 Okay, hello everybody. Good morning. Good afternoon. Good evening. Welcome back to the second lecture in this series by Hadim Waal of Impa in the on the geodesic flows on surfaces without conjugate points So as you can see we are in a new lecture room today We are in the main lecture room of ICTP the boot in each lecture hall, which I think many of you probably most of you are familiar with and Just a couple of announcements before we start the first one is that the last lecture is now available on the YouTube ICTP mathematics Channel so if you search for ICTP mathematics, you'll find a YouTube channel and the whole mini course will be available So the first class on Tuesday was available is available now and within a couple of days we will put this lecture on and then just to Say again that next week there will be two more sessions of this mini course So one of them will be Tuesday at four o'clock Italian time Exactly the same time as last week and there is a change of time for Thursday I will send an email also and I will post on on Facebook the change So the new change is just an hour before the previously announced change So it will be a two o'clock Italian time two till four o'clock Italian time on Thursday will be the final session to our double session of the mini course Okay, so thank you everybody for joining us and let me Pass the word to had him who will continue his lecture. Thank you Oh, sorry, I forgot to unmute the microphone. Yes, so Yeah, I said in the abstract I have which is online Today's lecture is supposed to be on major of maximum entropy But I think we still need some geometric background and stuff before we get to that So I'm not sure if today we will So I will be still talking about geometric Background and if time allows I mean we might talk about partisan Sullivan measures. We will see So let me just write in this corner recall of what we did basically from last lecture So recall and I will just leave it here. So so recall the definition of Jaco be filled So a way to think about it in two-dimension is just a simple second-order ODE Okay, J equals zero. So this is in two-dimension and Looking for a transversal Jaco be filled. Okay, Jaco be fit that is transversal to the to the geodesic and and Remember the definition of conjugate points. So this was just this picture. You don't have a Geodesic where you have a Jaco be filled that is zero here and start growing and then decay to zero So in two-dimension think of the Jaco be fills are just you know, you you fix a transversal direction to the geodesic You're just looking at the magnitude like a function that you multiply by That direction to get like that function should satisfy this equation So just think of it as a as a simple equation and that captures some geometric properties of this geodesic So these two points are conjugate if this happens and remember that We proved that non-negative non-positive curvature implies no conjugate points so this is a kind of a generalization of of non-positive curvature and also Remember that no conjugate points Imply that the exponential map is a covering map is a covering map So yeah, by the way last time at the end of the lecture somebody asked about an example of Local differ which is not a covering map and you can just have this on the like the exponential map on the two sphere You know the two sphere is simply connected So the universal cover is itself if the exponential map was a covering map You know that means that the universal cover is R2 which is not so the exponential map is a local difere morphism But it's not a covering map for the two sphere Okay So, yeah, so we will this is basically recall from last time So from today, I want to start with some notation that I will be using Notations So remember we said that from this property the universal cover of the manifold is a copy of R Rn or R2 in the case of surface. So we denote by let X denotes the universal cover cover of m So we also write sx as a unit tangent bundle. So this is a space of vectors in Tx Since that So gvv equals 1 so see that I mean Tx I'm in the unit and I'm in the universal cover now So see that I'm just lifting the metric on the manifold to the universal cover and I call it g again. So I'm looking at the tangent space of the today. We'll be just talking about universal cover not the base manifold Okay, and then so this is a unit tangent bundle and One more notation is that given v in sx. I denote by cv x the geodesic starting By v like cv prime at zero and also a lot one last notation is that given P Q in x So C P Q is a geodesic connecting P to Q with This parametrization that CP Q at zero equals zero It was P. Sorry So these are some notation that I would be using all over today's lecture so For today's lecture the first section that we're going to see is a very important property In this study, which is a divergence property the definition that We say that x the universal cover with G has a divergence property as divergence property if for any pair of geodesic C1 C2 we See one zero equals C2 zero We have that this guy they diverged in the universal cover we have that the limit When T goes to infinity of the distance between C1 T C1 and C2 and then C2 T Is infinite it's plus infinity. Okay, so this is a very Yeah, yeah, so he's asking is it yes, yes, yes So it if you're working in a compact set this this this doesn't make