 All right, so today we enter the realm of nonlinear dynamics. There are three kind of well-studied classes of examples. This is not meant to be an exhaustive list of nonlinear dynamical systems, but behaviors that are seen in these three classes appear in higher dimensions and in much more general settings. Kadeem will say a little bit more about that in his lecture, the third lecture today. So the first class of examples are diffeomorphisms of the circle. Now, I will say nothing about this, but there's a rich theory of classifying, for classifying diffeomorphisms of circle. It goes back to Poincaré, someone named D'Angois. The field's metalwork of Jean-Christophe Yokoz was essentially on diffeomorphisms of the circle. It's closely related to starting with Poincaré. It's closely related to a topic in smooth dynamics that is not mentioned at all this week, but is very important. That's the theory of Komagorov, Arnold and Moser, or KAM theory. This belongs to the realm of what we sometimes call elliptic dynamics, or parabolic dynamics, elliptic dynamics. The second kind of class of examples, which I will discuss today, and these are well understood, are expanding maps, and in particular, expanding maps of the circle. You can look at expanding maps of other manifolds. It's not a huge class of systems, because most manifolds do not admit self-covering maps. So these are maps that are not diffeomorphism, but whose derivative is sufficiently large everywhere. So for example, well, we've already seen f of x equals 2x, which is if we want to write this multiplicatively on the circle, z equals 1 in the complex plane, this is the map f of z equals z squared. Well, this is a linear example, but we can perturb it quite easily to get many nonlinear examples. Here's an example that I will actually discuss in your homework exercises. So if I consider the map, g sub a of z equals z times z minus a over 1 minus a bar z, where a is a complex parameter of modulus less than 1, and g sub a, you have to prove this, this is in the homework, is a map on the circle to itself. So it sends the circle to itself, and this is an example of an expanding map. Note, when a is very small, g sub a, the modulus of a is very small, g sub a is very close to the doubling map. If a equals 0, this is the doubling map. So this is what we would call a perturbation of the doubling map f, when a is very small. So by this, I mean that in this example, the values of g are very close to the values of f. The derivatives are very close to each other. In fact, in this particular example, all of the derivatives are extremely close to each other if a is small. And third, the third class of examples are the Anosov Diffie Amorphisms, another richly studied class of nonlinear dynamical systems. Like the expanding maps, this turns out to be a very special class. It only conjecturally, these types of Diffie Amorphisms only live on certain manifolds, like tori. That's fine. The methods that we will use to analyze some Anosov Diffie Amorphisms, again, are quite general and can be used for systems that aren't Anosov, but share some of the properties. Now, I haven't defined Anosov, and I won't today. I might tomorrow. But I've given you an example. So the map f from t2 to itself, given by f of x, y equals 2x plus y plus epsilon sine 2 pi x plus y comma x plus y. So for epsilon small, this is a real number. For epsilon small enough, this is an example of an Anosov Diffie Amorphism. And Kadim will give you an idea of what are the key properties of this example for epsilon small. And he will start talking about the dynamics of this example. I will not talk about the dynamics of this example today. But note that this is a perturbation of the cap map. Because when epsilon is 0, this is the cap map. So today I'm going to focus exclusively on the dynamics of expanding maps on the circle. Hannah talked about these maps quite a bit last week. I'm going to summarize just a few properties of these maps. I could go on and on and on and on and on about these maps. But I'm going to focus. So I'm going to focus on the property of ergodicity. OK, so this, as I said, these maps, g sub a, are examples. Well, state of theorem, suppose that a map, or I should call this f sub epsilon, suppose that g is a map from r mod z to itself with the following properties. The first is that g is c2. So I need twice continuously differentiable, because I need that the derivative of g is once continuously differentiable. And if it's not twice continuously differentiable, if it's only once continuously differentiable and nothing more than that, this theorem does not hold. There exist counter examples. So this is a very important hypothesis. And it's going to play a role in the distortion estimates that I referred to yesterday. Remember, we were looking at ratios of derivatives of high powers of a map and trying to bound those? This will play an important role. Next, I