So, today we will prove the following proposition. Any two invertible hyperbolic linear maps conjugate. So we have A of x equals A x, B of x equals B x. So, probably one of the most important exercises in the first exercise sheet was to think of all the possible dynamics related to different linear maps, right? So it's fairly straightforward depending on whether B is less than one or bigger than one or positive or negative, you have different dynamics, okay? It's very useful for you to have thought of all these possibilities to understand both the statement and the proof of this, okay? So first of all, let's see if we can remember what invertible means in terms of the properties of A and B. Do you remember? Exactly, okay? So invertible means that A is different from zero and B is different from zero. They're both invertible. What does hyperbolic mean? Exactly, so they're also different from plus or minus one. So what we're saying is that any two linear maps, as long as A and B are not zero or plus or minus one, are conjugate. So they're equivalent in some sense. So before we look at the proof, let's discuss this a second. So why do we... So do we remember what conjugate means? Definition of conjugate. There's a bijection. So these are defined on the real line, right? So conjugate, A and B are conjugate. If there exists a bijection H from R to R, because this is where these two maps are defined, and what property does this bijection need to satisfy? That's right. So A composed with H is equal to H composed with B, okay? Which means that if you... Well, which means that it maps orbits to orbits of these two maps. So why do you think we need to exclude non-invertible and non-hyperbolic linear maps? Why can I not say that any two linear maps are conjugate? What is the problem with non-invertible and non-hyperbolic linear maps? For example, how can you immediately know that if A is equal to one, and B is different from one, these two cannot be conjugate? Why? What fixed point? Exactly. If A is equal to one, all points are fixed points. And why is that a problem? It must map to fixed points. Exactly, okay? A conjugacy must map fixed points to fixed points. Periodic points to periodic points. And also all other orbits to orbits. So you can immediately see that if this is one, all the points are fixed, okay? If this is different from one, then there's only one fixed point, which is the origin, and therefore these two cannot be conjugate, okay? Of course, if both of these are one, then they're conjugate to each other in a kind of trivial sense. What if this is one and this is minus one? That's right. So here, if it's minus one, every point is periodic of period two. But it's not fixed, so you still cannot be conjugate, right? Because one has all fixed points, one has all periodic points cannot be fixed, okay? So what we're saying, and of course the non-invertible case means having zero here. What happens when you have zero here? When you have zero here, it means everything maps to zero, after one iterate and then stays in zero, okay? In the non-invertible case, nothing maps to zero. Things may approach zero, but they do not map to zero. So you cannot have a conjugacy in that setting, okay? So it's important to understand, to think about why we have these two assumptions, okay? As we shall see, if you've done the exercise, you know, even within the class of invertible and hyperbolic linear maps, there may be different kind of behavior. And I want to emphasize here that this conjugacy that we're going to build does not see this different kind of behavior, okay? And afterwards, we're going to show that if you require these to be topologically conjugate, then this will not be true anymore, because topological conjugacy distinguishes these different kinds of behavior. Hopefully today we will be able to see both cases. So we're going to assume that these are both invertible and hyperbolic, and we're going to construct the conjugacy. So how do we construct the conjugacy? At the last lecture, we gave a general method, okay? It's not a coincidence that I gave a general method. What was the general method for constructing conjugacies? Do you remember? I gave a general lemma that says, if this, then the two systems are conjugate. I gave a technique. Mariam, do you remember? You don't remember? Do you remember? Ba? The omega limit. Yes? Yes? Is that correct? What does everyone think? It's saying if two maps are conjugate, then the omega limits are mapped to each other. Is that right, Agni? I see you've forgotten more than I had. I confess I had forgotten exactly where we were in the last lecture, but I see you've forgotten even more of the contents in the last lecture. At least before the lecture, you spent half an hour looking to the notes of the previous lecture, you know? So you've at least a little bit refreshed your memory about what's going to happen. So this question about, is it true? First of all, this is an interesting one. Is it true that this conjugacy maps omega limits to omega limits? We just said it maps fixed points and periodic points to each other. Does it map omega limits to omega limits? That's right, only if H was a homeomorphism. If H is a homeomorphism, so it means it's continuous both ways, because the omega limit is defined as some limit points, right? So if this is continuous, it will map the limit points to the limit points. But just as a conjugacy does not necessarily do that. And anyway, that is not a technique for constructing the conjugacy. That is a consequence of the conjugacy. How do we construct such a conjugacy? Nobody can remember. Ah, fundamental domains. Okay, magic word. What is a fundamental domain? Anybody can tell me what a fundamental domain is? Fundamental domain? Do you remember? Yeah? Okay, you're kind of remembering what I said last time, but this is the construction I said. What is the definition of a fundamental domain? This, you're starting to go into the way I use the fundamental domains. But what is the definition? Do you remember the definition of fundamental domain? Do you remember the definition of fundamental domain? Oh, la, la, la, la, la, la. Okay, what's the definition of fundamental domain? Yes, this is what we were saying before. Yes, so we have a space. Okay, as you correctly remarked this, I could just start with this