 Good morning. We have a busy day today. Three lecturers, three different subjects, and then also a Q&A session at the end of the day, where all the lecturers are supposed to be present and to answer the most burning questions about the modern or condensed matter. Do we still have a volunteer to moderate the afternoon Q&A session? The person who was doing it yesterday? OK, we don't need to decide it now. But let's say after lunch. And with that, we have our first lecture of the day, Professor Alex Kamenev. We'll talk about SYK model. All right. Can you hear me? Cool. Thanks for coming and thank to organizers for inviting me. So they asked me to talk about SYK model, which is, as you can notice in archives, is a popular subject last couple of years. Some people say it's a model of a black hole. I'll talk about it a little bit later. Some people say that it's a model of non-fermic liquid state or incoherent metal. I'm not sure it's any of this. But in any case, it's a cute model. And there is a lot of interesting physics inside of it. So maybe it's not a complete waste of time to learn about it. But I'll mention all these applications. So I'll do it as a blackboard presentation. Otherwise, I will go too fast. But I will need few slides just to show you numerical data and some history. So as you can guess from the name of the model, it was suggested by, oops, it doesn't work too far away. Yes, it was suggested not by these people, but by completely different people. It was suggested by Oriel Baigas and Flores in back in 1970s. And then these two gentlemen, French and Wong, wrote a very similar paper. So their motivation was coming from nuclear physics. But nevertheless, the model which they considered was precisely equivalent to what is now known as Sajidif Yekitaev model. Now, Subir, Sajidif, and Jean Vuillet had no idea, apparently, about those old papers. So they had completely different motivation. They thought about spin-fluid states in a random Heisenberg magnet. And that's how they came with the model. So while I'm mumbling here, you can read their abstract. So apparently, their main result is that susceptibility of this random Heisenberg magnet behave logarithmically, and that's a signature of a spin liquid. So if I forget to mention how this logarithm would come about, remind me, I'll show you. It's one line calculation at some point. All right. So I'll tell you about the model. I'll tell you about symmetries and infrared behavior of the model. Maybe tomorrow I'll come to tell you a little bit about gravity side of the model. We'll see how it goes. No, right. So the model is extremely simple, and that's why it's probably worth thinking about it. It's just interacting Hamiltonian for fermions, chi i, chi j, chi k, and chi l, interacting with some matrix element, j with four indices, and no kinetic energy. Just only interactions, and that's it. Now you should say something about this matrix element. And the simplest thing you can say is that it's just random. Now what means random? Let's say that in average it's 0, and it's Gaussian distributed with a second moment, which is given by a constant j square. So this j is the only energy scale in a model as it is written. There is nothing else. Now another important parameter here is capital N. That's number of our fermions. And you will see that good idea is to scale this variance of j with n cubed. I'll explain in a moment why it is a good idea. Now let's see. So you can think about these guys being Majorana fermions. And that's what Ketayev suggested. Before him, all the people I mentioned thought about them as being complex fermions. And in that case, you probably want to say that it's c dagger, c dagger, cc. It's almost equivalent. At least I don't know any significant difference. Yes? Yes, so the question is, if I use complex fermions, whether it's the same j, it's almost the same. Now in Majorana representation, this j has to be anti-symmetric with respect to any pair of indices. Anti-symmetric with respect to ij, kl, you know. If it is complex fermions, then it's anti-symmetric with respect to c dagger, c dagger, cc. But it's not necessarily anti-symmetric with respect to. Well, from the point of view of ensembles, they are, again, almost equivalent. I'll mention in a sec where the difference may be. So if it is not clear in five minutes, then ask again. All right, so again, it's not a big difference. If it's complex fermions, then it's conserved number of particles. If it is Majorana fermions, it's conserved parity, but does not conserve number of particles because you may have here, c dagger, c dagger, c dagger, c dagger, or three daggers and one c. Now in a modern quantum computing world, you can think about this model as being sort of a quantum dot with Majorana quantum wires coming to this quantum dot and there is a Majorana fermion sitting at the end of each wire, and they somehow interact through whatever interaction you can imagine. Why it is random is another question, but again, since we don't have anything better, let's think about it. OK, so before I come to analytics, let me