 Okay, so the second. Okay, so that as always is our real chamber picture. We had our analysis of element here and not. We already showed that every element here is an analysis of element. And using that, so that if we have your lift of the conjugacy to R3, and we write this as H1 tilde plus H2 tilde plus H3 tilde. With respect to this splitting into the first, the second, and the third Eigen direction, we were able to prove that H1 tilde is, was our painful last off. Okay. And why H1 tilde? Because we were able to move toward this boundary here of this veil chamber C. And as we have said, we could have go to this kernel of chi2 and prove that H2 tilde is also smooth. That's fine. You can do it. It's not a problem. But what we are looking for is for the H3. But for the H3, we want to continue our argument. We need to go all the way up to this kernel. And to go to this kernel, we need an analysis of element here. And that's what we are going to do. We are going to find one analysis of element there and N1. And the advantage is that now we know who is H1 tilde. Okay. So let me repeat one picture we have used yesterday. But now it will become even more important. H tilde of the stable manifold always equal to H tilde of X plus the stable space for the raw N0, which is nothing but H tilde of X plus this one. That's E2 because the N0 is down here. So the stable guy are the first two element. And the H tilde of the unstable manifold N0 of X is H tilde of X plus EU raw N0. So this is the stable manifold for the linear action. It's the translate of the stable space. And this is the translate of the unstable space, which is H tilde of X plus E3. Okay. So that's what we have. This has implications of what is this H1, H2, H3. So this is the E1 direction. This is the E2 direction. This is the E3 direction. If you have H tilde X here, it has this value here, H3 tilde of X. And then it has this other value here, H2 tilde of X and this other value here, H1 tilde of X. Okay. What we are saying here is there is two important implications of this equality. It's one that you have, that this guy is contained here. And the other is that this guy is equal to this. It's not just contained. It's completely equal. So it's an onto map when you restrict to this unstable manifold. And the same for the second statement. So it means that when I take a point here, take its unstable of X, take the stable manifold of X. Of course, you have to leave everything to universal cover to apply the H tilde. If not, you apply the H. This guy, this line is mapped into this vertical line here. Okay. So that's what says this second statement. So this is not the line, this is the curve, but still it goes onto this vertical line. And it means that the H1 coordinate is constant. The H2 coordinate is constant, but in the H3 coordinate you are onto. Okay. So I have that this expression here means H1 tilde of the unstable manifold. So let's put this way. Of Y is H1 tilde of X for every Y in the unstable manifold of X. Alpha naught always. So I'm not moving alpha naught. Similar thing. And finally, H3 tilde restricted to the unstable manifold is homomorphism. So this maps of the Y tilde X naught into the E3 space. Heldar continues homomorphism with Heldar continues inverse. Okay. And similar thing I can say up here. Okay. From here I get H3 restricted to the stable manifold alpha N naught of X. I should make H1 tilde plus H2 tilde from stable manifold alpha N naught of X into 1 plus E2 is a homo H3 tilde of Y is H3 tilde of X for every Y stable manifold alpha. All this is an exercise so you can do it by yourself. Okay. Have all these nice properties. Okay. Now here so there is some asymmetry here. So here I have this H3 along the unstable and it's one dimensional going to one dimensionals. And here I have to put H1 plus H2. And I cannot speak directly just of H1. And the problem is because I don't have a splitting of the stable manifold into one dimensionals. I have a splitting for the linear but not I don't have a splitting for the nonlinear. If I had a splitting I would expect exactly that I can then split these two business or to have a a foliation attached to one and another foliation attached to the other. In principle yeah you don't know really so even if you have a splitting you have to prove something it's not that it's for free. And even more here because you have the whole action. But if you are just with one individual system let me maybe it's an important very important observation. So if you just have one guy. The perturbation and also of linear map real. I can values. Then you will have that the tension to the three split as he has the strong direction plus a weak direction. So if we are in this setting. Like a weak stable direction. Plus an unstable direction so you will have this splitting there if you are only doing perturbations. But this and you have the concurrency because with one and also you already have the concurrency and you will have. A strong stable for the Asian week stable for the Asian unstable for the Asian so all these things you will have now this for the Asian will be mapped by the concurrency to the unstable for the Asian that's easy. These two guys joined the grade today. Stable for the Asian and it's going