 That's where we ended last time. We were looking at a minimal area disk diagram, d in x. And remember, that meant that we have a diagram. We have a diagram. And well, we know that our diagram is built from squares. There's a lot of squares. Inside, I'll draw part of a diagram to remind you what's going on over here. And we said we were going to look at dual curves. These were little kind of hyperplanes in the diagram. And we're going to be interested. So there are two dual curves. And we're interested in looking at how the dual curves are going to be wandering around in the diagram. And well, there's a few pathologies that cannot occur when the diagram is minimal. What does it mean for the diagram to be minimal? Well, among all disk diagrams, with a particular boundary path that's that orange path that goes around the disk diagram, we can look at the disk diagram that has as few cells as possible. Usually, it's as few squares as possible, but as few cells as possible is fine. So the minimal diagram or unminimal diagram, they're not necessarily unique, will be one which has as few cells as possible, keeping that orange boundary path fixed. And when the diagram is minimal, the dual curves, and it's a diagram mapping to a non-positively curve cube complex, the dual curves cannot exhibit any of these pathologies. So a dual curve in particular cannot self-cross. What's your question? When you say minimally, do you mean it's an image in the cube path, what does it look like, the D itself? D itself. We're looking at how big D is itself. So when we say a diagram in X, we mean a disk diagram in the plane, together with a combinatorial map from that disk diagram to the two skeleton of X. So let's quickly sketch why this theorem holds. So the idea was hexagon, certain moves, so the method of proof relied on two transformations that allow us to maneuver back and forth between different diagrams with the same boundary path. So this one we like to call a hexagon move, sometimes righto-meister move. And what you're seeing is pushing across a 3-cube. You have three squares. You know that there has to be a 3-cube over there where they map to inside of X. They're going to surround the front half of a 3-cube. And you could just push across that 3-cube and change your diagram. So this is a move that happens inside, maybe could have happened right over here. There's an opportunity to change, to modify that diagram by this hexagon move. And then there's another move, which is a bit more drastic. And it's actually not going to occur inside of a minimal area disk diagram. And it looks like this. We have to be conscious of it. So over here we have two squares in the diagram that are meeting along a pair of edges. And those two squares have to map to the same square. That's why I'm drawing these labels over here. They have to map, they fold on their way to X. Because the cube complex X doesn't have two squares in it that meet along an entire corner. Because what would that mean in the link? Well, it would mean you would have a bygone. If these two squares map to different squares in X, you would have a bygone in the link. But the link is a flag complex. So it doesn't have any bygones in it. So what we can do, if we saw a god forbid, if we saw a pair of squares like this inside of a disk diagram, what we can do is we can remove those two squares and replace the pair of squares by this little arc of length 2. And that reduces the area. So of course that would never happen in a minimal diagram. The diagram couldn't be minimal and have this property. And have a pair of squares like this. In the combinatorial group theory, this sort of thing is called removing a cancelable pair. Those two cells kind of cancel with each other. All right. So now, how does one prove that theorem? Well, the main point is if any pathology exists, if one of those four pathologies existed in the diagram, then there exists a bygone. This is what we would call a bygone. Why is that? It's pretty clear. For instance, if you had this inside the diagram, then you could just choose a dual curve starting somewhere, see where it goes. And either it would self-cross, in which case you'd have one of those pathologies, or you'd find yourself a little bygone somewhere. So an innermost pathology would actually be a bygone. And then what we do is we now transform. So what we're going to do now is we're going to take a bygone. And we're going to show that if there is a bygone, then the diagram wasn't minimal. So let's see how that goes. It's pretty simple. You see, I'm just going to sketch it out. If you had a bygone, we're not saying what's on the inside. It's more squares maybe, or maybe not. If you had a bygone, then with a little bit of work, there's a little bit of extra effort over here, what you could do is you could find a little square corner. I'll give you away in a moment of being sure about that. But you could find a little square corner inside this bygone, just like this. And then once you see that square hanging over there, what you can do is you can push it out of the bygone by using a hexagon move. Get this right. Let me do the following. So you all see that the yellow dual curve, for instance, that was over here, got replaced by that yellow dual curve over there. And what happened is we've now found a new disc diagram with a smaller bygone. And well, you keep doing this until there are no bygones to push out at the bottom or to push out at the top. And at the last stage, you might perhaps keep doing this. At the last stage, you'll have a situation like this. This is also a bygone, isn't it? You've squeezed all of those squares out, though. And now at this point, you can see that if you visualize it the right way, you're looking at the outside of a column of cubes stacked one on top of the other. And you can replace this by this picture over here. And when you do that, let's get this right. 