 So I put up on the board over there a reference. I think, essentially, everything I covered in the first part is also covered in a nice survey paper by Ettinges, called Groups Acting on the Circle. It's a lovely little expository paper. It starts very gently and ends up very complicatedly. But in particular, it starts very gently. In L'Encinema Mathematique, and there are lots of beautiful mathematics contained in there. So I highly recommend. This second half is going to be somewhat disjoint. I will use some of what I talked about before, mostly the definition and properties of rotation number. But if you got lost, you can sort of start over right now. And because for the first half, I was talking about general properties. What can we prove about theorems generally about groups that act on the circle, or about particular homeomorphisms? Here I want to prove not a theorem about all the groups that act, and what do they have, but about a particular action of a particular group. So this is the advertised theorem about surface group automorphism acting on their boundary. And let me start by just describing this. So this is where we are. I want to just draw a different picture of what this action looks like. So some of you might know that if I take my surface and forget that it's fundamental group for the moment, I want to look at the mapping class group of the surface. But I also want to sort of see what happens to the fundamental group. So I'm going to fix a base point. I want things not just up to conjugation, but I want to fix a base point. So this star will be with a marked point. There's a map from mapping classes of homeomorphisms up to Ipsotopy to automorphisms of the fundamental group, where my fundamental group is now with this base point. This is, in fact, an isomorphism onto an index 2 subgroup. And the only reason it's index 2 is we look at mapping classes that preserve orientation. So this is an isomorphism, what do you do to the fundamental group, onto an index 2 subgroup, which I'll write as a plus. Sitting inside of here, I have inner automorphisms. That's isomorphic to the surface group itself, and that inclusion has quotient out. What is this surface group sitting inside the mapping class group? This is what's often called the point pushing subgroup. What do I mean by point push? Here's a piece of my surface, and I have a marked point. And one thing I could do is I could just take my finger. Here's something that's isotopically trivial if you didn't require me to fix that point at all times. I could just take my finger and go like push this around a little loop and put it back where I started. Or in other words, in this annulus, I'm going to do something that looks like a sort of a dain twist here and then a dain twist on the other side in the opposite direction. That will move this transverse curve before to the picture after where it goes like this. And that's my marked point. So I know maybe this was a closed curve you drew. So that was pushing around the curve gamma in pi 1, where gamma was this thing that I pushed my marked point around. It's a based curve. Yeah? Can you go as an example with an isotopically simple? Yep. I mean, it's harder to draw a picture of what happens. But I can tell you what the thing is, you do an isotope of your surface by pretending your surface is made of a very stretchy material and dragging your point along until it crosses over. So here, how would I point, push around? Or I mean, if you really believe that this is a group homomorphism, you can just compose everything out of simple closed curves, concatenated, and you just compose. But yeah, this is not what you want. So let's see if I, what would it look like to point, push around? I mean, so the point is that if you let's do a curve that maybe goes like this and then crosses itself and goes down there. I'll try and point, push around that. OK, so in my color scheme from before, I'm supposed to sort of point, push around like this in this neighborhood. OK, so I have my, now my base points over here. It's a new picture. And I had a transverse thing like this. OK. And oh, sorry, that's, if I'm going to go this way, I should be on this side of it. Sorry, I'll be, yeah, I'll be pushing it along. So I'm going to follow this loop, which will make this curve look like that. OK. And then I'm going to hit here, and I'll see this other piece that I was crossing. Right, so this will now get sort of moved along for the ride. I'll have to have nudged this over. And then maybe I'm crossing this image of a white thing again, and it will continue. OK, and then this whole tube of things will end up, this will get nudged up, and it'll end up sort of capping off over here with my point. OK, that will be the image of a transverse thing if I loop over myself. I'll do another picture, I'll do another picture slowly. We'll actually mostly care about the picture of what happens on simple closed curves. But the point is that if you do this point push, OK, so there's that white curve. And then I do another one, say, in this curve, in this direction. What will it do? It'll now pick up these two little white strands and move them around, both of them at the same time, and come back. And you'll see a doubled up picture of this one. So that's what I was trying to draw here. If you cross yourself, you see the two strands moved again. OK, so this is a famous inclusion here. And there's