 Thank you. So I've redrawn the pictures from last time. We'll come back to them later. But first, I wanted to say a couple of things. So I must apologize for an error in the second lecture. In the second lecture, I put down, which was called a number 3i. That's the scalar curvature of the horizon would be constant as a consequence of the Gauss-Goudatz equations. This was part of exercise 7. And I got carried away. This is true for the photon surfaces I'm going to talk about today. But it's not true for the black hole horizon that this follows from Gauss-Goudatz. So it's, of course, true after we proved any of these theorems that we proved in the last two classes, but it's not true a priori. Not so easy to show a priori. So I hope no one wasted enormous amounts of time on this exercise. OK. This being said, let's go back to very, very briefly reviewing the long proof that we did last class by Banting and Masudo Lalam of the static black hole uniqueness in three dimensions because we'll generalize it to high dimensions today. So what we did is we took our static vacuum system and we doubled it through its horizon. Then we made a conformal transformation and inserted a point and infinity and convinced ourselves that all the assumptions of a weak regularity version of the rigidity case of the positive mass theorem were satisfied, which told us that everything was actually R3 in this case. And then I very briefly mentioned to you a method using the cotton York tensor developed by Robinson and his collaborators that allowed us to recover Schwarzschild. And somebody asked me later, how did we make potentially several black holes disappear and just remain one black hole to remain? And this was actually hidden in the recovery procedure with the cotton York tensor implicitly, OK? Somebody else asked me after the lecture, well, but this argument with the cotton York tensor even requires third derivatives of the metric. And our metric regularity along these doubling horizons and also at this point was only such that we had C11. And so why does the Eisenhardt theorem and all these things even apply in the recovery with the cotton York tensor? The answer is very simple. We're only interested in showing that this original manifold was Schwarzschild. So we just forget everything we did. Now we know this upper part is conformally flat. And we can perform the whole argument in this original manifold. And this will be a recurring theme in the different arguments we'll give today in higher dimensions, where the only difference from the procedure we studied will be in this recovery part anyway. So every time that we arrived at this point where we know we're globally conformally flat, maybe we'll do topological arguments on this level. But then whenever we do any smooth arguments, we'll do them in the original manifold so smoothness is not an issue. Then somebody said that in the Bartnick paper there was a condition on the Ritchie tensor. No, there's not. I read up. So the Bartnick positive mass theorem is a higher dimensional version of Witton spinproof positive mass theorem. And it is under the same conditions as the original theorem. So no extra conditions there, so we're good to go there. And it's actually, I read it up. It's theorem 6.3 if you want to check it up. And one last thing I wanted to mention because it came up in a question yesterday after the lecture is, in fact, yes, the Banting Masoudou-Lalam proof is strictly speaking stronger than the Israel proof. So the Banting Masoudou-Lalam theorem implies the Israel theorem. And I'll write that down. Theorem 2 implies theorem 1. So if you know that we potentially start with multiple black holes and we show there is only one and it's Schwarzschild, that implies we start with one black hole and we show it's Schwarzschild. And the technical condition in this theorem that the gradient of the laps doesn't vanish, it doesn't occur here. So even from that perspective, it's stronger. Okay, so today we're going to talk about generalizations. In generalizations, of course, you could think in many directions. One could be to include matter fields. And this has been done for electrovacuum and for many other fields. I'm not going to say much about them because I don't know much about them. Of course, I know the electrovacuum proof and, in fact, there has been an electrovacuum proof in the style of Israel by Israel and collaborators and also by Masoud al-Alam with the Banting Masoud al-Alam approach, which also works in the electrostatic realm. And then there have been dilaton fields and in high dimensions even other models like brains and whatever to which this has been generalized. And if you're interested in this, see Heusler's book that I gave the reference in the first class and for newer developments, Robinson's overview article. I also gave that reference in the first class. So they list, Robinson even has an update on his website of his article having a small section on higher dimensions with matter. Okay, so that, we're not going to discuss that very much. Another direction of generalization, of course, one could do and we will talk about is higher dimensions. And as I just said, of course, higher dimensions could include matter as well, but we again are going to ignore this today. So we're going to only do it here, only in vacuum and see all those for non-vacuum. One paper I want to mention though is an upcoming paper by my student, Sofia Jans on archive shortly. This is a Bunting Masoodu Lalam style proof in an electro vacuum in