 Well, thank you. It's a pleasure to be back here in Trieste. I was only here before once, 24 years ago, but I still remember. I mean, I was here with Alberto Verkhovsky. And of course, like always with Alberto, it was extremely vivid. But that was also the time when I met Sasha Retsnikov. He made a great impression on me. And I'm sure a lot of us miss him dearly. He unfortunately died too young. And another thing that I think it's very hard for me to give a lecture in this room. See, last time when I entered this room, the astronomer Fred Hoyle was speaking. And Abdul Salam was chair. So it's kind of a very high standard. OK, let me say what I want to speak about. I would have loved to be able to speak on the following problem. You have a covering. You have a homeomorphism here that you can lift for some reason here. And you would like to understand a little bit the recurrence property of H tilde in term of H. OK, so this is a covering. And by recurrence property, of course, you could look at fixed point. But I'm not going to look lifting fixed point or periodic point. But I'm going to look rather at the non-wandering point and the chain recurrent point recurrence. Unfortunately, it's difficult to do something for arbitrary covering. But what we can do is something about cyclic covering. OK, there are many reasons why you would like to understand this covering problem. One of them is that it's related to the Poincaré Birkhoff fixed point theorem for analysts. Another reason is about, well, you have the analogous problem trying to lift obri sets for the Lagrangian systems to covering. And this has some consequences on the solution of the Hamilton Jacobi equation on the universal cover or something like that. OK, and in fact, whatever I will say about this homeomorphism case can also apply to the obri set case. Now, I'm going to restrict to the case of cyclic covering. That means that the fiber is z. Or more precisely, I'm going to assume that there is a z action on x tilde, OK? So that I will denote it that way. If x tilde is a point in x tilde and n is in z, I'll denote it by x tilde plus n, OK? That's the action. And the action is, of course, free and proper, OK? And x tilde divided by z is just x. OK, so what kind of result I want to give for this? Maybe I need a, so you see x tilde will, if you have x, which is a compact manifold, x tilde will look like something like that with two n's. And I would like to say something about converging to one end or the other for the iteration of, for the orbit of h tilde n. So maybe for this, the best way is to introduce a function to measure how far you go into one end or the other. So like all infinite cyclic covering on a manifold, this covering x tilde x, in fact, is given by the pullback by some function, OK? So we can find a function into the torus or the circle such that this covering is the pullback is just eta star of the covering of r over t, OK? Like all coverings. The coverings are classified, sorry, into, yes, are classified by homotopic classes of x into the circle, OK? So you get something like that. So I have here t, x, here I have eta, and here I have a lift eta tilde, OK, into r, which satisfies the following property, eta tilde of x tilde plus n is eta tilde of x plus n, OK? Now, let me maybe just state the theorem and then explain things a little bit later on what things mean. So the first theorem, which is a folklore theorem, I mean, it's not so difficult to prove is that suppose the non-wondering set of h tilde is empty. Then for every x tilde in x tilde, so maybe I should say that when I put tilde, it means I am in x tilde. I don't put them, it means I am downstairs. So I'll try to be consistent with that. Sorry? Yeah, x tilde plus n. My tilde, like 2s, my 2s are like 1. My exponential is equal to the log, usually. So I mean, like all mathematicians, I am extremely dyslexic, OK? So for every x tilde, either eta tilde of h tilde n, x tilde tends to plus infinity, or eta tilde of h tilde n, x tilde tends to minus infinity, OK? So if you don't have non-wondering points, then all points go fly away to either plus infinity or minus infinity, to one end or the other. What about the chain recurrence case? So we have either one of the conditions, OK? So if the chain recurrence is empty, we have one of the following three possibilities, one. The chain recurrence set of h is not connected. So this is the chain recurrence set of h downstairs. It's not connected. Or for every x tilde in x tilde, eta tilde of h tilde n of x tilde goes to plus infinity. 