1, 2, 3. OK, I'm not. I don't know if I'm counting correctly. I hope I counted that right. When you do that, you'll reduce the area by 2. Did I get it right? Erase the. Oh, well, that's a separate issue. OK, excuse me, I had a question. What a question. Sorry, I don't get how the first clue occurs. How does what occur? The first clue on the bottom left of the figure. How does this occur? Well, I mean, it might have been that way to begin with. It just turns the image, or? Well, so at the first stage of this process, you find squares that are kind of hanging at the bottom, and you push them out, until there's nothing left on the inside. It might have been that way in the beginning, I don't know. I mean, that's also a bygone, isn't it? But after you've squeezed all the squares out, then you have this. And this is, in some sense, a generalized removal of a cancelable pair, isn't it? Wait, can you just say again what you do in this last step? At this last step. You know, I'm actually very combinatorial minded, right? I gave it to you in one big sweeping. I said, well, imagine this thing. It's a kind of column of cubes stacked one on top of the other. And you're looking at four of the sides of this column of cubes over here. And if you'd like, you can imagine pushing across to the other two sides of the column of cubes. And that's what's happening over here, unless I screwed it up. I screwed it up. OK, let's correct it. But another way of thinking about it, which is maybe better, is you can keep doing hexagon moves until you've pushed one of these squares all the way over to the other square, and you can remove the cancelable pair. OK? So let's do it that way, instead of doing it in, you know, I've done this. And then I have one, two, three more, right? And I've now created a smaller bygone, right? And now you all see, I keep creating smaller and smaller bygones until at the last stage, I can remove a pair of squares, OK? OK, that's better. Thank you for correcting me. Well, before I move on to begin the lecture, because this is really from last time, you can also, if you think through what's happening over here, you can also pick up the furthermore statement, OK? And the furthermore statement says that if it so happens that you can find a diagram that has a configuration like this to dual curves that kind of cross in one spot, and then kind of touch each other, come up right next to each other in another spot, then in the cube complex, there exists a square, S, which could have been positioned right there, OK? This is going to come up again later. But this essentially implies that, right? Because you kind of imagine what's happening over here. And you follow all of these steps, then this square kind of keeps pushing, pushing, pushing until it arrives at the other side. And that shows you that that square exists. Yeah, so I have a question about the first step in this process. What happens if this vertex has four squares meeting there? Which step, right over here? How do you find this? How do you find that? That's a very good question, OK? So you can either work a little bit, OK? Because I hid that. Or you can look at the following proposition. Let D to X be, but you can cook it up. You could think about this on your own and define what it means to be a kind of lowest. You could find the lowest square and you'll find it, OK? But there's another, this proposition, which you can then feed into. This is some type of big induction. Then you'll see that you could find it as well. Let D to X be a minimal area diagram with X a non-positively curved cube complex. Then either D is what we call trivial, right? What's the smallest possible disk diagram that you can have a single point, single vertex? Or D looks like, it's some type of pastry. Maybe someone will know the name of these pastries. Is there anybody here that eats a lot of pastries? So it's surrounded by three dual curves that are meeting at three corners. It's some type of triangle. It's a triangle, triangular. Or D is a triangular pastry, OK? Or the usual case, the number of square corners in D plus twice the number of spurs in D is greater than or equal to 4, OK? So what are those? So a spur in a disk diagram, so maybe I'll give you a picture of that. Here's another example, some more stuff inside. I don't know what. Well, I guess just a single square on its own, right? So a square corner would be like that. And a spur is one of those. So a spur in a disk diagram is when you have an edge sticking out, ending at a vertex of degree one on the boundary. That's called a spur, OK? So here you have a spur right over here. Maybe there was a little tail, I don't know. So the third option is that when you count the square corners, so here I have one, two square corners over there. And I've got a spur over here. Here I have four square corners. This square has contributed four of them, OK? And I leave that for the exercises to understand why this is the case as a consequence of no bygones, OK? I gave a little hint over there of how to do it by a quick induction. And if you don't want to work hard to understand this, then you could set up some induction in your head and say, well, I believe everything. So I believe that