 I don't know what to say. I feel very lucky to be here. I feel very lucky to know Alain, to have known him for a while. He's been a good friend. Once I lost a CD and he bought a new copy for me. What kind of a friend does that? That's a good friend. Exactly, Francis Cabral. Yeah. Right, so I want to talk about something which I Well, you may see in a little while why I chose this subject besides the fact that I actually have been thinking about it. So this is my title on the co-homological dimension of automorphism groups of right-angled art and groups. Can you actually see that? I mean the highlighting. Okay, so I started this project some time ago with Ruth Charney and we have some very recent results with Ben Millard, who's a student at University College in London. So I'm going to talk about right-angled art and groups. Most people here probably know what they are already, but let me just go through the definition. So they're finitely generated groups. What makes it a finitely generated group, a right-angled art and group, is that some of the generators make mute and there are no other relations. Meaning you don't have to say any other relations in order to define the group. That gives you the defining relations of the group. So the way you usually describe a right-angled art and group, a general right-angled art and group, is by drawing a graph. So if you have a simplicial graph, then the art and group you get, this is the rag, the generators are the vertices of gamma, and the relations, well, you say that two generators commute, if and only if there's an edge between those two vertices of the graph. Well, there's v and there's w and they're joined by an edge in the graph. Okay, so that way, by just drawing a simple graph, you can describe exactly what your group is. So I thought I would draw some examples on five generators. There's an example of a graph. It doesn't have very many edges. In fact, it doesn't have any edges. So it's got five generators and none of them commute, so that's the free group. I could, on the other hand, I could go to the other extreme, take the same five generators and draw an edge between every pair of vertices. Did I miss any? No, I missed one. Okay, so every two generators commute, so that's a free abelian group. I'm used to having a lectern that's a little taller, but I don't, yeah, but anyway. And then, but in general, you can have pretty much anything. You could have, for instance, these three generators commute giving you a little free group of rank, a free abelian group of rank three, and here's a free abelian group of rank two, and those aren't connected, so the top three don't commute with the bottom three, so you get a z cubed free product with a z squared, etc. Basically, you can do anything you want and you get a group, a gamma. Okay, so if I draw you a picture of a graph like that, then that gives me some, some groups, right angle art and groups. So these are amazingly simple to define, and if you've been paying any attention at all to geometric group theory or pre-manifold theory, you've learned that they have lots, in spite of the fact that they're so simple, they have an extremely rich class of subgroups, which include, for instance, virtually, they include all hyperbolic three-manifold groups, for instance. Sorry? No, finite graphs. These are finitely generated groups. Right, so what I wanted to talk about today is automorphism groups of right-angled art and groups. So if you start on the left-hand side, if you look, you get the group of automorphisms of a free group, on the right-hand side over here, you get the automorphism group of z to the fifth, z to the n, which, of course, is the same as the general linear group, and since I'm basically a topologist, instead of thinking about, I don't, well, I'm the sort of a topologist that doesn't like to keep track of base points, which means I'm not a really good algebraic topologist, but I'm a low-dimensional topologist, so instead of thinking about automorphisms of free groups, I want to think of outer automorphisms of free groups, and of course that doesn't change anything about GLNZ. So these are two classes of groups that have been, we've been studying for quite some time now, and we know a lot of things about them. What I want to do now is try to take the things that we know about these, things that we understand, about them, and try to see what I can learn about, in particular, the outer automorphism group of a general right-angled art and group. So here's a list of properties. There's actually quite a long list of properties by now that are both everything on the left and everything on the right share. There are, for instance, lots and lots of finiteness properties. They're finitely presented, and so that implies in particular that their two-dimensional homology is finitely generated and their one-dimensional homology, and it's also true that in every dimension the homology is finitely generated. Both of them have torsion-free subgroups of finite index. So that means the co-homological dimension of those, you can compute the co-homological dimension of those subgroups, and it turns out to be finite. And this is big news. This is basically Kaluba, Kilek, and also you have to put Ozawa in there. Okay, so if you read Alain's book about property T, there's a section at the end where he