 Okay, good morning everyone. So in the last lecture, we looked at the topological conjugacy classes of linear maps, right? So we have the space of linear maps, AX equals AX, A in R. So R is the parameter space of the space of linear maps, right? For each value of A, you get a linear map. And we looked at the topological conjugacy classes of invertible hyperbolic. So if you remember, there were four conjugacy classes, minus infinity to one, to minus one, minus one to zero, zero to one, and one plus infinity. So we showed that in fact, any linear map in this set, so any linear map except for A equals zero plus or minus one are conjugate to each other. But they're not all topologically conjugate. If you want the topologically conjugacy, the topological conjugacy distinguishes, separates, okay? Maps that belong to these four regions. So what we, interesting question now. So if you remember the difference between the conjugacy and topological conjugacy, it's just that the conjugacy is a homomorphism in this case. So question, what about C1 conjugacy? Can the conjugacy H be a different morphism? In other words, in the same way as when you go from simple conjugacy, in which H is just a bijection, to topological conjugacy in which H is a homomorphism, you distinguish more in detail the properties of the system. And so you split it up into four instead of just one conjugacy class. The question is, if you say, if you want this conjugacy to be C1, is that a stronger condition? Obviously, two maps which belong to different topological conjugacy classes cannot belong to C1 conjugacy classes, right? Because if it's C1, in particular, it's a homomorphism, so they cannot. So the question is, if you take two maps within the same topological conjugacy class, are they also differentially conjugate? Or does this create a finer division into more C1 conjugacy classes? Okay, this is the question we will address today. So I will give, the answer actually is fairly simple, and I will just give you the answer, which is that indeed in general, two linear maps, even in the same topological conjugacy class, are not C1 conjugacy, are not C1 conjugates. And the answer is the, and the reason is the following, the position. I will state it in a little bit more general form. So let F, R to R, and G, R to R, be C1 differentomorphism, which C1 conjugates. So there exists differentomorphism, H to R, such that H composed with F equals G composed with H. Let P, Q be fixed points for F and G, respectively. If they exist, if they exist with H of P equals Q, then the derivative at P equals the derivative at Q. So what is this saying? I stated it in one dimension, actually this is true in higher dimensions too. But basically one dimension, so this is F, R to R. This is G from R to R. Suppose they are conjugate by C1 differentomorphism, okay, this is the assumption. Suppose there exists a point here, which is a fixed point, which is mapped to fixed point Q by this differentomorphism. Then what is the statement saying? It's saying that the derivative of P must be equal to the derivative. Okay, so this is the assumption, right? So the assumption is that there exists the differentomorphism, C1 differentomorphism. And the conclusion is that if there exists two fixed points that map to each other, then this differentomorphism may have the same derivative. So what is the implication for our question here? Why is that relevant to this question? Excuse me, zero is a fixed point. So for all the linear maps, zero is always a fixed point, that's right. So what is the derivative of the fixed point at zero? A, okay? So each of these maps has a fixed point, right? You can actually draw the graph of these maps, right? X equals AX has a graph, okay? This is the graph of the map A of X equals AX. It's linear, so the derivative is always A, right? In particular, at the fixed point, the derivative is A, okay? So if you take two different linear maps, A of X equals AX and B of X equals BX, can they be differentially conjugate? Only if A is equal to B. So if A is different from B, they cannot be C1 conjugate because of this proposition. What if A is equal to B? Do we know that they are C1 conjugate? So this just gives a kind of negative, right? This says if this does not hold, then they cannot be C1 conjugate. This is how we use this. So what we're saying is any two linear maps cannot be C1 conjugate unless A is equal to B. What if A is equal to B? They're the same map, very good. I thought you should get that. So if they're the same map, does that mean that it's conjugate to itself? Is a map conjugate to itself? Yes, but we didn't, I mean this is a little detail which we should have addressed probably right at the beginning, sorry? Is a map in general C1 conjugate to itself? Exactly, you take the identity, conjugates the map to itself. The identity is C1 defilmorphism and it conjugates the map to itself. Sorry? That's right, we didn't, yes, we proved it. So I guess it's part of the definition. We, in principle, we didn't discuss it explicitly in the class, but we implicitly checked that the map is conjugate to itself and we proved that it's an equivalence relation, okay? So the consequence of this proposition is that how many C1 conjugacy classes do we have for linear maps? Infinity many, uncountably many, even, right? Every map is its own conjugacy class. So this shows what the comment I made at the beginning, is that the difference between conjugacy, topological conjugacy, and C1 conjugacy, right? General conjugacy is a