 So I wanted to make a few comments about last time. First comment was about Dane twists. So I said, you just suspend a Dane twist in an annulus. And yeah, you can think of it like that. But I prefer to think of it the way I originally drew it. So we have two spheres, concentric spheres, with S2 cross i in between. And the outside sphere is fixed. The inside sphere rotates by 360 degrees. And all the intermediate spheres, so first of all, you pick an axis, sorry. And first of all, and all the intermediate spheres rotate by some amount between 0 and 360, maybe 200, whatever. And I like that picture, because I think if you think of the suspension picture, well, it doesn't look very smooth at the cone point. But this, on the other hand, is definitely smooth. And to get you to think about this, I put this as the first exercise for today. So somebody asked in class, I think it was Joe, what the horizontal slices looked like. So you can figure out what the horizontal slices look like for your first exercise. So that was the first comment. The second comment is about why the cut vertex lemma gives you an algorithm for deciding whether something's a basis. And I'm just making this comment, because the picture I drew at the end of class last time wasn't really the right picture. So what's the point? So that the star graph having a cut vertex, so what is the star graph? The star graph is the union of the images of all of the generating loops. I should have written it in orange. OK, and so there's this big graph that you make. And suppose it's got a cut vertex, A. The point of having a cut vertex is I can draw two other spheres, one on this side and one on this side. And both of those spheres intersect the star graph in fewer points. The sphere I drew in class also had that feature, but it wasn't one of these spheres. It was just a random sphere that happened to intersect the star graph in fewer points. But the fact that there's two spheres, if you look at A is only one half of the sphere that I cut open on, so A bar is somewhere else. It's in one of these spheres, say S1. So the fact that S1 separates A bar from A means that S1 is a non-separating sphere in the three manifold. So you use S1 separating A from A bar. Is there anything else? S2. S2. Thank you. Yeah, use S2. Separate A from A bar. And what do I mean use S2? If you remember, that means switch S2 with A. So you have a new sphere system also, and separating spheres. It also cuts the thing into a ball, a punctured ball. And then the other question somebody asked was, well, so I use S2. What happened to an edge that was over here? I guess, right. So this edge doesn't intersect the new sphere at all. But think about this edge. This edge goes into A, so it continues out A bar. If I glue A, so I glue A and A bar back together, then I've cut open along S2 instead of A. So this edge that was coming from B to A keeps going out A bar and goes somewhere else to C. So in the new picture, this edge doesn't intersect S2 at all, but it does intersect C. So I just don't have that. So anyway, that's what happens to the edge. So I just wanted to answer the questions that I didn't answer last time. OK, right. So what about today? What are we going to do? So we're thinking about two models of the free group. I'm calling the rows this thing. And the doubled handle body I'm calling MN. And we know how to tell whether a map from the free group to itself is an automorphism. That's Whitehead's algorithm. And we also, if it's an automorphism, we know how to factor it into a product of elementary automorphisms. So they were the permutations, inversions of a generator, rho ij's and lambda ij's. So that's what we did last time. So what we're going to do today is start thinking about the entire group of automorphisms out of Fn. Or out of Fn, today actually I'll be thinking mostly about outer automorphisms. So this is pi 0, homotopy equivalences of Rn. And I'm not going to stick in the base point. If I wanted to think about automorphisms, I would be thinking about pi 0, homotopy equivalences of Rn that preserve a base point. And we also saw that this was pi 0 of homeomorphisms of Mn, modulo-dane twists. Yeah, pi 0 of homeo of Mn with a base point. Again, modulo-dane twists. OK, so we want to think about these groups. So how do you think about a group? This is a conference about geometric group theory. So geometric group theorists like to have a space with an action. Let's do out of Fn. And what they require of this space is that the action be free or almost free. In other words, they want it to be proper, meaning that point stabilizers are finite. We're thinking of these as discrete groups. They're discrete groups. And co-compact. So that's what you want to do to use geometry to study the geometric group theory. On the other hand, you might be an algebraic topologist. And there, what you want is a space x. Space x isn't good enough. You do still