all dimensions, starting at three. So we'll talk about that in a second. Other generalizations we'll talk about today is to photon surfaces and photon spheres. And I'll say it more about this later. And of course, and there's been lingering at the background of my lecture a lot. You could talk about non-static situations and most interesting at the moment, I guess, is stationary space times. And there's lots of open questions there. It's known in real analytic situations, but the full case is open. Okay, but I'm not an expert on that, so I'm not going to say anything on that stationary, except that there is a living review article on this topic by Piotr Kuchel and other authors. If you don't know, the Living Reviews is an online journal which you can access through the library of your institution, which is a review journal that posts updates on a regular basis, so they're always very up to date. So they change authors also a long time, so I'm not putting authors. But it's very good and very easily, well-readable living review article on stationary black hole uniqueness there. But before we go to higher dimensions or to photo on surfaces, let's go and look at other proofs of the three-dimensional case. And of course, if you're a physicist, you may wonder why it's even interesting to study different types of proofs of the same theorem if you already know the result by bunting in my pseudo-lalarmina as much generality as we like. And the answer is if we want to go and generalize through different scenarios, of course the more proofs we have, the better it is, because maybe other proofs will generalize better. So, and also we want to do a little bit of justice to the people who did the work. So there's one proof by Robinson from 1975 and the article again is in the book and in Robinson's review article. And it generalizes the Millard-Tomhagen, Robinson, Cyford paper I already cited was 1972. And it's for connected horizon. So a single horizon in our language. And it is not, and this is important, assuming that the differential of n is nowhere vanishing, okay? But still it's assuming that there's only one black hole horizon. And so if you remember Israel's proof, which was then generalized by Millard-Tomhagen, et cetera, to lower regularity, what we did is we wrote down the Einstein equations in a constraint and evolution equation system and integrated some of them on the level sets and in n direction from the horizon to infinity. And the idea in the Robinson paper is exactly the same, but instead of integrating inequalities as we did in the Israel case for densities, we will integrate a magical identity that I won't write down because it's very long. So he finds a magical identity, coordinate independent between new of n and the cotton York tensor, which is completely geometric and contains the cotton York tensor as norm c squared. And of course, this is non-negative, so this gives us a magical inequality. And then he integrates this inequality again over the surfaces and out to infinity. And again, like in the Israel proof, he gets inequalities between m and the area of the horizon and the normal derivative of n at the horizon. These are algebraic, algebraic inequalities. The mass, the normal derivative of the left function at the horizon and the area of the horizon. And again, these together with the information that we have that m is one on four pi the normal derivative at the horizon times the area. These are sharp, so they are saturated. So all the inequalities were identities. In particular, the cotton tensor had to be zero everywhere and we're back in the situation where Masoud al-Alam took us and bunting. So all saturated. And unfortunately, no one has found a way to write a magical identity involving the vial tensor instead of the cotton tensor in high dimensions. So no one knows how to generalize this proof to high dimensions either. But it's conceivable that the magical identity can be used for other things as well. So I still think it's worthwhile studying this proof. There's very recent work by Agostiniani and Masieri. Of course, this is Lorenzo who is here from 2017. And because this is after the review, I have to give the reference on the geometry, the level sets of bounded static potentials. And this appeared in CMP in volume 335, issue one. The page is 261 through 301. And their approach is in some sense reminding me of the Israel approach and of this Robinson idea. Only they don't do a magical identity and then integrate it out on the surfaces and all the way to infinity. Instead, they write down magical functionals on the surfaces, I'll write them down in a second, and show that they are non-increasing from the horizon to infinity. Which also then gives them inequalities between all these numbers, which also can be seen to be saturated. And then their functionals are such that if they're not non-increasing but constant, they are characterizing Schwarzschild. So it's a similar idea, but instead of integrating something from the horizon all the way up to infinity, you construct quantities where you do integrate over the level sets, but then you look at them and you discover that they're non-increasing. So, study integrals for numbers P bigger than three. And these integrals are called UP, and they go from zero one to R, and they go like this. You take a number T and you assign to it two M over one minus T squared to the power two P minus one in three dimensions. And you integrate over the level set where N is T and what you integrate, maybe I should