 Well, thank you very much to the organizers for inviting me to be here. It's just a great pleasure just to come here and learn all the wonderful talks last week and this week it was this very nice conference you put together. Okay, so I'm going to speak about Poncat Extensions Russian Maps. This is a joint work with Peter Macienko and Angier Mociendra, and my main motivation is the so-called solubence dictionary. Let me just put some entries here just to remind you what it is. It starts usually by so this is an interplay between climbing groups and iteration of rational functions. It starts by noting that there are similar objects and similar kind of language they're usually the first entry here to say to people is that you have the limit set for some gamma in PSL2C you have a limit set of gamma which is just a fixed loxodromic fixed points, a closure, and then you have also the Julia set for some R in the semi-group of rational functions. This you may note this is just a quotient of two polynomials usually it goes up. And here you have the closure of repelling cycles. With this you get the discontinuity set which is just the Riemann sphere minus the limit set. You get the Fatou set which is minus the Julia set. You also have as a group you have the so the algebraic structure you have here just have gamma is a group and then you lose you have the semi-group over the over the naturals semi-group. The theorem that made this celebrate was the famous Fatou conjecture which is the same the following. So if the entry is related to here you have alphor's finitus which is if you have gamma finitely generated then the discontinuity set of gamma module. The orbit set of the discontinuity set consists of finitely many surfaces. Over here we have non-wandering domain. This is by Sullivan which is all Fatou components pre-periodic or periodic. If you if you take also the the Fatou set mod by the action of the semi-group then you also have is finite. But this was already known by Fatou so this part is already with surface. I haven't speak about surface now. The surface are of finite type. Yes, there's a finite union, finitely many surface of finite type. Yeah, I will come back to this because that's why I'm putting this entries. That's the last part of this talk. Okay, so but then as every dictionary, this is not a perfect dictionary, there are many things that don't translate to well. Here you have group, semi-group, but there are others like here you either have a measure you have zero measure or the whole sphere for the limit set. In the rational case you can have intermediate cases as a result of Serita and Booth but we still like and the first thing the first time I met this Solivance dictionary struck me as something very strange. Well, it was wonderful to have such a thing, but in a sense these these two objects are kind of different. Well, yes, you have the Mavius group, sorry which is the group generated by all inversions on spheres, on circles, on the plane and classical Ponca extension put this Mavius group into the isometries of the hyperbolic real space and I'm going to miss the field here because I'm just going to use this hyperbolic three space and the usual construction is just very easy. You just if you have the circle, you just put a cap on it and then every make the same construction This is isomorphic to PSL to C plus C2 and in as we have seen before PSL to C is isomorphic to isometries H3 This is all the homographies. No, I can say it correctly. This is of this type which is of course contained PSL to C is the subgroup part of the semi-group of rational functions. Yes, this is this is of the form of two polynomials. So what is my problem here? My problem is to extend this Ponca extension defined on the semi-group on the group of PSL to C into some kind of extension defined on the semi-group of rational functions. This is my goal today and there are implicit many ways to do that. If I just restrict to this case well, it is amazingly huge the way of possibilities. A stupid one is just to take so the boundary of of H3 I can be identified with with the sphere S2 and then consider V3 which is the boundary of P3 and then you just radially extend the rational thing. But of course, this doesn't extend the usual Ponca extension here. It is nice in the way that this makes a so we are looking for representations of RATC into the semi-group of endomorphisms of H3. So our idea is this one. So we want to make it in a coherent way of the Ponca extension. So the idea is that all morphic covering between Riemann's surface is a maybe as map on the coordinates. So this is a very ambitious, yes, just wait a little bit. They're going to a head. Yes, I want it. I am doing dynamics. So I'm willing to do dynamics. But in a minute, you will see that I cannot do it. Well, so far we cannot do it. So this is a very ambitious entry on the Sullivan's dictionary by showing that a rational map is indeed a maybe transformation on their uniformization. So this may seem a little bit crazy to you, but you will see that it's not that much and we can manage to do it in a typological generality. That means for an opens and then set of the rational maps. But not in the way that we would like it, as Alberto mentioned it. It's not a semi-group extension. We would like to make it homomorphically, but we will show you that there is a semi-group the function case where this actually works. But just let me let me see what I'm trying to do. Because it's really, to me, it would really sound crazy that a rational map is actually a maybe transformation, but because while rational map have degree, maybe transformation don't. So what is the degree going? But here are two examples, some starting samples and you will see that what I'm trying to say here. So you have the map that goes