What's the word y'all? We got some more BR articles. I've done the last week basically the last seven days I reacted to a bleach report article and y'all absolutely loved it I didn't have any idea that y'all were gonna enjoy the article reactions more than the actual footage reactions But here we are now today We don't have a top 10 at this position this and that but we got the top story lines for every team in Orlando I'm gonna read through and I'm gonna tell you what I think should be the top storyline if I agree If this is a top storyline and answer some of the questions attached to these story lines Okay, so it should be fun set out to Dan for fail and uh and bleach report if you're new around here subscribe This is the way you find out if you're subscribed because a lot of y'all probably don't know You see this box when you're watching this video in this box right beneath it There's a little red button if it's red click it that means you aren't subscribed to the channel just yet and like 65% of the people watching these videos aren't subscribed so go ahead you watch it Anyways, you might as well subscribe to the channel and you watch it anyway You might as well like the video all right enough of that. Let's get into the top story lines for every team in Orlando I can say Then my favorite thing about Orlando so far is that we have so many players that are vlogging or going on Instagram I'll go into live like we get to see inside of the bubble. It's like we're kind of there in a way So sound to Evan for me a mentee stable. I just saw Troy Brown Jr. Just dropped the vlog Javail McGee These are all players that are vlogging their experience so far and I'm guessing we're gonna have more players too So thank you to y'all because we've been missing NBA content Okay, first question is for the bosses Celsius is Jason Tatum ready to spearhead a legit title contender that I would say That is the the big question because Jason Tatum is blossoming to the best player only bosses Celtics roster Is he ready to be the number one guy on a championship team? I don't know it might be too early But Jason Tatum is that special where maybe it's not too early and he can actually do it I saw in the comments of like one of my last videos one of the top comments was like they could see Jason Tatum having a Dirk like career where like he stays a part of the boss of Celtics and eventually somewhere down the line gets the Celtics back to a Championship and I could see Jason Tatum being that type of guy as well I don't know if they're ready this season, but I'm not counting them out either Yeah, I'm saying no matter who they're matched up in this first round second round in conference finals They legitimately have a chance because they are such a nice team They're one of the few teams that are I don't want I think it's top top 10 and offense and defense that are in the bubble And they have a way as a team of winning very ugly games And those are the type of games that the playoffs are really about you know I'm saying some games Both teams are gonna be off and the bosses Celtics have found a way to win a lot of those games and Jason Tatum Is gonna have to be the focal point of that so not a bad question not a bad question I don't even have the answer to it. We're waiting to see next Brooklyn that's cares the bird goes a loan ish I Also see people ask it like what's the point of them even being there? Hey, one thing I can't says that carries the bird can mess run up his trade value for the Brooklyn That's to eventually try to make a splash for a better player Not saying that cares the birds bad or anything. Obviously he's an above average player But I think the Brooklyn that's are trying to get their hands on that third star And if cares the bird goes to Orlando and you can see like man cares the birds got a lot better Then that can help them potentially put together the trade I'm just throwing a name out there because his name is always in these conversations of Bradley Biel I don't know I again I don't think the Wizards are selling Bradley Biel But if they were to try to sell him a cares the bird goes down to the bubble and wins the Brooklyn that's some games by Himself, you're like, okay, if I'm the watch the Wizards, I know I have to trade Bradley Biel Maybe cares the bird is a guy I want to replace him with so cares the bird is up in his trade value They would like to keep specific then when he there But obviously he's not traveling he could have helped this trade value at the end of the day the Brooklyn That's are gonna be a fan favorite team because they have Jamal Crawford and Michael Beasley right now They may add more players who knows next The Dallas Mavericks, how far away are they from contention? That's a very good question because as y'all know They had the league's best offenses right here, but they get in the playoffs or slowed down It's it's a lot different than the regular season game So you wonder how their league best offense will translate to like the playoff basketball The ring of Ron Mahoney wrote. Oh, that's a lot Well, coach stir up the yeah, I don't know. I don't know if they're ready to be contenders to check I think that these eight games in the bubble for them though I think they have to succeed at the moment I wouldn't be surprised they jumped all the way up to like four maybe even three because I think they have a Favorable schedule and of course they have basically their entire roster suffered to wipe out traveling with them. So How far away from our contention are they we'll find out we'll find out next We have the Divernuggets slim Nicole Yoke is right So we're trying to figure out if this guy this this guy right here that has definitely relied on his size a lot Him being slimmer. How much does