 We have S1, which can be seen as the quotient between R and Z, or can be seen as X mod 1. We have this projection from R into S1, which can be written as this. And we have said that for any continuous map, there exists a continuous map such that it makes this diagram commute. I want to do the details of this. I will give you some hints. The first hint is that this projection is a local homeomorphism. So if let's, well, this is okay, this is okay. If the distance between X and Y are less than 1 half, then pi is a homeomorphism onto its image. Okay? So there's a unique point, a unique pre-image, okay? And well, the idea is we will take any point in, we can remember we want that pi of F equals F composed with pi. And we take any of these, we will have a unique point, which is of this form. And so we can define, there are many choices for this, and there are many choices for this. Okay? But we are going to choose one, and we are defining this. These are hints. I want you to do it, okay? Okay, then another hint, this is one hint. The second hint is that F, small F and big F2 are not, this is continuous. And this is compact. So this is uniformly continuous, okay? So there exists some delta such that if two points are delta apart, then these are one half, their images are one half apart. And so remember that we said that this is a local homeomorphism. So if these are one half apart, we can choose a unique pre-image, okay? So once we have chosen a base point, then we can extend small F using the fact that pi has a unique pre-image over certain intervals. Well, these are hints, you have to do it. So prove that there exists at least one, so prove that there exists at least one F, which is continuous. Now why, I have said also that if we have two lifts, their difference is a constant integer. And why is that? So let's suppose that we have two lifts, these are two lifts of small F. Why do they differ in only one integer? Well, that's it. Great. We have this equals F composed with pi, but this is also, so by pi they are sent to the same point, which is this. So this implies that this belongs to Z, but this is continuous, so it has to be constant, okay? Perfect. So this allows us to define the degree. The degree is F of X plus one minus F of X. And we claim that this is an integer. And one of you has said that this is because both of them are lifts of F. But why is this a lift of F? It's very easy to check. We have to check this, okay? But F is a lift. So F is a lift. This implies that this holds for all X, but in particular it holds for X plus one. But now pi of X plus one is pi of X. So this is also a lift of F. And so since both lifts differ in an integer, this is an integer. Now the third thing we have seen about degree was that if F is a homeomorphism, this homeomorphism means that it is continuous and its inverse exists and is continuous, okay? If F is a homeomorphism, then the degree of F has absolute value one. And so in order to prove that, we have to prove that if the degree is zero or a modulus greater than one, then it cannot be a homeomorphism. But now we have two cases. If the degree is zero, what happens? It is not. It's not one to one. Why it's not? We have F X plus one equals F X, okay? So big F is not monotone, then small F is not monotone. Now if the degree is greater than one in modulus, what does this mean? What does this imply? This wraps around the circle twice. The image wraps around the circle twice. So what does this imply? Yes. But we have to prove that. We have a simpler. We have that this is greater than one. So we, as you said, but this is a consequence of Falsan, if you like. There is a point between X and X plus one that has the same value as F X, okay? So that this implies that there exists Y. Can you see why? We have F X here. We have F X plus one here. And the distance here is greater than one. So we have all values here, okay? So in particular, we have F X plus one is here. Let's assume that F X plus one is greater than F X, okay? So there will be one point that corresponds here to this. There will be one Y, there will be one Y such that F of Y is, is this, is it clear? There's, when you have two options, either F X plus one is greater than F X or F X is greater than F X plus one. But in any case, you will have, in the other case, you will have F X minus one. Yes? Pardon? Yes, sorry. Yes. Here. Yes. You, we will have, we will have, yes, pi X. Thank you. So we will have some Y such that F of Y equals pi F of X plus one, okay? And so this implies that this implies that F is not monotone. Big F can be monotone or not, but small F will be not monotone. In fact, big F can be monotone. And one, one example is this, okay? Small F, small F is not monotone, but big F is. The degree is two. Okay. So this is all about lift and degree, but I have some more comments. I am especially interested because the title of my course is somewhat misleading. I