 Thank you very much. So first, thank you for letting me talk here. And this is a joint work with Andrew Sayle. And what I'm going to talk about is auto-automorphisms of right-angle arting groups. And I will investigate which of them are vast and which of them are not. OK, so let me first remind you what's a right-angle arting group. OK, so you'll start with a finite graph, gamma, with vertex set x1, xn. And you construct a very simple group, very easy presentation. You put a commutation relation whenever there is an edge. So that's a rag. So rags are very important in these days because you can embed many groups in rags. That's not what I'm going to talk about at all. I'm going to talk about automorphism groups of these rags. And so the easiest examples of rags are, for instance, the free group fn. You just take a bunch of vertices, but no edge. You take z to the n, you take a bunch of vertices, and you put all the edges, but there are many more. I don't know if you can do a direct product of free groups. Take free product of a billion groups, but many more. And OK, so what I will be interested with is the group out. I need a notation for this group. We'll denote it by a gamma. So given a finite graph, the arty group corresponding to that is a gamma. And I want to talk about out of a gamma, which is just the group of all automorphism modulo inner. Examples include, I mean, that the extremes of the spectrum out of fn and gln is out of fn. And so in some sense, those groups here, this family of groups, interpolates between out of fn and between glnz. And I want to draw a frontier, a boundary, between groups we have a large behavior and those which don't. And that's the goal of the talk. So in which sense, larger or vast ways in which out of fn is larger, or I would say, faster. Then OK, so for example, those groups are not small. They contain three groups. But the way I want to, the kind of bigness or largeness I want to investigate is something that will not well behave with respect to subgroups, but with respect to quotients. So for instance, one way in which out of fn is faster than glnz is that there is an epimorphism. A canonical epimorphism just by looking at the action of an automorphism on its abelianization. So this is the first very basic way in which out of fn is faster than glnz. But there are deeper ways. So here are three other ways. More interesting. The first one is a profanite property. So it's about involving all finite groups. So here's the definition. A finite-generated group involves all finite groups if for every finite group f, while there is a finite next subgroup of g that maps onto f. Definition. And for this property, out of fn, ba is very differently than glnz. So here is, for instance, here is trm by read. And so that's for the mapping class group by Novald Botzkic out of fn is that out of fn and the mapping class group of the surface, probably sufficiently large, involve finite groups here. And on the other hand, the theorem I don't exactly know whom to attribute it. Probably it's related to the congruent subgroup property. So for n, glnz does not. So for n equals 2, things are very different. gl2z is actually the same as isomorphic to out of f2 and is virtually free. It does involve all finite groups. So maybe I should have said so. The basic example of a group that involves all finite groups is a free group f2, because if you take a finite cover, you can get three groups of arbitrary rank. So you can map onto every finite group you want. So that's the first way in which out of fn is larger or faster than glnz. Here's a second way called sq universality. A group g is said to be sq universal. So sq stands for sub-caution. So if any countable group embeds some quotient of it. This says that this group is very far from being simple. It has, if it's finite generative, it implies that it has uncountably many quotients. And here, again, there is a big difference between glnz and our fn. So theorem is that mapping class groups and out of fn are sq universal. Damani, myself, and Dennis Ossin. And on the other hand, wondering what's going on for glnz. Well, there is a Margolis normal subgroup theorem tells you that any quotient of glnz is either by a central subgroup. So you don't do anything. Or the quotient is finite. Any normal subgroup, pgln has finite index. Altrual. But in particular, it's very far from being sq universal. So that's for n, at least 3. That's a higher-ranked phenomenon. Here is a third way. Third way involves quasimorphisms. Should I define homogeneous morphisms? Is a map from G to R such that it's a quasimorphism, meaning that h of gh minus h of gh of what? That's not good. It's bounded by a constant c. And homogeneity is just this is rated to bounded homology. And let's say that for this talk that G has many quasimorphisms, so this is a real vector space. You can add quasimorphism. You can multiply by a real number. And so you want the dimension of this set of homogeneous quasimorphism to be infinite dimension R of this set. So if your group is finitely generated, there are obviously a morphism from G to the additive group R is one of those. But there can be only five dimensional of two morphisms. So if there are an infinite dimensional set of quasimorphisms, then they are kind of exotic ones. And here again, there is a huge difference between