 Shall we start? Okay. Let's start now because we are all very tired. I took 10 minutes more last class so I expect to take less this class. I expect to finish sooner. I hope so. Well, let's finish with this construction. If you remember, we built for any expanding map F with degree of F2, and we showed that there was a semi-conjugacy from the full shift of two symbols into this F, okay? And so this HF composed with sigma was F composed with HF. And the same thing happened with HG. And both HF and HG were defined by this. This we showed that it was continuous, objective. This was, everything was okay. So now we are planning to do a conjugacy between F and G for any two expanding maps of degree two. And the idea is to take any point in the circle, take its pre-image, its pre-image if it is unique, this will be uniquely defined, that will be fine. But if it is not uniquely defined, then we have already shown that if this is not uniquely defined, it has only two pre-images. And so our job is to see that these two pre-images go under HG to the same point. After we do that, we only have to prove that H is continuous, and that will be all. Because since G and F are interchangeable, this will automatically show that H to the minus one exists and it is also continuous. So it will be automatically a homeomorphism. Do you understand that? So we only have to show, our generality will be first see that H is well defined we were almost there, and then to see that H is continuous. And that will be all, because H to the minus one, H to the minus one by definition will be H F composed with HG to the minus one. Okay? But now there was nothing special about F or G. So we already know that this H to the minus one is well defined and continuous. Okay? So this implies that it's well defined plus continuous. And so this automatically implies that H is a homeomorphism. So all we have to do is to see that H is well defined and that H is continuous. So if we're in, we have a point that it has a unique pre-image, then this is naturally well defined. So our only problem will be when we don't have a unique pre-image. In that case, we have already seen that when H F to the minus one, when we had this situation, two different sequences went to the same point and the sequences were different, then we already knew that there was some iterate of this point that was, that went to the PF, PF is the fixed point of F. We have shown this yesterday. Okay? So, but we also have shown that what we can show it. It's very, very simple. Let's do it right now. We have my hypothesis that these are different. And so there will be a first symbol in which they are different, okay? Let's assume that this first symbol is the symbol N. Then since we need that FN, PP, we have to have that from a certain point on every will be eventually constant. Why is that? We can only have this once because we have that FN, P is one. But after that, we cannot no more have this because we have that FN of X is P. And so FN of X is P. This implies that F of X does not belong to zero one nor to one zero for all N greater than N. Why is that? Because P does not belong to this. Remember that this, this is delta zero, this is delta one, this is delta zero one, and this is delta one zero, okay? And P does not belong to this interval. So it cannot have this in its representation after N. And so this implies that this has to be all one and this has to be all zeros, okay? So we are assuming that we have this, we have exactly this representation. We have to show, so there are no other possibilities. If we have two points that go into X, we take the first N such that this happens and we will have that this is the representation. All we have to see is that these two points go to the same point, right here. Okay, so we already have this. This is what we want to show. We want to show that HGX and HGY go to the same point, okay? But now we have the following, HG of H of X and HG, sorry. Yes, they are not equal yet. But we know that they have the same representation. So by definition, by definition, they will be in the first, they will have the same first N plus one symbols, plus N symbol, sorry, N minus one symbol. Look at this delta depends on G. It's not the same for F than for G, okay? Delta depends on P and Q and P and Q depend on G, okay? But this will belong to this because they have this representation. On the other hand, we will have that HG of X will be, well, this we have called it A N minus two, B N minus two, remember? And so this will belong to A N minus two, B N minus two, intersection G to the minus N plus one delta one, and all the rest will be zeros. All the rest will be zeros, N, N, okay? See if I have written it there. And on the other hand, we have that HG of Y will also belong to A to the N, B N minus two, but now it will belong to the G, to the minus N plus one of delta zero, and from this on, this will be all one. Okay, we have that already. And remember that we have already proven when we were defined, when we were proving that this was an interval for each fixed N, we have proven that J N minus one, restricted to the open set here, was injected. Remember, we have done this. So there's a unique point that goes to Q. There's a unique point here that goes under G N minus one to Q. Let's call it R N minus two. So G N minus one of R N minus two, exactly Q. Why do we care about this? Well, we care because now we will have the pre-image. We will have that A N minus two, R