 OK, good morning, everyone. So I'm going to continue today. And let me just recall the sort of structure that I'm using for this series. So it may have seemed a little artificial yesterday because I considered three cases. In cases two and three, we're really not that different. But the reason I separated them is because the extensions to higher dimensions are quite distinct. And in fact, one of the things that I would like to do is unify them a little more. And so remember, the three situations I considered were the first one was the complete case, so asymptotically flat. And so we were interested in, again, studying manifolds with non-negative scalar curvature or Ricci scalar in physics language. And in particular, we're viewing them as a special case of the constraint equations or the constraint inequalities for initial data for the Einstein equations. Can we explain that last time? So we considered the asymptotically flat case. And I showed using a gaspanae argument and also another more interesting argument that generalized to higher dimensions, how to show that the behavior at infinity is affected in a very explicit way by the local inequality, the point-wise inequality that the scalar curvature is non-negative. So we discussed this case. So last time we did this for n equals two and three. And then I'm interested. So the focus of the lecture is going to be on local or quasi-local versions of this. And so I separated that into two cases. One was the case where I had a region omega with a boundary surface omega, which is smooth. And in the two-dimensional case, we gave a rather explicit restriction that occurs on the boundary, particularly the rotation number or the total curvature of the boundary is restricted by the condition that k is non-negative. And then the third case that I considered is similarly the case of a domain where sigma, instead of being smooth, is polyhedral or piecewise smooth. And in the case of surfaces, I looked at the case of geodesic triangles. One could consider also other polygons where the edges are geodesics. And actually, as I pointed out, you really only need the edges to have non-negative geodesic curvatures. So you need them to be locally convex in the two-dimensional case. And so we discussed these last time for n equals two. And so what I'm going to do today, so in today's lecture, I want to give extensions of these to n equals 3. So I want to do 2 and 3, what we know about it, at least, for the n equals 3 case. Excuse me. And then my plan for tomorrow, so tomorrow I plan to discuss how these ideas, some of the ideas, extend to higher dimensions, n greater than or equal to 4. And then in my last lecture on Friday, I'm going to discuss the more general, what's called the spacetime case, or the general dominant energy condition. So remember, the condition of non-negative scalar curvature is only a special case of the dominant energy condition. And some of the ideas, although not very many so far, do extend to this more general case. So I want to talk generally about that on Friday. So assuming things go as planned, that's what I expect to do. OK, and so I'm going to start with the three-dimensional version of case 3, which is the polyhedral case. And actually, it really starts with some ideas or questions and conjectures of Grumov. And I'm just going to state two of them here. So we're going to consider three-dimensional manifolds now. And we're going to consider case 3, which is the case where sigma, the boundary of omega, is, in this case, a polyhedron. OK, so when we say a polyhedron, so I'm going to really discuss, so of course there are many different polyhedrons. So a polyhedron is just the convex hull of a finite number of points in R3. And so the polyhedra have faces in R3, which are flat faces. And they have edges where two faces join. And the angle at which they join, so let me draw a polyhedron. So an example is a cube, cubicle or rectangular polyhedron. I didn't do that very well. I used to be really good at drawing cubes. Let me try that again. OK, so let's start here. And then let's put this one here. And then let's do that, dotted here. OK, so that's a cube. So it's really the convex hull of eight points in R3. And it's made up of the faces, which are flat squares or I could look at rectangular solids or parallel pipeds also. And then it has the edges, which are very important for us. So the edges are the one-dimensional parts here. And then it has the vertices, so the eight vertices, which are there. OK, and so that's a polyhedron in R3. But we can similarly look at polyhedra in a curved manifold. So this could be in a curved manifold. The difference is that each of the faces would then be a smooth surface. So it's a curved surface. And the edges would then be arcs in the manifold. And the vertices again are point. So I could look at a polyhedral region in a three manifold. And that's what we're interested in doing. And so the faces are curved. And the important thing for us is going to be, again, we're looking for an extension of the Toppenagov theorem, so the angle comparison theorem. So the important thing for us is going to be what are called the dihedral angles. And so those are, so if I take a point p in an edge, then I have two surfaces that join there. I have this one and the back surface. So the edges are just the common boundary of two surfaces. And those surfaces meet at an angle. So if you like, you can think of the angle between the normals or you could draw it here. So this angle theta in general is a function of p. And this is called the dihedral angle. OK, and if I did this in a curved manifold, then the dihedral angle would be a function along the edges. It's a function of a point along the edges. Well, it may sound a little weird to do this in a manifold, but after all, when we consider geodesic triangles in a surface, that's exactly what we're doing. We're considering a collection of three curves which are joined at the vertices. And then the triangle comparison theorem allows us to compare the angles or even the distances of points in that triangle to points in a corresponding Euclidean triangle. And so that's the spirit of what Gromov was looking for. So these are proposed by Gromov. You might ask why Gromov would care about this. And so let me explain that a little bit. And so the reason I mentioned the Toponogov theorem yesterday, so in addition to being an important tool in sort of smooth Riemannian geometry, the Toponogov theorem also allows you to define the condition of a lower bound on curvature. So sectional curvature is bigger than some constant, bigger than or equal to some constant for metric spaces. So because everything in the Toponogov theorem depends only on distances, well, once you have the distances, you can talk