 OK. Thank you, Pepe. So I announced that my lectures will be on character varieties in complex dynamics, but in fact, I switched a little bit the subject. It will still be on character varieties, but there will not be a lot of dynamics in these lectures. In fact, what I would like to talk about is about some recent works on the question of dominations in the character varieties of surface group with values in the group PSL2R. So we already heard about these groups this morning. We studied lectures. It was not PSL2, it was just SO21, but these are almost the same groups to finite index group. So let me begin with speaking a little bit about this group, PSL2R, which will be the main subject of the study in the three lectures I'm going to give. So this group is the group of matrices, squared matrices with real coefficients and with determinant 1. But we want to identify a matrix with its opposite. So this is a P here. So we are cautioned this group by the subgroup consisting of the identity matrix and its opposite. So this group will really be the main subject of the lectures. And more specifically, we are going to be interested in the representations of the fundamental group of a closed-oriented surface, S. So S is a closed-oriented surface. So we know the classification of topological classification of surfaces. And surfaces, they are classified by their genus. So genus 0, we have the sphere. The echelogen sphere in three space. Genus 1, a torus. And all the surfaces are obtained from the sphere, like the torus or iogenus, by adding some handles. And so the classification tells us that the only invariant, so what I wanted to say is that the only invariant for understanding a closed-oriented surface is an integer invariant, which is the genus. This is the number of handles you have to add to the sphere to get your surface. So we'll take such a surface. And we will look at its fundamental group. And we look at the set of representations of the fundamental group of that surface into the group PSL2. So that will be the main, what we're going to do in the three lectures. So today I would like to speak about some really particular and really important representations, which are coming from uniformization theory of Riemann surfaces. And these representations are called fuchsian representations. And the whole lectures today will be devoted to define them and to prove the uniformization theorem for closed Riemann surfaces. I mean, prove this is not true. This is a hard theorem, so we will try to have some indications of proofs of this theorem. So that will be the goal for today. And tomorrow we will see what will happen. OK, so let me first erase this for some moment. And let me first introduce extremely important object associated to this group, which is its symmetric space. We have seen this morning in third lecture. This is the helper half plane. So this is a set of complex numbers with positive imaginary part. And the group PSL2R acts naturally on this set by a homography. So if you have a matrix A, B, C, D, and you have a complex number of positive imaginary part, you make this matrix on this number by this formula, AZ plus B over CZ plus D. And it is really easy to see that the imaginary part of this number is still positive. In fact, if I am not mistaking, the imaginary part of AZ plus B over CZ plus D is, let's see, so the determinant of the matrix, so it's one here, or mine, so it's one, and times imaginary part of Z over the normal of the complex, the modulus squared, the square of the modulus of the complex number CZ plus D. So you see that this group PSL2R acts on the helper half plane. And in fact, it is quite easy to see that the action of PSL2R on the helper half plane is transitive, namely if you take two different points in the helper half plane, you may find an element of PSL2R sending the first to the second one. And so you can identify helper half plane with PSL2R divided by the stabilizer of a point. And if you take the point to be the square root of minus 1i, this positive imaginary part, then the stabilizer in PSL2R of this point is just the set of rotations as O2R. So this is really easy to see that. And so this gives you an identification between the quotient of PSL2R by the subgroup SO2R with helper half plane. And so here, we see that these subgroups of the point i is, in fact, so it is a compact group. And in fact, it is a maximal compact subgroup of PSL2R. It is a maximal compact subgroup. And so this is the symmetric space of this semi-simple group. So it carries naturally a Riemannian metric. In this parametrization of this symmetric space by the helper half plane, this is quite easy to write this metric. So the idea here is that if you take the stabilizer of the point i, we see that this is a group of rotations. And if you compute the derivative of an element of SO2R at the point i, you see that it is a complex derivative, I mean. This is a complex number of modulus 1. So in particular, it preserves a gradient metric at the point i. And so you can push this metric by the whole group PSL2R to get a metric on the helper half plane. And this metric would be invariant by the whole group PSL2R. So this is exactly the metric of the symmetric space I was speaking of before. So this metric can just compute it. It's really easy. So this is, if you write your complex number