 OK, great. So we'll continue where we left off last time. So just as a quick reminder, while we all wake up, last time we saw three definitions of a translation surface. And all of those two are going to be the most important. The first is that it's a Riemann surface with a holomorphic one form on it, a.k.a. anabelian differential. And the second is it's a collection of polygons with edge identifications up to some cut, paste, and translate equivalence relation. And we saw lots of examples last time, square-tiled surfaces, the slip-torus construction, and unfolding of polygons. So today I want to talk about the modular space of all translation surfaces. So the moduli, of course, just means parameter space, so just the space of all translation surfaces. And to start, let's just do a thought experiment. So say we have the regular octagon with opposite-sides identification, opposite-sides identified. You can think, what can I do to this to get some other translation surface? Maybe a nearby translation surface. And the answer is pretty clear, actually. You want to wiggle the edges. So sort of perturb this to get a new octagon in the plane. And let's think about how many ways there are of doing that. So I have eight edges, but I actually can't change all of the edges independently, because they have to be identified in pairs. So this edge better stay the same length and in the same direction as this edge, et cetera. But what I can do is I can fix four of the edges, and I can sort of wiggle those edge vectors. And then the opposite ones are wiggled in the same way. And then I get a new translation surface. So oh, it'll close up because I'm keeping the opposite sides of the same length. So if you want here, I've got minus v1, minus v2. If you want to sort of put an orientation around them going around. So it looks like I have local coordinates for a modularized space, and the local coordinates look like v1, v2, v3, v4, and c to the fourth. Now, these are only local coordinates for the following reason. If I change these enough, this picture will actually stop making sense. So let me do that before your eyes. Let's get rid of these. So maybe I'll take this vertex and sort of move it in here. So there's a different surface, and then I'll move it a little bit more. So I guess here, this was v1, v2, v3, v4. So I'm sort of changing v3 and v4. And I'm not even changing v1 and v2. But let's just take this a little bit further. It just doesn't make any sense anymore. So the local coordinates, because eventually the picture will start, will stop making sense. And furthermore, there's sort of not canonical in that there are sort of many different local coordinates. So this is sort of the nicest one if you pick the argument, the regular octagon. You can argue that it's sort of canonical. But in other situations, it's not so clear what the nicest coordinates are. For example, maybe I can do this cut and paste thing. So maybe I'll cut here and then sort of delete the top and put it over here. This is the same surface. And then maybe I have sort of local coordinates. This was v3, v4, I still have v1. And then I have this sort of new coordinate, w. So I could have a new local coordinates, maybe w, v3, v4. And there's sort of infinitely many ways I can cut and paste and reglue. And so I could pick different edges and I could get different local coordinates. But you'll notice very quickly that all the different local coordinates are related to each other in a very nice way. So in other words, you can think about the transition functions. In here, the transition is quite easy. This w is a linear function of the v's. It's something like v. So this is v4 minus v4 plus v1. So it's just the sum of these edges. So change coordinates in gl. OK, so does this example make sense? Because if you understand this example, you can understand the general picture. It's worth pointing out there is some content to saying these are local coordinates. There's something you have to believe, which is that when you just nudge the edges a little bit, you can't cut it up and reglue it into what it was before. So that's something which I don't think should be obvious to you. But it's true these are local coordinates. OK, so let's sort of state the general version of this. So recall from last time that we had a Gauss-Bonnais type theorem. So there were sort of two versions of this fact. One was the number of extra 2 pi's of angle was minus the Euler characteristic. This was a version of Gauss-Bonnais. And an equivalent fact was the sum of the orders of the 0's of omega. And these were equivalent because we saw that a 0 of order k, so that was z to the k dz, we observed this was up to scaling the pullback of dz under z goes to z to the k plus 1. So this corresponded to a k plus 1 times 2 pi. OK, anyways, because of this sort of Gauss-Bonnais theorem, we're interested in partitions of 2g minus 2. So let kappa be a partition 2g minus 2. So it's just some integers that add up to 2g minus 2. So for example, if g is 2, we have kappa equals 1, 1, or just a single number 2. And these corresponds to the two possibilities we talked about last time that in genus 2, you can have two cone angles that are 4 pi or one cone angle at 6 pi. And in genus 3, you have a bunch of more possibilities, but it's quite similar. Did I get them all? Anyways, we have some possibilities. And if I'm going to talk about modulite spaces, I need to say what the modulite space is going