 So, good afternoon. Today I want to complete this chapter on piecewise expanding maps. So we've studied piecewise expanding full branch maps, right? We've studied the coding, and now we come to the, as in the other topics, to the notion of topological conjugacy and structural stability, right? We've studied the topological conjugacy between the symbolic system and the full branch piecewise expanding map. Now we want to discuss when two piecewise expanding maps are topologically conjugate. And when a piecewise expanding full branch map is structurally stable. So let me start with a lemma, which should be fairly straightforward, is that these two maps, let's consider the map f from i to i, g from j to j. I could write the formula, but I will just draw the pictures. Suppose this is zero one, zero one, and let's draw, let's suppose one of the maps. Is this one 2x mod one? And the other one is just the tent map. So this is an issue we've described before, which is the fairly subtle issue about the orientation of the various branches and how that affects the topological conjugacy class of the map. So are these two topologically conjugate? So we know they both have a semi-conjugacy to the shift again, right? Because we have constructed, in this case, it's a semi-conjugacy because of the boundaries and so on. Does this semi-conjugacy match up? So this semi-conjugacy is not a bijection, it's not injective, and this is not injective. So in principle, it's not clear whether you can go through the symbolic dynamics to construct the conjugacy. Maybe you can or maybe you cannot, right? If the points where it's not injective here coincide exactly with the points where it's not injective here, we might be able to actually use this to construct the homomorphism between these two. But in this particular case, there is a very simple reason that you can see nothing to do with the conjugacy. There is a very simple reason that shows you that these two maps are not topologically conjugacy. Are not topologically conjugacy. Can you see why that is the case? So there is no homomorphism that conjugates the dynamics here. Can you give me what the several reasons? But there's one very simple one. Very simple obstruction. What is the most fundamental property of topological conjugacy or even just a conjugacy? What does it preserve? Fixed points. Exactly, OK? So there's two fixed points. In both cases, there's two fixed points. One fixed point is here, so this is the diagonal. And the other point is here. So this has two fixed points, 0 and 1. This also has two fixed points. 0 and this fixed point here, p. And a topological conjugacy must map fixed points to fixed points. And a topological conjugacy must be homomorphism between 0, 1 to 0, 1, between the spaces. So homomorphism from the interval i to the interval j, which in this case are the two intervals, must map the end points to the end points. It's a homomorphism of the interval. Any problem with that? OK, so your question is that one is discontinuous and one is continuous. How can they be topologically conjugate? This might be an intuitive reason for which you might say, I don't think it's topological conjugacy. What we want is a clear mathematical obstruction that proves that they're not topologically conjugate. Yours is an observation that seems to think, well, I don't think it doesn't look like it. So you're saying that if we, that's right. OK, that is a good observation. Yes, so if we could find a homomorphism, we would be able to write f composed with h equals h composed with f. And if this was a homomorphism, you'd be able to write f equals h composed with g composed with h minus 1. And these are all continuous. And so this should be continuous. And it's not. Yes, I think that is a reasonable argument. My argument was, in some sense, even more elementary than that one was just this observation that this map has two fixed points at the end points. Any homomorphism, if it's a homomorphism, h is supposed to be a homomorphism between i and j. So a homomorphism of i to j means that it maps the interval i to the interval j. And in particular, it maps the end points of the interval to the end points of the interval. On the other hand, it must also map the two fixed points to the two fixed points. And it cannot do that, because these two end points, which are the fixed point for the f, must map to the end points here. But one is not a fixed point. The fixed point, in this case, is in the interior of the interval. OK, so I think this is a reasonable argument. This is also a reasonable argument. As I said, there's many ways. There's many reasons for which these cannot pitopologically conjugate. And these are two reasons for which it's true. So the question is, what is exactly, in which cases, can to full-blanch piecewise map pitopologically conjugate? For example, if this, instead of this, if we had, for example, the tent map, but we had these partition elements was not in a half, but we just took some constant C here, and we constructed, and we defined this map like this, are these two topologically conjugate? That is what the topological conjugacy should do. Yes. So we need an argument for that. We need to actually prove that there's a conjugacy there. So we will do that. I will sketch the proof because most of the steps are the same. We've done similar calculations now before. But basically, yes. So let's try to state. So, OK, good question. So my argument about the fixed points, so here we have, which fixed point we have? We have a fixed point at 0 and another fixed point here. So of course, the position of the fixed points is not preserved. This is another fixed point Q, let's say. So if you were to construct a conjugacy between here and here, this conjugacy would need to be a homomorphism that maps 0, 1, to 0, 1. And it would map the end point 0, to 0, 1, to 1. And of course, it would have to map P to Q. But that's fine. Well, this is just a homomorphism. It's not even saying that it's a linear rescaling. In fact, in general, it would not be a linear rescaling. Just because these are piecewise affine maps, that doesn't mean that the homomorphism would necessarily be just a linear map, right? Or affine map, or affine rescaling. But all it does, it's a homomorphism. And this is perfectly possible for it to map P to Q. The distance as a homomorphism can change distances completely. So as we will see in the proof and as you're mentioning, it is almost immediate that if you're going to construct this homomorphism, probably this homomorphism is