 To je všeč vzgledat po vrste nino, ki smo zelo v rovnih vzgledaj. Nelj smo zelo povrste na rovnih vzgledaj. A potem gaustvaj način in celo vsega. Danes smo nekaj tega vzgledati, ki smo vzal. Nelj smo zelo vzgledati, da je tudi vzgledaj. ker je zelo oclubila zelo. Ko se razkupavamo karado in zeloいくlje. Tudi mi to se pričo je, da je zelo bira sample, sebezoj renormalizacijo v dynamicah, je zelo izprutniji barrier, sebezoj renormalizacijo in zelo idle reormalizacijo. S tem kot, da je bilo lačen ljudi, so v svoj poslednjih večkot vse zelo izgleda. So se nožno izgleda, da površemo za taj moj nekaj, bo pa nekaj dobroj neskot, da so bilo odmahok. Cos nekaj dobroj nekaj izgleda, nekaj geometrik in nekaj neč zelo izgleda. To ne znam, če so briti, kako vidimo hradenštih hal normalezatičnosti? A počne, variables ne znam, ki je nema vživljenje vodne in življenje. Znači, zelo, da všeč ga je všakhova aplikacija, bo, da ne znam, da je všeč in življenje. Postoje, da imo priče priči. Nelj postojo nega nekampa vstvedu, del,コ je tresna hradenštika kaj je? Zdaj sem tukaj zelo, da nekaj idej, da se poživajte z dve vrste, da entropizirujemo dinamikalne sistemov, kaj je renormalizacija. Tukaj, kaj je filosofija renormalizacija? Tukaj, kaj je spas, nekaj nekaj nekaj nekaj nekaj nekaj nekaj nekaj nekaj nekaj nekaj nekaj nekaj nekaj. Poživajte, da se dokon愈ače je dinamikalne sistemov, se nekoj, boš začivajte, da jste tez nekaj v terakšen spas tr!. beltaj je da se gledaš. Tukaj, da se prišljaš s dinamikalne sistem, sta se požiš? na tem, da se gledaš v te Herbic, do denamikalne sistem, nekaj nekaj nekaj nekaj nekaj nekaj nekaj nekaj nekaj nekaj nekaj nekaj nekaj nekaj nekaj nekaj nekaj nekaj nekaj nekaj nekaj nekaj nekaj nekaj nekaj nekajnek. Zato to je buzina. Zato se skupit u nekaj dynamojnjih systems, in u nekaj dynamojnji system, ki pozaj nekaj dynamojnji system. Če bi se pozavati, nekaj dynamojnji konfettiči na zelo, napravna, jo prihleda, vlasti sputni, in nekaj, kako živite hunične vrste. The interest is in your system, but you are normalize to understand it. In in our case, it may be not the deepest case, but let me just see it in our case. In our case you take the rotation and this randomization, which is inducing and zooming, to je vse, ki se lahko videli, skupaj gausti na različenje... nega, zato si gausti na različenje podjel sem različenje na različenje. Zato nekaj sem všem, na različenju imam zelo, nekaj sem všem, zato sranjo renormalizacij. Zato da sem te glasbite, da te vse almost nekaj. Zato da gausti mači na zelo, da je nr. se mači na zelo, nekaj sem tega začel na različenje? Tako, pa kaj se vse, Se bozda, da je idet način, da je identitiva. V svoj početnih početnih, da bomo početnih od 2 interva, 1 interva štriči do 0, in da bomo početnih od 1 interva. In zrča, da bomo početnih, kaj je pačen, da ima začin perijodik. To je vse odrčen perijodik, da si početnih, da je pačen, da mi renormalizacija vsega način. V svoj početnih. Kako je alpha vsega? Tudi, ti je to početnje zjavljenje. Toko to, da je izvečen, da se je našličen. Ako mi je zapisnila, da je to z vseh zelo, kaj bo bil, da ti so izvečen. So, nekaj, da se je izvečen, da je za vseh zelo, da se je izvečen našličen, da je začnila. Tako, da je to početnje, da je zelo, da je vseh vseh izvečen. Vseh vseh izvečen je začne našličen. If I zoom in difference scales, smaller scales, I see the same picture than the bigger scale. This is idea. Eventually, 0 is periodic,lah normalization, periodic means that my map is subsimilar. Then, what about for typical alpha, what you see for typical is that the v sej tog. Music bo notateljno izgivatve. Pistoha ima testoviti, da tega izgivati plantovine, da se poživajte, da sarenje veče bo tega čom všet네 začnev. Nezavodne take, če je napisela, da Heart of the Continue. Tako bih lahko poprati, če bih take vsebe zame adagi. Rejo državce. Rejo državce, da poč smite začve drugi vse birbe, kako bi se začeli, da tudi se čutnočuje ispraviti. Zato je, da bi se tačno se razpravili na izvečenju, in ay bom nekada se tudi pačne informacije o tega teželku do rotatorji kod nekočnega vsega skke. Pač ni se zelo se po pozitivno izgleda. To je je Filosofija. Pač je to, da bom se začal v sezal nadej na druga kontekst. Zelo, da se zelo zasušila, izgleda so mnogo. Znamo, da je bilo, nekaj bilo v pošli v pošli, v pošli verčnih. Zdaj, kaj smo pripravili? Zdaj smo pripravili pripravljenje. Zdaj smo pripravljenje pripravljenje, kaj smo pripravljenje pripravljenje, kaj smo pripravljenje pripravljenje, kaj smo pripravljenje ok. Zdaj smo pripravili algorizm, kaj smo pripravili pripravljenje pripravljenje pripravljenje bo啱ih pošli. Zdaj smo se načinili, zvano se prosto izgleda kaj ide jev believingo z vsem, da smo pripravljenje pripravljenje tako, da tega potreba potrebač vziošaj, v čestni dinamijnške tašne, od roklini in kakutani tašne? Z слов vstavil je obtilitega večične kirjeva. Ta je nekaj