 Okay, let's continue. So today probably we are going to speed a little bit up. So let me just draw again the picture we had there. This is say the kernel of Chi 1, Chi 2. This Chi 1, Chi 2, and Chi 3 are the diapunov exponents of these two matrices. So we have A and B in SL3C as diapunov. So eigenvalues, lambda 1, lambda 2, lambda 3, and all sorts. Same for B, and we are assuming all of them are positive. You can take the square of the matrix, then the eigenvalues will be positive, and we are assuming they are real. So it's our standing assumption. So the same with the others. And then we define this Chi i of KL as being the logarithm of lambda 1 to the k of A, lambda 2 to the k, 1, lambda i. These are these linear functionals, which we can extend naturally to R2. And these are exactly the pictures. So these are the kernels. And let's say the Chi 1 is positive here and it's negative here. The Chi 2 is positive here and negative here. And the Chi 3 is positive here and negative here. So always you will have a picture like that. Modern interchange in the Chi 1 with Chi 2 and Chi 3 and so on. Okay, so let me put the sign. So on this cone you have that the first is plus, the second is plus, and the third is minus. Then if you jump here, you are jumping the first. So this plus is changing to a minus. So you have this. Then here, the guy that jumps is the third. So you have a minus plus plus. Here it jumps the second. So you have a minus minus plus, which naturally is the opposite cone to this. So whatever was plus has to become a minus. And then you jump the first again. So you get a plus minus plus, which again is the opposite of this. And you finally switch the last one. And you get this. Okay, so these are the sign of the first, the second and the third exponent in each of these cones. And over the lines, you will get that the guy, one of the guys vanish. And the other two respect the corresponding sign. Okay, now we have alpha from C square morphisms of T3. It's just a homomorphism. So it's an action. And I assume that, well, let me first define here, rho from C square into SL3C by rho of N equals to A to the K times B to the L, where here the N is the pair KL. So it's this homomorphism is injected by essentially assumption. So you have this nice homomorphism and this homomorphism rho induced an action on the torus T2, where for each N you apply this matrix on the torus T2. Now we have another guy. And we are going to assume what is known as the linear data, which will be an associated rho, which is this. So each alpha of N will have a lift alpha N tilde from R3 to R3. There are several such lifts, but for what I'm going to say now it doesn't matter. Eventually I will have to be more careful with that. Such that alpha of N tilde is some matrix A of N evaluated at the point X plus some function phi N of X. We just solve this. This is the linear part and this is the periodic part. Okay. And so this N belongs to SL3C and assumption equals exactly to this rho N. So that assumption, it will happen, okay? So it will happen. So these matrices here will generate the homomorphism. It's some easy homotopy or homology theory. So you can do it by yourself or just put the composition here and you will get at once that this A N generate the homomorphism. And so you will have one generator which is A, another generator which is B. And you can show that you can assume that the eigenvalues are real. So this is unpositive. This is not really big deal. So there is no loss of generality on the assumptions I'm doing here. Okay. Say it again. No, it's not right. Of course, there is KK. It should be an L here. Good. Thanks. Okay. So now I will make my next assumption which I don't really need to do this first step but let me just do it to make things easy for me. So next assumption, there exists N not where such that alpha of N not is an awesome. Okay. The linear guys are all an awesome because no eigenvalue has modulus one but I'm assuming that one of the nonlinear guys is an awesome. Okay. So says this N not here where the guy is an awesome. Let's assume we have this guy. We don't know anything about all the other guys but this guy is an awesome. We go to our previous lecture. We know that there exists H from such that H compose alpha of N not is draw of N not H homomorphism. Homotopic. There is one more property you have which I will leave it as an exercise. It follows essentially, well, the statement will be two statements. One is easy follow from the formula I gave you. The other is a little bit harder. H is heather continuous and H inverse is heather continuous. So H and it's inverse are heather continuous. So the fact that H is heather continuous is just follow from the formula I gave you. The fact that H minus one is heather continuous is a little bit trickier but the geometry behind is the same as the fact that H is heather continuous. Okay. So I want to conjugate not just this and not. I want to conjugate all Ns with the corresponding row N. Okay? So my first step will be to show that H consuates all of them. You will do it. I will not do it but I will state how to do it. To this N it's very important to be able to lift the action to the whole T three, to the whole R three. So here I took a lift. So for each N I can take a lift but I may make some mistake and this lift may be not matching correctly. So if I take this lift at random this family of diffeomorphisms of R three will not form a group action. Okay? I have to be careful to choose the lift so that I really get a group action. I will now do it carefully. I'll find not as a fixed point. And this is obvious because the linear guy fixed the zero and H is homomorphism so the pre-image of zero will be fixed