 So maybe we'll start with an announcement. So I think you should have received yesterday I sent an email to the mailing list. Did you all get it? So I recent for people who were here last week. I recent to the link to the virtual part So I hope you all know what I'm talking about. I thought I will show it to you But I don't have the link ready now. So this is this special screen where you can write Anonymous Lee questions so we'll have a panel discussion today at 430 and Yeah, we are happy to address or try to address or Questions that you might want to ask Through this virtual pad, but I wanted to show you. I think in Lucia is not here, right? Okay, so that was the announcement. So I'll go to the virtual pad and try to write questions you might have them and We'll tell you maybe later about the Thursday event and the information fair. So Okay, so today it will be in some sense a connecting lectures. So I think I hope You could follow them Sturmean sequences characterization the characterization of these cutting sequences of a line with a square grid Which also called the linear flow on a square or the linear flow on a torus and So first of all, I want to spend maybe the first 10 15 minutes by telling you something which is Closer to kind of What my it's actually my research and closer to what people have been doing more recently inspired by the square and Then I want to connect to true Translation surfaces and the world of IT's where we will see yet another example of renormalization So the first Whatever I do on the board I try to do slowly and this what I really want you to understand Whatever I will do on this lights is more like a picture of what's going on Closer research. So I don't if you if you get lost in the this part Don't worry. It's just to give you an idea of What's next and let me tell you also the story of this. So I'm going to tell you something that we did with John Smiley few years ago and There's done a prosecution that I will mention which is also joined to work with the rena pasquinelli who is here and another postdoc Diana Davis and It all started because when I was an undergraduate student I actually read this little paper by Caroline series, which I try to explain very much in detail to you And when I read it, I don't know I feel like but when I read it as an undergrad I was like, you know, wow, this is beautiful. It's this picture of renormalization I really fell in love with it. And so I kind of I had it in my mind and I kind of and at some point we were discussing with John Smiley and We were looking at the octagon. So So you could ask yourself, okay, we started with a square, but Can I do it with other polygons? So let's see What happens if I have a regular octagon and Again, imagine that you glue like we did in the torus you glue opposite sides you identify opposite side by a translation and If I have this octagon like in the torus when I go out I have to come back to the opposite side So I can't define a linear flow And let me show you a picture So what do I do? I start from a point of my octagon and I travel along a straight line with a certain slope and When I hit the boundary when I hit a side like in the torus. I come back To the opposite side of the opposite side, okay So you go out come back and you keep doing like this when you go out here B is glued to the opposite side B So I come back, okay Okay, so I travel along a line and use the gluing of the opposite sides to come back and What happens if I hit an angle if I hit an angle I stop so the process is not well defined at angles But I'm only going to care about lines which are by infinite So I'm ignoring those which hit the corner. I'm just looking at my infinite trajectories Okay, and you can prove those which hit the corner are actually countably many directions So I'm ignoring a small set And I can do the same thing we were doing with the line in the square We can code it by recording the sequence of sides which are hit So here I had for the square I had two pairs 0 and 1 Here I have four pairs and the alphabet here is a b cd. You could use 0 1 2 3 Just I had the slides a b cd So we can record the cutting sequence. So I travel along my line. Let's do an example I travel I cross the side a I record an a I cross the side B next I record the B then I cross a again C D I can record in a by infinite sequence There's pairs of glued sides that I hit Okay, clear the problem and now you can ask exactly the same questions Those sequences will not be random sequences in a BCD Can we characterize the sequences which appear and if I tell you that a sequence is a cutting sequence of a line Can you find the direction of the line? And what we did with just my name we gave a full answer to these two questions exactly in the spirit of Caroline series Presentation of Sturman that we did yesterday and today. So really doing some geometric and combinatorial Normalization and I was really happy because when I was a student like I love this paper And when I kind of find out I can we can do something similar that was really kind of enthusiastic that yeah But so my I already told you this to people who are here last week when you learn something if you learn some little tool Try to which is important to you learn it. Well, right understand it really deeply understand it So sometimes tools which are yours are those that then you can export in your own research So spend