another easy to define map where you just forget that this point existed. OK, so any homeomorphism up to isotope of a surface where isotopies and the homeomorphism have to fix that point defines a homeomorphism up to isotope not fixing the point called forget that you had to fix the point. So that's a mapping class group. And this sequence is called the Berman exact sequence studied by Joan Berman. And the Dey-Nielsen-Barr theorem is what's telling you that these are isomorphisms. So this whole story is just to tell you that if you like mapping classes better than automorphisms of groups, there's a way to see this. And there's a way to see this boundary action on d plus pi 1 sigma g. And that says follows I could fix a hyperbolic structure on my surface. So now it's universal cover is the hyperbolic plane. And I can take the, say, the Poincare disk model for the hyperbolic plane as a universal cover. And I'm going to fix also I have my base point for pi 1, my marked point on the surface. I'm going to fix a particular lift of that. That star maybe I'll call this guy lift of star. If I have a homeomorphism of this surface, there is a unique lift of it to a homeomorphism of the universal cover that fixes if my homeomorphism fixes this point, there is a unique lift of it that fixes my chosen lift of the point. So if f of star equals star, there exists a unique lift to a homeomorphism of the hyperbolic plane fixing the lift of star. And this homeomorphism of the hyperbolic plane induces a homeomorphism of the boundary. And that map on the boundary depends only on this guy up to isotope fixing the point. So only on what f represents in the mapping class group. If I took some other thing that was isotopic to this, I could just lift that isotope. My isotope, because it came from downstairs on a compact surface, it's only going to move points of most distance 25 or something. It moves points of bounded distance. So up here, lifting the isotope, well, it'll move points around, but each point will move distance at most d, whatever the bound was. In this model, the metric makes distance d look smaller and smaller and smaller as you go to the boundary. So I will have moved no points on the boundary by performing this isotope. So that's why it depends only on the isotope class of this map. So what does this say? This gives you a way to lift mapping classes to homeomorphisms of the boundary of the hyperbolic plane. So this gives a map from the mapping class group your surface with a marked point to homeomorphisms. Here, if I preserve orientation, I'll preserve the orientation of the boundary of the hyperbolic plane, which is the circle. And under these usual identification, this is the same thing. This is the boundary of your group. If you want to think of your group as the translates of this point under deck transformations, it's a good way to sort of see that these two are the same action. So you may, as you prefer, think of this in either way. You want to put them both out there. This also sort of gives you a nice way to see what's happening with the push subgroup of pi 1. So you can see this directly from the group perspective. Or if you see what happens if I push this point around. OK, well, one way to do that is I could lift this push. And if I lift the thing supported in this annulus, it'll lift supporting on a bunch of strips. And it'll move this base point over to the next one. OK, well, that wasn't very good. That broke my rule. I was supposed to choose a lift that fixed the lift of my base point. OK, so that was the wrong one. I should now post-compose with a deck transformation that takes this back to there. What's that deck transformation? It's exactly, you know, this was the axis of that curve. That deck transformation is exactly, though its boundary action is the boundary action of that isometry coming from this deck transformation. So that's gamma. And here is its lift. The boundary action of the push subgroup is just the usual action by deck transformations. So I will record that for posterity here. OK, so you can think of that as the PSL, if you want. If you put your hyperbolic structure on your surface, you can think of that as coming from the usual subgroup, discrete subgroup of PSL to our defining your surface subgroup acting on the boundary of hyperbolic space. So here gamma is acting by translating along this axis. So on the boundary, it's got two fixed points and it's moving points. So in other words, the automorphism of the mapping class group, it's a much bigger group in which I see this surface subgroup acting on the boundary of hyperbolic space. And I want to talk about a question and its answer. So here is a question that was in a problems list called Problems on Mapping Class Groups written by Benson and Farb in I think 2006. So he asked, here is a group and a very natural sort of coming from geometry action of this group on a space. It's the boundary of the group. Is this the only way this group acts on its space? So here's the question. Suppose you have an embedding, a faithful action of this group on the circle. So I'll call this homomorphism, if you will, from this group to the group of homeomorphisms of the circle. Now, yeah, if you wanted this to be injective, so a faithful action, the question is, is this the only one there is? So is rho necessarily conjugate to this thing I described here? Let's call it the standard boundary