3 for every x tilde in x tilde goes to minus infinity. So in that case, only one of the two possibilities happen. And in fact, the fact that r of h is not connected, by the way we do the proof, because we will show that if it's not connected, we have a kind of good decomposition, dynamic decomposition. Then in that case, in fact, what you can show from Conle's theory is that if in case one, if one happens, then h has a non-trivial factor in x. So this follows from Conle's theory. OK. So let me just state this in the case of the annulus. So the annulus is, this is the case, let's say, 0, 1 cross r. And of course, x tilde, sorry, cross t. And of course, x tilde is 0, 1 cross r. And so this is, as a corollary, we have something which is a theorem which is due to John Frank. So it's a Birkhoff Poincaré version due to John Frank's. OK, and it says the following thing. Suppose h tilde is a lift of h on 0, 1 cross t, preserving orientations, boundaries, and maybe I should add the homotopic to identity. So if the picture now of h tilde is like that, so h tilde, so h tilde moves all points to the left on one side and all points to the right on the other side, then either h has a non-trivial attractor, h has a fixed point. OK, if you try to apply this theorem, in fact, the proof goes like that. You prove a theorem like that one, more or less, in the case of the annulus. And then if you don't have a non-trivial attractor, and now you cannot have these two conditions because the difference on the boundary, then it means that your chain recurrence set is not empty. And in dimension two, you prove that by Brouwer's theorem, plane translation h tilde, not empty, implies h tilde has a fixed point. OK? So now, of course, this is not exactly the Poincare-Berkoff because Poincare-Berkoff proved there are two fixed points, but it's under the condition that h preserve, well, let's say the Lebesgue measure or something like that. OK? Because we remove the condition of preserving the Lebesgue measure, there are cases where we can have only one fixed point, of course. I mean, you can just go around like this, and here you just have one fixed point. OK? And I mean, it seems that you go around like that, but if you go to the universal cover, you can have it in the other way. OK, you can have only one fixed point in that case. OK, now let me come back and state what is a non-wondering set, prove quickly theorem one, and explain how to do for theorem two. So the non-wondering set, so let me just take a dynamical system on some topological space. I say that z is non-wondering if for every u neighborhood of z, there exist n larger than 1 such that gn of u intersect u again. OK? So this is the usual picture. OK, maybe you don't intersect, but after some time, you come back and intersect. OK, so the set of non-wondering point is denoted by omega of g. It is always closed. If you pick up some point z in z, then the set of omega limit points, so these are limit points of g and z as n tends to plus infinity. So this set is always contained in omega g. Therefore, if z is compact, omega g is not empty. But if z is not compact, it's not difficult to give an example where the non-wondering set is empty. Take, for example, a translation on a line. OK, and now let me come to the first theorem. So I will denote by omega h tilde as a set of non-wondering point of h. So I suppose it's empty. OK, so this is the proof of the first theorem. Now, if I look at the set of points of this map, in fact, because of this type of thing here, it's proper. So the inverse image of an interval is always compact. So I know that the set of, let's say, y tilde such that eta y tilde is between minus a and a is compact for every a finite. Therefore, if I start with an orbit of some point and I assume that the non-wondering set is empty, the limit points for h tilde and of z tilde. There are no limit points. So this implies that eta tilde of h tilde n of z tilde goes to plus infinity in absolute value. OK, it cannot be bounded along some infinite subsequence. OK, now what I have to do is to show how to remove the absolute value to go either to plus infinity or minus infinity. And this is a standard and easy argument. Let me remind you how it goes. The main point is that if you look at eta tilde of h tilde of z tilde minus eta tilde of z tilde, in fact, this is a function defined on x. Oh, sorry, x tilde here, I should say. It's not z, it's x. OK, on x. Because if you just try to compute eta tilde of h tilde of x tilde plus n minus eta tilde of x tilde, I forgot to say that I am assuming, of course, that h tilde commutes with translation. OK, for example, if it's homotopic to identity, there is other case where the action of h tilde is h tilde of x plus n is x tilde minus n. OK, it's a homomorphism, so it might be one or the other. OK, so if I just try to compute that, I get here h tilde of x tilde plus n to which I apply eta tilde. And therefore, I get here eta tilde h tilde x tilde plus n minus eta