want what's called the expanding property. So this is a regularity property. We call this regularity. The more times you can differentiate, the more regular a map is. The next is that the infimum overall x in the circle of g prime of x is greater than 1. Now, since this circle is compact, this infimum is realized. So in fact, this infimum is greater than or equal to some number bigger than 1. And this is the expanding property. Those are the two properties. That's all I need. Regularity and expanding. Then for any subset, any Borel set in R mod z, x is invariant under g. The Lebesgue measure of x is 0 or 1. Notice I haven't given a hypothesis that g preserves Lebesgue measure. This statement's true, even if g does not preserve Lebesgue measure. But if g does preserve Lebesgue measure, so in particular, g preserves Lebesgue, such as the example f of x equals 2x. Those examples, g sub a, for a less than 1, this is an exercise. So if g does preserve mu, then g is ergodic with respect to mu. Meaning that, for example, Birkhoff's ergodic theorem applies. So this is the main theorem I'm going to discuss today, a remark that I won't write down. But this line is kind of actually not a very, this can be generalized. So if g preserves any measure mu that's equivalent to Lebesgue that has the same 0 sets as Lebesgue, then it will be, this condition implies it'll be ergodic with respect to that equivalent measure. And there's a fact. Won't have time to prove this. But any expanding map preserves a probability measure that is equivalent to Lebesgue. So in fact, any expanding map has an invariant probability measure that has the same 0 sets as Lebesgue and which is preserved and which is ergodic. That's a conclusion. But I'm just going to focus. I'm not going to stick with Lebesgue measure. Other questions, I should say. Mu equals Lebesgue. Excuse me, I heard. So some people define ergodicity for systems that don't preserve a measure but preserve what's called a measure class. So any map of this form is going to send Lebesgue measure, push forward to a measure that has the same 0 sets as Lebesgue. And you could define ergodicity for such systems in this way. But our definition of ergodicity with respect to a measure assumes that the measure is invariant. And since I haven't assumed invariance of the measure, this is a statement more general. Why do we need an invariant measure? Because suppose this property holds but you don't preserve a measure, Lebesgue measure. Then you have no Birkoff ergodic theorem for Lebesgue measure, but you still have this property. Other questions, that property can be useful in other ways even if you don't have the Birkoff theorem for Lebesgue, that's still a useful property. So with this, I'm just going to do this for perturbations of the doubling map. So for example, EG for G sub A, A close to 0. Now, in the exercises, I kind of outline, depending on your background, you'll be able to prove this for all maps. But let's just focus on this example. So perturbation just means the derivative is pretty close to 2 everywhere. And the values of the maps are very close to, like G of x is very close to 2x mod 1. That's all I mean. So if G is close to the doubling map, then the following properties hold, and we're going to use this. So there exists some constant lambda greater than or equal to 1. Well, such that d prime of x is greater than or equal to lambda. So that actually just follows from the definition of an expanding map. But you see that if you perturb the doubling map, it will be an expanding map. The second, and this is not so hard to see if you're close to the doubling map, the map has degree 2. So the number of pre-images of any point is exactly 2. And in fact, any expanding map is going to have a degree, some integer m, at least 2, so that the number of pre-images is m. Some of this should be reviewed from Hanna's lectures last week. There exists a unique point, x0, in the circle, such that G of x0 equals x0. So these are the properties we're going to use. We're also going to use that at C2, but I haven't stated that quite yet. So first thing we do is to build a partition very much like the partition in the sequence of partitions, very much like the partition into dyadic intervals that we used in analyzing the doubling map. Except we're going to use the dynamics to construct this sequence of partitions. This is a very common theme in smooth dynamics. You want to look very close to a point. You use the dynamics to construct a small ball. So I'll show you what I mean. So construct partitions. So nested sequence, like in the homework problem, exercise 2 from lecture 2, nested with the following properties. The number of elements in the mth partition is exactly 2 to the m, number of atoms. The elements of PM are half open intervals. And if I take an i in the mth partition element, then the length of i is less than or equal to 1 over lambda to the