 OK. Prvam, da se organizacijo, da se vsegače. To je druga razlika, da sem tudi. To je veliko prav. Zato sem tudi prav, da se vsegače. Vsegače, da sem tudi prav, da sem tudi prav, da sem tudi prav. Zato sem tudi prav, da sem tudi prav, da sem tudi prav. Zato se se circumstavimo v svečenju različnih različnih earrings ljud. Odpovršnje je zelo pristelje pristeli. So n데 powdera svečne je. Zato tudi prav, da sem danaoviti na njom ljudve različnih Stock. So, da so zelo tega se večeho. Ok. Zdaj mi revoj, kako je geometrične strukče. Zdaj se vlasti vse definitivnih. Zdaj sem bil, da sem bil, da sem bil, da sem bil, da sem bil, da sem bil, da sem bil. izgledaj na spasih X. M je vzgledaj. G, X structure, M, je vzgledaj na m, kaj je vzgledaj vzgledaj. Vzgledaj na m, vzgledaj na m, je divitano na m vzgledaj na m.° In vzgledaj na obroženjem n. Res compte vzgledaj na m. g, vzgledaj na kajšljanje konec. Odstvarnanja moranj. isportsn Sahne da mu te becamereek z vstellj, zakovač je že po para naz piping. Volim neki neko sprav, da le medevkaja enom. Junior. In nekonek. One že no ambition. are hyperbolic structures, this is the most famous, maybe, example, and conformal structures, this is another example, real projective structures. So, a question is fixing a g and x describe the gx structures on m. So, a problem, which is more general, is just describing representations of the fundamental group in 2g, but you don't have a geometric structure, but at least you have a representation in 2g. So, this can be seen as a first step to trying to find a geometric structure on m. So, I have to define what is a representation variety. So, the variety, I would choose an algebraic group, so it could be any of those, SL2R, or SL2C, or SL3C, et cetera. And I'll call m of gamma g, the set of all representations of gamma into g, and as g acts on m of gamma g in the set of all representations by conjugation, I can take the quotient. The quotient is non-housed in general, so there is a trick to define an algebraic variety, which is called algebraic quotient, which I won't describe, but you can just forget about these two parallel bars. So, why study character varieties of fundamental group? There are many reasons. I think during those two weeks, you've seen many examples of deformations of geometric structures, since those unknown people, until there is this phenomenon of rigidity, theorems, ve, mostl, which was mentioned today in the morning, and interesting, more interesting recent results where the representation variety, or the character variety can give you informations about the topology of the manifold itself, so the work of color shaling. So, I get to the point, so pgl3c. So, I want to obtain geometric structures from pgl3c. So, pgl3c is too big. The dimension, complex dimension is 8, but there are interesting real forms in pgl3c. That means subgroups of pgl3c whose complexified algebra gives you the algebra of pgl3c. So, I am enumerating them. There is pgl3, pgl3, pgl1, which we love most, some people here, and sl3r. So, pgl3 is not related to any geometry, to any geometric structure, in fact, on three manifolds, but those two pgl1 are related to Cauchiriman structures and sl3r to real flex structure, I'll explain in a minute those structures. So, I don't want just to find representations into pgl3c as the title of the talk pretends. I want really to find geometric structures on three manifolds. So, that's why those three groups, especially the two non-compact groups are very important. So, the main idea behind that is that actually those two geometries, pu1 and sl3r, could be related to a geometric relation of contact structures in dimension 3. So, I'll just briefly describe those two geometries, and they are related to contact structures, each of them. Moreover, there is a natural embedding of pgl2c into pgl3c, which is the reducible representation of pgl2c into pgl3c. So, in a way, apabolic geometry is somewhere inside this group. So, what I'm going to do is the following. The whole technique is, I'll try to triangulate a manifold. Here it's a toy model in two dimensions. So, I'm taking the punctured torus. The punctured torus is triangulated by two triangles, and here I want to realize that topology geometrically, and geometrically means realizing each triangle as a geometric triangle. So, I think this picture explains what I'm going to do in three dimensions. So, you have to think the rest of the talk is an analogy of this drawing. So, let me recall for those who didn't see what is CR geometry. No, CR structure is pu1 acting on s3. S3 is the boundary of a complex ball of dimension 2. So, this group can be viewed as the isometry group of the complex ball, as in the talk of Martin yesterday. But it can be seen also as a conformal, sort of a conformal acting on the boundary of s3 by analogy to real epabolic space, and the isometry group acting on the boundary of cp1. So, it's the same idea. So, it's a geometry which is not Riemannian. So, it's not in the scheme of the talks in the first, in the morning, because in the morning we had a volume form which should be invariant under the group. So, it's not Riemannian, or pseudo Riemannian. And the real flag structures are, it's a g-x structure, s3 r, everybody knows. X is the, it has to be a three-dimensional manifold. And this three