about angles, but you don't even have to do that. So you could look at triangles in a metric space and you could hypothesize that all triangles can be satisfied the Toponogov theorem. And that condition for a smooth manifold is equivalent to the condition that the sectional curvatures are non-negative. And so another important application of Toponogov is it allows you in some sense to complete the space of smooth manifolds. So you might say, suppose I had a sequence of smooth manifolds where the curvature is non-negative, sectional curvature or bounded below by some constant. The question is, can I take a limit of that sequence? And so if you study geometric analysis or analysis more generally, it's a very important idea to have that your space be complete. So you can take the space of smooth functions and you can complete them in lots of different norms and you get very important spaces, sobolev spaces and things like that. And so that's the same idea that is used here. So you would like to be able to say that the space of manifolds with curvature bounded from below by a constant is compact in some sense. And so you have to attach the limit. And the limit in general will not be a smooth manifold, would be a metric space. And so for curvature bounded from below, sectional curvature, that's actually a very useful idea. And it leads to what's called Alexandrov geometry. So it's a very important area. And so this is an initial attempt to try to do something like that for scalar curvature non-negative instead of sectional non-negative. And so that's obviously much harder because it's a much weaker inequality. But nonetheless, there are some interesting things you can say. So notice if I took a cube in R3, the dihedral angles are all 90 degrees. And we just have all the dihedral, let me call it theta naught, is pi over 2 for all dihedral angles. Of course, if I took a cube in a general curve manifold, then you can't say anything a priori about the dihedral angles. OK, but I'm going to state two conjectures, which actually are now theorems. They were proven recently. Well, Gromov himself sketched a proof of one of them. And the other work I'm going to describe is a recent work by a former student of mine named Chow Li. And actually, the one case is really very close to the proof of the positive mass theorem that I gave yesterday. But it's a local theorem. So let me state what Gromov claims here. So of course, you need some conditions. So we're going to compare a cube in a manifold M, a 3-manifold G. We assume Rg is greater than or equal to 0. And then we'll consider a cube. Let me call it sigma in M. And we can assume it's the boundary of a region. So we have a cube now with curved faces. And then you have to make certain hypotheses. So remember, for the topenaga theorem, the edges were either geodesics, or they had non-negative curvature. They were locally convex. And so the corresponding assumption here, so the assumptions are, assume one, is that along each, on each face, the mean curvature is non-negative. So these surfaces that form the faces have to be mean convex. So that means that a sphere is mean convex. So it means the mean curvature vector points into the region omega. So that's one assumption. And in fact, of course, the Rg non-negative is important. Let me write that down as a condition. And we're assuming certain smoothness conditions. We're assuming that the edges are smooth and the surfaces are smooth up to the boundary. So these are piecewise smooth surfaces. In order to do this, I won't write that down, but that is required in this. Then I think that's all I need to assume. And then the theorems, let me state two theorems. So theorem one. So this, again, was sketched by, there's a proof sketch, which I'm going to describe, which is actually a rather interesting proof. It's somewhat different from the one that I'm actually going to give, the one that Zhao Li gave in his paper. So the first theorem is that assuming sigma is cubicle, then under these assumptions, the statement is that if I take the maximum of the dihedral angle theta p over the edges, p in the edges, that has to be at least pi over 2. So in other words, under these conditions, you cannot have a situation where all of the dihedral angles are smaller than pi over 2. So at least some dihedral angle has to be bigger than or equal. And in fact, so in the three-dimensional case, he also conjectured that equality is true if and only if it's equal to pi over 2. That was also proven in three dimensions in this paper of Zhao Li, so equality only in the flat case. OK. And then, of course, there are many other, just like there are many polygons, there are triangles and quadrilaterals, et cetera. There are many polyhedra as well. So the other interesting case, or maybe you could argue the most interesting case for polyhedra, is the case of a tetrahedron. So a tetrahedron is a simplex, so it's a generalization of a triangle in the plane in two dimensions. And so, yeah, question? Yeah, so actually, in Gromov's paper, he showed that if there's a point on the manifold with negative scalar curvature, you can take a very small cube at the point which violates this condition. So you can find a small cube with mean convex faces and for which the all dihedral angles are smaller. So that means that this condition recovers the non-negative scalar curvature condition for smooth manifolds, at least. And I'm going to explain this in terms of the positive mass theorem also in a minute. But let me just first state, yeah, question. Actually, this is purely local. So we just assumed the metrics defined a little bit outside the, it doesn't have to be a complete space. So there's a corresponding theorem for the tetrahedron, which was only conjectured in Gromov's paper. His proof sketch didn't only work for the cube, but it was also proven by Chow Lee in a way a little more complicated, so I won't be able to go through the proof other than to say generally what's involved. But the theorem is, if sigma is tetrahedral then, again, under these conditions, the maximum theta p over p in the edges is greater than or equal to, let me call it theta 0, or maybe theta t, theta tet, which is the dihedral angle for the regular tetrahedron. It turns out you can work it out. It's the inverse cosine of 1 third. It's something around 70 degrees. So the nicest tetrahedron in R3 is the regular tetrahedron, and it has the same dihedral angles at all points on the edges, and that number is the dihedral angle. So the statement is that in this curved setting, if the scalar curvature is non-negative, and these conditions are satisfied, then you can't have a situation where all of the dihedral angles are smaller than the one for the regular tetrahedron. And again, there's inequality only