going to map 0, 1, 1, to 0, C. And 1, 1, 1, to C, 1. Because really, all we need is basically this homomorphism will map the combinatorics, right? So each point here has a certain combinatorics. If you call this i0 and this i1, and if you call this j0 and you call this j1, in this case, the j0 and j1 have different lengths. But all the construction that we've done before, remember when we did it, in our constructions, we didn't say that all the intervals i0, i1 have to have the same lengths. They will all have different lengths in general. Everything works in exactly the same way, right? So here we get points that have all possible combinatorics of 0s and 1s. Here we have also possible combinatorics. And the homomorphism will just map all the points with the same combinatorics to the map to the points with the corresponding combinatorics. So let me state it, and then we'll look a little bit at the sketch of the proof. So proposition. So let f and g be full branch piecewise expanding maps with the same number of branches, with the same orientation, such that corresponding branches have the same orientation, right? Exactly like in this case. So in this case, all we care about is the number of branches and the orientation on each branch. So on this branch, the orientation is positive. On this branch is positive. And on i1, the orientation is negative. And the orientation is negative, which is not in this case. Then f and g are topologically conjugate. So just to finish this observation about the counter examples, why do we need the same number of branches? If you have one with two branches and one with three branches, exactly, right? Because full branch maps, clearly, if you just look at the picture, they have exactly the same number of fixed point as the number of branches, a fixed point in each branch. So you have to have. So as you can see from counter examples, you kind of need both of these conditions, same number of branches, and the same orientation for the reason that we have done before. However, the good thing is that as long as they do have this, then they are topologically conjugate. So the proof we've already started discussing the basic idea of the proof. So just to have some example in mind, let's suppose we have, say, three branches. In the pictures I've been drawing so far, I've been drawing the branches about equal size. But maybe that's a bit misleading. Remember that nowhere in the assumptions is that, right? So now let me draw them just to remind you to emphasize this fact. I draw them on very different sizes. So it doesn't matter which how I choose them. But let me just draw them like this just to show that it doesn't matter whether they're continuous or not continuous. So here they have to have the same orientation, which means that the graph basically looks similar like this. So this is F, and this is G. So basically define, so we construct a map H. Let's suppose F is from I to I, and G is from J to J. We construct a map H from I to J, mapping points in I to points in J with the same combinatorics. So what does that mean, and is this well-defined? So first of all, we let, so more specifically, since the branches have the same orientation, then the end points of F and G have the same. So what are the combinatorics of the end points here? What's the combinatorics of 0 for F? 0, 0, 0, because this is a fixed point, right? What is the combinatorics of 1? Not 1, 0, 0, 0. In this case, 2, 0, 0, 0. In the general case, it would be L minus 1, 0, 0, 0, 0. So this depends only on the orientation of the first branch and the last branch. But this determines exactly, if you remember, the orientation of the first and the last branch determines exactly the combinatorics of the two end points. So as long as these two maps have the same, the orientation of the first and last branch, then what would be the combinatorics of 0 here? It's the same, 0, 0, 0, and the combinatorics of this point is also the same, which is L minus 1, 0, 0, 0, 0. So of course, to construct this homomorphism H, this map, this bijection, we start by mapping the end points to end points and notice that these have the same combinatorics. And then we will continue this way, matching up the combinatorics slowly. So what is the combinatorics of these end points here? So we will map clearly, then we will map then, sorry, of the intersection point, then map I star to J star in the obvious way by mapping corresponding intersection points to each other. I think that's fairly clear what this means, right? So it means that because we can order all these intervals, even if there's 1,000 intervals, we have them all in a row. We map the interval has an orientation. We map the first one to the first one, the second one, the second one, and so on. And so what is the combinatorics of this point? Remember that each of these points has two possible combinatorics. What about these points? Also I have two possible combinatorics, but they're the same two possible combinatorics, right? So this point here has a combinatorics that is given by the fact that here, so this is I0, I1, I2, in this case, and this is J0, J1, J2. So this point here has a combinatorics that can either start with 0, and then it will have 2, and then it will have 0, 0, 0. Or it is 1, followed by 2, followed by 0, 0, 0. You see what are the two possible combinatorics in this case? You see that? This point here, you can either think of it as belonging to I0, or you can think of it as belonging to I1. So the first symbol of the two possible combinatorics of this point intersection point can be either 0 or 1. But after that, it's fixed because whether you think of it as belonging to I0, belonging to I1, its image is 1 at this point here, and we already know that the combinatorics of this point is just, in this case, 2 in the general case, L minus 1 if we had the general picture, followed by 0, 0, 0. And this has exactly the same combinatorics, this point. So this combinatorics also has the two possibilities. You may consider this part of J0, or part of J1, but then its image is 1, which has exactly the same combinatorics. So the combinatorics, given the symbolic coding of these two points, is the same for the two maps, and therefore it's natural to map. That is why we map it to each other, because we're matching the combinatorics up. And the same thing for this. So we take this as an opportunity to revise. This is what we studied in the previous lecture. But this is an opportunity to try to make sure that you review and you understand what I