tilde x tilde plus n. So it's indeed a map on x. And therefore, it is bounded. It's a continuous map on x, and therefore it is bounded. So I can find, if I put soup of eta tilde of h tilde x tilde minus eta tilde of x tilde, it's equal to some m, which is finite. OK, soup over all x tilde here. Now, if I fix x tilde and assume that n is such that, for all little n bigger than large n, we have, well, eta tilde of h n tilde x is larger than 2m. OK, because it goes to infinity in absolute value, I can assume that from some n on. OK, and now if you look at, you try to look at the distance between this point and this one. OK, this is always less than m, because here you just replace the x tilde by h n of x tilde. OK, so if you look here for n large at 2 consecutive integers, the difference is at most m, and the distance to 0 is at least 2m. So this implies that eta tilde h n tilde plus 1 x tilde, and eta tilde h tilde n x tilde have all the same sign for n, little n larger than big n. OK, so you get the convergence. OK, now I want to prove the second theorem. So let me remind you what is the chain recurrence set. So let me get back to the dynamical system here. And for this, I also need a metric delta, metric defining topology of delta. So things depend on this metric now. And let me remind you that the chain is a sequence with at least two elements. I mean, if it's only a point, it's a point, it's not a chain. And we say that the chain like that is an epsilon g delta chain. So it depends on epsilon, g, and delta. If for every i equals 0 up to n minus 1, the distance of g of zi to zi plus 1 is less than epsilon. OK, and then you say, so you say also that the chain z0 zn is from z to z prime. If you start at z and you end at z prime, OK? So z is chain recurrent for g and also delta. Depends on the metric. I'll make some comment on that. If for every epsilon, there exists a chain starting at z and ending at z, which is epsilon g delta chain. So the set of chain recurrent point is denoted by delta. Now it depends on the metric delta. In fact, it only depends on the uniformity class of the metric delta. So if z is compact, it does not depend on delta. But if z is not compact, it does depend on delta. So in my setting here of x tilde and x, I'm going to use the chain recurrent for both the lift and the map on x. Therefore, I have to tell you what is the metric essentially on x tilde. But what I do is I put a Riemannian metric on x and take a distance on x associated Riemannian distance. And on x tilde, pull back Riemannian metric. And then take the distance associated to Riemannian distance associated to Riemannian metric, I should say. So this has some properties, which of course it has some equivalence properties. So x tilde plus n, y tilde plus n, is equal to the distance of x tilde to y delta. It's invariant by translation on the cover. And also it has the following property if x, well, OK. The distance between two points in x, which are projection of two points in x tilde, is the infimum for n and z of the distance of x tilde to y tilde plus n. OK. And the infimum is at n. So these are going to be the metrics that I'm going to use in this situation, x tilde x. Now I have to give you a good way to understand the chain recurrence set. And for this, I'm going to tell you about the Pajot barrier. So this was introduced a few years ago by Pierre Pajot. So it has appeared in topology and its application. So let me get back to a general dynamical system and introduce the barrier. First, I defined an action between two points in the following way. P sub delta from z to z prime is the distance of g of z to z prime. OK. And for a chain, I propagate this to chain. There are several ways to propagate it. You could take the sum of the PD, but here we are going to take the maximum of PD of zi zi plus 1, i equals 0 to n minus 1. OK. And this is maximum i equals 0 to n minus 1 of the distance of g of zi to zi plus 1. OK. And if you have two points, you define the Pajot barrier between the two points in the following way. So the Pajot barrier, while it depends on delta, of course, and on g, but from z to z prime, is the infimum of the little p delta of chains for arbitrary chains starting at z and ending at z prime. OK. So it is not difficult to see that p delta of zi is 0, if and only if z is a chain recurrent point. OK. But this is kind of a very nice, quantitative way to understand the chain recurrent set. Let me tell you what are the properties of p delta. So first one is it's always non-negative. And it is less than the distance of g of z to z prime because you have always a chain from z to z prime, which is just two points. Note that this implies that p delta of z to g of z is always 0. So you can kind of think of this as some way to