 Thank you very much. So as Katie said yesterday, I am going to speak about bending of representations in the group of homeomorphisms of the circle. OK. So the first thing I'd like to comment is on semi-conjugacy. So there was a definition given by Katie. And there are actually several ways of doing that. But what I want to relate right now is I want to relate this with minimality, as this is one way of thinking about it. So there is a basic fact, an ancient fact, I think, by Poincaré, is that if you have a discrete group and acting on the circle, so that's to say we have a representation of the surface group in homeo of the circle. OK. Then one of the following holds. So you may have a finite orbit, or you may be a minimal action. So this means that every orbit is dense. And finally, you can have a minimal counter set invariant by the action. OK. Another thing is that maybe it is an exercise with a definition given yesterday to check that in the first case, if you have a finite orbit, then your action is semi-conjugate to an action by rigid rotations on the circle. And in the last case, the representation is semi-conjugate to a minimal action. So that's basically the idea that the action is really what you see on a counter set or on a finite set. And so you're going to just contract the complementary by conjugations. And to the limit, you get a semi-conjugation. And yes? Yeah. So OK. So this is a semi-conjugation in this group. And indeed, with the definition Katie gave, oh, sorry, oh, sorry, sorry. OK. Maybe I give a quick alternative. Or maybe take that as a definition, maybe. OK. And if you have an action with a fixed point, then it semi-conjugates to the trivial action. And that's what is here. OK. And now, so to really what's going. For some definitions, you want a lot of content. But what you said doesn't mean that by fixing a point, then you're not semi-conjugated. Yeah. Mm-hmm. No, they don't have to. And then everything is OK. Well, so if you take the relation generated by what Katie did, then yes. OK. But I mean, there is a very direct way which was really used by Gis in the article in which he defined this directly for maps from R to R. And that comes with another fact. So for this definition of semi-conjugacy, is that for minimal actions, semi-conjugacy is nothing else than conjugacy. OK. And one last thing on this theme is that, obviously, if you have an action of a surface group to the group of homomorphism of the circle, which has a finite orbit, then it's never going to be rigid. You can change the angles of your rotations. And so really, we are interested in minimal actions and deforming minimal actions among them. OK, that's one thing. And also, just Katie said that yesterday, but maybe I'm going to insist just a little more. So you have the map rotation number from this group to this one, which measures, in average, when you iterate how much the maps rotate in the circle. And you have its lift from upstairs. So I don't need that. OK. And so now you have this thing, the translation cycle from a homeo S1 to the square to R, and actually, just to minus 1, 1. That's where it takes its image. So you can take a map F and map G. And you can lift them and compute the translation number and compare it with the sum of rotation numbers. OK. And this, obviously, is not going to depend on your choices of lifts. And so that's a well-defined map. And here, the theorem of Matsumoto, using a lot of that paper of this, tells you that the data of a semicondigacy class is exactly the data of rotation numbers for each curve, as well as translation cycle for each pairs, of course. OK. And then I'm going to enter to the banding deformations of representations. So this is quite classical in the Leigh group setting. Oh, yeah. So this is a theorem that says that you have a map from homeo from gamma. You take a discrete group, any discrete group, to homeo of the circle. OK. And to each representation, you can assign the function on three variables. Yes. That given a representation, you are going to look at the map, gamma 1, gamma 2, gamma 3, which you're going to look at the rotation number of rho of gamma 1. OK. And you're going to look at the translation cycles of rho of gamma 2, rho of gamma 3. OK. And Matsumoto says that rho 1 and rho 2 are semicondigates if and only if m of rho 1 is equal to m of rho 2. So once again, you can use that in this setting. Yes, yes. OK. OK, OK. OK. And then let's do bending. So we are going to bend representation along simple closed curves. So actually, that's only along non-separating simple closed curves that we are really going to work. So this here, this is going to work only for representations of surface groups. OK, so we have preferred presentation. And when I write the product of commutators, I want to think that I'm using maps of the circle. And so I want to compose from right to left. So I'm going to write a1, b1, a1 inverse, b1 inverse, compose with until bg inverse. And this is equal to 1. OK, so this is the presentation of a surface group. And now whenever I have two curves that look like that, which are at the beginning of such a system of generators, then if I have a representation of a surface group into actually any group so far, then I can make up another one. So I'm going to bend along a1, and that's kind of, I'm thinking of the dent twist along the curve freely homotopic to a1. And so the bending along these curves is going to, I'm going to take some element in the centralizer of a1. And I'm going to cook up a new representation rho prime by sending a1, et cetera, ai, bi to rho of ai, rho of bi. So basically you don't change anything except for b1. Except that you take rho prime of b1. This is going to be, first you do rho of b1, and then you compose with