 Gabriela Mondello from the University of Roma sapienza, and he will be talking about spherical matrix with chronic singularities. Thank you and thank you for the invitation. I will use mostly the blackboards and just a little bit the projectors for some pictures. Vstaj sem pričel, da bo napet je počet na sferigulnih metričnih na vsefazici, in v koniglu singularitih. Prvno, izgledaj, da splatijo, da je veliko vzutim, z Dmitri Panov, v KWL. Všeč. Prezaj mi odlično ta motivacija. Na neko, da je zelo vse objevaj. Vse je, da je, da je svoj zelo vse, kam je, da je, ki je. Zelo, da, da je, da je, da je, da, konektin, zelo vse, ili vzlemati stvari s in vzlemati, tako tako vsih udovodnih relastih. then, for every conformal class, there is exactly one remarkable metric, which is one metric with constant curvature. and because of the Gauss-Baun-Neckl strain, this must be a negative curvature. So in every conformal class, there exists a unique metric age tako, da h je minus 1 v kvavji. The result is that one basically obtains a correspondence between conformal structures, say complex, so complex structures on s and hyperbolic matrix on s Tudi, kaj se je, koncentruja, in s težem, da ni so vzelo odrečenja isotopia in to, ki smo je, na kratku, je to, Taikmula spes, oveč. So, zelo je, da na spesu konformačnja struča potrebeno vse spesial metriče, kaj je kompletivno kanoniko. Tako, zelo smo všeč všeč všeč, kaj je zelo. Vseč, da je vseč vseč in genus 1. Zato, da je toro, z vseč metriče, zato vseč vseč in genus 1. in da se preddajte mačne metri. Anatom 250 metri, in nama je spet v cagesenje, a medija je občas, da tudi počusti, da počusti za metri vse in v怎麼辦i za ise ozvore, da je čil한테, ali na nekaj kompak 232. Kompak 232 je, abo kaj za nimi ne tukaj. in ne zelo nekaj interes. In zelo s večim kontekstem, še je zelo zelo zelo zelo, zelo zelo v počke, nekaj počke, zelo v počke, zelo zelo, zelo v počke, zelo v počke, zelo v počke, zelo v počke, zelo v počke, zelo v počke, zelo v počke, zelo v počke, zelo v počke, zelo v počke, zelo v počke, zelo v počke, zelo v počke, zelo v počke, zelo v počke, zelo v počke, zelo v počke, zelo v počke, zelo v počke, večOFFČ which are totally geodesic, В Oracle, in also in the hyperbolic setting you can ask this as a kasp, but you can also ask that this looks like a cone point. So we can assign a number at each conical point, and we can want to look for metric of constant curvature in konegal singularitaj na konegal 2 pi theta i. Poživam tvoj 2 pi za pratega raznega. Na konegal singularitaj, zelo smo imeli, kaj je. Sveško v tle pivih kaj je. Zelo, da je bilo, na pripivku, zelo, da počekajte, to je, nekako je zelo zelo, počekajte, kaj je, zelo, Teta, i, square, to je tukaj, kaj je tukaj. In v hyperbolici bomo vsega vsega vsega vsega in v sperega vsega vsega vsega. Veseli smo zelo površniti tudi različke vse. Zato, kako se počeš? Zato, če je zelo počeš? To je početno, da je zelo početno v 99,91. Mikovan, Mikovan, Interjanov, design singularities tudi, da je to nekaj postavljenje, da je tudi genus, tudi je to nekaj postavljenje na karakteristiku in vzvečenje z njegov. Zato, kaj ste zelo, 1 theta n tako, da je 2 minus 2g plus sum of theta i minus 1 less than 0. In vse konformačne klasi, da je to samo, da je konformačne klasi, da je vse unike metriče, da je k equal to minus 1, in da je tukaj, da je tukaj, da je tukaj, da je tukaj, da je tukaj, da je tukaj, da je tukaj, da je tukaj. Čest, ta izmene isto ima tukaj, da je to neprostno, da je to neprostno, da je to neprostno. Čest, ta izmene igra to neprostno, da je 2 minus 2g plus sum of theta i k minus 1 neprostno. Čest, je to neprostno? So one can then wonder what happens in the case k greater than one, sorry, greater than zero. So what about k equal 1? Okay, so let me call this quantity, just for praks, reason delta j, let me call it beta j minus 1, it's called the fact, so in fact it's zero, exactly when the point is smooth, above the angle is exactly to pi, and then it could be negative or positive. This basically represents a delta at the curvature minus delta j, so if delta is positive there is some point of negative of negative curvature concentrated there. ... in pa, da bilo načas, da se sem spremno pravno... ... ja bilo pa včas, da najteželji možno vz㈆u pošljanje iz nekor spasternja... ... in tudi zelo si, ki če prejazeljamo... ... da srečnem, da bote veliko pošljenja pred pانom... ... na stran Setu. ... da spremnejo izpadiček... ... zvrb na Vincentav. Zazam, da je tudi tref eti tudi. za moment, da bomo izgledati. Prvoj set je sp. teta, kaj je set metrič h na s s dotom s kurvajšem 1 in teta 2 pi teta j at pj to je vzljajte teta vzljajte teta teta teta teta s kaj je teta kaj je teta vzljajte s modem teta s modem teta teta h vzljajte h vzljajte teta teta teta teta vzljajte teta vzljajte teta teta teta vzljajte teta kaj je teta dva teta vzljajte metrički metrički Treba je, da je bilo gaustbo nekos trajn, nekaj to bila nekaj poživljeni, zato se nekaj, da je nekaj poživljeni, da je vse sega metrička. Zato je tudi početno, kako je to konjunkt, ovo je homotopitaj, da je druga, klasička problema je vsega, je, da je to smutno, tako, če je dimensja, v tem spasih, in, da je dobro, nekaj simpletik, in nekaj volum form, nekaj volum, nekaj volum v tem spasih, tukaj je za Teta, nekaj, potem, izgledajte, v tem, da se vse zelo, nekaj, potem, da se, da je, smutno, in nekaj, nekaj, nekaj, nekaj, je, nekaj, nekaj, nekaj, nekaj nekaj, nekaj, nekaj v ešli formučnosti, tato je razču, boste, kaj