 Thank you very much for the invitation. It's a pleasure to be here. It's 933, so I'll finish by 1033. I promise I won't go over time. So I'm going to talk about, I'm going to define some objects which are interesting in the context of a certain set of problems. Some of these problems are solved by means of studying these objects and some of the problems are still open and so hopefully I'll give you a flavor of what goes on in this little corner of geometric group theory. So the primary focus will be on groups of piecewise projective homeomorphisms of one-manifolds connected one-manifolds such as the circle or the real line or a closed interval and I'll start with a question which goes back to the 1950s. It's called the Von Neumann Day problem. So it's a question that was asked by Mahlon Day in the 1950s and he attributed the question to Von Neumann. The question asks are there non-amenable groups that do not contain F2, that do not contain the free group of rank 2 and so this notion of amenability was defined by Von Neumann in his study of the Banak-Tarski paradox. He understood that the reason behind the paradox was the non-amenability of the free group of rank 2 and he observed that well, the free group of rank 2 is non-amenable and that any group that contains the free group of rank 2 as a subgroup is also non-amenable. So let me give you a quick definition of non-amenability. It's not the original definition of Von Neumann. It's a reformulation due to Tarski and this is the definition. A group G is non-amenable if there exist sets A1 through An, B1 through Bm, subsets of the group that are pair-wise disjoint, pair-wise disjoint and elements F1 through Fn and G1 through Gm of the group such that the group can be expressed as a union over i of fi times ai and as the union of Gj times Bj. So this is simply the translation of by group multiplication on the left of the set and so the classical this is a so-called paradoxical decomposition and the classical paradoxical decomposition that appears in the in the Banak-Tarski paradox is that of the free group of rank 2. So for instance, if you take the free group of rank 2 generated by A and B, freely generated, then you take the words that the reduced words that start with A, the reduced words that start with A inverse, B inverse, you have these four sets, you can perform such a paradoxical decomposition. It's a very interesting exercise to do that. And so this is the notion of non-manability. The number of pieces in this is interesting. How many pieces do you need? So what is this? So this is M. What is the smallest number of pieces N plus M you need for a group G? This is the so-called Tarski number of the group and very little is known about Tarski numbers of groups and that's an interesting topic in and of itself. So the question, the von Neumann-Day problem, was solved. The answer is yes. It was solved by Olchanski in 1979-1979, constructed Tarski monsters, Tarski monsters. These are finitely generated infinite torsion groups, hence they cannot contain free subgroups and Olchanski proved that they are non-manable. However, these examples are not finitely presentable. They do not admit a finite presentation. And so the question was, the question that followed was, do there exist finitely presented examples and the answer is again yes and it's due to Olchanski and Sapir from 2002. 2002 they constructed torsion by cyclic monsters torsion by cyclic finitely presented groups and the way the construction is performed, they don't contain free subgroups and they start with one of these Tarski monsters. So the resulting group is non-manable because amenability is inherited by subgroups and so non-amenability is inherited by overgroups and the construction is quite difficult. It's 110 pages long and Sapir estimates that the number of relations in this presentation is more than 10 to the power 200, which is more than the number of atoms in the universe. So it's a very unwieldy presentation, even though the construction is a great achievement. So people were still hunting for examples which are more down-to-earth and some examples were constructed recently and I'm going to describe some of these examples. So Mano in 2012 constructed the following groups. For A a subring of R, the group H of A is the group of piecewise PSL to A homomorphisms of the real line with break points in the set P of A. This set P of A is the set of fixed points of hyperbolic elements of PSL to A. So recall that PSL to R acts by Mobius transformations on the real projective line, which identify with the reals together with a pointed infinity. So this action is this action by fractional linear transformations and so you can consider piecewise PSL to R homeomorphisms. So for instance, you take the real line and you chop it up into finitely many intervals and you define a homeomorphism, the restriction of which to every interval in the partition is a restriction of a Mobius transformation coming from PSL to A such that the two restrictions agree on the common point. So it's a homeomorphism