 Okay, thank you. I'd like to thank the organizers for the invitation. So what I would like to discuss today is actually a little more general theorem than what was in the abstract. So it's joint work with Ciri Lecure. So it seems that we are able to prove the full quasi-isometric rigidity of the class of clenion groups now. Let G be a finitely generated quasi-isometric to a clenion group. Then G contains a finite index subgroup, isomorphic to a clenion group. So what I would like to try to do is to give the main ingredients for the proof of this result and tell you the difference when the group is converse co-compact. So the strategy is exactly the same, but each time we work with a general case it's more difficult, almost at each step. So I would like to explain where it comes from and why it was not true to start with. Let me first remind you what I mean by a clenion group. It's a discrete group K of PSL2C. You can see it first as acting on hyperbolic three-spaces by azimetry. So the action is properly discontinuous. It also acts on the Riemann sphere by modulus transformations. So the action usually is not properly discontinuous on the sphere, but there is a natural partition of the Riemann sphere into the limit set and the omega. So the limit set where the dynamics is chaotic and omega, which is the ordinary set, where the action is properly discontinuous. So these two actions actually can be seen simultaneously. For instance, we see H3 as the three-ball in R3 and the sphere as its boundary. In this case we get an action and the limit set is the accumulation point of any orbit at infinity. So is that drawing something like that? So usually the limit set is a factor. So you pick any point and it will accumulate. So one point of interest for clenion groups is its tight relationship with three manifolds. So here I will assume all the time, because I'm just interested in what's going on up to quasi-izometry and up to finite index that the groups are torsion-free. So if it's torsion-free, then you can define the clenion manifold. M sub k, it will be. So I look at the action on the hyperbolic three-space and in the ordinary set. So since the action is properly discontinuous, it can take its quotient and it makes me a manifold. So it's a three-manifold which is orientable and has a complete hyperbolic structure in the interior. So the group k is convex or compact if the manifold is itself compact. Another way of seeing that is to consider the convex hull of the limit set in H3. And so because the limit set is totally invariant by the action of the group, the convex hull 2, and to be convex or compact it will be equivalent to saying that the action is co-compact on the convex hull. And so this is an equivalent definition for convex or compact. So the problem with the general thing is first year we have almost, the only thing that we know on the clenion group is that it is finitely generated. We have no structure somehow on this thing. But we have a lot of freedom because we can try to deform it by isomorphism. And through the right group we want to look at to use the geometry of the clenion group to find a subgroup which is isomorphic to it. So the first step is to choose a good k in its isomorphism class. So it turns out from Thurston's theory that we can always assume that k is geometrically finite and all power value points are rank 2 or have rank 2. So what it means is that the clenion manifold here has a compact piece and you have finitely many ends which are all total structures. You have a compact piece and finitely many tori which goes to infinity. And of course when the group is already convex or compact then it's already a good model. So we can use the geometry of the convex or compactness. So here what does it mean for k? So in particular it means that if you pick a collection of representative of the conjugacy classes of the parabolic stabilizers then this is a relatively hyperbolic group all p in pk which is isomorphic to this square. So the thing is it will be to work with this setting. So what are the first consequences that we get from that? That the group G itself will be relatively hyperbolic. And also because the z-squares are not relatively hyperbolic groups themselves we know that we will have a good structure. So the consequences from the assumptions maybe the first that I should have said is that the group G is finitely presented because it is quasi-isometric to a finitely presented group and second G is relatively hyperbolic and with peripheral subgroups are isomorphic which are virtually so in the convex or compact case we know that the group is directly word hyperbolic which is much stronger and provides much more information. So from this kind of things what we know is that there is a canonical boundary that we can define for a relatively hyperbolic group which is called the Baudiche boundary which consists in gluing. So you have your calligraph. I don't know how to draw a calligraph. So you have a calligraph. What you do is you glue, you can manage to get an action by azometries a proper hyperbolic space in the sense of rolloff. Each peripheral subgroup is the stabilizer of the point and all the other stabilizers which are not conjugate to one of these are finite. I mean cyclic or finite, elementary. Okay, so it's exactly what's happening for the for the Clinton group K that here we have the action, we have some cusps and each time we have the cusps the stabilizer is exactly this way. And so the boundary at infinity of this space so maybe I should give it a name very space X is what is called the Baudiche boundary. And so what we obtain here is an action, yes? P, the little P? Yes, yes. I'm sorry, I did it backwards. So the action