je, čudov, spasih, taj je, vsezodajte da je,spasih, kaj je, pravda, je at least one metric in each conform a class. Asking that F is finite, or with finite fibers, actually, so that there are finitely many metrics in each class, or one could also wonder if F is proper, or if it branches and where it branches. And OK, so we are close to the end. Is there a complex structure on this S? This is, I mean, in the hyperbolic case, this is one can use the bijection with a cumulus space and say, OK, yes, there is a complex structure. Here it is. But here it's not clear exactly because it's not a bijection. And the last one is how the generations of spherical metrics look like, which is somehow, in some senses, studying what is a good boundary for this space. OK, let me see. I would like to just discuss some of these problems. Like this one, which seems the easiest, and then also smoothness and dimension of the presentation variety and of the space of spherical metrics and also the fact that when is the monodromy a local differ. And here we cannot treat most of the questions except for this. And the upshot is that in the end we will discover that these two spaces are both except somewhere. I will discuss are both smooth of the same dimension and this map is proper. And actually one can also orient the space so that indeed there is a degree. So in each connected component, once we chop off some walls, there is a degree and this is the number of expected metrics in each conformal class, which says that counted with the suitable multiplicity on the general fiber there is a fixed number of metrics no matter what conformal class we take and no matter how we move the angles, provided the angles remain in a fixed chamber. The space of angles is chopped off by some hyper planes and then we have some kind of number in each class. Probably there is also some kind of one can study what happens when you cross the walls, but we haven't done this. OK, so let me see something about the first problem, which seems not particularly interesting, but actually most of the solution arrived a few years ago. By the way, so let me first let's use this projector and show what we know in the case of lower genus. So there are some cases which seems very simple, but still case of genus zero with n equal one point. In this case there is not much to say, this is just the round sphere and the angle must be exactly to pi. Already when we have n equal to two, there are two examples, and we show now in the picture, this was worked out by Toriyanov and for n equal to three, this is even less elementary and was classified by Jeremenko and in this case there is exactly one metric in each conformal class. So now I will show you how they look like. They are not very difficult to understand. OK, so the case is in genus two, sorry, with two points, in genus zero, there are these cases, so this is just a portion between two meridians and then you glue the two meridians so you have surface with two conical points, same angle, there are pi apart one from the other and this you can have for any angle. Or there is this special case, so you have two angles which have angle multiple of two pi and the distance between the two points can be arbitrary. So basically you produce this just taking the sphere, taking any two points and then taking a cyclic cover of order, or integral order, or order k, there is a double cover. So here you have angle four pi at these two points. So these are the only two cases in genus zero with two conical points. In genus zero with three conical points, this is equivalent to classifying triangles and actually classifying triangles, well, one can think that you have basically maybe this and maybe this, but there are many more other triangles that you cannot embed in the sphere. So there is, this is quadrilaterals, so there is indeed a constrain, sorry, so for triangles or which is equivalent to three punctured spheres, there is already some constrain, we will see now. So I said most of the result of nonemptiness came in 2011 by preanalysts, Bartolucci, De Marquis and Malkjordi, who showed that in genus at least one, of course we take angles that theta, that satisfy the Gauss-Bonnet constrain and then they showed that in each conformal class there exists at least one metric age with the assigned angles to pi theta, but indeed they gave they gave exactly an estimate from below on this number, so they showed that this n theta with this n let me call it m, this m is a function of g and theta and actually it increases as theta increases. So actually they gave examples where this goes really, we have an estimate from below on the number of metric, but this leaves out the case g equal to zero. I must say also that both them and Trojanov also showed that if the angles are all less than one, then there exists a unique metric always, so for small angles we always know and also them is, also their work implies this result. So what we did with Panoff was trying to fix the case of genus zero and see what are the possible constraints to the existence of a spherical metric in genus zero. This is last year. Here, no, no, no. Here is energy. Well, that's, well, if theta is smaller, yeah, it must be g equal to zero in this case. So it's really, it is really, well, what it does is a bit more than this. So in the differential equation you can allow some angles also to be a bit bigger than two pi and most of them to be smaller. So there is a critical constant that tells you that