 Who thinks that I speak too fast and then the lectures are too fast? Oh, come on, some of the different questions. Do I speak too fast? OK. Are the lectures too slow? OK. Are the lectures too fast? OK. Too slow. Good. OK. No. Oh, are the lectures right? No, you see, it doesn't match. There are too many undecided people. This is not correct. OK, so I think I'm going to start. So today, I think it's a key lecture because I think we really want to connect this rotation and Gauss map. And it's going to be our first example of renormalization. And it's really the basic idea of renormalization. But I think even though someone thinks it's too slow and almost all of you know continuous fractions, I decided that I made a mistake in skipping. I want to have on the board a couple of the properties that you almost all already know about continuous fractions because I'm going to need them today. So let me first start with this quick recall. So recall for almost all of you. Recall from continued fractions. So I need the notation. So we have alpha irrational. And we have, hence, alpha has a unique infinite continuous fraction expansion where these entries are positive integers. And I want to, probably you've already seen, I can define truncation. So I want to call pn over qn finite continuous fractions obtained by terminating the fraction at a n. So this is infinite, but I stop it at a finite level. And then I can clear out the nominators and I get the rational. So these are called convergence of alpha. And we are going to use them today. Actually, the denominators we are going to use. And let me recall you some facts about this convergence. So just three facts. So first of all, this convergence, you may well believe, will converge to alpha. So as n grows, you will tend to the infinite continued fraction. And actually, this convergence is exponentially fast. This I don't need, but it's just for your own. They converge exponentially fast in the sense that you can bound the difference between alpha and the n truncation. And this is less than some constant over something like square root of 2 to the n, so exponentially small. Maybe we'll go here. And then the key thing that you probably have seen, because it's one of the things you do in continuous fraction, there are recursive formulas. So especially the Qn and the Pn denominators and numerators both satisfy this set of recursive equations. So Qn plus 1 is An plus 1 Qn plus Qn plus Qn minus 1. This is a recursive relation involving the denominators and the entries. And also you have initial condition. So Q0 is actually, you set it 1. And Q1 is A0. And the Pn satisfy actually the same recursive formula. So Pn plus 1 is also equal to An plus 1 Pn plus Pn minus 1. The same formula, but different initial conditions. So you want to take P0 to be A0 and P1 to be A0 Q1 plus 1. Doesn't matter so much. Just a recursive formula for actually, I don't need the P's, I will just need the Q's. Have you seen this when you did continuous fractions? OK, great. So if you want, in principle, you could try to prove them. So it's just a manipulation of continuous fractions. So if you haven't seen them, you may want to play with them. But I want today, we will see a dynamical interpretation. So this formula will have a dynamical meaning. So I don't care about manipulating continuous fractions. I care about giving it a dynamical meaning. And similarly, let me tell you the third property that you also know, probably, that these are best approximations. There are something called best approximation of the first type or the second type. I don't remember which one. But I say I want this one. So if I look at this distance, this is actually less. It's the best possible. So this is the best approximation among all rationales with denominator bounded by Qn. So this distance is less than, let's say, put less than or equal to have to be safe. But I think less is correct. Q alpha minus P for every P integer and for every Q1. One less than Q, less than Qn. As I said, put less or equal. I guess I want to put less or equal. So in any case, they minimize the distance from integers. So maybe you can also use this notation. You can write norm of x, z to be the minimum of x plus n as n is an integer. And then you can kind of read this as saying Qn alpha less than Q alpha for Q less than Qn. So this is like distance from integers. So this is all I wanted to remind you. And first thing I want to remind, I want to give dynamical more than geometric. What does this tree means? Dynamical interpretation of property 3. So what does this tree say about the orbit of the rotation? How do I find Qn if I look at an orbit of the rotation in space? They are best approximation, but dynamically. So I'm going to draw a plot a large circle. I have to do many pictures today. So I'm going to put 0 on top of my circle, just because I like 0 on top. So if I look at the orbit of 0, if I look at the orbit under the rotation by alpha of 0, so this is just basically a fractional part of n alpha. Well, on the circle is whatever. On the 0, 1 is fractional part of n alpha. But say that I plot my, what is special about the Qn? If I look at the Qn iterate of the orbit of 0, how will I recognize that I'm special? You're just shy? Maybe my question is clearly unformulated. OK, maybe. So this property tells me that when I have the Qn, the