would have called it expanding maps on the circle, not piecewise expanding maps because of the following example. Remember that last class we had proven that if F is expanding on the circle, this implies that the degree of F in absolute value is greater than one. But what, what happens with this following example? I assume that all of you are familiar with the tenth map, but if not, I will throw the tenth map here. Check the tenth map on the circle. It's called the tenth map because it looks like a tenth. So here the derivative is two and here the derivative is minus two. Okay. There is one. This is a tenth map. Okay. And this map piecewise expanding. So it's expanding except for two points in the circle. And what's the degree of this map? The degree of this map is zero. But this is because this, this map is not expanding. It is piecewise expanding. And this result is for expanding maps. Expanding requires that the derivative is continuous. It, it exists at every point and it is continuous. Okay. If it fails in just two points, the result no longer is true. Okay. Okay. So we, the result we are going to prove in this part of the course is just for expanding maps of the circle, not piecewise expanding maps. Okay. So let's go back into what we started last class about coding with Corina today has mentioned. And oops. So we want to code a number between zero and one or a number, a point in the circle. And we're going to use its itinerary to code it. We will see later that this code is not unique. It is easier to think of 10 branches instead of two, but I have done the drawings for two branches. So I will explain it first with two and we are going to see it with 10. So we have this point, the red point, the red dot here. And in the, when it begins, it is on the left interval, there are two intervals, the left interval, which we are, we are going to call zero and the right interval, which we are going to call one. Okay. So in the first iterate, in the zero iterate, it's in the zero interval. So its symbol will be zero. Okay. Now we iterate it. Still in the left interval. So we will put a zero. Now we iterate it again. Again, zero. We iterate again. Again, zero. But since this is a doubling map and if X is positive, eventually it will change sides or it will be a fixed point. So now it's one. Okay. Again, one, zero, zero, one. And now it begins to be periodic. So at one point it will be repeating one, one, zero, zero, one, one, zero, zero. And that will be its itinerary and a type of coding. In this case, the coding is unique. We are going to come later to this. But in this case, the coding is unique. Let me, the middle point is exactly one, one problem we have. What happens with the middle point and with the, with zero and one? What do we do? When there's, when our red dot falls here or here or here, this is the same point, we have a problem of which notation we should use. Okay. And it's, in fact, we will use both. It's easier to see it when we have ten branches. It's harder to draw. But when we have ten branches, we have 0.9999999. It has the same symbol as one, zero, zero, zero, zero. Okay. So we will use both symbols for the same point. It's the same as happens here. We, if we want to mark this point, we can put zero point. Well, the sequence will be, will be in the left. Well, let's, let's choose the left here. And then it will be in the right, in the right, in the right, in the right forever. We first, from these two, we choose the left, the left interval. But then once we have chosen the left interval, it is in the right interval of the left. And on the right of this one, and on the right of this one. And then this is one possible symbol for this point. But on the other hand, we could have chosen the right interval. And then it will be on the left, on the left, on the left, on the left, on the left. So these two symbols will be useful to mark the itinerary of this point. The itinerary has two possibilities. In, in this case, we will have more possibilities. Okay. Well, not more possibilities. We will have also this situation. The situation will come with 0.999. But this can happen, this can happen in any, in any of these branches. Because we can have 0.3999. We will have this situation at each separating point. So for each itinerary, we will have a unique point. But not every point will have a unique itinerary. Some points we'll have and some points we won't. Okay. Okay. So this is essentially coding. Coding is assigning to each particle in the interval a sequence of symbols. In the case, by means of its dynamics. So we, we have chosen the, the map f equal 2x mod 1. And it's itinerary will give us a symbolic, a coding for it. So we will define the space of symbols. We will define a metric in the space of symbols. We will see that this space of symbol is particular. And we will come later into it in more situations. So we, we call this space Sigma. And there are