 OK, I'm going to pretty much pick up where I left off. So as I said, I want to tell you about some geometric techniques we've been using to try to understand some, answer some of these questions about art and groups. And the first geometric object we want to play with is called the Deline complex. So let's review, let me recall what that is. So we have our defining graph gamma. And we take D gamma to be the, it's a cube complex with vertices, our cosets of special subgroups. Remember, T is a subset of S, and we look at the subgroup generated by T, where we require that AT is finite type. So remember that just because the, oh, by the way, you should assume we're working with A gamma is infinite, that the full group is infinite type. Otherwise, these aren't interesting. We're trying to deal with how do we answer some of these questions for infinite type. So I'll be mostly interested in the case where the full group is infinite type. But it could have finite type pieces in it. For example, any edge generates a finite type. Art and groups, it'll have subgroups that are finite type. OK, so vertices are that. Edges are simply where we get from one coset to another by adding a single generator. And cubes are where we have a get, well, that is a cube. That's a one dimensional cube. Higher dimensional cubes are what I call intervals where we can add more than one thing. So AR, where T is contained in R. And it just adds one thing. We just get an edge. If we add two things, we get a two cube. If we add three things, we get a three cube, et cetera, et cetera, et cetera. All right, so that's what the deline complex is. And we're interested in this because, as I said, originally we introduced it because we were trying to prove the SOFAC, which I'm not going to prove, is that D gamma is homotopy equivalent to this hyperplane complement, the universal cover of the hyperplane complement that we would like to prove is contractable. So the question is that we want to ask is when is, so question, we have this cube complex. We want to know is D gamma cat zero? OK, and why do we want to know that? Let's just recall some basic facts about cat zero. Actually, maybe I should say, let me say instead, I mean, we do want to know if it's cat zero. But what we're actually going to check is whether it's non-positively curved. Is D gamma non-positively? OK, so let me recall some basic facts about x is a geodesic from this morning, geodesic metric space. Then the point is if you have x simply connected and non-positively curved, that implies that it's cat zero. And if it's cat zero, that implies that geodesics are unique. So what do I mean? I mean, given two points in your space, there's one and only one geodesic connecting them. That's what I mean by geodesics are unique. So given two points, there's a unique geodesic. By the way, I'm using the cat zero metric here, not the taxicab metric. Yes? OK, so straight line. So if I want to get from here to here, I go this way. I don't go around the outside. All right, I'm not a taxicab. What is the definition of non-positively curved? So now it is metric space, not a cube complex. No, no, cube. Oh, I'm sorry. I'm sorry. You're absolutely right. No, I mean, there is a non-positively curved for general. I'm going to be interested in a cube complex. It just means locally. I mean, this makes sense for any geodesic metric space. It just means locally cat zero. It's a local version of cat zero. Yeah, yeah. In my particular case, it's going to be a cube complex, so non-positively curved will be proved using the flat. We'll get to that, how to prove that. But this is actually true more generally if you define this to just mean locally satisfies the cat zero condition. OK, it's not important. We'll get to this in a second. So it's cat zero, geodesics are unique. And here's the point that I'm not sure whether Danny said this morning. Once geodesics are unique, x0 is contractable. And this is the point that we want. Why is that? Well, if I have any space, x, and geodesics are unique, then I can pick any base point whatsoever. There's a unique path to any other point in the space. And I can just contract the space along those paths. And you could show that that's continuous, and blah, blah, blah, blah, blah. So it's just general nonsense about cat zero geometry. Once you have cat zero, you automatically have contractable. And that is actually what I want, is contractability. All right. So I would like to prove this is contractable. That's the k-pi-1 conjecture. So it suffices to prove that this is contractable. So it suffices to prove that this is simply that d-gam is simply connected and non-positively curved. I already know it's simply connected. That follows from this. All right. It's automatically simply connected if it's homotopy equivalent to a universal cover. All right. So I already know this. And what I need is this. So we're reduced to showing. So need to show. Or we want to know that d-gam is non-positively curved. And