 And at the end of last class, I was asked a question that I want to share in the whole audience. And that question was, didn't you say you were going to be talking about high-dimensional versions of black hole uniqueness in the last class? Yes. So why are you saying it's not true this class? So maybe I need to clarify what I said by the end of last class. I didn't say black hole uniqueness was false in high dimensions. It's, in fact, true. And we'll see that next tomorrow. But Israel's proof of static black hole uniqueness is not known to generalize to higher dimensions. So while the claim is still correct, the proof doesn't generalize. That's what I was trying to say yesterday. And so the proof we're going to see today, in the way we're going to see it, also does not generalize to higher dimensions. But we'll see tomorrow how you can alter it so it does generalize to higher dimensions. So today we're going to talk about a multiple black hole situation. So what was question two and answer two yesterday. And we're going to discuss a proof by Banting and Masur Alam from 1987, which is 20 years after the proof of Israel that we saw yesterday. And it's much more geometric. It's one of the most beautiful proofs, actually, I know, in mathematics, which is why I wanted to talk about this topic because I really like this proof. And it has had a tremendous impact on mathematical relativity throughout. So many proofs are based on it, or imitate aspects of it, or actually use the results in the way you get it from this theorem. Hugh Bray's proof of the Penrose inequality, Romanian Penrose inequality, for example, wouldn't go without this proof. So that's why it's important for anyone who wants to do mathematical work in general relativity to know this proof. So with this motivation, let me state the theorem. My Banting and Masur Alam, 1987. And let me say up front that the key ingredient, sorry, the key ingredient in this proof is the Romanian positive mass theorem that you've seen last week. So again, we start with the static system, three-dimensional, the static, vacuum, asymptotic to Schwarzschild, meaning that the lower order terms of this form, this is the decay terms are like one over R squared, faster decay faster than one over R squared. The first derivative is like one on R cubed and the second derivative is like one over R to the four. This is what this notation means. This is just for those people who want to know the precise decay of the lower order term of mass M. And again, they assume positivity of that mass to begin with, but there are more modern methods where you don't need to assume it to begin with, but today I'll gloss over those with inner boundary, which is now potentially disconnected. So it consists of finitely many sigma two I, sigma two I connected static horizon and I bigger or equal to one, then I is one. And the corresponding spacetime, this isometric to Schwarzschild of the same mass. So we can potentially have multiple black hole horizons to begin with, but eventually we'll only have one and everything will be Schwarzschild as we discussed last time. Oh, I'm sorry, this would be an S, of course. This was gonna happen. Okay. And I'll write the remark. One can drop the assumption that M be positive by a maximum principle argument, which I'm not gonna explain here because I'm not sure everyone knows maximum principles. Okay. And recall from last class that we already know that sigma two I is totally geodesic, has mean curvature zero by assumption and also the stress free part of the mean curvature is of the second fundamental form vanishes and also the normal derivative of the lapse is a constant for each I, not necessarily the same constant for each I and actually is a positive constant, again by the maximum principle for those of you who know what that means. We discussed this last class for a single black hole horizon carefully. Okay. So now comes the idea of the proof and I'll do a lot of it in pictures. I'll only do those computations explicitly where I think one can learn something from the computation or otherwise the proof is not clear if one doesn't do the computation. Okay. So here's the idea. The first step is that we double the manifold. So we have our asymptotically flat, oh, there's not enough room here. We have our asymptotically flat manifold with potentially several horizons. So this will be sigma two one and this will be sigma two two and this is where S goes to infinity here as well and this is M three. So the first step, what we're gonna do when we did this explicitly in the case of Schwarzschild before is we're gonna reflect it through the horizon and have a second copy and call this M three tilde and we're also going to copy the coordinates that we have near infinity and calling them X i tilde and S tilde, the radius. So here we have coordinates X i tilde and radius S tilde that goes to infinity. Up here we have our asymptotically flat coordinates X i. Okay and because the surfaces are all totally geodesic so the second fundamental form vanishes completely we can do this here without having a crack. Okay, the first derivatives actually align. This is due to this condition, this assertion