distance from 0 on the circle is less than all the previous iterates. So I want to say, if I look at the distance between alpha Q of 0 and 0, this is nothing else than the Q alpha, this integer norm. It's the distance from integers. And Qn are successive closest returns. By this I mean, so if I keep iterating my rotation, at some point, well, I start from 0. I go far from 0. At some point, I get back close to 0. And this is actually going to be the first orbit at Q1. At Q1, I'm very close to 0. Then if I keep moving, well, I will again go farther from 0. And maybe when I come back, I actually come back but not as close as before. So I might have to keep going, keep going. And I come back, and maybe I come back again close, but not as close as Q1. So if I wait until Q2, I will actually be on the other side. And I will be, maybe I should draw it a little more. I would be actually Q2 is the first which is closer distance from 0 is less than the distance of Q1. And so on. If I keep going, it will take me longer and longer and longer number of iterates. But at some point, this will be the Q3 iterate. It will be even closer and then so on. And actually a fact that you can convince yourself and it also comes with a continuous fraction is that remark is that r alpha to the Qn of 0 alternate between left and right. So now I will get it wrong. Let's see Q1. So this is the left of the origin to the left of 0. If n is odd and it's to the right of 0, if n is even. So the closest return is once to the right and then to the left and then to the right and then to the left. And they accumulate towards 0. Great. Now I want to understand. So this is the dynamical interpretation. So I want to understand what happens on small scales. So in some sense, I want to zoom in around the neighborhoods of 0. And I will use these special closest returns to pick a family of neighborhoods of 0. OK, so let me define and let me set. Let me give a notation. Set in to be, this will be an arc to be an arc. But sometimes I will think of it as an interval. So we always identify the circle with an interval. So arc, but I will straighten it up and think of it as an interval. But for now, I want it on the circle because I want an arc with n points. N points are alpha to the qn of 0 and are alpha to the qn minus 1 of 0. This morning I was trying to get my indices consistent. And let me check my hook one. If it can be off by 1. But I think I checked it in the morning and I want it with n and n minus 1 to be consistent. Let's see. I lost my notes. Yes. OK, so I cannot tell you which one because some ones it will be one to the right, one to the left. And the other time it will be the other way around, right? So in my picture over here, this would be, for example, between 1 and 2, this yellow interval will be, did I get it wrong? I want it to be 2. I guess it's correct, I too. And i1 actually will be q1 is actually, sorry, q0 is 1. Is it right? q0 is 1, so alpha. So I have to take the point alpha. So if this is alpha, and this would be actually i1. i1 has n points at q1 and q0 and q1. Yes, that's right. Sorry, did I confuse everybody already? The n points are n and qn and qn minus 1 iterates to 0. So if I want the n points of i1, the n points of i1 will be q0 and q1. But q0 is actually 1, so this is actually alpha because 0, after 1 iterates, maps to alpha. And this one is this point that I have over there. So this is the first interval, and the second will be smaller. Essentially, what's happening is that I have an arc, which is my i1. Oh, gosh, this chalk is not good. There is 0 somewhere in between. And then at some point, one endpoint I keep, and one endpoint I keep, and the other endpoint I shrink. So I get another interval which contains 0, but it's smaller on one side. This would be i2. And then I will keep this interval and wait for the next closest return, which will happen on the left, and so on. So I'm getting a sequence n plus 1 contained in in. So these are nested arcs, and each one has one endpoint in common with the previous one, by definition. And this will be the interval on which I want to zoom. So I want to study the dynamics of the rotation on these smaller scales. And I will define in a second inducing. But before doing that, let me make a convention, so an easy remark. So by convention, I'm going to decide that i minus 1, which is not defined here, i minus 1, I'll take it to be, I write it. I take it to be basically the whole circle. Let me write it like this. It's s1 cut open at alpha. So if I have my whole circle, this is 0. This is alpha. I'm going to take my pair of scissors and cut it open at alpha. I can draw the little scissors. And so the interval looks like this. It's 0, contains 0. And here it's alpha. And here this length is 1 minus alpha. So this point is minus 1 plus alpha. This is my starting interval. And of course, it's just a different choice for opening up the circle. So how does the rotation look like when I cut open my circle? Well, we draw the graph. We said it's x plus alpha mod 1. In this case, it would be x plus alpha modulo this interval. So we had a plot, you remember, of the graph of the rotation. But I want to think dynamically. So dynamically, how does the rotation look like? I have two intervals, one of lengths alpha, one of lengths 1 minus alpha. And I want to just say that if you cut open a rotation, it