action. So that is the statement of the question. Maybe this group has a remarkable property that it only acts on the circle in one way. So this statement needs an edit that I think is just a typo. You can do something silly to any group acting on the circle. And this is not just conjugacy, but something called semi-conjugacy. Jair Minsky's picture of replacing an irrational rotation with this thing where you thicken some points and whatever and you get something that's no longer an irrational rotation, there's an invariant canter set, is an example of a semi-conjugacy. In general, you can imagine a picture where if you have a countable group acting on the circle and you look at the orbit of a point and just enumerate this orbit, OK? I don't know, x1, x2, x3, whatever. You replace point xi in your circle with a little interval of length, I know, 1 over 2 to the i. That'll give you a bigger circle, diameter 1 bigger, OK? And it tells you how your group is still supposed to act. You permute these intervals, just as you were moving the point. Move 1 to the x by the unique affine map. That'll give you an action that's not necessarily conjugate to the original one. If the original one had dense orbits, now this one is in case 3 where there's no longer all orbits dense, I have this interval that goes along forming the complement of something that will look like a canter set. So there's a trick that you could do called semi-conjugating that I don't want to worry about too much, but it's how you move between the canter set picture and the non-canter set picture. You can go back and forth. But if you don't like that picture, you can sort of think conjugate without moving too much of the spirit. And the reason I want to gloss over that part is I want to spend this lecture telling you the answer to the question, because it's a good way to showcase some different tools. The answer is a theorem that I proved with Maxime Wolff last year. Yes, and in fact, even better than this. Any action of this group on the circle is semi-conjugate, the best you can hope for, to the picture we just saw, the boundary action. It looks as best as you can like the group acting on its boundary, or is trivial, where trivial means literally trivial if genus is at least 3. And there is another silly way you could be trivial in genus 2, when g equals 2, the abelianization of this group is a finite cyclic group. It's a Z mod 10 Z. It happens to be, I think that's a theorem of Bermann. So you could map to the abelianization, and then you're like a cyclic group of order 10. You could act by a rotation. So for g equals 2 can factor a non-trivial to show that the abelianization is that. So yeah, so this theorem says that it's either injective or it's badly, badly non-injective. It's trivial, or factors through a finite group. And I want to just spend this whole time sort of outlining all the steps in the proof, because it shows you how to use a bunch of the basic techniques that we used before. This is one of those things where you prove a theorem and you find a very complicated way to do it. And you're like, yes, we answered a hard question with a hard proof. And then you're like, well, that step could be easier. And you sort of clean it up. And you're like, oh, that only is a very basic fact. And then you're like, well, what about step two? Oh, actually, that wasn't so bad either. And at the end, it's something that is, I think, appropriate for this mini-course and a good way to teach people about how to understand groups that act on the circle. OK, so let's prove it. I mean, the point is you don't need them to know a lot to get started. What do we have? I have a mystery action of a group on the circle. And I want to show that it's not so mysterious. It's either extremely boring, or it's one that we can understand as much as we want. It's group acting on its boundary. So that's my setup. Suppose, and I'm going to assume the genus is at least three, so I don't have to worry about two different cases. I'll point my finger at the one place at the very end where we're going to actually use that assumption. So I'll assume genus at least three, but I don't need to use that until the end. And I have some mystery action of this automorphism group on the circle. And the goal of the first two parts here, the first part, is OK, well, this group is big and complicated, but there's something I know that sits inside it. There's that inner automorphism group. There's the surface subgroups. And I know very well what I'm aiming to show that action looks like. I want to prove this theorem. I want to prove that it's either the usual deck transformation action, it's looking like that, or it's totally trivial. So let's study the surface subgroup first. So the restriction of rho to my surface group, that's the inner automorphism subgroup. And what I need to prove some facts about this that show it looks either like it's trivial or it's like the usual one. Well, one property shared both by the completely trivial everyone's the identity action and the usual boundary action is that these push or deck transformation, they all have fixed points. In PSL2R, they're all hyperbolic transformations. And in the trivial action, they're trivial. So our first lemma is to say that, hey, that's true even under the mystery one. And how do I say something has a fixed point? Well, it's a fact that you have fixed point if and only if