 It's fun to be here. Um, so I think a geometric group theory we enjoy, we like to talk about groups that act nicely on nice spaces, where the nice spaces are usually kind of combinatorial objects, you know, cube complexes, curved complexes, graphs, uh, things that kind of have a combinatorial flavor. I'm going to talk about groups that act on the circle, and this is not a kind of discrete object at all, except secretly it is. Um, and although I won't kind of highlight why secretly it is in my talk, um, because that involves co-cycles and things like this, um, I do want to highlight some parallels between groups of homeomorphisms of the circle and sort of some of the other groups that we've seen so far and approaches to- to- to talking about both. Um, so my plan is, here's the plan. Uh, in the first half, so that's this hour, I'm going to talk about some basics, let's call them fundamentals. These to prove a theorem, um, that would be the end of this half of it. Um, I'm going to prove in quotation marks, a Titz alternative for groups that act on the circle, or for the whole group of homeomorphisms of the circle. Okay. This- this says orientation preserving. I like oriented things. So, uh, this is the group of orientation preserving homeomorphisms. No- no flips. Um, and, uh, I'll say in a second what that is, and then in the next hour, I'm going to prove a theorem about a particular action of a particular group. Um, um, about the group of automorphisms of the fundamental group of a surface. So I want G to be at least two here. So a hyperbolic surface, nice surface group. You can look at the group of all automorphisms of that group. Great, this is an object we're all good with. Oh, look, and- and this is even a hyperbolic group, so there's some geometry going on there. Um, so this group being a hyperbolic group has a boundary. Um, and the group of automorphisms acts on this boundary. Hopefully that's something you can dig out of your mind from last week. Uh, and what is this boundary? It's a circle. It's a circle. Topologically, it doesn't have any further structure than just being a topological circle. So this action is by homeomorphisms. Okay. And so I'm going to prove a rigidity theorem about this action. It turns out, uh, for this particular group, any action of that group on the circle is- looks like this one exactly. Okay. Uh, so that's going to be an exciting theorem and a way to teach you sort of some- some- some tools for talking about these objects. Uh, and hopefully relate it to things that we've thought about, uh, last week. Um, the Titz alternative. So you- many of you or some of you might know, uh, what this theorem is for linear groups, right? It says that- Titz's theorem is that if you take a finitely generated subgroup of GLN, some linear group, then there's a dichotomy. Either that group contains a non-Abelian free group, or, um, virtually up to finite index it's solvable. Okay. So it sort of looks free or it looks belian. Uh, this is- and- and so in generally we say that, you know, a group G satisfies a Titz alternative if something like this is true. Finally generated subgroups are either looking like solvable or contain free groups. This isn't quite true for groups acting on the circle generally, but something very close is in- due to Margulis. Uh, so he showed that any, not even finitely generated, but any subgroup of homeomorphisms of the circle, um, either contains a non-Abelian free group, either contains an F2, um, or the amlog being solvable or looking kind of like your Abelian or in this case amenable. Uh, the best analog you can have here, um, is that there is a measure on the circle invariant under your group. Okay. So, or there is a gamma invariant probability measure on the circle. This isn't a dichotomy. It's- it's possible that you could have both happening at the same time. But I'll explain sort of what all these words mean. Uh, when we get there. Uh, well, let's start right at the beginning. How about some examples, uh, of groups of homeomorphisms of the circle. I'm going to give you some examples to convince you that, uh, this is a- this- this big group is a rich and interesting object. Okay. Uh, so we saw one already, uh, odd pi 1 sigma g. Uh, that's- that's a pretty complicated group. Uh, let's maybe try and go, um, for- for- for other things. Uh, how about like just the group of rotations? So I can think of the circle as like, I don't know, you know, uh, unit vectors in the plane I can rotate. Um, another thing we've already seen so far, let's see what have we seen. Uh, last week we saw PSL2R acts on RP1 by Mobius transformations. And I mean RP1 is just a circle. So that gives you, uh, a nice action of this group on the circle and the SO2 subgroup inside here is acting by rotations. What else have we already seen? Uh, oh, this morning we saw Thompson's group, uh, group F. This acts on the interval by piecewise linear homeomorphisms. And if I take my interval and I just glue the two sides together, I get a circle. So this acts on the circle with a fixed point. Uh, but there's even a- there's a- there's a slightly bigger Thompson's group, uh, that acts on the circle. This is contained in a group, um, T, uh, which is in between F and the group V advertised before. Okay. Uh, T is what happens. It's a