 going to do today's lecture. Maybe it might be obvious for most of them, but at least for me, when I was seeing it for the first time, it was not obvious. So for the people who are seeing it for the first time, feel free to stop me any time. And we can discuss, we can repeat things all over. So during the whole lecture, I will consider dynamics on the torus, T2. All the property that I'm going to list, they hold for any, if you replace this by any sum of money for, but just for the lecture, I will keep it for the torus. And I have B is a borrow sigma algebra, and mu is a volume, which is low back. I will be given a sum of my acting on T2, and that mu is F invariant. First of all, I would like to discuss the relationship between Ergodi city and some other properties like transitivity. And to do that, I would tell you how I think about Ergodi city, like the intuition I have for Ergodi city. So you have seen the definition of Ergodi city saying that for every a in Ergodi city. So F minus 1 of a equals a implies the measure of a equals 0 or 1. So this is the definition of Ergodi city. So the way I like to think about it, which is a good intuition for what is coming next, is that if you think of your space where the dynamics is acting on, Ergodi city is kind of related to what some people might call irreducibility of the system. If you have two sets, if you can reduce your system into two sets, A and B, where A is F invariant and B also is F invariant, one of these decompositions is not meaningful in a measured theoretic sense. Like the measure does not see one of the set. So in general, if this is possible, you can think of your dynamics having a decomposition, like a reduction. You can say your F1 is acting from A to A, and F2 is acting from B to B. You have a reduction. Ergodi city is telling you that if this happened, you can ignore one of the dynamics. Like the measure doesn't see what is happening in F2, for instance, or in F1. So that's the intuition. So Ergodi city really, for me, it means whenever you have a decomposition, the maps will take points here and map them there. Like it will make the two sets by the action of the dynamics. That's how I see Ergodi city. So this is closely related to, you can think of it as having a relation with the notion of transitivity. Transitivity as you define it, having a dense orbit. Now I will give another definition, an alternative definition which you will see is closely related to this. So I should finish what I was saying here for people who are taking note. So F me Ergodi should imply that me of A equals 0 or me of B equals 0. Like the system doesn't see one of the dynamics. You can just think of your dynamics as being one. You cannot reduce your dynamics. Definition, this is the definition of transitivity. F is said to be transitive V open upsets of T2. By the way, am I writing big enough? OK. Yes, thank you. Non-empty open subsets, there exists and this is that F minus N of U intersect V, not empty. So you can see that it's saying even more that like whenever you have two open subsets, you have one which is going to intersect the other one after some iterates. So there is this first exercise that we will probably discuss in the afternoon. Ergodi city is respect to volume implies transitivity. Now to do this exercise, I will just give you a hint to recall what you have done last week in an exercise by Hannah that you have seen that recall. Ergodi city implies that for any A, B measurable, not this. So you have seen this, I mean, you can see easily like how this property follows, the property of transitivity followed from this implication of ergodi city. So one other question that one might ask is like are there equivalent? Like is topologically transitivity equivalent to ergodi city? And the answer is not. The answer is no. So in general, transitivity does not imply ergodi city. And there is an example which is due to first and third. So this example is on the torus. We have F from T2 to T2, which is defined by F of theta 1, theta 2, because theta 1 plus alpha, theta 2 plus I of theta 2. So I mean, I'm not going to discuss the exact condition on theta and alpha. But if you are interested in it, you can just type on Google the first and very example. There are conditions that alpha should not be defined by any rational. And there is a condition, there is a choice on the map so that F is, there is a choice, say, says that F is transitive or very respectful. So now for the next, I want to, I want to, so there is another characterization of ergodi city also that I want to talk about, which is, if you have, sorry, so this is the definition of transitivity. Yeah. Yes, yes. Yes, I'm thinking of writing that as an exercise later, yeah. Yeah. There is this other. Yes, yes, yes, yes. Thank you, yeah, yeah. So now there is this other characterization of ergodi city. We use this two fact, continuous function, NL2 plus the fact that the projection map, if you remember you have the conditional expectation, just to record BF is a subset of the sigma algebra, the tariff invariant is continuous. You use these two fact, prove that ergodi city is equivalent to the fact that for everything in the, for every continuous function on the torus, your constant function. Yeah, this follows from these two facts. You can use the continuity to deduce that. Already from the book of ergodic theorem, you have for NL2, but using the denseness, you can have this way and the other way also. So the rest of the course, I just want to discuss examples. Yes, yes, yes, this converges. Yes, yes, yes, yes, yes, yes, yes, yes, yes. Yes, yes, yes. Exactly, that's a very important point because here this convergence is almost everywhere, but one might think that it depends on the function, but it doesn't, you can have a full measure set for which every function you take, you take a point in that full measure set you have convergence here. So let me write that maybe more clear. If there exists a set, how do I call it? A0 with full measure such that for every continuous function, you have that. This is for every x in A, in A0. Like this convergence is almost everywhere, but for every function, for every continuous function. Thank you. Yes, implies. Yeah, you have to. You don't take the. It's not whatever the difference. Yeah, thank you. Yeah. Yes, it's positive measure. Yeah, yeah, yeah, yeah, yeah, yeah. That's also another property that you prove from Ergodis City, which is more than this because this, you have open sets. You can take two positive measure sets and f inverse of one intersect the other, f minus n for some n that has positive measure. That also flows from Ergodis City. Yeah. So it was the, so you know what bf is. So the conditional expectation was defined to be that the integral of eb prime is in bf. Bf are the elements in the sigma algebra such that f minus 1 of b. So for the rest, I will discuss examples. And our