z as x plus i y, where x and y are real numbers, then you may consider this metric, which I am going to denote gp, because it has been introduced by Poirier. And this is a metric on the helper half plane, which is invariant by the whole action of PSL2R. And in fact, it is the unique one up to multiplication by a positive constant. In fact, I almost gave the proof before. So this metric is going to be really important. And I will call it what kind of metric in the lecture, or maybe sometimes I will call it the hyperbolic metric. And so in fact, we can see easily with the kind of arguments I just gave that the isometry group of the helper half plane equipped with this metric is almost the same as PSL2R. In fact, this is the same if you ask that the isometry preserves orientation. So to conclude the discussion, what we just have shown now is that you can see this group, PSL2R, as the isometry group of the helper half plane equipped with this Poirier metric. So it's a positive isometry group of this helper half plane. So in terms of Riemannian geometry, this metric has very nice properties. So namely, it has constant curvature. It is natural because the isometry group acts transitively. So it has constant curvature. And we can compute it. It is minus 1. So the curvature of GP is equal to minus 1. And in fact, the helper half plane, it is quite easy to see that this is a unique, simply connected surface, Riemannian surface of curvature minus 1, which is complete, which means that geodesics run for all times. So OK. So now what I would like to do today is to speak about a very well-known dictionary, which works for surfaces, namely that there is a dictionary between from one side hyperbolic geometry. So the geometry of this metric, which I think we will heard more precisely in Francois Gerito's lectures this afternoon. And on the other side, complex geometry in one dimension, namely Riemann surfaces. And so I'm going to speak about this dictionary today. This is quite really classical files, but I think it is very interesting. And it will be very important for the second and third lectures I end to give. So the first, how to say that, maybe the first result, which is quite simple, but which is extremely important, is Alema, which. So this dictionary has been, in fact, known from maybe 19th century, maybe the beginning was Schwarz and maybe before that, maybe Riemann already. But here is Alema from this mathematical Schwarz, German mathematical from the 19th century, which says that for any holomorphic map on other half-planes to itself, such a map has to decrease the hyperbolic, the poincare metric that we just defined. So let's write it like that. So holomorphic maps between hyperbolic planes from the hyperbolic planes to itself decreases this poincare metric. OK, let's prove this, Alema. So the proof of this, Alema, is, generally, is done by looking at another model for hyperhalf-plane, which is the unit disk. So if you consider the unit disk with variable w in this unit disk, and you consider the hyperhalf-plane, so there is a conformal map, the projection between these two open subset of complex plane, which is just given by w. OK, you see that the modulus of z minus i is less than the modulus of z plus y if z is in the hyperhalf-plane. This is closer to i than it is from minus i. So in fact, this homography maps hyperhalf-plane objectively to the unit disk. So you have this identification. And so in fact, if you want, so you can push the poincare metric defined here to a conformal metric on the unit disk. And then what you want to prove is to prove that if you have a map from the unit disk to itself, it will decrease this metric. So this metric, in terms of w, I have not verified there is a constant. I suppose this is 4. It is given by the following expression in terms of the w variable. This is the expression of the poincare metric in the unit disk model. So what we have to prove is that if, as I said, if we have a map, homomorphic map from the unit disk to itself, then it decreases the poincare metric. So now we know that we can, so if we have a homomorphic map, we can compose it by an isometry of, so hyperhalf-plane, or unit disk. It depends where we are working. And because this isometry is homomorphic, so we don't change the problem. So we can assume, so all we have to do is to prove that, so using this isometry that we just described, so the group PSL2R, the H model, we just have to prove that if we have F from the unit disk to itself, we can up to composing F by some isometry. We can assume that the 0.0 is mapped to itself. Then we have to prove that the derivative of F at the 0.0, so this is a complex number, is less than 1. This is exactly the expression of the fact that the metric at the 0.0 is decreased by the map F, right? So we need, so this is, so using this identification between H and D and the action of the isometric group PSL2R, this is exactly the reformulation of Schrodinger. So how do we do that? This is quite easy. We just express F as a series, so I recall the proof. So F is a homomorphic map from the disk to itself, so in the W variable, this is just a series. So it maps 0 to 0, so the first coefficient is 0, so we have the second coefficient corresponding to the power to the W term, and we have a series expansion like that. And we want to prove that alpha 1 is