 Ok, so let's start as always with this real chamber picture. I hope you remember it. End of course. So we have here, this is chi 1, this is chi 3, this is chi 2, so these are the kernels. We have here the size plus plus minus minus plus, minus plus plus, minus plus minus. Then you can finish. Ok, so this corresponds to the sign of the first direction of the Lyapunov exponent along the first direction, the second direction and the third direction. So we keep with two matrices A, with this row from C square, SL3C. We have these three Lyapunov exponents. We have the splitting E1 plus E2 plus E3. And the first, the chi 1 is the Lyapunov exponent along the first direction, the chi 2 along the second direction, the chi 3. This Lyapunov exponent along the third direction. So that's the picture we need. So we know, so we have our alpha from C square into different morphisms of C3. So we are going to assume infinity different morphisms. We are going to use the DRC infinity. C1 plus epsilon is enough, but you need to modify the technique. If you want C1 plus epsilon, you want to use this technique. You will need CR with probably R equals to 10, or maybe a little bit more than 10, maybe 200. Some finite R. So we were able to take a finite in the subgroup gamma of C square and try to build H from T3 to T3. H equals alpha n equals to rho n equals H for every n in gamma. H was Heller continuous, and the image of H was also Heller continuous. So it's a homeo. And moreover, we had explicit formulas for the H. And you can look at the proof, and you can see a little bit more. So some u of x, the alpha n, we were able to lift it to our n. So this was rho n x plus some function phi n of x. These are the periodic functions. And we gave explicit formulas for the H, but let me write them there with some more care. So we can write ux as u1x plus u2x plus u3x. Know that R3 splits at the sum of three sub-stories. So I can take this u of x and split it as the sum of three vectors, one on each space. And the same I can do with this phi n. And write it as phi n1 of x plus phi n2 of x by n. If you remember what we did when we show the existence of this H tilde, you can see that if phi i of n is positive, let's write the exact formula so that I don't make a mistake, this is negative. Then I of x is minus sum from k equals to 0 to infinity e to the chi i n times k and i on alpha k times n. Pause with the second. This is positive. This should be negative here. Pause alpha minus n. Okay, so you have this number is negative, so this is an exponential conversion sequence and this is a bounded function. So this series is conversion by the wire stress n test. And in the case the chi i n is positive, then you have a similar formula. Oops, x equals to sum from k equals to 0 to infinity. Oops, there is a factor in front. e to the minus chi i n times the sum from k equals to 0 up to infinity of minus chi i n k i n i alpha k n. Okay, now you may complain in principle. There are too many indices here. Okay, so the point is when we made the proof, we picked this n naught here and we did all this computation with this n naught. Okay, so the n was equals to n naught. But now I can look exactly the same with any n. So this expression makes sense for whatever n. So in principle these UIs will depend on the n, but I proved before that the h was the same independent of the n. Okay, so I have these expressions independent of the n. Okay, so these very same functions Ui have a lot of way of being written for each value of n I can write it in this way. This will be very important, improving smoothness of the guy. Did you can see just from solving this equation, you can just, it's a matter of rewriting this equation, you can see that you have formula U1 prime from which one is deduced very easily which says that the Ui of x is minus some from k equals to 0 to n minus 1. So it's essentially the same formula, but I'm only adding up to n minus 1. Chi i n times k by n i on alpha minus k n plus compose alpha minus n of x plus e to the n chi i n times the Ui compose alpha minus. So we have this. So this is, okay, where? Here? Yes. Minus k times n. And yes, so these two guys together is minus k plus 1 times n, I agree, so you could put them together. But I want to make them separate because there is some kind of symmetry here. So you go k forward and then k backwards. And then this is some dirt that I don't want to see. So this formula is exactly the same as this formula, but only treated n times. And then if I take n to infinity, this goes to the total sum, and this guy is killed because once I know Ui exists, this guy is bounded, and I have some exponential conversion term here. So this will get any bounded terms. So you have C0 conversions of these partial sums to the total sum. And there is a counterpart to prime. I will not write it, but it's essentially the same counterpart because you have here. In the case, the guy i n are positive because if the guy n is positive, then here this goes to infinity, so you cannot say much. You want to use the other formula. Okay, so we have this formula, and these are crucial formulas to prove smoothness. Let me see if I'm forgetting anything or that's all I wanted. Okay, that's