 We're beginning with expanding maps on the circle. Let's begin with something. I thought that this was something of us mathematicians. We care too much about classification. But in fact, this is a thing of scientists. We all care about classification. We classify insects. We classify even mental illness. Is that, can you read this? Yes? OK, so in general, we classify things. And we aim at classifying things. And in order to classify things, we need a criterion. So in math, you will know. In algebra, for instance, you have isomorphic. If you have a differentiable structure, you have diffeomorphisms. So the criterion focuses on what is important for us. Well, in dynamics, we have a criterion. We focus on the dynamics. So we will say that two objects belong to the same class if they are what is called conjugate. I think that you all know what a conjugate is. Does anybody here not know what a conjugate is? Let me just remind you what it is. So we have two dynamical systems, f and g. And we will say that f and g are conjugate if there exists a homeomorphism such that it makes this diagram commute. So h composed with g equals f composed with h. So this produces classes of dynamics. So we aim at classifying expanding maps on the circle. This is what we are going to do. We are going to classify expanding maps on the circle. But sometimes we don't have all the information. We cannot tell whether two dynamical systems are conjugate, but we can say that some dynamics is contained into another dynamics. In that case, we're going to speak about this is conjugacy. Sometimes we will have this, the same picture. But instead of h being a homeomorphism, it will be only a continuous surjective map. In that case, we are going to say that g is semi-contrugate. Also over here should have set y to x. That makes the diagram commute. It is never clear to me what does this mean. Who is the one who has more information? The one who is up has more information. So we are going to say that f is a factor of g. I presented both definitions. I will tell you why. Here, as Corina has said, important objects of study are the orbits. So if there is a conjugacy, the conjugacy will take orbits into orbits, periodic point into periodic point. Orbits into dense orbits. And the other way around. But in this case, you will take periodic orbits into periodic orbits, dense orbits into dense orbits, but not necessarily the other way around. You could have a dense orbit that does not lift into a dense orbit. Can you think of an example? I will put lots of exercises. You just pick a couple of them. But I will put lots of exercises. So one exercise. You can do it with very simple dynamics. Not necessarily the circle. We can think of even simpler dynamics. So think of periodic point p for f such that it's not periodic. So it's a pre-image is not a single periodic point. And does not contain a periodic point. Can you think of a periodic point p for f such that pre-image does not contain also? Can you think of a dense orbit for f such that its pre-image does not contain a dense orbit for g? You can do lots of questions. Even at this point, many of you are familiar to these notions. And there's some question I don't know the answer. Perhaps it's known. I don't know. You can think of the semi-contrugacy as an order relation. Because you have that, well, it is transitive. If you have this is a factor of this, this is a factor of this, then this is a factor of this. But and of course, every map is a factor of itself via the identity. But what about the symmetry? Is it true that if f is a factor of g and g is a factor of f, is it true that f and g are conjugate? Are conjugate? It's a question. Anyways, we have this conjugacy and semi-contrugacy are will be useful to classify maps. And so let's begin with Corina has begun with the description of maps on the circle and on the interval. So forgive me if you are already very familiar with this, but I want to introduce the lift of a map in the circle. You remember that the circle, as Corina has said, you can see the circle as this quotient between the rears and the integers. And you have this projection. And this projection can be what I put it like this. But you can look it as this, or x mod 1. This can be seen as a semi-contrugacy between a map that commutes with this and the map in the circle. For instance, Corina has presented three maps. One was let's put f1. The notation will be the same. But you can define this in the circle or in the rears. And let's call it f tilde when it's in the rears and f when it's down. And then you will have that this commutes. This is a factor of this. It's a non-interesting factor. But let's call it like this. And in the same way, she has defined the expanding map, which was 2x mod 1. And then, sorry, 2x mod 1. And then you can lift it to the rears like this. And then you will have this 2. And the same with the