looks like an exchange of two intervals. So what happens when I rotate the big arc? Oh, sorry. This point is minus 1 plus alpha. This is the length. So when I add alpha, this is just a shift. So I'm going to draw the image by the rotation of these intervals geometrically. And this is just shifted. And it goes exactly to the end of the interval. And the other one is just what happens to the other one? I'm adding alpha. And then I have to take the result modulo what I have. So when I do it, it just goes back exactly to the other point, right? So this is how a rotation looks when you cut it open. It looks no matter where you cut it open, it will look like an exchange of two intervals. So maybe let me write this. It has interval exchange, interval exchange map of two intervals. So the rotation is equivalent to this. In general, so if you have, and this is also called two IET, interval exchange map, in general, you could have a D IET. And this would be a map where you have an interval and you chop it up into these subintervals of any lengths you want. And then you kind of permute to them. In this case, you need to specify how you want to rearrange them. But you can kind of permute them so that each has the same length in the image. And you rearrange them so that they again form an interval of length one. I'm telling you this because I will give you an exercise later, which involves an interval exchange of three intervals. OK. And actually, I can tell you what the, OK. No, no, I cannot tell you yet. OK. And now I just want another easy thing. And then we can do, maybe I'm going too slow. So first I'll also, no. I want to tell you what does it mean to induce. So what does it mean inducing? Inducing. So if I, let me say in general, if I have a dynamical system from x to x, from x to x, and I have a subset of my system, y, and assume that I know, assume that for every y in y, or maybe not for every, but for almost every, with respect to some measure, for almost every y in y, the point returns to my set. And you just did one career recurrence this morning. So this is not such a big assumption, you will believe. So assume that y, for every y in y, there exists an n such that fn of y belongs to y again. So this is a return time. For example, this is true for our rotation for any interval, just because orbits are dense. So if I pick any interval, points will come back infinitely often, actually, to them. But for one career recurrence, this would also be the case for measure preserving. OK, assume that you know this. Then you can define the induced map. Then you can define, I will call it fy from y to y. This is the induced map. Sometimes we can also call it first return map. Sometimes we can also call it one career map. For one career, maybe I'd like to use it for sections of flows better. I'm not sure if it's. Let's say it's an induced map. And what is this? This is just you wait until the point comes back and define a map from y to y by setting fy of y equal to f to the r, let's call it r y of y of y. So let me explain this. So what is the r y of y? It's an integer. It's a power. So this is not a power. This is a composition. This is r y of y composition. It's the usual notation of f composed with f that number of time. And r y is the first return time. So it's the minimum n greater or equal than 1, such that f n of y is back in y. So this is first return time. I have all the statements to state what I want to show you and the algorithm I want to explain. OK, we'll leave over there. I want to see or I'm somewhere that I don't erase. So OK, maybe I'll do it here. I keep the n point. I keep the intervals. And what I claim is that maybe I'll take the way I wrote it. So the main proposition. So let me say the induced map of r alpha. Maybe let's give it a name. Let's call this induced map t to the n. I'm calling it t because it's OK. The induced map, this is the definition. The induced map, t to the n, t to the n is just the name of the induced map. The induced map of the rotation on this interval, i n, contained in S1, is actually an exchange of two arcs. It is an exchange of two arcs. And I'm going to call these arcs delta n and delta n minus 1. Again, I hope my indices are right. Let me just check my indices. Delta n and delta n plus 1, correct. And for n greater than 1, and for n greater than 1, I want to say, so let me let ln lambda n be the length. This is the length of this arc. This is the arc length. So we have that lambda n divided by lambda n minus 1 is actually the ratio of the length of these two arcs is actually given by the nth iterate of the Gauss map. So I see the Gauss map as ratios of these induced lengths. Moreover, you can also say what are the return times, the first return times are respectively qn minus 1 and qn, respectively. So and it's mismatched indices. So qn minus 1 is the return for delta n. And qn would be the return for delta n minus 1. Voila. And this is what I want to prove today and show you. But I want to first explain what this means. So what am I saying? I have my rotation. I have this sequence of nested intervals. And I claim that I look at this nth nested interval. The dynamic which I see as the first return map is exchange of two intervals. So the dynamics of the one-care map or the first return map will be something like this. I will have two intervals which are exchanged. So