 Just to recall, we used density points, Lebesgue density points, to prove that transitive isometries are agotic with respect to Lebesgue. Not sure. And there were two key takeaways. For an isometry, balls are exactly preserved. They're sent to balls, and in fact, balls of the same radius. And secondly, isometries preserve density. So meaning that if I have a set and it has a certain density in another set, and I apply f to the n, the density of the image in the image set is the same for an isometry. All right. So that was example one, if you like. Here's example two. I'm going to give a proof that the doubling map, x goes to 2x, or let's say f of x equals 2x mod 1 on r mod z is agotic. In fact, it's agotic with respect to any invariant measure. Well, it's agotic with respect to Lebesgue measure. OK. So proof. So either using the Lebesgue density theorem or using the exercise from yesterday, you can show the following. So LDT, Lebesgue density theorem, or exercise 2 from yesterday. We can show the following. If x is the subset of r mod z, well, the Borel measurable set in r mod z with positive measure, then there exists a dyadic interval in the circle. Let me draw a picture. Here, x would be any set of positive measure. But somewhere, let's say this is the origin, which is equivalent to 1 under the identification of the circle with r mod z. And somewhere, I can find an interval i whose endpoints are of the form 2 to the m comma k plus 1 over 2 to the m for some k between 0, 2 to the minus 1. I can take the circle, and I can partition it such that. So you can partition a circle into intervals of this form. And you can let m go to infinity. And we get smaller and smaller such intervals. And the exercise yesterday shows that there's some interval somewhere where this set x, this positive measure set x, has very high density. So I'm going to add a quantifier here. So given epsilon greater than 0, there exists such an interval such that the density of the set x in that interval i is greater than 1 minus epsilon. Now suppose that x is f invariant. So suppose that f of x equals x. Or f inverse of x equals x, but that will imply that f of x equals x. OK. So what we're going to do is we're going to look inside this interval, and we're going to apply the dynamics. Now, the dynamics is not going to move this interval around. But it's going to do something else very nice, which is if we apply enough iterates, as we apply iterates, it expands each time by a factor of 2 until when we've applied m iterates to this interval, we get the whole circle. So note that f to the m of i, so I made that interval half open, we note that f to the m of i equals r mod c. Because if I take f to the m of i, I'm going to, if I don't do this mod 2, I get the interval k comma k plus 1. I'm sorry, if I don't do this mod 1, I just keep multiplying out. Multiplication commutes with taking things mod 1. If I do that m times, I get the interval k comma k plus 1. And then when you take that mod 1, you get 0, 1, which is the whole circle. All right. And so what happens to the density of f to the m of x? On the one hand, the density of f to the m of x over f to the m of i, this density, by definition, is the measure of the intersection divided by the measure of f to the m of i. Now notice here is not an invertible map. So the measure, if I apply, well, this is correct. And this is equal to the measure, now I'll make my note, of f to the m of x intersect i, because it is, divided by the measure of f to the m of i. That's because i is a very small interval. So why can I go from here to here? Because i is a small interval on which f to the m is injective. So going backwards, there's no issues. This is, in fact, the image of the intersection is, in fact, the intersection of the images. So it's this ratio. Now, what I want to say is, well, the measure of f to the m of this intersection is not the same as the measure of this intersection. Because even though f preserves measure, that f inverse, which doesn't even exist, does not preserve measure. This is not f inverse. This is f. And so in fact, the measure of this, because we're looking at Lebesgue measure, the measure of this set is going to get scaled at each step by a factor of 2. So this is 2 to the m times the measure of x intersect i. And again, in the denominator, it's 2 to the m times the measure of i. And now, beautifully and quite remarkably and unusually, these two numbers cancel. And so this is equal to the measure of x intersect i divided by the measure of i, which is this conditional measure, which we've chosen i, so that this is greater than 1 minus epsilon. So on the one hand, we have this. On the other hand, f to the m of x equals x. f to the m of i equals the whole circle. That conditional measure, well, the measure of x is equal to the conditional measure, the measure of x conditioned on the whole circle. That's always the case. And that equals, because of this, the conditional measure of f to the m of x in f to the m of i. And so what have we shown? We've shown that the measure of x, I think you maybe can't