here. So the regularity at the gluing here will be C one one where C one one means it's C one. So with the continuous first derivatives and the second comma one means the highest order derivative is Lipschitz continuous. If you don't know what that means it's not so important for the rest of the proof. If you do know what it means, that's the regularity we have on all these surfaces. Apart from the surfaces, we're smooth. I mean the manifold you can think of smooth anyway and in addition we have the metric and we also just copy the metric down here and then everything has this nice regularity here. The second derivative could jump but the first derivative is actually continuous, okay? Which is because the second fundamental form vanishes so in some sense this could be used as a reflection symmetry plane or surface. So that's the first step, we double M three and G and then we glue them together so we have M three hat to be M three together with M three twiddle and of course we identify the boundaries and we only count them once and we have G hat to be G on M three and G twiddle on M three twiddle and now we want to do something similar for the lapse function. Here we also have our lapse function N and down here we want to have yellow the lapse function N twiddle so that together we can combine them to a lapse function N hat and our definition of N twiddle of a point P twiddle will be minus N of P. What does that mean? If we have a point P twiddle down here because this is a double this means it's a reflection copy of a point P up here. This we can plug in into N, take minus of it and call it the lapse of this point P twiddle and then we combine them to get N hat to be N on M three and N twiddle on M three twiddle. So that means now on the whole thing we have a manifold, we have a metric which is smooth except on these surfaces where it has this regularity and we also have a function and now let's think about this function. Of course the function is smooth up here and smooth down there. What does the function do across these? Well it becomes zero there so changing it from plus to minus doesn't matter makes it continuous and if you look at the condition that new event is a positive constant here you can actually make this equivalent interpret this as being equivalent to being able to merge this function N and N twiddle C one one here as well. So the conditions here are exactly what we need to merge these two pieces together along the horizons in the C one one fashion which is exactly smooth enough for what we want to do later. And this is also exactly what happens we didn't talk about the left function in detail in the isotropic Schwarzschild picture where we double Schwarzschild this is exactly the same thing that happens there on the new on the copy we have a negative left function in the Schwarzschild case as well. So if we hadn't started with Schwarzschild in isotropic coordinates with half Schwarzschild in isotropic coordinates we would now have complete Schwarzschild in isotropic coordinates. If we had Schwarzschild just a second if we had started with Schwarzschild in different coordinates which are only asymptotic to the isotropic coordinates we would still have double Schwarzschild just in funny coordinates. Yes. It could be see infinity on the boundary also I think that's something one typically assumes yeah but here all we need is that it matches C one one so then we could have assumed that from the beginning. We don't know anything about the second derivative across the surface in this reflection procedure end of story we just don't know a priori. A posteriori we're gonna know it's Schwarzschild and then of course it's smooth but a priori we don't. But that's all we're gonna need to prove it's Schwarzschild. Okay? Good. So now let's think about the asymptotics for a second. Here we know that it's asymptotic to Schwarzschild as S goes to infinity then the metric here is going to look exactly the same as S twiddle goes to infinity and the left's function is gonna look like minus one plus M over S instead of S twiddle instead of one minus. Okay? So that's natural and the next step what we're gonna do now I'm gonna erase this here hoping that everyone already copied it so I have room for more pictures. Now we're gonna perform a conformal change from this scenario where now we have here M three hat G hat and N hat to scenario here where we'll have M three I don't know how to call that wedge maybe. G wedge and N wedge. No N wedge M three hat wedge and G hat which will be conformally related to this in the same sense that if I started here with Schwarzschild then double it to double Schwarzschild and then perform this conformal change I would get flat space only I would get flat space without the origin in Schwarzschild, right? So I want to do something that if I started if this is Schwarzschild then this would be R three without the origin and the Euclidean metric. So what was the formula in Schwarzschild? The formula was that this metric here was a function phi to the power of four delta. So to go here I want to multiply by phi to the power minus four to get delta, okay? So I want G wedge to be some function phi to the power