 Well, I'm really pleased to be invited here and I think all the organizers and here in Palo and anyway, great crowd and thank you very much for the opportunity. So I'd like to talk about a set of topics that I find very appealing and I'll spend a certain amount of time today motivating what I want to do and I'm not quite sure how far I'll get to by the end of the week, there's sort of an infinite amount to say, but I hope to give you a bit of a flavor of several different approaches. So I'll describe the problem in some detail as we go along, but what I find appealing about it is that many different parts of mathematics come in. So I mean there's really input from analysis and geometry and also geometric analysis, synthetic geometry, algebraic geometry, low dimensional topology. I mean it really has a lot of lovely connections. So I'll try to impart some of that but I'll sort of obviously have my own bias perspectives and so today I'm going to give, I think it'll be some background and I'll try to talk about the more classical parts of the subject and so it'll be, well, we'll see, a rather different flavor than what I'll be talking about later in the week but I think it's important to cover various different grounds on this and I'll try not to lie too much about how I describe things but of course if I lie too much, the sun's going to go out. Okay, so let's see. So I want to say a little bit what I mean by a metric with the chronic singularity so fortunately Alex gave me the perfect introduction and so you have a bit of intuition about these but I'll describe what I mean a little bit more specifically and as he mentioned I'll be interested in surfaces with more general chronic singularities. So okay now I got to get used to this chalk which I was told there's some pricks too. So let's see, so hold on. No, Scott you're wrong. I like corners but that's still not working. No, here we go. Okay, sorry. Okay, so chronic singularities. So one can talk about these in sort of generality in higher dimensions and in surfaces and I'll define what I mean by a chronic metric initially and then I'll specialize to surfaces. So chronic metric. So the model cone is just a warp product so it looks like g which looks like dr squared plus r squared times h where I have a cross-section zh and this is a compact let's say compact smooth manifold and what this looks like is just exactly this. Here's a copy of z and it's whatever Riemannian manifold you wish and you just sort of cone it off so as r gets large this is going out and as r gets small this is coming down to a conic tip. Okay, so nothing too mysterious about this. More generally we'll talk about, I'll just refer to them as conic metrics, but asymptotically conic metrics it'll be some sort of perturbations of these. So there'll be some sort of metric perturbation and I might think of this plus higher order term so if you like just adding on extra tensors that decay as r goes to zero. Okay, so that's sort of the general class of things. Now when we talk about surfaces there's only one choice so of course if this is supposed to be n dimensional then z is n minus one dimensional and if the ambient space is two dimensional the cross-section is called the link is a circle and the only invariant of it is well you know all circles are isometric once you fix the length so there's exactly one real parameter which is the length of that circle which is the cone angle. Okay, so in one dimension in two dimensions for surfaces I can write these in a special way I can write them as dr squared plus and maybe I'll write this as alpha squared r squared d theta squared where here alpha is any number greater than zero. Okay, so theta is on the regular circle of circumference 2 pi and alpha is greater than zero and so the cone angle here is 2 pi times alpha so I'm allowing cone angles of any any positive number. Okay, so this picture looks exactly right so a cone angle that's strictly less than 2 pi I've made the circle a little bit small and when I take the cone over it it looks sort of like a decreased version of Euclidean space when the cone angle is exactly 2 pi well of course Euclidean space itself is a conic space it just the cone singularity is kind of invisible and it's doesn't really play real it's arbitrary and then I can let the cone angle be greater than 2 pi and then you can't visualize it anymore you can think of it as something which is wrapping around a bit but in general I'm interested in in real cone angles and so Alex talked about some of the milestones where we have cone angles maybe pi 2 pi 4 pi and so on those play very interesting roles they actually one has a little bit more rigidity there but those are just special cases of this general construction okay so that's okay so now here's my general problem so given let's say M and I'll maybe put a conformal class on it see this is a compact Riemann surface okay I want to choose a divisor I'll just be a disjoint collection of points p1 through pk be a disjoint collection of points set of points on M and I'm going to choose parameters alpha 1 alpha k and then what I want to do is I'd like to find a metric G so this is in the conformal class C such that G has a conical seniority at pj with angle 2 pi alpha j j goes from 1 to k okay so in other words whatever the surface I'm always looking at compact surfaces by the way I have a collection of points and I want to find somehow metrics that have own angles well if I just specify this problem that's not very interesting or hard namely it's easy to find metrics in a given conformal class and to modify them a bit to have a certain cone angle is not very interesting either not very hard so I want to say G has constant curvature so in other words I can think of this as the uniformization problem in the category of of these types of cone metrics okay so that's a problem it's been around for 150 years I mean in some sense people thought about this for a long long time I'll talk about what's known and and so for certain cases we're going to see that there's a radical difference when the cone angles