sense It's yeah, it's it's void like you don't have you don't have divergence property if you're working in a compact set You know, yeah, if you are not in the universal cover and yeah, it's yeah So, so this is just this property So it tells you that if I have two geodesics emanating from the same point P. They are going to diverge okay in the universal cover and Remark that in negative curvature Remark Negative curvature implies divergence property why the The proof being that if you have a geodesic here Who geodesic C1 and C2? You can fill this space with geodesics. Okay, you can parametrize this space You know, I'm just saying that you can have a geodesic variation Let's say this is V and this is W. You can define the geodesic variation as T as I said last time exponential P at T V plus SW and if you fix S it gives one geodesic, okay, and as I said last time in an exercise that J T equals D alpha over D s at s equals zero is a Jacobi field and in negative curvature You can see that just looking at this equation K prime plus K J because zero J prime double prime you can see that Jacobi fields are growing exponentially because of this negative curvature such a Jacobi fields J T will be bigger than like exponential of T Absolute value of K Of minus K if you want a negative curvature here Yes, just from this simple equation. You have exponential growth of the Jacobi fields So you will see that these two points are going to and and and I mean this is just the vector field here But these two points are going to diverge and then yeah, so actually today's lecture is mainly about You know recovering the picture we have in the Poincare disc in the in the hyperbolic Poincare disc, you know you have this Hyperbolic Poincare disc where you have the geodesics are given by this guy So in today's lecture we are going to define what is this boundary here in the in this case of no conjugate points How to define the boundary and to recover the picture we have in the Poincare disc With assumption that we don't have conjugate points. So before that to define the boundary To define this boundary we would need We would need the divergence property and then yeah, so there is this theorem Actually, it was proved by green so that if mg surface without conjugate points mg Conjugate points as divergence property. So it's true. So every surface Without conjugate points has this divergence property and this is an open Question in higher dimension like it's not known whether this holds true in higher dimension. So Is this true in higher dimension? There's no contra example yet if this was in the I think early 60s Yes, this is quite old in the early 60s. Yeah, or end of end 50s early 60s Yeah around that time and Yeah, and everything actually I'm gonna do To study the geodesic flow it highly Realize on this divergence property. I will show you Like all over how we use divergence property is key. Yeah, yeah, yeah, that's yeah You can say M is a is a has divergence property. Yes. Yes. It's a yep if the surface Yes, yeah, probably. Yeah, it's not be very clear. You're right So, yeah, I mean that the surface has divergence property if the universal cover has divergence property That's what I mean. Yes. Thank you so Yeah, so this is open question and And It's yeah, I should say that there are some cases where we know it's not too like in the in the finseler case in the finseler case there is an example by Howard Mezzel that You have a higher dimensional Moneyfall that doesn't have divergence property. It uses tech military and stuff, but in the remandian setting. It's not known There's no contra example and there's no proof or yeah So basically the proof of divergence property You can get it when you have growth of Jacobi fields see this proof that I was just sketching here If you have growth of Jacobi fields, you can have divergence property So what we are going to prove is that we have growth of Jacobi fields like a Growth that doesn't depend on the geodes. That's kind of uniform. So that's what we are going to prove first So yeah, so we will just prove a proposition that gives a growth. This is exponential growth. It's good but yeah We don't have it in general in no conjugate points So proposition Is it okay? Okay, I go the other way Proposition this was again first proved by green in Two-dimension so this is about okay growth of Jacobi fields. Okay, so let Mg compact Surface without conjugate points conjugate points and see if from R to M Geodesic and that j is in Gamma remember this notation TM the Jacobi field so the proposition say that given a positive and little a or so There exists T here So here T the points that T here depends only on big a and I'm saying that there is a general version of this proposition where this T depends on the geodesic But for the two-dimensional version It doesn't depend on the geodesic this T depending on the geodesic is not enough to prove To prove divergence property. So there is this T such that If you start Or I start Jacobi fields with j of zero equals zero. Okay, J prime is bigger than a imply that j of T is Bigger than capital a for all T bigger than T So this is a resource about growth of Jacobi fields actually There is a Generization of this an improvement of this results by Geha Kniper who proved that you can have like something like T here linear this time. Okay, but the proof I'm gonna Show you it just