measure if you are along the same orbit or not. It's not exactly that because you have this infimum and all these things, but it's kind of that. OK. And the other properties, the second one is the ultrametric property. OK, this is because to go from z to z double prime, you get an infimum, including all chains, going from z to z prime and z prime to z double prime. And then you, let me just make sure I don't forget any property. No. And then when you give a definition like that, of course, it does not seem clear at all that it is continuous in z, z prime. But it is indeed the case because you have the following things. p delta of z prime of z z prime minus p delta of z z double prime is less than the distance of z to z prime. This is quite easy because if you have a chain that goes from z to z prime, you can deviate it into a chain to z double prime by just changing the last moment. And if you look at the difference between the p, little p delta of the chains, you see that it's bounded by this quantity. Delta of z prime, z double prime. Thank you. And here it's the same thing for the first one but with a little twist. No, I cannot make p delta of z z prime small because that would mean that I am very close to the chain recurrence set. Going from z to z prime with small chains is this big indication. Sometime I cannot. It cannot be 0. p of z z is not always 0. So you have to be careful about that. I mean, I'm telling you it's continuous in the second variable. And now I'm telling you it's continuous in the first variable. And here you have to take g because if you do the argument at the beginning, you see that's what you need to do. So note that it is continuous. It is uniformly continuous. If g is uniformly continuous. Now let me just say a few things now in the case of my covering. I will have two barriers, a barrier downstairs on x. OK, now the distance are fixed. I'm not going to mention. But so I have p of x, y on x and p tilde of x tilde y tilde on x tilde. So I have to compare these two barriers. It's not difficult to see that p of pi of x tilde of the projection, pi of y tilde, is less than p tilde of x tilde y tilde. But in fact, you can also get, because you can always lift the chain downstairs to a chain upstairs with the same p. You get that in fact p of pi x tilde pi y tilde is equal to the infimum over n in z of p tilde x tilde y tilde plus n. So this suggests to introduce kind of three functions. OK? So if I look at p of x, x is 0, and I pick some point x tilde over x, I can attain the infimum either at 0 or at n larger than 1 or n less than minus 1. So this suggests to introduce the three function. Rho tilde plus of x tilde is the infimum of p of tilde of x tilde x tilde plus n, n larger than 1. Rho tilde 0 of x tilde is just p tilde of x tilde x tilde. OK? And the last one is rho tilde minus x tilde is the infimum for n minus 1 of p tilde of x tilde x tilde plus n. OK? Now it is not difficult because of the property. In fact, I should have said here that p tilde of x tilde plus n, y tilde plus n is the same as p tilde of x tilde y tilde. It is not difficult to see that these functions are, in fact, functions which are defined on x. Because if you add the same k here, it doesn't change anything. OK? Moreover, because as I said, the barrier is continuous, is uniformly continuous if the homeomorphism is, h tilde as a lift of h is uniformly continuous. Therefore, the barrier is uniformly continuous, and this function r continues. And infimum is not necessarily continuous, but in that case, it is the case. And so I have these three functions, rho plus, rho 0, and rho minus, which are defined on x. And what happens is that the chain recurrence set is precisely rho 0 plus equals 0, union rho 0 equals 0, union rho minus equals 0, but rho 0 equals 0. It means that the point above, any point above, is in the chain recurrence set of h tilde. So in fact, this is the projection of r of h tilde. So I have written my chain recurrence set below as the three sets, one which is the projection of the chain recurrence set. They are all closed because the function are continuous. And here is where the z property comes in. You can prove that rho plus equals 0, intersected with rho minus equals 0, is included in rho 0 equals 0. OK? So if you make the hypothesis that rho h tilde is empty, you get a decomposition in two disjoint sets. And it's completely natural decomposition from the dynamics. So either you get the decomposition or one of them is empty. So in fact, if rho 0 equals 0 is empty, then you have conversions for every x tilde in x tilde. You can show that eta of h tilde n of x tilde goes to minus infinity. And if rho minus is empty, it's opposite. OK? So this gives you exactly the theorem I mentioned. I'll stop here.