mean by the combinatorics. So what is the combinatorics? You tell me, what is the combinatorics of this point here, the possible combinatorics of this point? There's two possibilities, right? Yes. Yes. So it's either 1 and then 0, 0, 0, 0, because that's the combinatorics of 0, or then 2 again and then 0, 0. So this point has two possible combinatorics, either 1 followed by all 0s, or 2 followed by 2 followed by all 0s. And this is completely determined by the orientation of these branches, and therefore this point has exactly the same two possibilities. OK? So let me write this observation down. So notice that points in i star have the same combinatorics. By this I mean the same two possible associated sequences, the corresponding points for j star. Now notice that all the pre-images of, recall that i star hat and j star hat are the sets of pre-images of i star and j star. And notice, of course, that because so this is a crucial observation, so since all the branches, since corresponding branches have the same orientation, then the points of then corresponding points, i star hat and j star hat, are ordered in the same way in i and j. So what does this sentence mean? What do I mean by they're ordered in the same way? And what do I mean by corresponding points? Yes, I'm not going to try to repeat that question because it was, but you're kind of floating in the right direction. So first of all, the meaning, just the meaning of ordered in the same way means that if I take two points and one is to the left of the other on this interval, then the corresponding points will have the same ordering here, one will be to the left of the other here. And what do I mean by corresponding points for j star? What do I mean by corresponding points here? I mean points that have the same combinatorics. All the pre-images, what does a pre-image, what are the points in here look like? Well, the points in here, the points that belong to the interior of one of these elements. For example, you take a point here, x. And after some time, they fall onto one of these intersection points. So before falling into one of these intersection points, they have a certain combinatorics given by the sequence of elements of the partition that they fall into. So the corresponding point here would be simply the point that has exactly the same combinatorics. So that before falling onto j star, it has exactly the same combinatorics. So if this point here goes 1, 1, 2, 0, and then it falls on the intersection, there is a unique point here that has the same combinatorics, 1, 1, 2, 0, and then it falls on the corresponding intersection. So that's what I mean by corresponding points. If I take here two points nearby that belong to i star hat, and both of them fall onto some intersection after some point, and one is to the left of the other, and then I take the corresponding points here that have the same combinatorics as this, then they will be also the corresponding point of the one that's on the left will be on the left, and the corresponding point that's one on the right. And of order preserving, the combinatorics, the order of points on the combinatorics preserves the matching combinatorics is all well aligned. If you remember, I mentioned this already in the previous lecture when I talked about what happens if the branches are not the same, then you see that the symbolic shift is, the symbolic sequences are embedded in the line with a twist in some cases. Remember I mentioned that? In this case, you do not have that, right? If you had a different orientation here, then what you'd get is you'd get two points with the same combinatorics, but because they both land in the same place after a while, they might be twisted. Here they might be one to the left and one to the right, and here one to the right and one to the left, if this had the different orientation from this. But because they have the same orientation, then this is two. So when you revise this, think carefully exactly about what this means, and you see that this is two. And the same is true also for all other points that do not belong to I star. So the same is also true points in I at star or J minus J at star, okay? What I mean is that corresponding points, IE points with the same combinatorics have the same order on the intervals. So what is the conclusion of all these observations is that there is a well-defined bijection. So the bijection actually we proved already before this observation, as soon as we proved that the combinatorics of points in I star match up exactly with the combinatorics of points in J star, that already allows us to define a bijection between all the points here, because if a point is in I star hat, then it has two possible combinatorics, but they match up exactly here and here. If a point is not in I star hat, so it's here, then it has a unique combinatorics, and therefore there's another point here with the same unique combinatorics, and therefore we can map a bijection that matches up the combinatorics exactly, okay? The additional information, the reason for all these observations about the order is that therefore H from HI to J, which maps points to corresponding points with the same combinatorics, is an order-preserving bijection. It's well-defined, and it's an order-preserving bijection from I to J. Do we know that H is a conjugacy? So now we need to show that H is a conjugacy. Why is H a conjugacy? We've got a bijection, we need to show that it's a homomorphism, and that it's a conjugacy. So that it maps the dynamics. It preserves the dynamics. How do we know that H is a conjugacy? By construction, H maps the points with the same combinatorics to points with the same combinatorics. That's right, exactly. So that follows exactly, yes. So it is automatic by construction, right? Because we are mapping the combinatorics to each other, and we know that the dynamics is conjugate to the shift on this combinatorial information, basically. So if you take a point X here, and you map it to a point Y here that they have the same combinatorics, then the image of f of X, we know what the combinatorics of the image of f of X will be, and we know the combinatorics of the image of f of Y, of G of Y will be, and they will have the same combinatorics, and therefore H will map the points to each other. Right, so it's clearly, H is a conjugacy because it maps, because it preserves the combinatorics, because it preserves the combinatorics, okay? You can make sure you understand this. If I write down all the details, then I don't have any