to the stable for the Asian for the linear this week for the Asian. It's a little bit harder to prove but you can prove this week for the Asian will go to the send to the intermediate for the Asian for the linear but the strong for the Asian. Will essentially never go to the strong for the Asian of the linear so unless you have some very specific type of system the strong for the Asian is not going to the strong for the Asian so these things. Do not happen strong for the Asian don't go to strong for the Asian is a perturbation so it's because it's a perturbation of the linear so it's it's it's here to show theory yes yes so it's it's just a perturbation. Yeah if it is not a perturbation then in stealing this type of setting a lot of things are known but. But if it is a perturbation is this all here to show theory so it's and just write down formulas and. So yeah even so according to this even if we can split this is it's not at all obvious. That the corresponding for the Asian will go to the corresponding for the Asian but it will be true so we will see if the coast was smooth this will happen and we are trying to prove this. Okay and I already know that this H1 tilde is smooth okay. So let's move the next step so I need I have only these two bundles and I need to jump into the next place and in the next place I need to find. Another two bundles okay so the way it should be is. Just you need to follow the the recommendation of the linear so if I take this and one there so the stable for this and one will be just the e2 direction and the unstable for the and one will be the e1. Okay that's for the linear now I have to produce something similar to the nonlinear now the e3 correspond to the unstable in them and not not case so this and not will still be. So this e3 corresponding direction will still be in the unstable for then one so I have one piece of unstable for then one. Okay now I want to produce the e2 which is this table. It will not be easy to produce this e2 and then I want also to produce the e1 so and I have just this H1 so let me just tell you what you do. We have this linear map so one observation before. One more property for this H1 map so if you write an integer in e3. So take this integer. In e3 you can write it as n1 plus n2 plus n3 where the nI are in the eI and the fact that the H tilde X is a map that goes to the torus it will give you that this H1 tilde of X plus an integer is the same as H1 tilde of X plus this n1. If you look what happens in the integer translate it's just doing that so it's not doing more than that. And so let's define some subspace by f of X is the kernel of the H1 tilde. This is for X. I just have this corner. What are the properties of this f of X? Property 1 n is in c3 then f of X plus n is f of X. Okay so this is just an application of this property here. So it's really this f of X so f of X finds vector bundle. It's not exactly vector bundle but it's on the torus t3. Okay because it's independent of the choice of the guy. Take a point to the torus t3 you take an element of f3 that projects to it and I have this vector space. Okay and if I take another guy I have the same vector space so it's well defined on the torus. That's the first thing. The second thing follows from the property of the H tilde. So since H1 tilde composed with f is rho n composed H1 tilde. Okay so it's only for the first piece but it's still if you want you can even put here e to the guy 1 n times this. So this implies f is alpha n I'm sorry for every n where this implies that d alpha n of f of X is an invariant so to speak bundle. Okay so I have a new invariant bundle that's all I wanted. So what is this kernel? So you always have to imagine this H1 tilde as the projection to the first coordinate because that's what it is. It's projection plus some periodic guy so it looks really like projection. So this kernel in the nicest case will be giving you this e2 plus e3 direction. Okay this f you have to think this f as the e2 plus e3. Of course e2 plus e3 is useless so I want to have something like e2 or something like e1 plus e3 no e2 plus e3. But that's what you get what to do. So you use what you get you don't just produce what you want you just get something you use something. So that's what we get. Okay so now I can take a couple of lemmas. Lemma 1 mention f of X is 2 for every X so it's d3 and hence f of X is a continuous. So the first statement is contrivial. The second statement is just some trivial application of implicit function tilde. Okay so this kernel is the tangent space to the solution to the level surface of the H1 equals to constant. But I need to know that these guys are two dimension. This guy could be three dimensional. It could even be three dimensional at every single point. Could it be? So if you have a function from R3 to R and the kernel of the derivative is everything for every single point. What you can say? First of all you have a linear map from R3 to R and the kernel of the linear map is everything. What you can say about the linear map? That's easy. Zero. Exactly. Now if I know the kernel of the derivative everything for every single point I know that the derivative is zero for every single point. What can I say about the function whose derivative