 This is what we call, sometimes the get-together, the reception or whatever, how you want to call it, on the terrace of the Adriatico, but it's a dinner, okay? I've been asked by many people, it's just some kind of spritz or whatever. No, it's a complete package, liquid and solid, so don't take any other commitment for the evening, okay? But it's at the Adriatico guesthouse, okay? Not here, not in Galileo, it's down there, on the sea, okay? And second short communication, this afternoon at 4 p.m., we will meet here, everybody interested, this is what in the program is just called the discussion session, which basically we are meaning that we are asking the lecturers to be here with us and just be ready to answer questions to whomever wants to ask, okay? So, prepare your questions and your, I mean, questions or opinions or topics of discussion, it will be completely open. I mean, eventually we might end up in the bar taking just a coffee if there is nothing to discuss, but it's a free moment of discussion and things, okay? So that's it, so we are ready to start and I'm really very, very happy to introduce you Rick Shane from Irvine and one of the most important teachers of my mathematical life. And he's talking on some geometric properties of space-time. Okay, well thank you, Claudio, can everyone hear me, the microphone? Yeah. So, right, so I'm going to give four lectures mostly related to geometric properties which are connected in some way with gravitational energy and mass. And so, I think yesterday for giving an introduction to the ADM mass, so I don't have to repeat that. But I'm going to be interested in, so from a physical point of view there are ideas called quasi-local masses, which there are many different proposals for quasi-local masses. And from the geometric side of view, from the geometric point of view, the mathematical point of view, there's the idea of comparison theorems and comparing geometry of curved spaces with a flat, not necessarily a flat, but a simple background space, like a Euclidean space or a constant curvature space. And these two ideas are really very similar and there have been some interesting proposals in both directions which have been kind of divergent, but I think they're obviously related. And so, I'd like to present kind of both sides of that story and discuss what might be possible relations between them. Well, let me just say that my background is a mathematician, so I did my PhD in differential geometry and geometric analysis. And then I got into relativity a few years after I graduated because of connections with the Einstein equations with geometric problems, in particular the positive mass theorem. And then over the years, I've kind of been a want-to-be physicist, I've learned more physics, I always enjoy hearing real physicists speak, but I do, I think, I've learned some of the language which is quite different in some cases from mathematical language and I view general relativity and also string theory to some extent as a very rich source of natural differential geometric problems and also, in some cases, ideas for understanding geometric notion. So, I think if you're working as a differential geometry, it would be a very good idea to learn the relativistic or the geometric physics side of it as well because you may be able to do something interesting there. Okay, and so, I'm going to begin with, so actually, I'm giving four lectures. My first two lectures are going to be on three-dimensional manifolds, so these will, for the Einstein equations, be space-like hypersurfaces in space-time as Piotr discussed yesterday, so you can think of them as initial data sets and I'm going to be looking and so Piotr wrote down the constraint equations which are conditions on the initial data, so remember if we take initial data for the Einstein equation that consists of a triple, a three-manifold, a Riemannian metric G and a symmetric zero, two-tensor K, which will be the second fundamental form, so these correspond to the initial position and velocity for the Einstein equations, the gravitational part of it, and a very important constraint on these arises from the set of four constraint equations and so let me recall those, so these come from the Einstein equations and so in case there are matter fields present in the space-time, then the first one expresses the mass density of the matter field as observed by an observer moving normal to the space-like hypersurface in the space-time and this Piotr showed yesterday or didn't show but wrote down yesterday is related to the scalar curvature of G, so it's the scalar curvature of G and then plus a quadratic term involving K and that term is minus the norm of K squared taken with respect to G and then plus the trace. And then the, this is the scalar constraint equation and the vector constraint equation, I'll put also the 8 pi here, I'll write it as J i, which is the divergence, so it's D J of K i J minus trace K. Okay, so those came from, those are the constraint equations, they come from the Gauss and Kodotzi equations or actually Piotr had another nice way to motivate the fact that there must be constraints among the initial, the initial data and for positive energy theorems, it's very important to have the dominant energy condition and so what that gives us in terms of mu and J, so this is, these are conditions on the matter fields present and so they say not only is mu positive but mu should be greater than or equal to the norm of J, so J as I've written it is a one form and this is again the norm with respect to the metric. So it corresponds to the fact that the mass density is positive for any observer, so that's a natural condition which is satisfied by most matter fields that one considers and I believe that this condition is absolutely essential for proving positive energy theorems, they're probably not true without that. And so I'm going to look, in my first two lectures, I'm going to think of a special case, so there's a, this is the case, so remember G is the initial position, K is the initial velocity, so it's perfectly reasonable to look at a gravitational field which is initially at rest, so in other words K equals zero or actually the same analysis would work even if trace K is zero. And so in this case we can forget about the vector constraint equation, this inequality just says mu is greater than or equal to zero and if the trace is zero then this being greater than or equal to zero implies that, so let's forget K here is zero or actually some of the analysis also works when trace K is zero and so we would be, the constraint equations would tell us that the scalar curvature is non-negative. And so this is already a very challenging geometric problem to understand three manifolds with non-negative scalar curvature, so it is known, so if you look at compact three manifolds and you ask which compact three manifolds have metrics of non-negative scalar curvature, it is actually known but it's quite a hard theorem, it ultimately comes out of Ricci flow ideas for three manifolds but there's a lot that can be said by other