want x to have a proper action, but you want a contractable space x with a proper action. And actually, you don't really care whether the action's co-compact or not. The reason is that topological invariance of this quotient space are topological invariance of the group, like cohomology, Euler characteristic, things like that. Here, quasi-isometry invariance of this space are quasi-isometry invariance of the group. So you may be interested in. So you want this for more geometric applications. You want this for more topological applications, algebraic topological applications. So we're going to get a space x. We're studying the group of outer automorphism. So we'll start with outer space. And outer space will satisfy the second one. We'll make algebraic topologists happy, but not geometric group theorists necessarily, not yet. Satisfying algebraic topologists. And but after we get this space, it turns out there are closely related space. Let me give outer space a name other than x. Actually, this is how outer space got its name. I was giving a talk many, many years ago about outer space. And I was calling it x. And Peter Shailin yelled out from the audience, you should call it outer space. And it was a really good suggestion, because everybody is hurt, even if they don't know what it is. Outer space, that sounds cool. OK, outer space, making algebraic topologists happy. But x has closely related spaces. One's called, I guess I should write. I was going to give it a name. How about on? It has closely related spaces. One's called k and that's called the spine of outer space. Then there's also on hat called the boardification, outer space. And both of these make everybody happy. They're both contractable. They're still contractable. The action's still propertor. And the action is co-compact. But first, we have to describe this space, outer space. So we define free groups in terms of graphs and in terms of three manifolds. Right. This is the third time. Yeah? When does it have a metric? This is true. But if. So we're not that happy. Well, it does have a metric, of course. It's got a non-symmetric metric. Anyway, this one will be a simplicial complex. So you can just make every simplex a regular simplex. And then you have a metric space. And then it's quasi-symmetric. You happy? No. No, you're happy. Good. We've talked about graphs. We've talked about this three manifold. Actually, the quickest definition of outer space doesn't use either of those ideas of a free group. What else is a free group? Uses the fact that a group is free if and only if it acts freely on a simplicial tree. So a free group is definitely free. So and it acts on trees. Right. So what do I mean? So for me, a tree is a one-dimensional simplicial complex. My trees, I'm not looking at real trees at the moment. I'm only looking at simplicial trees. So I won't bother to say simplicial all the time. For me, a tree is just a one-dimensional CW complex. And it's contractible. And what else? Action. The kind of actions I'm going to look at, these are going to be simplicial actions. Or in other words, they're going to be graph morphisms the way I defined them last time. So they send vertices to vertices and edges to edges. And what does it mean to be a free action? Every group element moves every vertex, well, every point. So one way to get an action of a free group on a tree is take a graph with the fundamental group, a free group, and look at the universal cover. So this is G. This is T is G tilde. And then the fundamental group of G acts on G on the universal cover, which is a tree by deck transformations. Right? So to cut down this, so we're going to make a space out of these actions. And to cut down the size of this space, I'm going to make some assumptions. I'm only going to use minimal actions. In other words, I'm not going to allow any invariant subtrees. I'm also not going to, in this picture, when I take a graph and I look at its universal cover, and I think about the action of the group on this universal cover, I don't really care where the, I don't want to care if I sprinkle bivalent vertices in here. That doesn't really change the basic picture, the tree or the action. So actually, this can be technically phrased so that it automatically does this, but I'm just going to say it for emphasis and no bivalent vertices. So I'm going to use, I want to make a space. Now I want to make a space of such things. All I've done is shown you an example. I've shown you how to get lots of examples. Take any graph with the right fundamental group. Look at its universal cover. We have an action on a tree, but I want to make a space of such things. So in order to make a space, I'm going to make these trees metric trees. I want, in order to have, I want the action to be by isometries. So I'm going to make the metric invariant under the action. So now I've got points, and at least I can sort of imagine what a nearby