group- subgroup generated by F acting on, uh, S1 which you think of as this- this interval with the two endpoints identified. Okay. By piecewise linear, that gives you a nice linear structure. Um, by piecewise linear homeomorphisms plus, I- now I'm gonna allow you to move the endpoint that I glued. So you can take this group to be generated by this and there's an order two rigid rotation. Okay. So, uh, silly way to write this if you like more- more- this kind of notation is the transformation x goes to x plus, uh, a half mod 1. Okay. Where I think of the circle is, uh, this interval mod mod 1 or r mod z if you like. Um, and this group is, uh, interesting and special because, we thought that this group was not simple, although it's commutator subgroup is. This group is another example of a simple group. Um, good. So I've got some particular examples. Let's do a nice, uh, uh, very hands-on example or a couple. Um, uh, free groups fit inside of here. Okay. In lots of different ways. In fact, you can prove if you decide you want to challenge that the generic two in the sense of bear, even generic two homeomorphisms of the circle generate a free group, but I want a particular example that we should hold in our minds for later. You can do this even in PSL2R, but I'm gonna forget that I know anything about PSL2R. Um, I'm gonna produce, uh, free group generated by two homeomorphisms say f and g. Okay. So, uh, what do I want these to do? Here's my circle. I'll give some instructions to f. So I want f to be anything that takes, let's take a little interval here and stretches it out. So I want f to take this interval and stretch it out to being something like this. Okay. So f of i is going to be really long like this. And so what's left over, I wanted to take the complement and shrink it. Okay. So this interval here, I can call f of circle minus i. And g, I'm gonna have to do the same thing to two other intervals. I know maybe I'll call this j. And I'll do the same kind of stretch and shrink. Just take your elastic circle and pull it this way and smooth it that way. So that this interval over here is g of the complement of j. Okay. I claim that these two things generate a free group. Okay. Just as I've described though, I don't need to give you any more information. And there's only one way that we really know how to prove, okay, there's only one widely applicable way that we know how to prove that things generate a free group and this is the ping-pong argument, right? So actually what I claim is that if I take a point, I don't know, say this point x, that is not in any of these colored in regions that I drew. I have a, definitely I have a map that takes a point, that takes a word in my free group. Okay. And sends this to, so it's some abstract word in f and g and f of inverse and g inverse. I just take that word and I apply it to my point x, that this is now a homeomorphism, it sends x somewhere else. Okay. I claim that this map is injective. So not only is this group free, this point has trivial stabilizer. So for some of you, there's a one line proof or there's a one word proof and it's the word ping-pong. And for others of you, here's how you do it. The proof is by induction on word length, okay? And I'll do the, okay, the base case word length zero, that's trivial word, okay, I don't know. Let's do the case of word length one. I want to show that f and g and f inverse and g inverse all send x to different points. So let's check. If I apply f, I'm supposed to take this guy and shrink it into there. So I would go here, somewhere in this region. If I apply g, I'm supposed to land in, I'm not in j, so I'm supposed to land in that region. Okay, so these are different. If I apply the inverse of f, the inverse of f, what takes this and stretches it down, so it'll send x into here. So f inverse sends x into this region, and g inverse, same kind of logic, sends it into that region. And so these regions, the four of them distinguish the image of x under words of length one. And, okay, so that's the base case. And to do this longer, you assume that you sort of have this already. And I can actually take further nested intervals, right, inside of this picture of this interval. If I draw the image of these guys under f, I'll see some little sub-intervals in here. I'm supposed to take this and smoosh it. So I'll see a picture like that. And you can use these kind of refinements, these level two, to distinguish the image of x under words of length two, et cetera, et cetera. So this is a great exercise to do if you haven't had to do one of these before. And I'll leave it as proof by audience. Great, so that's a nice example of how to make a free group. And actually, if you are an expert at this, there's one thing here that people don't often notice. If you use this very by hand induction proof to do this, you don't only recover that this is injective. You specify the location of these intervals, i, j, and then f of the complement, g of the complement, is enough to determine the cyclic order of this point x. If I want to know if some first word of x, if I start reading around to the right, do I encounter a word or a second word, first or second in what order? That's recovered just from this combinatorial data. And that will come out of the proof. I mentioned that