N minus two, goes under this into delta zero, and A R N minus two, B N minus two will go under G N minus one into delta one, okay? So in particular, we will have, you write it, G N minus two, R N minus two will be delta zero and G N minus one of R N minus two, B N minus two will be precisely delta one. Well, this implies that this is R N minus two, B N minus two, okay? G N minus one of this is delta zero. So G N minus one to the minus one of delta zero will be this interval. And this, let me put some colors here. A N minus two, R N minus two. So this is smaller, so we can forget this. We will have this equals this interval intersection all these pre-images, and this equals this interval intersection all these pre-images. Are you following me? So now, now we have, let's zoom this situation. Zoom this situation. We have A N minus two, R N minus two. This goes to P, this goes to Q, goes against to P, and all this goes to delta zero and all this goes to delta one, okay? But now we have, this A G X belongs to this, let me write it in blue, belongs to this right interval, but after that, it always is on the left side, okay? So it is here, but at the next iterate it is here, in the next iterate it is here, and we know, because we have proven that if a couple of points stays forever in the same interval, they are the same point, okay? But we know that R N minus two belongs to there. So it will be precisely in this point. On the other hand, we have this belongs to A N minus two, R N minus two, and it is always on the right side. So this is this, we know that this consists of a single point, okay? It is easy to see that this consists of a single point and this single point will be precisely R N minus two. Both X and Y are going to go to R N minus two, R there in the slide, but then this implies that H is well defined. So in the case, in the case that we have only one pre-image, it is obviously well defined, and in the case we have two pre-images, it will go to the middle point, okay? It will go to the middle point. So now we have to prove that it is continuous, and then we have two situations. Either it has a unique pre-image, or it has two pre-images. In the case that it has a unique pre-image, it will be more or less simple. It will be very similar to the proof of the continuity of this, because all we have to do, we have that if it has a unique pre-image under F, in case one, if this is unique, then we know that no iterate of this will be the fixed point, okay? No iterate under F, of this will be the fixed point. And so this will imply that X is in the interior of all, is not in the boundary of any delta Xn, okay? Will be in the interior of all its Xn for all n. So it is simple. All we have to do is to choose the size of the interval a n, b n, that we want, and then choose a delta such that all points are in the same interval. If all of them are in the same interval, then they will have the same sequence representation, and so they will go to the same interval. Let me do it with more detail. Let me do it with more detail. There's a unique point, okay? There's a unique point. It's in the interior of all this. And so how can we do this? Let me erase here. Okay, so we want to show that hg, we have that hf to the minus one is some X, okay? And so all that I am going to do at the beginning is to show that if two points X and Y are sufficiently closed, so they have the same entries, okay? Let me, if these are closed, we want to see that the hg of X and the hg of Y are as close as we want. This will do it, but in principle, we have to show that this, if these two points are closed, then we have this, okay? We can do this because we know that this an, b, n, the intersection from n to n minus one of g have that this, have proven this, its length goes to zero, okay? So we, if these two points are close enough, then they have the same representation up to n. And so it is enough to choose n such that this is less than epsilon and we will have that if these two points are at distance one over three to the n, then they will be at distance epsilon. This is not the problem. So now we have to show, we know that these are in the interior of here. So we know because h is to the minus one is unique, we know that h, we know that x is in the interior of this, let's emphasize it, it's in the interior of a n, b n. So h n is in the interior of a n, b n. So we choose in the interior of a n, b n and we will have that they will have the same representation and so their distance is that clear. If we choose two points in the interior of this, of this interval, then their first n symbols will be precisely this. If their first n symbols are precisely this, these are close enough and so this is less than epsilon. So if h is injective in this point, then it's easy for us to show that it is continuous. I want to show you some things extra so I will go a little bit faster here. We are very close. Let me just give you the idea. So let's assume that we have two pre-images. I don't want to go into too much details. Let me just tell you how we do it. I will leave you the details because I want to tell you an application of all this. So let me just show you. If we have two points, we know that fn of x is going to be p and then we know that precisely this is going to be x0, xn minus two, zero, one, one, one, one. y is going to be x0, xn minus two, one, zero, zero. And the situation is going to be the following. For some n, we will have this. You already