 Thank you, I want to thank the organizers for the invitation. It's been a great week. So this is joint work with Hayroboki And Rafael Potri So just so we know we are talking about the same thing. Let me recall the classical Morse Levin So when you have a negative record of space You have some quasi-sodesic So provided the quasi-sodesic is long enough it's going to be about the distance from a true ishodesic This is the classical Morse Levin negative curvature. So this is obviously false when When you look at flats, right? Because when you're in R2 some spiraling thing is actually a quasi-sodesic, right? So it turns out that recently Kapovic The Ivan Porti in 2014 I think So they stated they proved a very nice version of the Morse Levin for symmetric spaces of higher rank. So symmetric spaces of semi-simple groups of non-compact type So what we did With Hayroboki, so we actually found a new proof of their theorem I believe this is essentially a different proof So this is what I would like to talk about So I'm going to actually talk about the proof of the state of the theorem and not why it is important Or why is this is this relevance? So let me so the actual statement requires some Knowledge of the symmetric space. So the statement is going to come later in the talk not at the beginning So let me start by saying By explaining the main tool we're going to use This is called the what's called the mediated splitings So what is this? So when you have some fiber band on Over some compact base and you have some bundle of tomorphism So this is so this is linear on fibers and this is over some dynamics on the on the base So T here could be in could be discrete time or continuous time could be a flow or So we say that it has a dominated splitting if well the bundle splits Into two continuous invariance of sub bundles and what else so we're we're going to ask some conditions So which is vectors here are contracted more than vectors here So this is some sort of very weak condition in some sense. So I'm not requiring that vectors actually contract They could both be contracting they could both be expanding. I'm just saying that here goes It's contracting more than here So let's just put it this way Okay, so if I pick a vector on E if I pick any vector on E and any vector on F if I flow long enough So this is this gets smaller So this is actually a given this is very strong, but it's actually equivalent So actually just the limiting going the limit going to zero So this is what I'm saying. So these vectors here are contracted more than vectors So the so the basic example of this is when you have some some sort of an also some hyperbolic dynamism So this is flow was something negative record. This is actually what's called hyperbolic, which is This is contracted and this is expanded Also, but this this comes in the family was called the weaker versions of hyperbolic So for example, so that's that was one example of the main splitting other another example Which we're not going to use but I just gonna I could just state it like that so that People are talking about So if you have some representation of some hyperbolic group into PSL DR say So this is going to be an also in the sense of Francois if only if The associated flat bundle has a dominator Okay, so this is another characterization characterization of being another representation in the sense of Francois If you don't know what this means, it doesn't matter where I'm going to use it Okay, so what I wanted what I'm going to try to use now is there is some very nice characterization of having a dominator splitting like this Which is due to Bucky and Grumbelon, so let me introduce some notation So when you have a She has a map going from some linear map Well, it between two to norm vector spaces, we're gonna call Sigma one of a These are called the the singular values. Yes, so this is what is this so? Your matrix a takes a ball and sends it into some ellipsoid So the singular values are just the size of the axis of this the match of it, right? Let me just introduce another notation, which is I'm going to look at here So for example, if say the piece Similar value is greater strictly People's one for example, if you have some sort of gap which is singular values Then I'm gonna we're gonna then all by you P say of a Which is the the vector space by the by the first P axis of the ellipse Greater the greatest P. Yeah, yeah, yeah, yeah Metric spaces in metric in a product of B and a product on W So what is the characterization that Bucky and Grumbelon found it's actually W some day much Yeah, so I know thank you for so when we're gonna also consider Like the image of this the pretty much of the smallest axis sometimes the pretty much is like this And here we're gonna have some other space We're gonna call it SD minus P So the statement Is the following so you have some Some Automorphism familiar automorphism of some vector bundle then it has a dominated splitting Say of dimension P say say for example that one of them has a mission P if I don't leave Then when you look at The pay plus one singular value of your matrix of your automorphism and you divide by the The pace the P1 This is contracting. Okay, so if you have if you have some some gap on the singular values after time T That is contract that is contracting with tea Then you you will find some dominated splitting you will find some invariant sub-bundles So the The straight direction is actually sexually sexually forward is nothing to do It is the backwards direction. That's a little harder. And so let me just say a few words about the proof So what they actually what they do is they actually find who is the splitting is going to be So if you want to find a splitting like this, right? inviting splitting So How is it? How is it gonna? How do they found it actually they show the following? So you want to take this you P say of Psi T Into the time minus T. Yes, so this is this is some vector that's living on the fiber over X, right? I go backwards t times and I look how this map expands the