to be. And it's going to be determined by one of these partitions, which are going to describe the cone angles, or probably the 0s of omega. Stratum, which I'll use the notation h kappa, is a set of all translation surfaces x omega with cone angles 2 pi times kappa i plus 1. Or equivalently, omega has 0s. So the kappa i are just the numbers that appear. OK, and the theorem that I was alluding to in this example, or maybe I'll call it a proposition, is a complex orbifold. You don't know what an orbifold is. Just pretend I said manifold. You won't be harmed of dimension 2g plus the size of kappa minus 1. Let's just ignore the orbifold points. So away orbifold points, it has an atlas of charts 2cn with transition functions in gl. Sorry, n is the dimension 2g plus the number of things that appear in kappa. So here n would be 2 and here n would be 1 minus 1. The same thing, you just have to keep in mind the orbifold points. So this is a very mild kind of orbifold that has a finite cover, which is a manifold. So if you add a little bit more data to what's called rigidify the modulite problem, then the orbifold points go away, and then it's just a manifold. So the orbifold points are not at all scary. You can sort of wave them away very easily. Other questions? The absolute value of kappa is the just number of terms. So here, that's 4, 3, 2, 2, 1. As will be clear in a moment. So one point of view is that these local coordinates are provided by edges of the polygon. But I want to try explaining it also from a different point of view. So coordinates, usually, edges of polygons. In other words, they're complex lengths. But I want to rephrase that. So if I have an edge gamma, I can think of gamma as giving rise to a homology class. But the two vertices at the different ends of the edge aren't necessarily the same point. So in fact, it's a relative homology class, sigma equals set. And the complex length, the edge, that's sort of a complicated, well, a needlessly complicated way of computing the edge of a line. The length of a straight line segment in C is to compute the integral of dz. So this is omega. So dz is the Abelian differential. So these are what's called relative periods of the Abelian differential. Relative because I'm using relative homology classes, as opposed to absolute homology classes if I was using the usual homology groups. So in fact, what you can do is you can pick. So now that we've made this transition to this more abstract point of view, we can abandon the polygons if we want and sort of give another description of local coordinates. So we can pick any basis, h1, x, sigma, z. And usually the way you do this is follows. So you start with a basis gamma 1 up to gamma 2g for each one, z, the absolute homology group. So these are actually loops that close up. So the usual picture will pick some curves like this that close up. And now you add to that, add a curve, a fixed point of sigma to every other point of sigma. So in other words, I have a bunch of points in sigma. And I just pick one to be my favorite. And then I just pick any curve I want. So these relative periods, in a sense, measure the distance between the cone points. And then altogether, I get a basis for the absolute homology group. And the coordinates are just these relative periods. And this explains why the dimension is what it is. Wait, why the dimension is what it is. Because I have 2g absolute homology classes. And then I need to add the number of cone points minus 1 relative homology class. Any questions about that in terms of this picture? Yes. So this isn't a very good example, because here there's only one cone point. So you can't distinguish. But let's do another example. Oh, maybe let's do one of these examples with the slit torus construction. Two tori, and as usual, I'll glue them together. So now let's illustrate this basis. So maybe I'll pick that at gamma 1, gamma 2, gamma 3, gamma 4. So that's the basis of absolute homology, which would have four curves in genus 2. And now I have two points of two cone points. So there's one cone point. And there's another cone point. So now I'm going to pick one more curve, gamma 5, which joins them. Oh, if I pick, no, A and B are the same length here, because it's a slit. Or they actually, A and B here give the same relative homology class. They're homologous. So it looks like this. Here I have gamma 1, gamma 2, gamma 3, gamma 4, and then two singularities, gamma 5. That's sort of this A or B. And the other one of A and B is just sort of going around the back and is homologous. Because A and B together divide it into two tori. Does this answer your question? Oh, because I have two cone points, but I only need one relative homology class. Because the relative homology class is, indeed, measuring the relative positions of the two zeros. So what I was trying to indicate over here is, since it's relative positions, I can sort of fix one to be the reference point. And then, in a sense, I'm measuring the distance of all of the other cone points from that one. Although you have to be a little bit careful with this word distance, because I'm not necessarily picking this to be a GDC, but somehow the holonomy of the integral of one of these curves. Well, these gammas aren't even curves. They're homology classes. They're the integrals of omega over these homology classes. Well, it depends on how I pick the curves. So yes, in this example, it's the length. But what I'm sort of warning you is