all I needed. So let's try to prove smoothness. I will make my first attempt. Probably I will fail, but then I will introduce more here, Ian. Hopefully we will succeed. So I have this explicit formula, say here. Okay, so I told you yesterday that I happily could move along any place here and the guy will be an ossef. Okay, I will have stables and I will have contraction along these stables for whatever n I choose there. Let's see if I take even the very n naught. What could I do? So I want to have, say, there the... I will try to prove smoothness of the u1, smoothness of u1. So the u1 here is the first direction. So here the first direction is contracting, so I'm using the first formula. So I want to prove smoothness of this and if you have a series, if you want to prove smoothness, what you do is you take derivatives. Okay, you take derivative of these terms. If these terms, the derivative of these terms are summable, then you are in good shape. Okay, we just take derivative then. Derivative of ui is... will be more specific of how I'm taking derivative soon, but let me just write it down. This is just a number. Then I'm taking derivative of phi and i, pose alpha minus k, and compose alpha minus k. And, okay, here is my first step. So this guy is independent of the k and is smooth, so I can't forget about this guy when I'm going to do my computation. Okay, so let me not add it. It's very painful to have this guy. That's what we have. So I have some exponential convergent guy here and then I have the derivative of this piece. Now some chain rule is in order, so you can do derivative of phi and i, compose alpha minus k, and times the derivative of alpha minus k. So that's what this guy is. So if I'm taking first derivative, that's what it is. Okay, this is just a function composed with homomorphism, which is what the difference is, which is onto, so the c0 norm of this is bounded. So this is not a problem. But now I have the norm also of these guys. Okay, now these are matrices and the norm of these matrices are growing to infinity. And indeed you can be a little bit more specific. I could take a vector here in the unstable direction. Okay, if I take n in that cone, if I take the vector in the unstable direction, then this is exponential conversion. I'm in a very good shape, but we will see in a second that this is useless. So I would have been able to prove it faster before. So the interesting case is that I take vectors along this stable direction and when I take the vector in the stable direction, this grows exponentially. This is the hyperbolic condition that will tell me that this will grow exponentially. Now I have an exponential decay, an exponential grow, and the problem is that most likely this guy will win to that guy. And if you take higher order derivative, indeed, what you can do, so you can use just more fancy general formulas, and so the Fadi-Brino formulas and essentially what you get is a bound like this, but then you will have a power k here. Okay, so that's what you're paying, but it's even worse because you have exponentially increasing to the power k, so it's k times this exponentiality against only this term, but maybe k is not a good quantifier. And this is the degree of infercibility. Then I have here this guy, and I have to kill this guy. So this is not working, so how you fix it? How you fix it is you fix it in the following way. First of all, I want... Yes, that's what I want to do for higher smoothness, exactly, but this is not working because the bound I can get here will explode, so I cannot really control this at all. So then what should I do? What I will do is I will, first of all, try to fix a little bit this expansion that is appearing here. Try saying, okay, I don't want really to use the formula I used because this is really bad what is happening there. Okay? So what I want to use is the second formula. I want to use this formula because if I use this formula, forget about this very bad guy in principle because I'm... the end is still down here, so it's not up there, so I am down there. So this guy is bad now, so I'm using the second formula in the wrong case, but this guy is very nice because when I do this very same argument I did there, so I will have instead of the minus here, let me write this down what we'll do with the second formula. So the ui the second formula will be the guy i n so this minus guy here, I forgot the minus sign here, but that doesn't matter, and then I have the sum from k equals 0 to infinity e to the minus guy i n times k, and then I have derivative pi n i negative compose alpha k n now I do the same argument there and this I can bound by the norm of partial n of pi compose alpha n pi and then the second guy I have to bound is this d alpha k times n then I want to see what happens along stable directions along unstable will not be a problem I will tell you later how to do and I will need to control this bound and now this is beautiful because if I'm taking positive iterates this has exponential decay and this exponential decay essentially will win to this other guy also this one has to be careful this will not be 100% true because I don't know how much faster, I don't know this exponential