very weak notion of conjugacy. All these linear maps are all conjugate to each other. C1 conjugacy is a very strong definition of conjugacy. None of these maps are conjugate to each other. You don't gain anything, you've not really grouped them in any way, they're all different conjugacy classes. So you might as well have the original maps, you know? You've not gained anything in this particular setting by taking C1 conjugacy. Topological conjugacy seems to be a very interesting compromise because it groups them in just four classes, so a small finite number of classes. Each of these classes has some obvious characteristics that they share that seem to be fairly natural and it seems fairly natural that you don't need to distinguish really between different maps in here, okay? So of course there are certain situations in which it's important to know whether maps are C1 conjugated to discuss that. So I'm not saying this is never relevant, I'm just using this example to really highlight the difference between these three notions of conjugacy, okay? I think this is a very good example and we will come back to these three notions in different settings. Yes, the topology on what? So there's two topologies involved here, right? There's the topology on the spaces. So a topological conjugacy means that H is homomorphism from this space on which one dynamical system is defined. And this is a space on which the other dynamical system is defined. So these need to be topological spaces for you to be able to say that this is a homomorphism, right? So implicit in the definition of topological conjugacy is that two systems are conjugate if there's a homomorphism that conjugates the two dynamics. So in particular the two spaces on which the dynamics are defined need to be homomorphic. They need to be homomorphic topological spaces. I'm not sure if that was your question or not, yes? Yes? Okay, so let's prove this. This is really a very simple calculation. I will do it, but I could almost leave it as an exercise, but let's do it. So it's really just an application of the chain rule. So using the conjugacy, we write by the definition of conjugacy, we have F is equal to H minus 1 composed with G, composed with H, right? Because H is invertible. In fact, here it's a diffimorphism, so the inverse is also C1 diffimorphism, right? And so by the chain rule, we just differentiate, right? So what is the derivative F prime of X is equal to the derivative of all of this in the point X, right? And you know what the chain rule, how the chain rule works. So this is equal to the derivative of H minus 1 prime in G of H of X, right? Times G prime in H of X times H prime, okay? So the chain rule just says that you differentiate each map in the point at which it is applied. So now we use the fact that the P is a fixed point. So since P is a fixed point, P is a fixed point, and H of P equals Q, then we have that H minus 1, so we're applying this at P. Okay, so let me write this. F prime of P is equal to H minus 1 prime of G of H of P times G prime H of P times H prime of P. So this is equal to, so H of P is equal to Q. H of P is equal to Q, right? So this is Q and this is Q. So this is H minus 1 prime in G of Q times G prime in Q, H prime in P. But Q is a fixed point for G also. So this is H minus 1 prime of Q times G prime of Q times H prime of P, okay? And now this H minus 1 prime in H of P is equal to, but, okay, so H minus 1 prime in Q is equal to H minus 1 prime, H minus 1 prime in H of P. Q is H of P in H of P, which is just equal to H prime of P to the minus 1. Okay, this is just the inverse. The derivative of the inverse is 1 over the derivative in the image, okay? And so this is the inverse of this, okay? And so this is just equal to G prime of Q. Excuse me? Ah, yes, you're right, you're right. But it cannot be 0 because otherwise it would not be a defilmorphism, right? If the definition of defilmorphism, a defilmorphism cannot have 0 derivative at a fixed point because otherwise the inverse would have infinite derivative at that fixed point, which is not acceptable, right? If the defilmorphism at a fixed point has 0, in fact, not just at a fixed point, at any point it cannot have 0 derivative anywhere because otherwise the inverse would have infinite derivative, okay? So defilmorphism cannot have 0 derivative. Thank you, that's a good point. Okay, so this is just a simple calculation that shows this. Okay, so this completes our discussion of the different kinds of conjugacies. So now to wrap up this section on linear maps, we want to use, apply all these ideas of conjugacies to study the structural stability. So if you remember at the beginning I said one of the most interesting applications of this idea of this kind of equivalence class is to talk about structural stability for one-dimensional linear maps. So do you remember what structural stability means? Anyone? In the same conjugacy class, that's right. So structural stability is the following. I take a system and I say what happens if I make a small perturbation of this system? Does it change the conjugacy class or not? Does the system change or not? Okay, so there's two crucial ingredients that are required to formulate this. One is what do we mean by changing the system a little bit? And the second, what do we mean by the system stays the same? So in other words, let Ax equals Ax, okay? Does the dynamics change under small perturbations? So there's two notions that are involved here. Two notions. What do we mean by one does the system