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3/6 Outer Automorphisms of Free Groups

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3/6 Outer Automorphisms of Free Groups
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A group is a mathematical object encoding natural notions of symmetries and transformations. Geometric group theory is an area in mathematics devoted to the study of discrete groups by exploring connections between algebraic properties of such groups and topological and geometric properties of spaces on which these groups act. As a distinct area, geometric group theory is relatively new, and became an identifiable branch of mathematics in the early 1990s. Geometric group theory closely interacts with low-dimensional
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ZweiInnerer PunktMinkowski-MetrikMultiplikationsoperatorKlasse <Mathematik>RangstatistikVorlesung/Konferenz
Transkript: Englisch(automatisch erzeugt)
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, OK? 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 the class last time wasn't really the right picture. So what's the point? So 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 Sposun'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,
it's kind of 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.
S2, thank you. OK, yeah, use S2. It separates 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 n 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. 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, yeah, it 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 double-handled body, I'm calling mn. 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,
row ij's and lambda ij's. OK, 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, 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 homomorphisms of mn modulodyne twists.
Yeah, pi 0 of homeo of mn with a base point. Again, modulodyne 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 of, 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. I'm 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 group, 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 contractible space X with the proper action.
And actually, you don't really care whether the action's co-compact or not. The reason is that topological invariants of this quotient space are topological invariants of the group, like cohomology, Euler
characteristic, things like that. Here, quasi-isometry invariants of this space are quasi-isometry invariants of the group. So you may be interested in, so you use this, yeah, you want this for more geometric applications.
You want this for more topological applications, algebraic topological applications. So we're gonna 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, okay? 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 Shalin yelled out from the audience, you should call it outer space. And it was a really good suggestion,
because everybody, even if they don't know what it is, outer space, that sounds cool. Okay, outer space, making algebraic topologists happy. But X has closely related spaces.
One's called, I guess I should, oh yeah, I should, right, I was gonna give it a name. How about O-N? O-N has closely related spaces. One's called K-N, and that's called the spine of outer space.
Then there's also O-N-hat, called the bordification, outer space. And both of these make everybody happy. They're both, they're still contractable,
the action's still proper, and the action is co-compact.
Okay, but first we have to describe this space, outer space. Okay, so I wanna define, so we define free groups in terms of graphs, and in terms of three manifolds.
Right. Yeah? O-N doesn't have a metric. This is true, but if. It's 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 put a simplicial, make every simplex a regular simplex, and then you have a metric space. And then it's quasi-symmetric. Okay. You happy? No. No, you're happy, good. Phew. Right, so we've talked about graphs, and 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 contractable. And what else? Action. The kind of actions I'm gonna look at, these are gonna be simplicial actions,
or in other words, they're gonna 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 a 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 the universal cover, which is a tree by deck transformations. So we're gonna make a space out of these actions, and to cut down the size of this space, I'm gonna make some assumptions.
I'm only gonna use minimal actions. In other words, I'm not gonna allow any invariant sub-trees.
I'm also not gonna, 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 wanna care if I sprinkle bivalent vertices in here, that doesn't really change the basic picture, the tree or the action. So, actually this does have, this can be technically phrased so that it automatically does this, but I'm just gonna say it for emphasis,
and no bivalent vertices. Okay, so I'm gonna use, I want to make a space.
Now I wanna make a space of such things, all I've done is shown you an example, shown you how to get lots of examples, take any graph with the right fundamental group, look at its universal cover, you have an action on a tree. But I wanna make a space of such things, so in order to make a space, I'm gonna make these trees metric trees, I want the action to be by isometries,
so I'm gonna 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? Okay, so what's the equivariant Gromov-Hausdorff topology?
If x and x prime are metric spaces, then x is close to x prime if,
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 in s are within epsilon of the corresponding distances
of points in s prime. I'm just gonna write it like that. Okay, so that's the regular Gromov topology on metric spaces. What makes it, but we don't just have metric spaces,
we have metric spaces with actions, right? 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 needed 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 s,
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, right? And by ejection, and what do I want? I want the distance between two points, s1,
not just the distance between s1 and s1, 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, okay? 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 x and x prime, to s and s prime. Okay, so that makes it so, since I now have actions on metric spaces,
there is an echovariant topology, and right, so now I could define definition one. Is there any way to draw what invariant sub-3 would look like?
What an invariance, this is one of your exercises. Yeah, well, I'll just, for one thing you could do, supposing you had a perfectly nice action on a tree and you put a little spike in the middle of each edge,
okay, 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 sub-tree, okay? So, right.
Definition one, the space-free actions of f n
on metric simplicial piece. Okay, 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. Oh, I see, it has two advantages. First of all, it's very short, succinct, someone asks you what outer space is and you don't really want to talk to them, you say that's the action of f n on. 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 f n on O n, as described. So what's an element of O n? 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. Okay, so this is a point in outer space.