minus four times G hat. So in case I started with Schwarzschild and I picked phi to be the function I wanted in Schwarzschild then I get the delta back. But now I need to say what phi is in general. And actually even if I was in the Schwarzschild case I don't know if my Schwarzschild is written in isotropic coordinates. So then I don't know if this function takes exactly the same form in the variable S as it did in Schwarzschild. So I need to find a way to write this without appealing to coordinates if I can. And the idea is the following. Look at Schwarzschild, maybe I'll do that here. And this is a cool idea that's been used in other scenarios as well. But I think it has been tremendously underused. Is there a question? Do you have a question? There's something you can't read? Or what is this one? Oh this is the trace free part of the second fundamental form. So if we were in Schwarzschild we would know in isotropic coordinates and we would know that phi of S is one plus M over two S and N of R was one minus two M over R. And R was phi of S squared times S. And this allows us to look at N of R of S. Smart Seller suggested to call it. And you'll call this N hat of S in the Schwarzschild case just as over there. And this you can compute to be one minus M over two S divided by one plus M over two S. So the lapse function that reads like this and the R variable reads like that and the S variable in conformal coordinates, isotropic coordinates. And now the cool idea that they had was to introduce a function phi of N hat. Not of a variable, not of a coordinate, but of this function as over one plus N hat. And if you do that, then what you get is that in Schwarzschild phi of S is nothing than phi of N hat of S. So this is rewriting the same thing over and over again in different ways. So I'll repeat, we have this conformal factor that we computed for Schwarzschild. We have the lapse function that we wrote down for Schwarzschild. We write this lapse function in isotropic coordinates. It's the same physics, it's just written differently in different coordinates. And then instead of expressing it in terms of S or R, now we express this function little phi in terms of a function of N hat, which then is a function of S. So then we can express this phi of S in the Schwarzschild scenario as this capital phi applied to this N hat. And this we can do here, where phi is this capital phi applied to N hat. Or in other words is two over one plus N hat. This is just an algebraic trick in the Schwarzschild example, but in general it allows us to write something which completely is independent of coordinates. So even if we worked in Schwarzschild but in different sets of coordinates where the phi would take a different form, this would still be true. So we've peeled the physics out of the coordinate expressions or the math or the geometry or whatever you wanna call it. We've peeled it out of the coordinate representation of everything. Now we managed to write this conformal change in a Schwarzschild scenario like this. And of course no one stops us from just performing a conformal transformation like this and see what happens, okay? So that's what we do. We define our G wedge to be phi to the minus 4 G hat and actually our M3 wedge to be just M3 hat. So we don't change the manifold at all, we just change the metric on this manifold and in that in a conformal way with a conformal factor that we can compute from the lab's function. Okay, in order to do that of course we need to verify that this phi never becomes zero which is obvious but it also never becomes infinity which is fine because N is between zero and one and then N hat is between minus one and plus one and that's okay. Yes. Because I want, this is the manifold with boundary and this is another copy of the same manifold with boundary. Now I glue them along the boundary so I want to have a name for the manifold without boundary for the total manifold, okay? And in fact, one thing you can observe about this manifold is that it's geodesically complete, okay, this here is geodesically complete. Not important for the proof but, okay and now I perform this conformal change so I don't know how to draw it yet because I don't know the asymptotics yet so I'm not gonna draw it before we've figured out the asymptotics. But before we figure out the asymptotics or let's figure out the asymptotics first, why not? Asymptotics. So as I said here we just have the asymptotics of M3G and N has the asymptotics we assumed, yeah? G and N, G, Gij is like one plus M over two S to the power four delta ij plus lower order terms and N is like one minus M over S plus lower order terms as S goes to infinity. Now let's look at P to the minus one because we're gonna need P to the minus four so I'm starting with P to the minus one. That is one plus N hat and in this case one plus N over two. One plus N over two with this asymptotics is like one minus M over two S plus lower order terms. So then G wedge ij is five to the minus four G ij is like one minus M over two S plus lower order terms to the power four times one plus M over two S to the power four delta ij plus lower order terms. Okay, I've just multiplied the formula