 So I'm going to be talking about automorphisms of free groups. So basically for the next six hours or so, I'm going to be telling you things about one particular group, namely the group of automorphisms of free groups, except that as I go along, this will be morph into various related groups. So automorphisms of free groups, but this is an old, old subject, basically been studied since the beginnings of group theory. Free groups are kind of the most basic, infinite, discrete group. So if you want to understand an object, you want to understand its symmetries, so you want to understand its automorphisms. So the plan is to introduce some, here's the plan. Introduce some basic objects and techniques that are used in the modern study. The groups, automorphism groups of free groups. Now I say we're going to introduce stuff that's used in the modern study of automorphisms of free groups, but these objects and techniques, turns out, are not necessarily so modern. So I'm going to start today with some techniques that were invented in the 1970s. And in fact, I'll go back to some techniques that were invented in the 1920s and 1930s, so almost 100 years ago. But it turns out that they're still relevant and still useful. And if you want to understand what people are doing today to study automorphisms of free groups, you have to know this language and these techniques because it's kind of part of the undercurrent of what's happening today, from the 1920s on, but still useful. And I'm assuming in this course that I'm talking to people that don't already know a lot about either automorphisms of free groups or the spaces we use to study them, like outer space, for instance. I look around and see faces of people that do know a lot about these things. So you have to be patient. It'd be good if you go to these groups and explain to people that haven't already seen before, which you know. So when we do the exercise groups, you'll be the stars. But you may find this a little pedestrian at the beginning. OK, so what are we going to do? We want to use topological and geometric methods. These ideas aren't new either. The techniques I'm going to talk about from the 1920s and on also use topological and geometric methods to study automorphisms of free groups. So you need a topological space. And this space is going to model the free group. So my free groups in this course will all be finitely generated. And I'll try to remember to always call the generators a1 up to an. OK, so you're going to use topology. You're going to think about groups. You want a space that models the free group. In other words, you want a space with the fundamental group of x isomorphic to the free group. So you can probably off the bat think of several spaces that might work. So let's try so candidates. Oh yeah, I should have said you want a space to models the free group, but we're actually supposed to be studying automorphisms of the free group. So then maps from x to x, well, they induced maps of fundamental groups. And if we've identified this with the free group, get maps on the free group. In order for these, we're interested in automorphisms of the free group. So we want these maps to be isomorphisms. So in other words, we want to model these. We want to think about maps that are homotopy equivalences. So we need a space. And then we're going to study, actually, a homeomorphism is an example of a type of homotopy equivalence. We're going to study maps, in other words, on x that induce the isomorphism on the fundamental group. And we might wish for more than that. But we'll settle for that for now. OK, so what are the candidates? Well, the first one, you probably see in algebraic topology, is a finite graph. And in order to have the fundamental group fn, we'll make it connected, so that we actually have a good fundamental group. And we want the Euler characteristic. And we want it to have fundamental group fn. So we want the Euler characteristic, which is vertices minus edges to be 1 minus n. So for example, there's an example of a graph. It's got fundamental group f3. But that's not the only common space that has fundamental group, the free group. Another space that you encounter a lot in this course. And in geometric group theory, you might have a punctured surface. These are punctures. It's got some genus. Then the fundamental group is a free group. So here, pi 1 of x is isomorphic to fn. Here, pi 1 of x is isomorphic to the free group on 2g plus s minus 1 generators. Another space that you commonly encounter if you're thinking about free groups is a handle body. So this is a solid thing filled in surface. And another way to write it, it's a connected sum of s1 cross d2's. N copies of them. And again, here, we have pi 1 of x is fn. And finally, maybe slightly less obvious and not the first thing you think of, is if I take this handle body and double it, meaning I take two copies. These are both solid handle bodies. And I'm going to glue them together by the identity on the boundary. So what do I get? I had a connected sum of s1 cross s2's here. I have it now. I connected some of n copies of, well, if