 OK, thank you. So what have we done so far? We've defined outer space. That was for n, at least, 2. And this is a picture of it for n equals 2. So we defined it in three different ways, right? And we also defined the variance O and S. I defined those in two ways. So I'm only going to think really about these two ways. So what did I do? I defined it as a sphere complex, a part of a subspace of a sphere complex. And I defined it as a space of graphs with leaves, if S is positive. So that's what we've done so far. So in particular, in terms of sphere complexes, we had this manifold M and S, which is a connected sum of S1 cross S2 minus S balls. The sphere complex of M and S is defined the way you define the curve complex. It's isotopic classes of spheres. Complex of isotopic classes of spheres. And so let me just, I'm going to talk about this model for a while. Oh, sorry. And then O and S. I'm just reminding you of terminology and notation at the moment. O and S was a subspace consisting of basically of complete sphere systems. Now, this is the complete ones. Incomplete ones were called S infinity. So I'm not going to redefine all these things. I'm just reminding you, since you've had other things to think about over the weekend of what we called all these things. But I will show you, remind you in this picture. So here we have for n equals 2, here's a simplicial complex. It's got some infinite valent vertices and some bivalent vertices. So A here and B are each vertex is a single sphere in the manifold. So there's two different pictures of my doubled handle body. There's this kind where I've drawn just the handle body without doubling it yet. I should write times 2. And that's the sphere A. That's the sphere B. They form an edge because they're disjoint. And this is the picture where I've taken this picture, opened it up, so I get a ball with four patches on it. And then when I double, I get a three sphere with four two spheres on its boundary. And so since I've cut open along those spheres, each of them corresponds to two spheres in this picture. A, A bar, B, and B bar. And then where's the sphere C? Well, here's a sphere C. I need a sphere that makes the three simplex with this so I could use this for C. But there's another sphere I could add to this picture. Namely, yeah, let's not use that one. Let's use this one. I'll again draw only half of it. So I don't know whether you can see that. It kind of, here's a disk. This part goes in the back of the picture and this part is in the front of the picture. It's a little bit hard to see in this picture that C prime. In this picture, it's kind of easier to see. I should have drawn C first. This is C. It goes around A and B. And where's C prime? Well, it goes around A and B bar. So that's the picture over here. And then there are these little, these guys here, these bivalent vertices. I need orange chalk. So they correspond. So there's one other sphere in this picture I can draw. In this picture, it looks, well, yeah, OK. There it is. That's X. And in this picture, it looks like that. So depending on which kind of picture you like to draw, it's easier to see the spheres maybe in this picture than it is in that picture. I don't know. So that's a picture for n equals 2. It's hard to draw these pictures, right? But we know, especially after we assiduously did the exercises, that there's this dual picture. Instead of a sphere system, I could draw a graph. But the downside of that is I have to mark the graph. I have to mark the edges that aren't in a maximal tree with a basis. But it is easier to draw a graph than a handle body. And I could mark these like that and these like that. And if I, yeah, so the edge here is marked like that. And the orange guy looks like, well, maybe I should make it an orange, looks like that. So you were supposed to draw this picture as part of your exercises. So if you didn't, I just did it for you. Anyway, that's outer space in rank 2. And you can do that in any dimensions. So I wanted to make some remarks about these complexes, these sphere complexes. Let's start with no punctures. So in the literature, people have studied this often using the graph or tree model of outer space. And it's also called in the literature the free splitting complex of the free group. And why is that? It's because if I have a sphere, if I take any single sphere, it either splits the manifold into two pieces. And Van Kampen's theorem says that the fundamental group of the whole thing is a fundamental group of that, pi 1 of the left, amalgamated along the trivial group, which is the fundamental group of a sphere, pi 1 of the right-hand side. So that's the splitting of the free group. If I used a non-separating sphere, this is using the sphere x. If I use a