 Thank you. So on Sunday, I was preparing the exercise sheet. It was like 10 PM. And I had to put labels as easy, medium, or hard. So I fell for that trick. I shouldn't have. So everything was hard. Really think about it. Like, the first time you meet some mathematical topic, it's hard. There is no way around, right? So I wanted to start this lecture by discussing a little bit the exercise that I saw you yesterday, very fast, just to show the rule was don't write any formal proof. And I will not write any formal proof. But I will just tell to the good people that you are who worked hard on some exercises. I'll just tell my spin on it. But OK, probably you all looked at the first one. I don't know if you have the exercise sheet with you. I'm reading it out loud. Show that SL2R acts transitively on the set of unordered triples of r union infinity. So I should say, yeah. Let me, oh, something disappeared. Yeah, so let's see that. So we have a few matrices that we prefer here. We have those ones. Those ones, this one, I suppose, and this one. And one should do using only that, actually, because anyway, that's enough to have everything that is interesting. So I draw the picture. I take a, b, c. And I want to put them to minus 1, 1 infinity. So I first use this one, which is the translation. So that b is now at 0, or c is now at 0, anyway. So then I use this one, which takes 0 to infinity. Oh, I'm happy now. So using this one and this one in order, I'm taking some of those three points to infinity. Now using this one again, which is fixing infinity, I'm taking the two others so that they are centered at 0. And now using this one, which is a dilatation, I'm just pushing them apart until I reach minus 1, 1. So this shows that I can reach from any triple. I can reach minus 1, 1 infinity. Good. And after that, there was a question about how the ideal triangle, minus 1, 1 infinity, was delta thin in the sense that any side was in the sum delta neighborhood of the two other side. And I know that some people are really attached to find the optimal value of delta. And I should say I don't care the optimal value of delta here. So for someone like me who doesn't want to do any computation and who doesn't care about the optimal value of delta, I cut the triangle in three parts above some parts here, where it is easy to see using the explicit metric, which is divided by the height. That distance from there to there is small. But now I can just say that this point is in the orbit of that point. So there is a cutoff here from which the sides are close to each other. And same here because of the group action. So I'm left with something here when I need to check whether those three sides are at bounded distance from each other. This is a compact situation. Of course, they are at bounded distance for each other. So there is some delta. And if you like, you can compute the optimal delta, which is something like I don't remember log of 2, log of 3, something like that. And now if I want to do that for all triangles, well, I take an arbitrary triangle. OK, can I draw that? Well, as I say, I can certainly put this side by isometry on the vertical here. So now what is the picture of it? I have an arbitrary triangle whose one side is vertical here. And the two other sides are something like I don't know. Really, I don't know. OK. So they are the whole triangle fits in this one. Is that correct? Yeah, I think it is. This one we know is delta thin. So there is no room inside. So the whole triangle has to be. OK, what was next? Zn is quasi-isometric to Rn, an explicit quasi-isometry. Then the integral part of, oh, sorry. That version was, Mr. Moussa, what was that? Yeah, there are two versions of it. Rn, qi, to Zn, which is a stronger version of the exercise. So show quasi-isometry for Rn to Zn. Well, if I take a tupper, I can certainly look at the integral parts of each of them and see where it goes. OK, let's now check that it's good. I suppose it was. So show that I'm just scrolling through the exercises and it's not official formal solutions, as you understand. Show that two paragraphs about finite generating set are quasi-isometric. Well, so there is a word metric associated to S1, a word metric associated to S2. And a good idea is to check what is the maximal value of the word metric of S1 for elements of S2. So look at max for S in S2 of the word metric dS1 from 1 to S. That's one way to control it. And the other way is replacing the two roles. And that tells you how to send one edge here. What is the length of the path there? And then when you know what is the maximal factor, you're basically done from one way to the other. And from the other way to the one. That's not English, but write a proof of Schwartz Milnor Lemma when you have time. Let G be a finitely generated group so that if H is finite index subgroup of G, then it is quasi-isometric to G. That's already exercise four. So H, finite index in G. Well, H acts on the calligraph of G by isometries, left translation, properly discontinued G. It's just a subgroup, so freely, actually. And with finitely many orbits, so co-compactly. So we know that it means that H is finitely generated