 Thanks to the organizer for the invitation. It's great to be here. I was here 14 years ago in 2002 when I was still an undergrad. Great to be back and see that nothing has changed really. It's same. Okay. So the plan for these talks is to give an overview with maybe one or two of the special cases of main theorems. Developments in homogeneous dynamics which has happened in past maybe 40 years or so. So let me start with the setting. The setting is going to be G is going to be a locally compact group, but this is of course well too ambitious and the examples to keep in mind are G is going to be Rn, Zn, or the group of n-by-n matrices with determinant 1, really 2 by 2 matrices with determinant 1. And I'm going to have a manifold X on which G acts. And my actions are going to be continuous actually almost all the time real analytic, but let's say continuous. And the question that's going to be somehow the driving force of lectures are described two things. The orbit closure for every X in X are all ergodic invariant measures. For the action of G, of course, I stated it vaguely enough that so what do I mean by the word describe? Already at 9.30 this morning we were told that this is a hopeless problem because if I take the action of Z on the shift space then we know that there is just awfully many measures. So in this generality, this is hopeless. But I want to present cases that this can be done and actually has applications to number theory and geometry. So in the examples I have in mind X has a lot of symmetries and G is a subgroup of these group of symmetries and a lot of symmetry forces for rigidity. Hence we can really describe all the orbits and all the measures in certain special cases. So let me maybe begin with some examples, okay. First the baby example is irrational rotation of a circle. We know that every orbit is dense, so I described all the orbits and the only invariant measure is the rotation measure. So this is a uniquely organic system, meaning the only organic invariant measure is the rotation and depending on your point of view, this is interesting or really very uninteresting. Notice the very extreme behavior. Every single orbit behaves the same way. Every single orbit is dense. So as soon as I have a system for which one orbit is periodic, but not every orbit is periodic, then I am beyond this realm and so uniquely organic systems are good for what they are built for, but they won't really help you in systems that are more complicated. The examples I want to provide are examples that you have really different behaviors, but still you have complete understanding. So this is complete rigidity and as I said in the morning, we saw that the shift on two letters. I am just taking the non-invertable version. No classification is possible. Basically dimension is the only obstruction. You give me a dimension, I can construct a measure for you that has this house of dimension and impossible. But a celebrated conjecture still open. A first number says that, okay, this shift on this space, of course, you can think of it as times two map on r mod z. So this is the same as times two map from r mod z to r mod z. So times two map from the circle to itself. First number conjectured that the only ergodic non-atomic measures on s1, which are invariant by times two and times three map. So the setting that Vertigo was talking about, you have two commuting maps from s1 to itself, is the rotation. It is the Lebesgue measure. So this was conjectured, I think in the early 70s by Firstenberg and still open in this generality. Spectacular progress was made by Rudolf where he assumes if you put some positive entropy for one of the elements that's acting, then you have this classification, but without this positive entropy input, this is still open. Topological version of this was proved by Firstenberg in the same paper that he made this conjecture. So this is sort of the type of questions that I want to answer, but of course this is too difficult. So I'm going to look at easier examples. We are bigger group x, which I'll make it more specific, what bigger means and then the space also has structures. Okay, s1 has some structure, but space x has structure and then I try to answer these questions. So something we know, again continuing examples is Oppenheim's conjecture, which is no longer a conjecture. It was proved by Margulis in 86, but still this is the name one goes with. So this is clearly a difficult question. It's I'm asking for measures invariant by times 2 times 3. Now I'm going to give an example which comes from number theory and then see how dynamics can be used. So suppose b is a quadratic form in three variables. So that means it looks like pij x so x1, x2, x3 bij equals bji. So I have a quadratic form, examples 2xc minus y squared. Then Oppenheim made the following conjecture. Suppose p is indefinite, meaning that p of some vector, p of some non-zero vector equals zero. Non-degenerate, this is okay. The non-degenerate form, the inner product that it introduces is non-degenerate. Then, yeah. If I look at all the integral points, evaluate my quadratic form at those integral points. This is a dense subset In R, if and only if b is not a multiple of a rational form, a multiple of an integral form. And what do I mean by an integral form? An integral form is a form where all bij's are integers. So this was the conjecture that he made in 1930s, early 1930s. And was proved by Margules 86 using dynamics. So I want to spend maybe a few minutes translating this to a dynamical question and that will set up the stage for what is the manifold I'm thinking of and what are the groups that are acting. Is the statement clear? So