 Thanks for coming to the last of these four lectures on the DOZZ formula. So today I'm going to give a sketch of the proof of the DOZZ formula that I explained. So I explained the DOZZ formula in the first lectures, but I'm going to re-explain exactly what we proved. There are some new faces, so I think that if I give a few reminders, it will be useful to, well, I guess, to all of us. OK, so let me first start by, I wrote it already, I cheated. Let me recall what the DOZZ formula is. OK, so it's a rather complicated function. So remember, so what we did is we introduced, in the first lecture, a probabilistic expression using the Gaussian free field for the endpoint correlation functions of viewville conformal field theory, and what we're trying to do is we're trying to prove that the three-point correlation function of viewville conformal field theory is given by the DOZZ formula, that our probabilistic construction matches this formula that physicists derived in a non-rigorous way. And so let me recall, so we see it at least once in these lectures written on the board. So what is the DOZZ formula? It says that the three-point correlation function of viewville, so we saw, and I'm going to recall in a few, well, maybe before stating the DOZZ, let me recall what we did in lecture one. So in lecture one, we defined the three-point correlation, the endpoint correlation function for n greater or equal to three. We define, so if I take points here belonging to the complex plane, and I take weights alpha k, which are going to satisfy some bounds that we call the extended cyber bounds. We defined an endpoint correlation function by two, so mu. So mu is a positive parameter which plays no role in the theory. It's just something, it's a theory that is scale invariant if you wish. So for the power minus s, gamma of s, so I'll tell you in a moment what is s with respect to the weights, I'll recall this. So expectation of, so I think I'm going to, I wrote it like this. Integral over the complex plane of some function which defends on all the points to the power minus s, and let me say what these things are. So s is related to the weights, it's the sum of the weights minus 2q over gamma. So the viewville theory depends on two parameters, the essential parameter gamma, which controls all the conformal invariance properties. And mu, the so-called positive cosmological constant, so gamma belongs to 0, 2 in these lectures. We could take it up to 2. Mu is some positive parameter, whatever, okay? And what is this, and I forgot one term, which is, sorry, just make a little bit of room, product for i smaller than j, 1 over zi minus j, alpha i, alpha j, okay? And times this expectation, so what we're taking is a moment, so a fractional moment, of the exponential of the free field integrated against a function, so recall what the measure was. It was nothing but the exponential of the free field x, okay? That I integrate against, so some background measure. Remember that this stands for the maximum of the Euclidean norm of x and of 1, okay? And what do I get, I integrate against this exponential of the free field? I integrate some function which has singularities on the points here, zk. And the magnitude of the singularity is given by these weights, alpha k. So f of x, so z is the product for k equal 1 to n. x plus, sorry, x plus divided by x minus zk to the power gamma alpha, okay? So this was the probabilistic expression. Now, of our gluville correlation functions, and these correlations exist provided the alpha k, so zk are distinct points in the complex plane. And provided alpha k satisfies some bounds which are written in the notes, which were in lecture one, which are called the extended cyber bounds. This quantity exists is non-trivial and is our clean probabilistic definition of the gluville correlation function, okay? Now, in particular, so the condition n greater or equal to 3 came from the fact that the extended cyber bounds give constraints on the alpha k's and these constraints force n to be greater or equal to 3, okay? So I'm not rewriting the extended cyber bounds. I'll write them probably when I look at the three point function. Now, from this, I can look at the three point correlation function when one point is sent and it's at infinity, right? Now, if I do this, so what I do is I introduce what is called the three point structure constant or the three point correlation function where one point is infinity. So I can introduce, so this was my three point structure constant or maybe an even better notation which is not in the notes, but it's to write this, okay? And it's defined as the limit when Z3 goes to infinity of Z3 to some power for delta 3, the three point correlation function, Z3, okay? So I take the correlation function with points 0, 1, Z3. I renormalize by some factor, so this was the conformal weight, okay? So delta 3 is alpha 3 over 2, Q minus alpha 3 over 2. Let me, for those who, okay, I see there's a new face, so let me just recall that Q and gamma are related by this special relation. If I take this scaling limit, I get something non-trivial and I get a probabilistic expression and what is my probabilistic expression? It's 2 mu, so I'm going to set some notation into stone here. I'm going to introduce alpha bar, which is nothing but the sum of alpha 1 plus alpha 2 plus alpha 3. So I just take the limit in my general expression, okay? Minus alpha bar minus 2Q over gamma, okay? And I get expectation of, so I called it in the notes, rho alpha 1, alpha 2, alpha 3, to the power minus alpha 2Q over gamma. Where, so here I am just copying basically what I have there. Let me write it properly, alpha bar is the sum, okay? So this will be my notation during these two-hour lectures. And my random variable here, rho alpha 1, alpha 2, alpha 3, it's nothing but to the power alpha bar, the sum. And I have my singularities, and I'm integrating against the exponential of the free field, okay? So let me, okay, so this looks correct. Okay, so I can define my three-point correlation function. And I get a clean probabilistic expression and provided, okay, alpha 1, alpha 2, alpha 3 satisfy some bounds. This exists and is a non-trivial definition of my three-point correlation. Now the DOZZ theorem says that this guy here, so this is, okay, this is an easy term, this is the gamma function, so this is explicit. So if I take this times this expectation with respect to the exponential of free field that I'm integrating, then this thing is equal to the formula that is upstairs, the DOZZ formula. And what is the DOZZ formula? So I wrote it explicitly at least once in these lectures. So you see it's a, it's a rather intriguing and mysterious formula when you look at it at first. So it's not, so it's pi mu times so L. So this L function is also set in stone in these notes. You know, it's the ratio of gamma over gamma 1 minus the variable. So I take this L function, so this power here, and it's, and especially there's, there's the ratio of four epsilon functions. So these are called the Zamolochikov's epsilon functions, divided by four epsilon functions. And these epsilon functions of Zamolochikov have an explicit expression. It's exponential of this, you know, this, this integral here. With a, so, whereas that is a parameter. And so today I'm going to try to explain how you prove that this guy upstairs, which is, you know, just some fractional moment, essentially, is equal to this rather complicated formula. Okay? So that's the plan. So it's the main, okay. All right, so let me, so I've written this formula. So they're in the notes, and I'm not going to, I'm going to erase it now. If you allow me, it was just to throw it once. And I'm rather going to explain why this formula is so special, okay? Essentially, I'm going to explain that this formula, it satisfies a unique system of functional equations. And that's by, and these functional equations that, that this formula