 Just a second, a second. So as Amy has told you in the last lecture, we are going to see the proof of the following theorem for the case where the manifold is two torus. So if we have, oh, this theorem uses the Hopf argument, goes back to 1939. And the proof is by Anosov. So some people also say that it's Anosov-Sinai. So let's play safe, and I will cite both. So what we are going to see today is that if we have a map on some object which is called manifold for those who know what it is, otherwise, just think of the case T2. And if this is T2, Anosov, the theomorphism, which preserves the volume. I have called volume M in my last lecture. But just to be consistent with Amy's notation and with Kadin's notation, let's call it Mu from now on. This is volume. So if we have a C2, Anosov, the theomorphism, which preserves the volume Mu, then F is ergodic. This is a very strong result. F is ergodic with respect to this Mu. We are going to see in the case that M is a torus T2. M is compact. In general, for those who know, M will be a compact and close to a manual manifold. But if you don't know what it is, don't worry. We're just proving for the case that M is T2. So in the very special case that M is T2, so this what I'm going to say holds only in the special case, in fact, what you need is that M is two-dimensional. If you have a two-dimensional manifold, then I think you have seen this with Amy today, that if F is a C2, Anosov, the theomorphism, then the stable and unstable fluctuations are C1. And that is key in what we are going to prove now. What does this mean? It means that the charts, the foliation charts, in fact, what I'm going to write is even stronger. So there are foliations charts that are C1. So there exists C1 foliation charts, C1 foliation charts that take unstable leaves into horizontal lines and stable leaves into vertical lines. So let me take some colors. And what we are going to have is that if we have a C2, Anosov, the theomorphism, we will have that the foliation by stable and unstable leaves will go under a foliation chart into straight lines simultaneously. So there is a C1 foliation chart that takes simultaneously this into horizontal lines and this into vertical lines. So this is going to be very, very strong in our proof today, because this in particular, let me do a remark, which is going to be very important, a consequence of this theorem is that Fubini holds, because this is C1. So Fubini holds. So as a consequence, so if we want to measure what I said A here, we can just take one unstable leaf and integrate this A along stable leaves. Why? It's because there's a C1 foliation chart that takes this into straight lines and horizontal lines. So Fubini holds. This is going to be very, very important. So we are going to prove Anosov's theorem and what are we going to do? We are going to proceed. Remember, in my last lecture, I did this for the 2111. And I told you that there were three main ingredients. And one was the fact that the bulk of sums in the future are constant along stable leaves. In the past are constant along unstable leaves. This was one ingredient, which we are going to have here, too. The other ingredient was Fubini, which we already have. And the third ingredient was accessibility, which we are going to have because, in fact, we only need a stronger case of accessibility, which is local product structure, which you have already seen. So if you have local product structure, then you have accessibility. Why is that? Because you can cover. For each point, you have a neighborhood where you have local product structure. And then you can cover the manifold with these charts having local product structure. And then since it is compact, you take a finite sub-covering and then you will have an S2 path from any point to any other point. So having local product structure in a compact manifold is stronger than accessibility. And these three elements is the only thing that we are going to use today. OK, so we are going to proceed in the very same way as we proceeded in this case. So what we are going to do is we are going to show that for every continuous map, the conditional expectation is constant almost everywhere. And remember what we have called and this limit existed almost every x. And almost every x take all inside. In the points where they don't exist, you just can take lim soup or just take zero or whatever. Ignore it. We are going to consider the map just almost everywhere. And so we know almost every x, this limit exists and are equal to the conditional expectation. We are going to call these points the set of good points. Just to find this set of good points, g2, such that this limit exists, let me put it like this. And if we take the volume normalized, then we would have that this set as measure 1, total measure. Just remember the good points. And also remember, it was constant on stable manifolds and phi minus was constant on unstable manifolds. So we are going to call this t because it was trick 2. So let's define the boxes with which we are going to work. Remember that in the previous proof, we have used squares around certain points. So now we are going to define square. Remember, I told you that we just wanted to take squares so that the boundary of the square was made of stable and unstable leaves. We are going to define squares. It will be defiomorphic to squares. But they will have boundaries made of stable and unstable leaves. So for each point, torus, just take the stable manifolds of p. Take them very, very small. And let me do a drawing. The red one, there's some colorblind person here. So