I could have picked a curve that sort of went. I mean, I could have picked any basis that I want. So I could have picked a curve that sort of. And then, well, so first of all, just a homology class. But I could pick the actual curve with the shortest geodesic length in that. And I can look at the geodesic. But then, that might be a union of many saddle connections. And so the integral wouldn't be directly related to the length. You can still use it as a coordinate. I mean, let's put it this way. Say I was stupid and picked some awful basis, which I shouldn't. And you're wondering, is this basis still a coordinate? Well, you could pick a more reasonable basis, where all of the curves are actually geodesics, if that makes you happy. And I think it makes me happy, too. But if I know all the integrals from my basis, then just sort of a linear coordinate change will say all the integrals for your basis. And so with integer coefficients. So you know all of the periods. I mean, once you know the periods of a basis, you know the periods of every single homology class. So if you picked A, you mean if you picked A instead of B, you actually get the same. You would exactly the same thing. You literally get the same number. Because despite the fact I drew this as curved. Yeah, one way or the other. So those are homologous. So in other words, I can look at this whole loop. And I could integrate omega over that loop. But Stokes' theorem, I mean, since omega is closed, I get 0. The integral of omega over this is equal to the integral of d of omega over this surface. So I get 0. So that's somehow manifesting the fact that the two sides of the slit have equal length. Are there questions? I really like getting all these questions. You guys should keep asking them. OK, so I want to take a minute to explain the word stratum. So let's fix G. And I want to look at the union over all of the partitions of 2G minus 2 of this strata. So this is the set of all x omega or x has genus G and omega is not 0, with no restrictions on the types of the zeros. Because I'm sort of taking the union over all possibilities. And this is what's known as the Hodge bundle over mg minus the 0 section. And so here mg is the moduli space genus G Riemann surfaces. Another beautiful, beautiful moduli space. OK, so the Hodge bundle is a very classical object. And I have it sort of broken up into all of these pieces. And these pieces are of different dimensions because the number of terms in the partition change. And when you break up a space into a bunch of nice pieces that sort of have different dimensions, you call it a stratification. So this is a stratification of the Hodge bundle. And that's why you call the pieces strata. OK, and just because people often ask about this. So the smallest dimensional stratum has dimension 2G. That's when sort of there's only 1, 0. mg has dimension 3G minus 3. So the smallest dimensional stratum doesn't even have a chance of projecting onto all of mg. The biggest dimensional stratum, so maybe I'll just write this out because it's just not strictly speaking necessary, but it gives us a feel for what's going on. Smallest stratum dimension 2G. So that's when kappa is just a single number. And the biggest, as dimension, well sort of the biggest kappa can be, like the most terms it can have is when it just has a bunch of 1s. So then it has 2G minus 2 1s. So the dimension is then OK, so I have 2G plus 2G minus 2 minus 1. 2G plus the number of cone points minus 1. So approximately 4G 0s. And just to complete the sort of general picture, mg has complex dimension. All of this is complex dimension 3G minus 3. So the hodge bundle, the biggest stratum is open in the hodge bundle. It's like the bulk of it. So it's really big. And the smallest stratum is sort of small compared to mg. No, the fibers are quite complicated. So the stratum lives over mg. And the sort of projection map, as you just forget the differential, the fibers are sort of complicated. Both the absolute and relative periods can change in the fibers. Yeah, and the fibers, even their topological type, can vary. Yeah, I think so. It's sort of, well, the real answer, we don't have a good understanding. It would be quite interesting to get sort of a good description, even just of the image. But also of the fibers. It's unclear what you should hope for. Oh, no, the open stratum, that map is onto. Yeah, so you're missing this sometimes called the discriminant locus, the sort of multiple zero locus. And I don't think we have a great understanding of exactly what that looks like. It's some sort of sub-variety. I'm not sure exactly what you would want to know. But anyways, I personally can't tell you much about what. Yeah, it's somehow moving the zeros of the form. But everything about the form changes as you do that. Yeah, everything in sight is an algebraic variety. No, it's not a proper map. So the image is not a 100% sure about that. But the map is not proper. Sort of zeros can come together and coalesce into a higher order zero. Or in other words, cone points can coalesce into sort of a cone angle, cone point with larger cone angle without the surface degenerating. OK, so I just want to get sort of the most basic picture of what a stratum is. And probably the most basic question is whether strata are compact. And the obvious answer is no, because you can scale. I can look at two times omega and three times omega and et cetera, et cetera. But I can also look at the