satisfies, well, they stem from this special property of the epsilon Zamolochikov epsilon function. So remember L, L of z is gamma z over gamma one minus z. And you want me to leave this formula? Okay, so you, so I leave it. So I'm going to leave this two, okay? So these two, these two guys, these two guys, sorry, are the same. Okay, so what I'm going to do, so okay, I'm going to, I'm going to, so you see when I shift, this is a shift equation. If I shift my variable in my epsilon, Zamolochikov's epsilon function, I get the same thing times a ratio of gamma functions, okay? Essentially up to some power. And same thing for this guy. I get the dual equation, okay? And this is really what characterizes Zamolochikov's epsilon function. And it, and so if you use these two functional relations, then you're going to deduce functional relations on the DOZZ. So this is what I'm going to write, okay, I wrote it here. Now, one has the following, one can show the following thing. So I take DOZZ and I shift the variable gamma over two, okay? And I get the following equation, minus one over pi mu. Some function, I'm going to, so it's a curly A or a matkal A, if you're doing minus gamma over two times the same thing, okay? And I'll say in a moment what this function is. And I have the same dual equation. Well, not quite, so there's some subtlety here. So I have that when I shift my variable, so I have to introduce the dual cosmological constant, okay? Where, so the dual cosmological constant is not mu. So remember, mu, it's just my parameter up there, which is in factor. It's mu pi, my L function, my ratio of gamma functions, okay? And so if alpha zero belongs to minus gamma over two, minus two over gamma. So it's one of the two values here. What is worth A? So it's a rather complicated expression, but I'm nonetheless going to write it. So it's a ratio of lots of these L functions. So I'm writing it really explicitly. Of course, in this business, that maybe is a monochic buffer. I don't know if someone knows these formulas without notes. It's a rather complicated expression. So I'm sorry, I really want to write down these expressions. So things, so I don't do a too abstract discussion. I really want you to see at least once the formulas, okay? So L is the ratio of two gamma, okay? So DOZZ satisfies these two shift equations here. Okay, so how does the proof go? The proof, well, the idea of the proof is to show that this guy here, satisfies the same shift equations up there, okay? So of course, by conformal invariance, the three point function, it's invariant if I do permutations of the variables. If I see gamma alpha one, alpha two, alpha three, it's the same thing as C gamma alpha two, alpha one, alpha three, etc. Or you can see it in the DOZZ formula. I mean, it's something which is permute. I mean, if I do a permutation, okay? If I take a permutation, of course, I have that DOZZ or the three point correlation function we've constructed. Of course, it doesn't depend on the permutation, okay? So take a permutation, sigma, it's invariant by changing the coordinates. It's not a minus here. Does this use the fact that any three points are conformally equivalent? So it would not be obviously true for a higher number? No, no, it's going to work. I mean, you mean this permutation? No, no, we're going to see that. Is it a matter of permutation symmetry or the fact that- No, it's a conformal invariance. It's a conformal invariance. It's a conformal invariance. So if you have, so okay, so DOZZ is the unique solution of this and these two shift equations. So if I prove that my three point correlation function satisfies these two shift equations, and this I'm done. And it's because of a theorem, and that's what's funny about this story. It goes back to L'Euvil's theorem a long time ago, as I'm going to explain. So imagine I show that this guy satisfies permutation and variance in the two shift equations. So what do I do? I fix alpha two, alpha three, and I look at this function. I look at the ratio, say, of DOZZ divided by my three point correlation function, okay? And so this function here, well, since they satisfy the same shift equations. Well, so I suppose my three point satisfies these two shift equations. Well, this means that this function, it has two periods, right? Okay, and so it also has this period, two over gamma. So I have a function which has two periods. Now here I have to make an assumption. I'm going to suppose, and this is something that people know in the conformal field theory business. There's a difference between, when the central charge is a rational number and not a rational number. So if I want my argument to work, I first have to suppose that gamma squared does not belong to the rationals, okay? Because essentially what I'm saying is that these shift equations, they don't uniquely characterize your three point function if gamma squared is a rational number. It hasn't appeared in these lectures, but let me just mention that the central charge of uville is 1 plus 6 q square. So if I write q, that's gamma over 2 plus 2 over gamma. And so you see if this guy is rational, central charge is rational. And so we suppose this, and then we can deduce the general place by a continuity argument. Because everything we're doing is continuous in gamma. So if we suppose this, then by uville's theorem, I have a function which has two periods and the ratio of the periods is irrational. So it means that it's a constant function, okay? And of course the constant a priori depends on alpha two and alpha three. But then I do the same thing. I look at this constant which depends on alpha two, alpha three. By permutation invariance it's going to have a shift equation in the second variable, in the alpha two variable. So it's a constant which only depends on the third variable. And I do the same thing and then at the end I get that this constant in fact doesn't depend on any variable. You don't really need to repeat the argument, it's symmetrical. So it's also respect to one for the constant. Well, okay, yeah, but it's basically what I'm saying. You just repeat the argument on each variable. Okay, and then at the end I get a constant. And so all I have to do is find the constant. And if there are lots of values you can take to identify the constant. And at the end you find one because I put a two here. Because if there was a four then it would be two times d of zz. Okay, so that's the idea of the proof. So the question is how do you prove that the three point correlation function satisfies these two shift equations which uniquely characterize? Now I must say that, okay, when d of zz first derived their formula for the three point question in Uville, it was really, even in the physics standards it wasn't very, it wasn't considered rigorous. And I wrote to you a citation on this d of zz formula. Yeah, so Zamorochikov, Zamorochikov, they said, okay. So they're talking about their derivation of d of zz. They say it should be stressed that the arguments of this section have nothing to do with the derivation. These are rather some motivations and we consider the expression proposed to the d of zz formula as a guess which we try to support in the subsequent sections. So I showed how they derived it in the first lecture and it was really about doing some analytic continuation guess. And so soon after the derivation, physicists, they were looking for, let's say, a more convincing derivation. And let me say that, okay, some of our work is inspired by a beautiful idea by York Teshner. So soon after d of zz Teshner, in the conformal bootstrap approach, he explained how, in some sense, using special four-point correlation functions, we can recover these two shift equations, okay? So Teshner was working with axioms of the conformal bootstrap. So basically he was saying, okay, if UVIL theory exists, it's going to satisfy axioms. And if I gather all these axioms at the end, the three-point correlation