and the break points are these points where you partition the real line and they must lie in this PSL to A invariant set P A. The theorem of Mano, please stop me to ask a question if this definition is not clear. Finitely many intervals in each homeomorphism, yes. The theorem of Mano is that if A is not equal to the integers then H of A is non-amenable and does not contain a free group, does not contain F2. So these are really beautiful counter examples to the, well, which was originally stated as a conjecture that there shouldn't be any non-amenable groups without free subgroups. So these are really beautiful examples, very concrete, easy to understand. However, they are not finitely presentable. In fact, they're not even finitely generatable. And so I defined, together with Justin Moore, the following example. So I'm going to define three piecewise projective homeomorphisms of the real line. The first is simply translation by one. The second is the identity if T is less than or equal to zero. It's T over 1 minus T. If T is between zero and half, 3 minus 1 over T if T is between half and 1. And T plus 1 if T is at least 1. And the third one is C of T is the identity outside the unit interval 0, 1. And it's 2T over 1 plus T if T is in 0, 1. And define the group G0 as the group generated by A, B and C. And we proved that G0 is non-amenable, does not contain F2, is finitely presented with three generators and nine relations. So this was an improvement from that. And then subsequently I proved that G0 is of type F infinity. So it's finitely presented in a very strong sense. Type F infinity means that a group is of type F infinity. If it is the fundamental group of an asphherical CW complex which has finitely many cells in every dimension. So there are these so-called finiteness properties type FN which are topological generalizations of the classical finiteness properties, finite generation and finite presentability. And it's a very beautiful topic and this is particularly interesting in the context of groups acting on the circle. Because groups acting on the circle tend to have interesting finiteness properties. And for every group that you construct it seems to be sort of an independent challenge to find it's the right finiteness properties of the group. And so this is the first example which is type F infinity non-amenable and does not contain free subgroups. And then later I showed that G0 admits a paradoxical decomposition with 25 pieces. This paradoxical decomposition is reasonably explicit. It probably can be made even more explicit and it provides an estimate for the Tarski number of this group. The Tarski number lies between 5 and 25. So yes. Those are the three generators, yeah, that's it. And so regarding Tarski numbers, Tarski students Johnson and Decker and Johnson proved that a group has Tarski number 4 if and only if it contains free subgroups. A group cannot have Tarski number between 1 and 3. And whenever you have a non-amenable group which does not contain free subgroups, it's interesting to understand its Tarski number and very little is known about these Tarski numbers. And also I proved that the same holds for H of A. The same holds for H of A. So groups of piecewise projected homeomorphisms have been studied before Mano. There's a famous group called Thompson's Group F which was discovered by Richard Thompson in the 1970s. And his original definition and description of the group is a group of piecewise linear homeomorphisms of an interval. And he visited Thurston in Princeton. And Thurston, I've heard people claim this that Thurston immediately recognized the group and gave the following equivalent definition to Thurston. This is the group of piecewise PSL2Z homeomorphisms of the real-life with breakpoints, the rationals. And Thompson's Group T, again due to Thurston, is replaced Z by Z together with... I'm sorry, let me just write the definition. So this is the group of piecewise PSL2Z homeomorphisms. The real-projective line are together with a pointed infinity with breakpoints in the rationals together with infinity. So basically, so F is the subgroup of T that fixes infinity. And a result of G since her jazz crew shows that these actions are C infinity smoothable. This means that they're topologically conjugate to a group action by C infinity to few morphisms. There's a homeomorphism which conjugates these actions to the diff infinity of R and S1 respectively. And so when these groups were studied by Mono and then by me and Justin Moore, this question was raised by G's and Navas. Are these groups C infinity smoothable? These new examples. And we proved the following with Christian Monati and Michele Triestino. Fortunately, Christian is not here. So one, H of A is not C once smoothable even for any subgroup. And second, if A contains nontrivial units, so units other than plus minus one, then H of A does not admit any C1 action on the real line. Does not admit a C1 action. And thirdly, the same as two 4G naught, the same as two 4G naught. So this settles this question. So