write it consistently like this, with the area measure that we've always looked at, you integrate this, which is the normal derivative also, if you want, to the power P. So for P equals one, this is just one, and this is just the integral over the normal derivative that we've studied that gave us this identity down here by the divergence theorem. But now we introduce this additional parameter with a normalizing factor, and in fact I should mention that the proof assumes that the mass is positive. To begin with, it even makes sense of this factor, and to get a definite sign on this thing. And then they work really, really hard with method from potential theory and geometric measure theory to show that these integrals as a function of T, so as a function of the level set, value N are non-increasing. And also, U P equals constant, and I'm glossing over a lot of details here, of course, if and only if we are in Schwarzschild of mass M. So now again, instead of having something that we integrate all the way out from zero to, from zero, which is the horizon, to one, which represents infinity, we look at functionals that are now non-increasing. So, most likely decreasing unless we are in Schwarzschild. And this, of course, then gives us, again, inequalities between the normal derivative of the lapse at the horizon and the area of the horizon. And these are saturated as, again, M is one on four pi, nu of N, N equals zero times this area. And this implies that U P is actually constant. And then this implies that we Schwarzschild. So we got rid of the cotton tensor. So in principle, you could think that this method works in higher dimensions if you put their dimensional power in the correct way. And in fact, I think their monogenicity properties are very nice also in higher dimensions. However, you cannot, as far as I understand, get these things to be saturated for the same reason that the Israel proof doesn't generalize to higher dimensions and that is that you would need a Gauss-Boune on the way. And if you're in higher dimensions, you can't do a Gauss-Boune and then you cannot get saturation and then you cannot conclude. So the problem again is not in the analysis but in the geometry. But I think it's a very, very nice proof and it's also a nice new aspect to this theme of looking at the level sets, integrating something over the level sets and then getting inequalities between the quantities on the horizon and the quantities at infinity, now not by integration, but instead by monotonicity properties of some functionals. So the next other proof for N equals three I want to discuss is a proof by myself which is not quite yet on the archive. Maybe only for baby reasons. She doesn't like me to sit on the computer, okay? So it's in progress but that's not a problem with the proof but just with the typing. And it's a Banting Masulu-Lalam argument up to here and then it offers a different suggestions of how to recover Schwarzschild without using the cotton tensor, without using anything that's inherently three dimensional. So in particular it will allow everything to be done n-dimensionally later, okay? And here's the idea. We found, so we found, I'd write it three dimensional to be consistent with last class that this manifold M3 wedge union P infinity. So M3 wedge was this guy. No, there wasn't any wedge. P wedge was actually the same thing as R3 delta by the positive mass theorem. But I mentioned yesterday that we also found that the Laplacian with respect to this added metric of the conformal factor phi was zero on M3 wedge. Be careful, it's not true that this is vanishing at this point that we glue in because the conformal factor has a singularity at this point. And recall also that the definition of phi was nothing else but two over one plus n hat where n hat was the lapse function that we defined here just to be the lapse function and here to be minus the lapse function. And now we argue as follows. One, we think about the topology, meaning we prove that I is one. This goes like this. We know this guy is R3 globally, globally isometric to R3. So if there were several connected components of the horizon to begin with, then this would have non-trivial topology. For example, second fundamental group or something, whatever you like. And this has trivial topology but they're isometric globally, no. So just for simple topology reasons, there has to be just one black hole. Then the second aspect is the geometry of the horizon and now it's a singular horizon. So I mean a single horizon, not a singular horizon, luckily. So the argument goes like this. We showed in the beginning, or I claim that we could show in the beginning that this horizon is ambilic. This trace phi part of the second fundamental form vanishes. Then we perform the conformal change. It's well known that being ambilic is invariant under conformal changes. So it's also ambilic in this manifold. But this manifold is R3. So now we know the horizon is ambilic in R3. So sigma two, ambilic in M3. Sigma two is ambilic in R3. But it's a very well known fact which can be verified by a computation. And if you've never seen it, I suggest you do that. That any conformal, any ambilic surface which is closed in R3 is nothing else but a round sphere. There are no other ambilic surfaces. The only closed ambilic surfaces in R3 are spheres and spheres of the form radius S0 and center C. So C is in R3 and S0 is a radius. I'm calling it S again recall because we call the coordinates in R3S, the radial coordinate S and not R to remember