m. So we have some exponentially small upper bound on the length of these intervals. For the doubling map, that would be 2. That would be precisely, or 1 half. The length would be precisely 1 half to the m. But here we just have some upper bound. It's not a half, necessarily. If I take i in PM, then f to the m of i equals r mod z. So this is the sense in which this is a dynamical partition because the elements exactly injectively map onto r mod z. And in fact, it's a diffeomorphism that is a C2 map with a C2 inverse on the interior. So those are the three properties. Let me explain how this construction works with a picture. And let me then explain especially why this property holds. The other one should be obvious. So the construction goes as follows. Let's start with constructing p0. So this construction, I believe, I'm going to describe, because this is the construction that allows you to code orbits of g using the two shift, using 0's and 1's. So these little intervals kind of play that role. I think she might have called them delta. So here's the construction. Here's, well, let's do p1, p0's trivial. Just going to be the whole. So I have, let's just put it here. This is the fixed point given by the properties I stated before. This point is fixed. So x0 is a preimage of x0. Well, x0 has two preimages. So there's another point, x0 prime, that gets mapped under g to x0. I'm going to use these two. To construct two intervals. And you know what? Just to be safe, I'm going to put m minus 1. It's not going to make a difference. Or else I could call this p0. Oh, I'm so clever. No, I'm not going to. This is p1. OK, so note, so I see here's my two elements. Well, what happens to this interval if I apply g? This point's fixed, and this goes over to x0. So I get the circle. Similarly, this point's fixed. This is a little trickier to see. But this is going to also go over to x0. So this is a, this goes to this. Wait, well it has to. OK, so these are the two sort of branches of g inverse of these two intervals. Now how will I construct p2? Well, my sequence of partitions will be nested. This is x0, and this is x0 prime. I already have sort of p1 constructed. Now to construct p2, well, this is x0 prime. I already have found all the pre-images at this point, but this point has two pre-images. Let's call them x1 and x1 prime. And I can divide the circle into four pieces now, and I get a partition. I don't know why I did it this way. Here we go. So I can divide the circle into now four pieces, and you can check. Well, what do you need to check? You just need to check, sorry. So now we have four intervals. The way I've done this, if i is in p2, then f of i is in p1, or g of i, sorry. So these are the two pre-images under g of x0 prime, and so on. So you continue. You take the pre-images at this point, and the pre-images at this point. That'll further divide your intervals into two. So continuing inductively, continuing inductively, we get that for i, we construct these intervals using pre-images. For i in pm, g of i is in pm plus 1. And the number of elements of pm is 2 to the n. So the n points, before I say this, the n points of elements of pm, if I said pm minus 1, sorry, it's very important. Previous to the higher level. The n points of the elements of pm are the pre-images of n points of elements of pm minus 1, pn points, yes, pm minus 1. So inductively, the number of elements of pm is going to be twice the number of elements of pm minus 1. And so the number of elements in the number of elements in pm is therefore 2 to the n, since the number of elements in p1 is 2. Finally, what about this estimate 2? So this estimate 2 can also be shown inductively. So for i an element of p1, clearly the length of i is less than or equal to 1. It's just length, but this denotes. And now, note, if I take i in pm, g restricted to i is edifio on the interior, and it's on to i prime in some pm. Now, the length of i by change of variables formula, that's equal to the integral, because this is edifio morphism on the interior, this is the same as the integral over g of i, right? That's i prime, the derivative of g inverse. But g inverse is greater than or equal to 1 over lambda, because g inverse prime is greater than or equal to 1 over lambda, less than or equal to 1 over lambda, because g prime is greater than or equal to lambda. So this is less than or equal to the integral over g of i, 1 over lambda. And this is equal to 1 over lambda times g of i, the length of g of i, which is by induction less than or equal to 1 over lambda times 1 over lambda to the m minus 2, which is less than or equal to 1 over lambda to the m minus 1. Yes, you have to check that. But basically because you chose the end points to be the unique pre-images, the map has degree 2. So you want to use that. Every point has two pre-images. I have to ask, well, where could those two pre-images lie? Could they lie in the same interval? So