 Welcome back. Now that you're all so comfortable with graphs, we can move on into three dimensions. We'll skip two dimensions, go straight from one to three. So I'm going to talk about the doubled handle body model, which is, remind you, up is up here. This is one picture of it. This is another picture of it where you're supposed to think this is in the three sphere, and I've glued A to A bar, B to B bar, and C to C bar. So this is picture one. This is picture two. It's the same object. Oh, OK. They're different objects, A, B, C, A prime, B prime, C prime. This is A union A prime. Yeah, B union B prime, C union C prime. So this is kind of A, and this is kind of, that's the front of A, and that's the back of A. OK. I'll just call them A, B, and C. OK, so that's the model. Let me, so we had these automorphisms, which, let me just remind you, there was a sigma permutes the A i, iota i flips, sends A i to A i inverse, rho ij, maps A i to A i aj, and lambda ij maps A i to A j A i. OK, so those are the automorphisms expressed in the generators. How do I model those in this picture? So let's do the sigma ij's first, sigma first. Well, it's pretty clear what you do. You just take a homeomorphism of this picture that permutes the A's, A's, B's, and C's. So if I put A over there, B over here, and C over here, you should be fairly easy to convince yourself that what you have is homeomorphic to what you started with, yeah? So there's a homeomorphism that does that, that sends this guy to this guy, this guy to that guy, and that guy to that guy. And similarly, flipping A to A inverse, let's put A over here, then it's really easy to think about that. So on the handle body picture, you just twist the handle body, twist both of them. Yeah, so that'll send the bottom to the top, twist both of them and glue by the identity, and that's a homeomorphism. And that sends A i choose a homeomorphism that permutes these spheres A sub i. Oh yeah, I guess in order to do that, I should explain how I'm identifying the fundamental group of my manifold with my generators. I've got, I'm going to pick a point that's not in one of the spheres and draw a loop. That's A, that's B, that's C. So those three loops, if I choose, if this is my base point, I should call it P maybe, those three loops generate the fundamental group of my handle body. So each of these spheres is dual to one of these loops. It, the loop punctures the sphere exactly once. So if I permute these spheres, it's going to permute A and B. If I flip this sphere upside down, if I flip this handle upside down, it's going to send A to A inverse. OK, is that clear? I can, those things are supposed to be obvious. One that's maybe not so obvious is these row ij's, but it's actually not that hard. Let me take, let me draw my handle body a little bit funny. Let me draw the ai sphere here, the aj sphere here. And then I have all these other spheres down here. And I've got generators ai and aj. OK, so in order to, I need a homeomorphism of this thing that's going to send the generator ai to the generator ai times aj. So I claim what I can do is take this sphere. I'm basically going to take the connect sum of these two spheres. So I'm going to draw a little tube between them. And after I straighten out this tube, it will look like this. So I'm going to draw a little tube between them. And then I'm going to mess around with that tube until it looks, till that. So that's, you know, if I take two spheres and I connect them by a tube, I still have a sphere. And then I'm going to push it out until it just intersects this half of my handle body in a disk. And I claim the disk will look like this. So this is a picture of ai connected sum with aj. So I've taken those two spheres, joined them by a tube, and then made it look pretty for you. OK. Now I want to take a homeomorphism that interchanges the a-j sphere with the, yeah, the a-j sphere with the, interchanges these two spheres. So maybe I'll draw this in pieces. Let me first cut this open. Let's solve this mess down here. OK. So now I've got, these are my, now I'm just going to switch these two spheres, switch these two spheres, do a little twist up here, do a little twist down here, switch these two, and switch these two. And so this one, ai, ai connected sum aj. We'll show up over here. And aj will show up over here. And ai hasn't moved. I'll glue it back the same way it was before. And I won't do anything at all to these. OK, so that's, that's a homeomorphism. Now I'm going to glue them back together. I just, I just do it apart because I thought it was easier to see what I was twisting after I glued it apart, but I didn't really have to glue it apart. I could just say, interchange these two spheres. But I guess I think it's sort of easier to see if I take it apart, then twist here and twist here and glue back together. Anyway, so now I have a picture that looks pretty much like it used to look, but these spheres have different names. This now is ai plus aj. This is aj, and this is ai. But let's look what happened to my generators. Of course, nothing happened to these guys, but what happened to this? What, this ai, it goes up to, at first it hits the ai sphere, then it hits the ai plus aj sphere, then it comes back to the base point. So over here, first it's going to go up to hit the ai sphere, then it'll hit the ai plus aj sphere, and then it'll come back to the base point. Similarly, this one, what did it do? The aj sphere went up and hit the