because that secretly lives in the proof of Ethereum, I'll quote later, but it's not necessary for understanding everything today. Okay, so there's free groups. Maybe just one more example of very pervasive things. All right, so I went from complicated groups down to simpler ones. Let's go for maybe one of the easiest groups. How about just an infinite cyclic group? It's very easy to produce infinite cyclic groups. Here is one way to do it, where, for example, f could be specified as follows. So I take my circle, I'm going to choose some closed set. So it might be finite, or it might contain some intervals, or it might have a part that looks like a canner set, or some accumulation points, I don't care. Take any closed set, there's k. And on every complementary region, so every interval in the complement, like here's an interval in the complement, I can identify this interval with the real line, it's topologically the same. And under whatever identification I looked, I chose, I can have f in these coordinates, I can make it look like x goes to x plus 1, just the translation. So this closed set is going to specify the set of points fixed by f, and the complementary intervals are going to be the non-fixed points. And for each of them, I have a choice of whether I thought infinity was this way, or infinity was that way in the real line, which way I thought was positive or negative. And that'll determine whether f is shifting points to the right or to the left. So I could choose a function positive or negative to choose right or left, maybe positive there on this complementary interval, maybe I wanted it to go the other way, minus. So this data is really an assignment, or a little more specific, than an assignment of complementary intervals to pluses or minuses, telling you the orientation of the line, which way points are moving, clockwise or anticlockwise. So what's a fancy way to write that? I don't know, pi not of the circle minus k, all right, point regions in the complement to the set containing plus or minus. So that gives you lots and lots of choices of ways to produce homeomorphisms. All of these will be infinite orders, so they'll generate a z subgroup. And the fact is that this data, the closed set in the plus or minus assignment is a conjugacy invariant. So the pair k and this assignment up to conjugacy. So obviously I could rotate my set k and rotate all the places I put pluses and minuses, and I'd have the same kind of looking thing. I'd have a conjugate homeomorphism. Okay, is a complete conjugacy invariant. So it distinguishes homeomorphisms up to conjugacy in homeomorphisms of the circle if k is not empty. Okay, so provided you know you have some fixed points, you can always write it down using this recipe and the recipe. The topological data of this is enough to specify up to conjugacy. So why I like homeomorphisms rather than diffeomorphisms, you can kind of draw a picture of them. Can you draw any homeomorphism like that? No, like here's a bad example. What if I take rotations? Okay, so here's a remark, not generally. Okay, for example, how am I supposed to draw a picture distinguishing a rotation by angle pi and one of angle, I don't know, pi over three or something like this? Both of them are just saying move all your points around some amount. So I don't know what to do. For example, rotations, non-trivial rotations I guess. I'm sorry, does it mean that every homeomorphism of r conjugate to the substitution? Yes, with no fixed points. Every fixed point free homeomorphism of r is conjugate to translation. I mean that is the proof of this fact basically. It's conjugate to translation by an orientation preserving or reversing homeomorphism. And that you can actually do by hand. You don't need a sophisticated theorem to build conjugacy. So that's a nice, didn't put it on the sheet, but that's a nice exercise to do. Okay, so actually let's solve this problem here and try and write down something that looks like, I don't know, a conjugacy invariant that will pick out different rotations, okay? My goal being to come up with a number or something easy to write down that will solve my problem. And I don't have to invent this, this is secretly hidden in work that goes all the way back to Poincare. This is Poincare's rotation number. And it's something that will play a giant role in the second half of this in hour two, okay? So let me set up some definitions. So what does the rotation number do? What I want to do is I want to pick a point on the circle. I have a homeomorphism. I want to capture on average, as I iterate, how far does this point go around? Okay, so you want to capture the average amount, you move around the circle under iterates of f, okay? So to do that, rather than, this guy doesn't have a fixed point of imagining I'm really rotating a circle, rather than remembering every time I crossed over and went around a time and then another time and then a third time. An efficient way to do that is to lift to the universal cover and unwind the circle, okay? So I'm going to start by working there. Okay, so let's set some notation. Here's the definition slash notation. If I think of my circle as r mod z, I'm going to lift it up and get all of r. But if I remember the circle is there, right? I have this deck transformation which