manifold is the set of flags in rp2. So, flags in rp2 is just, so rp2, if I make a drawing of rp2, here is rp2, and I just take a point, p, and a line. So, it's the set of all those pairs. So, it has to be pairs p, l, p in rp2 and l, I think a line in the dual space of rp2. This is a three manifold, this space. And you can think of it as s3 r divided by the upper triangle matrices. So, the upper triangle matrices, so, just to, this here, so if you have a line in rp2 belonging to, so this is a line in rp2, belonging to a plane in rp2. If you want to fix the line and the plane, then you need a matrix, which is like this, upper triangle matrix. So, this is borau group. So, those are the two contact, homogeneous contact structures, which are supposed to uniformize or to uniformize all contact structures on three manifolds. So, I need to talk about flags in order to put those two geometries together. So, instead of talking about real flags, I'll do all the computations with complex flags. So, I'll put the two geometries together. So, the space of flags in cp2 is just one point in cp2, and a line passing by that point. And there is an action of SL3C on the space of flags. And the space of flags is a homogeneous space, then SL3C here divided by the borau group, the same thing, but in the real case, complexified. So, it's interesting to observe that if you have cp2, you can obtain a conic. So, let's see, this is a conic, cp1 inside cp2. And each point of a conic determines a complex line, which is tangent to that conic. So, the embedding of cp1 into cp2 defines an embedding of cp1 into the space of flags. So, here, I'm writing, I just defined a map from cp1 to the space of flags. And the same thing for S3, if I have S3 contained in cp2, I just take a tangent line passing through this, a tangent complex line passing through one point. So, I obtain a map from S3 to the space of flags. So, there are those two cases, which are related to CR structures, and hyperbolic geometry. And there is a further map, if I take a real conic. Yeah, I'm taking any conic, a quadric, yeah, a quadric cp2. And this defines a natural map of cp1 into the flags, just take the tangent line. So, I have all those maps. So, I want now to describe triangulations. And in order to describe triangulations, I would describe triangulations of flags, by flags. That is, I have to take configurations of flags in cp2. So, the action of SL3C is transitive on the set of generic pairs of flags. So, we do a drawing and everything will be clear. So, if I take two flags, like this, in cp2, there is only one, up to SL3C. But if I take three flags now, if I take three flags, there is one complex invariant. So, this is zero dimensional, this is one dimensional. And if I take four flags, that's what I need to define tetrahedra. Then you need, so, if I take four like this. So, any configuration of four flags, this is four dimension. So, this is very classical. People in the 19th century were dealing with this. And we are done to define triangulations. So, in order to parameterize those configurations, there is a classical way. It's by cross ratios. So, all those things are very classical. There is a way to, if you have four points in cp2, you can imagine all the lines passing through the first point. So, one. So, here. So, I imagine lines between the points. And take all the lines passing through the first point, right? One. In fact, as I'm dealing with flags, at point one, I have also a line passing through it. So, actually I have four lines passing through one point. And all the lines passing through one point is just cp1. So, I have four points in cp1. Once you have this one for flags, and this defines cross ratio. So, x denotes the cross ratio of those four lines passing through the same point. So, in this way, you get a lot of numbers. So, not a lot of cross ratios associated to many configurations. So, there are too many numbers. Because it depends on how you choose the order of those lines. So, it's huge, there are all those numbers. But I said that the dimension is four. If you have four flags, you just have four dimensions. So, there are relations between those numbers. And I can write down the relation, it doesn't matter. So, I'll skip those relations. There are classical relations. We don't care. So, let me start again with the same picture of the tantra torus. But now in three dimensions. So, the most classical example is the figure eight knot. It's an example of non-compact three manifold. And this is a famous triangulation of the complement of the figure eight knot that nobody understands. So, I won't explain. So, there is this configuration with gluins of sides that you have to spend some years understanding. So, we have what we call an ideal triangulation of M. That is, I have M as a truncated gluing of tetraedra. T, there is our tetraedra. So, if you see here, you have just two tetraedra. So, I will decorate this triangulation. The decoration means that I'm going to put at each vertex of the ideal configuration, ideal triangulation. I associate a flag. So, that's a decoration. So, I put a flag at each vertex. And if I do that coherently, then I get for free a representation into PGL3C of the fundamental group. So, you can think of this as a black box. But you can imagine how to do because everything, all