 Okay, so today I would like to speak about some concepts which has been developed in the beginning of late 90s, probably, which is a concept of domination between representations of surface grouping in PSL2R. So let me give the definition. So this is a way of comparing representations of surface group in groups of isometries. So the definition is the following. So let G, rho be two representation of the fundamental group of, again, S is a closed oriented connected surface of genes larger than or equal to two. And we take two representations. One is called G and rho is, the other is called G and rho. So let's, we say that J dominates. Let's put the definition in that way. If there exists a Lipschitz map from a half plane to itself, Lipschitz map, which is all equivalent, meaning that, so this terminology means that for any X in H2 and for any gamma in the fundamental group of our surface, we have phi of J of gamma X equals rho of gamma phi of X. So this is a kind of order between representation, a way of comparing representations, the various representations of the same fundamental group of a surface in PSL2R. And in fact, there is an alternative way of, I mean, a more strong version of domination, which we will call strictly, I don't know, maybe this is not the right term to use here. So we'll say that G J strictly dominates rho. If there exists a Lipschitz map, so, but with a constant of, a Lipschitz constant, which is strictly less than one. So Lipschitz map means that the distance between the image of two points is decreased by the map. So this should be true for any couple of points in a half plane. And for strict domination, we will ask for C Lipschitz map with C, a constant between 0 and 1. And this means that we have this inequality for any couple of points, X and Y in a half plane. So what we are going to do today, so yesterday what we did, yesterday we defined the volume of representations and we saw that it can take only a finite number of values and we announced the result that the representation of maximal volume in modules are the frictional or anti-fictional representations. Today we are going to prove, in fact, something more than that. We are going to, in fact, so yesterday we proved that with respect to volume, frictional representations are maximal. Today we are going to prove that this is also true for the domination, for domination. Namely, that if you take, so, friction representation, maximal representation with respect to domination. So in general, so J will always be the representation which dominates, so this will be the representation. And in fact, we are going to prove that friction ones are the maximal ones, so J will play the role of the friction representation. So here, in fact, this action you can just think of this action as being the action of the fundamental group of the surface on its universal cover once you have hyperbolized your surface with the hyperbolic structure corresponding to the J. So this will be the plan for today and as a corollary, we will get of the proof that friction representation are maximal with respect to domination. We will get Goldman's theorem that this is, they are also maximal with respect to the volume. So here, you see that this notion of domination gives you another in the set of representation. Namely, you will say that one representation is bigger than another representation if it dominates it. This is another. In fact, it's really, it's transitive and it's, so if J dominates J, let's say rho and rho dominates J, then rho is equal to J. If J dominates rho and rho dominates, let's say, rho prime, then J dominates rho prime. This is an order, total, okay? This is not a total order, but it is an order. So maximal with respect to this order. This is not so clear to prove this kind of thing. Up to conjugation. Yeah, right. Sorry. Up to conjugation. So those things are not so trivial to prove. I will try to not speak about this order, but there is an order which comes from domination. And in fact, what we are going to prove today is that the maximum element with respect to that order are fixed representations. No, no, this is not a total order. So sometimes you cannot compare, represent it. Is it a tree or a lattice or something? I don't think so. Have you understood the question? So for the, that's right. Good. So it's not, yeah, right. So we will see that we cannot dominate. So a fixed representation is maximal. So if you take two different, you cannot dominate. There are many maximal elements. So we can wonder why we are doing, we are caring about this kind of order on representations. And in fact, so I would like to say some words. So in fact, it is related, this concept of domination to the construction of space time in three dimensions, which are quite different from the space times that Todd was speaking about in his two lectures. Namely, these space times are curved space time of constant negative curvature. These are called ADS manifolds. So these are manifolds equipped three dimensional. So these are three dimensional manifolds equipped with Lorentz metric. So a family of metrics on the tangent space whose curvature has one, two, is one, two, curvature, Lorentz metric. And we ask that the curvature of those metrics is negative. So the curvature of form is minus the metric. So in fact, I decided not to describe exactly the structure of these manifolds. But what I can say really briefly is that there is a model for this ADS-3 geometry. So it's a rigid geometry which can be described by locally by a standard model. And the model I can tell you