unit area locus, which is just the set of x omega, where the area of omega is 1, which is the sum of the areas of the polygons. Or if you don't like that, the integral of omega which omega bar times i over 2. But even the unit area locus is not compact. Well, shearing doesn't always sort of degenerate. The main point, though, is that when you write things in terms of polygons, some of the edges can reach zero length before the actual surface becomes degenerated. So for example, here, I could let the slit length go to zero. That would degenerate the surface. Or I mean, there are more complicated things you could do. You could somehow let the slit length go to zero and shrink this torus, but increase this torus at the same time. So there are all sorts of interesting ways you can degenerate a translation surface. Oh, yeah, a translation surface by definition requires the differential to be non-zero. Not identically zero. No, no, no, I'm saying the hodge bundle is by definition the locus of all x omega, all of them, not just the non-zero ones. The zero section is just the set of x comma zero. So what I'm saying is the union of the strata is the hodge bundle, which is all x omega, minus the zero section, which is the x comma zero. So but the unit area locus in a stratum. No. So for example, if I shrink the slit over mg bar, then you can do something. But if I shrink the slit, I'm sort of pinching this curve. And I get a nodal Riemann surface, not over mg. Yeah, so what I'm saying is I could shrink the slit to zero. In the limit, I get a picture that looks like two tori glued together at a point. This is not a point of mg. This is not a Riemann surface. This is what's called a nodal Riemann surface. This is not in the hodge bundle. Yeah, this is not in the hodge bundle because it's singular. So there is something called mg bar, the DeWien-Mumford compactification. And this nodal Riemann surface lives in mg bar, but it doesn't live in mg. Yeah, that would be the quotient by the S1 action. So there's an S1 action on each stratum where e to the i theta acts on x omega by just rotating omega. And so this rotation obviously doesn't change the flat metric, but it sort of changes the choice of north. So if you quotient by this S1 action, you get the space of cone metrics with trivial hoonomy, which is in quite a translation surface because you don't have a choice of north or you don't have sort of an atlas of charts to see with transition functions being translations. How is this related? Yeah, it's different. This is still different, right? Because this is even smaller than the bundle of abealing differentials. You're asking about the bundle of quadratic differentials, which is even bigger. So it's sort of in the other direction. Maybe we can talk about that later. OK, so Mazer actually gave a compactness criterion. So a closed subset of a stratum is compact if there is a positive lower bound of saddle connections. So remember, a saddle connection is just a straight line statement during two of the cone points. You think of it as an edge and some polygonal representation. So this is a fairly concrete compactness criterion. And it also clarifies that you can think of there being two sources of non-compactness. One is the saddle connection might join two different cone points, at which point when that gets small, you're not necessarily degenerating the surface, but you're sort of degenerating the differential because two zeros are colliding. But also, the saddle connection could join a zero to itself. And if that's getting small, then it's really the whole surface degenerating, not just the abealing differential. There are special types of geodesics. A general geodesic is a union of saddle connections. OK, but just to round out our picture here, Mazer and Beach, so I'm not going to talk about this a lot, show that the volume of the unit area locus in a stratum is finite. So a word about volume, there's a natural Lebesgue measure on each stratum because the transition functions are in GLNZ. And the determinant of an integer matrix that's invertible is 1 or minus 1. So there's a natural sort of volume form. The whole stratum under that is infinite because of the rescaling the surfaces. But there's a way of modifying that to get a measure on the unit area locus. And with respect to that measure, the stratum are finite volume. So you think of a stratum as being some sort of object, some sort of manifold like this. Somehow it's not compact, it has this end. You think of it as just having one end. It's not like a cusp in that the diameter of the end is not going to zero. So you can have two degenerate surfaces that are very far apart from each other, which is why I didn't draw it like this, which sort of says there's really only one way to degenerate. But somehow, despite the fact there are all these different ways to degenerate, somehow they don't use up a lot of volume. It's somehow wide but not very thick. It gets a bit challenging to visualize it. I don't think you should take it too far trying to visualize it. I'm just trying to give you some sort of starting picture to think about what a stratum looks like. Well, so first of all, very easily you can get SL. And then you just say one final comparison. So you should compare the unit area locus in a stratum to some homogeneous space like SL and R mod SL and Z. But probably you want to take n bigger than 2 if you want a representative picture, at least if you're looking