the flags are coherently glued. So, when you follow a path along your manifold, you make gluins of flags. And gluins of flags are done by this PGL3C. And you obtain this map. So, this gives you a holonomy, sort of a holonomy representation. But I didn't have any structure. I just have very coarse data to define this map. So, when I have gluins. So, for instance, when I have a gluin of a face, I have a triangulation. So, I have one tetraedra here. And here there is another one. So, let's suppose that that face here, it's glued with this one here. But for that face I have three flags, F1, F2, F3. For that one I have, let's say F1, let me be F1 prime, F2 prime, F3 prime. So, I have three flags. And three flags have to be glued together. So, you have to be careful that three flags forms a configuration of three flags form a one parameter, has a one parameter space. So, you have to be at the same point of this one parameter space. Yeah, yeah. You have more equations. I will come back. Yeah, I will come back to this. It could be just, sorry. In fact, it's C minus a couple of points. So, it's a very simple space. At the same time you have a boundary holonomy. This manifold has a boundary. So, you have a representation of the boundary of the manifold into PGL3C. And I will suppose that the boundary has just one component and it's a torus. In that case, I will impose that the boundary holonomy preserves one flag. To preserve one flag, I have to have the boundary holonomy, which is Z plus Z. So, it has two generators. It has two generators here. It's an abelian group. So, I will impose that the two matrices corresponding to this representation are in abelian group, the preserver flag. So, another way to think about this, a more formal way to think about this decoration, is to think of all the cosps in the boundary of three manifold. Let's say it's a hyperbolic for that sake, for that example. And I associate to each cosp a flag in an equivalent way. So, I have a map from cosps to flags in such a way that it's equivalent. So, I want, it is a very formal definition of a decorated representation. I won't use this definition much. I just want to stress that if you have representation of the fundamental group of a manifold in pgl3c, you can associate a decoration, a decorated representation. And this, the boundary holonomy will fix a flag. So, again, I repeat bigger the matrices that should be, they will be very important in the following. So, those coordinates using the decorated representation were used by Falk Goncharov to analyze higher tašmila space in the case of surface groups. And they also studied in a very similar way by Garofallidis Thurston and Zikert. And they were generalized to pgl3c by Garofallidis Thurston, Zikert, Dimov Tegelberg, Goncharov. So, there are many activity on those coordinates because they give very useful parametrizations of representations. So, let me explain some applications of them. So, the first application is a definition. So, there are mathematicians. But Gabela is a physicist, and Dimov Tegelberg is a physicist. So, the first application is the definition of the volume of representation. Here the volume is not compared to what was explained in the morning because you don't have the invariant volume form. So, you have to do something different. I was surprised that there was a volume in this case because the geometry is not remanual. So, as Ruth explained before, the dialogarism appears usually in the forms for the volume in three dimensions. And if you have a gluing of tetrahedra, you just sum the dialogarism function on each of those coordinates, cross ratios that I introduced before. It seems to be very at ok, but it's very natural in a way because of the properties of the dialogarism. So, this volume is of the find, and it was defined by many people in several generalizations. There are many ways to do it. So, I listed all the people because I know I'm filmed, so it's a bit dangerous. So, a second application is combinatorial proof of this result of Menard Ferrer and Porti, which says that if you start with a hyperbolic manifold, M, and you have a representation, the geometric representation of the fundamental group into PGL2C, that is the discrete faithful representation, which gives you a hyperbolic structure, then you can compose with the irreducible representation of PGL2C into PGL3C, and you get a point in the character variety. You get a point in the character variety here, and the theorem says that it's a smooth point of complex dimension two. So, this parameterization, and in fact the eigenvalues of the boundary alonomy parameterize the character variety. So, if you choose two eigenvalues in the boundary alonomy, this parameterize locally the character variety. So, for those who know the representation into SL2C, this is very natural to think. Yeah, I didn't say this, yeah. Sorry, yeah. So, let it be the manifold with boundary atoros, just one. So, for PGL2C, you generalize this result, and may not party other methods. Okay, so a third application is a generalization of a formula of Neumann's AGA. So, this formula says that don't be afraid. So, first you have to just see this. This is a funny notation for this. In fact, you have two functions, f and g, and I just want to say that I multiply log of g times the differential