I found for this to the power four with this formula. Now, if you have something like this you can compute it like a binomial and then you get one minus M over two S to the power four plus lower order terms but one minus M over two S times one plus M over two S is also binomial and gives you one plus something of the order one over S squared, so no one over S term. So this completely reads like and I'll write it in a suggestive notation one plus zero over two S to the power four delta ij plus lower order terms. Okay, so the G wedge, the new metric I get up here in this infinite end looks like asymptotic to Schwarzschild of mass zero. So I can now try to draw that. So after this conformal change M three well the northern half here G wedge I'll not write anything there, I'll write it on top. So it would still be asymptotic to Schwarzschild but now of a different mass namely mass zero. Recall if this was Schwarzschild that would be Euclidean space so mass zero is very consistent. Yeah, so in some sense asymptotically for any such metric the same magic happens as in the Schwarzschild case only just asymptotically we kill the mass. Okay, so what happens at the other end? So this is what they call the northern half. So now let's look at the southern half or the double. Of course the metric has the same asymptotic as before because we just reflected it but the left function now has the opposite sign so it will be like minus one plus M over S. So G twiddle ij looks like one plus M over two S to the power four delta ij plus lower order terms oh S twiddle and N twiddle will look like minus one plus M over S twiddle plus lower order terms as S twiddle goes to infinity. Now then if I compute my little phi minus one that'll be one plus N twiddle over two. This doesn't look as nice anymore because the one and the minus one cancel. And instead you get M over two S twiddle plus lower order terms. That means that this conformal factor goes to zero not to one anymore in this end. So it will do something tremendously different. So here in this end the conformal transformation just pulled everything even more flat. In this end it will do something different it multiplies it with something that asymptotically becomes zero. That's not so surprising either because if we started with Schwarzschild then the lower end would fold in and would be R three around the origin. So we can expect that something like that happens too. And in order to observe that it's a useful idea to introduce a different set of coordinates that allows us to study something near a point rather than near infinity. Okay but let's first write down what would G ij look like. G wedge ij would look like V to the minus four G twiddle ij. And that is it would look like M over two S twiddle to the power four which is what we've just computed delta ij plus lower order terms. Okay so it looks like something that goes to zero. That's not good for a Romanian metric, yes. Because that's what happens in Schwarzschild. If you look at the isotropic or the conformal Schwarzschild in isotropic coordinates then the left's function is positive up here and negative up here. We call the Kurskal diagram I drew last time, right? On the inside the black hole on the other side on the other exterior it has the opposite sign because time changes, I mean. Yeah? That's why I'm doing it here because I want to do exactly the same thing that happens in Schwarzschild because I want to recover Schwarzschild, right? If I did anything else I'd be doomed. So here we are faced with the problem that the metrics looks like this. If we were exactly Schwarzschild there just wouldn't be any lower order terms. They would also look like this. So now we introduce new coordinates and like this we take the coordinate x i to be m squared over four times x i twiddle over s twiddle squared. So what does that mean? Here we had the coordinate x i with radius s and we decided to call them x i twiddle and s twiddle down here. Okay, that's what x i twiddle and s twiddle are. So I take the i th new coordinate is the i th old coordinate divided by its length squared and multiply by this number. Geometrically what this means if you're in Euclidean space and you have a sphere of radius m over two, we call in isotropic coordinates that's the radius of the horizon. Then this is an inversion in the sphere in the following sense. If you have the origin here and you take a ray and you take a point here then there is something called an inversion in the sphere which is like a reflection in the sphere that you can take a point from the outside to the inside and vice versa. If you perform this twice, you get the identity. So it has all the features of a reflection in some sense. And if you then instead, if you know this then it has this formula if you take the unit sphere but now we want to take the sphere which is the horizon. So the sphere of radius m over two and then you get this formula. So this is the inversion in the sphere. And if you think of this as being exactly Schwarzschild in isotropic coordinates then we said the northern half and the southern half are isometric and there is a reflection and this is exactly the coordinate change