 So, first of all, I'd like to reiterate that I really encourage you to all ask questions during the talks and after the talks and whenever the more questions, the better, really. My goal is really to talk about dynamics on modular spaces, modular spaces of translation surfaces. But for today, I'm going to start out and just make sure we're all comfortable with what a translation surface is. And so I'll start by giving you three different definitions and I'll sketch the proof that the three different definitions are equivalent and then I'll give you a bunch of examples. Okay. Can people at the back read this? Yes. Okay. Let me know if it gets too small. Okay. So, a translation surface is a collection of polygons in the complex plane with pairs of edges identified via translation up to a equivalence relation that I'll cut, paste, translate equivalence relation. Okay. So, let me give you an example. And through the examples, I'll clarify exactly what I mean by some of the things in the definition. Okay. So, the first example of a translation surface is the square with opposite edges identified, okay, which does indeed give you a surface. And so, let me say what I mean up to cut, paste, and translation. So, and this will sort of come as no surprise. But first of all, I'm allowed to cut a polygon in two, giving two polygons whose edges are identified via sort of the identity translation. Okay. So, having cut the polygon in two, I'm now allowed to translate one of the pieces elsewhere in the plane, keeping the same edge identification. Okay. And here to sort of illustrate the situation, I'm just going to translate it over, over here. So I have this and the opposite edges are identified. And then, so the opposite of being allowed to cut is I'm allowed to paste back together. So if I have two polygons sitting next to each other, sharing an edge, I'm allowed to sort of just delete that edge and consider it to be one bigger polygon. Okay. So those are the moves. But I just want to remark that rotation is not one of the moves. Okay. So if I take the square torus and rotate it a little bit, then I actually get a different translation surface. Although if you pause to think about it, that's actually not so obvious. It's a good exercise for you to do if you're looking for something to think about why is it that you can't, you know, cut up a little slightly rotated square into millions of pieces and reassemble them into the big square. No, you don't. You can cut anywhere you want. So you can sort of make new vertices if you'd like. How does this work? Okay. So another example, everybody's second favorite example is a regular octagon with opposite sides identified. And one of the first things you should note is that when you identify edges, various vertices get identified. Okay. So let's just follow through. This edge is identified there. So this vertex is identified there. And then this edge is identified here. So this vertex is here. And when you follow that through, you'll see that there's actually only one vertex. And then you can, yeah, it'll be a consequence of the definition that the quotient of these polygons via the edge identifications is a compact smooth surface. So what you can do is you can just add up the angles around the vertex. And what you'll find is that there's six pi angle. And you can also compute that this is a genus two surface. What you should not try to do is try to think of it the way you usually think of a genus two surface. Okay. It is homeomorphic to this because it's a genus two surface, but it doesn't help to try to visualize it. It's sort of possible to visualize, but it's very hard. And in other examples, it can't be visualized. Rather, the genus is something you compute by computing the Euler characteristic, vertices minus edges plus faces. And that's what you should do if you want to check that this is genus two. I also want to give a non-example. Okay. It's getting a bit small. Let me rewrite some of these. Maybe I'll rewrite them in capital letters. Is that easier to read? Thank you for letting me know. Okay. So this is a polygon with edges identified in pairs via translations, okay? But it's not allowed due to various details I didn't write here because it actually takes a while to write out all the rules quite formally. But the problem here is that, say this edge B. You can't translate these and have the polygons be on the same side of the edge. Okay. So the problem is when you translate, the polygon is on the same side of this edge. So one of the rules that I didn't write was if you have an edge identification, then when you translate to actually glue the edges, you have one polygon on each side of the edge rather than two polygons on the same side, as is the case here. So this one is not allowed. Non-example. No, I don't think it makes a non-Hausdorff. It's just somehow not the structure we want to deal with. You could be interested in that. I mean, it's, you know, but we're not going to. Okay. So I want to revisit this cone angle. I said there was a six-pi cone angle here. And