that affect his game? for the most Wait Bleacher poor use this article isn't his fan flipping him off and Beard decided to use this article they wild for that one Yeah, like Nicolio's the guy that used his body a lot to baby smaller defenders and everything and now he's thinner and in most cases In most cases a player getting more fit has benefited their game and basketball and Probably pretty much every sport because you're just more agile you just faster you're quicker But a guy like Yoke it's doesn't need he didn't need to be faster quicker or things like that So I'm curious. I am also curious about slim Yoke itch and how much That will change this overall play. I'm guessing he still will be great. Will it be better? Well, he's just be the same, you know next we have Silly without Victor depot I'd well for good thing for them is that they have a good sample size of playing without Victor The depot right because Victor depot had just come back and only played 13 or so games or so So they had already played so many games without him Obviously, they're also still missing Jeremy lamb who had an injury to What is their ceiling without them? I I can't for the life of me look at this team and say that they are contenders Without Victor the depot and without Jeremy lamb, they're gonna give some teams some hard times in the playoffs But I'd be very very surprised if they get somehow got to the conference finals I'm like, yeah, you sup you should surprise me. You surprised me. Yeah, I don't think I Don't think they're ceiling is that that high without those two players next Houston's fadence in micro ball the the main thing that confuses me about the Houston Rockets is the fact that they just decided not To have a singular big on the roster they have ties to channel But he's basically 40 and he hasn't played other than like garbage time minutes But even if if I was the the general manager if I was there or more of the Houston Rockets I love the idea of the small ball obviously But I would always have that one player my tenth eleventh man on a roster who may not get PT Unless the situation calls for right my seven footer that's just sitting there.  They just signed Bob Mute Do they got somebody else too? None of those guys will even come close to what click I mean, I click appellate but PJ Tucker can do for this roster So there's just one injury away from the whole season being scrapped and I'm not even talking about an injury to one of their elite players And I don't also don't know if they run it to Yoke it's in the second round. Who is guarding Yoke it's Who is guarding the other big so they go against Anthony Davis, but anything is probably gonna have a 40-point average in the series. I Just wish that they would have kept one big on the roster just in case their security blanket But they didn't need it. They decided they didn't need it. Next. We have the Clippers. What's the closing line of yeah? You have so much talent. I would say that my question for them mostly Is I mean, we didn't have a lot of time of PG and Kawaii basically the whole roster playing together The games that we did have their whole roster healthy and playing together They were superb, but it steals a small sample size And now we're going to the bubble where these guys don't probably don't have the chemistry of some of the other teams That have played 60 so games together while healthy. So that'll be my thing. They close the lineup I'm sure that they have so many different Variations it's just gonna be situational.  They're gonna have to replace Rajan Rondo who's one of the better playoff players in recent history It sucks. It definitely sucks if I was a Lakers fan, I'd be pretty bummed out about these two players missing time But now they have JR DR waiters as Caruso is gonna have to play a lot of menace and obviously LeBron It's gonna have to take the the crown of being the playmaker and one of the only playmakers really on the roster next Justice Winslow's fit. I forgot Justice Winslow even got traded here, bro That's how long we have been without basketball. He actually this this jersey just fire I feel like this jersey would look great on anybody, but what is his fit? I have no idea I haven't seen that boy Justice playing so long. He played in just 11 games to see is that I don't know what he's good at anymore I don't know what he's bad at anymore So I'm just gonna skip this and hope that he works out I would say that that trade is a worthy trade though for Justice Winslow because I mean When he was playing well a couple years ago, he was looking really good and the fact that they traded Iggy a guy who wasn't playing for them anyway, and I think it was Jay crowd was also in that trade Who was actually playing really nice for them? But you get a younger guy that fits the timeline a little bit more and we know that his potential is still pretty high next Here same thing integration of Jay Crowder and Andre Iguodala so that was the bad thing about the whole shutdown of the season is that it happened like right After basically the trade deadline or maybe a month after the trade deadline So we didn't really get to see Iggy play He only played four ten games and not a lot of time alongside some of the other guys Eric Sposier didn't really have that time to figure out rotations with these new guys Both of these guys are good enough to be rotational pieces on on the team that's trying to compete But they they didn't really have the time to be like, okay Jay Crowder plays best when he got Jimmy Butler here or he plays best Maybe bam out of bios in the best because they need a little bit more space away Iggy on the court They really have a chance to really toy with all of that because of the suspension of the season And it's only eight games