has a list of open problems. Number one on the list is does, out of FN and out of FN, do they have property T? Okay, so I'm actually not going to talk. I wish I could talk about property T. Actually Alain invited me to Neuchâtel in 2002 and tried to convince me that property T was interesting. He convinced me it was interesting, but I had no idea how to do anything about it. So I wish I could be standing here right now telling you the proof that out of FN has property T, but you have to invite those guys instead. So, yeah, so the exciting things are happening in the world of out of FN. Excuse me, what did you say about property T? The outer automorphism group? Yes. So does the automorphism? Yes. For N at least six, and at least five, and at least five. So does the automorphism group? I just, since I'm talking about outer automorphisms in this, I mentioned out. So those are properties, that's just a small part of the list of properties that these groups share, and there's lots of other things like they have finitely many classes of finite subgroups, there's homology stability theorems, and the list goes on. But they're definitely not the same group, and there's a lot of ways in which they're different. For one thing, GLNZ is a linear group. Out of FN is not. I mentioned that they both have finite virtual co-homological dimension, but there's a big difference. For out of FN, it's linear in N, and for GLNZ, it's quadratic in N. Another difference, big difference, between out of FN and GLNZ is that every solvable subgroup of GLN, out of FN, is virtually abelian, and in GLNZ, it's definitely not true. You could say it's virtually polycyclic. Property T distinguishes the two groups, at least for N equals 3. For N equals 3, of course GL3Z has property T, but out of FN does not. So that's been known for a long time. It has a subgroup of finite index that maps on to Z. N equals 4 is open. Right. Other things that distinguish these, kind of related to the ... Yeah, so the Dain function, which measures how hard it is to solve the word problem in a group. For out of FN, it's exponential, and for GLNZ, it's polynomial. Actually, there's other big news in out of FN that is not so ... hasn't maybe received as much press as property T. So there's something you can define for a group that has torsion-free subgroups of finite index, called the rational Euler characteristic, which is just the usual Euler characteristic of the subgroup of finite index divided by its index. For GLNZ, Burrell, I believe, proved a long time ago that the rational Euler characteristic is zero. I think of N as at least three. And it was just proved, it's actually not even on the archive yet, that for out of FN, the rational Euler characteristic is always negative and it grows more than exponentially with N. The rational Euler characteristic? Yeah, yeah, I'm sorry. This is greater than three. No, but for N equals three for GL. Yes, it is exponential. Yes, GL three is that you're right. It's GL four and up. It's it's polynomial. This, well, I have to ... so this is not the subject of my talk, but I have to say this. This is proved by somebody named Michael Borinski. He's a physicist and he proved it. So a long time ago John Smiley and I made this conjecture that the Euler characteristic is negative and grows more than exponentially with N. And we did some work on the Euler characteristic. In particular, we just we used a certain function which we called tau to do a calculation and we found some sort of a generating function. But we couldn't tell whether and we generated lots of numbers. We generated the Euler characteristics for rank up to ten and then Don Zagie simplified our generating function and got it up to a hundred and they seemed to do this so we made a conjecture. But what Borinski realized is that this function tau can be thought of as a character on the con-crimer hopf algebra of graphs which physicists have been studying and using to calculate Feynman diagrams and so he used some things he knew about this, the number theory associated to this hopf algebra to prove this. Okay, I'm done. But lots of exciting things are happening in out of FN. Right, so the general question that I wanted that I want that I'm curious about is how does the shape of gamma actually affect the group of out of automorphisms and its subgroups. In particular these properties, you know, which are true for all out of a gamma. How do you tell when one property holds in some set of graphs and not in others, etc. So what I wanted to address today, what's in my title, is the virtual co-homological dimension. So I want to study that. So in order to study this group, I need to know like what's in it. So we have to start by, I'm going to start by talking about some basic elements of out of a gamma. So let's start with a graph, any old graph, I just, I'm going to use that one as an example throughout the lecture. So there's a couple of obvious kinds of automorphisms. If I take a generator V and send it to its inverse and send everything else to itself, then that gives me an automorphism. All I have to do to make sure I've gotten an automorphism is preserve the commuting relations are satisfied. So if I just invert a generator, it still commutes with everything it used to. Another obvious type of automorphism is a graph automorphism. So if I have, for