are less than 2 pi or greater than 2 pi things behave very differently geometrically analytically and so on the story is understood and a little bit surprising in some cases when the cone angles are all less than 2 pi when the cone angles are greater than 2 pi then there's well like in all of these problems you have to specify are you looking at negative zero or positive curvature negative in zero curvature is in fact easy there's still some interesting aspects but the existence problem is pretty easy and I'll give you the full solution today the spherical case is hard and in some sense the answer is still not quite known in fact it's rather far from known surprisingly so I'll be describing that this week you know what's known and what are the various approaches that people have used to study this but anyway there's the problem okay now let me just say a little bit about motivations so why spend so much energy well it's you know there's an infinite amount you can say about surfaces and this is you know yet another whole class of problems I want to motivate it a couple of different ways so first of all motivations so one is is that as I already mentioned the words there's sort of a general class of you know problems of finding canonical metrics and I want to think that this is a problem of finding canonical metrics on singular spaces okay so a canonical metric I mean that's a little bit of a vague term but you know we're interested in constant curvature metrics or you know Einstein metrics or something like this and you can look in every dimension in every category of space you know what's the most reasonable natural metric that you expect to just find a few of them or you know a finite dimensional family of them now you know why look on singular spaces and you know I think the argument my kind of message is that singular spaces are pretty natural they come up in some sense just as often if not more than smooth manifolds and so even if you're just interested in smooth manifolds you often encounter singular spaces as sort of and when I say singular I'll I'll specify what sort of singularities I'm not looking at sort of candor sets or you know wild singularities I'm looking at spaces have stratified sort of well regulated singularities so you know these come up just as often as smooth manifolds and for example if you take a you know a more stratification take a Morse function you look at its various levels most of its levels are smooth manifolds but every once in a while you have to cross a singular level and it has a nice stratification and understanding that stratification is you know classical well-known thing but you get into a new category of spaces so you know there are many similar constructions in topology where you just sort of naturally led to singular spaces algebraic geometry obviously algebraic varieties compactifications of a locally symmetric spaces you know I can give you many many natural examples of singular spaces and so part of the goal is to you know develop tools of geometric analysis that work on singular spaces just as you know we have a lot of familiar tools that work on smooth compact manifolds okay so that's one sort of thing now this is kind of the lowest dimensional version of a of this problem so singular spaces I mean well there are actually some interesting singular one-dimensional objects namely networks of curves and there's some nice problems there but this is sort of you know in terms of bringing in a surface geometry this is the first interesting case now this particular problem actually has two interesting extensions you know let's say it's part of sort of several interesting problems so one is is it's the lowest dimensional example of a problem a well known problem in a complex or algebraic geometry of a problem in complex algebraic geometry which is a finding Kaler-Einstein metrics and so many of you may know this has been a subject of you know great interest in and work over the past many decades and there was a big success not long ago sort of finally completing the last case and it's you know unlike some other problems you cannot always find a solution and we're gonna see that here too so there's gonna be a certain and there was emerged as a called stability condition algebraic geometry case stability and it turned out that you can only find a solution of that problem when you're a case stable so this is actually the very lowest dimensional case where you see that phenomenon and from that point of view it's interesting there's a complicated obstruction okay so here's another example that goes back to certainly the 70s with Thurston so a class of what Thurston called a cone manifold and so I know it's kind of bad terminology because he wasn't thinking of just isolated conic singularities but he was thinking of constant curvature spaces which he built synthetically sort of this GX type construction and you have spaces that have stratified singularities so he was thinking obviously of the three-dimensional case primarily but one can define this in higher dimensions where you know you might have a hyperbolic structure away from some singular locus and there's some sort of branching or some sort of conic behavior as you go around these co-dimension to singularities and you might have higher co-dimension singularities okay and it turned out that these were a rather important tool in sort of his approach to sort of geometrization and putting hyperbolic metrics on certain manifolds and so he had an approach which again was synthetic but these have emerged as sort of interesting tools and then slightly more generally you can just ask for some sense of a very natural problem of you know find singular Einstein metrics and when I say singular I mean sort of with very specified class of singularities these often arise as sort of boundary points on the space of smooth Einstein metrics okay so I don't want to belabor this too much but just to say that the types of techniques and ideas I'm going to introduce for this problem can be used in all of these things or generalizations that you can