začelti všeč. Eli z dvečkih pristjih mači, kaj jeFFa dissolve in izložiteva tega, da je tko ne umelim, da bo delil bil još, da je to z vse mene všeč. Zobaj, da pristjem malo hubštva in iznosim vseč, kaj ta je to napravima. in tega zelo skupaj s tem, ki je neče nekaj zelo. Ok, zato taj moj sem spreviti, da se zelo početite. Tukaj, so ti početite. Tukaj je tegem, ko je tukaj 3-gupi tegem, ko je tegem na rotatorij, je to Stinehaus tegem. Vse je tukaj zelo početit, da se izjemno zelo pripočaj. Zelo se početite in zelo. Tako so početiti, da so da morali, ki so najbolj, že je zelo na kvaliteti, je to na kvaliteti, že je bih pa je začetit, in je dobro vse zelo početne početne, ki sem razumila vse. Tudi tudi je klasika rezačne obrženje, kaj je prišlo vse s te vših pizdi. Zelo bi me vse videl. Vse sem prišlo vse, ali je vse tezda z homoemorfizimom z vsev, a z nekaj zvomorfizim. Vse je vsev zelo, vse je vse, da se vse vse. vse je tudi poslutno, da se stavimo homomorfizm in difomorfizm, zato se odrpunili z normalizacijami. Vseh, kaj je z normalizacijem, moramo, da smo zelo v ročniku. Však se početim, da se prišli, da se početimo za difomorfizm, nebo da se je početim z normalizacijem srednjem. Vzelo, da sem težko početim. Zvoljte, da se razproma v drzci, Mrkvamo in Herman'a preforiti o konjučajnej palette več ročnej saj vrste. Tako je občas, ki puno petr, ki jen prepočiti iz toga algorizme. Kaj je bilo dar jažno presednja na srečce, in zasredno je je čapter, kaj je tega algorizme za srečce. S vseho zelo se dolgo izelo po bočce. Lekčr, recently, had notes in ga squares the references, there is book topics in there go theory by Tsai Waycinnah. And lega, zredno normalizaciji. And let me go back to just reach it the rotation now. Let's go back to really the rotation. There are still things that people are proving about rotations, and even, let me tell you two examples where I prove something about rotations using exactly these partitions. And one, I think so, for a sample, I have a joint work with Senay ki smo pričali, da je vseh počak, ta je vseh počak, pa je naštavno dobro moram, ki tako je vseh, in nekaj vseh pričak je vseh, da je vseh pričak, da je vseh pričak, pa se zelo počak za vseh pričak. In, da se počak vseh počak, tako, da sem razgavljala, da se pričak vseh pričak, ljub, ljub, ljub, ljub, ljub, ljub. Vespoj, da se vseh pričak vseh naštavno pa počak, sentično, kaj bo, nekaj ni pričo si, ne. Še bom se najemnačnja začeta, da so čakela, če so začati,不žoveči vse nekaj ni odličen, zato ga sem hvetnimo na patičnih noutov. Tire postaqu 순št, odzaj sem vsi, da je piločen, zato možemo in seništaj počkih potečnih tudi, grezil si takšno, ne razdin. Nelast sem odličen, da počanjamo jea stvarnost, in zrečen sem naštenje, da se zrečenimo, da se zrečenje in polači, moj so vse predstavil, do vzgledaj, zato je zrečoval, da je zrečen... Daj je zrečen do vzgledaj, že nekaj pogleda se na dančno zelo. Pa vrte se to takovpo beg summonem. Zelo, je gena, da gač 한 strano in ja vsealskim imacnju. Kaj je, da忘em. Z SO that this symmetry is really key in this paper, so there is a key, so if useful to know your tools and play with them well, and even more funnously, like just few... In last year actually I have a... Sorry. I have postdoc Michael Bromburg, some of you know him, he is a postdoc in Bristol and we just prove, I will try to tell you show you a slide in end, vzvešli smo neko, da je to pravda, če je zelo učenje z vseh zrpavljogov, da so dobrozvane, da je to želite pojeliti, da je to nekaj dobrozv, da je to želite, da je to involves na vseh zrpavljogov, da je to je dobrozv, da je to dobrozv, da je dobrozv, da je to dobrozv, da je to dobrozv, da je to dobrozv, zato sem experienično waters. Laševa staj vzredi, tako vzredi do tudi možete v'' tregup'' tudi. Fah, kaj jo ne zelo. Sve se večrbi da si ne bo. Ok. So, ok, zapravi, vzredi, je to izgledal. Vzredi, je to tudi tregup' tudi. Vzredi, stejn' house'' tudi. Vzredi, raising over and over. To bude tko hodnje lukov. To je tako neko zelo ravno zašložitva. In vse mojto loče načo odličovali, tudi nekaj je bil su onj деньги, zašli se kaj lajila in tdi možete taj zeložiti. Zdaj si dishese. Moje, tako, kaj je? Tudi za dej lira. Periodično bi je dobro. For any alpha and for any point on the circle and for any n I'm going to plot the rotation orbit up to n. So consider the initial piece of the orbit z r alpha of z dot dot dot dot dot r alpha to the n of z in my orbit. So I just plot a finite segment of my orbit. So I'll say this is my z, and I plot plot plot plot plot plot plot plot whatever something. And now I have, I don't know, n plus 1. I have n plus 1 points and I want to look at the gaps. So what are the gaps? So these points are not ordered. So as I apply my rotation jump, I don't follow the order. But I can look at the natural ordering on the circle of these points and I can look at the distances between neighboring points on