proposition. And so show me where the spurs or the square corners are inside of this sub diagram. And there'll be either a spur, which will kind of tell you that you can close it up a little bit somewhere at the end. Or there'll be a square corner, at least one, along the top or bottom. And you'll be happy, OK? I'm more interested, though, in conveying the culture here than elegant systematic proofs. Excuse me? In the further most statement, you said that if there is something in the diagram, then there is a square s you mean in D. The square s is in, thank you, in X at the images of E1, E2. Or in other words, you could have put a square there and mapped it to X. Is it clear now? It's a little wishy-washy? Or it's clear. But how do you know that it exists? You said it's your. If you follow this kind of sketchy proof, then you kind of imagine first pushing things outwards until there's nothing on the inside. Ignore that this was here. We never used it. And then start taking this square and imagine pushing it further and further to the right. At the last stage, after you've done it, after you've pushed it all the way to the right, it will be right alongside these two edges. And that's the square s. What arrives there after pushing, pushing, pushing is the square s that the furthermore is insisting on. But the pushing is building a second or more. Yes, that's correct. But this means that you changed the. It's not this square. In the cube complex, it's a square that's parallel to this square. It's witnessing that there is this column of cubes. And you're just looking at the other at the top end of it instead of the bottom. OK, so you check. OK. OK, so let's start today's lecture now. I want to talk about convexity. So I'm actually going to focus on sub complexes. So y tilde contained in x tilde is convex if gamma is contained in y tilde whenever gamma is a geodesic in x tilde with n points in y tilde. So a remark, we're going to really be very combinatorial over here. When y tilde is a sub complex, not just a subspace, it turns out that y tilde is convex, provided that geodesics gamma joining vertices of y tilde must lie in y tilde. And also, one other thing, a cube is contained in y tilde whenever its one skeleton is contained in y tilde for each cube of dimension greater than 0. OK, so you have to think about this a little bit. But what I'm saying is that maybe I'll draw an example of a convex complex so that you'll have something to have in mind. It's the one skeleton of y tilde. Thank you, dear. It's just a very simple example. So that is y tilde over here. And it lives inside of x tilde. And the statement, you should have the remark in mind, I'm really focusing on the main point are geodesics joining vertices in the one skeleton. So geodesics in the one skeleton joining vertices. Excuse me. So convexity is saying you can't have a geodesic, like this orange one, which starts and ends inside of y tilde and leaves y tilde. OK, yes? In the first part of the remark for convexity, if I take a three cube, I don't think it's one thing. I remove it. It only takes three of the two cubes in its boundary. It's convex how long it has to be the entire cube. Let's see. So that's correct. Did I make a mistake? Let me understand what you're asking about. Three cube, that is three of the faces. Now, not the entire one skeleton is inside that sub-context. OK, so here's my three cube. And what are you asking about? Essentially, if I take only three of the faces, three connected faces of them, the convex hull of that is going to be the entire cube. That's correct. But the one skeleton is not going to be the concrete. Yeah, you're right. Let me fix the statement. But it still has to be the entire cube because you need every geodesic joining two corners. So he's complaining that I made a mistake. OK, so let's correct it. It should say whenever a corner of a cube lies inside. So how am I going to crack that? It's a little painfully. And by a corner, you see what a corner is. Thank you for saving me over there. So what's our mean in the picture? This one, yes? It is convex. And the orange path is not a geodesic. So this is convex. And this is not a geodesic. It's a picture of failure. Now, probably the main point over here is the following definition. A map from y to x between non-positively curved cube complexes is a local isometry if it is a combinatorial map. So it sends cubes to cubes if it is combinatorial. And for each 0 cube, little y, the induced map from the link of y to let's name little x, where y maps to, the induced map from the link of y to the link of x is an embedding as a full sub-complex. So the, well, when you imagine, here's a simple example. Maybe I'll draw it horizontally, stacking up a bunch of cubes. So this is R2 mapping to R3 over here, very poorly drawn. And what's happening at the level of the links? Here's the 0 cube that I'm interested in. That's little y. And here is its image, little x. And at the level of links, the link of little x looks like this red octahedron. So this is the link of little y contained in the link of little x. And to be a full sub, so there's two things going on over here, there's the embedding. So that's really saying that this is a locally injective map. So the vertices of this link, the vertices of the link of this point are mapping to vertices of the link of that point. But it wouldn't necessarily be an embedding. So we're requiring that it's an embedding. And we're also requiring that it is full. So whenever a simplex, whenever a set of vertices in the image of link