and I don't know really how much faster this is so I have to work a little bit more even to say that but at least it's a little bit more hopeful than that the big problem is that even if I can take say imagine I take higher derivative then the point is I can play a little bit and in case this I don't have a function to take derivative to begin with if chi i n is negative this doesn't make sense this is not a function so what you do and that's essentially the idea of fissure, cullinan but here it's not a function, it's true but this is a distribution so as a distribution it makes sense one has to prove it and then if it is a distribution you can take derivative of a distribution and if you eventually show that taking enough derivative you get a nice function then you can say something so that's the claim that's the rema u1 take n to be chosen carefully c is this cone chosen carefully write it, this is formula 2 then formula 2 defines ui of x i, i, e for any function psi p the integral of ui times psi is equal to the sum from k is equal to 0 up to infinity of the minus i, i, n there is an e to the minus i, i, n also here there is a k here integral of pi, n, i composed alpha n alpha k, n times psi so this is yes, for some expression to be chosen carefully I didn't tell you exactly what does it mean I will tell you later I don't have even a function to differentiate yes, that's why I don't have I could take formal derivative of this I could do that and I will have a serious but even if I do that which is what is written there still not obvious that this exponential contraction here will win to this exponential expansion there it is not, no it is not it is not so this will not win to that because yes absolutely you have to make very careful things but even if you were in the best of the cases the only thing you may hope is that this number is exactly the same as this number so in the best of the cases you will have balance so this will not explode but it will not go to 0 it will be used yes so I still need to control this so the point is that what I can show essentially here is that this guy doesn't grow exponentially but not too exponentially so super exponential it at most grows only exponential with some explicit balance on the exponential I will have now this will be enough first of all to get this statement and the next statement which is much stronger because this time I can get it without this discussion but the next statement I will need this discussion to get it but you only discuss things in the distributional sense so I need to modify a little bit all my statements here to make it work but I think that's where everything comes from so the way you realize that this is the right thing to do is by looking at this term so if you want to get conversions in the distributional sense of this series you need to know something about what you are summing now what is this let me just translate this in a very crude way so your say your alpha n of x so your alpha n is a function f phi n i is just a function phi which is a sinfinity function your psi is also a sinfinity function so what is this is the integral of phi composed with fk times psi let's say instead of dx I put a d mu now maybe you start realizing what is this term if I take out the integral of phi times the integral of psi so this is the decay of correlations for these two functions I may be very lucky and have that this is less than e sum tau times k minus tau times k is norm of phi curler norm of psi I may be lucky enough to have exponential decay of correlation for curler continuous function and also systems used to have that ok so I only need to do it with respect to the right measure and constants of course will not kill my problems up there I will not have problems with the constants ok so that's why you hope to have some conversions and observe that when you want to say that this happens in the distributional sense it means that I need to have conversions of this series but I need to have bound on the c r norm of psi for some r and here I am getting respect to the curler norm it's a distribution but it's not even of very high level let's state let's discuss a little bit this issue for the one that exists the measure mu equals to some cell function eta x x where eta is in infinity and mu is alpha n invariant we have an absolutely continuous invariant measure for this moreover h star mu is hard measure I'm sorry measures go from push back the measures you don't pull them forward or you pull them back you define a measure there also it's ok it's hard to measure on this you take the h which maps on your t3 with the proaction you have here the 3 with your alpha action you take the measure mu here and it's sent into the most natural invariant measure there which is hard measure this in non-trivial statement is a consequence essentially of Lipschik theorem or Katokkonenko or more so the best form is a consequence here but it's not that hard to prove in this context you have to play some homological equations it's not really that hard in this context so that's the first thing to do so I have this measure this dx and I have this eta which is fixed so this statement will tell me that it's not harm