space. I don't need to take a subset, but we have some space. So we say that some u is a fundamental, we have a map, an invertible map, so bijection from the space to itself. u is a fundamental domain if every orbit passes a unique time inside u. Every point, right? Because, remember, because this is invertible, okay, I have to revise a little bit, but I don't want this to become a habit, okay? So if this is invertible, I can take forward and negative and pass the images like that. So the orbit of x is this set here. This is the orbit of x in forward and backward time, right? You iterate the map, you have a map. So you take the point x, which we can call x0, okay? Sometimes it's convenient to just slide this as xn. And then you look at x1. Which is the first image of x0. Then you look at x2, which is the second image of x0 and so on. And you take the orbit. And then you can take the orbit in backward time. This gives you a countable set of points here. And we say that u is a fundamental domain. If this orbit stops in u just once and only once, okay? There is some point, some time xA, where this orbit has one point inside here. This is a fundamental domain. As I said, it's a bit like a gate. Every orbit in going wherever it's going has to go through this gate. And it only has one or like a gate or just like a little bridge, okay? And it just stays on the bridge once. It doesn't stay there many times. It goes through this bridge. And what I proved is a lemma is that if a construction, okay, is that if you have two spaces and each of these has a fundamental domain, then it is enough to define a bijection between these fundamental domains. And this bijection extends to a conjugacy on the whole space, right? Because you use this bijection to... So how do you use this bijection? You start with any point here, x. What you want to do is you want to construct a conjugacy between f and g, right? So all you do is you take this point x and you wait until it enters the fundamental domain. This can be from positive or forward time. There's some tau. So tau of x, okay, is the time when f tau x of x belongs to you. There's a unique tau for every x because by definition of the fundamental domain, there's a unique time that you fall in there, okay? And then it's easy. You just take this point. You wait till it falls in there. And then since you've assumed there's a bijection between the fundamental domains, you look at the image of that point in this fundamental domain, and then you go backwards. You take minus tau iterates, okay?  And you define this as the point h of x, okay? If tau was negative, if you had to go backward in time because the full orbit intersects the fundamental domain only once in forward and backward time. So tau x may be negative, then it's no problem. You just hit here, and then you go back, and then you still iterate by minus tau, which means in this case it's forward time for j, okay? And then we did the calculation in class, and we showed that this is a conjugacy between these two systems, okay? This is a conjugacy between these. And this is the technique that we're going to use here, exactly. We're going to construct the fundamental domain for a and the fundamental domain for b, and show that there's a bijection between these fundamental domains, and that's finished, okay? We already have a nice lemma that we can use to apply in this case. So this, if you remember, is the technique of fundamental domains. So let's construct the fundamental domain for a. So let's suppose, just to make things easier, that a is, so we choose a specific number. Let's say that a is between 0 and 1. It's positive and between 0 and 1, okay? Just to make things a little bit intuitively easier. And then as an exercise, you can try all the different cases. So what do these points do? How do you think we should construct the fundamental domain? Let's take a point x here. This is our point x0. Where does it go? So we have an a and of x is just equal to the a to the n of x. So as n tends to plus infinity, what happens to this point? Yeah, we choose a between 0 and 1. It goes to 0. How does it go to 0? Does it jump around back and forth on the other side of 0? Does it just go like this? It goes like this. x0, x1 is just a times x0, x2, x3, and so on. Yeah? In backward time, it goes back here, okay? So how should we construct a fundamental domain? How do we find a region that this orbit will fall in once and only once? Where could we think to put this region here? I need to get some colored chalk. So for example, if we take this region here, then this orbit will fall in this region once and only once. Yeah? But we need this to be a fundamental domain for all points. So the first observation is that since these points never map here and these points never map here, our fundamental domain will have to have two parts. We'll have to have one part with two connected components or two components. One part will take care of all the points on this side. The other part will take care of all the other points on this side. So now we need to make sure that this is also a fundamental domain for some arbitrary other point, y0 here. And what does y0 do? Well, y0 maps to y1, maps to y2, maps to y3. Ah! Missed it. We jumped over. So it was not good. How can we guarantee that we're not going to jump over? How do we choose this fundamental domain? That's all right? We cannot define it. It cannot be connected. Why not? So let's try this. Let x tilde greater than 0 be an arbitrary point and let i plus a be the set a x tilde x tilde. So I suppose that this is some point x tilde and this is the point a x tilde, okay? And I take it open here and closed here, this interval. I claim that this is a fundamental domain for positive x, okay? Claim i plus a is a fundamental domain for x prime equals the set of points, positive points, okay? This is why I introduced x prime in the previous thing because I want to apply this to some invariant subsets. So is this a fundamental domain? How can I be sure that wherever I start, I will fall in here once and only once? Okay, let's look at different cases. Suppose I start in here already. So tau equals 0. Can I guarantee that no other iterates will belong to here? Only if I start in here, I want to make sure that I leave after one iterated both in forward time and backward