read this little bit, plus the measure of x equals, what I wrote over there, f to the m of x in f to the m of i equals, by what we've said over here, the density of x in i is greater than 1 minus epsilon. So mu of x positive and invariant implies that mu of x is greater than 1 minus epsilon for all epsilon. And so therefore, mu of x is 1. And that shows that there are no f invariant sets, measurable sets of positive measure other than those with full measure. So therefore, f is ergodic. So what are the key things we used here? Questions about the proof? Again, we're not using any smooth. You're not seeing a derivatives pop up here. Well, secretly you are. These are derivatives, these 2 to the m's here. Someone's messing with my lights. Does that make it easier to see? Just choose something, though, because it's having some kind of epileptic seizure here. So what are the key properties of f of x equals 2x that we used here? Well, one is that f maps dyadic intervals onto the circle, 1 to 1, and onto r mod z. So the fact that under any interval, if I wait long enough, we'll map onto the whole circle. The other key property is that f preserves the shape. Well, I guess that's kind of implicit in what I've said here. The other key property is that f preserves density. And I said this before for transitive isometries. Could I have either both set all of them on or all of them off? I don't know who's controlling the lights. This way is better. Bad. Could someone please turn these lights off? Thank you. If you want, you could turn overhead lights. No, no. This is worse. This is OK. It's really strange. I feel like I'm lecturing in the middle of the night somehow. Where is that? OK. And so the second is that f preserves density, i.e., the measure or the conditional measure. Who's doing this? Can you please? No. Is this just happening? OK. The conditional measure for any m. And this is key if I take any m. If I look at the density of f to the m of some set x inside of f to the m of y, that's the same as the conditional measure of x and y. And now let's just for a minute review why that is the case. So this equals the measure of the intersection. So because f is injective, I should say, f preserves density on injective domains. So as long as y, f restricted to y, is injective, this is a true statement. Because we have the change of variables formula, which I'm going to write now. So the conditional measure, mu of f to the m of x. Sorry, I already wrote that. This is equal to the intersection right at this way, since it's injective. And now the change of variables, these are Lebesgue measurable sets. We're in dimension one. So if I want the measure of the image, and my thing is injective, what do I do? I integrate f, the derivative of f to the m prime of dx over the set x intersect y. And I divide by the integral over y of f to the m prime. Or so I should say d mu of x, the beg measure of x, d mu of x. Now in the case that we're considering, the linear case, this is constant and equal to 2 to the m. And this is constant and equal to 2 to the m times the measure of x1. And those cancel, and we get this density. Now, what's m? Oh, it's m dyadic. It's 2 to the m. It's OK. No, it's OK. It's always good to clear it, because I bet other people have thought that. OK. And this is the conditional measure. Everything cancels. OK, but now suppose that we just changed f of x equals 2x just a little bit, right? And we wanted to see what happens to density. Well, let's suppose, on the other hand, if f prime is not constant, and the measure of x and y is very small. So in general, in our proofs, we're actually interested in controlling small densities, which is the same as kind of getting large densities. If you can show that a density of x and y is very small, then you know that the complement has high density. And it's typically easier to show low densities are preserved than high densities are preserved. So on the other hand, if this happens to be small, that does not say less than epsilon than the best naive on the ratio or the density of f to the m of x and f to the m of y, given what I've got here, is just this is less than or equal to, well, the soup, the absolute largest value that I can get, overall x in, well, overall x in x intersect y of the absolute value. This should be this. It doesn't matter. I'm assuming the derivative is positive, but I'm assuming orientation preserving. But in general, if I had just a map from the real line to the real line, an upper bound on the numerator is the soup, and a lower bound on the denominator is an inf. And if I have a bound like this is less than epsilon, then this will be less than epsilon times this ratio. And what I want to impress on you is that if we think about our proof, we were not allowed to choose m in advance. We chose epsilon. Epsilon determined an m in which we had density bigger than 1 minus epsilon. And then we had to iterate m times, where m could be extremely large. And we wanted to show that density, high density, was preserved, or equivalently low density was preserved. And here, the