less than 1 in modulus, so we just write it like that. We write this series expansion as a multiplication of W by a certain function. This function is this one, and this function I'm going to denote it by H of W, right? So we have F of W is equal to W times H of W, right? But now, so we need to prove that H of 0 is less than 1. This is exactly what we want to prove. This is H of 0 is exactly alpha 1. So how to prove that? So we will use the mean value. So it is a kind of maximal principle. We will use the mean value theorem for a homomorphic map, so namely, if we, so if I consider a real number R, which is less than 1, positive real number radius, which is less than 1, then I can write this formula. So here, I forgot to divide by 2 pi. So namely, the value of the homomorphic function as a 0.0 is just the average value of that function on any circle around 0. And I push the circle to the boundary of the unit circle. But now, is it right? Is it correct? Yeah. So now I know that, so what is the proof now? OK. Let me try to think of a problem. Right, I have to use that. OK. So now, so let's do some proof. Because later, there will not be some proof like that. So let's do that right. OK. So we have this inequality. And now, this is exactly H is equal to F over W. So the modulus of W on this circle is just R. So we just get this inequality. And now, we know that F takes values in the unit disk. So this is less than 1. And so we get that for any R bigger than 1, the value of the function H as 0 has modulus less than 1 over R. So taking the limit, when R tends to 1, we get our desired inequality. That absolute value of H of 0 is less than 1. OK. So we get this Schwarz lemma. And in fact, this Schwarz lemma is extremely important. So one of these is first corollary is the fact that if you take a complex automorphism of the upper half plane, then in fact, it is an isometry for the poincare metric. Because the map decreases the poincare metric. And the inverse of the map also decreases the poincare metric. So the poincare metric has to be invariant by the map. And so you get something really important that I am going to write. I'm going to write what I just said. Namely, complex automorphisms of H. So by this, I mean the set of biomorphism of upper half plane is the same as the positive isometries of upper half plane. This is the same as, I mean, it identifies with the group PSL2 are acting by homographism. OK. So this set of identity, so this identity is really what I meant. So one of the instance of this dictionary between here, Taishmula theory, at this level, hyperbolic geometry. And maybe we could put another instance in this dictionary, which is here, the group hacks by homographies on upper half plane. So we have here projective geometry. But this instance of the dictionary I will not really speak about in these lectures. OK. So let's move and state one of the main results. So as I said, this dictionary has been really well-known since the 19th century by, of course, Poirier. And during the whole 19th century, it was, this dictionary was developed until it's, I don't know how to say that in English, abutismo, maybe. Anyway, until mathematicians understood that statement, which is now called as the uniformization theorem, I am going to state part of this theorem. So there are many mathematicians here, Klein, Poirier, Schwarz, and many others, Kober. But the theorem, I am going to state it in the following way. Any closed Riemann surface of negative other characteristics, or of genus bigger than or equal to 2, carries a unique conformal metric of curvature minus 1. So in fact, if you have any closed Riemann surface of genus bigger than 2 or equal to 2, then you can think of this surface as a hyperbolic surface. And in fact, this theorem can be made more precise. So I wanted to state this theorem like that because this is one of the form of the uniformization theorem, which we will use later. But in fact, it is more precise. So here is a statement that, in fact, Poirier proved in, I think, in 1898. It is more precise than that. So in fact, for any negative function, so let's call the surface S phi S to R. So it is negative. It takes negative value. And it is a smooth, in fact, of class C1. There exists a unique conformal metric, curvature, this function. So we can prescribe the curvature of a conformal metric on our Riemann surface. And in fact, as we will see, it's really important also. And also for the theory of representations of the fundamental group of a surface in PSL2R. So this version has been proved by Poirier in an article. It's kind of interesting what happens here. So this is an article Poirier wrote in this year. And in fact, it was completely forbidden during many years. And it was reproved in 71 by Dergé, who gave a different proof 70 years later. It was reproved by so using. So at the time of Poirier, he hasn't. So this is a theorem about partial differential equations. So Poirier really gave a proof by hands by solving this PDE by hands. And the proof of Dergé is much more modern. It uses all the tools that we know to solve PDEs, electric PDEs, namely sobolef, sobolef spaces, sobolef embeddings, et cetera. But what I would like to do today is to try to describe for you what Poirier did. It's kind of dirty at some moment. So I'm going to skip all the technical details. But I