and these eight games matter It's tough. It's tough next Postseason Eric Bledsoe has been very very disappointing so far and maybe this is the year that he is like the X factor for me They can't win a championship if Eric Bledsoe decides to be subpar in Multiple series. That's what I'm saying next Ziya Ziya Ziya we all know big surprise big surprise. I know Zion is the X factor for the Pelicans They were such a good team when they had the lineup of Drew Lonzo BI D-Faves and Zion Wilson that lineup was destroying teams and now we're probably gonna see that lineup full-time So, yes, biggest storyline is that Zion Williams in his playing basketball again, and he looks fitter than he ever has before Can you play themselves? Can the Thunder play themselves into a different timeline? Okay? I got to read what you mean by that Um that the present core lasted longer than the surprise Okay, I understand what they're saying. I think I Don't know. I just I'm just happy to see them in the playoffs at the end of the day They got my favorite up-and-coming young players say they got my favorite vet in Chris Paul So I just want to see them play. I don't even care about the timelines. Yadda yadda yadda I just want to see them boys play and give somebody some some hardships because that team is actually really nice Next team is your lendo magic roster to cross roads Yes, hopefully this is the season if they don't get out of the first round Which we can all bet that they won't because they're gonna be a lower scene They're gonna have to go against Toronto Boston or Milwaukee Bucks They can decide to make some changes to their to their team I saw in every four years vlog that John Isaac is in a bubble. So we don't even know if John Isaac is playing I hope he is because that gives me a reason to watch your lendo magic But we don't know if he's playing and hopefully this is the season. They're like, okay Maybe we'll try to record it. Okay, maybe we'll try to try to boost a bit in nature Not hit a reset because they have some pieces here But they try to re-orchestrate this roster from just being the six seven eight seat every single year because there's no fun in that There's no fun in being mediocre. It really not next Wait, what did I say? We're planning on life without Jonathan. So he's there, but it doesn't like he's gonna play Oh, that's tough. That's tough. Next is the 76er. So like if this roster work That is the biggest question They're the biggest Question mark team in the entire thing because obviously on paper this team should be amazing But they haven't figured that out But they had these moments these stretches when they look like the best in basketball But then they also have these stretches where they like, you know, I'm saying so cut this teamwork that that is They are the biggest X factor team if this team went on and got eliminated in the first round I would not be surprised if this team went on and got to the conference finals. I would not be surprised That's how big of a difference this team can look from day in and day out 76ers man 76ers fans Y'all y'all have it tough, bro Y'all have a real tough because you just don't know which version of the 76ers are gonna show up Okay, my boy, Mikhail get us some love in the article Mikhail Bridges is arrival now I see the Phoenix Suns they're not playing for that much. You know I'm saying we don't I don't think anybody expects them To to really make that jump to the 18. I'm not saying it's not possible But I think people are prioritized and dame and the trailblazers or pelicans and of course the grizzies who already have the Spot, I don't think people are really thinking that the Suns are gonna make it But without Kelly Ubrey what they're saying is that Mikhail Bridges probably gonna slide into that spot And he's gonna have room to grow and just overall as a team. You know, they're very young roster Overall as a team they have time to grow inside this bubble get some playoff It's not playoff experience, but see what the playoff atmosphere could kind of be like without fans next the return of Zach Hollins and use of nerca Jim and that's one thing that Terry Stotz has not had a chance to figure out his lineups and stuff because of the suspension of the season You should know kids is back and and Zach Hollins is back and when old Trevor Riza That means you're probably gonna get some mellow at the three Again, like it was old times and Zach Hollins and use of nerca's I don't know how they figure out the big man rotation because they have so many of them now It's not white side also be in there But it's something that we're gonna have to wait and see the Marvin Bagley third place in the pecking order I somebody said in my comment section I always skim over the the Kings and that's not by design. It's not like I dislike the Kings or anything There's just a very confusing team overall Because Marvin Bagley is confusing to me and their coach Luke Walton is confusing to me So yeah, I would love to see where Marvin Bagley places on the pecking order I would guess he's probably number three Maybe even number two on some nice. Then we have the end of the postseason streak. I would guess so We have no Lamarcus Aldrich But if anybody could do it is Greg Poppins, bro If anybody can do is Greg Poppins and line Walker cut off his hair I'm expecting big things from line Walker. Maybe not this bubble But it's like in the general of the course of his career. Next the state of the half-court offense Oh, they better get a system advanced stats They bring 18th and point score per half-court possession below the board. Okay bench trends transition attack So yes, they're great when running But obviously the game slows down the playoffs So how can they do that without having a go-to guy