be used and this is sort of a good test case for that okay okay so now let me point out that you know some examples of these things just off the bat and in fact well Alex gave us a lot of examples in the last lecture let me give you one other sort of general class of examples which is very closely related to what he did so let's suppose I take m2g2 to be any smooth compact surface compact constant curvature so it could be flat spherical or hyperbolic right and so obviously with the hyperbolic case there's a lot of examples if I take f from m1 to m2 which is a branched cover and I pull back the metric I let g1 the f pull back g2 then I'm going to get a metric on whatever m1 is so m1 might have very high genus because it's a branched cover and it'll have conic points exactly the ramification points of the branching and if I am winding around k times then the cone angle here is 2 pi k so in other words my my 2 pi maybe I should call it a l okay so if I've wound around if I've ramified l times then it's opened up to a cone angle 2 pi l okay now this already gives you a little bit of a hint that there's going to be some complexity so if I if I've asked over here can I find a constant curvature metric with specified points and specified cone angles right already we know that that's going to be a bit dicey because if I were to say I want to sort of find a hyperbolic metric with arbitrary points and arbitrary integer multiple of 2 pi cone angles I'd be asking for a branched cover with sort of a you know specified ramification points you know and so that's not such you know that's a constrained problem okay so okay so I'll mention various people that have worked on this as we go along but so I'm going to now go to the you know sort of the classical uniformization theory so first I want to remind you classical uniformization from a PD point of view and so it's exactly this one which I'm going to be using so I'll be talking about other perspectives as we go along as I said synthetic geometry even has sort of a a very important role to play here and has provided some answers that we can't quite get other ways in some cases but I want to talk about this problem in the case where I've no cone angles and this is you know really ancient history but just to remind you how the how the PD perspective goes trying to solve the uniformization okay so the problem here is that I have mg this is smooth compact and I want to find a metric g which is conformal to g not so remember I said I wanted to be in the same conformal class so g looks like e to the 2 u g not the only reason for writing the conformal factor as an exponential is a I can be sure that it's never negative or zero just by writing it this way and secondly it makes the equation look nicer right otherwise I could write this in I could write it as you to the fifth or you to the you know whatever log of you or you know you can write anything here you get a worse equation anyway if you write it as e to the 2 u then there's a classical formula which is not very hard to derive that if I take the Laplacian with respect to g not of u minus the Gauss curvature of g not plus the Gauss curvature of g times e to the 2 u that's equal to zero this is just a relationship between any two conformally related metrics it's pretty easy to derive and it's called the the Louisville equation okay so okay there we have it and from our point of view what we're asking is we have a smooth compact surface I want to find you such that k is equal to a constant okay so in other words one way you can think about this is somehow find to you so that that coefficient is constant which if I write k is equal to e to the minus 2 u times this stuff then that's like solving some PDE the other way to think about it is I can sort of fix k and then I can ask well can I find a solution and you'll pretty pretty quickly see that you're gonna run into some constraints so I'll just remind you there's a constraint which is the Gauss-Benet theorem which just says that you know the integral so whatever this curvature is so the integral of k times da so this is case of g da sub g which is the same as k times e to the 2 u da g naught this this has to be 2 pi times the Euler characteristic of m okay so very classical formula and all that tells you is that if you are in a surface with negative Euler characteristic then this k it better be a negative number okay similarly Euler characteristic zero or positive you correspondingly have to choose that to be zero or positive now the exact value of k is not so important so in this case if I replace u by u plus a constant I can rescale this so u plus a constant the Laplacian is the same u plus a constant it gives you an extra factor out here a positive factor so in other words I can always assume that this k if it's going to be constant is negative one zero or plus one okay so okay so I know what constant to put in there now I can ask can I solve that problem you know once I put in the right sign here okay so there's this old theorem of course that was obviously really a theorem in complex analysis you can think about that about this many different ways but I want to state it purely this way as solving this equation I can ask you know can I can I solve this so the well-known answers of course yes this always has a solution if the sign of k is chosen appropriately and I'll just want to remind you how you do that okay so there are three cases so the easiest case maybe I'll call case zero is k is equal to zero why is that easy well simply because the nonlinear term so anytime you have a nonlinear equation you're gonna presumably have to work harder when k is equal to zero that exponential is gone and so I'm just asking to solve the equation Laplacian Laplacian of u plus interview is equal to k not now one thing which somebody should a squawk to notice of course I certainly expect questions is which Laplacian to my do I mean so there's always a normalization and here I mean the Laplacian which in the standard Euclidean cases d squared dxj squared some j goes from well in this case one to two