famous, our favorite example is the two on one map to prove Ergodis City. Later on in the course, you will see a more general proof that works for general system, not only for two on one, but here we have a proof using Fourier series that works perfectly fine. Like an example, we take our favorite. F a is when you use this map on the two torus, you donate it by F a. So we will prove that the back, F a is as good as respect to the back. So I, so you have seen already the proof for 2x mod 1 using Fourier series last week. And here I will use a slightly different notation that makes life easier later for me. You can think of it first for the 2x mod 1. For example, if you're given f of x equals 2x mod 1, you can represent it in the complex plane by f of, by considering that s1 is a subset of c, and you consider the map f of z equals z squared. If you write it in the coordinate on the circle, this just is equivalent to this guy. So z squared is just doubling the angle, which is a doubling map. For the two on one map, we can have similar by considering that t2, yes, my parametrization goes to e to the 2 pi ix. Oh, yes, yes, yes, yes, yes, yes, yes, yes. Yeah, yes, yes, yes, yes, yes, yes, yes. You have here the point on the unit circle, which is denoted by e to pi ix. And you just double the angle. This is e to the 4 pi ix. Yes, yes, yes, yes, yes, yes. Yes, this is z squared. So I can do the same thing for the two on one map by thinking this as a subset of c2. And the map that we will have will be f of z1, z2, z1 squared, z2, and z1, z2. So here you can, again, see the same thing if you use the change of coordinate of z1 being e to the 2 pi ix and z2 being e to the 2 pi iy. Here you will see exactly what you have is e to the 2 pi i 2x plus y. And here is e to the 2 pi ix plus y. Which is exactly the, this map is giving xy to 2xy and x plus y. It's just a change of coordinates. And it will simplify the calculation that I will do later. So to prove ergodicity, we are going to prove that invariant function under the dynamics are almost everywhere constant. So to do that, we pick an invariant function. Take it, l2 says that c composed with fa was phi. So here we use the Fourier expansion of phi. We can write it almost everywhere as phi of z1, z2. These are the Fourier constants. So if I look at the Fourier expansion of c composed with fa, what do I get? This is exactly phi of z1 square z2, z1 z2, z1 square z2 to the power n, and z1 z2 to the power m. Some more space here. So if I do some arrangement, I will have that. What I get is the sum of nm z2, I will stay. And I have z1 to n plus m. And I have z2 n plus m. And here you can observe very well that this is exactly the two integer that I have here. Observe that nm is exactly 2n plus m. So this I can rewrite it as something I forgot to say that these coefficients, they go to 0 when the norm of this goes to 0, obviously. Like here, cnm goes to 0 for the convergence as, and this we can just think of it as. Here you can rewrite it as just cA minus 1 of z1 n z2 m. You just see that what you have here is just the pre-image of the couple that you have here under A. And this should be equal by the invariance of phi under the dynamics. I have the Fourier sequence. The Fourier coefficients should be the same. Yes, yes, yes. I will have that cnm equals cA minus 1 nm. So similarly, you use the invariance. So using c compose fA square equals c compose fA, that's cA minus 1 nm is cA minus 2 nm. Recursively, we have that now cnm equals cA minus 1 nm equals cA minus p nm and so on. So now here we have to, OK, I will need some space. Observe that if n and m are not 0, using the definition, yes. Thank you. Yes, thank you. So we have to observe that for m different from 0, 0. From 0, 0, you have that the 2, 1, 1 map will take any point to infinity. If we take any point to infinity, then minus p will go to infinity. This implies that using this property here, we will have that cnm equals 0. So we are left with c0, 0. So this proved that c is exactly c0, 0 almost everywhere. c0 almost everywhere. Which proved the ergodicity. So there is a generalization of this, which we will do later as an exercise. Is it clear for everyone? So this is something we will do also later. I don't know before I write something. So you can prove that in general, if you have a n by n integer metric determinant of a being plus or minus 1, then you consider the map fA from Tn, which is n modulo zn self. So we can show that fA is ergodic with respect to Lebesgue. Respect to volume. If and only if A has no eigenvalue, which is, if no eigenvalue is a root of unity. You think still the same approach, the Fourier series, we will discuss this probably in the afternoon. So this is an example of a linear map that is ergodic. So I want also to show you an example of a map that is not ergodic. Probably many examples, but for those who don't know. So you can consider this map f from the tree torus, which is still a tree mod z3, which is given by f of x, y, z is equals to, you have 2, 1, 1 applied to x, y. And here you have just identity with respect to volume. So to see this, it's kind of simple. You can just think of the two torus being t3 being t2 times s1. So you have your two torus here. And you times s1 when you do the identification of these two by just identity. So this is s1. You do the identification of these two torus by identity. You can see what is this map doing. If you take a point on these torus, let's say this has a coordinates t2 times 0. This map is just taking a point here, map it inside the torus. Because if you look at where we put here 0, it just moves inside these two torus, which is the base here. So you can see this guy is invariant. If you want, you can now take something which has a positive volume, like if you take this set a, which is given by t2 cross 0.5, write it here. You consider this set. This guy will be invariant, because each of the torus is invariant. This guy will be invariant. And the volume is half of the volume. It's not 0, neither 1. But you can do, OK, still time, yes. Wow. Thank you. Yeah, yeah, yeah, yeah. That's very true. So A is a matrix, B are the measurable sets. So you can see really that as I was telling you at the beginning, this space, you can have many, the composition of the space into many meaningful dynamics in a measure theoretic sense. Like, you can have the composition for which both has positive measure. So but you can do something else to make it ergodic, which is a little bit, it looks like a little bit like the Fusenberg example, but it will be ergodic, by just adding here something that depends only on x. So that if you take the torus, you don't map it to itself, you just move it in a different level. Every point will go to different level. So if you consider this map, F, y, z. So here you can prove that F is ergodic back to the back. I think this is all I wanted to say for this lecture.