would like to show that if you are really brave, you can solve these kind of nonlinear partial differential equations by hands. So brave and generous. But it's really not. So it's really not. There is nothing deep in the Poirier approach. So who will see? So just first, before I go into some parts of the proof, I would like to say why this theorem is related to representations of fundamental group of surfaces in pièce à trois. I don't know, in fact. Marcel probably, yeah. But there is another one. I can tell you that. So if you let me one minute. So actually, I'm going to make an advertisement. So here is a book that we wrote with many people. So we are maybe a group of 15 people that we wrote to retrace the history of this uniformization theorem. So there are many proofs of this theorem. So here is the expression of this theorem, this uniformization theorem for closed surfaces of negative error characteristic. But in fact, the uniformization theorem is valid for any Riemann surface. Even if it is not closed. And so there are many proofs. And we wrote a book about all these proofs and the history of this problem. So this is called uniformization de surface de Riemann. And there will be an English version, I think, available soon. I have a question. So you're looking for that? Yeah, right. So OK, so Berger, you're right. Marcel, M.S. Berger, so probably Marcel. Journal of differential geometry, 71. Riemann's structures of prescribed Gaussian curvature for compact two manifold. You're up early. Sorry? Maybe it's going to be. Do you want the pages? They are there. M.S. Berger. M.S. Berger. Berger, OK. So that Berger. You're the first person. You're the first person. But you don't believe me? I don't believe you. OK. I don't know. My question is to you. Berger, so my question is to you, Berger. OK, maybe. You did that. You know, you know, it's crazy. It's crazy. OK. OK, so here is a. So I'm going to reformulate the first part of the uniformization theorem. And so the thing is that if you have, so if you take a Riemann surface, so maybe I should recall. So who knows, who don't know, what is a Riemann surface? Please, hands. That's OK. OK, I don't believe you. Sorry? I don't answer the question. Do you want the definition? The definition? OK, do you want the definition? Yeah, this is the definition. OK, I was just thinking about the definition of what is a Riemann surface. So everyone knows. OK, so a correlation of this, of the first part of this uniformization theorem, is that any closed Riemann surface with negative Euler characteristic, right in this way, is the homomorphic to the quotient upper half plane by a subgroup, which is discrete and torsion free. So let me give some comments. So if you take a subgroup of PSL2R, which is discrete, in fact, using the fact that PSL2R acts by isometries on the upper half plane, you can show that this group acts properly discontinuously on the upper half plane. So third explained this morning what is the definition of properly discontinuously. And so in fact, you can make, so this means that you can make the quotient of the upper half plane by your group. And in this quotient, you will have a nice topology, which is house off, right? So this is, and torsion free means, so it is easy to see that it is equivalent to the fact that the action is free. You have no fixed point unless, for any element of your group, this element has no fixed point unless it is identity. OK, so in fact, what it means is that if you have a discrete and torsion free subgroup of PSL2R, if you make the quotient of the upper half plane by this group, you get a Riemann surface, right? And so a corollary of the uniformization theorem is that you can see any closed Riemann surface negative alert characteristic in this way, right? And so this is just, so the proof from the uniformization theorem to the corollary is just look at the universal cover. So look at the Riemann surface of negative curvature. Take this unique conformal metric of curvature minus one. So now you have a hyperbolic surface. So the universal cover of this hyperbolic surface, it is a simply connected, complete surface of curvature minus one. So it is positively isometric to the upper half plane. And so you get this color. So here the group gamma in that proof is going to be the covering group of the, so the fundamental group of your surface acting on its universal cover, okay? So it is interesting to say that unless it is quite trivial, but it is interesting because you see that you get, in fact, so if S denotes our, my original Riemann surface, you get an identification from the fundamental group of S with the group gamma by which you have the upper half plane. And it gives you, in fact, a faithful representation. You can think of that as being a faithful representation of the fundamental group of S to a discrete subgroup of PSL2R. And in fact, let me give the definition. So the representation that you get in this way is called friction. Those are the representation I want to speak today. So there are friction representations which comes naturally with hyperboleization of Riemann surfaces. Okay, so now I want to give some ideas of proof of this uniformization theorem. I will try to, as I said, I will try to skip the technical details, but still this is analysis. So maybe for most of us, let's say, it can be quite