I would like to stress on the fact that this group T is a very interesting group. So this was the first example of a finitely presented infinite simple group. And so there are several examples now of finitely presented infinite simple groups, groups that act naturally on the circle or on the canter set. And I construct. Oh, I'm sorry, I'm sorry. I'll just write it. I think it's faithful C1 action. That's different from being not C1 smoothable. Yes, yes. C1 smoothable is a stronger requirement. It requires that you admit such an action. But in fact, that action is topologically conjugate to the given action. So C1 smoothability is a condition on the action. And whether a group admits a C1 action is a requirement on the group. So if you give me an abstract group, does it admit a C1 action? And if you give me an action by homomorphisms, is it topologically conjugate to a C1 action? So these are two different but related questions. This in the early 1970s by Thompson, Richard Thompson. Sorry. Okay, I think everybody has read it so I can move on. So I played a little bit with the Thompson's group T and the smoothability aspect. And I wrote down this homomorphism. This is the identity if T is not in 0, 2. It's 2T over 1 plus T if T is in 0, 1. It's 2 over 3 minus T if T is in 1, 2. And I define a group S as generated by T together with this little s. And my theorem is that S is finitely presented, simple, and does not admit a non-trivial C1 action on any one manifold. So this is an interesting contrast to Thompson's group T. There's another question about group actions on one manifold, which is whether there's a property T group that acts by homomorphisms or by C1 to few morphisms even on the circle or the rail line. So question, does there exist an infinite property T group that admits a faithful action by homomorphisms or the circle? Now these groups are not, they don't have T because they all have infinitibilization. However, you can consider groups that act on the circle, for instance this one. So the ones that fix infinity have infinitibilization, but for instance this is simple. And you can construct cousins of this, you can consider cousins of this example. And you can even consider examples of groups of piecewise projective homomorphisms acting on the circle with infinitely many breakpoints. So all the examples I've defined so far, every element has only finitely many breakpoints. But you can also consider actions with infinitely many breakpoints. And we call these countably singular piecewise projective homomorphisms. So they're outside a countable closed set in the circle, they are of the desired regularity, which is C infinity. So the question of, I guess Navas has written about this question and he asked me this question, what about these examples? What about the piecewise projective groups that act on the circle? We were able to partially resolve this question. So the theorem of myself together with Matevon and Triestino, if G acts by piecewise projective homomorphisms with finitely many breakpoints on the circle, then if G has T, it is finite. And for the countably singular case, we were able to show the following. If G admits a faithful action by countably singular piecewise projective or even C2. So you can choose any C2. So this is Kajdan's property T. So this has many equivalent formulations. One of them is a fixed point property for actions on Hilbert spaces, affine isometric actions on Hilbert spaces. Every such action has a fixed point. And it's sort of very much an opposite property compared to amenability. The only groups that are both amenable and have property T are finite. And really the property that we are using in these theorems is not property T. It's a consequence of property T, which is called property FW. And that property emerges naturally in the setting. We consider actions of these groups on the groupoid of germs. And if a group has property T, then it has FW. And I mean, if I have time, maybe I'll talk a little bit about FW and how we do this. But really in this context, the right property is actually the consequence of property T, which is property FW. FW. FW, yeah. It's a property of concerning group actions on sets. It's very down to earth. It says that whenever you have a group action on a set, there's a notion of a commensurate set and there's a notion of a transfixed set. And it says that for every action on a set, every commensurate set is transfixed. So I can state it at some point, yeah. So singular difumorphisms on S1. Then if G has property T, the action has a finite orbit. The action finite. So what that does is it reduces the study of this action on a compact manifold to a non-compact manifold. And in that setting, we do not know how to solve this question. And actually this argument is quite robust. So it works in more generality. So theorem, and again the same people as here. Let G be an C2 difumorphisms. In fact, we can even get away with C1 and a half. B and aperiodic action by countably singular CR difumorphisms on a closed manifold M. If G