mention numerical things. Now since it's a Majorana fermions, they obey this Clifford algebra. So they all anticommute with each other, and if it's the same i, then it's just square to one, or one half. Now you know that if you want a matrix representation of a Clifford algebra, that's called Dirac gamma matrices. And if you need four or five of them, you want to have matrices four by four. If you want two or three, you can get away with matrices two by two. In general, you need matrices with the size of two to the power n over two. So all my operators are going to be represented by matrices if n is not even, then it's an integer part of it. All right? So you can form these matrices. That's very easy to do on a computer. You can form a Hamiltonian, and then you can brute force diagonalize this Hamiltonian. So you will have a spectrum, which is a many-body spectrum. And this is a density of states of this many-body spectrum. So now let me walk you through it. So if you don't have any interactions, that energy of all states is just zero. There is nothing. So all energy levels are completely, and there are two to the power n over two of them. So in this case, n is 32, so there are two to the 16 energy levels here. Yes? Yeah. In the previous slide, what is the class T bionara wire? Class T in the bottom? Class D bionara wire. Oh, class D. Doesn't matter. Doesn't matter. It will take me too far away. Sorry, it's left from our talk. It's a myrana wire without time reversal symmetry, which is what the experimentalists do anyway. Right, so you diagonalize it. Now they repel each other with many-body levels. They repel each other, and now they are not all at zero, but they form a band around zero. Now that looks more or less like Wigner-Dyson semi-circle, right? There are two difference here, more than two actually. So one is that in the case of Wigner-Dyson, we typically try to think about what happens in the middle of the band. That's where Fermi energy is apparently sitting, and we somehow focus here. Now in this problem, the many-body ground state is here. It's just the very last eigenstate of my Hamiltonian. So at zero temperature, I basically interest only, my interest is mainly in the very last eigenstate. If I have a low temperature compared to this ultraviolet scale, then I'm interested in a corner of the spectrum. Yes, there is a square root singularity at the edge. I will talk about it. Yes? Well, I mean, there is a standard routine in your computer which allow you to generate the set of gamma matrices, whatever dimension you want. Yeah, it's a standard thing now. But still, going much beyond 30, people do 34. Some people manage to do 36. But you probably can't go much farther than that. Now, you can do some analytical theory, which is, I will not do it here, but it's done by Garcia, Square, and Verbersche. So then you will see that depending on how many fermions you take, your density of states changes. So this is 32, which we already saw. But if you would be able to take 64, nobody can. You will see that it's a pretty good Gaussian. Still, at the very end of this Gaussian, there is a square root singularity. And the physics is around here. What else should I say here? Yeah, there are exponentially many energy states here. So the density of states is exponentially large. And that means leads to finite entropy, even at zero temperature, unless you resolve energy scales which is comparable to level spacing. I'll mention it again. No, right. So the next thing you can do, you can ask what's the level statistics, namely, you can take the distribution of consecutive level spacings. And what you will find that it's perfect Wigner-Dyson. It's one of the best Wigner-Dyson distributions you can imagine. Nobody quite actually understand why it is. Because if you think about this huge exponentially big matrix, it's an extremely sparse matrix. There are that many matrix elements in this matrix. That's a matrix with a size of 2 to the power n over 2. And only very few elements have this J i, J k l, n of order n to the 4 matrix elements which are not 0. So you deal with exponentially sparse. Yes. Yeah, very good question. So question is, this level statistics is taken in a bulk or in a tail. So I think this particular picture taken in a bulk. But actually, if you look for the distribution between ground state and the first excited states, it's also perfectly Wigner-Dyson. Two. Full disclosure, you have to be careful. Because the many body space actually splits into even and odd subspaces. So this Hamiltonian can source parity. So therefore, part of the many body space, so there is 0 here, 0 here. If you properly select your basis, the actual size is 2 to the power n over 2 divided by 2. Now, if you take spectrum within one of these parity subspaces, you will have this Wigner-Dyson, including the very bottom. All right. No, chiral symmetry will be off diagonal. OK, you can ask what kind of Wigner-Dyson. So in this case, it's GOE. Now, the answer is sort of funny. It modulates with the period 8. So it's all GOE, GOE, GU, GS, symplectic, symplectic, unitary, orthogonal, depending