that you obtain on the boundary is an action which is called uniformly quasi-mobiles. Okay, so what does it mean? It means that we have a natural family of matrix which are so-called visual matrix and the group does not distort cross ratios. We have a boundary on them. So there exists a distortion function which is homomorphism such that each time you pick a cross ratio four points and an element of the group then you look at the cross ratio of the image it's controlled by the cross ratio. And here what is the cross ratio? It's a matrix cross ratio. So you can have several normalizations but they are all equivalent. So I'm used to using that one. What is important is that you use the four letters each time you make the product of the distances and divide them in another way. In particular what is interesting is that when the cross ratio is small here it is small there. And this is enough to characterize the matrix transformation. And so in our setting what we obtain is that the limit set of my Clanian group is also the bondage boundary of the group with the same geometry. So we obtain directly an action by uniformly quasi-mobius action on the subset of the remand system. G is quasi-isometric. G implies that G acts on the limit set. And the action is a conjugate to the action that I had here on the boundary direct. And so now there is a famous theorem which says that if a group that a group of uniform quasi-mobius maps on the remand sphere is conjugate to a group of mobius mappings. And so then if the limit set is the whole sphere then we are already done. And so this is corollary which is due I guess to Canon Cooper. Maybe I should quote Schwartz. If lambda K is a whole remand sphere it's not really G it's the action of G which is conjugate to that of the Clanian group. So what I claimed in my theorem it was something a little stronger meaning that G contained a finite subgroup which was itself Clanian. Not a quotient. Okay, so I'll come back to that later on. I don't want to mix everything but there is a general difficulty actually to prove that. I don't have any simple proof going from finite quotient to subgroup. So one strategy for the general case would be I have this action which is already nice on the limit set and I could try to extend it to the remand sphere to keep it uniformly quasi-mobius and then apply this theorem. But you have to realize that in general it's hopeless to proceed like that because the topology of the limit set can be very complicated. For instance, if it is not connected there's no reason why you should be able to extend even one element to a global homomorphism. And even when it is connected there are many difficulties because you have local cut points and you can change the way the action acts has nothing to do with the action of G on the limit set of K has nothing to do with the action of K on its limit set, a priori. So we need to add some more structure and work differently. Maybe before I explain how I want to proceed what I should do is provide other cases where the theorem was known for a long time which will enable me to rule out these cases in the sequence. So for instance, we know that the theorem is known for the known cases. K is free due to stallings, I guess. When K is a function this is due to Cassandre and the Rice and Gabay and some many cases were treated by Toukiah before. And so in this case, the limit set of course is I think that is homomorphic to the circle. I want to go to other cases. So since it's complicated to look at the whole action of the group because we have all these cut points and stuff like that and all the components the first idea is to try to cut the group into pieces and then study each piece and show that they are themselves closer to the Clanian groups. And from that we'll get some more structure to work with. So the first step will be to split G to pieces. Then for each piece, each piece is virtually Clanian. And then once we have these pieces we can associate the three manifolds get a collection of manifolds, MV, the Clanian manifolds that we want to piece together back and obtain a manifold M such that the fundamental group hopefully will be isometric to G and then we can apply the first one's theorem. So this is how we want to prove the theorem. In both cases, confess the compact or general case or geometrically finite. So the splitting of the pieces I will use the splitting of elementary groups. So these are very well understood when the group is hyperbolic. When it's not the case we have to use more powerful results for finitely presented groups because it's much more delicate. So this is the first instance where there's a big difference I think. And then to show that each piece is virtually Clanian it won't be that hard once we... it will be more or less the same pattern for both cases. And then we have a problem here because there's no reason why these manifolds we will be able to glue them together to get a three manifold with exactly the right fundamental group because here the only thing we can get is to get something up to finite index. And so it's not the way that we want to glue them back and all that and stuff like that might not give us a manifold at all. So here there's a lot of work. In the case of word-hypolic groups it was well explained by the work of Y's on cat zero cube complexes in our setting when the groups are only relatively hyperbolic he doesn't cover exactly the cases we need. So we had to play a little bit with that to be able to get something of finite index there. And then to some theorems it's the same, doesn't make much difference. So let me try to detail a little bit all these steps and say what are the main... So the first thing we can do is since