I want to give you a very concrete way of thinking about that cone angle and more general cone points. So a cone point of angle two-pi times k plus one looks like. So I should say that this is called a cone point. It's a singularity of the flat metric induced by the metric on the plane. And what it looks like is very, very concrete. You just take some half planes and you glue them together in a cyclic order. So this is a good way to visualize a cone point. You just take, so if I want 2k plus 2 times pi angle, well pi is just a half plane. So I'll just take a bunch of those and I'll glue them together cyclically. And then a neighborhood of that cone point looks like a neighborhood of this point here. Okay. So this brings us to the second definition of a translation surface. Translation surface is a closed surface x with a finite set of points in x, sigma subset of x, such that x minus sigma has an atlas of charts to see whose transition functions are translations, translations, and such that a neighborhood each point of sigma looks like the above k plus 1 construction above. Okay, so you have this finite set sigma of singularities. Away from that, you have an atlas of charts to see so that when you change coordinates between two of these charts, it's just a translation. And at each point of the finite set, it's locally isometric to a neighborhood of the special point in this construction. Any questions about the definition? Yeah? So it's a consequence that it's oriented of that definition. It'll become clear in a minute, actually, why it's oriented. Other questions? Yeah, yeah. So if k is equal to 0, there's no point in including it in sigma. So if k equals 0, you just take one upper half plane and one lower half plane and glue them together cyclically, which is like this. And it just looks like the plane. So at k equals 0, it's just sort of a regular point, 0, which is equivalent to just getting c. So it's a smooth point. Let me write non-singular point. Yeah, I want to emphasize the surface is smooth even at the points of sigma. It's the metric that's singular. OK, other questions? OK, so I want to talk at least briefly about why the first two definitions are equivalent. And then we'll have a third definition also. So proposition definitions 1 and 2 are equivalent. So proof. Well, really it'll be a proof sketch. So let's start with definition 1 implies definition 2. So definition 1 is a surface built out of polygons. And really this is fairly easy. We're most of the way there. If I take a surface built out of polygons and I take sigma to be the vertices, then it's evident that away from sigma, I have coordinates coming from just the coordinate in the plane. OK, so the coordinates charts come c. Come c, which is where the polygons live. OK, so I have this atlas of charts, and really it remains to see that what happens at the vertices. So the main point is why are the cone angles 2 pi times the natural numbers? So I gave you this construction. You only get cone angles here that are integer multiples of 2 pi. The next speaker, Rafe, will take great exception because he's going to let the cone angles be anything he wants. So it really is a question. We have a flat metric. It's singular. There are some cone points. But why are the cone points integral multiples of 2 pi? And the answer is true because the surface has what's called trivial linear halonomy, which I'll explain now. The surface has trivial linear halonomy. OK, so what does that mean? This is the largest eraser I've ever seen. It's pretty awesome. So what that means is you take some sort of loop on your surface, and then you take a frame, so an orthonormal frame, and you parallel translate it. OK, and when you come back, the frame is rotated. So in general, the halonomy is a map from pi 1 of the surface to o 2, the orthogonal group. And it just says, how does the frame come back? So here I'm saying the frame comes back trivially. And the reason it comes back trivially is at every point of the surface, there's a well-defined choice of north and a well-defined choice of east, because that's defined on the polygons and I'm only gluing the polygons via translations. Every point of x minus sigma, et cetera, because these are defined in the plane and they're preserved under translations. So now let's think about the consequences of this for cone angles. So the fact is that cone angles not in 2 pi z have non-trivial, just going around the cone angle. You don't even need to do anything global. So what's an illustration of this? So this is the Pac-Man figure. OK, so here's a cone angle that's not an integer multiple of pi. Where I glue this via a rotation. So now I can sort of do the same thing. I'll take a loop and I'll sort of, maybe I won't drill the whole frame. I'll just draw a choice of north. So I'll take this vector and I'll parallel translate it till I get here. And then this is glued via rotation. So it comes out over here and I get non-trivial halonymy. So that's a sketch of the proof. If you had cone angles that weren't integer multiples of pi, you