time. I leave. Is it true? Why is it true? So I take a point here x, okay? Indeed, if x belongs to this interval, i a plus, then a of x, which is a of x. Does it belong to that interval? Why not? That's why to get open on one side and close on the other, okay? It's a good observation. So why not? Because clearly if x, then ax is less than ax tilde because it's less than or equal to ax tilde because x is less than equal to x tilde, clearly. So because I've chosen x in this interval, then x is less than or equal to x tilde and therefore a of x would be less than or equal to a of x tilde and therefore it's outside the interval. And in the other direction is also the same, right? So a minus 1 of x is equal 1, 1 over ax and this would be greater than or equal to x tilde, okay? Because x is greater than or equal to ax tilde, okay? Because x is, sorry, this would be strictly greater than, because x is strictly greater than ax tilde, okay? So in this case it's very simple because you're literally just multiplying by constant. So it's clearly the way I've chosen this interval is by taking x tilde, multiplying by this constant. And this is exactly the image of x tilde, right? This is equal to a of x tilde. So I use the dynamics to define this interval, okay? And because of the way I use the dynamics, it's clear that if I take a point inside this interval, it maps outside in forward time and it maps outside in backward time. In fact, more than that, look at the image of the whole interval. Where is the image of this whole interval, i plus a? Moreover, a of the whole interval, what can you tell me about the image of the whole interval under this map? Where does it lie? x tilde maps to a of x tilde and ax tilde maps to a squared of x tilde, which is somewhere here, right? So the image will be from a squared of x tilde to a of x tilde, sorry, so here it's open on this side, okay? And where is the backward image? So this is equal to a squared of x tilde, a of x tilde, okay? And the backward image will be equal to x tilde, 1 over ax tilde, which will be... So the backward image, of course, in backward time, this is invertible map. So this endpoint will just map to this, a of x tilde times 1 over a is just x tilde. This endpoint will map to here, right? So it will map to something like this, this is 1 over ax tilde. So there will be exactly a jation to each other because of the way I've defined the original interval. And what happens if I continue iterating? I will fill up the whole space, right? So all iterates, a n of i plus a for n in z are disjoint. Do you agree with that? They're all disjoint, right? This interval here is disjoint from this because I take it closed here and open here, and this one here, I take it open here and close here. And the union of them, what's the union of them? All of r, will the union include some point here? Exactly, positive numbers, right? So this is exactly the set of positive numbers, positive real line. And why does this show that this is a fundamental domain for the real line? Exactly, so why does every point eventually map to this? Because every point belongs to some image of this, right? So if you take this point here, x, this will belong to some iterate of this.  And if this is the nth pre-image of this, then it means that exactly nth this will fall in here. So because the union of all the images and pre-images covers exactly all the real line, then it means that every point belongs to one of these images or pre-images and therefore after some time it falls in here. So that proves that every orbit contains a point in here. And the first observation we made is that if you're in here, you must fall outside, shows that there's a unique time that you fall in there. So this is a fundamental domain, okay? So this shows claim, this proves claim. So what about a fundamental domain for the negative real line? Exactly the same construction, right? What about if a was bigger than one? Can we do the same thing? Okay, this will be exercise, okay, to check for all of the cases. We did it for a between 0 and 1. You should check that it works. Remember, as long as a is different from 0 and different from plus or minus 1, so you should check. We've not, this construction did not rely on choosing a specific value of a, okay? To do this, we only use the fact that a was between 0 and 1. If a is bigger than 1, then you have x tilde. And then a of x tilde will be on the other side. But everything else, if you think about all the arguments we have, they will work in exactly the same way. If a is negative, it's a little more subtle, but you can think about it, okay? Very important exercise. So if a is negative, in fact, the images will swap from one side to the other each time. So you have to think how to adapt a little bit the construction. But I think it's much more useful if you try to do it yourself as an exercise. Okay. Okay, so now we have a fundamental domain for a. We have a fundamental domain for b because we can do exactly the same thing. So we have a fundamental domain i a plus here. We have the other part of the fundamental domain i minus a here. Okay, this is supposing a is in 0, 1. And then let's do the same thing for b. Let's suppose for simplicity also that b is in 0, 1. Okay, again, you should, in the exercise, you should generalize and look at all the other cases. But they're exactly the same thing. So here we take some other interval i plus b. And here we have some other interval i minus b. This is a fundamental domain for the positive real values. This is a fundamental domain for the negative real values. Notice that we cannot find a fundamental domain for everything, right? Because 0 is a fixed point. 