like Hawaii later like last year? Of course, they have Pascal Siakam who they're who they're relying on But we just haven't seen Pascal do that just yet And maybe this is the time he does because there is times in the playoffs last year where the Raptors were bleeding And they needed Kawaii to stop it and Kawaii hit the shot to stop it And they I don't know if they have that in the half-court right now So that is yes the biggest story last last question mark for them. Look at Cal Lyre, bro Then we have the Utah Jazz the really go beard dynamite your dynamic We've talked about a plenty of times here on this channel before but yes We got to figure out how they're gonna fair together and then lastly Who emerges without be old bartons and wall? Hopefully is Isaac Banga Hopefully it is Troy Brown, Jr. And I love watching Rui Hachimura. It's gonna have to be one of those guys Overall cool article cool questions and cool story lines if you enjoyed it leave a like if you knew subscribe. I'm out


Thank you very much for the organizers and you to be here. So let me start quickly with the topic of the course. So the idea is to analyze in this first lecture and probably in the next one what happens with instead of iterating one different morphism, you iterate two of them, which commute with each other. And in this start with the nicest situation, then later probably if we can we go and move on to do some more complicated stuff, probably making the group more complicated, but let's start with this simple case. Okay, so to introduce the objects, let me just start with one guy. Okay, which I guess a lot of you are more accustomed to. Different morphism from M to M. In general, M will be a compact manifold, but it could be something different. Ifomorphism, when we study these guys, there is one notion which is called hyperbolicity, which is very useful to understand them. So there is the ultimate hyperbolicity assumption, which is the analysis of condition. So let me define it is an Ossoff if half a splitting into a contracting space and an expanding space of the attaching space, so that this means in each point have the invariance condition. This is the first condition and the second is the contracting and expanding property. Okay, so this is land less than one positive such that C positive DXFN of B is less than C lambda to the N, B for every B. S and similar thing, but when you go to the past.  Okay, and the ultimate example of this guy is the matrix. This which has eigenvalues lambda, B plus, this should be the other guy, plus minus, plus minus quarter of five over two. You have the eigenspaces ES plus U, as it is to these guys, which you can compute quite easily. You guys, this is the ES guys here. You have the unit cube. This is not supposed to go through this diagonal. This is supposed to go, let's try again. Taking this guy is supposed to go through. So these are the eigenvalues. So this is the one, one, the zero. This is unstable, this is the stable, eigendirection. Okay, so and since the entries of this guy are integers, we can look at it as acting on the torus T2, which has used the same letter for the corresponding guys. So here T2 is R2 mod the integer. So if you take a point, plus the integer lattices will be mapped to a point, plus the same integer lattice. So it goes to this quotient space, and this is just the torus. This space is stable and unstable. You can put it just by translating over every point on the torus, and you will have your decomposition. You are looking for here, and you will have the contraction. You can take the lambda here to be exactly this lambda minus appearing here, which is the contracting guy. And one over lambda minus exactly the lambda plus, so you get exactly this condition, beautiful condition here. Now, this is a good example to understand, and you have a lot of properties. So it has a lot of imperial measures, you have a lot of periodic points. So for example, you can show that periodic point of F coincides with the integer points. F is A, the rational points of the torus are exactly the same as. As the periodic point, you can put it as an exercise if you want. It's a simple one with a perfect exercise. So this already give you a lot of imperial measure because you have a periodic orbit, and you can put equal weights of one quarter, one quarter, one quarter, one quarter here, and you have an invariant measure already. Also, the fact that the determinant of this matrix is one makes it this map area preserving. So it's area preserving on R2, and you have the natural area measure on the torus T2 that is also preserved. Okay, so this is also nice. You can prove ergodicity of the area measure using Fourier analysis, which is very useful. We are going to discuss very little of it later, not for this map, but for a little bit more general than this, so higher dimensional analysis of this. But one problem that it has is the following. This is the second exercise. It says that if B is here, so these are integer matrices, have another such matrix here, and I know that this matrix commute, I want to commute in matrices. As I said, the course is about commuting matrix, commuting different or fism, so I want another commuting different or fism. If this matrix happened to commute, then there exists a K integer, which is not zero such that B is plus or minus A to the K. So it could be minus this matrix, but this is not that much. Means that, of course, an iterate of different or fism will commute with a different or fism, but this is not really fun, so you're not adding any other guy there. And indeed, you have this exercise and you can try to go further and prove a similar statement for any matrix in SL2Z, okay? So if you are in dimension two in the torus T2, you will not have two