rotation. The issue is that you might be already familiar with this result, but let me state it anyways. When you have a continuous map on the circle, you can always lift it to the rears as a continuous map and in a unique way. So if you have a continuous map on the circle, then there's a unique map in the rears, which is continuous and unique modular integer translations. I will tell you what it is in a minute. But such that they are semi-contrugated. Such that f is a factor of this continuous map. Small f is a factor of big f. And f is unique up to integer translation. I will tell you in a minute what this is. And this is called a lift of f. What does it mean that f is unique up to an integer translation? Let's keep the notation here. So that means that if we have two lifts of the same map, then their difference will be an integer number. Will be a constant integer number. This allows us to define the degree of a map. Roughly speaking, the degree of a map is the number of branches. And the map has, when you draw it, as a map of the interval. But we can define it. If we have a lift, the degree of the map will be that integer. I don't know how to, here. This is an integer. Can you tell me why this is an integer? You know why? x plus 1 minus f of x belongs to the integers. Do you know why? Yes, both of them are lifts. Because if you have this as a lift, then immediately, this will be a lift, OK? You can check it yourselves. f of x plus 1 will be also a lift. Will be a continuous map. It will commute with this. We'll make this diagram commute. And so since the lifts are unique up to an integer translation, this will also belong to the integers, OK? And this is called the degree of f. And roughly speaking, it's the number of branches that small f has when you draw it in the interval. So for instance, what is the degree take? And this is independent of the lift. If you take another lift, it will have the same number. And if f is a homeomorphism, then the degree will be 1 in modulus. So what is the degree of this map? Can you tell me? You can see it as a map of s1. You can define this as s1, as a map of s1. 0, isn't it? Because it's 0 in 0 and 0 in 1. So the degree of this is 0. And what's the degree of this one? Can you see why? Well, you have a lift of fx2x mod 1. Well, a natural lift is just 2x, OK? This is a natural lift. And so if you calculate this, this will give you 2. And what's the degree of this map? Well, you have it. If it's a homeomorphism, it's in modulus, it will be 1. But this will be 1 because you take out this mod 1 and you will get a lift, OK? So the number of, and as Corina has drawn, the degree is the number of branches, OK? This has two branches, has degree 2. This looks like if it had two branches, but it has only one because the branches are horizontal in some sense, OK? So this has only one branch. And this has zero branches. This is hard to do. It makes more sense when it is expanding. So the number of branches is the degree, OK? So we will classify expanding maps according to their degree. We will see that all expanding maps of the same degree are all conjugate. This is what we are going to see. This I'm not going to, if you're interested, it will be there. I will not enter into that. So let's go into linear expanding maps. These are what are the rigid models. The linear expanding maps are the rigid models. So first of all, what's an expanding map? What's an expanding map? We are going to say that this is an expanding map. If it is continuous, plus C1, and the derivative is greater than 1 for all x. And since s is compact, we will have, in fact, that there will exist some constant doing this, OK? You will have a constant and uniform bound, because it is C1. OK, so I will go first into two examples of linear expanding maps. You are very familiar with the two of them, but one was suggested to me by Stefano and I find it a great idea. So the first linear expanding map we're going to see, we have seen it already, is this one. 2x mod 1, the one with two branches. But probably it is a good idea what Stefano suggested, because you're more familiar with that. It's this one, each 10 of x, which is 10x mod 1, which is a degree of this, OK? It has 10 branches. Well, it will be hard for me to draw them. No, that's OK. That's OK. Sorry. 