how do we do that? Now we really are working in a cube complex. We genuinely look in a cube complex. And we know that non-positively curved, to show things non-positively curved, we need to show that the links of vertices are flag complexes. All right. So d-gamma non-positively curved if and only if links on links of, put it here, if and only if links of vertices, of all the vertices, are, I'm going to capitalize this for a reason, flag complexes. And now we need to remember what it means to be a flag complex. Because we're going to try and check. We're going to get in there and look at it and see if it's a flag complex. So what does it mean to be a flag complex? It means, so the link is a, is a, is a, is a suplicial complex, right? Because the link inside, so we're at a point. And inside every cube, the link sees a little, a little simplex, all right? So the link's a simple complex. A flag complex, remember, is a suplicial complex. Here's how I like to think about it. This is in a formal definition. Simplicial complex with no, what I call, empty simplices. What do I mean by that? So what I mean is whenever I see the one skeleton of a simplex, that simplex is there. It's filled in. You never see, for example, a triangle with no filling in it, all right? Whenever I see a one, the one skeleton of a simplex, the simplex is there. Instead of talking about one skeleton of simplices, I'd like to talk about, so the one scale, so the, so the, so the one, the one dimensional space is, is, is a graph. I mean, we're looking at a graph. So instead of saying on the one skeleton of a simplex, it's what I call a clique. Have you ever heard the word clique? A clique is a graph where every vertex is joined to every other vertex by an edge. So another way of saying this, that is. So what do I mean by this? What I mean is every clique in the one skeleton spans a simplex. Everybody say k clique. Clique with k spans a k simplex. OK? All right. OK, so that's what we've got to check. We're going to look at the, at, at cliques in the link and ask whether they span simplices. All right? Let's do it. I thought I'd actually do something like get in there. And, you know, I'm hoping that, I'm hoping some of you looked at some of the questions on this. All right? So we need to understand the links of vertices in d gamma. And I'm not going to do a complete proof. I'm going to do a link of a particular vertex where you already see what can go wrong. And it turns out that's the picture, basically, for all of them. But let's do, let's look at. So consider the vertex. So I'm going to look at a vertex in d gamma. Just call it x is the vertex corresponding to a empty set in d gamma. All right, so a vertex in d gamma, where are they? There they are. They're cosets of finite type subgroups. All right? A empty set is the trivial group. It's as bad as finite type as you can get. I said those things can be, those t's can be empty. I'm taking t to be the empty set. And that is a vertex in d gamma, all right? It's not the only kind of vertex. But let's look at that one and study it carefully. So what is the link in d gamma? So I'm at the a empty set. And I'm going to look at a little kind of sphere around it and ask what I see. Well, the vertices in the link come from edges that emanate from here, all right? They hit this thing in a vertex, right? So the vertices come from edges. Well, what are the edges attached to a empty set? Well, any generator generates a finite type. Any single generator is finite type. So I have something for S1. I have one for AS2. I have one for every generator, et cetera, yeah? Each one of those is a vertex. So there's a vertex. So link on x has a vertex for each SI, for each SI. Each SI gives me a vertex, namely the edge that goes to ASI. Everybody understand what I'm saying? For each SI, I have one of these edges. So there's one for each SI. When are two of these joined by an edge? Well, they're joined by an edge in the link. In other words, when does this exist? It exists precisely when this is some cube, precisely when there's a cube there. So when is there a cube here? Well, it's got to have another vertex up there. And what does that vertex have to be? It has to be AS1, S2. So in order for this to be here, it has to be AS1, S2. Well, is that allowed? It's allowed if and only if those two generate a finite type subgroup. In other words, if and only if the MiJ is not equal to infinity. So two of these, two such, are connected by an edge in the link, if and only if. So I have an SI and an SJ, if and only if a SISJ is finite type, namely if and only if MiJ. So if and only if, let's make our lives easy, spans an edge in gamma. That is, MiJ is less than infinity. In other words, there's an edge, if and only if there's an edge in gamma between them. Yes? Because otherwise, if there's no edge, if MiJ is infinite, then this cube is not allowed. It doesn't exist. All right? OK. Well, guess what? We just constructed the one skeleton of the link. We know