at genus G greater than 1. So if you're familiar with this, you should think of it as some sort of non-homogeneous analog of that. So that's sort of the introduction to strata. So now we get to the star of the show, which is there is a GL2R action on each stratum. And it's the easiest action there is. So if G is in GL2R, then G times x omega is obtained by acting linearly on polygons. So just some examples of this. So the easiest surface is this square torus. And I could maybe act on it by 1, 1, 1, 1, 1. And that has the effect of shearing it over a little bit. And you keep the same edge identifications. Similarly, I could take 2, 0, 0, 1, 1, 1, 1, 1, and act on the square torus and get sort of a stretched out square torus. So this is really, really an elementary thing. All you need to know is that if you apply a 2 by 2 matrix acting linearly on R2, parallel vectors of the same length go to parallel vectors of the same length. So you can keep the same edge identification. That's the only thing you need to know to define this action. Question is the definition of the action clear? A lot of the time you just look at SL. The scaling is not a big deal. You could look at the SL action. So let's start having remarks. So 1, SL to R acts on the unit area locus of the stratum, which maybe I'll write H1 of kappa. And if you're doing dynamics, that's what you study. So if I look at cos theta, all this does is it rotates. Typically, if I act on a translation surface, I'll get a new translation surface on a new Riemann surface. So usually it's changing both the Riemann surface and the differential. And the final comment is you should think of this as some sort of transcendental thing. The strata are varieties. They're complex manifolds. This action is somehow not respecting the algebraic structure or even the complex structure. Yeah, it's only the conformal matrices, which are just the rotation matrices. Oh, and the scaling matrices. So if you act by twice the identity, that just multiplies the differential by 2. Oh, you mean why is it on the stratum and not just the whole hodge bundle? Well, I get an action on the stratum because when I act, I get another translation surface. And you can check it doesn't change the cone angles. So it's somehow just very, I mean, that's a philosophical comment, which is that it's not respecting some algebraic, yeah. So let me tell you what I mean by this more precisely. So maybe you'd present the x omega using the way an algebraic geometry would, sort of an algebraic curve with some algebraic one form on it. And then I say I want to act by 2, 0, 0, 1, 1, 1 half. Tell me a new algebraic curve with a new one form. You just forget about it. I mean, maybe you can write a program to spit out something to 20 decimal places if you're really energetic and smart. But one of the things that makes Riemann surface interesting is there are many different perspectives. It's the same thing that if you have a Fuchian group, you can't give the equations of an algebraic curve, except to 20 decimal places using a, sort of, involved computer program. Wouldn't help at all. You still couldn't compute new, sort of. I can act on R2 by any matrix in GL2R. It's really just very elementary. Yeah, well, that's why it doesn't give you the same translation surface back. So you're thinking about there were two situations where we have atlas of charts. We had an atlas of charts on a translation surface, and we had an atlas of charts on the stratum. But this is just, sort of, orthogonal to both of those discussions. This is, I give you a matrix and a translation surface. You get a new translation surface. Yeah, that was about coordinate changes for coordinates on the stratum. But actually, good lead-up question to what I was going to do next. So I want to think about the stabilizer in SL to R of the flat torus, the 1 by 1 flat torus. So let's think about this. So if I do 1, 1, 0, 1 times the flat torus, well, that just shears it over by 1. And this was one of our original examples. I can sort of cut it and move it over there, and I get the same thing back. And similarly for 1, 0, 1, 1. So actually what you see is that this is SL2Z, sort of unrelated to the SL2Z in your question. But OK. So what's the reason I'm presenting this example? Because this is really important to internalize. You might think you apply a big matrix. Now your polygons become very, very sort of stretched out, very sort of degenerate polygons. Like you apply the matrix 1,0001 over 1,000. Like you're sort of looking at the x-axis, basically. So you're like, oh, I have a really different translation surface now. But there's the cut and paste, which is somehow what makes this whole subject difficult. That's a very degenerate picture of a translation surface. But maybe you can cut it up and get a new picture. So here I could act on this by a matrix of size of billion. And it would give me some very degenerate picture, but I could cut it up and get this picture back. And for this reason, the action is sort of a lot more complicated than you might think at first. Let me put that question aside for now. And similarly, so I'll just make this as a remark, for any square tile surfaces, filed surface. So for example, we talked about this one yesterday. But just as surface you get by gluing together squares, the stabilizer is a finite index subgroup of SL2Z. So you can prove that as an exercise if you want something to do. And by the way, another exercise