that we are in the isotropic coordinates in Schwarzschild. Okay, so again, we were ambilic to begin with. So we are ambilic in R3 and the ambilic surfaces in R3 are classified and they're all round spheres. So that means that topologically, now we know we are round sphere horizon and we know more. We know in R3, the image of the surface and I'll suggestively draw it like this is a round sphere intrinsically. If it's a round sphere intrinsically, we can verify that it has to be been a round sphere in here because the conformal factor, and I'll write this down in a second, the conformal factor on the horizon. If you look at the formula and had a zero on the horizon, so the conformal factor on the horizon is a constant. So the metric on the horizon induced in the original manifold and induced in R3 are related by a power of, by two, by a power of four, so by a constant. So it's a round sphere. So phi on sigma two is constantly two. That means that sigma two in M3 was already intrinsically round. We knew it was extrinsically totally geodesic. So round, but also now we know it's intrinsically round because the conformal factor is a constant. And three, it's in fact, the whole thing is four-chilled by the following argument, which is much simpler than the argument using the cotton tensor or the explicit computation that Israel gave. And the argument goes like this. We know that phi is a harmonic function outside the ball of radius S0 around the center C in R3 because it was harmonic with respect to the new metric, but the new metric turned out to be Euclidean space. And if we stay away from this one point, which is inside this ball, we get this. We now just computed that phi on the boundary of this ball is two, which is a constant. We also know from the asymptotics that phi goes to one as S goes to infinity, S being the radial coordinate in R3. Well, I can solve that, right? So that means that phi of some X in R3 is nothing else but one plus, we check S0 over X minus C. So here's the picture for any point X out here. I compute, and I'll now write this as C. I compute the distance from X to C, plug it down here, S0 is the radius of that ball and one is the value at infinity. So there's, the only thing I don't know is the center and the radius. But now I know more than the asymptotics. I know that N goes like one minus M over two S, M over S. So I can do a compare asymptotics N being one minus M over S plus lower order terms. And hopefully not too surprisingly, get that S0 is M over two, which is the radius of the horizon in the isotropic coordinates. So this is indeed one plus M over two X minus C, which is almost the conformal factor from Schwarzschild except that we have the distance to some constant point C instead of just having S there. But that's a matter of translating the coordinates on R3. So translating coordinates on R3 says this is Schwarzschild and we're done. So that's a method of recovery that doesn't ever switch into the Schwarzschild coordinate picture like the others do, which is why it seems so much simpler. And instead of using anything about the geometry of the original manifold, we now stay entirely in the R3 and exploit this equation that we just computed as a side computation for computing that the scalar curvature transforms to something non-negative. And then we forgot about it in the other proof. And if we don't forget about it, it gives us the answer immediately because it really is already the conformal factor for Schwarzschild. And this clearly generalizes to higher dimensions and explains why there's an N minus two here in higher dimensions because that's the seed solution of the Laplace equation in that dimension. It also gives you a very nice insight why there's no Schwarzschild for N equals two because then you would have a logarithm and fundamental solution which would grow and you could never get something asymptotically flat. Questions? The C is the center of mass. That's a good question. So if you think about your ordinary Schwarzschild solution and you just translate your coordinate, then that's what's gonna happen in the isotropic coordinate picture. It's not so easy to do that in the Schwarzschild coordinate picture because it exploits the spherical symmetry much more. And of course, geometrically, you can't distinguish the two Schwarzschilds because they're just written in translated coordinates. That's why in this proof that exploits the geometry very much, it comes up naturally. Because I'm not exploiting the spherical symmetry, instead I'm exploiting this conformal picture which still has this translational freedom there. So somehow you could have anticipated that you catch the center of mass term, right? So usually we center our Schwarzschild at the origin even in the isotropic coordinates, but in the isotropic coordinates there's no reason to do so except that it's simpler. So this shows how this works. So let's move to higher dimensions explicitly. So as I said, it's not known how to, or if it's possible to, to generalize to higher dimensions any of the proofs, Israel, Robinson, and Agostiniani and Masieri. Agostiniani, Masieri, plus their matter versions because they use Gauss-Bohne. In 1989, 1998, one, I already mentioned this yesterday, gave a generalization of this part of the Banting Masoud-Ullalam proof and then a recovery mechanism using the vial tensor instead of the cotton tensor for higher dimensions. So this is, and you find the reference in the Heuser book and, or in the Robinson paper, n bigger equal to