Okay, so in this, in this second lecture, I will still be speaking about the representation of a finite group of a surface in this group PSL2R. And I would like to first recall what we did yesterday. So yesterday we introduced the so-called friction representations, which are the representations coming from a hyperbolic structure on the given surface, namely if, so we take, as yesterday S is a closed oriented surface of genus G, which is larger than or equal to 2. And if we have a hyperbolic metric on that surface, namely a Riemann metric of curvature minus 1, then we have an identification between the surface S and the quotient of the upper half plane by some group gamma of isometries of positive isometries of the upper half plane, which is our group PSL2R. And this gamma in fact is, so acts freely and properly discontinuously on the upper half plane and it identifies with the fundamental group of our surface. So this gives rise, so any such identifications, any hyperbolic structure on the surface gives rise to representations from the fundamental group of S to PSL2R, which is just the identification between the fundamental group of S and the group gamma, this friction group. So in fact these, so these representations are called friction representations and in fact a friction representation from the fundamental group of a surface in PSL2R is function if and only if it is faceful and with discrete image. The proof of that is that in fact the fundamental group of a surface of a closed surface determines the topology of the underlying surface. So you can detect for instance the genus of the surface by knowing the fundamental group of the surface. You just take the abenianization for instance of the fundamental group and you get the first homology of the surface which is just a free abenian group of rank 2G, so you recover the genus of your surface just looking at the fundamental group. So in fact let's say another characterization of friction representation is the following as I said representation is friction if and only if it is faceful discrete. So by discrete I mean that the image of the representation is a discrete subgroup of PSL2R with natural topology on this group PSL2R and as I said yesterday we didn't really prove that but it's not really hard to prove it is equivalent to be discrete it is equivalent that the action of the group gamma on upper half plane is properly discontinuous. So we have defined that yesterday and we also have seen that in fact any complex structure on the surface as gives rise to a unique hyperbolic structure and so to a unique representation of the fundamental group of S in PSL2R up to conjugation. But today I will not really, I will try to, I mean unless for really small moment I will forget the Taish Milo theory I will really try to speak about hyperbolic geometry and representations of surface group. So what I want to begin with is to say that if we look at general representations of a surface group in PSL2R there are no reason to be friction of course so for instance if you take the trivial representation which maps any element of the fundamental group to the identity matrix of course so this is one example of course it is kind of easy to build representation which are not faceful for instance just I give you can for instance do the following procedure to build some representation which are not fictions. You think of your surface S as being the boundary of a handle body of genus G so you feel your, you think of your surface S as being in three space like that and you look at the interior region delimited by the surface this is a handle body of genus G and in fact it is, so let's call H the interior region in three space and in fact it is quite easy to prove that the fundamental group of H is a free group on G generators and so it is the first thing and that if you look at the natural map coming from inclusion of S as a boundary of handle body so this map is a surjective so this proves that you can always surject the fundamental group of a surface onto a free group on G generators where G is the genus and now it is really easy to define representations of the surface group into PSL2R just by defining a representation from fundamental group of the handle body and pre-composing by this surjection you will in this procedure build a representation of the surface group and now why it is interesting it is that to define a representation from a free group on G generators in PSL2R is extremely easy you just have to specify G matrices which will be the images of the generators of this free group so you will have G times the number of parameters of this free group namely three G parameters for defining a representation from this group to PSL2R so what I just wanted to say here is that by this procedure you can construct a lot of representations from a surface to PSL2R this is just a kind of easy example just to say that there are representations from a surface to PSL2R that are not functional so in this lecture we will try to make a little to organize a little bit this space so the space of representation of the fundamental group of our surface in PSL2R and so this is quite classical so the main thing I am going to do is that I am going to to to organize this space as as a disjoint set of connected component there will be a finite number of connected component and each of these connected components will be detected by an invariant which is called the volume of the representation so I am going to to define this volume I am going to explain some obstructions that the volume of representation can that the volume of representation satisfy and and at the end I would like to try to to speak about how we can come back to geometry from a representation which is not necessarily fuchsia namely how we can try to geometrize representation so this is more or less what I would like to speak about today so first let me try to define what is the volume of a representation and so this kind of really classical invariant which goes back maybe to the 70s so so what you try so I was I say the word geometrization that will be an important word for the the end of the talk but but from now on still we will see that it appears in some ways so we we we knew we know from yesterday that fuchsia representation we can associate to them a hyperbolic metric on the surface and so in general for a non-fuchsia representation of course this will not be the case as as I said but still we can try to to appeal to geometry and how we will do that in really great generality what I would like to say what is a geometrization it is going to be a map so so what I claim is that there always exists a map let's say for for now continuous and even smooth if you want map but I will consider also continuous map so I will be I will stay in this category for for this generality map from which it's called the developing map or which I will think of as a developing map from the universal cover of our surface to uh upper half plane so it is kind of a local identification of our surface with a upper half plane so as in the case of a hyperbolic structure