ai plus aj sphere, then it hit the aj sphere, and then it came back to the base point. So first it hits the ai plus aj sphere, then it hits the aj sphere, then it comes back to the base point, okay? So can you see that? So this orange, when I did this, and did this twist, and then did this, this is what happened to the orange graph. So if I think of this as the same manifold as this, with those things switched, then what happened is the ai sphere got mapped to the, well, the ai plus aj sphere. And the ai loop used to be over here, it got sent over here in this manifold. So it used to be over here, now it goes like that. So it sent ai to ai plus aj. Aj, on the other hand, got sent to its inverse. Okay, so that's a homeomorphism of my original manifold. This is what it does to the fundamental group. It sends ai to ai aj, and aj to aj inverse. It's not quite rho ij, that's, if I compose this with iota j inverse, this goes to ai aj inverse, that goes to aj, and this is rho ij inverse. Okay, so if I'd done this slightly differently, I would have got rho ij instead of rho ij inverse, composed without iota j. But this is the basic move. You take a sphere ai, so that's the basic move, that's basically a diffeomorphism of the three manifold that realizes one of these rho ij's. And let me just point out what I did in this picture. I had, well, this was an ai. So I took these two spheres, ai and aj, and I took their connected sum. So I do a little tube, and I got this sphere. So in this picture, of course, the ai is glued to ai bar, aj is glued to aj bar. And what I'm doing is switching this sphere with that sphere. That performs that automorphism up there. So I take a homeomorphism of this manifold, I just got two spheres in the manifold, I'm performing it, taking a homeomorphism that switches the two spheres. And that's how I model the rho ij's. And I realize that this is hard to swallow if you haven't thought about it before, so I want you to think about it a little. So an exercise on the exercise sheet is to explain how you would get, that was a right transfection, rho ij, explain how to get lambda ij. What's different? What do you do different? Or to get rho ij instead of rho ij inverse. Okay, so after you've struggled with that for a while, you should be completely comfortable with these pictures. So, whoops, we used folds, excuse me. So we used folds to prove that these guys generate odd of fn. I should point out here that I'm being, I haven't, yeah. I guess I have been drawing a base point in my pictures, but I really don't think about these homeomorphisms as necessarily preserving base points. So I usually think about outer automorphisms instead of automorphisms. But maybe it's easier to see the automorphisms. So these generate odd of fn, so they also, so their images generate the outer automorphism group with just the quotient of odd of fn, modern inner automorphisms. So what these pictures prove is that the map from homeomorphisms, this is a homeomorphism that I've described over here. Pi not of homeomorphisms of mn to out of fn, is onto. So that's theorem. So you have to combine that with the fact, so we know all the generators of out of fn, and we can model each of them by a homeomorphism, so that means that map is actually surjective. So that's the beginning, so Whitehead knew this in the 1920s, 1930s. So Whitehead used this fact to study automorphisms of free groups. So, What do you mean? They're non-trivial automorphisms. So if a function has a, this is a non-trivial, oh they're not inner? These are not inner? Yeah, abelian eyes. Abelian eyes, what happens to an inner automorphism? It's the identity on homology. And when you abelian eyes, these guys are not the identity on homology, so they're not inner. Good question, thanks. Whitehead in 1930s used this picture to give an algorithm to decide whether a map phi is an automorphism. Now I'd like to point out that we already have an algorithm to decide whether a map is an automorphism. Namely, Staling's folding gives another one because if you take this map and that I described and try to start folding it, if that's an automorphism, that's a homotopy equivalence and every time you fold, you'll preserve the fundamental group and yeah, if you get stuck someplace you can't fold, then you don't have an automorphism. I don't wanna, actually I don't wanna, I don't wanna dwell on this because I really wanna concentrate on this. This is a, so Staling's folding gives another one and Staling's fold also tells you how to factor your automorphism into a product of rows, lambdas, sigmas and iotas. But this one is a different algorithm, very different and it's nice and easy to program on a computer for instance and gives you an easy test, immediate test to see whether something's an automorphism and the point is that if it's an automorphism, so if the ith generator goes to some word, w i, that's a, being an automorphism it's the same thing as saying these w's form a basis. So what Whitehead did is give an algorithm to decide whether a set of words is a basis or part of a basis. Okay, so the algorithm, yeah, so the algorithm is gonna tell us whether a set of words is a basis. So how does that work? So given, so the idea is that we've got this, let me back up a bit. We have an automorphism, we have a map from FN to FN, we wanna understand whether it's an automorphism. We know from what we proved over there that if I have an actual automorphism, I can model it by a homeomorphism. So