 Alors oui, je voudrais remercier les organisateurs de cette éventue pour la possibilité de parler ici. J'espère que je vais avoir des choses intéressantes pour vous. Et donc, let's start. Et le titre des différentes lectures est, le titre global est hyperbolicité et généralisation. Et j'aimerais commencer par des exemples de plus en plus. Et l'exemple de plus en plus que j'ai dans la main, c'est le point carré space-model pour le plane hyperbolic. C'est très simple, c'est juste de déclarer que l'H2, le plane hyperbolic pour ce modèle, est le plane à l'extérieur et avec des métriques. Donc, à ce point, x plus iy, on met cette forme quadratique sur le plane tendant, on va dire ça. C'est-à-dire que si j'ai un parc, c'est assez régulièrement, sa longueur sera simplement l'intégral sur le domaine, je dois dire sur lequel c'est défini, et sur l'intervalle de, bien, la norme de sa vitesse. Donc, la norme de x'2 plus y'2, sorry, la route square root de x'2 plus y'2, divisé par y of t. Ok, donc, qu'est-ce que ça ressemble à ? Donc, quelques facts en ordre. Donc, la ligne verticale de la géodésique et les circonstances en r, comme ça. Donc, ici c'est r, ici c'est le plane à l'extérieur, et nous avons la ligne géodésique, donc la route de la petite longue entre deux points, qui sont comme ça ou comme ça. Donc, il y a beaucoup. Mais, même deux points, il y a seulement une géodésique entre elles. Et la groupes isométriques de l'orientation de l'orientation de l'H2 est isomorphique à PSL2R, actée par une homographie. La action A, B, C, D, appliquée à Z, sera élevée à AZ plus B, ou à CZ plus D. Et c'est PSL2R, parce que la identité minus est triviale. Donc, c'est exactement le kernel de la action, et c'est pour ça que l'action s'actualise au centre. Ok. En fait, vous avez déjà prouvé ça avant de trouver la géodésique. Et la action est conforme. Pourquoi parlez-vous de ça ? Parce que les objectifs principaux de ces discussions seraient les groupes, les groupes hyperbolicités et les généralisations, qui sont les plus importants. Et l'H2 est un bon espace pour quelques groupes. Donc, qui sont-ils ? Donc, le premier, peut-être le premier et le plus fort, c'est le groupe modular PSL2Z, qui a une bonne fortune, je ne sais pas si c'est une bonne fortune, pour préserver des très bons objectifs dans les planes de l'opportunité. Donc, chaque fois que j'enlève cette ligne horizontale, je veux dire que le plane de l'opportunité est quelque chose d'intéressant. Donc, ici est 0. Ici, c'est le circle de l'unité. Ici, je pense que c'est de l'unité. Et quelque part, ici, ici, c'est l'exponential de l'unité par 3. Et, pour moi, let's draw this arc here and call it A. A will be the geodesic, but for the geodesic for H2 between i and exponential i pi over 3. What happens to A when I apply PSL2Z ? First, I have this matrix, which is 0, 1, minus 1, 0, which flips A over i. And then I have 0, 1, minus 1, 1, which turns A of angle 2 pi over 3 around the other endpoint. And those two matrices generate SL2Z, so PSL2Z as well. So I claim that PSL2Z, A, this is very classical, the orbit of PSL2Z applied to this segment A is a tree, the modular tree, or the serre tree, sometimes I heard that. I believe. It is completely explained and detailed in Serre's book, cours d'arithmétique, non, cours d'arithmétique, which I don't know the title in English, and it's a course in arithmetics, maybe. And that's probably why we call that serre tree. So what do I want to say ? Yeah, I was going to write something PSL2Z of A, the orbit is a tree, meaning this is a connected graph without cycle. Oh yes, it's a modular tree, PSL2Z. It looks like that, let me finish the drawing. So I told you this matrices flips it here, and then this matrix flips unfold three times, three folds. We still have those arcs that continue there, and each time we arrive to a vertex here, it's unfold three times. But of course, here also, because here is the image of I, but below is the image of this point. So there is a three fold here, and so on. So this is the kind of pictures when you start, you don't know when you finish. Ok, maybe I finish now. So it's very interesting to have a space on which a group act, that's what groups are made for, acting on spaces, that's why they are interesting. So here we have an action on a tree. Another group for which H2 is interesting. Yeah, I'll come back a little bit to this picture. Another group for which H2 is interesting is a fundamental group of surfaces of genus, I need a base point, of genus G, closed orientable surface of genus G larger than 2. So it looks like that, for genus 3 for instance. So what happens here? I claim that H2 is a good model for universal cover of sigma G. There are several ways to see that. One, for instance, is through uniformisation, Riemann uniformisation theorem. But I'd like to