everything works when the angles are not too big, but certainly this works if the angles are all small than two pi. This happens in this sphere case. So this is some kind of case, sort of trivial, when it's already done. But the problem is when you are in genus zero, but the angles are very big. This is what you don't know. Yeah. You must have a, well, yeah, you must have at least three because the other cases are already done. Okay, so the, what are the constraints in this case? So the constraints, certainly you must have, okay, these angles are zero, then there is a Gauss-Bonnet constraint which is two minus two g plus sum of delta. Let's express in terms of delta. This must be positive. This Gauss-Bonnet. And then this is not enough. There is a third condition which is the following, is the distance. So let's take rn with the l1 distance. And inside rn, let's take this set of points, integral points with odd sum. So this is points m in zn such that m1 plus mn is odd. So the third condition can be phrased in this weird way. So the l1 distance from the vector of the fact delta to z to dn, odd must be at least one. So let me state more precisely what are the implications. So this is a necessary condition. We will see also why. And we proved also almost the converse. So what we proved is that if s of sp theta is non-zero, non-empty, then this condition must hold. We are in general zero. And the converse holds if we have strict inequality. So here if we have the distance delta, zden zero is strictly greater than zero. And of course all the other conditions. So there is slightly tiny bits of uncertainty when this is exactly zero. And we will see now when I will motivate the proof why this is slightly uncertainty. So where does this come from? So let me show also what this looks like. So the domain of angles. So let's take A, the possible deltas. Let me call this condition A, B, and C. This, sorry, let me call it C and B. And we put C only in genus zero. Because there is no constraint in the heterogeneous. So this is the set of the facts that satisfy A, B and possibly C if we are in genus zero. This is somehow the space of admissible angles. And the picture of this strange set is OK, so these are quadrilaterals. We don't care. OK, so it's something like this. So I took A and then I chopped just a unit cube. Here these are the integral points of Zn. Of course, OK, this picture is just schematic. So I drew something three-dimensional, but actually in three-dimensions it doesn't look this way. In three-dimensions is just a tetradron. So this is just a symbolic picture of what happens in higher dimension. So in higher dimension this is, so the chopped domain is this light brown. Then this dark brown, darker brown, are the boundary of this domain imposed by this condition. So you see this condition imposes that this delta is far from odd integral points and this dark, this dark black, these are the odd integral points. So it must be far from this. This wall that is inside we will see later what it is. OK, so why, so let me just give the idea of why this, of why say this implication holds. So here I have to motivate just this. So the fact is that if we have, so the obstruction lies in the monodromy, in the sense that the problem is not exactly creating H, but if we have an H then the monodrom in order to exist must satisfy like this, why? Because if we have H then we have seen that we produce a representation inside SO3. Now we can lift this representation, this is also classical in the closed surfaces that we can lift this representation to SU2. So of course here one can say, OK, that's not very surprising because that's a free group here because we are punctures. So you can always lift the representation to SU2. But indeed we want to lift the representation that behaves in a prescribed way. So the proposition is the following. So that if we have H, varical matrix with angles eta, which is the vector of angles, then the rho H lifts to some rho H hat in SU2 such that at gamma j is the loop that winds simply around the point pj. So this has eigenvalues, the exponential plus or minus i pi delta j. So why this? So you see that when the point is smooth delta is zero, so this must lift the identity. When the angle is 4 pi, you want it to lift to minus the identity. When it's 6 pi, you want it to lift to the identity and so on. So it's this kind of standard lift that you want basically that you want that the 2 pi is lift to the identity and then that if you start increase in the angle, then the lift varies continuously. Now here just one, but there are 2 to the 2G lifts, like in the close case. But if we are in genus zero actually this means that it's canonical. So lifts turn in 2 to the 2G ways up to conjugation. So this means that in the case of genus zero it's canonical. And now the obstruction is in the existence of this lift. One can understand the obstruction very geometrically. So if we have a sphere, we have the base point, we have pj and we have this which is gammaj. And let's call cj, which is a wrong hat of gammaj, that's an element of SU2. And because we know that gamma1 gammaM is the identity in the fundamental group, then we must have that c1, cn is the identity in SU2. Now this is the equation that gives us troubles, because suppose that so identify SU2 with a sphere just placing a base point and sending just g to g times v0. OK, so now using these elements we can we can describe on this S3 a broken polygon so we have v0 then we apply