a. And since a, since in the centralizer, since a commutes with rho of a1, when you compute the commutator, you see you didn't change anything. And so this does give a representation, a new representation, which is a bending of rho1 along a1. A bending of rho, sorry, along a1. OK, so this is for a non-separating curve, a1. I could do the same along a separating curve. Just the same, I would have a curve c, say, and conjugate everything on one side and do not change on the other side. And that would define another type of bending, but that's really the same philosophy. And so in order to understand what I can reach by bending, that boils down to really understanding how are the stabilizers of individual elements in the group. And so let's look at centralizers in the group of S1. So I take a map F, and we have different possibilities. And the most accidental possibility is when F is a rotation of infinite order. So if F is a rotation, OK. So when I say rotation, what I mean is conjugate to a rotation, to a rigid rotation. So if F is a rotation with angle of rotation, which is not rational, then it's quite easy to check that the centralizer of F is nothing more than the other rotations that look the same. OK. And the point is that in any other case, the centralizers are very much bigger. So first thing, if F is a rotation of a rational angle, then I can think of this rotation as a covering of the circle onto a smaller circle. And so the centralizer of F then is just going to be an extension of homeo plus the circle. OK. But now if I have something more interesting than a rotation, so if F is not a rotation, then it may have irrational rotation angle. And in that case, it's going to preserve a counter set. So let's say it's K of F. And if not, if it is a rational, then I'm going to consider, I can consider the set of periodic points of F. And actually, it's going to be a little more interesting to consider the boundary of this set. And so in both cases, this thing I'm going to call K. And so what I can do is consider the circle. And I remove K. And this is a union of intervals. OK. And F permits these intervals. OK. And so I'm going to choose a representant for each orbit. And then I'm going to define some map from that interval to itself and use the map, F, to extend it everywhere. So if this rotation was irrational, then in this action of F on this set of intervals, there are no stabilizers. So I can do really everything I want. And if not, then there is just a cyclic permutation of F on each orbit. And then I just have one map to commute with. But so I really have a lot of freedom there. And so that gives me tons of one parameter groups in the centralizer in each orbit. And what we are more interested in is to do this in a specific way. So it's going to be useful to do it so that if I have a parameter group S, T, so I can build this way one parameter groups in the centralizer that have positive dynamics in the sense that any time I have such an inter-inter-inter-inter all I alpha, then for every T positive, I'm just going to push the point to the right. And for T negative, I'm going to push to the left. And so this I'm going to call a positive one parameter group commuting in the centralizer of F. So in particular, we have for all T non-zero, I'm going to have the fixed set of ST, which is going to be exactly K, this set I was starting with. So that's something useful. In every centralizer, I have these maps that push things in a way that we can control. And so now, just as a warm-up of what comes next, I'm going to have a warm-up of what comes next. So that should be four, I don't know, or three. Maybe three. Let's use this to prove that there are no irrational rotation numbers for a rigid representation. So more precisely, so take a rigid representation and then for every gamma, which represents a simple non-separating curve, we have that its rotation number is going to be rational. And also we are going to see that rho of gamma cannot be rotation. So this is the easiest part, and let's see this right now. So suppose I take a curve alpha, like this, or A1. And suppose that for contradiction, that some curve is mapped to a rotation. So it's a rotation, so it lives in some group conjugates to SU2, and in this group conjugates to SU2. So I have rotations r theta. And now I just claim that the map that assigns theta to the rotation number of rho of B1. So I should draw this more properly, maybe. Always the same picture. Rather A1 is going to be B1. So if I look at this composition, then the rotation number is not going to be constant. And one way to see this is that I can just lift all these elements to as homomorphisms of r. And when I add a full turn to theta, this is going to add one to the translation number here. And so this is an element of the circle. And if I do something, it upstairs increments by one. And so downstairs it cannot be constant. Does that make sense? So this is not constant. And so the representation is not rigid. So now let's assume for contradiction that I have a curve mapped to an element with irrational rotation number. So now suppose that the rotation number of rho of A1 is not rational. And here I'm going to use a minimal representative of rho. So suppose rho is minimal, which I can do. So if rho of A1 has an irrational rotation number but is not a rotation, then it fixes some counter-set. So it has a minimal counter-set. Let's call it kA1. And then I claim that I can choose B1 properly so that rho of B1 does not stabilize kA1. So why is that? That's because actually the whole group is generated by the curves of B1 that look exactly like this in the surface. And so if this was not the case, then kA1 would be invariant and so my action would not be minimal. So there is some B1 