0 goes to 0. So if you try to include 0 in a fundamental domain, you cannot have a situation where you have a unique point of the orbit inside that fundamental domain, right? Because the whole orbit is equal to 0. If you have one element of this orbit in the fundamental domain, then all the points in this orbit will be in the fundamental domain. That is exactly why the other day when I introduced the notion of fundamental domains, I restricted myself to a subset of the whole space. Because that's exactly how I do it here. I construct fundamental domains for these subsets. But it doesn't matter because now we construct the conjugacy subset by subset. So now we construct conjugacy h from r to r. And how do we construct the conjugacy? Well, first of all, we know that this conjugacy has to map fixed point to fixed points, right? So we know that 0 must map to 0. It's the only fixed point that we have both for a and b. This is a and this is b, okay? So we define h of 0 equals 0, okay? And then how do we define h in the other parts? How do we use that technique? We already, I just said at the beginning of the lecture, what we do when we have fundamental domains. Honey, what did we do? Okay, and what do we need for that? We need some relationship between the fundamental domains. What did I just say? In abstract setting, if you have two systems and two fundamental domains, what do you need to be able to define a conjugacy between these systems? Just a bijection between the fundamental domains, okay? We've already proved the lemma that says if you've got the bijection between the fundamental domains, you're okay, okay? So can we construct the bijection between these fundamental domains? This is just an interval. This is an interval. Can we construct the bijection between two intervals? Yes, we can construct bijections in infinitely many ways, okay? It's very flexible. All we need is a bijection. This bijection can just be even just a rescaling of whatever the size of these two things. That's a very simple bijection, okay?  All of this gives h r to r conjugacy using the method that we built before, okay? So what you do is you just take any points, but this we've already covered in the abstract setting, but in practice to define the overall conjugacy from r to r, you take a point, you iterate it until it falls in the fundamental domain, you use this bijection that we had defined, and you iterate it backwards by the points by the same number of iterates and this defines the correspondence between this point and that point, and that gives you a bijection that is a conjugacy which we already proved, okay? So this proves, except from the fact that we haven't explicitly checked all the different cases, okay? We've checked it for a between 0 and 1 and b between 0 and 1, but all the other cases are just variations. You just have to think about how these fundamental domains work and everything else works. So this proves the fact that any two hyperbolic invertible linear maps are conjugate. So this leads us, as I anticipated, to the question of is this good enough, okay? So for example, let's suppose we consider the map A of x equals 1 half of x and B of x equals 2x. Both of these are invertible hyperbolic linear maps and so they are conjugate. What is the dynamics of these maps? What is the dynamics corresponding to this map here? What is A n of x? So here we have A n of x equals 1 half to the n of x, okay? B of x equals 2 of x. Actually, let me use 3 of x, otherwise there's a symmetry which I don't want to include. Exactly. So in this case, what happens is in forward time everything is converging to 0. 0 is attracting, okay? It's an attracting fixed point. In this case, what happens? A z goes to infinity. It goes to infinity. So points are moving away, right? If x is positive, it will be moving in this direction. If x is negative, it will be moving in this direction, right? If x is negative, you still get 3 to the n minus 3 to the n. Some point, it becomes very big, okay? So what is the omega limit of x here? Remember what the omega limit is? The omega limit, the accumulation points of the future orbit is 0. Exactly. When you have an attracting fixed point, that means that nearby points have that point as a fixed point. In this case, it's not just nearby points, but all points, in fact, have the omega limit equal to 0. In this case, the omega limit is empty because it just goes to infinity. There's no accumulation point. What is the alpha limit of these two? Alpha limit are the accumulation points in backward time in the past, right? So here in the past, points are going to infinity in the past in the sense that in backward time when n is negative, right? Then this is like 2 to the n, okay? So they're moving away to infinity in backward time, right? So the alpha limit is empty. What is the alpha limit here? It's 0, okay? Points are moving to 0 in backward time. So they're kind of inverses of each other. We have shown that these are conjugate, okay? So what do you think this conjugacy, how does this conjugacy work? Because what does it mean to show? We've just did this construction here, okay? And I claim that these are conjugate, which means that we have this bijection H, which maps orbits to orbits, okay? So how does this bijection, what do you think this bijection looks like? What is the image of some point here very close to 0? So, how is it a scaling? So this bijection will map some point x to some point y here, okay? Because that's what a conjugacy needs. It maps some point x to some point y. And it maps the orbits to the orbits, the forward orbits to forward orbits and backward orbits to backward orbits, okay? That means that the forward orbit of x is going to 0, and where is the forward orbit of y? It's going to infinity, okay? So where does this point map to? This is a n x. Where does the bijection H map this point here? It must map it to b n of y, because that's what the definition of this conjugacy is, right? And where is b n of y? b n of y is way out here, b n of y. Is that possible? That this bijection is mapping points that are very close to 0 to points that are very close to infinity here? I mean, it better be because I just finished proving that these two are conjugates, right? I'm just pointing out what this actually means. Can you have that? Can you have a map that maps the positive real line, switches things around? Of course you're going to have that. It's not a problem at all, right? Like 1 over x, okay? It's not a problem at all. But I'm pointing it out because you should think that that is what's happening here, okay? So remember that the conjugacy we just constructed here maps 0 to 0. It maps the fundamental domains to the fundamental domains, okay? And then it maps the forward orbits of the fundamental domains to the forward orbits of the fundamental domains, just what I've described here. Now, in this case that we did here, we took a between 0 and 1 and b between 0 and 1. And so it didn't