to C to the end and the lattice family. Here are the holomorphic coordinates, the map downstairs is your rational map and leaves to a very neat maybe transformation on the complex plane. Also for the lattice family, you can see an elliptic function. Here's the vice-trans function, but you can take any other and then you have a nice family of lifting of rational maps that lift as a maybe transformation. And now that you have succeeded on doing that, then you can. So what do we want? So what do we want to impose to our to our extension? Well, first has to make sense to speak about it on this conference. So this is the first thing I need to do. So it has to be geometric. So I forget to mention that in the static way, I mean static, there is no dynamic. If you have S, it's a maybe surface. Sorry, I just learned from Adolfo the other day that this should be called projectives surface, but I'm one dimensional guy, so sorry, maybe surface. Then there is a canonical way to attach a trim manifold to S. So if you have your surface and you have some conformal disk here, you just do what conquer edit for this thing. And this is this is a Pulkarni pink. So it seems that in Mexico, we are putting a lot of your papers. So this is this is done by the current because it is a canonical way. If you think of the C star, for example, and you do this construction, but you get this is the whole hyperbolic think minus the C taxes. So you just have to remove this thing out. And then this is the the the manifold that you get by attaching bigger and bigger balls. So that's the category I'm going to move in. It's a category of movers maybe maybe surfaces. And we consider a whole morphic covering and assume that so there are many teeth now. I'm sorry to there are a lot of assumptions to be made and a lot of assumptions to be meet, but I hope I can manage to do that. So assume that there are two planning groups and two components W1 and W2, such that the discontinuity sets of these two planning groups have that the stability that the orbit on the stability of this thing. It's exactly the given given two surfaces. So now assume that there's a maybe transformation such that the structure here live to the to the corresponding components of the omega limit set. And this is an inducing homomorph from gamma one to gamma two, which is just the inner automorphism induced by alpha is just composition by that. So if you consider the three manifolds with boundary, sorry, this is a type of this shouldn't be there, but just consider the high three hyperbolic space union, the closure of this. Then it induce a maybe morphism between these two manifolds. And that's what we're looking and if the map in this, so the map of this slide actually extends the original automorphic transformation between the surface S1 and S2 and without our tilde the punker extension of R whenever it exists. So here's where the degree goes. The degree is encoded by the degree of the new map is encoded by the index of the image of gamma one into gamma two. And in fact, you have that degree of the original map is less or equal than the other degree. This degree can explode to infinity. But the quality happens exactly when the stability of the corresponding client groups is exactly the client group itself. So you need an invariant component. So this makes a restriction of what kind of uniformization we can consider for the client groups. Sorry, so this is just to come. The previous was an abstract construction that managed to build an abstract three manifold. But we wanted to put it inside our favorite manifold. So we got some manifold M1 which is H3 union of gamma one divided by gamma one. And we wanted to put it inside the hyperbolic three space with the problems that there may be counter remaining geodesics that we have to remove. In order to do this. And then you have a map between two of these guys. So this is our tilde. So this is the map f star, f star tilde R. And since we want a concord extension, we don't want to have so many punctures. We actually want to have the original map here. So we want that this map continually extends to the geodesics. So we have a very, so we actually have that f star R extends continually, extends to a map. There's a nice right and left action PSL to see acts on the rational set, the rational maps by left and right. So I did, I was in the setting of holomorphic mapping. Now, R is a rational map. Usually you just remove the critical values, the critical set and the critical values to have a nice holomorphic map, which you can make this work. Here I changed the word for projectives. So here's the equivalence of under many sections. So we get some set in the rational maps and assume that there exists a map that extends the set into the endomorphics of the hyperbolic plane, hyperbolic tree space. And that this map is an extension for every element of A. Then we can make it equivalent just by defining this new map where G and H are elements of the Navier's loop and make it, we just force it to be. But this may not define the map, maybe a multi-valued map. Why? There is a chance that there is a, there exists a Q and R to rational maps and for Navier's transformation, say, gamma 1, gamma 2, gamma 3, gamma 4, such that satisfies this equation. In this, if we have such a thing, then there is no reason that this, this, these two objects may be the same. If we just define it the way that this, that way. And in fact, the situation is tricky. Even in the case when A consists of a single point. Because, well, you might clean this together and then have some H1, H2 thing. You may have solutions for this equation. Just to get one, you consider C to the K, C to the K is equal to E 2 pi over K, 2 pi I over K over C to the K. So, here we have such a, such a, such an example. C to the K is a