1, 2, 3, 4, 5. OK, so here I will not draw them all, but you will have branches like this. There will be 10 branches like this. It will expand by 10. OK, so let me show you an example of how this works. This is the dynamics of a point. Have you ever seen this? Have you ever seen the dynamics of a point with the green line y equal x, like this? Yes, OK. So you have this, the line y equal x, and the two branches here. And then you will have the dynamics of a point. We'll go here, here, here, and so on. OK, so in order to classify these dynamics, we will do what is called symbolic dynamics. In the case of 2, and I will do first in the case of 2x mod 1, but then I will do it in the case of 10x mod 1, because you will find that you are very much more familiar with this than with this. How do we do this? Oh, sorry. I will do this later, but let me first tell you how we are going to do symbolic dynamics. If the point is in the first half, we will assign them 0. If the point is in the second half, we will assign them 1. Then you iterate once. And if it is still in the 0 part, we will assign them 0. And if it has changed, we will put 0, 1. And then the trajectory of the point will give us a sequence of 0's and 1's. But this is not hard. However, you will be more familiar with this in the 10 expression. Let's get symbols for this sequence. So this symbol, this sequence, so in the first, we have to put them inside one of these 10 boxes. So it will be not here, not here, here between 3 and 4. So the first symbol will be 3. This is a symbol 0, symbol 1, symbol 2, symbol 3, 4, 5, 6, 7, 8, 9. Then we apply the map, then we will have pi. And then we do it mode 1. So the second symbol will be which one? 1. The second symbol will be 1. We multiply by 10. We do it mode 1. And the third symbol, 4. And then, in fact, the symbols will be just its expansion. So it will be very simple for us. The symbolic sequence of this orbit will be just its decimal expansion. And in the case of this, it will be its binary expansion. So we will go get back into this later. Let me first just see a relationship between the periodic points and the degree. So we know what periodic points are. And one thing we want to know is the number of periodic points some linear expanding map has. And in fact, we are going to calculate the number of periodic points each expanding map has, not necessarily linear, whatever expanding map has. This will come as an easy consequence once we have classified all expanding maps. But we can do it in advance. We can calculate the number of periodic points each expanding map has only knowing their degree. So let's call this the number of fixed points of fn pn of f. We are going to calculate this pn of f in terms of the degree. The linear map for linear expanding map 2x mode 1, the number of fixed points of e2 to the n is 2 to the n minus 1. The minus 1 comes because we are identifying this and this, the 0 and the 1. So we lose 1. As a consequence, it's not as a consequence. They are what Corina told equidistributed. So since they are also equidistributed, they are going to be dense. So I leave this as an exercise. This calculating exactly for the linear expanding map is not hard. You can calculate it. This can be a hint, but if you find any other way to calculate them, I will be happy with it. So let's do it with a general map. Well, first with other linear expanding maps. But then we are going to do with any expanding map. So what happens when we have like this 10x mode 1? What happens in this case? Can you tell the number of periodic points in terms of the degree n to the n minus 1? OK, let's see when this will absolute value. In fact, no. I'm sorry. It depends on the absolute value of m on the sign of m. And the periodic points of em are dense. In fact, this will happen with any expanding map, not only for the linear expanding maps, but this will be a consequence of our classification. So we are going to show that these are dense. And we are going to show that this is the number. And in fact, it's not going to be hard. But the main part of the proof, you are going to make it, because it's an exercise. Let me tell you how we are going to do it. Well, I have already told you what an expanding map is. It's there, and we already know what the degree is. This is the main part of the proof. If we have two maps, f and g, on the circle, the degree of the composition is the product of the degrees. The degree of the composition is the product of the degrees. And this is going to be an exercise. In particular, the degree of the g composed with f is equal to the degree of f composed with g. And also, we have this for the iterations. Degree of f to the n is the degree of f to the n. And this, which is the key part of the proof of this formula, is going to be an exercise. OK, so let's prove why we have this. I don't know if I will have enough time, so I will give you some hints. So if it is an expanding map, the degree is greater than 1. Can you tell me why this is? Not necessarily linear. For any expanding map, the degree is greater than 1. One hint is that if f is a lift for an expanding map, we also have that f prime of x is greater than 1, because the derivative is something local. And the lift is locally the same as the small f. Pardon? Yes. No, the degree is general. The degree is global. Yes, we can calculate it at any point. It will be constant. So the degree will be x at any point. Yes? No, it's strictly greater than 1. Yes? Yes. And it will be