when I wait for my rotation to come back to this small arc, I see two intervals which are exchanged. And the iterates that I need to do of the rotation to come back are different for the two intervals. So in one case, I have to wait for one, and we have to wait qn iterates to come back. For the other, I have to wait qn minus 1 iterates to come back. But I just thought I spent some time to say at the beginning that the rotation, you can think of it as an exchange of two intervals. Because now if you want, I can glue back, glue to circle. So if I glue to a circle, this map, these two intervals, and I rescale the circle to unit 1. If I rescale the circle to unit 1, then this induced map rescaled is actually you get a rotation by Gauss to the n of alpha. So basically, this inducing is also giving me a map from rotations to rotations. And basically, this map is renormalization. So this map, we'll call it R, renormalization. So renormalization, if you want, is R of my rotation is actually obtained by R induced on R alpha induced on this interval, in, and then rescaled. And this is actually equal to R of Gauss to the n of alpha. So on the parameter space of rotations or rotation numbers, this renormalization procedure is described by the Gauss map. Just to understand my time, I have only 20. How much time do I have now? Now we have to work. This is kind of all the introduction. OK, I made also some pictures because I'll try to make a picture on the board, but it's not clear that it will be as nice. So maybe I keep also pictures on top. And I would put these pictures on the web. So I would suggest that maybe you look and don't try to write because you can get a different picture and then you get confused. So just look at the pictures and listen, and then, OK, OK. And also, I will post some lecture notes. In the other days, we went from a course which I was teaching in the introduction to the dynamic system. But this part is not written in a lecture notes form. But I have some handwritten notes that I think are quite nicely written. And I scan them, and I'll put them online. So you can find this procedure described with all the details in the notes also. OK, so we consider our rotation by alpha. So this is our rotation. And the first interval I want to look at, I want to set delta 0 to be 0 alpha. Actually, all these intervals, you could make them semi-open if you want, so then you have. OK, so this is my red interval. It's red. No, no, no, no. It's the first green interval in this picture, OK? Maybe it's bigger in my picture. And maybe I'll do it green also here. So the first thing is how many iterates can I fit before overpassing 0? So how many full copies of? So remember that if I take a 0, a 0 by definition is if I divide 1 by alpha and take the integer part, this gives me a 0. So a 0 is actually how many times I can fit an interval of size alpha in a unit circle. So iterate, so I want to iterate delta 0, a 0 times. So this is what I did in my picture, 1, 2, 3, 4. I iterate a certain number of time. And this number of time, a 0, is exactly how many times I can fit a full interval. And then what is left is going to be a reminder. What is left is supposed to be, that's why I like the picture on there, that what is left, this is going to be my reminder. And I'm going to call it delta 1. So delta 0 was the original arc. And delta 1 will be my first arc. And let delta 1 be the reminder. And notice, notice that alpha is r alpha to the q1 of 0, because q1 is equal to 1. So this endpoint is r alpha to the q1 of 0. And that endpoint, what is it? And the left endpoint, the left endpoint of delta 1 is r alpha to the a 0 of 0. But a 0, if you remember this recursive formula, a 0 is q1. So this is r alpha to the q1 of 0. So the endpoints of these two intervals are exactly r alpha to the q, sorry, r alpha to the q0. And r alpha, did they write it right? q0, this is q, sorry, sorry, sorry, sorry. This was q0, q0 is 1 and q1 is a 1. So what I wanted to say that what I defined to be i1 is the union of delta 0 and delta 1. And I claimed that now I want to induce. So this is my first inducing interval. I'll do it yellow. The union of these two is exactly what I want to induce. And I guess I have a pointer, yes. OK, so this is where now I want to focus. It's my first induction. So what is the Poincaré map, the first return map of the rotation on the yellow interval? So let's see what happens to the red interval. So 0 is mapped to alpha, because I'm rotating by alpha, right? So 0 is mapped to alpha. So the other endpoint will drag along. And the red interval will just move to the end of the yellow interval, OK? So 0 goes to alpha. And the reminder goes, well, whatever it is. And this is already back. The red interval in one step is back to the yellow. So the return map just takes one iterate. So maybe I can say, well, if I want to write everything, because what do I want to say? I want to say that r alpha of delta, what is it called? Delta 1 is already contained in delta in i. So the return time is 1. But notice that 1 is also q. The first iterates are confusing, because 1 is also q0. So we proved that 1 returns in q0 iterate. What about the green bit? So the green bit, now I wish I had that. We can use think of the pointer. Does the pointer point? No. Oh, gosh, where is the pointer? Here. So this green, let me move it again. So