out of fn epsilon there. So theorem. So here, a fein, a virginia fugiora, out of fn, and the mapping class group of the surface have many morphisms. On the other hand, this is not true for g and z. So I think this is due to, then there is a more general version, which is that g and z. Well, in fact, I think it has absolutely no non-trivial homogenous. All these three properties, these three largeness properties like, so involving all finite groups, being ex-universal, having many quasimorphisms, they are inherited by quotient. So if you map onto a group that has one of these largeness properties, this vastness property, then the group you started with also inherits this property. And that's a nobulous fact. There is the strongest form of vastness, which has a name. It's called being large. g is large. If it virtually maps onto a free group, it has a finite index subgroup g0 in g, such that 0 maps onto the free group f2. That's largeness. And this largeness property implies all the three properties I proved. I mentioned, sorry. And what is, sorry, examples, it's known that, well, out of f2, which is the same as gl, z is large. Just because it's virtually free, it's known that out of f3 is large. This is due to Grinwald and Lubotsky. This is not known out of fn, or n larger. And it's known that gl and z is not large, greater, at least three, maybe because it has property t or because of all those other reasons. This largeness property is the strongest vastness property in some sense. Strongest, it implies all the three properties. So maybe I should write notes. These properties, so the picture of the goal of my talk is the following is the word of rags. On the left, or on the right, I'll put the vast behavior. So we're out of fn. I will see also n is here. On the left part, I'll put the non-vest, so the skimpe part. And here I have gl3, greater than 3. That's here. And the question is, where does the frontier go? What's the boundary between those two behaviors? And that's out theorem. We describe the boundary following. So it says that the following are equivalent. The first thing is a combinatorial condition on the graph. I'll make it precise right after. Then if this condition on the graph holds, then out of a gamma, well, if and only if, out of a gamma involves all finite groups. And equivalently, this also holds if and only if, out of a gamma, is sq universal. This also holds if and only if, out of a gamma, it actually has. So I can describe the precise combinatorial condition. So it means that for the group to be vast in this way, gamma has 3 equivalence class. I need more space. What is the condition star, if and only if? So I need to explain the terms I'm using here. So if gamma has a free equivalence class at least 2, or gamma has a nabellian equivalence class of size exactly 2, or gamma has still separating sections. So these are features that were already used in the study of automorphisms of rags. So what are them? So to explain, so I need to explain this. What are those equivalence class? And what's a separating intersection of things? Well, this is rated to a generating set of the group of automorphisms, which involves several kinds of automorphisms. And the first one, this kind is called a transvection. So what is a transvection? That's a particular kind of automorphism. So take two vertices, x and y in gamma, which are distinct. And that's an automorphism that sends x to xy, and it fixes all the rest. So this does not always define an automorphism of the rag. It depends on some combinatorial condition. This is automorphism, if and only if. So either the link of x is contained in the link of y. So what is this link? So we are in this graph gamma. The link of a vertex is a set of neighbors. So you want that every neighbor of x be a neighbor of y. That's because you want that if you send x to xy, x commutes with many elements. And then you want that xy commutes with the same elements at least so that this defines a homomorphism, either. So that implies that x and y don't commute. And when they commute, the right way of putting it is a variant, just that the star of x is a star of y. The star of x is just a link together with x. So the link is a sphere of radius 1. The star is the ball already. And so you see that this is actually an ordering, or at least a preordering. And so we say, in this case, let's say x less or equal to y. And corresponding to this partial preordering, there is an equivalence iteration. And that's the same as either they have the same link or they have the same star. Maybe I don't write this. So examples, maybe I put x1 or x2, x3, x4. And here I want to add some more edges, y1, y2, y3. And those guys form an equivalence class. They have the same link. And those guys form an equivalence class. They have the same star. And this is a free equivalence class because the elements don't commute with each other. They generate a free group in the rag. And here it's a Nabilian equivalence class because they generate a Nabilian group in the rag. So that's the definition of the two words which are here. A free equivalence class is something like that. It's a set of elements which don't commute with each other, but you can transvect the element onto the others. And a Nabilian equivalence class is the same, but the element commutes together. That's a feature. Now to