 It's like before. Before again? Okay, here. Whether it is hence G as this shape, with A plus square root of this stuff. Yes. And then in the slide after, we write that G to the minus 1 of A is gamma T, and gamma T is A to A plus PB, AB. Yes. And B is this value. But it seems to be that gamma is G of A, not G to the minus 1 of A. Because if you substitute A in Z in the previous form, in the slide before, you obtain A to A plus IB. Okay, let's try this out, because I know that gamma has to be G minus 1 of A. Okay. So something else has gone wrong. Okay. Let's see what has gone wrong. So this is the Lovener evolution. It becomes singular when the denominator is zero. So it is singular at G equals to A. Therefore, if I invert this at this point, Z, which is equal to G minus 1 of A. I have singularity. And this point by definition is gamma of T. This we cannot change. It happens here. Now, the problem which is possible I haven't checked is if this map somehow has a problem. Let's see. G of T of Z is equal to A square root of Z minus A 2 plus 40. Do you think it has to be plus 40, I think? Is it 40? Minus 40? Yes. I can see what you mean. Minus 40. So let's say G of A is equal to A plus 2i root T. So which is right because as T goes up, it goes on the imaginary axis. So this is A plus 2i root T. And this is A. This point is A. Now, what we want to see is G minus 1 of this guy. How can I do that? So what I want to say is G minus 1 of A is equal to A plus 2i root T. G to the minus 1 because you wrote G equal to G to the minus 1, right? You are saying here? Yes. Here I only put instead of Z, A. So from here it follows that A is G minus 1 of A plus 2i root T. So this immediately implies that to be true. This is true. So that is not true. Something is strange here. Always easier to look at here. When you teach, you realize that next to the blackboard, you are at a disadvantage to the student. Student looks at it from here. It's not correct. So something is wrong here. I start from this equation. This one is correct. I set B equal to 0, which is the origin of time. So I end up with this equation. And that equation is solved to give G equals to A plus A squared plus 4T. But I have a boundary condition that at T equals to 0, I need G of 0 to equal Z. In this case, it is equal to A plus square root of A squared. So the solution is to change this constant here into Z minus A squared. And if I take the positive root, A goes out and I get Z. So the correct G of T and Z is equal to A plus Z minus A squared plus 4T. So something has gone wrong here. I don't understand why I've got it wrong. Something is wrong here because I have a Z. So this is correct. This I know to be correct because I've worked with that. Maybe it comes out of this solution. This is now correct. Now to answer the problems from there. GT of A now becomes A plus 2 root T. And it means that you started from A, but G is mapping it to somewhere here, which is A plus twice root T. I read the tip, gamma of T satisfies that G of T of gamma of T must equal A, which is, in fact, this expression. So if I put in there, I must have A plus gamma minus A squared plus 4T. So it means that gamma minus 1A squared plus 4T is equal to 0. So I get gamma equal to A twice I root T. But to get this answer, I have to choose the positive root, the negative root. So only when we have the A of T equals A constant in the equation? Yes. And the problem sheet I gave you was for you to attempt solving it for any other function. So a simple function like just T, and you see how difficult it is. So, in fact, I tried very much myself over these years to find another function of T which I can analytically solve, but I couldn't. It would be nice to have one. Okay, now? Okay, another question? Okay, let's do a problem then. Let's do a problem. So in percolation, S of P is equal to probability a connected cluster of size S at RQ at unprobability exists. So it is, you say, I turned the squares on and off with P, probability P. What is the probability that I suddenly find a cluster of size S? At the moment, it doesn't matter to me. Okay. So then it's the probability to find that? No, not at least. That you find, you just randomly find a cluster of size S. Now, all these clusters, some of them are size S. Divide them by the total, that will be NSP. This we know is of this shape. So it has a power law, but big clusters are reduced. And theta tends to zero as P reaches PC from below. So here P is smaller than PC. As you reach PC from below, theta goes to zero so that at the critical value, you have a scaling behavior. Here is the problem. Do this for one-dimensional percolation. Calculate this guide. This is the two-dimensional version. This is playing with his phone. I'm a terrible professor in Iran. Here you have democracy. We don't have democracy in Iran. So students can be subjected to extreme force. Do the partial differentiation. Let me see the final answer. You don't need to. For any P, you can do it. This can be done for any P. No, that way is in two dimensions. I apologize. That form is for two dimensions. In one dimension, completely different shape. This I gave you for knowledge. But when you