you would perform to the isotropic coordinates to swap the southern and the northern half. So that's where this idea comes from. And then we do a computation and namely that AI and we introduce capital S as capital X I squared plus capital X two squared plus capital X three squared. So the Euclidean radius in this new R three. So now we wrote R three without the origin in different coordinates and this is our new radius. It's and we can relate the old and the new radius by seeing that the new radius is m squared over four times S twiddle. So this is how the length of these vectors changes under this inversion. And then we can compute the differential D X I twiddle as being nothing else than m squared over four D X capital I minus two capital X I over S squared. Wait, there is an S squared missing here and S four X K D capital X K. This is chain rule, okay? This is chain rule applied to the inverse of this formula. And now that I have this, let me write, sorry, I want to understand this metric. So I write D wedge as what we've computed m over two S to the power four delta I J plus lower order terms D X one I twiddle D X J twiddle. This is what this really means to write it like this. Yes. And now I've got a formula that allows me to take this S and transform it into a capital S via this formula. Oh, this should be an S twiddle, sorry. And take the D X I twiddle and D X J twiddle and replace them by this, okay? To transform this metric into these new coordinates. And I'll need to erase something before I can write this anywhere. So you do the computation and what you get is delta I and J plus and now it's not lower order terms anymore about higher order terms because now we're inverting the radius. So instead of taking S twiddle to infinity, we're taking capital S to zero. So it's higher order terms D X capital I D X capital J and there's a D wedge when we started. In other words, if I take G wedge and plug in D capital X I D capital X J, then I get delta I J plus higher order terms as capital S goes to zero. Okay, so let's look at this formula once more. There was an S twiddle in here. We take this S twiddle and replace it through this formula by a capital S. So it will be like capital S to the four times some M's and so on. And all M's go away, capital S to the four and then this S squared here and this S squared four actually kill this S to the four. Okay, if I did this precisely exactly in Schwarzschild and isotropic coordinates I wouldn't get any higher order terms and I would have recovered the flat metric. So let's draw that. So what happens is that this asymptotic Schwarzschildian end gets bent in the same way, the southern end of the Schwarzschild band gets bent in to something which looks like almost flat space with one missing point. If it was Schwarzschild, it would be just our three without the origin and we can glue in a point P infinity to close this off geodesically completely. So we glue in this one point and in fact one can do this glue in as was verified by this computation if you track the higher order terms more carefully also so that the regularity here is C11. And now the new thing, so then M3 wedge union this P infinity together with the metric G wedge extended by the flat metric into this point is geodesically complete. Just like R3 which is Schwarzschild R3 minus the origin after you glue in the origin is geodesically complete. Yes, it is capital S going to zero, yes. So the important thing we've achieved by this gluing in of this one point is that our resulting thing is geodesically complete and if we were Schwarzschild, we had started with Schwarzschild it would just be R3 delta. Any questions for this computation before I raise it? No linear I think. This magic with the one term that's missing that's happening in the other end doesn't happen here because of this cancellation of the one and minus one. So we lose one order. Okay, so now we've produced a geodesically complete manifold which has the same topology as before the conformal change except we glued in this one point and it has an asymptotic one end it's asymptotic Schwarzschildian with one end that has mass zero. So M wedge three united with P infinity G wedge is asymptotic to Schwarzschild or asymptotically Schwarzschildian of mass zero. We've already computed that and I want to write it down in words and now if we were Schwarzschild we would also be flat completely, yes? Of course in general we won't expect to be able to show that a priori that this is flat because then we would be pretty done. We need more steps. Instead what we want to do is we want to use this magical positive mass theorem in the Romanian case for positive energy theorem or Romanian positive mass theorem whatever you want to call it that says if you have a geodesically complete manifold we do which is asymptotically flat of mass zero we have even more than that asymptotic to Schwarzschild of mass zero and it has non-negative scalar curvature then it has to be Euclidean space. There's a little bit subtleties because it says ended a smooth but let's discuss that later. Okay so now let's compute the scalar curvature of this to see whether it's non-negative. BmT is the positive