function should satisfy these two shift equations. So it has to be d of zz. So on our side, what is difficult is we actually have to, and this is kind of our program, we have to implement rigorously in a probabilistic setting all these axioms that physicists suppose in the bootstrap approach. And so this is what I'm going to try to explain in the rest of these lectures, is how somehow we're going to build all the axioms of the conformal bootstrap, in some sense, with our probabilistic expression. And then at the end we're going to be able to derive these two shift equations, okay? But okay, the problem is, and it's still an open program, there are lots of axioms of the bootstrap that we do not know how to justify from probability theory. And so what I'm going to do today is I'm going to focus on trying to explain to you how we get this equation here. So this equation is based on, so the proof of d of zz, to summarize a bit, it's based on two papers that we did with Antique, Couppier-Nen and Rémy-Haud. So let me call the first paper KRv1. So in this thing, we derive what are called the BPZ equations. So we kind of set all the stage of the conformal, we reveal somehow all the conformal structure behind our probabilistic construction. And as an application, we prove, so let me call this shift equation one. So S1, shift equation one. This is shift equation two. And in the second paper called the Integrability of Uville, proof of the d of zz formula, we show, so here we show shift equation one. And this paper is focused on showing shift equation two. So I will not discuss in these lectures shift equation two, it's much more complicated to derive rigorously. It's why it's a separate paper because there are lots of, this is really a technical paper. And so I'm going to put under the rug all the technical difficulties in deriving this one. But I'm going to try today to show you how we can show shift equation one. Okay, so that's the plan. So shift equation two will not be a BPZ equation? It will, but the problem is, as I'm going to explain in the sequel, there are two ingredients. You have to use the BPZ equations for degenerate field insertions. And then you need to do operator product expansions rigorously. So you need to do Taylor expansions on your correlation functions. And the Taylor expansions are rather nice in this case. But if you want to do Taylor expansions with the second BPZ equation, it's much more involved. And this is also where the two-point function appears and but I don't have time to explain. So I'm really going to explain this part of the proof today. So the first of the two papers, which establishes the DOZZ. Okay, so I'm going to replace this expression for the general endpoint correlation function by the four-point correlation function. Because I'm going to be working with the four-point correlation function. So the idea, as I said in lecture one, is I'm going to take special four-point correlation functions, which are going to satisfy differential equations. And I'm going to try to use them to get information on the three-point correlation functions. That's the idea. So let me introduce the four-point, so I'm going to replace this thing. I'm just going to take four points in this thing here, with one at infinity. So I'm going to take, so alpha zero, say. And I'm going to look at a four-point correlation function with weights. So the first variable, I'm going to call it Z zero, well Z actually. I'm going to define, so G alpha zero is the four-point correlation function with one point at infinity. So I take my expression, three, and I get a four-point correlation function with a point at infinity. And I get an explicit expression, okay? Using my, I get the following expression, so let me not write it. So it's one over Z, yeah, alpha zero, alpha one. One over Z minus one, alpha zero, alpha two, okay? So the trivial part, so there's a two. Mu to the minus alpha zero plus alpha one plus alpha two q over gamma. The gamma function, so usually I call this S. And some, so I'm sticking so to the notations of my lecture notes. I call this, so this is a random variable. And what is this random variable? Well, it's this two, and I integrate against my free field measure. My exponential free field volume measure. Okay, so I'm just writing with four points, essentially what I wrote with N, okay? Now, I'm going to call this piece, so this is kind of a trivial piece, okay? This thing here. I'm going to call this piece, all of my notation, tau alpha zero of Z. In this random variable, do you take absolute values or complex values? No, so, okay, I should have maybe recalled this. Our probabilistic constructions in all these lectures, I take real alpha zeros, okay? Is that your question? Yeah, and also the absolute values. Why would there be an absolute, there's no need. Yeah, no, just wondering about herbal structure, never mind. Okay, of course the DOZ is valid for complex gammas, alpha zeros, but by the probabilistic door where we're trying to understand it, we need to take real weights, essentially. Okay, so I introduce a four point correlation, and I think it. And so, so let me first start with a remark. You can notice that if I take Z equals zero here, well I get the three point correlation function. With, sorry, where I replace alpha one by alpha one plus alpha zero. So let me, let me put infinity here, so my other notation for the three point. So, so you see, if I, if I look at what happens, so this is the link, okay? We're starting to see these shift equations arise. Because, you know, I, essentially when I look at this function around the point zero, I, I, I'm starting to see the three point correlation function which I've shifted. Okay, so. It is z one, then z again, then z three, in the, in the G, your definition. Ah, I'm sorry, I'm okay, excuse me, zero, sorry, what did I do? One, okay, so this is the four point correlation function with one free parameter, zero, one and infinity. And a trivial observation here, just plug in the definition. You'll see I get the three point. Now what's the idea behind the proof? The idea is that for special values of alpha zero, which were predicted in physics. So, so the two values, alpha zero equals minus gamma over two and minus two over gamma. This thing here, this tau alpha should satisfy a PDE. It's called the insertion of a degenerate field. So for alpha zero, for alpha zero belonging to the finite set, minus gamma over two, minus two over gamma. Okay, physicists expect, and it's in their bootstrap axioms. They expect that this tau here I introduce, so just the four point correlation function where I've removed a, you know, a trivial function in front. It just should satisfy a second order partial differential equation, which is the hyper geometric one. That's the second order BPC equation. And so what we proved in this paper, the first one, is exactly that. So proposition if you want. If I take this, so this is taking partial derivative with respect to the complex variables, so I hope this is clear that the partial derivative is, you know, this, sorry, minus, okay, I'm taking notations of. Well, this guy, so if alpha zero, to write it, alpha zero belongs, okay, so I'm going to put my proposition here. So if alpha zero, if alpha zero belongs is one of these two values, then this correlation function satisfies a finite PDE, which is the hyper geometric one. So this is what we proved in the first paper. So the BPC and the dual BPC equation. Where A, B, and C are related, of course, to the coefficients. So it's kind of, you know, tedious. But let me write the full conditions very clearly on the board. So Q is 2 over gamma plus gamma over 2. Okay, I think this is correct anyways. I mean, the important thing is to get the idea. Okay, so I have this finite PDG. And so the interesting thing is now that, okay, this came, this one magic thing about this thing is that if, what we proved also, so it's not very difficult, that