largest value of f to the m prime, it might be, suppose I'm very close to the doubling map, but I'm not the doubling map. This could be something like on the order of 2 plus some small number eta to the m, whereas this down here could be on the order of 2 minus eta to the m times epsilon first c1 map. And you'll notice that as m goes to infinity, this goes to infinity. So this proof that I just gave you uses very heavily the fact that the derivatives were precisely on the nose constant everywhere. And a main theme in smooth dynamics is that one needs to be able to bound quantities like this for nonlinear maps. And if you can do this for all m on some interval, what you've achieved is what's called a distortion estimate. So a distortion estimate says that if I choose an interval, well, if I choose some image, I can go back as far as I want. And the ratio of the max and the min of the derivatives up to the m is bounded, independent of them. All right. So that's getting a little bit ahead to the nonlinear setting. But I will come back to this, but I wanted to emphasize that this proof still needs some work if we want it to do anything more than just f of x equals 2x mob 1. OK. So now let's go to example 3. That's the cap map. So if we let a be 2, 1, 1, 1, then the map, f sub a on t2 is ergodic, which we know. We already know this because of Fourier series, so from Kadeem's lecture. But let's see if we can try to do a proof with density points. Well, what happens? Yes, I'm using, where am I using f of x equals x? Both are true. Yeah, it works out fine. Yeah, I know. That's something that's like, wait, what? No. It's because you're looking at the intersection of x with a little interval on which f is injective. So you do get that f to the n. Yeah, but it looks like, I know, it's a good point. I think about that. Yeah. In fact, I should have said f inverse of x equals x, because that's actually assumption that you need to prove it is not f of x equals x. But that's OK because f inverse of x equals x implies f of x equals x. So I should have made this my assumption. So thanks. I really used that. Anyway, so how would we try to prove this with density points? Well, we could do something very similar. So suppose here f is invertible, so we can do it either way. Suppose f of x or f inverse of x equals x, and this is f sub a. Well, if we remember our picture of what the cap map does, if this is the torus, suppose mu of x has positive measure, so x is some invariant set with positive measure, then somewhere, yes, indeed. I can find a very small ball in which I have very high concentration of x. OK. So, well, what happens if I iterate under f? What happens to this ball? Well, if you remember the picture from my first lecture, what happens is we have these two special eigen directions for a, and these two eigen directions, the derivative of f sub a is a, call this x naught, is everywhere a, right? It's constant derivative. Just like f of x equals 2x, I constant derivative 2. This is constant derivative of a. So everywhere, if I look close to a point and I look in its tangent space, I have these two eigen directions. And this eigen direction, so this point is mapped somewhere else. Maybe it's mapped here. I'm more likely to be mapped, say, over here. And this direction here is contracted. So if this is some arbitrary point x, then here's f sub a of x. And the linear map takes this eigen direction to this one and contracts it by some factor that I'll call 1 over lambda. And this direction is expanded by factor lambda. What is lambda? Well, you can solve the characteristic equation. And it's the larger of the two. I have to always write down the characteristic equation. x squared minus 3x plus 1 equals 0. And so the larger solution is 3 plus root 5 over 2, which, by the way, is the square of the golden mean. That's the square of 1 plus root 5 over 2. And actually, the slope here, just for fun, the slope is 1 over the golden mean, which is an irrational number. And here the slope is minus the golden mean. OK, so now I've really cluttered my picture. My point was simply the following, that if I take a nice ball in which I have high density and I iterate it forward, I no longer have a nice ball. I have an ellipse. Well, if I iterate it a lot, I'm going to get, well, you saw what happened to the cat, right? I'm going to get something that's elliptical, but alarmingly thin and not small at all. So even though f is a diffeomorphism, and indeed, what would be the formula for the density, the change in density for f sub a? We could do the same calculation, maybe I'm going to remind you what it is. So here the density of a set of f to the m of some set x and f to the m of another set y. F is a diffeomorphism, so it's injective, so I don't have to worry about a lot of things. Well, in this case, this is going to be equal to the integral over x intersect y, not of f prime, because after all, this is a f prime as a matrix, but it's the absolute value of the determinant of the derivative. It's this. It's this. And by fortunate accident, because, again, we have