if sigma is the regular tetrahedron in R3, and the region inside is also flat, so the trivial situation. OK, so actually it turns out that either of these statements imply the positive mass theorem. So these are local theorems which imply the positive mass theorem. Let me just make that as a remark. It also may be relevant to Piotz's question. Either, OK, so these are local theorems which imply the global theorem, the positive mass theorem. So you would expect that if you can prove the right sort of statements for bounded regions in your manifold, then those should be stronger statements than the asymptotic, the global positive mass theorem. And so let me explain that. So again, I'm again looking at the case of the positive mass theorem where I assume the asymptotics I assumed yesterday. So assuming the metric is conformally flat near infinity, so let me justify this remark. So let's look at the setting we looked at yesterday. So we assume we have a three-manifold. So the outside of compact set K is dypheomorphic to R3 minus a ball. And I can choose coordinates in m minus k. So there exists x1, x2, x3, where the metric gij is conformally flat. So it's u to the fourth, u of x to the fourth times delta ij in these coordinates. And so, OK, it's a special asymptotic form, but it's possible actually to show that you can approximate arbitrary asymptotics for which the mass is defined by one of this type. So this form is actually sufficient to prove the positive mass theorem with some preliminary effort. So we're just going to look at this case. And so let's look first at, say, theorem one. And so the point is, so this metric is conformally flat. So what we're going to do, so remember the basic idea of the proof yesterday was to show that if the mass were negative, so again, we'll prove this by contradiction. So assume it's negative. OK, so again, m is the leading order term of u. So it's 1 plus m over 2 mod x, and then plus terms that fall off faster. OK, and so if m were negative, what we showed yesterday is if you take a plane, I took the x3 equals constant plane, but I could take any coordinate plane. If I move it far enough out in either direction, then the region on the, so if I move it to forward the region below the plane, if I move it up, the region below is mean convex. And if I move it to the right, the region to the left is mean convex. So you can control the sign of the mass controls the convexity, or the mean, actually, even the convexity of the coordinate planes. And so what we're going to consider is so what we want to do is, assuming m is negative, we want to make a counter example to theorem one. m1 negative implies there exists a counter example. And so that will show that if we could prove theorem one, which is really a statement for very general domains, then in particular it implies the positive mass theorem. OK, so how do we get that? Well, we just take a very large cube. So we take just a cube where the boundary of the cube is out in the asymptotic region. So these are, so this is very big. So the entire boundary, so this is a region that contains the entire set k, the compact set. And it lies as far out as we like. Choose it very far out. And then the calculation we did. So let me make some observations about this cube. So first of all, because the metric is conformal, the angles of the cube are the same as the Euclidean angles. So this would be a cube in R3. And its dihedral angles are all 90 degrees, or all pi over 2. So the first point is that the dihedral angles are pi over 2. Because for a conformal metric, the angles for the two metrics are the same. Conformal change doesn't change angles. It only stretches the lengths of vectors by a common factor. So the dihedral angles are all pi over 2. And what I showed yesterday is that if m is negative, then the faces, in fact, so m negative implies h is strictly positive on the faces. So I get something which is borderline for theorem 1. I'm getting a cubical region in my space where the dihedral angles are all pi over 2. So this is exactly pi over 2. And the faces are actually strictly mean convex. And so now what you can do, well, OK. So it already violates the equality part of theorem 1. So technically we're done. On the other hand, since h is strictly positive, you can actually, let me draw a two-dimensional picture. So since h is positive, I can deform the faces a little bit, keeping them with positive h and reducing the dihedral angles. So if I looked at this picture, I could replace this by a spherical cap or something similar here. And then I could keep h positive. And I could now make the, so when I do this, the dihedral angles will be smaller than pi over 2. So it's a little hard to see. That's a two-dimensional picture. So if I pushed the faces in a little bit, then I'll reduce the dihedral angles along the edges. So you could either use the equality case of theorem 1, which is a little harder than the first one, or you could argue that because h is strictly positive, I can deform the faces a little bit and make all of the dihedral angles strictly smaller. So that's a little exercise. You could do it, for example, by replacing the plane by a very large sphere centered far out which approximates the plane in this region that we're considering. So that would then still have positive h. And if you do that, you'll reduce the dihedral angles. So you'll be pushing in this face and pushing in that face. And so that will cause the dihedral angle to be smaller. So we could either use the equality case or we could deform slightly to make the maximum dihedral angle strictly smaller. However, too, so that's the argument. So in particular, the statement does imply the positive mass theorem, so that already makes it quite interesting. And similarly, it's true for the tetrahedron. If I took a regular tetrahedron, I won't try to draw it, but if I took a regular tetrahedron which lies out in the asymptotic region, then its dihedral angles would be exactly the theta-tet, the angle for a tetrahedron. And again, its faces would be pieces of planes which are far from the origin. So they would also be strictly mean convex and then I could deform in to reduce the dihedral angles. So either one of these theorems implies the version of the positive mass theorem that I proved yesterday. Okay, and so as I said, this theorem was sketched in Gromov's paper except not the equality case and theorem two was conjectured in the paper. And both of the theorems together with equality were proven by Chow Lee. And I'm just gonna describe today the proof of the cubicle case and I'll just make some comments on theorem two. It's a little complicated, so a little more complicated. So let me first give the Gromov sketch because it's pretty interesting. I mean, it's certainly