so in other words it's little plus in with the plus d by dx squared not minus so that's my normalization I use here okay well when does this problem have a solution so this is classical fact this has a solution if and only if the integral of k not is equal to zero but the Gauss-Benet theorem says that if I have any metric on here five any metric on the surface g not then its Gauss curvature has to integrate to zero so that just sort of built in if that's the Gauss curvature function for any metric whatsoever even if it's a variable function it has to integrate to zero and that's a necessary and sufficient condition for this this problem to have a solution okay now you can modify you like I did before I can modify by a constant you up plus or minus a constant well that would correspond to just scaling my surface okay so there's not a natural way to normalize in zero curvature but in the other cases there is okay the minus one case is more interesting okay so let's assume that Kyve m is less than zero and what I know now is simply that the integral of k not is negative and I need to solve that a problem so I need to solve Laplace not of u minus k not minus e to the 2u is equal to zero please sure well why would I compute it with respect to a different well I mean so I didn't mean this being I mean I just mean that I mean the Laplacian with a positive sign here but the curvature I mean so what other normalization for the curvature do you mean well okay so I'm of course this is not the Laplacian with respect to a general metric I've written the Laplacian with respect to a flat metric and you know when I write Laplacian sub zero I mean the Laplacian with respect to the metric g zero okay which looks like you know one over the determinant of gd by dx i root gd by dx j okay yeah so you're absolutely correct there okay so I need to solve this and so I want to introduce you to the method that's been used and I'm going to be able to solve this conic metric problem with exactly the same method so I want to go through it carefully it's not a very sophisticated method in fact if you're trying to solve elliptic nonlinear elliptic equations it's as they say the most boneheaded method you could try and you know you're in luck if it works it rarely works but in this problem it works okay so this is called the method of sub and super solutions and it feels like cheating because you don't have to do very much work so the method of sub and super solutions okay so you may have seen this so bear with me I just wanted to review it here because we're going to want to modify this in the singular context and something that I maybe didn't go you know there's a whole yoga a whole methodology of trying to do elliptic theory and parabolic and hyperbolic theory in on singular spaces and you know this is sort of the first example where you see sort of the new features that come out when you try to sort of understand the Laplacian on iconic manifold okay so the method of sub and super solutions is well let me look at the equation in the following way Laplacian you is equal to f of x and you so here x is a local coordinate so that just kind of a symbolic for local coordinates so this is some it's called a semi-linear elliptic equation and so the method says the following suppose that I can find what are called sub and super solutions so definition is that you lower is a sub solution if Laplacian of you lower bar so there's this stupid sign convention which is what leads people to sort of talk about the Laplacian the other direction is greater than or equal to f of x you bar and it's called you upper bar is a super solution if Laplacian of you upper bar is less than or equal to f of x okay so that's just a definition for function so maybe you can find them maybe you can't obviously if you found an exact solution then it's both a sub and super solution okay so this method is based on just the maximum principle which is a very powerful technique which again rarely works but when it does it gives you very powerful estimates okay and I say it rarely works I mean if you have a system if you have you know I mean all sorts of examples it doesn't all sorts of cases it doesn't work okay so the theorem in this general sense is that if there exists you bar which is less than or equal to you bar so sub and super solutions then there exists an exact solution a true solution which is sandwiched in between them you lower bar is less than or equal to you is less than or equal to you okay so in other words if you can somehow find a sub and super solution and they have to satisfy this equality this sub solution is less than the super solution then you're done okay now you know these functions don't need to be very complicated you can just build them out of you know functions that you know and love and that's what we'll do here but it turns out that that's enough so the claim is is that we can do that here so what I want to do just briefly is first of all apply it to this case to this theorem this level equation and then secondly just remind you of the proof of this theorem okay this will look very different than the sort of analysis I'll be doing later on but it's sort of a good basic technique which is very worth knowing and reviewing even if you know it well okay so apply this to our problem well so the functions that we know and love the best are constants and we'd like to say that we can just apply constants as sub and super solutions right which is what we want to do however the problem is is all we know about k not as it's integral is negative we don't know that itself is negative so there are various ways of handling this but the first thing we can do is prepare ourselves by finding an initial conformal transformation namely an initial function so that we can transform to a function that has strictly negative curvature okay so here's my first step so my preparatory step so I wanted to find e to the 2 u0 times g0 I'm going to call this g1 and I'm going to choose this so that k g1 is strictly negative everywhere in the surface okay now how do I do that well what I'm