before doing that let's make maybe the first exercise. Three is the best number you can put here. So n equals two, this is wrong. Okay, hint is look at the quadratic form square root of two x squared minus xy. So for n equals two statement like this cannot hold basically because quadratic integers cannot be very well approximable. This is the statement that square root of two you cannot make it, you cannot approximate it better than, there's certain c that you cannot make it make the approximation better than c over q square, where q is the denominator. And if you try to write that then you see that you cannot actually get close enough to zero. And slightly more complicated exercise, which is again not too difficult if let's call this star, if star holds four and now then it holds equal to and now. So really the hardest case is n equals three. If you prove it for n equals three, then you proved it for quadratic forms in n variables for every n bigger than three and for this you need to, okay, I give you a form that is not a multiple of an integral form. Let's refer to it as an irrational form. So this is the definition. So for this what one needs to do is that I give you an irrational form. You need to find some rational subspace of dimension n minus one, where the restriction of the form is non-degenerate, indefinite. These are quite easy to do because these are generic condition. You need to make it also irrational. That one requires some. So n equals three is really is the hardest. And indeed the history also exactly moved in that direction. When n is bigger than or equal to 21, this could be proved using circle method. I don't think number theory. And for certain special forms of lower dimension as low as nine or something, this was proved by Davenport and co-authors, but there is no non-dynamical proof for n equals three. Even the simple diagonal form square root of two x squared minus yz. Even if you take this form, there is no non- dynamical proof. Well, as of yesterday of Oppenheim Conjecture for this form. So now what was the idea? The idea is to translate this to a question in dynamics. This in some sense can be tracked back to a very influential paper of Kessels and Sonata Dyer, but explicitly was written by Raghunathan and is referred to as Raghunathan's conjecture. And that's what I want to do here. So translating to dynamics. First, I need a group. I'm looking at values of quadratic form at some vectors. So I am free to change my vectors by something that leaves the values invariant. What is this? This is the group H. I should, strictly speaking, put an index P. Wait, let me do that. These are matrices, three by three matrices with determinant one so that my form is invariant under their action. So the value of the form doesn't see the action weight G. This is a big group. HB is a rich group. Depending on how we are doing in time, this is an exercise or I will hint at it. It's a group that locally looks like NSL2. But forget about what it is. It's a group. Certainly anything that satisfies this you can take product and then you can take inverses. It's a group. It's closed. So it's a closed subgroup of it. Now in order to understand my quadratic form at integers, as I said, I am free to move all the integral points in the space. So I have the integral point sitting here and I want to know what my value of quadratic form is at all these points. I am allowed to take Z3 and shift it by any element in this group and then compute the values because by definition the value does not change. So now already it looks like an orbit. If you're thinking of it in R3, it's orbit of infinitely many points that are moving in R3. But R3 is a space that doesn't have a natural finite measure. So I'm going to think of Z3 as one point in a certain space. So what is the space? X3 is the space of lattices. So this is lambda subgroup of R3. So this is lambda is discrete subgroup. The volume of R3 mod lambda is 1. So space of 3. We have normalized volume. We certainly know one point in here. That is Z3. But we also, given Z3, we know every other point. Why? Because if you give me a guy like this, that means I have a basis vector. A basis vector needs to satisfy volume equals 1. That means the determinant of a matrix that I put these three vectors. So you give me lambda. This is the proof of something that, so you give me lambda. Lambda gives me a basis V1, V2, V3. I put them in a matrix V1, V2, V3. And this condition precisely is that the determinant of this matrix is 1. So I can identify this space with SL3R, because that's exactly the space of matrices with determinant 1. However, points have non-trivial stabilizer. So let me take one particular point, namely Z3. Then the choice of basis is not unique. It is unique up to 3 by 3 matrices with determinant 1 entries in Z. So this space is naturally identified with that space. The map that I just gave you. Given a lattice, you can send it to any basis. This space is not unique, but it's unique modulo of the stabilizer of the lattice. So this space is identified with that. And a difficult exercise, so it's really a theorem, is that SL3R modulo SL3Z has a finite invariant measure. We'll prove this at least for three replaced by two, either today or tomorrow. But it's true for any n. This space has a finite invariant measure. Another fact, which is this one is an exercise, is that this is X3 is not compact. Let me maybe hand wave, and then you can think to make this more precise. Why is it not compact? Given a sequence of lattices, I need to give you a sequence of lattices which could not possibly have a converging subsequence. What do you do? You take a torus and a carbonate and