calculate NSP in one dimension, you get a very different form. But you can write it in that way. So write it in that way. The next step is, after you've done this, which you have, what is the characteristic cluster size? So do this now. The answer is NSP is equal to 1 minus P, a squared P to the power s. If somebody doesn't understand the answer, tell me. No, I don't mean average size. Characteristic has a meaning in physics. Have an exponential decay. Exactly. Questions are easy. I said easy questions, but still everybody is thinking. So we write this as 1 minus P squared e to the power s log of P. So the characteristic size of a cluster is minus 1 over log P. Unfortunately, it is positive. So you have a percolating problem. You choose one site which is on. What is the probability that it is the member of a cluster of size s? That's this guy. Calculate it. It's a little algebra. One minute. No, because if you sum this over s, it should get 1. So it's itself in the dominant. Sorry, I didn't mean. This probability, if I add all clusters of size s must come to 1. Yeah? Yes, but it's the same thing. Suppose I write it like this. Then I say that s omega s of P is equal to 1. Yeah, right. Thank you. So you want to sum this? OK. We move to differential geometry now. Proof that all two-dimensional, the Emanian manifolds, conformally flat. Matthew is sitting in the back. He doesn't know I'm doing differential geometry. So what is a conformally flat metric? A conformally flat metric is one that you can express in this form. So it is flat, but not completely flat because it has this coefficient. So if you have a metric like that, you say it is conformally flat. So how do we prove this? I give you one minute to think and then I explain. What is a Riemann manifold? Riemann manifold is a mathematical space which has a metric, a Riemannian metric, defined locally on it. Simple. But when you say a theorem like that, it's a very important theorem. All Riemannian manifolds are flat. Definition of Riemann manifold? Riemann manifold is a topological space which is equipped with a metric on at least locally. That means you can hope that it's global, but usually you cannot. You just define it on a coordinate patch. And this theorem also only applies to a coordinate patch. So the correct way is all two-dimension Riemann manifolds locally are conformally flat. When I was a teenager, there was a cartoon called Ikyu-san. Did you see it here? Did they show it in Italy? It is a Japanese cartoon. The character was called Ikyu-san. Japanese pronounced that Ikyu. San means master. So it's master Ikyu. And so if they had any problem and asked him, and he goes like this, and I solve it. Yes. Nationality. Linear. No. Rho is not small and it's not linear. It's Rho is any function of z. Any function of x and y. So z and z bar. So to solve this, remember that the metric can be written like that. But you can also write it like this. So the theorem says you can find mappings which takes these to zero, gz and gz bar to zero, and you are left with, because this is now a flat metric, and this is a factor. So there are conformal transformations which takes you from there to here. Why is it? Yes. The question is find them. So if we know that they exist, but this question asks you to find the transformations. The proof that they exist is mathematical proof, but we don't need it. We know that they exist. There exist mappings which can map this form into that form. So in fact, if you like, I need to find w, allow z to go to w, such that this goes into w, w bar, dw, dw bar. Expression depending on z and z bar. That is, this move from here to here is simple because what you do is that you take, if I expand this, this is g11 dx squared g22 dy squared plus g12 dx dy. And what I do is that I define z as x plus iy and put it in there. And then what happens is that I have dx squared twice i dx dy minus dy squared g11. Sorry, this way is better. x is z plus z bar divided by 2. So it is dz divided by 2 plus dz bar divided by 2 s squared. So this comes out into one quarter of g11 dz squared plus one quarter of g11 dz bar plus one quarter of g11 dz bar s squared. You do that for all three terms and collect them. Eventually you will end up with the bottom shape. But gzz is some linear combination of g11, g22 and g12. Is there two factors in the third two symmetric terms? You have to find the mapping which makes g1, you see it makes this combination g equal to 0. You find the map, you see this under conformal mapping. G maps like this. So there is a mapping. x goes to x mu. So dx mu will be dw mu dx mu dx mu. Therefore g mu nu equals dw dx g dw dx. Find this mapping. This is like a matrix. So if you take g as a matrix, it's m, g, m transpose. And I wanted to look like, sorry, and I wanted to look like that. This, if you like, is a function. This is a matrix. Function is not the row xy. Just the row xy, yes. Is the row xy? They should write a row, if you like. It's the row xy. The line is how the matrix transforms under this transformation that we are looking for. Yes. I'm sorry. Omega is an epitome transformation. Omega is a transformation of the coordinates. The statement of