four, generalize Banting Masoud-Ullalam up to Schwarzschild recovery and then produces an involved argument using methods including the Bohner identity, Bohner formulas, the Moritz theory, and the vial tensor to be construct, to recover Schwarzschild. And as they use the vial tensor, these methods only work for n bigger or equal to four and not for the case already covered by Banting Masoud with the cotton York tensor. Then in 2002, Gary Gibbons and Iida and Shiro Mizo, letter two from Japan, I believe, in 2002. And again, the reference is in the Robinson article, citation 1112 in Robinson because there are several ones by these authors in that year. They do the same as Fouang, so meaning they generalize the Banting Masoud procedure all the way up to the Schwarzschild recovery to higher dimensions. And then they try but recover Schwarzschild maybe a little bit a-historically under the condition that Dn is nowhere vanishing by a rather explicit method, similar to that of Israel actually. And in fact, they argue that in principle, at this point of the proof, we should actually be able to show that there is no zero in the gradient of the lapse, which is one of the things that Wang actually does. He shows that using the Bohner formulas in the Moritz theory that actually this doesn't happen after you already know you're globally isometric to globally conformally Schwarzschild, globally conformally flat. So they're right, but they don't show it. Both these proofs are for spin only, for spin manifolds only, because at the time, the only available positive mass theorem in higher dimensions was Wittens, and Wittens proof with spin manifolds, and then the Bartnick's regularization thereof, or low regularly version thereof. Using Bartnick's weak regularity. Neither of the papers actually gives an explicit explanation why the fact that a manifold is spin, it survives this construction. If you think about it, what the assumption would be is that this northern manifold be spin to begin with, and then you would have to verify then doubling preserved spin structure. And I thought about it and I'm convinced myself, but I can't reproduce the argument of the top of my head. And then that this conformal change doesn't affect the spin structure, but then you glue in one point and that could very well affect the spin structure again, and they don't say anything why, but I convinced myself that yes, but that's something, if you're interested in spin structures, I think you should think about why you get a spin structure here. Okay, and neither of them says anything about it. So then, what do I raise? I raise this. The argument that I'm currently erasing goes to higher dimensions easily because all the arguments were completely geometric or about fundamental solutions of the Laplace equation on R3 outside of ball. All those go easily to higher dimensions. There was no Gauss-Ponnet. And in fact, now we know that there is a positive mass theorem in higher dimensions, also without the spin condition by Shane and Yao, and I'll give you the references. So, to higher dimensions, dimensions, and of course, this applies to these other theorems too. Well, I'll say that later. And the positive mass theorem story, I also already said this yesterday, but I still need to give you the reference. Is that Shane and Yao in 2017 in an article called Positive Scalar Curvature, Minimal Hypersurface Regularity, The Singularities, and I couldn't find it published anywhere yet, but on the archive, it's 17.04, 05.04, version one. Together with this paper I mentioned yesterday that my Mac Farron and the Hedy from 2012, which is called On the Positive Mass Theorem for Manifolds with Corners, and appeared also in CMP, Volume 313, number two, pages 425 through 443. So, Shane and Yao prove Positive Mass Theorem under smooth conditions in all dimensions. Sorry, without a spin condition, and Mac Farron, Shackle, Hedy in all dimensions prove that if you have a non-regular, non-smooth manifold, which is smooth enough to start a Ricci flow, then, and has zero mass, then it keeps zero mass under the Ricci flow and becomes instantaneously regular in which case then the Shane and Yao rigidity case tells you you're smooth and thus you're globally isometric to RN. So using this, instead of the Bartnick version of Witten's proof with spinners, we get away from the spinner condition by now. And of course, this can also be inserted into Huang's argument and also into Given's Ida Shiromitsu and removes the spin condition in those works as well, a posteriori. So that means we now have proofs and several versions of proofs for a higher dimensional, static vacuum black hole uniqueness. So we know in all dimensions, and bigger or equal to three, if we start with potentially several static vacuum black holes, which are asymptotically, sorry, Schwartz Hildian. So suitably isolated, then there is only one black hole and it is actually Schwartz Hildian of the same mass we started with. And this mass has to be positive. None of these theorems that I just mentioned today in the higher dimensional section needs to assume positive mass, they all get it out as a consequence, as I explained yesterday. Sorry. But however, if you think about the proof again, let's think about the proof again under the aspect of have we ever used the part of the static equations which involves the Hessian? We ever used this equation. We doubled, okay, first we derived properties of the horizon and when we derived properties of the horizon, we did use this equation and