time to understand fully something in details until you have a clear picture in your mind because those pictures You can carry them with you as your weapons to attack future research Okay, so we start by looking at First of all by some symmetry consideration, well, maybe we can actually just look at the upper upper Half plane and again, you remember for the for the square We were looking at the upper quadrant and we had two types. We had elementary restrictions according to whether I was a Bob in zero pi four or in pi four spy, right? And this restriction had to do with which sites can follow each other Here I can only have after a zero I can have another zero and okay You can remember so we first lose these elementary observations I look at my geometry and they make elementary observations So let's assume here there will be eight types not two types But eight types say zero to seven and each type corresponds to a small sector of of Angle pi over eight so if my trajectory is in this blue sector, so horizontal to pi over eight So you can kind of if you want you can use the symmetries of the octagon to bring you back to this small Sector it comes from symmetries of the octagon If my trajectory is almost horizontal and if it hits a side a I Cannot go up if I'm almost horizontal. I'm bound to hit D next Do you see after a I have to eat hit the D if my trajectory is not very more than pi over eight I cannot go up so I can only hit from a I can hit the D And I record this by putting an arrow from a to D after a I have to see a D Let's do another another one after the D my trajectory remember is Horizontal to pi over eight. I don't know exactly where but after a D I can hit a B or I can hit a a But I cannot go down because I have to go in this sector So after a D, I can hit a B or an a okay, and You can think what happens after a B. You can hit a D or a C and After a C you can hit another C or a B, but you cannot hit anything down there so I record this as a little diagram and An elementary restriction is that my trajectory my cutting sequence has to Give me a pass on this diagram. So if I read through my sequence I After an a I can see a D after a D. Maybe I can see a B. Maybe I can see an A but I have to Reading my sequence gives me a pass on this graph Does it make sense So this is true is my slope between zero and pi of rate. This is type zero so definition of type zero is being giving me a pass here and You have to do the other types you can just use the symmetries of the octagon to perm to permute it If you want you can do same argument for the other eight octagon eight sectors and you get a similar Diagrams if the direction is here. You have a similar graph the zero. What these are eight possible types and the first definition is Sequence is admissible if it is one of these eight types so How does it work? This corresponds to the two types and the octagon and the first lemma which is elementary in just indeed Staring at your picture. What we just did is that an octagon cutting sequence is admissible So if I have a cutting sequence the slope will be in one of these eight sectors and for that sector I will have to leave in that diagram. Okay, so this is admissibility It's more a little more sophisticated than the square But it's still elementary by looking at the picture and making some geometric consideration So cutting sequences are admissible in this sense Is this enough? No, we need an operation like derivation We need to bring it to the next stage by asking this admissibility on many scales on many levels And this is again, you need a derivation operator like we did for sturmium and I'll tell you what this is kind of Maybe make magic not magic. It's more surprising. I didn't we didn't expect it when we started Okay, let me tell you just is a combinatorial definition So let me say that the letter is sandwiched a letter is sandwich like see is sandwiched If before and after I see the same letter So see is like the ham in a sandwich of bees So this letter is sandwiched And now here our derivation which is particularly beautiful in some sense, so it's Given a cutting sequence the derived sequence omega prime the derived sequence Is obtained by the following rule keep only sandwich letters So let me see. This is an example This is a sequence and as I scan through my sequence. You see a is sandwiched great I keep it D is not sandwiched has a and B. So I drop it B is not sandwich So I drop it the next sandwich letter is a bee preceded and followed by C and So on this you scan it the next sandwich is a Then see sandwich this be the sandwich Okay, you scan and you keep only letters, which are sandwiched, right? and Now we say that an admissible sequence is derivable If the derivative by this if I take my sequence I keep all the sandwich letters and the new sequence of sandwich letters is admissible So again, it lives in one of these eight possible types and The sequence is infinite the derivable if I can do this infinitely many times. Yes So we are kind of ignoring the place we start so we are also yesterday You are looking at sequences up to shift So I think a by if by infinite sequence you're right So I don't know how to where to put the center and I'm ignoring it So if yesterday also if you wanted to do a more detailed analysis with where the center is you need to work a little Harder here we have never done it So I just think of my