questions to ask you in the exam, right? So, and finally, that H is a homomorphism, H and H minus one are continuous. I will leave this as an exercise because it's also quite simple and very similar to the arguments we've used before. So again, you use the combinatorics. If two points are close, then the combinatorics is the corresponding sequences are close for a long amount of time, and if they're close for a large number of digits, then these have the same combinatorics, so these also are close because they have sequences that have a large, whose sequences, their corresponding sequences are the same for a large number of digits, okay? It's a by now standard argument that we've used before several times. Okay, very good. Any questions? So we have now that we have a criteria for the topological conjugacy between full branch expanding maps. So what do we usually do after we know, after we have characterized in some sense the topological conjugacy classes, right? So what we've done now, if we know that the topological conjugacy classes of full branch piecewise expanding maps are basically determined by the number of elements, number of branches, and the orientation of these branches, okay? Which means that you can more or less classify. If you fix number of branches, then you look at all the possible orientation and you get all the possible equivalents, all the possible conjugacy classes. So often what we're interested in is the notion of structural stability. Remember what structural stability means? Yes, we want to perturb these maps. So the question is how do we perturb such a map? It's not completely obvious what we mean by a perturbation of this map. And in mathematics, the nice thing about mathematics is that you have to write in some sense to control your definitions, right? So we could, if we wanted to decide what do we mean by perturbation? So for example, given this map, what is a small C1 perturbation of this map? Well, I could, for example, decide that I want to fix these three intervals and just inside each branch, I can perturb the map in a C1 way, okay? Would we get structural stability with this definition? So my definition is I'm saying, okay, a small perturbation is one in which I keep exactly the same partition. And then here instead of this, so let me actually draw the picture a little bit bigger because I want to discuss this a little bit more. So my question is what do we mean by a small perturbation? Suppose I have this I zero, I two. Let me just, as an example, take the same map that we had before. Okay, now again, I'm sorry if I always draw these as straight lines, I don't mean them to be straight lines, right? The theorem, these are just examples, but the theorem full branch piecewise expanding only requires the derivative to be bigger than some lambda, right? We don't need them to be straight lines. So now I take a small C1 perturbation of this, which means I can make this, I can change the graph a little bit, but I always work within these branches. So I still assume that the same interval I zero is that this is a full branch piecewise expanding with the same intervals. So I still suppose that this endpoint and this that I zero maps to everything and so on. Is it structurally stable? Exactly. We have the new map has the same number of branches, the same orientation, and it's still expanding, right? Because if the derivative here is strictly bigger than some lambda bigger than one, if I take epsilon small C1 perturbation, the derivative is still close. So it's still bigger than one, uniformly bigger than one everywhere. So I get a new map that is still expanding and so on and satisfies, so it's structurally stable. So this is a little bit restrictive to assume that we keep the same ones. What happens if I allow these also to change a little bit? If I say, okay, I can perturb a little bit and also I can change the position of these boundaries, then also it will be okay, right? So if I say, okay, a C1 perturbation means that I have still the same number of branches and maybe the new interval I zero prime is close, the boundary points are close to where they were and I still need to have the new boundary point to map to everything. But you can see the definition is not starts getting not so nice, right? I mean, there's no real solution to this. I'm pointing out the fact that it's not a very nice class of maps on which to define the notion of a C1 perturbation, but there's various ways to do it, okay? You can fix the intervals and say that you perturb inside the intervals or you can allow the intervals to vary a little bit and then you still say it's a small C1 perturbation. The simplest setting, so the way I will formulate it is like this. I will apply it to a slightly different class of maps. So I will say let F be a map from the unit circle to itself and suppose this is C1 and suppose this is F prime of X is greater than equal to lambda greater than one for all X and C in S1. I will say that then F is structurally stable. Is C1 structurally and how is this related to piecewise expanding full branch maps? Because hidden in the condition that F is a map from S1 from the unit circle to itself and the fact that its derivative is bigger than lambda implies that it can be viewed as a full branch piecewise expanding maps. So let's look at this relation. So we take the circle S1 and we have a graph of S1 to S1, a map from S1 to S1. So how do we represent this? Well, we can take an arbitrary point here and we can cut, we can write this circle as the unit interval. With the end points identified, this corresponds to this point here on the unit circle and we just draw it like this, okay? And then we draw the graph of the map as a map from the circle to itself. We draw it here, this is zero one and we just remember that this point is the same at this point. What does that mean that this point is the same as this point? It means that the graph, so we now draw the graph of F, right? So the image of zero might be any other point, it might be this point here, okay? Then what do we know about the graph? Well, we know that the graph is increasing because the derivative is everywhere lambda strictly greater than one. Therefore it can never fold back on itself, okay? It needs to increase. Now this point here, so it must reach the top at some point, but this point here one is identified as zero because really we're working on the circle where zero and one are identified, right? So that means that when the graph reaches this point, really it is the same as being on