in the world what it is. You take this group, PSL2R, this is a three dimensional manifold, and you equip this, so you think of this as being a three manifold, and you put on it the, what is it called? So we have the killing form. Thank you very much. The killing form, you were describing yesterday, maybe yesterday talk. So you consider the killing form which is of signature one, two on this lead group, and so this is a Lorentz manifold with lots of symmetry. Namely, if you multiply by an element of PSL2R on the left, it's going to be, to preserve this killing form. And also if you multiply by an element on the right, it's going also to multiply, to preserve this killing form. And this is a model for this geometry. And in fact, there has been many works beginning with, in the 80s, by Ramon and Kulkarni. And many people proved that in fact, such manifolds, if they are compact, up to finite coverings, they are all of the form PSL2R divided by following relation, X, an element of PSL2R that I call X, maybe add here, G. If you take G, an element of PSL2R, you are going to identify it with rho, gamma, let's say, G, gamma inverse, where G and rho are two representations from a certain fundamental group of a certain surface of genius larger than two. In PSL2R, such that G strictly dominates rho. So there are many non-trivial things to be seen, so maybe one thing that is not so hard to see and that I let as a, let's say, hard exercise, it's an exercise, to see that if you have a strict domination, assume that rho is strictly dominated by J, then this action, so the action of the fundamental group of S on PSL2R given by this formula, acts properly discontinuously on PSL2R. So this is my exercise that I propose, and in fact, you can also prove that the quotient of PSL2R by this, such an action, is compact. This is not so difficult to prove that. So in fact, those three manifolds are quite easy to define, but it is quite very difficult to prove that this gives you up to finite index as a list of all closed ADS three manifolds. So maybe to put some names, so Kulkarnir and Raymond, I don't know how to pronounce his name in English, Raymond, proved that quotients of the universal cover of this group that are compact, that are compact, that are given by such a form, where G and rho are some representations. Maybe at this time we didn't know that rho had to be dominated by J. Then there is a theorem by, I hope I will not, by Klingler, saying that in fact, any ADS closed three manifold has to be covered by the universal cover of this model, so essentially this model. And then the fact that the representation of rho has to be strictly dominated by G is quite recent theorem by Fanny Kassem. I hope I don't miss names in this classification. So, I think that the fact that the representation of rho is, sorry? Everybody has to start with K. Yeah, that's right. Yes good remark, there's three K, three dimensional. I don't know. Okay, so I won't speak more, sorry, anti-deceit manifolds. anti-decider, anti-decider manifold. Okay, I won't speak more about these space times and three-dimensional geometry. I will really focus on this notion of domination for representations of surface group in PSL2R. So let me try to see what I want to say now. Okay, so maybe I will state more precisely some results about dominations and then we will go to the proofs. So the first result, so maybe there is a recent result which in a sense closes this story of classification of ADS-3 manifold. Sorry, I said that I didn't spoke more about that, but we should put a name here. Nicolas Tolozon, who recently proved the following theorem and next I am going to give some particular case of this theorem, which is a theorem which can be stated only in terms of representations of surface group, but in fact, it really closed the story of the classification of those closed three-dimensional ADS manifolds. So the theorem is the following for any representation of volume satisfying, so not maximal volume. The set hyperbolic structures, so of fuchsian representation that dominates Ho is homomorphic to a copy of Taishmuller space of genus G. T of G, this is Taishmuller space. So we know that in fact, this Taishmuller space is a ball six G minus six real-dimensional ball, so I could say this is just homomorphic to a ball, but in fact, the way Nicolas do is, he gets a really natural parametrization of the set of fuchsian representations that dominates Ho by Taishmuller space. The natural map from Taishmuller space to this set of representations that's, so I forgot the word here. Sorry? Taishmuller space of S? Taishmuller space of S, right. This is homomorphic, this is well known, this is homomorphic to a ball of dimension, real dimension six G minus six, right. So in fact, today I will not be able to give the proof of this theorem. It uses in fact too many materials, namely it uses harmonic maps, equivalent harmonic maps, but I will give a weaker statement, which is the following, so which is just the fact that if you take any row with the same hypothesis of non-maximal volume, then you can find a fuchsian representation that dominates this row. And in fact, this is interesting because if you consider, sorry, right, you're right, thank you. This is, yeah, in fact, I chose to not speak about conjugation, but so at each step, you are completely right. So the set of conjugacy classes. So there is something which is completely evident. This is that if you have a representation that dominates another representation, then if you take some conjugates of the two representation, the same