difficult to prove, but, is that if you take a real valued solution to this PDE, it's a 1D space. I mean, it's a, the solution space of this is one-dimensional, okay? So this means that my four point correlation function, it's a one, it belongs, it satisfies this PDE for these two values. And so I can completely characterize it because it's a 1D, 1D space. So let me describe the solutions. So the general solutions of this, so remember I have, so I have A, B and C which are related to all the parameters of my theory. So if I introduce the hypergeometric functions. So usually the GUS, I think they're called the GUS hypergeometric functions. So remember this thing here, I hope I'm not going to miss it. But I think it's something like factorial n, factorial with respect to b, set to the power n, you know, something like this. Where these factorials, it's a, a plus one, a plus n minus one, I guess, like that, okay? So I introduced these, this hypergeometric function, I introduced the second one. And I'm going to describe my solution space with different parameters. Well, all the real solutions to this guy is nothing or are given by, so I can write tau of alpha zero of z is equal to lambda, so this is the free parameter, times my hypergeometric function squared. Plus my second hypergeometric function squared. Where this guy is a special, this guy has an explicit expression, okay? So this means, so let me, let me write this, so let me write this special, this expression. So remember A, B, C have these special, these relations. So if I say this guy is in R, I have this is equal to, so it's very, it's also here, once again, rather complicated, so. So for people who don't like formulas, I think this is not the right lecture. I mean, never mind the exact expressions, but so, okay, so to summarize. What did we prove? So we proved two things, we proved that tau satisfies these PDEs. And we also proved that the solution space of this, all real solutions of this PDE are given by lambda times a hypergeometric function squared. So this is the appearance of the conformal blocks, conformal field theory. Plus some coefficient here times the second hypergeometric solution squared, this one here. And the coefficient here is given by this rather complicated expression with gamma functions. So that's what we proved. So how does the proof go in some sense now? Well, of course, we know what lambda is for tau of alpha zero. Because, okay, if I take z equals zero, this is zero, okay? Because this is worth zero, and this is worth one. So lambda is nothing but the value of the function taken at z equals zero. And we know by our probabilistic expression what it is. It's this thing, so c gamma alpha one plus alpha, alpha two, alpha two. So we know, so we know this function what it is now. So how do we prove, how are we going to prove this shift equation, right? Well, we're going to try to find a second expression for the coefficient lying in front of f plus z square. And this s, this second expression, is going to, we're going to see, oh, sorry, it's the other way around where it's written, sorry. This guy is going to be lambda here, it's the value z equals zero. And remember alpha zero is worth minus gamma over two, or minus two over gamma. And so lambda is nothing but this guy here. Now, I'm going to do what is called an operator product expansion, or just a Taylor expansion in math, to try to find a second expression for the coefficient lying in front of f plus z squared. And this second expression I'm going to get is going to make this guy appear. And then by unicity, this will give me a relation between this guy and this guy, which is nothing but this shift relation. I don't know if it's clear, but what I'm trying to do is to find two different expressions in front of this guy. And by unicity they're equal, well, I get the shift up there, okay? So that's the program. So right now, what I did is essentially for alpha zero, I'm going to now focus on the case alpha zero equals minus gamma over two. So here in my, so now I'm going to focus on this. Okay, so let me give you- This equation right, the same as the tetanus, I guess. Excuse me? The equation in the second order equation is the same as the tetanus, right? Yes. So then he expands on hyperparametric, right? So it's different, what they do in physics is they, yeah, or, okay, but tetanus, what he does is what they do in physics is, he supposes that the correlations they factorize in a conformal piece in a bar conformal. And then he says the two conformal blocks, I put them together, but we don't have these two independent holomorphic and anti-holomorphic parts. We only have one PDE, and the fact is that what we say, what we prove, is that you don't need to do all these factorizations of holomorphic, anti-holomorphic parts because you have a unicity statement over there. What is, for instance, new is the fact to say that you see the conformal blocks appear just with a PDE. They're sufficient to show this unicity statement here. Whereas physicists, they cut it into holomorphic and anti-holomorphic parts. So it means that the question was the unicity statement? No, because he's working with a straight D. He's working with a holomorphic and anti-holomorphic part. Of course, he knows that physicists know, of course, that the real correlation function which I'm working with is going to satisfy this PDE. But they work with holomorphic and anti-holomorphic. Okay, so let me explain. So let me take out this equation and just, because okay, this leads to this and this summarizes all we need. Okay, so the proposition is the following. So I'm going to take alpha zero equals minus gamma over two. And when I'm going to focus in the second hour of these lectures, I'm first going to state what I'm going to prove in the second hour of these lectures. If I take alpha zero equals minus gamma over two, then I can show the following, okay, and sorry. And I'm going to suppose alpha one plus gamma over two is smaller than q. Then I have this. And if I do a Taylor expansion of my correlation functions, well, what do I get when z equals zero? Well, I said it, it's the three point correlation function. This is trivial to prove. And I can prove this. So this is a proposition, if you want. So it's stated, not as a proposition, but it's relation 5.14 in the notes. So I still have my relation here. So now I'm only going to work with alpha zero. It's worth minus gamma over two. So I can actually, I can actually replace it here, okay? Plus little o. So if you prefer the coefficient here is, so I'm looking at z to the power gamma q minus alpha one. So this is a proposition I want to prove in the next hour of these lectures. So proposition, I have this Taylor expansion where B of alpha one or B of alpha is minus mu pi the product of three L functions. So these gamma functions. Now you see, if I have this Taylor expansion, well, what happens when z goes to zero here? The first order term is going to be, as I say, it's going to be lambda is this guy. And the second order term, when I do an asymptotic expansion, it's going to be lambda times this times z to the power two one minus c, okay? So this means, so this unicity statement on my partial differential equation along with this proposition that I'm admitting, you know, right now. It implies, okay, so maybe I'm, wait, what am I going to erase? I think I want to erase the DOZZ for now, can I? Okay, I think I, I like it as much as you do, maybe even more, but I have to erase it. So, so this means that B of alpha one by unicity B of alpha one c gamma alpha one plus gamma over two alpha two alpha three is equal to c gamma. So this is lambda times my complicated function, you know, alpha of alpha of okay. So alpha zero now is worth minus gamma over two. Now if I simplify everything, I do all the algebra, there are tons of gamma functions all over the place. Well, this, this relation is equivalent to shift equation one. I