 Thank you very much, and thank you again to the organizers for having me participate in this wonderful affair. Indira was concerned that my last lectures would be dry, but that will not be the case. So, let's get to it. First, let's talk about wall spaces. My goal today is to tell you how to get from a group to a cat zero cube complex. The main theme is going to be Mikhail Sagiev's Ph.D. thesis, which has been developed upon a lot in the last 20 years or so. So, wall spaces coincidentally are a notion that arose in Orsay, I believe, because of Frédéric Palin and Frédéric Arglond. So, a wall space is a set S with a collection script W of walls. Each wall in our collection is a partition of S into two subsets. We'll assume that the subsets are disjoint from each other, and we will also assume that no two walls are the same partition for technical reasons. Although a lot of what I'm going to describe has great flexibility to it, you'll see as we proceed. So, this definition, which is the most elegant form of the notion of wall space, is suggestive of quite a lot of generalization that will become apparent as we proceed. So, we're also going to require that the number of walls separating any two points in S is finite. So, this is the number of walls separating P and Q. A wall separates P and Q if P lies in one of the half spaces of the partition, and Q lies in the other half space of the partition. If you exchange the two parts. If you exchange the two parts. Yes, is it another one? We're not going to assume that each wall is just a partition, so it's a collection of two subsets. So, it's the same wall. The order doesn't matter, even though the notation suggests that it does. The order does not matter. It's not fixed, which is left, which is right? No, it's not. It's just a convenience of notation. And now, what's known as empty? We'll assume that they're both not empty. Otherwise, it makes a little bit of a mess for certain subtle points, but don't worry about it. Okay. You missed me on one empty set, by the way, that I was wrong on last time. So, what's the picture? The set is a set, some object, maybe. Maybe it has some additional structure to it, some geometry, maybe it's a metric space, who knows? We will see. And then a bunch of walls. There's a yellow wall. So, look, I'm drawing the wall as if it were a wall. But we all are thinking, oh yeah, it's dividing it into a left half and a right half, or a half and another half. Never mind the order. And then there are other walls. There's a third one. I'm not going to be so generous with the chalk in the future, just this one time. Okay, so here's a wall space. It has five walls. Okay? And, you know, I guess if this really is going to be a partition, I suppose we would have to decide where the points on the, what I'm going to, what I tend to call these walls, right? The yellow things, this yellow thing, I tend to call that the wall, right? And you'll see why soon. But if you're being very careful about this definition and being faithful to the definition, you have to decide for these yellow points, are they going to be on this side or on that side? Okay, so let me choose a way. This definition is flexible in that fashion. And it's actually quite a bit more flexible than that. So let's give some examples. So some examples that you're conscious of, perhaps. I mean, that was a perfectly good example over there. But what if we took, if we're being very, very faithful to the definition, and we took the zero, the vertices of a cat zero cube complex, and the walls correspond to the pairs of half spaces associated to the hyperplanes, right? So each hyperplane in the cat zero cube complex, when you just consider the vertices of the cat zero cube complex, that hyperplane cuts the vertices into the ones that are in one of its minor half spaces and the vertices that are in the other minor half space. So the hyperplane gives you a wall. I suppose another example that the two Fred Rieks were motivated by was, I'll just say, coxsweter group or coxsweter complex, and the walls are associated. So that's going to be the set S is the coxsweter group, the elements of the coxsweter group, the walls will correspond to the reflection, the reflection walls, right? So each reflection, when you have this coxsweter complex, each reflection, or even if you imagine a coxsweter group acting in one of its usual ways, each reflection kind of give you a wall, okay? And so you can cook up a wall space from that. If we were being really, maybe I'll be particular about this. So for instance, in a dihedral group of order, so why should the, let's see, of order 3 or 6 depending on who you are, degree 3 I guess. You can see 3 walls over here, right, associated to 3 reflections, okay? And, you know, I personally, I'm not so picky about the idea that the walls should be, should be an actual partition. I'm okay with them being just some, a union that adds up, a union that is equal to S, and, you know, I allow the two half spaces to kind of intersect along what I like to think of as the wall. And that's a, that's a workable definition for what we're going to