explain the second point in the definition, which is separating intersection of links. This is related to another type of automorphism, which is called a partial conjugation. What is a partial conjugation? Well, it's an automorphism defined as follows. For all x in x, x is mapped to a conjugate, and you conjugate by y. And for the other elements, you just don't do anything. That's a partial conjugation. It's a conjugation on some piece and identity on the rest. So here again, for this to work, for this to be an automorphism, you need some conditions. And the condition is x is a union connected components of the graph minus whatever commutes with y, which is star. And actually, you could also add some more elements in the star of y because if an element is in the star of y, it's exactly the elements which commute with y. So those two definitions agree on this set. So that's an exercise to check this. So star of y, which doesn't change the definition of the automorphism itself. And now separating intersection of links is some condition that appears when you want to understand when those guys commute or don't. So here you say that x, y, z is a separating intersection of links. If the following holds, the link of x intersection y separates x, y. And you also want that x, y, and z don't commute. So the picture is as follows. You have x, y. They have something in common in their link. Let me have some other stuff. And there is a connected complement of the complement of this intersection. That's the intersection of the links. And here you have the z. And what you can easily check is, in this case, if you take big z as a component, see the x, z, and see y, z. Don't commute and generate a free group. So in fact, there is a theorem by Pérez and Gotz-Androhan saying that, in fact, that the group generated by all those partial conjugations is abelian, if only if partial conjugations. Butte, abeliana, if only if there is no seal. So that's the only source of non-commutation. All right, so I think there is, for instance, a theorem that can state with those definition that's due to today, which analyzed which of those automorphism groups of rags contain free groups. And that's if and only if there is a seal or there is one of those equivalence classes of size two, or at least two. So here, there would be a two, because then you have also many transactions that don't commute with each other. All right, the point with these particular automorphisms is that, in fact, they form a generating set of a finite index group of the group of automorphisms. So theorem by Lorentz, transactions, and partial conjugations index, psi will denote by r0. In fact, you can be explicit about what's missing. What's missing is just that there are automorphisms that may come from permutation of the graph, automorphisms of the graph. And then there are also automorphisms that fix every generator, but just send one generator to its inverse. That's what's missing. One may wonder, so in the theorem I just raised, skillfully, one may wonder what those three properties which are rather different, like involving all finite groups of profanite property, sq. universality is something related to quotient, quasi-morphisms is something different. So they, in general, they are not related. But here, something happens. I mean, there is this coincidence. The boundary between the behavior for those three properties is the same boundary. So this can be explained by the following trichotomy, which we prove. So one of the following holds. So 1a is that out 0, a gamma, maps onto out of fn, or sl. That's the first possibility. A second possibility, 1b, is that out of a gamma large. In fact, this one implies this one. So I, well, not exactly, but I could have forgotten about it. And there is 2, which says that out 0 of a gamma fits in a short, exact sequence. So here, you have a polycyclic group. Here, you have a product of ni. Ni's are at least three. So either, so here, that's what happened if the combinatorial condition holds. And this is what happens if it doesn't hold. And here, it's fairly easy to deduce the theorem from this. Because, well, if your group maps onto out of fn or into sl to z, or if it's large, then it satisfies all those vastness properties that I mentioned. And if it does not, then you have this short, exact sequence. And because all those, the group sln z that appear here, are higher-ranked, so they have this n greater than 3. And it's also an exercise to show that, because those ones are not vast. They are skimpy. And this one is just polycyclic. And then this property lifts to this property, for instance. So it can't be, for instance, I don't know which one is your preferred. It can't be sq universal. Yes? No, 1b implies 1a. Well, it's not, because it's a nor. It's not known whether, yeah, they take out of f4. It's not known whether, so it fits here. But it's not known whether it's large or not. So large, meaning it virtually maps onto f2. But why? OK, so the proof is really combinatorial, and that G-brake doesn't evolve. So there is a sequence of work by Charlie and Vokman for constructing outer spaces for these groups. And we don't rely on that at all. So the method we use is more algebraic, as you see. It's about mapping onto other groups, or about proving this short exact sequence here. So let me maybe explain more precisely