and quasi-isometric to the space on which it acts. They use that all finitely generated non-imbellion free groups are quasi-isometric to each other. So a free group of rank K is a fundamental group of the rows with K petals, meaning the graph with one vertex and K edges. So what do I want to say? The free group of rank 2 is a fundamental group of this simple thing. And now if I want to find finite index subgroups, that means I want to find finite covers of this topological space. But there are easy finite covers that I should ban the word easy from my vocabulary from now on. Because what is hard two minutes ago is now easy once you understand it. And I should have used colors to show the covering. This is sent to that as an unfold cover. And this is sent to that each time. OK. Now this is a rank whatever free group which corresponds to a finite cover of the rows of two petals. So it's finite index subgroup of F2. So if you never saw that, it means that all three groups, non-imbellion free groups, can be seen as finite index subgroups of rank 2. So they are all quasi-isometric to each other. On the contrary, it shows that z squared is not quasi-isometric to z cubed. And I think that the right way to see that is to know in advance that the growth function is a quasi-isometric invariant. And check that they do not have same growth function. The growth function say how fast that the ball of radius r in the kilogram of this group grow. So in one situation it's quadratic and the other it's cubic. And this property is preserved by quasi-isometry. So they are not quasi-isometric. Do you know all that free group? Other than this one, no. I don't know. Well, maybe I know other proofs, but not element. Doesn't mean that the number is known. OK. OK, that's it for the, if you allow me, that's it for the sketch of the exercise. OK, so let me resume then what I was saying last time. Maybe I should just erase everything here. So last time I was speaking yesterday, I was speaking about hyperbolic groups and relatively hyperbolic groups. And with two definitions of relatively hyperbolic groups, one using hyperbolic fine graphs, so angularly locally finite graphs. And the other one using proper, so yeah, proper. I say again, that means closed balls are compact. By opposition to this one, which is not locally finite as a graph usually, proper hyperbolic space with invariant system of four balls. And I found that, OK, I didn't check that. I think so. Is that the picture in the poster? Yeah. Yeah, so and oh yeah, we all have it on our badge. A picture of a hyperbolic space with an invariant system of four balls, so I'm very flattered too. I struggled with the technical definition of horrible yesterday. You remember that it was virtually impossible to apprehend in two minutes. That's really hidden here is a point at infinity, some ray that I don't even want to draw it, but OK, let me draw a ray pointing directly at that point. And on the ray, I take larger and larger balls, et cetera, take the union, and that's a horrible. That's horizon, I suppose. OK, so much better way to explain what it is. So I erase construction drawing. So this means that our ugly subgroup in G, so we have a group G and P, a subgroup that is the reason for non-hyperbolicity usually that are ugly, I mean ugly. Non-hyperbolic subgroup or parabolic subgroup will fix one point at infinity, and its conjugate fix another point at infinity. So this will be fixed by P, and this one will be fixed by, well, a conjugate of P, G, P, G. And what's the morality of this situation? In both cases, it says the same thing, but in different languages. I want to say what it says. Here we see that the conjugate of P fix convex sets in my hyperbolic space that maybe sometimes are close to each other, not for very long. If I travel inside my convex set, I must dive through the point at infinity and here through another point at infinity, so very far away from the part. So the thing fixed by P and the conjugate of P have virtually nothing to do with each other. Like in a free product, in a free product A star, blah, blah. The conjugates of A have nothing to do with A. And so I was saying this is visible in the picture. I tried to make it visible. This is also visible with a finest condition that says that when I have a cone point over a coset and I choose an edge E, well, I can only come back to this coset in bounded time in finitely many ways, serious ways. Of course, I could do crazy things. Come back without saying it, go somewhere else, come back. But that's not what I want. If I want just a simple loop, only finitely many possibilities. That's the definition of fineness. Only finitely many other edges with given angle. So let's have an example that I discussed with some of you yesterday of a cone of graph that is not fine. And that very easy that counter example, z square, or z plus z, in which the second factor will be P. So here is the first factor. Here is the second factor. And of course, well, you all know z plus z. And what do I have to do? I have to cone off the coset of the second factor. So the vertical lines. Coneing off the vertical lines means that I have just the