G is finitely presented it is accessible so we can split it over finite groups. We obtain a first action on each tree, T1, it's a simple action with no edge inversions finite quotient such that the stabilizers of edges are all finite and the stabilizers of vertices are either finite or one-ended subgroups. So in our setting it corresponds to look at the stabilizers of the connected components for the one-ended subgroups. So the one-ended subgroups correspond exactly to the non-trivial connected components. And so now this group, G is quite isometric to K I want to relate the action of the vertices to the action of the same splitting that I can do with K when I cut along compression disks my manifold. This is a result of Papazoglu and White which says that the quasi-isometry between G and K gives us quasi-isometry between the one-ended. So in some sense I have reduced the general case to the case where I can assume that at least lambda K is connected right now. This is still not enough to try to use the Sullivan theorem. And so when the group is hyperbolic it's not very hard to check that this is true. But here, because the structure is not really on the, I don't know how to say it, the geometry coming from the relative hyperbolicity is much weaker in some sense because it's not exactly on the KD graph, it's on a graph that you have transformed to get the hyperbolicity. And if it's easy to go from the KD graph to the new space X that I have talked about earlier it's very hard to go back actually. So now that we split our finitely many finite groups, I want to split our two-ended subgroups so I apply a JSJ deformation. So the version that I'm using so in the case of word hyperbolic groups is due to Bowditch and it was adapted to relatively hyperbolic groups by a graph. So the idea is that we can read splitings of two-ended groups from the topology of the limit set of K. And so what we obtain for G for instance is that we get a new action so now I assume that G is one-ended and so K also and lambda K is not homomorphic to the circle. So we get an action on the second tree with no edge inversions, simply sure of course. And so what we have is that the edge groups are all two-ended and the limit sets disconnect the whole boundary so lambda K. And for the vertices we have at least four types of them. The first time GV is also two-ended and as here it disconnects also the limit set. Or it could be so-called surface group so this means that GV is virtually free and its limit set has a canonical cyclic order. There is a third type which is called rigid meaning that a theory we don't exactly understand what's going on. So it's relatively quasi-convex, not one of the above cases. And the third case is GV is one of these parabolic subgroups or conjugate to it. And this corresponds to cut points on lambda K. So in the case of convex co-compact groups we don't have any cut points. So this latter thing does not appear. And also what we know from Bowditch is that the graph is always finite, the quotient. Whereas in general we can say nothing about it. Also it's not even clear that if you do the same, we have also an action of K on the same tree with the same properties. In the case of word-hypology groups it's not very hard to check that the vertex groups will all be quasi-isometric one to the other. In the general case it does not come for free. So here we have an action of K also. Exactly the same. And each time GV is to ended, KV will be to ended. Each time it is a surface it will be a surface and so forth. And so to understand first for instance the quotient is finite in the case of G. In the case of K it is well known because it comes from the geometry but in the case of the group G we had to go through the Bowditch paper and check how he proved the finiteness of the graph. So it seems that for the surface groups and the rigid, for the two-ended and surface groups he already proved the finiteness of the number of orbits of these things and of the edges which takes out. And we had to work to understand what was going on for these stabilizers of points. So one way, so we had to make a local study of the parabolic sum groups and prove that the action was exactly the one that we thought exactly as if it was a clenion group somehow. Even if it's only virtually this square and not this square. So the thing is that if you send the point at infinity what you get is a collection of components an infinite collection of components which are permuted by the stabilizer K. So here I have two actions on this. So the point P, the point infinity is a parabolic point for both G and K. And here I'm looking at the action of the stabilizer inside K. So let H point at infinity. So which is isomorphic to Z-square. So we know exactly what is this action. So we have an action by a translation like that and another one like this. Which permutes the colors. And so all these sort of lines which 200 lines which go to infinity they are well distributed. And what we know is that the action of the stabilizer of the point at infinity for the other group has to preserve this orbit and it is also co-compact on the limit set. Take off the stabilizer inside G. So from this we can see that all the stabilizers are two-ended for each component and they are all commensurable, two by two. Each component is virtually cyclic and they are all pairwise. So this means that the stabilizers inside G it acts essentially exactly as the one from K. And this shows that we have only finitely many edges, orbit edges attached to a parabolic point, for instance. And so this is the step that one of the pieces which we are missing to prove that the graph will be finite, the quotient graph, to have only finitely many pieces