 Okay Today we are going to Go on on the classification, but first I will work out the details that I had left yesterday I I will tell you what some of the things I left as an exercise we will do it with details now Because this will allow us to understand better the proof of the classification So in at this point, we are trying to find a semi-conjugacy between the shift of two symbols and any F which is expanding this degree two and Well, basically what we want What we want to do is that we have a sequence Let's write it like this So we will want we will only have the following criteria It is the only key step in the definition of Our semi-conjugacy is the only thing that we will require the rest will be just Even by the fact that we pretend we expect H composed with Sigma to be equal to F Composed with H I'll do it in general, but I'll first Work out the details for the two for the special case so This is the only thing that we this is the definition essentially we are going to only require this and Of course, we want this to be a semi-conjugacy and From here it follows the definition Why is that? Because if we want this for every X We want that for every X then Implies that this is going to be So at the same time this is going to be F of H of this We expect this to be F of H of X But this is going to belong So if we Get all these things together We will have that H and so We will have H of X This is because of this this implies But this automatically implies and this in terms implies that a checks But this is automatically This follows automatically from the fact that we expect this to happen and expect this to happen This only two things yield us What is going to be the definition? So just Just requiring that this Happen we automatically have its definition Well, we have to prove that this is a definition So first we're going to do it Let me let me first do it for The two X-MOD one map and then we're going to do it in general case So let's prove first Let us first prove That this happens that this is a unique point and Well We have to prove that In order to a way to prove this is the following is to prove that this is an interval of length One over two to the n plus one and we can do this by induction I will first do it for two X-MOD one because this allows us to Do it for the general case F So how do we do this? well One way is to do it by induction Delta X zero Is an interval of length? One half Okay, remember have this This is Delta zero. This is Delta one. This is one half. We can prove by induction that Delta X zero X n equals in fact K over two to the n plus one K plus one Two to the n plus one for some K and that if risk X e2 To the n Plus one. I have this some dyslexia with if this is n or n plus one Let me let me put it like this then we will check that this restricted to the interior of this set is Injected so we can prove this by induction with This holds for n equals zero This For n equals zero zero is Yes, it is n plus one for n equals zero this holds Okay, we have that Delta X zero Equals K over two K plus one over two for some K and also that e2 restricted to the interior is Injected it it takes the interior Into not only it is injective it is Injected and is it equals As one minus one point it takes it takes the interior of Each of these into the whole interval except the fixed point Okay, this is trivial for each one. I'm doing a lot of calculations because this will be This will help us to calculate the general case the general case is not that that trivial But well now assume Assume that we have proven this for some n Assume that this is true for some n and let's prove it for for n plus one Okay, so we have that Delta X zero and also we have that this is injective But since this is injective and takes this interval into the whole Circle minus a point This implies that there will be a pre-image of one half Inside here. Okay this implies that There exists some point plus one which is the pre-image of One half and and it will be long but now this is This is injective and we know that e2 to the n plus one Is exactly two This is exactly this okay, so They breathe if we intersect this Delta X zero X n with e2 and Then this is this is injective and there is only one Here in the middle we will have that this is either when plus two some 2k Plus one over two and That's two this It's easy to calculate it. This will be 2k plus one over two and plus two okay so When we iterate when we iterate this n plus one times we will get one of these branches this will be very very Short this will have lengths one over two to the n and this will take this small interval into the whole interval zero one then We will have either here Delta zero Delta one and this will go This will have this will partition our small interval into one half So one over it one of the two halves Will be this but this is Delta X zero X n plus one And so we have obtained This statement by induction for the next case and it is injective it is injective because We have that e to the 2n plus two This has a length one over 2n plus two and so if we multiply it by two and by two We will have an interval of lengths one and it will be injective when it will not wrap twice Okay, it will wrap only once Well, the same thing. Oh, this is This proves That age is well-defined for the case e to to the end It to sorry We will do it for general F in a minute, but first we have to prove two more things So we have to prove first age is well-defined is well-defined this we have proven for e to Yes, we have proven it for e to I will prove it for F and then we have to prove That age is continuous and then we have to prove that age is on top And then we're done. Okay so Let me do it for general F So for general F Have this we will not care about the length because we have already proven What we have proven yesterday is that if we have two points inside? This inside this they're then they are equal. So yesterday we have proven that if this is non-empty it is unique