do, okay? And it's a little bit easier in a way because that way there is a thing that we actually call a wall, if you actually manage to get the two half spaces to intersect. But needless to say, over here, in the most elegant version of this definition, S would be these six points, right, the six vertices for instance, and we have our three walls. So now we talk about Sagiv's construction. It is going to be a process which, so this is going to be a construction which takes as, takes input a wall space and gives output a cat zero cube complex, which we call the dual to that wall space. So let me describe Sagiv's construction and really what I'm going to be doing is I'm going to be describing the cubes of the cube complex. So the main thing is the idea of a zero cube. That's the hardest thing over here. A zero cube in this output, in this dual cube complex, so maybe we, let me be a little formal here, a zero cube of this dual cube complex is a choice of half space for each wall. So what do I mean by choosing half space? Well, if you've got yourself a wall, then you could say I'm going to choose the left half space for the wall or you can choose, or you could say I'm going to choose the right half space for the wall. So you're going to need to choose a half space for each wall and this choice is going to have to satisfy two conditions. The most important one is that such that no two choices, no two chosen half spaces are disjoint. So what that means is that heuristically this picture is not allowed. Because I've chosen, it's kind of nice little communication device to put a little normal vector on your wall to indicate what your chosen half space is. This is going to be a choice for each of these two walls, but it's problematic because this chosen half space, the yellow chosen half space and the orange chosen half space are disjoint. And the other condition is a little bit, you can ignore it for now, but let me spell it out. All but finitely many choices, all but finitely many chosen half spaces contain some particular and hence any some fixed point little s0. And so here let's indicate the sort of behavior that we don't accept, although it can fail in other ways. So here's our point s0 in our set, the underlying set of the wall space. And then maybe there's a sequence, it doesn't have to look quite like this, but that's the easiest way to get the idea across. This is not allowed. So these are not allowed, forbidden. S0 is one of the points, so my set s probably it has a bunch of points. And you just choose some point, it doesn't matter which one, maybe this one, that's one of the points. And then we require that, well maybe there's a few of the walls that are, a few of the half spaces in our zero cube, maybe a few of the half spaces won't contain s0, but almost all of them do. But the question is quantified, do you say there exists s0 such that you also hold s0 equals? There exists for some fixed point, so you choose s0 ahead of time. Okay, and then, thank you, you choose s0 ahead of time. And then you want, and then it's also true because of this requirement over here that it doesn't depend upon your choice. Okay? Okay. Yeah, thank you. Okay, yes, please. How does two correspond to your picture? It looks like your choice of half spaces, interestingly many of them don't contain that first point. Right, so I don't like that. Okay. I like to draw the things that were not allowed. Okay. Huh? Yes, so, and they don't have to be in this kind of descending chain. They could be, you know, they could be often in all sorts of different directions. Right? They could have been a forbidden, it could have been a forbidden picture which looked like this, et cetera. And our point is over here, but there's more and more walls that are... That breaks the wall as well, right? Yes, it does. That's a great point. Who said that? So, I like this definition because I like it that a zero cube has a lot of stuff. So, I don't know, it's a lot of information. Now, we need to say what the one cubes are. We need to say what the one cubes are and it's actually pretty simple from here. I should really illustrate this. Let me say what the one cubes are and then I'm going to illustrate it, excuse me. So, there's going to be a one cube. It's going to join two zero cubes if they differ, if their associated choices differ on exactly one wall. So, I'm going to illustrate that in a second. And then, we're basically done. You then, we've built the one skeleton, add an n cube for larger n whenever you see its n minus one skeleton is there. So, you add higher cubes wherever you can, starting with the squares and then the three cubes and so forth. So, let's breathe. So, it turns out this is going to be cat zero. It's going to be simply connected and non-positively curved. Let's breathe some life into it with a few examples. So, maybe the most traditional example goes back a few hundred years, maybe 100, 150, something like that. So, this is a situation where no two of the walls cross each other, so to speak. We say the walls two walls cross if all four possible intersections of their half spaces intersect, have