 All right, so thanks to those who came back. So the second lecture will be at least formally unrelated to the first lecture. But if you remember something from the first lecture, then perhaps some of the techniques or the questions will look familiar. So I'm going to start with some motivating questions. Yesterday we looked at how lattice points in R2 distribute themselves with respect to spheres of large radius. So now we're going to consider the hyperbolic plane. So gamma is going to be some group which acts cofinite volume on the hyperbolic two-space. And what we can take is we can take a point at O for 0. And we can take its orbit under gamma. And we're going to get some picture that looks like this. So you get some periodic picture. And this corresponds to gamma acts cofinite volume. This means that this picture really is the universal cover of a fixed surface of some genus. So it looks something like this. And here we have O. And maybe you have a fundamental domain, which looks maybe something like this. Maybe you have a fundamental domain. And you would like to understand, again, the type of question. So Br at O is the ball of radius R around O. You would like to understand the number of orbits of O intersected with this ball. And you could ask, is it proportional to, for example, the volume of this ball? And so this is analogous to the same problem in R2. But now the difference is that there's an issue. If you try to do even the simplest error term, so remember, we could try to somehow try to draw these fundamental domains like we did for the circle, something like this. And we could try to analyze the boundary of the circle. But the issue is the following. The issue is that the length of the boundary is proportional. Well, it's comparable to the volume of Br0. So both quantities are exponential in the radius. So they're very large numbers. And so they're of comparable order of magnitude. So if you try to do the base accounting using the same techniques as in the R2 case, then it's not going to work unless you have some extra ideas. And OK, so this is one type of question. But it turns out that the techniques that I'll explain, they allow you to treat more general types of questions. So here is another type of question that you can consider. So you can take O and you take gamma across geodesic. Here's gamma. And then this is covered by a picture that looks somehow this. This picture is not very accurate, but then you have stuff inside. This is the universal cover of this type of picture. And here's your point O. And you can again can ask if you take a sphere of large radius. So again, Br0, you can ask for the number of geodesics that this sphere meets in the universal cover. So what does it mean that this geodesic is a distance R less than R from this point? It means that if you drop the perpendicular, that this is perpendicular, then this length is less than R. So you see from this point, you might have a perpendicular that goes this way. You might have some perpendicular that goes on the other side. And you could have some other perpendicular that comes on this side. So from this point, you can draw many perpendiculars to this geodesic. And they would just correspond to the universal cover pictures like this. And you can ask how many such perpendiculars there are of length less than R. So you can ask for the number of, I'm going to call gamma tilde the lift of gamma intersect with Br0. This is asymptotic to what? And we'll see that it's asymptotic to the length of this geodesic times the volume of this ball. So we can treat the techniques that I'll explain. They can treat these kinds of more general questions. And the way to treat them is going to be by first establishing some equidistribution result, then see how it's related to counting. So in the case of, I'll just explain the case of this point count. So let's say that you take, again, the hyperbolic plane. Here's O. Now you consider BrO. And now, instead of considering the ball, let's consider the sphere. So the boundary of BrO is SrO. And we would like to say that when we project this to the surface, this curve that you get on the boundary, it lies quite uniformly distributed on the surface. So the type of question that the type of statement that you would consider would be the following. So suppose that you have a function from H2 mod gamma. So a reasonably nice function. And then you lift it. This is the lift. And you consider the integral of f tilde over this sphere. And let's say normalized by the volume of the sphere. Then you would want to say that this converges to the integral over H2 mod gamma of f. You would like this kind of equidistribution result, which you can get from some dynamical techniques. And once you have such an equidistribution result, then you certainly get, so if sears are equidistributed, then the whole ball is also equidistributed. Because if you have a result like this, then you can integrate. You can integrate. And you will get that the ball itself satisfies a similar type of integral property. You'll just replace the volume of the ball. And here you'll integrate over the entire ball. So if you have a distribution of sears, you have a distribution of balls. And if you have a distribution of balls, then how do you get the counting? Well, you can take your f to be a bump function around O on the surface. And then you'll see that the frequency with which the ball projects to a neighborhood of O is going to approach the volume of this ball. And so you get this counting result. So I'll do this in a little bit more detail in a second. But I just wanted to explain, at least vaguely, why such equidistribution results can give you the counting results. Are there questions so far about the idea? No. All right, so now I'll state things in the more general setup and how these techniques work. So I should say the following. So these types of questions about, for example, counting lattice points in the hyperbolic plane, they go back a long time. So I'm not sure exactly who were maybe