instance, this graph has an automorphism that kind of flips along the diagonal and it changes V and B, C and W, U and A. Again, if edges were connected before, they're going to be connected afterward. So the graph automorphisms actually induce automorphisms of a gamma. The subgroup generated by inversions and graph automorphisms is finite. So that's just a little tiny bit of the group of, in general, of the group of automorphisms of a gamma. So how do you get infinite-order automorphisms? Well, you think about z to the n. How do you get an infinite-order automorphism of z to the n? You take an elementary matrix, right? You send one of the basis vectors to EI plus EJ. So those are elementary matrices and they have infinite order. For the free group, you do something that looks almost the same. You take a generator. So this is a1 up to an. You can send ai to ai times aj. Call that rho ij. That's an automorphism. You can undo it by multiplying ai by aj inverse. And you don't do anything to the other guys. But notice that that's different from multiplying on the other side. So those are two kinds of automorphisms, right and left. Those are called right and left Nielsen automorphisms. Okay. So, so that's really easy and extremely well known. But for a gamma, for a general graph, you need to be more careful. You can't send Yeah, you need to preserve the commuting relation, right? If you do something to a generator, you can't just arbitrarily multiply it by another generator. If you do that, you better make sure that the result still commutes with everything it used to. So what do you do? How do you get infinite order automorphisms of a gamma? Well, here's the general idea. You choose a vertex. I'm going to choose this one. And I'm going to look at the star of this vertex in the graph, which means the full subgraph spanned by everything that's by it and everything that's a distance one. So here's the star of M in this picture. And then I'm going to look at the component. I'm going to take the star out of the graph and look at the components of what's left. So there's one component. And there's another component. So I'm going to look at these components. In the first case, you can really see that. Okay, kind of. Should use a darker color. Never mind, I'll stop playing. Okay. Right, the component on the left, it's a singleton. It's only got one guy. Everything that commutes with you also commutes with M. So I can actually take you and multiply it by M and do no harm. I've still preserved all the commuting relationships and I can do that on either side. Okay. On the other hand, if I've got a component with more than two vertices in it, like C2 over there, then I can't do that. I can't multiply A by M because then it would, it used to commute with B and it no longer does. So, but what I can do is conjugate the entire component by M and the fact, the rest of the graph won't notice that I've done that because it's kind of isolated from M by, as far as it knows I, yeah, I haven't done anything. So in fact, that also gives me an automorphism of gamma. So I can't get away with just these multiplying one generator by another. I have to either, sometimes I can multiply a generator by the other, sometimes I can conjugate a component of this gamma minus the star by the generator. But anyway, these types of automorphisms, they're all in infinite order and they're called, whoops, I didn't want to highlight that. I wanted to highlight this. They're called gamma-whitehead automorphisms. I should probably call them gamma-neelson automorphisms, but and then the theorem is that the group of all automorphisms and therefore the group of outer automorphisms, too, is generated by inversions, graph automorphisms, these elementary gamma-whitehead automorphisms that I just told you about and one more type of automorphisms called twists. So what's a twist? Well, it too multiplies one generator by another, but in this case the generators commute. What I did up here, I multiplied u by m and they don't commute, but if I multiply, so the twists, yeah, I'm multiplying v by m, but v and m commute, so that's different. So notice that if I started with the free group, then I don't need the twists. In fact, I don't have any twists because new generators commute. On the other hand, if I have a gamma z to the n, then I only need actually one in inversions and twists. I don't have any of these gamma-whitehead automorphisms. I conjugate stuff by an element of z to the n, nothing happens. But in general, this is what you get. You really need all those different kinds of automorphisms. So for general agamas, the twist subgroups is actually very well understood. So the answer is that it's the subgroup generated by those twists is actually injects into the general linear group and you can say exactly what its image is. The image lies in a parabolic subgroup that you can write down specifically and you can, anything you want to know about this, you can just read off from this representation, including for instance, its virtual co-homological dimension. So the twist subgroup we understand, so what I'm going to talk about today is what we call the untwisted subgroup. So it's the subgroup generated by everything except twists. So a couple of remarks about these two subgroups. First of all, every automorphism can be written as a product of a twist