is zero at every single point? Constant. Can H1 tilde be constant? Was somewhere here? Yes well something like this. Maybe this one? Yeah this one will do that for us. Thanks. Yes it cannot be constant. Indeed it's projection plus bounded so how could it be constant? So the bounded should kill the projection and a bounded guy cannot kill the projection. This projection is periodic not only bounded but periodic is bounded. So it cannot be constant. Okay now next question. What does Sartre's theorem tell me? Almost every point is a regular point. Regular means exactly this kernel is minimal possible. Okay so I know that this guy has dimension two a lot of times but that's something to start so at least I can play with something. So I want for every point but I got a lot open at least because certainly it will be an open property to have kernel two. Sartre's theorem tells me some regular points plus I have this property here. Or if you want, yes this property here. So this f of x is invariant. So proof of lemma let k be a set of point x such that dimension of f of x is three. So this dimension could be just two or three. There is no room for anything else. Okay so property one k is compact. Property two alpha n of k is k for every n is invariant by the whole action. We have a compact invariant set by the whole action. This almost killed this. They are so close when you get this. There is a result by variant. It will be useless for us but still I think it was stating it. This is essentially the counter byte of f of x times two times three for compact invariant set that Amir was talking about a couple of days ago. We says that if rho n, so for the linear of k is equals to k for every n in c square, k compact, k equals to the total space. That's good. We already ruled out k being the total space. k is empty. That's good. We want to show it is empty. Three to three k is finite. And since it is invariant, it's just finitely many periodic orbits. Okay the problem is that variant doesn't rule out k finite because he cannot. So you have compact invariant set which are finite. So you have finite orbits. And having a finite orbit where this phase is useful for us. And it's not even just not useful. It's useless if you want to look at the generalization to higher dimensions. So there is no variant theorem in higher dimension. And it may fail in a very bad way. So it's quite bad. There are some examples by Sirin one and Ellen Lindenstrauss about how these things can fail for minimum. It's complicated. But even if it were just a periodic point. So we need to do something about this periodic point. And will you still use the theorem to make my life simpler in the exposition? Because I only need to assume the case just some periodic orbit. So some amount of periodic orbits. In the general setting what you do is you have a compact invariant set. It supports a measure. And you play with the measure instead of the periodic orbit. And then instead of using stable and stable manifolds for periodic point, you use spacing theory with respect to the invariant measures. Not harm. It's okay. But anyway, here you have learned. So still we know something about this compact set. Now we can do something about it. Okay. P and K, the gamma equals to the stabilizer of P. So all the n such that alpha n is of P. That's it. And we know we have already discussed this. We have always that this gamma is finite index. So we can assume always the point is fully fixed. Okay. For all the elements of the axon. Or maybe I didn't comment that. So this theorem is for the linear action. But since the linear and linear topologically conjugate, the same is true for the alpha axon. So it's the property which is invariant by topological conjugances. Okay. I take this P. Here my fixed point. And I know that for this fixed point, I have Lyapunov exponents. And I know they almost match with the Lyapunov exponent for the linear. So it's much up to a constant. But that's good enough for me. So this point will happen. The three of P. Two of P. This you have. And the stable manifold of P is this plane down here. Okay. So let me concentrate on this plane. E1 of P. Now this plane is the stable manifold for the n naught guy for the point P. So it's mapped to the E1 plus the h map this map plane to the other plane. And indeed it's not just h. It's this h1 plus h2 tilde that are making this map. And of this I know the h1 tilde is smooth but has derivative zero here at this point. Okay. The next statement is the following stable manifold. Something very similar to this. For this I need to pick my right element, my correct element. So I pick this guy here. Let's call it n3. This I'm picking a guy whose guy one, the f1 is very close to zero. The guy two is reasonably large. The guy three who cares. So I know that for the guy I'm taking, so for this n3 guy I'm taking, this is a strong contraction and this is just a weak contraction. And I can take this as close to zero as I want. And this will become huge indeed in this case. So there exists a foliation. The lip sheets, when I see one foliation, uu of, well if it is s is ss of uu of this guy. This fixed point, this is the same as the stable manifold alpha is not. So it's stable manifold doesn't change. Such