wouldn't sort of have a consistent choice of north and east on the surface, but you do. So that's just one direction for the proposition. Now we should talk about definition 2 implies definition 1. So just a bit of terminology, a straight line segment joining two points of sigma is called a saddle connection. And the two points are allowed to be the same. So this is just a straight line segment. The terminology comes from dynamics. Maybe I'll explain more about that in a minute. Anyways, to see that definition 2 implies definition 1, you should think that given a definition 2 surface, you cut it along a maximal set of disjoint saddle connections. Okay, so first of all, it's not entirely obvious there is a saddle connection. But it's not actually so hard to prove. You can pick an isotopey class of paths from one point of sigma to any other point of sigma and then tighten it to a geodesic. And that geodesic will be built out of saddle connections. Okay, so you find one and then you cut it and then you find another and then you cut it and you keep doing this until you can't find anymore. In the flat metric, the only metric we have right now. So I keep finding these saddle connections and I cut them until I can't anymore. And it turns out the only way you can fail to find another saddle connection is if your surface with boundary is a triangle. Because if you have a triangle, any path joining two of the cone points is just, you know, you can tighten it to a saddle connection but it's just one of the boundary ones. Okay, so this gives a collection of triangles and these triangles can be embedded in C. Okay, so you cut, you get the triangles, you put them in C, you keep the edge identifications. Now you've expressed your surface as a collection of polygons with edge identifications. Sorry? Yeah, because of the trivial linear halonomy. Yeah. You embed them so that the north on the surface is the north in the plane and the east on the surface is the east in the plane. And then when you identify two edges, if that's going to match up, it has to be via translation. Yeah, good question. Okay. So now, another question? So now we'll have our last definition. Definition three. A translation surface is a pair omega where x is a compact Riemann surface and omega is a non-zero Abelian differential. Okay, so an Abelian differential, by the way, it's a complex valued one form. Okay, so it's just a differential one form such that when you write it in a complex coordinate, when you write it in a complex coordinate z, it looks like a holomorphic function times dz. Okay? So in other words, it's a global section of the canonical bundle if you like that language. Okay, so now I want to sketch to you why the third definition is the same as the second definition. And then we'll be able to move on from there. Before I do that, I want to give a lemma that says that actually you can always take this function f to be of a particularly nice form. And the idea is you can sort of integrate this to get it especially nice coordinate where f is, you know, typically f is just the function one. So proposition, for any Abelian differential, any Abelian differential on a Riemann surface, every point of x, there is a local coordinate z. So z equals 0 corresponds to that point on the Riemann surface so that in that coordinate, omega is z to the k dz for some k greater than 8 equals to 0. Okay, so this k is the order of vanishing of the Abelian differential. Okay? And there's only finally many points where the Abelian differential vanishes, so at most points, omega equals dz in this coordinate. And this is purely local, as I said, you get this coordinate by integrating. So by the way, if you want to see some of these details, these lectures are loosely based off of a survey that I have available on my web page, which is called translation surfaces and their orbit closures, a survey for a broad audience. So I do lemmas like this in significantly more detail in the survey. So moreover, k equals 0, the coordinate, the local coordinate, is unique translation. And that's really just the fact that if two functions have the same derivative, then they differ by a constant. Okay, so using this, we can now show that the proposition definitions 2 and 3 are equivalent. So let's start with definition 3 implies definition 2, sigma to be the set of 0s of the holomorphic one form. The local coordinates z in which omega equals dz give the charts on x minus sigma. You take every local coordinate where omega equals dz, and it's unique up to translation, so if you take two different local coordinates, they just differ by a translation. So now we just need to discuss what the metric looks like at one of the 0s. In definition 2, we should have this atlas of charts and we should understand something about what's going on at the 0s. In particular, we want to know that the flat structure you get from z to the k dz when k is bigger than 1 looks like all these half planes glued together. But it turns out that's actually pretty easy to show. So the differential e to the k