and an untwisted subgroup and untwist, and yeah. Yeah, you can put all the, put all the twists in front of all of the, of the untwisted guys if you like. So you can write this as a product. The intersection of these two groups is very small, it's just inversions, the subgroup of inversions. It's an interesting group. It really kind of measures the difference. Well, the kernel of the map from out of A gamma to GLNZ that you get by abelianizing A gamma, kind of measures the difference between out of A gamma and GLNZ and that, that group, which is an interesting group, is contained in this U of A gamma. And of course, if there aren't any twists, then it's equal to U of A gamma is actually, actually equal out of A gamma. So this is a group we're going to study. We're going to study this untwisted subgroup. So to explain what I, how I do this, I want to go back to the case of the free group. When we wanted to study the group of outer automorphisms of the free group, we built a simplicial complex called case of N. What are the vertices? Yeah, what are the vertices? How can I say this efficiently? These are graphs, oops, G, together with isomorphisms from the free group on N generators to pi 1 of G. No, so I haven't put in a base point there because I don't care about base points, so this is right. But I have an isomorphism from the free group to pi N of G and what's an edge? I get an edge from G to G prime. So these are the vertices of my simplicial complex. I get an edge from G to G prime. If G prime is G and I'm going to collapse some edges. In an acyclic subgraph. So you call an acyclic, so that's a subgraph with no cycles, so we'll call that a forest. So edges are, we call them forest collapses and a k-symplex is a chain of k forest collapses. Okay, so the group, that's a simplicial complex. Easy to describe. And how does the group of outer automorphisms act? Well, one vertex comes with this isomorphism from the free group to the fundamental group and the out of FN just acts by changing the isomorphism. So I've got an isomorphism with FN to pi 1 of G. If I take an automorphism, that gives me a new isomorphism from FN to pi 1 of G. So that's, that's a simplicial complex. The group of outer automorphisms acts and the theorem that Mark and I proved in 1986 was that this the simplicial complex is contractable. The action of out of FN is proper, meaning it has finite stabilizers and co-compact. So let me just draw some pictures. That there is a picture of one of the vertices of KN. It's called a rose because it's supposed to look like a rose and the edges are marked with elements of a basis of the free group. That gives the isomorphism with the free group. Okay, so that's a point. That's a vertex. What does a neighbor of this thing look like? Well, here's a neighbor. Okay, there's another graph and that's supposed to tell you, you can see some loops in there. They go up an edge and down the tree and that gives me, yeah, that's a point. And I claim that these two are connected by an edge because if I just collapse this edge, I get that. On the other hand, okay, so another, here's another neighbor. Basically, I can draw any graph I want, draw a maximal tree in the graph and label the rest of the edges by these A's. Then I can collapse one edge to get that picture and I can collapse two edges to get this picture. This is what, this is E and this is F. Et cetera. Right? So there's a little picture of a neighborhood of a couple of vertices in this complex, in this simplicial complex that are next to my to my rows. So what's the precise construction when you say you take all the graphs? Of course there are. Yeah, yeah. So I've been- Both of them classes or you have to rigidify them in some way or do you take- I take, I've been a little cavalier here. I take, I only take graphs, all of whose vertices are at least trivalent and they're connected and I take isomorphism classes, yes. Right, I guess so I've been talking about this since 1986 and it always surprises me that everybody doesn't, I mean it's like when you teach calculus. You kind of think they ought to know this by now. So I'm sorry if I'm a little bit glib. Right. So that's, that's this, that's a space. I want to give another kind of a description of what the neighborhood of a rose looks like in this complex and to do that I have to move to partitions. So there's a very well-known correspondence between trees and partitions, namely if I have a tree and I have an edge in the tree, then that naturally gives me a partition of the other vertices. So for instance, this edge in this tree gives me a partition of the vertices of the tree into two pieces, the vertices that are in here and the vertices that are out here. If I have two different edges, that gives me two different partitions and they're compatible in some sense. The sense is very precise. They're two, so the partition has two sides and I can choose a side of one and a side of the other so that they're actually disjoint. So enough or another way to say that as you can draw circles representing the partitions that don't intersect. And in general I can describe my whole tree as a set of partitions. I have a partition for every edge in my tree. Okay, so there's there's a well-known correspondence between partitions of sets of elements and finite partitions and finite trees. I guess you don't