infinite sequences not as I do not place the zero Okay, so this is kind of surprising derivation. You don't know if you Comes from but so what we proved is that octagon cutting sequences are infinite at the rival in this sense so you see it's a theorem similar to in the flavor of the Sturmean and It's not only for the octagon It's for any regular to end gone. So that cagon dot the cagon whatever and You can also do it for odd gongs like the pentagon. You have to take two of them. Okay. This is Diana Davis didn't her PhD thesis and Okay, you remember yesterday we said it's almost an if and only if so it's infinite a derivable if and only if It's in the closure of cutting sequences This is not if and only if not even for the closure But we also have a second theorem which is if and only if and this is in the spirit of the substitution Version of Sturmean. So we also say we can explicitly tell you a recipe to write substitutions such that cutting sequence Sequences in the closure of octagon cutting sequences if and only if you can write it as this Expression by substitution. I hope You remember this is kind of what we did at the end of yesterday So there exists some indices from one to seven and some letters or such that omega I can write it as a limit of Applying sorry, sorry, sorry applying the substitutions to some periodic sequence You remember this is the second formulation. This is an if and only if yes Yes Yeah, so there is this little story. So actually so exagons are the same than squares They are actually square with a marked point You can cut and paste a square into an exagon and our sandwich letters work well for the exagon And gives you a different way to characterize, but actually is the right here. Yes So this is the right I came to Bristol to do a summer research project And I think the first thing I asked you was indeed to check what happens for the exagon and the square And she started doing that and then she read the paper by smiley and then I'll tell you what happened next in maybe maybe in 20 minutes or so, but and then we did we had the research paper which started from Generalizing this octagon picture to Something else which will come next. I will tell you okay, but so you can ask She knows everything she can give you her stage thesis. Yes Okay, so I'm lying something I'm Hiding some details under the carpet in reality the substitutions are not on the same alphabet or in a different alphabet and You need to change to convert them back to the standard alphabet at the back And actually this theorem does not appear in our paper of that time. It's something that I realized Maybe it appears in your in your thesis. Yes, so she wrote in here and I wrote this form Which I realized only some years later that you could do these substitutions. Okay Okay, and I want to tell you one key idea So the key idea of this result is all is the same is the same What was the key idea? the key idea was to prove that and the derived sequence of an octagon cutting sequence is Again an octagon cutting sequence by giving it a geometric meaning that was the key lemma for the torus The derived sequence of a square cutting sequence is a square cutting sequence And here we want to prove that the right sequence of an octagon cutting sequence is an octagon cutting sequence How did you remember what we did it? How did we do it for yesterday? We compared The square cutting sequence with the parallelogram cutting sequence, right? And we proved that maybe in k steps, then you get the derived sequence and We also used kind of randomization So we need but then this each parallelogram I can map it back to the square Restart the process So what do you need to do the octagon you need a geometric map to normalize the octagon and Luckily such a map exists and it's another kind of magic or beautiful thing which was known To beach bill beach who died recently who's also a very important figure in the study of Interval exchange maps and translation surfaces. Okay, so I will just show you one map. So So if I take let's take an octagon and let's apply to it this matrix So this is yesterday. We used you remember 1 1 0 1 or some This is minus you can ignore sorry say 1 then you have this 2 1 plus 2 2 0 1 This is the linear map and if I apply my octagon the image is sheared So this is a horizontal shear of a certain amount Okay, so I have an octagon and I can and the image is this all prime is this image by this shear and You can see that I wrote to my octagon in a grid made by rectangles and trapezoids and So this is just for ignore the grid for a second. Just the octagon is sheared here and The magic of the octagon is that I can cut and paste like a little puzzles I can cut and paste the sheared octagon back to the original octagon So beach Remark to this this all prime octagon you can draw it on paper and you can cut it along these lines that I drew and Your eight nine pieces will reshuffle back to make an octagon. Okay, you can make a little puzzle like you make with the squares So this sheared octagon can be cut and pasted back to the original octagon and Okay Okay, so the key lemma is that that the right sequence is keeping the sandwich letters actually gives you the cutting sequence of a Certain line the same line the same trajectory, but with respect to the sides of this sheared octagon Like yesterday, it was for the parallelogram here I can actually draw my