the problem was for a local problem or it was universal? It's not global. It has to be local. So I don't want to teach Romanian geometry here. I'm avoiding your question. So when you have a Romanian geometry, the Romanian manifold, usually you cannot cover it with one coordinate patch. For example, a sphere at least needs two coordinate patches and they translate to each other on their overlaps. So this is only true in one coordinate patch, only true for two dimensions. There are manifolds in three and four dimensions which are conformally flat, but they are special. You have to make an assumption on the metric to become conformally flat. Here you don't have to make any assumption. It automatically... The solution of that problem solves it. Yes. I mean... Yes. You have to find a w which gives you the m such that g in general goes to this. Okay. You have to solve... You set up a set of differential equations which does this. I have notes to do it but it's really messy. It takes a lot of time. I can give them to you later. You just find the equations and then you solve the equations. I mean for each value... Yes. So for each couple of indices we have a differential equation. So we have a set of differential equations that we have to solve basically. Okay. The beauty of it is that it becomes similar to Cauchy-Riemann conditions. But it needs a few lines of mathematics. Okay. Thank you. Give me... So we can play karaoke now. All right. You're playing a system differential equation for the component of omega in the new variable. Yes. Okay. And we should keep all the odds back. Yes. Okay. Let's go to the next problem. Fractals. This is called a Cantor set. Cantor set is a fractal that Cantor did not discover. Somebody else discovered it. But we all call it Cantor sets because he proved a lot of theorems about it. So what you do is that you take the real line, let's say 0 to 1, you take half of it and throw it away and repeat it. So this goes from 0 to 1 third and this goes from 2 thirds to 1. You then repeat and then repeat until infinity. And also of course on the other side. You know this is true about the Ising model as well. Ising did not discover the Ising model. It was constructed by his supervisor whose name was Lenz and he gave it to Ising as a PhD program, PhD problem. And as most PhD students do this, after six months, Ising went to Lenz and said, no face transition in it. Wrote his PhD and left. The right Ising model was solved by Ansager after the war. And what did Ising do? He opened the shop selling textiles. Okay, Cantor set, calculate the fractal dimension. Don't find it on the internet. Solve it. I don't get the sentence at all. Log 2 by log 3. Log 3. Yes. Log 2 divided by log 3. Gabriel, you finished? Okay. So how I do it? I have to cover this by lines. See here because this fractal is smaller than the line, I cover it by line segments. So line segments at each level sk is 1 third. So at k equal to 1, a segment of length 1 over t will cover this guy. But how many do I need two of them? Because there are two line segments at each stage. So nk is 2 to the power k. And fractal dimension is defined as minus log of nk divided by log of sk limit of k tending to infinity. So log 2 divided by log 3. Approximately 0.6. Does anybody has anyone done the numerics? What is 0.6 or something? Does anybody know? The next digit, the next 3? 0.63. Is equal to 1 over the fractal dimension of, of what? Serpinski Triangle. This is the action for the scalar field in arbitrary dimension. What we want is the energy momentum tensor. It is related to the variation of s with respect to g mu nu. So what is it? So of course one term is this. It's easy. It's just this determinant which is a little difficult. No, determinant of g. This is determinant of g. So variation of this with respect to g first term gives this one. The problem is this next term which has variation of determinant of g. It's a very, very important rule. The teacher is not confined to that area. You can walk out. Have you done this in the exercise classes? No. It's an instrument which I use very effectively in class. What do you do? I walk in. I walk in and then I go, I. I got a little stuck. I got a little stuck. I asked her if she knows how to do it. So you have to get the variation of g with respect to g mu nu. So you can write it like that. So it becomes something like this. I don't know how to resolve that. So you can do that in two dimensions. But in d dimensions, which is I'm doing now, it's a little difficult. Yes, it's like a derivative. But it's a variation because this is not a parameter. It's a function. Yes. Then it's the variation multiplied by some function. And this function has to be some function of g. There is no other choice. Yes? So it's just that what exactly is true, I don't know, but I know that this is the right answer. So the question is what is the variation of the determinant of g? It doesn't help. But anyway, one, check that d mu nu is zero. From here to here, g now is a constant. So I do this calculation. The way we do this calculation is that at this stage, we take g to be a function of x and y because we want to do this calculation. But