methods as well but the geometry is much more complicated than the geometry say of three manifolds of positive sectional curvature or even Ricci curvature, so in Riemannian geometry there are different curvature notions and the scalar curvature is the weakest of those, so this is quite a, it's only a single function constructed from a three metric which has six components locally, so it's a, in principle quite a weak condition. Okay and so in fact it's sort of remarkable that you can say anything about such three manifolds, if you reverse the sign and you ask which three manifolds have metrics of negative scalar curvature the answer is all of them do, so there's no obstruction at all to making the scalar curvature more negative, so it's very special to this situation and it actually I think in a certain sense is the structure of the results in this area are very much coming from relativity or related to ideas about gravitational mass and so I'm going to motivate three different, I'm going to talk about three different settings or three different sort of ideas and I'm going to motivate them from the two dimensional case, so this is, this obviously we went one dimension down we would get a much simpler problem so I'm going to start the motivation with the case of surfaces, two dimensional surfaces and I'm going to look now at the Gauss curvature Kg which is non-negative, so this is a much simpler class of, it's a much simpler problem to understand these manifolds and I want to talk about three different properties of these spaces, so I'm going to take property one, this guy's idea one which is the case when I have a complete surface which is asymptotically flat, so I talked a lot about asymptotically flat in n dimensions for n bigger than two but you can also look at asymptotically flat surfaces and so let's say we have an asymptotically flat surface, well what would it mean, so it could mean the curvature falls off to zero at some rate, now the thing about surfaces is that the model at infinity so if the curvature falls off to zero, the model at infinity is unique up to a finite number of parameters, namely if you think of flat surfaces then all you have are the cylinder is a flat surface, non-compact ones and you have cones and so the idea, so we're going to look at the situation, so the cylinder is a kind of degenerate case, you can think of it as a degenerate cone, so let's look at the situation where m outside a compact set looks like a cone, okay so what is a cone, well it's a, you can draw a cone in R3, that's a flat surface or you could draw a complicated cone, so this is a convex cone and the cone angle in this case is smaller than 2 pi, so it has smaller angle than the Euclidean cone and you could take, well it has a singularity but you could smooth it out, you could just say cap it off like that and then you would get a smooth surface which is asymptotic to the cone, you could make it equal to the cone at infinity if you wanted to and have non-negative curvature, so you can have examples like that and so asymptotically flat surfaces would be surfaces that outside near infinity look like cones, okay and so what's a cone, well the cones are determined by a cone angle, so a cone in coordinates looks like this, d rho squared plus a squared rho squared d theta squared, think of rho and theta as polar coordinates in the plane and a here is a positive number and so the period of theta is 2 pi, so this has cone angle, this would have cone angle 2 pi times a, okay and so I wrote that down because I'm going to come back to it a little later when I talk about the three-dimensional case, but so that's what cones are and the cones, really the cones together with the cylinder are really the only flat, non-compact flat exterior regions that you can have and so they're parametrized by one number which is the cone angle, the cylinder would have cone angle 0 I guess, okay and so what can we say about this question, well if the curvature is not negative then and let's assume also, so actually let's assume the manifold is simply connected, pi 1 of m is trivial because actually the only non-simply connected manifold of this type is a cylinder, so there's really no loss of generality and assuming it's simply connected and then what can we say about the relation between this condition pg non-negative and a and so the basic fact comes from the Gauss-Bonnet theorem and that would tell us that if we take a large region, if we take some sphere, some large region in s sigma and we integrate the Gauss curvature then assuming we take s sigma also to be simply connected then this is 2 pi minus the integral over the boundary of s sigma, the oriented boundary of the geodesic curvature, kg with respect to arc length this is the geodesic curvature, okay and so in particular if the Gauss curvature is non-negative this is non-negative and when sigma goes to infinity for large sigma out in the cone region the total geodesic curvature is just the cone angle, so this is the cone angle, so this would be 2 pi times a, we take cone angle a and so what this tells us is that this is non-negative so that implies the cone angle a is less than or equal to, well a is less than or equal to 1 or the cone angle is less than or equal to 2 pi, okay and moreover you can see that equality holds only if k is identically 0 and if k is identically 0 the surface is isometric to r2 so it's a theorem that tells us that the local condition that is kg non-negative has an impact on the asymptotic behavior at infinity and so what it says is that assuming kg is everywhere non-negative then the cone angle in the flat region at infinity must be less than 2 pi, so that's property 1 and so I'm stating these for surfaces because each one is going to have a generalization to higher dimensions and I'm going to talk about, so the generalization of this one of course will be the positive mass there, okay so this is the first setting and the second setting I want to consider is again the same two dimensional setting but where I take now a region just an arbitrary region omega smooth, say with smooth boundary contained in m and again for simplicity I'll assume the region is simply connected, so in other words I have my surface and I just take some region omega in there and then again I've written down the Gauss-Bonne theorem if I apply the Gauss-Bonne theorem on omega then what I get is that the integral of omega of k with respect to the area element is 2 pi because again the other characteristic is 1 because it's simply connected and then minus the integral over the boundary of omega of the geodesic curvature, let me just call it k ds, so I get that from Gauss-Bonne and again the fact that k is greater than or equal to 0 tells me that that is greater than or equal to 0, now I want to give a slightly different interpretation of this so I want to understand the 2 pi in the following way, so I can think of taking this region omega in my curve manifold and comparing it to a region in R2, so here I have m, here I have R2 and I want to do it in such a way that the boundary geometry is