matrices commuting, unless one is essentially a power of the other, or half a common power or essentially half a common power. So what we can do is we can just go to higher dimensions, so let's go to dimension three. Because, well, here I exclude zero because it's obvious that this, I don't want to delete the matrix. So, you're right, I shouldn't exclude zero if I want a correct statement here. For this statement zero, it's okay. It's not, I should say that it's not identity and they have a common power, thanks. Okay, so now I take a matrix in SL3Z and I looked at these matrix, so stream by three integer matrix with a terminal one, so I can make it act on the torus and I have a different fism of the torus and, well, you will have essentially, depending on what is the matrix, you will have a same picture as this one, but I can make it more delicate explanation of what it happens and that's what I'm going to do. So I don't want A to be the identity matrix, so identity has very trivial dynamics to discuss. So what you will have always is three numbers, which are essentially the modulus of the eigenvalues, so we have dimension three, so we have these three numbers and need to do some discussion here, which is, yet again, some exercise. You may have double eigenvalues, but then you will need to have modulus one if it is not diagonalizable. If it is diagonalizable, then this will be one and it's complex conjugate and still this will not be fun enough and the nice case is when the three guys are real and different, okay? So the exercise is to formulate correctly what I'm going to write here and solve it, okay? So unless lambda one positive, lambda, let me just state which are the lambda i's, so lambda i are modulus of eigenvalues, then the statement is this lambda one, larger than lambda two is larger than lambda three, lambda i different from one. All of them, we are in a situation this or completing non-hyperbolic. So completing non-hyperbolic meaning the eigenvalues are all of modulus one, indeed plus or minus one. Not interesting for us. Okay, so this exercise I will tell you that if you really want to compute in guys, you want three real eigenvalues, the three of them of modulus different from one. Okay, so that's the implication. If you have a complex eigenvalue, you need to conjugate, so you will have an equality here. So you need really real eigenvalues there. So there are such matrices, you can find them. It's not too hard. Here is one example, which is yet one more exercise to show that this example works, like this matrix. It has a determinant one, I hope, and it's eigenvalues are all reals, two are larger than one and one is between zero and one. And you have the matrix B. You can put two identity minus A. This will commute with A trivially. It will have a determinant one, and the property, if I make my computations correct, if not, you can try to fix them and find the correct guy. A to the K times A to the L is the identity. K, L different from the zero, zero. I'm sorry, K, L in Z, then K, L equals to the zero, zero. The only way they are a power of the other, and A, that's absolutely. If they have a common power, so to speak, then it's because you are doing the trivial thing. Then I have, generally, a map from C2 to SLGC, which is taking K and L into this power. And what this exercise tells you is that this is a one-to-one homomorphism. Let's call it wrong. And they have two different morphisms of the total T2 that commute with each other. Okay, now, now since I have three different eigenvalues here, I have the composition of R3 as E1 plus E2 plus E3 into the corresponding eigenspaces. And since the eigenvalues are different, you will get that B preserve the same splitting. So it's jointly diagonalizable with A, and it's just because it preserve this very same splitting. We have that A of EI is EI, and B of EI is EI. Also, it means that if I iterate, let's call lambda I of A, the eigenvalues of A, and lambda I of B, the associated eigenvalues of B. So if I consider this iterate of these matrices on a vector B1 on the direction E1, then what I will get is lambda one of A to the power K times lambda two, lambda one, oops, of B to the power L times B1. I get multiplied by these numbers. That's how this guy is acting on these spaces. So this is for every B1, E1, and the same happens with the E2 and E3. Now, since these numbers, this is yet another exercise, is these three numbers are real and different from one, you can show that not only this power is never the identity, but also this number here can never be one, another exercise. We are not going to use it right now, but it will show up later, I'm sure. Okay, so since these numbers are different from one, you have that you can put together the two numbers which has models larger than one, or if there are two of them, if there is only one of them, you just put it by itself, and the two numbers which are smaller than one, if there are two of them, and you get this splitting ES plus EU that you wanted. Okay, good. So this is the type of examples we want to study. So, well, this is the linear version. We want to study the non-linear version of this guy. Move forward and define what are the leapons of exponents in this setting. So we have this splitting of R3. Well, you can do it in whatever dimension you want. I'm just doing three dimensions, make life of everybody simpler. So I have this splitting to three spaces, and I can define a sheeted functionals from C squared R, which is chi i of the point in C squared will be the logarithm of the absolute value of lambda one to the a to the k times lambda one of b. So it's the logarithm of the absolute value of the number I am multiplying there on the EI direction. And this is what is understood as the leapon