is translating by integers. So we'll say homeo of r superscript z, okay? Just like in many areas of math, you write this as the invariance. This is the set of homeomorphisms of the real line which commuting, which commute with x maps to x plus one, integer translation. And hence they commute with all translation by integer amount, okay? If I have something that commutes with this, well then it defines a map on the quotient of r by this map, right? So there's a surjection from here to the homeomorphisms of r mod z. I want everything to preserve orientation. I think you do automatically if you commute with this, but this will make it clear, which is just homeomorphisms of the circle. So this surjects, and the kernel of this map is just this translation. So this actually lies in some short exact sequence like this. Where this one is generated by x goes to x plus one. So lifting to the line, let's me think of elements there up here and that counts how far I move around or how many times I've wound around. Okay, I'm gonna define the rotation number upstairs first. So for what's good notation for a lift, maybe with a little squiggle? For one of these homeomorphisms of the line we'll define the lifted rotation number of f to be, what do I want? I want the average amount, it translates a point, okay? So I'll take a point, let's choose zero. That's a good point in the line. I'll take f, I'll iterate it n times and to get an average I'll divide by n and I want the limit. I didn't leave myself as much space as I hoped for. I want the limit as n goes to infinity of this quantity. One of your exercises is to prove that this limit exists. And in fact, zero wasn't a special choice. I could have taken any point x and I'd still get the same number, okay? So this is, it's annoying to do on the back, blackboard, but I've given it to you with some hints. This is a well-defined number, it's just some real number. That is the rotation number of f. And if I looked downstairs, everything makes sense, mod z now. So for a homeomorphism of the circle, I'll say the rotation number of f is, I'll take a same kind of limit. I'm gonna choose any lift I want, apply it to any point I want. I picked zero, apply it n times divide by n. And this will give me a number, but if I took different lifts, that's like composing this with a translation by one, okay? That will change my average translation amount by one or by two or by whatever power I chose. So this is only well-defined mod integers. And that is the rotation number or the lifted rotation number. So let's do a little mental check, I really, if I have a rotation by, I don't know, maybe that one that's like x goes to x plus a half mod one, that order to rotation, hopefully I would get a half out of this. Well, let's see, let's pick the lift that's really x goes to x plus a half. If I do that n times, I'll add n times a half to zero. I'll divide by n, I'll get a half. I don't even have to pass to a limit, this is great. So the rotation number of a half mod z is a half. And indeed, that gets rotation number a half. So in general, the rotation number of rotations is exactly what you think they should be, rigid rotations. This has lots and lots and lots and lots of nice properties. So let's write some of them down to use. And again, these are things that I'm asking you as exercises to check. I'm going to clear on the definition. Okay, so properties. Okay, so I already said that rotations or their lifts, translations, if I take any real number, this is alpha. That's pretty immediate from the definition. It has a homogeneity property. Meaning that the rotation number, the lifted version of a power of f, k would be any integer, is k times the rotation number of f. Including for inverses. Better than this, so this is a particular abelian subgroup, right? It's a subgroup generated by f. And I'm saying it looks like a homomorphism, an additive homomorphism that are on this abelian subgroup, right? This is true in general for things that commute. That's all you need to prove this. So it's a homomorphism to some subgroup of real numbers as an additive group. When restricted to abelian subgroups, but not in general, all right? As you may even know from say multiplying matrices in PSL2R, you can write a product of two things with six points to hyperbolic matrices. You can write a rotation, a non-trivial rotation of a product of two hyperbolic things. Hyperbolic guys will get rotation number zero, they have fixed points. Your rotation is whatever you wanted it to be, okay? So not, not in general. But it is generally a conjugacy invariant. Meaning that the rotation number of conjugate of f is the same as the rotation number of f. And although it's not a homomorphism in general, it's up to bounded error, which is all we care about in geometric group theory, right? Homomorphism, namely it's what's called a quasimorphism. What does quasimorphism mean? It means something that is bounded distance away from a homomorphism in the following sense that satisfies that if I compare the rotation number of a product, here my product, my, my operation is like function composition, right? If I compare this to the rotation number of f and the rotation number of g, well, if it really was a homomorphism, if the sum of these should be the same as