it just Just shows Yeah, it just showed that you have this grow and this growth is enough as you can see You just want to create the separation this growth is enough. All right, so Let's start the proof No, just for any Give me yeah, they are not related. It says that given a and a You can find the time There is no relation between these two. It says that for any a and big a and little a there is a So I should say maybe for every maybe For every there's no so the proof the proof is just Just analysis just uses the two-dimensional nature of the prop of this of this problem. Okay, it's just analysis We will see so we have J equals zero So J prime is not zero we can suppose because this is a linear This Jacobi equation is linear so we can suppose that J prime is one J prime at zero equals one So when I talk about Jacobi fields, I'm talking about vector field But in these two-dimensional thing I can just talk about a function as I said in the beginning I'm just finding a scale of function in the transverse direction so that that vector field Scale by that function the function will be J is a Jacobi field So I'm just thinking of J now as a function. Okay. Now you can suppose that let yt Be another solution that is linear independence with J be another Jacobi fields that is linear independence in In the sense of wrong skin like if you look at this guy the wrong skin so linear independent V is J. I Don't know if what I'm writing is it's correct English, but So I'm just saying that this guy is linear independent this set of solutions in In terms of wrong skin. It just tells you that you take this function The wrong skin between yj Which is just you take the determinant between, you know, you take J J prime y y prime the determinant is K I think I should Yeah J prime J y prime minus y J prime this guy is not zero by And at zero this guy is minus y zero. We can just suppose that it is negative Oh, we just switched the two so from this we can suppose that because the wrong skin doesn't you know We can suppose That y of zero is positive because if you have a solution that start at zero the non-linear The other one which is non-linear should start with non zero We suppose that and then look at this function now H of t Which is j of t over y of t if you look at H prime of T this is just The wrong skin of j Y a Yj I think in this way. Yeah, j y like this over y square which is Strictly positive so this means that this function is strictly increasing and And and you can see that H is strictly increasing Everything I'm using now is just OD. I haven't used yet. No conjugate points. I will show you where I use it is strictly Increasing so now let's suppose by contradiction that this is false like this is false like you have a sequence xn so suppose Proposition suppose there exists xn going to infinity since that this guy is Converging you can suppose that j at xn is converging to a number Which is C in R? Just the negation of the of the proposition and define this quantity a n this number Which is one over h xn So this guy will be bounded so you can suppose that this is converging because this is strictly increasing so The inverse will be decreasing so you can suppose that this is converging to a number a which is positive And you define this other now Function which will be a Jacobi fields. Okay minus a j t Because if you have a Jacobi field you take this is a linear equation. Okay is a Jacobi field and see that by this construction z of xn is converging to zero So there is this claim That is going to use the no conjugate points that z at t Is greater or equal to zero for all t? positive see that if this is not true like if if this claim is not true if if There exists t naught such that z of t naught It's strictly negative and I know that z at xn is positive Is always positive so that means that means that you will have to zero because at zero Z is zero and xn It is positive here. So this is z xn So if there is a negative value it should be doing this you would have It should not at xn You should go like this at least yeah, if there is one point where it is zero. I know that these xn They are infinitely many of them and they are always like z of xn is strictly positive So this implied that this this function is is non zero is is non non negative so Because he has no conjugate point. Do you see what I'm saying? If this was true if there is a negative value, I will have to zero for this field Which is a contradiction of no conjugate points. So another claim Is that z is strictly positive now? for t positive Why is this true? This is true because if you if you just look at so if There exists t naught such that z at t naught equals to zero This would mean that z prime at t naught equals zero because z cannot be negative So if it touches this axis, it should be something like this, okay? So this would imply that z prime at t naught equals zero and In particular if you use a definition of z it will contradict the fact that y and j are linearly independent So this would imply that this would be equal to y prime t naught minus a z prime t naught Yeah, so this would imply that y and z linearly independent So this Jacobi feels it's always positive 40 positive and then now You can define these functions you one which is z over z prime over z and you two which is y prime over y and If you're familiar with this you can see that these you they are