this point here. And so the graph continues and it's still growing like this, right? And then what happens? It grows, suppose it hits the roof again here and then it grows and then what do we know about the image of the graph at the point one? It needs to be the same as the image of the point zero because really this is a map of the circle to itself and these two points identified, right? So this must hit if you want the end point, the image of the end point one must be the same as the image of the end point zero. Now this doesn't look like a full branch piecewise expanding map to me or is it? It's not full branch, right? It's got three intervals, but unfortunately two of them are not full branch. Do we have fixed points here? Okay, so how do we turn this into piecewise expanding full branch map? Here, can anyone see this? How many pre-images does every point have? This is not a bijection, this map, right? It's a non-invertible map. How many pre-images does every point have? Two, at most or exactly? So this point here has how many pre-images? Two, and this point here? Two, right? This point here has also two because this is the same point as this point, right? So it has this pre-image and this point is a pre-image. Zero goes to this and this point here goes to this. It also has two pre-images. So every point has exactly two pre-images. So this is a two to one map on the whole circle. So maybe this is not the right graph. This is not really the right picture. What is wrong with this picture? So what is wrong with this picture is that we cut it, we cut the circle in the wrong place. This was not a very clever choice of a point to cut the circle because suppose that we cut the circle in this fixed point here, okay? So somewhere here there's a fixed point, P. Let's say this is the point P. So we could have used this. So we could have used this as the place to cut the circle and to draw this as an interval with these two points identified, these two end points identified, okay? If we draw on this picture here, then what we would have if we'd cut the circle at the fixed point, then here we would have the fixed point. So that means that the graph, so this is now the new interval, let's say, these are the two end points. So this would be the fixed point. So then the graph would start from this point here, right? And then what would it do? It would increase by lambda and then it would increase. It would look like this. So we are really, we are now drawing the graph because we are shifting this, we could cut it anyway. So now we cut it here, right? And if we cut it here, this is the new initial point. So zero goes to the fixed point, goes to the fixed point. And then if you think about the graph, what it must do, it must still be two to one because it's exactly the same graph and this is the way it's going to be two to one. It's going to go up and it's going to go all the way once down the circle when it gets here and then it's go, the other side is gonna go, the other part is also going to go down. Notice that it still only has one fixed point. Why? It looks like two fixed points. It looks like it has one fixed point here and one fixed point here. But yes, but they're the same fixed, they're the same point, right? So it just has one fixed point. So another way to say it, a more kind of topological way to say it is that every uniformly expanding map of the circle is a covering map. This is what a covering map is. A covering map of a space of a topological space or a manifold is a map that maps the space to itself a certain number of times exactly. And you can see that you can play this game with any number of branches, right? If instead of this picture here, you had got, this might have been a covering map with many more branches. So I just do it like that. What I know, remember that the only assumption I start with that is C1 and it's uniformly expanding. So I could have ended up with something like this, okay? This is another example. Could have been like this. It has to have this shape because it has to have exactly, it has to have some number of full branches. And then it has to have one, the first branch and the last branch must match up because the image of zero must be the same as the image of one, right? So here you have I zero and then you must have a full branch followed by a full branch because the derivative is always bigger than one. So it cannot fall back on itself for those strange things, right? It has to be a branch that is a different morphism onto the whole thing and then a full branch and then a full branch. And then the last branch, it must close up exactly at the same point here. So here we have several fixed points. This is exactly, this is one, two, three, four, five, six branches, right? And each point has exactly six pre-images, right? Because any point here has one, two, three, sorry, five pre-images because it looks like a six branch but really it's a five branch map, right? Because these two should be, if we'd cut it in the right place, we would have five full branches. We just made a mistake or we didn't know at the beginning where to cut the circle. So each point has exactly five pre-images. You have one, two, three, four fixed points, okay? And now if I cut the circle, instead of cutting it where I cut it, which was completely arbitrary anyway because the circle does not have any origin as such. I can just decide to cut it where I want and I decide to cut it exactly at one of these fixed points. So if I cut it at one of those fixed points, then the graph, because this is now a fixed point, the graph is going to start from here. It's going to come right out of there, okay? And then therefore, if it comes right out of there, then it's going to have one, two, three, four and then five full branches like this. This is the graph of the same map but you just identify the interval with the circle starting from a different origin. And it's exactly the same. Okay, so what's the proof? So the first step of the proof is that every C1 expanding map of S1 can be seen as a piecewise expanding, sorry, first I'd say is a covering map. So every C1 expanding map of C1 is a covering map. This is a standard notion from Algebraic topology or topology is a covering map. And so can be seen as a piecewise expanding full branch map. And why is this therefore structurally stable? Now take a small C1 perturbation of this map of the circle. Why is this structurally stable? Because if you perturb this map a little bit, you will still get, you see this is a map from S1 to