what the vertices and what the edges are. What is it? It's gamma. It's exactly gamma. It's one vertex for each SI and one edge for each edge. All right? So that is so the one skeleton, that is thus the one skeleton, the link of x is exactly gamma. It can exactly be identified with gamma. A vertex for each SI and edge whenever SI, SJ. So if I look at the link, I know what its one skeleton is. It's exactly, I just see gamma. That's it. All right. So what we want to know now is whether, so now we know what the one skeleton is and what do we need to check. Well, we need to look at cliques in gamma and ask whether those get filled in in the link. Whenever I see a clique in gamma, that's a clique in the link. I need to know whether it gets filled in. All right? So let's figure out if it gets filled in. So we need to check whether every clique in gamma is filled with a simplex in this link. All right, so let's draw one and try and construct it. All right, so maybe let's just draw with three, but it could have more. Let's say I have an AS1, an AS2. I mean, it doesn't matter what the numbers are, but I have three of these, right? And they generate a clique. What does that mean? That means that these two, there must be a cube here. So I'm looking at the link, right? And I'm drawing part of d gamma here that's around this. So there's a cube there. There's a cube here. And then these two also join a cube. So I have three faces, but that's not good enough. That just forms this. And I want to know is that filled in? So to fill that in, this whole picture has to be in some cube. There has to be some cube that completes that this whole thing lives in, yes? There has to be a three-dimensional cube spanned by those, right? We've got these three edges, and I need a three-dimensional cube to fill it, all right? What's going to be at the other end? I'm not drawing this very well. So what's going to be at the other end of this three-dimensional cube? Yeah. What's going to be the, that's not drawn very well. It's the diagonal of a cube. What's at the other end of the diagonal of this three-cube? I'm having a hard time drawing this cube. Can anybody see what has to be there? It's got to be, just like it was there, at the other end of the cube has to be S1, S2, S3. Right? All right, it has to contain them all. It has to contain this, this, this. I haven't drawn this very well. It's supposed to look like a cube. So I have these three edges, and in order to complete that cube, the other end of the cube has to contain all three of those, all right? So it has to be A, S1, S2, S3. So what are we saying? We're saying that whenever they're pairwise, so in order for this to be true, we would need this to be finite type, otherwise it doesn't exist. So the issue is, if I have a bunch of things which are pairwise finite type, is the whole union of them finite type? And the answer is, sometimes yes, and sometimes no, and then we're in trouble. So here's the condition. I will now write out the condition for this thing to be. So here's the claim. Is that that the length of this x is flag, if and only if, for every clique, for every clique. And I'm just going to write t in gamma. What I mean is a set of vertices which span a clique is what I really mean. For every set of vertices which span a clique in gamma is what I really mean. At is finite type. Every clique, at is finite type. I'm going to give you examples in just one minute, but I want to introduce terminology, definition. a gamma is, we say a gamma is fc type, is a fc type if this condition holds, if every clique, if star holds. That's it. I simply would refer, sometimes, as I will show you in examples in just a second, sometimes this holds and sometimes it doesn't hold. And we call it fc type if it does hold. Why fc flag complex? The fc I was going to underline it, but it disappeared up there. fc stands for flag complex. It's exactly what we need to guarantee the lengths of flag complexes. You just guarantee that the length of that vertex. Yeah. So you have to prove this for all. I haven't written the statement, so I'm going to write the, after I do the examples, I'll write down the theorem. It turns out this is all that goes wrong. So it turns out if you look at, so what happens if you look at any other vertex is you have what's called the upward link, things bigger than it, where exactly this problem occurs. And then you have the downward link, which are things smaller than it, where no problems ever occur. So if there's more, you're absolutely right. I haven't checked all the vertices. You have to check all the vertices. But the issues all show up right here. Turns out they all show up right here. The rest of it is you can, you know, takes a little bit of lemmas, but it's no problem. All right? So if you're going to see any problems, you're going to see it right here in this link. Okay, so I'll write down the theorem in a minute, but let me do some examples so I make sure everybody understands what fc type