by alpha N not. Okay? So you have a fixed point. So then we can lift alpha N not alpha tilde N not touch that alpha tilde N not fixes this point. Fixes A point. Okay? I just pick this lift. Now I take another let's call this point X not. So C three X point X not tilde in R three such that X not tilde plus C three X not. So these are the consulate the classes of points in the total. So that is this means this guy projects to X not. So you can just pick one and make a lift that fixes this point. That's the easy way of taking lifts indeed. Now I take another alpha N but what happens with alpha N of X not? I wish this guy were also X not. So these were also fixed by X not. This may fail. I don't know if this is true but if I apply alpha N not to these then these two guys commute. So this is the same as alpha of N times alpha of N not of X not but X not is fixed by this guy. So this is alpha N of X not. I get another fixed point for alpha N not. So for whichever N I'm picking alpha N of X not is another fixed point. An alpha not is a loss of different efficient. It doesn't have infinitely many fixed point have a lot of periodic orbits but not fixed points. Fixed points are only finitely many. So it means that there should be so the amount of N's where I will start getting different things are finite. Okay, so then I can take gamma which is the stabilizer of this X not which is all the N's such that alpha N of X not equals to X not. So this is subgroup and this observation I just said implies that X square mod gamma is fine. Because if you have infinitely many co-classes here then it will mean that you are doing infinitely many different things here which implies that alpha N not has infinitely many fixed points and this cannot happen. It's just an exercise, you can do it. Well, this is the gammas. If you remember the theorem I stated I stated there exists a gamma subgroup of finite index where I have really concurrency. Well, this is the gamma I was stating in the theorem. Okay, so from now on I'm working only on gamma. Forget about the other elements of the action. It doesn't matter, I have the same picture. Okay, once I have a fixed point X not and it's fixed for every element so I have that by definition alpha N of X not is X not for every N in gamma. I can fix the leaves such that alpha of X not. Of X not, I define these leaves so that they have this fixed point. And now this will form a group action. Have that this alpha tilde, so with this definition. Exactly, so all these arguments to be able to pick this very same X not tilde for all the guys. So I can have the point X not here in the torus T3. I have the point X not tilde R3 and it has all the other translates. So here is the X not tilde plus the one zero zero, for example, so it's a translate of this guy. So all these guys project onto X not and I can fix the leaf and it will be a unique leaf that fix this point. I can pick another leaf that will fix this point and I can even take another leaf that sends this point to this point. Okay, but this will be all different leaves. I choose the leaf that fixes this point. Exactly, so that, yes, yes, yes. And that's, I'm choosing this leaf here so that this same X tilde not works for everyone. And you can do it. So as long as you know that it's a fixed point here, you can leave it as a fixed point there. So this is my definition, this alpha of N tilde. I'm defined up there, say it again. No, no, you get it from the H. Indeed, X not, yeah, because here you have to fix point, which is zero. So indeed, X not is nothing but H minus one of zero. It's exactly this guy. Okay, now I can leave everything to R three and I can now write down this formula for these very specific leaves, okay? And I have a family of smooth map, phi N and these linear maps here. And they really make some nice game. Only on gamma. Yes, absolutely, absolutely right, thanks. Yeah, probably we forget about gamma and not being the whole C square. So gamma is isomorphic to C square in an event. So don't worry if I just renormalize everything. I assume that gamma is already C square, okay? So I'm only working on this gamma. Okay, so you have this global fixed point. So now the next lemma says alpha H compose alpha N is raw N composed H for every N in gamma. The same H not only work for N not, it work for every other guy. Indeed, more is true. The way we solve this equation, and this is very important, the way we solve this equation. The H corresponding to a lift because of the form X plus U of X. And we show that H tilde compose alpha tilde N not was equals to raw N not pose H. So that's the way we started to solve this equation. So everything in R three. And that's what we get. And what you prove is that the same is true for the H tilde. And this formula implies this formula. H is homomorphism. So let me just leave this as an exercise. So the observation is that there is a unique H tilde solving this equation. Okay? You can solve it in very different ways. So you can assume there are two of them and then get it or the point is that if you define take an N, oops, pose H tilde, call this H tilde N. So the claim is H N equals to H. So that's the claim. That's how you prove it. So you define this guy here, this H and tilde, and then you go and prove that this H and tilde coincide with H tilde. And then you get there should be minus one somewhere. Yes, thanks. This is an exercise. So it's really, there's plenty of proofs of this. So let me not enter into that. Okay, so now I know already that the homomorphism works for everyone. Okay, so this is the easy part. Now comes the hard part to prove that this homomorphism is a different morphism indeed. Okay, oh, to lift, just to lift, just to lift