going to do is I'm going to solve Laplace in with respect to g0 of u0 is equal to so this k not now unlike in the flat case I can't solve this problem right because this does not have integral zero so what I want to do is I want to subtract off its average so k not bar is just the integral of k not divided by the area of the surface okay I know I can solve this problem by the same general criterion I talked about and so then if I take this solution so now I'm going to define g1 by this formula here and now if I look at Laplace in u0 minus k0 and I want to write this is plus k1 times e to the 2 u0 we know that this is always the transformation rule between the Gauss curvatures of g0 and g1 right but we know that this thing is equal to well by the way I've done these things this is equal to k not bar right so therefore k1 the true Gauss curvature is equal to k not excuse me yeah k not bar times e to the 2 u0 k not bar was negative so that's strictly negative okay it's variable it's not a constant right we don't know what it is but it has the advantage of being strictly negative okay so that's a nice preparatory step okay and now we can find sub and super solutions really trivially namely we just take very large negative or positive constants consider u lower bar which is equal to minus a and u upper bar is equal to plus a so if I plug this in I'm going to get you know Laplace in not of minus a minus k0 minus e to the minus 2 a well this part is gone this is strictly positive by how I chose things and this is extremely tiny if a is big enough so that's going to be greater than zero everywhere okay so there it is there's my sub solution and then similarly you can check that this is a good super solution okay okay so very easy I've done no work at all doesn't seem like I'm doing analysis and everything kind of rests in this this theorem here okay so let me sketch the proof of that so that's a you know a very useful tool classical fact it's you know finding this sort of the proof written down is not everybody refers to it without writing down the proof so let me just give you the proof okay so this is done by an iteration method okay now what I want to do is first change the equation a little bit so I have this equation Laplace in a view so what I'm trying to do is proof of this general theorem general PD though this thing over I've written over here okay so what I want to do is I want to rewrite the equation as Laplace in minus u Laplace in minus lambda view is equal to f tilde of x and u which by definition is just f of x and u minus lambda u okay now why do I do that because what's not to make this thing invertible though when lambda is a positive number that's invertible right so that's sort of a nice fact but in fact it's going to turn out that you want a certain property of this so what I want to choose is I want to choose lambda so large so that this function f tilde of x and u is monotone decreasing in you in other words I want du of f tilde of x and u to be less than or equal to 0 okay so why well first of all how do you why should we be able to do this because of course you can take on any value could be up to plus infinity minus infinity well in fact all I care about is that this happened for values of you which are in between the sub and super solutions okay so suppose that we have our sub and super solution so me just do this over here so I have these sub and super solutions which are you know let's say c2 functions so that I can define this point wise so I have u bar and u upper bar that gives me bounds on where f is and so it's certainly true that you know just by compactness of the surface and so on that I can make this function I can choose lambda so large so that this is true for any value of you in between you lower bar and you upper bar because I'm just dealing with a compact set of values there okay so I so I want du f tilde is less than or equal to 0 if you lower bar let's say the infimum of you lower bar is less than or equal to you is less than or equal to the supremum of you upper bar okay so that just a bounded range and I can easily arrange up by choosing lambda large enough okay so now what I do is I'm going to define a sequence I'm going to let let's say u0 be you lower bar now I'm going to define uj plus one iteratively is equal to Laplacian minus lambda inverse f tilde of x uj okay so namely I just solve the equation Laplacian minus lambda uj plus one is equal to something that I already knew at the previous step okay so that's called an iteration and see let's go back here so the fact that we can always do this is just the statement that Laplacian minus lambda is invertible because lambda's a positive number and so the real interest is inequality so the claim is that all of these things stay sandwiched in between you lower bar and you upper bar please okay so it doesn't really matter here because so and I'm definitely sloughing over the regularity issues so you know I can assume everything is smooth so I work in you know ck alpha so let me defer that for a little moment but let's say you do it in LP or L2 or something like that it doesn't matter very much the important things here of course is that at the end of the day because this is an elliptic equation if let's suppose that I found you lower bar and it's smooth right then the solution of this is good and as long as capital F is a smooth function then this guy will be smooth so you solve it in some space and then apply elliptic regularity okay now I'm assuming and I'm happy to sort of talk a bit more about general elliptic regularity I'm going to be talking about it specifically in this context of conic singularities but anybody want me to say a few more words about that be happy to okay so so you do this and these all follow by the maximum principle so let me just sort of do one of these just for for instance right so if I take uj plus 1 minus uj right this is going to be Laplace minus lambda inverse of f of x uj minus f tilde of x uj minus 1 now this thing here so I carefully arrange things so that's monotone decreasing by induction so