is. So examples, here are three graphs and the question is are they fc type or are they not fc type? Three, three, three, three. This is gamma I'm drawing. This is gamma, so eg. Here's some graphs, gamma. There's one, here's another one. Three, three, three. I'm putting a lot of threes. We're playing with threes today. Three, three. And the last one is any graph. Right angle, right angle, right angle. Meaning any graph where everything's labeled two. All the edges are labeled two. That's right angle. Right angle means every m, i, j is either two or infinity. Yes, the right angled ones, the only relations are commutators. So right angle means any graph you want as long as all the labels are twos. All right? Okay, is this fc type? So the question is fc type or no? Yes, why? Well, the only cliques are edges. There are no other cliques in this graph. Cliques are sets of vertices all of which are joined by edges. So the only cliques are edges and a single edge labeled three. Well, that's the braid group on three strands. That's definitely a finite type. That one, any single edge is a dihedral. Is a dihedral. Okay, so yes, no problem here. What about this guy? Well, this one has a clique right here, right? Those three guys generate a clique. And that 3, 3, 3 thing was the coxer group for that was that affine reflection group that I said was infinite and I used it as an example of a case where you get something infinite even though there's no, the m, i, j is or not infinity. This was that affine coxer group and it's infinite, bad. So no, it has a bad. In fact, the minute you see a 3, 3, 3 triangle, forget it. You're lost, you're gone. Okay, right angled art and group. So if I see a clique, it means I've got a bunch of generators all of which commute with each other and oh, I guess we need to know what the coxer group for that is. What's the coxer group for that? If I just take a bunch of generators all connected to each other and put twos everywhere. Anybody know what the coxer group for that is? Yeah. It's just a direct sum of bunch of Z mod 2. Yeah, it's just, they all commute. It's just a bunch of Z mod twos that commute with each other, so it's finite. Z mod 2 plus Z mod 2 plus Z mod 2. They all commute with each other. So a bunch of generators of order two. So the coxer group's finite. So the answer is yes always. We mean no conditions whatsoever. On rags are a special case of FC type. FC type, all right? Just a special case, all right? Okay, so, okay, so what's the point? I claim we've actually sort of more or less, we haven't proved it because we haven't checked everything but we've kind of more or less proved the following. Let's see, here, theorem. D gamma, the Deline complex is cat zero or non-pi's of the curve or equivalently cat zero if and only if a gamma is FC type. So the answer is there are lots of nice infinite type art and groups for which it is and lots of nice infinite type art and groups for which it isn't. I mean it's just, that's the way it goes, all right? Okay, so once we know it's cat zero, we have the fact that I discussed at the beginning, it implies that it's contractible and hence K pi one conjecture holds. So this immediately applies K pi one conjecture holds for a gamma by the argument I gave up there that if we could show that it was cat zero then it's contractible and hence this is true, all right? That was the argument I gave at the beginning. Okay, so that's great, I mean that's kind of, as I said, this was, Mike Davis and I did this and this is what we were setting out to do was to prove the K pi one conjecture. There I should say, I'm not gonna talk about it in this talk, but there are some other metrics one can put on D gamma and there is one called the Musang metric which is conjectured to be cat zero for all gamma but nobody's been able to prove it. We were able to prove it when it was, D gamma was two dimensional but we can't prove it for anything bigger than that so it's possible that there are other metrics which are cat zero but the cubicle metric is cat zero if and only if this FC type condition is satisfied, all right, okay. So, okay, so now what? Oh yeah, I wanna say it's better than that. So it turns out, and let me put a few names on this because there are a bunch of people involved in this. So first Mike Davis and I, then one of my students Joe Altabelli and then Eddie Godow actually proved that if A gamma is FC type then all of the conjectures, the old, that list of conjectures, the conjecture, the star conjectures, all that list over there hold, or true, hold. And I'm gonna say a little bit about how you prove the others because they all sort of depend on this one. They all sort of, all but one of them kind of follow once you have this. So let me say a little bit about this. So I'm gonna work my way up words so we know, so we say okay, we know this is true, all right. So why is there a finite k pi one? And for that I need to tell you what the self-edit complex is which is interesting