hard sometimes. But let's try to understand that. I will keep on. So first, I would like to say some words about curvatures and to express the PDE that we will have to solve. So what we will try to do is the following. So S is our surface, which is closed and of genes larger than or equal to 2. And we want to find a conformal metric on S, which has curvature of minus 1. So what we can try to do is the following. We can equip that G, so G naught, be a smooth conformal metric on S, for instance, using a partition of unity. We can find such a metric. And what means conformal here? It means that, so the conformal structure induced by this metric is our Riemann surface structure. But in more concrete ways, if you take Z, a holomorphic chart. So this is a biomorphism between an open set of S and an open set in complex plane for the Riemann surface structure on S. Then the metric G naught is going to be expressed as exponential phi naught, phi naught is a function, times dz modulus to the square. So this is an expression of, a local expression of our metric, where phi naught is a function, a smooth function. And so this is what the meaning of being conformal. And so I want to, so I'm not going to define what is curvature. I would like to just to say that, so if we have a Riemann metric on a two-dimensional surface, we have a curvature induced by this metric, which is just a function in two dimensions. And if the metric is conformal expressed in this way, we have a formula for the curvature. So the curvature of G naught is kappa of G naught is equal to minus one-half of the Laplacian in the Z variable of phi naught over exponential phi naught. Okay, so this is an expression that you can derive. This is just compute. So when you just look at the definition of curvature, you compute and you find this. So what we will do now, we will try to multiply this metric by a function, so to consider another conformal metric on the surface S, in such a way that it becomes of curvature minus one. So we look, so we are looking a metric which is expressed by the following expression, right, where sigma is a smooth function from the surface to the ring numbers. And so that the curvature of this new metric, G, is identically equal to minus one. So that is the differential equation which will be satisfied by our function sigma. So let's just write this differential equation. So, okay, so let's do some computations. You can start sleeping for a while. I will tell you when you can wake up. So there is three lines, maybe five minutes. Okay, so this is for the first part of the theorem, but what we want to do is to prescribe our curvature. So here, in fact, it is some given function kappa, which is negative. This is the only hypothesis. And we want to solve this equation. So kappa is given. We want to find our function sigma so that the curvature of G is equal to kappa. So, okay. So the local expression, so here. Okay, so G locally is just exponential sigma plus phi naught times dz squared, right? So the curvature of G locally is just minus one half laplacian of sigma plus phi naught exponential sigma plus phi naught, right? Okay, so I'm going to rewrite this equation. So we have, so here, this is a laplacian with respect to the z variable, right? So what we have is that we have laplacian in the z variable of sigma plus phi naught is equal to minus 2 times exponential minus 2 kappa, exponential sigma plus phi naught, right? This is the laplacian in the z variable of sigma plus laplacian in the z variable of phi naught, which is by this expression given by curvature, so this is minus 2, curvature of G naught, exponential phi naught, right? And this is equal to minus 2 kappa, exponential sigma plus phi naught, right? We divide everything by exponential of phi naught and we find the nice expression, so it's, do you see that? So we find 2 kappa of G naught minus 2 kappa, our function, exponential of sigma. So we find an expression which involves only our original metric G naught, the function sigma, the density sigma that we are looking for, and here there is this operator, exponential minus phi naught times laplacian in the z variable, so it's kind of bad in a sense, but this is, I mean, in fact, so this is the density, so the inverse of the density of the G naught metric, and the laplacian in two dimensions is conformally invariant, so this is nothing but the laplacian in the G zero metric of sigma. I'm going to rewrite the equation, okay, so maybe I'm going to rewrite it here, just so the equation that we want to solve is just minus 2 kappa, sigma plus 2 times kappa of G naught, right? So this is the equation that we want to solve, okay? So maybe we can say, okay, we are done. We have written the equation, like in the t-shirts that we can buy. This is the equation of the universe, and this is, but we will try to solve it. So the proposition is the following. So the proposition is for any, so let S be a closed remand surface of a negative characteristic for any negative function of class C1, and for any, no, no, sorry, sorry, sorry, this is a positive function for any, no, sorry, sorry, sorry, sorry. Okay, I made a mistake, sorry. Let's write this, okay, and I'll do it again. So you take two functions which are, let's say, smooth, but in fact of class C1. So assume that theta is positive and phi is positive in average