very convincing and it probably can be made rigorous, but it's not so easy to do. So Gromov had a sketch of the first part of theorem one for theorem one and you'll see that the argument really works only in this cubicle setting. So the idea is the following. And so this uses a little information that I haven't actually stated, but I hope I can just make it plausible by drawing a few pictures. So let's start with our cubicle region, cubicle polyhedron. And then what we're gonna do, so remember the hypotheses are that H is positive, so we're gonna prove by contradiction. So we assume all of the dihedral angles are strictly smaller, so the maximum over P in the edges is strictly smaller than pi over two. So those angles are less than pi over two. And then the idea is the following. So the idea is to construct a metric on the three torus with non-negative scalar curvature, and which is not flat. So it uses a theorem, which is the fact that the three-dimensional torus does not carry a metric with positive scalar curvature, and the only one with non-negative scalar curvature is flat. Okay, so he's gonna compactify it. And so the idea for compactifying it is to double this. So you can take this, the cube, and we could double it in this direction to begin with. So we take another copy here, and then we double it. So we take another copy, and we just reflect across the face. Okay, and so then we get a metric. Well, when we do the doubling, we get a metric which is piecewise smooth across that face. It's not, however, so it's continuous, but it's not C1 across the face. So we get a, so it's not a smooth metric. But nonetheless, the advantage of doing that is now this side here is isometric to this side. When I do one doubling, I create a situation where this side is isometric to that side. And so in particular, I could identify the two and get a metric on S1 cross a disk by doing that. Okay, and so, and then I can repeat the argument. I can also double down in the direction here. So I take these two and I double again. It's a little hard to draw all these things. So I double it again, and I get a little sketchy about it. So this is a doubling, and I get four. I then get four copies of this same fundamental domain. And when I do that second doubling, I'm making the upper face isometric to the lower face. So I could identify those faces. And then I repeat it in the, so I do that in the board. I repeat it in the third direction. So let me not try to draw it, but I double it in all three directions. And then I produce a new bigger cubicle polyhedron where now the big one, so I can draw a big one here, which consists of eight copies of the original one. And this big one has the property that this side is isometric to its opposite side. This side is isometric to its opposite side. And the other one, top, bottom, front, and back. It's a little hard to draw, but. So each side is isometric to its opposite side, so it defines a metric on the torus. So I can identify those sides, and I get a metric on T3, the three-dimensional torus. Okay, now, of course, the problem is, the metric is highly non-smooth. It has a lot of singularities. And so we haven't used any of the hypotheses so far. So what you have to argue is that this three-torus metric, so this is a metric on T3, you have to argue that it has non-negative scalar curvature in some weak sets. Or, so in particular, you might hope that it could be approximated by a metric of non-negative scalar curvature. And it's possible that it can be, but it's rather tricky to do that. So let me explain. So the mean convexity condition is the condition that when I do a reflection across the face, the resulting manifold has non-negative scalar curvature across that face. And so again, that's been studied quite a lot, piecewise smooth metrics. The question of whether the scalar curvature is non-negative or not depends on the jump of the mean curvature across the boundary. So we can draw a nice and easy picture. So let's do it, let's look at n equals two. And let's say we take a piece of a spherical cap which is convex, so it's less than a hemisphere. Now if I take that and I double it, then I get this piecewise smooth surface, it's S3 with a non-smooth metric. Now, because this was convex, the new doubled surface actually has positive K. So the sign goes the right way. So I could smooth this out and I would get, so I get some positive distributional content for K, the Gauss curvature across that boundary. Whereas if I took a concave, I took a bigger piece of the sphere. So I say I take a piece like that, which goes more than halfway around. And then if I double it, this has K negative. If I smooth that out, I would get saddle points. I'd get a point where K is negative. You can see it from Gauss-Bonnet, too. The contribution of the total curvature from the smooth part is strictly bigger than four pi. And so that means the contribution from the singular part must be negative, right? So the total curvature of any smooth metric is four pi. So the condition on the faces tells us that at the singularities at which two faces come together where the face is doubled, that those points have non-negative curvature. And then you have to worry about the other singularities, where there are also singularities along the edges. So if I look at an edge, then at an edge, there are four faces that come together. So at an edge, I have a little bit hard to draw, but there will be four faces. And the assumption that the maximum dihedral angle is less than pi over two tells us that if we go around the edge, then the total angle is smaller than two pi. So in other words, along an edge, the metric looks like a product. So we could just cut it in orthogonal to the edge. And then that surface that we get in the orthogonal plane, orthogonal surface, would be a cone surface. And the cone surface would have cone angle less than two pi. Now cone surfaces with cone angle less than two pi are positively curved. So that's a cone that looks like that. And so I can smooth it out to make the curvature positive. Okay, and so this assumption tells us that the scalar curvature at the points, well, okay. So that's the curvature in the transverse direction. But because it's roughly a product, that will dominate the scalar curvature at these points. You should be able to smooth it out so you get enough Gauss curvature in the transverse plane to dominate any negative stuff that comes in along the edge, because the metric is varying smoothly in the edge direction. Okay, so you can see that this is would be, I mean proving this would be kind of a nightmare to write down, but on the other hand, it certainly looks very plausible and likely true. Then there's another singularity course which are at the vertices. But now at the