non-separating sphere, I get a representation of the free group as pi 1 of the remainder, m minus s, as an h and n extension. And anyway, in either case, it's a splitting of the free group into pieces. So that's why it's called the free splitting complex. It's a little bit more complicated to define if you do this. I think this is pretty clear what it is. But people have worked with that. And in particular, the big result on that is in 2012, Handel and Mosher proved that S of m is Gromov hyperbolic. So they did that by thinking of it in terms of these free splitting, in terms of graphs and trees, in 2013, Ilyan and Horbez gave a proof using spheres. So it doesn't deviate much from this proof. But I think it's easier to understand, because it's got the definition of the complex is simple. And well, I'm obviously biased. I like sphere complexes. So this is a delta hyperbolic space. So people often think of studying outer automorphisms of free groups as they want to imitate things that have been done for mapping class groups. Well, one of the big results in mapping class groups is that the curve complex is delta hyperbolic. And this gives you an analog for out of fn. The sphere complex is delta hyperbolic. So that was remarks on for n equals 0. I also wanted to make remarks about, well, on S for S bigger than 0. First of all, when I defined outer space and rank n, I only talked about n at least 2. I did write it up there. But for n equals 1, it already makes sense for n equals 1 as long as S is positive. Yeah, I already said that. Because what is the handle body in this case? Well, it's a handle body of genus 1. That's just S1 cross S2, a single doubled solids torus. And I've got a hole in it. So there are spheres, non-trivial spheres. There are spheres that are not homotopic to the boundary. There's one. And yeah, I could leave. So I think I asked you to think a little bit about this space. It also makes sense for n equals 0 if S is bigger than or equal to 3. Well, let's make 4. If S is at least 4. So what is the doubled handle body if n equals 0? Well, you can think of a handle body as a three sphere with some handles attached. You attach n handles. If n is 0, you don't attach any handles. So m0 is just the three sphere. And so if I have a three sphere and I take at least three balls out of it, I wanted to take four balls. If I just take three balls out of it, then any sphere I try to draw in, this is going to be parallel to one of the boundary components. But if I take four balls out, then there are interesting non-trivial spheres. So this makes sense. As long as I've got at least four, if I remove at least four balls, I have an O is 0 S. So there is one subtlety about when I'm talking about the sphere complex of m and S, which I guess I already alluded to when I was drawing this picture for S bigger than 0. I don't use spheres. There's another kind of trivial sphere. I said before that trivial spheres are spheres that don't bound balls. But I also want to think of a sphere as trivial if it's parallel to a boundary component. So I remember in terms of graphs, supposing I have, let me do a surface of genus 2, if I've got at least one hole in it. In terms of, so a sphere system is dual to this. Here's an example of a sphere system. If I didn't have this hole, then this sphere would be the same as that sphere. But I got a hole there. It blocks that. So I'm a little bit right. There's an example of a sphere system. And dual to this sphere system is a graph. It's got three components, one, two, four edges. And remember what I did with this boundary component was I drew an extra edge from the boundary component to the graph. Now that boundary component doesn't have a sphere going through it. If you remember, I thought of a point in this space as a point in the middle of a simplex, which I gave barocentric coordinates to. So I'm thinking of this as a sphere system with weights on it, but the leaves don't have weights. So the leaves don't have lengths, just the internal edges. So in fact, other people are studying this sphere complex. Well, more precisely, they're looking at the quotient. Oh, yeah, I should have mentioned. We got these sphere complexes. And we also have these groups, A and S. Well, it's diffeomorphisms of this guy up to or homeomorphisms, diffeomorphisms, whichever, modulo-dane twists. Or it's also homotopy equivalences of a graph that looks like that. So those are the groups A and S. They act on these spaces. If S equals 1, this is a group of automorphisms of a free group. If S equals 0, it's the group of outer automorphisms. And in general, it's a group. And if I look at the quotient of this complex, the whole complex, the whole sphere complex, this is a space that other people have been studying