know that because we have already proven it. We will have here this lambda x0, xn minus two, lambda. Put it like this. It's not precise, but you will understand it. Okay, so we are in this situation. We want to prove that this is continuous. The situation is that we have two codings for this point. So how do we proceed? Very simple. We show that it is continuous on the left and we show that it is continuous on the right and then it will be continuous. Why is that useful? Because here we have an easy representation and here we have an easy representation. All we have to do is to see that on the left it converges to rn minus two and on the right it also converges to rn minus two. But then we just have to take, how close do we want to get to rn minus two? Okay, we want to get epsilon close here on the left. Perfect. So if we want to go on the left, as many symbols as you like on the left. And you will be able to prove that this is as close to rn minus two as we wish. If you want to go epsilon close on the right, you take n symbols here and any y that has this coding of n symbols will be epsilon close to rn minus two. Is that clear? I leave you the details. This is not hard. It's just show if y belongs to r, if y belongs to this set, I'm sorry, you, yes, we have that, we have that h e of x is rn minus two. So h of x rn minus two. So if we take y in this sub-interval, then it will be close enough on the left. And if we take y in this sub-interval with n enough symbols, it will be close enough on the right. And that's the proof. You have all the details in the slides. So let me finish because I want the next 10 minutes I want you to give an application. So we have, if you run the details, you will see that h is well-defined and h is continuous. But then, since there was nothing special about g and f, this implies that h to the minus one is well-defined and h to the minus one is continuous. So h is a homeomorphism. Now, as a conclusion, we have that all, we have that all expanding maps of degree two are conjugated. And of course, well, we have this. We have to prove this, but since it is a semi-contrugacy, we have this and then we have that h composed with f is g composed with h. And this proves that it is a conjugacy indeed. So this finishes the proof that every couple of, every couple of expanding map of degree two are conjugated. Just work out the details, because I want you to show something related to what we have been doing. So I wanted to tell you a little bit more about our Godicity. Just, it will take me, I hope that it takes me only 10 minutes. So with this, we finish that every couple of expanding maps are conjugated. But now I want to go into what you have been doing with Oliver, Irene, Lucia, Stefano, and Davide. I want to go a little bit into our Godicity and some true application. An application that is used in real life and it's not used correctly, but it's still used. So, and it has to do with our Godicity. So I wanted you to tell you a little bit. This has to do with the method that is the mark and recapture method. It's, this is being used actually, you can Google it, you will find it. It is used to measure populations of species and also in epidemiology. It's used to count how an infection spreads. In general, it's used to estimate populations. This method is used actually, currently. What does this method consist in? So let me explain it by an example. Let's suppose that we have a lake and we want to count the number of fish in the lake. So this is a lake and the dots are fish, okay? So what do we do to count it? Because we cannot count all the fishes. They are moving. So how do they do? So they capture, say, 1,000 fishes, okay? And they attack them and then they release them. And then they let them get mixed. And then they recapture them and count again, okay? And then they conclude the estimate, they estimate how they capture again 1,000 fish and count how many of them are tagged. And so they estimate the same proportion and in this form they have the estimation of the fish in the lake. Well, this is fantastic. It's very clever, but it can fail in so many different ways. It has a lot of assumptions. And so let me tell you about the assumptions we are doing in these methods because they have to do with what we have been learning this week. So this is the method. We have the recaptured fish. This is the sample of 1,000 that we have recaptured. This, the yellow square is the proportion of tagged fish that we got. And so the sample had 1,000 fish. Let's say that the tagged ones are 10. So it's the 1%. So what we can conclude is that the 1,000 fish were 1% of the population. So the total population is 100,000. This is the system, how the system works or how the system pretends to work. The first assumption that we are doing is that the amount is not altered. What happens if when we tag, the tag is toxic to the fish and the fish begin dying? Then the system is not good. The second assumption, oh, sorry. Second assumption is that the fish move and get back because we could tag them and they never get back. So when we recapture, we have zero tagged fish, we cannot measure. Thing is that they will get mixed because if they were all together and come back and then when I recapture them, I recapture 1,000 and certainly that's not the whole population. So an animation here, so bad. I had an animation here. There was a movie of the fish and well, I can only explain it. The fish can get dying. This was the