But we have proved that we have to prove that it is non-empty So the hints I gave you yesterday. We are going to do it We have to prove that for all n we have this. This is an interval We don't care about the length, but all we know is that f to the n plus one of a n Equals f to the n plus one of b n equals p remember p was the fixed point We already know that if an expanding map has a fixed point p and F to the n plus one restricted to the interior of a n b n Is injective? And so we will proceed in this we have to prove this by induction for the first iterate. It's obvious remember remember that we have There exists some q a unique q Such that q is different from p and f of q equals p This is this q is the one that plays the role of one half okay, so For the first iterate you will have that delta x0 will be just pq or qp In any case we will have that f of q And f of p are both p and f of Restricted to the interior is the whole interval minus The minus p. Okay, so let's assume that we have this for some n and buying The induction hypothesis we have that this is injected It's injective, but it takes this whole interval into the whole s1 so in particular there will a there will be a unique rn in the interior such that Fn plus one of rn Equals q okay, there will be a unique one And now by the definition of degree we will have that fn well Then we take a n plus one bn plus one to be equal to either a n plus a n rn or Rn bn, okay One of these two depending on if whether it is delta x0 or the other one and then we will have that this No fn plus one Delta xn plus one this will either yield us but now fn plus two of Am plus one Will be a n so or or rn so this will be fn plus one of a n or Fn plus one of rn Yes, no That's two so in any case it will be of q or f of p and the same this will be fn plus two of either bn or Fn plus two of rn In any case this will be f of p or f of q Which is p so we will have that Fn plus two takes each of these extremes to p But since this is degree two This is degree two and we have this This is a n plus one Fn plus one here Will take this into this and so it will take if fn plus two will take each of these Fn plus two will take each of these into the whole interval so we will have that fn plus two Restricted to either a n rn or rn bn so we have That for each end this is an interval and we have proven it by induction. We are ready for the next step Okay, so all all we have to do is to identify The middle point such that it is the pre image of q So in this way we have proven that this is well-defined Yesterday we have proven that it is unique we have proven that it is an interval and Yesterday we have proven that it can have only one point Now we have to prove that it is continuous and this is not hard to prove Let me we have to remember the metric sigma two okay So first of all we have not proven it, but now maybe it's a time to prove it. We have for general for general Expanding we will have that this can you this is we have denoted Despite this or by this can you give an estimate of the length of this interval? Remember that we had that f is expanding so Well, we can estimate it. It's we have this because f is expanding But s is continue s is compact. So we have a uniform bound for the derivative, okay, so If we take any two points inside this set If we take any two points inside this set We know This fn is injective, but we also know that This is onto Okay, so If we take two points we have that if x and fn of We have done this calculation yesterday this belong to the same x delta xn for all n From zero to n But we have done it if you remember we have done this calculation, this is equal to f prime of some intermediate point times the length of Fn minus one and this is bounded from below by lambda But we can do this n times and so we will have that this is greater or equal than lambda n x minus y But we do this for all x and y and so if we take the If we take the maximum here We will have that for all x and y. This will be lesser equal than this and So we will have that the length of a n Bn Will be less or equal Than the maximum length of this which is Less than one because they are in the same delta x zero. So in particular they are in the same interval So this will be less or equal than one over lambda to the n Okay, you take The maximum here with x and y Variant into a n b n and so this will be less or equal than one over lambda n So in general we will have that this The length of this is less or equal than one over lambda to the n Where lambda is lower bound for the derivative of f so Why is that important to prove the continuity? Okay, so now We want to prove that h is continuous in order to prove that h is continuous We have to prove that this x is close to y then h x close to h y but We have a measure for the Closer closeness of x and y is They are close if they are equal for enough amount of entries. Okay Y equals zero if these These are equal then the distance of x and y Is less than one over three to the n plus one Okay, on the other hand if this happens then For sure x h x and h y Since these two entries are equal for n iterates They will both belong Because this is going to be equal you get it So if these two are equal for n plus one entries then they both will belong to this So they were both will belong To a n b n so H x and H y Will be less than one over lambda n apart Okay So if you want this to be less than epsilon you it's enough to take large enough n Okay, so h is continuous and now we have to prove that h is untrue Do you have an idea of how we do it? We have x in s one and how would do we get Some x H to the minus one of x if any How do we get it? I think we have shown it That's that's it. That's it. We have to follow the itinerary of x. Let's put q here. So delta zero and delta one Zero and delta one are not The same size in general, but in any case we will For any x we take We can decide where it belongs to delta zero or delta one And if it belongs to both like in this