has T, then the action is conjugate to an action by CR difumorphisms to an action CR difumorphisms. So this is a smoothing criterion. So again countably singular here for a general manifold means the same thing. Outside a countable closed set, you are of the desired regularity. And no, R is just any real between 1 and infinity. And if you consider the recent result of Brown, Fisher and Hurtado concerning Zimmer's conjecture. They partially resolved Zimmer's conjecture. If you have a high rank lattice acting on a lower dimensional manifold by C2 difumorphisms, then this cannot happen. It's impossible to have such an action. And this statement generalizes that to actions by countably singular C2 difumorphisms. Because whenever you have a property T group acting by countably singular CR difumorphisms and the action is aperiodic, then it's topologically conjugate to an honest to God CR difumorphisms action. And that you can apply the statement and obtain the contradiction. So there's some progress on this question of now us. However there are examples and I'm going to tell you some more examples where we still do not know if they have property T. And they could be actually interesting candidates to consider for property T. They could be actually interesting for some completely different reason. I hope I'm not erasing what I just wrote. It's not going to be difumorphic but it's going to be homomorphic. So there's some moving around of charts that has to happen in this. So if you have an aperiodic action of your favorite high rank lattice on a lower dimensional manifold by countably singular CR difumorphisms, then by this result the action is topologically conjugate to a faithful action by CR difumorphisms. And if r is at least 2 then you obtain a contradiction because Brown-Fisher and Hurtado prove that this is impossible. So slightly tangential but very much related is this question. So it's a question that goes back to 1980 and is due to Remtola. Do there exist finitely generated simple groups of homomorphisms of the real line? So equivalently do there exist finitely generated, so I should say infinite here because Z mod 2z is an example. But besides Z mod 2z there's no finite group that acts on the real line faithfully. So do there exist finitely generated simple left-orderable groups? If you like to study group actions on the real line then you know that being left-orderable is equivalent to admitting a faithful action by orientation preserving homomorphisms on the real line. A group is left-orderable if it admits a total order which is invariant under left multiplication. So there are lots of examples of simple groups of homomorphisms of the real line which are not finitely generatable. For instance various commutator subgroups of groups of piecewise linear and piecewise projective homomorphisms. And with Kim and Kuberta we studied these chain groups of homomorphisms of the real line and we proved that whenever the chain group acts minimally on its open support then the commutator subgroup is simple. And a corollary of this is that there are continuum mini isomorphism types of countable simple groups acting on the real line faithfully. However none of these examples are finitely generatable and the real trouble is that there is a lack of method of tool to prove simplicity for a finitely generated group of homomorphisms of the real line. And there was basically no candidate until last week. We posted a paper with James Hyde last week we solved this problem. There exist continuum mini isomorphism types finitely generated simple groups of homomorphisms. In fact our construction is really explicit. We produce a machine that takes an input and pops out an output in explicit group action. I don't think I can give you all the details but let me try to give you an idea of how this goes. So a quick remark. These are groups of countably singular piecewise projective homomorphisms of the real line. So groups of piecewise projective homomorphisms of the real line where you allow infinitely many breakpoints. In fact every element in this group has infinitely many breakpoints. And so without let me let me try to give you an idea. I don't know if I'll be able to present the construction. There's not enough time but we start with a map row from the half integers to the set a b inverse b inverse. So this is a set of symbols and we call such a such a map quasi periodic. How do you spell quasi? If we require that the integers are mapped to the set a and a inverse. The translation by half of the integers are mapped to b and b inverse. And for every so we treat the image of this or the evaluation of this on the half integers as an as a by infinite word labeled with the with half z. And whenever you have a finite subword then its inverse appears somewhere in the by infinite word. And whenever you have a finite subword then for every finite subword there is an n a natural number n. Such that whenever you have a block just a subword of length at least n then that contains your starting subword as a subword. So this is