the rotation number is 0. So that rotation number we defined before picks out whether points move at all or not. So lemma 1 is that if, let's do it for nice curves, if gamma in pi 1 sigma g represents non-separating. So I don't want it to cut my surface in half. Simple closed curve. Gamma's supposed to act on the circle in some way. That's some homeomorphism of the circle. Well, lemma says its rotation number is 0. That's certainly true for things that do nothing and for hyperbolic, for the usual boundary action. OK, prove the lemma by looking at what happens in the mapping class group. So this is where I need to use this picture. In the mapping class group, gamma, my push around gamma. Oh, good, this was the nice picture where it's a non-separating simple closed curve. OK, I'll draw another version of it here. So there's my gamma. And the thing that's supported on this annulus and takes this white curve here, the one that just goes around, the one I drew, is equal to the composition of two drain twists, a usual drain twist in this curve on this side of my mark point, and then a drain twist in the opposite direction in this curve. A usual drain twist in this curve takes this arc to something that looks like up, right? And then one going the opposite way will take the other half of this arc to one that looks like the mirror image of this, so like this around and then here. And that's exactly the picture I had drawn there. OK, so the point is that I can write gamma in the mapping class group, which is the automorphism group of my surface group, as the composition of two drain twists, let's call it twist one, and twist two, where the direction of twist two is the opposite. OK, so I'll pretend twist two is going the usual way, but I'll write it with an inverse. And in the mapping class group, these drain twists around simple non-separating simple closed curves are conjugate. Where T1 conjugate to tau two. OK, so now here's where I just want to use facts about rotation numbers. So these are conjugate. Oh, and there's, I guess, another fact I want to use. These are supported on disjoint little annuali, right? My two blue curves didn't cross. So from what we know about mapping class groups is that T1 and tau two commute. OK, so now let's try and study what this looks like for rotation numbers. So that means rho of gamma is rho of T1, T2 inverse, which is rho of T1, rho of T2 is a group action composed with rho of T2 inverse. Great. And so the rotation number of this circle homeomorphism is equal to the rotation number of these because they're the same thing. Oh, but they commute. And rotation number is a homeomorphism on a billion subgroups exercise. So this is the sum of these two. Well, I don't know why I switched to a color. And this is a group action. So this is a homeomorphism or representation so I can put the inverse out there. Oh, and homogeneity of rotation number means the rotation number of the inverse of something is the minus of its rotation number. So I can take this minus and move it over here. So that was homogeneity. And now what else do I have? Oh, I have conjugacy invariance that was also on my list of properties that follows easily from the definition. So this number and this number are equal because these two elements are conjugate. They're conjugated groups, so in particular, they're conjugated as homeomorphisms of the circle. So this is 0. They just completely cancel by conjugacy and variance. And that's the proof of the level. So just by using things I know about rotation number and mapping class groups, we already have a good step along the way to this looks like how it should. This argument doesn't work because your Dane twist won't be conjugate. And I don't need to know that yet. So I mean, the answer is yes, because the theorem is true. But I don't have a three-line argument that does it. OK, great. Next, Lama. So I'm sneaking this in because I wanted to define a very important tool in studying groups acting on the circle. And this is the bounded Euler class. And I'm going to give you a very hands-on definition of what this is. So let me state step number 2, or Lama 2, which is another statement that's going to say in a fancy way that this has some property either in common with the trivial action or the usual boundary one. So I'll use a word we haven't defined yet. The Euler number, this action of a surface group, is either 0. This is what the trivial action would give you, or up to sign the Euler characteristic of your surface, 2g minus 2. This is what the usual boundary action would have. And now I'll tell you what this Euler number means. So postpone proof for a definition. And I'm going to give you not the usual definition. Even if you've heard this word before, you may have not heard it. But what I particularly like, and actually I learned it out of this way of thinking of it, out of a paper of Anna Wienhardt on the higher-tech Euler theory of all things, where they study actions of surface groups. So I'm sure other people have shared this perspective. But that's where I happen to learn it. OK. So this definition is for an arbitrary action of a surface group on the circle. Suppose I forget I was this row from before. I just sum up row action of a surface group on the circle. All right. I'm going to take a pants decomposition of my surface. Just pick anyone you want for now. And I'm going to say, what does each pant contribute? And