continuous. So it will be all the time about 1 or all the time below minus 1. And there you go. Then you will have that this. Either this is greater than 1 or less than minus 1. So the degree will be in absolute value greater than 1. And so why is this? Now remember what we have said, that the degree of fn, we have this. We have just said this. So let me. This we have already done it. But let's see that we only need to prove this. Why is that? This is the number of fixed points. So this is, in fact, p1 of fn. If we prove this, we will have that p1 of fn will be, if this is true, we will have this. This implies this. But this is equal to this. OK? So we only need to prove this for p1. We only need to prove this for p1. So let me see if in this 10 minutes I can do this. So is it clear why is this? If we prove this for any f, in particular, we will have this. But this is equal to this. So just to prove it for fixed points is enough. So we have to count the number of fixed points in terms of the degree. We begin by taking a lift of f, call it big f. Let's consider g. We need to count in some sense the zeros of fx minus x. Or what is the same, the number of times this cuts z. OK? Between 0 and 1. And so we have to count the number of zeros or integers of gz, gx. OK? And but this equals the degree of f minus 1. You know why? This is f of x plus 1 minus x plus 1 minus fx minus x. OK? So this is the degree of f. This goes with this. And this is minus 1. OK? So this implies that there exists at least amount of points such that g belongs to z. Can you see this? What happens if the first one and the last one are the same? You still count it. You will not have that because the degree is greater than 1. But you will not have it because the degree is greater than 1. So you will have the map is expanding. So yes, all this. But this you are counting twice. That's why you subtract 1. You are counting this twice. You have to subtract 1 because you are counting the end points twice. When you have this, well, you will have this gx plus 1 and gx. And you count all the times this cuts. How many integers are there in between? And then you will have degree of f minus 1. Because this is this. It's because of continuity of g. It's at least, at least, in fact. Because you will have, by the Bolsonaro, you will have at least degree of f minus 1 points that belong to z. At least. But on the other hand, we have that gx is greater than 0 in absolute value. OK? Because gx is fx minus x. And this f prime is greater than 1 in absolute value. So this will be different from 0. So it will be either increasing or decreasing. So we have that this cuts at least degree of f minus 1 points in absolute value. But if it is because of Bolsonaro, but if it is strictly increasing or decreasing, it will cut each of them exactly once. And so the number will be this, the number of fixed points of f, which is the same as the number of zeros of this, will be exactly degree of f minus 1. The number of times these cuts and integer coordinate will be exactly the number of zeros of g, which is the number of fixed points of f. And this is exactly proofs this. So we have, we counted, this is not for linear, it's not for linear expanding map. This is for any expanding map. The only thing that we have used is the derivative of f. We have used that the derivative of f is in absolute value greater than 1. So all expanding maps have degree f minus 1 fixed points. And the number of periodic points are degree of f to the n minus 1 in absolute value. So we could count the number of periodic points just by using the definition of degree. We will later see that, in fact, we have another path of calculating this because it is very simple to calculate the periodic points of a linear expanding map. And then we will see that all of them are conjugates. So they will have the same number of periodic points. But this is a direct proof of this. OK, so to finish, I have three minutes. But I will define, you already know this. So let me just, it's just a remark. We will say that f is topologically mixing. If for any two open sets, we have this, we have this. And in particular, there's another exercise. If f is a homeomorphism on the circle, then f is not topologically mixing, is it? The definition is up here. It's for the future, OK? So from one end on, you already know this definition, don't you? And you have, OK, rotations are not topologically mixing. Isometries are not topologically mixing in the circle. But let's put it some more difficulty. Homeomorphisms on the circle are not topologically mixing. You can get a flavor with the rotations. And so expanding maps on the circle are topologically mixing. This is not hard. I have it in the following slide, but I will end it here because I want you to do it as an exercise. Expanding map on the circle, are topologically mixing. OK, and then I finish here. Thank you.