but here I don't ask that it is a locality for morphism I just have a continuous map which is equivalent with respect to the representation or okay um so I claim this is always possible to find such a map so here's a proof so the proof is take a triangulation of s tau s um and uh no no no now I try to to define the volume of any representation right um so take a triangulation of the surface I denote it by tau and lift it to the universal cover of s so I denote by tau tilt with tilt I denote all the objects at the level of the universal cover so we get so if this the interior of this disk is the universal denotes the universal cover of our surface we get a triangulation which will be invariant by the action of the fundamental group of our surface right get a kind of tessellation uh which is invariant by this action and so what we will do we we just uh take for any vertex of our original triangulation we lift this vertex so for any p we denote by p tilde uh a vertex of the lifted triangulation which uh which is the lift uh of p so we have a finite number because we have a finite triangulation on the surface we have a finite number of points of vertices here which form a kind of fundamental domain in for the action of the fundamental group of the set of vertices at the level of the lifted triangulations and now what we will do we we will define a map so we want you to define a map from the universal cover of our surface to the upper half plane and uh we will do the following so we uh for any p tilde uh that I just defined for the finite numbers we uh prescribe any value so arbitrarily any value the map we want to construct and uh now we uh we extend we extend uh the the the definition of our map to the set of all vertices by using equivalence okay since we have taken a fundamental domain for the action on vertices there is a unique way to do so so this is maybe we prescribe this is the first step second step we extend d to the set of all vertices of this lifted triangulation using row equivalence so if for instance I consider a vertex here so it is a vertex that I denote by 2 tilde let's say so this q tilde is some element gamma of the fundamental group uh the image of some of some elements of some vertex p tilde that I in my fundamental domain uh by an element of the fundamental group and I just define the image of q tilde is just row of gamma times image of p tilde that I already defined in the first time okay so there is a unique way to do so and uh now we have defined our map d on the set of vertices of the lifted triangulation and we extend this map on the first skeleton so on the edges by sending any edges edge to the geodesic between the image of the extremities of the edge I consider and then the interior of I mean the two the two skeletons so the the faces of my triangulation I extend the map so that it maps the triangle to the triangle in hyperbolic space right so to do to do this I again I have to I will not make this really precise I just say that so you to to make to to do that really completely you you have to also take a fundamental domain in the set of all edges in the lifted triangulation define extensions of our map so that it becomes the image becomes geodesic and then extend by equivalence and do the same for the set of faces okay but this is very simple procedure to prove the existence of such a developing map okay so this is extremely simple and in fact uh it is also simple I will not do so to smooth such a map so this map I just defined is kind of continuous only but you can smooth such a map uh to get a smooth equivalent map to uh from from the universal cover of your surface to to the upper half plane and so I'm going to to erase this proof and I want to say something before continuing which is the following so this is a first step towards the geometrization of our representation in the following sense so here for instance I so if you think of the the proof uh before so we have uh in fact founded for each triangle of our triangulation in our surface so we can leave this triangle to the universal cover and look at the image of the this triangle by our developing map we just constructed for and we get a hyperbolic triangle so in fact for any triangle of so it is well defined because it does not depend on the lift because of our triangle because uh the map is equivalent so uh the image of two lifts differ by an isometry um of of the upper half plane so in fact uh if we follow the proof we just done so we had a triangulation of our surface and each triangle uh is now equipped with a hyperbolic I mean a structure of a hyperbolic triangle and more than that uh we can also we have another an extra information namely that this triangle uh has an orientation because our surface is oriented so we can look at how it is mapped to the upper half plane it is either mapped to the upper half plane with positive orientation or with negative orientation and there are some uh degenerated cases where the triangle could be mapped to us a degenerated triangle where we could not define orientation but uh we will try to just don't think about these kind of degenerated cases which are not which are which are not generic so in fact uh we have kind of geometries uh geometries our representation in the sense that uh we uh we have geometry on these triangles plus signs a sign to this triangle so this is what I so this is the first attempt to get a geometrization of our representation so of course here uh this geometrization will in this generality will not be uh really um how to say efficient for many problems and all the game uh that we will try to develop in this lecture and tomorrow will be to try to find uh better geometrizations which enable to say something interesting on our representations but anyway this is really the idea of what is geometrization okay so now let me define what is uh the the volume no no no it is not at all canonical so this map is not at all canonical it depends first it depends on the triangulation second it depends on the first uh so you had left uh the vertices and you might uh send the vertices anywhere you want so there are there are these finite number of parameters uh which are not canonical okay so this is not canonical but uh but it is canonical in a certain way let me let me say a property which which is very important uh and which will serve for for for what will follow is that uh the map there the developing map is well defined up to homotopy within the set of row equivalent map so what i am going what i am saying is that the set of row equivalent map continuous row equivalent map or smooth equivalent map is connected in this sense it is canonical and the proof is extremely easy also it's just if you have two maps so here is the universal cover you have a first map d uh and a second map d prime which are both row equivalent right so if you take a point here you look at so you have two images one by d and one the other one by d prime and so you can define a map a family of maps so dt for t in the interval zero one and uh in such a way that dt of p describe the geodesic segment between d of p