I double s1 cross d2, I'll get s1 cross s2. The d2 is doubled just along its boundary. So this disk becomes a sphere. And I'll get a connected sum of n copies of s1 cross s2. So those are candidates for a group that models a free group. I want to say a little bit more about this model, because in fact, this is a model that we'll use a lot. And maybe you're not so used to thinking about this. Let's look at another way to visualize. I want you to get used to these pictures, because I tend to draw lots of pictures in my talks. And if you don't know what I'm drawing, there's no point in coming to the talks. Let me call this is a doubled handle body. So I take my two handle bodies. And I'm actually going to pick n spheres in them so that the complement is simply connected. So I need to take n of those to make that happen. Here I've got three spheres. Let me call three disks a, b, and c. And a prime, b prime, and c prime. And as I said, the manifold I'm talking about is I take this guy and glue it to this guy. But I only glue it by the boundary. So this disk together with this disk will become a two-sphere in my manifold. So here's another way to think about that. Let's cut this guy open along these spheres. Here's a. Here's the other half of a. Here's b. Here's the other half of b. Here's c. Here's the other half of c. And do the same thing over here. This is easier on the iPad. You can just cut and paste. But if anybody's taking notes, that's not nice. OK, so I'm supposed to glue these by the identity on the boundary. Right, so now what have I got? I've got these two three balls with some patches on them. And I'm supposed to glue these two three balls together, but I don't glue the patches. I'm only gluing the boundary, which is not. The patches are not part of the boundary. So what happens when you two glue two three balls together? Well, you get the three sphere. So after I glue this together, I'll get the three sphere, which is kind of hard to draw. It's the three sphere. It's everywhere. But I've got these little spheres. a, a prime, b prime, sorry, a prime, a. a and a prime make a sphere. a bar and a bar prime make a sphere. b and b prime and b bar and b bar prime, et cetera. c and c prime and c bar and c bar prime. So after I glue them along the boundary, I have a three sphere, which is all a space with these six little spheres cut out, removed. So this is s3 minus 6 three balls. But of course, this isn't my space because I cut this apart. So in order to get my space, I have to glue these back together. But still, this is a good picture of my space. I just remember that I have to identify these guys together. So this is one picture of my space that I'll use occasionally, maybe a lot. And usually when I'm drawing this picture, I'll only draw half of it because the other half just looks exactly the same. But here's another picture of this space, just six balls hanging in three space. And they're labeled in pairs to remind me that my manifold is really what I get by gluing those guys together again. So is this picture clear? You have to understand this picture if you want to understand the rest of the lectures. So those are my spaces. So as I said, any homotopy equivalence between these spaces induces a map on the fundamental group. Your first exercise, which is on the exercise sheet, so if I take a homotopy equivalence, a couple of remarks, any homotopy equivalence induces an isomorphism over there. If I have homotopic homotopy equivalences, they give me the same map on fundamental groups. So I really have a map. So every homotopy equivalence gives me an element of, well, it gives me an automorphism, gives me a map. I'll write something wrong. Why is that wrong? There's no base point, right? If I really wanted automorphisms of the free group, I would have to have a base point. What this is, I'm thinking of this as automorphisms of pi 1 of x. But pi 1 of x is not a group. In order to talk about pi 1 of x, I have to have a base point. And so in order to talk about homotopy equivalences that give me automorphisms of pi 1 of x, I have to talk about homotopy equivalences of x that preserve the base point. And then homotopic homotopy equivalences give me the same automorphism. And so pi 0, that's a path of homotopy equivalences. That's a homotopy of homotopy equivalences. This group maps to this group. If I don't want to worry about base points, and I'm a topologist, so I don't like to worry about base points, so I'm a low-dimensional topologist. Algebraic topologists really like base points. And if they don't have one, they throw one in. But I don't a priori have a base point. So when I take a homotopy equivalence, it might move whatever I decided to choose as the base point over here. And in order to get a map on fundamental groups, I have to choose here if maybe this x gets moved over here to f of x. In order to get an actual map on fundamental groups, I have to choose a map path from x back to f of x. So a loop here might get sent to a loop here. So I have to choose a path back to think of it as a loop at x. So I don't actually get an automorphism because I have