first P vectors Look at what I get. This is what it is Awesome ellipse some ball here Look at the image here. It's gonna give you some ellipse. You look at the biggest The biggest P axis of the ellipse So what they show is that this is this is convergent and This and the limiting is going to be some invariant bundle but to show that this is convergent. It's not actually very hard Some tricky computations and very clever computations using this actually give you that this is a Cauchy sequence That that's actually the distance in T is decreasing exponentially on T This actually is this actually is the problem was hardest when you when you look at the other guys Well, if you want to find the other one, you're gonna look at the other space that's standing here What it contracts the most to the past and what they show okay? This is also conversion just because because of the same but the hard part to show is this is that these guys are in the next Some that these guys actually don't intersect So this is this is the actually this is the to show this is the non-trivial input of the statement and What they do is They use some some old machinery what's called which is called the city that's theorem Which is such actually a measurable a measurable theorem. So in some sense using some Measurable theorem from the 60s called the city that's they can show that this is true almost everywhere for anybody a measure That's always true for every point something like some sort of argument Like that Okay, so let me explain Let me know we're gonna leave that theorem there We're gonna come back to it like in half an hour or so And let me explain you Let me say something about the metric spaces of let me just always stay in SLDR. So it's simpler if we have time We can say something about the general case afterwards, but let me explain something about the symmetric space of SLD So this is gonna be very basic so that everybody follows So what are we gonna do? We're gonna define the symmetric space I'm gonna be consider the space of inner products on our D up to a home of three Okay, so this is gonna be our space It's just a point here is an inner product up to a home of three So this is a contractable space, of course. It's a convex space convex set some some vector space So so this this guy has some action, right? So SLD acts here or PSL DX here. So it acts transitively So we're gonna this will give you some way for example to understand what what is the tension space to this guy? Yes, so for example, if you pick some inner product up to home of three then And then this induces some involution only as you're right just the edge of the pollution transposing So this involution pleads the vector space into eigenvectors and vectors that are sent to its opposite. So this is Again, okay, so this is very simple So it turns out that okay, so this is this is what is this is the algebra of of the compact group is stabilizing So if This is some compact group stabilizing all and there the other is what so the other is just a set at the space Matrices are going to lisable on a all or turn on set. Okay, so this gives you the tension space to this guy Just because you have XD Here you have some points Disacting here KO The derivative of the stabilizer of this point, so this goes to zero and PO goes with the tension space and So the thing is that here is LDR and you have this you have the killing form This is what the killing form of AB is simply the trace of AB Okay, so some so this is very silly, but if you consider some some metric which is actually on P so if So this matrix is diagonalizable So this trace is just the sum of the of the eigenvalues, right? so This is just a sum in particular, but this is positive definite and not a generator, of course, right? So the so the metric The metric on the symmetric space there are many methods we're going to choose is This is killing form restricted to this space Yes, so in some sense in some sense the metric is in changing what changing is the space what you what you measure so it turns out okay, so Okay, so what's so you have some point say you have some point on it on XZ you have some times on vector This is just PO This is what this is just some matrix which is diagonalizable on a All-Orthogonal set so you can look at for example the set of all matrices that are Deagonizable on this same orthogonal set So for example consider and we can consider the set of matrices Deagonizable on this special particular set. Okay, so what happens with this set? So this is an abelian the algebra And so what is this so when you what happened when you look at the orbit? Oh, this this is simply the set of All inner products that make this set orthogonal. Okay, so There is something this some very easy computation that you can do actually if you restrict the metric of this of the symmetric space to this set Trivial computation tells you this is a metric to a In fact, but this is this is a little harder is that this is actually a totally scholesic flat so so the intrinsic the intrinsic geometry is flat right because it's just a vector space with some inner product but The fact is it is scholesic. So it's actually in a better flat so maximum in dimension and the scholesic flattas flat so So for example, what you get from this is that this time from all to each of the some vector It's simply the sum of taking values. Okay But so what happens for example? You have some some matters like that. So So this is one flat that's going through through P and this and this tension vector, right? This is just some flat there, but there are probably maybe there are more flats That goes through this point and this vector, right? So what could happen? So what do we do? So we pick T. We looked at the basis where it was diagonalizable and mattresses diagonalizable Inner product that make this special set orthogonal are the flats. It's a flat through that goes through this point So maybe there are more because maybe T could be a good diagonalizable on several other bases The fact that they could be going on many