and d prime of p uh parameterized by a length one interval right so you take the geodesics uh between the these two maps and it defines for you uh so a homotopy between d and d prime and this homotopy is made of row equivalent maps because we have used geodesics which are invariant by the representations and so these the maps uh dt are all um row equivalent okay so in a sense there in this sense there exists a unique uh row equivalent map of course uh as we will see uh we will try to to find uh more I mean in this class of row equivalent map in this class of developing map or geometricizations we will try to find more more interesting ones but um so so now let me uh define what is the volume and uh for this I will uh I will need to uh to specify uh to maps which are either uh I mean so I will need to some regularities so I'm going to to use uh smooth maps or the maps that I have constructed in this uh in the in the in the proof that there exists some uh developing map so so what I claim is that if you take a developing map uh you can look at uh the pre-image uh by uh our developing map of the so so maybe I should say what is the idea here so the the here you see we have uh we we have constructed in the proof a geometric object made of uh the union of these hyperbolic triangles some are positive and some are negative and in fact uh we want to say that this geometric object has a volume and this volume is just the the sum of the volumes of the positive triangle minus the sum of the volumes of the negative triangles right and so this is perfectly it makes perfectly uh good sense uh and in fact uh what is going to be true is that this uh invariant does not depend on the choice of the developing map and uh in fact uh more it will take uh uh almost an integer value if you you divide by 2 pi um but uh so this is for this kind of uh developing map I have just defined here but if you take any d so more formally if you take a any d uh which is smooth or I would I would like to say piecewise smooth then uh you can look at uh the pre-image by d of the volume form associated to point parametric right and uh because this map is row equivalent and that the volume is invariant by the action of the isometric group this is going to be a form which is invariant by the action of the fundamental group okay and so in fact it is a form which lives on the surface s so you can just integrate this form on the surface s so now s is compact so this is a real number and uh we uh the the claim is that uh this v which actually will depend on d does not depend on d it uh it only depends on um on the representation row so this depends on the representation row this is why I denoted by v of row and um I I would like to to also write some properties of this volume moreover uh we have that one over two pi times the volume belongs to a finite set of integer values so so all the values so all the integer values uh which in modulus are less are equal to two g minus two so the absolute value of the polar characteristic of the surface um and um so what can I say so this is called so this this is called the middle wood inequality and uh it takes extremal values uh if and only if uh row is fuchsian or anti-fuchsian so what is the definition of uh anti-fuchsian representation we have defined what is a friction representation so anti-fuchsian representation is the same as a friction representation but you just uh change the orientation of uh the surface so change the representation this one right right this is the opposite of the volume so this is the opposite of the volume okay so this is I will not really really speak about this other class but uh so to answer your question of course this is the opposite and so this uh this fact so right so so in fact uh um so I don't really get the question so so this so I repeat what I said maybe to Alberto so this one over two pi times the volume is just opposite of the other class of the secret bundle that you obtained by suspending the action of the representation on the circle projective action there is a negative sign right because you know for instance for a fuchsian representation we know that this bundle has a class 2 minus 2 2 minus 2g and the volume by definition by my definition is positive but really I want that the fuchsian representation has volume 2g minus 2 so is it clear for everybody that the volume of representation so okay so let let so let me just say a word so here this is a theorem by proven by Goldman in his thesis in in in the beginning of the 80s or end of 70s I don't know exactly but um so these two theorems belongs to the 70s let's say a little bit extended to them so anti-fuchsian representation is the following so you take uh a surface and you consider a fuchsian representation so it is an oriented surface right you take a hyperbolic structure on this surface and you get a fuchsian representation but now you uh you you take the same surface but you exchange the orientation take the opposite orientation and you you keep the same representation the same representation is no more fuchsian in fact it is anti-fuchsian and one way to to to see that it is that this volume takes the opposite value so if you exchange your orientation on your surface and keep the same representation the volume changes size okay sorry because an anti-fuchsian is different from a fuchsian representation you can see that by just looking at the volume so are we are we so yeah so it seems like the same representation but in fact uh right that's right this is the same right right this is have you heard uh so what was said is that I don't know if you so what was said is that uh another definition for anti-fuchsian is that you take a fuchsian representation and you take a conjugate of this fuchsian representation not by an element of ps l2r you would get a fuchsian but by an element of an isometry which reverses orientation so an element of pgl2r and then you get an anti-fuchsian representation so pgl2r adds also for me right just forget about pgl2r an isometry which reverses orientation now I have a doubt if just forget right so yeah yeah right so so yeah so my doubt was uh was uh was true anyway so is this okay so so what I uh so so just to to understand well I mean the the easy part in gunman theorem is the fact that uh if you have a fuchsian representation then uh its volume is 2g minus 2 right so the volume of rho uh is equal to 2g minus 2 if rho is fuchsian well this is really easy because we have by definition we have uh hyperbolic structure on on our surface and this hyperbolic structure gives rise to an identification of of the universal cover of our surface with hyperbolic plane which will be equivalent with respect to our fuchsian representation and this uh different morphine fission between uh universal cover of the surface and upper half plane is a positive different morphine so then uh the volume of the representation is just the volume of our hyperbolic surface by gauss bonnet is just 2g minus 2 so sorry 2g minus 2 times 2 pi by gauss bonnet so it is uh this this statement is just gauss bonnet theorem and uh but the opposite