if is an automorphism by a homeomorphism and I'm getting bollocksed up with automorphisms versus outer automorphisms again, I'm really talking about automorphisms. And if you notice my picture up there, I didn't move the base point. So actually I've proved that that map is onto. I can do all these things without moving the base point. That's what I've actually proved. So let's just talk about automorphisms for now. So if is an automorphism, model it by a homeomorphism, fixes the base point, okay? So here's my MN, I'm only drawing half of it and I got some homeomorphism. I wanna draw the picture I've been drawing up there, namely I'm gonna redraw exactly the same picture, I guess. I'm gonna choose N spheres and choose a dual basis. This picture is different because I'm calling them A1, A2, A3 instead of A, B and C. So good thing I do a different picture. Okay. Right. And the other picture for this, I'm also gonna draw, those are my spheres. Here's my base point, P and I've got a sphere, A1 goes in one side of A1 and out the other side. This is A1, A2 goes in one side of this and out the other side and A3 goes like this. There's one picture, there's another picture, okay? Okay, so now I'm gonna apply my homeomorphism and watch what happens to the orange graph. I'm not gonna, I'm gonna keep the A's the same place they were before. I have a homeomorphism from this manifold to this manifold. I'm not looking at the moment at what happens to the spheres, I'm looking at what happens to the orange graph. What happens to the orange graph, well, P goes to P, but it gets all messed up, it does stuff. I shouldn't make it too fancy. Yeah, it does stuff, it gets messed up. This is a homeomorphism. It becomes some other graph, right? So now let me cut this open again and look at this picture. So P went to P and this orange graph here got messed up. It goes all over the place. Maybe it looks something like that. Whoops, actually I'm gonna prove that it can't look like that, but okay. So now, yeah, now squint, I can do this and make all those spheres become just dots and you get a graph. I'm not going to draw it accurately. This thing is called Whitehead's Star Graph. If you think about carefully what happened, you'll see that there is an edge from the sphere A i to the sphere A j, if and only if A i, A j inverse occurs in some W k. So you can check that if you want, but topologically the picture's clear. I have a manifold, I take a homeomorphism, I see what happens to the orange graph. Gets messed up, to the orange graph gets messed up and then I cut open along my canonical spheres and I get a new graph called the Star Graph and here's Whitehead's Lemma. If W1 up to Wn is a basis, then the Star Graph has a cut vertex, which is not the base point, okay? So a cut vertex is a vertex, so that when you remove it from a graph, the graph divides into two pieces, okay? I love this proof. Yeah, so that's what we're proving. Yeah, proof. Why is the code left like something that's standard basis? Where's the what? I'm gonna graph for this stuff. Star Graph has an outer vertex or is this. Thank you. If you got this graph, you're happy. You got the basis. Right. So why is this? What time am I supposed to stop? 11.30? Okay. Okay, so in that picture, all I drew was what happens to the orange graph. Well, something also happens to the spheres AI. Proof. Did I quit a name? I guess I called it fee, which I shouldn't have. Okay, this time it's gonna be more convenient to look at this picture. So under homeomorphism, a sphere is still a sphere. The image of a sphere is still a sphere. Make. And by general transversality theory, we can isotope these spheres to be transverse to the spheres AI. Okay, so now I have a sphere running through this picture. F of AI is some sphere. It's got a different color. And it's transverse to this picture. And I claim that that means that, yeah. So here's a piece of it. Maybe here's another piece, another piece, et cetera. I've got some surface here. It's a sphere cut into pieces by the circles of intersection. So here's my green sphere. It intersects the white spheres in a bunch of circles. The white circles cut the green sphere into planar surfaces, right? But the green sphere is still a sphere. So I've drawn here, you know, the complementary components are the planar surfaces. Here I've drawn them as they would appear inside this three manifold. There, yeah. Here's a pair of pants here. Here's a pair of pants here, et cetera. But nevertheless, they are planar surfaces. And since this is a sphere, there have to be some disks. The innermost pieces are disks. So F of AI, a bunch, is a union of planar surfaces, including at least two disks. Take innermost things here, they're disks. Can't put anything inside a disk. So there's at least two disks. Here's one in this picture. And there has to be another one, which I haven't got yet in the picture, and maybe it's here. So that's what F of AI looks like in this picture. Now let's bring the orange graph back into play. When it started, A1 intersected A2 or A3. And it didn't intersect A2 or A3. So I'm going to continue over here. A1 intersects A1, not A2 or A3. So I'm supposing this is a picture of F of A1. Then F of little A1 was this guy. It intersects the green sphere in exactly one point. So it has to miss one of those disks, A1 in exactly one point. So F of A1 intersects F of A1, not F of A2. Or F of A3. Similar F of B1 doesn't intersect F of A2, doesn't intersect F