non-empty intersections. So, let me first draw the dual for you. Is there enough contrast between what do you guys like? Orange or red? Or orange or red? Red. I mean, I'm not talking about what you like. What do you see better? Red. Okay. So, this is, here's my wall space. My walls are the W's and my dual is this cube complex C. Okay? And let's look at one of the zero cubes and try to decide what... Let's look at one of the zero cubes and decide what choice of half space for each wall it's associated to. Okay? So, are you able to see the purple or not? Okay? So, that's a nice one. We're simply orienting everything towards this point right here. Okay? So, it's very easy now to start imagining other zero cubes. And how do you know that, how do you know the axioms are satisfied? Well, this is not a concern because any two half spaces contain my big point. Right? So, we're okay. And this one also. Let's now ask ourselves, well, what's the difference between the purple zero cube and the orange zero cube? With the orange zero cube, everything's going to be the same except for one small difference. And the difference is right here. I've taken this half space and I flipped it so that it goes in that direction. Okay? So, this one cube over here between the purple and orange zero cube is kind of associated specifically to this wall. Okay? So, now you see a one cube. Right? Well, now you know the one skeleton. It's going to get more complicated. There weren't any, there weren't any higher cubes to add. Right? Because we don't see there one skeleton. It's a tree. It's actually, it's not the most traditional usage of the word dual, but it's actually dual. Right? I think it's dual. It's a slightly more conventional dual, right? That's where we're even more familiar with the usage of the term. Okay? So, here again, these are my walls and I'm going to draw the dual and it's actually going to be even more familiar than the case of the tree. In this case, because there's a very simple, there's something very simple about this type of wall space that I've drawn. Each zero cube is actually going to correspond to a point in the complement of all of the yellow walls. And I suppose we should play, we should play the same game. Let's choose, let's focus on a particular zero cube. And it's again the case that it's associated to the choice of, it's associated to the choice where we're directing all these normal vectors towards that zero cube. That type of zero cube is called the canonical zero cube. So, and then, you know, if we flipped it over to this orange one, you know what the change is going to be. It's just that change right over there. So I'll just put the change that way. And if we flipped it to this green one, then there'd be this change here. Green is not so good. Let's see what this does. To check that those are the only zero cubes, it's, of course, it's a finite problem, but is there a way to do it, it's hard going over all the possibilities. What do you, I don't know what that means. So you define what a zero cube is in terms of the choice of half spaces. So you have some evidence, zero cubes correspond to the point you've drawn. So maybe there's a surprise zero cube that we didn't see. Yeah, we're going to do that. So let's, maybe I should move forward. So in a way, all this picture is exhibiting. It's, you know, you're looking and you're seeing that if you, you could flip this one, you could flip that one. If you flip them both, you get here, right? You get to this white zero cube. If you flip, if you flip the, if you flip this one, maybe the white one's more interesting. If you flip that wall and you also flip this wall, right? And you leave all the other walls just like the purple zero cube wanted. Then you get this zero cube. Okay, so we need to do more exciting example. So I'm going to do that, right? Because otherwise this isn't so interesting because it's just like something that we all know about. We learned about when we were, you know, when we were in high school, I guess, right? These little dual pictures, yeah? So, so what's a more interesting example? Okay, so we already know the game and we know about these canonical zero cubes, right? Where, where you just, where you just orient all the walls towards some point and that's going to satisfy the axioms. And we found seven, we found seven and maybe, maybe it's a good idea just this one time to, to do this. Okay, and what I'm going to do is let's call our walls the, the first, second, and third walls. And actually let's direct them. Let's make the, make this choice so that we're starting at this zero cube right over here. Okay, that's going to be our sort of neutral choice. And now let's, let's name all of the other zero cubes, right? Because we can make a little naming system now. I suppose maybe you'll forgive me for calling neutral one and let's, let's name, let's name them 111. That's all of these and forgive me for, for the number one being that, that just means neutral. So 111 and let me, let me proceed. 