integral over the surface of the function phi with respect to the volume form induced by the metric G naught, okay? Is positive, right? Then the following equation has a unique solution. And so you see that from this equation to that equation you just, so you just put theta equal to minus 2 kappa. So kappa is our original function that we want to, for which we want to prescribe curvature. So it is, by definition it is negative, so minus 2 kappa is positive function. And this function, kappa of G naught, this curvature of the metric G naught, if you integrate the curvature you find up to 2 pi is a large characteristic, so you find a negative function. So if you put minus phi is equal to this function, then you get that phi is positive in average, right? So you get here that integral phi dv of G naught is positive by cos bonnet inequality. And so in fact you just apply this proposition to get the function sigma which satisfies this equation and getting back you will get a metric given by this expression whose curvature is exactly kappa, okay? So we just need to prove this proposition, so we need to solve these differential equations, okay? So, okay, so in fact as you see this differential equation is hard to solve for various reasons. So the main reason it is difficult to solve is that you have this nonlinear term exponential of sigma involved in this equation. And the way Poirier solved this equation is quite amazing. So what Poirier says, he says, okay, this is a very difficult equation. So what I am going to do is I am going to solve simpler equations. And in the space of all differential partial differential equations on my surface I am going to move and successively solve this differential equation step by step and at the end I want to end in this equation. And so it's a kind of continuity method in the set of partial differential equations in the surface S. So here I am going to just list all the steps. All the steps are four steps, I think, and you will see all the differential equations that Poirier is able to solve successively. And so the first equation that Poirier solved is Laplace equation. So now it's extremely well known, extremely easy to solve, but in his paper Poirier gave a new proof of that. So let me take my notes. So the first, so Poirier says, okay, this one is too difficult, but I know how to solve the following differential equation. So I am going to skip, so for now on just to simplify notations I am going to note by the Laplacian, the Laplacian with respect to the genot metric. So our original conformal metric. So let's write, okay, so let's do it in this way. Okay, so Poirier erased the nonlinear term, he said, okay, let's erase that one. I know how to solve this equation provided that, so it means, what does that mean? It means if you give me phi, then I can find a sigma whose Laplacian is minus phi. Okay, of course this is not true in general. We have to impose some condition. Namely, we know that the integral, the average value of the Laplacian of a function is always zero, but this is the only obstruction. So this is Laplace equation. So if integral of the function is zero. If phi is given, we can find sigma such that this is, so that its Laplacian is the function minus phi. And in fact, we have some, I will give you some, later some, we have some estimates on the norms of sigma, on the norms of the function whose Laplacian is phi in terms of the norm of phi. But let's try to just skip all these things about norms and et cetera. So the second equation, so Poirier moves to another kind of equation. He says if I am able to solve this equation one, so the Laplace equation, I will be able to solve the following equation. So delta of sigma is equal to lambda sigma minus phi. So here eta is given is positive. So it is a positive function. Phi is given. It is any function and Poirier says I can solve this equation. I can find sigma such that delta, the Laplace of sigma is eta sigma minus phi. And so this is, here it is, now it is more complicated than the Laplace equation because we have added this linear term. But still it is linear. And then he moves to the, directly to the non-linear equation. So he moves to our equation, delta of sigma. But in fact, so this is done by, in Poirier's paper, it is done by hands. I will maybe say a word about how this is done, but it is quite classical to solve this equation, Laplace equation. And for the second and third step, they are divided into several steps again. So this equation, in fact, there is two steps. Poirier proves that if you fix eta and phi, then for small parameter lambda, you will be able to solve this equation where lambda is a real number, a positive real number, which is close to one. And then, successively, he says that if he is able to solve this equation, then he is able to solve this equation for bigger lambda and he is able to increase the value of lambda to get to this equation. And the same is true for this third step. It is divided into several steps to get to this equation. What I do, I don't want to bother you too much about this or so. What I want to do is to try to indicate very briefly how Poirier can do to solve, let's say, this equation assuming Laplace equation is solved. Just that you can see what is the method of Poirier. We will just do that and then I will finish to bother you too much with