that, right? So let's see what happens if I take this and I subtract off these guys, okay? Well, the claim is that this is uniformly bounded. And in fact, in this case, you can show that it's bounded by one in absolute value. Excuse me? Yes? Are there any non-trivial, obedient subgroups of the group of homomorphisms? SO2, z, z to the n, all kinds of things. That don't eat within things with fixed points, right? You could think of just some little interval of the circle. You could identify with the real line and make a group of translation. Okay, but if there are no fixed points, then there are weirder examples than things in SO2, yeah? And I'll actually, how about I show you one right now, making sure that in a second, once I make sure I've told you all the properties I'm gonna use later on, yeah? When does the quality hold? Good question, if f and g don't commute, typically not, okay? But you can also construct examples where f and g don't commute and the quality does hold. So, oh, sorry, when does the quality hold? Sorry, I meant, I was answering some version of this question where I thought, when are these, when is this zero? Do you mean when are these, when is this one or when is this zero? When is this one? Yeah, when is it zero? Okay, so I answered the zero question. Actually, both questions are interesting. And in general, if I give you f and I give you g and I ask and I tell you what these are and I'm like, what are the possible values can this guy take? That is an interesting problem. And in certain cases, for this particular question, there's a complete answer which I can show you a picture of. There is a graph you can draw, what are the possible values? It has this crazy self-similar fractal structure. This was done algorithmically by Danny Caligari and Alden Walker. And they have a computer program that sort of answers this question in many cases. So yes, you asked actually a very subtle and interesting question. What values does this take when, based on what data? Let me take a slight detour for a minute which might help us over there. I wanna talk about an example we saw later last week. Jair Minsky had a picture where he was trying to describe how you make a lamination that's not a foliation on the torus or on the punctured torus. And I'm gonna be just very slightly less sketchy than he was. He started with, if I take, you wanna understand that the torus are in annulus with foliation of irrational slope. So one way to make an irrational slope foliation on a torus is I'm gonna take, here's my circle, so I'm gluing these two sides together, cross interval, all right? And I'm gonna mod out by an equivalence relation where I'm gonna glue top and bottom of the interval by an irrational rotation or translation. If I do that, well, that'll take this vertical stripes under my gluing. What am I supposed to do? I'm supposed to shift them all over before I think of them being attached back down, right? So that's like, I've really identified my, taken like a slope like this and I've used this to identify my stripes, okay? And if I sort of shear this back over to say, hey, just glue them up and down normally. You'll get an irrational slope foliation of your torus, okay? Then he's like, well, here, instead of doing that, I want one where it's not all dense. And in fact, I only want these sort of going on a canter set of points. I'm gonna like take one of these leaves that goes wrong and I'm gonna thicken it up a little bit and then thicken it, but not so much over here and something like this. If instead you glue by not rigid rotation or translation, but by some homeomorphism F that, say, has rotation number alpha, but not conjugate to a real rotation, okay? The result is that you'll produce something that has a canter set somewhere here identified with itself by kind of a shifting over. And this picture will look exactly like the picture we saw before, okay? Where there's stretched out and thinned out lines following some canter set. So if you did the, if you got interested in his lecture and the Dangeois counter example, as was suggested then, the Dangeois counter example is exactly constructing a function that satisfies this property that lets you build one of these strange relations. I wanna explain sort of what's actually going on there, but I wanted it to tie it to his lecture first. Okay, so consider that maybe motivation for what's about to, what's about to come now. And this is a very useful kind of dynamical trichotomy that says that pictures either look like this or look normal. So here is a theorem which I'll need for what I wanna do next. And so as the following, if you take any group acting on the circle, so any group of homeomorphisms of the circle, then either, this is an exclusive either or. Okay, one of three things can happen. Okay, one, it could have like a fixed point or more generally a finite orbit. A finite orbit. That's one possibility. Another possibility is I don't even have like a very rich large group like PSL2R or something like this or like a surface, co-compact service group in there. Another thing that could happen is that all orbits are dense. This also happens if your group is just like a single infinite order rotation. And the claim is if neither of these two things happens, then something weird goes