on purity of N, capital N, modulo 8. Now, I will talk only about even N. So for even N, you see it's a period 4. Artogonal, unitary, symplectic, unitary. And that's because this Clifford Algebras, that scientific notation for them, has additional symmetries, which play a role of time reversal symmetry, and crumples the genus. That's a good question. I don't know. My understanding is no, but I may be wrong. The question is about both periodicity of topological insulators with dimensionality. Yes? Look, for any capital N, or at least for any even capital N, your Hamiltonian splits into even and odd subspaces. This is not number of particles. This is size of your matrix. So number of particles is a different thing. And it's not conserved even in Majorana case. In complex fermions, it is conserved. No, right. So that's observations. And there is symmetry understanding why it may be the case. But actually, nobody knows why Wigner-Dyson is applicable here. Some people say they do, but actually, I don't think so. Anyway, let's see what. Right, so you can look for what is called sigma 2, namely variance of number of energy levels within a window of size epsilon. And this epsilon is measured in mean level spacings. So according to Wigner-Dyson, it should go like a logarithm. This RMT stays for logarithmic prediction. If you do numerics, you see that it follows this logarithm very precisely, but at some energy, it starts deviate and shows something else. In mesoscopic physics, this is known as Alshulev-Schklovsky statistics. And this energy is known as Stavlis energy. What is it for Svk model? Again, pretty much nobody knows. If you go to very large energy, it will be Poisson, and it will be linear. So the question is, what's the structure of the weight functions? Yes, very good. So Volodya is the best expert in that. That's a good question. Somebody must have done it, but I can't recall it. Yeah, so the statement is that if it go to linear at large energy, then the slope of this linear dependence has to do with fractal dimensions of the weight functions. Yeah, I don't know why nobody. Anyway, good question. OK, let's see. Yeah, that's probably not what I want to show you. No, right. So I will go now to analytical theory of this story, and I switch off my computer. So if you have any questions about numerics so far, please ask. As far as I understand, there is basically no difference between complex. So Felix Israelov and company, yes. Yes, right. Yes, so the remark was that in nuclear physics, people still continue to think about this thing, and they have many results, maybe even some which I forgot to mention. OK, so we want to do a theory of this thing. The Hamiltonian, once again, is 1 over 4 factorial, sum over i j k l up to n j i j k. And I will think about them as being Myrana, but that's not very important. j in average is equal to 0, and j square in average up to symmetry properties is 3 factorial. j square over n cube. Again, I promised you to explain why n cube is a good choice. I will do in a sec. Now, some people generalize the model taking instead of four interactions, taking cube fermions. So in my case, q equal 4. And the common notation today is Sykq. So I will mostly talk about Syk4 in which notation. But practically, all the results are easily generalizable to any q. I will try to mention it if I forget, ask me how to do it. No, right. So now, in terms of energies, the only energy scale is this capital J. So there is a scale which is 1. And this is my ultraviolet energy scale. Now, down to smaller energies, there is a whole zoo, probably. So for one thing, there is an energy scale which is 2 to the power minus capital N over 2. That's where the mean level spacing is. And that's where numerics, or some of the numerics which I mentioned, is done. What we will see in a while is that there is at least another energy scale which is 1 over capital N, or maybe 1 over capital N log N. There are the quantum effects. So this is mean level spacing. So this is quantum. Physics starts at this energy scale. So here it is classical. You will again see in a sec why I say so. And possibly there are some other energy scales in between. Again, nobody quite knows. Cool. OK, so now, how do we go about the theory? So let's say we want to calculate a partition function, which will be e to the minus integral from 0 to beta Hamiltonian d time. Time is imaginary. Now, it's still a random thing. So we want to average this partition function. Now, you may know that averaging partition function is not such a great idea. And you better average logarithm of the partition function. So we will need replicas to think about that. So we will have z to the power N and product of A from 1 up to little n, h a. Now, eventually, this n will have to be taken to 0. But for a while, I will not worry about this too much. So that I can write as a functional integral over my eranas, which has index a and index i. i goes up from 1 up to capital N. A goes from 1 up to little n. And then e to the minus integral z tau from 0 to beta i z tau. No, I'm too