even have to say finite. Okay, and the nice thing about this is, yeah, I already did this, I can actually reconstruct the tree from the partition. So if I have a partition like that, then there's a tree. I just put a vertex in every component of the partition of the complementary sets and connect the dots if there's a circle between them. Okay, so trees and partitions. Now, when I was talking about the neighborhood of my rows, I was talking about graphs together with a maximal tree. So if I have, if T is actually a maximal tree in a graph, so I've got, you know, edges. I don't know, I have too many edges. You don't have that much time. So there's a maximal tree in my graph. Then, and all of these edges, remember, were labeled AI. Then if I take one of these partitions corresponding to an edge, whoops, I wanted a blue partition. It actually, not only partitions the vertices, it partitions the vertices of the tree, but I can also think of it as a partition of the half edges of my generating set. Right, and I look at the, maybe the thing, the way to do this is cut all these half edges into two pieces. AI and AI inverse, and put AI and AI inverse in the same piece of the partition if they end at the same, same vertex in the tree. So what am I saying? Not only could I construct this tree from this set of partitions, I could also reconstruct my marked graph from the set of partitions. I could, right, I know, I know how to glue in the extra edges if I know these partitions. So for instance, in the two graphs that I had in my, the neighborhood of my rows, the first one, I have this one edge and the front, actually in this one the front edge of all three generators is on one end of the edge and the back edge is there all at the other end. So that gives me a, this graph corresponds to that partition and the second graph, I've got two edges in the partition, two edges in the tree, two partitions, and that tells me how to reconstruct that graph. So, yeah, and the punchline is that I can reconstruct the graph from this partition, this set of partitions and each PE partitions, the set of generators. Okay, so what this boils down to in the end is that I can describe the link of a rose in my simplicial complex as actually just the geometric realization of the partially ordered set of compatible sets of partitions. A compatible set of partitions gives a tree. If I throw out one of the partitions, I collapse an edge in the tree. If I throw out all the partitions, I get the rose and so the dimension of this space is the number of partitions in a maximal collection of compatible partitions. Okay, so that's a picture. So why did I go all through all that? Well, I claim that these gamma-whitehead automorphisms I was talking about, things that are actually automorphisms of a gamma, can also be described in terms of partitions. In fact, partitions of the generators of my right-angled art and group, which remember the vertices of the graph. So how do I do that? Let me remind you of the kinds of automorphisms I was talking about. For instance, I had this picture and I said I could multiply u by m. Well, the link of m doesn't play. So I've written down the the vertices, which are the generators and their inverses in a list and they kind of separate into sets. There's... I shouldn't don't actually want this part in the set. There's the link of m, which just consists of v and w and their inverses and then there are these... there's m itself and its inverse and then there's u and its inverse, which I treated separately, and then there's these components, c plus or minus. So how do I draw a picture of... how do I draw a partition that represents, for instance, the white-head automorphism that multiplies u by m? Well, I just partition u and m from the rest of the components. How do I draw the partition that... Yeah, so that was... you goes... yeah, what about if I want to do... I want to multiply on the left instead of the right? Well, I claim that this does that and I want to think of that as multiplying u inverse on the right by m. So it multiplies u on the left by m inverse. And what about a picture of this partial... this conjugation? Well, I just take m and m and separate it from n inverse and put c inverse in the same way. So these are examples, as I said, of elementary gamma-white-head automorphisms. There is a much more general thing called a gamma-white-head automorphism. So it's a combination of these basic ones. So it multiplies some generators on the right, some generators on the left by u inverse, and some components it conjugates. So here's the general picture. You have... you start with your vertex m and you write a list of all those components and all those singletons and their inverses. And then you partition that set into two pieces. You have to separate m from m inverse. So those are the kinds of things I've done. And then you define the automorphism by you multiply something on the right by m. If it's inside, if it's in one piece of the partition in the side containing m and you don't do anything if it's outside the side containing m. So let me just do one more example. So there's a partition. It separates m from m inverse. It contains two guys, one of the singletons and one of the components. And what do I do? Well, if something's inside p, so I have to conjugate every a, b, a and b by m and I