bases is saying that T has some condition is an eigenvalues, right? What are we going to do? We're going to consider the following. So if you take U and V on E I'm going to consider the following function linear function the eigenvalue minus Okay, so if this is zero then there are several flats through T because this is diagonalizable on on many on on many bases that That are contained in the vector spaces by U and V, right? Okay, so this is a finite set of of linear forms So this finite set of linear forms is called the root system of A. Okay So what do you have? You have A and you have this finite set of linear form and when the forms vanish It's saying that there is some trouble in the vector So you're going to look at all the possible kernels of all these possible linear forms So if you have a vector that doesn't lie on any of them Means that there is only one flat through that point on the vector because it is only one There are no coincidences in the eigenvalue. So we're going to choose an only one Orthonormal set So the so connected components of this of this complement is that's called a white chamber It's picking one of these connected components. Okay, so now if you pick some white chamber just pick one pick this one for example So this is picking what about chamber is exactly the same as picking some order on the set E So just fix one of them So now now that we've picked one we can choose an order So we're on right now. Now we can write this as this Yeah, and the white chamber is going to be vectors such that all of these linear forms are positive which means that the vectors are A Okay, so what happens now? Let me just record for later that this is special now that we have this order of this special linear forms here are called the simple roots Okay, so what happens when you when you have two points now? We just study what happens when you have a point and a vector So what happens when you have two points in the symmetric space? You want to understand if there is some flat going through them and you say so what do you have to do? So these are two in the products and you want to find a set which is orthogonal for both That's a flat through them So if you write so So how do you do that? So We're gonna write Q as some the image of some matrix times all so direction of the group is transitive And so we have we have this method and also we can look at this carton the composition of the matrix with respect to So this these two guys are okay, and this guy here It's a it's a it's an element of the on the white chamber. Okay, so which is which is now the orthogonal set Who's gonna be now the orthogonal cell which is? The cell which is gonna be orthogonal for all and Q is gonna be simply the image of the of E by this guy So this this set of D lines is going to give you exactly the flat That's good. That's going to that's gonna go through all and Q Okay, so again as before There could be many flats through on Q and that depends on whether this guy some cohesiveness on eigenvalues Because if this guy some has some cohesiveness in eigenvalues means that there are many choices of this vector of this element So so what you're gonna do? Let me just write some definition We write K of P as destabilizer K of this of the a vector space Be so remember we had a preferred basis here, and we had an order on it. And so let me write down See all the simple roots. Let me put it this way to see the numbers P That's that So this is this is this is this is telling me where there are no coincidence of the eigenvalues of of a of g and so Let me just find this is a set. Let me just find a defined K of theta as the intersection of all of these Let me write this as depending on G or more better depending on all and Q Okay, so finally what's happening what I'm saying is that any element any element of H a K theta So all of these or all of these sets are simultaneously orthogonal for all and Q So this is telling me all all the plots are going through all Okay, just just for the record So we instead of considering the the basis We can we can stick with it with a little less information which is called we're gonna go We're gonna consider the flag the flag But this is a complete flag touch that the which is So I'm just considering the flag which is This vector then the sum of the first two than the sum of the first three Yes, this is telling me what where it can jump So this is defined up to up to the moment you jump it you can choose anyone So this is exactly saying so you consider this complete flag And so so now what I want to do is instead of considering all these bases all these different bases I'm going to consider the flag So this was a complete flag, but this may this may kill some some vector So this is now going to be some partial flag partial flag So why is the interest is the interest of this partial flag because I'm gonna think of this flag as a pointed infinity on the symmetric space from For the flats that go through all and she and she so let me let me make this picture So in some sense was what's gonna happen is if I look at flats that go from oh and she all if I look at Not plus but by the cons I mean The image of the instead of looking at the whole flat I'm looking at the image of the by chamber only if I look at all this all this flat that go through all on this guy Do in some sense that they're gonna have this flag at infinity They want to share this flag at infinity in the following sense. So let me write it this way So this flag is contributed flag which was gonna call you you of she I'm gonna call this flag you she of she because exactly This flag is corresponding to where G has its biggest axis as we had at the beginning And so what I what I'm going to say is that all these guys share some this flag at infinity. So I'm not I'm not defining a boundary I'm not looking at the visual model. I'm just saying we have to think think about this in the following way So when you in in H2 for example, if you have a point Inside an appointed infinity you can you have only one should I see great