just mention more down to earth way, maybe. Ok, let's do that for sigma 2, or let's sketchly do that for sigma 2. How do I get sigma 2? Sigma 2 is very simple. It looks like 2 torrées glued along a common boundary and indeed it is. So it comes from 2 torrées, so rectangles, but with a boundary, so pentagons with sides glued like that. So if you ever glued the sides of a rectangle to get a torrées, it would be without trouble seeing that this is a torrées with one hole and the same on the other side, maybe with colour. Ok, so the colours mean something. I'm not identifying this to that. Green arrow is identified, so I should do that. So now if I want to understand the universal cover of sigma 2, well it will be tiled by those hexagons. So, octogons. Sigma 2 is tiled by octogons, or by the way some people manage to draw where the sides go in this picture. I can try, but... Ok, you can try. This somewhere. How is it tiled by octogons on the vertex that is the image of all the vertices of my folded octogons? Around it I see eight different octogons. So with valence 8, meaning that at each vertex there are eight octogons. But fortunately H2 comes to tell us how it looks like. In H2, there exists one regular octogon such that the angles at vertices are 2 pi over 8 so that I can put 8 of them around the vertex. So I know this is pi over 4. So how hard to see that? Maybe I just want to show you where it is. So I'm just drawing H2. I'm choosing I and I'm choosing 8 4 in rays, geodesics, 8 rays, 4 geodesics the right regular angle here. So how it looks like? 1, 2, 3, and I did the fourth one. So it's like a spider, I don't know. And I told you angles make sense because the action of PSL2 is by conformal transformation, so preserving angles. And I want to try some regular octogons. So I travel the same length on each of the rays, the same hyperbolic length and I join them by geodesics. So I don't know how they look like. Ah, come on. Something like that. That's a geodesic hyperbolic octogon if I'm not mistaken. And then I'm trying to see what's going on on the angles here. Well, let's call that this parameter T. If T goes to infinity and you see like very close to an ideal octogon angle is then very close to zero. If T is close to zero you see something that lives in the tangent space essentially. So angles are cloned to the Euclidean situation and you can check that in the Euclidean situation angles of octogons are I don't know how they are but larger than 2 pi over 8. So in between there is somewhere there is a good octogon. So I can take this octogon as a model and build the tiling so that I have a tiling of the Euclidean of the hyperbolic plane actually with the right number of octogons around each vertex. Now I have to check that the action of pi1 of sigma g matches this tiling so there is a little exercise of label checking and it's not very complicated. So the end of this so hence so there is a little bit of checking here but h2 can be endowed with an isometric action of pi1 of sigma g2 that is free and with compact sigma2 that is universal covering action. Ok. So here in these two examples we have two nice examples of actions so psl2z acts on the modular tree pi1 of sigma g acts on h2 should have been more careful. Both actions are by isometries properly discontinuous and co-compact. I just say what means properly discontinuous and co-compact so properly discontinuous means that if I take a compact subset of either of the spaces and I look at the elements I should write it so I write it there so properly discontinuous so for all k compact the set of elements g and g such that gk inter k is non-dempty is finite. co-compact means there exists a compact that meets every orbit or you can cover your space with a single compact. Ok. This is the best possible situation by isometries the one that tells you the most about the group so let me just state define and state what is needed for Schwartz-Mille nor Lemma so I need the concept of quasi-isometries so let me define that if x and y are metric spaces a map from x to y is a lambda mu quasi-isometries so here I should say lambda is greater than 1 mu is greater than 0 so if two things really for all x for all x prime in x for the distance of f of x to f of x prime is controlled both ways by lambda distance of x x prime plus mu and 1 over lambda distance of x x prime minus mu so it's like bilips sheets but with error and there is a second condition which I was not careful enough to put on the same board but that will go just below that it is almost surjective every point is at mu distance at most mu from the image of f ok so I write indistinctly the distance in x and y as d if you prefer I should have put d y here d x there d x there I hope d y there but it makes notations heavier ok and shorts mu normal amount say that if x1 and