the first people to do this. But I think maybe Hubbard or Huber and Selberg and some other people had some results about this counting. But the general setup that I'm discussing is following Eskin and McMullen. This is a little bit more recent. And their approach was one of the first ones after Margulis that used dynamics to study these kinds of counting problems. So some of their results were covered by Margulis, but not all of them. And most of the countering results before that were done using spectral methods like in the first lecture. So the setup is going to be the following. G is going to be some semi-simple Lie group. So if that is too much, you can consider SL2R or SLNR. Gamma is going to be a lattice in G. So again, SL2Z or SLNZ. And we're going to have a subgroup H side G. So I'm going to put a technical, this is a technical assumption, and I'll explain in a moment what it means. So such that G mod H is what's called G mod H is a fine symmetric. More precisely, this means that there exists sigma from G to G in evolution such that H is the fixed points of sigma. And to just have some examples, you can take SLNR and H to be SONR. And for SL2, you can take H to be the diagonal matrices. So this also works. And there are some variants of this. What I want to say is that, yeah, so this is going to be the setup. And so we're going to have these equidistribution results and counting results. Sorry. So the key assumption on H, so assume that gamma intersect H is a lattice in H. So this example when G is SL2R and H is this group where some group conjugated to this will correspond to the example of a geodesic. So remember, the geodesic flow corresponds to the diagonal subgroup. A closed geodesic is a closed orbit of the diagonal subgroup. So this means that if you have a closed geodesic, you have gamma intersecting the diagonal subgroup in a Z subgroup. And so this example will be a non-trivial example, which applies to this geodesic counting problem. OK, so this is the setup. And then the first theorem is that these guys proved this is an equidistribution theorem. It says the following. So we define Y to be this group, this thing that we assume is a lattice. So it's gamma intersect H, module H, and G, and X. So it's contained in X, which is gamma mod G. So you have a picture like this. You have your X. Maybe it looks like this. And then you have some subset. So both of them have finite volume, because then they have a natural volume. And Y sits inside X. So the equidistribution result says that for any F, let's say, smooth and compactly supported, it's enough to be continuous on X. So you have a function. You can average it over X, or you can average it over the orbits of Y. So you can do the following. So you can consider the integral over Y times G. So you act with G. So you have G mod gamma, and G acts on G mod gamma on the right. So you can act on the right by a big element in G. And you move around Y. So you apply an element, and Y moves in some complicated way. And you divide by the volume of Y, G, which is the same as the volume of Y, if you normalize it correctly. And you integrate F. So in this approach is the integral over X. Again, the normalization factor has to be the volume of X. The integral over X of F. And this happens as G goes to infinity in G mod H. So this quotient G mod H is some variety which is going to be important. So this is the equidistribution result. So then for a similar setup, we'll have the counting. Yes, well, you have to conjugate it. So you can take a conjugate of H. So everything I say is the group H should be this group up to conjugation. So there will be a conjugate of this group which will intersect as L2Z in a geodesic. And so this group will indeed not intersect as L2Z. But there will be plenty of conjugates of this group which will intersect as L2Z in a geodesic. And it's about those groups that we're talking. So does this clarify it? So I should say that, yes, the group H, I'll usually write it in some simple form like this. But it's always assumed that you can conjugate things to be in this form. It doesn't have to be of this, necessarily of this form. Where equivalently, you can always conjugate gamma. So you see if you have the lattice gamma, you can always conjugate the lattice. So you can take g, gamma, g inverse. And it's always going to be, again, a lattice. And then you can assume that H is of that form. But the lattice is a little bit different. So now you also have the counting theorem. So the counting theorem works in the same setup. But now what you consider is the following. So again, you have your g mod H. And you have here a point, let's call it O. And you can consider the orbit gamma times O. So this is the orbit of O. And so the assumption that gamma intersect H is a lattice is essentially equivalent in this case to the fact that gamma times O inside g mod H is discrete. So the discreteness of this set is equivalent to gamma intersect H being a lattice. So I'll try to come back to these two examples in a second and try to explain what these theorems are saying. But now, if this is a discrete set, we can start trying to count things. So we assume that Bn is a sequence of reasonable sets. So the condition that they introduce is called well-rounded. And I'll explain in a second what well-rounded means. So Bn are contained in this g mod H. And so the theorem, the counting theorem is that 1 over the volume of these sets, Bn times the number of elements in this orbit which intersect Bn is asymptotic. So if you divide the left side by the right side, you get 1. This is asymptotic to the volume of the Bn's. And I assume that the volume of Bn goes to infinity. So the sets get larger and larger. So let me do first the picture with the geodesics. So let's say that G is SL2R. G is SL2R. And H is the diagonal subgroup. Then so as you know, SL2R has a natural representation. So if you think about the adjoint representation, it acts in R3 of these guys. SL2R, so if you want this as the adjoint representation on R3, where