relation holds. So you can, so if you have G that dominates whole, if you conjugate J by some element of PSL to white, the conjugation still dominates whole. Okay, so thank you. Thank you. Sorry, is it clear that this is clear that you can give yourself a row? Yes. And then anyway you can, why does it not get there? Okay, is it clear that, okay, this is clear that you cannot have trig domination by a friction with another friction. So, okay, so you cannot have G dominates strictly. Hello, okay. So I, sorry, I didn't get your question, so. Right? Something I'm asking, I guess, in all the conversation, if you point it not whole, like every fixing of the row. Ah, okay, okay. So if you take a row which is not trigger, you might find, I mean, a simple, let's say a simple closed geodesics with, which is mapped to a lo psodromic element. So it has a certain translational length. But then you can find an element in dash mirror space which pinches this geodesics, right? So this element will not dominate whole. Okay, is this clear that the set of friction representation, so the question was, is it possible that fixing a row, every friction representation dominates whole? This is never true. I guess it's always a set of fixed representation that dominates another one. The part maybe for trigger representations and maybe some stupid examples. But for most representation, all these set is trigger in dash mirror space. Right, so the space of representation of given volume, this is how to say that. So this is homotopy equivalent to a certain symmetric power of the surface where the power depends on the volume. This is two g minus the volume, I think. Minus the modulus of the volume, okay? It is a theorem by Ichin, okay? It uses Higgs bundles. No, in particular it's connected and this was proved originally by Goldman using different metals. No, no, it's never bounded. In fact, it is never bounded. It is never bounded. So in fact, as far as I understand, so concerning, so we have, so Nicola gave a parametrization of this set of dominating representations, but the geometry of this set is still to be understood, I mean, as a subset of dash mirror space, okay? But we know that it is never bounded, okay? So what I want to say now, I wanted to say something that I had to say at the really beginning of the talk, so when I made this definition of domination, so in third stock, we had some domination problem also to construct this Margulis space times and this domination in his context was formulated in terms of translation length or Margulis invariance. So in fact, this domination in this context of representation of surface group can also be given in terms of translation length. It is, so we will say that, so an equivalent statement, this is not at all trivial, it was proven again, Fanny's thesis is that a representation dominates another representation if and only if you can see the domination at the level of the length spectrum. I mean, it's the length spectrum of the first one dominates the length spectrum of the second one. For those who don't know the length spectrum, just forget about what I just said and but I just wanted to point out that in fact domination is something that you can read on the length spectrum of the representations, on the characters of the representations. Anyway, so now I want to, as I said, to prove a very weak version of this theorem which were proved independently by, so the, Gérito Casel-Volfe in the work I mentioned yesterday and by myself and Nicola. So the following results so were previously proved. This is that for any row with volume strictly less than 2g minus 2, so of non-maximal volume, there exists a J-function strictly dominating. So it is not a trivia, this statement and also in fact there are other kind of statements. For instance, in their paper, they prove that if you take any fixed representations, then in any earlier class, I mean any volume which is non-maximal, you can find a representation of that volume which is strictly dominated by your fixed representation you started with. So there is also other kind of opposite statements for this theorem which was also originally, I mean maybe guessed or partially proved by François Salon. But today we will concentrate on this theorem, on this statement and I will try to give quite complete proof of that statement. So maybe let's just analyze what does it mean. So I would like to try to remind you that yesterday we spoke about geometrisations and in fact this kind of statement is going to be a consequence of efficient geometrisations of given representation rho. So in terms of a geometrisation, I recall you that a geometrisation or maybe it could be called, I don't know, a pre-geometrisation is just a map from the universal cover of the surface to upper half plane which is a rho equivalent. And in fact yesterday what we did we just out of such a map we just extract the volume of a representation by taking the pullback of the volume form of the Poincare metric by such a map which gives in fact, which is invariant by the fundamental group of S and which gives rise to a volume form on the surface S. But in fact what you can do is you can really try to think of S as being a geometric object out of this pre-geometrisation by just pulling back the remanent tensor of the Poincare metric. So you might, instead of taking the volume you might think of your surface together with the pullback of Poincare metric itself, right? And in fact it is a tensor which is not a remanent metric because a priori this map D is not a summation, is not, or even, is not