as going As I say to these two So here you can do something very similar, which is if I have this flag if I have a partial flag and a point in sign I Can choose a whole family of flats that that go through this point and at infinity go to the flag This is simply want I have my flag. I have my inner product. I Apply Grammys meet to this flag so obtain an orthogonal set and then I complete this orthogonal set as I come with Orthogonal sublines this this is going to be all the flat is going from this point to this to the point at infinity So this is actually very this is actually very important the fact that The flag at infinity that you get from all to see all is the biggest axis of the ellipse of how Is the biggest axis of the ellipse when you when you look at the inner product? So let me just say again a quick Formula that follows from the captain the composition which is Distance from oh to see oh This is going to be if I play the captain in composition. They both stabilize. Oh, so this is going to be exactly The norm of she and this is going to be the square root of the log So when you have oh to see oh This distance here is going to be something like the log of the norm of she For the inner product. Oh, it's going to be comparable to that So the distance from oh to see oh is what is the norm of the matrix with that inner product? and at infinity what what what is it associated flag at infinity is The flag that that explains you how the biggest axis of the ellipse of the image ellipse are So just one one last statement and we can we can actually state couple which lives for this theorem The following following observation which I personally like very much If you have say now say now that you have two flags Is a is some partial flag? Index on some numbers It's on some numbers bi and you have some other flag in general position again a partial flag So so they're in general position are in general position if when you take complementary spaces they They don't intersect So now you can do something very similar to what we did right, but we didn't know so if we had a point and the flag at Infinity we could reconstruct This is the image of this while chamber here Simply by so this is this is a flag with the inner product you order gonna you apply garbage meet and you obtain an Orton and set and then you consider all set of lines that expand the same sets at the same time So now you have two flags What are you gonna do they enjoy a position so now when you consider? You can consider all inner products that make these two flags or tunnel This is what's called a parallel set So this gives you some sort of sets inside It's a union of flats If you have some more time I said and then what and then you look at all the lines that Expand the same guys and you look at all those when a person and make this nice orthogonal. So this is a union of flats so actually you using Using the order I had in my set. I can actually interpret as this as Going somewhere like starting here and going there Okay, so what what is the name I want to state for you is that if you have any a product here? Yes, and you want to understand what is this distance for the statement is That this distance is what you're gonna do. You're gonna you have you have these two guys, right? So you have these two spaces which are complementary You're going to measure their angle with this inner product And you're gonna take the smallest of that and the log of that and this is this So what do you do you take the angle between e? I You take the scenes Hello, you take the minus you take the mean take the minus you get something positive And this is the detail is going to be comparable to this comparable in the following sense That you will have some numbers some constant numbers L and C and the same on the other side So if you want to understand that this I'm from this this point all to this Spell of the set of what you're gonna do is you're gonna see you gotta measure the angles of Distro guys from this point for this inner product What's the statement? So let me just let me just take the the simplest version of the statement so Yeah, otherwise we won't have time to do just playing how the proof finishes so so what you're doing you want to consider some I'm gonna consider some cone say that we have picked this one for example consider some cone inside this by jump Can't touch the walls. I don't know. It's just some cone in there Consider some position some position is So let's make it by infinite so to simplify the statement, but we can also get the other statements Stronger statements So we're gonna say that the position is is C regular if what if When I look at the carton of them this carton projection from one point to the other right, so this was what this was this was You pick a matrix has to go from here to here and you look at its singular values. This is this vector If this belongs to C, okay, so you have this question. So desic and then So this is saying something about the distance So it goes further away in distance comparable to n for example, but also the vectors that we had you get a Relay inside this some some given comb This can be make or see that's that's that's not a problem So what's the statement? this one of this one of the statements, so if xn is a C regular position desic Then what there exist? two flags partial flags In general position So the type of these flags depends on where on the walls the cone intersects or not such that Is it bounded from the parallel set? Okay, so actually it's the actual table is a little bit stronger because you have this position desic Have this parallel set here, but actually you can pick a point and you look at the white chamber that goes At infinity to this guy and so infinity to this guy So this is not actually the whole parallel set you can look at some images of course So let me let me try to give you the proof that we found So what you're gonna do is you're gonna consider sequences of matrices that explain you how this guy is moving so Say that you fix them all here, right? So