I'm going to do this by induction by induction uj is greater than or equal to uj minus 1 which means that this is less than or equal to 0 okay and then well here's the point is that Laplace minus lambda and its inverse are you know what's called well positivity preserving but of course that's a misnomer it makes positive curvature and positive signs and negative signs so I'll remind you why that's true in just a second so this whole thing so I claim that this whole thing is greater than or equal to 0 and that's the inductive step okay so why is that true so what I'm really claiming I can see from two points of view and I'll remind you both points of view so I'll remind you two points of view the first is that if I have let's say Laplace minus lambda v is equal to f so if I have Laplace minus lambda v is equal to f and let's suppose the f is greater than or equal to 0 right then I claim that v so I was the hard part here is remembering what happened so if f is greater than or equal to 0 then I claimed that this means that v has no which way do I want to get here so it's a sub solution so v has no man so v does not reach a positive maximum v does not reach so the met the super v is less than or equal to 0 okay so that's the maximum principle well so that's just noticing that Laplacian v is greater than or equal to lambda v right and if the supremum were strictly positive then lambda v would be strictly positive but the Laplacian at a maximum has to be less than or equal to 0 okay so that's one point of view the other point of view is that if I take the integral kernel for Laplacian minus lambda so suppose I write Laplacian minus lambda inverse applied to any function f is equal to the integral of g maybe I'll call it g lambda of x y f of y d y okay so this is the green's function at energy lambda or however you want to call it then this g lambda just point wise is less than 0 of course it has a singularity when x equals y but point wise it's less than 0 okay well you have to prove that one way of proving that is sort of an analog of this result or you can relate it to something that's arguably even more elementary which is the heat kernel so I can write g lambda is equal to minus the integral from 0 to infinity of e to the minus lambda t times well right sort of provocatively e to the Laplacian t dt so this is the heat kernel h of t x y okay that's the integral kernel if I put a dollop of heat at a delta function of heat at y and I wait for time t then that's how much heat persists at the position x okay so again there's a maximum principle at work this is a strictly positive quantity I'm integrating against a exponentially decreasing but strictly positive guy from zero to infinity and I have to take a minus sign because the zero is on the lower limit of integration so this just looks like thinking of the Laplacian as a number so this is really a functional calculus statement but it's true okay and it's true pointwise in fact you know if I put in arbitrary values x and y not equal to each other this is a this is an equality yeah please yeah well so what I'm doing is I have a different right hand and left hand side right so you know I'm solving this equation Laplacian minus lambda uj plus one is equal to f tilde of x uj of course this is the same as what I just wrote down but I assumed that I've already figured out what uj is inductively and then I find a new function uj plus one okay so they're different and I keep on stepping up and I find new ones okay okay good okay so I find these functions and they satisfy this chain of inequalities so what well yeah okay so they satisfy this chain of inequalities I'm just applying arguments like this again and again it's an easy exercise and so what I get is that you which is equal to the pointwise limit of uj so I'm doing the most unsophistic kind of kind of thing that we teach our real analysis students not to do with functions right it just take the pointwise limit that's usually a disaster right because we know nothing about the pointwise limit except for that it's bounded okay it's an l infinity function that's all I know about it because it's bounded between u lower bar and u upper bar okay however what I know about it is that so Laplacian minus lambda of this u is equal to f tilde of x and u well this right-hand side it just I mean f is a smooth function so I can just take the pointwise limit of f of x uj and it converges to that how can I differentiate this thing well you know if you're like me think of this as a distribution and you can differentiate anything right so you can think of this as just a distributional equation okay so I think of you as this as a distribution on m and the right-hand side a priori is just an l infinity and then I apply iterative elliptic regularity so the first thing I do is note that this is going to tell me that this equation says that any distribution which satisfies that equation has to be better than it looks it has to have two derivatives in l infinity so so u is in w to infinity okay so that's kind of a weird function space it just means two derivatives now infinity but in particular that means that and in fact let's just be safe call this two derivatives in LP for all p less than infinity right this is going to say that u is in c1 alpha the holder space c1 alpha right and now the right-hand side is in c1 alpha and u has two derivatives as it's Laplacian and c1 alpha that's an a space in which I can apply regularity so now I'm saying that you know Laplacian of u is in c1 alpha that's going to tell me u is in c3 alpha etc so I keep on bootstrapping and I get eventually that you is in c infinity so I found a solution okay and moreover the solution is unique so if I have two solutions Laplacian u is equal to f of x u and Laplacian of v is equal to f of x v and maybe excuse me I want to do this with the monotone guys just to be safe Laplacian minus lambda and what I mean Laplacian minus lambda v is equal to that the difference Laplacian minus lambda of u minus v right is equal to f tilde this is f tildes now of x u minus f tilde of x v and