vertices we can say, well, the vertices are co-dimension three, so they shouldn't make any difference. And so on the other hand, again, if you wanted to smooth the metric, you have to worry about those points. So it's a very interesting argument. And so the way the argument works is you construct, so assuming you had a violation to the first part of theorem one, you would be able to construct a metric on T three, a non-trivial metric with non-negative scalar curvature. So that's the proof that's sketched in Grumhoff's paper. And actually, Chow Lee initially tried to make that rigorous. In fact, he wrote at least one paper with Montelides trying to argue that you can smooth metrics like that, but they never really were able to handle the complexities of this together with the vertices and that sort of thing. So that was the original sketch. On the other hand, it's a very compelling picture. And you believe that if you work in the subject, you believe that it should work. And so it's very strong evidence anyway for theorem one, at least for the first part. And so what I'm gonna do in the remaining, well, probably in the next 15 or 20 minutes is give a rigorous proof of theorem one, okay? And I'll say a little bit about what's different for the tetrahedral case. So notice this reflection argument, the idea of compactifying it wouldn't work in the tetrahedral case because it simply isn't correct combinatorially. So it's an argument which is specific to the cube, which suggests it's also not quite the right argument. Anyway, and so I'm gonna give a rigorous proof, so maybe I can erase the sketchy proof. Okay, so actually, the rigorous proof is really very close to the proof I described yesterday. It uses, so the proof again, we assume for contradiction, and I'm not gonna describe the borderline case, but I'll just do the first part. So we assume for a sake of contradiction that we have a cube, so sigma's cubicle, and we assume the maximum theta of P in the edges is less than pi over two. Okay, then the idea is very similar to what we did to prove the positive mass theorem yesterday, so we have a picture, so again, these pictures are a little hard to draw, but we have this picture where we have, so let's think of the top face and the opposite one, the one on the bottom, and then we have the other faces. So let's think of this, so let's think of the top and the bottom faces, and now we're assuming that the dihedral angles are smaller than pi over two, so let me draw a two-dimensional picture here. So the picture is something like this, so the pi over two. So we have the upper face and the lower face and the dihedral angles along the upper and the lower faces are strictly smaller than pi over two. Okay, so what we could do, let's say if we were in the two-dimensional case, we could construct an interesting geodesic in this picture, namely we could look at the shortest distance from this side to this side, so we look at curves which separate the two faces, the top and the top from the bottom, and we could minimize the length. Because the dihedral angle is less than pi over two, that curve is not gonna hit the top or the bottom or the vertices, cause if I push down on the top, I reduce the length, if I push up from the bottom, I also reduce the length, and so what we're gonna get is this special geodesic in here. So this geodesic is what would be called a free boundary geodesic, so it minimizes the distance between these two arcs and so it has the property that it's a geodesic which is orthogonal at the boundary, and it's stable in the sense that it's the shortest geodesic which does that, which separates the top from the bottom. Okay, and so that's in two dimensions, but in three dimensions, I can do the same thing. The only difference is that this geodesic is replaced by a surface, so I can construct a free boundary minimal surface. So I draw it in here, let me call it sigma. So it's the least area surface which separates the top from the bottom. Okay, so the dihedral angle condition, the dihedral angle's less than pi over two gives me existence of this very special surface. It's a surface which has least area, it's called free boundary because the boundary isn't specified, it's allowed to move along the side, and you would worry in general that the surface might escape to the top or the bottom, but the mean convexity and the dihedral angle condition guarantee that doesn't happen. And so just like in the geodesic case, so again we're assuming these are geodesically convex, and so the curve also is not gonna escape in the interior of the top or the bottom. Okay, and so we get this very special surface, and again as in, so this is the analog of our asymptotically flat surface that we considered yesterday, and then what do we do? Well we looked at the second variation condition, so the fact that the surface locally minimizes area, and we then chose a special variation and used the Gauss-Bonaille theorem. So we're gonna do exactly the same thing. So we're gonna take this surface sigma, and then, well we have to work a little bit, we have to think what the second variation formula is. It's a little more complicated because now we allow variations which also move the boundary. So if we take variations that fix the boundary, then it's the same as the one we wrote down yesterday, but what I claim is that there's a boundary contribution in the second variation, so this has to be worked out, but the claim is that if I take my sigma and I vary it, so I think of the normal vector, new, and actually on the boundary because the surface is orthogonal on the boundary, the normal vector is tangent to the boundary, so I'm again allowed to think at least infinitesimally of taking a function times the normal and varying the surface, and so I do a variation which is phi times the normal vector is an infinitesimal variation, I push it in this direction, I push the surface and then the condition that the second variation of areas non-negative is the following condition, so this is d second area sigma t, that turns out to be the integral over sigma of the gradient of phi squared, same thing I had yesterday, gradient of phi squared and then minus one half scalar curvature of m minus scalar curvature of sigma plus the square of the second fundamental form, and let me now give some names here, let me call the second fundamental form here h, so that's the second fundamental form of sigma and m, the reason I worry about notation is because it will also involve the second fundamental form of the boundary, so I'm gonna call the second fundamental form here k, so that's the second fundamental form of the face in m, okay, and so the formula here is this, this would be norm of h squared sigma times phi squared, so that's the same thing I wrote down