fast. No, no, no. Sorry, sorry, sorry. One second. Yes? Yeah, I have finite temperature. Well, yes. Yes, temperature, of course, is on our scale. So in essence, you should think where the temperature is to understand what's the physics. No, right. So now I want to average this stuff over my disorder realizations. So I will have to average this on disorder if I have e to the minus integral or plus d tau j i j k l chi i chi j chi l with the same replica index a. And I average this like this. Then I will find e to the minus j square. I will need to look what's the coefficient. n cubed over 8. Then I will have double integral d tau prime. And sum over i j k l of the capital N chi chi chi chi a a a a chi chi chi chi b b b i a k l i j. And this is a time tau. And this one is a time tau prime. So I'm doing it a little bit fast. So if it's too fast, then slow me down. All right, so that's a beast. We can make it a little bit more user friendly if we introduce notation, green function, which is a matrix in replica space. And then time space is equal to 1 over n sum over i from 1 to capital N chi i of tau and chi i b tau prime. So this is just an notation, which is reasonable to call green function. And in terms of this green function, what I'm going to have is e to the minus here, j square over 8 d tau d tau prime j a b tau tau prime to the power 4 sum over a and b. In more general model, instead of this 4, I will have q. You can easily see that. All right, cool. So now what I need to do, I need to tell my partition function that I want to use this definition. Now how do I do that? I will multiply this by 1. That's my constitutional right. And this 1, I will say that this is integral over this matrices j of a delta function nj minus sum i. Integral of a delta function is 1, so nothing bad happens so far. And then I will write it once again. I will say that this is integral over dg and d sigma. Sigma is a conjugate matrix. And I will have an exponent. And how do I want to write it? 1 half integral over d tau, d tau prime sigma tau prime tau a. Yeah, I don't do anything high tech. I just write this delta function as an exponent. Now with that, my partition function will look as an integral over dg and d sigma. And then now look from now on, all my fermions are purely Gaussian because the only thing which depend on fermions is this, right, and this. There is no more fermions because my interaction Hamiltonian is now written in terms of g. And then I have this guy in terms of g and sigma. So sigma and sum look similar, so I'm sorry for that. So anyway, my fermions are now purely Gaussian. I can integrate them out. That give me determinant actually since it's my run of fermions. It's a Phafian because this matrix should be purely anti-symmetric. Well, let me be loose about it. Let's say that my sigma is integrated in imaginary direction. But in principle, you're right, of course. It's better be i here. OK, so that will be s, which depend on g and sigma. And now what do I have? s now will be n independent because look, n is sitting here. n is sitting here. And there are n my run fermions, so this n will be sitting in front of my Phafian. So long story short, this s sigma is given by 1 half. Norit, so that's already not bad because now we have a field theory written in terms of two matrices, j and sigma. I have an action. It doesn't have capital N at all. There is this energy scale, j squared sitting here. So this is actually AB. That's what comes from my interaction. Now you see that n plays a role of inverse Planck constant. So what is sitting here, if I would be careful, is actually, of course, a Planck constant. So n to infinity limit means classical physics. Planck constant goes to 0 to infinity. This is classical. Now you see that this happens not by accident, but because I carefully chosen this scaling of the variance. If I would put here a different power of n, then it will not be such a nice scaling. Question. Your line for the action looks pretty much as Latinx word functional. First term is trace log g minus 1 sigma, sigma g. So does it mean that there is a way in this model to effectively re-sum the skeleton diagrams? To do what? To re-sum the skeleton diagrams. Oh, yeah. That's next line, of course. Norit, so yeah, probably if there are no other questions, I go exactly to this. Right, so if n is large, and we're looking interested in a classical limit, then we can look for equation of motions of this action. So ds dg is equal to 0, and ds d sigma should be equal to 0. Now ds dg will give me sigma plus or minus. I will be losing about factors, j square g cube. And ds d sigma will give me 1 over d tau minus sigma is equal to plus or minus g is equal to 0. Now what is written here? Are there two equations of motions? Let me rewrite them in a slightly more user-friendly way. The second one will have form d tau minus sigma times g is equal to 1. And 1 means delta function tau minus tau prime, and delta function AB. And this is a good matrix equation, and this is nothing else but Dyson equation. g is a green function, sigma is a self-energy, and that's why I have