look like there was anything strange going on. But if we had taken b bigger than 1 here, then what this conjugacy would be doing, what this, when we extend, we have this bijection between the fundamental domains. But when we extend it, we take a point here, we move it forward until it takes here, okay? And then we map it to here. And then we need to apply backward. This map backwards, which means in this case since b is greater than 1, backwards means it's going back here, okay? So it's taking points that are very close to infinity, points very close to 0 and points very close to 0, very close to infinity. But there's nothing wrong with that, okay? I'm just kind of emphasizing that. Does it preserve omega limits of points? It does not preserve omega limits of points, right? Because x, the omega limit of x is 0, the omega limit of y is empty. It does not map omega limits to omega limits. Is it a continuous conjugacy? It is continuous. Well, is it continuous from the real line to the real line? So for example, it maps 0 to 0. If it's continuous, it must map points that are close to 0 to points that are close to 0. It's not a homeomorphism. Exactly. If it is the homeomorphism, it must preserve omega limits. So you work by contradiction and say it cannot be a homeomorphism. And I also want to point out more directly that it cannot be a homeomorphism because 0 maps to 0, but points near 0 are mapped close to infinity here. So it cannot be continuous at 0, this particular bijection, this particular conjugacy. So this conjugacy that we've constructed in general will not be a homeomorphism because of these particular examples. So what we want to do now is we want to say, well, what happens if we look at the topological conjugacy classes? So we are interested, we think, okay, this conjugacy is all good, you know? But in some sense, you know, if a map has an attracting fixed point like this one and the map has a repelling fixed point, we would like to think of them as different, not as the same. Okay, we would like not to put them in the same equivalence class. We would like to distinguish these two situations because we think it makes a big difference if in the future you end up at 0 and something happens at 0 or you end up at plus infinity, which is a completely other thing. Okay, and this is why we're interested in topological conjugacy. So topological conjugacy will ensure that the omega limits are mapped to omega limits so two maps can only be topological. If two maps are topologically conjugate, in one everything tends to 0 and in the other one also everything will be 10 to 0. So now we want to investigate the problem of topological conjugacy of linear maps. Okay, so let me write down the result, but I think you can guess what the result is, right? So what is, when are two linear maps going to be topologically conjugate? So in this case you can see clearly that they will not be topologically conjugate. What if I put here one third, will this be topologically conjugate? Do they at least have a chance to be topologically conjugate? They have a chance, right? We need to prove that they are, but they have a chance because at least they have the same omega limits. So every point is converging to 0 here, every point is converging to 0 here. They look more similar from a topological point of view. One is going much faster than the other, okay? And that would make a difference because after the topological we're going to ask whether this conjugate can be differentiable conjugacy, okay? We're systematically trying to understand what all these different conjugacies mean, okay? So, proposition or maybe even theorem here. So, when are two, so, two one-dimensional invertible hyperbolic linear maps topologically conjugate if and only if they are fixed points, okay? If and only if, one, they're fixed points which means the fixed point the origin, 0, are either both attracting, repelling. They are either both orientation reserving or orientation reversing. So, let's just make sure we understand what this means. So, they either both attracting. What is the condition on the linear map that ensures that the fixed point is attracting? So, that's right. So, the absolute value of A is less than 1. That's what this means, right? It can be negative also. Then everything goes to 0 in forward time, the fixed point is attracting. And similarly repelling means A greater than 1. I'll just use A to indicate generic. Orientation preserving and reversing. I don't know if I actually formally defined this, but it's clear. Orientation preserving means the derivative is positive. It means that the order, it preserves the order, right? So, I will use perhaps as a definition the fact that A is positive and the fact that A is negative can be the definition here of orientation preserving and orientation reversing. So, what exactly does this mean? I take two linear maps, A and B, okay? How do I decide if they're topologically conjugate? I look at the value of A, I look at the value of B. What has to happen? What exactly have I said here? Either one or the other or what? So, for example, if A is equal to plus 1. No, A is equal to plus 1.5 and B is equal to minus 1.5. They're not, okay? Because they need to satisfy this and this, okay? So, they need to be both, no less than 1 and both positive or no less than 1 and both negative or both no bigger than 1 and both positive or both negative and so on, okay? So, how many different equivalence classes do we have? Four, exactly, okay? So, linear maps, four topological conjugacy classes, okay? These conjugacy classes are minus infinity to 1, 1 to 0, sorry, minus 1 to 0, 0 to 1 to infinity. So, what is the statement saying? The statement is saying that if you take two linear maps, AX equals AX, BX equals BX, then these two maps are topologically conjugate if and only if both A and B belong to the same one of these four conjugacy classes. If A and B belong to two different sets, then they cannot be topologically conjugate. If they belong to the same set, then they are topologically conjugate. So, we will prove this now. 1.5 and 1.3, they both belong to this conjugacy class so they are topologically conjugate. So, again, the difficult part is to show that they are. The part in the direction in which you suppose that A and B belong to different conjugacy classes. Let's