I apply the same sort of ideas that if u is ever bigger than v then I look at the supremum of u minus v this right hand side is positive and then that contradicts the same sort of argument that I did before that the Laplacian has to be less than or equal to zero at a supremum of u minus v okay so u minus v is both less than or equal to zero and greater than or equal to zero okay okay so that's like I say a old argument going back 50 years or so but it's still a very powerful one when you can make it work so we've completely solved the uniformization theorem in in a in the zero and negative cases now okay in the contact case okay so let me say a few words about the spherical case but quickly move on to the iconic situation which is more a bit more challenging so the spherical case so the reality which we know from complex analysis is that any metric whatsoever on the sphere is conformal to a constant curvature metric there's only one conformal class okay and you can prove this by you know finding a certain sort of holomorphic function from so if I have M what I really would want to do and there's this is sort of a task thing is if M is looks like the sphere then well you see what do I want to say so I'd want to find a holomorphic function from this on to the complex plane so holomorphic that has a simple pole at P and you can find such things okay I mean there's sort of a standard tricks to do if we were to look at this from the PDE point of view we'll plus in you minus k0 plus e to the 2u so this is one of these things that you know it's sort of good pondering you do something incredibly minor looking like change a minus sign into a plus sign and it just screws everything up okay namely this method of sub and super solutions just fails you know you can try to apply it and things don't quite work and you think well I could just work a little bit harder and make them work somehow it just won't work and the reason is is that there are families of solution well it's a little bit contradictory but there are too many solutions here is the reason it doesn't work so in fact it turns out that there are solutions that so here's the sphere it turns out there are solutions to this problem which kind of bulge at one point they kind of bubble at one side and you know get very small at another side I'm not going to say too much about this point of view but the fact that you have a non-compact family of solutions so where these solutions come from is pretty easy I just take a conformal transformation just a dilation on the sphere that sort of sucks the south pole to the north pole or you know any pole to any other pole and I pull back the round metric with that I get that's always conformal to the round metric and I get a family of functions which get larger and larger as I take a stronger and stronger dilation they all satisfy this equation and though that's not directly contradicting the fact that I might use sub and super solutions that turns out to be you know really it really screws things up and so the method just doesn't work now I'm not going to sort of give you full details of the proof that you can just solve this equation by hand for any function k not that integrates to the right thing namely 2 pi times order characteristics namely 4 pi and but the way that you do this is that you do this variation way so you use the calculus of variations so variational ideas so namely what I want to do is I want to look at the energy one half times the integral of grad u squared so then I'm going to have to do plus minus the integral of k not times u and I want to find the I want to find you such that this thing is minimized but I want to do this subject to the fact that the integral of e to the 2 u is fixed let's say is always equal to one this is the case so this is a variational problem so in other words I look at this functional here and I look on a nonlinear constraint set so this is some big infinite infinite dimensional manifold in the space of all use in fact it has co-dimension one so it's this huge hyper surface in the space of all use and somewhere on there I want to find the minimum of this functional now why does it do the right thing well if I take the Euler Lagrange equation of this which I'll compute for you tomorrow and use this is Lagrange multiplier I get exactly that equation okay how does the plus sign come out well it's this minus sign here and that minus sign means that this is possibly unbounded below and you run into all sorts of problems about why can you even find a minimum and so on whereas if this were a plus sign which would be that being a minus sign the negative case this would be easy okay so there's a whole line of work that sort of goes here and this turns out to be a somewhat delicate variational problem precisely because there are too many solutions right it's always this sort of a normal thing if you have this big non-compact family of minimizers it's very hard to find them variationally because you know you take a minimizing sequence and it can go off to infinity okay so there's sort of a whole line of work there which I am not going to be pursuing very much in these lectures but I just want to mention this okay so the last couple of minutes let me just set these problems up now for the chronic case what I'm really interested in and why I'm you know so the first thing is a theorem kind of completely settles the problem I stated in one specific case so here's a theorem by Bob McEwen and it was proved about 1981 I think and what he proved is that the the main problem is solvable if the corresponding curvature is less than zero or equivalently there's a number which is the chronic Euler characteristic I'm gonna call this kind of a less than zero okay so there's a condition which is just like an Euler characteristic condition and provided that simple number is zero as negative this is just a combinatorial thing then I can always solve this problem for any cone angle okay and the method of proof is going to be just that we have to set it up properly and there's some new details which I want to describe