on. There, an invariant, so a gamma invariant, closed subset. How do I wanna state this so that I don't, let's give it a name, k. It's contained in the closure of every orbit and so that the restriction to k has all orbits dense. All orbits on k, dense in k. It doesn't tell you very much what k looks like, but I can say a lot more. k in case three is homeomorphic to a canter set, and it's unique. It's the unique set with the properties listed here. closed set contained in closure every orbit with the action has all orbits dense. This is often called the action being minimal. There is a gamma advanced closed subset k containing the closure of every orbit with all orbits k, dense in k. So, let's say this in English words with the action on k is minimal. Yeah, so I can forget the rest of the circle ever existed. I now have an action on something I claim is a canter set and all of the orbits of this are dense. So I could make it bigger and not have something like this happen. If I take a minimal such example, I'll get like a smallest possible set that satisfies these, I'll get this property, but it's not necessarily... For example, the whole circle is a gamma invariant closed set contained in the... Oh, contained in... Yeah, you're right. You're right. If I put it this way, containing closure every orbit. Yes, that's true. This is a constant. I didn't realize what I had wrote. You're correct. Let me give you a quick proof of this because it's good to prove some things. I mean, one way to approach this is you say, well, suppose neither of these holds and like let's find k. But very generally, we could start right from the beginning by being like, okay, what am I aiming for? Let's look at... I'm trying to understand closures of orbits. So let's look at the set of subsets of the circle consisting of orbit closures. So this is a set of things that look like gamma on a point, the closure of this where x is some point in the circle. This is partially ordered by inclusion. So I can take a minimal element. So you show that it has the finite intersection property or whatever you want. Take a minimal element. Under this ordering. If that's finite, you just showed that some orbit is finite. Great. So if your minimal element is a finite set, you get case one. If your minimal element is the whole circle, what does that mean? That means that the closure of every orbit is the whole circle. So we're in case two. And if it's not, well, what is a minimal element of this? In general, it will be some gamma invariant set because orbit closures are gamma invariant. It's going to be a closed subset. And I claim it will look like a canter set. So every point should be an accumulation point. So it has some accumulation points. And the set of accumulation points will be invariant on your group action. Your group is acting by homeomorphisms. So if I didn't want this, if there were some non-accumulation points, I could throw them out and I'd get something smaller. So every point is an accumulation point. It's perfect. And the same kind of argument says it has an empty interior. Otherwise, I'd throw out the interior and I'd get some smaller closed invariant under my group set. So the closure of an orbit in that point would be contained in it, would be smaller. So it must have empty interior. So these last two just come from the fact that I took a minimal element. So that says, oh, I already proved sort of my agenda that this is homeomorphic to a canter set. And the contained in the closure of every orbit is just because I could take an intersection, right, if I had something that was smaller. OK. So that was a little quick and sketchy at the end, but I promise there's nothing too complicated going on. So check that this has the properties that you want. Is the existence of a minimal element guaranteed by compactness? Yeah. You just check that descending things. Yeah. Yeah. OK. Can you give us the measure of space? Nope. You can do one of Yahya's examples if you believe that this really works to make fat canter sets on which you do a shift kind of thing. So things are pretty pathological if you don't, if you just have actions by homeomorphisms. OK. Let me in the remaining 16 minutes or so tell you about what I really wanted to kind of get to, this proof of the, of the, of the Tetz alternative. The rotation number I need, maybe next time, but not right now. And I'm going kind of quickly, my aim is to give you a flavor sort of how some of these kind of arguments go so that you can play with them and imitate them and sort of have a toolkit to play with some problems in the problem session. OK. So let's prove the most general kind of structure theorem that we have for this group. This one, I have a subgroup of homeomorphisms I want to find a free group or an invariant measure. So for instance, if I was acting literally by rotations, my invariant measure would be like Lebesgue measure on the circle. That would be nice. More generally, maybe I could, I won't write the Lebesgue measure one down, but if I'm in case one of there, if I have a finite orbit, I can just put point masses on all of those finitely many points. My orbit has cardinality five. I'll put a point mass of one over fifth on each point. Those points will get cyclically permuted around. That will be an invariant measure