the same so well a curve of course doesn't have much geometry, the 2 curves are isometric if and only if they have the same length, if I can choose an arc length parameter, so there are many different such domains but the simplest one would be a circle, so in other words I could take this curve, let me call it gamma which is the boundary of omega and I can isometrically embed it into R2 and so the simplest way to embed it would be as a circle where the circle has length equal to the length of gamma and then I can isometrically embed, I just map the circle in there and I follow around, I correspond the arc length parameters, so I move a certain arc length along gamma and I map that point correspondingly distant along, so I would fix one point which goes to some point and then when I want to map any other point I just follow around in, so this is oriented, I just follow around so that I move the equidistant along the circle and then what I can do, so remember the famous theorem of Hopp for embedded curves in the plane which is that for any embedded curve in the plane the winding number or the total change in the tangent is, total change in the angle of the tangent is 2 pi, so in particular for any embedded curve, in particular for a circle, the integral of let me call the curvature of this curve K0, so this is 2 pi, so let's call this curve gamma 0, this is my region omega 0, so I want to compare omega to omega 0 and they have the property that their boundaries are isometric and now there are many different choices of omega 0 in this case, but what I want to then do is I then have a correspondence, I have a map which takes a point here to a point on the circle and we call this map say phi and then I can write and the arc length parameter is the same, so I can write this 2 pi as the integral of K0, where K0 is the curvature, I can write it as the integral of K0 well really composed with phi minus KES on gamma, so in other words I can view the Gauss-Bonnet theorem in this case as giving me a comparison between a domain in my curve surface and a domain or really a collection of domains in R2 and then the curvature non-negative condition tells me that this is greater than or equal to 0, so in other words the total curvature on the flat space is bigger in general than the total curvature on the curved space and that's a reflection of the fact that the Gauss curvature is non-negative. Gauss curvature non-negative tends to make the rotation number or the total curvature smaller than it would in the Euclidean case, so that's the second property and again I write it in this way because it turns out there's an exact analog of this in three dimensions which I want to discuss which comes from really from physical ideas and so this is the second case and then the third is really a special case of this one in two dimensions but it's where we take our region omega to have polygonal boundary so we consider polygons and I'm just going to, we could do this for any polygon but let's just take the simplest one which is a triangle, so I could look at a special kind of region omega whose boundary consists of three geodesic arcs which meet at some angles so this is my region, let me call it T now because it's a triangle and so each of these is a geodesic that means K equals 0 and then if I use the Gauss-Bonne theorem for this region then what I get and so you can see this by sort of smoothing out the edges so if you remember your undergraduate differential geometry you can actually get the formula I'm going to write down from the one there by approximating by smooth domains in particular the integral over K of T, P a is equal to, so let me write down the answer and then I'll just explain why it, how it comes from that so what it is if I consider the angles of the triangle the interior angles theta 1, theta 2 and theta 3 then it's the sum theta i, i goes from 1 to 3 and then minus pi okay and so let me just explain how that comes from here so the Gauss-Bonne theorem so actually the curvature contribution when I, so the curvature measures how the tangent vector changes and so if I'm moving this direction then the way it changes the amount it changes at each vertex is the exterior angle so this is the exterior angle which is let me call it alpha 2 and then similarly here the exterior angle alpha 3 so if I do this directly if I put this in yeah expand okay yeah the drawing oh yeah the right okay so let's take a okay so we have let's call this theta 1 we have the exterior angle alpha 1 we have theta 2 so we're going this way alpha 3 so actually the when you smooth out the domain and approximate the triangular region this term here becomes the sum of the exterior angles so you get the integral of k so from the formula there you get 2 pi minus the sum i goes from 1 to 3 of the exterior angles alpha i but now you can see the exterior angle plus the interior angle adds up to pi and so this is equal to 2 pi minus the sum of pi minus the interior angles theta i goes from 1 to 3 and so this is 2 pi minus 3 pi which is the which is the minus pi part and then I get plus the sum of the interior angles so that's where that comes from okay and so and so the the the gaspanae theorem for these domains with polygonal boundaries give you formulas relating the the interior or the exterior angles to what they would be in Euclidean space so in Euclidean space the sum of the interior angles is pi and so and so what this tells us geometrically is that is that if we take a triangle in a surface with non-negative curvature then the sum of the interior angles would be always strictly bigger than pi unless unless k is 0 inside the triangle now now actually I want to point out that there's there's actually a much better statement than this which is true and and again it's going back to this idea of of constructing comparison triangles so so actually what I can do is I can take so let me call so if I look at my triangle here let me doctor this picture a little remove the exterior angles so if I take my triangle in M and I call the side lengths here say this is L3 L1 L2 then I can because just by the triangle inequality because because L because one side length is less than the sum of the other two I can draw a unique up to congruence triangle in R2 so I can look in R2 and again I can use this same idea and I can draw a R2 triangle with the same side lengths side length L1 L2 L3 those have the same side lengths and then I can look at the corresponding angles so let me call this one say theta 3 bar 1 theta 2 bar theta 1 bar so those are the the comparison so there's there's a comparison triangle which is really an isometric embedding of this singular curve the triangular the triangular curve into R2 okay and then I have corresponding angles and there's actually a stronger theorem which is true and this is called the top of comparison theorem what it says is that each angle is bigger than the corresponding angle so this is a very important tool in differential geometry it says actually not only is the sum of the angles bigger than or equal to pi but each corresponding angle