110, 100, 101, 001, 011, 010, right? So, so 010, that's this, that's this guy right over here, right? That's 010. And why is it called 010? Because we've flipped the, we've flipped the first wall. We kept the second wall in the, in its, in its original position so it hasn't changed. And we've flipped the third wall, right? So, so, right? So we, you got from 111 to 010 by flipping two walls. Now what about, so, so now we see that there's another zero cube that we left out before, right? And that's, that's 0, 0, 0. We like 0. So, so that's the, the type of zero cube, a sort of secret, secret zero cube that was hidden. And well that, that corresponds to, that corresponds to this orange cube. It, it corresponds to reversing all three of these. And you look at these three orange chosen half spaces and you see that their, their, their triple intersection is empty. So the 0, 0, 0 cube is, is not a, what I call the canonical cube, right? So, excuse me, canonical zero cube. Thank you. Okay, I, I said what a, what a canonical zero cube was. Shall I, shall I say it? Shall I say it again? I'll put it over here. But it really belongs right after we defined what a zero cube and a one cube was and we said what the cube complex was. It's an observation that zero cubes exist in the dual. Just orient all half spaces, orient all walls towards a point of s. And you'll get, let's name that point, little s, obtain a canonical zero cube, cs. What, what was your question? If s is empty, there's no such canonical. If s is empty, there's no, there's no, okay, good. Okay, you're fine. So, yeah, so the truth is, the truth is, I'll give you a private confession now. I too love the empty set and I normally don't even, I, I, it was just as a public service that I told you about the set s. Because I, I only imagine the walls, right? And so I got a bunch of walls and, you know, you can wonder what, what, what they mean even, but it never bothered me. And, and then I just make sure that I don't have two walls pointing away from each other and I don't have, you know, certain, certain terrible things. Actually the second type of terrible thing doesn't bother me very much either because it's really some, it's something at infinity which I'm actually quite happy with. And points, points in s, I never give them much consideration, okay? So, oh my goodness, I'm super, super slow today. Exercises, well, started at 9.15, all right? Time flies when you're having fun. So let's, let's just call it an exercise that C is non-positively curved and C is simply connected. Both of these are actually easy. One of them I even put in the exercises even though I'm sure it's one of the ones that you're not going to get up to because I put too many. And let's move forward. Let's discuss the notion of a co-dimension one subgroup. So I guess let me sort of invite you to exercises. So one of the things that I will ask in the, in the, in our exercises is, you know, to compute the dual to a system of N families. Of parallel walls in R2, for example. So you can have some orange ones, some red ones, right? And I'm imagining that each of these families is infinite. So you get some nice wall space like this, right? And I want you to figure, figure out exactly what the dual cube complex looks like. You will be able to, okay? Let's now talk about co-dimension one subgroups. Yes? Those two pictures, have you, have you actually drawn all of those there? In the first two pictures, there are no non-canonical zero cubes. So exercise when, when W does not have three pairwise crossing walls, there are no non-canonical yuck. Every zero cube is, is canonical. Thank you. Where were you when I was writing, making my exercises? Huh? That's, that's good. So, and that's very well, very well observed. And that somehow things get interesting as soon as, you know, as soon as you have a bit of a mess, right? And it's, it's exciting because there's a, we love dual because a dual means a secret, right? That there's a secret and then we feel that we're in the game and other people don't know. So if the secret gets exciting as soon as you're starting off with a wall space that seems kind of tame and you look, and there's a dual and the dual is something gargantuan, something maybe could be infinite dimensional. And, you know, with a very, very rich local, local complexity. So that's actually what happens. So let's, let's see where these come from. So I'm not talking about co-dimensional and subgroups. G be a finitely generated group with caligraph epsilon. A subgroup H of G is co-dimension 1 if there exists r more than 0 such that the caligraph minus the r neighborhood of the subgroup, right, which is just a subset of the vertices of the caligraph has at least two H orbits of components, K that are in the sense that K is not contained in any finite neighborhood of H. Okay, I need to, it's a little bit of a mouthful. So what you should be, let me spell out examples before we try to express the definition a little bit more humane form, because really we should be giving an incorrect definition which is more humane, and then we'll think about it a little more carefully. But just an example quickly to have in mind, any copy of Zn in Zn plus 1 is co-dimension 1. Any cyclic