you're gonna consider hn of all gives you xn and you're gonna let gn one say So what's gonna happen is that this this the sequence of sheen So these are all bounded this this is a bounded This guy has bound the norm just because of the quasi-scholesic relation. These guys have bound the norm bounded norm Respect to this this whole point here and more over So what happens when you multiply them? So this is going to give you the map When you multiply all of them you multiply some fix the amount of them Sorry from n to m. This is going to give you the map that goes from here to here and this The fact that this lies in the cone And the fact that the norm is growing As my n-m because it's a quasi-scholesic It's gonna tell you that you're gonna have the exact same relations work for the again for the singular values that that depend on them On the for the walls that for the walls that doesn't intersect the cone C So let me just write this So this is very straight. This is very very straightforward, which means that This is very straightforward I literally follow from the definition of quasi-scholesic and being away from certain walls And this is this is this whole for the P, right? Okay, so now let's go back to our usual sort to our first statement When we're going to build ourselves some co-cycle and apply the dominated splitting condition. So it turns out that Actually very strange You consider the set of sequences Perhaps they have bounding norm fix some norm And that verify that condition over there So as long as you fix this contents L C and Nanda This is some compact set and you have you have the shift on this composite because it's some set of sequences So you're gonna be the co-cycle, which is just this set times V And who's your who's gonna be your? your bundle automorphism just so this is a bundle automorphism and And by definition of the actual of the base they verify Bokey-Gormelon's criterion, they have the hypothesis And so it splits. So the Bokey-Gormelon criterion is saying exactly the same century one when you look at the This u vector is going to converge to something look at the complementary guy when you do it backwards This is going to converge to something. This is the Bokey-Gormelon criterion and they are in the exam But these are the sorry so in particular, so then the exam and the base is compact. So in particular the angle between these two guys is bounded below Because the base is compact. So what's happening is you have your O and you're supposed to do this going through O So when you look at this flat from O to XM This is going to be some u of these guys This is what we did at the beginning and what you know is that this sequence is going to converge and the same to the Opposite in the opposite direction, you know, they're in the exam and you know that the angle for them is bounded below by some constant independent of the sequence Because this is this constant here. This is a base. This is a compact set But the angle is just the distance On this point to this flat. So the fact that this The fact that this distance doesn't depend them So you have your uniform amount that doesn't depend on the sequence that you picked It's going to tell you that if you pick some other sequence some other point on the sequence You can pull it back here You can pull it back back here and look at this what this distance is and you're going to have the same distance So this this this argument actually gives you okay You can get much stronger results as a scope of which level 42 So for example, you only need the sequence to be long enough You don't need to have a bi-finance sequence You just need it to be long enough and then you have the same statement and this is just because this is again holes here because Because this convergence here was exponential in some sense so You don't need the whole sequence if you multiply enough matters as this is going to be close to somewhere already and the same here some easy Contradiction argument gives you the some direct some afterwards So let me let me just say one last word on what happens in the general for a general assembly simply group Because there is there is there is some non-trivial statement To be made it doesn't follow exactly exactly like this the idea will be the following So the actual statement that that's missing for semi-simple in group is this statement is what does this mean? Can you do you have a statement like this this distance compare to this two guys? So what you're gonna do you're gonna put your semi you're gonna put your symmetric space into some other symmetric space of some matrices So say you have x she semi-simple group non-compact type etc etc So what you're gonna do you're gonna consider some representation for example into SLD Let me reduce your representation say so so that you get some totally shoddistic embedding of the symmetric spaces So you have some point here So when you have these two flags at infinity in symmetric space of sheet, they're going they're going to go to two flags two flags Sorry partial flags in the boundary of the metric space of SLD So what you want to understand is a distance from your point or to this parallel set of the city to these two partial flags So the problem here is that you know how to do that in SLD in SLD It's just so just the angle from this here from this guy. The problem is that the parallel set It could be much bigger now Parallel set in SLD associated to these two flags is much bigger than this So maybe the distance is attained here and this could be very far away So this distance is going to be compared is comparable to the angle between these two guys But you don't know what happened to this distance. Okay, so there's something to show there that it actually remains bounded from this guy So the actual statement is Consider some orthogonal vector here. It's gonna make some fix it It's gonna it's gonna make some fixed angle with the parallel with the bigger parallel set So this is gonna give you the the problem the bond between these two distances here. I think I'm