sort of quasi periodic. And the second input is is a real alpha in 0 1 minus the dietics. And for every pair of these this datum you have an explicit group action 0 alpha. There's a subgroup of the group of orientation preserving homomorphisms of the real line. And what the labeling does so this group is generated by well so it's probably two generated but it's the smallest generating set that I mean the generating set we describe in our construction has a total of eight generators. And all of these so these generators are separated into two subsets and for one of the two subsets the labeling on the integers determines how the generator acts locally around that integer. And for the other subset the labeling determines the labeling on half plus z determines how any generator from that set acts locally on on a neighborhood of on a unit ball around that half integer. And so this is a very explicit group action and the the the atoms of this consist come from the group of piecewise linear the orientation preserving piecewise linear homomorphisms of 0 1. This is a very beautiful group and it's been studied a lot in the 1980s in particular by Matt Bren and his students. And this we use subgroups of this group and we use them to build these zero alphas and we I mean by definition they are finitely generated and we prove that they're simple. So and these groups are candidates for property T groups acting on the real line because well I have no idea how to prove or disprove property T for these groups at least then the methods I know for disproving property T fail for these examples. And so it just makes me curious whether they can have property T and perhaps one can use some computer programming like what was recently done by Osawa and Novak and third person. And you and ensure and say something about whether these groups have property T. So so if I start describing the construction I think I'm going to go over time by at least 15 minutes. I'm tempted to do that but I'm not going to do it but I'm happy to tell you the construction privately if if you want to hear about it. And I'll also say that these groups they all contain free subgroups there's a very simple ping-pong argument that you can you can clearly see. I mean am I going to be sent to jail for doing that because I am I'm going to go over time if I I mean OK I'll try I don't even know how to summarize it. So so we start with there's a subgroup H alpha of PL plus 0 1 to define H alpha. I'm going to have to define Thompson's group F not in the way that Thurston defined but in the way that Thompson originally defined this is the group of piecewise linear orientation preserving piecewise linear homomorphisms. Of the unit interval with breakpoints in the dyadic rationals such that the slopes wherever they exist are powers of two. So what we're going to do is we're going to define a subgroup H of F which can consist precisely of the elements of F. For which the the germs at zero and one have the same derivative. And so this is a nice subgroup of F it's finitely generators. And given any alpha outside the dyadic in zero one it's that we explicitly construct an element which is supported on the on some interval on some compact subinterval of the open interval zero one. So for instance we chose one over four comma three over four such that the derivative at one over four of this homeomorphism is is alpha. Or if you like to do the alpha and V the group H alpha is generated by H together with this homeomorphism. And so these are four generated groups. And it's not too hard to show that they contain uncountably many isomorphism types. And we use these groups as building blocks to construct zero alpha. And so let me actually maybe I can do this. I can give you a very quick construction without writing any notation. So let this be an integer. Let's call it call it X or N say N. If row of N is a then on the neighborhood N minus half and N plus half on this interval. The generator of the generator which is associated to one of the generators of H acts exactly like it acts on the unit interval except that the unit interval is translated to this interval. So it acts exactly like that. And if row of N is a inverse then that generator of H alpha acts like it's flipped. So it's conjugate. The way it acts is conjugate under the unique orientation reversing isometry of this interval. It's the mirror image. And the similar thing happens for the other four generators which are also associated to the four generators of this one around the half integers. I mean the integers translated by half. So whenever you have a point N plus half you look at N and N plus one and the respective generator acts like the generator of H. If row of N plus half is B and it acts like that generator of H flipped if row of N plus half is B inverse. And so you might ask well what if I take the trivial labeling. What if I map every integer to A and every half plus Z element to B. Then well this group is not simple but this is actually a nice group. It's the lift of Thomson's group T. So Thomson's