is larger so theta i in the curve in the curve manifold is bigger than or equal to theta i bar and again that's provided k is non-negative so in other words triangles in surfaces with non-negative curvature or fatter they have they have larger angles all angles are larger than the corresponding angles of a comparison triangle in R2 okay and so those are the three settings that I want to look at and what we're going to be looking for is generalizations of each of these three situations to the three dimensional case with with non-negative scalar curvature and it turns out remarkably there are partial generalizations for all of them and so let me also before I stop let me make two remarks so so so first of all so the first remark is that in in the triangle comparison theorem it's not really necessary that the sides of the triangle that the edges be geodesics the comparison theorem would still work in the same way that this sum is bigger than pi if so so to get the sum of theta i greater than or equal to pi we only need of course k greater than or equal to zero and of course k is greater than or equal to pi we only need of course k greater than or equal to zero and the and the geodesic curvature of each k little k greater than or equal to zero on each edge okay and you can see that again from Gauss-Penet so here I wrote the the formula when I write Gauss-Penet the the the integral of minus k on the boundary comes in so so in particular I get the contribution from the vertices but I also get the contributions the edges and if I assume k is greater than or equal to zero on the edge then this is a negative term and so I could take it to the other side and I get the course the same sort of the same comparison theorem so it sort of fits with your idea that that this is a geodesic well if I moved it out a little bit then then that would have k greater than or equal to zero and that would enlarge the angles okay and so and so the same is true just under the assumption that k is greater than or equal to zero I don't really need the side the the edges to be geodesics that's the one remark and then the second remark is that this this triangle comparison theorem actually works much much more generally so so the so Toppenagov works whenever whenever we have sectional curvature inequalities Toppenagov is true for any n greater than or equal to two and but it requires sectional curvatures so these are sectional curvatures bigger than or equal to some kappa where kappa is a constant and then you can you can make the comparison of a triangle in your general space with curvature bounded from below by some constant to corresponding triangles in the space of constant curvature kappa and so and so this is really just a very very special case of a much more general theorem which is very very important in in Riemannian geometry so the Toppenagov theorem has been used to do lots of lots of important things in Riemannian geometry so it really has a natural n dimensional extension but the required curvature condition is much much stronger than we're assuming so the sectional curvature condition is a much stronger condition than the scalar curvature being non-negative so you couldn't expect triangle comparison theorems for scalar curvature under scalar curvature conditions okay so those are a couple of remarks so those are the three sort of general settings are there any questions on this I'm going to now move to the three-dimensional case I assume everybody remembers their undergraduate geometry okay so so now we're going to consider three manifolds with non-negative scalar curvature we're going to ask in is there any sense in which in which these these kinds of theorems generalize okay and so so I can erase this part well so Piot already discussed the notion of the notion of asymptotically flat in in the higher dimensional case and so in higher dimensions one of the main differences is that the the model at infinity that is the the question of what what the asymptotic behavior should be at infinity is is not determined really there are many many different possible asymptotic forms that you might you might assume in fact you saw that Piot really only required fall off to the Euclidean metric at a relatively weak rate and so and so it allows it allows rather complicated rather complicated asymptotics on the other hand if you assume that if if we assume that the asymptotic model that is the model at infinity is rotationally symmetric then there's again a finite parameter family namely the Schwarzschild metrics and so I'm going to look at a special case and it turns out actually to be sufficient to handle this case but but it's a it's not so hard but I won't I won't go into it here so I'm only going to consider a special class of asymptotically flat manifolds we're going to take the following special case we're going to take a three manifold now the metric G scalar curvature non-negative and we're going to assume it's asymptotically flat in a rather strong sense and so and so this the the sense I'm going to assume is I'm going to assume that at infinity it's asymptotic to a rotationally symmetric model which would be a Schwarzschild metric and in fact I'm going to make the assumption that G near infinity so so outside some compact set so picture of an asymptotically flat manifold is like this I have some topology or geometry inside so actually in three dimensions it's it's no longer true that you should assume the manifold simply connected there are many many possible topologies for for this class of RG non-negative they are restricted in in in a strong way but there's still lots of lots of different possibilities and so so I'm going to assume that outside some compact set so I cut out some finite set K then out in this region near infinity I'm going to assume the metric is conformally flat so I'm going to take the special assumption that this is you to the fourth time the Euclidean metric in some coordinates out here okay and then and and and I'm going to assume the scalar curvature is zero or falls off very quickly and what that implies is and I'll talk about conformal deformations of metrics actually in the next lecture in more detail but what that turns out to imply is that you as a harmonic function or very close to being a harmonic function outside a compact set so as P out said for the Newtonian potential that that forces good asymptotics on on you and so and so what you can prove from the condition that RG is essentially zero is that you and so of course you has to go to one but you is one plus a constant and I'm going to call the constant m over two and then there's a one over R term and then plus terms that fall off faster and this first part one plus m over over two mod X gives us the short shield metric so if there were no tail here then this would be the short shield metric written in conformally flat form so the short shield metrics rotationally symmetric so you can you can choose coordinates which are which make it conformal to the Euclidean metric okay and so and so if you do that then then the conformal factors one plus m over