then I'm going to sum them all up. So for a pant, here's my pants. So there's my pair of pants. Its boundary curves represent elements of the fundamental group. And if my base point was in here going around this, shoot, I should orient this this way. So it's the induced orientation from my surface from the pants. So this is like a curve that goes like this and comes back. And this boundary is like a curve that goes like this and comes back. Let's call this guy A, this guy B. And then my induced orientation here is this way, right? Yes. And so this last boundary component, I'm just trying to write its boundary as elements of the fundamental group if my base point were here. This boundary component is now going to be what I should do, B inverse followed by A inverse. I'll write it like function composition. OK. So those are its boundary components. And the Euler number for the pant is going to measure how badly rotation number fails to be a homomorphism on this pant. OK. Sorry, I don't understand your picture. So you're right here. Yes. So I have my pants on the surface. And I just want to, here's my pants. I just want to say, forget I drew a base point. I want to say, what are the boundaries of my pant in the fundamental group? OK, so if this is A and that's B, that's A inverse times B inverse. My computation was I had to draw a base point and be like, all right, going around this loop is like I go over here, and then I go around, and then I go around the back, and then I go back. I'm slow, and I always get orientation wrong. So I was just drawing it very carefully. But the information to record is that this is what I've written down. OK. So I'll steal some space over here. Actually, I guess I can go over to this board, lots of space. I want to measure how much rotation number is not a homomorphism on the pant. So the Euler number, the pant part, P of rho, I define to be the rotation number of the action of A plus the rotation number of the action of B minus, sorry, I'm just going to sum over all the boundaries, plus the rotation number of rho B composed with rho. So I want literally that boundary on A inverse, rho B inverse. And if you like, if it's nicer to you to understand it this way, this is the inverse of B composed with A. So this you could think of as minus rotation number of rho B composed with rho A, if you liked. What is this independent of the choice of pants? Well, I haven't even told you that you have to sum them up yet. But yes, OK, that's one of your exercises, is to show by elementary pants moves that my definition is going to be independent of choice of pants. All right, but there's another problem I have to fix here. These are all things mod Z, and I actually want to get a real number. So I'm going to just lift these like we did to the line and take their lifted rotation numbers. And so then I'll take the lift of this. I'll take its inverse. I'll take the lift I chose. I'll take the inverse of that. And I'll look at this composition. And here it's easy to check on this level that the number you get here is independent of choice of lifts. If I move this by an integer translation, I change its inverse by an integer translation in the opposite direction. So I would add one and I'd minus one. So this thing I've got is well-defined. And this, up to mod Z, is measuring the failure of this guy to be a homomorphism. So this is no longer part of the definition. OK, and so then here's the part that you're like, why is this not depending on pants? I define the Euler number of rho to be sum over all P in my pants decomposition of the contribution from the pant. And exercise doable by changing pants one at a time is this is independent of the pants decomposition. And if you don't like that, there's another exercise which has proved that this is the same as this other number. And the other number, it has no reference to pants whatsoever. It was something you could just sort of produce out of nowhere. OK, but here is a very concrete thing that I like this version because it's very concrete. You can choose whichever pants decomposition you like. And you're just looking at rotation numbers of some nice simple closed curves. And a fact that comes out of the quasi-morphism property I said, OK, so this was actually basically on our list of properties before, so that this number is bounded in absolute value by one. This is a property I called before, called quasi-morphism. And if you prevent me a second of digression, this actually, this statement, what I called fact, is really part of a famous theorem in mathematics. This is the Milner-Wood inequality. Hinge is a statement about Euler numbers or it's really a statement about Euler classes of flat bundles. And it boils down to this observation once you frame everything in the right way. That is just to say that there's some kind of interesting math hidden in here. OK, so let's see what I can extract. Let's go back. So I gave you all the definitions. It was just a statement about rotation numbers that added up. And the claim is that this is either zero or this Euler characteristic looking thing. So let's prove the lemma. And at least to sketch it. So I can choose whatever pants decomposition I want after you do your homework. And I'm going to choose one that looks like this, like a really nice looking one. This is my surface. I'll draw it in kind of a round way. And I'll put pants curves like this. So