under my group action. OK. So you know, so I don't need it to act by literal rotations. All I need to know is that this point, these three, these five points are always, they all stay in their order, right? I'll elements my group, permute these. I can put point masses there to produce an invariant measure. OK. In particular, if I have a single fixed point, then this tits alternative is very easily satisfied. It could have a billion fixed points or uncountably many. I don't care. Just put a point mass at the single fixed point. It never moves. That's an invariant measure. So we're really interested in what happens in two or three. All right. So when this kind of thing doesn't happen, I'm expecting it to look like rotations with Lebesgue measure, some version of that, or to what if I want to prove there's a free group, I better imitate my ping-pong picture, which got erased from before. OK. So I'm looking for one or the other. Well, one thing that would put me in the invariant probability measure would be if I had a group that acted by homeomorphisms where those were equicontinuous, where the action is by some family of homeomorphisms. These are all equicontinuous. I'll remind you what that means in a second if you forgot your analysis. Then the Arzela-Skoley theorem means that your group is compact. It's a fact or an exercise, which you can't do, and I put on the sheet. It's a fun one. Any compact group acting on the circle is conjugate to a group of rotations. So gamma is conjugate. So compact implies gamma conjugate to a subgroup of SO2. So then that conjugate of your Lebesgue measure, putting it back under that conjugate, gives you an invariant measure for the action. You're just acting by homeomorphism. Gamma could be countable or uncountable. Oh, maybe I want it to be countable in what I'm thinking here. No, this is fine. I think fine just in the contact of entomology. Do you have a sort of example that's bothering you? It's not just a finite quantity of compact, but it will be countable. You mean that the closure of it should be... I see. Right. Well, wait. So if the closure of gamma is compact, then its closure is conjugate into SO2. And so gamma itself is a subgroup of SO2 being a subset of its closure. So that's good enough. Right. Then gamma... Yeah, so that's fine. Then the closure of gamma is relatively compact is fine. So its closure is some subgroup conjugate into SO2. That'll do it. Thanks. Okay, so if you fail to be equicontinuous, if not, okay. So by the book definition says what does it mean? That means that there's a bunch of intervals of length going arbitrarily small. So that elements of your group stretch them to have a big enough... to have a... bounded from below length. Okay, so let me write that down. So there exists some bounds, some epsilon. There exists intervals where the length of IN goes to zero. And there exists some sequence of elements of your group where the length of GN of IN is at least epsilon. All right, so this is the case that we're worried about and we want to hopefully produce some ping pong out of this and it's looking pretty good because I have like tiny little intervals that get stretched. And that's what I was using in this ping pong picture, stretching and shrinking lengths of intervals. All right, so to make my life easier, I'm going to assume for a little bit that I don't have one of these crazy canter sets that seem like a pathological thing. I'm gonna assume that my action is minimal. All orbits are dense. And then we'll deal with what happens if not else was. So let's assume for a moment a little bit of time that gamma acts minimally. And what I want is to run my ping pong argument. Let's draw this picture of what I want. I want to find something like this and some F that takes this interval to like this big one and then some G that has something like this and it takes this thing and smooshes it down or it takes this one and stretches it over like I had that before. And then I'll have a free group and I'll feel happy. But the problem is that this picture, the way I drew it, doesn't always happen. All right, I might have to draw a more complicated picture. All right, so here's a problem. This picture doesn't happen the way I drew it even in like, I don't know, like SL2R acting on like raised the origin or like the two-fold cover of PSL2R if you like, right? If I wanted to try and do this picture, an SL2R, the thing about SL2R is everything looks the same on both sides of the circle, right? Everything commutes with an order two rotation. It's like an action online. It's not the projectivized version. So I can't have something that takes this giant interval and swooshes it down. Whatever I do up here, a squish is also done down here and I'd have to have another squish. So I wanted to do this in SL2R. I'd have to draw the following picture. I wouldn't, you know, I'd maybe need more of these. And where my F is taking these two points like here and taking these two points there. And so my domains that I was stretching or squishing are no longer connected. And I would want it G to do something like this. And then I could run the same argument, but it would look different. But this is just silly. You know what did I draw? I drew the two-fold cover of this picture. Everything commutes with a