the circle. So this is what is called a gap. So a gap is just the distance on the circle distance between closest to successive points on the circle. And again, let me stress that unless some special cases, these are not successive in the orbits, but they're just successive in space. So I just look how I reorder them and look at the spacing. And I don't know, if you have some n alpha, you can also write it as fractional part. So you can look at fractional part of an alpha, say that this is your n alpha and then you can look at the minimum between n alpha and k alpha as k ranges between 0 and is different than n. And maybe also you want to say that so something like this, you look at the closest neighbors and the conclusion well, it's called three-gap theorem. Can you guess what the conclusion is? There are only three gaps. So there are three gaps at most three gaps. There are only at most three gaps. And actually there are two, there are two, there are two, there are two, two if and only is, and is equal to two if and only is, and is equal to qk for some qk denominator of alpha denominator of convergence denominator. So there are maybe two or three. So it's a very, very simple proof and you can see it very clearly in the towers and you can also compute what are the gap lengths and what is the gap frequency and get a lot of information. So let me just as my point I can pick 0 because in any case everything is homogeneous so I can pick 0. What do you want to do? I just want to, oh yeah, so I want to remark that if I look at the sequence qk plus qk minus 1 this is increasing. So there exist unique k such that n is in between qk plus qk minus 1 and qk plus qk plus 1. So I'm just saying these numbers are just growing so my n will be between two of them. A which picks my n. And now I hope you all did the exercise. If any of you who did the first exercise yesterday for my course I ask you to plot the orbit of 0 up to this length and I ask you to plot the orbit to 0 up to that length. Just for curiosity, who has done it? A few people have done it so I'm not going to do it if you haven't done it you have to try or trust me. So if I plot if I plot the orbit of 0 up to n this is what I want to study I want to plot these points and see what are the gaps. So there are more points than this and less points than that so let me see let me give you the solution of yesterday's thing if you plot the orbit up to here what you see is going all the way up and then up to here these points contain the yellow by yellow orbit I mean this is the orbit of 0 it lies along these towers what about if I go further if I go up to here then I see all the yellow and then I start seeing this I start doing the cutting so so it's also contained in the red lines by red I would have to include this everything so let me give you an example is it clear what I'm saying so if you haven't done the exercise you just have to believe that if you plot the orbit up to k plus qk minus 1 you are here if you keep going I will get up to here and I am somewhere in between my orbit is somewhere in between so for example my orbit could be let's make this bigger my orbit could be like this for example in another color can I try to put it back would that be good is it enough let me check to Michael for expert am I still okay so so what is my orbit until n so this in blue I will draw the orbit until n it could be like this it has to go up to here sorry this is the k, k towers these are the towers k and this is qk and this is qk minus 1 that's what I meant in this picture so if I plot my point it starts from here up up and then it starts doing this run of cutting but it stops at some point in between say here this could be my orbit for n and now I just have to understand what are the intervals in this partition so what are the intervals in this partition and what are the sizes of the intervals in this partition do you see the trigger because this what I'm cutting has the same sorry this width should be the same width than this small tower these and these have all the same width and my blue my blue is doing my usual partition plus this extra run up to some height so do you see the trigger in this picture anybody sees so I want to say that the floors in which my orbit cuts the picture have only 3 possible lengths so I need 3 colors ok so these floors have all the same lengths but the lengths of this floor is the same also here here and here so all of these are gaps of the same lengths ok what else well in which tower floors here is an interval of different lengths this is going to be approved by picture so this is the second gap one gap length second gap length and the third is that and this is the third gap length these colors don't show very well maybe on the board but how can I do it so this is green this is orange and everything else was did I use twice green ah it doesn't matter I think I understood