of A1, and F of A3, doesn't intersect F of A1. So let's just look at F of A1. I've got my graph, wherever it is, so this is F of A1 union F of A2, union F of A3, that's my graph in this picture. It intersects at most one of these disks. So this whole thing misses D1 or D2. Say it misses D2. So whatever my graph looks like, it doesn't intersect any of this green sphere. I have to be very careful how I draw it so I don't make it intersect the green sphere. Could intersect that disk in exactly one point, but there's no other points of intersection, right? I'm not using anything deep here. Okay? Yeah? Explain again what this D1 looks like. Are we saying that this third sphere is completely inside of this D1 in some sense? No, okay, so this is my, this is my sphere F of A1. I'm just looking at, I'm taking one of those spheres, my favorite sphere, say A1, and I'm looking at its image under this homeomorphism. So this is my sphere under this home and I made it transverse to those original spheres up there. So in particular, it intersects my original white spheres in circles. So now I've drawn these circles in this picture. Those are circles of intersection. In this picture, here's the circles of intersection. I haven't drawn enough of them, but they're there. Maybe I have drawn enough of them. I don't know. So this planar surface here, this is a surface of the green sphere. It's, this is a pair of pants, so it might be this pair of pants, for instance. But there's at least two discs in this picture. So I've drawn two of them. Here's one of them. Here's the other one. Now, so this is a picture of the green sphere sitting inside my manifold. And now I want to look at my star graph. The star graph is this. F of A1, union F of A2, union F of A3. All three of these things intersect F of A1 in exactly one point. So there's at least one of these discs, either this one or this one, that doesn't contain that point. And in this case, it's this one. Okay, so there's no point of intersection of my star graph with this disc. So I'm done. Because, so that, so if the boundary of this disc, what did I call it? D2 is contained in that, this is AI bar. Then AI bar is a cut vertex, right? So I have this disc. I have this star graph. This disc may have other things in here. I have this star graph. Some of the star graph is outside this disc. Some is inside the disc. I don't know where the disc is. The disc might go up here. Some of it's outside, some of it's inside. This, this vertex cuts the graph into two pieces. So it's a cut vertex for the graph. Is that clear? Is that, does that, have I answered your question or not? Okay. How do you know the stuff that's stuck inside? How do I know there's stuff, because there are vertices inside, because there are spheres inside. And if I missed a vertex altogether, I wouldn't have an automorphism. That would mean I had a set of words that didn't use a certain letter. So that's not gonna be an automorphism. But good, good question. I remember you from classes. I was asking these good obvious questions here. Okay, then AI is a cut vertex. So that's the end of the proof. So the reason I said, pointed this out, this is just keeping track of what I did. Point is, if I have a set of words, it's, this is just thinking about what I did. This makes it easy to draw the star graph without drawing three manifold pictures. The three manifold picture gives you this really cute proof that there's a cut vertex. But in order to actually draw the star graph, you don't have to draw the three manifold picture. So, so how do we turn this into an algorithm? Yeah, so how do we turn this into an algorithm to decide whether something's a basis? Well, first you take a set of words, to draw the star graph is if there's no cut vertex, then you don't have a basis, that's easy. But what if there is a cut vertex, what do you do? So let me tell you how to do the rest of the algorithm. And you get to do this in your exercises. So the algorithm, draw the star graph, no cut vertex, then the base point, then it's not a basis. What if there is a cut vertex? So here's the cut vertex. This one's a cut vertex. Cut vertex is my, I erased the wrong picture. So if this, remember these points or dots are really spheres, supposing there's a cut vertex. Then I can draw a sphere that separates, draw a sphere that separates, oops, I forgot to draw the base point, oh well. Cut vertex A that separates A from its opposite number up here. Now I actually have two choices here. I could have drawn this one or I could have drawn this one. I think I wanna use one that contains the base point, but I can't remember. If I'm thinking about outer automorphisms, I don't need a base point, and that's what I usually think about. But yeah, draw a sphere that separates A from A bar, S. And now I'm gonna do, so this is like the picture I had where I wanted to model the rho ij. I had an ai and an aj, and then I took a sphere that had, that separated ai from ai inverse, ai from its opposite sphere. And then what did I do? I exchanged those two spheres. So take a homeomorphism, exchanging a with S. Okay, so then I'll get a new star graph. So this one maybe I will try to do a little bit carefully. So I'm gonna replace A by S. Everything else remains the same. So look what the orange graph does. The fact that this is a cut vertex means that there are some edges inside, according to Joe, and there are some edges outside. But the sphere S intersects the orange star graph in