a linear map, it is the case. Well, this is just the determinant of a to the m over the determinant of a to the m times the density of x and y. So the density of x and y is preserved, because these are the same. This is one, right? In fact, even if this wasn't one, as long as f were injective on that little ball, these would cancel. So we do indeed have that density is preserved. But the problem is, balls are not preserved. And so while we know that the density of this invariant set does not change in its image, its image is useless, the image of the ball is useless from the perspective of the Lebesgue density theorem. We can't get this density point to a density point for the complement while keeping it a density point. So density points are a priori if, for example, if we tried to employ the strategy for transitive maps or transitive isometries, which recall, that means I want to take a density point for my invariant set x to a density point for the complement of x, which we called x prime. Well, we can do that. So I mean, I haven't proved that this is transitive. But the transitivity part works. So we can do that. We can take x close to x prime if we wanted. F sub a is transitive, and that's much easier to prove than ergodicity, which I hope to impress on you. I hope we're, well, it's impossible to impress on you how much easier for an area preserving map, it's much easier to establish transitivity than ergodicity. So we could get x close to x prime, but the point is the picture looks like this. Huge amounts, huge amounts of x, huge amounts of x prime. But if we apply f to the m, this will still be huge in the image. But maybe the image looks like an incredibly flat ellipse. And so here's f to the m of x. And indeed, the green has huge density in this highly elongated ellipse. But that's completely useless. It could just occupy it. So skinny, it could occupy that 1% of this ball that is not x prime. So this argument fails. And the reason is it fails because, well, f does not preserve the shape of balls. It doesn't even roughly preserve the shape of balls. All right, so we need a new idea. And the new idea is very closely related to Corina's lectures. There's another dynamical system that's lurking here, a type of dynamical system that's lurking in the map 2111. And we need to use that other dynamical system to help us prove that invariance, that must be trivial. So he new idea. f sub a has what are called stable and unstable filiations. So recall my little, well, you don't have to recall. Just look over here, that little infinitesimal picture where the derivative of f everywhere just acts like a. That's a microscopic or an infinitesimal, not even microscopic, but an infinitesimal picture. But in fact, since that picture holds everywhere, we have a global picture. So suppose for starters, I take this line. Is that a square? Remotely a square. So suppose we take this line of slope. I believe it's 1 over the golden mean. So we take the line in our do. And we do as Karina did. And we project that line onto the torus. And we want to stay parallel, which is hard to do when you're close to the board. And so what you've seen is that, not so parallel, is that in fact, because this slope, if we call this theta, then I think using Karina's notation, tangent theta, which is less than 1 in this case, is irrational. Consequently, this line, you can go backwards with it, too, this line winds densely through the torus. OK? Now any point, I can do the same for any other point. So here I've associated this line that's immersed in the torus. It never intersects itself, but it's dense. So I'm going to call wu of the origin. That's because this is 0. But through any other point, I can do the same. I can just translate the origin wherever I want to any other point. So maybe I start here. I wish I had two different shades of green. For the colorblind in this room, I guess that wouldn't be so useful. Actually, I guess that would be better than using red. OK, so here's the lighter shade of green. But I could do the same. I can take this line. So if this is the point just x, I can project this line to the torus. And I get a line that's distinct from the other line. And it, too, winds densely through the torus. And I'm going to call this wu of x. And as a set, it's just wu of 0 plus x, like literally the set sum inside of, well, you could either do it upstairs in projector. You can project and do it downstairs because addition commutes with projection. Torus is a group. And so this collection of lines partitions the torus. I'm going to give it a name. I'll call it wu. It partitions. It's a partition of the torus, a hideous one in some sense. But locally, it doesn't look so bad. And so it's a partition. So the union wu of x, just trivially, is t2. In fact, it generates a nice equivalence relation. Two points are as any partition does. It partitions the torus. It's invariant. So you can check these properties in the sense that if I take any x in t2. So by the way, I mean, if I took an x prime, I should say this. If I take an x prime in wu of x, then wu of