to choose this path. And I can choose lots of different paths. What I get instead of an automorphism is an outer automorphism. So I could also write pi 0 hg of x. Gives me not an automorphism, but an outer automorphism of pi 1 of x. And then I don't need to really write the base point because I'm not worrying about the base point. So there's this map. And your first exercise is to prove this is an isomorphism. So you have to prove it's a homomorphism just by the way these groups are defined, these things are defined. But you have to prove it's surjective and injective. It's not hard. Second one. So that makes an extremely nice model for my free group because homotopy equivalences in my space give me exactly outer automorphisms of my free group, which is the group I was trying to study. What about a punctured surface? Punctured surface, it's certainly true. Actually, even it's true for surfaces that homotopy equivalences are homotopic to homeomorphisms. So instead of pi 0 of he of x, I could think of pi 0 of homeomorphisms of x. So I get a map like this. Turns out that this is not surjective. The basic problem is that if you have a surface, there's a certain intersection form. Yeah, if you have two curves on a surface and you apply a homeomorphism, if they intersected, then they still intersect. If you count the number of intersection points, the number is the same. So there's a certain intersection form on the first homology that's preserved under a homeomorphism of a surface. That implies that the eigenvalues of the map on the first homology come in pairs. And this is explained. Well, this is said in more detail in the exercises. And you can find homeomorphisms of the free group that don't have this property, that the eigenvalues come in pairs when you look on first homology. OK, so that's not so good. So we don't want to think about this too much. What about number three? Again, you can homotopy equivalents to a homeomorphism. So pi naught of, you could look at, so there's a map like this. And that is surjective. So that's good. We made progress. The problem is that it's not injective. In fact, it has an infinite kernel. And again, for the exercises, I'll tell you why there's a kernel. So let's go back to, so now we come to candidate number four. There's a theorem by Ladenbach that says this is a good model for automorphisms of free groups, namely the map pi naught homeo of mn to out pi of fn out pi 1 of x. Well, it's surjective. And the kernel, well, it's not quite an isomorphism, unfortunately, but it's got a very small kernel. k is a finite two group. And it's generated by Dane twists in two spheres. So let me just explain what a Dane twist into two sphere is. You may know what a Dane twist in a surface is. You take some annulus in the surface. There's an annulus. The surface is all around here. Maybe there's some genus in the middle. So you take your annulus, and there's a homeomorphism of the surface that's supported just on the annulus, which doesn't do anything to the outside circle. Twist the inside circle by 360 degrees, which doesn't do anything. And gradually twists more and more as you go in, so that the image of some hapless little curve that passes the annulus after the Dane twist looks like here nothing happens to it. It gets twisted more and more and more and more and more until it goes back to there, where it the same place it entered. So that's a Dane twist. You're all probably familiar with this. Now, what's a Dane twist in a two sphere? All I do is suspend this picture. So instead of an annulus, I'm going to have instead of two, an annulus is two concentric circles that are filled in. Instead of two concentric circles, I'm going to have two concentric two spheres that are filled in. But I'm going to do exactly the same thing. I'm going to do nothing to the outside. The whole homeomorphism is supported in this area between these two two spheres. On the outside two sphere, it's the identity. On the inside two sphere, it's the identity. And in between, it twists a little bit. So this is a suspension of what you're used to thinking of as a Dane twist on a two sphere. OK, so I say this is generated by Dane twists in two spheres. When I do this rotation, so I've got some axis, and I do this rotation, I'm rotating by 360 degrees. That's D gives a path of rotations, in other words, a path in SO3. That's a group of rotations of three dimensional space. Pi 1 of SO3 is Z mod 2z. And this path is a generator. So when I square it, in other words, when I do D twice, I get the identity. So the same thing happens here. D squared is the identity. And in fact, this kernel, each of these generators is only an order two element of homeomorphism of this manifold. So it turns out if you do Dane twists in those particular spheres that I've put up there, that generates the entire kernel of this map. Sorry? The kernel of finite two groups. It's a finite group, and every element has order two. Yeah, it's Z mod 2 to the n. And the generators are those Dane twists in those particular two spheres. Yeah? Can we always choose this path so that the rotation axis remains