two are where this m is the short shield mass that described okay so I'm going to make a simplifying assumption that my asymptotically flat metric near infinity is of the special form and again you can justify you can reduce the general case to this by by by some methods which are not so complicated really but I I don't want to go into it here so I want to just consider this this this class of metrics okay and so and so I want to think of so notice what I'm what I'm thinking of is replacing the the cone in the two dimensional case which is the unique rotationally symmetric flat metric by the short shield metric in the three dimensional case which is the unique rotationally symmetric scalar flat metric okay and so I've written down that the basic theorem here which came directly from Gauss Bonnet in two dimensions the corresponding theorem in three dimensions is that this is called this is the first case of the positive mass theorem in this case where rg is non negative what it says is that if the scalar curvature is non-negative then then the adm mass m is non-negative and m equals 0 only if mg is isometric to r3 so that's the first case of the positive mass term and I want to explain this in a geometric way which will lead to a kind of localization okay and so so that's my goal here and so I'm going to explain the proof of the theorem or a proof of the theorem in the remainder of the lecture today and I'm going to indicate how it it leads in quite a natural way to to a a local conjecture and and really a family of conjectures which have been verified in some cases so so this is this is the theorem so so again you should compare it to the the case of an asymptotically flat surface where the condition that a is less than or equal to 1 corresponds to the positivity of the mass that's the idea so again it's the same idea that that the point wise condition scalar curvature non-negative has has an effect on the asymptotics at infinity so so it's it tells us that in fact you could not at infinity have a negative mass Schwarzschild metric which which has a filling which is a smooth manifold of non-negative scalar curvature so so in order to motivate the proof I'm going to give it I'm going to give a proof of this theorem so that's why I wrote down this this metric here the cone metric so okay so we could prove this theorem using the gasp on a but of course gasp on a is not going to work in three dimensions that not there's no gasp on a theorem involving scalar curvature in in 3d so so so it's it would be interesting an interesting challenge to give a different to give a proof of this which doesn't use gasp on a and I'm going to describe a proof which is actually going to generalize almost verbatim to the three-dimensional case okay and so what we're going to do so so notice I've written the cone metric here in a slightly different form but if I want to I could I could I could write it in conformally flat form so the first step so I'm going to prove this theorem this is the proof n equals 2 and really the same set of ideas is going to work but it'll be a little more complicated because I'm going to have to replace geodesics by surfaces okay so so for n equals 2 we can give a proof in the following way we can first of all do a change of variables and write the metric in conformal form so just like we've done here we've written the in the short-shield metric in conformally flat form I can rewrite so this is my metric g is g near infinity I can rewrite g in the following form I can make a change of variables and I can write g as I'll replace r by rho I can write it as r to the 2 alpha times the Euclidean metric dr squared plus r squared d theta squared and it's rather easy to do that because you see how do we do that well this metric has to be equal to that I'm not changing theta so it has to be true that r to the 2 alpha plus 2 is equal to a squared rho squared okay and since I want everything is positive here that means taking square roots it tells me that r to the alpha plus 1 is equal to a times rho okay and then if I differentiate this if I take d of this then I get alpha plus 1 dr is equal to a d rho alpha plus 1 to the alpha is equal to a d rho and so if I choose alpha plus 1 equals a or if you like alpha is a minus 1 then then then these two guys cancel out and I get r to the alpha so I get d rho squared is r to the 2 alpha dr squared so so it's a simple change of variables the relation between alpha and a is that alpha is a minus 1 so in particular when a is less than 1 that means the the cone angle is less than 2 pi the alpha is negative so this is a negative alpha but that's okay it's still a complete metric as long as as long as alpha is bigger than minus 1 so a is positive so this alpha is between minus 1 infinity so those are all complete metrics they're just different coordinate they're they're the conformally flat descriptions of the cone metrics okay and now I want to consider something about the geometry of these metrics so so first of all I can I can reintroduce Cartesian coordinates here and I can write g near infinity is r which is mod x to the 2 alpha times dx squared r dx 1 squared plus dx 2 squared euclidean introducing euclidean coordinates I can write the metric in that form and then and then I want to do the following thing I want to consider the geometry of lines so I want to consider the line say x 2 equals lambda where lambda is large and I want to consider x 2 equals minus lambda okay and and and now I want to I want to compute the curvature of these lines okay so there's a useful formula which I'm again this is part of the conformal deformation theory so there's a conformal formula I read it down in any dimension says if I take a metric g which is a conformal factor let's call it v squared times g bar g bars some other metric and if I take a hyper surface so these are these are surfaces in general I can take a hyper surface sigma in there then I can I can express the mean curvature h of sigma with respect to g in the following form so there's a simple formula v to the minus one this is h bar that is the mean the mean curvature with respect to the bar metric and then there's a term involving a derivative of v so that term is n minus one times the normal derivative so let me choose a normal vector here new so I take d d is the derivative it's just the ordinary directional derivative no metric involved new is the unit normal with respect to g bar so new v over v okay so there's a simple formula actually in general the second fundamental form of a hyper surface if you can formally change the metric it changes by a by a trace term and the trace term involves the derivative the normal derivative of a log of that that metric so this is a calculation so h here is the mean curvature but when n equals 2 it's just the it's just the geodesic curvature so it's just the the quantity we're looking at and and so I want to apply that in this setting so this is my g bar which is the Euclidean metric and I'm going to look at these two surfaces I'm going to look at a line which is pretty high up in a line which