anyway. So in addition to this Deline complex, by the way, by the way you might say doesn't this show that A gamma is cat zero? And if you did the problems that I suggested, it doesn't because the action's very much not proper. It's co-compact but it's definitely not proper. It has these huge infinite stabilizers. So it doesn't, so the group itself is not a cat zero group. The action's not nice enough. Okay, but yeah, so how do we get a finite k pi one? Well, for that we constructed a different complex which is also on the top equivalent to H gamma tilde. And once we know H gamma tilde is contractible, it turns out that this complex gives us a k pi one space. So that complex is called the Salvetti complex because it was first constructed by Salvetti in the finite type case. But we did the general construction for the infinite type case. But he first introduced it in the finite type case. So let's talk about that. Oh, I can go over here. So let's see, so I guess we're, so I'm not gonna do the full proof. I'm just gonna say a little bit about how some of these others follow. So, yeah. How many rags of FC type? Is it, they're not rags, asking you if it's FC type? If I drew a random graph and put a label that I'm not expecting it, is it or is it unusual? Yeah, no, no, it's some, there's, so I actually have a student who's got some actual, basically you have to avoid too many threes. I mean if random, so what do you mean by random graph? So this is a whole, there's a whole paper on this I can tell you about that where you have to have a probability for the labels as well as the graph itself. It's not enough to know the graph. You have to have a probability for the labels. And essentially, if the probabilities of threes are too high, you're gonna see one of these three, three triangles and you're done with. But if you can avoid, you know, if you keep the probability of threes reasonably low, then yes, you can get a high probability of getting ones that are, so I mean, that's not the only condition you have to get. But essentially that's the problem is these threes. So it's an interesting question. I can tell you about where the paper, tell you, look at the paper. So, yeah, oh I should say, I think somebody asked me this in between and for those of you who aren't that familiar with Coxer groups, you know, in order to know whether something's a finite type or infinite type, art and group, we need to know whether the Coxer group for that graph is finite or infinite. Well, Coxer groups for finite graphs are completely classified. You go to any book on Coxer groups and you'll see a list. You'll see a half page long diagram showing you all the finite type Coxer groups. They're one of these and if it's not one of these, it's infinite, all the finite Coxer groups. It's not one of these, it's infinite. I mean, we know exactly which Coxer groups are finite and which ones aren't. So that's not a difficult problem. It's a known, it's a completely known fact and they're almost all infinite. They're only very few that are finite. Okay, so, all right, so where are we? We are, oh yes, I'm gonna just write this as a remark but it's sort of a remark about the proof of this, about some of what goes into the proof of this. So, how do we get the finite, so the finite k pi one that we get from conjecture two. So the finite k a gamma one space is a generalization of the self-edit complex which Danny, you did this morning, right? You've described the self-edit complex for right angle art groups. So we're gonna generalize this. Self-edit complex for regs. So here's what it looks like. All right, so I'm gonna construct it for you. Let's call it S gamma. So let's S gamma for self-edit. So I have my graph gamma and I'm trying to construct a finite k a gamma one space. So here's what I do. I'm gonna start with a single vertex. I'm gonna put on a loop for each generator just like you would for S one, S two, one for each generator. And in the right angle case, what you then had to do is start gluing on tori for commuting elements, right? More and more tori the way you have collections of commuting elements. Well, we're not just interested in commuting elements now, we're interested in any collection of elements that's finite type, that generates a finite type subgroup. So if I have a collection T that generates a finite type subgroups, that means that the WT, so let's say T is contained in S and let's say the coxsiter group is finite. So the art and group would be finite type, all right? Then there's something, for any time you have a finite coxsiter group, you have something called a coxsiter cell and I drew you one of these earlier. So if you have, for example, a two generator thing, you just look at all the, you look at its action on Rn, you look at all the reflection hyperplanes, all right? It chops up your space. You take a point in the interior of one of these regions and you take its orbit and then take the convex hull of that orbit, all right? You can do that in