subgroup in the fundamental group of a surface of an orientable surface is co-dimension 1, right? So this is sort of suggesting the terminology. What's the picture? Well, let's first look at a very explicit example of Z inside of Z squared. So here's the caligraph of Z squared and here's a, here's a, like orange, like red. Here's a cyclic subgroup, maybe a nice diagonal cyclic subgroup. And the first part of this definition is, well, the subgroup should cut the caligraph in half, right? It should cut the caligraph. It should coarsely cut it. We want the caligraph minus the neighborhood of the subgroup should have more than one component. That's the first thing that we want, okay? So, well, I guess it already cut it right now. Maybe this wasn't an exciting example. Maybe I'll leave it as it is. I'll change it here, here. Now I chose a slightly, I chose an index 2 subgroup of the one that I had before. Okay, so now my subgroup does not quite cut the caligraph when you remove the subgroup from the caligraph. It doesn't cut it because you can sort of still travel through, right? There's a little gap over here. But you could take a neighborhood, right? And so is it a closed neighborhood or an open neighborhood? It doesn't really matter. The definitions don't really matter. You can, let's assume it's a closed neighborhood, okay? And let's take a two neighborhood. Let's take a two neighborhood of this subgroup. What does it look like? One, two, one, two, one, two. I think that's what it looks like, right? And now, as you can see, this yellow neighborhood of my subgroup. So I have the caligraph of the group. I have my subgroup. I have the, what, two neighborhood of the subgroup, okay? And it cuts, that yellow neighborhood cuts the caligraph into two, it cuts it into two components. But we want, we want those components, there's two things that we want to say about the components. One is that we want those components to be deep, right? We don't want, we want those components to be non-trivial, to be sort of impressive. So the way to state that is to say that they get very, very far from the subgroup. And that's the case over here, okay? So these components are deep. The two white components that you see are deep. They go far from the subgroup. And the other thing that you see is that, this is a little more subtle, has at least two H orbits of components, right? So actually when you look at how the subgroup is acting on this situation over here, it's quite nice. It keeps, there's actually only two H orbits, in fact, right? The top white component and the bottom white component don't get exchanged by the action of the subgroup. Now, if this had been maybe a Klein bottle, right? This could have been the caligraph of a Klein bottle, for example, of a Klein bottle group. And maybe the subgroup might have been acting by a glide reflection, then how many H orbits would there be then? Right, if H were a glide reflection, there'd only be one, and it wouldn't count as a co-dimensional one subgroup. Okay, so those are, and if you look in the literature, you'll see people make mistakes about that, myself as well, many times, especially in the early literature on this topic. I mean, so the notion of co-dimensional one subgroup is a very old notion, but it kind of became really very important after Sagiib's work. So let's see how this is now related to wall spaces. Oh, I should say this. Here's an alternate and elegant, alternate definition of co-dimension one is the following, comes from the following. So H is co-dimension one, if and only if, so this is a, G is this finitely generated group with caligraph, epsilon, H is co-dimension one, if and only if the coset diagram, the coset graph, people sometimes call it the Schreier graph, has more than one end. Okay, so if you have a graph, the number of ends of the graph, having more than one end means that you can remove a compact subgraph and get several infinite components that are, and get several, and get more than one component, which is deep in the sense that it's far from this compact subset. Okay, so there's a way of defining the number of ends. This graph will have three ends. This graph over here, a kind of train track, will have two ends, right? Because you're kind of cutting just this one point, doesn't do anything, but cutting a little more, you've got two ends over there, and so forth. On the other hand, this caligraph over here only has one end, because no matter what compact subgraph you remove, there'll only be one infinite. So if it's a locally finite graph, let's just stick to a locally finite graph, there'll only be one infinite comp-metric component when you remove a compact subset. Okay, so the number of ends is the maximal number, or the largest number allowing infinity, or the limit of the supremum of the number of infinite complementary components as you remove larger and larger compact sets. You've uncountably many ends. If you define the space of the end as the inverse limit, then if they are all finite, but in some cases you could get like a control set of ends. Well, okay, I don't want to stop and think about that. So