of exponents. In general, when you are lucky and you have eigenvalues, the logarithm of the eigenvalue is the leapon of exponents. And here you have this nice eigenvalue and these directions and the logarithm of the eigenvalues is this leapon of. But now the leapon of exponent is not just a number, it's a function. Because I'm playing with C squared guys, so I have really a function. So let me put here my R squared and inside the R squared is a C squared, of course. And I have three functionals. The one corresponding to the first direction, the one corresponding to the second direction and the one corresponding to the third direction. So I can naturally extend it, it's the linear map, then to map the linear map from R squared to R. So just instead of putting K, L, you put TS or allow K and L be real numbers. Then I have this linear function and the linear function is essentially determined by the kernel of it plus some normalization. And here the normalization will show up later, but the kernel is very important for us. No? Of the guy one, say here's the kernel of the guy two, say here is the kernel of the guy. It's a one. It's also one, yes. So the subindex, well it's an I, I'm sorry. This should be an I and this should be an I. So the subindex correspond to the direction on your applying your dynamics. In this splitting here, E1, E2, E3. So then once you have these kernels, another very important information is to which side of the kernel your eigenvalue, your level of exponent is positive or negative. Let's say I should have done this computation before. So let's say this kernel is positive to this side and negative to this side. Then I hope I can do this second guy positive to this side and negative to this side. Then if I do this, this should be probably negative to this side, positive to this. So the very important observation here is guy one of N, so for every N in C square, guy one of N plus guy two of N plus guy three of N. Okay, what happened with the sum of these three guys? This is the logarithm, the first eigenvalue plus the logarithm of the second eigenvalue plus the logarithm of the second eigenvalue. So it's the logarithm of the product of the eigenvalues. The determinant is one, so the product of eigenvalues is one. The logarithm of one is zero, so the sum of these three guys should be always zero. It means that if two of them are positive, the third has to be negative. So you cannot have the three of them positive and I hope I made things in the right way and you cannot have three positive. So a point here where the three guys are positive. Now what happens if you sit in one of these lines? Where is the kernel? So if this is the guy one, the kernel of the guy, if you are in the kernel of the guy one, it means guy one is zero. Guy one zero means that the eigenvalue is one or has modulus one. Here are real numbers, positive real numbers say. So if the modulus of the eigenvalues is one, so the guy is one, which shouldn't be possible unless you are irrational. So this means that this line will be totally irrational. Okay, so it's not a rational line. This needs to be proved, but it's yet another exercise. There is no integer point on any of these kernels. It's not that bad, this is, you can do it. Okay, so these are called, so you have here the termine, three different, several cones. Let's say, oops, we should go all the way up to here. So this comes with the associated guys. This is three cones and these are called veil chambers. So whenever you pick an element here, you are getting a lot of information, but in particular, you are getting the information that on which direction is unstable and which direction is stable. If you are a plus plus minus, so meaning the first is positive, the second is positive and the third is negative, then you have that the this and this are unstable and this is the stable guy. And in this way, you can look at these combinatorics and distinguish what will be the unstable space and what will be the stable space for these guys. Okay, what we do with this? So this is all the linear picture. Now, what we do with the non-linear picture. So let's discuss a little bit different morphisms of the torus T3 or TN for that matter. Of a map F, it's TN to the torus TN. A diffeomorphism. Doesn't need to be diffused, but since we are going to work mostly with diffeomorphism, it could be a homeo or it could be just a map. It doesn't matter. So some very basic algebraic topology, so it's covering theorems, will tell you that you can lift F tilde, or F to F tilde, lift is a tilde, is a map from RN so that the natural diagram, here you have the canonical projections diagram commute. This lift is not unique, but let's not worry about the non-uniqueness now. You can put some conditions to make it unique or you can study what's the non-uniqueness or it's not really big deal. Now this lift will always be of the form AX plus pi of X where A is an N by N, so I say it's a diffeor, so invertible integer matrices where the inverse is also an integer matrix. Okay, so essentially it determines this plus or minus one. So integer matrices with a terminal plus or minus one. And the phi function that is y that phi of X plus N is phi of X for every N, is ZN periodic function. So you can always write this diffeomorphism, so the lift as a linear guy plus a periodic guy. So being periodic means in particular this bounded, okay, while the linear map is a linear map, so it's not bounded at all. So it means that if you look RN close to infinity, you are moving by a huge linear map and then you are moving a bounded amount. Okay, so it's a bounded perturbation of a linear map. Okay, so we have always such lift. So let us assume A is hyperbolic. So it's an analysis of matrix. So it means that RN