have to multiply u inverse by m. Yeah. Multiplying u inverse by m is like multiplying u on the left by m inverse. And a is in p, so I have to multiply it on the right by m, but a inverse is also in p, so I have to multiply a on the left by m inverse. So that gives me a conjugation. Okay, so it's a fairly simple rule. There's a partition. The partition has to be the right sort of... it has to keep these components together. And it has to separate m from m inverse. And given any such picture, I get a gamma-whitehead automorphism. And I call the partition a gamma-whitehead partition based on m. Okay. So now we get to how do we study this group? So Ruth and student of hers Nate Stambach and I in a paper that just came out in 2017, we defined a simplicial complex, like the one I defined for a free group. But now the vertices, instead of being graphs, they're they're certain special cat zero or NPC, locally cat zero, cube complexes. So a graph is just a locally cat zero cube complex. The cubes are one-dimensional, but they're cubes. Yeah, but they have to have with the right fundamental group, and you have to have an isomorphism of pi one with a gamma. And what's an edge? You get an edge between x and x prime. Well, for graphs, we collapsed an edge. Here we have a higher-dimensional thing. Instead of collapsing an edge, we're going to collapse an eye bundle. So a product of a graph with an interval. x prime obtained from x by collapsing a certain type of eye bundle. And again, I'm being a little vague. They're they're details here. So what we did is we defined this complex, and then we proved it's actually contractible, and this subgroup of untwisted automorphisms acts properly and co-compactly. The mineral vertex, so for the free group, it was this rose. So what's the analog here? It's a so-called salvedi complex. So it has a generator for a loop for every vertex of gamma, and it has a torus, t to the n, for cliques in gamma. So for instance, if two vertices are joined in the gamma or joined by an edge, you have two generators of the a-gamma, and you sew in a torus to make those two generators the meridian and the longitude, and you do that in higher dimensions, too. So that just by construction gives you a space with the right fundamental group, and it turns out to be, and it's locally cat zero, non-positively curved. Okay, but we actually don't need to understand what kind of spaces these are because we have a description of these spaces in terms of partitions, just as we did for the free group. In fact, if you take one of these minimal guys, these salvedi's, and decide what's around it, it's exactly the same as before. It's every vertex in this link is given by a compatible collection of partitions. Now you have to be fairly careful about what you mean by compatible here, but there's a definition which works. So you got this cube complex, and it's, sorry, this is a simplicial complex, and the vertices are collections of compatible partitions. So as before, the dimension of k-gamma is the number of partitions in a maximal collection. Okay, so where have we got? I said I was supposed to be talking about virtual co-homological dimension. Well, I have a proper action of a group on a contractable space, so the dimension of that space gives me an upper bound on the virtual co-homological dimension. And actually when we wrote this paper, we conjectured that this would be exactly the virtual co-homological dimension. But story's not quite over. It'll be over in five minutes because that's how much longer I have. How do you get a lower bound? Well, an easy way to get a lower bound is find a free abelian subgroup rank R, of some rank, rank R, then R, then the VCD of u of a-gamma is at least R. And of course, it's less than or equal to the dimension of k-gamma, which as we've seen is just the number of partitions in this maximal collection. Okay, so where do we go from here? We have this nice dimensional space. We would like to find a free abelian subgroup of the same dimension as the space because then we would have shown that the VCD was exactly that dimension. So the idea is I'm going to be very sketchy now. You take one of these compatible collections and so given any of these partitions, you've seen that they correspond to automorphisms, these gamma-whitehead automorphisms. So now you think about, well, when do the two of these automorphisms commute? They don't always commute, but sometimes they do and it turns out you can decide exactly when they're going to commute and if there's a partition in your set that you don't like, you can often get rid of it and replace it by another one, which will commute with more things. So this requires lots of sort of cute combinatorics. Yeah, which I shouldn't tell you about. And if you're lucky, you can replace all of the partitions in your biggest collection by ones corresponding to commuting automorphisms. So let me just state the theorems that we can prove. Not yet on the archive. We can find, we can tell you the the biggest possible free-abilian subgroup you can build that's generated by these gamma-whitehead automorphisms. We can tell you precisely, we can calculate it, give me a graph, I can tell you what this dimension is, what the, this rank of this free-abilian subgroup is. And, so the next question is, well, when is this free-abilian subgroup exactly