this. Sorry? Eta is positive. Eta is positive everywhere. In any of these steps, eta is positive. It is really important. OK, so just to say, I think the Poirier's paper is maybe 45 long, so it's kind of really long. There are many steps to, but each step is extremely elementary. So let's see one of these steps. OK, so I assume that if there is a function on the surface whose average is 0, then I can find a function whose Laplacian is my original function. I assume I can solve the Laplace equation. So let's see how we can solve this equation here. OK, so we want to solve this equation here. So eta is a positive function given. Phi is a given function which is smooth but nothing more than that. And lambda is a small parameter. And for each small value of lambda, we want to find sigma which satisfies it. So of course it is natural to think that the solution, if it exists, it will depend on lambda analytically. So let's express sigma. The solution, we are going to express it as a series that I am going to express in this way. You will see why in a moment we have to do this. So we are going to look for our solution in this form where sigma k are smooth functions and ck is a constant for any k. And now the function phi is a given. But for any parameter lambda, I can write phi in the following way. Phi naught plus lambda phi 1 where the average of phi naught is 0. There is a unique way of decomposing a function as a sum of two functions. And so now let's just write the equation that we are looking for. So I will just erase the other kind of steps that 4k has to solve. So here what I assume is that I can do that. I assume that I can solve this equation. Namely, if you have a function whose average is 0, then you can see this function as just another function. I assume. No, no, no. In 2 and 3, yeah, it is not. So for 2 and 3, we don't need to assume that phi has average 0. Yes, this is maybe a bad thing about notation. So the function phi may be different for each of these steps. So here phi has nothing to, so if it is of average 0, it's not necessary to make this transformation. So let's just write, so we have the Laplacian of that. So if I write this equation here, let's write it here. So we have Laplacian of sigma. So it's going to be a series. So the constant disappears because we, we derivative, we make a derivation. So lambda sigma 1, lambda squared Laplacian of sigma 2 plus et cetera. And this is, so this is minus Laplacian of phi. So this is minus Laplacian of phi 0 minus lambda Laplacian of phi 1. Plus lambda eta times sigma. So the first term is going to be lambda eta sigma 0 plus c0 plus lambda squared eta sigma 1 plus c1 plus et cetera. And so we, we identify the terms and we get the following equations. So Laplacian of sigma 0 is equal to phi 0 Laplacian of sigma 1 is equal to minus Laplacian of phi 1 minus plus eta sigma 0 c0. And then we get, so the sequence of equations. Okay, for any k. There is a minus here. Oh, sorry. Like that. Oh, no, no. I am wrong here. Sorry. Yeah, right. There is no Laplacian here. Thank you. Yeah. You are not sleeping. Yeah, yeah, right. So here this is like that. Yeah, of course. And here this is like that. Good. Okay, so we are almost at the end of that through, I mean of the demonstration that I want to do. So you see we have chosen the decomposition of phi as a function phi not plus lambda phi 1 in such a way that the average of phi not is zero. So we can solve the first equation. Okay, we can solve find sigma not by Laplacian equation. Right. So now now we at the at the second and third, et cetera, steps we have equations which for which you see a constant which appears. And in fact, the value of that constant is given. You can compute it once you have. So what I mean is that the constant C cap, you can compute it once you have find found the functions sigma k. Why? For instance, for this first equation, if you integrate here, you get zero because this is the integral of Laplacian. And here you get integral of phi that you know, minus 5, sorry, phi 1 and plus integral of eta sigma not, but you know, sigma not, you already get get sigma not from the first equation plus C not times integral of eta. So C not you get it from. So you put C not is equal to the value that you have to do. And then you can solve this equation because then you have chosen C not so that the integral of this function is zero. So you can solve this equation and et cetera. You do at each step you so you find sigma k by solving the Laplace equation that gives you the value of the constant K and you continue. And so in fact, formally, you can solve this equation. The whole question now is to prove that convergence occurs and so you have to work. But this is not hard. I mean, you have to just know what you get in the Laplace equation in terms of the norms. And then you have to estimate in all this process the norms and to prove that the series you are constructing converges and then you will. So the statement is that you do that and for small values of lambda, the series you get converges. And so you can solve this equation. And so I don't want to say more today. I mean, I think this is quite incredible that Poirier was so brave to do this procedure. I mean, it's completely crazy. But you can read the article. It's amazing to see all these pages of computations. OK, so thank you.