see, I will leave it as an exercise, okay? To show that if two maps, if A and B belong to different conjugacy classes, you need to show that they are not topologically conjugate, okay? But we've already discussed this in some length. For example, if A belongs to this conjugacy class and B belongs to this one, why are they not topologically conjugate? Exactly, because in this case, the fixed point is attracting and in this case, it's repeller, same as we did before, okay? So, you should think as an exercise, you should think of all the various possibilities and make sure that you understand that if A and B belong to two different classes, they cannot be topologically conjugate, okay? And what we will do now is show that if they are, then they are topologically conjugate. If they're in the same class. So, we will take advantage of all the construction that we've done so far, the construction of the conjugacy. All we need to show, we've already shown that all of these, notice that all of these contains exactly the union of these is all the invertible nonhyper, invertible hyperbolic linear maps, right? Because the only points we're excluding here is the points minus 1, 0 and 1 from this set, okay? So, what we've just shown is that all these maps are conjugate to each other. So, just by normal conjugacy, we have only one conjugacy class, which is the union of all of these. And we're showing that if you have a stronger sense of conjugacy, you differentiate these different types. But we've already constructed the conjugacy between all these. And what we just showed is that if you belong to different class here, there's no way that this conjugacy can be a homeomorphism. And now we will show that in the case in which you belong to one of these, you can prove that your conjugacy that we already constructed can actually be chosen to be a homeomorphism. And that's all we need to do, right? So, proof. So, we have AX equals AX equals BX. So, again, we need to kind of choose one case. So, for simplicity, let both A and B be, say, in 0, 1, okay? Again, you should make sure you check the other cases, but the argument is almost exactly the same. So, we already have our conjugacy. Remember, what we did is we had here a copy of R. Here we had our fundamental domains, IA plus, A minus. Here we have a copy of R. Here we have our fundamental domains, IB plus, B minus. This is 0. This is 0. And we constructed H. First of all, by defining a bijection between these fundamental domains and then extending it to the whole set, okay? So, we use exactly the same construction, except that now, of course, we can assume in the construction that the bijection between the fundamental domains is actually a homeomorphism at least, right? So, H is constructed, H is defined from a bijection H tilde between the fundamental domains, okay? I plus or minus A to I plus or minus B. So, let's at least, since we are hoping to get a homeomorphism, at least let's start with this H tilde, which is the original bijection between the fundamental domains to be a homeomorphism. Can we do that? Can we define a homeomorphism without any worrying about the dynamics at all? Does there exist a homeomorphism between A, I, A plus and I, B plus and between I, A minus and I, B minus? Of course, they're just intervals, right? We can have infinitely many homeomorphism. We can take, for example, as we said before, just the linear rescaling of this, okay? It's very simple. So, we can choose H tilde to be a homeomorphism. And then, do you remember how we defined H starting from H tilde? I've said it in words several times already today, yeah? H tilde, that's right, that's right, okay? So, we define H of X to be equal to, so we take any point X here. So, we start by applying A tau of X, then we apply H tilde, and then we apply B minus tau. This is our definition where this is the tau of X, right? So, it's the time that this X needs to enter the fundamental domain. So, in the calculation I did in the previous lecture, we showed that this gives you a bijection and gives you a conjugacy, okay? Which is what we've just used to prove the conjugacy. Now, using the fact that H tilde, sorry, this is a tilde here, using the fact that H tilde is a homeomorphism, we want to show that H is a homeomorphism. And we define H of zero. So, this doesn't give us a value at zero because zero does not hit the fundamental domain. So, we just define H zero equals zero, okay? So, this is the definition of H. We've shown that this is a bijection. We want to show that it's a homeomorphism. So, we have a lemma H and H minus one are continuous. That's what it means to be a homeomorphism. Because we've already shown that it's a bijection, so we just need to show the continuity. So, we will show the continuity for H, and H minus one will be exactly the same argument, exercise. So, how do we show the continuity for H? How do we know that H is continuous? Remember that somewhere we're going to have to use this assumption here. Because as we showed before, this is not true if this is not, if b is bigger than one, right? So, somewhere we'll have to use this assumption here. So, continuity is, as you know, a point-wise property, right? So, what does it mean for a function to be continuous? It means it's continuous at every point. So, the definition of continuity says H is continuous at a point. That's the definition of continuity. We need to show that it's continuous at every point. So, let's consider different points, okay? So, suppose, for example, H, so we need to show H continues at every point in R. So, let's look at different points. Suppose, for example, H belongs to IA+. Is it continuous? Is H continuous? Let me write IAH+, explicitly here. Let's suppose it depends. Remember, IA+, was of the form AX tilde, X tilde, okay? Is H continuous at X? Tau is zero. Tau is zero because it's already, the point is already in the fundamental domain, right? So, you do not need to iterate the map. You're already in the fundamental domain. So, tau is zero. So, A tau of X is just X itself. Tau is zero here, too. So, this is just the identity. So, we have these two are the identity because tau is zero. So, you just apply H tilde. So, H of X is equal to H tilde of X. And so, H tilde is a homomorphism. Almost. Well, almost, right? What about continuity at this boundary