a nemation, sorry. So a priori there could be some direction in the tangent space of S where the distance is going to vanish and in fact so as a geometric object but you can still measure lenses of passes and define a kind of pseudo distance on your surface and get a geometric object. So the kind of geometric object that you will get by this procedure can be some, so S is say like that but it would be really possible using such a map to get as a geometric object something which is degenerated like that for instance because that would correspond to the fact that some curve, on some curves this tensor is vanishing and so in terms of geometry that all this curve is going to be contracted to a point. So for instance you can get some degenerate, so some some collapsed things but still you have a geometric object and we will really try to think about this geometric object in the rest of the talk, okay? So what really we will try to do to prove the statement, this statement here, okay? Is the following problem. So we will try to think to find a developing map. So a pre geometrization, let's say that can be dominated by conformal, sorry, by a hyperbolic metric on the surface S. So the problem in fact is find a good geometrization D and remanual metric G, let's say G on S with which dominates this remanual tensor. So of course if we dominate strictly dominates like that then we can find a certain constant C such that C is less than one and we have this inequality and if we have such a thing then we can, this is completely evident to see that this D is going to be a C-lipschitz map between the universal cover of the surface equipped with this metric to to a half plane and this map is going to be raw equivalent. And so in fact, if we think of the action of the fundamental group as being given by a fictional representation that I call G, we will have that, we will have exactly that always dominates, it is strictly dominated by J, right? So the problem is to find, so this is, I just shifted a little bit, I didn't say anything deep here, I just say that in fact you can translate, this is not a real translation but you can try to find a certain geometrization of your surface in this sense in that you can dominate by a Riemannian metric of curvature minus one and this will give you the answer. Okay, so and now my claim is that if you find some efficient geometrization you will get that. So let me give the construction. Yeah, so maybe we have saw yesterday two examples of geometrizations and I would like to describe what looks like this kind of metrics for those geometrizations. So the first geometrization that we saw is in the proof that there exists some developing map. So I recall you, we took a triangulation of the surface and we lifted the vertices of this triangulation to the universal cover. We assign any value for this lift and then we extend the map equivalently and we extend on the one and second part of the triangulation by using barycentric coordinates. So in fact, in this geometrization, so we had, we started with a certain triangulation of our surface and recall that that was exactly the same. I mean, what we had is that we had replaced each triangle of this triangulation by a certain copy of a hyperbolic triangle in H2 and there was a sign to the triangle. So the sign, there was some signs which correspond to the fact that the triangle was mapped by D with positive orientation or negative orientation in H. Okay, so in fact, what is this tensor in this case? So this is really easy to see what we get as a geometric of this. Is this just a hyperbolic metric with some conical angles? So we have some conical angles at the vertices. So in fact, if you glue two triangles along an edge, you don't have any problem. I mean, you just get a hyperbolic metric even on the edge where the remand tensor is not, I mean, you could have some problem to glue, but in fact, you have no problem. You get really a hyperbolic surface but what happens to the, at the vertices. So in the vertices, what happens is that you can, you can glue several angular domains and it is unclear what is the value, total value of these angles. It might be different from 2 pi. So you get a kind of conical singularity coming here but this is only the only singularities that may happen in this case. So we get conical hyperbolic surface. So by that, I mean a surface with a metric of curvature minus one apart from a certain finite number of points and at these points, the geometry completion of the surface is just given by, defined by a conical domain which could be defined by, you take a conical domain in H2 for instance, with an angle alpha and you glue the two boundaries of this angular domain by an isometry. And so when doing so, you get, you get the call. So of course here you get only cones with angles less than or equal to 2 pi but in general, you can also define conical angles of angles with angles bigger than 2 pi just by taking, instead of taking some angular domain in a half plane, you just take an angular domain which is how to say above edge, something like this. So the angle here is making 2 pi and then a little bit more and you can make another number of turns. So what I, yeah, exactly. So what I want to say is that the angle can be any positive real number, not negative, no. Because there is no sense of, yeah, the thing is oriented, so no, there is no, I mean, those are things that are isometric if you would just invert the, okay. So that was examples of, let's call them baby geometrisations, let's say. And next we saw another kind of geometrization