I didn't show this to you, so I have to tell you. So in order to show it's ambilic and to show that the normal derivative of the laps is a constant, you needed this equation, okay. But after that, we doubled, didn't need this equation. We did a conformal transformation, we didn't need any equations for that. Then we computed the scalar curvature of this new guy. That involved only the Laplace of, so we used the Laplace with respect to g of n is zero, for sure, in the computation that the scalar curvature be non-negative and we used that the scalar curvature is zero to conclude that the scalar curvature of this new guy is zero. But really, for the positive mass theorem, we only needed that the scalar curvature of this guy is non-negative and to show that, we would only have needed that the scalar curvature of the original guy be non-negative and we could have used only that the scalar curvature of the original metric is non-negative. Then we did a positive mass theorem that didn't, of course, need the static vacuum equation for the Hessian at all. Because it's just the positive mass theorem. And then the recovery method I erased over only used Laplace phi is zero. And Laplace phi is zero was a disguised version of Laplace n is zero, okay? So we didn't use this either. So we didn't ever use this, except when we derived that the horizon was ambilic and had constant normal derivative of the labs. For the proof, I put Zenerbaum in progress so you know which proof I'm talking about. Except that we use Hessian n equals n Ritchie, which then implies scalar curvature is zero to derive that the trace phi part of the second fundamental form vanishes and that nu of n is constant, which we actually never used in this proof anyway. We only use that it's non-zero. Okay, so that's the only point where we use this. So we actually proved a much more general theorem. We proved a theorem saying that if, if m and g is Romanian with smooth boundary and the boundary has vanishing mean curvature and vanishing trace free part of the second fundamental form. So the boundary is totally geodesic. And if n on m3 to our extent smoothly to the boundary vanishes on the boundary and the normal derivative of n does not vanish on the boundary and n is positive in the interior and m and g n is asymptotic to Schwarzschild or asymptotically isotropic as we always also call it of mass m. And for those who are interested in the details, the error term is Sn minus one and such that the scalar curvature is non-negative, which is also known as the dominant energy condition in this context. And the Laplacian of n vanishes in m3, mn, then Schwarzschild of mass m. Now this could be rather surprising because in the beginning we started with all this rationale, yes? Oh, it's the three, thank you, more. So in the beginning we started with all this rationale that we have several black holes and they gravitationally attract each other and therefore they should accelerate and shouldn't sustain in static equilibrium. And now we got rid of the static in the thing, right? Instead we said let's have things that are black holes and have more properties like black holes in static space times, but let's forget about the static other than that and let's just think of a function n that takes the role of the lapse if it was static, but doesn't have to be static and the whole result sustains. There really, there wasn't so much statisticity in the proof and at least if you modify the last bit of the proof in the way I showed you, there isn't much statisticity except that the horizon components are priori, they're potentially multiple components, have the properties that they would have in a static vacuum environment, right? So from a PDE perspective, this says instead of looking at an old highly over determined system of elliptic PDEs with highly over determined boundary conditions, you don't need the highly over determined PDE anymore. Actually it's sufficient to look at a harmonic equation with so many boundary conditions. Why are there so many boundary conditions? On the inner boundary we put the directly condition of vanishing laps plus non vanishing normal derivative which is almost a Neumann condition but at infinity we impose that the function goes to one that's a Dirichlet condition with two derivatives. So the first derivative is also prescribed which is another Neumann condition and then the second derivative is also prescribed which is even stronger, okay? So we have hugely over determined boundary conditions for this lapse function. So it's not a good idea to think if I have a metric that on MN which has non-negative scalar curvature and I'm totally geodesic boundary, then I'll just find some N like this and then everything is Schwarz yield. That would be too much to hope for because of course there are manifolds with boundary that is totally geodesic that are not Schwarz yield. But if you can solve this highly over determined system of boundary conditions for the lapse then you'll be able to solve boundary conditions for the lapse then you're Schwarz yield which just means that it always never happens. Okay? So this could be slightly surprising, hopefully. So the error, the lower order term in the definition of asymptotically isotropic I never said how fast the lower order term needs to decay. And this is saying in dimension N the lower order term would be at least of the order 1 over S to the N minus 1 and then the first derivative like 1 over S to the N and the second derivative S to the N plus 1. It's the same