statement is more difficult I I hope I mean we will we will give a proof of this of the opposite statement which is really called man's theorem tomorrow uh but uh today I would like to to give a proof for all the other statements namely that uh the one over two pi times the volume of our representation can only take this set of values and uh that we uh and and that's it to get what is possible to get representation that I think every right uh right this is so I will not write it but uh each of these values uh occurs as a one over two pi times the volume of our representation certain representation okay right we will see that right zero is attained 2g minus 2 is attained so we will see in a moment that uh that we will we can we can do any value okay so uh let's go let's go uh to prove uh uh first so so what I am going to to do uh to prove this theorem is uh to prove that first that one over two pi times the volume uh is always integer is always an integer and uh to do so I am going to use so in fact now we don't really know that it does not depend on uh on the equivalent map D but uh what I am going to do is that uh I am going to prove that for any of this map uh this integral takes an integer value and so because they are all homotopic uh this volume is going to be constant and not depending on the representation and to do so I will only give the proof in the case of the developing map I already constructed in the proof that there exists so namely in this kind of um um geometricization by uh positive and or negative triangles so so here is uh here is our surface and this surface is triangulated by hyperbolic triangles I am going to to assume that none of these uh triangles are degenerate and so there are some parts which are positive of the surface so some some triangles are positive and some other ones are negative so it defines for me some regions of the surface that I am going to to call uh so in the red this is say positive ones in the blue these negative regions right and uh I am going to denote s minus s plus uh the the union of the triangles which are positive and s minus the union of the triangles of our triangular relation which are negative and uh now what I want to do is to mimic uh what I just say that the volume of the fiction the proof that the volume of the fix and representation is uh over 2 pi is 2 g minus 2 so we will uh to do so we will just apply Gauss-Bonet theorem to each connected component of the regions uh s plus or s minus okay so so what I have to to to to do is to apply Gauss-Bonet to some so if I take a connected component of s plus or s minus it is so so in fact to be really precise what I do is that I I consider the a connected component of the interior of s plus or s minus and then I take the geometric completion but let's just forget about these things uh what if I have a surface with boundaries uh and the surface is a hyperbolic surface and the boundaries is piecewise geodesic so how do we apply Gauss-Bonet theorem for such so let d be a hyperbolic surface hyperbolic meaning remain surface of curvature minus one so a compact hyperbolic surface with a piecewise geodesic boundary but so let's say it is a connected or it is not really necessary here so what is so if if we look at the boundary then it is going to look uh as something like this so we will have a curve so assume say our surface leaves on this side of the boundary so we have our domain here so this domain uh it it it could it could have topology in our setting but uh so what will um um be important in the Gauss-Bonet theorem is to consider the external angles if we take p a point in the boundary which is a corner we have an angle let's say alpha of p which is the uh I don't know the angle of the of the surface so the conical corner corner angle maybe I don't know what is the terminology corner angle of the surface of the domain d at p and so Gauss-Bonet theorem says that if we consider minus uh the volume of d plus sum of this angle alpha of p minus p for any p in in the set of corner points in the boundary then we get that this is equal to 2 pi times the electric characteristic of our domain d right I'm I'm is it is okay or so this is this is a curvature term of the boundary okay we have no curvature channel coming from the point which are not corner point because those segments are geodesic ones but we have only this kind of terms which uh occurs okay so now I can I can uh write this formula um use this formula for the two surfaces s plus and s minus so I get um so volume of the developing map I I was just constructing is by definition of the volume is just the volume of s plus minus the volume of s minus and I will try to to make the proof here um so this is um so 2 pi times the electric characteristic of s plus minus the electric characteristic of s minus right so this is an integer so we are okay and now there are the boundary terms coming from the corner points um and uh so let's analyze what happens there so we have so let me okay we have some at the corner points positive corner points of minus the sum at the negative corner points right in this way okay so now in this sum we can try to block this sum into all the points a sum on all the vertices of our triangulation of these quantities so this will give so we will so if we try to to look at all the these corner points um coming from a single vertex of our original triangulation we will see successively a corner on a negative corner point then a positive one and then a negative one etc so we so if we so if here we have a vertex of our triangulation in this double sum I will see for each um a conical domain start getting out of this point I will see some some terms appearing and what are these terms well if I don't know this is going to count by a negative term so it will be something like alpha of p minus if this is so let's so let's maybe call this point p1 plus p1 minus no this is p1 minus p1 plus p2 uh minus p2 plus etc there will be an even number of domains because they are uh because this is clear I don't know how to say and uh and then um uh in the sum we will have so I'm sorry it's a kind of uh we will have so sum over alpha of alpha of p uh k plus minus pi minus alpha of pk minus minus pi so the pi are going to cancel out and we get the as a result we get the sum of the alpha of pk plus minus the sum of alpha of pk minus right here so and and then to get this double sum we have to sum these sums over all the vertices of our triangulation but now what I claim is that this quantity for any vertex this quantity is an integer and this is this is really clear because around this vertex we have a developing map local developing map which is defined which is going to map the red region positively and then the blue region in negative sense so we have to so if if we are in the red region we we turn the anti-clockwise orientation and then in the blue region clockwise orientation but we have to come back at the at the same value because the map is well defined so in fact the the sine sum has to be an integer right so each of these is an integer and so this ends the proof