but that's it the whole solution okay the k equals zero case turns out to be even easier as usual it's just a linear problem and the k positive case is the one that's still open and unsolved and it has a lot of subtleties okay and that sort of relates to what I'm talking about over here now I want to sort of conclude today by talking first of all what this chronic Euler characteristic is and then secondly sort of why they're still interesting things to say even in the negative case okay so what is so suppose I have any conic surface so maybe there's some handles in here but there's a finite number of conic singularities okay I'd like to compete the conic Euler characteristic so what that means is I want to take the integral so I want to take m minus the union of balls j goes from 1 to k and I just want to take epsilon balls around these points pj case I'm just excising balls here okay now on this surface I want to apply the ordinary garden variety gas beneath there I'm on a man on the surface with boundary okay so I'm going to have the integral of kda on this m sub epsilon I call this m sub epsilon and that's going to be the integral over the boundary of m sub epsilon of something plus the Euler characteristic of this thing so 2 pi and then I'm going to have chi of m and every time I've removed a disk it subtracts one from the Euler characteristic so I have 2 pi of chi of m minus k okay so so the whole issue is what is the boundary contribution now the proof is you know the various ways of doing this but so this this boundary contribution the integral of boundary vm sub epsilon so this thing so there are various ways of how you might write this what this thing does so this is going to be asymptotically the same no matter what this conic metric is to the corresponding contribution on a flat cone okay so this is going to be a sum j goes from one to k of integrals around little circles and I'm going to have this corresponding so if you like this is a circle with cone angle 2 pi alpha j okay now there are various ways of writing down this contribution but it's essentially the whole ennemy around that it's the parallel transport around there so this is just going to be the sum of 2 pi times alpha j okay so what I'm left with all together is that the integral of kda is going to be 2 pi times the sum j goes from 1 to k of alpha j minus 1 plus 2 pi chi of m and so I call this number chi of m beta so I'm going to switch parameters now is equal to chi of m plus the sum of the beta j's where each beta j this just turns out to be a more convenient parameter alpha j minus 1 okay so it's a way of writing it and so the theorem of macoans is that this is a necessary condition if I'm going to have negative curvature you know if k is everywhere negative right then this right hand side better be negative and he says that's a necessary and sufficient condition okay proof is same kind of boneheaded stuff though we have to worry about how do we solve laplacians and conic manifolds and I'll be telling you about that in negotiating detail okay so that's one thing and then and then I promised you that there are some interesting questions here and so here are the questions that I want to get to by the end of the week so I can find a solution with singularities now sort of the the frontier of this problem is the frontier of what happens here namely what happens when the points coalesce so what we're after is sort of a very accurate analysis of what happens when you have clusters of conic singularities that that vanish I mean that that coalesce into one point okay so we know that there's always solutions by macoans general theorem but something horrible happens to them I mean they're gonna converge to something with cloning singularities I'd like very precise information on how that happens right now why do I care about that so much well it's emblematic of a lot of problems where you are have clustering singularities and you're interested let me give you two examples and I'll stop there one example is suppose you have a holomorphic quadratic differential on the surface and you look at a family of such things where the zeros are clustering together okay how do you sort of understand the fine asymptotics of that now there's old work on this that actually Alex went out to me was solved by my colleague Halsey Royden you know forty years ago but in fact we have a very interesting new approach on sort of how to understand those asymptotics okay another problem this comes out of quantum mechanics really suppose you have a bunch of Coulomb potentials you have a shorting or operator with Coulomb potentials you know one over mod X singularities and they're all clustering you know you have a bunch of particles which are very close together what happens how do you analyze the fine spectral theory of that so the tools I'll be describing give you a way of understanding that and this is just sort of the easiest kind of model case to understand things like that okay so I'll stop there so yes I mean so the the the right question is given a collection of k distinct points in the surface how do you compactify the space of k distinct unordered points in the surface now there's complex ways of doing that I mean you know complex geometric ways of doing that which I don't want to do so what I want to do is define a real manifold with corners which describes the compactified configuration space so that turns out to sort of describe all possible ways that these points can come together right and then sort of the theorem and after is saying that on that compactified configuration space I have a universal curve which is over every divisor I have the curve M you know and it's sort of changing as the points move around and that itself you know as the points cluster you know that that has itself kind of a complicated resolution is a real manifold with corners so I have some very complicated geometric object it has a lot of boundary faces and so on but it describes all possible ways that points can come together and I'd like to understand this family of metrics on that space