line here if the line. Sorry. Actually. It's a lie if the line is in this horizontal if the line is in between 0 5 8 I Can draw my line here or draw my line inside there By this puzzle I can draw the line here and look at when it hits these Slanted sites and it only hits the slanted sites where the original letters were sandwiched. This is what you want to Check this is the key lemma that I'm not going to tell you But once you have that you can normalize renormalize you can apply the inverse of my matrix and Line here by applying the inverse of the matrix will map back there. So Sorry, we do it again If I have my sheared octagon and I apply the inverse the sheared octagon goes back to the octagon and the line changes so The derived sequence is actually again an octagon cutting sequence in a new direction and Then from here you can prove that cutting sequence are infinite the rival ball and also all the other results That you want and bow the octagon So I don't want to really explain this in detail, but I want I hope you see the analogy I'm doing something very similar to what I was doing for the square and it works for Regular polygons also and I really need this idea. Okay, and Now I want to move on and I want to move to surfaces. So we already did this for the square So you remember when I started with the square? I told you if I glue opposite sides of my square. I'm actually on a torus and My line here is actually a trajectory of some flow on the torus So I want to do the same for the octagon to start with Okay, so if you got lost in the octagon part Ignore it. It was just for fun and to tell you something related to my research and Think that people are still doing I will tell you related to this. So so if you are lost Don't panic wake up and listen from now. Okay, so it's ketchup point Okay, so this should I want I think some people know maybe not everybody know so what happens if I look at my octagon and glue This opposite side. So I glue a with a B with B C with C Do you know what I have? Ah, so you're right. Good question. I'm gluing by translation So I'm good. I'm taking the unique translation which maps a C in to C So I have a surface if I glue, you know, what is the genus of the surface? Huh, okay, so I don't know if everybody has seen this I think not so let me give you a movie a Movie so if I have this octagon, what do my mathematician do they reduce themselves to the previous case? So so first of all I want to claim. Okay, that's my picture. I claim that this with glued sides is that it's a pretzel It's surface of genus too. So how do we reduce to the previous case? I can fit my octagon in a square piece of paper Okay, what is an octagon? It's a square without with cut out corners It's a square where I cut out the corners. Actually all this this thing is I think it's I learned it from Anton Zoric So I think it's good to Draw the picture, but he has some pictures in his flat surfaces Survey, okay. So what do you do once you have a square for the square? We have to glue a with a all the and C with C So I hope you all believed yesterday that the square gives me a doughnut a torus What happens of these corners these missing corners the missing missing corners glue to each other to make a little rhombus here Does it make sense? So if you look at his picture this little Triangles Form a like a kite a little kite So I have a square is the torus a square a square with cut corners Is a torus with a rhombus hole? So I need to cut a rhombus hole from my doughnut And now what do I have to do of this rhombus hole here? I have B and here I have D so I need to glue B with D and D with D I still have to do the gluing of the corners of the other four sides So if I glue B with B, I'm left. I'm squashing I'm squashing my Rhombus, and I have two holes with D as a boundary and now you have to glue the two holes And now it's a little movie tick tick tick when you glue let me show you that when you can deform your surface and Glue them like this and that's where you get the second handle and the pretzel, okay? Okay, and your Trajectories here will actually be a line the in by infinite trajectory will be a by infinite line on this Genius to surface it will be a trajectory of some flow as we will see now Okay, so let's generalize this So let me let me give you so take a Polygon and maybe take even a collection of polygons. I will stick to one But you can take more than one a collection of Polygon polygons I write it with one but you can take two three But your polygon has to have the following property such that For every side for every e for edge side of P There exists another side e prime of the same parallel Parallel to e and of the same length So this is what we had for the octagon. So we had each side has a parallel side of same length and And in addition actually you want them. I will not write formally this but you want So this is good if you have a side like this with that Which is to the right of your polygon and a side which is to the left of your polygon. This is okay This is e prime, but you don't want You can have a polygon which has I don't know two sides like this This is no you don't want this you want them with with on opposite sides of the of the of your polygon You can say this by orienting your polygon and then you want the boundary to have let's see Opposite orientation same only I mean, okay, the picture is so you want them to