rotation of order two in SL2R. Here's the projectivized version. It's the picture I had before. And more generally, if you have some kind of group that acts on the circle, listen, the circle happens to be a K-fold cover of itself. So I could look at all the lifts to a K-fold cover and I'd get a more complicated looking picture. So this is just to say that I might have this silly issue where I had a kind of a complicated action. I could have just taken a quotient and then I could draw my original ping-pong game from before. So this is nice picture. And this is more complicated. So I want to simplify my life right off the bat and be able to detect if secretly your whole group actually just came from one of these lifting tricks. If I could have just been like, listen, you were really acting on this circle and then you took all the lifts to this cover. So the outline of the proof is to first produce that covering map if there was one and then say, all right, if I didn't have this covering map going on then I have exactly this picture going on. I can find intervals in the configuration I wanted before from my step zero of this lecture example of free group and show that there's a free group in it. And I don't have much time left in this hour so I will maybe outline how this works and we can either take it up next time by request or you can sort of fill in details as you like. Okay. So let's see, if I act minimally, I can massage this statement into something else. So let me massage it right away. Put it right there beside, remind us. So if I pass to a subsequence, these G-N-I-Ns, it's a sequence of intervals in the circle that will house dwarf converge to some set after passing to a subsequence. So I could pass to a subsequence so that this condition will say that, you know, all of these contain a very slightly smaller interval, maybe call that J, and then the inverse images of J under G-N are getting really small. Okay, so this says that there is some interval J of length more than epsilon over 2, I can certainly guarantee, with the length of after passing to my subsequence G-N of J, the inverse of J, I guess, going to 0. So I have a fixed guy that gets shrunken. So let's call this interval a contractible one. If my action is minimal, I can move any point to any point. So minimal implies that I could put J wherever I wanted. That says that for any point in the circle, there exists some point Y, okay, so that the oriented interval between X and Y is contractible in the same sense that J was. There exists a sequence of elements that shrink it to arbitrarily small length, okay. Such that there exists some sequence H-N, depending on X, with the length of H-N of this guy, XY, going to 0. I will pick out if I was secretly covered by a cover by the following function. So I'm going to define a function on my circle that assigns a point X to the biggest possible Y where I can do this. So the supremum preceding clockwise around the circle, so that XY has this property. For example, if you're in PSL2R acting by Mobius transformations, this function happens to be the identity. If you give me a point X and you give me an arbitrarily large interval, there's some very, very, very strong hyperbolic element that takes this to being as short an interval as you want. And the claim is that in general, this might not be the identity. I'll state it as a fact that you can check. This guy is a finite order rotation. Or a finite order homeomorphism. And I've cooked it up so that it commutes with the action of your whole group, meaning that phi of GX is G times phi of X. So these are easy to check from the definition. It's saying, when are you in this picture instead of that picture? Here, phi is order 2. But it commits with your group, so I might as well look downstairs. And to say that phi is the identity map says that at every point, I can take as big an interval as I want. Maybe I took this really giant one here. And I can contract it to intervals as small as I wish. And similarly for one containing some point over here or bound with a left-hand point like this, I can contract it as small as I wish. That is exactly what I need to make the ping-pong argument work. So I'll just summarize and say, so passing to a quotient, phi is the identity in which case you can play the ping-pong argument. Okay, that's a good place to stop. Perhaps that's too much. I should say, if you weren't acting minimally, then we have this Cantor set picture. And I can just pretend my Cantor set was the original circle. You imagine, you know, like Cantor's staircase function that takes your interval with a Cantor set in it and collapses each complementary region to a point. This is some continuous function that makes your Cantor set disappear, right? My Cantor set is invariant, so I could apply the collapsing function and I'll get a new circle on which my group acts by homeomorphisms. I can run this case, all right? Either I was equicontinuous on this new circle and so I can pull back an invariant measure to my old one or on this new circle I could play ping-pong and that must mean I contain a free group. Okay, so in very short, the fact that I'm assuming that gamma acts minimally is no big deal because my other option is this Cantor set. And I think that's a good place to end this first bit.