the same, or can it vary? This is just an example of a Dane twist. OK, I mean, I'm saying what I mean by a Dane twist. So I choose an axis, and I've got two spheres, and I choose an axis, and I rotate by 360 degrees. That's what I'm calling a Dane twist. I'm not sure I understand. I'm curious what happens if the path is exactly on this equator plane. What happens if the path is a little bit higher off? Well, up here, oh yeah, up here nothing happens. Let's see. I'm not sure I can draw this. That's a good exercise. So yeah, good question. So you're thinking I should have drawn my suspension sphere like this. Is that what you think? Yeah. But that's homotopic. So yeah, maybe it would have been better to draw it like that. Then you can, it's still a bit straight through. It will go, OK, just do this and make this a little higher. Sorry? Just the suspension of the original Dane twist. It's a suspension of the original Dane twist, yes. That's all. I'm just trying to draw a picture of it. Yeah. OK, so double-handled body, blah, blah, blah. Let's see. OK, so let's go to, so those are going to be my models for automorphisms. Let's go back to automorphisms. Let me show you some. So remember, my free group is about generators A1 up to An. If I want to tell you an automorphism, all I have to do is tell you what happens to the AIs. So example, I can send AI, let me call this, actually call it iota i, send AI to its inverse. And don't do anything to AJ if J isn't equal to i. OK, that's clearly an automorphism. It's clear how to undo it. You just send AI, it squares the identity. Here's another one. If I have a permutation, I can permute the generators. That, too, is clearly an automorphism. Just undo it by taking the inverse permutation. Here's another one. It's called rho ij. I can take the ith generator and multiply it on the right by the jth generator. And AJ goes to AJ if J isn't equal to i. And then it's inverse. I just, I multiplied AI by AJ. I can multiply AI by AJ inverse. That's an inverse automorphism. And there's also, since this product is not the same as the inverse product as the AJ AI, there's also a left by lambda ij that sends AI to AJ AI and doesn't do anything to the other generators. OK, so those are some easy examples of automorphisms of free groups. Let's see what they look like in. So those are our candidates. These we didn't like. These are left in the running, one and four. And those are, in fact, the ones we'll use. Do I have three boards or just two? Oh, I have three. But if I use that one, how do I get it back to? So let's look in the graph model. Let's take a graph. Let's take the easiest graph with fundamental group Fn. It's got n petals. I'm going to label them a1 up to an. And it's pretty easy to see how to model i0 to i. It just takes the ith petal and flips it. And it's pretty easy to see how to model. So this induces AI to AI inverse on the fundamental group. It's pretty easy to see how to model sigma. Just permute the petals. What about how to model rho? Well, let me do lambda ij. I want to model that on this graph. I've identified the fundamental group of the graph with the free group by making every petal one of the generators. How do I model lambda ij? Well, let me draw the ith petal like this and the jth petal like this just for good measure. And then all the other petals are down here somewhere. What I want to do, I claim, is divide the AI petal in half and fold the first half of it over the aj petal. So I'm going to do this in stages. So first I divide. Well, I can divide the ai petal in half right now. And then I'm going to, the aj petal isn't going to do anything. It's going to stay exactly where it is. These other guys aren't going to do anything. The ai petal is going to slowly get the first half of the ai petal, or the second half, I guess, is going to get folded over the aj petal. The end I'll have, it'll be folded all the way over. Let me just take this and put it out here again. Now let me look at what I've done. So this was, let me look at the ai petal. That's this, goes like this, goes like this. And now, again, I'm back to things where I'm supposed to, where I, we're back to where I started. I want to think of this as the ai petal and this is the aj petal. But the ai petal goes over both the ai petal and the aj petal. So this is an example of a homotopy equivalence. It sends, it's the identity on all the white edges. And it sends the orange edge over the ai petal and the aj petal. So it's clearly the automorphism that takes ai2. I should be careful of my directions here. So it goes over the ai petal in that direction and over the ai and it goes that way over the aj petal. So it's really that map now. So it goes, it sends ai to aj ai. So did I do that right? I have a feeling I didn't. You can't see from the drawing at the end what order the two leaves go. Right. This is aj always. Nothing happened to that. Yeah. Yeah, yeah, it goes to ai aj. The first half still goes, doesn't get folded on anything. The second half goes to aj back in the right direction. Yeah. So that's right. So that's actually not lambda aj. That's rho