which was given by foldings. What happens for a folding? It's interesting to see what happens. So the geometry of the underlying tensor is just nothing but a hyperbolic metric without conical angles. So in the case of foldings, you just get a hyperbolic structure without, so without conical angles. Why do you do so? Well, this is really evident. So if you have, you have something, so if you have a map which folds, so if, so suppose here you are, so suppose the map D, so here is H2. Assume that this region plus is mapped to the plus region here, here is the geodesic. And assume this region minus is also mapped to the plus region in H2. So you have a folding. You have folded the minus part of S tilde to the plus part of S tilde, right? So then if you take the pullback of the Riemann tensor in hyperbolic tensor in H2 by this map, because the folding you can think of this as being just a reflection in H2, which is an isometry. Then in fact, here you really get a hyperbolic metric of curvature minus one. Okay, so a model for that is just you take H and you take a geodesic and you take the map which on one side of the geodesic is just the identity and on the other side is just a reflection with respect to the geodesic, right? And because those maps are isometries, what you get as a tensor on your surface is just hyperbolic metric, okay? So the reason here is just because this, the fold is a geodesic. But this was the case in four foldings. So of course here, this is not so evident to see if you have laminations which are not just simple closed curve, but this is also true in this generality. Okay, so in fact, with the two examples I gave, it's quite easy to see what is the geometry for this tensor. And now what is interesting is, and what we will try to do is to try to find some geometrization which are going to produce some conical angles. And in fact, more than conical angles, we want that the angles are big, namely bigger than two pi. So why we are going to look for that? This is for the following reason. Let me write a very classical lemma which we have seen yesterday. I mean, in my first lecture in some form and in the romance lecture yesterday, which is Schwarz, so a generalization of Schwarz-Lemma. I don't know if that can speak and I'll force, I mean, I was trying to make something chronological but I'm not sure about these two guys. Anyway, there is maybe, here maybe, sorry, I have to change, so you, right? I have to, ah, here, okay, right, thanks. Okay, so what is this lemma? So the lemma tells the following, let S be a closed human surface, so of the characteristic less than zero. G, G prime, two conformal metrics with the curvature of G less than the curvature of G prime. Then, and everything is strictly negative, so curvature of negative, so metrics of negative curvature. Then G is dominated by G prime. So you have to write the formula that we approved. So recall that in the first lecture I gave a formula which relates the curvature of two conformal metrics in terms of the conformal factor between the metrics. So let's introduce this conformal factor. So proof, let sigma, the function, so the conformal factor, so namely what I mean by that is that let's write G equal to exponential sigma times G prime, okay? There are conformal metrics so you can find such a function. So sigma, let's say, is smooth. So let's say the two conformal, two here, we are going to consider smooth conformal metrics. And so what we proved in the first lecture is that we have the following formula. So this factor is satisfying a certain partial differential equations that remember we solved in the first lecture. So the Laplacian of sigma with respect to G prime is just minus two times the curvature of G times exponential of sigma plus two times the curvature of G prime, okay? So we proved this. And now what we want to prove, we want to prove that sigma is fast, is non-positive. So we want to prove that exponential of sigma is less than one. Sigma is non-positive. So assume the contrary. So let P be such that sigma attains its maximal value at P. And assume sigma of P is positive. It's really trivial application of maximal principle. Let's do it. Assume sigma is positive at P. So then, because sigma has a maximal value, it's Laplacian with respect to G prime is non-positive, right? So we have, so this gives that the curvature of G prime at the point P is less than or equal to two times the curvature of G at the point P times exponential of sigma of P. Okay, so let's recall what we had as an assumption. We had an assumption on the curvature. So the curvature of G at the point P is bounded by the curvature of G prime at the point P. So we get a complication. This is what I note here, which goes back from yesterday night, 11.30. So maybe there is a mistake. Right? I think it is with respect to G prime, as I told you, so we are not going to verify this again now, but we can discuss it later, but I think it is with respect to G prime. Anyway, it doesn't change the argument here, right? So anyway, for any Laplacian, the Laplacian of the function attending a maximal value is less than or equal to zero. But we can discuss this point later. So I'm trying to see if I have time to explain. So here, in fact, we will try to use a complement version of this lemma, which is due to Minda, which is a case of equality. And we will have to really use that for what is going to, so because here you see that this lemma, so we are really happy with such a lemma because this is a way for creating dominations, right? But we want strict domination, and there