many notations I have chosen this one. Sigma plus 2. I will explain it. The plus is because we are going to choose. I'm sorry. The two is because it is two symbols. And the plus is because it's n and it's not the integers. Okay. We will see later that there are also space of symbols, the my sided sequences. But we are going to use only one sided sequences for this space. And this is, this is an example of a sequence. We will have sequences of zeros and ones. Many of them. We know that this, the cardinal of this is Aleph 0, Aleph 1. And we can define a metric on this space. We can define many metrics. But we are going to define one which is useful for us. And which is the following. We want to measure how distant, how distant two points are. We are going to measure the distances between each entry. But we are going to give different weights to each of these distances. Okay. The closer, the closer to the first position, the more weight it has. Okay. And so we are going x, each of these will represent the sequence. Okay. These are going to be xn with n in n, yn with n in n. And this is the metric we are going to use for this. Okay. So the distance of this is going to be 1 over 3, plus 1 over 3, plus 0, sorry 2, plus 1 over 3 to the 4, plus 1 over 3 to the 5, et cetera. So if the difference is 1, we weight it by 1 third to the power of the position it has. Okay. So since this is a convergent series, it is a well defined metric. I leave you as an exercise to check that this is a good metric. Okay. It could be infinite and still be a good metric, but this is a finite metric. Okay. And we have the following, we have the following properties. This is a compact metric space. Well, we have to check that it is a metric. I leave you, I leave it to you as an exercise, but it is also compact. And also it is a very practical metric because we can measure exactly the distance by means of how many entries coincide. Okay. So if, if two sequences coincide up to the entry force and then they are different, then its distance will be, in fact, will be 1 over 3, 5. And then, so this has an interesting consequence. Well, we have to check it is compact. I don't know. I have not much time. I have not much time. But this has an, it, yes, it is a product of, this is a, yes, this is a short proof. It's the product of, countable product of compact spaces, it's compact. So this is a short proof. Well, this, with this metric, but it's okay. Thank you. And this, this other equality, inequality, I leave it to you as an exercise, but it has an interesting consequence. I will leave it, this will be the first exercise in the afternoon, but this has an interesting consequence, this situation. The, the consequence is that this is a totally disconnected space. What does this mean? It means that every point has a neighborhood basis of cloven sets. What's cloven? Cloven means closed plus open. They are closed and open at the same time. Okay. And these both are closed and open. This is also an exercise, but it's easy to check because we, you have that, the distance is less than one over three to the n plus one. Then it's easy to show that this is an open set using the triangular inequality because it is symmetric. On the other hand, this equality, that the fact that XI equals YI for I equals zero to n is a closed property. It's easy to, to check that this is a closed property. So this is the, this, the set of poles is the cloven basis. And this is interesting because we are going to show that, well, we have, we are going to prove that there is a semi-conjugacy between one map we are going to define in a minute and expanding maps of degree two. We are going to define a shift map in a minute and we are going to show that there is a semi-conjugacy between this and any expanding map of degree two. But this is crucial. It is just a semi-conjugacy because this is totally disconnected and this is connected. Okay? So it cannot be a homeomorphism. But this semi-conjugacy closes the gaps. It is totally disconnected. You can imagine this as a cantor set. Okay? A cantor set. You are familiar with the cantor set? Well, the cantor set is a totally disconnected set. So you can imagine that this semi-conjugacy closes the gaps. And the gaps are produced exactly here. You have these two different symbols for one point. Okay? So let's define this what's going to be our, oops, our map. Okay? Here, points that are in the ball of sequence, in the ball one over three to the sixth of center one, one, one, one, one. It's all balls that its first six entries are one. You have infinitely many. Okay? Okay. So let's define this map which is going to be semi-conjugate to the all the degree two maps which are expanding on the circle. Which is the shift transformation. The shift transformation is defined by shifting. We shift, this is a sequence we have here x0, x1, x2, x3, etc. And then we shift all the sequence. We erase the first