solutions of the Riccati equation So because these are Jacobi fields, so u1 and U2 are solutions of the Riccati equation Which is just u prime plus u square plus k equals zero and Yeah, there is this thing. I am not going to prove here, but You have these solutions. They are uniformly bounded in no conjugate points So I will not prove this, but you can find it in the literature u1 and U2 are Uniformly bounded and then now if you if you if you write this the wrong scan of the wrong scan of y z yj over z t You can see that this is equals to u1 t minus u2 t times y t And this wrong scan is constant is a number and and see that if you do this at xn for instance at xn xn You can see that by definition of xn. This is u1 xn minus u2 xn times this is xn z xn Right and this guy being uniformly bounded A n sorry this is A n by definition of xn and this guy being bounded it imply that this is bounded Uniformly with n which contradicts the fact that this guy goes to infinity wait. We have it goes to zero. Sorry Which contradicts? so this is just Using ODE's like this solution of the ODE the two-dimensional nature of the problem Which contradicts that of xn goes to zero, right? So So this gives if you agree This if you use this you have divergence property visa sketch the same sketch I have in negative curvature I have growth of jacobi fields imply divergence actually uniform growth in terms that this time should not depend on the geodesic As I said, there's a version of this of this prop of this Proposition where T in higher dimension it depends on the geodesic. So that's not enough. So this implies this implies this implies divergence property and then At the next step, I want to talk about a more slimmer or So 2.2 So we are asking is compactness used in the proof here compactness is not used in the in this proof But I think in the higher dimension in this two-dimensional version. You don't need compactness because everything we are doing is We we are doing it in there in the universal cover Is not compact so So we are going to see now how divergence property is used to define the ideal boundary so before that So for the rest, I'm supposing that I have now mg is Surface is without conjugate points Without conjugate points and I'm supposing I'm supposing that either I'm supposing. I mean, I'm supposing it's of high genius What I want to say is just I want another metric on my manifold which has negative curvature Of genius at least two at least two. So this implies that in particular particular There exists g0 on m of Negative curvature this is quite general and then yeah So there is this more slimmer That is also very much useful here Is that it says that there exists an R? positive its universal so that for every CG so sometimes I write CG. I mean geodesics with respect to the metric G I have two metrics on my on my manifold But the metric I want to study is a metric without conjugate points and the way to study it is to rely sometimes in the in the metric of Negative curvature CG geodesic CG CG zero CG geodesic Like I said G geodesic already there exists CG zero G zero geodesic does that The house of distance between these two geodesics is uniformly bounded. So everything is in the universal cover here is bounded by R Where the house of distance think of just you know having a Like a R thickening of this geodesic will contain this one Okay, the distance is less than R but this is a very useful and I will just sketch the proof how to Prove this this fact Basically, it I mean if you're used to this stuff It just means it just can't follow from the fact that geodesic with respect to this metric are quasi geodesic with respect to this one or you can prove it this way Proof So I have this geodesic CG This is CG and I want to find another geodesic with respect to the metric G zero that bounds Let's suppose that this is CG zero the way to do is is the following so so take The this this is CG and okay and is large and is generic and is So for N in N You can define this geodesic that connect these two guy. This is a geodesic in CG zero. So this is CG zero The geodesic with respect to the G zero metric connecting this point CG zero zero and see E N So what I'm saying here is that the claim is that so so this will depend on N. Okay? It depends on N. Okay So what I'm claiming is that this geodesic is converging when N goes to infinity It is converging to a geodesic Let's say at a bounded distance to this guy Why so you look at the how the picture is going to be if I do this Going to be something like this. You might think that this distance is going to get bigger and bigger but No, because so see that the end points of these geodesics Are nearby. Let's say this is just distance one. Okay, this is bounded distance. So if I want to look at the angle Between so the angle between CG zero and Prime this geodesics at zero and CG zero and plus one prime, you know the two consecutive guys This guy is exponentially small So this guy is exponentially small. Why because again of exponential growth of Jacobi fills in negative curvature So I can always fill this space with Jacobi fills and I see that this Jacobi fills is not growing up to time N So what does it mean? It means that this angle must be very small Okay, so this is exponentially small In N Okay, just two consecutive guy since they stay at a bounded distance to time and that means that you don't