point would look like. I could just change the edge lengths a tiny little bit, and I should get a nearby tree with the same action. But to make that precise, I should say the set of these things makes a space. To make a space, put a metric, and then use the equivariant Gromov-Hausdorff topology. So I'm assuming that not everybody knows what that is. Is this true? OK. So what's the equivariant Gromov-Hausdorff topology if x and x prime are metric spaces, then x is close to x prime. For every finite set in here, you can find a corresponding finite set in here, where the distances are within epsilon of each other. If for every finite s and x, there exists an s prime and x prime, also a finite set, and a correspondence by ejection s to s prime, so that the distances between points and s are within epsilon of the corresponding distances of points and s prime. I'm just going to write it like that. So that's the regular Gromov topology on metric spaces. But we don't just have metric spaces, we have metric spaces with actions. So to include the action, so this is the Gromov-Hausdorff topology on metric spaces. How do you make this equivariant? Well, x with its action and x prime with its action are close if, well, you need finite sets in x and finite sets in x prime. And you also need finite sets of group elements in x and x prime. And you want if for every finite s, s and x, there's s prime and x prime for finite s and finite a and g. There exists s prime in x prime and by ejection. And what do I want? I want the distance between two points s1, not just the distance between s1 and s2 to be approximately the distance between s1 prime and s2 prime. I want that. But I also want if I move s2 by a group element, it's still true. This is supposed to be true for all g and a. So basically, if you understand this, then this isn't much harder. You just throw in the group action to make sure the group action does approximately the same things to s and s prime. So since I now have actions on metric spaces, there is an echo variant topology and right. So now I could define definition one. Is there any way to draw what invariant supreme would look like? What an invariance. This is one of your exercises. For one thing you could do is supposing you had a perfectly nice action on a tree and you put a little spike in the middle of each edge of every edge. So the group still acts on this tree, but these spikes are completely unnecessary because the tree without the spikes is an invariant subtree. Definition one is the space free actions of fn on metric simple trees. So that's a nice definition. It has the advantage of being succinct. It doesn't really tell you a lot about what the space looks like. I say it has two advantages. First of all, it's very short. Someone asks you at outer spaces and you don't really want to talk to them. You say, that's the action of fn on the other hand. I'm sorry. Yeah, it also has many other advantages, one of which is that it's easy to generalize to other situations. Like if you're looking at if your group is not the free group, but a free product of groups, you can make similar spaces. And a lot has been done with that. But anyway, that's the first definition, right? But I want to go to the second definition, which was actually historically the first definition, which is using graphs. Oh yeah, so I didn't actually tell you what the action is. Maybe I should tell you what the action is. Maybe I'll put it here. The action of fn on on as described. So what's an element of on? It's an action on a tree, which I could think of as a homomorphism from the free group to the group of isometries of a tree. That's of, yeah, sorry, no, an action of a free group. So a point in this space is an action. So this is a point in outer space. It's an action of a free group on a tree. So how does the group of outer automorphisms act? Well, if I have an outer automorphism, I can lift it to an actual automorphism. And then I could just compose, this was rho. I could just compose fn with this actual automorphism. And I'll get, this is phi hat, from fn to fn. And I'll get a new composition, a new map from fn to the isometries of t, rho composed with phi hat. So that's a new point in my space. It's the same tree, but it's a different action on the tree. Because I've twisted the action by this automorphism. So that's how f, I'm sorry, yes, you're absolutely right. I'm asleep. Action out of fn on on, right. Do you have fn twiddles? No. No, I'm sorry. This is b, yeah. Right, a point in outer space is an action, which is a map into isometries, from the group, the free group to the isometries. And if I change this action by an automorphism, I get a new action. And one of your exercises is to show that, oh, yes. I guess I'm really not done with this definition quite yet. Only use minimal actions. And I don't really want the action to depend on how I drew the tree in the plane. Say two