is pretty low down okay well this this surface is aligned so in particular h bar or k bar of course is zero because it's just a geodesic in the in the Euclidean metric so that's zero so the only the term that comes in here is when I compute k so so I can write k what I get is k is equal to v well v would be then mod x to the alpha so be v to the v inverse so x to the minus alpha and then n is one now I get a minus and then I get the normal derivative so the normal derivative is just so again it depends on which normal so it turns out the right normal to take is the downward one so I want to write this as minus this the yeah I want to write this as minus this and then and then divided by v again so this would be minus 2 alpha and then I get the d dx 2 that becomes a minus d dx 2 of mod x to the alpha the v is mod x to the alpha and when I do that calculation what happens well if I've done it right this should be positive if alpha is positive so so let's assume let's assume alpha is positive if alpha were positive then this would be an increasing function of of x2 and so I seem to have the sign I'm like p owed I can't get the sign right let's see right oh okay there's no minus here so I had a minus sign that took the normal to be down so that's a plus and mod x to the alpha it doesn't matter exactly you can do the calculation but mod x to the alpha is obviously an increasing function in the x of each of the coordinates right mod x is just the square root of xx1 squared plus x2 squared and so in particular this is this is obviously positive similarly if I did the calculation down here with the upward normal I would also get that it's positive it is positive and so and so what that means is that the the region the region between those two lines is a convex region so the the both the upper and the lower boundaries are convex it's a convex region and so what does it enable me to do well I can look now at geodesics if I take a point here and a point here this is on a surface so I just take two points which are far out in this strip then I can solve the geodesic equation I can minimize length but what could happen well it could happen the geodesic goes far away but that won't happen if alpha is positive because because this is convex so in other words when I take the least length geodesic it will lie in the strip and then I can let the points go to infinity and I can construct a doubly infinite geodesic so I can construct a geodesic line called a geodesic line that's a geodesic which realizes the distance between any two points okay so in particular if alpha were positive so we're trying to prove so the fear the what we're trying to prove is that a is less than one that means alpha is negative right so the theorem so so a is positive so the theorem here I erased it so the theorem is that a greater than or equal to zero implies that a is less than one and that means that alpha is negative and so this tells me that if alpha is positive the geometry would be such that I can find this slab so it's the region between two lines which is in fact convex and that would give me a geodesic line in here and it's a it's a theorem in Riemannian geometry it follows you can you can prove it from the the the second variation of arc length but it's a it's a theorem in Riemannian geometry that you cannot have a geodesic line in a manifold with non-negative curvature even Ricci curvature in in higher dimensions unless the manifold splits so this is a contradiction unless unless k is zero but but k is zero would would would mean that alpha zero so so alpha positive is not possible so this cannot occur it's called the splitting theorem says that you can't do that so so in other words if if you have a surface with non-negative curvature you won't you can of course minimize distance between any two points but if you let the points go to infinity that minimizing geodesic will run away it will go off you can never find a doubly infinite minimizing geodesic and it's right so it uses a theorem it's actually quite a bit easier in this case I don't want to belabor the point because I'm mainly focusing on the higher-dimensional case but but but again I just want to emphasize the strategy so so the the condition if we assumed that our cone angle were bigger than 2 pi that we correspond to alpha positive and the geometry at infinity for alpha positive is such that you have this convex strip and the convex strip is incompatible with k non-negative because because you if you had a convex strip like that you could construct a a minimizing geodesic which is doubly infinite and so and so that's a proof that's a proof of the two-dimensional case which doesn't use the gasp on a theorem and in fact the proof generalizes almost exactly to three dimensions except so so the main strategy is the main analog so let me write a little table here what's what's that well right so yeah so this this would be a slab in our two except the metric inside here is something we don't know right some k non-negative well the metric can't be flat because because alpha is positive right yeah but of course it is flat outside it is flat outside a compact set but these curves are not geodesics in the in their geodesics in the flat are too they're not geodesics in the cone metric in fact the geodesic curvature strictly positive and so those those are not geodesics in the in the metric we have on the other hand since since the region is convex we can construct a minimizing geodesics in that region then we could get this infinite doubly infinite geodesic and that contradicts the k non-negative condition right so so basically the idea is that the the curvature being non-negative causes geodesics to be unstable just like a if you take a the equator on a sphere the the fact that the sphere is positive curvature means that it's unstable you can push it off and decrease the length so it's the same general kind of general idea in in fact it's quite easy to prove but I just didn't want to go through the the the argument but but again I just want to emphasize the strategy so the strategy is the behavior at infinity gives us this region in which we can we can solve the geodesic problem and then this doubly infinite geodesic contradicts the assumption that k is non-negative because k non-negative tends to make geodesics unstable so you you cannot have a doubly infinite minimizing geodesic okay so that's a that's a proof of course it's you know somewhat more complicated than the Gauss-Ponnet proof but the Gauss-Ponnet proof proof has no hope of working in higher dimensions okay so I want to just close today by explaining how these same ideas generalize to three dimensions in fact this was the original proof of the three-dimensional positive mass theorem in this case and so there's a there's a basic table if we look at n equals 2 equals 3 then then we're going to generalize curves so curves in the three dimensions will be surfaces so instead of looking at geodesics in our three manifold we're going to look at surfaces and the curvature of a curve the Gauss curvature will generalize to the mean curvature