and have an action here. So it's important in this proof that you still have an action here, so the commutativity. So it's important that this guy here commutes with the alpha and not. If you are not lifting correctly, then you will not have the commutativity. You have the commutativity plus some error term and then it makes everything very nasty. Comes the app of exponents, the nonlinear action. Okay, so we discuss the app of exponents for the linear action but now we have a nonlinear action, the alpha. And we want to discuss the app of exponents for these guys. Now to discuss the app of exponents, you need to have a measure. Okay, and since measures are a delicate subject here, we are going to play with the most simple measure which are periodic orbits. Okay, periodic orbits are easier to deal with. Take a point P, periodic by alpha and not. They're dense of such guys. The same argument as we did there will give you that this gamma P, which are all the ends such that alpha N of P plus two P. So this is subgroup, so this is stabilizer of P and you have that C squared mod gamma P. Because once this guy is periodic for this guy, then you will not have infinitely many periodic points of the same period. So you need this to be finite. It's exactly the same argument. Which means that I can concentrate on this gamma P and assume that this point is periodic for all the elements. On the gamma P, this point is fixed for all the elements on the gamma P. So now I take N gamma P and consider this derivative which is a map from P, P to three, and map from R3 to R3. So three dimensional vector space to three dimensional vector space. It has eigenvalues and it has what is called the upon of exponents, which are the logarithm of the modules of the eigenvalues. And this is for each N. So it's a very same discussion as we did here. It will happen here. Okay, so you will have is dimension three. You will have three liapons of exponents, subscript P. So you have these three liapons of exponents as it is P. So just think of, you have two commuting matrices and you have the whole C square action here. So you have the commutativity. So you have two commuting matrices and they are jointly, not essentially not diagonalizable in principle. They may have the short and blocks but they have the same short and blocks. Okay, so this is just basically in Azure. So these two guys may coincide here. Okay, but still I have, I won't have to go through all the possibilities but you will have in the nicest case, you have three directions and this is liapon of exponents along the first direction. There's liapon of exponents along the second direction and liapon of exponents along the third direction. And that's what I go, we are going to prove. The next lemma says that that's really the case. There are three directions for this P and these liapon of exponents, the three of them are different and indeed the three of them are exactly, not exactly but they really look like this. I will like to prove that they are the same. I will not be able to prove that they are exactly the same but close enough. So the lemma depends on some ordering. So in principle I make some arbitrary choice of this chi one, chi two, chi three. I will need to make some more careful choice of who is one, who is two and who is three but by such choice you have this lemma. There exists a positive constant indeed. There is a r shouldn't one indeed. R shouldn't be equal to one, it could be one. Such that for every P gothic by alpha and not, there's three numbers, alpha i of P, i one, two and three. This is a positive real number. Indeed it's more than positive, let me just put the i. This guy is larger than one over C inverse and smaller than C, so it's bounded and bound away from zero independent of P. Such that alpha i of P times chi i P is equal to chi, maybe let me emphasize a row here for the chi i of the row. So the liapon of exponents of the nonlinear are positive multiple of the liapon of exponents of the linear. So I want this guy to be a one. So if you imagine that these two guys are smoothly conjugated, then this will be a one. Because if the guys are smoothly conjugated, the derivative of the conjugacy will make these matrices conjugated to the row matrices. And if you have conjugated matrices, then you have same eigenvalues hence same liapon of exponents. Now I have a weaker property than this. I only know they are positively proportional. But I know that the rate of proportionality is bounded above and below. So this has to do with the Helder property, oops, H. The fact that this rate of proportionality is uniformly bounded above and below is the Helder property of H and the inverse. But I will leave this part as an exercise, but what I want to explain is why they are proportional. Okay, so let's try to argue of how to prove they are proportional. Now you have to play a little bit between the linear and nonlinear dynamics, the proof of lemma. Look here, and this is lemma two. Not both at the same time. Okay, so here, yes, it depends on the point, the periodic point P and also it depends on the direction you are taking. So this is not the number, this is a linear function. That's exactly the point. And the positivity, so good. Now that's a very good observation because this tell me exactly what I need to prove. So I have two linear functionals and I want them to agree. So what's the first thing to do to show that the kernel agree? The kernels of the functional agree, then the functional agree up to a multiple. Then the second thing I need to prove is that they have space agree. So when I have positive for this guy, I have positive for the other guy and vice versa. Because then I will know that they agree up