last time, and this is integrated with respect to the area measure, and then I get another term which is minus the integral over the boundary curve of the second fundamental form of the face applied in the normal direction, new, new, so when I do, so when I'm varying in the normal direction, so the shape of the boundary is gonna matter, and this is again, this is times phi squared and this is integrated with respect to arc length, so it's a calculation that that's the second fundamental form, so just to explain this boundary term, suppose I were doing it for an equatorial disc, so I consider my boundary surface to be a two-sphere and my sigma to be this disc, then of course sigma is just flat, so if I do a vertical variation, so I think of pushing it upward infinitesimally, then in the interior I don't change anything, but on the boundary in order to do that I have to squeeze the surface, and so this boundary term, so in this case the second fundamental form of the sphere is positive, so that gives a negative contribution to the second variation, so there's that boundary term. Okay, and so now what we can do is we can fiddle with this a little bit, so we can write, so we have to use the mean convexity of the boundary, so what we can do, so at points on the boundary we have the normal vector, which is normal to the surface and tangent to the boundary surface, this is new, and we also have the tangent vector, let me just call it T. T is the unit tangent to the boundary circle, the boundary of sigma, and so if I take, so that's an orthonormal basis, so I can write the mean curvature as the sum of the second fundamental form in the new direction plus the T direction, so I can replace this part, so right, so the condition is this is greater than or equal to zero, that's stability, and so I can rewrite this as the mean curvature H at a point on the face and then minus the second fundamental form in the T direction, like that, and then H is non-negative, so that's good, so that's a good term, I can throw it away, gives me a negative contribution on the left, and so I can replace this term by plus K of TT. Now, the other observation is that K of TT is the geodesic curvature of the boundary, so the claim is that K of TT is equal to the geodesic curvature K of boundary sigma in sigma. So why is that? Well, the reason is because the, what is the geodesic curvature? Well, it's the second derivative, it's the derivative of, it's the second derivative along the curve, so it's del TT, and then the normal component of that, so the point is that the, that is the normal in the surface, and so the point is that the surface is normal to the boundary, so if I take the inward normal here, let me call it eta, the co-normal, then that is exactly the normal to the, that's minus, I'm choosing the inward normal here actually, so that's the normal to the boundary phase, okay, and so this K TT is nothing more than the geodesic curvature, because again the co-normal or the normal to the surface is exactly normal to the boundary phase, and so the normal curvature is exactly the geodesic curvature of the boundary, and so in particular I get this inequality so I can throw that term away and I get that again this term also is positive, non-negative, this is non-negative, so I throw all the terms away, what do I end up with, I'm gonna take phi to be one, like I did last time, I take phi to be one, and then what's left is, so minus a half, so a half of our sigma is K is the gauss curvature, and so what I get is the integral over sigma, this becomes a plus K, K sigma, and then I get from here the integral of, so I get two minuses as a plus, the integral of K dS on the boundary of sigma, and I get that this whole thing is greater than or equal to zero, okay, and now I use the Gauss-Bonnet theorem, so the Gauss-Bonnet theorem would tell me this term is two pi times the Euler characteristic of sigma, and then minus the integral over boundary sigma of K, almost, right, except I also have the angles, right, so the point is, so there's an issue here, so it's important that the surface be C1 at the corner, so that I want the angle, the corners, the interior angles of sigma on the boundary to be exactly the dihedral angles of the, at the corresponding point of the polyhedron, so I get a sum of dihedral angles coming out here, and so this is two pi, and then let's see, what do I get, I get minus the sum of the exterior angle, so that's the sum over i goes from one to four of pi minus theta i, so remember the contribution to the geodesic curvature at the, at a corner is the exterior angle, so that's pi minus the interior angle. Okay, so putting that all together, let me erase this one, what do I get? Well, I get the two pi times the Euler characteristic of sigma, and now this term cancels that term conveniently, and so what's left is plus, I get two minuses as a plus, so I get four pi from there, and then I get, sorry, I get minus four pi, sorry. So I'm looking at this term, I get minus four pi, and then plus the sum of the, it goes from one to four, and I get that this is greater than or equal to zero. And let's see, so let me write this as, let me write it as two pi times the Euler characteristic of sigma minus one, and then minus two pi plus the sum of theta i, i goes from one to four. Okay, and then remember, each of the theta i's is smaller than pi over two, as we've assumed, right? Our contradiction's coming from the fact that the dihedral angles are too small, so the sum is not negative, but this term here, minus two pi plus that is less than zero, less than or equal to zero, and moreover, the Euler characteristic of any surface with boundary is less than or equal to one, so I don't know my surface as a disk, if it were a disk, then this would be zero, but it could have a handle, but in any case, the Euler characteristic only becomes more negative, so this term also is less than or equal to zero, and that's a contradiction, so that's the proof. So the proof uses the dihedral angle bound, the fact that dihedral angles are less than pi over two in two ways, one is to construct this separating, minimizing surface, and the other is we use the dihedral angles on the other faces in the Gauss-Penet argument, okay? And so it's really very much an analog to what we did yesterday for complete asymptotically flat surfaces, but somewhat more intricate because we have to handle the boundary terms, but everything works out really beautifully, and it gives a proof which is rigorous, so the rigorous proof is given in Charlie's paper, so that's the proof, and then the borderline case is somewhat harder, I'm not gonna go through that here, but so that's theorem one, and so theorem two, which I erased, uses a similar sort of idea, except you can't use, so if you