actions are equivalent. If there's an isometry, so fn acting on t and fn acting on t prime. If there's an isometry from t to t prime, making the commuting with the action. So for instance, let's just take the free group on two generators. Say a maps this way, b maps this way. So a moves everything to the right, b moves everything up. You're probably familiar with this picture. So that describes an action on this tree. But I could have also drawn it like this. So I could have made b go down like this and a go across. Same action of a and the flipped action of b. Well, I can give you an isometry from this tree to this tree. My hand just did it. Does that? OK. And that commutes with the action. So those are the same tree. I just drew them differently in the plane. Right. So I should say free actions on equivalence classes. Is it not space of equivalence classes or free action? Space of yes. I think your hour just needs to go somewhere else. Do you make all the assumptions minimum? Yeah. And the metric is not necessarily the metric coming from the special can be? The metric on the tree. This has to be the metric of the special complex or can be. So the trees have metrics. We don't have a metric on the space yet. We just have a topology on the space. But for a given tree, we have more than one possible metric? Or we're using only the? No. A given tree has, I think of it as a metric tree. So a tree has a metric. Comes with a metric. So yeah, fn acts on a metric tree. And the metric is part of the data for this point. So if I made this edge a little shorter, it would be a different point. Is that clear? So another thing I could do is interchange the a-axis and the b-axis, do a diagonal flip. If these two edges weren't the same length, that wouldn't be an isometry. So it wouldn't be the same action. But if they were the same length, it would be an isometry, and that's the same action. Yeah, this might have been more than a five minute elevator talk, describing the space. Um, yeah. But anyway, so that's definition one. That was supposed to take no time. Let's get to definition two. So this is going to be a definition in terms of graphs. We want this space. So definition two in terms of graphs. So as I said, historically, this was definition number one. So supposing you have an action, fn acts free action on a metric tree, then the quotient is a graph. In fact, it's a graph with fundamental group, graph g with fundamental group, fm. So I have, yeah, I already drew this picture. I have a group acting free, free group acting freely on this tree. The quotient is going to be some trivalent graph. So an exercise that you have is to check that if this was a minimal action, no invariant subtrees, then the graph doesn't have any univalent vertices. No, it is finite, first of all. Has no univalent vertices. And in fact, I also didn't allow my trees to have bivalent vertices. So I won't have bivalent vertices either. So if I'm looking at my space, it's a space of minimal actions. So the quotients by these actions will be graphs with no univalent or bivalent vertices. So instead of looking at the trees, I want to look at the quotient graphs. But I have to incorporate the information about the action. So how do I do that? If I had a point here, then in this case, my free group is on two generators. I can look at the image of this point under all of the generators. Maybe this is a1x. Maybe this is, I don't know, a2x. I don't know where a2x is. And that gives me a map from a1, a2, the standard rows into, well, after I go to the quotient. So a2, this path will get wrapped somewhere around here. And this path will get wrapped somewhere around here. They may be long paths. This gives me a map from the rows into the graph and exercise this map as a homotopy equivalence. In other words, the induced map on the fundamental group of the rows to the fundamental group of g is an isomorphism. So I'm going to call this a marking. I've identified the fundamental group of the rows once and for all with the free group. So what I can think of is my action gives me a way of identifying the fundamental group of the quotient graph with the free group. Gives me an isomorphism of the fundamental group of the graph with the free group. So what is the marking? What information is the marking? Actually, it's just the iso, well, depends on what you read. Today, it's just an isomorphism between the free group and the fundamental group of the graph. And notice I'm being sloppy. I'm not talking about base points because I'm talking about outer automorphisms. So the marking, g star is a marking. And if I have an isometry between two trees up here that commutes with the action, I get an isometry. So equivalent actions give me isometric graphs. The markings commute. So the isometry commutes with the action. So that's going to mean that the marking commutes in the sense that I've got this isometry from g to g prime. And I