if I put the measure mu here instead of this measure dx when I'm discussing this monic analysis that's the first statement second statement by Gorodnik and Spachier but in this in this framework has an understandable proof in the general framework the proof well of course there are at least two persons that can understand and more has a referee and one can understand it but use heavy machinery so you use green tau tuition results but in this setting of the torus it's really some Fourier business which is quite understandable so it says the following which is this decay of correlation so of course you will say it's decay of correlation for the nose of map so this is a very old subject let me just be careful on the statement so given theta which is the Heller exponent there exists tau positive the row action is given here so these guys depends on the row action but only on the row action did you know less than that but it doesn't matter now such that if phi theta psi is also in c theta and that's it then the integral of phi that composes alpha and times psi x, alpha not rho so it's for the linear action this is minus the integral of phi the x times the integral of psi dx smaller than or equal to some constant yeah there is a constant here positive e to the minus tau norm of n times now it needs epsilon more space norm of phi c theta norm of psi so the decay is uniform for all the elements of the actions so that's the key business and since the con sugan c h is Heller continuous I can take this decay of correlation exactly the very same with the same exponent tau there maybe a modified theta here 2 decay for the alpha action and this is exactly what I needed here exactly what I need there and as I told of course there appear the integral of phi and the integral of psi but this is not really big deal but this is not a problem this decay when you change the alpha instead of the rho is exactly the proof of this level yes and I guess I get the conversions so here you have to do the following things so when you go to the proof assume first the integral of psi is 0 and then this thing here is exactly the same as this so this guy decays exponentially but also with bounds of the norm of psi c theta indeed let me just put it here what I know is that ui but it was continued to begin with so what I'm going to write here is useless but it's important this is c theta star that's the proof that this guy is in c theta star of course the fact that ui is in c theta star is back so I know that ui is Heller and Heller will be in its dual c theta this is not fun what is important here is that this formula holds in the distributional sense okay and indeed rather than using exactly this sum you can use this formula here and get it even faster but let me not enter into this now comes the second lemma which is even more important this heavily so the key point here is being able to use the wrong formula okay so that's the key issue oh yeah maybe I should say something more probably that's why this is complaining yes I'm sorry yes so here I have some clear bound tau this is more than tau I have tau mod of n k this decay now I need to win to this guy here which is positive so that's this beaches in California so the proof of lemma 1 is just the following there exists n indeed for every eta positive there exists n such that I want two statements one is chi i n is negative that's just because the guy belongs here but it's larger than it minus eta that's times n and this is just essentially irrationality of this guy that's one thing and then I will need other more things but let me skip until the next step where I need to prove the next lemma but once I have this then I can choose really this n so that this is in absolute value much smaller than the tau I am winning and the tau is fixed independent of the n second lemma positive n derivative along the s direction of ui this is n derivative along ws direction belongs to c theta so there exists theta so I am taking nth order derivative but only for vectors along the stable direction for this guy I have and I know they are all always in the same distribution space of the same degree c theta so this that happened here is not only true just for the ui which was used it's also true for the derivative and I don't need to move the theta and that's the crucial business because of course from here you will get that dn ui is in c n plus theta star so this is almost a triviality okay but you can do it without modifying the theta the prove is a little bit more complicated and I just realized I have only 5 minutes but follows exactly the same spirit here okay and there is one very important step that I'm not putting here but it's it's what I said so it's this guy now since I'm getting very to the boundary of my n of my c this guy now is not anymore exponentially contracting well it's exponentially contracting but I don't have good control on how fast it is but at least I can play this eta business and create some control so I have a counterpart of this eta for the non-linear action also which comes in the same flavor as what we did yesterday when we show that all the elements here were an awesome so once you know that for every measure the Lyapunov