mass theorem. Okay, so this is the really the rigidity case the positive mass theorem. So how do we compute that? Well we have a formula for G wedge, right? So G wedge and G wedge was phi to the minus four of G hat. And G hat was really just G in two copies. So we can compute our wedge and so this is a conformal transformation and we can use we can use the formula for conformal change of the scalar curvature to deduce exactly what is the conformal what is the scalar curvature, okay? And the formula there is such that I never remember it is R of G hat. And this way round turns out to be more useful it's phi to the minus, let's be typo I think seven or something like that. R G wedge minus eight the plus in G wedge phi over phi. So the scalar curvature of the old metric can be computed or expressed in the scalar curvature of the new metric and this is the Laplacian of our conformal factor phi in terms of the new metric G wedge. But this we know is zero from the static equations. One of them was scalar flat. Okay, so now that means we know that our new scalar curvature our wedge is eight times the Laplacian of phi with respect to the new metric over phi. And that doesn't tell us much yet about the sign of it. Okay, but we can compute this because we know what phi is. We call that phi was really two over one plus N hat. Now the same way that there are formulas for conformal change of the scalar curvature or the transformation of scalar curvature under conformal change, there's also formulas for the transformation of Laplacian upon a conformal change of the metric. And the formula reads like this for any function F, which is smooth enough. We have the Laplacian with respect to G wedge of F can be computed as phi to the four, Laplacian G hat of F minus two over phi Laplacian, oh, no, sorry, D phi applied to the gradient of F with respect to the metric G hat. Okay, so the Laplacian with respect to the new metric of any function F is the conformal factor to the power of four. The Laplacian with respect to the old metric of F minus two over F, D phi applied to the gradient in the old metric. Now we need to apply this to phi and then we get Laplacian of G wedge applied to phi, is phi to the power four, the old Laplacian of phi minus two over phi D phi squared in the old metric. Then another nice formula is, in fact, that this part here in the bracket by a straightforward computation is nothing else than the Laplacian with respect to the old metric of phi minus one, which is one over phi and not the inverse function, okay? If you compute the Laplacian of one over a function, what you get is Laplacian of that function minus two over phi times some power, phi to the minus two, sorry, phi to the minus two, okay? So now what we got is that Laplacian of phi with respect to our new metric, which is what we need to understand the scalar curvature of the new metric is minus phi squared, Laplacian of phi to the power minus one in the old metric. Recall that phi was two over one plus n. So that means that phi to the minus one is one plus n over two. The one this is is minus phi squared, Laplacian of g hat, sorry, n hat, one plus n hat over two. Now this is really nice because of course Laplacian of a one is always zero, yes? Where's the minus sign coming from? Probably I missed one here, thank you. Everything is going to be zero in a second, so it's not that bad, but thank you anyway. So if I have the Laplacian, Laplacian is a linear operator, so then I can compute it separately for the one half and the n hat over two. The Laplacian of a constant is of course zero. So what I remain with this here is really one half of the Laplacian of g hat of n hat. Now recall on the northern half g hat and n hat are just g and n, so this is the Laplacian of n and n was harmonic, so this is zero. On the southern half g hat was g and n hat was minus n, so the same argument applies. So by the static equations, this is zero. I'll go through the computation once more in a second. So we get that the Laplacian of phi with respect to g wedge is zero. Now look at this. The scalar curvature of the new metric was proportional to this Laplacian. So that means the scalar curvature of the new metric vanishes as well. So it's scalar flat. So we write here m wedge union p infinity g wedge this has r wedge equals zero. And now I've lied to you the whole time because everything is not smooth on the gluing surfaces and then this point p infinity. So now I'm writing where smooth enough. So away from this one point where we may not have enough regularity and away from these two surfaces were finitely many surfaces where we may not have enough regularity to even compute the scalar curvature which involves second derivatives. Here and here, so to say, we're scalar flat. So we do satisfy the assumption of the positive mass theorem except being smooth. Now if we were smooth, which if we had started with Schwarzschild we would be smooth. The positive mass theorem would now tell us, okay then this is globally isometric to Euclidean space. There is a weak version and this positive mass theorem as you've heard last week is proven by Shane and Yao in 1979 and 81 and by Whitton in 71. 