can be written as this U and after the norm of A to the N is less than Z lambda to the N, oops, restricted to this stable part, lambda less than one, positive for every N larger than zero and norm of EN E minus N in the unstable is less than Z lambda to the N, same things and A of S is yes and A of U is the same as with an also. So the transformation of the total city with the linear is an analysis of the morphism and there is your splitting. So what can we say about the F? Can we say anything about the F just from these conditions? So this very condition we start with. Indeed, we can say a lot. We can say that F has a fixed point. This is yet another exercise. Prove that F in this situation where A is hyperbolic, F will have a fixed point. It, you will have a huge amount of periodic orbits also. And indeed you have a little bit more. This is a theorem by Franks from Franks, the 70s or 60s, 60s probably, that says that in this framework, there exists H from the N to the N and continues on the topic to identity, which essentially on the topic to identity means that this matrix that appeared here corresponding to this H is the identity matrix and H compose F is a compose H. And the proof in this is so he never called it here and indeed it's a simple computation that we are going to do. Worry, but the point is that this linear map is a factor of F. That's the meaning. It's a topological factor. This being homotopic to identity implies automatically that the map has to be onto plus continuous. And it is not hard to prove it in this setting. So what we are trying to discuss then is whether we can make this H better than just a continuous map homotopic to identity. Okay, so can we make it a homeomorphism? Can we make it a different morphism? We have contraction and expansion. This lift X plus some phi of X. And let's look for H of the form identity plus some U of X where U of X is the same as U of X plus N for every N. Identity plus periodic. So we put the question. H compose F is A compose H. This is the same as very few, I will do it maybe a little bit faster. You can do us an exercise the slow motion.    There is a functional equation but it's linear in the variable U. So A is a linear map. So this is a linear in the variable U. So I can write down a phi here.  Go from RN to RN but I can split them since RN split as ES plus U. I can split this RN as ES plus U. So I can look to phi S from RN to the ES space and US from RN to the ES space and the same thing E U U 1 phi U. Okay, I will do just this part. The other part will be exactly the same. So I can put S's here. AS means to the stable part of the A. Let me just multiply this by this vector. Maybe let's remove this, not really this. So this is your functional equation. So let's move there and rewrite the functional equation there. S compose F inverse is minus here. Is US minus AS U compose. So that's the same equation. So how I solve this equation, I can solve it explicitly. If you think that this operator L of US equals to AS S compose F inverse. Composing with a map here in the C0 topology is an isometry. As soon as this function is onto. And AS is a contraction. So this is a contraction. And what I'm having here is identity minus L times US. So we know how to take the inverse of this guy. So I minus L inverse is sum of L to the K. K larger than or equal to zero. I can just write explicitly the solution. So I will get that US will be minus some K larger than or equal to zero. I don't want to put this here. A to the K, pi S compose F to the K. That exercise show that this is a convergent sequence in C0 topology. Okay, the bias trust M test. And you will have, analogously, that UU is sum K larger than or equal to zero. A to the minus K, UU, pi. There should be some minus one somewhere. Yes, here. And this is a plus one here. So this is the formula for the U, essentially. And again, you can check that this is a convergent sequence. And since the function pi S is periodic, the F will preserve the integer lattice. This will also preserve the integer lattice. So you have that this function, US is periodic, this function UU is the end periodic. So there is your solution. That's the proof of the theorem. Now how to make this H better than just a continuous map. The continuum, which now is Franks and Manning. Then it's really a hard theorem. I'm not going to prove it. That says that if is an anosoph, from TN to TN is an anosoph, if you, from TN to TN during that is a homomorphism. So the age we just built is indeed a homomorphism. There's one to one. So it was already continuous and onto. Now it's one to one. So this is very important. It has more than just this piece. This Franks, oops, Franks Manning. I didn't wrote Manning. So part of the statement, which we will not care much is that the matrix A is indeed a hyperbolic matrix. If F is an anosoph, different from this one, then the matrix A is indeed a hyperbolic matrix. This is part of the theorem. Just not only assuming that. Okay, now comes the next question. Can we make H a different morphism? Even can we make it absolutely continuous? Meaning it takes Lebesgue measure zero to Lebesgue measure zero. And the answer to both questions is most of the cases, no. You cannot make these guys not even absolutely continuous. And you can see easily that you cannot make it a different morphism. Just have that A of zero equals to zero. So zero is the zero vector. And if you look at the formula, you will have H compose F at H minus one of zero is the same as A of zero. So this is the same as zero. So this means that F of H minus one of zero is H minus one of zero. H minus one of zero is also a fixed point. Okay, I'm using this homomorphism. So there is such an H minus one of zero. And if you take the derivative of this