y span a simplex of the link of x, that simplex has to lie inside of the link of y. That's what a full sub-complex means. So you know the notion of a full subgraph, take a bunch of vertices, throw in an edge. If you have a subset of the vertices, you throw an edge in whenever its vertices are in the set. That's going to be a full subgraph. A full sub-complex of a flag complex, or of a simplex complex, actually, a full sub-complex will have the property. What it means is that whenever the vertices of a simplex are in the sub-complex, the simplex is in the sub-complex. So there's a similar variant version of this, which I find quite convenient. Maybe I'll call this an exercise. Equivalently, the map is a local isometry if it is a combinatorial immersion. So now I'm asking that it be locally injective. This just means locally injective. And it has the following property. There are no missing corners of squares. What does that mean? What it means is that whenever two edges, E1 and E2, this is inside of y, whenever you have two edges, E1 and E2 and y, at some vertex of y, maybe I'll call that little y, actually, that's what we were calling y before. Whenever the image of this pair of edges, let's call them now E1 bar and E2 bar, this is inside of x and v bar is the image of v, whenever these two edges at a vertex bound a square, whenever they bound a square, let's call that square s bar inside of x, they actually had to bound a square inside of y. So here's the square, maybe I'll draw it in yellow, here's the square that actually has to exist. So let's stare at this for a moment. What this says is that if two edges at the vertex v in y map to two edges at a vertex in x, which form the corner of a square, which I'll call s bar, then those two edges actually already form the corner of a square s, which of course maps to s bar. So there's no missing corner of squares. If there could have been a square there, there was a square there. Now, this is somehow much more specific. It's much more specific than this definition over here. It just seems to be focusing on one small tiny little point. But it turns out when you think it through, they're equivalent and very heavily depending upon that these were not impossibly curved cube complexes. So a quick example, any immersion, really combinatorial immersion of graphs is a local isometry. And in some sense, we're always generalizing things that work for graphs. So when you have a local isometry of graphs, people usually call them just an immersion. People usually just say an immersion of graphs. It's a local isometry. Well, there are no squares around, so that's automatically the case. I will describe, so in the exercises, there'll be more discussions of local isometries. Let's state the theorem. Let phi from y to x be a local isometry with y and x connected, of course, all cube complexes over here are connected, non-positively curved cube complexes. Then number one, the induced map between their universal covers is an injection. Number two, consequently, the map phi induces an injection from pi 1 of y to pi 1 of x. And actually, a bit stronger than number one. In fact, the lift of the universal cover of x to the universal cover of y to the universal cover of x is actually not just injective, but its image is a convex sub-complex. So let's sketch why these are true, and it'll give us a chance to see our disc diagrams in action. So I think let's start by proving injectivity. Suppose that A is not equal to B are two distinct points of y tilde, and suppose that they are mapped to same point of x tilde. Since these are complexes, it's enough to show that this map is injective, it's enough to show that it's injective on zero cells. So we will assume that these are vertices. And now we will choose a path P to y tilde joining A to B. This is a combinatorial path. So it's a path in the one skeleton. And actually, we're going to choose that path such that P is equal to the boundary path of a disc diagram D, where D to x is a disc diagram. And moreover, let's do it so that D is minimal among all possible such choices. Well, yes. But right. But it's closed in the disc diagram, which is mapping to x. So here's A, here's B, here's P. Right now, they're different from each other. A and B are different from each other. In x, of course, A and B are going to map to the same point because they're the same point in x tilde. And so if there's a path from P to x tilde, which if there's a path from P to x tilde, which is closed, so the path from P to inside of y tilde, of course, A and B are different from each other. But inside of x tilde, they're actually the same. The images are the same. So I should always say A bar and B bar. That would have clarified things, sorry about that. And so since this is a closed path in x tilde, which is simply connected, there's going to be a disc diagram. And well, we like minimal disc diagrams, so let's choose the smallest possible disc diagram. And let's even, we don't really care which choice of P we're going to use from A to B. Whoops, from A to B. We'll choose whichever one would give us the smallest disc diagram. You mean D to x tilde or D to x? Well, it doesn't matter, of course. If you have a disc diagram in x tilde, it projects to a disc diagram in x. If you had a disc diagram in x, it would lift to a disc diagram in x tilde. Since D is simply connected. x tilde. That's fine, thank you. And now what do we know? We know that disc diagrams have spurs or corners or they're trivial, right? So if D is trivial, then A equals B because P is the trivial