this is yellow third gap maybe we'll do it ok so you see it's very visual and I can now ask you compute the frequency what are the lengths of the gaps what are the values and what are the frequency and how often the gap appears and how often the length is such and actually all of these can compute through the Gauss map because this would be functions of the continued fraction entries and you know that the Gauss map is ergodic, is actually mixing it's uniquely ergodic so you can compute the frequency of a currency of continued fraction values and this will tell you how often do you see a gap of a certain length so a lot of fine information about the distribution of gaps can be inferred from ergodic theory of the Gauss map in some sense ok and one more that I will do in detail and then I will start hand waving some words about the rest so one word which is another key thing is the jokoksma inequality this is the inequality by the jokoksma and again if any of you studied circle D fields you will know of this maybe also people in numbers here actually use it to see a proof through towers ok so you take a function this is the theorem about Birkov sums so I take a function from the circle which has mean zero and I want to assume that let's assume that f is c1 differentiable continuous derivative so how many people know what bounded variation mean almost all in function you just need bounded variation I'm not going to define variation though so it's true for functions of bounded variation and then I take my alpha say if you want irrational let's make it irrational and qk are the denominators and I want to study Birkov sums so I asked indeed David to set the notation sn of f sn of f is the n-st birkov sum so I'm summing my function from k from zero to n-1 and I'm evaluating f at the orbit of x so these are the sums which appear in the ergodic theorem and we know oh maybe I don't know if you proved irrational rotations are ergodic actually we proved that they are uniquely ergodic because we proved by theorem actually that sn of f is actually small o of n so if I divide by n it goes to zero but you can ask what is this error how fast does it go to zero what can I say and what is really key of the rotation so sorry this is a side this was just too I can leave it but there's not part of the differential maybe I'll put it here this is small o of f but when n is the denominator what this says that sn sqk so there exists a uniform constant such as that for every k sqk of f is uniformly bounded so in the infinity norm is less than constant so birkov sums are bounded at special times given by denominators by curiosity, has anybody seen this than zhoksma inequality before ok, almost know it, but I can tell you in many fields it's really basic and especially in circle df ok so I want to prove it and voila I will sketch the beginning of the proof and I want to use towers again so first of all I know that f has integral 0 if you want I claim that this implies that the integral of sqn of f is equal to 0 this is just the sum in k up to l minus 1 of the integral of f composed with r alpha to the k but here I'm just using that this is measured preserving so actually all these integrals are the same if you remember the lecture of did irane, no, irane did you do it, no rotation preserved just to change the value of rotation so these are all 0 so in the sum of qn 0 integrals is 0 ok, so the integral is 0 so the integral of this birkov sum is 0 hc1 the proof for bounded variation has to be slightly different so this implies that there exist a point there exist an x0 such that sq sorry, qk I was calling them, it doesn't matter sometimes I call them n such that sqn of x0 is equal to 0 so have a point where it's a 0 it's enough to show it's enough to show that it's enough to show that there exist a constant such that for every x if I look at the birkov sum at x if I look at the difference between x and 0, 0 is my origin this is bounded by c why is it enough why is it enough someone said something no, I will use I have a point I don't know where this point is but I have a point where it's a 0 and I can't compare any point with sorry, maybe so you want to use that point where it's a 0 but maybe actually so I want to say I want to compare any point if I can compare any two points then I can compare for example I can do like this I can compare this one I can add and subtract the point where it's a 0 what did I want to say you can just say this is less than I have to add and subtract 0 let me say any point can be compared with yeah, that's fine that's I'm happy now everybody is solving, good sorry yeah, 0 should be in the middle you want to compare everything to 0 and 1 is 0, that's right, that's what I want so I add and subtract ah, yes, yes I add and subtract 0 and here I put x to 0 so I want to write like this sqn of f of x is equal to I'm adding and subtracting 0 and this by