that one n1 principle only but also will be two for others. But I only need then one of the lip. Y is Y and invariant foliation. But it's only inside this stable manifold. It's not in the ambient space. And one is this guy n3. I'm sorry, n3, n3 euro. And this is also n3, I'm sorry. All right, n3 is the same in the same big chamber. Here the condition on n3 is guy one of n3 is negative and is much, much larger and closer to zero than the guy two. Second, the tangent space at the point p of fss at the point p is this e2 of p. So it's an almost vertical. Foliation, also you have that the third property is that fss of x is transverse to this. Well, I should put a name. So there is some center manifold here, wc of p. Let me not write it. There is such a center manifold here. So you can show it naturally well defined center manifold. Let me not enter into this. Well, maybe I would have to enter into this because I want to say one more thing. So calling it c, let me put a one just to emphasize the tangent to the first direction for h tilde of this w1 of p is h of p plus the e1 direction. So it's a guy that is mapped there. This w1 of p is a smooth manifold. You can see, I can tell you from here who is w1 of p. So I take a point here. If I take a point here, the e1 direction is the stable and the other two guys are unstable. So the first direction is contracting. So if I take this point here, this picture I have e2 plus e3 is unstable and e1 is unstable. So then it has a stable manifold for this other element. This stable manifold is exactly this w1 p. But I can say it only for the point p. I cannot say it for other points. It is fine. So I have this foliage, a smooth foliage. Now what happens with this smooth foliage? This is the very next step. So this is the smoothness comes because of very strong domination. So then you can even make it cr. But liptiousness, which is the only thing I care, only comes just from this inequality plus the fact that this is one-dimensional. And then you can get liptious and even c1 plus further. This is just some computation. But that's the only thing you need. If you want more smoothness, then you need much more smaller. My action was infinity. I can do c1 plus alpha. But then I will need to play more. Now the next is the crucial lemma, which was not true there. This general case. Has to do with one more property that has this strong stable foliage that I will state. So let's say this is lemma 1. That was lemma 2. Lemma 3. H tilde, which you only care about the 1 plus 2. But still, of fss of x is equal to the h tilde. Indeed, I only need the contain. But it's indeed equal to h tilde of x plus this e2 direction for every x. The stable manifold of p for this element. So inside this strong stable manifold, I know that these guys are mapping where they have to map. And the proof of this lemma, followed from the following adendum, lemma 2, y is in the fss of x. There exists an epsilon positive such that if and only if distance alpha n3 to the k of x, alpha n3 to the k of y, k is faster than e to the, so it's chi2 direction, chi2 of n3 plus epsilon. So these strong stable are characterized by the strong speed of conversions. And the point is that in this plane, essentially you have two possible speeds, the chi2 speed or the chi1 speed. And in the linear, it also has only these two possible speeds, the chi2 speed or the chi1 speed. So I should put a p here, because the chi2 is just for the linear, here is for the nonlinear chi2. There are some constants, but everything is fixed here. Now here comes a trick, and that's why I really need to finish the lemma 3. I need really this difference to be huge. Okay, I need this to be true here and I also need this to be true for the points p. I also need some theta held constant appearing in this inequality, because the point is that if I apply my conjugacy h to this inequality and since the conjugacy h is held or continue, I just pay a times theta here. Okay, so since I pay theta here, I know that the image of hx and hy are going to zero faster than what the chi1 direction is telling me. So they need to be on the chi2 direction, on the same chi2. Okay, so that's the proof of the lemma. So I, as always, run out of time, so let me just say this about this proof. So I have lemma 3 that sends the strong stable to the strong stable. Okay, now that's essentially the end of the proof. I take this box here, I have color chalk, I never use color chalk neither, so it's not a problem. I take this box here and I map by h. It's mapped into some box. The verticals go to the vertical, the horizontal, this horizontal goes to this horizontal. Okay, now the size is this length is one and this length here is epsilon. Okay, so what's the area of this box here is epsilon. Okay, there are some constant because these are not vertical lines, these are just curves, but it's essentially epsilon. Okay, now this box is mapping to something which has uniform length in the vertical. Okay, so the vertical is larger than, I can even make it larger than one half, but let's say some constant c0. And what happens for this horizontal? I have this segment here and I'm applying h1 to this segment. Because if I restrict to the one direction, I'm just applying h1. I