81, sorry, 81, little bit ahead. And there was a version of Whitton's proof for weak regularity manifolds already in 1986 by Robert Bartnick which is one year prior to this. Still they gloss over this and they just say, okay, now we apply the rigidity case of the positive mass theorem by Shane and Yao and by Whitton and then everything is Euclidean space. But in fact, they could have just cited Bartnick with his low regularity spinner proof, so Whitton style proof of the positive mass theorem and said, okay, this is enough regularity to apply Robert Bartnick's result of the rigidity case of the positive mass theorem and conclude that this is actually really globally isometric to Euclidean space with the isometry being smooth here and here and of C11 regularity here and here and at this point. So I'll write some of this down. Oh, I said I walked you through this again so I'm not gonna erase it. So I'll write it here. Rigidity case positive mass theorem in weak regularity by Robert Bartnick, 1986. And I wrote down the citation somewhere like this here. This is the mass of an asymptotically flat manifold and it appeared in communications on pure and applied mathematics 1960. In 1986 on pages 661 through 6931. Okay, so let me walk you through this computation once more just in case I lost you previously. We've computed before that the manifold is asymptotically Schwarzschildian of mass zero up here and that here we can glue in one point to make it closed and geodesically complete. And then we looked at the scalar curvature. We used a formula from conformal geometry that tells us how the scalar curvature of a manifold is related to the scalar curvature of really conformally changed manifold. This allowed us to conclude because we knew from the static equations that the original manifold was scale flat that the new scalar curvature is proportional to the Laplacian of the function phi with respect to the new metric. Then we computed or I told you a formula how to transform the Laplacian of a function with respect to a conformally transformed metric in terms of the old metric. And we plugged in F equals phi and realized that the term we get here is nothing else than proportional to the Laplacian of one over phi, which is very nice because one over phi is a constant plus something proportional to our lapse function. And because the Laplacian is linear, we can compute it as just the Laplacian of the lapse function or Laplacian of minus the lapse function on the southern end of our thing. And this according to the static equation is zero, which means we've concluded that it's scalar flat. The new metric is also scalar flat, okay? Yes, I would think so. And I think it's better to discuss this after tomorrow's lecture. I'm not saying I'm gonna prove that tomorrow, but I'm not the weak case. But so maybe it's worth a comment now that you asked. So we've got this equation and all that we've done, that we're doing now using this equation is to conclude that the scalar curvature vanishes of the new metric. Tomorrow, I'm going to show you an alternative proof of what's going to come now of the closing argument of the Banting-Masrula-Lam proof where instead of using only scalar flat, we also exploit this equation much more, okay? Which allows us to generalize it to higher dimensions. But today, somehow Banting and Masrula, they don't explicitly write it like this. They somehow use it implicitly just in the computation of the scalar curvature. And then they say, okay, now we know it's scalar flat and they forget about this fact. At least that's how I read it. Okay, so now, thanks to Bartnick and Bartnick's generalization of Witten's proof of positive mass theorem rigidity. We know that our resulting manifold is precisely R3. So positive mass theorem rigidity in lower regularity. Let's say low enough regularity. I'm not claiming that you can't prove it in lower regularity. Regularity tells us that M3 wedge union P infinity, G wedge is globally isometric R3 delta. Just for those of you who are interested, so far we have not made use of dimensions except that we're using here the positive mass theorem, but actually Bartnick's version is end dimensional. So up to now, everything works fine and dimensionally. Yes, did he? Really? Okay, I'll check for tomorrow. Remind me after the class please. I don't know, but definitely there is higher, by now we have higher dimensional versions of Witten's proof in low regularity by Lee and the flock for sure with no more than the necessary assumptions. And also of course we have a high regularity, Shane and Yao proof in higher dimensions, which I'm sure you've heard about last week, or I think so, but so we do have the theorems and I thought that this was fine, but I'll check again. So this globally isometric, so now we're globally isometric. So now this weird thing turns out to be flat Euclideans, okay, so this guy is in fact globally isometric to R3 delta where the isometry could potentially be of low regularity here and here as well as here and it's smooth everywhere else. So now we need to recover it, why is that helping us to know that we've been schwarzield to begin with? Okay, now we're in the same situation where we