I think I'm going to continue, and then I'll talk to you about it. I'd love to talk to you about it afterwards, okay? So to think about this lemma, what's happening actually is that the neighborhood, when you think about what's going on with that lemma, the neighborhood that you're removing, the neighborhood of the subgroup that you're removing, is going to correspond to this compact set that you're removing in the coset diagram, right? This is the graph that we're interested in. We want to say that it has more than one end. So the neighborhood that you're removing, the neighborhood of your subgroup that you're removing, is this compact set that you're removing from the coset diagram. And you want to have at least two infinite complementary components. Well, the complementary components are going to correspond to the H-orbitz. The complementary components over here in the coset diagram are going to correspond to the H-orbitz of deep components that we're interested in. That's the sort of connection to this lemma, and I think it's a much more elegant way of stating it, but maybe not always the way we think about it. So let's now conclude this by explaining the work. Let's now explain how we're going to get from a co-dimension one subgroup to a wall space. We already know how to get from a wall space to a cat zero cube complex. We've discussed the co-dimension one subgroup. Let's talk about how to get from a co-dimension one subgroup to a wall space. So what did Sageev do is he, the co-dimension one subgroup, here's his construction. Let H be co-dimension one subgroup. Choose R such that the R neighborhood of H has at least two H-orbitz of deep components. In particular, it has some deep component K, and then let W be the wall where one part is K and the other part is its complement. And then I'll draw you a picture of all this in a second. Our collection of walls just be the set of all translates of these partitions. So we obtain, so this is going to give us a wall space on G. So S is going to be G. Well, I guess you can let S be the caligraph if you'd like as well. That's fine. This yields a wall space with a G action. So not only do you get, so what will happen then is that you obtain a dual with an action by your group. So this is the game. So the picture is the following, very simple picture. Here's our caligraph and here's our subgroup and it has a little neighborhood, which is cutting our caligraph up in an interesting, in an essential way. And then, well, we know that there's lurking over here, there's a way of deciding, right, just maybe choose K to be one of these, choose K to be one of the deep components. There's a lot of leeway when you're playing the game over here. K would be one of these and then the complement of K, so that's going to be left W. And then the complement of K, if you're being very particular about this, this might be right W over there, so that's the complement of K. And then if you corsify your vision and you're not so particular about this, then you might just say, okay, I got this wall over here, and there's this side and that side, that's my wall. And then when you take that wall in the caligraph, so I'm just looking at it in a kind of more summary, heuristic fashion, and when I take that wall and I translate it around by the group action, that's that entire collection of translates of that wall, so there's going to be this one and this one and this one and this one, those are just translates. And, well, probably a translates of this wall is not going to cross itself. So when you translate this wall, well, maybe it crosses itself, so you might get that one and that one over there and this one and so forth. And now you have a whole bunch of walls, and the group is acting, permuting these walls around. And so the dual cube complex, which we now know how to define, we're already prepared for that, is likewise going to get acted on, I think I switched my conventions, but there's this dual cube complex, and it's a very concrete example of here, it looks like everything is canonical, but it's not usually so simple, okay, it isn't really such a big deal either way. Well, this cube complex C is our goal and the action of the group on the wall space gives you an action of the group on C. Okay, so G-ax on C. All right, well, of course this construction works with several co-dimension 1 subgroups and many things about the setup can be flexed a little bit. The one thing that you might consider, it's an interesting exercise, why is the number of points separated by walls finite? So you kind of imagine your points, it's pretty convincing that there'll be finally many walls that separate P and Q because there'll be finally many walls whose yellow neighborhoods even intersect this white arc, this white geodesic. Okay, so that's more or less the reason. Another thing to note, and then I'll stop there, is that as you can maybe guess from the construction, the stabilizers of hyperplanes of the dual cube complex are pretty much end up being commensurable with conjugates of our co-dimensions 1 subgroup H. Okay, so I'll stop now and I have to stop. Good.