chosen these notations, of course. So this equation is just formal Dyson equation, nothing else, nothing more. This equation is slightly more interesting and says what sigma is in a classical limit. And what it says is sigma is j square times g AB, in general, this is q minus 1. So that's my set of classical equations. Now what it does for me, it does precisely what Mitya said. It resum a certain subsequence of diagrams, namely, can I raise that? Yes. Well, sign is important, but my signs are probably wrong. Right? So it's don't ask too much from me to follow. Yeah, so pretty much all factors of 2 pi and plus and minus are wrong to the first approximation. If you want to follow them, go to the literature. All right, so the second equation what tells us that the self-energy sigma is actually given by free green function, which is natural. So there is a random coupling constant here and a random coupling constant here. We do a Gaussian average. So I should denote it as a dotted line. And there is a factor j square associated with this Gaussian averaging of a random coupling constant. That's this statement. And what my Dyson equation, together with this definition of self-energy, does, it resum all the diagrams with the topology like this. OK? And what I'm stressing is that that's sort of an exact result in n to infinity limit. And this is actually a classical physics of my model, not a quantum physics. Yes? What happens with what? So the question is what happens with this guy and the answer is here. Now, that's still not a piece of cake. So it's a formal resumption of this diagrammatic sequence, but go solve this equation. That's not that easy. OK? Now, what these smart people understood, Subir probably was the first, although I'm not sure exactly, that there is an interesting limit where you can solve these equations exactly. So let me tell you this. So this limit is known as conformal limit. For now, it's just not a formal notation. And what you're formally doing, and I'll explain in a sec why it is a good idea, you say that d tau is equal to 0. So you neglect this. So at the moment, it's just a brief guess, but you will see why it is a good guess. So if you do that, now you can solve these equations. That's already much easier. And the solution is the following. I'll give you 0 temperature solution, and then I'll comment about finite temperature. So at 0 temperature, you will find that GAB of tau and tau prime is equal to sigma AB of tau and tau prime will be BQ 3 half. So in general, this is 2 divided by Q. And you can easily guess what this is. I guess it's 2Q minus 1 divided by Q. Now, that's not a big deal, actually. You can guess these powers just from power counting here. That's one line. And then you can convince yourself that actually things work. For that, you need this sigma because it's a time-ordered green function. So this is the solution. Now, if you go to energy representation, then what you say is G of epsilon, again, is just power counting, 1 over square root of tau. You do Fourier transform D tau. So it's tau to the power 1 half. And you go to energy, you find that it's proportional to 1 over square root of epsilon. And sigma, again, power counting, 3 half D tau. So this is 1 over square root of tau. You go to energy. You get that this is square root of epsilon. So now you understand that this conformal idea is actually not such a bad thing. So this thing, it's actually epsilon times J. And that's also epsilon times J. Question on the left. Question. Question? Yes. That's what I said. That's a conformal limit. Time derivative is equal to zero. Yes, time derivative of everything. So I neglect this, right? And then this is a solution. If I bring back this, then it is not a solution. That's what I hopefully will discuss in five minutes. But for now, it's a guess. So within this classical limit, I do another approximation. I neglect this, then I solve it. And that's what I get. Now, why it is not such a crazy idea is that because this is in energy representation, this is epsilon minus sigma of epsilon. And sigma behave like square root of epsilon J. So this is square root of epsilon J. So if epsilon is much less than J, then this guy is much larger than this guy. And then maybe it's not such a big crime to forget about this. So therefore, this conformal limit is only justified at 10 of this, which is much, much less than 10 questions. Cool. Okay. Now, that's actually the place where physics should start. But unfortunately, that's the place where 99% of the papers stops. Basically, pretty much 90% of what you can see in the literature is about classical physics of this model. So they stop at this level. So in particular, this logarithmic susceptibility which Subir and Ye had in their abstract, I showed you this paper is one line calculation from here because spin is two fermions. So spin susceptibility, chi of omega is given by this diagram. Of course, this is omega, this is epsilon, this is epsilon plus omega. So one line calculation, this is d epsilon square root of epsilon, square root of epsilon plus omega. That's my green