are in schwarzield, but why does that help us? And there are actually a number of different arguments how to do that, and I'm not gonna go into much detail, but in the original, so Bunting and Massoud, at this point, the site work by Robinson were actually already present in this paper of Miller, Thimhagen and Robinson and Seyford that I cited yesterday, Robinson and Seyford, should be 1972, where they use the cotton York tensor, which I'm gonna write down in a second from conformal geometry to conclude plus the static vacuum equations to conclude with schwarzield that we started with schwarzield, okay? So this argument is not due to Bunting and Massoud, and it's been there before. Just to give everybody proper credit. So the cotton York tensor, and this is for n equals three, this is the point where this becomes a three-dimensional proof, okay? The cotton York tensor is a tensor defined as a derivative of a Ritchie, I'll write it down in a second, so it's a third-order differential operator, and it sort of plays the role of the vial tensor in higher dimensions. So you know probably that from dimension four and higher, the vial tensor is zero if a metric is conformally flat. In three dimensions, the tensor that plays this role is the cotton York tensor. And in fact, there is a theorem by Eisenhower saying that it's conformally flat if and only if the cotton York tensor vanishes, but we're not gonna need that. So the cotton York tensor C is defined at that easy, most easily defined in abstract indentation. So it's two nabla Ritchie j k plus one half d k, of course not like this, k i nabla. R is the scalar curvature, rick is the Ritchie curvature, and C is the cotton York tensor, and the brackets are anti-symmetrization with a factor of one half, okay? Okay, that's the cotton tensor, so it's a third order differential operator, and the cotton tensor C vanishes if G is conformally flat. And actually, I forgot to say that if you didn't know any of these conformal geometry formulas, it would of course be a good exercise to verify them, the scalar curvature. And also this is a nice exercise to do if you want to get more familiar with conformal geometry. So what Millard, some Hagan, and Robinson, and Zyford do next is they take the cotton tensor, which involves the Ritchie, the scalar curvature in the metric, and of course covariant derivatives, and plug in the Ritchie equation from the static equations. So we had n times the Ritchie was the Hessian of n. You can write this as Ritchie ij is one on n, Hessian ij of n, and we know that the scalar curvature is zero. This was our static equations, and you can plug them into the cotton tensor. And then you get a formula that the cotton tensor is such and such and such in terms of the metric and the labs and derivatives of the labs. Okay, so C is then C of dn, probably up to third derivatives, and G, and a zero, because we already shown that we're conformally flat. And then what they do is go back to this Israel idea of looking at the level sets of the labs function. Of course, the tensor that zero everywhere is also zero on each level set. And then they conclude from this formula here that is lengthy, that each level set has constant n to begin with by definition, and constant mean curvature and vanishing teres free part of the second fundamental form. So it's ambilic. It has constant Gauss curvature and so on and so forth. So we cover all the identities we did yesterday in a different way, and then close the argument like Israel did by explicit computation of Schwarzschild. So we have the cotton tensor. This implies that each n equals constant, which yesterday we called them sigma two n is as yesterday, I put it in quotes round and mean intrinsically and extrinsically and then the same explicit computation as yesterday and it shows you that you're Schwarzschild, okay? So this is where this proof became three dimensional because we use the cotton tensor and the cotton tensor doesn't have the same nice conformal property in higher dimensions. However, there is an argument by, it plays the role of the vial tensor but the vial tensor has a different formula, completely different formula, yeah? It's a part of the Riemann tensor, right? So it has two derivatives. This has three derivatives. So it has the same spirit, but it's not the same thing, okay? So but you're thinking in the right direction because one, and I'll write down the reference next class, I forgot to bring it or I can look at it. No, I cannot. I'll bring it next class. Wang in, I think in 1980, in 1998 or something like that proved the same thing in higher dimensions. So all the first part I said it was immediately higher dimensional except all the numbers four get replaced by four over n minus two and so on. And he found a way of using the vial tensor here instead of the cotton tensor to make it higher dimensional. I'll mention this again tomorrow, okay? So there is a version of this recovery of Schwarzschild argument using the vial tensor which is a little bit more involved than the original cotton York tensor argument but has the same conclusion in higher dimensions by Wang. And maybe I'll just close here for today. Thank you.