then all my pants look the same. There's like a rotation or like a reflection that moves one to the other. In fact, you can prove that even in the based mapping class group, there is something that takes the triple of curves bounding one of these pants to any other one of these pants triples. So my mapping class group moves these around. And you see I also cooked it up so that all of the boundaries are non-separating simple closed curves. So we're going to pick this pants decomposition. Pick these peas. Now what do I know if I try and understand the Euler number of one of these pants? OK, it's a sum of rotation numbers. I knew something about rotation numbers. That was from lemma 1. If I have non-separating simple closed curves, then they get the number 0. So their lifts are, I don't know, they're some integers. Everything is just like 0 mod z. All right, so that's great. This is something that's 0 mod z. So that's 0 mod z by lemma 1. And our fact, or what I said with some famous theorem called the Milnerwood inequality, says that this number that's 0 mod z, there's only three options. It's 1 or minus 1 or 0. So it's actually equal to 1 or plus or minus 1 or 0 by the clausomorphism fact. So that's something that's true for individual pants. And there's a little work to do, but the fact that I chose these all so that there was some element of some homeomorphism of your surface that I can think of as a mapping class or as an automorphism of pi 1 that took any one of these pants to any one of the other ones, like if I want to go through there, I should just rotate the whole thing around like this. Tells me that these are the same. I get the same number on each pair of pants. Conjugating by that homeomorphism tells me that I should conjugate rotation numbers and they don't change. So every pant here can be taken to every other by an automorphism or a mapping class. And rotation number is conjugacy invariant. So you get the same value. All right, so I'm just summing up the same number how many times, how many pants in a pants decomposition? 2g minus 2 many. So I get 0 times 2g minus 2 or plus minus 1 times 2g minus 2 for the total. These boards a little bit. Great. OK, so we sort of accomplished our first goal here, which was studying what happens with the surface group. That was still a setup for my proof. OK, so the reason this Euler number, I just told you some recipe that you don't even believe me is well-defined because it depends on the pants composition, except I swore to you that you didn't. This number that takes an action of a group on the circle, the surface group, mind you, and spits out a number, is an amazingly rich invariant. And here's one theorem that I'm going to quote that tells you how much it just knows. OK. So if we're in the case where the surface group, the Euler number of your action of the surface group, the one I was thinking of, one we had, but this is true for any action of a surface group, if it agrees with the Euler characteristic of your surface, it turns out that's as big as it can possibly be, then there is a theorem. So maybe I should state this is the theorem, and this is due to Matsumoto, Shiginori Matsumoto from about 1990, I want to say. He proved that this number tells you what the action looks like in the case where it is the Euler characteristic. Then this action is, you guessed it, the standard boundary action. So it is semi-conjugate to the usual. And somehow, secretly, what's going on in his poop of this is he wants to recover the fact that you were a surface group acting like by-deck transformations on the boundary of hyperbolic space. And he uses kind of ping-pong-ish arguments to say that when I had elements of my fundamental group that corresponded to curves in your surface that crossed and played ping-pong, then the fact that this number is really big means that they'd have to play ping-pong on the surface. And so it's like they had crossed axes. And he sort of rebuilds the surface out of this. It's a beautiful theorem. But I'm just going to quote it and be like, look, in half the cases were basically already done. So it's not super hard to go from this. This is the surface group. This is the inner automorphisms. It's a normal subgroup using the fact is a normal subgroup in my automorphism group. It's easy to show that if this guy's acting in the usual way, then actually the whole group has to come along for the ride and be acting in the usual way too. All of row, not just the restriction to this normal subgroup, but the whole thing is the usual boundary action. And for a minute that it was actually genuinely conjugate, not semi-conjugate, then this means that everyone in pi1 sigma g is acting with sourcing dynamics with an attracting fixed point and a repelling fixed point. Automorphisms of the group have to, an automorphism will send some element of a fundamental group to another one. So I know where this point, where it's attracting point went. It went to the attracting point of its image under that automorphism. That's enough to determine what you do, I guess, on a dense set, all of the attracting points. So that determines what your whole action is. And that's the kind of argument you use here. OK, so we just won. OK, so we only have one more thing to do, and then we proved the whole theorem. We want to know that if I get zero, then my action is trivial. We have the Euler number. And for this, I'm just