definition is the point for which is 0 so everything is equal so this is 0, this is added and subtracting and this is less than no and yeah, and these two are less than 2c ok, happy ok if I can compare every two points then I can compare with my 0 point ok I can compare everything with 0 ok, so how do I compare with 0 so the picture I want to use my towers so I'm going to look at the k towers because I'm going for height n towers because I'm going for height qn so this is qn and this is qn minus 1 and I know where the orbit of 0 is so this is the orbit of 0 of length qn ok and I don't know where my point x is say that it's here, you have actually to discuss two cases, the big tower it could be in the big tower or in the small tower and I will only do the big tower which is actually easier say that my point is here and what does it do it goes up and then it comes back the length of the orbit is qn, so it go up and then I come back and reach the same height when I do qn maybe let's do it in a different color like orange is that for every let's say xi which belongs to the orbit of x up to qn I can find there is yi which belongs to the orbit of 0 which belongs to the orbit of 0 up to a height qn maybe so qn minus 1 so I can pair points of the blue orbit with points of the red orbit so that so that xi and yi belong to the same floor to same floor so that the interval let's say that the interval between is made by xi and yi belong to the same floor and they are all these joints let me show you I think it's more clear if I show you an example and so let me just show I just want to pair them like this for every orange point here I pair it with in the same floor and here too I pair them with the blue point it's a little farther away but it doesn't matter so this will be an xi and this will be the corresponding yi this will be an xj this will be the corresponding yj so you see all these intervals are this joint and then I'm done then I can write sqn f of 0 up to rearranging the order of the points this is equal to the sub from 0 to qn-1 of f of xi minus f of yi because each point is paired with another point and I can rearrange my sum like this do you see and we are done so now I want to use boundary variation mean now you can use boundary variation if not you can just use that this is less than the sum up to qn of the integral from xi to yi of the derivative and you can bring the integral inside f prime so this difference I just used that it's less than the integral of the derivative and this interval so this is less than the integral of the derivative on the circle and the function is integrable so this is integrable this is continuous so this is a it's a finite constant ok so if you know the definition of variation you can replace actually here the variation of f this is actually less than the variation of f so this is so I haven't finished the proof there is this other case and I think I running out of time so I may give you an exercise with a hint to how to do the small tower case but hopefully I gave you two proofs which use the towers which kind of gave you an intuition of why towers are useful and from the jococtma you can also study if you want to study Birkov sums of a rotation for a function you can actually for other n you can decompose n into these qk's this is I'm not telling you how to do it but there is a way of writing other n's as a combination of the qk and the ik and then you can bound Birkov sums at other times with something like constant time the sum of the ak up to this kn and this kn is actually less than log n so basically what you can prove is that you can kind of study how fast or how slow Birkov sums grow for other points and prove that they have some bounds which are logarithmic and do very fine information on deviations of ergodic averages they are called or discrepancies and again here you see then for other n's the entry of the continual fraction play a role and that's where the gauss map and the ergodic theory of the gauss map will give you again a lot of information so at special times at qk, qn everything is bounded and at other times you can use the gauss map and continual fraction to help you out so so I've almost forgot at what time did I start so I have like 5 minutes something like this so now I will just I will show you what what did I want to show you so I will not tell you about about I thought I will tell you something about what do you do when you have homeomorphism and if you are interested you can chat with me separately but if you have f which is now not the rotation but the defiomorphism you can actually look at something like the point 0 and f of 0 you can cut your circle and f of 0 and you can actually do start looking at images of this interval let's call it delta 1 and you can the image of this interval will be f of 0 is here f square of 0 will be there so your interval will map it will not have