for other reasons. Yeah. Well, let's take N at least 1 is called the modular space of tropical curves. So I guess what I'm doing at the beginning is telling you different names for these spaces and the ways different people study them. So algebraic geometers are interested in this as a kind of degeneration of the modular space of Riemann surfaces. It can be seen as living at infinity of the modular space. And in fact, it's isomorphic, homeomorphic, to the curve complex of the surface of genus, the appropriate genus, whatever the appropriate genus is, modulo A, I guess it's 2g, something like that. So this thing is definitely not the same as the curve complex, the sphere complex. However, when you mod out by the group action, it actually becomes the same. So it's the same space. So that's the modular space of tropical curves. If N equals 0, well, then what is this group? If N equals 0, then a graph is just, there's my graph, S. And I'm looking at homotopy equivalences of that graph that fix the leaves. So there aren't too many up to homotopy. So I'm looking at the whole sphere complex, M0S. The tropical geometers call that the tropical grass manian. Actually, it's not very hard to prove that this thing, we know pretty much everything you could possibly want to know about this. There's a symmetric group that permutes the leaves. So the homology of this space is a representation of the symmetric group, and we know what it is. In fact, we know the homology of this space. We know, in fact, it's only got homology in one dimension. And in fact, the whole thing is homotopy equivalent to a wedge of spheres of dimension S minus 4. So that space is well understood. OK, so this is outer space. I wanted to find, in the first lecture today, I want to define a couple of related spaces that are pretty easy to define. And then in the second hour, I'm finally going to tell you. So I've defined all these really pretty spaces, which I just like because they're pretty. But it turns out they're actually good for something, too. And so that's what I'm going to do in the second hour is kind of tell you how we've used these spaces to prove things about the groups. But I still have some more pictures to show you. So that's what we're doing. OK, so we were remarked last time. So what? This is the sphere complex I've drawn a picture of here. The outer space in this picture is just interiors of the simplices that aren't orange. All these guys are orange. I'm going to make them all orange. Yeah, I guess I shouldn't have made these orange. So it's just the interiors of the two simplices. It doesn't include any of the vertices because if I just have a single sphere, the complement isn't simply connected. And actually, if I have a and x, its complement isn't simply connected either. So the line from a to x is not part of outer space either. And certainly x isn't. So a single sphere or a pair of spheres, one of which is separating, is actually not in outer space. And that has the unfortunate consequence that when I take the quotient by the group, it's not compact. And so in particular, this space is not quasi-isometric to the group I'm studying. And as a geometric group theorist, I don't like that. If I add in the orange stuff, then I do get something that's compact. The quotient is compact. But the stabilizers aren't finite. And as a geometric group theorist, I don't like that either. So how do I fix that? It's almost compact. You know, there's just these stupid stuff at infinity that shouldn't be there. So what I'm going to do is just throw that stuff away. And the right way to do that, this is only true, of course, for n, at least 2. I'm going to take the barocentric subdivision, ns. OK, so let me do that. Do this in two stages. I'll take the barocentric subdivision. It'll start to look kind of messy in my picture. It'll definitely start to look messy in my picture. And I can keep going, but you get the idea, right? I divide every triangle into six triangles. And now what I do is throw it. So I now have a simplicial complex, which is a refinement of the simplicial complex I started with. And now I'm going to throw out all of the vertices that aren't in outer space by simplices n, o, and s. So in this case, let me look at this triangle here. So I'm going to throw out all of these vertices. So in this triangle, let me not, this picture's too messy. Let me just draw it over here. I've got a, c, and b. And I've taken the barocentric subdivision. And this vertex is not in outer space. This isn't, and this isn't. So the only vertices that are are that vertex, that vertex, and that vertex, and that vertex. And so I'm just going to take the sub-complex that's spanned by those vertices. Yeah? You don't want that not there? Or is k and s the ones