dz is proportional to the pullback of dz under this cyclic cover goes to z to the k plus 1. Okay, and that's basically the whole proof. This half plane construction is just a way of visualizing the cyclic cover from c to c branched at the origin. So in the other direction, definition 2 implies definition 3. I'll just say set omega equals dz on x minus. The charts are well-defined up to translation, so dz is well-defined away from sigma, and then you have to check what happens at sigma. And I'm going to skip that, but I think you're probably getting the idea now. And it's, as I said, it's in my survey if you want more details. Yep? No, they're different. So the question was what about is x omega the same as x lambda omega, and the answer is they're not. Okay, so somehow all of these different definitions are one of the things that makes the subject interesting. The definitions that are more elementary connect to various problems in dynamics that I'll talk about later. And of course, this final definition connects to, you know, Riemann surfaces and Teichmeler theory and the modularized base of Riemann surfaces. Yeah, yeah, that's a really good point. So you could in particular just think about e to the i theta. And if you rotate, it's probably not the same translation surface. Okay, so let's go back to examples. So we started with this example. And let's say this is a one by one square. Well, this is one of the few examples where you can write down in a very easy way the Riemann surface and the differential. So the Riemann surface is c mod z a joint i. And there's a differential dz on c, which is invariant under translation. So that descends to the differential on the Riemann surface. Oh, and I should say one of, you think of this differential as giving the flat metric. Okay, so like if you want the length of a tangent vector in the flat metric, you should feed it to the appealing differential and see what the absolute value is. So you're probably a little bit more familiar with giving a flat metric in terms of a two tensor. But you could take the square of this if you want, and then you have a two tensor instead of a one tensor. You do get the same metric. Yeah, but a translation surface is more than a surface with a flat metric. And it's even more than a surface with a flat metric with special cone angles and trivial halonymy. Because you have this atlas of charts that's well defined up to translation. Or equivalently, sort of a fourth definition, I'm going to stop writing definitions, is a translation surface is a cone metric with trivial halonymy and a choice of north that's defined everywhere. North, that's sort of a variant under parallel translation. So that's a very closely related thing. Sometimes people call that a half translation surface. Yeah, if you had a section of the square of the canonical bundle, a so-called quadratic differential, then it has a double cover, which is given by an appealing differential. And then the main difference in terms of the sort of elementary definitions is that then you'd be allowed to glue with rotations by 180 degrees, as well as translations. Other questions? OK, so another very similar example to this. A whole class of examples are called square-tiled surfaces. So an example is something like this. And I'm just going to glue opposite, sort of perpendicularly opposite edges. So you just make a surface by a bunch of unit squares. Or a bunch of, they don't have to be unit, but a bunch of squares of some size. And then you see very easily this covers the square torus, which, as we said, was c mod z adjoin i. And so this means we have this map f, and here the appealing differential is the pullback of dz on the torus. So I want to think about this squared-tiled surface in particular for a moment. When you do the game of sort of identifying the edges and seeing which vertices are identified to which, you'll find that there are actually two cone points. So this particular example has two cone points, four pi each. Again, you compute the angle just by sort of adding up the contributions. So it's very easy to compute the angle. And you can check that it's still, this guy is also genus too. And again, you do that vertices minus edges plus faces to compute the Euler characteristic. So this guy's a little bit different than our earlier genus two guy, because our earlier genus two guy had one cone point, and the cone angle there was six pi. So we know a surface of higher genus can't literally have a flat metric because of gazpone. Gazpone says the integral of curvature is given by the Euler characteristic. So actually a version of that formula is still true here, but the curvature you should think of now as a point mass at the singularity. And the rule is each extra two pi angle counts as one of negative curvature. So each extra two pi. So here I have two points, each with one extra two pi. Because four pi is two pi, which is the amount you're supposed to have, plus two pi extra. So this is Euler characteristic minus two. And before I had one point with six pi cone angle, which