equivalently it's locally isomorphic to SO21, so SO21 acts on R3 in a natural way. And so you get these three types of orbits. So the orbit, so hopefully if you've seen the hyperbolic plane before, you know that the hyperbolic plane corresponds to one of these hyperboloids. So this subgroup, this is also a hyperboloid or whatever this outside figure is called. So this is going to be G mod A. So the stabilizer of a point here is going to be this subgroup A. And you get G mod A. So this is the space of geodesics in H2. And H2 is this guy. So this group is G mod K, which is H2, and K is SO2R. So if you have a point here, then you get a geodesic on this plane. How do you get it? You take the orthogonal, the perpendicular, which is going to be a plane, and it's going to cut this thing in a geodesic. And so what does it mean to take a closed geodesic? So gamma inside SO2R acts on this. And so a gamma orbit on this outer hyperboloid corresponds to a geodesic. So let me try to draw some kind of orbit close to live on this outside hyperboloid. So if this orbit is discrete, then this corresponds to a closed geodesic. You see, if you take a non-closed geodesic, a non-closed geodesic is typically going to be dense on this surface. And its orbit in this space is going to be dense under the action of gamma. So gamma doesn't act co-compactly on, even if it acts co-compactly on the hyperbolic space, but it doesn't act co-compactly. In fact, most of the orbits are dense. But there are some points, the special points corresponding to closed geodesics, which will have this discrete orbit. So the counting theorem over here refers to the following fact. You take, essentially, imagine a sphere of large radius, and it will intersect this hyperboloid in some set. And the volume of that set with a natural measure will be proportional to the number of green points that you see in this picture. So this is the counting theorem for the case of geodesics. And for the case of when H is SO2R, this is just a hyperbolic plane. And we're doing the, so this assumption that gamma intersect H is a lattice is trivial because H is compact. So it's true for any gamma. And then we're counting lattice points in the hyperbolic space. OK, so are there questions about these two examples, how they work? OK, so now I can move on to the proofs. I have to explain first what are going to be the main tools that we'll work in. Oh, I didn't explain what well-rounded means. So well-rounded, a set is well-rounded. So this is referred to a family of points BN. So it says the following. It says that for any epsilon bigger than 0, there exists U epsilon neighborhood of the identity in G such that, so you want a set as well-rounded if you can approximate it from inside by smaller sets, by reasonable smaller sets. And you can also approximate it from the outside. So you want the following inequality to be true. The volume of BN, you want it to be bounded by y minus epsilon times the volume of the following set. So you take the intersection, the lower bound probably want the union, yes. So you want the union over all G in U epsilon of the translates G times BN. So you have your set and you move it around a little bit and you take the union of all of the sets that you get. And this should not be much larger than the volume of the set that you started with. And you also want a lower and upper bound, which should be 1 plus epsilon times the volume of the intersection where G is in U epsilon of G BN. So this G times BN is when you intersect them, you get a smaller set than BN. And you want this to be roughly of the same size for any epsilon. OK. So now the setup is as it is here. I'm just going to keep it. And so let me mention the tools that will be used. So main tools. So just as a reminder, so this appeared in Amir's lectures. So we have however more mixing, so it says the following. It says that if you have alpha, beta, two functions in L2 of G mod gamma, oh. So I'm going to put this start to mean this is, so this means average 0. So it functions that average to 0 on this group, on this quotient space. Then the following holds, you get that the integral of G pullback alpha times beta over this. I'm going to start abbreviating this space by X. So these functions, they asymptotically become orthogonal as G goes to infinity in G mod gamma. Sorry. As G goes to infinity in G. So you act on the function, you move one of the functions and it becomes asymptotically orthogonal to. Yes, sorry. Yes, thank you. Gamma inside G is irreducible. And this is probably an assumption that wouldn't be bad to make here just to be safe. Yeah, G is non-compact. Yeah, this theorem is certainly false for it. And yeah, it's false for G compact. OK, so now let me formulate it slightly equivalently. So now for any alpha and beta in L2 of G mod gamma, we're going to take, so we're going to do the. So this is just the same theorem, but now you see what happens if the functions have integral 1. And the claim is that this will be approximately the integral of alpha times alpha beta plus a quantity that goes to 0. So this goes to 0 as G goes to infinity. I am sorry for writing here. But does everybody see at least a little o of 1? Let me write it here, plus a little o of 1. So we have this asymptotic equidistribution, so this is just the same statement as over there. So then the other key input which we'll need is the following geometric assumption. So this is where the assumption that H is a fine symmetric comes in, so we have the wavefront lemma. So it says the following. So let me draw a picture on this side. So let's say that this is G, and inside you have H. So what the wavefront lemma allows you to do is it allows you to thicken H a little bit, such that if you move it, so here is GH, then, sorry, I want to act on the right hand, sorry, so HG, then, so if you move it a little bit, then this group is in the neighborhood. So this new neighborhood is not too distorted. So let me say the following. So for any neighborhood U epsilon of identity in G, there