of log of f, and I doing this, so just forget this thing, and think you have this formula. So, the formula that we obtain says that the derivative of the volume is equal to this. So, see the important thing is that it depends only on the eigenvalues of the boundary alonomy. S is the number, the cusps, so here I'm summing over each cusp, and each cusp I have those eigenvalues, and I do this. It's a fantastic formula, because it says that, for instance, if a, if the diagonal is 1, so if it's unipotent, if you know that the eigenvalues are 1, then you see in the formula you have log of 1 always, which appears, so this is 0. So, for unipotent representations, the volume is a critical point. So, the volume, there is a notion of volume of representation. If the representation is in PSL2C, it coincides with the hyperbolic volume. If not, it's something, it's a volume of representation. It's defined combinatorially, using the dialogarism, you sum over the trade. There is a notion of volume, which behaves well. PGL and C in general. For compact manifolds, it's a pullback of a homology class, but if it's an open manifold, it's not. You cannot do this, but that's what I was going to say. Bohere, Bohere, Yodzi, they managed to define as in bounded homology, this volume, which is, I think, it's the best definition of the, of this volume, and they obtained this wonderful result that says that the volume is maximal, if and only if the representation is the geometric one. So, if you have a three manifold, hyperbolic three manifold, then you'll know that the unipotent representations, bounded unipotent representations are critical points, and what they prove is that there is a maximum, which is precisely the one which comes from the geometric representation. The geometric representation, so I go back a couple of, is this one. So, if you have a hyperbolic manifold, you have a natural representation in PGL to C, and then you compose with, so you call this geometric representation, this is the maximum volume. So, this is very interesting, but we don't understand really how this volume, the volume is a function, how it behaves in the character variety. So, in particular, we can show that the volume is zero in representations in PU to 1 and SL to 3R. So, I'm saying too many things, so I'll continue the slides. So, those were applications now, how to compute explicitly the character variety, how to do that. So, it's very hard to compute, you can imagine that you have to do some tricks, if you are lucky, sometimes you are lucky, then you can get explicitly the representations. So, let me show you how to do it with brute force. So, you use coordinates, the cross ratios. So, you do what Thurston has done, he writes cross ratios, and he imposes compatibility conditions. The compatibility conditions are phase equations. So, when two phases are glued, the three ratios have to coincide, so that's what it means, this equation. And you have edge equations, as you said, when you do a full turn around an edge, you have to go back to the same point. So, it's for those who have seen real hyperbolic geometry, you remember those equations in Thurston's notes. But if you didn't see that, you just have to write those equations, and I will write for you in the case of the figure eight knot. So, I return to the figure eight knot, and I write all the equations I need, I put the parameters, and I write them. It's too much, right? It's a lot of equations, but here are the edge equations, the phase equations, and here it's a computation. You can compute the eigenvalues of the boundary polynomial. So, there are two generators, so there are four eigenvalues in total. So, let's compute. So, this is a diagram of what you should do. You should triangulate the manifold, I repeat what we are doing. You should put the coordinates zij, you write the equations, you solve them numerically or with Grobner basis. And if you do this, what I told you is this black box, I get representations. And then after the representations, we could go further and try to get geometric structures. So, it's really brute force. Let's try to do this. So, the first case is when the eigenvalues are one. So, it's the unipotent case. It's not the full character variety, but it's part of it. It turns out that there are very few solutions. So, there are the hyperbolic solutions, which were obtained by Riley and later by Thurston. And then there are some, a finite number of CR solutions. CR solutions meaning that the autonomy will have values in pu to 1. So, here is the group. This is the fundamental group of the figure eight knot. It has this presentation, there are two generator. And I will not write the hyperbolic solution, real hyperbolic solution. I write only the CR solutions. There are three that you can write down. So, probably you don't care. Just matrices, you get at the end. You get representations. What to do with the representations? Well, that's what Pierre Ville was saying. Once you have representations, he described the beautiful parameterization of the white-head link representation variety. And the question is, what representation corresponds to geometric structure? Who knows? So, let me briefly say what we know about this case. So, for the first one, I analyzed many years ago, but it's a very strange representation. It's