to work with. And the fact also that we will share a little early but that they are all, the components are all commensurable two by two is important when I will want to build a manifold because I will have to create a torus and glue the torus to some finitely many pieces to be in the same homotopy class. Because these cut points come from accidental parabolic. Yes, because the stabilizers are, I don't know if they are billion or not, at this point. But the action is, you can change the metric on the limit set to make it by isometric. So where am I? And so what I wanted to prove is that... Okay, so the next thing is what I claim, or proposition, if you want. The quasi-isometry between K and G builds quasi-isometry vertex groups. It turns out that because the perishable subgroups are all virtually a billion, the vertex groups are undistorted in the calligraph of G and... So this is, I think it... I learned this from a paper of Rosca on the relative quasi-convex subgroups. Maybe it's from somebody else, I don't know. So I think that the vertex groups are undistorted. And what we know also from the work of Papazoglou is that the image of a vertex group is bounded essentially the vertex group of the image. Inside G are mapped. And the fact that they are undistorted and the image is exactly what we want tells us that we get a quasi-isometry, vertices by vertices. So what is even better is that because of the structure of the tree, it has to map also the edge groups to the edge groups. What we obtain is that for the non-elementary vertices, at least for the case of surface and rigid edge, we obtain quasi-isometries between the vertex and the collection of edge groups which are incident to the same here. So as relatively hyperbolic group, they will map the peripheral structure close to the peripheral structure there. And this is much better because now we have the quasi-isometry which was just on the groups, now they are decorated because we have these edge groups added to them. So in our quest for clenion groups, of course when it's too ended, we know that it is virtually a subgroup of PSL2C. You just take the finite index group which is cyclic and you can realize it. It's the same for surface groups because we have this canonical order and then we can apply two case theorem. For the case of Abelian, so it's virtually Abelian, so we also know how to realize it as a convex compact group if you want. Not a convex compact. This is a clenion group like a lattice in Z2. The essential problem is how to deal with the rigid cases. And so in this case, we have this quasi-isometry which preserves the edges. And so one way to understand how this is done is to pinch, so here, this is a genuine clenion group so we have a manifold underlying this group. And these curves, these subgroups, corresponds to Anoli and provides an acylentrical pairing of the manifold. So this data provides... ...period... ...period... ...period. So what is quite standard then is to pinch the manifold to create casps to send all the edges at infinity and because it is an acylentrical, so what we'll get is... ...there exists a new clenion group K'V which will be isomorphic to case of V where all the parabolic points, all the edges have become parabolic. And because it is an acylentrical, I can assume that it has a total geodesic boundary. So what does it mean on its limit set? Limit set... ...it is the complement pairwise disjoint round discs and of measure 0. It has measure 0 because it is a geometrically finite. And the quasi-isometry that we obtain here exactly as in the beginning will give me an action of G sub V on lambda prime K by quasi-mubius mapping. So from this, actually, there is some sort of miracle which happens. There's a theorem of Bonk-Clayner and Merenkoff which tells us exactly how does such mappings look like. In a particular case where you act on the complement of a collection of round discs of measure 0. There are always restrictions of Mubius mappings. So it means that the action of GV on this set is already an action by Mubius transformations. Or any quasi-mubius map, self-map. Lambda K prime V is the restriction of a Mubius mapping. So it means that these rigid pieces, they are also clenion groups. So they are not exactly the clenion groups I was expecting because I have created many parabolic points by making the pinching to start with. But then it's easy to kill all the rank 1 parabolic points that I have created using, again, Thurston's theorem. You have a manifold which is, with rank 1, you can modify it. It's something that I already used implicitly in the first set. So maybe something that I should have said is the fact that we obtain an action by quasi-mubius maps here that we get the same geometry. It follows also from the work of Groff when he works on the JSJ decomposition. First thing that he does is he shows that when quasi-relatively hyperbolic groups are quasi-isometric, then the quasi-isometry remains a quasi-isometry when we add the whole balls to make them parabolic in the standard sense. So before I raise that, what I try to convince you is that each vertex group now is virtually a clenion group. Now what we have to do is to try to glue them back together and make sure that we'll get a manifold. Of course, since at each step, we are taking finite index subgroups, we don't know how they look like. So there's no reason why it should be the case. One thing that we have is the planarity of the limit set, which tells us at least that the boundary curves, for instance, they are simple for each vertex group. But we don't know if they are primitive and of course it's very hard to see how to... There's no reason why it should be the case. So there are some cases