this is the mean curvature of the surface and in particular k equals zero curves which are geodesics will generalize to minimal surfaces so minimal minimal surfaces are surfaces of zero mean curvature and you get them by minimizing area so if you if if so just like for points on a surface we take two points we can connect them by by a minimizing curve if we in a three manifold we take a closed curve we can span it with lots of surfaces and we could look for the surface of least area which spans it and that surface will exist and under reasonable conditions and and it will be a minimal surface that's called the plateau problem so so so the the the proofs are really an application of the plateau problem the idea the the geometric condition at infinity is very analogous in fact it's precisely the same as this one and so and so let me now so there's a there's a point yeah so there are a couple of so I want to give the proof of so there are a couple of points that I a couple of preliminary points I want to make and that is so I've as I said there's not a unique model at infinity I've taken my metric near infinity to be conformally flat so so one point is I can do a conformal deformation to make the scalar curvature everywhere positive we may assume and again I'll talk more about conformal deformation later but actually there's no loss of generality in assuming the scalar curvature is positive at all points so that just gets around this issue so you might have worried here that it might happen that this geodesic lies entirely in the flat region in the flat part and and so then I mean the splitting theorem doesn't care whether that happens or not but but in the three-dimensional case that's something we'll have to worry about so so we can get around that by assuming the scalar curvatures everywhere positive so so if we had a in order to prove the positive mass theorem it's actually sufficient to look at metrics which not only have non-negative scalar curvature but we can make it strictly positive and it falls off at infinity rapidly okay so that's the first point it's one one bullet here okay and so the proof is going to be the same so so how does it generalize well I now have three coordinates so my metric g near infinity is u to the fourth times dx1 squared plus dx2 squared dx3 squared okay and so what I'm going to do is I'm going to look at the slab x3 between x3 so I can draw this here this is say x3 equals lambda large lambda and x3 is minus lambda okay and then and and again the same reasoning if I assumed if m were negative so I have I have the m which occurs in this metric so my my conformal factor u is 1 plus m over 2r I'm trying to show m as positive so let's first show it's non-negative so if it were negative we again use the same formula to show that these are mean convex so if I choose the downward normal I get that h is positive here if I choose the upward normal here the inward normal I get h is positive so so the the region the slab is a mean convex slab and again that just follows from putting in but plugging in the form of the metric there so again h bar is zero so the v is v in this case would be u squared and the point is if m is negative then then this function m over 2 mod x is is a is an increasing function in each coordinate because m is negative so 1 over mod x would be decreasing in any coordinate but since m is negative that would be increasing and again just using the same formula you can see that if m were negative you would get a a slab which is what's called mean convex so in other words the mean curvature points in top and bottom and the consequence of that again is is if I choose a curve so I could take a circle here the circle could be x1 squared plus x2 squared equals sigma squared where sigma is large if I take a circle and say x3 equals zero so that lies out out in the out in the the the exterior region and then I don't know what the metric is in here but it doesn't matter what it is I can always minimize area so I construct a surface here we call it sigma sub sigma which is a least area surface and now because because the because of the the mean convexity of the region that surface will remain in the slab and then when I let sigma go to infinity I'll be able to trap that surface and I'll be able to produce a complete area minimizing surface in lying in the slab so so I in the same way I could produce a a geodesic line in the two-dimensional case for n equals two in three dimensions I'll be able to produce a a complete area minimizing surface so I let sigma look at the limit as sigma tends to infinity sorry I'm writing too small so I look at these surfaces and I take the limit well I take as sigma tends to infinity while I choose a sequence of these and I'll be able to get a limiting surface sigma so these these minimizing surfaces have a good compactness property so you you can bound the curvature and such things and so and so you can if I look at the surfaces in any compact set then I'll I'll get a compact family of surfaces and I can take a sub sequential limit and this sigma will be a complete least area surface okay and moreover it's again there's some there's some minimal surface analysis involved in this but as you might expect that surface is asymptotic to a plane it's bounded in height and so it's it's asymptotic to a plane so sigma itself is asymptotically flat so outside a compact set the limiting surface is a nice graph and the graph the graphing function goes to a constant and the derivatives go to zero at a at an appropriate rate so we can produce this least area surface which is asymptotically flat okay and then so what well so that that has to be that has to be then measured against the geometry so the the geometric condition is that the scalar curvature is non-negative and so and so we have to understand sort of the analog of the splitting theorem and we don't really have a splitting theorem for quite in this setting but what we have is is stability or second variation so the basic formula which we'll use we'll use again in later lectures so let me write it down carefully is if I take if I take a surface and this is an n minus one dimensional hyper surface in some n manifold and then so in here is if I take this hyper surface if I assume that there's if I assume that things are oriented so there's a there's a normal a choice of a unit normal vector then if if sigma is a stable minimal surface so what that means is that the mean curvature of sigma zero and sigma locally minimizes area in other words the second variation of area greater than or equal zero so what that means is if I do a small compactly supported perturbation of sigma so I I consider a nearby surface say they agree on the boundary I consider a family of surface this is sigma equal sigma not and I consider sigma t so I think of varying sigma then when I do a variation I require that the area not go up in other words the the sorry that the area not go down in other words it should it should go up and that's the condition that the surface is a minimizer or a local minimizer and and that that is a geometric