that uh this volume takes an integer value so it is constant and does not depend on the developing map since uh the sets of developing maps are connected okay okay so after this painful proof we will go on to uh to prove that one over two pi times the volume is bounded in absolute value by two g minus two so this is uh what is called the middle wood inequality I have to hurry if I want to to speak to give some examples so I will try to so uh proof of middle wood inequality so the proof of middle wood inequality starts with the following observe very simple observation is that so we will try we will use uh so we we now know that uh volume does not depend on the geometrization we have chosen so we can try to uh to to use the geometrization I I defined um at the beginning of the talk namely those made of hyper positive or negative triangles um so now suppose you have such a geometrization by positive or negative triangles uh what I claim is that the volume of the representation is always so let's let's do this so yeah so the volume of the representation is just in absolute value is just the absolute value of the sum of the uh a re of the positive triangles minus the uh sum of the re of the negative triangles okay so this is certainly bounded by the sum of the re of all the triangles and uh as you uh probably know or probably we will see uh uh in france was lecture maybe um is that the rear of a hyperbolic triangle is just uh pi minus the sum of the angles of the triangles in particular it is bounded by pi so we get a bound for the volume which is uh pi times the number of triangles which we need to uh triangulate the surface s so this this bound here does not depend on the on the representation so this is the first step of uh of in miller wooden equality so what is a kind of uh good triangulation I mean an efficient triangulation of a surface is uh let's say reman's triangulation I don't know if it is really him but anyway so this is uh you know you you consider this uh these curves starting from one point so this is you first consider a point in the the surface and you consider um uh these curves um so this is for in the case of genius two and if you cut along this curve you get as you probably know this octagon okay right so to to to go back to the surface you have to identify this and this with uh negative differ morphism this and this and uh right you probably uh all know this uh this construction and so now uh you can get the triangulation of this surface by just triangulating this region and so here in this genius two example you get six triangle right so you get that the volume of representation in genius two is always bounded by pi time six okay but uh in general in if you have a genius g surface you will get a four g gone here and you will get if I'm not mistaken four g minus two triangles so this uh this bounds gives you um that the the norm of the volume is less than pi times four g minus two so this is not what we want because we want four g minus four right so the trick here is to do the the following construction to remark that um the volume of representation has a multiplicative property with respect to taking a covering of a surface a non-ramified covering of the surface so this is a kind of extremely classical trick uh but which will give the result here so what I claim is the following is that if you take if you take your surface and you take a d21 uh that I am going to call p non-ramified covering finite um above our surface s okay uh then you can consider the representation row prime from the fundamental group of s prime to p sl2 r which is just uh the composition uh of the map from the fundamental group of s prime to fundamental group of s and which goes then to p sl2 r via the representation our original representation right so you can uh you you can restrict in a sense this is an injection you can restrict your representation to um the fundamental group of uh covering of the finite covering of the surface and what I claim this is evident is that uh the volume of row prime is just degree of the covering namely d for the degree of the covering times the volume of row so uh I can apply uh this this bound for this um uh for for the for the restriction of the representation on the fundamental group of the covering namely volume of row prime is less than pi times 4g prime minus 2 where g prime is the genus of s prime right but if you consider the quantity 4g prime minus 4 this is this is a two times a large characteristic of s prime and this is multiplicative so this is d times 4g minus 4 so we get here that this is uh pi times d times 2 4g uh sorry sorry sorry sorry sorry yeah up minus 4 plus 2 right and so if I want uh is it okay this is okay I think right and so uh I am not interested in the volume of row prime I am interested in the volume of row so so this is just this divided by d by this formula so we divide everything by d so here's the d cancels we get this bound so and taking d to infinity we get uh the miller wood inequality that we wanted to to prove okay so so this is the proof of the miller wood inequality and um I have uh approximately 10 minutes to uh left uh to talk more precisely about uh geometrizations so maybe efficient geometrization uh of surfaces so let me uh give two examples of geometrizations which uh as we will see tomorrow will be very efficient for for certain problems of dominations uh so first first example uh is uh geometrization by folding in a sense we already know what uh so our geometrization was really made by folding so we had positive and negative triangles and uh between a positive and a negative one the developing map uh folds uh is really a folding so so in a sense we have already seen that you have to think that that's right I will try to to make a picture so so I'm going to try to to make a picture of the developing map so if you have in our surface uh two triangles one is positive the other one is negative what is doing the developing map it is just mapping the red triangle isometrically and positively into uh upper half plane and the the blue one it maps negatively so you see a kind of overlapping of the two triangle namely a folding uh appearing for the developing map okay so we already saw that but there are some foldings which are really much uh more uh interesting and efficient um which are foldings uh which only occurs along geodesics so what uh what happens here is that you see the folding I just uh draw uh occurs along a geodesic but what happens uh at a vertex it's much more complicated because you have successively positive and negative regions and you fold uh a certain number of times so here uh you have a complicated combinatorics appearing at the vertices so these kind of things we don't want in the geometrization I am going to describe and uh so I am going to introduce um some some geometrization which has been um developed uh by first on in the context of uh in fact geometry I mean representation of surface groups in um hyperbolic spaces in general which is the following so the definition is the following uh this is a developing so this is a pro-equivalent