change? And two, what do we mean by small perturbation? So what do we mean? In this case it's particularly clear. So what do we mean by does the system change? Means is the new system conjugate to the original one, okay? And as we have seen, there's various ways to define, you know, there's various conditions we can use to say they're conjugate. So depending on whether we use a simple conjugacy or differentiable conjugacy or topological conjugacy, we will get a different answer as to whether the systems are the same or not, okay? And the second is what do we mean by small perturbation? In this case, it is fairly natural. There's an obvious topology on the space of linear maps. What is the topology? What is the space of linear maps? The space of one-dimensional linear maps is just given by the parameter a. So there is an obvious way to say what it means for two linear maps to be close is just a and b should be close, right? So small perturbation means a small neighborhood of a. That is a small perturbation. So in this particular setting, the question is to formulate the question more precisely than this. So given a in R, does there exist... does there exist a neighborhood, let's call it u of a, such that for all b in u, a of x equals ax and b of x equals bx, r, say, topologically, or c1. This is the notion of structural stability. So I take r. This is... I take it as the parameter space of linear maps. This is 0. This is 1. This is minus 1. So my question is this. I take a parameter a and I say, is the linear map ax equals ax structurally stable? What is the answer? Is it structurally stable for this parameter value a? It is. So can I find a neighborhood of a such that all the maps in here are all conjugate to a? Yeah? Does it depend on where I chose the point a for every a, minus 1 and 1, what? Then what? Then it will be stable if it's between minus 1 and 1. And what about here if it's bigger than 1? Is this structurally stable? Awesome. Let's make sure that everyone is on board here. Annie, do you know where we are? Is the question clear? What about the answer? So what do we know about whether two parameters a and b like this are conjugate or not? Mariam, that's your name, right? What do you think? Are these two maps corresponding to these two parameters are they conjugate? a of x equals ax, b of x equals bx. Are these two conjugate? Do you know? Was that a kind of, OK, 50-50 chance? Why are they conjugate? That's the definition of conjugate. But why do they belong to the same conjugacy class? Did we show that? And so, the same conjugacy class. What kind of conjugacy? Topological conjugacy, c1 conjugacy? Topological conjugacy. What if I'm interested in c1 conjugacy? A and B conjugate? Exactly. So if I'm interested in the c1 conjugacy, does that exist a neighborhood of A such that all the parameters in here are conjugate to A under the c1 conjugacy? No. So is a structurally stable under c1 conjugacy? It's not structurally stable. Structurally stable means you change a little bit and things do not change. According to the conjugacy, you have chosen. So the notion of structural instability depends on which notion of conjugacy you have used. If you choose topological conjugacy, then all these points, as you said, except for 0, minus 1, and 1, they're all structurally stable. Because if you take any point, because remember, these are exactly the topological conjugacy classes are exactly this. This is one topological conjugacy class. You can use different colors. This is another topological conjugacy class. This is another topological conjugacy class. And this also is four topological conjugacy classes. They're open in the standard topology on the real line. Which means that if you choose any point inside here, you can find the neighborhood any point. However close to 0 you are, you can find a small neighborhood that remains inside the same conjugacy class. So every invertible hyperbolic linear map is structurally stable with respect to topological conjugacy. Exactly. When I say invertible and hyperbolic, I use exactly to exclude those cases. What about 0 and minus 1 and 1? It's structurally stable. This map at minus 1. Is it structurally stable? There is a map at minus 1. a of x equals minus x. It's a perfectly well-defined map. No, it's not, but it's still a linear map. So my question is, is that structurally stable? Exactly. And remember, we know that for minus 1, for example, every point is periodic of period 2. When you take a small perturbation, you have only one single fixed point at 0 and no other periodic points. So this map at minus 1 cannot be even conjugate. We don't even have an anthropological conjugate. It just cannot even be just conjugate to any of the maps that have arbitrarily close to it. So the points minus 1, 0 and 1 are not structurally stable in any sense whatsoever. Because if you make a small perturbation, you will get something that is not even conjugate to the original map. But if you take anything outside these three points, then they will be structurally stable with respect to topological conjugacy, but not structurally stable with respect to C1 conjugacy. Because no map is structurally stable with respect to C1 conjugacy. Because with respect to C1 conjugacy, every map, every arbitrarily small perturbation, is a map that is in a different C1 conjugacy class. Because the only C1 conjugacy class is just individual points. So maybe let me summarize this. So proposition. Every invertible hyperbolic linear map is structurally stable with respect