It's an action of a free group on a tree. Right, 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 f n with this actual automorphism and I'll get, this is phi hat,
goes from f n to f n and I'll get a new composition, a new map from f n to the isometries of t, rho composed with phi hat. Okay, so that's a new point in my space.
So same tree, but it's a different action on the tree because I've twisted the action by this automorphism. So that's the, how f, I'm sorry. Yes, you're absolutely right. I'm asleep. Action out of f n on o n, right.
So it's f n twiddle, no. No, I'm sorry. This is, 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 f n acting on t
and f n 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. Okay. And that commutes with the action. Okay, 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, free action?
Space of, yes. I think you're out of your seat. Yeah. Do you make all the assumptions, the minimal? Yeah. And the metric is not necessarily the metric
coming from the supercharger, it can be? The metric on? On the tree, on the tree. This has to be the metric of the supercharger complex or can it be? So we don't have a, right. So the trees have metrics. We don't have a metric on the space yet.
We just have a topology on the space. Right. 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. Yeah, so if I made this edge a little shorter, it would be a different point. Is that clear? Okay, 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. Yeah, describing this space. Yeah. But anyway, so that's definition one. That was supposed to take no time.
Let's get to definition two. So this is gonna 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 Fn. So I have, yeah, I already drew this picture.
I have a free group acting freely on this tree. The quotient is gonna be some trivalent graph. So an exercise that you have is to check that
if this was a minimal action, no invariant sub-trees, then the graph doesn't have any univalent vertices.
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 is 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 wanna 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 is 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 gonna 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. It 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, 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, okay? So the marking, G star is a marking.
And if I have isometric, 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 gonna 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 gonna get a quotient graph and I'm gonna 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 one as the marking. That's unique up to homotopy. So here's definition two.
O-N is the space of equivalence classes of marked graphs G-G.
And now I have to say some more things. I have to say G has no one or two bailant vertices
and G is finite, has no one or two bailant 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, okay?
So that's the second definition. That was, as I said, the original definition although the original paper also talked about actions on trees. I wanna, 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. No, that's the wrong one.
Yeah, okay.
Let me just pick, let me show you how to, so remember when we were talking about Stalling's Folds, I showed you how to think about a Stalling's, 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 wanna 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, not a base point in the graph but well it could be if you want. So then label all the extra edges.
Well this is, 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. Okay. So I'll label the rest of the edges by a basis for the free group. That completely describes the way I've identified the fundamental group with, so it describes the isomorphism of the,
describes the isomorphism of the marking. Now one reason I wanted you 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 rows 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 Stallings knows, 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,
you know, some intermediate graph that's about there. As you travel along this path, the identification with the fundamental group is gonna 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. Stallings folds gives you a path from this arbitrary graph,
the arbitrary marked graph that you started with and collapsed the tree. That's certainly a path in the space. And then you start this folding process and it'll eventually land you at this base point of your space. So that will imply that ON is connected by Stallings 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. Right. 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 wanna do? I'm stuck.
Okay, 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 FN on the tree. Same thing's true here. The quotient tree isn't gonna, the quotient graph isn't gonna change. The only thing that will change is the identification of the fundamental group with the group. Okay? Yeah? Can you explain again what is the topology
of all this simple definition or? Well, it's the same space, so it's the same topology. But I'm gonna 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, same topology. So in a minute, I'll tell you. But basically the idea is I've got this graph and it's got, it's a marked graph, it's got a metric. Things close to this marked graph are things where the edge lengths,
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 metrics be arbitrarily big? At the moment, yes. I haven't normalized. That's the next statement. It's often convenient to normalize. 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. So, but in order to make, be able to draw pictures of a reasonable size, I wanna 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, the metric, 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'm gonna, I can normalize graphs so that the sum of the edge lengths is equal to one. 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 the 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 wanna keep them, the sum equal to one.
Equals set of G prime G, such that the edge, well, yeah, equals 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. Okay?
So this is sigma of that. If you haven't seen this before, yeah. So why is it just an interval?
Because I wanna keep A plus, the sum of the lengths equal to one. 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 one. 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.
Yeah, we only label the complement of a maximal tree, say U and V. And then, so I've got three lengths to play with, L1, L2, and L3. But the sum of the lengths always has to be equal to one.
So I'm going to get a simplex sigma of UV is that. When I collapse, say, the middle length to zero,
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.
and what yeah 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 um let me just talk a little bit more about this so I've got this UV so U goes that
way and V goes that way now the way I happen to draw this as I point it 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 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 places like in outer space of rank two and yeah the exercise says explain the picture I
drew in class class of all of outer space of rank two but I didn't draw the picture I have other things to do in the second lecture yeah so draw a
picture of outer space and rank two if you can put five simplices together you should be able to figure out the whole thing looks like so yeah I'm out of time for this morning so I mean for this one