exponent has disbound it's true uniformly and then you play exactly the same game here and for all Lyapunov exponents you will have a bound like this and then you will have this type of bound uniformly for the non-linear just by playing the Lyapunov exponent so you need this type of bound to play this let me not enter into this because if not I will need today, tomorrow and another week just to prove lemma 2 but that's the the philosophy of lemma 2 is essentially you have some plus some interpolation techniques where you bound what you can bound in c theta star and then the rest you allow to go to c and plus theta star and then you do the rest ok so that's what you get the next lemma putting this plus this together so it says that if action w from t3 r is theta so it's a further continuous function and the partial derivative of order s of these w functions are in c theta star for every n so exactly what I'm having here so this partial derivative makes sense in distributional sense and as distributions belong there then there exist alpha positive which is little bit smaller than theta so theta as you wish but it's a little bit smaller such that this dns w belongs to c alpha these partial derivatives are actually functions and are further functions so this is just an interpolation technique so it's a very standard fact in harmonic analysis if you have a function you bound the c r norm and you bound the c l norm you can bound all norms in between ok and that's what you are doing here ok so you have that the very high derivatives are bounded here so this has some implication on how this w norm and then I can bound all the other guys in between so this interpolation plus all of embedding is just some very basic harmonic analysis so it's the only subtly that you have to work a little bit harder is that everything here is subordinated to the stable direction not to the total space so then you have to do some harmonic analysis subordinated to the stable direction but that's doable, that's not really it's not even painful it's nice ok good so I have now that my function u i is differentiable along the stable direction let's analyze u i along the stable direction so I have the same way I split the u i I could split the h tilde as h tilde 1 plus h tilde 2 plus h tilde 3 ok and I know that h will take the stable manifold of n naught of a point x into the stable of the linear guy which will be h of x plus the stable space of n naught which is nothing but h of x, there is tilde here everywhere plus the stable space which is the e1 plus the e2 direction so we take this stable manifold onto this translated space now what is this translated space so here I have say the e1, e2 e3 so I have my h tilde x and I have this now I have the unstable manifold correctly I have my unstable manifold of this point x so these are curves and this is just a plane now this plane is mapped into this plane this line is mapped into this line through the point x so it should be into this line ok now this line by definition so you have the counterpart here so h tilde of the unstable manifold n naught of x is h tilde x plus the third space which is the unstable manifold for this n naught so this the fact that the image of this manifold is this line means that this first coordinate is not moving and this second coordinate is not moving ok so what I get is that h1 tilde restricted to the unstable manifold of this point x is constant it has to be constant it cannot move it's a constant function it's constant h1 of x ok the only variable that moves is the third guy so the first is constant the second coordinate the third is moving well constant function is moved now u1 is h tilde minus the projection of the first coordinate projection is moved so u1 is moved as well along the unstable and then there is to finish a theorem of Schurme h1 minus the projection of the first coordinate so h because h tilde is x plus u so h tilde 1 is projection of the first coordinate plus u1 ok so it's the natural projection from r3 into this e1 coordinate so if this is constant and this is constant and this is smooth this is smooth it's just this statement so Schurme theorem if w from t3 is h tilde and this here we need continuous let me formulate the way I want to use it have dns w in c alpha and dnu w in c alpha for every opposite fly for every n then w is same thing ok so he has much more statement like that it's smooth ok then we have to go to u2 and to u3 now let me just comment one very important thing we needed this eta here and to get this eta it means that I have to take dn very close to this line where the chi1 is very close to 0 I needed to be able to take this this point here ok and that's why I was able to handle the e1 direction and handle the e2 direction by taking the guy here but I cannot handle the e3 direction because to handle the e3 direction I need to make this chi3 of n very close to 0 so to handle the 3rd direction I need to shamp this wall and approach here but to do that I will need to know that the guy is here not an awesome and that's the next thing to do cross the wall and get an awesome guy there and once I get an awesome guy there then I'm done that's what we are going to do tomorrow