where we know we can conclude is what I would like to call a regular JSJ decomposition, so what does it mean? It means that for instance, G is torsion-free, which is not bad to create a manifold. What I want also is that the edge groups here are all the same, for instance, at parabolic points, edge groups. Another condition is that there's a problem also when you have pair of cut points, because you want them to be all in the same homotopic class when you want to glue them back to analyze. So you want that for any loxodromic edge group, G7E, it's action on the set of connected components. Then the K take of this is a tree. So what I'm saying is that if all these properties are satisfied, then I know how to create a manifold. So the idea is what is that for each, so of course I've erased the thing, for each vertex that I had, so when the vertex was an elementary, a cyclic group, then I will consider a solid torus, and what we know from the JSJ decomposition is that you have only infinitely many edges attached to them, and so they will all be parallel inside the torus. And because the action is trivial here, I'm sure that the action will be primitive and simple because of the planarity. When I had the surface case, so the group is a free group with infinitely many holes, so you can multiply it by an interval to make it a three-manifold, and each hole corresponds to edge groups also, and they are an ally for which I want to glue them together. In the rigid case, so this was the surface case, so maybe here I should have added that there's z squared stabilized. In the rigid case we have this pairing here, which is already given by the fact that we have the manifold, so I don't know how to do that, so we have a collection of an ally, I guess I needed these two holes, and then we have the tori, so of course I'm not used to making this talk, so I'm probably going to try the vertices, which I want to see as something like that, and similarly I will have, because of my condition, a collection of curves on the boundary, which will all be parallel. And now I can use the graph C2 over G, which will tell me to which analysts I have to pair the analyte together to build my manifold, and then it will have exactly the fundamental group that I had started from, because that's what Bastar theory is telling me. So in this case, the way I'm doing it, you can check that it will be irreducible at or at the end of that, so I made it all sketchy here, because essentially this is over thanks to Thurston's Uniformization theorem, if we have that. So in the word Hi-Public case, it's easy to see, well easy, it follows from the work of Weiss and Hegel that the group has a special action on a Casio-Cube complex, because it's word Hi-Public, and we can employ all the theory. And in this case, we know that quasi-convex subgroups are separable, so we can clean up, get rid of all the elements that, everything which did not make it a regular JSJ decomposition. So then we'll be able to find a finite indexer group. In the general case, it's not word Hi-Public, and I don't even know if I can find a torsion free, if I can find an action on a Casio-Cube complex or the whole group. But what I know is that each piece is acting on a Casio-Cube complex, so not co-compactly, but co-sparsely, so meaning that it is geometrically finite in the realm of Casio-Cube complexes. And Weis has a certain number of results concerning these issues, but of course, it's never exactly the right setting, so we had to adapt what he did. And so maybe one idea that actually he explained to me, because of course we were kind of stuck. And I think it's interesting in its own right, and I will conclude with this, so how to use the fact that on each vertex, so we have a graph of groups, G is written as some sort of graph of groups inside, we have a collection of stuff like that. And what you know is you have very good properties on the vertices and on the edges. And so what he told me to do, was to say, well, just take quotients, vertex by vertex, and make sure that they match on the edges. If it's the case, then you'll get a first quotient to a hyperbolic group, or maybe even you can manage to make it virtually free. So we get some normal subgroups for each of those, in such a way that when an edge is written VW, then the intersection of your normal subgroup corresponds exactly to the other one. And so for instance, if it has finite index, then each quotient will be finite. And we can create a new group, which will be the fundamental group or the graph of group with the same graph, but now what we gain is that, maybe I should do it like that, GE over N of, that this one will be virtually free. And using this kind of ideas, and to do that it's not terribly hard because we know that we have a lot of freedom of these vertex groups because they act on cut-zero-q complexes nicely. And so we can use what we did at this stage, okay, essentially. And then we make sure that everything matches, and once we're here, we can separate anything we want because it's a virtually free group. And then we can build a finite index subgroup where we'll get rid of everything. So of course, when we build this graph, we have to be careful not to put into the kernel the elements that we want to separate afterwards. Okay, so, but for this, we can use done-filling theorems and stuff like that. So it's a game between the manual special Gaussian theorem and the standard done-filling theorems of Osin, for instance, and make sure that everything goes exactly as we want in the end. And so here there is some work, but it essentially follows from what Wes did, but we still have to do some work to get it. So I will stop here, and I thank you for your attention.