this green bar, it's my green arc. So I can follow it, follow it, follow it, follow it, up to here. And the last time it comes back, it will actually follow this bit and exactly fill the spot to which I had. So delta 1, so what I want to say, sorry, delta 0 returns. In how many iterates did it return? Eh? What did you say? Who said? You said the green one. No, the green one moves, rotate by alpha every time. 5 in my picture, in general. What is the number? So it fits a0 times fully. So I start with one copy. I have to move to the second, to the third, to the a0 ones. And then I have to do one more iterate. So it's actually a0 because it's a0 minus 1 plus 1. So I return in a0 iterates, in a0 iterates. And a0 is q1 by definition. So we proved the first step of our assumption. So we proved that we come back in q0, q1 types, and then the induced map is an interval exchange of two intervals. What happens next? Ah, this is too fast. What happens next? So now I need to induce on the yellow, sorry, I need to induce on a smaller interval, which I don't know yet what it is. But first, let me remark that if I look at returns remark, if I look at returns of the rotation to this I1, every time I return to the yellow interval, I can forget what happens in between. When I return in the yellow interval, I'm actually doing one iterate of the induced map. So returns of I1 are iterates of t1. t1 is this induced map, this exchange of these two intervals. So let me just try to understand what happens of the red interval. I'll draw it here. What happens of my red interval when I apply my exchange of two intervals? So this is green. And my map t1 maps it here. And this is red. And my map t1 maps it there. So if I look at the red interval, the first iterate is here. What happens then? Then I have to think of it as a part of your original interval and drag it along with the green. So you have to plot some orbit of some interval exchange, especially if it's next week it will be good. So my map takes it here. Then if you want, think of it as part of 0, 1, and then you drag it along. So I want to say that this is t1 of my interval delta, what was it, delta 1? This is delta 1. And then if you look at t2, this is actually going to be a copy of Joshian to the left. Sorry, t1 squared of delta 2 is here. And then if I look more, basically every time that I'm applying t on the red interval, it's shrinking towards 0. And I can look how many iterates it takes me until I can fully stay inside. So for how many iterates? So for, ah, we didn't do the main thing. I'm so sorry. I missed one bit. I missed one bit. Sorry, I need to go back. I haven't proven the full proposition because I didn't show you the ratio of the lengths, which was an important part. So let's look at, sorry, I'm going to go back to the first step. So lambda 1 was the length of delta 1. And lambda 0 is the length of delta 0. What is the ratio? So lambda 0 over lambda 1. What do I want to prove? Lambda 1 over lambda 0. I want lambda 1 over lambda 0. So lambda 0 is alpha. It's a length alpha. What is the length of the reminder? The reminder is 1, everything, minus a0 times alpha. You agree? I have to remove a0 times my big interval. But now simplify. What is this? 1 over alpha minus a0. And a0 was the integer part. A0 was the integer part of 1 over alpha. So this is fractional part of 1 over alpha, which is the Gauss map. So this was part of the important part of the proof of the first step. And similarly, how many copies, the cardinality of copies of delta 1 in delta 0? How much are those? I need to take the length of delta 0, lambda 0, sorry, the length of delta 0, and remove this cardinality of copies of delta 1 in delta 0. It's just I have to take lambda 0, divide by lambda 1, and take the integer part. So this is what the definition I divide and take how many times it fits. Lambda 0, I hope I get it right. So I prove that lambda 1 over lambda 0 is Gauss of alpha. So this is integer part of 1 over Gauss of alpha. This is the opposite ratio. And what is the integer part of 1 over Gauss of alpha? No? So Gauss of alpha, the continue fraction was a0, a1, dot, dot, dot. Yes, a1. That's right. So what is the length? So now I claim that this point is actually, this point is actually, what is this point? I think I'm going a little bit too, let's see where I go. I think I can erase here. I want to keep the theorem. So I want to know if I do this a1 iterate. So if I apply my induced map on the red interval a1 times, what is the endpoint? So maybe let me draw it again. I have my, here I have r alpha to the q1 of 0. The left endpoint, I want to look at the left endpoint of the red interval. So the red interval, I have to cut it. Iterating t1 cuts the red interval as many times. Here there are a1 copies. And I want to know what is this point. I claim that this point is obtained by taking r alpha q1 of 0. Taking this point, first I have to map the interval here. To map it here, I just need to apply the rotation once, because this is the first time. So then I need to apply the rotation once. But then to go from here to here, I'm actually dragging along with the green. So the green takes a0 times to come back. So I need to apply r alpha to the a0. I hope I'm going to get it right. So sorry, I have to apply it a0 minus 1 times. Because I'm already in the first copy, and then I have to