to a multiple. And that's what you are going to prove here. That they agree up to a multiple. So let me now draw the picture. Pro guy, this is the one direction, this is the two direction, this is the three direction. Now I have this P point there. I know that the alpha I'm not, I don't really need it, but I know that the alpha I'm not is an Ossoff. So I will have, and let's say the N not was in this place I draw. So I have a picture like this. Okay, well this is expanding and these are contracting. Even though I don't really have these two lines so far. I only have a two dimensional stable manifold and one dimensional unstable manifold. In principle, this eigenvalue here could be complex. So which means that I don't have these two lines. I have to prove that I have these two lines. Okay, let's go first to the third direction. I will leave the hard part, which is the other two directions for you. Okay, that's an exercise. Let's show that these two kernels match. This is indeed very easy. Well, I need to make a choice here. So you have to be a little bit careful. You have to fix all the problems I'm doing here. But it's not that bad. I know that these unstable manifold go by the consistency to the corresponding unstable manifold, consistency sends unstable to unstable. So I have a way to number in these three with these three. And now I'm numbering the Lyapunov exponent so that this third direction match with this third direction. And I have a topological background for saying that. Okay, so one, we said that I want to prove this kernel. So what are the possibilities is that these two kernels do not match. Okay, but if these two kernels don't match, then there exists an N such that chi three P of N is positive and chi three rho of N is negative. So I can make one positive and another negative. So if they don't match, I can do that. This is, they don't match by a positive multiple. I can do that. But then this will imply that when I take alpha N, it will expand this direction. But you're conjugated to a guy that is contracting this direction. This is not possible. Okay, so contracting is mapped by a homomorphism into contracting and expanding is mapped to an expanding. So if two points are positively asymptotic by the consistency, they remain positively asymptotic. And if they are negatively asymptotic, then they remain negatively asymptotic. So I have these two points that for the alpha N, for this N are positively asymptotic. And this will be mapped to two different points because H is homomorphism and they cannot be positively asymptotic. So this is a contradiction. That's how you prove it. And that's it. The same thing you will do here a little bit more. But that's why it's very important that I know already that the H works for everyone. Okay, the very same H works for everyone. Okay, that's it. That's the proof that the kernels match. Not only the kernels match, they have spaces that should match because if not still, you can get the same inequality. And then you do the same. Okay, so let's move on because I don't have infinite time. So what is next step? Next step is to answer question of one of you. You, but anyway. I have these guys and also, it's just by itself. It's very few conditions. So I want much more guys. So what I will prove is if I call this cone C, so the next lemma will be that I could pick whatever point here and all these guys will be an also as well. Not just this guy, every guy here is an also. And it's a little bit more delicate, but it's not that bad. So this is the veil chamber C. So the lemma, which is the one which contains the original and also element, lemma, for every N, P, alpha N is an also. All these guys are an also. Okay, how do we prove that? So you see what we are doing. So maybe look a little bit random, but it has a direction. But still what we are doing is the following. If you know that H were smooth, all the statements I'm doing are all trivial. Okay, because if H is smooth, then I have full matching of this leapon of exponents. I will have then also property trivially satisfied, just by using the smooth consequence H, but I don't know H is smooth. So what I'm recovering is all the properties H will give me. Okay, and at the end of the day, with all this property, I will prove that H has to be smooth. But well, you can start doing infinitely many properties and not getting really what you want. This has a direction, so it took a while to realize that this is the type of property you have to recover. This is just periodic data. It is enough. It's periodic data, but with uniform bounds. Yes, because the point here is to prove this lemma, I really need the periodic point to become fixed for all elements of the actions. So I needed to assume that the measure associated with this point was invariant by the whole action. And then this take us to Amir's lectures. So what are the measures invariant by the whole action? Alpha action. Well, alpha is continuously conjugated to the linear action, so I can just restate the question. So what are the measures for the linear action? Well, this is the same as times two times three as a concepture. So you have the periodic measures, good. You have the back measure, good. Is there any other measure? Not of positive entropy. So I could state this for the back measure, well for the entropy maximizing measure, which is the counter part to the back measure. And for any invariant measure, I could state, but there are not many measures, so it's useless. But I have it for periodic points. And these are a lot. The periodic points are a lot of them. And that's what it enters here. Okay, so how you prove that a guy is an Osof is you use the definition, okay? So we need a splitting, so