map D so again from universal cover of the surface to upper half plane um which is uh a isometric so which is continuous and is uh a isometric embeddings um in restriction to any uh exterior component of uh the lift of an invariant geodesic lamination so a p1 pi1 of s invariant geodesic lamination so this geodesic lamination is taken with respect to a certain hyperbolic structure on s and consequently on the universal cover of s so let let me give some some comments on this definition so what is so if you have a hyperbolic structure given on your surface uh so the so universal cover of the surface is just identified with a hyper half plane and a geodesic lamination is just a closed subset which is a disjoint union of geodesics or maybe something like this so there is an infinite number of geodesics maybe even uncontrollably many geodesics but this is a closed subset uh i'm going to denote it by maybe lambda tilde so this is uh uh the the the lamination and so what we will uh we'll ask is that uh in the complementary region of this lamination which is a certain domain with geodesic boundary uh we want that uh the map D is just an isometric embedding so it is a restriction of a map of PSL2R okay but the maps can uh so sorry uh an isometric embedding so I I say the mistake of course so we we we want it is isometric but we don't ask it is it preserves orientation it can uh it can fold of course it's called a folding so there are some regions which are mapped isometrically but with positive orientation other ones which are mapped with negative orientation exactly as uh we described here and uh in fact there there is uh so those are uh important geometrisations of surfaces and uh in fact there is a recent result by uh Gerito, Castel, François, Gerito, Fanny, Castel and uh Maxime Wolff uh building on works of maybe Gallo Capovic-Martin uh which shows that uh any representation apart from um representation uh can be geometrized I don't know what would be the good word in English I think you understand so by by uh folding so any representation it is not always true for instance identity representation will not work in fact any representation which is different uh to uh um an Abelian a virtually Abelian representation Abelian representation means a representation with Abelian image in PSL2R and virtually means up to a finite index above so this is a um a very nice theorem which is uh quite recent maybe 2013 maybe and it builds on works by uh Gallo Capovic-Martin where they proved uh a similar statement for representation with even a lack lesson uh so we prove the same statement for um um uh non-elementary representation with even a lack classes I mean with even volume with even v of o over 2 pi okay but but in the in the approach they really get something which is fold only along the finite number of simple closed geodesics um but you can prove it's a kind of free easy exercise to prove that uh you cannot uh geometrize uh a representation whose volume over 2 pi is odd by uh folding along simple closed geodesics it's not possible it's a kind of easy exercise uh that you can do by looking at the positive and negative regions on the surface and uh let me just finish by giving the example um maybe I will I will do that tomorrow because I will be out of time if I do so so tomorrow I will begin by another kind of geometrization which is much more in uh in the spirit of uh Taishmuller theory and uh in the spirit of what we have seen yesterday that will answer your question I love for yesterday so I will begin with this and and then I I'm going to speak about domination of hope connotations thank you


Today, I ask the angels to visit you in this greeting, a divine appointment of God's salams here at Maristans Healing Journey. Let's invite Allah's light into the brushing of our shoulders. Let us set our intentions right. Let's make the ummah's noor grow bolder. When I cast you a salam, wrap it in rahma and send it back.  A salam from a muslim? That's a gift from the heavens. And you never know, it could have been sent down from the seventh. A salamualaikum. May God's peace be upon you. May the noor of a noor shine down on you. May the angels twirl around you if you knew it would astound you, what is packaged in our peace. This is a greeting that sets the spirits free. And I don't know about you, but the world outside does not necessarily screen peace. And yet the barakah of this greeting can turn our usra into ease.  You would not believe me if I told you it had the power to ease a muslim's strife. Sometimes it's a salam in the streets to remind you that you are not alone. And sometimes it's a salam to someone in need to remind you your mouth is a hole. It is a sign of the end times when we pretend not to know one another. Like a family full of strangers, no sense of sister or brother. But this ummah is equipped with a vocabulary to carry each other's sorrows. To ensure God's blessings are all widespread, to ensure people spread peace tomorrow. So let's turn our mouths into gardens, orchards full of trees from which we give to our ummah all this. And a salamualaikum that you give me. Salamualaikum warahmatullahi wabarakatuh to the beautiful audience all over the United States all over the world. My name is Amal and my name is Hope in Arabic.


So today was a significant day in this offseason because it was the official press conference for the Liars luncheon. And that is of course the Baltimore Ravens pre-draft presser where they answer a lot of questions about the roster, about the draft. In this particular presser they did both of those and answer questions about Rashal Bateman and Adafi Away's fifth year option, Andrew Voorhees with the potential that he has to start on the offensive line. And they covered a lot of different stuff that we are going to get into in just a few seconds. Before we do, make sure you subscribe and turn your notifications on and run them likes up on this video. I appreciate y'all coming through as always, but without further ado, Eric DeCosta and John Harbaugh, they had a whole lot to say and I had to take my notes for this one. So, Eric DeCosta said yesterday, he said from 9 to 6, so 9 a.m. to 6 p.m. So all day, a full work day, he said they had been looking at offensive linemen. They had really been doing their homework on there. He said that by the end of this week, they'll know, well, they'll know who they're going to draft at pick 30. So again, this is the last luncheon. So a lot of this talk that they do at this press conference is business is business. There may be some truths in there, but a lot of it is just putting certain stuff out there, giving certain responses, because they're not going to come out and say, oh, yeah, well, this is what we're going to do in the draft. And this is our plan for the draft. No, nobody's going to come out and say that because you got to strategize this thing. So with this first part, when he said that by the end of this week, they'll know who they're going to draft at pick 30. To me, that let me know like, hey, and Eric, as y'all know, he's always open for