case or this case Then we choose either one, but in any case we will be able to choose one symbol Okay, so we take x and just decide whether x belongs to delta zero or delta one and Then we take f of x and we decide whether it belongs Delta zero or delta one and so on We will have at least one choice and then so we take x Itinerary of x and we get a pre-image Okay, so H is onto but the key Problem that we will face in a minute is that in general It will be only onto it will be not Injected okay, okay, so let's Let's begin with the classification. So now we have a well-defined We have proven the details and we know that the the problem the problem for our For our age will appear Precisely in these endpoints and we have this The drama that these endpoints are dense in our interval So we will lose injectivity in a dense set This is very strongly non-injected But even so we will be able to classify them all we will be able to construct Conjugacy between any two expanding maps, that is what we are going to do now We're going to start So what we are going to prove is that any expanding degree two map Is a factor. This is the first theorem and then this one this is the theorem will have proven it already We have proved just proved this but the theorem is that any two expanding maps will be Conjugate, this is a consequence of this But let us so let me state what we are going to prove now We will have now any two expanding maps What's next so there exists some H Which is a conjugacy Okay, so let's first study the points of non-injectivity this is Because it's key when in the points of injectivity we will not have a problem but in the problem in the The situation is that we are going to compose Remember that we had some hf and he and we are going to construct The conjugacy by doing this But we have to We have to show that this is a well-defined map because this In general will not be a single point. And so what we have to do is that every time that this has a Couple of green images we can put them back into the same image This is the whole proof But in order to do that we have to study the points of non-injectivity So I am Confusing h and h f and hg, but we have this this is going to be our age. So Let me put it in in the blackboard how We should work with this So to begin with we are assuming that if it's an expanding map of degree 2 and We are going to call hf The semi conjugacy we have just obtained and let's assume that we have two points Two different points that go to the same x What we are going to prove is that this is a pre-image of the periodic of the fixed point So the points of non-injectivity are the pre-images of the fixed point So this if these two points If this has two different pre-images, so there exists and such that All the points of non-injectivity are contained in the pre-images of p But we will see that pre-images of p are dense. So this is not an easy an easy problem Yes p is the fixed point So how do we do this? Okay, so first we have seen this in our Induction hypothesis we have proven that f is injected in the interior of the delta x zero F is always Injective in not only we have f is injected and in general we have that f n plus one is Injective on a n b n So Have proven this a minute ago This on the this is in tech. Sorry. This is interior This small zero This is not a standard notation. So Let me write it. It's injective on the interior on the end points. It loses injectivity but we have also proven it a minute ago that in the boundaries This goes to p Indeed delta x zero Either one has an an end point as an end point p and q so f of p is p f of q is p so both Go to p so If we have x in the interior of either delta x zero or delta zero or delta one We don't have any problem for the first symbol If they if they are if they if x is in the interior of delta zero or in the interior of delta one Then the symbol the first symbol is unique Okay, this is not there's no ambiguity We have either to choose zero or one and the symbol is unique However, if if we are in the boundaries, we have two choices if we are talking about p or q We we can choose zero or one. So there we lose injectivity So in the interior, there is no ambiguity as we have just said the problems began in the boundaries So, let me see if I have so Here we have if if we are in the situations of the red dots We're in the situation of the red dots Then here the first symbol is zero here the first symbol is one, but here in the black dots We can choose either one For the second symbol we will have this situation If we are in the situation of the red dots The second symbol will be zero, but if we are in this situation The second symbol could be either zero or one because it's in the boundary of the second pre-image of delta x zero Okay, and so well here they are evenly distributed, but in general they will not be evenly distributed However, they will be dense because as we have seen the length of each Interval goes to zero Okay, what about the points of non-injectivity we are supposing that we have two different sequences that go to the same point okay, and So let's let's take the first integer So that they are different H of x is x H of y is x and then we will have that x is x zero xn minus one and xn, let's say zero And whatever and then y will be equal, but here we will have a one Okay Well, what I claim is that if you reach to this situation Then what comes next has to be unique. You don't have two choices or many choices You have only one choice if you want that these two points go to the same points. Who's next here? If if H if you have that H x and H y goes to X What has to come here? 1 1 1 1 1 0 0 0 0 is the only possibility Why is that we are going to prove it? This is intuitively clear, but we can prove it ah We have this and so this implies that H x and H y pose belong to the Intersection This is because the same the first n symbols are equal So we have this and we have that the first is different so Since the first is different