that's an exercise. If you want, you can think of drawing the graph of g. And you might find this easier in the iterates of g. You might find it easier to picture all this by looking at the graph. So the graph of g is going to kind of go like this, but it's not going to be linear. It's going to have two branches. All right. So we have these intervals. And now let's prove ergodicity. So that was kind of step one, was constructing these partitions. Step two. Now we're going to use the sort of density point argument. The exercise that I gave you turns out to be much more convenient to use in this setting than the actual Lebesgue density theorem. So we're going to use that exercise. So suppose we have a set that's invariant under g. And suppose that this set has positive measure. So this exercise, as I just mentioned, lecture two, says that there exists a nest. So we have this nested sequence of partitions into intervals. The lengths of the intervals in these partitions go down to 0. In fact, they go down to 0 exponentially fast. And so the exercise shows that there exists an m. So given epsilon, there exists an m. We have no control over what m is. It could be huge. And there exists an interval i in PM, such that the density of x in i is greater than 1 minus epsilon. The Lebesgue density theorem, we would have almost every point. And then you'd have to take symmetric intervals. And those might not really coincide with these intervals. But that exercise is beautifully adapted to this dynamical sequence of partitions. Now let's let x prime, as usual, let's let x prime be the complement of x. Well, first of all, x prime is also invariant, clearly. Forward and backward invariant. Well, the density of x prime in this interval is less than epsilon, because it's a complement. So now if we apply g to the m, as we did last time for the doubling map, we can blow the x prime using an invariance of x prime. We can blow it up to x prime sitting inside the entire circle. And using that gn versus of x prime the invariance of the complement, we get that the density, well, we get that the measure of x prime, which is of course the same as the density of x prime in the entire circle, is the density of g to the m of x prime, the side of g to the m of i. And now let's use change of variables again. As I illustrated or I discussed a little bit in the last lecture. So using the change of variables, well, I'm going to write out more than what I wrote here on my page. So this is the integral over x prime intersect i of g to the m prime divided by the integral over i of g to the m prime. And that's because g to the m restricted to the interior of i is a diffeomorphism. So I can use this formula for the measure of the intersection over the measure of i. And these are d mu. And now this is less than or equal to the soup over all x in this intersection of g to the m prime of x divided by the inf over all x g to the m prime of x times the conditional measure of x prime in i. And so this is less than or equal to, well, I can take the soup over all x in here. But let's just really, since we know nothing about x prime, we might as well consider this supremum. And then this is less than epsilon. Now we come to the key theorem that we need to finish this proof. So I'll state the theorem, finish the proof, and then prove the theorem. So I'll call this the distortion theorem. And it says that there exists some c greater than or equal to 1, such that for any m. So this m is arbitrary, and that's key, the order of quantifiers. And for every interval i inside of pm, this ratio, the soup over x and i of g to the m prime of x divided by the nth, is less than or equal to c. So remember last time you could do a naive sort of estimate one could do a naive estimate on this ratio. The naive estimate would say, OK, well, we know that the infimum of g to the m prime is greater than the nth power of the infimum of g prime by the chain rule. So this we could get is greater than or equal to lambda to the m. So we could get, we're not going to use this, but I'm just so we can get less than or equal to lambda to the m times epsilon. The supremum, on the other hand, I have no hypotheses on the supremum. So it's just some other number. It might be very far from lambda. It's some number bigger than lambda. So some eta to the m, where eta, the supremum, is definitely greater than lambda. And if eta is bigger than lambda, which is what you expect, you don't expect the derivative to be constant, this is going to go to infinity as m goes to infinity. So it's useless. And this is what we might call a c1 estimate, and it's useless. But this is an estimate using that g is c2 in an absolutely crucial way. And once we have this, we're done. Because then, using this theorem and going back to our proof, we now get tracing backwards that the measure of x prime is less than or equal to c times epsilon. But epsilon was arbitrary, so