entry and we shift all the sequence to the left. This is why it's called the shift map. Okay? So the shifted sequence we have as entries, as entry zero, what was the entry one of the previous sequence. Okay? And maybe you can see it better with this sequence of sequences. This is x. So this is x. And we shift the sequence. Now we shift it again. And look at the zero how is balls one. Okay? The shift. Erases the first entry. So the shift map is not invertible. Okay? Because each point has two pre-images. There are two possibilities for each sequence. You can put, if you want to come back, here you can put either zero or one. Okay? Okay. So the shift has many, many properties. We are going to, the shift is the example of chaotic dynamic, chaotic dynamics. Okay? Everything that can happen in a chaotic map happens in the shift. Okay? And for instance, it has dense periodic points. But at the same time, it is transitive. It is topologically mixing. All of this, we are going to see it today in these 10 minutes. And in the exercise, in the problem session. Okay? So what is the property of a fixed point by the shift? How is a fixed point by a shift? One sequence? One constant sequence. Okay, yes. A constant sequence is a sequence, is a fixed point. Okay? If you have, if you have two different ones, you no longer have a fixed point. Because when it shifts, it's not equal to the previous one. So you have the next, the next one has to be equal to the previous one. So that implies that it is constant. Okay? And so we can do all the calculations just by the definition. If the shift equals x, then that, that implies this. And this implies that xn plus one equals xn for all n. And so once we have defined x0, all the other ones will be automatically defined. We only have freedom to choose the first entry because all the rest will be automatically defined. So we have two possibilities, either x0 is zero or x0 is one. And then we have the sequence of zeros and the sequence of one. So these are how fixed points look like. Well, then you can imagine how periodic points will look like. Periodic point, you know what a periodic point is, a point with a finite orbit. So at some iteration, we will go back to the initial orbit, to the initial point. And well, let's see what happens with periodic two points. We have this, then we have this. And this implies this. This is a recurrence of degree two. And for those who have worked with recurrence, you know that you have two degrees of freedom to choose the initial conditions, which are x0 and x1. And then you have four possibilities for x0 and x1. But then after you have chosen these four possibilities, then all the rest has to follow. And you have no more freedom than that. You will have four periodic points of period two. Well, you can easily guess how many periodic points of period n you have. But then it's a good exercise to prove that periodic points are dense. You have to put in any ball a periodic point. That is a good exercise. Well, I have so many exercises that I don't know what to choose this afternoon. Okay, that is a good exercise. And so they are dense. They are dense in the totally disconnected space. You have your, in any ball, you can plug in a periodic point. You can plug in many periodic points, in fact, infinitely many, because they are dense. And this I like it very much. The shift transformation is transitive. But there are many ways in which you can do this. You can prove, in fact, I will leave it as an exercise to prove that it is topologically mixing. And we have not proven it, but topologically mixing implies transitivity. If a homeomorphism, if a continuous map is topologically mixing, then it has a dense orbit. But another way of proving that a transformation is transitive is to show a dense orbit. And then this is more interesting. The shift transformation is transitive, and we are going to construct dense orbit. And why, how do we do that? I have three minutes, but I would like you to think about this. The balls have two variables. You have the number of entries and the length. Because you have, if they coincide in this n entries, then it's radius. If you want this to be in this, with radius 1 over 3 to the n plus 1, it has to coincide in the first n plus 1 entries. And that's all. So we want this orbit to eventually enter any ball. So it will have to contain all the strings of zeros and ones. So we will want to construct a sequence that contains all finite sequences. And how do we do that? Well, we start with zero, then one, then the sequence is of length two. Zero, zero, zero, one, one, zero, zero, one, one. Then the sequence is of length three, one. And eventually you will have Romeo and Juliet written in binary code inside this sequence. Everything will be inside this sequence. Your favorite book will be inside this sequence. And it will be transitive. Okay, and I think this is enough for today. Thank you.