horizontal lines remaining, well, infinitely many. And slices here that are all conned off by some special vertex. And since there are infinitely many, you see that cycles of length 6 from here going down to the horizontal, going right, going up. I have infinitely many paths of length 3 that come back. So it's not fine. It's not fine because cosets remain parallel to each other. Because conjugates intersect. This is ultimately the reason. So it's not fine, and there is no hope to have conjugates well separated like that. OK. So let me move on now. And I wanted to talk to you about a theorem about relatively hyperbolic groups that tells you somehow that you're never that far from hyperbolic groups when you are in a relatively hyperbolic group. That's a dense feeling theorem. So a little bit of historical commons here. So what is dense feeling or dense surgery? To begin with, it has to do with three manifolds. So if you have a three manifold where the boundary is a torus, so for example, think about a knot complement. But instead of removing just the knot, you removed a small neighborhood of a knot. That was dense motivation. Three manifold whose boundary is a torus that you can be tempted to fill in the torus by a solid torus. Solid torus is just the disk times circle. So I don't know how to draw it. Maybe suggesting that this is full. So the solid torus has a torus as boundary. Your manifold M has a torus as boundary. Maybe you want to glue the torus on the torus to fill the hole. And then you realize there are many ways to do it. You have to choose how you glue some framing to some framing here. And if you think a little bit about it, what is important is the slope of the meridian in the framing of your torus. That's a homotopy equivalent invariant. So yeah, that's a dense feeling. Choose a slope in T. Glue the solid T so that the meridian goes on the slope. And much later, Thurston, who was interested in hyperbolic manifolds. But this is just topological, right? If Thurston was interested in hyperbolic manifolds, among other things, I suppose, prove the following. That there are only finitely many slopes that give you something weird. Other than that, all other slopes will give you a compact hyperbolic manifold. Did I say an assumption here? No. So there is missing assumption. So I say that again, Thurston. So if you start with m, complete hyperbolic manifold, well, yeah, finite volume, hyperbolic manifold with one cusp of dimension 3, for all but finitely many slopes, the result of what you glue, so is a compact three manifold. This is just topological. But not any compact three manifold. It's a hyperbolic compact manifold. Is the statement clear? Because I had to say it twice the first time this part was missing. Is the statement clear? So you start with a hyperbolic knot complement. You remove a little bit more than just the knot itself to get a full torus. You glue a solid torus along a complicated slope. And then you get a hyperbolic manifold again. Well, there is a statement in the group theory saying this in the realm of relatively hyperbolic groups. So the theorem, which is due to Ossin, there is no other version by Groves and Manning, is the following, is an adaptation of this setting. So if G is hyperbolic, is there a question? There is a missing word. Yeah, so m union st glued on the slope. So I should have said choose a slope. And let me write it like that. This means st glued on m such that little m goes on the slope s. If G is hyperbolic relatively to p, there exists a finite set, f, I call it, inside p minus the identity such that for all normal subgroup of p avoiding f, the group G quotiented by the normal closure of n in G contains the faithful image of p mod n. n is hyperbolic, relatively to p mod n. OK, is this really the same? So G would be for the fundamental group of m, hyperbolic relatively to z squared. There is a finite set in z squared minus the identity such that whenever I take a cyclic subgroup of z squared killing a slope not enough, then the group, the quotient, is hyperbolic relative to z squared mod some cyclic group here. That has to be virtually cyclic, actually, to get a manifold. So hyperbolic relatively to z means hyperbolic. So it tells the result of Thurston at the level of fundamental groups. Of course, to have a true hyperbolic structure on the manifold, it's another story. But at the level of fundamental groups, it gives the den-feeling theorem. So it says also that assume, for instance, that p is residually finite and choose n to be finite index normal subgroup. So blah, blah, blah, blah, blah, relatively to p mod n, relatively to a finite group. So it says that G mod n will be hyperbolic. So if G is, again, if G is, too many long words, if G is residually finite and if n is deep enough finite index subgroup, then the result is hyperbolic. So I'd like to see it as we are nowhere near, we are nowhere far from hyperbolic groups in this setting. You said it's possible to write this to z. Yeah, there is, yeah. So the respect that I use twice in two sentences, so it's fair that I addressed. If G is hyperbolic relatively to p and p is hyperbolic, then G is hyperbolic. I believe