point here? So, it's a little bit different saying that H tilde is a homomorphism between this interval and another interval. And saying that this map is continuous when you think of this interval as embedded in the whole line. What is the difference between these two things? Remember, continuity means continuity from the left and from the right. That, if you think of it just as a map from the interval to another interval, then you just need continuity from the left. But now, this is a point in R. We're trying to look at this map from R to R and saying that it's continuous at this point. And then we need to say that it's continuous from the left and from the right. And H tilde is not defined on the right of here. Okay? So, we need to consider this point separately. So, if X belongs to this open interval, AX tilde, X tilde, okay, then H is continuous. This is a special point and we'll have to worry about this. Anya, do you understand this comment I made about this continuity? H tilde is a homomorphism from this interval to this interval, okay? So, as a homomorphism of intervals, if you include the endpoints, then the definition of continuity, you just need the continuity from the left because there is no right at that point, okay? But now, this H tilde is a homomorphism from this interval to this interval. But really what we're interested in is the continuity of H, which is a map from R to R. So, if we're interested, which is a bijection from R to R. So, the continuity at that point means we're interested from the left and from the right. We need to prove both. From the left, it follows because H tilde is a homomorphism. From the right, we have to think about it. Before we do that, let us look at zero. So, we've proved it inside the fundamental domains. Of course, this also, by the same argument, we get it inside the interior of this fundamental domain because, again, H tilde is defined there, okay? So, this is one case. So, two. If x equals zero, let's try to prove it at x equals zero. So, we know that H zero maps to H zero. What does it mean to be continuous at zero? That's right. That's right. So, we need to say what this means is that for all epsilon, there exists a delta. Such that if x is less than delta, in other words, x, the distance between x and zero, right? So, x minus zero, which is the same as x is less than delta. Then H of x, which is the same as H of x minus zero is less than epsilon. This is the definition of continuity, right? In other words, if you take points very close to zero, they map very close to zero. Is this true? First of all, convince yourselves. Why is it true? Once you know why it's true, you'll know how to formalize the argument to prove that it is true. Yes? It was a minus variable.  We know, so this is where we're going to use this property, okay? Of course, because if you remember, it's exactly at zero that the continuity failed for the other case, right? Remember when we said because in the general conjugacy case, if A was between zero and one and B was bigger than one, then we had that points close to zero were mapping close to infinity, right? So that's exactly at zero that the continuity failed if A and B belong to different conjugacy class. So here we need to use this as you correctly say. And what this means is that all the points are converging to zero here and the B and all the points are converging to zero after A. But how do we, why does that imply continuity at zero in terms of this definition? Good, good, good, good, very good. So let's try to, let's try to go step by step here. So what is the heuristic idea, first of all? Is that if you start very close to zero, what can you tell me about tau of x? It's negative and small, large, it's large. Crucial observation, it's large. The closer you are to zero and the larger this is, you agree with that? Yeah, because it means, you know, you are as, if you look at the point from the, if you look at this fundamental domain and you iterate it forward, it's getting closer and closer to zero. If you're very close to zero, it means you took you many iterates to get there and so it takes you many iterates to come back. So this is large. So you take a point very close to zero, you map to the fundamental domain with a large tau.  So what is this large tau? Where is this large tau going to take you? Because tau is large, it's going to take you back close to zero, okay? And this is the fundamental intuition behind this. The closer you are to zero, the larger tau is and therefore when you come back, you will go back by a very large tau which must take you close to zero, okay? And now we formalize it, which is exactly how you said. So how do we formalize this? So what is the epsilon and the delta, right? So the epsilon exists here. This is an epsilon neighborhood, okay? So we want to make sure that the image of these points will be inside the epsilon neighborhood here. So how can we guarantee that the image of points here will be inside this epsilon neighborhood? All we need to guarantee is that tau is big enough, right? Because we always start from here. So we can looking just on this side, we know how big tau needs to be so that all this point falls inside the epsilon neighborhood, right? So for any epsilon, there exists some n epsilon, okay? Such that bn of i, in fact, both. I can just take the union. I can do them at the same time, the positive and negative. Such bn is inside minus epsilon epsilon for all n greater than equal n epsilon. You agree with this statement? Yeah? So given any epsilon, I can fix the n epsilon that I need. That will guarantee me that all these points here after n epsilon interest will fall inside here and will stay obviously inside there for a long time. So now it's very simple. How do I choose my delta on this side? Now I need to choose a delta neighborhood here, minus delta delta, that guarantees that the time it takes to get here is bigger than n epsilon. That my tau here is bigger than n epsilon. Because as long as this tau is bigger than n epsilon, then when I move to the other side by tau by h tilde, then I need to iterate backwards by tau epsilon. So maybe this should be, no, this should be positive, right? So here I will get a negative tau, so I need to go backwards by a very large number. So moreover, for all n epsilon greater than 0, there exists