condition appearing in the Banting Massoudan. I'm not claiming that this is the lowest assumption that you can make it's just an assumption under which this works and under which the gluing of the P infinity point is straightforward. I didn't want to glue it in I think, but I didn't want to go in that direction. So let me spend the last 15 minutes on explaining a little bit about this photon sphere business and I'll continue writing this theorem later so leave some gap in your notes if you're taking notes. Oh, in fact maybe let's do it immediately. You can generalize this theorem also remains true if additional components of the boundary are allowed to be of the following form. N is constant so the components like yesterday we will call them sigma 2i. The lapse is constant the N there exists a constant C i bigger than N minus 2 over N minus 1. I'll explain what this means in a second. And the scalar curvature of the induced metric sigma i is C i times h i squared and h i is the induced mean curvature of course is constant and 2 times the normal derivative on sigma 2i is C i minus the special N minus 2 over N minus 1 times h i times N i is constant. Let me say this in words now we are allowed to start with the northern manifold so a Romanian manifold with the lapse function that's harmonic and non-negative scalar curvature with some components that have mean curvature 0 lapse function 0 and are ambilic and some components that have this other quantity I forgot to write that they're also ambilic so the lapse function is a positive constant the normal derivative of the lapse is a constant the scalar curvature is a constant the mean curvature is a constant and they're ambilic and all these constants are related via this parameter constant C i and of course h naught needs to be 0 as well and we could have some components of one kind and some components of the other kind if you take C to infinity then you can think of the horizon components as arising as limits of those anyway and I'll say a few words later what the C i means and then everything is also Schwarzschild so the boundary conditions you can make them different from just horizons conditions this way and before I'll say what this has to do with photon surfaces and photon spheres let me explain the only two pieces where this makes a difference in the proof we cannot double because if we have okay let's have this a horizon component so h equals 0 and so on and so forth but the other components we cannot double through them because they have positive mean curvature and so if we doubled they would get a crack but we need C 1 1 so instead what we do and we first did this with Greg Galloway in a special case of this and then I did it on my own we glue something in and we glue something in which has a horizon and it's end and then we double and the thing we glue in is in the work that I did with Greg Galloway and I'll give the references later to whoever is interested and write them in the lecture notes and they're all on my website in the case with Greg Galloway we glue in a piece of a schwarzschild neck so therefore the static equations hold on this glue in piece also and then all the rest just continues the same way in higher dimensions we only have that the surface has positive constant scalar curvature if you look over there and in higher dimensions this doesn't tell us even a sphere which was the whole problem why Israel's proofs and so on didn't generalize so we glue something in which looks like a schwarzschild but the base is not a round sphere the base is something else and this means that the static equations won't necessarily be satisfied on this neck which is why it's great that we didn't need the static equations anymore and in fact we can do that in a way that the scalar curvature remains non-negative and the lapse function we will just continue to call the lapse function will be harmonic on the sky so that then the rest of the argument goes through because we didn't need the Hessian equation anymore okay and I can like it will be on the archive soon with the details and then the recovery just works exactly the same only you have to first forget about the glued in regions and do the whole argument I gave just in the original manifold as before okay so much for the strategy for the proof now what's the role of the CI if you think about so this is positive if you think about a surface of this type of course a spherically symmetric cross section in schwarzschild would be of this type and CI would tell you that it fits into schwarzschild anywhere between strictly speaking the horizon and infinity the threshold is exactly what you get if you try to fit it into Euclidean space and if you go below the threshold you would be in a round surface in a negative mass schwarzschild so this characterizes the round surfaces in positive mass schwarzschild again this is the threshold for Euclidean space and if you go below you would be in a negative mass schwarzschild so it tells you you're one of the surfaces that could a priori sit in a schwarzschild as a round surface so here's the story with photon spheres and photon surfaces if you're in a space time a photon surface is a time like hypersurface in any space time which is ambilic and you can think about this for a while and the reference will be in the lecture notes and this ambilic turns out to mean nothing else than if you take a photon so a null geodesic started tangentially to the surface somewhere then it will remain in the surface okay that's why it's a photon