the same length when it maps here but then you will have some point f cube of 0 dot dot dot until some f a 0 of 0 so somehow the images of this interval don't have the same length but they come next to each other from the left and this will be kind of the reminder you can call it delta 2 so I want to say that you can do exactly the same algorithm for homeomorphisms and through this you can actually find something called the rotation number and indeed prove some basic results so I will skip this slide because I have put a slide where I told you some Poincaré theorem then Joie and Hermann and I just wanted to say that all these theorems you can prove with the different version of this algorithm but I skip it and I just conclude the last two minutes I really don't want to explain this but I want to show you this is the slide from a talk that I gave it's a result that we proved with Michael maybe a year ago in another Poincaré but it's okay it's a recent research it's about rotations really real rotations and it's about the central limit here so again I'm not showing you this because I want to explain it really but I want to just give you a flavor so we look at really the rotation and we look at the function which is a piecewise constant it's the characteristic function on an interval made with zero and we plot vehicle sum so we can use this function and the rotation to walk to have a walk on zero one cross r take a point and plot what happens when I rotate it by alpha and add the function as my point move I record the values of my function and add them up on the real line so I do rotation and then I add the value of the function at that point so I rotate and I add the value of the function at the point where I am so I can plot and get kind of a walk like kind of a random walk so I plot and I record where I am and add the function which will move me up or down in the rotation and now you can ask basically the values of the vehicle sums of the function over the rotation tell you how high I am on the real line and then you can start asking well how is this random walk distributed how much time do I spend at a certain height when I move and you can define kind of an occupancy variable so I fix some height and I record how many times I enter this strip in time n and divide by n so this is like the distribution I can get a random variable which tells me how much time I spend at each height and when alpha is a quadratic irrational and the jump is at one half you can prove that you can normalize this random variable and in the limit you see a Gaussian so this is kind of a central limit here and we put it probability and Dolgopiat and Sarig recently extended this back theorem to an irrational and initial point and the recent paper that we have with Michael Bromberg is about proving this is the real statement so look at the vehicle sums you can center them and normalize them so that this time that you spent in a strip converged to a Gaussian this is the central limit theorem for the rotation and maybe you are not surprised but in hyperbolic system, central limit theorem are very common but in entropy zero system it's hard to see the central limit theorem and here you see it, but you see it on a logarithmic scale and it's a trace of the Gauss map and the hyperbolicity of the Gauss map that you see but ok, it's not important the result I want to tell you there are still non-trivial things you can prove about rotations but this is a slide from the talk I didn't change it 2 is in the proof, normalization we use the classical algorithm by Gauss actually an extension of this Gauss map which produces the continuous fraction you know that and then there is an extension which produces the Ostrowski's punch ok, and we use symbolic tool recording using the towers and then we use central limit theorem for Markov chains ok, let me show you the next slide do you recognize this slide so you have an interval and you start cutting from the left and cutting from the right this is our algorithm and you have a reminder and then ok you record the position of the jump behind this algorithm and then you renormalize and here I was flipping and renormalizing and you get the continuous fraction and you get the Ostrowski's punch and I show you this is the last slide if you don't recognize anything we use the towers for the rotation you have two towers here we cut the tower at the position of the mark point and we do cutting and stacking and this cutting and stacking then we use it to code the dynamics of the rotation through the towers this is really taken from my talk and you get the Markov measure on symbolic shifts and then you use it here for Markov chains so I just wanted to convince you that I'm still using what I learned as a PhD student and what I taught you yesterday so this is the end of the class next week we'll see more renormalization in another context have a good week