have growth of Jacobi That means that the angle that you have is exponentially small That's one way to because Jacobi fills. We know they grow exponentially in negative curvature and Another way to prove it would be to use some convexity of the distant function in negative curvature You can just use convexity of the distance function in negative curvature or so so using Convexity of distance function so distance function in negative curvature Distance function meaning if I have two geodesics if I look at this function fd distance between c1 t and C2 t and This is a distance with respect to g0 and this function is a convex function and you can prove that You know if if basically what I'm saying is that if you have two geodesics where the end points are nearby In between they cannot get far apart very much So if two geodesics they start and end in some balls in between they cannot stretch too much That's what this is saying and you can use this to prove To prove this most lemma, so this is just a sketch and Yeah, now I'm gonna Go to the definition now of the ideal boundary 15 minutes so Definition so we first define an equivalence classes of geodesic rays. So let C1 C2 ray. I'm just saying assuming that they are defined in this. I just take this x To geodesic ray to geodesic rays So we say that C1 is equivalent to C2 if this guy the distance between C1 C1 0 Infinity and C2 0 infinity is bounded. So this is C2 like the geodesic ray. They say at bounded distance is Finite is finite. Yes To geodesics are a key to geodesic ray are equivalence if in the future. They say at a bounded distance. They're not going to So I define the boundary the ideal boundary of X is defined to the set to be the set of equivalence classes of geodesic ray with respect to this equivalence relation set of equivalence classes so This is just a set. Okay. Now we want to put a topology on this set So there is this lemma that follows from the divergence property and most lemma that for every P in X the map Fp which is defined from SP X to the boundary of X Which takes V Gives C V the equivalence class of CV This guy is a bijection. This guy is a bijection so proof That I really Under estimated the time I saw this I would have done it in the first lecture. So Okay, so Yeah, this is the proof is just Simple see that the injectivity is given by is given by Divergence, right? So injectivity from Divergence property because because if you have two guys two vectors If these these two are going to be far apart. That's what divergence property says So which means injectivity and The surjection is given by the most lemma the most correspondence or most lemma. Yeah, I call it most lemma. So Such action follows from most lemma How is that if I take one equivalence class? If I take this equivalence class and P I can look at the You know the picture you would think of is I have here P. I have here the ideal boundary, okay? Now I don't know if it is nice like this and here is is one points in the ideal boundary Xi And what I want to connect? Xi to P with a geodesic What I can do is that in negative curvature? I know this is this is something you can do in the Poincare risk You can always connect this guy to this guy by G0 geodesic in the Poincare risk This is a picture you have and what you do is you find the most Correspondent to this geodesic. This is G geodesic given by most given by most So in particular it defines this guy it gives the It gives the this ejection And what you do now the topology you put here you put the topology so that this map is a homeomorphism Okay, so now define topology on this guy says that FP is a homeomorphism So in particular the boundary I define is like S1 is homeomorphic to S1 because this is S1 This is a unit circle Okay Yeah, so that's the definition of the ideal boundary Yeah, now I want to talk about Boozman functions and stable manifolds and stuff So I want to talk about invariant manifolds 2.3 This is about invariant manifolds invariant sets because I'm not proving that these are manifold invariant sets and So by the way see that from this definition I have a nice there is a correspondence between I can have a correspondence between sx and X times the boundary of X By just if I take V here I Associated to the foot point P V so P Pi V pi is a projection from sx to x and Then the second components is just the equivalence class of CV and this is a bijection Okay, and I will be using so given V in sx We have the Boozman function at V so Boozman function V So is B V Q so this is a function as defined from X to R That's just B V act B V Q is the limit when T goes to infinity of The distance between CV T. I will draw a picture for this Q minus T So this is a Boozman function. So what it's doing is You take I have my V here and I have the geodesic. So okay going to end at the boundary so this is CV and then And then I have another point Q So I take this distance and this distance The difference so this is V So the difference between these two distance is going to converge to the Boozman function At Q attached at V for the point Q. Okay, so you do this Going to infinity and then you take always T because T is just this distance. So this is CV T okay, and Clearly you consider this function is a lift its function. So just by triangle inequality if you take B V Q Just triangle inequality here will be less than limit of Distance between CV