pinch it. So you take lattices and make one of the vectors short while keeping the area equals one. This means you make this short and the other two vectors long. This couldn't possibly converge to a lattice in any sensible way, as this space is not compact. So now let's go back to our problem. We wanted to understand the value of the quadratic form at integral lattice Z3. We are allowed to move it with this group, and this one gives me orbit of one point in the space X3. So I have the point Z3 HP applied to Z3 in the space of lattices. First, an easy observation. If B is integral, meaning all the coefficients are integers, then HP Z3 is closed, which you can think of this as HP times SL3Z. This is a closed subset of three by three matrices with determinant one is closed in X3. So already you see there is a relationship between the form being irrational and the orbit. This is the easy relationship. If I know the coefficients of the form are integers, then the set of vectors that I see by moving vectors with HP in R3, this forms a closed subset. But the hard direction is what theorem of Margulis and generalizations by Danny and Margulis are, which is if B is irrational, then this orbit is dense. Danny Margulis is that if B is irrational, then HP Z3 closure is the whole space X3. And of course, this finishes the proof of Oppenheim conjecture because I am the whole space of lattices. In particular, I get arbitrarily close to any vector in R, so B gets arbitrarily close to any real number. Let me pause here for a second. See, something that is quite particular about this theorem, which makes it difficult, is the fact that you gave me a group, HP, and you gave me a point Z3, and I needed to understand this particular orbit. Usually in the usual sense of ergodic theory, what you know is that the orbit of a typical point is nice. And this is certainly the case here also. For the action of HP on this space, orbit of almost every point is dense, orbit of almost every point is equidistributed, everything you want, even with rates. That's not too difficult to prove. And actually, we will prove today that the action of HP on this space is ergodic. So I understand orbit of typical points. However, the question that comes from number theory does not allow me to take a typical point in the space. The question comes with the point. The point is Z3. And ergodicity tells me that if I perturb Z3 to a nearby point, then I have this density. But that does not tell me anything about the point Z3. So it's quite important that I have description of absolutely every point in the space. Almost every statement has nothing to say when it comes to a question of this kind. So that's the type of rigidity that... Okay, that's a good question. So Margul has proved exactly what I cheated here a little bit. This is not really what Oppenheim conjectured in the beginning. First of all, he conjectured for dimension 5, then Davenport conjectured for dimension 3 and above. And what he had conjectured is that you get arbitrary close to 0, not that you become dense. You get arbitrary close to 0, and that's what Margul has proved. That also proves that for forms over R3, this also proves this density statement. However, Danny Margul has proved this orbit closure statement, which is what he built on really ideas that we've developed in probably. So this orbit rigidity statement is stronger than Oppenheim. And that's where the non-compactness of the space was actually used in Margulis' proof. So now that he was mentioned, let me maybe state the weaker statement. The weaker statement is that for every epsilon positive, this was the original form of the conjecture. There exists a non-zero vector in Z3 such that B of V is less than epsilon. So this is what Margul has proved. And in order to show this, what he needed to understand is whether this orbit, HBZ3, in this space stays bounded. This is a non-compact space, and the cusp corresponds to precisely lattices that they have short vectors. In order to get this, Margulis needs to find a short vector in this orbit, HBZ3. And that's what he proved. So he didn't go all the way to get the density. But before I erase this, let me go back here, another exercise which related to this. This is a somewhat harder exercise. What is the difference between the isometry group of this and the isometry group in three variables? So this is the key difference between the questions I'm asking and the space of shift. So I'll give you the answer once you need to move it. The answer is that this is a one-dimensional torus. This group is a three-dimensional group that is locally like SL2. And that's what the main difference is. This is a very rich group, as I said here. This is locally isomorphic to SL2R. It has many elements in it, but this group you can easily see what it is. It's just a multiplicative group. So this is the one-dimensional group that's a torus. This group is much bigger, and that's why this can be done. Else there was no hope. Maybe I will quickly mention another example, which is okay, that is proved. So let me mention something that is not proved, and this is little with conjecture. This one goes back to 1929. And the following is the statement. Limf of n, so for every alpha beta in R, limf of this product equals zero. This is the distance to closest integer. This problem also translates to a question in dynamics. And this is exactly the setting that Frederico was talking about. This becomes two commuting maps acting on the same space, actually. SL3R, module SL3Z. Let me maybe write the first line of translation, and the rest I will leave to your imagination. So what is it that I want to do? I want to take n, n alpha, n