ij. OK? So I can describe this whole thing with one picture. I've divided the ai petal into two pieces. I sent the first half over ai and the second half over aj, et cetera. This is one picture. But when I draw this picture, I'm making various assumptions. This one picture doesn't actually tell me what the map on the fundamental group does because I haven't identified this picture, the fundamental group of this picture with my rows yet. So implicit in this picture is that the petals correspond to the ai. And I have to say which petal corresponds to which ai if I'm actually going to specify the automorphism, as somebody was pointing out a minute ago. So the petals correspond to the ai, but it doesn't really matter too much which petal I call which ai because if I called the wrong petals ai and aj, I would end up just with a permutation of the petals. So this might be rho ij. Well, this picture is either rho ij or rho ij combined with some automorphism, some permutation of the petals, permutation possibly inversion. So that's what it is. OK, so this is an example of a Staling's fold. So this is an example. So this idea of folding one edge of a graph over another was originally due to Staling's in a paper in the 1960 called The Topology of Finite Graphs. I love the title of this paper. And it's become a technique that you'll see over and over in the literature on automorphisms of free groups these days. So I used it to illustrate how to model a particular automorphism. Turns out you can use it to prove that the automorphisms theorem, the automorphisms eotis of i sigma rho ij, and actually you don't even need lambda ij, generate odd of fm. OK, so I want to sketch the proof. So what I've just done drawn up there from the start to the end is an example of what's called a graph automorphism. So I should probably tell you what I mean by a graph, but this is a sketch proof, so maybe I'll be a little bit vague, x and y graphs. They have vertices, edges, and they've got an initial vertex map that takes an edge to a vertex and an involution that takes an edge to an edge oriented in the opposite direction. So that's the combinatorial data I need to describe a graph for you. And a graph morphism, well, it's just what you think it takes, vertices to vertices and edges to edges. But in order to be a morphism, it has to play well with the initial vertex map and with the involution. Plays all means commutes. So that's what a graph morphism is. And x and y are graphs. So the edges come, these are oriented edges. Maybe I should have said that. So you have two edges for every place you see a little line segment. So I've only defined an initial vertex map, so an edge has an initial vertex. It also has a terminal vertex that's the initial vertex of x bar. And that's the common, yeah. So, and here's a little lemma. I'm not using the sphere model anymore until the next hour. So I'll erase this. Do you allow your involution to have fixed points? No. Thank you. I posted a paper yesterday morning on the archive where I do allow my involutions to have fixed points. But we're still in 1960. So we don't have, not right now. What would it mean for the involution to have fixed points? Did you have some edges that were under-represented by a single edge instead of two? It means that you'd have leaves on your graph. Let's not go into that right now. There's a slightly different combinatorial notion of what a graph is, which involves half edges. And yeah. Right. Let's not get distracted. Yes. So here's a little lemma. Supposing you have a graph morphism. What, how do I want to say? f is not locally injective. This is really more of an observation than a lemma. Then there are two edges with the same initial point. f of e1 is equal to f of e2. In other words, the picture is, if you have this graph morphism, x to y, and you've got, then it's not locally injective. That means there's some place, single vertex, two edges, initial and yeah, e1 is an e2 bar. So e1 and the map sends both of these to the same edge of y. As I said, this is more of an observation than a lemma. Why do you have e1 isn't e2 bar instead of e1 isn't e2? No, I guess you're right. I did want to rule out loops, but maybe the lemma is still true. I think you're right. I'm going to leave it like that. So the point is that if you've got a graph morphism, it sends edges to edges. All I've done is describe it combinatorially. But if you want to think about this topological, an edge gets mapped to an edge, you can think of that as just a homeomorphism. So it's certainly locally injective on the edges, on the interior points of edges. The only place it has a chance of not being locally injective is at vertices. And not locally injective means that, well, in wherever this vertex goes, there have to be, if all these edges go to different places, then it's locally injective. So there must be two of these guys that are going the same place. That's all. It's just an observation, OK? Yes? Yes. My graphs could have loops or multiple edges. That's fine. If, right. OK, so what's a stalling's fold? Let's see. 10, 0, 8. I've got five minutes to do