infinite index subgroup on aphalitis. So, but the limit set is S3 itself. So, it's a very strange. And we don't know if it corresponds to a autonomy of a geometric structure. On the other hand, the other two, so this was analyzed by Derot and Jian Wang. It's a very beautiful representation. In fact, it's a triangle group of type 334. And in fact, it's uniformizable, as Pierre explained in his talk. Remark, so in fact there are other structures branched and maybe not non-branched, which have the same holonomy. We don't understand why. Another remark is that Derot continued and he analyzed other deformations, he analyzed uniformizations of deformations of this representation. And this was for the figure, but historically Schwartz found, as Pierre Ville explained, the first geometric SU21 structure on a complement of a knot, a republic knot. But what I want to say here is that there are some examples which are being studied, but not many, very few. So it's really annoying, because we don't know what's going on. As for geometric structures, there is a website curve, which has this wonderful computer understanding representation varieties efficiently, which computes representations. So this was worked by the implementation of those things, due to Kozalev, Rouiller and Gornet. Gornet implemente in snappy, the Ptolemy model, which computes representations, not geometric structures. So now I can go back to this table, which is not so bad. So here is a list of manifolds. This list are hyperbolic manifolds. Ordered by complexity. Complexity means the number of tetradors you need to obtain them. So they are the first two tetradors. Sorry? So all of them have only one cusp, except the last one, which is the white-head link, two cusps. So here you should see all the... This list was done by Kozalev and Rouiller. And they are all the unipotent representations obtained by this method. So they are all here on PGL3C. But you can see that there are some representations, which are in those preferred subgroups of PSL3C, which give you geometric structure on three manifolds, maybe. So in PSL2C there are always the hyperbolic ones, but you see that there are much more in PU21. So you can think that CR structures should be much more important in life, three manifolds, than hyperbolic geometry. There is something wrong going on. So that's a mystery. Those in the table were unipotent. Can you imagine that it was hard to get them? Now we want to get the whole character variety. To get the whole character variety is a bit difficult, and this was done recently. So there are three irreducible components. So the character variety is an algebraic affine variety, and I'm saying that there are three reducible components containing irreducible representations. So this was proved independently by Heuzener, Munoz e Porti. So let me make a drawing of them instead of explaining the technique to prove it. So the drawing that I would like to make is this one. So this is one example. It's essentially the only character variety of a hyperbolic method computed up to now. It's very hard to compute. So this... Let me... So it's sort of... There are three components. I drew these three components. So those two are hyperbolic. So there are two... There are two unipotent hyperbolic, and we know by Bvlger, Boucher, Yodzi that they are maximal volume. So that's why I drew them like in the extremes here. And these two... This is one component. It contains this row one and row one bar. They are always couples. You can take the conjugate of it. And they have zero volume. That's why I put here. And then there are two other components where the representations are... So row three. Those are the unipotent representations that are points. There is a picture of what should be the character variety of irreducible representations ordered by some volume function. We don't understand exactly what it's going on here. OK. So the last general result that the only general result that I would like to mention is once we understood that for the figure eight knot you have a character variety where each component is two-dimensional. We could imagine that this is a general phenomenon. So we start with m, orientable three manifold with a torus boundary. So compact means that... Well, it has boundary here. I have to have a non-empty boundary. And suppose that the representation is the risk dense. It's irreducible. And I suppose also that the boundary is regular. Meaning that the centralizer is the minimum possible. Then I can say something. So you take an irreducible component of the character variety and then this irreducible component has dimension bigger than 2s. So 2s... S meaning the number of torus components of the boundary. So this was an important theorem in the case of SL2C where you substitute instead of 2, 1. So let me... In the case of SL2C you can read this theorem in substance notes. Here you have the dimension is bigger than s. So it's an important theorem because you want to make sure that the space of deformations is non-empty. So this allows you to do den filling, for instance, in your manifold. This theorem is more general than what I stated. So we can... If you substitute n here so you can guess that you get n minus 1 here. So let me give a proof of this. No, I won't give a proof of this and I'll stop here.