equal to the dimension of your k gamma, and we have a condition? So here's a theorem. Yeah. So we say a vertex is principal if its link isn't in any bigger link. So, and then we look at the, here's the criterion. Supposing I have a vertex that's not principal, so its link is in a bigger link. Here's the link. Suppose it's in some bigger link. So, that guy has a bigger link than that guy. And supposing there's also something else in this bigger link. If it's true that whenever I have this situation, then these two m's are actually in the same component somehow of gamma minus the star of u, then it's always true that the VCD is exactly equal to the rank of this free-abilian subgroup. So, I'm pretty much, right. Okay, so that's nice. This is something that's, here's an example of a graph in which this is true. There are lots of examples. But here's the theorem, too. Sometimes it's not true. Sometimes the dimension of this space is strictly bigger than the rank of this free-abilian subgroup. It's strictly bigger than the rank of this free-abilian group you can build. If, in addition, there's another condition that these non-principle guys, almost everything within distance, two of them, is bigger than them, then, in fact, this K gamma has an equivariate and invariant deformation retract of strictly smaller dimension. And, in fact, you can, in many of these cases, you can retract this space down onto a space of dimension equal to the rank of this subgroup, this free-abilian subgroup you can make. So, this is a condition under which this happens. And there's an example where it happens. The dimension of the space is six and the VCD is only five. So, let me just state some conjectures from everything. So, for out-of-fn, the rank of the biggest-abilian subgroup you can build inside of out-of-fn is exactly the VCD of out-of-fn. And my feeling is that should be true for this untwisted subgroup as well. So, that's the first conjecture, that the VCD is exactly the rank of this abelian subgroup. Right. So, it's conceivable that you could find some other free-abilian subgroup that wasn't generated by gamma-whitehead automorphisms. But I don't think, I think the second one is just a true conjecture, so I won't say anything. Another thing, as I mentioned, for out-of-fn, all the solvable subgroups are virtually abelian. And I can prove, well, Ruth and I have proved for many classes of graphs that this is true for these untwisted automorphism groups as well. Every solvable subgroup of U of A gamma is virtually abelian, but it's not a theorem yet. It's only true in lots of cases. So, those are some conjectures related to this. What is the second conjecture? The rank is just the rank of the largest free-abilian subgroup inside of U of A gamma. So, this guy M of L is the rank of the largest free-abilian subgroup in U of A gamma that's generated by these gamma-whitehead automorphisms. But there could be some weird other free-abilian subgroup somewhere in there. I don't think so, but there could be. So, let me just say, in summary, maybe this is the point. If I want to study this group, in particular the untwisted subgroup, which is the part of this group that I really don't understand, I should replace finite graphs as in the study of free groups, the outer automorphisms of free groups, with finite locally cad-zero cube complexes with the right fundamental group. And that will give me, well, will give you then a contractable space with a proper co-compact U of gamma action. And, right, and most of my talk was about trying to use that to get a better handle on the VCD. That was the rest of my talk. But I should say that there, as I said, there are many properties shared by, I listed at the beginning, some properties that are shared by fn and out of z to the n, right? And it turns out that many of those properties are also shared by all out of A gammas. And there's a list of people that have proved theorems, some people, some of the people who have proved theorems in this way. So here's a question for Alan. When does out of A gamma have property t? So maybe this space, I mean, we've got a geometry now. So it's conceivable that could help understand this question. And that's my last slide. The conjecture about standardly subgroups, it depends for GLNZ, right? Right, I was talking about the untwisted subgroup. So, of course, for GLNZ, it's generated by twist. It is the twisted subgroup of itself. So, yeah, I guess it's not quite true to say that it's a parabolic subgroup in that case. But, yeah, no, I'm just talking about the untwisted subgroup. Can we expect that every solvable subgroup of out A gamma is fanatic generated? Yes, I think that's true. I think it's in a paper by day, matte day. Yeah, so we know a fair amount about solvable subgroups. We know lots of interesting examples. And as I said, in many cases, we know when the solvable subgroups are either solvable subgroups of GLN, a little GLNZ inside of out of A gamma, or they're virtually abelian. So, yeah, we know a lot about solvable subgroups in general, but this question we don't know. And can we expect that the cohomological dimension is always achieved by a need port and subgroup? That's a good question. I would think so. But based just on examples, evidence, evidence of examples, I think that's probably true. But I can't prove it. Maybe you can.