two dimensions, but also in higher dimensions if you have more generators than two. You take the convex hull of the orbit of a point, all right? That thing is some nice Euclidean polytope, polyhedron, all right? You get some nice Euclidean polyhedron and not only that, it turns out, let me draw this a little nicer. It turns out that the one skeleton is exactly the Cayley graph of Wt. So if this is the identity, then I've moved, I've taken the orbit, so every point in here corresponds to a point in Wt and getting across one of these is exactly doing some, can be done by a reflection. So maybe this is S, T, S, and this'll be T, S, T. And notice that if I read around here, it says that S, T, S equals T, S, T, which is exactly the relations I want. So I fill this in, this is called a coxidor cell for Wt. And when I do this one for each T, one for each of these, I make a coxidor cell and then I glue it on. And how do I glue it on? I just glue it according to the labels. I glue the S's to the S's and the T's to the T's, and that's the salivary complex. Now in the case of a right-angled art and group, these guys are just tori. All I get is S, T equals T, S, they're just tori. But now I get some sort of bigger structures and I simply glue one for each of these T's, gets glued onto this, and it's finite. There are only finitely many subsets and I glue on finitely many cells and it's almost obvious from what I said that it has fundamental group. What's the fundamental group? Well the fundamental group is first generated by these with some relations added. What are those relations? They're the art and group relations. In other words, it's basically a no-brainer, an easy exercise to show that this is true. What's not clear is that the universal cover is contractible. This is easy. So the issue is, is this a K, a gamma one space? And to be a K, a gamma one space, we need this to be contractible. And so the proof of four was basically proving that the universal covering space of this is homotopy equivalent to, so to prove, to prove, projector four show what we showed was that the universal cover is also homotopy equivalent to H gamma tilde. It was another space, homotopy, yeah. Is this without restriction on like a gamma being XC type? I guess this homotopy equivalent's always true. This homotopy equivalence is always true. Yes, this one's always true, so does it. And the other one was true too. You know this D gamma homotopy equivalent to H. This was always true too. These two homotopy equivalences are always true. The question is, can we prove any of this stuff is contractible? And we needed cat zero to prove this was contractible. This is always the case. This is always the case, all right? So the point is once we've proved conjecture five, we've also proved conjecture four. So once we know that this guy's contractible, we get that this guy's contractible, and then we're done, all right? So once we know this, we get that conjecture four, five implies conjecture four. Okay, but conjecture four implies conjecture two. Why? Because if you have a K pi one space for a group, then you could just look at the action of a subgroup, just go to the universal cover and just look at some subgroup acting on it, and you get a K pi one for the subgroup, but torsion groups can't have finite dimensional K pi one spaces. So a space with a finite K pi one space cannot contain torsion. This never cannot contain torsion. So that comes out for free, that a gamma is torsion free, and it leaves us with only two conjectures to prove, and I don't have time to tell you much about this. I'll just say that the first one, the solvable word problem, used the cat zero structure to say, well, there's sort of canonical paths between, well, it uses the cat zero structure to reduce the problem to looking at stabilizers, but the stabilizers are all finite type, and we already have the word problem solved for those, so we can just sort of reduce to stuff we already know. It allowed us to reduce it down to chunks that we already understood. So again, the D gamma, the cat zeroness of D gamma was absolutely crucial in proving the word problem for that, and the last one, I can't say anything about, totally different methods. That was Eddie Goodell's work about the center. I'm not gonna say anything about that right now. Okay, so anyway, you actually get all the conjectures. This was just telling you a little bit about why some of them follow pretty quickly and easily once you've got that one. All right, so yeah, let's keep moving. All right, so great. So we have this fairly good sized class of infinite type Barton groups called FC type for which we can answer the questions. All right, now what? All right, in fact, the whole picture sat there since that was in the 90s that most of this was done, and there was no progress at all for quite a long time after that. And so I wanna tell you about some more recent work that is beginning to produce some new results. All right, so now we suppose