to topological conjugacy, but no map. But every linear map is structurally unstable with respect to C1. Structurally stable with respect to topological conjugacy, structurally unstable with respect to C1 conjugacy. I'm not going to write down the formal proof because we've discussed it. It's pretty clear, but you need to make sure that you understand how to prove this. This could be a good candidate for an exam question. I'm saying it's a good candidate. It collects results we've been doing until now. It tests understanding of the concepts that we're talking about and the specific results that we've discussed until now. Complex parameters. Yes, depending on the family of dynamical systems that we have, we will have different topologies on the space of dynamical systems. So in certain situations, there are certain families of maps which are naturally parameterized by the complex numbers. In other families, as we shall see in fact now, we will talk about nonlinear maps and we don't really have a parameter space, but we have a topology on the space of all maps, like the C0 topology or the C1 topology. I will talk about it either today or tomorrow. I don't know if you've done these topologies on the space of continuous maps or C1 maps, no? Okay, I will introduce them. But that's right. So implicit in this is because we're looking at some very simple examples, and the family of all one-dimensional linear maps is parameterized by the real line, then the most obvious topology on the space of linear maps is just the standard Euclidean topology on the real line. But of course, this depends on the topology we choose on the real line. Because the notion of neighborhood depends on the topology. So what you want is a neighborhood such that everything in there is conjugate to each other, all the maps in this neighborhood are conjugate to each other. We will see many examples of this kind. Yes, yes, yes. Try to formulate your question a little bit better. So you see these, the conjugacy classes, each conjugacy class is an open interval. One, two, three, four conjugacy classes. So if a conjugacy class is open, then automatically any element in that conjugacy class is structurally stable. Because by definition of open, any element in there will have a neighborhood that's contained in that conjugacy class. Okay? So in this case, this is another way to see this, is that if the conjugacy classes are open, clearly every element in one of these conjugacy classes is open. In general, this is not necessarily the case. You might have conjugacy class that has a boundary. In this case, the boundaries of the conjugacy classes do not belong to that conjugacy class. But you might have some situations in which the boundary belongs to that conjugacy class, but it's still the boundary. So it means that the boundary belongs to the conjugacy class, for example, on the left, but not the one on the right. You might have situations like that. So if you're in the interior of your conjugacy class, you're structurally stable. If you're on the boundary, you're not. This is another way, equivalent way, to think about structural stability. So is your map in the interior of its conjugacy class? And to define that, you need a topology on the space of system to talk about being in the interior of your conjugacy class. You need a topology that tells you what it means to be in the interior of your conjugacy class of maps. Okay? And we will see several examples now. Okay, I think maybe this is a moment to take a very quick two-minute break and have a glass of water. Okay. So now we are going to remain in a one-dimensional setting because it's a simpler setting to illustrate some concepts. But we're going to take the step beyond linear maps to nonlinear maps where you don't have such a simple and obvious parameter space of families. So we're going to now consider interval, study interval different morphisms. So we suppose i is a compact interval. So generally you can just think of i equals 0, 1, for example. But it really doesn't matter. Everything will... Sorry? Yes, yes. So it doesn't really matter. We will study the dynamics of maps on compact intervals. Okay? So if you have a map on one compact interval, just up to the scaling is the same as any map on any other compact interval. It doesn't make a difference. And we take f from i to i, a different morphism, c1, different morphism. So this is interval i, interval i. This is just a map from i to i. Remember, different morphism in particular needs to be a bijection. In general, the different morphism will look something like this. The map will look something like this. So just a few observations about interval different morphisms. First of all, notice that the end points are mapped to the end points. So end and points. Also, since it's a different morphism, by the same reason we said before, we have that f' is not equal to 0 for all x and i, because otherwise the inverse would have infinite derivative and that cannot be. It's not part of... It's in the definition of c1, different morphism that that cannot happen. And so this means that either f' of x is greater than 0 for all x and i, or f' of x is less than 0 for all x and i. So you can only really have one of these two cases. In this case we call it orientation preserving, and in this case we call the map orientation reversing. If a map is orientation reversing, then the second iterate of that map is orientation preserving. So in some sense the orientation reversing maps are kind of special and there's a very