T if you put a P here minus distance between P Q Minus T So you just use triangle inequality here. You see that this is a is a Lipschitz function Okay, so so B V Q minus B V P is less than the distance between P and Q and more There's a theorem by Eschenberg in no conjugate points. So this was in I think 77 by Eschenberg That BV is C1 is a C1 function and the gradient of BV as one like is not one Because one so I will not again talk about proof of this so From the boost one function now you can define the these horospheres that are just So by the way from this correspondence So if I talk about BV, I always also use this notation. So B V Q can be seen as B Pi of V like Q and like CV You know I sometimes use this notation This is a base point the food point of the vector and this is n point So given these two it uniquely determines V and given V or so So this is another notation that I will be using sometimes You just say B P Q Xi for P Q are in the universal cover and Xi in the boundary So Xi in the boundary and P Q are in X Okay, so now given P and P and X and Xi in the idle boundary I Would have the horosphere the horosphere attached at Xi Through P So H P Xi is the level set the zero level set of the boost one function This is that P P Q Xi equals zero Okay, these are the zero level set and think of them as just now I can have the picture of Of the compactification I Want to give some exercises by the end So see that this is what you have so this is now the compactification of the universal cover So let's say this is X union the boundary Okay, and see that if you have a point Xi here, this is what you have of H P Xi This is just usual or is here, you know in the point card. It is and From this you define the stable manifold at V So the stable set W SS V You see sometimes I use P Xi sometimes I use V but thanks to this identification Because when I talk about stable set is better to to talk about the Vectors in the unit and your bundle. Okay, so this is a set of points Q Gradient of B V Q Says that Q is in H P V and C V the equivalence class of V of this geodesic right, right? So You have this this is a subset now of the unit tangent bundle This is a subset of sx because this is a point and this is a vector with unit norm by this By this theorem here So up to now I haven't defined the geodesic flow and now it's time probably So and then I will post some exercises and stop Just two minutes so definition nation so the geodesic flow flow So feet II is a flow from the unit tangent bundle to itself Which says feet II at V? Equal CVT prime You just take the vector V So this is CV zero and this is CV T So these are unit speed parametrization. Okay, so and this guy will be fee E V So few exercises one is to prove that this set is invariant under the geodesic. Oh some exercises That is one Is that oh I forgot to define the unstable actually this is stable. So you define the unstable at V to be just the stable of minus V. So when you have V you have minus V Okay, so it's just that you would have this is V and it is also it correspond to another Horror sphere here. This is this will be Minus V is here and this is V Right, so Yeah, exercise one. Oh just on time so check V T W SS V equals W SS V TV and Number two is that in negative curvature Implies that we have transfer so it is between these two guys W u u V equals V and Another one is that if W is in W SS V Then there exists are positive such that the distance between FET V and FET W is strictly less than or equal to R for T positive Okay, this is why they are called stable because the the the guy the vector that I'm going to diverge to infinity and the last one is that in negative curvature that there exists Lambda in zero one is that if W is in W SS Be everything here is in the universal cover. Okay, I don't need to take local like I don't need to take Local manifolds here for some people who then the distance between FET V T W is less than lambda T then distance between VW All of these things you can use them you can prove them you can do this exercise is using Jacobi fields, okay using properties of Jacobi fields in negative curvature for instance and this one using like Boozman Boozman functions here and Yeah, this is almost the proof of In negative curvature that the geodesic flow is unknown sort of except that I won't prove yet that these manifolds are smooth But it's almost the usual unknown sort of flows in negative curvature so Yeah, I think I stopped here for today Yeah, okay. Thank you very much. Hadim There any questions? Yes Yes Yes Yeah, that's a good question Yeah, he's asking why the function BV the Boozman function is well defined what I said is just that it's finite It's well defined here because geodesics are minimal like if you take two points you cannot find two Because see that the way it's defined that I have key here and I'm taking the Geodesics, but the fact that this geodesic is unique Makes the function well defined it follows from the fact that no conjugate points imply that geodesics are Minimal in the universal cover like if you give me two points. I cannot find two geodesics connecting them just from this picture here Yeah Thank you. Any other questions? Okay, well Thank you very much. Hadim for the second lecture. Thank you everybody So the third lecture in the series will be on Tuesday Next week at four o'clock Italian time Thank you very much