so we want to define v1 just as c1 inverse v0 so this is c1 inverse and then taking vk to be ck inverse vk-1 then in the end vn must be cn inverse c1 inverse v0 and so it must be v0 so this polygon that lies on the sphere must close and you can compute very easily what are the lengths of these sides and the lengths of these sides because we know exactly how we lifted so the length of the side from v k-1 to vk if this is delta j and this is lj the length on the sphere is something like this so here this is 1 this is 2 this is 0, this is pi actually this is the distance between the two points so basically you see that this function is delta j this function is 2-delta j this function is delta j-2 this is 4-delta j OK, so now in this it is exercise you see that you must have exactly where is this so I am claiming that this is just an exercise starting from this and just show you an example so why should this be true example suppose that we are in case of ok we are in case of 2 points and suppose that we take at 3 angles which are pi sorry let's take 3pi 3pi and 0 OK, 4pi, 4pi, 4pi this corresponds to delta which is 1, 1, 1 and so actually from z and o from delta to 0 and in fact if you go and try to draw this this diagram then every angle 4pi corresponds to multiplication by minus the identity so if you multiply 3 times minus the identity you don't get the identity the picture on the sphere is you start at v0 then you are sent to minus v0 this v1 then you are sent again to v0 and this is v2 and then v3 you are on the other side, this doesn't close up and of course you can see that if you perturb a little bit this then it's still difficult but you are back to v0 you need to perturb a lot to be able to be back to v0 you have to perturb at least by 1 in total so that's why you need that this distance is at least 1 OK, this is just an easier argument so let me discuss just 2 other points so the monodroming map is to to say what structure what wants to give the space so because we said that attached to every metric we can attach representation and a developing map so basically we have attached what is called an S2 SO3 structure which is somehow a particular case of a Cp1 structure which is a kind of Cp1 structure where the monodroming is in SO3 or actually in a conjugate of SO3 so basically you can define this as a subset of the space of Cp1 structures we take this space of Cp1 OK, so let's take another thing, so let's take first Cp1 structures on S dot and then we want to impose something like the analogous of the angles at the punctures so that the Schwarzson derivative is has residue Schwarzson derivative is quadratic differential quadratic residue at Pi which is minus T i square half at Pi with respect to the Poincare projective structure so we put on the surface the hyperbolic, the Poincare hyperbolic metric with casps that defines a projective structure we respect to that we take just the projective structure that have this quadratic residue as asking a conical behavior of angle T i at Pi so this is what we call projective structures with moderate singularities and then so this is the boundary behavior then we have to ask that the monodromy is in a conjugate of SO3 so what we do we look at monodromy map so here this is a monodromy map the same as in the case of spherical surfaces but here is Cp1 structure so it has a monodromy map inside PSL2C and here we can map, we can find representations into conjugates of SO3 actually this SO3 is like PSU2 which is inside this and we take the preimage so projective structure that have monodromy is here so the definition of this is monodromy is in PSU2 intersected well in p-mod intersected p-mod SP so that's one way to define this space and so the result in 93 by Luo was that this space is smooth so in 93 just analyzing the deformation theory of this space Luo showed that S of SP theta is smooth of dimension real dimension because it's a real up to now real manifold dimension S6g minus 6 plus 2n if the angles are not integral and so there is this this constraint and also if h sorry, if rho h is not rho h is not a billion so if we take a point which corresponds to a metric then this is smooth at h near h is a manifold if this is not integral and this is not a billion this is just a little deformation theory and the same holds for rep same statements when you have representations in SO3 now this is not true that this is when the angles theta these are rho such that the boundary is exactly the one of a rotation angle theta which means that the trace of rho gamma j must be twice the cosine of theta j pi so the this is not true in general that this is smooth when these angles are not integral there are two kind of problems so this problem is just because if this is a billion then there are a lot say actually coaxial so if it lies in a subgroup one parameter subgroup then when you take conjugation you make a mess so that you want to exclude this case this really is kind of technical problem it depends on the fact that this representation variety in some sense why as a problem because you will think that this is almost very close to this when you deform a little the representation you can follow with the developing map which is in fact what happens in general but when the representation has an angle which is integral the monodromi at the point does not see the axis of rotation so a rotation of any angle determines an axis on the sphere except when the rotation is multiple of 2 pi in that case it doesn't