that maybe should accompany that one is show that at least if you're allowed to vary the size of the squares, so they don't have to be 1 by 1, but could be x by x, square-tiled surfaces are dense in every stratum. So the combination of those two exercises will give you a little bit of insight. The next very basic thing I want to do. So first of all, why did I say an SL2R? Well, essentially, it wouldn't have made a difference. Because when I act by a matrix, the area changes by the determinant of the matrix. And if I have two translation surfaces that are equal, obviously they have to have the same area. So the determinant had to be 1 or minus 1, anyways. And to be honest, if I was less lazy, I would write gl2 plus to say the positive determinant matrices in gl2, because that's usually what you want to work with. But proposition, again, with SL2R bits, the SL2R orbit of any x omega is unbounded. It's related to the number of tiles, but the index is not the number of tiles. Can I describe the index? The index is at most the number of square-tiled surfaces there are in that stratum with the same number of tiles. But now I'm giving you hints for the exercise. So let's get back to this. Does anybody have an idea about how to prove this? Why should the SL2R orbit of every surface be unbounded? So it's not true that the e to the t00 e to the minus t orbit of every surface is unbounded. Surprisingly enough, by far most of them, to an extreme degree. But you can't always just use just it's not always true that the e to the t00 e to the minus t orbit is unbounded in the stratum. And remember, we had this Maser's compactness criterion, which says that it's unbounded. That means that there are saddle connections of small length. Yeah. Yeah, so then if I rotate and scale, why does it make it obvious the? OK, so that's really close. You're saying the right things, rotating and scaling. But really what I want to know is I want to get small saddle connections to know that I'm not compact. I want to show that for all epsilon, there's something in the SL2 orbit with a saddle connection of length less than epsilon. So how can I do that with rotating and scaling? Yeah, so rotate so that you get a vertical saddle connection and then shrink that saddle connection. I mean, essentially, just take any saddle connection, you can shrink it with GL2R. So proof thick theta, so r theta x omega. This is the rotation by theta. I won't write out the rotation matrix, but that's the rotation by theta, has a vertical saddle connection. And then consider, let's say, of length L. So I'm just letting L denote the length of that. So there's not a unique theta. There'll be many, infinitely many, even. So then e to the t 0, 0 in the minus t, r theta x omega. So then I scale the vertical direction by e to the minus t. So has a saddle connection of length e to the minus t. So that makes it clear the orbit is unbounded. Ah, so that's very good. So the sort of obnoxious answer, that's what Maser's criterion says. But then the question is, how do you prove Maser's compactness criterion bounded below? So here, the length is not bounded below of this set. Is it if and only if and Maser's criterion? So it's if and only if. Yeah, it's if and only if and Maser's criterion. But no, there's a deeper question here, which is sort of like, why is Maser's criterion true? And the answer is that on every translation surface, there are only countably many saddle connections. And there is a lower bound, which will be familiar to any differential geometers in the room because there should be a lower bound for the length of the shortest geodesic. So it's sort of comparable to that. So one more proposition to just sort of help us get a feel for this. So the stabilizer gamma in SL2R of any x omega is discrete, not co-compact. OK, so a subgroup of SL2R is discrete. If there's no sequence of matrices in it that converges to the identity matrix. And the proof of discreteness is pretty easy. So discrete since if g is sort of close to the identity matrix, then x omega and g times x omega have different coordinates. Edges of the polygons give local coordinates. And if I act by a matrix, I have different coordinates. The reason the matrix has to be small to some degree is that you want to make sure you stay in a single coordinate chart that's injective. If I act by a big matrix, OK, well, I have different coordinates, but somehow I may have gone around and come back to the coordinate chart. OK, so that's why discrete, so not, so that was discrete, not co-compact since the orbit is isomorphic to SL2R mod gamma and is not. So by the orbit stabilizer theorem, the orbit looks like SL2R mod gamma. And it's actually homeomorphic to SL2R mod gamma. And but we know the orbit is not bounded, so it can't be compact. And the definition of co-compact is that this. So somehow the stabilizer can be pretty big. It can be SL2Z, for example, which is a lattice in SL2R. But it can't be too big in that it can't be dense in SL2R. And it also can't even be a co-compact lattice in SL2R. OK, so that was our introduction to the moduli space in the GL2R action. Next time, I'm finally going to get to tell you what a lot of the motivations are. I'll tell you the connection to billiards, sort of some dynamical problems. I'll also tell you the connection to the teichmeler metric on MG. And then we'll actually prove some things about this action. And the hodge norm will make an appearance. OK, thank you.