S1, smooth map. It does not intrinsically come with any partitions or any full branches like it just comes as a C1 map of the circle. If you perturb a little bit in a C1, you will still have a C1 expanding map of the circle. And since every C1 expanding map of the circle is a covering map, the perturbation will still be a covering map, right? And clearly it will still be finite to one with the same, it will still have the same number of branches. So it's a covering map. And so every covering map is L to one has a certain number. A covering map means that every point has the same number of pre-images, L pre-images. So clearly if you take a small C1 perturbations, you cannot change the number of times. Think of it as this map takes S1 and wraps it around itself a certain number of times. That's exactly what it does. So if you take a small C1 perturbation, you cannot change the number of times it wraps around itself. You can just change a little bit of the derivative of this. And of course you will change, because you're changing the map a little bit, you will change the specific partition. So you will change the place that you need to cut the circle to be able to see the full branch piecewise expanding map. So you will get a slightly different piecewise expanding map, full branch piecewise expanding map, but of course it will still be a piecewise expanding full branch map. It will still have the same number of branches and it will have all the branches have the same orientation. And so we're in the setting that we had before and so it would be topologically conjugate. Yeah, that's a very good question. So this of course cannot have infinitely many branches because this you cannot, if you have a C1 expanding map of S1, you cannot not have infinitely many branches because the derivative is okay. I guess I should assume in this case, well it's part of this fact that it's C1 that it cannot, in this particular case, this kind of map always has a finite number of branches, but your question is perfectly legitimate in terms of can we study by similar techniques maps with infinitely many branches, okay? And you can. So in that case, you will get a symbolic coding with an infinite number of symbols and you get the shift map with an infinite number of symbols and that is also an interesting class of maps. And in fact, maybe some of you have seen the Gauss map, the Gauss map, one over X fractional part of one over X, F of X equals one over X mod one is a very nice example of a full branch piecewise expanding with an infinite number of symbols. And we will look at that map in the ergodic theory course, actually. But in this case, you can only have a finite number of branches and these number of branches cannot change. And so you get, so for C1, for C1 perturbation, G, the number and orientation of each branch is the same. And so this implies by the previous result that it's topologically conjugate to F and therefore we have structural stability. Okay, so we still have about 20 minutes left. I want to give one last example, but I think maybe we'll take just a couple of minutes break and then come back. Okay, so I just want to complete now this section on expanding maps and I want to introduce a map that is an expanding map in disguise. So define a map. So I'm going to introduce a map that is part of the quadratic family. So the quadratic family of maps. So just this map here, minus 2, 2, 2, 2 plus 2x squared. So what does this map look like? So the interval of definition is minus 2, 2. So draw it like this, minus 2, 2. This is just a simple quadratic map that you know well. So you can see that 0 goes to minus 2. Minus 2 goes to minus 2 squared. 4 minus 2 is 2. 2 goes to 2. So you can quite easily see that it's a simple parabola that is exactly like this. So this is a full branch map. Right, what are the two partition elements? The obvious branch is here. We have this branch here. We can call it I zero. This we can call it I one. And it's full branch. I zero maps exactly onto the whole of I. In this case we have the interval I is not the interval zero, one, but the interval minus 2, 2, but it doesn't matter. We have an interval. But is it expanding? It's not expanding, why not? The derivative here is zero. So obviously it's not expanding. So what goes long if we try to apply our whole construction that we did before for piecewise expanding maps? Even though it's not expanding, sometimes you do some construction, you prove some theory with some assumptions, but the strategy is much better than you thought and you can actually apply it in much more general situations. Can we apply? Can we do our symbolic dynamics, our coding and so on in this case? Where did we use the expanding property? So before that, just to construct the symbolic coding, yes? To get what? Why? Exactly, to prove that I is a single point. So when we do, we can start doing this construction here. So we can, if you see, so this branch I zero maps to all of this. So we can construct two. If we wanted to reconstruct I zero one here, right? This is, sorry here, we do the construction here. So we could construct I zero one here because this is the interval in I zero that maps to I one, right? Because this is I zero and this is I one, right? And then here we get I zero zero because this is the interval in I zero that maps to I zero. And we can do the same thing here and we can start constructing these intervals, okay? But the problem is, when we try to apply the mean value theorem, depending on the dynamics, there will be some regions in which the derivative is very small. We will start having intervals where the derivative is very small and we don't know that the derivative is bigger than lambda to the n for lambda bigger than one and we cannot do this, okay? And we really cannot except that in this particular case, we have a magic, okay? And we will be able to construct the topological conjugacy. This is one of the very few cases in which we can construct an explicit topological conjugacy. And we will construct the topological conjugacy with the following map, which you know this is a map from zero one to zero one, which is defined like this. This is the point one half and this is the 10th map. So this is the map G of Z, which is defined as two Z, which of Z is between zero and one half and two minus two Z if Z is between one half. And of course, using this conjugacy, so our theorem that F and G are topologically conjugated. And one of the interesting thing, this is actually a very remarkable result but it's also very