function. Right. So you immediately see that I have a logarithm of j divided by omega. So this logarithmic susceptibility is just this. And we will see tomorrow that pretty much all what people say about incoherent metals and things like that is not more complicated than this observation. Okay. But being sort of positive, you can say, well, this is non-theoremic liquid. Why is it non-theoremic liquid? Because what you find is that your green function, function of epsilon like this up to scale j when it does something else. And this is one over square root of epsilon. So this is to be compared with a theoremic liquid. We know that green function has a peak at epsilon p, which is p square over two m minus chemical potential. And the peak is narrow and blah, blah, blah, right? So here it's obviously completely different. So again, the way to think about it is that without interactions, all the energy levels were strictly degenerate at zero. Now you introduce interactions, energy levels start to repel each other. Now actually, single particle energy levels are not defined even because they all mixed within this huge many body exponentially large space. But effective single particle green function still has this huge peak at zero and then it has a long tail of decay. Right? So quasi-particles you can say are completely lost and another way of saying is that self-energy is larger than the energy. So that's non-thermalic, yes, Valeria? Nothing. Nothing. So this periodicity is seen in exponentially small energy scales, two to the minus n, okay? So these energy scales have no representation. In the limit n to infinity, you cannot even discuss this stuff. So we will see in a moment that there are other energy scales, namely more notable there is energy scale J divided by n. And remember, n is inverse plant constant. And what we will see is that what green function does actually, it does like this. Now whether it's, so here it's square root of epsilon. Whether it's a Fermi liquid or not is a sort of semantic. But what it is in a mesoscopic language, this is a zero bias anomaly. But we'll have to work a little bit harder to get to this. Sorry, just following on the semantics. So even if you go to energies which is much larger than J where presumably the self-energy is much smaller than energy. It is not a Fermi liquid in a sense that there is no Fermi surface in this model. No, there's no momentum. There's no momentum. It's a zero dimensional thing, yes. But still, I mean, if you have a set of levels, you could expect green function to have peaks near these levels, but it doesn't. That's of course a miracle of favoraging. Right, so in some sense, even for finite n, in some sense I have a thermodynamic limit because I have an infinitely large ensemble of my systems. It's a tricky stuff. Now, how much time? I still have 15 minutes to go. Any questions so far? Conformal limit green function, what is the definition of p, no, left side? The conformal limit case, maybe I'm missing or... P, p, g equal to p over j, a scalar to j. Oh, yes, sorry about that. P is one over four pi to the power one quarter, but in my universe it's one. As I told you, two is equal to pi and equal to one. Yeah, sorry about that. All right, now you may notice, and that's again probably Subhir was the first who noticed that this is actually not the only solution. You can find our, so this is not the only solution of these equations, even in the conformal limit. There is a whole family of solutions, and that's important. And this family of the solutions is given by reparameterizations of time. Namely, you can say that your time goes to f of tau. Now, time goes from zero to beta. Your f should also go from zero to beta, so you should understand that it's in general something like this. So this is my function, f, right? It better be monotonic because time is probably monotonic and we cannot go back in time, but otherwise it's an arbitrary function. And if you do that, you will discover that you have a whole family of solutions which are given by JAB tau tau prime, again B square root of J, Pignum f of tau minus f of tau prime, f prime of tau to the power one quarter, f prime of tau prime to the power one quarter divided by f of tau minus f of tau prime to the power one half. And again, this power is actually one over q in a more general setting. And sigma has a similar form, tau, tau, tau, f prime of tau to the power three quarter, f prime of tau prime to the power three quarter, f minus f to the power three half. Now you can easily check that with this substitution, my equations are still valid, this is trivially true because that's just the second line is just a cube over the first line. And the first line is true because I can make change of variables from tau to f and then this convolution operation will go through. Okay, so there is a family of solutions. My, the one which I just told you about and I sort of called nonfermic liquid and blah, blah, blah is not unique solution. There is a whole family of them. So there is a soft manifold of my action, right? So notice that that was my extremal