is this, so this complement, which says the following. So if furthermore, if there exists a point Q such that this inequality is strict, or let's say, so here is a complement. So this is the same statement then, but the conclusion, so what proved Minda is the following conclusion, which is Minda. He proved that under the same conditions, if, so either you have strict inequality at any point, or the two metrics are in fact equal. So the proof of that is not completely trivial. You have to do some work. So in fact, you have to analyze what happens. So if you assume that, so what we have to prove is that if the maximum value of sigma is equal to zero, then in fact, sigma is identically equal to zero, and so we have to do some analysis close to a point attaining such that sigma attains its maximum value at zero, which is the boundary of these sets of points, and to, I will not do that, in fact, I will not have enough time to describe that, and this is quite technical, so I would like to try to skip this, but I mean, it's elementary, but I didn't manage to recall the argument by myself. I had to come back to Minda's paper, so with the help of Nicolas, so it's not so easy, I think, but let's skip that. So we have this stronger version where we can strictly dominate G by G prime if there is a point where the inequality here is strict. Okay, so here is another complement of this first half force peak lemma, which is the following. So maybe I should say that if you have, so if you have a conical hyperbolic surface, as I defined before, it gives rise to a Riemann surface structure on the whole surface, so including the point where you have conical angles. This is very easy to define this conformal structure. So for instance, if you take a domain, so suppose here I am going to treat the case where alpha is less than or equal to p, which is not the case I would be interested in, but it would be simpler for what I am concentrating for what I say now. So suppose you take such a domain, you have a coordinate here, which is suppose it is a domain in a graph plane. So to create the conical angle, we identify these two boundaries by a hyzometry. And how to find the coordinate which is going to define Riemann surface? Well, we have just to take a certain power of the variable z, which is going to be something like two pi z to the power of two pi over alpha. If I am not mistaken, it's going to describe, when z describe our angular domain, so if variable w is going to describe the whole disks, so we have multiplied the angle by this factor using this kind of transformations. And so we get really here the coordinates, which is going to identify a neighborhood of the conical point to a disk. A disk in her path plane. Complex numbers. So these coordinates is going to be the coordinates that defines our Riemann surface. So we have a Riemann surface given a conical hyperbolic surface. And in the opposite direction, if you have a Riemann surface, so to go in this way, maybe assume you have a Riemann surface and to get let's say a conformal conical hyperbolic surface. So it's going to be a metric, conformal metric, which is going to have an expression like this. So in some complex coordinates for Riemann surface. And we want that this function, so this function is going to be smooth apart from some finite number of points, which corresponds to the conical points of our structure. And this function has to satisfy such a partial differential equation, which is this expression that the metric is of curvature minus one. And at the conical point, G takes the form Z to the power two alpha minus two times some smooth factor, which does not vanish, times of course the Eklien-Riemann tensor. So in fact, the conical points in the expression of a chart of our Riemann surface structure had some singularities, which are just given by powers of the modulus of the coordinates. And you can read the conical longer in the power you are interested in. So you can just, to prove that, you just look, you just write that in the variable W's, the conical metric you are interested in is just given by Poincare metric, and you write what does it mean, and you can prove that it means exactly that. So where phi is a function which does not vanish and which is smooth and continuous at the conical point. Okay, so now what is the version of the alphos, Schwarz-Alphos-Lehmann compliment that we are going to be interested in? So if G, so let's say G and G prime are conformal metrics, curvature or minus one on the same Riemann surface I was considering with G prime, so the, so, sorry. Let me, let me rephrase the thing differently, sorry. I'm going to say it like that. So, so let S be a Riemann surface of negative other characteristic. G prime, the Poincare metric on S, and let now G be a conformal metric with curvature minus one and conical singularities of angle strictly bigger than two pi. But by this I mean that there are some conical singularities, a finite number and there are one, at least one which is strictly larger than two pi. Then G is dominated by G prime for a certain constant C less than one. And the proof of this is exactly the same proof as I explained before. So we just write G, we consider the density of G with respect to G prime. So this function sigma, the log of the density and we write the partial differential equation that it satisfies everywhere apart from the conical points of the metric G. And we see that this density because of the condition on the angles has to tend to minus infinity at the conical points. As you see, when alpha, so here, sorry, I