proof. We need to create a splitting, which is invariant by the alpha n. And once we know that the splitting is invariant by the alpha n, we need to know that this guy is contracted by iterated of alpha n. This guy is contracted to the past by iterates of the alpha n. That's what we need to know. Okay, so the most difficult part in general is to find the splitting. But here the splitting we have. You can see what happens in the linear. I never change signs. So the stable for every linear guy here is always the same stable. It's the E1 plus the E2 plus the E2. And the unstable is the E3 direction. So for the linear, that's what happens. And I will do the same for the nonlinear. So I will choose, maybe let me emphasize here that this corresponds to the nonlinear. So I will define this stable space as being the stable space for the n naught and this unstable space as being the unstable space for the n naught. Just we'll define it this way. And for n naught, I know because n naught is an OSF. Now the fact that alpha n commutes with alpha n naught will make that these bundles are invariant. So the invariance is not a problem. So now I need to gain contraction. How we get contraction? So I take logarithm of the norm just to make my life a little bit simpler. f equals to alpha n. Okay, so it's just one different of this one here. So now we can forget a little bit about the whole action. I take function a of x, logarithm of the norm of derivative of x at x on the stable direction. I take this function. I need not just this function, I have the whole family of functions. So if I want to show that the guys, and also what I need to prove is that claim there exists some n positive pressure such that norm of dx fn on this stable direction naught is less than one for every x. I know that this contracting for every single point x at this iterate, then the guy will be an awesome. But then you can just write down formulas and get that the constant c and the lambda is strictly less than one. So it's a compactness assumption. So if it is less than one at every point, then it is less than one minus epsilon. At every point, this one minus epsilon is the contracting rate on n iterates. Then you, but this fix be capital M, but then you will have, that it is also true for every small n. So now what does this mean? This means that what I need to prove is that there exists some n such that a n x is for every x. So it's the same thing. That's what I need to show. Because I take logarithms and that's what I get. That's the next lemma, I erase this too bad, too late. Lemma says the following, family of continuous, these are continuous, indicative of cycles. Okay, so that means n plus m of x is smaller than a n composed of m plus a for every n, m x. So it's multiplicative ergodic theorem, guy. Now if integral, more than the integral, the infimum of one over n integral a n x d mu x negative for every infimum in n positive, for every mu variant by f, then there exists n positive such that a n positive n of x is less than zero for every x. And that's what I want over two times so I will not write anything else but lemma just comment, I will leave this as an exercise. It's a non-trivial exercise but it's still an exercise. I have, I even have done my own proof of this but this I have seen probably 10 different proofs of this, maybe not that very different but by different people thinking that they were the first proving it. So it's this type of lemma which is very basic, it's very important and I think the first ones were some guys in the 90s that maybe, maybe it appeared even earlier or not. Anyway, so let me just comment on the proof. So what does it mean? So the for all here is some kind of unique ergodicity. So imagine that you instead of having a sub-additive co-cycle you have an additive guy. So this would mean taking a Birkhoff sums of functions. Okay, so if you know that the integral of the function is exactly the same constant for every invariant measure then what you will get is that Birkhoff sums converge uniformly. Okay, so this is the same as when you play with unique ergodicity. Now here is essentially the same assumption but with inequalities and with multiplicative ergodic theorem. Okay, so if you know that the integral is negative then you can prove that it's less than a constant for every measure so the space of measure is compact so you can always put less than a constant here, a negative constant here and then you use the sub-additive property to show that you will have this going to minus infinity with n in a uniform way. So indeed what you prove here is that there exists constant c negative such that a n of x is smaller than minus cn for every n larger than n or even better, constant d positive. That's what you prove and again it's not really that difficult. Okay, so once you have this, this is very important for us not just in this context, it's in general will be very useful for us in what comes next time. So you get this claim which implies this claim which means that this table guy is contracting you do exactly the same unstable. Now, and I finished with this, how I get this integral to be negative? Well these are the Lyapunov exponents. If you take this logarithm of the norm, this limit here is the Lyapunov exponent on the stable direction so the largest Lyapunov exponent on the stable direction as c to the measure mu. Now the measure mu can be approached by periodic orbits and on periodic orbits I have that this is less than a constant, less than zero because on periodic points I know the Lyapunov exponents. That's why I prove the lemma that I erase exactly there. Okay, so that's it, next time I will use this to prove the smoothness of the guy, thanks.