is two extra two pi's, and that was also genus two. So let me just write this down. So gazpone says that there are two g minus two extra two pi's of angle. And another perspective on that from Riemann surfaces or algebraic curves is that omega has two g minus two zeroes, counted with multiplicity. A zero of order k, as we saw, gives an extra k, two pi's of angle. So they both give the same answer, which is good. Excellent. Okay, so that was a few examples. Now I want to give you a bunch more. I want to give you all translations or vices in genus two. So I'll give you a construction which is sufficiently flexible to give you all surfaces in genus two. And it's a very intuitive thing coming from the first time you study topology, and you learn that the way to get a higher genus surface is by looking at a connect sum of a bunch of tori. So there's a version of connect sum that works to give you translation surfaces, and it's called the slit torus construction. So I'm going to start with two tori, but as translation surfaces. So they don't have to be the square torus anymore, but they're just given by any parallelogram with opposite sides identified. So there's one, and then there's another. So I have two tori. And now as the name suggests, I'm going to make a cut in both of them. And the cut is going to be in the same direction, and it's going to be the same length. And actually now the way it's helpful to draw the cut is sort of like this. If you cut a piece of paper, then bend it a little bit. You see it as two sides. And then I'm going to glue them together. So I'm going to glue this edge over here, and this edge over here. And when you do this, you get a translation surface. And it has two singular points, one for each edge, each end point of the slit. And the cone angles, well, there's two pi, and there's two pi. So it has two cone points for a pi angle each. And so this is enough to get you, so this gives, let me just write this, gives all surfaces with, maybe I'll just call it four pi singularities, or cone angles. So two cone angles of four pi each. So by the way, just to give a proof of this, how you would prove this. You triangulate the surface, and then you look for a triangle that has vertices at both singularities. And then you use the fact that every genus two surface has a hyperliptic involution. And you show that, so you take, so you show you can find an edge of the triangulation joining the two singularities that's not fixed by the involution. And you take that, and then you take its image under the involution, and those are the two sides of your slit. And then you cut it and show the pieces are so simple they have to be tori. So that's an argument that, frankly, you don't need to know, but it's nice. You can try to reconstruct it later as an exercise, or you can look it up in my survey, or you can just forget about it. But that actually doesn't give you all translations of vertices in genus two, because, for example, it doesn't give you the octagon with opposite sides identified where you have a cone angle of six pi. So this has a variant where, again, you take a torus and you cut it to make a slit. But now, instead of gluing on a torus, you glue on a cylinder. So the way you're supposed to think about this, you glue those two edges, so now you have a cylinder. And then you have a cylinder and you sort of bend it around, and then each circle gets glued onto one side of the slit. And this, again, gives you a genus two surface, but this time with a six pi cone angle. And this will get you all genus two surfaces with a six pi cone. Yeah. Yeah. So what Scott's saying is that this locus on the surface looks like a figure eight because you have two circles, and then when you sort of glue them in, they sort of have a point identified. So it's sort of a little bit looks like, although, as I said, you shouldn't actually try to visualize things like this, but it looks something like that. There you can see the slit torus, but it's like you had the slit, you glued a pair of points together on the slit, and then you glued in this handle. Other questions? This chalk is fighting me a little bit. OK. So another example, or another class of examples, is given by what's called unfolding polygons. And in a later lecture, we're going to talk about how problems of billiards in polygons ends up relating to dynamics on modular spaces, and this construction will feature very prominently. Today I'm going to give this a sort of, not especially well-motivated, but relatively easy construction, just a way of building surfaces. OK. So you start with a rational polygon. So that means rational means angles are in pi times q. So all of the angles are rational multiples of pi. And then we're going to do what's called unfolding it, which really you have to imagine that you have a surface and it's been folded up like origami onto this, and then you're going to sort of unfold the different copies of the polygon until you get a surface. So I'm going to start this with the square, and to make things easier, I'm going to draw