once it is finished and we arrive at t, I now take g to be a constant, which is one minus one minus one minus one. So given that g is a constant now, show that the energy momentum is conserved. This conservation of energy and momentum is the foundation of physics. If we don't have it, we have nothing. So this is conservation of energy and momentum. Yes? In the last line, the derivative d mu, shouldn't it be d nu? No, I think it should have a factor of one-half in it. In the first term, the nu is missing. Okay, this term vanishes because equation of motion is d mu d mu phi. And this term exactly can sense this term. I just have to redefine my indices for it to become nice. So I bring this down, call it beta, call this alpha, put the g alpha beta here. And they become exactly equal. But this is a factor of one-half here, which in my derivative I had it wrong. Yeah? So phi is a massless scalar phi? Massless scalar phi. If it's a massless scalar phi here, you see there is no mass. It's equation of motion. Do you derive the equation of motion from here, variation in respect to d mu of phi? You just get Laplacian of phi is zero. Or in fact, this is more than Laplacian, it's box in any dimension. Number two, work out the trace of t. The trace of t is one minus t divided by two times something. So it is equal to zero at d equal to two. I don't understand your question. Let me just repeat. T mu nu is this. I take the derivative with respect to d mu. It comes to d mu, d mu phi. And then I have... Then I have this guy. Yes? Then I have this term. Here these are two terms because the derivative of each one of them, a factor of two comes out. Two terms are exactly the same, so I just cancel the factor. Now these two terms are exactly the same if I rewrite it with a metric. So you're okay now? So, yes. Those two terms are equal, yes. You have to just rewrite and rename the indices for them to look the same. So the derivative of t mu nu is zero. This is conservation of energy and momentum. Now, next step, take trace of t. Trace of t is t mu, t mu. In other words, you multiply with g mu nu and sum. The answer is 1 minus d divided by 2 d phi squared. See if you can get it. This is minus, this is plus. Here? Or in front of everything. So this shows something is special at two dimensions. It doesn't happen in other dimensions. So at two dimensions, let's go back and rewrite the action. Same action in two dimensions. Is there anybody who did not write this? There's many. Because you didn't want to. Yeah? I don't get the factor one out in front of d. The trace of t is d. The trace of t is g mu nu, d mu nu, d nu phi. Sorry, the trace of g. The trace of g is d. So I don't get the one out. I get 1 minus d. There is a one-half in front of g. I don't touch. Well done. So we see that this term comes out, 1 minus d over 2. So if you set d equal to 2, the trace of t vanishes. If the trace of t vanishes, then we have conformal invariance. So in two dimensions, I rewrite the action to see that it has obviously got conformal invariance. I write it in terms of complex variables. Now, the variation of s with respect to phi gives you the equation of motion, gives you d d bar phi equal to 0. Which means that phi is a function of z plus another function of z bar. This is the power of complex analysis that you solve with differential equation by choosing an ansatz. You don't need to solve anything. Just any function of z is a solution. Now, what I want is the green function. What is it? So you know how to find the green function? No. Nobody knows how to find the green function. Take a differential equation like this. I take this to be two-dimensional problem so that it is similar to our problem. But I put in an extra constant here to make it a little harder and a negative sign here. I want to find the green function. So what I do is that I take a Fourier transform. Take a Fourier transform. You get k squared plus m squared. Fourier transform of g is equal to 1. Fourier transform of delta function is 1. So this equation is now algebraic, easy to solve. So you get g of k equal to 1 over k squared plus m squared. Now, inverse transform. This is the Fourier transform of the green function. So if I inverse transform, it gives me the green function. i.e. the green function x minus, let's say just x, is e to the ikx over k squared plus m squared e to k. If I want x minus y, I will simply transfer here. So if I want to calculate this, now I can do it by taking it to the complex plane. I have a pole at plus minus im. So I take it to the complex plane and there is a pole here and a pole here. So what I do is that I close the contour on the upper half plane. So this pole is picked up. So I have to set k equal to im. It gives me a minus mx divided by 2im. And then this, of course, can be written as k plus im, k minus im. One of them is the pole, jumps. The other one is 2im. So modulo sum constants. Calculation is finished. However, what is interesting to me is the limit of m tending to 0. Because this is what I want, which is only the first part of this operator. So in the limit of m tending to 0, see this goes to 1, but this goes to 0. And so this is singular