that you're removing? Or am I just going to? In. Yes. Thank you. Simplices that are actually in outer space, yes. So that's what it looks like here. On the other hand, the simplex a, b, and x, well, x isn't in outer space, a isn't in outer space, and b isn't in outer space. But on the other hand, the simplex spanned by, so remember, what is a complex in the barocentric subdivision? So simplex is a chain, or vertex rather, is a chain, s not contained in sk. That's a k simplex, yeah, vertex. Right. So this is a, this is a and c, and ac is in outer space, a, c, b is in outer space, a, b is, c, b is. So I have those vertices. Here I have a, b is in outer space, a, b, x is in outer space, but nothing else is. x, b isn't, that doesn't cut my thing into simply connected pieces, so it isn't, and all I've got is this. OK, that's the spine of outer space. It's the spine KNS is the spine of ONS, and it doesn't contain any simplices that aren't allowed. It lives inside of ONS and misses stuff at infinity, yeah, misses bad systems. So it misses bad systems, so in particular, it's KNS mod ONS is compact, is a, no, so the action, what am I trying to say, the action is proper. The only place I got into trouble in this sphere picture is when I had a vertex here or a one cell where the associated sphere system had a non-simply connected complement, and then the stabilizer of that system was infinite. But as long as I've gotten rid of all those simplices, then the stabilizers are all finite, and the action is proper. And and the KNS mod ANS is compact. So it's everything I want. ANS, it's got, I now have, oh yeah, well, I haven't said quite everything I need to say. I also need to say and ONS deformation retracts onto KNS. And that's easy to see, in particular in this picture. In the bare centric subdivision, I have some faces, some simplices. And these simplices have one face that's in KNS and one face that's not. Every simplex has at least some face that's not because there's a vertex. None of the vertices are ever in outer space. So I can just linearly retract from the face that isn't in outer space to the face that is. Here too, here most of the simplex retracts onto the vertex that is in outer space. And of course, the bare center of every maximal vertex is in outer space because, yeah, that's how I defined outer space. Simplex is a sphere system that's complete. Question? Yeah? Does the dimension always go down exactly like I want? The simplisher? No. This is an exercise on your sheet. One, two, I've got another board somewhere. Yes, it's an exercise. Well, you already did the exercise computing the dimension of ONS. We had dimension of ONS is 3n minus 4 plus s. That was if n's at least 2 to get n's at least 2. And the dimension of KNS is 2n minus 3 plus s if n is at least 2. And the answer is slightly different if n equals 0 or 1. And that's also part of the exercise. Right. So that says KNS is contractable. It's this says nothing. And that says the action's proper and the quotient's compact. So let's see. Right. So I want to understand a little more about the space KNS. It's a space that's, as we'll see next hour, is used a lot to study the group. So let me first talk about the local structure. So as I've defined it, KNS is a simplicial complex. It's a sub-complex of the sphere complex that's, yeah. I was going to draw it in this picture. But this picture's too messy. But I'll draw it anyway. I'll get rid of all of the extraneous stuff and just keep the spine. So what does a spine look like? Well, we saw it looks like that in these simplices, like that in those simplices. So it looks like a trivalent graph with little spikes sticking up out of it. I guess I'm not quite done. I should have drawn it in a different color. That's why it's called the spine as it looks kind of spine-like. I should have drawn it in a different color. OK, so what does KNS look like? Well, what's a simplex in KNS? First of all, it's in the barycentric subdivision. So it's a chain of subgraphs of systems. S0 contained in S1 contained in SK. This is a K-simplex. Each of these systems gives you a simplex in the original space. And what's a simplex in the barycentric subdivision? It's a chain of simplices. So that's a chain of simplices of systems in the, you know, on my iPad I could do this and just erase the stuff I didn't want and just leave the stuff I do want. It's really nice, but I don't have my iPad. It's much worse on the chalkboard. OK, so for instance, right. So this vertex here corresponds to the system ABC. That's that vertex, is that system ABC. This vertex is a system AB. So we have AB contained in ABC. That's this one simplex in the spine. So a simplex in this barycentric subdivision is a chain of systems. And they're all complete. So