map with symmetries. But this is not so bad if we can, if we have the following thing. So, if you have a rational map satisfying this equation, then this actually is an, this is an equality for this equation. If and only if H1 and H2 are various automotions of the corresponding three manifolds. In that case, everything works fine. Okay, so I think I, I, now I'm set up to, to say what we are looking for in a Poncat extension. So, first we wanted to make, to make it geometric as I explained before, to have the same degree. That means that it has an invariant, invariant component. We can restrict the question of Albert in many different ways. The first thing is just to consider what happened to the semi-group of a single map. This is the dynamical extension. So, if you just take the map and all the, the extension of this map, give a, are coherent with the dynamics of the rational map. You can make it even better if you consider semi-groups of the rational maps and consider homomorphic extensions of the semi-groups. So, you have the following equation. Composition goes to composition. And another, another, another way to, to, to restrict ourselves is just to ask for equivalence on the radius action. This is sometimes when it's equivalence over some, some, some side. For instance, for the right is called, what is called conformal, conformally natural extensions. Which is, which has been exploited before by works of, well, it is based on socket, socket theorems, but then it was worked by Duadi and Aaron. And later on for rational maps like Castor Peterson, of course, Macmuller. And then it was extended for, by the song Cortuan Gallo. But the problem with the barycentric extension is that it doesn't define how it's not a, it's not a homomorphic. It may be, and also we have very, very little control of what kind of maps we are getting on the geometry of these maps, although it may be useful to compute volumes of, in the static case. So I think that's, that's. So the semi, as in the client group, the case which is easy, in some sense, is the function case. And the corresponding entry for the function case in the, in the rational maps is the so-called Blaschke products. These are all rational maps that leave the, the unit disk environment, the unit circle environment, and have the following, the following thing. So each one of these maps is a automorphism of the unit disk. You may recognize that. And some, some rotation. This is the theorem. There is a extension defined on the semi-group of Blaschke maps, which satisfies the four conditions, except the fifth. The fifth may be satisfied if you don't, if you don't ask. That is a variant on the, the whole PSL2C, but only the set of maybe transformations that leave the disk environment. So it's, it's purely the friction case. But you may wonder what, what, what if I wanted to make it actually a, a variant under the maybe, maybe section on the whole, and the whole thing. Well, you can do that on subsemi-groups of the Blaschke maps, as long as the orbit, the bi-orbit of maybe transformation doesn't hit this back point. So this is inside, inside the, the rational maps. This, this map is a very bad point of having so many, so many, so many symmetries. Okay. So I think I'm way ahead. So maybe I can, I can say something about the proof here sketch. So the key facts are very simultaneous uniformization theorem. Secondly, that the Blaschke maps are symmetric. These are the two key facts. Start with the Blaschke map with the critical points removed. And then you consider two surfaces. You consider delta, delta one. This is the star with delta two. Delta two, which is S2 intersected with unit disk. And delta two star is S2 intersected with the C minus. So you're intersected with the, you have here your, your mirror. You have some critical points inside here. And some other critical points inside here. And then you consider two pieces, the outside of this piece and the inside. And then you apply Bernoulli's uniformization theorem for the blue, blue part and the red part together to get. So we get, we get gamma one and gamma star uniformizing. And then you see that, and then you see that this, this group actually should be the same. Because of the, because of the, the reflection, please. So this, this, these groups are acting in the whole space. And then we have a nice homomorphic thing. Uniformizing the, the, the sphere S2. And once you have uniformizations, then you can, then you just consider. You do the same thing for gamma one, for S1. And then you have this. Because this is a map preserving the homomorphic structure. This has to preserve the homomorphic structure. So you have here a nice maybe transformation. So this is how you succeed to make a, a Poncares extension. And once you have it on the level of surface, you do the classical Poncares extension to get the. The hyperbolic tree manifest you want. This, this is a, this is the way we, we do it. Okay. Now, so far you have been managed to avoid this word quasi conformal. I couldn't manage to do, not to do it. I hope it wasn't a forbidden word in this conference, but here, here we have it. I need to introduce this space, for this space, which is the class of, for a given rational map. We say that R and Q are always equivalent. If you can get two quasi conformal maps, phi and psi, so just make this diagram commute. And then given a rational map, we consider all, only all the rational maps that are always equivalent to the given rational map. This is a nice, nice space inside, inside the, the, the rational maps. Everything is quasi conformal in the, in the quadratic family. Everything is in this. This is exactly, thank you. This is exactly the way, the, the place when we lose dynamics. Okay. Thank you. Thank you very much. So, so far we have done it in a dynamical