just, you know, simplifying this guy with this guy and with all these gamma functions at the end. You spend one page of algebra and, and you find that you get shift equation one. Okay, so I think I'm, yeah, I'm going to take a, a break maybe in a minute or two. And I hope that, I hope that you understand. Okay, never mind the formulas, just the general philosophy. The general philosophy is to combine the unicity on a PDE and identify the coefficients in two different manners and combining the unicity statement, it implies a relation on the three-point correlation function and it gives shift equation one. And then what I'm not going to discuss is in these lectures and then basically what I'm going to do is I'm going to look at alpha zero equals minus gamma over two, but I can also do the thing with two over gamma. Do a, use a unicity statement in the PDE, do Taylor expansions and identify the coefficient in two different ways. And it leaves to a second shift equation and then I'm, it characterizes the, the OZZ. Okay, so two minute break. Okay, just following questions, I, I, I'm just going to say again what I said, maybe it was, it was not clear. So I, I have a unicity or uniqueness. People told me I have to say uniqueness. I have a uniqueness statement on my partial differential equation. So this is really analysis if you want. Which says that tau minus gamma over two of z, okay. I wrote, I wrote over there unique, the uniqueness statement on the PDE. And this uniqueness statement says that tau of the z variable, so this is a, the t is a tau, in fact, that's a, or a minus gamma over two z. It's nothing but the three-point correlation function where I replaced alpha one by alpha one minus gamma over two, the first hyper geometric function squared. Plus the same three-point correlation function times this ratio of gamma functions times the second hyper geometric function squared. So this second hyper geometric function, in fact, it's, it's not the hyper geometric function per se, it's times z to some power. And so, as a straightforward corollary of this uniqueness statement, I get that when z goes to zero, I have this Taylor expansion. Is that this function, this four-point correlation function is the three-point, plus, so the same three-point times the coefficient, z to the power two one minus c plus little o of z to the power two one minus. So, now what I do is I try to derive this expansion directly in a probabilistic setting. And I find this proposition, star, say, and this proposition I'm going to prove now in the next hour says that when I do an asymptotic expansion around zero, I get the three-point correlation function with shift minus gamma over two plus the three-point correlation function but with shift plus gamma over two instead of minus gamma over two times an explicit function and you'll see where this comes from. Plus little o. So by unicity of a Taylor expansion, the coefficient that is here is nothing but the same thing as the coefficient that is here. And so this leads to this relation and if you do the algebra and you're a bit patient, it's exactly equivalent to shift equation S1 up there. Okay, so that's, that was the idea I, I, I, I want, I hope that it's rather clear what's going on, okay? So it's really playing with partial differential equations and look at what happens at the boundary of their, you know, their domain of definition. So here it's z equals zero, it's, you know, it's not really defined and you do, I mean you do, it is defined, sorry, and you look at what happens around special points, sorry. All right, so to, so now what I'm going to do is I'm going to prove this proposition that I can do this Taylor expansion. Excuse me, so just the fact that tau satisfies the PDE, that's easy to show? No, that's not easy to show. It's, it's, it's, it's, no, it's not easy, it's, it's, it's rather what, essentially what you do is Gaussian integration by parts because you have some kind of Gaussian structure behind your, your definition, right? It's the correlation, it's nothing but the expectation of, you know, so z, the variable's here, but this is the exponential of the free field. So you can do lots of Gaussian integration by parts, but the problem is you have lots of divergences. And so it applies, you know, introducing burling transforms. So it's a mixture of, you know, probabilistic estimates and, and complex analysis tools under the rug. I mean it's, it's really, it's really one of the big parts of our first paper, is to show these PDEs are true. But I have to choose what I'm going to, you know, to present. So I don't want to present how we derive the partial differential equation. It's, it's, it's, it's rather technical and there's lots of tricks involved. Rather, I, I, I want to emphasize, you know, the, the, the general scheme behind the pool. But in a way, it's known because it, it, it, it generates seeds in the C. F. T. Power, as well. So what, what corresponds to that knowledge in the Gaussian free field approach? Where do you, how do you see it? Well, you see it, it's the same, same, same thing I'm going to say to Pascal is essentially you, you start taking a, taking derivatives. You do integration by parts and at the end you see that there's only special values, well the two special values that I wrote, so minus gamma over two, minus two over there, where the, the terms cancel miraculously. And you, you, somehow you have faith in conformal field theory. And you believe you've construct, yeah. And you believe you have the right probabilistic definition. You believe it and then you say, okay, if I, if I use the Gaussian structure, I should see these, these PDEs. In a way, if you have not known that it should work, it would be difficult to spot. Yeah. That's why our, our projects somehow it's kind of, of course it's inspired by, by what physicists did. You, it's, you're importing this bootstrap language into probability. Oh yes, of course it's difficult to spot. I mean, why would it, why would it work? I mean, I mean, when you look at this expression, how can you guess that these two special values are going to work? In SLE, in SLE, you know, you recover, you recover the VPC equation from the, on the boundary, the properties of, you know, the SLE. So anyway, so, but here it's somehow more difficult to spot from the start. It's, it's, yes, it's, it's kind of difficult to, to, to, but somehow this is what, you know, my student, Petun Anzu is working on, is, is trying to kind of, you know, exactly kind of identify what kind of structure probabilistic structure is behind the fact that you have all these BPC equations. It's, but it's kind of a, it's still a bit mysterious for us, but I think that Tounan is doing progress on this. Well, we're not yet there. Okay, so let me, let me prove, so this is now I'm, I'm going to jump for one hour into pure probability. Okay, so maybe I'm going to erase, well, I can erase nearly everything I, so I'm going to erase the Schiff's equations, which determine the OZZ, okay, I don't really need these anymore. Okay, and so as I said, so though I can work in the dual case, I'm, I'm not going to do it in, in these lectures, so everywhere I'm going to replace alpha zero by minus gamma over two. Everywhere, minus gamma over two, it's a bit long, but I'm doing it professionally, over two, minus gamma over two. And, so r, my random variable, which I'm taking the expectation. I'm going to say that it's, so alpha one plus alpha two, so this is alpha bar in my notation, it's minus gamma over two. And here, I have x minus z to the power gamma square over two. So this is my, I replaced, theoretically, I think I managed to replace all the alpha zeros by minus gamma over two. Okay, and let me, let me put some, some notation, so it's going to be clear. So I'm going to set, so if you look at my, my random variable, it's the integral over c of some kernel, which depends on z and x, times the exponential of the free