forget to roll. So alpha, so here the formula is two times alpha divided by two pi minus two. So you see that when alpha is strictly bigger than two pi, this exponent is positive. So the log, so sigma is going to be given by the log of the modulus of the times a positive number which decreases to minus infinity at the conical point. Okay, so you get that your sigma is going to infinity at the conical point so you can just make exactly the same argument. You take the maximum of this function, it has a maximum, so, and you do exactly the same argument and you get strict domination of G by G prime. Okay, so this is extremely powerful because if you just apply this, you get the following statement that if you have a geometrization of representation which is a conical metric and where the conical angle are bigger than two pi, then you can consider the underlying element surface structure, you uniformize this surface and you get a metric which is going to strictly dominate your conical hyperbolic metric which is exactly what we wanted to get, okay? So in fact the problem is how to find efficient geometrization, namely conical matrix, so let's say hyperbolic matrix with conical angles strictly bounded below by two pi. So I'm going to indicate two ways of doing so. I'm going to say, so here I'm going to try to mix all the techniques of all the works I have spoken about. So assume, so first I want to say the following. So just, I make a remark, if you have a conical metric whose angle are strictly bounded from below by two pi, then in fact the volume of this metric is strictly less than the volume of hyperbolic metric on the surface, namely strictly less than a large characteristic times two pi, I mean two G minus two times two pi because of Gauss-Bonaire formula, okay? So in fact you see that in a sense it is efficient in the following sense that the volume of the geometrization has to be small. So in a sense we are trying to get to get geometrization whose volume is small and in some moment I'm going to indicate how do you do, I mean a general question, I mean conjecture concerning this, the problem of finding something, geometrization with minimal volume. But anyway, so here is two steps. I'm going to just indicate the first one because I will not have enough time. So the first step is to take the geometrization by folding, so we start with the folding. Assume it is a folding along a set. So you have a hyperbolic structure, a folding is given by a hyperbolic structure and the data of some disjoint simple closed curve and at each time you cross such a simple closed curve you fold, okay? So of course here as I said the geometrization gives rise to a hyperbolic metric. This is exactly the hyperbolic metric we started with. So we are not happy, we cannot dominate this by another hyperbolic metric, okay? We have no con angle but how we can make a con angle it's extremely easy. Assume, so look at the developing map here in this region, there is one region which is, so say this one and this one, two semi-disc, assume the wide semi-disc is mapped by D to some semi-disc in her para plane, so here this is her para plane and the blue one is also mapped in the same region, right? So now what you can try to do is you just move, you choose a point here and you are going to replace your folding by another folding which consists of moving, so the image of this point by D is here, so you move this point in the direction of the folding. What you get, and so you are going to get a folding in the following form, so here you are, so the folding is going to, sorry, it's not exactly the way you have to think about that, okay, sorry, so let's make a picture which is equivalent, so if you make a picture which is equivalent, you have, so I'm going to represent the universal cover of some neighborhood of the geodesic, here on the left you have the white side, on the right the blue side, of course this is invariant by some element of the fundamental group, okay, and all this picture is mapped by the developing map by two something which is only on one side of one geodesic, okay, so what I want to do is to replace the image, so the fold in the image by something which goes in the region of the fold, namely I'm going to do something like this, so this point is going to move here, you just, in a sense I want to say you eat the fold, you just erase the part of this fold and you just get something with here, a conical angle which is the double of this point, of this angle here, why, because what you have done is that if you look at the pre-image by the developing map of the yellow curve, you get these two curves and what we just, what you do is you just erase this part and you glue these two parts together, here you have conical angles which are bigger than pi on both sides, so when you glue you get some conical angle of angle bigger than two pi and you win, so you just win by doing this procedure, okay, and let me just say two words, so there is a second way of doing things which uses equivalent harmonic maps, so there is a theory which tells you that you can always find an equivalent harmonic map given a high-pabric structure on your surface to a half plane and then if you use this map you can show that the folds in fact have a convexity property and in fact when you do this procedure you will find that the metric that you get, you can apply, you can uniformize that metric with appeal to uniformization theorem, apply as fast lemma and you get a domination also, so you can, this was the way we did with Nicola and so I think I will stop here, yeah.