going to enlarge my surface group a little bit. My surface subgroup, the inner automorphisms, sits in a finite index subgroup of some slightly bigger group of automorphisms, which I can realize as the fundamental group of a hyperbolic orbifold, if you like. And specifically for this, I'm going to study instead, a subgroup that I'll call delta. Claim there's one that has the following presentation. It's going to look like the fundamental group of something that's a sphere with four cone points. So I'm going to get all geometric topology on you for a second, so hang on for the ride. Order 2, 2, 2, and 2g. If that doesn't mean anything to you, I'm going to write down a presentation for it. So this is generated by four things, where a squared equals b squared equals c squared equals d to the 2g equals the product of all of them. And these are all just trivial. Those are your relations. So this, you can show, sits in this automorphism group. You can think of it as coming from special finite order symmetries of your surface if you write your surface in a particularly nice geometric way. And it really is a finite extension of the surface group. It contains the surface group with finite index. You can write a surjection of this group to the dihedral group with 4g elements that has as kernel a surface subgroup. So this is one way to show that you can realize this group is a group of automorphisms. I can write down exactly what this surjection is, but I don't know if it's super enlightening. I need to send a to some element of order 2. So if my dihedral groups generated by an s and an r, this is going to s. This order 2g1 is going to r. And these go to other reflections that you cook up so that their product is the identity. So you can press me to write down exactly this before. But trust me, we've worked out the details. This contains a surface subgroup with finite index. And the point is that groups of this form also get Euler numbers. And the Euler number is multiplicative under covers. If you're good at Orbefolds and Orbefold fundamental groups, this is a sphere with four special points. So its fundamental group has a and b and c and d are corresponding to loops around these points. I can make a pants decomposition of this sky by slicing it like here and use the same kind of recipe to get an Euler number. So if you don't like that explanation, you can just accept this fact that Euler number applies with the same kind of formula to actions of these kind of groups too. And we'll take a value here is a multiple given by the index of the Euler number on the surface group. Now rotation number definition just tells you that it is, if I think of these as my pants and loops around these as my generators, then it's going to be of the form rotation number of row A plus rotation number of row B. That's for like two of the pants. And then here's where they match up. So I'll get a rotation number of row of C and a rotation number of row of D. And I didn't take any lifts or anything. So this is just like plus some integer is the best I can say. But it's always of this kind of form. It is of the form from the same kind of definition. Great. OK, so let's see. I know some things about these guys. Namely, this is order 2, and that's order 2, and that's order 2. And so these all have to look like literally rotations of order 2. So this, let's circle with the color, this, and this, and this are all elements. They're all half integers. OK, or maybe they're 0 or whatever, but they're all in half z. And I guess this one is 2. It's an integer, it's also in half z. So the only way that this guy is supposed to be a multiple of this, what I mean in the remaining case, with this guy, this is supposed to be 0. I know that 0 is equal to some half integer plus, I guess, the rotation number of row of D, whatever that's doing. Well, this implies that the rotation number of row of D must also be a half integer. And since D was finite order, it's acting by some finite order homeomorphism of the circle. Those are all rotations. So if it looks like a rotation with rotation number, some multiple of a half, it's either an order 2 rotation or it's trivial. But in particular, if I do it twice, I'll have done nothing. So using the fact that I know I'm finite order, this improves me to I'm actually an order 2. And now we're done by the finaling fun exercise. Basically every finite order mapping class, like D squared, for example, here or finite order automorphism of a service group, here D squared is finite order, it's order g. It normally generates the whole mapping class group or automorphism group. Here's your end of the proof exercise. D squared, it's normal closure or it normally generates odd or you can write mapping class. Maybe I'll write mapping class because it's more suggestive for how I think about proving it. So what did I find? I found something that normally generates the whole guy in the kernel of my action row. So the whole thing must just be trivial. And that is how you end the proof, modulo exercise. All right, so that was a lot. But I wanted to sort of outline how one goes about proving a whole theorem using some of these techniques, like rotation number. And this particular exercise is not on your sheet because it's about mapping classes, not about circle homeomorphisms. But if you like mapping classes better than that, I won't be offended if you decide to prove this one. Thank you.