point, saddle point of my action, right? That's these equations. I told you this is a solution, this is the mean, the minimal of my action. But now I tell you that there are other solutions so I can go from one solution to another solution by changing tau to f of tau. So there is a whole family of solutions which minimize the action provided I forget about this d tau. So this is a soft mode. If I forget about this, if I forget about d tau it's actually zero mode of my theory. Nothing depend on this f. If I recall that I still have d tau then it is a soft mode because that's derivatives and that's what we are going to discuss. But before I do this, so let me just mention in passing that there is a very special choice of f namely if I choose it to be what? I forgot. Tangent of pi tau over beta. So this is a strange choice because it maps the interval, the infinite interval of tau to a finite interval zero to beta in f. So with this particular choice of my reparameterization I can write down finite temperature solutions of this guy. So this is a standard trick that in conformal limit finite temperature and zero temperature are different by conformal transformation which is this. If you worry about finite temperatures then just use this f, substitute it here and you will have finite temperature solutions. So they will look actually very simple. They will look like sine of tau minus tau prime pi over beta to the power one half. And here it will be sine to the power three half. So doing finite temperature and zero temperature is pretty much same physics. I just need to have this special conformal transformation. So therefore I will mostly limit myself to zero temperature for now. And if we will need finite temperature I will just jump to finite temperature results. Cool, okay. So in a sort of scientific language this symmetry is called diff s one which stays for diffeomorphisms of unit circles into unit circle and which is represented by this figure and my action provided I forget about this is symmetric with respect to this diffeomorphism transformations of circle into a circle. This is infinite parameter group of symmetries because you can expand this function f in a Taylor series and each coefficient in this expansion will be a parameter, okay. And the generators will of course form a verisoroalgebra as usual. I don't need it but hopefully you know what I'm talking about. Now in our observation which is also crucial is that the actual symmetry is not diff s one but actually diff s one slash sl two r, okay. And that's what I will explain in a sec. Namely there are special transformations conformal if I choose f of tau to be a tau plus b divided by c tau plus d such that determinant of a b c d is equal to one. This is called Möbler's transformation. You can check that if you use this f and substitute it here your green function doesn't change at all. This particular transformation does not introduce a new solution into my saddle point equations, okay. Any other f of t does create a new solution. This particular form of f does not create a new solution. So therefore I have to separate, factorize these transformations out of total group of my soft modes. So my soft manifold is actually a coset space of diff s one divided by a sl two r. So you see why it is a sl two r because it's completely parameterized by two by two matrices with real coefficients and unit determinant. Okay. So since determinant is unit there is one constraint where are actually three parameters in this group. So in the language of Virasora algebra I have to take out three lowest generators of my Virasora. Which is just, yeah, so which I'm given here. Okay, all right. So now what the next step will be I will just announce it. I will not do it today. I will do it next time. But the velocity is this. So we found the saddle point. That's good. Now we found a soft manifold around the saddle point. Great. Everything else is presumably massive modes but this particular degree of freedom is a soft mode. So what we want to do, we want to integrate out the soft manifold and forget about massive modes. Like what we usually do in the sigma models. Okay. So what we want to do is to, in my theory, instead of having this huge integral over two matrices I want to reduce it to the integral over this reparameterization modes with some action which depend on this reparameterization modes and then I will calculate some observable. Observable depend on reparameterization mode. For example, if I want to calculate green function that's an explicit way how the green function depend on reparameterization mode. So that I know. What I don't know yet, I don't know what is this action in terms of soft mode and I don't know what's the measure. Okay. So what I will start doing tomorrow I will tell you what's the action. I will state what the measure is. I will not try to derive it here but we know what it is and then I will show you how to do this integral explicitly and then we'll be able to obtain something which I already erased but I will repeat it again. All right. I probably finish here and let's do questions.