special result and what we're going to show is we're actually going to have an explicit formula for this conjugacy. And so we're going to actually define H from zero one to minus two two by X goes to two cosine pi, sorry, I'm calling Z, I'm using the symbol Z here, two cosine pi of Z. So what does this picture look like? This is zero one, this is minus two two. So what does this graph look like? H, so when Z equals zero, this is just cosine of zero, which is one, two times one is two. So this is two, right? And so this, and if you, when Z is equal to one, this is cosine of pi, which is minus one, so this gives minus two. So this is exactly, this is the graph H, right? So I wanted to give this example for several reasons but one of them is just to show you in this course and in the first part of the course and the second part of the course, what I've emphasized is not only the results but the techniques that we use for constructing topological conjugacies. And this is very interesting because this is almost the only situation in which we really have an explicit map for conjugacy between the two maps that in some sense look alike, but it's not at all clear that they should be topologically conjugate to each other. Okay, and so what we do now is we already have that H is a homomorphism. Okay, in fact it's almost a C1 diffeomorphism, right? It's not a C1 diffeomorphism simply because the derivative is zero at these points. So the inverse would have infinite derivative at these points, but on the open interval zero one to the open interval minus two two, it's a C1 diffeomorphism, right? It's almost a C1 diffeomorphism. So they are almost C1 conjugate. Remember how I emphasized how difficult it is for two maps to be actually C1 conjugate. It's a very strong condition and we will come back to that later. They are almost C1 conjugate. Okay, and the proof is very nice and easy because we just need to check. We have the formula for F, we have the formula for G, we have the formula for H, we just check, right? So to check that this is the conjugacy, we just calculate F of H, so F composed with H of Z is just equal to F. So what we are doing is we are looking at a point here. This is a point Z, right? We are going to check the two possibilities. One possibility is that you apply H and then you apply F. The other possibility is that you apply G and then you apply H and they need to give you the same point here. That's the conjugacy, right? So this is just F of two cosine pi of Z and this is equal to, so this is the formula here. So we have two cosine pi of Z squared minus two and this is equal to four times cosine squared pi of Z minus two and this we can write as two times two cosine squared pi of Z minus one and then for those of you who know all your trigonometric equalities, identities which I never remember, I always have to look them up but this is a standard trigonometric identity that this is exactly equal to cosine of pi two Z, right? So this is just equal to two cosine, sorry cosine of two, yeah, two pi Z, okay? This is a standard trigonometric identity here. And on the other hand, let's see, where can I erase, okay, we don't really need this picture and then just checking instead on the other hand. So for H composed with G of Z, we have actually two possibilities. If Z is in zero one half because the map is defined differently in these two cases, right? So here we have H of two Z and this is equal to, so H of two Z is just two cosine, two cosine, two pi Z which is the same. It's what we want, right? If, sorry, this is a bit, so if Z is in this. Okay, and otherwise we have H of two minus two Z if Z is in one half one, okay? Strictly speaking, I should only decide whether to include the one half above or below but it doesn't make any difference and so H of two minus two Z, so I plug it in here. So this is equal to two cosine pi two minus two Z and this is equal to two cosine two pi minus two pi Z but as you know, cosine is two pi periodic, right? So this is just the same thing as two cosine minus two pi Z and this is just also cosine is symmetric so this is the same as two cosine two pi Z which again is exactly the same thing. So this proves that F and G are topologically conjugate by direct verification of the fact that this is exactly a conjugacy of these two maps. Any questions? You're happy with that? Okay, so what, sorry. So what is the corollary about the dynamics of this? So all everything we know about the tent map, the topological dynamics of the tent map, I'm sorry, that transfers to the topological conjugacy of this which otherwise we will not be able to do, right? Because for the tent map, we did all the symbolic dynamics and everything and so on. For this map, we cannot do it directly because it's not expanding. We're lucky that we have this conjugacy which allows us to transfer this information. So for example, what is the topological? Do you remember what are the basic features of the topological dynamics of the tent map? Corollary. Periodic points, what can we say about the periodic points now of this map? Is that a periodic points is dense, okay? So periodic points of F is dense in minus two two. There's an infinite number of periodic points and the set is dense. Also something about transitivity, dense orbits, right? So F is transitive. So there exists X such that omega of X equals minus two two. So this is one corollary which does not follow immediately from here. Also we get things like sensitive dependence on initial conditions because the sensitive dependence on initial conditions we got basically from the tent map, we know that any two points will eventually fall on opposite sides and therefore because this is topologically conjugate and the midpoint one half maps to here, these intervals, IZ and I1 maps to each other, we know that any two points here will eventually fall on opposite sides here so you still get a kind of sensitive dependence initial conditions, okay? But just the last thing I want to mention is something a little bit extra that we get here using the fact that this conjugacy is essentially differentiable, right? Remember what is the additional property of differentiable conjugacies? You remember? What additional property does it have? If it's differentiable, what additional structure does a C1 conjugacy preserve? Sorry? That's right, the derivative of the fixed points. And at the periodic points. So, let me redraw the tent map. So, what fixed points does this have? It has two fixed points, okay? Now notice that what this, this is a