that implies that the measure, did someone fix this? The measure of x prime is 0. And therefore, we assumed x was invariant, had positive measure, and therefore the measure of x, which is the complement is 1. And therefore, we're done. We've proved ergodicity. So now, let's talk about the distortion theorem. And we prove it. So we're given g. Let's define a function c from r mod z into the positive real numbers. Well, first it would be a priori into the real numbers. By c of x is the log of the derivative of x. Notice this makes sense, because the derivative of x is bigger than 0. Well, in fact, it's bigger than 1. So c satisfies the following properties. So first of all, c is greater than 0. But in fact, it's actually greater than or equal to the log of lambda, which is greater than, strictly greater than 0. Secondly, and this is key, c is c1. It's continuously differentiable. That's because g is c2, which is twice continuously differentiable. That means that g prime is once continuously differentiable. But log is as smooth as you want. So if I compose, and I'm bounded, this is bounded away from 0, so this function is c1. And the derivative is bounded away from 0. But the key point is that c1, in particular, it's elliptions. So this is, if you haven't seen this before, it's just the mean value theorem. So the mean value theorem in calculus implies that there's some constant m greater than 0, such that, if I take any two points, x, y in the circle, c of x, the values are within this constant related to the derivative of c times the distance from x to y. And third property, and this is the chain rule, is if I take any m, so you might have to think about this a little, then the log of g prime, g to the m prime of x, which is the key thing we're trying to estimate, is equal to the log of g prime of this product. And log is additive. And so this equals the sum of the logs and the logs of g prime are c. So we get this estimate. So it's the sum of the values of c from 0 to m minus 1. That's just chain rule plus additivity of log. And I guess I shouldn't have erased what I want to prove. So let's use g. Sorry. OK, so now, if we're given i in PM and we're given two points, x, y inside of i, we want to show that there exists a c such that g to the m prime of x, divided by g prime of x, divided by g to the m prime of y, is less than or equal to c. That would show that the soup over the inf is less than or equal to c. So you take a point that, if you want, you take a point that achieves the soup. You take a point that achieves the inf. And you take the ratio. So let's take a log of this expression. So we want to show that the log of g to the m prime of x minus the log. So now I'm using the additivity of the log again for ratios. This should be prime is less than or equal to. And let's just be safe. Have to. But anyway, so we would like to show, I guess we have c less than or equal to 1. So we want to show that this is less than or equal to some constant. Let's just call it k. Now we use property 3. So what we're going to try to estimate is the sum from j equals 0 to m minus 1 of c of x minus the sum j equals 0 to m minus 1 of c of y. So I'll just put this inside the sum. Equals. But this is less than or equal to by the Lipschitz. Well, this is less than or equal to the sum. This is not x and y. A little flustered here. So this would be g to the j of x. Sorry, minus c of g to the j. Well, I'm just going to save time. I'm just going to take these absolute values inside. And now I'm going to use the Lipschitz property of c. So this is less than or equal to m times the distance between g to the i or g to the j of x and g to the j of y. These points lie in p sub m minus j, because it's a little interval. And we have an upper bound. I'll go a minute over. We have an upper bound on the lengths of those intervals. So this is less than or equal to, again, I'll pull out the m. And that's less than or equal to the sum of the length of that is bounded by 1 over lambda to the m minus j minus 1 by the bound on the lengths of intervals. Well, this is from j equals 0 to m. Of course, I could just re-index this. This is the same as the sum from k equals 1 when j equals m. I'm going to get 1 over lambda to the minus 1. OK, put a lambda here. k equals 1, certainly less than or equal to this. I don't need to put a lambda here. It's fine. Less than or equal to this. And this is equal to m over 1 minus 1 over lambda. And because this is less than 1, so 1 over lambda is less than 1, this is a perfectly well-defined constant. And this is our k. So to summarize this estimate, which looks like maybe it was very special, you see this again and again and again and again and again in smooth dynamics. Whenever you have some Lipschitz quantity that you need to bound for arbitrary, whose variation you need to bound for arbitrarily many iterates, and you use hyperbolicity to make this bound. So I've gone over time.