beta, but then find the distance to closest integer. So what does that mean? I want to find to look at a product that looks like that, that, and I want to show that limf of this over n is zero. So I find m1 that realizes that number, m2 that realizes this other thing, and I want to take the limf. It's product of three numbers. Whenever you have a product of three numbers, you're allowed to multiply this by something, multiply that by the other thing, and divide this by the product. So this, any element of this form, where m1 and m2 are integers, this is the same as I multiply the first one by that, multiply the second one, and multiply the third one by e to the s, multiplied by two numbers divided by their product. Now let me write this in matrix four. You gave me two numbers. I'll put them here, alpha and beta. I have three integers, n, m1, and m2. I'll put them here, m1, m2, and n. So, and I have my group, which is e to the t, e to the s, e to the minus t minus s. This one, I'm allowed to choose any m1, m2, n, which means this is z3. That's a point in the space given by alpha. So I have a point x alpha beta, which is 1, 1, 1 alpha beta z3. And this is the diagonal group. So I'm asking whether the orbit x alpha beta, whether I know what this orbit closure is. If this orbit closure has arbitrary short vectors, then I get local conjecture. So it's a, it's a different group acting, though. It's a group which comes with two commuting automorphisms that are diagonalizable. It's a much harder question. This is still open. It's really conjecture. And spectacular progress was done by, was made by Einstein, Ketok, and then Neuschraus, which built on earlier work of alone on quantum unique rigidity using entropy and entropy methods. So sort of an overview of the type of questions that you want to understand. Any questions? So really what you want to show here is that if an orbit is bounded, then it's periodic. That's the statement you want to show. You want to show that bounded orbits must be periodic. That's enough. That's enough. And difficult. But it's in, the spirit is very similar to times two times three. However, in times two times three, we understand the topological conjecture. In here, the topological conjecture is quite difficult to understand. And their approach uses measure. So they classify measures with positive entropy. Okay. So maybe in the last seven minutes, I want to prove the following proposition. Maybe I'll only be able to prove the first part of it, but that's enough. Suppose SL2R acts on a probability space and it's ergodic. Then part one, the diagonal of SL2R acts ergodically. Part two, the group of upper triangular matrices, the unipotent group also acts ergodically. This goes by a name. X mu is a probability space and the action is ergodic. Journal. I mean it's preserving the measure. Probability space. And the action is ergodic. For this, first I need an exercise, which is an exercise in linear algebra. SL2R can be generated with the two groups. The group generated by upper triangular unipotent and lower triangular unipotent. Elementary operations are allowed. Hence, you have the exercise. So that's, this means group generated. Now I'm going to prove part one. So I have an action of SL2R in a space. So SL2R acts on the space. That means it acts on L2 of X mu. And this action is isometric. And ergodicity means I don't have any invariant function. So if equals F for OG in SL2, then F is constant. Now I want to show A acts ergodically, so I should be able to show the same thing. But AT, which is my notation for this guy, F equals F for all T. I would like to show that then F must be G invariant, which would finish. So in order to show that, I will show F is invariant under any element in here and any element in there, and then exercise will finish. So is the proof or not? Okay. The proof is the following. e to the t, e to the minus t, 1, 1, s, e to the minus t, e to the t is e to the 2t, s, 0. This is really the proof. Why? I want to show this equals that. So I take their differences and calculate the altunar. If I want to show they're equal, I better show the altunar of the difference equals 0. This F was invariant under that group. So and this action is isometric because the action is isometric. I can put that minus altunar does not change because I'm acting by same vector. This is because isometric action. Yes, this is because the measure is invariant. Exactly. So it's isometric. Mu is SL2 or invariant. Now F is an invariant function. There I did not use the fact that F is an invariant function, but now I can use it. So I'll put inverse of this element there. I'm going to have a minus t at F minus. Here I use the fact that F was invariant. And I just get rid of that. Now I go back to what does this tell me? This tells me that the altunar is 1, 1 e to the minus 2t, s, F minus F. But the action is continuous. This is going to identity. So this is arbitrary small. This is less than epsilon for all epsilon. You get this one converges to 0. But that did not depend on t. That means 0. So I showed that same way. This exercise tells you that the function is SL2 or invariant. Now part two is a little more computation. It's a computation of the same sort. And I dare to leave this at 30. I need to stop, right? I dare to leave you this as an exercise. And hint is the following computation. That 1, 1t, 1, 1c can be written as 1, 1c, some diagonal, some upper triangular. You choose this to go to 0. Maybe I will let it be epsilon. So this is the hint for part two and similar computation. So what you need to show is that a function that's U invariant is A invariant. After you know A invariant, you can use part.