and this means that this calculation is no longer valid. It is because in the limit of m tending to 0, this comes to the real axis. And I have a pole on my contour. So I have to avoid this problem. And so what it does is that I have a pole on the contour and I have to make a little circle like that. And I now have to calculate this little circle here around the pole. So an integral like that has to be done for this pinching. When you look at this integral, you see that you have a divergence at k equal to 0. And if you scale x away, if you define k prime as k as k prime over x, this x disappears. And between these two, you lose k prime. You observe that you have a situation like that. So it cannot be that this is independent of x. Something is going wrong. And that is because when k is very much near 0, which is on this circle, actually you can take this equal to 1 and this is a logarithm. So the answer is a log of x. So I suggest to you that the answer is minus log of z minus w squared. You can see that if you differentiate this once and then twice, you get 1 over z minus w and it is singular at z equal to w like it should be. And it is regular everywhere else. So behaves exactly like a Green function is a solution of this differential equation and it is just not a solution exactly at z minus w. These things are called Green functions. When you deal with the problem of differential equations, finding the Green function is very useful because it solves your boundary condition problem. But here for us it has a far more important use. What is the use? Phi of z, phi of w, expectation value. So if you take this as now a quantum field theory, these become operators and phi will have an expectation value and the expectation value is the Green function. This is something from quantum field theory. I prefer not to prove it now. So it is equal to minus log of z minus w squared. However, it cannot be a conformal field theory. It is conformal, it is quantum, but not a conformal field theory. Why? Because we know it is not dropping. So what do I do? Dropping. In axiomatic field theory they show that if your field theory makes sense, you must always have this thing minus this thing to be a decreasing function of x. This is absolutely necessary because you want an effect here, an effect there interacting with each other as they go away from each other to become independent. Maybe in the usual standard situation they become independent exponentially, but at critical phenomena they become independent by power law, but still they become independent. They must be decreasing. Log is not acceptable. Yes, it's negative, but it depends on the length of scale you use here. You have to use a length of scale because this now doesn't have right dimensions and this length of scale. So you do something smaller than the length of scale, it becomes positive. The negative sign is not important. It's not bad. Playing a game? Okay, it's 4.30, but just give me five more minutes, I finish this argument. A little bit more energy. So what I do? I take exponential of phi. Only in two dimensions. This is allowed to do. Taking exponential of operators in higher dimensions is not well defined, but here we can do it. And this is called the vertex operator. And mathematicians call this theory vertex operator algebra instead of conformal field theory. You call it vertex operator algebra because of these vertex operators. Now what I want is, okay, now that I have defined this vertex operator, work out for me the correlator of two vertex operators if phi behaves like that. The alpha alpha. Do you know this theorem in quantum mechanics? You don't know. You said yes first. Exactly, so you know. So if you have two operators that don't commute, multiplication of exponentials of them gives you A plus sum plus A commutator B. I think one half. You have to use it to calculate that. I remember some BCH series which is in NX series. Meaning arguments of exponential. If this is zero then it finishes here. So when you have two operators whose commutator is a C number, that's all you need to write. Okay, the next step is that if you put phi here, this commutator of phi is related to the expectation of phi. So the alpha z, the alpha w is equal to e to the power of minus 2 alpha squared phi of z, phi of w, one half. So equals to e to the power of minus alpha squared, log of z minus w squared. So equals to z minus w to power minus alpha squared, z bar minus w bar to power minus alpha squared. So the alpha is a conformal field theory with conformal dimension, alpha is squared, alpha is squared. Yes? The cumulant expansion gives the same result. Why is that? The cumulant expansion doesn't give the same result. Gives the same result. I expected it to be only approximately the same. Which cumulant expansion? What are you talking about? Replace that with the exponential of a series of expectation values. Okay. Is that because we despise the other terms in the expansion of the commutators? I don't know. You mean to write it like this, one plus, write it like that. I have to see it on paper. I cannot imagine. No, what I didn't explain to you, that here you have commutator of two phi's.