simple classification of these, and I'm going to leave this as an exercise. So the orientation reversing dynamics of orientation reversing maps is really simple. There's not much that can happen. We leave this as an exercise. Whereas for orientation preserving there's a much richer structure. So from now on we're going to study always the orientation preserving case. All the iterates are always orientation preserving, so it's a kind of a closed class of system. What do we want to say about orientation preserving maps? If p is a fixed point, if f of p is equal to p, we say p is a hyperbolic fixed point. If f prime of p is not equal to 1, which is completely consistent with the linear case. We do not include the minus one here because we're assuming that we're orientation preserving the derivative is just 0. The derivative is positive. So if all fixed points of f are hyperbolic, then we say that f is hyperbolic. So what is the dynamics of interval diffeomorphism? So we start with the following proposition. f i to i, orientation preserving, c1 interval diffeomorphism. For all the limit sets omega x and alpha x, the alpha and omega limit sets are fixed points for f. Shall I prove it or leave it as an exercise? I will sketch the proof. It's a fairly simple exercise actually. But the conclusions are non-trivial. The conclusions are fundamental to the theory of interval diffeomorphism and show also that interval diffeomorphism have a particularly simple kind of dynamics. So here we're starting with simple systems where not much can happen. And then we will go and we will see that the systems with this very complicated omega limits. So basically what this says is that the only thing that a point can do if you choose a point x and you iterate it, it must converge to a fixed point, both in forward time and in backward time. That's what that says. So it cannot do strange things. It must converge to a fixed point. And that basically follows from the fact that this map is monotone, monotone, monotonically increasing. And so the sequence of points along the orbit is monotone increasing or monotone decreasing. And a monotone increasing sequence must converge to some point and the only thing that really needs to check is that that point is a fixed point. So we'll just sketch briefly the proof here, but it's very simple. So let x in I, if x is a fixed point, then what? Then there's nothing to prove. Then it's obvious because for a fixed point, the omega limit set of a fixed point is itself and the alpha limit set is also itself if it's an invertible map. So if f of x happens to be already a fixed point, then this is clear. Then alpha limit of x equals the fixed point itself, which is equal to omega limit and nothing to prove. Well, and that's the proof. Okay, that's fine. So suppose, so if f of x is different from x, then either f of x is greater than x or f of x is less than x. Those are the only two possibilities. If f of x is greater than x, then f of x minus x is greater than 0. By the way, what do these possibilities correspond to geometrically? It's always useful, especially in these one-dimensional maps, where you can gain a lot of intuition from looking at the graph of a map. So every different morphism of the interval, orientation preserving basically looks like this. The derivative is positive. It has to map end points to end points, which means that the end point, if this is the interval a, b, or 0, 1, it must map a to a and b to b. In the orientation reversing case, a would map to b and b would map to a. So it would go like this. It still has to map end points to end points. In the orientation preserving cases like this, a maps to a, b maps to b, the derivative has to be bigger than 0 everywhere, and so it has to do something like this. It can be like this, or it doesn't have to intersect the diagonal, of course. It could be just something like this. It could also be something like this. In fact, basically, these are the three kinds that we will consider. In these two cases, these are the only two fixed points, right? Because remember that a fixed point, okay, crucial observation. When you look at the graph of an interval map, a point is fixed if and only if the graph intersects the diagonal. This is clear, right? Because the diagonal is the line where y equals x, so it's a fixed point precisely means that f of x equals x if the graph intersects the diagonal. So you can immediately see from the graph where the fixed points are. Okay? Here it intersects the diagonal and here it intersects the diagonal. These are fixed points. It could be that these are the only two fixed points. For example, if you have this situation, or this situation, or it could be that you have several fixed points. Okay? It could also be that you have an infinite number of fixed points. What is an example of an interval different morphs with an infinite number of fixed points? Identity. Identity. Not of course. The identity is the best counter example in so many situations. Okay? Always try with the identity first. Whenever you're looking for some counter example or something, try the identity. Okay? Because otherwise you start constructing very complicated construction where the identity works. Okay? The identity works. Or you can have just something that is the identity in some piece. The identity means that it exactly coincides with the diagonal. That gives you an infinite number of fixed points. So you can try to construct something that oscillates lots, or you can just construct something that