the first time this fact is due to Ossian in this level of generality. So I'd like to sketch a proof of the den-feeling theorem of the theorem. And it's using so-called rotating families. But I will say what's going on. So take x to be one of two favorite spaces for G. And I will take the model with horribles. How do they say? OK, the proper hyperbolic space. So the model with invariant horribles for G relative to p. So this is this picture. Maybe I'll do it again. OK, all those horribles are in one single orbit of the one preserved by p. And for some, OK, I will deal with quantifier later. I will try to say what I want to do and see what's going with quantifier and the order later. I will choose a system, a much deeper system of horribles. So we'll see what much deeper means. While they are still convex, they are still preserved by the conjugates of p, it's just they're much farther away from each other. Much farther away from each other. And I will cone them off. I will remove them and replace them by a coning off. So here I have a new vertex. And now it's coning here. The advantage of doing that is about angles. So when you are at point of infinity, p acts on the whole sphere by turning the whole sphere, but it fixes the point. And so it permutes all the geodesics that go to this point. So it indeed does some kind of rotation on the whole sphere. But if there is any angle on it, it's 0, always 0. Because any rotation deep enough, close enough to the point at infinity, will do almost nothing. If you think about it in the upper half plane, the matrix 1, 1, 0, 1, 1, 1 million, 0, 1, it's just translating a lot on the whole sphere. But if you are ready to move up, it's translating almost nothing. So angle of rotation would be 0 in the yellow rubles. But in the blue, just coning off, well, it's something. It's non-zero of some angle. And now if I have sufficient resolution, finiteness, or if I am free to choose n as deep as I want, I can choose n so that for this rotation, it turns with large angles. Oh yes, VP will be this one, the new vertex that used to cone off the orosphere fixed by P. What do you mean by rotation in the general algorithm? So first, it's an elliptic element. But now I have also this concept of angles that I defined previously, that the angle between two edges would be the length of the smallest path going to their endpoints without going through the apex. So this elliptic element, I can say that, OK, it's a little bit abusive, probably to say it's a rotation. I don't know. But I'll finish this sentence to make it, to give some sense to that, that it fixes VP and it sends every edge to another edge that has a large angle with it. And I don't know how to call a transformation that fixes a point and make a large angle between edges, but rotation, I don't know. Is it fair? So that any element, any non-trivial element, sends any edge adjacent to VP to an edge making a large angle. OK. Picture zooming in to this VP, which is way too small here. This is calling off the orosphere somewhere here. And I hollow only elements in N, then I turn a lot along this orosphere. OK. And the main claim. Sorry, so the edges connect VP to the geodesics with N point P. Is that right? What are the edges here? So the edges will be, yeah, they connect this new vertex VP to an orbit, a P orbit, in the yellow orosphere. It doesn't have a name. I should say it's yellow HP. Do you want to test the horosphere or the horosphere? The horosphere. The horosphere. Because if I was to cone off the whole orobol, maybe that's what is written, cone them off. So only the orosphere. Thank you. If I was to cone off the entire horobol, for every N, I would find some edge, which is small angle. The same problem that the original orobol. But the horosphere is not fixed by P, is it? You look at the geodesics, which are? Yeah. The horobol is more or less the boundary of the orosphere. It's more or less the boundary of the orobol. The orobol is fixed by P. So the orosphere is preserved by P either. This is the small one, it's fixed by N. It's preserved by P. OK. One second. The yellow orobol is preserved by P. The yellow orosphere is preserved by P as well, as boundary of something preserved by P. And being preserved by P, it's preserved by N. And I take a P orbit in it, and I cone off the P orbit. So this is. The horobol is fixed. Does it mean that P is metaparabolic elements? Yes. Well, OK. Formally, no. I would like to think it is made of parabolic elements. There could be some elliptic elements in P as well. There could be some elliptic elements in P as well. But they fix the point at infinity. So, yeah. Imagine there is a finite normal subgroup fixing everything in the picture. Its elements are elliptic, because they fix points inside the space. But they belong to the group P. But as individualizometries, they are not parabolic. They are elliptic. Yeah? We're working to such end. OK. In general, you cannot. In general, you cannot. If your group is relatively finite, then you can. The action of P on the horosphere is proper, because in the complement of the horobol, the action of G is properly discontinuous. So if you are allowed to