what's going. Well, yes, the proof of this trichotomy, and the relation with the combinatorial condition. This, OK. Yes, so that's a direct product of SLNZ. And the Ni's, all the Ni's are at least three. And you can be more specific, in fact. In fact, let me add some color comments. So in fact, here, Ni's, that's the size of the equivalence classes. OK, maybe this is not understandable, because I didn't explain the relation with this. So 1a will hold if there is a free equivalence class of size at least two, then you will have a map here. You'll get this thing. And if there is an abelian equivalence class of size two, then you will get this part. OK, that's this and this. Now this will happen if there is a seal, and 1a does not hold. So what we prove is that in this case, if there is the seal, and if we don't have any free equivalence class of size n, essentially, then we prove the largeness of the. And so in the remaining case, what we know is that all equivalence class, they are abelian, and they have size at least three. So the Ni's are the size of this equivalence class. They are all at least three. And we have this short exact sequence here. So the proof consists in saying there is a first proof 1a. It's a simple observation. It's the fact that if gamma has an equivalence class of size, say n, then this group out of 0 of a gamma, maps on to either out of fn, or on to slnz. According to the fact that the equivalence class can be free or abelian. OK, so a similar observation had been made by Charney and Vogtman, for instance. But in a more restricted case, that's very easy observation. And it proves part 1a of the alternative. Now, about 2. So you have to prove that if there is, so that's in the remaining part. So you have to prove that. So you assume equivalence classes. And then what you do is you look at the representation corresponding to the abelianization. So you have a map from out a gamma of 0, size of gamma. So just abelianize the writing of our thing and look at the action of the abelianization. So this has a kernel, which is called aAA. And it's a result by weight and by the day. And weight, this is generated by two things, by partial conjugation and some commutator transactions. And in this context, you just check that because all equivalence classes are abelian, these are all trivial because of this. And so you remain to understand that. So the kernel is regenerated by partial conjugations. And so in fact, and I told you this result, which is skillfully here by Gutierrez-Pigotin-Reyn, that this is abelian because there is no seal. So this really is abelian in this case. But what's the image? The image is very easy to understand. The image here, it's made of block triangular matrices. And on the upper blocks, that could be something or nothing, depending on whether, OK, I didn't say the sizes. I mean, all these blocks correspond to the equivalence classes because that's where the elements can transfect on the other. So these are the size of the equivalence classes. And here, some equivalence class that can be transfected on the other, but not conversely. And so it's about whether some equivalence class is larger than the other or not. So you really, the image is described completely explicitly, it's a group of triangular matrices. And so those n i's here, so here you have a nil-potent or unipotent part. And the diagonal part gives you this product of the SL and IZ here. So that's really contrary to it. And so the main point maybe is what happened here. If there is a seal and 1a doesn't hold, so if there's no free equivalence class. So here, in case 1b, there is a seal, equivalence classes are abelian. This is enough. So then what we prove is that there is another, so x prime, y prime, prime, on which the group, we have to find some morphism to a large group. And that's what we're going to do. So such that our zero Bay gamma is mapped onto, well, not onto, to a subgroup of the group that they generate. So that's generated by the equivalence class of x prime, the equivalence class of y prime and z prime, which is out of some z a, z b. So in some sense, we have to do some combinatorial analysis to see what can happen. And we see that if we take, in some sense, a smallest seal, then we can construct this map here. And then, so here the map is not onto in general. So here we use two things. So it's not onto, but we essentially understand the generating set from the generating set we have out of a gamma. And so there are two options. So either we can completely describe as a free product, an amalgamated free product, as an explicit. And from which we are. And in the other case, so we use a representation that's related to the representation used by Grinwald and Lubbocki to prove that out of F3 is large. So here we have three factors. And that's important. And it's the same three as out of F3. And so we use some Fed Podsky station. And we get, so that's on the homology of index 2 subgroup of this group. And this homology splits in two factors. And one of them, because of the action of this group of order 2, so there is a fixed space and an anti-fixed space. And this one has dimension 2. And from this, we get a morphism from a finite index. And so the image is large enough. It will be virtually free. And when it's not, then we can rely on this case. That's all I wanted to say.