business, especially when it comes to getting even more draft picks. That let me know that one, they could know who they're going to pick at 30. But at the same time, he's like, hey, if y'all come with the right offer, if y'all come with the right trade offer, hey, I'll be willing to trade down, but we'll see what happens anyway. Hey, action. What do you think you are at outside linebacker as far as veterans, as far as the young guys, because you got some first or some excuse me, some second and third year players. And he said, we were excited to bring back Kyle Vanoy. And he said, it's great to have a veteran in the room. And he said, the younger guys, they got potential. And he's excited for them to reach their potential this year. And that is a big thing that the Baltimore Ravens do have right now. He described it perfectly potential. We haven't quite seen the potential from all the young guys. Some of it is due to opportunity. Some of it is due to injury. And some of it is due to, just hasn't clicked all the way yet. Opportunity, that would be Tavius Robinson. Tavius Robinson, he's just behind so many people on the Dept chart last year. So it was understood why he didn't get many opportunities. Injury, obviously, David Ajabo, hopefully he can stay healthy this year because when he's healthy, he's produced. And just for it, just not clicking all the way yet has been adopted away. And we're going to talk about him in a little bit. But anyway, it's nine picks, a number that you want to be at. How much value do you put on the fifth year option? Now, EDC said he likes the idea of having more picks and says that there is an advantage of having that extra year when it comes to players with the fifth year option. Now, he was asked about picking up Bateman and the way he's fifth year options. Like, what's going down with those? Because the deadline's coming up because it is coming up in, I want to say either, I'll forget when it's coming up, but within the next couple of months. But Eric Dacasa, he said that we're about a month away from making that decision. So soon, soon. But he said, we'll talk more about that after the draft. Now, it's hard to really gauge where he's going at with that because we remember with Hollywood. Remember with Hollywood Brown? They were like, oh yeah, we anticipate picking up Hollywood Brown's fifth year. I will never forget. They said that we anticipate picking up Hollywood Brown's fifth year option. Yeah, we anticipate doing that. Then boom, traded on draft night. So I don't know. This is something I just, I really don't know. I mean, for a lot of this stuff, we just don't know. We'll have to see it for when it happens. But for this, I don't know what to think. I really don't know what direction these Baltimore Ravens are going to go in with either of these two players' fifth year options because it's just, it's super hard to tell. But continuing, EDC said that this is his 20th year running the draft. That's a long time. And then when he said this, he said his first year was the Mark Clayton draft. And I was like, oh man, a lot of us are old. We are old. Because I remember Mark Clayton. But anyway, he said that Joe Ortiz had been with him for 19 of those years. Because somebody asked him like, oh, how is it without Joe Ortiz in the building? Now he's, of course, the GM of the Chargers. He said that he doesn't see, Ericka said he doesn't see a challenge with picking early versus picking late in the draft. And somebody asked, are any offensive linemen currently on the roster ready to step in right now? Because that's been a question that a lot of us have been wondering.  But Hobbs answered this one. Hobbs stepped in. And he gave one of them hardball answers. He said, absolutely. The best player is the player who plays the best. Lele John Hobbs over here giving us one line is giving us them quotes. But he said those guys are going to be competing with whoever comes in here. Now back to his one line, the best player is the player who plays the best. What he was saying, because he broke that down, which I appreciated. He said it's not always about the best player, but it's about the one who's playing the best. I mean, it's pretty self-explanatory. But still, I did appreciate him going a bit of extra mouth to explain exactly what he was saying. So basically, you could have somebody who may be expected to be, it could be a veteran. And this veteran is expected to be a starter at whatever position it might be. But then you may have a rookie or second year guy, young guy come in, and they may be playing better than that veteran is playing. So John Hobbs basically saying like, oh, no. Then the second year guy, they're better. They're the better player than the guy who's expected to be the better player. But anyway, continuing. He said, they were asked, where is Andrew Voorhees at right now? As far as his status, like, what's going on with him? Eric DeCosta said that he sees potential for him to be a starter at guard. And he went through the whole story about how he drafted Andrew Voorhees. He said, I did it for y'all. Those of y'all that covered the team and stuff, he said, because y'all thought y'all were done for the day. Y'all thought that we had signed off. Y'all thought it was a wrap. But I was watching and I said, you know what? Let me give them some more work. So he jumped back in the draft and made a trade with Andrew Berry from the Browns. And he said, usually we don't do many indivision trades like that, but Andrew Berry was more than happy to do it. And I was happy to do it. And we made it happen. And they went and drafted Andrew Voorhees. And he talked about how had he not been hurt, then he would have been drafted a lot higher. And I know all of y'all more than aware of that already. So let's see. We'll see. Time is ticking. EDC also said the draft is the best way to build your team, especially from a financial standpoint. And he said last year they had tremendous depth. Yes, they certainly did. And he said, now the challenge is doing that again and a lot of the young guys really stepping up to help feel those depth roles. And then of course possibly some starters too. He said they love to add a talented corner at some point in the draft, whether it's in the first, second or third round. Now this question I didn't really appreciate because this has been a question that a lot of us as Ravens fans have wondered at different times throughout the tenure of the Baltimore Ravens, especially under Lamar Jackson. But really a lot of other times too, but certainly since Eric Dacosta, he's been the general manager since 2019. There have been a lot of times where we've questioned are the Baltimore Ravens really this? What is that? Well, the question was with you being one game away from the Super Bowl last year, are the Baltimore Ravens really all in as far as 2024? And that all in question, I love it because actions speak louder than words because Eric Dacosta said if you're not all in, then you're all out. Another one line, a shout out to him. He said we're all in every year and that's what we all