exists a neighborhood U delta of identity in G. So this is the type of statement that you see in continuity. For any epsilon, there is a delta such that if you take the group G and then you thicken it by this U delta, this is H times U delta, and then you act by G, any element G, so for any G and G. So you act on this, you act on the thickening, you move it, then you're still contained in this single orbit times U epsilon. So it says that if you've moved it, you haven't really moved. So nearby orbits of H look similar. Sorry, can I say again? Neighborhood, sorry, and BHD, sorry, any neighborhood. So with these preliminaries, we can prove the equidistribution result quite easily. OK, so proof of equidistribution. So first we're going to prove equidistribution, and then we're going to prove how equidistribution implies counting. OK, so what's the setup? We have X is G mod gamma. It contains Y, which is H intersect gamma H. And we want to show, so we'll show that show for B beta continuous, compactly supported, on X, that the integral over Y times G1 over volume of Y, beta, approaches 1 over volume of X, the integral over X of beta. And the way to do it is basically using the two things that we know how to do. So you see the problem with how more theorem is that it applies to functions, not to geometric objects, not to submanifolds. So what we're going to do, we're going to make a function out of our manifold. And we're going to make a function using the thickening that we can produce using the wavefront lemma. So what we do is we thicken Y. So fix an epsilon, which is very small. So we'll fix an epsilon neighborhood, which is very small, and how small it is will depend on beta, which is uniformly continuous. So you can assume that it's not just continuous, maybe smooth. Or if it's compact support, then it's uniformly continuous. So we thicken Y. And we're going to show that this is true up to any epsilon. So if we show that this limit holds up to an error of epsilon for any epsilon, then this is true in general. So how are we going to thicken Y? We consider Y times U epsilon. Sorry, U delta. U epsilon is our precision. And we thicken Y. So this is an open set, X. So if you draw a picture, this is X. Maybe here's Y. Then we've kind of thickened it a little bit. So now you consider the indicator function. So take alpha, let me call it alpha delta, to be the indicator function of this thickened orbit. And the point is that this, by how we more, we know that this will be what we want. How we more it tells us 1 over the volume of Y, sorry, 1 over the volume of Y delta. So let me call this set. So Y times U delta, let's call this Y delta, times the integral over Y delta times G of beta approaches 1 over the volume of X of beta of X. Because we're just integrating. So integrating against this function, alpha delta, essentially amounts to integrating beta over this open set. And we divide by the volume because, so you use the how we more theorem in this form. So you can move the volume of X to the side. So if you work out, you'll get exactly, if you apply just how more, you'll get exactly this statement. So now, but we want to compare. So what do we want to say? So I want to say, the claim that we're basically done is that 1 over the volume of Y delta, the integral over Y delta times G beta, is equal to the volume of Y delta. Yeah, sorry, I was confused. So the integral over Y delta of G times the integral for Y of beta plus something that goes to 0. So I claim that the integral, so this integral over here, so you can move the volume 1 over volume of Y delta over the other side. So this integral is basically the integral of beta over Y times the normalization factor plus some small error. So this follows from the wavefront lemma. The claim now follows from this wavefront lemma because, what is the picture? You see, so I'm going to draw maybe a caricature. So let's say that this is GY here. Can people see in this corner? So if this is GY, so yeah, YG, G acts on the right, then the wavefront lemma tells us that this Y delta, right? Y delta is the Y U delta thickening of Y. So the wavefront lemma tells us that if we act on the left by G on this set, then we're in the epsilon neighborhood of the, so we're in this epsilon neighborhood. By the wavefront lemma, we have that Y delta times G is contained inside Y times G times U epsilon. And beta is a uniformly continuous function. So essentially averaging, so the correct thing to be would be to just move this Y delta, so let me, so the average, so 1 over the volume of Y delta, the integral over Y delta of beta is approximately the same as the average over just this volume of Y, sorry, G. So because these, so they're approximately the same up to an error, which depends on epsilon and the uniform continuity of, and the uniform continuity of beta. So we get that these quantities are essentially the same, which is exactly this equality. So we just move one of the volume of Y delta to the other side. And this is what we wanted because, you see, here we have this integral over the thickening. And we just proved that this integral over the thickening is essentially the same as the integral over just the orbit itself up to some small error. And we proved this for any error, for any epsilon. We want this as small enough. So this gives us the equidistribution result. So OK, so I don't think anybody ever complained if I end early, so let me just say that next time I'll finish proving how equidistribution gives counting. And I roughly sketched the idea at the beginning. So if you have that these orbits equidistribute, then you can obtain some counting results. So the sphere equidistribution already gives you the counting for a lot of points. But I'll explain how to essentially kind of transfer these integrals in general. And so this idea of kind of transferring an integral from one homogeneous space to another is going to be useful, and we'll see it hopefully in the next lecture in another context. So I'm going to stop here. Thanks. Thank you.