have this marking. Well, here I just find it as a map from a rose to g. And there's also a marking from the rose to g prime. And the fact that the graphs are isometric and the actions are equivalent mean that this map commutes up to homotopy. So I call such marked graphs equivalent. So when it's equivalence, classes gives me some notion of equivalence of marked graphs. So let me repeat. If I have an action on a tree, I'm going to get a quotient graph. And I'm going to get a homotopy equivalence from the rose to the quotient graph, which induces an isomorphism on the fundamental groups. So I could either think of this as the marking or the induced map on pi 1 is the marking. That's unique up to homotopy. So here's definition 2. On is the space of equivalence classes marked graphs g, g. But now I have to say some more things. I have to say g has no 1 or 2 valent vertices. And g is finite, has no 1 or 2 valent vertices. And g, well, you can either think of it as a homotopy equivalent from the rose to g or look at the level of fundamental groups. Right. So that's the second definition. That was, as I said, the original definition. Although the original paper also talked about action. So I'm going to get a homotopy equivalence on trees. I want to, this business about markings can be very confusing. Maybe it's not so confusing if you realize where it comes from. This is, let me leave that up there. Nope, that's the wrong one. Yeah, OK. Let me just, let me see how to show you how to, think about how to think about one of these maps in a single picture. So let me give you a picture of a marking. So here's my graph. There's my graph. And I want to think of that not as a graph, but as a marked graph. So what do I need to do? I need to identify its fundamental group with a free group. So one way to do that is pick a maximal tree like this. There's a maximal tree. Then every other edge kind of defines a loop in the graph. And not a base point, but what could be if you want. So then label all the extra edges. Well, I'm imagining I have an identification of the fundamental group of this graph with a free group. So each of these loops is an element of the fundamental group of the free group. So each of them has, it corresponds to a word in the free group. And these elements form a basis for the free group. Is that a maximal tree? Yes. It's not a maximal, yeah, this one, this is a maximal tree. Maximal tree. OK. So label the rest of the edges by basis for the free group. That completely describes the way I've identified the fundamental group. So it describes the isomorphism of the marking. Now, one reason I wanted to look at this is think what happens when you collapse the maximal tree. So that the whole maximal tree becomes a point. And you have edges labeled by u1, u2, u3, u4. So this is a picture we've already seen when we were talking about foldings. Now, one point in outer space is just to rose with edges labeled a1, a2, a3, a4. So the isomorphism with the fundamental group just sends each generator to a petal. And I know, or stalling's nose, you all know now, how to find a path of graphs from this picture to that picture. So somewhere along the way, you'll get a path, some intermediate graph. It's about there. As you travel along this path, the identification with the fundamental group is going to change. We know how it changes because we did this folding business. And so what that tells you is that on, described in terms of graphs, is connected. Stalling's folds gives you a path from this arbitrary graph, the arbitrary marked graph that you started with. You collapse the tree. That's certainly a path in the space. And then you start this folding process. And it will eventually land you at this base point of your space. So that will imply that O n is connected by Stalling's folds. Now, of course, there are many choices in how you take this fold path. I mean, you might have two edges here that have the same label. But you also might have two edges over here that have the same label. And which do you fold first? It's up to you. And there is another problem with this picture. Anybody see another problem with folding to get from here to here? These are metric trees, right? What if I run out of edge before I do all the folds I want to do? I'm stuck. So actually, before I start any of this process, well, I have to go down here, and then I have to change the edge lengths till they're big enough so that I can do all the folding I want. And end up here. But that's a big idea for how you prove the space is connected. Right. And as before, what does the action do? The action just changes the marking. So remember when I described it in the space of trees, the tree didn't change. The only thing that changed was the action of F n on the tree. Same things to here. The quotient tree isn't going to, the quotient graph isn't going to change. The only thing that will change is the identification