animation of a torus and the fish were just a square that was getting thinner and thinner and disappeared. So this is the possibility one. If there is no, if the measure of fish are not preserved, then this method fails. So the first assumption is the idea of the animation. If we have this, then the method doesn't work. We cannot estimate. And so the first assumption is what you have been saying, what you have been saying. We are assuming that the measure is preserved by the process. Well, we are assuming that the process is a homeomorphism. We're assuming it is invertible and it's inverse is measurable. That's not harm, but we're assuming that it is measure preserving. If it were not measure preserving, then the fish could disappear and this would not be good. The second possibility, again, this does not work. Pity, I don't know why it doesn't work. Okay, the second possibility is that they, let me do a drawing of what could happen. The second possibility is that you have the fish and the fish don't get mixed. They go all together and they do this movement, this motion. And so this is not good to measure the population because we are only, they are all together. And so when we recapture our recapture samples our first nothing. The next time half of the sample is tagged. The third is all tagged. The fourth, so it does not have a limit and it's not representative of what happens in the lake. So this is another possibility. The third possibility, okay, I don't have the animation. So let's just, the third possibility is that we have the equivalence of an irrational rotation. They don't get mixed, but they still go around the whole lake all together. And so their trace is dense. It's dense and not only dense. It's trace has measure one. Well, there's still hope when this happens, okay? And so the limit does not exist, but still we have hope. I will tell you how we can overcome this situation when we have the situation that they don't get mixed, but still their trace is odd. And the fourth possibility is what we are really assuming is that it gets completely mixed. I'm sorry, I don't have the picture, but they go completely dense. They get absolutely mixed and then you have in the whole lake, you have a proportion, like the lake, the proportion in the lake. This has a name and it has to do with what you have been studying this week and it's mixing property. And mixing property that means that any two samples get eventually independent. Probably you are familiar in probability that two processes are independent, one, the probability of the intersection is the probability. Well, mixing is one, they are not, I'm assuming that this is invertible, that is why I put this, but I'm assuming that this converts to be independent. This converts to be independent. This is stronger than ergodicity. If this happens, the system is ergodic. This is one of the exercises of today, okay? This is called mixing. If it is mixing, it is ergodic. If this happens, then have this, if the measure of the sample is positive, we have this and then we have, in fact what we are doing with the mark and recapture method is this. This is the assumption we are using. We are using that if we let enough time pass, this will be close enough to the measure of the sample, okay? Okay, so we are using that, in fact, what is being used in the mark and recapture method is the mixing property. And this method is being used all over and they don't check that the mixing property holds, okay? They don't even care about this, but it could fail because if this doesn't work, it could not work. What happens if we don't have mixing property? So I will go faster here. If it doesn't work, we still have hope. Have this situation like in the irrational rotation. Maybe this limit does not exist, like in the irrational rotation, this limit in general will not exist, but it will exist on average. If it is ergodic, this limit exists on average. So we can still use the mark and recapture method, but instead of just looking at the sample, we will look at the average of, we will take many samples and we will average them and we still can get the mark and recapture method. So I leave you as an exercise. This I will later put it as an exercise that if it is ergodic, this happens and it is an if and only if. So we can still measure the lake population if we know that the process is ergodic. And this is how we do it. This is another exercise for today. I hope you have not made it. This is easier. And, okay, let's just give a flavor of what we are going to do next week or what we aim to do next week and let me finish with some open problem. Ergodicity is fragile, okay? It's a very simple method. You take an irrational rotation on the circle. You perturb it a little bit and you get as close as you wish a rational rotation which is not ergodic, okay? So ergodicity has a problem that it is fragile. So one thing that we want is to study under which conditions ergodicity is robust. It's stable. I will not explain in detail, but you will tell that something is stably ergodic. If you take close processes and all of them are ergodic. So you have a neighborhood of ergodic systems, okay? All the perturbations are ergodic. So this is an open problem. It's not solved. Does stable ergodicity imply mixing? All the known examples of stable ergodic, the filmophisms are mixing, but is that the condition? This is not known, okay? This is not known. This is an open problem. And this is all. Thank you.