Okay, and now you want to so glue glue pairs e e prime by using The unique translation Like we did from the top from the octagon the unique translation in R2 Which maps e2 e prime. Okay, I just to I identified them by a translation and Maybe let me call it like this s. I'm going to call it P Module of tilde where this tilde is just the gluing of the sides P is equal to the tilde is a surface and Actually, it's what is called the translation surface I will tell you in a second a more formal definition, but let me say it's a translation surface say translation, it's a surface and it's a translation surface because They say gluings are made by translations. So this reminds you that doings are by translations and Let me add or not give you any another example You can for example take four sides, whichever vectors you want Let's call them a b C and e I'm going to make a non-regular octagon with pairs of glue So draw your favorite four vectors like this and then say that we repeat them in the opposite order So this is a B this is You have to make them parallel C and of course my picture is not good, but it will close out if you take four vectors the sum of four vector will be So well b see Yes, a is maybe too big. Sorry. What am I saying? I doesn't matter I mean, so, you know when I want to see when I'm gluing a with a I want to make a piece of a plane So if I glue two sides like this I get a piece of the plane if I go to sites like that I get a folded piece of a plane. I want my gluing to give me some piece of a plane That's what Okay. Okay. Okay. Okay. You just want this arrow here. Yes. Yes, absolutely. You repeat the same. Yeah, that's what you Oh, thank you. Let me clarify. I haven't understood the question Thank you. I thought it was about this gluing here, but yes Okay, so this is again. It's a surface of genus, too But it's a different Okay, let me say let me make some Let me make some remarks or maybe some facts so first of all We say that so so if P1 and P2 if you have two polygons which can be obtained Can be obtained From each other By cutting and pasting I will give you an example in a second by cutting and pasting by translations Let me give you an example that you already know Say that your first polygon is P1. It's a square and your second polygon is the parallelogram P2 is 1 1 0 1 of the square Okay, so I imagine these two polygons P1 and P2 Each with opposite sides identified You'll see that you can cut and paste one into the other. So if I take If I take this side I can if I take this triangle I can cut it cut it along here and Move the triangle here by using the gluing by translation I can identify this diagonal with that diagonal and I get exactly the same polygon P1 so if two polygons differ by cutting and pasting where I move things by translations in the plane then I Claim that the I get actually the same surface The glue if I glue opposite sides, I will get a surface I cannot distinguish from which I came from I get exactly the same translation surface And Then so I don't only get actually a surface what I really get when I glue a polygon It's more than a surface. I get a surface with an Euclidean locally Euclidean structure I get the surface which looks like the plane in almost all neighborhoods of points so So to This take is that as I'm going to give you in a second the definition of translation surface But this is kind of you can make a definition by polygons. We're taking this as an axiom and two is as is locally Euclidean Or it's flat locally Euclidean So it has a flat metric Euclidean metric outside outside outside vertices Versations of P and what do I mean by these? So if I take a point in the interior You all see that it has a neighborhood which is a piece of a plane This is a Euclidean neighborhood But and if I take a point on C Which is also here You can also find a very nice Euclidean neighborhood. These two things are Identified and when I glue my edges, I see a piece of a plane Okay, so everything is like every neighborhood on my surface every point in my surface unless I pick a vertex Has a nice Euclidean neighborhood. You can kind of build the surface by gluing pieces of paper Okay, and the consequence of this and the gluing by translations. It's also that There is a well-defined notion of Direction of direction tita tita. Well, there is a well-defined notion of direction tita So if I kind of go out here, I know when I come back Sorry, I know what does it mean to move in direction tita moving in direction tita is well-defined. So I can define There is a flow a Flow which I'm going to call feet t tita which moves At unit speed Along lines in direction tita in direction tita and this is called the linear flow or directional linear flow of Translation surface s Okay So this is what I was plotting for the octagon before Okay, I don't want to do anything. I don't want to write at the board anything more than that I really want you to think you have a yes sorry Luckily Euclidean means that me I have neighbor of goods which are really like pieces of plane Okay, let me let me read it if I say it now So I don't know so I'm very happy for you to think I have a polygon with opposite sides of equal lengths and parallel glued together Okay, so how many know the definition of a manifold in general with charts and transition with charts Okay, quite Many so let me just tell you I have one extra slide