think of the regular, so it's again modeled on the regular tetrahedron, and of course you don't have, so of course in a cube you can take the planes that are between the top and the bottom, and those will be surfaces like the one we considered there, these orthogonal, orthogonal at the boundary, in the tetrahedron you don't have any natural planes like that, so what you have to do for the tetrahedral case, so let me just see if I can draw a tetrahedron, yeah, so that's the convex hull of four points in R3, so in the tetrahedral case, if we took a regular tetrahedron we would have the natural planes that are parallel to the base, and those separate the opposite vertex from the base plane, and there are minimal surfaces, of course they're planes, and they meet at a constant angle, so there's an angle here, there's what's called a contact angle, which in this case is not 90 degrees, but it's an angle that's explicitly computable, so you can compute the angle, and so what Chal Lee does to prove theorem two is he constructs minimizing surfaces with the contact angle boundary condition, and he shows again that there's a corresponding second variation in the terms again work out beautifully so that you end up proving it, so it's a little harder, and in particular the theory of minimal surfaces with contact angles is less well worked out in terms of the regularity at the corners, and so he spent quite a bit of time in the paper just justifying the fact that the surface you get is actually C1 up to the corner, so you can make the argument, so there's some sort of analysis involved, which in principle could be in the literature, but it doesn't seem to have all been worked out, so that's a sketch of theorems one and two, and see these are local theorems that imply the positive mass there, I mean they give you information about explicit geometric information about regions in your space time. Okay, and so I wanna close in the last 15 minutes by discussing the analog or what we know of the analog to case two, which is the case of a region with smooth boundary, and so this goes by the name, so this idea comes not from geometry, but from physics, and it's called the Brown-York mass, and so let me tell you a little bit about the Brown-York mass, so this was defined in a paper in 1992 of Brown in York, and actually there was 10 years later, there was the proof of positivity of the Brown-York mass, which is done by Xi and Tam, it's a very beautiful paper in 2002 in JDG, but the situation is as follows, so we consider again our three manifold M, and we consider a region, and then as always our G is greater than or equal to zero, and then in order for the Brown-York mass to be defined, so this is analogous, so this is case two, we're gonna assume smooth boundary, in order to define the Brown-York mass, you need a comparison surface in our three, and so it's like what we did for in the two-dimensional case, the just using gaspanet, we interpreted the two pi as the total curvature of a comparison circle in our three, and so the problem is it's very easy to isometrically embed a curve into our two, but it's a lot harder to isometrically embed a surface into our three, so if you take a surface with a metric and you ask does that metric arise from an embedding into our three, that's an extremely difficult problem, and of course it's not always true, certainly not true globally, even the local problem is very hard. On the other hand, there is one great theorem where you have existence and uniqueness, and that's the case when the boundary of omega has positive Gauss curvature, so the boundary of omega is a smooth surface, and so it has an induced metric from M, and so we assume the Gauss curvature K of sigma is positive. So in this case, there's a theorem, it's called the violin bedding theorem, which asserts that we can isometrically uniquely embed this surface into our three, so we can find a phi from sigma into our three, and let me call the image sigma naught. Phi of sigma is just a convex body, a convex sphere. Okay, so this always embeds, this is our sigma naught, convex sphere, and we had here, this is an R three, and we had phi, which goes from whatever our sigma, and this is our omega, and so in this case, we can isometrically embed into our three, and the embedding is unique up to a rotation, up to rotations of our three, so this is called, it's a hard theorem, but it's known since about 1950, it's called the violin bedding theorem. Okay, and so using this theorem, it gives us a flat model to compare with, and so the Brown York mass is defined in the following way, so what we can do is the following, so again, it involves mean curvatures, so the connection between mean curvature and scalar curvature is always there in this theory, and this one unfortunately also requires a condition on the Gauss curvature, which is probably not natural, I mean one would like to remove that somehow, but the point is if I take a point P on the boundary surface here, then I have, let me call H of P the mean curvature of the surface in M, and then I also have the image point, which is somewhere over here, phi of P, and then at this point I have the mean curvature H naught, which is the mean curvature of the surface in R3, of the convex surface, and so I can use phi and pull these back, and I can just think of eight, so I can just think of H naught of P and H of P, and now the two surfaces are isometric, so the pullback is an isometry, so the phi is an isometry, and so the area form is the same on both surfaces, they're identified via phi, and so the Brown York mass of omega or really it's just a function of sigma, the geometry of sigma, that is it's a function of the induced metric G and the second fundamental form, or really the mean curvature of sigma, and what it is is one over eight pi, the integral over sigma of H zero minus H, the mu with respect to the area measure, so that's called the Brown York mass, and they motivated this from physical considerations, and I guess they conjectured that it would be positive assuming that the scalar curvature inside is non-negative. Okay, so notice this, I wrote down an expression yesterday which is exactly like this except for the eight pi for, and this was an integral over the boundary curve for domains, simply connected domains in a surface with non-negative curvature, so this is sort of the exact analog where the Giadesse curvature is replaced by mean curvature, so this is called the Brown York mass, and so you can fiddle with it, but actually I haven't given any exercises, I feel very guilty, I always make my students do exercises, but here's an exercise for you, so if you consider the Schwarzschild metric G, M, and we can write it, well actually it might be easier to compute in the other form, but