goes along the diagonal for a certain, you know, certain piece and that gives you an infinite number of fixed points. That is the identity even just on some piece. Okay? So in this case is the map hyperbolic. What is the derivative of these fixed points? One. One. So the map is not hyperbolic. These fixed points are not hyperbolic and the map is not hyperbolic. Okay? So I'm just anticipating one of the reasons to introduce this definition of hyperbolist in this case just as in the linear case is to exclude certain very unpleasant situations like this one. In fact, we will see that if this map, if all the fixed points are hyperbolic they can only be fixed points. Because if the fixed point is hyperbolic then it must cross and it cannot be accumulated by other fixed points. So it needs to be just a finite number. Okay, but that was just a little parenthesis. This is just to give some idea. So what does it mean here that f of x is bigger than x in the graph? For example, let's look at just this, this, okay, look at this graph here. Okay? For this point. This is x and this is f of x. Is f of x bigger than x or smaller than x? Bigger than x. How can you tell? Because the graph is above the diagonal. Right? So you can immediately tell for each point if you look at the graph whether f of x is less than x or bigger than x. So if instead you were looking at the pink curve let's take this point x here and then f of x is here and because it's below the diagonal this means that f of x is less than x. Okay? So here you can see exactly the situation that we described here that either x is already a fixed point or if it's not a fixed point clearly it must be either below the diagonal, the graph must be either below the diagonal above the diagonal which corresponds to those two cases that we saw before. Okay? So we suppose that in fact the two cases are exactly the same. So just for definiteness suppose we take this case what does this mean? It means that f of x minus x is greater than 0. So for example it could be let's just stick to one map maybe so this is the diagonal let's suppose the map looks something like this so let's suppose we take a point x here where the graph is above the diagonal so what's going to happen to the next iteration can we say something about what happens to the next iteration well where is this point here in your dynamics, in your dynamical space so this is the dynamical space a, b this is where initial condition x is 0 and this here is x1 the image where will x2 be what's the image of x1 excuse me that's right, that's right and how do we see is that a graphical representation so we're going to do it in two ways we're going to write down formally what we know but I think it's also very useful to understand how to look at it in the picture you want to know find out where x1 is here because when you know where x1 is here then you can look at its image by looking at the graph so how do you carry x1 from the vertical axis to the horizontal axis you just reflect it in the diagonal so geometrically you look at x1 and then you take a horizontal line all the way to the diagonal and then you take it down like this and this will be x1 and then you can look at the graph and you say ok what's the image of x1 but the image of x1 is here so x2 will be somewhere here but having x2 on the vertical axis is not much good for us what we want to know is where x2 on the horizontal axis so we can do the same thing we take x2 and I don't need what's the point of me starting from this point going all the way to the left and then going back all the way to the right so just from here I can just go directly down to the diagonal and what's the point of me now coming all the way down and then going all the way up again I can just start from this point and I can go up again so let me just draw a picture of what we're doing I'm just taking a little parenthesis to show you how you can analyze this graphically so let me zoom in to this region here the map looks like this and what am I doing I start somewhere under here and I end up with this point here which is this point here and then what I did is I said this is x1 now you come back and you reflect in the diagonal and then you go down reflecting in the diagonal means finding this point here and then I go down to x1 but then what do I use x1 for the next image which will be this point on the graph here which is this point here and now this corresponds to x2 so I take x2 and then I go back and I reflect it in the diagonal which means I move to the diagonal here then in principle I go down and I find x3 down here and then I look for the next image I come back up and go across to the graph again so really the itsets of the map you can just kind of graphically draw a picture like this which shows you that this point is moving from here to here to here wherever these images are from here to here and it is converging to this end point so this is a very useful graphical analysis it does not prove anything because it is just a picture but it helps you to get a feeling for what is going on so because the graph is always above the diagonal then the image, the next image is always bigger than the point that came before okay and so you have a monotone increasing sequence okay so repeating this argument so let's write f of let's write this as x1 minus x0 greater than 0 repeating this argument we get x2 greater than x1 okay and so the orbit xn n in n which is equal to n of x0 sorry n in n is monotone increasing is monotone in forward time so what