which is going to do that. And now we are going to twist B1 along A1 in order to ensure that the B1 is going to have zero rotation number. And once we do that, we are going to twist A1 along this curve of zero rotation number so that its rotation number has to change. So let's do the first. So here I know there exists some x in kA1 so that rho of B1 of x is not in the counter-set. So here I have rho of B1 of x. Somewhere I have x. And here I can use an interval of the complement of kA. And I have its pre-image here. Here it's going to be rho of B1 inverse of i. And so this I'm going to call j. And the thing is that x is in the counter-set. And so if I use this interval and apply rho of A1 many times then I'm going to get to some interval very close to x. And so here I'm going to find an interval of the type rho of A1 to the n applied to i. And this just tells me that if I look now at rho of B1 and then rho of A1 to the n I hope I'm correct here with the indexes. Then this homeomorphism of the circle is just going to shrink the interval j. So I start from j, I apply B and then I apply A many times and then this composition shrinks this interval. And the outcome is that the rotation number, this map has to have a fixed point in there. So the rotation number of rho of A1 to the n B is 0. Actually this right here is not bending. Sorry. So it's not bending so far but what it tells me is that without loss of generality the rotation number of rho of B1 is 0. And so B1 is a map that sends some points of the counter set to something else. So here I have now rho of B1 has fixed points. And so the outcome is that there are some fixed points of this map. And I'm looking at an interval complementary to that. And I can find an interval complementary to this set which contains points of the counter set. And there they are as many as I want. And so I can find some point Y and its image by some iterative rho of A1. Is it okay? So now I choose a positive one parameter group commuting with rho of B1. So let's write it B sub t. And then the claim is that if I apply rho of A1 and then Bt and this with t varying then this has a non-constant rotation number. And the reason for that is that in this picture I have some distance here delta. And so if I choose a lift to homeomorphisms of R then let's choose one lift and keep it. And so if we compute this to an element that actually lifts Y to R then this is going to be Y plus this delta and plus some integer that depends on the lift. And now the thing is that for t small enough then the B sub t is going to send me back there. Because B sub t has positive dynamics. So for t small enough B sub t does that. And then I have this rho of A1 to the n composed with B sub t. So this really means the lift in R that has fixed points. Then this is going to be less than Y plus m. And now the good thing with positive dynamics is that this thing I iterate a lot of times and the rest of times Bt I don't know what it does but it has positive dynamics. And so we know that Bt composed with rho of A1 tilde all this to the n of Y is less than this one. And this is because Bt just pulls things to the left. And now I'm in good shape because that's really this map that I'm using. So at time zero I have a point Y that is mapped next to this point. And at some time it is less than Y plus m. And so for some t not I have Bt composed with rho of A1 this to the n of Y. This is going to be equal to Y plus m. And this really tells you that Y is a periodic point. So you have the rotation number of Bt composed with rho of A1. Now here Bt not this is m over n. And if it was rational to start with and now it's irrational it was not constant. And so what's going on here? This trick of using positive centralizers enables me to really change some rotation numbers and that proves this proposition. Is it okay? And now let's go very much further to what Katie promised yesterday. So now from this we are going to go towards towards a condition sk. So this is really what Katie was telling us. So we say that f and g to homeomorphisms of the circle satisfy the condition sk. If they really look like what Katie suggested. So if f is conjugated to some lift of a hyperbolic element in PSL2R. So this is f. And the same for B and with points that alternate. And a big part of what we had to do is to prove the following thing. That suppose you start with a rigid representation. And we are going to suppose also that I have two curves A and B as usual. And suppose they don't share periodic points. And then we can use rigidity to increase it to that condition. So let me say just a few words to say that now this is extremely close now to geometricity. So just a remark on what we can do with that. Thing is this condition really tells us just about two curves A and B. So I have two curves A and B with some condition that we can hope I can find two curves with disjoint periodic points. And then actually they are going to have a finite number of periodic points. And from that I'm going to be able to do bending of one along the other to just push a little all the periodic points. And then this is going to tell me that if I have a third curve like C then I'm going to be able to push the periodic points of B so that they are going to avoid those of C. And once we've done that then I know that C also has 2K periodic points. And so this property is going to propagate along all the curves and is going to be satisfied by all the curves. And so if we know that we are really in good shape for geometricity. And so let me try to give some hints about how to prove that. So the hypothesis is I have these two closed subsets of the circle. And the hypothesis is that they are disjoints. And so in the compact circle they are actually very far away. So the