let's say we write it in the form I've been writing it in times the Euclidean metric, then when you do that the Schwarzschild horizon is at M over two, mod X equals M over two is the horizon of the Schwarzschild, and then you can verify that, so here's M over two, if you compute, so if we consider the spheres that is the mod X equals constant spheres, then those just have constant curvature metrics, and so we can easily compute the H naught, the mean curvature, we can embed them as spheres of a particular radius into R three, so you can explicitly compute the Brown-Yorke mass in this case, and what happens is at M over two, so this is M by, of say R, mod X equals R, this is the R axis, then what happens is at the horizon this is two M, so it's a little weird, the Brown-Yorke mass on the horizon of Schwarzschild is actually twice the ADM mass, and what happens is it's decreasing, so that also might be a little weird, but that's what happens, it's decreasing and it decreases to M at infinity, so what happens is that M of R is decreasing in R and the limit as R tends to infinity of M of R is equal to M, the ADM mass, so that's an explicit calculation, and I won't do it, it's quite straightforward, and so there's, as I say, this is the definition of mass, so the one over eight pi is just there so that this is true, so that for a very large sphere you're approximating the ADM mass, and similarly if I take a general asymptotically flat manifold and I take a very large sphere in it, at least if I assume reasonable asymptotics, it will be true that the Brown-Yorke mass will approach the ADM mass when it is large, that's not completely trivial because it's not so easy to calculate this, I mean you can't explicitly calculate the isometric embedding, except for a round sphere, but nonetheless it is true, okay, and so there's a basic theorem here, which you can think of as being the analog of theorems one and two for the smooth, for the case of smooth boundary, and so the theorem is just due to Xi and Tam 2002 is that assuming, again, scalar curvature is non-negative, the hypotheses we have, and also H is positive, so we assume K sigma is positive, we assume H is positive, that's also necessary for this, so we have positive mean curvature and positive Gauss curvature, then the theorem is that the Brown-Yorke mass of sigma is greater than or equal to zero, and equals zero only if omega is flat, in other words the, and the region, so omega would be flat and in fact you can argue that omega is just isometric to the region inside the convex surface in R3, so that's the positivity of the Brown-Yorke mass, and so I gave the argument there for the angle comparison theorem, so you might expect that there would be a similar argument for the Xi-Tam theorem, however the argument's totally different, so actually it's a real challenge and I think it might help to sort of unify these ideas to ask the question, is there an intrinsic proof that is which doesn't involve extension, so the idea of Xi and Tam, and I don't have time to go through it, but it's a very pretty paper, actually it uses an earlier ideas of Bartnick, Robert Bartnick, so the idea is to, if you like, is motivated by the Schwarzschild picture, so if I took the situation where I had actually the sigma was a round sphere and the mean curvature was constant, then I could take an exterior solution, which is actually a Schwarzschild, and I would reduce to this picture, and so they do that in general, so the idea is to take the region omega, this is omega in our three-manifold and to extend it, so the idea is to extend in a, well, by solving the constraint equation, zero scalar curvature, so there exists an extension, which is asymptotically flat, and so that, of course, it's only gonna be piecewise smooth, but it has the property that across the boundary, sigma, the mean curvature from the inside agrees with the mean curvature from the outside, so the mean curvature h out, h in are the same, so you find an extension to an asymptotically flat space and it's done in using this idea of Bartnick called quasi-spherical affiliation, so there's an interesting way of solving the constraint equation, in particular our scalar curvature, zero, and this is asymptotically flat and they get a picture like this, so there's a fulliation here, the natural way, the f sigma, sigma r, so this is sigma zero, r equals zero, and they do it in such a way that the Brown-Yorke mass of sigma r is decreasing in r, and then because the scalar curvature is non-negative and because the mean curvature doesn't jump across the boundary, so they have to extend the positive mass theorem allowing a slight singularity, so the metric is only piecewise smooth, but the mean curvatures agree across the boundary and so in particular this is decreasing in r and the limit as r tends to infinity is the ADM mass, okay and then using the positive mass theorem so they have to justify the positive mass theorem with this slight singularity, positive mass theorem tells us this is greater than or equal to zero, and then because it's decreasing this will be less than or equal to the Brown-Yorke mass of sigma, so that's the idea of the argument, so it's a very special extension which is fully aided by spheres for which the Brown-Yorke mass decreases to the ADM mass and then the positive mass theorem tells you the Brown-Yorke mass is positive, so it's a very interesting argument. On the other hand, it has absolutely nothing to do with this argument other than the fact that it uses the positive mass theorem at the very end, and so I intended to say a little more about the relations to these, but maybe I'll spend maybe 15 minutes tomorrow talking about how possibly these could be unified in some way, so as it stands there, so I pointed out yesterday that the case two, that is the case of a smooth boundary, implies the triangle comparison theorem just by approximating the boundary, well in this case you have this annoying restriction that the Gauss curvature is positive, and so that occurs for this case for the Brown-Yorke mass, it doesn't at all occur in the Gromov setting, so in this polygonal setting, all we have is assumptions on the mean curvature, which I think is the natural assumption to make, and so maybe I'll spend 15 minutes tomorrow just comparing these two, and I have a sort of wild conjecture, I wouldn't call it a conjecture, a wild speculation as to how these could be conceivably combined, but I have no idea how to prove it, so it's one of those things that is very appealing because it gives a very clean statement which kind of includes both, but it seems really hard to prove, so I'll finish this up tomorrow and then I'll talk about the higher dimensional case and how some of these ideas extend. Okay, so thank you.