field, and, so, notations of the lectures. And, and, and, and this kernel is, I'm just copying to, to make some room. The kernel is just this guy here, okay? So it's x minus z to the power of gamma square over two. x plus gamma alpha one plus alpha two, three minus gamma over two divided by x to the power of gamma. All right, so the idea is that my tau here, so let me put another notation, so it's, it's usually what I call s, but it's one over gamma, alpha one minus gamma over two minus two q, okay? So what is tau? So tau minus gamma over two z with my notations. I'm just rewriting to get things, you know, it's mu. So to the power minus p, I think, I can work this out. Minus p, sorry, gamma of p, so with my notation. My expectation of r minus gamma over two z to the power minus p. This is my four point correlation function. Okay, so this board is, is set into stone. These are my notations in this hour. Okay, I just re, I'm just rewriting what I wrote already, lots of times. The four point correlation function, it's mu to some power, gamma of p, the, the expectation of this variable. When z, and so I have to study what's going on when z goes to zero. So of course, when, and of course I have r minus gamma over two z, for z equals zero, it gives me, so right now, forget about this term and this term, okay, they're fixed in my problem, so. So I have to study this. So it's reasonable to write this, right? Expectation, so I, the first order term is, is, is, is given by this. And if you take this moment to the power minus p, etc., I get, of course, the three point correlation function with minus gamma over two shift. Now, what I want to do is go to the step after. So it's completely logical what I'm going to do. I'm going to write this minus, so if I want to study tau, I just have to study this moment. I mean, it's not always true, but here it works. It's reasonable to, to start by doing this. Minus p, expectation of r minus gamma over two z minus r minus gamma over two zero. My random variable taken, so I'm just taking the derivative. Of course, if this guy is getting close to this guy, then I should get this as an expansion. So I think everyone agrees with that. Okay, now, what is this guy? Okay, let me, I'm going to do it in steps. So what is this guy and this guy? It's just the integral of some kernel with respect to the exponential of the free field. So it's nothing but minus p, expectation of the integral over c kz of x minus k0 of x. My free field, my exponential free field, sorry. Times r minus gamma over two zero taken at the point zero. The power minus p minus. Now I use Fubini to take this integral out, integral over c. Now I, I, I think I can erase this board too because I recast the notations. So what is this? This is minus p, expectation of, sorry, the integral comes out. The integral over c kz of x minus kz of zero, okay? Expectation of my m gamma t to x. So I'm going to write it explicitly, okay? Have to watch out. Exponential gamma x of x minus gamma square over two. So I'm writing my, my measure explicitly. The exponential of the Gaussian free field we normalized times. So let me write it, so what is r zero? It's nothing but the kernel integrated against, okay. I write it in, maybe in, I'm going to call this variable u now. So I can match the, because I already have my variable x here. And let me write it completely. My m gamma is the exponential of the free field. So the power minus p minus one. So I have this expression, roughly the difference between the, the value. mz and n zero, it's, it's, it's this thing here where I, I integrate against this guy here. Now, now here I'm going to use Gersonov's theorem. Gersonov's theorem says that if I take the exponential of a Gaussian variable. And so this, so let me, let me recall Gersonov, okay? It says that, if I look at some function of the, so imagine I take a, a Gaussian variable. So when I, when I look at this, this is expectation one, this is a chi. This has expectation one, so it's just a change of probability measure. And if I look under this probability measure, a function of my Gaussian free field, okay? So you've seen this lots of times. Well, this thing is the same thing as this. It's the same, it's the free field that I shift by a deterministic function, which is in each point x, the function is worth the covariance of my free field at point x with the Gaussian variable. So I think everyone, this is in the, these, this is in the lecture notes. The Gersonov theorem, okay? So I really stated it. So here, okay, I'm doing this formally, but if you want to make what I'm doing rigorous, it's not complicated. You just work with cutoffs of your Gaussian free field everywhere, and you're allowed to take the exponential of the free field. And then you do everything I'm doing, and then you make the cutoff go to zero. So what I'm doing at a kind of formal level, because this is not defined point wise. You know, it's, it works the same. You just have to, to regularize x everywhere, say. Use Gersonov and then go to the limit. So I apply Gersonov, and this is, this is where you see how the minus gamma over two is going to become minus gamma over two plus gamma. And so plus gamma over two, how this guy is appearing. This is nothing but alpha one minus gamma over two plus gamma. And it comes from Gersonov. So I'm now going to erase. So I apply Gersonov, and what do I get? I get that this thing is minus p integral over c kz of u minus kz of zero. Am I, sorry, I think I'm wrong. Sorry, I'm integrating on the u variable and I'm letting z go to zero. Okay, that's, this is great. k zero of u times, so a function of u where I applied Gersonov. So what is f of u? f of u is equal to the integral of k zero of x. Exponential gamma square, this guy here. Sorry, there's an expectation, of course, everywhere, so. I apply Gersonov, and it's just amounts to shifting the free field. So exponential gamma x becomes exponential gamma x plus the shift due to the covariance. Minus p minus f of u, okay? So now you're starting to look rather well, because if u is worth zero here, then you're going to see that it corresponds to putting, to adding gamma to your alpha one minus gamma over two, and so it creates alpha one plus gamma over two. Also, another remark is p plus one, what is it? Well, it's one over gamma, alpha one minus gamma over two plus alpha two, plus alpha three minus two q, plus one. And so I write one as gamma over gamma, and that makes me one over gamma, alpha one plus gamma over two plus alpha three minus two q. So you see, Gersonov is going to create a log singularity of power gamma, which is going to add with the log singularity of power alpha one minus gamma over two, to create alpha one plus gamma over two. And you see it also at the level of the power here, because remember, the alphas, they appear in the power and in the log singularity, you're integrating in newville. Okay, so what I have to do now is state a theorem which is to study this integral when z goes to zero, and everything concentrates around the value u equals zero. So here's a lemma, the lemma is the following. So if I take a continuous function, say f, whatever the function, I'll apply it to the problem I have at hand here, but it says the following. The lemma says that if I take, so let me write it, let me write the explicit definition here of the kernels. So I'm just copying what's upstairs. If I take u minus z to the power gamma square over two minus u gamma square over two u plus gamma alpha one plus alpha two over two divided by u to the power gamma alpha one, one minus u gamma alpha two. I take any function f, so continuous function that I integrate. When z goes to zero, so it goes to zero, of course. It's equivalent when z goes to zero to z to the power gamma q minus alpha one, okay, where q is two over gamma plus gamma over two. Times f of zero times pi divided by what I want. So this is a purely deterministic lemma I'm stating, okay? So