We have but this inter this big intersection is the same Okay, remember this this is This intersection is the same So we will have that both There is an equal they will belong not only to this intersection, but they will belong To f to the minus n Delta zero here But since they are the same they also will belong to f to the minus n Delta one this is equal to f to the minus n delta zero intersection delta one and what is this? We know exactly what this is. What's the intersection between? Delta zero and delta one you are p so this is F to the minus n of p or q so This implies Automatically Let me put it here This implies automatically that if n of x is Either p or q So if n plus one here I should have put a big n so if n Plus one is p then we're done We could precisely prove That these are the pre-images of p and it's an if and only if it's easy to show that if it is a pre-image of p Then you have you can choose both symbols and we get what we want Okay, so here we have the theorem we have already told you about this if we have Two expanding maps with degree two then they are topologically conjugate in particular All are topologically conjugate to e2. We could do that prove that every is Contrugate to e2, but we are going to do choose any two Any two expanding maps and prove that they are conjugate conjugate or topologically conjugate is the same This is what I have just said So let's go into the proof as I have told you I We will have Let me do it we have here e2 plus and the shift map and For any expanding map we have a semi-contrugacy hf So here if we have a g a g But now hf is continuous onto, but it's not injective However, we are going to prove That we can define hf composed with he to the minus one and that this will give us a Homiomorphism this is what we are going to do so How will we do it? The idea is what will I have just told you We are going to prove We will take any in the injective points. There will be no problem Because this will have a unique pre-image. So that we have to prove it is continuous, but it won't be a problem but in the none Injective points we have just proven that this is the pre-image of the periodic point It's a pre-image The ends pre-image of the periodic point and so the periodic point as we have just seen It has exactly two pre-images There's no secret here. So the two pre-images will be of this form We will have the same digits at the beginning then one will have a zero the other will have a one and then They will go one one one zero zero zero. So we will take a point of non-injectivity We will have two pre-images and then we have to prove that for the HF These two pre-images go again to the same point Okay, this will be a well-defined map Then we have to prove that it is continuous and on to Okay So we have this I have already I have already shown this Let me write it F composed with HF equals HF composed with sigma G composed with HG equals HG composed with sigma and Well, I put it exactly the opposite of what we wanted The way we go to do HG composed with with HF to the minus one. Okay, so this will give us This will give us a map Here to here H Will be HG composed with HF to the minus one so Well, I will have only five minutes. So let's just prove that it is well-defined and then we will go on next class so If it consists of a single point, it's obvious that it is well it is well-defined. Okay It's not not a problem. So let's assume that it's Not that's not going have a Unique pre-image then we have this We have two points. Let's put it F here, F here And so we will have this we will have two pre-images for the same point and We will have that for some end This goes to the fixed point and so as we have just proved For some iterate of the shift map We will have as we have done here if we take shift n of X We will have all ones or all zeros This is mandatory because they have to go to the same point. Okay We have just proven this in fact We will either have to have all zeros or all one Because if if we didn't have all zeros or all ones we would have that Let me let me prove it here if we didn't have all zeros or all ones We would have one This will be part of the sequence or this will be part of the sequence. Okay, so this would imply that for some n For some n for some k for some k greater or equal than n This which is P Would belong to Delta zero one or to Delta one zero Okay But let me Throw this picture. We have Delta zero Delta one and we have Delta Delta Delta zero one is this And Delta one zero is this Okay, but P does not belong to this Is that clear I mean if we we want to show that shift n of X Is either a sequence of zero or a sequence of ones In order to do that we have to prove that this Subsequence is avoided It's not allowed And why is that it's because if it were allowed then we would have that this is either in this Set or in this set at this set P is not in either of these sets So it has to be Here or here and that this implies that they are only zeros or ones So let me finish with this so that we go on tomorrow is a contradiction and so We have the first n such that all are zero and all are ones and the previous one is zero or one and These all are I changed notation. So here I put it to put it to this is the situation Okay So I don't know if I want this will take some time. It's easy, but it will take some time I I think I will let it for tomorrow so that we can do it Easily, but let me just tell you we will have this situation and this the the idea will be that we will have Everybody will be equal up to some point and then at one point one will belong to here And the other will belong to here, but after the point falls here. All are ones After the point come here all are zero So they will converge to the same to the same point in the intersection of all intervals We will get the same point which will be either a Q or a P okay, and That will the Q the PF the PG or the QG, but let me do it More easily tomorrow. We go we go on tomorrow then. Thank you so much