we have an infinite type one which is not FC type. So this FC fails. So what if? Well, it turns out that we can modify D gamma. So that, to make it cat zero, for all gamma. Okay, what went wrong with D gamma? Well, what went wrong is you had things which were pairwise joined, they spanned a clique, but you couldn't finish it. You weren't allowed to put the AT at this end, that the AT that should have gone, completed the cube, sorry, AT was not allowed. It wasn't finite type. All right, well, let's stop worrying about whether it's finite type or not, and let's just fill in every clique and just say, I don't care if AT is finite type or not, let's just stick that cube in there, that missing cube. All right? So this is called the clique cube complex. Yeah, okay, so define, which we'll call C gamma. It's a lot like D gamma, in fact, the vertices are, well, again, cosets, so GANA is in a gamma, but now we don't put so much restriction on this. All we care is the T spans a clique. Before we required that AT was finite type, I don't care anymore, all I care is that the T spans a clique. I'm allowing a lot more vertices than I was allowing before. So for example, if I see one of those 333s, that's okay, that's allowed here. I'm allowing that, okay? So it's more vertices and the rest are the same. Edges and cubes defined as before. Edges means you add one guy, cubes mean you take an interval, so as before, same rules. So an edge is one where you just add a single generator and a cube is one where you add more than one generator. All I've done is added some extra vertices in here and of course some extra cubes come along with it when I do that, okay? All right, so theorem, and this was first proved by Paris and Nadelle, C gamma, we just got rid of the problem. This cap zero for all gamma, no matter how, for any of our finite labeled graphs. Okay, we seem to have fixed the whole problem, right? What's wrong? Sorry? So you don't want to be clear. Yeah, so exactly. So this is the good news, but then there's bad news. So unfortunately, there's two things to notice that we lose. The first is that C gamma is, well, we no longer know. We no longer know that C gamma, if C gamma is homotopy equivalent to H, we don't know what the relation now is with this guy. Remember we designed D gamma to understand this guy. Well, we've got a very nice complex now, but we have no idea if it's related to this anymore or how it's related to this. So we lost our connection to the K pi one conjecture, all right? Not only that, we're in really bad shape. If gamma itself as a clique is a clique, well, that's certainly possible, why not? So any two vertices and gamma are connected by some edge. Then this guy is totally uninteresting because if gamma itself is a clique, then I'm allowed, this could be a gamma. This could be everything. I could take T equal to S, and that contains everything. All right, so then the diameter of C gamma is finite. In fact, so in fact, every maximal cube contains the vertex, the vertex A, A, S, or in other words, A gamma. If I take T equal to S, if I take T to be everything, the subgroup generated by everything is A gamma, and that contains every other coset. So every cube contains this one vertex. This is vertex that every maximal cube lives in. Well, that's pretty uninteresting. For those of you who do geometric wave theory, you know that if a complex has a finite diameter, it might as well just be a point that's of no use for studying anything really, okay? So it's not much good in that case. All right, so what have we gained? We've gained something and we've lost something, yeah? Okay, so is it good for anything? And the answer is it's still good for some things. So here's a theorem, a recent theorem. Myself and Rosemar is right. It's in the audience. This is one person, two names. I should put prints. What do you, like this? Okay, so here we go. Well, actually, there's part of it is periscodal, but I'll explain which part. So if gamma is not the star of a single vertex, meaning there's no vertex that's connected to every other vertex in gamma, all right? So if it's not the star of a single vertex, then a gamma has trivial center. So this is conjecture three, all right? Basically conjecture three holds under some reasonable conditions. And two, if gamma is not a join, meaning everybody knows what a join is, meaning there are two subgraphs, gamma one, gamma two, such that everything in here is connected to everything in here by an edge. Every vertex here is connected, every vertex here by an edge. Then a gamma is a cylindrically hyperbolic. A cylindrically hyperbolic. And three, and this really was already in the Paris and Goodell. We have some different proofs, but this was already in the Paris and Goodell paper. Perhaps the most interesting thing is that all of the conjecture, so all the other conjectures all, other than the centralizer one, all the other conjectures star on that list reduce to the case where del is a single clip. That