of the fundamental group with the group. OK? Yeah? Can you explain your general knowledge of topology on this definition? Well, it's the same space, so it's the same topology. But I'm going to give you a different way of thinking about this space after I give the next definition that will give you a different topology, which is much easier to understand. And there's a theorem that says that this is the same topology. So in a minute, I'll tell you. But basically the idea is I've got this graph, this marked graph, it's got a metric. Things close to this marked graph are things where the graph looks like this with the same marking, but the edge lengths are slightly different. So that's the basic idea. Can the metric be culturally big? At the moment, yes. I haven't normalized. That's the next statement. It's often convenient to normalize. OK? So right. Yeah, so I claim that this definition in terms of graphs instead of trees makes it, gives you a much better idea of what the space looks like. But in order to be able to draw pictures of a reasonable size, I want to normalize my graphs so that the sum of the edge lengths, I should have put. I didn't write the word metric here, did I? I should have. I mean, there's a metric upstairs when I take the quotient. It's invariant under the action. So when I take the quotient, I get a metric downstairs. And I can normalize graphs so that the sum of the edge lengths is equal to 1. That turns out to be convenient. Then with a graph description, O-N, graph description, decomposes into a disjoint union, well, of open simplices, sigma of gg. So sigma of gg is a simplex containing g. And what is it? It's what I just told you a minute ago. I look at the same picture, the same graph, same labeling. But I let the edge lengths vary. But now I'm letting the edge lengths vary, but I want to keep them the sum equal to 1. Equals a set of g prime g such that the edge equals a set of g prime g obtained from gg by varying the edge lengths. So for instance, supposing I have this graph, simple graph, u and v. It's labeled by some elements of f2. Then what simplex is it in? Well, I can make a whole line of graphs that look just like that. On the one over here, the v guy is really tiny, and the u guy is big over here. The u guy is really tiny, and the v guy is big. And this whole open simplex is contained in the space. It's a space of marked graphs. Notice that the endpoints of this interval are not in the space, because if I collapse v all the way to a point, it doesn't have the right fundamental group anymore. So this is sigma of that. If you haven't seen this before, yeah. So why is it just an interval? Because I want to keep the sum of the lengths equal to 1. So if this is the length of the u edge, and this is the length of the v edge, then I'm just looking at the length of u plus the length of v is equal to 1. It's just a line. Otherwise, I'm looking at the whole cone. And let me do one more. Let's take a graph with three edges, l1, l2, and l3. They're labeled by words. We only label the complement of a maximal tree, say u and v. And then I've got three lengths to play with, l1, l2, and l3. But the sum of the lengths always has to be equal to 1. So I'm going to get a simplex sigma of u, v is that. When I collapse, say, the middle length to 0, I'll get this simplex up here. So this is a simplex in outer space. It's missing its vertices. This is a simplex. This is a simplex. And this is a two-dimensional simplex. So I have a two-dimensional simplex that's missing its vertices. I guess I don't have time to do that. I was going to do the three descriptions in the first lecture and then go on. But I guess I'll let me just talk a little bit more about this. So I've got this u, v. So u goes that way and v goes that way. Now, the way I happen to draw this, as I pointed out, if I just flip the direction that u goes, that's the same point in outer space. That's like flipping the tree like this, which is what I did in my example. So if I flip it, so in this picture, u goes this way and v goes this way. But if I flip it and expand it out again, then u will go this way. And v will, what did I flip? I flipped v. v will go this way. So in particular, so the u-loop hasn't changed. But the v-loop goes the other direction. So in particular, u times v, for instance, will go around like this. Whereas here, u times v went like that. So that's a different marking on my graph. It's a different simplex. And one of your exercises is to draw at least five simplex pleases in outer space of rank 2. And yeah, the exercise says, explain the picture I drew in class of all of outer space of rank 2. But I didn't draw the picture. Other things to do in the second lecture. So draw a picture of outer space in rank 2. If you can put five simplexes together, you should be able to figure out the whole thing looks like. Yeah, I'm out of time for this morning. I mean, for this one.