an interesting thing happens when you think of the dual graph. What happens if I look at the dual graph to this system? It's got three edges. What happens when I remove a sphere? Well, if I remove the middle sphere, this is A, B, and C. It collapses to the edge corresponding to that sphere, collapses to a point. I may be collapsing more than. So as I go down this chain, I'm throwing out spheres, and that means I'm collapsing graphs. Chain of collapses, G. Let me call it GK. But I'm not allowed to collapse any subgraph. I can only collapse. I have to keep the fundamental group Fn. And that means I can only collapse. C collapses forests. A forest is a subgraph with no cycles. So that's one way to describe. There's two ways to describe a simplex in the spine. One is a chain of sphere systems. The other is a chain of forest collapses. Of course, this isn't enough to say what the point in outer space is. These guys come with markings. And in order to be a simplex, the marking has to get dragged along with the collapses. So that has to commute. OK, so that's what a simplex looks like. What does the link of a vertex look like? Let's start with a trivalent graph, or in other words, a maximal system where the thing is cut into pairs of three punctured spheres corresponds to a graph, which is trivalent. To go down, I collapse an edge, or another edge, and another edge, and another edge, or maybe I could collapse the whole forest at once. So the link of G is graphs obtainable by collapsing a forest. So I can measure this by the geometric realization of the partially ordered set of forests in G. Once I've got a trivalent graph that's like a maximal system I can't stick anymore spheres in. So those are the only kinds of things in the link. So this is just maybe a fancy way of saying this. You look at all the graphs you could get by collapsing a forest. You get one for every forest. So that's a vertex in the link. And so then the link that you get is called the geometric realization of this partially ordered set of forests. And it's OK to erase this. What about the link of a, yeah. So what's the smallest dimensional simplex I have here? It's one where I have the minimal number of spheres I need to cut this into. So the complement's simply connected. If I have genus G, if I have genus N, if I take N spheres, I can cut it into a single simply connected thing. And the dual graph is going to be a rose. So we know what the link of a vertex is if the graph is trivalent. To think about the link of a vertex, a rose, it's easier to think in terms of sphere systems. So a rose is a minimal complete system. So if I have a minimal complete system, it cuts M, let me call it S, not because it's very small, M minus S not if I cut it open. I have a1, a1 bar, a2, a2 bar, N bar. Well, if I have a minimal complete system, I'm just looking at spheres I can add to it. So that's just the set of spheres I can put in this picture. This is an M0 to N. That's a ball with two N holes in it. So that's M0 to N. And I'm looking at spheres I can put in it. So the link, so this manifold is that. So its link of S0 is isomorphic to M0 to N. So that was easy. And by what I said, if I didn't already erase it, I probably already did. We know exactly everything about this. This is homotopy equivalent to a wedge of spheres of dimension 2N minus 4. And what in general, if, yes, and yeah, no, here it is. There's the SM0S, which of spheres of dimension S minus 4. And what if I'm in between if G is not trivalent rows? And that's saying that the sphere system is not minimal, not maximal. Right, I'm just talking about a vertex, sorry. This is not trivalent non-row. So that means that's not minimal and not maximal. So I can look at M minus S. And I get a MNS minus S. And I get every sphere that I could, everything I could add to this sphere system also is part of the link. Gives me a bigger system. On the other hand, everything I could collapse in G gives me a smaller system. So if I kind of look at everything, yeah. So I'm looking at everything I could get by collapsing a forest. I'm looking at everything I could get by adding spheres to S. I've kind of mixed the two pictures of what these spaces are. But the link is going to be the joint, the simplicial joint, of these two complexes. So there's a theorem that says this post set of forests is spherical, too. I'm not going to guess at the dimension. Number of vertices minus 4, yeah, some dimension. And so that implies that the link of any vertex is homotopy equivalent to a wedge of spheres. If I take a wedge of spheres and take the simplicial joint with another wedge of spheres, I'm going to get a wedge of spheres. So yeah, let me point out in this picture, the space I have is not a manifold. It's got these places