can you say about a monotone increasing sequence of points on the real line this is a sequence of points this is monotone increasing it's also bounded that's right it's bounded before it converges that's why we're interested here we're in a compact interval before this is an increasing sequence so every monotone increasing sequence xn which is bounded converges to some p in i it's monotone increasing and bounded so xn converges to p which is what we're seeing in this picture because you're just increasing increasing increasing there's no thing else that you can do except converge either you go to infinity or you converge to something what is left we need to show that this point p is a fixed point we need to show that it doesn't converge to something since we know that this is the only fixed point we need to show that it actually reaches the fixed point because it's monotone increasing but maybe it converges to some other point here that comes before the fixed point okay we just need to check that we need to show that it converges and it comes all the way here so we need to show that f of p equals p sorry? why by construction? it's that and why might that not be the case so your observations are correct so it's almost trivial but it's not completely we're using so you're saying it converges to p and now you need to show that the fact that it converges to p implies that f of p equals p is the upper bound and so in this particular case p is the end point of the interval but how do we know it does not stop here somewhere before that f of x but we don't know that f of p is f of xn is a subsequence of xn and so by the continuity of f okay this is what implicitly you were using this fact but this is a crucial ingredient here okay you need to use the continuity of f exactly so how do we write this out so the continuity of f plays the crucial role here so what we have is that f of xn the continuity of f we have that f of xn converges to f of p just by continuity okay but f of xn has the same limit as f of xn but f of xn equals xn plus 1 and so the sequences and so the limit of as n tends to infinity of xn is the same as the limit as n tends to infinity of xn plus 1 which is the same as the limit as 10 tends to infinity of f of xn okay and the limit here is p and the limit here is f of p it's just continuity okay so the integration is completely correct but the formula depends on the continuity of the map f okay so this allows us to study the conjugacy classes of these maps because this is of course our goal we want to classify all these interval diffeumophism and classification means understanding when are two interval diffeumophism's conjugates so what do you think let's see let's start with the simplest cases so conjugacy classes so first of all just heuristically of course what this proposition says is that the fixed point play a crucial role because diffeumophism has a certain number of fixed points every other point ends up in one of these fixed points so the fixed points determine the long term asymptotic dynamics of every orbit because every orbit in backward time and in forward time ends up in a fixed point so the fixed point play a crucial role in classifying these systems and we will see this more specifically so let's start by looking at two maps that have that both have just two fixed points so let f i to i and g c1 orientation preserving diffeumophism suppose both f and g have only two fixed points so what does it mean that I have two fixed points because we saw that the end points are always fixed then it means that one this and the other one maybe is also like this notice that you do not need to be defined on the same interval so this is going to be conjugate topologically conjugate differentially conjugate do they look conjugate to you could they be conjugate topologically conjugate makes sense that they're topologically conjugate why would you say how would you convince me without a formal proof the intervals are homomorphic as intervals but now the conjugate needs to conjugate the dynamics as well right it's not enough for the spaces to be homomorphic the homomorphism needs to conjugate the dynamics so we need a homomorphism that maps h composed with f equals g composed with h so we need to map orbits to orbits so you would clearly naturally map these two fixed points to the corresponding fixed points and then can you have a homomorphism that maps every orbit here to every orbit here that's right so what are the alphan omega limits of points in here and which one is it in this case why do you think who can be more sure yeah it's obvious you don't need to think it's obvious by the argument we just did you take a point x here its image is above the diagonal so it increases right this image is above the diagonal so it increases all the points are moving to the right and therefore they converge to this end point here and the alpha limit is this end point here and the same for this right all these points are moving to here and all the points are moving to here okay so the gut feeling is that this should be topologically conjugate because you think okay you know it's not exactly the same I mean you think that you should be able to take all the points these similar things a little bit like in the linear case when everything was going to zero you think they should be topologically conjugate and this is correct okay we will construct the conjugate here what about if this case here is is doing this are these still topologically conjugate why not why do you keep