Because an infinity sum of these very complicated objects called occlusive sum depends on these parameters. So it depends on the polarity of these states, mu and n. There's a sum over the polar terms, this part here. And then there's some integral, which is a Bessel function. It's very important. So it's just the integral representation of what is called a modified Bessel function of first kind. It's a very complicated formula. It has an infinite number of terms, but it's absolutely convergent to a precise integer that you can show that. So what is this close sum and sum? Just the close sum and sum is a very complicated formula, but has roughly the following expression, c. And then there's a sum. So this is for fix at c. So the sum over d and c, which are a co-prime. Then there's a sum over a, such that ad equals 1 mod c. And now you see it appearing here, this multiplier matrix. So it's a very complicated expression, but you can show that this multiplier matrix has a form like this, it's a sum over phases. So close sum and sums are just sums over phases of this form. d over c, area over c, very complicated. So now, why do I like this formula? Because you can do asymptotic expansion, or well, not asymptotic, you can study asymptotics of this integral. It's very easy to do. So if you have that integral, you can just, so dt, some t to some power here, a over t. And then it's just going to be this by telepoint approximation. What you'll find is that this is the exponential of 4 pi square root of a modulus b. This is a leading term, and then I can keep its corrections to that. I'll just put, like, order of 1 over a and b. But this approximation is valid only for a much bigger than 1, and fix the a much bigger than 1, and then it can expand that form. I don't want to erase these formulas, but. So for each, so you have a sum over Bessel functions. So there's a c, and there's a sum over the polar terms here. The spectrum, I'll call it spectrum, of polar terms is finite. That means what I'm saying is that n bar minus nu square for m is smaller than, there's a maximum value, and there's a minimum value here, which you can compute, it depends on the Jacobi form. And so in this expansion, in this regime, you'll have terms which grow faster than others. For example, it's easy to see that c mu of n is something like c nu of n bar, for which n bar minus nu square over 4m equals the maximum value, exponential of 4 pi n nu square, maximum value. And then you have a series of exponentials up to 4 pi. Now you can guess it for m times the minimum value. There's a maximum that starts decaying and decaying. These all grow exponentially, but the maximum starts decaying and decaying. And then you have the c contributions over c. So we start with 2, maximum value, plus dot dot dot. Again, so this is the c equals 2, case, minimum value, and on, and on, and on, and on, c to infinity. To say what? To say that this term here, which is the leading term, is nothing else than the black hole entropy. So this is just an area you can check. If you check some years computation, this will be area over 4g. OK, since I have five minutes. But now you have the leading term, then you just get a bit of corrections to this. It's very easy. Or you can check Wikipedia about how to expand the best of function. It tells you what type of corrections you have. But you also have all these terms here, which, though they grow exponentially, they grow exponentially, but they grow much slower. Because the exponential is smaller than this one. And therefore, it'll give rise to non-perturbed corrections. If you take the log of this term, then this will become exponential suppressant relative to area over 4. So these are called non-perturbed in all these terms. OK, I'll try to finish. There's some normalization factor here. So OK, this is actually in this limit. OK, so thank you for asking. In this limit for which I'm taking n minus mu square much bigger than 1 with fix of m in this limit, this thing does not grow. It's just a number, order of 1. And then you have that expansion. But there's a different limit to this expression. So there's a different limit, which is very important, which is the limit for which, if I rescale all these parameters, m lambda m. And that means that I have a family of Jacobi forms. There's like different CFTs. This is like the n of the gauge group of the CFT, roughly speaking. So if you scale all these terms like this, n, n, n, lambda much bigger than 1, what happens is that you still have these exponentials here and so on till infinity. But these coefficients that multiply the exponentials start growing. So here will be like order 1, for example. And then this thing, which was very subliving, can start dominating the expansion. So these coefficients here, there's some coefficient here in front that might grow exponentially like lambda to some power when you do that. So it's not true anymore what is perturbative and non-perturbative. So you have to redo your computation and so on. And actually, it's a very good question, is that, well, if in this limit, this term is area over 4, you may ask if in this limit you still recover the same leading answer. It could have happened that these terms grow so fast that they spoil completely this formula. So you have something like this. C u, bars, bar, I feel like 0, up a dot. Then you have this C u, bar, and bar will grow like lambda 4 pi. So the question is that how these numbers grow, such that this term is still the leading term. That's what you want area over 4. And this is an open question for a general super comfortable field here is, how does this grow? What I can tell is that for simple examples, OK, if I have how many minutes? One minute. OK, I'll try to. Now, it's nice that for T2 times T6 coefficients or T2 times K3, like what Samir might be talking about, these numbers can grow exponentially but not fast enough. C u. So grows can grow exponentially but fast enough. And because it doesn't grow fast enough, you have still the black hole entropy area formula. And this is the limit that I think Samir will be studying. Or it's Ashok's limit when he studies the quantum entropy is when you take this limit of charges. And in this case, these things don't grow enough. But for genical Abial, the problem is not answered. It's not known how these things grow in generality. OK, I think I'll finish here.