where it's got, yeah, that's not a manifold point. But on the other hand, the link of a vertex and the spine, so the spine isn't a manifold either. The whole space isn't a manifold. The spine isn't a manifold. In a manifold, the link of a ver, if you triangulated your manifold, the link of a vertex would be a sphere. In this space, the link of the vertex is not a sphere, but it's not so bad. It's homotopy equivalent to a wedge of spheres of the correct dimension. In fact, the dimension is the dimension of the spine minus 1. So that's, oh yeah, I've got 10 minutes. That's a little bit about what this space looks like. At least locally, it kind of looks sort of manifold-ish, even though it's definitely not a manifold. I wanted to tell you about one other space before I tell you what we use all this stuff for, because I really like this space. And also one of my co-authors is sitting in the audience. And it has the good effect of simplifying that picture. So let me call one more space. It's called the boardification. So this is another way of getting something which is proper and co-compact on which the group acts. So it's another way of getting rid of these nasty simplices that are out at infinity. The first thing we can do is get rid of some of them by only using spheres that don't separate MNS. So define O and S reduced to be the subspace. Well, I guess I could do it for the whole sphere complex. It is a sub-complex spanned by non-separating spheres. So I'm not going to use these silly x's, because they separate my manifold. So instead I'm just going to not use them. And then my space looks much nicer. All the orange stuff disappears. And again, this would be a whole lot easier with my iPad. OK, so what I did before is I kind of went on the inside of this space way down to, you know, I took the bare centric subdivision and threw out the bad simplices, bad vertices. Now instead of throwing out the bad vertices, I'm just going to cut them off. So I just want to cut off the bad vertices. You know, throw this part away. So how do I make that precise? So given a graph, a marked graph or a sphere system, S contained in S reduced. So this is right. So this corresponds to a simplex. Yeah, so this is a sphere system. It gives me a whole simplex sigma of S. So for instance here, I've got three spheres. That's a sphere system. That's a two simplex. I'm going to define J of S. It's going to be something sitting inside of sigma of S. And I'm going to get it by slicing off the faces of sigma of S that are opposite what I'm going to call core subgraphs. So what's a core subgraph? It's too hard to get the top one. I'm just using two ports. S0 is contained in S. That's a complete system. So this is core. Let S1 be the other spheres. So S0 is core. If the spheres of S0 are non-separating in M minus S1. So for example, my simple example, over and over and over again, A, B, and C. That's my sphere system S. If I look at just the sphere system A, then what's left over is B and C. And if I cut the manifold open along B and C, it looks like A becomes separating. So this is not core. On the other hand, if I started with A and B, the rest of it is C. And if I just take out C, then where were we? A and B. They're not separating. So this is core. I have one minute to tell you what to do. What I'm going to do is take all of these simplices corresponding to some sphere system S. And I'm going to slice off faces opposite core subgraphs. So in A, B, and C, A, C is core. So I'm going to slice off B. A, B is core. So I'm going to slice that off. And B, C is core. So I'm going to slice that off. And what I'm left with is just that. That's called J of S. In this picture over here, I'm just slicing off the stuff that I don't like in every simplex. So I'm throwing this stuff all away. And I'm left with just that. And that's co-compact. And let me just tell you what's true about this space. So if I slice off carefully, these things all fit together to give me a deformation retract of my space, I should say slice deeper if S not is smaller. So if I'm going to slice off this face and this face, I want to slice deeper on this face. So I make sure I make a new face every time I slice. The resulting space is homeomorphic to the Bestvena Fane Bordification. One more space. It's this space. It's homeomorphic to the Bestvena Fane Bordification of ON. That's for S equals 0. And that's in a paper by Kayuva Bukes, Peter Smiley, and myself, the fact that it's homeomorphic. Bestvena Fane made this Bordification in 2000. This paper is 2018. And I'm over time. So I'll tell you what you use this. Well, I'll tell you what you use all these spaces for in the next hour. But I don't know. I hope you like the construction of these spaces. I think they're pretty.