this is just, this is analysis, regular analysis, I have this thing. Where l is, remember my l function, l of z is the ratio of two gamma functions. What do you mean it's a distribution? Yes, you pick up the value, well, there's no need to speak of this. Yeah, you mean it concentrates around zero with, yes, as if you wanted to. Okay, so let me admit this lemma for a while, just to conclude my proof. So this means that when z goes to zero, let me finish. This means that with my problem at hand, I'm equal. I'm going to be roughly equal to minus p z to the power gamma q minus alpha one times f of zero. So if you do, if you look at f of zero, what is f of zero in my problem? Well, it's the integral of x plus. So I just plug in, I just plug in my value here, okay? And zero, this, if u equals zero, this is one minus x for u equals zero. This is log of one over x plus log x plus, okay? The covariance of my Gaussian free field. And so if I, if I do the algebra, well, I get alpha one plus gamma over two. Plus alpha two plus alpha three divided by x gamma alpha one plus gamma over two, one minus x gamma alpha two minus p minus one, okay? Now, if I work at the level of tau, this means that I find that that the p here. Now, I worked with, I did the asymptotic expansion on this guy, okay? So now if I look at my tau function, well, I can see that the p goes with gamma of p and it creates gamma of p of p plus one. And there's also a mu hanging, which, which I don't see in my expression here. But I write mu minus p as mu minus p minus one times mu. And at the end, I get the, the expression I, I gave you in the beginning, okay? So tau minus gamma over two z is roughly. So my correlation function like this minus, so p gamma p, which becomes gamma p plus one and mu minus p minus one times this expectation times pi mu over these functions. And I get my, I've derived my asymptotic expansion, expansion for one over two. And so this is nothing but there's a minus here and gamma p plus one mu minus p minus one, there's a two also everywhere I forget, okay? There's an error up there, there's a two missing, but never mind the two, it's, it's okay. So this guy is going to give me my, my, my three point correlation function. So this guy is going to be, so this guy comes here and I get c gamma alpha one plus gamma over two, okay? So I hope this, this is clear that the idea is that, okay. I may be rigorous, there's a two here, I hope. No, I forget the two once, once in a while, no, it's okay. I also think that I forgot, I'm sorry about these things, but I think I also forgot there's a gamma minus one. But they're global constants, I always, I always forget these guys, but they play no role, I mean, they're just global constants, okay? So all I have to do, if I want to conclude, is I have to prove this lemma here and see the, the gamma functions arrive. So that's not very difficult to, so I wanted to give you one case in kind of full detail. That's what I decided to do today. So my lemma, this thing here, well, it's going to concentrate around zero. When z goes to zero, the mass here is going to concentrate around zero. So, well, when z is going to zero, this is roughly the integral, let's say for u less or equal to, okay. It's going to concentrate around zero, so this guy is one, okay? It doesn't play a role, so I'm just going to look at this case, f of u. And so what do you do here? Well, you just do a standard change of variable, okay? You set u equals zy, so with a norm here. And what do you get? Well, by simple scaling with this guy and this guy and this guy, you see. So I'm going to write explicitly what you get. You get gamma q minus alpha 1. I wrote q explicitly times, so this is a simple scaling. Well, y minus z over z power gamma square over 2 minus y, gamma square over 2 y, gamma alpha 1, f of z, y, okay. And so now you, I think you get it, when z goes to zero, this goes to f of zero. This thing, it's the maximum of z, y and one. So of course, when to the power four, so when z goes to zero, the maximum goes to one. So this goes to one, this goes to f to zero. And then at the end, I'm equivalent to z to the power gamma. So q, f of zero times, well, this integral here. And luckily, these kinds of integrals, of course, were computed by people like, so, z over z bar is something that's isotropic. So you just put one if you want here, okay? And of course, it's not really a, it's not, it's not, it's not by pure coincidence that these kinds of integrals were computed by, by people like Fatih F. And well, Fatih F computed this integral. I think it's due to him, maybe it's due to mathematicians before, I don't know. But this integral, it's exactly equal to this. So you're really happy, you do your little expansion. At the end, you have this thing, and then you spend a few days looking in books, what is this worth? And at the end, if you find the right book, you see that it's exactly equal to that guy upstairs. Excuse me? Doesn't mathematical or maple know these integrals? Perhaps, yeah. I don't know. Okay, you say I've lost lots of time looking at what this is worth. Yeah, next time we have lots of these in our business, and next time we go see you, we ask you, okay, so that's the, so, you see, let me, let me conclude. I was able to do, sorry, the asymptotic expansion around z goes to zero, plus, you know, what I, you know, the three point correlation function with the right shift, so let me conclude what I did. I think it's minus mu pi, yeah, this is, I think it's in Selberg, yeah. So I showed that when z goes to zero, I get this minus mu pi over, you know, these, these ratio of gamma functions. And I did this, so I did this under this condition, okay? There's another one I don't, I'm putting under the rug, but essentially I did it under this condition. Remember, because what happens is when I use Gersonov and everything, I end up with uville, but with a singularity around zero, which is not alpha one minus gamma over two, but alpha one plus gamma over two. And if you recall things, the Cyberg bounds, the extended Cyberg bounds, we're only able to define uville conformal field theory provided the alphas are smaller than q. And so I need this condition for what I'm doing to work, okay? Because if this condition is not true, then this is bigger than q. And in fact, these guys are going to be infinity. What happens if you see exactly a q? You should get maybe a log correction. You don't know what you can do. Oh, you can do it at alpha equal q, I think. No, that's okay. No, the real problem is what happens when alpha one plus gamma over two is bigger than q. Okay, so this concludes what I wanted to say. Today, so I really wanted to go through full detail on these matters. So I have a bit of time left, so I'm going to, it's- Can I ask a question? So all of the calculations in these four lectures dependent very much on the choice of the background metric? No, nothing depends on the background metric. Well, it was, it had this sort of format. Oh, you mean I did everything with a background metric? You mean to just press- Oh, yeah, but look, look, look, look, if, if, imagine I have any background metric. Well, so remember my metric was this. What happens if I take any metric? Well, I mean up to some constant, this thing goes to g of zero. I mean all these conclusions, they, the metric never plays a role in everything when you go. Of course, I could have wrote everything with a, yeah? Where would it, yeah, I mean, okay, you, also, okay, in some sense when you do Gersonov and everything, the background metric plays. But it's only, everything simplifies. I mean really it's, it's really, up to a global constant called the veil anomaly, there is no issue with metrics. The metric plays no role at all. I know what you would expect that just. But here's an illustration. The relationship's not very transparent. Yeah, okay, but here's an illustration. This coefficient here, it's really universal, you