delta of n epsilon, okay? Such that if x is in minus delta delta, then tau of x has to be less than or equal to n epsilon, minus n epsilon, okay? This means that it's very big and negative. So as long as I take this delta neighborhood sufficiently small, clearly points very close, take a long time to come here. So I look first on this side because it allows me to fix n epsilon, okay? Once I know n epsilon, I can take delta sufficiently small so that any point that starts inside here must take at least n epsilon iterates in backward time to reach the fundamental domain. And that will guarantee that when I compose these maps, if I start inside the delta neighborhood here, I will end it inside the epsilon neighborhood, okay? Then, so this implies for all x in minus delta delta, strictly speaking I should put minus 0 here, okay? Because I'm not, h of x belongs to minus epsilon. And this proves the continuity at 0. Okay, so what about other points? Now I've done 0. I've done this fundamental domain. There's two more classes of points, okay? Third, if x belongs to the union for all n in z of a n of a x tilde, x tilde open neighborhood, union the negative of that, so union minus x tilde minus a x tilde open neighborhood. So in other words, if I look at this fundamental domain and I look at one of the images of the fundamental domain, okay? This is some a n i plus a. And I take a point in there. How do I prove the continuity at that point? Yes, that's already inbuilt into the definition of h. But look, what do we have? We have a composition of functions here. So h of x is given by composition of three maps. So to show continuity of h, I just need to show that each of these is continuous. And they are continuous. For these points, we just have, in fact, probably I should have made this the mark right at the beginning. It's kind of the first obvious approach. It's because h is defined as a composition, then as long as these three are continuous, it's continuous, okay? So this holds. The first case we did is a special case of that is when a and b are the identity map. And then you just use the fact that h tilde is continuous. But in general, even though it's not the identity map, this is a continuous map. So at this point, a is continuous from the left and the right. It's continuous. A tau is also continuous, okay? And then you apply h tilde which is continuous. Then you apply b minus tilde that is continuous. And you get continuous at this point. So for this point, then h is a composition of continuous maps. So there's only one other class of points left, which is the problem we had here, occurs at all the images and pre-images of this point, right? So here I put the open interval here because that's where it's a composition of continuous maps. The problem with being on the boundary point is that tau changes across that boundary, right? If you're on this boundary, on the left-hand side of the boundary, tau is zero. But on the right-hand side, tau is one. So tau does not depend continuously on x when you're on the boundary, okay? And similarly, when you look at the image of this fundamental domain and you take a point on the boundary of the image, tau does not depend continuously. So this is no longer a composition of continuous maps, okay? Because as part of the definition here, you have the tau that depends on x. So if you're in the interior, then tau is constant in the interior, and therefore tau is constant. So this is continuous, this is continuous, this is continuous, the composition of continuous maps. But on the boundary, it's not. So the problem we had here, on the boundary here, is exactly the same problem that we have at all the boundary points. Because what they do is you take these boundary points, you need to show continuity from the left and from the right. From the left, it's okay because from the left, it's just a composition of continuous maps. From the left, you have continuity of A, continuity of tau, everything is fine on the left. The problem is the continuity from the right, okay? But it's not a big problem because it's clear that this is a solvable problem because it's clear that if you take... because this fundamental domain maps to this fundamental domain, this image of this fundamental domain will map to this image of this other fundamental domain, and these points on the boundary will map to this point on the boundary, and it's continuous from the left, but also from the right. Really, this point here is just on the other side, and it's part of the image of the previous image of the fundamental domain, right? And it clearly will map nearby here because here also you have the previous image of this fundamental domain, okay? So I will leave it as an exercise for you to formalize this last case, okay? Which is a very interesting and important exercise to really make sure you understand the construction, okay? So the only cases left... so only cases left to check are when x is on the boundary, when x is on some a n of i plus or minus a exercise. So the exercise takes a little bit of thinking, but I have given you really all the kinds of ingredients, all the kinds of arguments that you need to use in these kind of things, okay? What we're doing here is fairly slow, is fairly systematic, but it's very important because these arguments we will use in other cases over and over again, okay? These are not special arguments. These arguments, these constructions are constructions that are more general and we will use in other situations. This is the simplest setting in which to really become familiar with them. So anytime you spend on this is absolutely worthwhile. Okay, if you get a chance in between preparing for all your other exams, also to revise this and to do the exercises, you can take the opportunity to think about what we're going to do in the next lecture, which is ask ourselves, can we do even more? Can H be constructed differentiable, okay? And you can think of what we did. So all we did, H tilde obviously can be constructed differentiable, okay? And the question is whether this composition is differentiable. A is differentiable, B is differentiable, so you can think a little bit in the context of this proof whether you can push it all through to the differentiable case or not, okay? So I think this is a good place to stop. Thank you very much.