an F on the square, which isn't really part of the construction, it's just to help you keep track. And the whole point is I'm going to do this with examples first, and then I'm going to give you the definition. But everybody learns it through examples. So the first thing I'm going to do is I'm going to reflect this polygon about this edge. And I'm going to get a second copy that looks like that. Now I'm going to reflect it about this edge. I'm going to get another copy like that, and then I'm going to reflect it about this edge. OK. So this may seem mysterious, but now I'm going to declare that I don't need to do any more reflections. And the reason why, so I could reflect this about this edge, and I would get this. But I already have one that just looks like that. That looks just like that. It's here. So the way the construction works, instead of doing that, I'm just going to identify this edges. And I'm going to say, look, I have reflected this. It is over here. And similarly, if I reflected this, I'd get a copy of what's down here. And I just got all edges identified. So this isn't the usual way of drawing the surface. If you reflect this, you get this thing. So this is identified there. Usually you would cut this square and put it right there, which would reveal that you have the torus with opposite edges identified. So you unfold the square. You get a torus of four times the area. OK. So now let's do another example that's a little bit more complicated. So I'm going to cheat and draw the unfolding to start. Because otherwise, my ability to draw polygons is actually very limited. So I'm going to unfold the triangle, the right angle triangle, whose smallest angle is pi over 8. And I'm going to start by reflecting it about this. And I'll get another copy. And then I'll reflect this one about this. And you could see where this is going really quickly. I get the octagon. And again, I could say, oh, I'm going to reflect this one in this direction. But I actually already have it. It's over here. So I'm not going to do that. And I'm just going to identify opposite edges of the polygon. So in general, you do exactly the same thing. You may sometimes reflect and find you have an overlapping copy. And then you can just detach it and put it somewhere else. You're building sort of an abstract surface, after all. But there is sort of a pleasing way of phrasing the general definition that lets you see what's going on. So in general, you can let G be the subgroup of S of O2, 2 by 2 orthogonal matrices, be generated by reflections in the edges of the polygon P. And when I say reflection, I mean like the derivative of the reflection, the linear part, to get something in O2 instead of in the affine group. And then the unfolding of P is equal to, well, you take a disjoint union of copies of your polygon with the elements of G applied. So you sort of translate them so they're disjoint in the plane. And then you identify edges when they differ by a reflection in that edge, up to translation. Okay. So by the way, who knows? What would go wrong if I allowed the angles to not be rational? Yeah, it would be infinite. That's exactly it. Okay? So this definition is perfectly fine, but this group would be infinite. And in fact, you can even, if you're bored, compute the size of this group in terms of like the least common multiple or the greatest divisor of the angles. It doesn't matter because it's all translations. Really, I mean like sort of all of this is up to translations. Yeah, so if you want, it's any other questions? So these are especially important in the theory in that they're related to the applications. They're also sort of pleasing from other points of view. So I'm just going to write this proposition just so you have a flavor of this aspect of the subject. So the triangle with angles p over n pi q over n pi over n pi unfolds to the differential form y dx over x times x minus 1 on y to the n equals x to the p times x minus 1 to the q. Okay, so this defines an algebraic curve, and really what I mean is you just take the normalization of that algebraic curve to get a smooth Riemann surface, and this defines a holomorphic one form on that smooth Riemann surface. Okay, so this, you know, yet another way of presenting a Riemann surface is with algebraic equations, and sometimes you can do that with the differential forms as well. So this is a very special Riemann surface. This is a cyclic cover of p1. The deck group is generated by multiplying y by a root of unity. And those deck groups, by the way, are sort of the action of g, or rather the action of g intersect sO2 on these polygons. So somehow unfoldings are very symmetric, and it plays an important role in the subject, or it's a difficulty in the subject, that the unfoldings are not at all generic. They're very special. Most surfaces don't come from unfolding polygons because they don't have such a big automorphism group. Okay, I think I'm happy to stop here, and the next time I'll talk about the moduli space and the gl to our action, and it'll really get started.