thinking instead of being sure yes and so what does not preserve what does not preserve the omega limits this homomorphism so you are saying correctly that if this homomorphism maps x to a point y then it needs to map the omega limit of x to the omega limit of y right so why does it not do that when you say well the omega limit of x is this point and the omega limit of y is this point so why can h not map this point to that point sorry this point to this point can it but can you have a homomorphism that maps this interval to this interval that maps this point to this point and this point to that point that flips the interval of course you can have a homomorphism it's an orientation of reversing homomorphism because it switches the order but it's not a problem it's just switch the order that's still a homomorphism it's not part of the definition of the conjugacy there's no particular requirements on orientation in general between two topological spaces you don't even have the definition the notion of orientation so it's just a homomorphism between these two topological spaces and there's no reason in this case why it cannot switch them around in which case it should be we need to check we need to prove this but there's no reason a priori why we couldn't imagine that these two are also topologically conjugate although we need to prove we need to check that but okay so keep this in mind so we don't have much time left so I don't think I will be able to finish the proof here okay so let me postpone the construction to the next lecture but let me first say that we're going to use the same construction as we did for linear cases we're going to use fundamental domains right so it would be a very good exercise if you want to try to do it at home it's very easy in the same way because every point here is moving from here to here for every point the alpha limit is this and the omega limit is this so you should be able to find which is a fundamental domain exactly in the same way as the linear case so any point in backward time will go here and it should spend just one iterate inside this fundamental domain and then another fundamental domain here in the same way and then you map by homomorphism the two fundamental domains and you extend it exactly like we did for linear case and we construct this topological cone so we check that it's what we get is a homomorphism so I will do some or at least I will sketch some part of the construction but it is exactly the same as we did for the linear case except for you don't have an explicit form when you need to explain why this is a fundamental domain it's not just multiplication by A or B you need a slightly more sophisticated argument to show that that's a fundamental domain that's the only difference excuse me sorry I didn't hear you yes it does work the definition is the same it's just to show that well the argument is almost exactly the same just to show that you cannot jump so here you take some X tilde here you take F of X tilde this is your fundamental domain closed on one side open on the other side and everything is exactly the same just to show that you cannot jump and that if you jump you land only once basically exactly the same when we did it for the linear case we used a little bit the fact that it was exactly multiplication by A but it's almost exactly the same argument so since we just have a couple of minutes let me say first what about C1 conjugates assuming we can prove that these are topological conjugates what about C1 conjugates are they C1 conjugates exactly so certainly if the derivative at these two fixed point is different then they cannot be C1 conjugate so for this to add there's not work exact in particular in this case it does not work so this is another example of how strong the C1 conjugates is so two maps even if they look very similar even if the so if this map is like this okay even if it's even if the two maps coincide almost at every point but at the fixed point they're a little bit different the derivatives are different the maps cannot be C1 conjugates okay just to show how strong it is so we shall see that in the space we still need to put a topology on the space of interval the film of this but instinctively you can see that if we will show that all the maps that look like this basically topologically conjugate then you will have a reasonable large conjugacy class which contains a lot of maps and it is also reasonable in the sense that these maps all have pretty much the same behavior all the points mapped one of these two fixed points on the other hand differentiable conjugacy you can still have two maps that are different but in the same differentiable conjugacy class so this is different from the linear case so in these two cases we will not prove it but you can prove that if they are if they do have the same derivative then you can construct a differentiable conjugacy between here so you can have maps that are different but in the same differentiable conjugacy class but you can also have two maps that just differ a little bit in the different differentiable conjugacy class so again the topological conjugacy class seems like a more natural and robust definition of equivalence okay so in the next lecture we will do this construction we will also look at maps with several periodic points okay although that again will be almost the same and then we will define the topology on the space of this interval the thermomorphism that allows us to address the notion of structural stability okay thank you