fewer points than the sphere A did. So A intersects the star graph in all of these points. S misses that point, star graph, the star graph. Intersecting S in fewer points well the star graph intersects S in fewer points than A. It's the same star graph, the star graph. Star graph intersects S in fewer points than A. So when you exchange A and S, you get this picture with S now replacing A and you have to draw the star graph. I don't know, these guys have the same valence. S now has valence one, two, three, four. Where A had valence one, two, three, four, five. I won't connect the dots. But that's what this new star graph looks like. And... Can you actually connect the dots? I don't really understand the process. Yeah, well, if you have a segment of your star graph going from A to B, you still have A and B over here, it still goes from A to B. If you have a segment of your star graph that went from B to A through S, then now it goes from B to S and out S on the other side. And yeah, so it goes, yeah. If you used to go from B to A, it now goes from B to S and out the other side. And yeah. And this guy... Yeah, what happens to this guy? You can draw a new star graph if you carefully look at what your old star graph did. So the intersections with these spheres haven't... So all these changes, you've switched one of the spheres with another one of the spheres. And yeah, it's still the image of a bunch of loops. It still intersects these spheres and a bunch of points, and you connect them when they connect to spheres. I claim it's the same picture. And I'm sorry, I can't... I've been talking for two hours. I can't be expected to do things like that at the end of two hours. Sorry. Yeah, I'm not sure if you said this, but do you need to specify that your sphere is separated from the components of the component? I wanted a sphere that separates A from A bar. So that guarantees that it's a non-separating sphere with people in this room who know too much. It seems to me you could draw other spheres on that graph that are separated A from A bar, but also what you want. But also don't do what I want? Yeah, no. I want to separate... There are lots of spheres that separate A from A bar. The important feature of this sphere is that since it goes around the cut vertex, it misses some of the edges coming out. So that means that it intersects the star graph, this image in fewer points than the valence of this vertex. So I claim that that means that a new star graph using this sphere instead of this sphere will have smaller valence at S than we had here. So the valence of all the other vertices will be the same, and the valence at S will be smaller. Is that okay? So I can't... I don't want to... The reason I needed a cut vertex was to get a sphere that intersected the graph in fewer points than the valence of A. You say it would. What happens to that? Right, most... Yeah. One that's inside of S, it doesn't intersect S at all. It will definitely be somewhere in the new star graph, right? I've done a... I have to see what... So in order to answer your question properly, I would have to actually think about what this homeomorphism does to this star graph. And I don't think I'm up to that at the moment. But you agree that... I mean, I had this picture. That's how I got the star graph, as I applied that homeomorphism. Now I'm looking at some different spheres up there that still cut that thing into a ball. And I'm looking what happens under that homeomorphism to that graph. And I get a graph that cuts. So I get a new star graph. So topologically, it makes sense. Combinatorially, it gets confusing. And yeah, I don't think I can do that in front of all these people. Okay. Does everybody agree that the picture is... All I've done is exchanged one of those spheres for a different sphere. A different, non-separating sphere. And so there's still... The complement is still a ball. I still have a star graph. It's just a different star graph. And it intersects my sphere in fewer points. Because, yeah. This is still the image of that rose under the homeomorphism. It just... And it intersects the sphere in fewer points than the old guys did. Right. So the claim is the star graph intersects in fewer points. So if it's... If the new star graph doesn't have a cut vertex, then stop. Otherwise, repeat. Until you get this guy. And then you know you had a basis. So the homeomorphism, of course, induces an automorphism of the free group. And so your basis will be transformed under this automorphism to a different basis. And what you're doing is transforming this basis over and over again until you get the standard basis by doing these things. So the induced automorphism that switches A and S is called a whitehead automorphism. So for obvious reasons. Stop means it's not a basis. If you ever get a set of words that doesn't form a basis. So you take a set of words, you transform it by an automorphism if it has a cut vertex. So notice that this lemma only goes one way. If it's an automorphism, if it's a basis, then there's a cut vertex. That's not an if and only if. You can have things that aren't a basis whose star graph also gives you cut vertices. So you might have one of those by accident, but then if you transform them by enough of this type of automorphism, you'll eventually get something that doesn't have a cut vertex. Right. I think I'm going to stop there. What I should do now is do some simple examples. That's what exercises are for. Sorry for that.