this. No. I have seven minutes to do this. Good. So what's a stalling's fold? E1 equals I of E2, yeah. And here I want to say E1, E2 is not equal to E1 or E1 bar. Then the graph morphism that you get by taking E1 and E2, and identifying E1 and E2. So this is x. This is, you get a new graph, y, is called a stalling's fold. So actually what I did up there is a stalling's fold. I took an edge, I had the second half of the AI edge, and I folded around the EIJ edge. So my picture, so all I have that E1 and E2 are different edges at this vertex. But there's nothing to say where the end of E2 is. That could be E1, E2. That's the example of what I did there. So that's called a stalling's fold, too. All you need is E1 is not equal to E2. Yeah, there are degenerate stalling's folds. It could be that E1 and E2 have the same endpoint. Then when you do this fold, you're going to get that. So a stalling's fold is called degenerate if it's not a homotopy equivalence. So I killed a loop here. Can't do that. That's degenerate. So these are the only two non-degenerate pictures. You fold an edge. You just have two edges. They end up at different points. You fold them together. Well, two edges, they end up at different points and you fold them together. I guess they couldn't both be loops, because then they would both end at the same point. So if they end at different points, then you can fold them together. So now how do we prove this theorem? Here's the theorem. Now the problem is, oh yeah, there it is. You can't find this. That's the theorem I'm trying to prove. I want that one to go up and this one to come down. So I'm going to start with an automorphism. So phi of a i is some word, w i. And I'm going to draw a graph that I want to represent. I want to draw a morphism of graphs that I want to represent this word. So represent this automorphism. So I just take the i-th loop and I divide it into pieces and spell out the word w i. So I guess as I said, over there I can model this whole graph morphism by a single picture where I take this guy. But then I'm implicitly telling you that I'm identifying the petals over here with the generator's a i. But anyway, there's my phi. Now if phi is locally, so the lemma, if phi is locally injective, then phi is a homeomorphism. I guess I'm calling phi both the map. I should call it something else. If f is locally injective, then f is a homeomorphism. So homeomorphism can't really do very much to this picture except permute the petals and maybe flip them. So that implies that phi is generated, is a product of permutations and inversions. So that's fine. And then if phi is not locally injective, if f is not locally injective, you can fold. That's that lemma up there. So there's some fold phi 1. I should call that. I shouldn't call it. I use phi all the time. Let's see. Fold, variation of phi. OK, so that means that there's two guys here. Somewhere that get mapped to the same edge over here. Maybe E1 and E2 get mapped to E, something like that. So I can fold E1 and E2 together, and I get that. Since E1 and E2 get mapped to the same place, my map factors like that, OK? And if it's not locally injective, I can fold again. I get something else. Who knows what I get? And I can keep doing that until, so I notice that the number of edges in my graph has actually decreased when I folded, because I folded two edges together. So I can keep doing this until finally I get to some graph over here where this map is locally injective. So actually, this works no matter what the graph doesn't have to be a rose for this to be true. It's still true. If I got a locally injective homotopy equivalence from this to a rose, that's a homeomorphism. So I'm really, I got a rose over here. So now I've factored my original map, F, into a bunch of folds like this. The last one's a homeomorphism. That just permutes and flips the leaves. So that's a product of permutations and flips. What about a fold like this? If I collapse that tree, I get a rose again, and I get a homotopy equivalence of the rose, that didn't actually do anything on Pi 1, nothing at all. So when I look on Pi 1, this is the identity. The only time I get a map that isn't the identity is one of my folds is that kind of a fold, where I'm folding a single edge around a loop. In which case, I get a product of, I get a right, I get one of these row ij's or lambda ij's. I may get more than one, but so a fold induces the identity on Pi 1, a fold induces a product of row ij's and left ij's. So when you look all together when I'm done, I started with this map. I factored it into a product of maps. If I contract the appropriate maximal tree in each graph I've got to along the way, every time I'll either get, I'll do absolutely nothing to the fundamental group, or I'll do one of these product of row ij's and lambda ij's. So I will have factored on the level of fundamental groups. I will have factored my automorphism of the free group into a product of identities and lambda ij's and row ij's with some permutations and flips at the end. So that finishes the proof of the theorem. So that's probably a good place to stop.