So, oh, sorry. Sorry. Sorry. This was a before I do that I had one example of two polygons which give the same and This other there is another example of two polygons Which are the same this octagon and that octagon can be cut it and pasting into each other by translation This is this puzzle magic that each discovered So if I glue this regular octagon or if I glue the sheared octagon by opposite sides I get the same surface the same translation surface Okay, that was my second example here for fact one, okay, so This is more formal definition So a translation surface is the closed dimension to manifold. It's a surface With a locally Euclidean structure outside finitely many points So what does it mean each point has a neighborhood isomorphic to our two and the changing of coordinates before? Neighborhoods are translations so you can write charts in our to and when you have you have to pass to one charge to another If you've seen the definition of manifold, you know that there are transition maps between charts This transition maps are translations over to that's a formal definition But this definition is not complete. Okay, this was my nice neighborhood. This was my Other nice neighborhood. The problem is I don't know what happens at this vertex So in the octagon you can check that all Vertices actually correspond to a unique point in the manifold in the surface I'm the one to spend time on this because I'm not gonna use it, but I'm telling you for culture So this in all the vertices are actually a unique point on the surface and this point doesn't have a nuclear neighborhood it has a strange neighborhood, which is called Conical angle singularity so it's and instead of having to pie around it this blue point has actually I think in this case, it's Six pie angle so a bit before closing up. I need to go three times around myself so if you want to flat model of this point on the surface it looks like Something like this where you have an excess of angle instead of having a piece of a plane You have to take six pine of planes and bend them around to close them up So that's a picture of a conical singularity. So the formal definition. It's a Surface where transitions of between charts are translations and finally many singularities look like this and And Also, I wanted to say this linear flow this going in a straight line. It's well defined Everywhere, but if you hit the vertex. So if you hit the vertex, I told you stop Why because if these points are all the same all these lines actually Enter these vertex. So, you know in a normal point I have one way to go out and one way to come in in direction theta Here you have many ways to enter this point And if you want to extend this linear flow on these singularities, you have to put saddles So what you the linear flow looks like it's a flaw with saddle points on the surface And that's reminiscent if you know about oil a carat if you know about Gauss-Bonne You cannot have a flaw without fixed points on a surface of genus, too You need to have some fixed points Okay, so this was a digression if you go lost Let's stick to polygons and maybe I think Amy had told me correctly How do you know if I just give you a polygon? What's the genus of the surface is? I will tell you Euler magic formula Euler magic formula. So if I take the number of faces Number of faces in my part if I have one polygon I have one face And I in the case a number of polygons if you want a number of polygons remove the number of edges the number of pairs of edges pairs of edges which are glued to each other and add the number of vertices But be careful the number the vertices vertices after gluing So in the case of the octagon it looks like there are eight vertices But you can kind of convince yourself that this vertex is the same than this vertex using the identification of A But D is actually the same than D. So it's also the same than these So you can convince yourself that actually all these points are glued to a unique point but if you compute number of polygons minus Number of pairs of edges plus number of vertices after Identifications this give you 2 minus 2g where G is the genus the number of handles in your surface So I don't know maybe I do have to check if it's right. Maybe I shouldn't check 1 minus 4 plus 1 It's right 2 minus. I should get genus 2 Okay, let me not check Okay Okay, so you there is a way maybe I'll give you an exercise. There is a way to compute I don't know the genus of your surface from the polygon. Okay Okay, so now I want to say Okay, so it was very special maybe I will skip something So it was very special to have if I take a random Oh, okay, maybe I'm sorry more if you want to know more about Surfaces I can give you some references The square and the octagon because they were regular but really because they are special had something very special. So they had They had some linear or a fine maps of the surface into itself so We saw this for so for the torus So if I take the matrix 1 1 0 1 and I apply to the torus You see I get this parallelogram But this parallelogram I can cut and paste it back to the square So I get a map of the square to the square exactly like the cut map This is a special case like the cut map in the cut map that Amy did very carefully in the first lecture If I apply an integer matrix to my square I can have a map from the square to the square by going using mod 1 mod z2 by Identifying