x prime is equal to wu of x. So it's not like, oh, I take an x prime. And this is part of being a partition. If I take an x prime, it's not like I get some different line. OK, it's invariant. So for any x in t2, we have the following properties. If I take f of wu of x, I get wu of f of x. The leaves of wu, by which I mean, so the leaf through x, I denote by wu of x. So that's called a leaf. It's an element of the partition. Our uniformly, well, they're uniformly expanded by f. But more importantly, in fact, they're uniformly contracted by f inverse. So here's a weak, something stronger is true. There's actually an exponential rate of contraction. But we'll talk about this later. If I have two points, let's just call it y, OK, and they lie in the same leaf, it's called an unstable leaf, that implies that the distance between fA to the minus n of x and fA to the minus m of y goes to 0 uniformly. I'm sorry, x prime, it's also true for y, but goes to 0 uniformly as m goes to infinity. Locally, if I forget what the leaves belong to, this might belong to the same leaf as this. But if I just draw like a little local box here, I can't really tell whether two of these leaves belong to each other. So locally, these are intersections. These are connected components over two. These are the connected components of wu intersects some kind of box of the leaves of wu. These are called plaques and denoted wu, if I have a point. This plaque is denoted by wu, loc of x. Locally, there's a smooth change of coordinates. There's a change of coordinates. In this case, it's affine, so that the plaques actually just look like little line segments. They look like coordinate disks, in this case. This property, plus this, funny I put this first and last, those two properties together are basically the definition of affoliation. Affoliation is a partition of a manifold into smooth submanifolds, like a torus, into lines. So that locally, the elements of the partition look like literally by some change of coordinates look like just horizontal, like a stack of pancakes. OK, so great. So we have affoliation. Its leaves are invariant, or the affoliation is preserved, and the leaves are contracted under f inverse. Similarly, and it also happens to have irrational slope, but actually, that's not super needed, for what I'm going to say. So also, there exists affoliation. So there exists something called WS, called the stable affoliation. It's orthogonal in this case to the unstable affoliation, it's orthogonal to WU. So this is called the unstable affoliation. And it has the same properties, except that the leaves of WS are uniformly contracted, but not under f inverse, but under f. These are parallel to the stable eigen direction. And now Hanan, the next lecture, is going to explain, or maybe not next, but the follow up to this lecture, is going to explain how we can use these two foliations to prove ergodicity. I just want to make a quick little observation that we can prove something even. It's almost an exercise, and maybe it should be an exercise. Ergodicity says that you have no invariant integrable functions, no invariant L1 functions. My observation is that once we have these two foliations, it's now easy to prove. So when I said f invariant, yeah, fA, good, this is fA, to show that there does not exist an fA invariant continuous function. And mostly, I'll just leave you to think about this. But briefly, the reason is, suppose I had a function that was invariant and continuous. Well, let's take a point and look at it I didn't draw the stable foliation, but here's the leaves of the stable foliation look like they're lines that wind densely. Take another point on the stable foliation and look at the two values of f, or phi, sorry, my continuous function. Well, if phi is invariant, then when I apply f, the values don't change. But I can get the values as close as I want by applying f. So for every epsilon, the value of phi at a point here, I keep saying phi, I mean phi, the value of phi here and the value of phi here actually are the same if I have a continuous function that's invariant. So phi is constant on this. Now if I want, I could just use the fact that this is dense. And I could say, well, I have a function, it's constant, and the dense set, it's constant. Maybe I don't know it's dense, but I could do the same argument going backwards now to show that c has to be constant along this leaf. And so now if I look at one of these little boxes, what I call these foliation boxes, I have a function that's constant along stable plaques and is constant along unstable plaques, which if I just look at the picture in R2, I have a continuous function that's constant along these lines and constant along these lines. It's constant. So that's a warm-up kind of idea how you can use these two invariant foliations to prove ergodicity of this map. And this modulo, a lot of technical details, is how you can prove, remember the example I gave with epsilon, sine, blah, blah, blah, that you can prove ergodicity for that map as well. OK. I'm gone over, so I think we should end.