x2 are proper length spaces and if g act on both properly discontinuously and go compactly then they are quasi-asométrique proper means closed balls are compact and length length spaces means the distance corresponds to infimal length of or minimal length of pass between points sorry oh yes I forgot to say that by isometrics oh and moreover and g is finitely generated that's the consequence of the lemma here I just give an idea of how it works so idea cover x1 by translates of a compact k1 build adjacency graph of a thickening of this cover and show that it is quasi-asométrique to some calligraph g so calligraph of g will be vertices are elements of g are of the form v g gs for s in some generating set which is which is well the the generating set corresponding to the link of k1 somewhere k1 and you have its translate and its link is the set of those that whose image intersect it and after that well it's only about x1 isn't it but there is an eddy part it's showing that given two generating sets both calligraph are quasi-asométrique ok so that's why we are we prefer actions that are property discontinues and co-compact basically because they are more or less all the same and they talk the most about the group in some sense now so I started with early examples of hyperbolicity now I should go to Gromov hyperbolicity which is an adaptation in the core setting of the world of quasi-asométries or of the nice geometry of H2 that we met so a definition so let me give the thin triangle definition of hyperbolicity so a geodesic space matrix x is delta hyperbolic if for all A, B, C, X and for all geodesics the triangle is thin so meaning that the side A, B for instance remains at distance less than delta from the union of two others so B, C union C, A and I'm sure a lot of you have seen this picture before so it means that A, B and C are like that so if I draw the geodesic first I should draw AC and C, B then it tells me that AB must remain at close distance from the two first so here delta is a number of course it is interesting for large triangles much larger than delta and it's there is no cost of thinking of delta as small by the way there are a lot of argument that goes through renormalization makes it gives a real sense of thinking of delta as small even sometimes I was going to 0 so there is nothing special about AB of course this definition is about any triangle so you can replace the rule of points so to illustrate it I started drawing AC and B and C, B but it's really symmetrical this quantity is interesting the distance from C to the branching point ok this is this can be defined as the Gromov product of A and B seen from C and it's formally one half of the distance of CA plus the distance of CB minus the distance of AB so this formula makes sense in any metric space but it's particularly interesting here because we see that there is from CA and from CB and I think of it as like a tripod I'm counting this distance plus this one plus this distance plus this one and I'm removing this one plus this one I'm left with this distance ok so a theorem due to Gromov there is something that I'm missing here how should I operate just the metro full pousse form et pas trop fort ensuite comment je fais pour le récupérer ok c'est ça merci so a theorem is that this notion of hyperbolicity is invariant under quasi isometry if two geodesic spaces are quasi isometric so x1 x2 and x1 is Gromov hyperbolic so it's x2 it's not with the same concern though but we don't really care about the constant the value of the constant so let me define the Gromov boundary of a hyperbolic space so in two steps first I need to define what it means for a sequence to go to infinity or in three steps actually a sequence so let a hyperbolic space a sequence xn goes to infinity if xn xk from x0 goes to infinity with n and k so here I have x0 I have a sequence of points I want that for large n and k they may be far away from each other but mostly their geodesic is far away from x0 and xn is equivalent to yn if the same thing if the Gromov product of xn and yk from x0 or from any point actually goes to infinity with n and k so with that we can say that the boundary of x is the set of equivalence classes of sequences and with the topology with the topology so it is a set of equivalence classes of sequences going to infinity with the topology that comes from the Gromov product so one can extend the Gromov product so this kind of sequences if two sequences go to infinity it makes sense of extending the Gromov product of one sequence with the other and so a sequence of sequences will converge to a point at infinity if this sequence of Gromov products goes to infinity so I hope it's enough to state it like that but the main observation is that so yeah, examples examples if t is a tree it is zero hyperbolic and