surface it catches photons and then a photon sphere we only know to define that in a static manifold static space time it's a photon surface with n equals constant along it photon surface p with pn with n equals constant along pn so what does that mean if you think about it for a while it means the color of the photon stays the same all the time from the perspective of the static observers so no redshift or blueshift the color of the photon or the energy level of the photon stays the same it's a way of saying that the photon is trapped and an example of such a photon surface is the cylinder excuse the stupid language this is a cylinder in Schwarzschild with spherical base r equals 3m and then up in time direction a cylinder and this is such a photon surface and in particular also such a photon sphere because n only depends on r so n will be constant meaning that if you stand if you were able to have a rocket that allows you to stand on this cylinder you could see the back of your head okay and it wouldn't be color shifted okay so that's a photon sphere by definition and excuse the language it could be in different spacetimes also of a different topology than just a cylinder over a sphere so now Greg Galloway and I showed that in Schwarzschild in all dimensions you have a funny diagram so this is Schwarzschild n bigger or equal to 3 and this is r equals 3m the photon sphere that's very well known and we found a whole family of photon surfaces in Schwarzschild which came to a surprise to us at least and I'll just draw a diagram so there are some which look almost like hyperboloids there are one sheet of hyperboloids in Minkowski space I'll also photo on spheres then there are ones that come in from infinity and asymptote to this 3m then there are ones that come in from infinity and asymptote to the black hole horizon which in our language is 2m then there are some that come from the horizon go to the photon sphere and return to the horizon then there's no more color but there's still one family coming in from the photon sphere and going to the horizon all of these if you take them and rotate you get surfaces in Schwarzschild spacetime which are time like ambilic so they're photon surfaces and they're spherically symmetric so these are all photon surfaces and just the one with constant radius is a photon sphere but they all have spherical cross section and you can if you look at the intersection with t equals 0 each of them is a sphere of some radius and each of these spheres we can show our priori if they are in a static vacuum spacetime we can specify all these conditions so as a consequence a theorem by Greg Galloway and myself which applies this abstract theorem is saying that if you have a static vacuum spacetime which is asymptotically Schwarzschildian and it has an inner boundary which arises as a photon surface such that at each cross section the level it's a level set of the laps and I won't have time to write this down it's Schwarzschild in particular if you start with the photon sphere that's a result we had for a while in three dimensions you'll have to be Schwarzschild so not only are black holes in static vacuum spacetimes unique and have to be Schwarzschild also photon spheres and now with the new theorem that allows for non-hashian solutions to show that also in high dimensions this holds true for photon surfaces in three dimensions actually you don't even need to drop the static equations so now instead of just knowing static black hole uniqueness we know something which is in some sense nicer and not clear whether this is a stronger or weaker statement we know this photon sphere or photon surface uniqueness why is it not clear that it's a weaker well on the horizon we have all these singularities of things where we need to be super careful that we don't compute anything that's not well defined especially if we work in low regularity in all spacetimes once we got out that far that we are at the photon sphere everything is super regular in static yeah so everything is super regular here and we don't need to look at static problems that there's no problems that any vector fields become null that we would like to work with so we're in the super regular regime of the spacetime not anywhere near the horizon and still we have a uniqueness and you need to be careful we only have the uniqueness to the exterior that means if you had you could potentially and this was something people in astrophysics believed and we're quite surprised when I first worked on this you could have had a star which is not which is static and it has non-homogeneous structure so it could potentially be non-round and it was expected in astrophysics that if the star is suitably compact it would form a photon sphere around it of trapped light this is saying well if it does then the photon sphere and everything exterior to it is spherical symmetric anyway but the static equations are real analytics so it should be then spherical symmetric all the way down to the star so this is saying if you have a non-sphere symmetric star which is static it won't form a photon sphere which is a surface it can still have individual trap light rays but they won't form a nice smooth surface they will do something more complicated maybe something even as complicated as incur we don't know but the naive picture that it'll form a surface is too naive and has to be given up and the same thing is true for these photon surfaces which I don't know if they have any physical significance beyond being pretty thank you