path, right? So that couldn't possibly happen. Otherwise, by the proposition D has a spur, well, it has at least two spurs or square corners, right? The proposition up over there, yeah, the triangle has three square corners, right? So there's one of them. Can't reach the others. OK? But what would that mean? Let's just look at the case of a square. Let's say that D had a square right over here. Since there are at least two, there's going to be one which is going to be away from this point right over here, where A and B kind of get identified. So there's going to be at least one away from this point. So that means that there's going to be a square kind of alongside y, alongside y tilde if you'd like. So that square is going to be kind of like this. But y tilde doesn't have any missing corners of squares. So this yellow square actually must live inside of y tilde, right? But that means that instead of using this diagram, we could choose a different this diagram. We could use this diagram over here, D prime, right? And of course, we'd be choosing a different, we'd be choosing P prime instead. And this is smaller, right? In the case of a spur, you'll do something similar. All right. So I guess we already have injectivity of the map between universal covers, which implies pylone injectivity as well. How about convexity? Yeah? How do you know that there is a minimal disk diagram that's in the image of y, sort of? How do I know that there exists a minimal disk diagram? There's boundary facts in the image of y. So we chose the path p inside of y, or inside of y tilde. That path p is closed inside of x tilde. Therefore, it's no homotopic because x tilde is simply connected. Simply connected diagrams have disk diagrams. We chose the smallest one. So let's look at item three now. So let's prove convexity. So that was three. Now we're going to do something very similar. Let gamma be contained in x tilde via geodesic from A to B. So y tilde already embeds. So I could just call these A and B now. And we're going to choose a geodesic from A to B. And let D be a disk diagram between gamma and a path, a combinatorial path sigma, such that D is minimal among all choices fixing gamma. So the picture is this. This is what I mean by a disk diagram between gamma and sigma, a disk diagram whose boundary path is gamma sigma inverse. And this is all the territory of y tilde over here. I'll draw it in yellow. That's y tilde. OK, so now it's quite similar. It's going to be a very similar style of reasoning. A little more elaborate though. But the basic idea is to choose this minimal area diagram and then use things that we know about dual curves now. So firstly, no dual curve starts and ends on gamma. So let me draw the picture of what can't happen. So here's a dual curve that kind of travels from gamma to itself. So this cannot happen. Why not? Well, if you imagine something like that, what you would do is you would say, well, consider each dual curve that travels through, that crosses the ladder carrying this dual curve. And where would it end? So I was kind of started over here. Where would it end? It has to end on this little part of gamma in between. Because could it come back and cross the orange? No, because a minimal diagram can't have any bygones. So this orange curve has to end on gamma. It can't come back there because then we would have a bygone and then d wouldn't be minimal. So all of the dual curves, I'm just going to draw some of them. All of them end on gamma. But what does that mean? That means that this part of gamma in between these two points is at least as long as this part right over here of the rail of the ladder of the dual curve. So what it means is that this path over here is shorter than gamma. Since then, gamma prime is less than gamma. It's two edges less. It's too less. Because it missed this edge, which is dual to the orange curve, and it missed this edge as well. And this part is no longer than this part. And the second step of the proof is the following. Well, if d contains a square, here's gamma, here's sigma. This is d. If it contains a square, what we want to show is that d doesn't contain any squares. So gamma actually lies, gamma will be forced to lie inside of y tilde. If d contained a square, well, we might as well choose the first edge which doesn't lie. We might as well choose the first edge of gamma which isn't equal to sigma. We might as well choose the first edge where gamma leaves y tilde. We might as well have assumed that gamma actually left y tilde immediately and arrives immediately at y tilde. So where does this square end up? It can't self-cross. We already said it can't come back to gamma. So it has to, if you look at the dual curve that sort of starts at this square, it has to end on sigma. I already said it can't come back. So it has to cut all the way through to sigma. OK? Well, now apply the proposition to this little sub diagram over here. We've got to have at least one more corner someplace. There's a, sure, maybe there's a corner over here and there's a corner over there, but there's got to be a corner or spur someplace else. Perhaps over here, because there's going to be at least three corners in this diagram. And then the same exact thing that happened before. This square would actually lie inside of y tilde, because y tilde doesn't have any missing squares. So the upshot is that d is trivial. Gamma and sigma are the same, and we're done. I have to stop. So that means you guys get to go and have your 15 minutes of coffee. Am I wrong? Yeah. Thank you. Thank you.