it's boundary it's totally disconnected if it's a local agipinite tree then it's boundary oh yeah I shouldn't say that it's a tree of SL2Z if it is for the modular tree it's boundary so the modular tree was essentially the regular trivalent tree or also called binary tree non unrooted binary tree so it's a counter set but for H2 it's a boundary of H2 oh, H2 is hyperbolic I think it's an exercise c'est un exercice the boundary of H2 will be the circle at infinity and you can identify it with R union infinity the upper half place model but with the topology of one point compactification here there is an important statement that goes with the setting over there that says what happens if two space are quasi isometric at the level of their boundaries proposition also due to Gromov at this stage everything is Gromov theory so if X is hyperbolic and proper then its boundary is compact and if Y is proper as well and if we have a quasi isometry from X to Y can you remind what proper it is closed balls are compact and if X is a quasi isometry then it induces a homeomorphism between boundary so I did not yet define X is hyperbolic group but let's do that and finish with a tentative application of a nice situation this one definition G is word hyperbolic group this is also definition due to Gromov as most of you probably know the word hyperbolic group if let me state it like that if G acts properly discontinuously co-compactly on a proper hyperbolic length space or for every day's definition it's enough to consider its scale graph for a finite generating set is hyperbolic so for instance psl2z pi1 of sigma g for g larger than 2 and I would like to use my 5 or 5 plus minutes 7 7 to sketch an application of that which I like which I like exemple assume you are given two hyperbolic structure on the same surface ok sigma g g means the genus but also g1 I'm sorry about conflict of notations will mean the Riemannian structure and sigma g the same g, the same genus g2 not the same Riemannian structure are two hyperbolic structure on sigma g ok the theorem it means that pi1 of sigma g admits two actions the first action by isometry which I call alpha1 on the universal cover the hyperbolic universal cover and the second one the first comes from taking the universal cover of this structure the second of this structure sending one orbit on the other gives an equivariant quasi isometry and therefore it's really sending the orbit on the orbit so it's equivariant by by choice and therefore it's a quasi isometry because orbit isometric to H2 and it induces a normal morphism of the boundary so boundaries we understand it's just a circle an homeomorphism of the circle cannot be that complicated can it so in particular it has to be almost everywhere differentiable as any homeomorphism of the circle and the theorem I would like to advertise here or advertise I don't know it's as I understand due to Kouzalo but inspired by mostore rigidity so and I think there is a mention in mostore's work so I will write it like that so if G1 is not isometric to G2 so if I really choose two different hyperbolic structures then this homeomorphism is almost everywhere of null derivative and I'm going to sketch a part of it is very simple conceptually but it uses something that is not here but so so assume that deep side the map and the boundary is not absolutely continuous so it means that there exist a subset E in S1 of positive Lebesgue measure such that deep side of E is sent as 0 Lebesgue measure so now I take the orbit of E for the first action I should say alpha 1 I don't know how to write that well the orbit of E is invariant it is sent to the orbit of deep side of E which is a countable union of set of 0 measure so here there is a a statement that the action has some mergodic property that the only invariant subsets of such uniform action are full measure or 0 measure for Lebesgue by Argodisity of course Lebesgue measure is not preserved but it means that E as full measure its image still as 0 measure so from that one can deduce the claim of the theorem that the derivative that F is singular I should have stated like that so F is no it's not F it's deep side so it makes it does not quite prove the theorem really because it says assume that this is not absolutely continuous then we have a singularity of this application and it's not so far from checking that derivative is null almost everywhere actually null everywhere it is defined and there is what about if it's not what about if it is absolutely continuous I should show you that G1 and G2 are actually isometric so since I'm out of time I will not do that but indeed similar arguments show that deep side satisfies some functional equations which makes it a mobius transformation which makes G1 and G2 conjugate by an isometre F2 thank you for this first lecture