condition so that's that's the same as the the analog of the condition that a geodesic be stable right so that's a condition involving the the curvature of the ambient manifold and so this condition is a condition on is a condition on the surface which is an eigenvalue condition it's related to an operator so let me write this down so this is a somewhat difficult calculation but the condition that the surface be stable is if I consider sigma t I can always parameterize variations by a function times the normal so I can think of sigma t is the exponential of t times some function fee times the normal vector I can think of getting the surface by moving in the normal direction by some amount and I and so if I the fee tells me how far I move positive or negative how I deform my surface and then the condition that the that if I compute the the area we call the area sigma t the condition that d second dt squared at t equals zero is non-negative turns out to be an eigenvalue condition it's a quadratic condition in fee it's an eigenvalue condition on the Jacobi operator so just as if you're familiar with geodesics this is just the same the analog of that so it's equivalent to the statement that we can write it this way the integral of we write one half over sigma and this is the scalar curvature of m minus the scalar curvature of sigma and then plus the norm of the second fundamental form let me call it k squared respect to g so k is the second fundamental form of sigma in m this times fee squared the mu is less than or equal to the integral of the norm of gradient fee squared on sigma and this is true for all fee of compact support so so a function with compact support gives me a way of gives me a deformation of the surface by moving in the normal direction at speed given by that function times the normal and then I can compute how the area changes to second order of course the first derivative the area is zero because it's minimal and so the the second derivative is a measure is telling you whether or not it's a stable critical point of area and so the condition that it is a stable critical point is the condition that this inequality holds for every function which is smooth with compact support so that's that stability and you can see the geometry in this well this is a positive term so you know it's good a good term so we want to take something like we want to take simple choices of fee like fee equals a nice function we took fee to be one for example then this would be this would be a good term and the scalar curvature is positive that's the that's the assumption we're making to begin with and so what it's telling us is is that it in in in some average sense the scalar curvature of sigma is also positive so the idea is here if we think of say fee is one then it's telling us the average or the integral of our sigma would also be positive so if it's scalar curvature is positive we take a stable hyper surface then it's also positive in some sense and that that's the basic idea of the way that's the basic connection between the between minimal hyper surfaces and scalar curvature and again it just generalizes this this this idea for for for geodesics okay so then how does the argument work well again I won't be too totally precise but so we first construct this this sigma and the sigma again is is asymptotically flat so it's asymptotic to to a plane at infinity an ordinary plane x3 equals some constant okay and so what we're going to do in the two-dimensional case is so we have this inequality for all fee of compact support right but in two dimensions so so step one step one is we can actually take fee to be one well that may sound preposterous because because I I'm requiring fee to have compact support but there's a special thing that happens for surfaces so the surfaces have area that grows quadratically so you can you can approximate the function one by compactly supported functions where this term goes to zero so that and this is called this is a standard but important trick in geometry called the logarithmic cutoff trick so so you can we can construct we can actually construct a sequence I which tends to one on compact sets with this is something special to two dimensions so this isn't going to generalize to higher dimensions d mu goes to zero so this is a two-dimensional surface and so it's because the surface has quadratic area growth so roughly speaking if you just take a cutoff function take a function that's one and then cuts off to zero then the derivative will be like one over the radius at which you cut off so so this will be like a one over r squared or one over radius squared on the other hand the area of the surface is like is like r squared is the radius squared and so it looks like it's bounded but you can fiddle with it a little bit and actually make it go to zero so so it's there's a so you can actually choose compactly supported variations which converge to to one and so in particular you can take fee to be one so in other words you can take that to be zero and so step two then is we then just write this down all these functions decay quite quickly so they're all they're all integrable functions and so we get that the integral over sigma so let me keep the positive terms on the left so I have scalar curvature of m that's positive and then I have the norm of the second fundamental form squared that's positive d mu is less than or equal to and then on the right I get the integral well actually one half r sigma is the Gauss curvature so so the scalar curvature of a surface is two times the Gauss curvature in the usual notation so this is the integral of k d mu or da on sigma so we get that inequality and now we can apply Gauss Bonet okay so the claim is this surface is so what does this surface look like what does this integral look like well this would be two pi times the order characteristic and then minus the limit of the total curvatures now because this surface is is asymptotically planar in a very strong sense the the total curvature of the boundary circle goes to 2 pi so this is 2 pi in fact minus 2 pi and now the Euler characteristic we don't know that sigma is simply connected but the Euler characteristic of any non-compact surface is at most one so this is less than or equal to one and so this is less than or equal to zero so so the total Gauss curvature because the surface is asymptotic to a plane in a very strong sense the total Gauss curvature is less than or equal to zero on the other hand these are positive terms this was assumed to be strictly positive and so that's a contradiction and so that's that's that's the proof of the the deposit that's this a proof of the positive mass there and I'm going to show next time that it's a proof that leads in a pretty natural way to a to a kind of polygonal comparison idea polyhedral comparison idea and so we'll talk about that next time so so as you can see the the positive mass theorem is very geometric it uses these ideas are all you know all sort of Riemannian geometric geometric ideas and so so I'll stop there for today I think I'm out of time