is to say, if we could prove them for the case, if we knew how to deal with this case where gamma was a single click, then we would know the conjectures were true for everything. So you can use this complex to reduce the problem down to this case. But we have no idea how to deal with this case. But it reduces us. It says, that's what we've got to look at now. We've got everything reduced to that case. All right, I had originally hoped to actually give a proof of the centralizer one. I think I don't have time, but let me make a couple of comments. This doesn't quite prove this. Yeah. Sorry, when gamma is a three-quick peak? Yeah. Oh, we don't know. It's just the easiest clip, right? Yeah. Well, we know some stuff about that for other reasons, because we know some stuff about two-dimensional. We know some stuff about that special case, but in general, we don't know anything about cliques. I mean, for an arbitrary case, we don't know anything. There are some special, some of the affine ones, John McCannon has some results on, and so there are some other known things for special cases, but no general theorems about the clique case. Okay, let me make a remark. Yeah, there's no way I'm gonna have time for this proof. Let me just make some remarks about this. This theorem isn't as good as we'd like, and the reason is, when I say not a star or not a join, the two conditions, gamma's not a star, gamma's not a join, I'm talking about the unlabeled graph. All right, so what do I mean by that? So the statement that gamma is not equal to, oh, that should be, sorry, is not a joined. It cannot be decomposed as a joined, is what that says, right? So the statement that gamma is not a join is, refers to, and the same with the not a star, refers to the unlabeled graph. So what do I mean by that? So all it's saying is that gamma, I can't divide gamma up, there's a subgraph gamma one and subgraph gamma two, and I cannot divide it into two graphs where every vertex here is connected to every vertex here. That's a join, right? What would we really like to say? We would really like to say that it's a cylindrically hyperbolic if gamma's not the product of two smaller things. We would really like to say, we'd like the hypothesis to read that a gamma is not equal to a gamma one cross a gamma two. By the way, if it's a product, it can't be a cylinder, it's, it can't, there's no possibility that it's a cylindrically hyperbolic. So this clearly has to be ruled out. That's what we'd like the hypothesis to read. So hopefully some of you did some of these problems. What would I need to add? What else would I need to know to turn this guy into this guy? I would need twos on all of these edges, all right? This thing says nothing about the labels on the edges. We would like to be able to strengthen it to say that we only have to worry about the case where they're all twos, all right? So it isn't yet quite exactly the hypothesis we would like, but we're pretty sure we just haven't done enough work that it's just a matter of working a little harder to get that. So I'm pretty sure that a cylindrical hyperbolicity thing is, works, it's just that we, where still our hypothesis is still not quite as strong as we want. And the same, by the way, for the star case, if it was a star with all twos, then it would be a product. And there'd be a center element. Then this element would commute with everything. So of course it would have a center, right? So you obviously have to rule that out, but we haven't quite ruled that out. We've ruled out a star without knowing what the labels are. So the hypotheses here aren't quite as strong as you'd like, but it's still progress. It's still good progress. So I would say, since I don't really have time to, I was hoping to actually do the proof of the center case. I'm done at six, right? Six is good. So then let me just say, instead, I don't have time for this proof, that I think that really, so hopefully you've got the idea that there are a lot of open and interesting questions here, and really the end of this story is, what do we do about the clique case? I mean, the end of the story is, we really are, have no tools at hand at the moment to deal with the clique case. We don't have any nice complexes that they act on. We really, that's the next step. That's the next big project, is if anybody has any brilliant ideas of how to address the case where the entire gamma is just a clique, all right? And I should also mention, because some people have asked, there are other special cases where things are known. There are lots of papers out there if the case where, the two-dimensional case, that's the case where the only finite type ones are edges. There's nothing else's finite type. The lot is known about those. There's some stuff known about some of the affine ones. So there are various other special cases where we have some results, but the ones I've given you are sort of the most general, the most general things we know. So I'm gonna stop there.