determine an axis it doesn't tell you well the axis the marked points of the conical point is moving on the sphere so this means that here we have more information than here and in fact what we do is constructing suitable resolution of this representation variety so that this is restored so basically how some kind of what you call a refined representation variety and a map hall at from the space of spherical matrix to the space of representation so that this is local differ at which monodromi so the resolution is basically very easy you just record at the loop which is the boundary loop you record the axis of rotation even if the rotation is of angle 2 pi or 4 pi so basically if you want to construct this wrap you first want to see what are the refined homomorphisms so what is a homomorphism from p1 of s to some group this is really an ennub of elements a1 b1 ag bg c1 cn in so3 such that we may satisfy the relation we know which is the product of ai bi times c1 cn equal to the identity now if you want to define a refinement of this you take a1 bg and then we put capital c1 cn and this is in the group times the algebra and you want that a1 the product of the commutator times the exponential of c1 exponential of cn is equal to the identity and that the norm of cj is exactly what in this case so basically g is the group so3 and g is the algebra so you see that in this case taking the exponential of this capital c you get the element of the group but when you have an angle which is these are different from 0 clearly when you have an angle which is 2 pi or 4 pi and so on here this element is non-zero whereas here this is 0 so then you take quotient by so3 and then you get the resolution you want so why I said this just one word how do you get the enumeration of the fact that the matrix are the degree is fixed in a certain chamber so we have a map from ssp so all the matrix with all titas, possible titas to taikimula space times a that space of angles now if you have a metric monodromi this means that the sum of so h a billion implies that you are all monodromis are in the same subgroup so this means that sum so that there exists a subset of n such that if you sum delta j and you subtract the ones not in this subset then you get something even so this is exactly the condition that the broken geodesic that really now stays on the maximal circle of SU2 closes up except the why you have plus and minus because if you go forward with some side then you get plus delta j if you come back stay on the same circle then you get minus delta i so these are exactly these determine some walls inside this space so if you restrict this map something to the projection away from the walls the proposition is that this map is proper is proper and not subversive sorry proper and the two spaces are as smooth of the same dimension so we know that this is the angle this has the same dimension as this so basically we know that this if smooth this has the same dimension as this and then the theorem I said before I said that if improving Luo we have got rid of this condition we have just proven taking this resolution that if the representation is not a billion then this space is smooth so basically this space because we have removed the walls when it can be a billion this is smooth and so the only thing that we have to prove is that this map is proper and really that there is an orientation here canonical orientation so this is you can use some simplatic structure to orient this to see that this is and then I stop to see that this is proper you have to show that if you have a sequence of metrics that diverge here then the sequence here diverges too now if it diverges here and it converges on t you have to show that it diverges in the space of angles so why this happens so I'll show you just a picture ok so the the picture is this so suppose that you have a sequence of metrics which are diverging so if the metrics are diverging this means that because the diameter is bounded the curvature is blocked that some conical points are crashing together so this means that there are some conical points that are getting closer and closer now because we are assuming so there is this cluster inside this box which is getting smaller and smaller now because we are assuming that to prove this that here the conformal structure is converging and we want to show that we hit a wall on this big things there is not much that can happen because if you have some topology some genus or you have more than one point and you are crashing here then really you have a cylinder that separates this from this and then the conformal structure becomes the conformal structure of a nodal curve so this means that on this big things you can have at most one point and no topology so basically here you can have just a sphere with some integral conical point or here you can have this kind of beams now if you look when this situation degenerates you look at what is happening here in this box then you can magnify a lot and you get something that is becoming flat this is becoming flat and with here you see something that is some kind of funnel that opens up and if you do the computation here for the using Gauss Bonnet on this flat surface what you get is exactly some relation like this where these positive delta Js are the points which stay in and the negative ones are the points which are away so basically you are hitting you are going to hit a wall so this proves it's proper thank you sorry for running over time