circle is going to look like that. So I'm going to have some periodic points of A and then I'm going to encounter some... I'm going to wait for some positive distance and then I'm going to encounter periodic points of B and then I'm going to hit some other periodic points of A, etc. And I have this pattern in some number. Okay, let's put the third one. So the circle really looks like that. And now from this information I can read a lot in the conjugacy class of the representation I'm looking at. So I have this number of things here. Of green things, say. And this actually is going to be odd. We are going to see this right now. And this is going to be twice K. And so from this picture we are going to take some information about the maps A and B. A has some dynamics. So A, no, A may not have fixed points. But A, so root of A is going to be something like P depending on A over Q, depending on A. And so the first iterate of row of A, which is going to have fixed points, is A to the Q of A. And this map A to the Q of A, row of A to the Q of A, this iterate of row of A, it has fixed points but it has no fixed points in between these two. And so either it has positive or negative dynamics. And so this gives me a sign. Let's say A, epsilon A1. And similarly, from here I have a sign, epsilon B1. And then I'm going to move to some other sign here. Epsilon A2, etc. And so what I get from the representation is a sequence of signs. So an integer N and a sequence, which is well defined up to cyclic permutations of signs. Epsilon A1, epsilon B1. Until the last, which is epsilon BN, which is the number of things here. And now the thing is that the translation, he knows of all this data in a very natural way. So you can set A to be the lift of row of A to the Q of A with translation number zero. And you can do the same for row of B. And now what's going to happen to me is that if I apply A to the epsilon A1 times a very big integer. And then B to the epsilon B1 times the same very big integer. And I'm going to do this whole composition. Then the translation number of this is going to be at least one. Because if I start from a point in this region and apply this A, I'm going to end in this region. And then I apply B to the correct sign. I'm going to be there, et cetera. And then I'm going to do just a little more than a full turn to this point. And the thing is that this is really the least complexity of what I can write in order to have this property. And so it really comes with the translation co-cycle. And as a result, if I can push my representation to do something and then this is different, then it's really not semi-conjugate. It's really that I did something nasty. And from this, I can conclude that the signs, epsilon A1, epsilon A2, et cetera. So I consider only those for A. Then they have to alternate. And the proof of that is that, suppose I have two consecutive regions where this thing has, say, positive dynamics. Then what I can do is push very much A by bending along B so that a point like here for A, yeah. When I apply this, I'm going to maybe to get, what I'm saying, okay. The thing is that when I apply a row of A to some power, then I'm going to get maybe in that region. But then twisting along B is going to help me to jump into that region. And so I have some periodic set that is going to disappear from this picture. And so that's really the idea of what's happening here. And so this is it. The number N has to be odd, has to be even, sorry, for the signs to cyclically alternate. And so I get my K. And so we are almost there with these dynamics, except that we may have periodic sets which are a little too big instead of points. And for that, maybe there is a good exercise for our students. Suppose you have a map F from R to AB with AB finite. And suppose you have a homeomorphism which preserves the orientation. I have a map like this, okay. And then there exists some T so that if I do F and compose with a translation by T, I can choose T so that this map has a unique fixed point. Okay. And this is really an exercise for a... This is really an exercise. And now what I'm saying is that, okay, suppose that K is 1, then that's really what enables me to shrink these two points, right? Because on the A side, I can conjugate so that I have A on this interval that's like a flow. And then B is going... Because of the alternating things with the signs, it's going to push things inside. And so that's exactly this setting. And so in order to push all these regions to points, then you just have to generalize a little this exercise to the case when there are a lot of them. And that's just doable. Okay. And maybe I... Thank you for your patience. Can you tell us a little bit about non-resistant representation of the banding deformation of synchronicity? Suppose you have a non-resistant representation. Oh, yes, yeah. Oh. Well, okay, there's a trick. So all these bandings give me a little, a slightly better property than just rigidity. It's really pass rigidity among representations. So all these... And we do have this... Actually, this is how we do the main result in the end. We actually do pass rigidity most of the time. And this pass rigidity is always through bandings. And so almost the answer would be yes. And... Contracturally yes. Contracturally yes. Anyone who does dynamics of any sort, because stable behavior comes from hyperbolic, theoretical fixed points. And so if you have something that's not conformable, it must fall off. Like that's got to be the source of it, right? And so all this technical work is to just realize the idea in a case where you have no regularity whatsoever and everything is only up to even semi-conductions. You have nothing to hold on to except for the translation that comes like always.