expect, coaches, staff, the players, the fans. He said that's what we expect. And there have been times under this current regime, under his regime as GM where it hasn't really felt like they were all in. Last year, it certainly felt like they were all in. 2021, that felt like they were all in too but everybody got hurt, so they were all out. But this year, and I've said it before, feel like Eric Dacosta, there's no room for error. There's no room for mistakes when it comes to building the roster this year because you have some significant holes on the roster. We'll see how he patches those holes up, but it's gonna be fun. Eric Dacosta, he talked about this too because a lot of regular fans have been frustrated about how the regime's been moving this off season. They said it's very, very slow, like what's going on. Eric Dacosta said that he said they're in the same place right now, roster-wise, that they were around this time last year. And he said a lot of the guys on the roster, were guys that from last year, a lot of the guys ended up getting signed in August. So he said they ain't tripping right now, basically. And he said the destination is September, not May. And that's a good point. I gotta do a little bit of fact checking to see if that's accurate. Because obviously he did get Jadaven Clowney late in the off season, got Ronald Darby late in the off season, got Kyle Vanoy after the season had already started. They were a couple of weeks into the season. So yeah, I mean, for enough guys, they did get signed later on. So yeah, basically just, that's a message to everybody from Eric Dacosta saying, chill out, relax. Even though I know us as Ravens fans, we don't chill out and we don't relax at all. But we did get the message. Now, he was also asked, where does the patience and that philosophy come from? Like, how you so patient as far as waiting till later on to sign different guys and whatnot? How did you get so patient? And Eric Dacosta said, I'm not a patient person at all. But he did say he learned it from Ozzie Newsom who was very patient. And he said that the Baltimore Ravens, I gotta learn to pounce when an opportunity presents itself for the franchise to get better. And it's funny, somebody also asked a question about, they were like, oh, well, last year, you made the announcement of Lamar Jackson's contract on draft day. But does this off season and going into the draft, does it feel like a lot lighter for you as far as like the status of everything right now? And Harbaugh was like, hey, well, we got two weeks. It could still make another big announcement. Something could happen. So I said, I'll see you Harbaugh. Don't tease us like that. Don't troll us like that. Even though Eric Dacosta, hey, he pretty much said the same thing because I think it was Jeff's rebuke. He asked, do you anticipate any moves being made over the next couple of weeks that could impact your prominent needs heading into the draft? And ADC said, you just never know. If the phone rings, maybe there's a move to be made. But he said that Kyle Vernoi was one of the last things on his checklist trying to get that done. I don't know that we'll do a lot over the next couple of weeks besides the draft board. Sliest luncheon. So, hey, there's gonna be lies in here. There's gonna be some truths in here and it's up to you to decide what you believe. But based on that, I think they ain't done yet. Shout out to Marcus Peters. But I can see them making, I'm gonna say, draft is in a couple of weeks. So we'll just like the 28th, something like that. I'm gonna go, ooh. I'm gonna go they make at least two more moves. You know what? I'm going big. Well, I mean, for this office, he's been really slow. So this really wouldn't really be big. But I'm gonna go, they make three more moves. They make at least three more moves, three more transactions, whether it's two signings in a trade, three signings. I'd say they making three more moves. Ooh, that sounds crazy saying that out loud. But y'all know I'm crazy. So that's what I'm rolling with. Three more moves. You heard it first here. It's there, it's documented in this video and it ain't going nowhere. Now, the last question. Well, last question that I took notes on. This was a good question too. And I think it came from my guy, Cordell Woodland. Shout out to him. He said, is the plan to develop Malik Cunningham as a quarterback or like, what's the plan with him? And Harb said that that remains to be seen because he's developmental as both a quarterback and as a wide receiver. That's interesting. I know that he's very raw right now, but those are two very different positions. Those are two completely different positions. And for me, I don't know what Malik Cunningham wants as a player. They obviously brought Josh Johnson back to be the backup quarterback and Malik Cunningham is there right now. There's some players that just, hey, whatever I got to do to get on the field, I'm willing to do it. There are other players that, I'm sure every player has that mindset, but then at the same time there could be a player that, hey, if you want to be a quarterback, if I want to be a quarterback, I'm going to be a quarterback. I can't play in all the positions. So I don't know what Malik Cunningham is or what he wants to do. So I can't speak for him, but I would hope that if he wants to be a quarterback, if he wants to continue to play the quarterback position and develop there, I would hope that the Baltimore Ravens would go that route. So, because when you start flip-flopping, then it just, it goes downhill quick. And I mean, recently the more, the quarterbacks that we've seen go from quarterback to receiver, the guy who was on the Texans. And I think he went to Ohio State. I cannot remember his name. I know that, I know y'all going to come through. I know y'all going to come through. He played quarterback at Ohio State, I believe, yeah. And he went to the Texans, he played quarterback for them. There was also, I can't not remember, none of these guys' names right now. There was also a quarterback on the Raiders. He started off as a quarterback, and then he switched to receiver. I cannot remember his name right now either. It's frustrating, but I know y'all got me. And I love y'all for that. There was also a quarterback, this wasn't recently anything, really none of these guys are recently, but there was also a quarterback on the Baltimore, Ravens. Kenan Reynolds from Navy, I believe. I think he went to Navy, but he was a quarterback in college, then they tried to transfer into wide receiver and it just, it didn't work out. So, with Malik Cunningham, I just hope that he gets to play whatever position that he wants to play. But yeah, this Lies Luncheon had some significant updates that we got, and we got to get them directly from the Baltimore Ravens, and it's always nice when we can hear directly from them. But what did you take out of this Lies Luncheon? What information did you gather from it? What do you think is the truth? What do you think is a lie? And what do you expect? Let me know in the comments section. Thank you, be clean.