apply it a0 minus 1 times. So what is this point? This is r alpha to the q1 plus, sorry, it's not a0, but it's a1. So this number of copies is a1, plus a1 times 1. You should think of it like this as q1 plus a1 times 1, where 1 is q0. And so what is this iterate? Well, this iterate is q2 by definition. So the moral of the story is that the endpoint of reminder is r alpha to the q2 of 0. So what am I saying? Then now if I want to induce on the next interval, so if I set delta 2 to be the reminder. So delta 2 to be the reminder interval. So this delta 2 will be where am I? Delta 2 will be the small interval here. So if delta 2 is this reminder, this one is my interval i2. And then I can continue the procedure the same way. So let me just say set delta 2 to be the reminder. And then I have that i1 is equal to i2 will be equal to delta 1 union delta 2. So maybe what I did, I'll show you the pictures. So this is actually how the iterate of the rotation on the whole interval looks like. Do you see the red interval? It goes along many times. And this is the red interval, you should really think dynamically. So the red interval goes along, come back here, then goes around, come back there. And left leaves my little reminder. And now I want to pick the red and the green and induce on this arc. And I have one more step. Now if I induce the red and the green on this arc, I get a smaller reminder. And this is where I have to induce next. So I don't want to do the, of course, this is the beginning of an induction. You could try to prove this whole statement by induction. But I think I want to summarize what is the algorithm before stopping. Because otherwise, I think we are all confused on what's going on, because we did too many steps. So let me streamline what's happening. Let me streamline the algorithm for inducing, which is now very simple. And it looks like this. I start with what was right, it was long to the left and short to the right. So it starts with, this is alpha, if you want. It starts with i minus 1, if you want, which is alpha, which is the whole circle, which is length 1. So what my algorithm tells me, my algorithm, you can think of it like this. Take this arc and chop it as many times as you can. This will be a 0 times from the left. You are left with a little reminder. Call the reminder, so this was your delta 0, call the reminder delta 1. This is our first step. Now we get i0. I might have messed up. Was this i0? I think I messed up a little bit. I knew I was going to mess up. Was i0 the one which I did I define with minus 1 or with 0? I think I might have messed up. I knew I was going to mess up. So I think it's, I should have defined the, is it still here? No, the trivial one where I cut open the rotation, I should have called it i0. Did I call it i0 or i minus 1? Okay, because I changed the indices in the middle of refining the class, which messed it up. No, and I think delta, we'll say it again. I erase this and i1 has n points. I think i1 is a 0 alpha, right? No, I forgot. I see, I think, okay. I think whatever was minus 1 should have been 0. Yes, so I said that i1 had, I know it was qn and qn minus 1, the indices. So yes, so I think i1 was, I'm not, I'm getting confused now. Okay, in this module, the indices, the next interval will be, I can check them up later and I think on the nodes there, right? Okay, so in any case, now I'll have, I start with large, large and a big interval. I chop the, here the small is on the right. I chop it from the left and point as many times as I can. I'm left with the reminder and now I start chopping the reminder. Now I start chopping the reminder from the right as many times as I can and I'm left with the reminder and then I start chopping the reminder from the left. So basically for, from n odd, let's see if I can get it right or not probably not because the indices from one end or the odd or even, you chop, you chop the small interval from the left end point as many times as you can and you're left with the reminder and for n even, you chop the small. For m even, the small would be on the opposite side. The chop the small from right, from the right end point as many times as possible. So each time you take the small, you go to the opposite end and chop it until you have a reminder and then you look small becomes large, reminder is the new small and you start over, chop the small and one time you chop from the right, one time you chop from the left. So this algorithm which you can take as an algorithm is actually producing the induced maps of the rotation on the smaller and smaller arcs and that's what we try to convince ourselves looking at the dynamics. So this algorithm, if you want, it's like a Euclid algorithm. It's a kind of geometric version of Euclid algorithm and it is indeed a way to produce the continuous fraction expansion. If you have two numbers, you have the chop one from the other and it is also going to be next week, we will actually describe an induction for interval exchange maps and this is kind of the basic case for two interval exchanges. It looks like almost like this partial sum convention. So I think I will start from here tomorrow. So tomorrow we'll rewrite again the algorithm and then I want to show you also the Rockling Tower pictures and then I want to show you some applications. What can you do with this? So okay, so hopefully.