group T acts on the circle not on the real line contains torsion cannot act on the real line. But if you lift the group to homeo plus R then this is exactly that group. And this group is perfect because it admits a quotient on to on to Thomson's group T which is simple and abelianization is hence trivial. And so this is interesting so this group is perfect and it contains a center which consists of integer translations. And the challenge is how do you kill that center and how do you well okay even if you kill that center then why should that be simple. So that's where the definition of cause a periodic comes in. That's where you you flip various things at various points to avoid. So basically one property that will hold for every element of this group is that it will fix a point on the real line. Every element in this zero alpha will fix a point on the real line if row is cause a periodic. And that's that plays a crucial role in the proof of simplicity. And so so actually I managed to give you some idea of do you have any questions about this. I mean did I was was this definition okay. It's literally it's actually very simple. You look at this these neighborhoods of these points and you act exactly like you act in H alpha except that you flip the action. If the image of row n is is the inverse letter. And so it's it's really sort of combinatorial and there there there are a few obstructions to this question. For instance the Thurston stability theorem says that if a group acts by C1 diffeomorphisms on a closed interval or a left close right open interval then it's locally indicable. So this group action when you completed to close minus infinity plus infinity then that should not be topologically conjugate to a C1 action otherwise you are locally indicable. In fact most of these groups tend to be locally indicable because there's this germ homomorphism. So if you look at the group of germs at plus and minus infinity then generally that homomorphism is non trivial. But for these groups because of this quarter cause a periodicity that homomorphism is actually isomorphism. And so and these groups are not bi-orderable because it's an old result that simple bi-orderable groups they do exist. Simple bi-orderable groups cannot be finitely generatable and so by result of David Morris amenable left-orderable groups are locally indicable. So every finitely generated subgroup admits a homomorphism on to Z. So in particular they cannot be simple if they're finitely generated. And so these groups are not amenable but then you can see that directly by constructing free subgroups inside. And actually if you can find a group which is finitely generated simple acting on the real line which does not contain free subgroups then that would be very interesting. In fact Nawaz raised this as a question in his ICM article. Not exactly in this way but an equivalent formulation of it which is if I remember correctly maybe Christopher can correct me if I'm wrong. But does there exist a left-orderable group which is not locally indicable and does not contain free subgroups? I think this is the question. This is one of the questions in his ICM article. So for instance if you can find such a construction of finitely generated simple left-orderable groups that do not contain free subgroups then you answer that question. But that's not true of these groups. They do contain free subgroups. Interestingly these groups not all of them but some of them are C1 smoothable in fact C infinity smoothable on the real line. So by G-syncergescu you can prove that these groups are the most basic one. The one where your alpha is actually you don't use that additional generator you just use this H. That group is topologically conjugated to that group action is topologically conjugated to an action by C infinity with a few morphisms. However thanks to Thurston stability theorem the extension of that action on the closed the compactification of the real line is not topologically conjugated to an action by C1 with a few morphisms. So there's some interesting subtleties around these groups and we are trying to prove that well I shouldn't tell you what we're trying to prove otherwise you'll prove it but I don't know it's still nice to share things. And we're trying to prove that finitely presented simple groups cannot act on the real line. We believe that should be true but we don't have a proof. It's not it cannot be periodic. Yeah otherwise it would have a quotient on to something that's that's literally I mean that's that has only finitely many breakpoints and that would be locally integral so it cannot be periodic. Exactly exactly exactly right so if you if you start with instead of H alpha if you just start with the H the original age without introducing that new element and you construct the group. Then that that is just built out of copies of Thompson's group F right and we know that F is C infinity smoothable thanks to G Sir Jessica so so on the real line it's C infinity smoothable but not on the closed compactification of the real line.