way. So the construction here, the, the black screen maps, oops, it's 10 meters. The black screen maps satisfied the four, the fourth was the, the homomorphic, the homomorphic thing. And even five, if you consider PSL to R. But now, now we lose it here exactly. So we can manage, we managed to do it in an open and dense set of rational set, but not in the way as we like it, which is dynamical. So we, we topologized this. We put a topology on HR, the whole space with the compact open topology. And so that's what I, so this is more than 100 years. So about 1870s clips and Rudolph proved that every two, every two regular coverings, branch coverings in generic position. That just means that the branch thing is not more than two. The branch points are, who is equivalent. And that's where, where we get the topological generative thing of our, branch points of work of the Riemann sphere in generic position. So in fact, there's an open dense, open and dense, since we have an geometric extension, we can prove in the rational maps, we can prove that if two maps are, are, sorry for make the hand wiggling. So if you have two rational, two rational maps, which are orbits equivalent, such a way that this has an extension, then this also has an extension. Then we have this open and dense condition, but, but we cannot manage to do it in the homomorphic way again. So now the next topic is co-ordinate rational maps. And this is a new phenomenon that, that we found out here. I don't know if was given before, but let me show you. So if gamma is a client group inside PSO2, are geometrically finite, then I would say then you get, then you get family is one until SN, finite family of surfaces of finite type. And there's a very natural question coming from this. So far I don't know the answer, maybe you know. If I give you any set of surfaces of finite type, finitely many surfaces of finite type, can you give me a group that uniformizes this finite set of surfaces? When can you do that? The beautiful uniformization here and say that you can do that for one Riemann surface. You give me two, you give me three, I don't know. Does anybody knows that this is already solved? Yes, yes, yes. Finitely. No. Yes. Thank you, sorry. Yeah, this is not the same thing. Sorry. So here's the question. The best answer I know about this one is the following. So for any client group, you have a co-board and family. This gives you a three manifold such that the boundaries of these three manifold are exactly these N surfaces. So these surfaces are co-board and this is part of the title. So this is a theorem by Masquit. If you give me, if S1, S2, La La La until SN is any family of surfaces of finite type, then there exists another, you can throw in another surface. Such that another Riemann surface, such that it's not until SN is a co-board and family. That's very nice. And the idea is that if you have such passport, then you glue some graph and in the graph you will go attaching the surfaces. And make some client, Masquit client combination theorems in order to get the actual uniformizing group. But this is, that's kind of the idea of the proof of this. And then as, then just looking at this, this, this construction, you see the possibility that they are rational after that co-board. Meaning that there are some hyperbolic three manifolds with planar surface on the boundary. So what I want, it could be that there are some hyperbolic three manifolds M1, M1 and M2. Some maybe transformation that goes from here to there. These are three manifolds with boundaries. So the boundary FF1, it has S1 and S tilde. And the boundary of F2 is S2 union S tilde. In such a way that the projections over each thing, so I don't know how to do it. But let me say there is R rational map. So these two are planar. One from S1 and two S2, Q from S1 tilde to S2 tilde, which are co-boarded. And this is our definition of co-boarders between rational maps. So it's exactly the definition. And then the question is, how can we make co-boarder co-boardisms between rational functions? And can we solve basket theorem? And first of all, this is a good category. And this, of course, map is co-boarded with itself. Thank you to uniformization theorem. If a map is co-boarded to other, then the other is co-boarded to the first one. So that we have a transitivity of co-boarded things. And this is, thank you to, thanks to the Thurston Geometry Session theorem. So we cannot manage without that. So the idea is, so you have some surface here, some three manifold, another surface. How do you name it? S1, S2, S3. And then you do the natural thing, which is attached. But if you attach two manifolds, there is no reason for these two manifolds to be hyperbolic. But then you just need to check that this is heteroidal. It's not, it's hacken. And then you can apply the Thurston Geometry Session. To this thing. And then you just clue and you get a nice hyperbolic three manifold. The next thing is the analogous to masking theorem. If you give me, this is just working, this is just working in the spirit of the Sullivan Sessionary. Going word by word and copy the proof in one way goes in the other. And then you have N maps, rational maps. Then you get another holomorphic covering. Actually you can't get it with no rational map, general holomorphic coverings, which form a family of co-boarder holomorphic maps. And finally, this is the last theorem, that if you have two conformal coverings that belong to the same Hurwitz class. And each of them, so if they belong to the same Hurwitz class that admit a question, I forgot to put admit a question uniformization each other. Then these two maps are co-boarded. In particular, there's an open and dense set of maps, a pair of maps in the pair of rational maps, which are co-boarded. I should stop here.