can't forget about the h2 direction and h3 direction, so these are 0. Now, h1 is smooth, derivative of h1 at 0 is 0. What I get in this direction is an epsilon squared or smaller. So I'm mapping something of size epsilon to something of size constant times epsilon squared. Much smaller. Now, if you remember, I told you 100 years ago that there is a volume measure here which is mapped by h to volume measure here. And this is not only true, what is true is that if I take these area measures along this table, it's mapping to area measure here. Times h1, times the Radon-Nichols derivative. The Radon-Nichols derivative is a continuous map. So if I have something of area epsilon here, it should go to something of area constant times epsilon, not constant times epsilon squared. So this is a contradiction. So this proved that derivative could not be 0. So I rule out, prove my lemma 1. I have my new bundle. I told you it's not the best bundle I wanted. What I wanted was the 1 plus h3 bundle and what I got was a bundle like the e2 plus e3. But that's all I need. So here, five minutes, I should finish. Now with f, with this bundle f, I will take this guy now and I show that this guy is an Osof. Okay, I need a criterion to claim that the guy is an Osof and this is a theorem by Manier which says from m to m, Diffio, you only need c1 mention of stable manifold of p equals to some constant s0 for any p of f. Of course, if there is no periodic point, it's even simpler for you. Similarly for the unstable manifold, so let me put a slash. If any points are not hyperbolic, it's even better for you. Second, for any b in the tangent of n or any x, and you have that the supremum of the derivative. Take this vector, you go forward, go backward. This is called the Quasi-Anossof property. Then f is an Osof and the nice advantage of this criterion is you don't need to find any bundles, nor cones, nor nothing. You just need to take a vector and send it to infinity for the future or for the past of both if you can for both. Okay, good. That's what I have to do. The first is okay in my setting. I take my alpha of n1 and I know that the Lyapunov exponents for this point is the same as for the linear for every point. Certainly then the dimension of the stables will be one of the n stables will be two for every point. So the first property is not a problem. So I have to do this second. And assume first it's not in this bundle I just built. Okay, if it is not in this kernel, then what does it mean? Well, it means that then dh1 tilde x of b is a non-zero vector. But now I use the properties of the h. So h1 tilde compose alpha kn1 is equal to e to the k chi of n1 h1 tilde that's the property of the n1 tilde. And then if I take derivative of this expression so this on the alpha kn1 tilde kn1 of x times the derivative of alpha kn1 tilde of alpha kn1 tilde apply to the vector b this is nothing but e to the k chi n1 derivative x of h1 tilde of b there is a k which is here. Now chi of n1 is positive this guy goes to infinity this guy is different from zero this guy the norm of this guy is uniformly bounded this guy then so if if the right hand side goes to infinity the left hand side has to go to infinity and the only way the left hand side is going to infinity is that this guy is going to infinity because this guy here is bounded so h1 is a smooth map so it's completely bounded so that shows that for vectors which are not in this f of x I'm fine they go to infinity in the future which is reasonable by f corresponding to the e2 plus e3 a vector which is not in e2 plus e3 has an e1 coordinate and a vector with an e1 coordinate should go to infinity that's what is written there so now I have to play with vectors inside the f of x but vectors inside the f of x are very easy to handle splits as what? as the f of x intersected with the stable manifold of alpha n0 plus unstable manifold of alpha n0 so it's f corresponds to the e2 e3 direction so I can intersect this here I intersect that there so this requires a proof but this isn't quite straightforward proof I hope ok, but now it's very easy, so I take a vector f of x then either v is in the eu alpha n0 of x or let's say I can write v maybe to write it more clear I can write v as a bs plus vu with respect to this splitting so either eu is non-zero in which case you come back here to your picture and see that the chi3 direction didn't change at all so it went from a plus to a plus so this unstable bundle is still uniformly expanding after I cross this wall so it will start having problems when I cross this wall so I have this theorem that says that have uniform bounds so then there you are fine this implies d alpha kn3 of b is larger than d alpha kn3 of b u which goes to infinity exponentially fast maybe some constant but that doesn't matter so the second case is if u is 0 then bs is non-zero well of course I should have said here I don't want the 0 vector there then the bs is non-zero and now I go backwards so here is k when to plus infinity now here I take k going to minus infinity and the point is that in this f intersection es is this eigenvalue which is directing the business and this didn't change so still I have uniform hyperbolicity along this direction same thing that's it thanks