I just put the names. OK, so if x is in the unstable manifold, in the local unstable manifold of p and y is in the local stable manifold of p, then the local product structure, the stable manifold of x and the unstable manifold of y intersect at a unique point. And we are going to call this unique point bracket x, y. So for any x, a local unstable manifold of p, and for any y in the local stable manifold of p, we are going to define x, y. And this point is uniquely defined. And now we are going to define the square exactly in the same way that we did define it last class. This is going to be a square center at p. And it's going to just be made by all x, y, such that it's in the local unstable manifold of p. And y is in the local stable manifold of p. So this is going, if we trace all possible points obtained in this way, we will get a small square around p. But now we can proceed exactly as Fubini holds. We can proceed exactly in the same way as we have proceeded in the last lecture. So let me just tell you the trick. We had the good points, g. And in the same way, we can define the bad points as the complement. Just we are going to take the bad points as the complement of the square and the good points. So since Fubini holds, now we have Fubini holds. So since Fubini holds, since the measure of the bad points is 0, then we have, and Fubini holds, both things imply that for almost every p, we are going to have, this is the Lebesgue measure of the, the Lebesgue measure on the local stable manifold of p. So this measure will be 0, because Fubini holds. If it were no 0 for mu almost every point, we could get a point, a set of measure, a positive measure where it holds. Then we integrate, and we will get that the set of bad points is positive. So we have this. For mu almost every p, we are going to have that. If we take, I will take any p. I will take any p into 2. And then I will consider sp. And b is in fact this, and this local stable manifold is so small as it is contained in sp. So I don't know if this answers. This is going to be just, we have p here, we have the unstable manifold of p, and the bad points here. Yes. OK, this, the measure of bad points in the local stable manifold of p is going to be 0. So as a consequence, we have that for almost every p in t, mu, p, u, almost every y in the local stable manifold is a good point. I do it twice because it's easier to work with measure 0 set because we have that this measure is 1, but then we have 2. So we have that for almost every point in the torus, if we take the local unstable, the local unstable manifold, almost every point in this local unstable manifold is going to be a good point. So it's a good exercise, though it requires some care that you can also assume that p is a good point, not only almost every point in the local stable manifold, but you can also assume that p is a good point. It's a nice exercise. You can also use fubini. We proceed as exactly as in this previous lecture. OK, I'll use white chalk. This can be seen OK? OK, then as a consequence of this, this implies that mu, almost every p in the torus, mu, p, u, almost every y in the local unstable manifold of p, y belongs to g. What this means is that for almost every point, almost every point in its local unstable manifold is good. It's a good point. OK, so now we are almost there. As I said, we can also assume. We don't need that, in fact, but it's just more elegant. We can also assume that p itself is a good point. OK, so now take any of these points, a good point, and p a good point such that this happens. And then we proceed exactly as we did in the last lecture. We have this up here, p. In fact, we only will use this unstable manifold. So we have this, then we have, since they are good, we have p is good, phi plus y, phi plus p equals minus p equals phi plus p. This is because it is good. OK, now the definition, this trig property implies that this equals y for all y, local unstable manifold of p. This is because p is good, and this is because of the trig. But now, this is because almost every y is good. This will hold almost every x. So let's erase here and set almost every y in the local unstable manifold will satisfy this. So we have almost every y here has the same value of phi plus. But now, we apply the trig again, and we have that this is constant along the stable manifold. So this equals this for all set local stable manifold of y. So for all set in the local stable manifold of y, and for almost every, let me state it like this, and for all y in w u log p intersection g, we have that this is constant, but we have the following. So we are going to finish. OK, so now, we have for all set in the stable manifold of y with y in the intersection of the local unstable manifold of p intersection g, this phi plus set is constant. So let's call this set square prime. For all set in square prime of p, phi plus of set equals phi plus of p. But now, remember, Fubini holds. So if we evaluate the measure of this set, we are taking all the points in the local stable, almost every point on the local stable manifold, and we are integrating the measure of the stable manifold along it. So we are going to have that a measure of S prime of p equals the measure of S of p. But what is the same for almost every point in the square, the value of phi plus is constant. So now, we can cover the torus with the squares. And since they can overlap, and so since they are constant in each of these squares, they will be constant all over the torus. And so with this, we have finished the proof that it is robotic. And that's it. Thank you.