homomorphism, but it's an orientation reversing homomorphism of these intervals, right? In the sense that you can see also just the direction of the tent map just the direction of the point here. So what this does is it maps this interval, it flips it over and maps it to this interval like this, right? So this point here maps to this. So this homomorphism, if you just look at the actual explicit formula here, this map, point maps to this, this point maps to this, okay? So this fixed point maps to this fixed point. So this also has two fixed points here and here. So the derivative of this fixed point is what? Two, because the slope here is two, yeah, minus two. The absolute value is two, okay? Or even the derivative is minus two. And this is the same, we could calculate it explicitly if you want to. You can find the fixed point, calculate the derivative. But what about here? What's the derivative here? The derivative of this map here, F prime of X, F prime of X, yes, it's two X. So here it's four. But here, so, so what's going on? Was this false? Sorry? Exactly, exactly. The comment I made before, right? So this is not actually C1 conjugate, okay? So at the bound, this is not true. But this is not the main observation I want to make. The main observation is all the periodic points that we have a dense set of periodic points here. So proposition, and then I will finish, okay? We're almost finished. So just, so proposition. For every periodic point P in the open interval minus two two of period N, we have FN prime of P equals two to the N. So the proof is simple of this. What's the proof of this is we've basically just said it. Why is this true? That's right, you can, well, a kind of simple way to say it is that this P is a fixed point for FN. If P is a periodic point of period N, then P is a fixed point for F to the N, right? And if the conjugacy is a conjugacy for F, it's also conjugacy for all powers of the two maps. So FN is conjugate to GN. F, the amplitude of F is conjugate to the amplitude of G. G also has a periodic point of period N, which is mapped to the periodic point here. And this periodic point here also has period N. And the derivative here is clearly two to the N because the derivative is always two here for all these periodic points. It's either plus two or minus two. The absolute value of a periodic point of period N here is always two to the N, okay? And this periodic point is a fixed point for G to the N, which conjugates F to the N. The derivative of this is two to the N, and so the derivative of FN is two to the N. So this, so the proof is simple, okay? Proof, let me just write it as that. Proof is that P is a fixed point for F to the N. So has the same derivative as corresponding fixed point for G. Okay, that's the proof. So all the fixed points for G to the N. And F and H is C1 different. On the open interval minus two to. So why is this remarkable for this map? Remember that periodic points are dense here. So that means I can find arbitrarily close to here. So I just conclude with this remark. So like you say, the chain rule is always very important in one-dimensional dynamics. Always remember that FN prime of P is equal to the derivative in the point P times the derivative in the point F of P times the derivative in the point FN minus one of P. This is the chain rule. The chain rule says that the derivative of the Nth iterate of a point is the derivative, the product of the derivatives along the orbits of this point. Which means that if P is a periodic point of period N, this is nothing but just the product of the derivatives with all the different points of the orbit. It's a kind of average derivative. It's the product of the derivatives. So what does it mean here? Notice that here you have some regions where the derivative is bigger than two. You have some regions where the derivative is smaller than two. You have even some regions where the derivative is very, very small. So if you take a periodic point here P and there are periodic points that are arbitrarily close to this point zero, okay? And then the derivative here could be of the order epsilon, arbitrarily small, okay? This means that the product of all these derivatives must be very large. So the product of all these derivatives is of the order two to the N over epsilon because the product of this times this is two to the N, as we've just shown. This is a big derivative if epsilon is small. In fact, how is it possible that the derivative can be so large if you only have N minus one terms and you only have, you know, the maximum derivative is four here. You don't have very places with this very big derivative. Well, if you think about it and you did the calculation, that means that this must be very long sequence. So the periodic orbit must have a very long period because to compensate the fact that you get a very small derivative here, it must spend a long, many iterates in the regions where the derivative is strictly greater than two in order to be able to compensate that and then the average derivative along the orbit is the average derivative two in the sense that this product is the same as if you were taking a product of twos which is what happens in this case. So what you have is a situation where this particular map, quadratic map has regions where the derivative is small and where the derivative is large but it's a very, very special map and because of this very special differentiable conjugates it's not enough that it's any conjugacy. It just turns out that the combinatorics of points is very, very special and if you have a periodic point then the time it spends in the regions where the derivative is bigger than two exactly compensates the time it spends where the derivative is smaller than two so that it's as if the average derivative was two at each iteration. It's just a very special structure. So in the last lecture tomorrow I will introduce just an last additional dynamical system which I think is interesting you should know and then I will spend also a little bit of time talking about the introduction of this quadratic map actually this map belongs to something called the quadratic family which is a one parameter family of maps which is maybe one of the most deeply studied families of dynamical systems and it's a very sophisticated and very interesting dynamics and this is one, the starting point in some sense of that and talk to you a little bit about some open problems and maybe even some recent research on this kind of family. Okay, so I think that's enough for today.