know, I mean this, I mean the newville, it doesn't depend on the background metric and you can see it up to a constant. When you, I mean, when you do these kind of fusions, these are called, you know, these OPEs, you, you, you're really zooming around a point so the metric becomes flat if you want. You don't see it. And this is an illustration. What makes this the problem much more complicated for the other value minus two of it? So, okay, let me, so let me spend ten minutes to conclude. So what, what is complicated? Okay, what is complicated is, is the following thing. So I'm not at all going to explain. There are some explanations in, in the lecture notes, but it, it took us a while to, to really do all this rigorously. The real difficulty, really what's difficult when you're doing these and all what I said fails, is when I look at the case alpha one plus gamma over two, bigger than q. And when I look at two over gamma, well, if I, I look at my parameters, I'm always in the case. Okay, I'm, I'm always in the case. No, bigger than. Okay, you're, excuse me? You're x1 and q. Yes, but what I'm saying now is, what I, what I did in detail today is I showed how you do a rigorous OPE, a rigorous Taylor expansion in this case. And now what is difficult, and that's our second paper. I mean, what it took us, you know, lots of technical details in here. What took us some time to really understand and do properly is to do Taylor expansions when I have this condition. When I'm looking at the other insertion, the dual of BPC equation, I'm necessarily, what happens is I, I, I don't shift with gamma over two, but with two over gamma, and I'm always in this case, in fact. So what I have to, I mean, what we have to understand is what happens when I'm in this situation. And then you just copy basically this case to this case for two over gamma. But the crucial point is, well, how do I do an expansion? Because this fails, right? Because at the end, I get, I get something that doesn't exist. And also, this integral here, you can see that it explodes, in fact, so. Well, in words, well, you, you, you, you look at, you look at what happens, you, you look at how this variable, you look at what happens to this variable when z goes to zero. Yeah, how it blows up. And so you, you, you analyze the guy around z equals zero. And what enters the day, the, the game is the tail expansion of, of my, my, my, my GMC, my Gaussian chaos measure. And the tail expansion of this, of the GMC around a point, it's given by the reflection coefficient of viewville, you know, the partition function of your, your theory. And so you need really to analyze so roughly what happens. So it's in the notes, but I, I, I can't, but I, I can, I can, I can state the, the, the proposition. So if alpha one plus gamma over two is bigger than q, then, okay. Let me write it. Then, so proposition, this is a proposition we proved. So let me write it clean, proposition. So imagine alpha one plus gamma over two is bigger than q. Then we can derive that tau of minus gamma over two z. The Taylor expansion is that my four point correlation function gives me same order, zero term, that's easy. Plus r alpha one c two q minus alpha one minus gamma over two alpha two, alpha three z to the power gamma q minus alpha one plus little o. Now the idea, so this is the lecture two. This is the two point correlation function of, of, of, of, of viewville field theory. And it's, it's, it's the, and you see what is called, kind of starting to arise, what is called the reflection principle. Namely that what replaces alpha one plus gamma over two, which I do not know how to define probabilistically, okay. Well, you see that you should replace it by some kind of, you know, reflection of your, your alpha one plus gamma over two times the coefficient, which is called the reflection for that reason, in fact. Now what you can, what, so how does this come in the game? It's because essentially when you're looking at Taylor expansions, you want to understand when z goes to zero what this variable is looking like. And so you have to analyze what's going on, say, around the singularity when z is going to zero. So you want to look at your kernel, okay, around this singularity. And, okay, it's, you can show it demands some, an amount of work. You can show that this variable somehow, you know, it's something which is scale invariant times a Gaussian chaos with a singularity. And then you have to analyze, you know, really precisely the tail behavior of this. And the tail behavior of this, we know it's ruled by the reflection coefficient. That was lecture two. And at the end, you can show this. And if you do this, you're not done at all. And then I'm going to conclude in, so I, I, I'm putting this, so here, I'm telling you in five minutes all the difficulties. So once you get here, you have to show, in fact, another thing you have to show that this is the same thing as this. Minus mu pi, so okay, this is going to be vague, but it's the last ten minutes. So it's just to give you a flavor of what's going on times r, okay? So, so you see what comes out of the probabilistic expansion is this thing. And what we show is that we, this probabilistic expression, we first have to show that it's equal. That, that there are shift equations which work also with the two-point correlation function of the variable. And so by working, you know, with tricks on using symmetry, we're able to show that this guy is worth this guy. And then we have really the reflection coefficient which is working. And so this means somehow that this guy, it's nothing but the analytic continuation of this function. So what, what we prove is that if you want to analytically continue the, the correlation functions beyond the cyber bound alpha bigger than q, well, you, you should, you should replace this by r alpha. So we show that this thing is analytic, okay? So if I take, sorry. If I take this thing for alpha less than q, and this definition for alpha bigger than q, this is an analytic function. So somehow what we do is we show that this thing is analytic, that the reflection coefficient is the right way to extend the probabilistic definition, because we have no idea how to extend it. But then by working with these equations, we're able to show that, okay, that this thing is analytic. And so we can analytically continue the three-point everywhere. And once we show that r, yeah, it satisfies the right shift equations, we see that this is nothing but, by definition, the three-point correlation function with this shift. And we do the same thing with two over gamma. And so with this, this analytically continued three-point correlation function, we can, we see also the reflection coefficient enter the game again. And since it's always the same function, it's these analytically continued three-point functions. At the end, we get the second shift equation. So, okay, it went a bit fast, but essentially the take home message is that if you want to understand a Taylor expansion or UVIL when you're putting a weight bigger than q, these equations we get from the BPC, they enable us to show that the right analytic continuation of UVIL is given by reflecting the weight alpha in this manner and multiplying by r of alpha. And on this analytically continued guys, we're able to show both shift equations and then conclude that they satisfy the DOZZ formula. Okay, so this was last five minutes. It's probably vague, but I think I gave a fair amount of details in my lecture notes. But it would be completely pointless to try the last five or ten minutes to show you how the Taylor expansions work. There are lecture notes for that and I don't want to. I want you to have a good impression on these lectures. And I feel that the first hour today was rather simple to follow. And if I start doing this guy, I try to show these expansions. It's going to be a mess in ten minutes. I think I'm going to conclude here. Thank you.