say, OK, there exist finite index subgroups as deep as I want, then you may say, I choose a finite index subgroup so that there is no small element in this proper action. But in general, you cannot do that. In general, you don't know if there is any finite index subgroup. Yeah? Did we assume that P contains no box of atomic elements? How do we know if someone had parabolic elements? I mean, parabolic elliptic. So P is assumed to fix the point at infinity of the horobol. And indeed, it will not have an atomic element. OK. Yeah, so what I'd like to recover where I was together with you. Oh, yes. I was here. So we choose n. Oh, I understand why it was a question. Are we allowed to choose n? We take n, if possible. Such that any non-trivial, OK. But now, so choose if possible. And now the main claim that makes everything work, and I explain quickly why, that now if I do, if I look at the quotient of the space, not of the group but of the space, I need a name for this space. So let's call the x dot the condoff space. So the final space, the condoff. So x minus yellow horobols union the condoff. I hope this is clear. This is catchy. So this is the result of all the construction. I take x. I remove the yellow horobols. I take the condoff of the horosphere. So the main claim is that the quotient of this space by the transformation induced by conjugates of n is a local isometry away from where it is not and where it is not is on the vertices. So of course, the rotation is not a local isometry at the vertices because it's folding something on the vertices. It's turning these two edges into a single one because they are in the same orbit. But everywhere else, it is a local isometry. And let me just sketch why. But this requires an argument. It's just an illustration of what's happening. And I have no time to develop the entire argument because I want to also make the conclusion. But when you take a convex and you make a rotation of very large angle around one point, then you have the image of the convex. It's another convex. And I claim that they are disjoint. They are disjoint because what if they were not disjoint? That means that I have this element and this element joined by this rotation. So here I have a path between them that avoids the cone point. But this is in the image of this one. So this element is realized by an element in N. So let me see what it does here. It does some rotation with a path that does not go through there. So it has to bond the angle. The angle is bounded by the length of this path. And we're working a little bit more. We realize that the angle cannot be as large as we wanted it to be. And finally, what to do with that? It's a localizometry away from the vertices Vp. And there is a general local to global principle in all the hyperbolic geometry around it. It says the following. So the question space locally will look like x dot. So it will locally be hyperbolic, Gromov hyperbolic. Sorry, but every space is locally Gromov hyperbolic. What do you do with that? Yeah, OK. I said I will do it with quantifier later to say how the quantifier comes into the picture. And maybe I will not have time to do that. But I will try to go through that. Stick with me for one minute and then I'll come back to that. So it's locally Gromov hyperbolic. Moreover, it comes from a space that can be thought as simply connected or throwing in the right cells could have been simply connected to begin with. And so you question it by elliptic element. So you keep this property. So it's still coarsely, so at large scale, simply connected. And there you have a difficult theorem. Carton-Adamar, but for Gromov hyperbolic spaces. So Carton-Adamar Gromov theorem that tells you that with good quantifiers, this implies that it is indeed globally hyperbolic. So here I state with good quantifier. So if a space is simply connected, let me state it like that. Let's say there. And 10 to the 7 delta locally delta hyperbolic, then it is 300 delta hyperbolic. And if you ask me, those constants are not so interesting, except that they are constructive. But I would have been happy with there exists L, such that, blah, blah. Now I can go back and see what's going on in the argument. I want these to hold for spaces of size 10 to the 7 delta. So I want the apices to be at least that apart. So now I can say what it means to take deep yellow horribles, so that apices are much, much more apart than this constant. Now I can say what it means for H to have large rotation, because I could not say it before knowing what were the yellow horribles. And so the N having a large rotation means on this faraway robot having rotation that assure that this argument will hold, that rotating this amount will put convex subset on this joint convex subset. And now I can say what is F in the theorem. F will be all the elements that I need to avoid to be sure that N does that. So it was a bit fast. I don't know if it cleared your question. But of course I'm not claiming that I'm using this theorem for space that I, like, won locally. We moved the rotation points far apart in order to have sufficiently room to have this locale pebble city. OK, thank you. I'll stop there.