by integers and this map you can think of it as an automorphism of the torus so there is a map from the torus to the torus which is linear it's actually smooth and it likes very much so and Actually, you also have this special affine map from the octagon surface to the octagon surface So if I take the octagon I share it by this special matrix The resulting octagon gives me the same surface. I can chop it back to the octagon So I have a map Diffuse from the octagon surface to the octagon surface, which is linear locally linear It's actually not smooth. Oh, I will not discuss the singularities, but okay And this is very special and just for your own curiosity These surfaces are called which surfaces the surfaces which have a lot of affine Diffuse which are rich of affine diffuse are called which surfaces and Reach means this is all digression Reach means that there are enough diffuse so that the linear parts generate a lattice in a cell to war so this Special this this if I draw a polygon at random like this it will not have a nice affine diffuse and People in the field of studying translation surfaces and in tech Muller dynamics have spent a lot of time trying to find Polygons, which are so special that they have a lot of affine diffuse and just slide only wanted to To to to have like a give you a taste of modern very current research So this is special surfaces are special dynamically for example for the torus or for the rotation we saw that trajectories of a rotation are either periodic or dense and uniformly distributed and This is a property which is quite rare in dynamics. Usually you have cases where there are trajectories, which are neither dense nor Periodic and all these special surfaces with lots of affine diffuse have this very strong dichotomy like in vile theorem Yes Yes, so prove it now. Yes. I actually can I have a very nice picture But it's sort of billiard, but I can I can show you a picture of a trajectory Which is not you can see that is dense I will show it tomorrow maybe and I cannot build it regularly, but yeah, but so okay So and I just said I mean this is like it's just fun. So you can actually You people have been listing so all all the surfaces which are special the square is special regular polygons Is special which discovered it and then there are some new families of special Surfaces and boom all their surfaces are others there are and people like Kurt McMullen and Alice Eskin Alex right so many people who are Big names in in like Muller dynamics are trying to discover new polygons, which have lots of affine diffuse So having a fine diffuse is rare and people study and search for this very special needle can you say needle in a In a haystack You're looking for this needle in a haystack and there's lots of research on this which is quite deep very deep And I'll give you one more example, which is beyond octagon because I promise to tell you what we did with the rena So this is another it's a collection of polygons This is one and this is another and you can glue them Following numbers one with one two with two and I told you you can get a surface by gluing not one by many Collection of polygons. So this is one surface and this is another surface and These are called boom all their surfaces They are surfaces which are glued out of Semi-regular polygons polygons where sites have two types and some rotational symmetry and okay so just keep this and this boom all their surfaces also have a fine Diffuse I have maps so they're very special and for those with with Iran and Diana Davis We we did the same thing that we did for octagon and and square So you can look at linear trajectories of this linear flow and try to characterize cutting sequences through geometric randomization and Okay, I'll just put it here because I wanted to advertise also Iran is what you can characterize these cutting sequences very With with the substitutions. There are substitutions and you can you can build the substitutions and build these sequences with substitution Okay, this was my digression Okay, and Okay, so now I just want I'm almost over but let me do one more picture before What did I start? I think I have out of time So let me just say one thing. I'll show you one picture and then I stop So what do we do when we don't have so maybe let me make this remark so typical or Random, maybe say if you pick at random your random Translations surfaces cannot be don't have Don't cannot be Normalized geometrically can can't be renormalized geometrically by renormalized Geometrically Maybe this is a line by a fine diffuse by a fine diffuse So what will we do tomorrow? So at this point tomorrow, so we will go to Poincaré maps Go to the Poincaré map to Poincaré map of the floor and I just put on this last slide The Poincaré map I put on I will explain this tomorrow. We start from here tomorrow. The Poincaré map will be Interval exchange transformation This is like interval exchange map We will we saw it briefly last week But we will do it carefully tomorrow. We'll define interval exchanges tomorrow and we will go to interval exchange And then we will define a renormalization combinatorial if you want a renormalization algorithm for it is and Okay, and and this is what you do when you don't have you can this renormalization is kind of a very powerful Tool for proving dynamic are results about these flows and these interval exchanges. Okay. Thanks