equation, you will get matrix Q, which is essentially the derivative of H at the H minus one of zero times the derivative of F at zero is A times Q. That means that these two matrices here and here has to be conjugated. And this is very strong condition. Okay, you can write easy perturbations that will kill this property. Most likely it's not. Now the theorem, still you can do something when you have the two commuting guys. So this theorem in the very final format is a theorem with zero and one. But I will just state it because I will not state the precise hypothesis here. So then it's by several authors. Alpha one to one, two commuting different or fission, but not being one a power of the other or some trivial thing. Assume this and not such that alpha of N not is an ossoff. So it's the F of that theorem. Then it will be topologically conjugate to the linear map. Now can we say more? Yes, we can say more. Then appears here on Frank's Manning. Okay, and indeed this row from C square into SL3Z which essentially correspond well GL3C which essentially correspond to the linear part of the corresponding matrices alpha and not. And if I want to make it correctly and there exists a gamma subgroup of C square, quotient is finite. Finite in the subgroup such that alpha of H compose alpha N is draw of N, compose H for every N in the gamma. If not there are fine guys which you cannot make linear. This is some subtlety but this is of minor effect. Probably it will appear clearly later. Okay, so I guess time is up. So the idea of the next lecture will be to prove this statement and the next two lectures we are going to in a sense apply this theorem to understand more general group acting. Okay, that's it, thanks.


Hello and welcome to the session. In this session, we discuss the following question which says, make a line graph for the information in the table given below. The table given below gives us the information of the scores in five English tests. We are given the number of tests that is the five tests 1, 2, 3, 4, 5 and the corresponding scores that is the scores obtained by a student in different five tests. Now we need to make a line graph for showing this information. Let's move on to the solution now. Now first of all, the number of each test that is 1, 2, 3, 4 and 5 is represented on the horizontal reference line and the scores are represented the vertical reference line. This way we have represented the number of each test on the horizontal reference line and the scores are represented on the vertical reference line. Now from this table, we see that in the test 1, the student scored 30 marks. So we get a pair of numbers as 130. Now in the second test, the student scored 50 marks. So the second pair of numbers would be 250. In the third test, the student scored 40 marks. The third pair of numbers would be 340. In the fourth test, the student scored 20 marks. So the other pair of numbers would be 420. And in the fifth test, the student scored 80 marks. So we have another pair of numbers that is 580. Now on this graph, we need to locate the points for these pairs of numbers. First we have the pair 130. We are going to locate this point 130 on this graph. We have the number 1. Now from this point 0, we move one unit to the right to reach up to the number 1. Now from here that is from the number 1, move 30 units up so that we reach up to the number 30. So we get this point, so this is the point which represents the pair of numbers 130. Now in the same way, we locate the point with the numbers 2 and 50. For this, for we will move two units to the right from the number 0. So we reach at this point. Now from this point, we move 50 units up so that we reach at this point which represents the number, we have the point 340. For this again, we remove three units to the right from the point 0. So we reach at this point. And from here, we move 40 units up so that we reach at this point and this point is 340. Next is the point 420. For this again, we move four units to the right from the point 0. We reach at this point and from here, we move 20 units up. That is, we reach at this point. So this is the point 20. The next point is 580. For this again, we move five units to the right from the point 0. So we reach at this point, move 80 units up to reach at this point. This point is 5. We located all the points on the graph that we got from the table. For the line graph, we will join the successive points by line segments. So as you can see, we have joined the successive points by the line segments to get a line graph. So from this line graph, you can easily make out the performance of the students in five different tests of English. In this session, hope you have understood the solution of this question. Have a good day.


Check out this chicken crescent reef I just made. Start out by adding your delicious canned chicken. One can of cream and chicken. Now let's cut us up some broccoli. Next we're going to add some red bell pepper. Kind of reminds me of Christmas. Some water chestnuts. Add a little bit of onion. About a cup of Monterey Jack cheese.  That's the green top. One more can of cream and chicken. Now you need two of these crescent ropes. Some oil on your pan. Roll out your dough. Separate your pieces. Just like that. Oh, like that. And we're going to lay them out. Now we start placing in our mix. Simply just fold over your little flaps of crescent dough. Now we bake at 375. All right, we're done. I got a special guest. He's going to give it a try and let you know if it's any good. So what do you think about that, Eric? Pretty good. Pretty good? I'd be lying to you if I said it was the best thing in the world. We use canned chicken. What do you expect? At least we got Dano's. Yummy, I'm going to get you some.