 And she will talk about the uniqueness theorem for static black holes. Thank you. So after last class, three questions were raised to me that I thought would be interesting for everybody. There were also other questions raised, but they were, I think, more specific. But these three I want to touch on today before I continue with the planned content for today. And one question was, why did we look at this conformal extension of the Schwarzschild spatial slice yesterday? Because there's this much better extension by Kraskal, the Kraskal extension. And why wouldn't we look at that instead, which is because it's much better? And the answer is this conformal picture will appear in the proof tomorrow. So we needed for the proof. We also already needed the conformal coordinates on Schwarzschild for defining asymptotic to Schwarzschild last class. And also we are interested mostly in the Romanian picture, so the Laurentian generalization by Kraskal is not so relevant. But for those of you who are familiar with Kraskal, I just want to point out that you can recover this doubled part of the spatial schwarzschild slice in the Kraskal extension. So if you look at Kraskal, then this part here, get colored as this part here, t equals zero and r bigger than two m is the part that we re-route in isotropic coordinates and then we doubled it. And in fact, when we double it to get a geodesically complete Romanian manifold in the Kraskal picture, this means we added the reflected line piece here. So this then would be t equals zero s less than m over two, where this was equivalent to s bigger than m over two. So the Kraskal picture contains this doubled part. But we will not make use of the Kraskal picture just for those who know. The second question I was asked was, are black hole horizons always compact in the spatial slice, the intersections with the spatial slice? And the answer to this is, in my talk for sure, otherwise I'm not an expert, so I'll not comment. And in my talk, all the spacetimes we will look at have asymptotically schwarzschildian Romanian time slices. And then we asked, in the definition of that, that they split into a compact, so mn can be written as a compact set c, this joint union with this end en, which was diffeomorphic to yn minus a ball. And if you look at this, now if this has a boundary, the boundary cannot occur here because this is going out to infinity. So the boundary will be a component of the boundary of the compact set that does not attach to c, okay? To e, sorry. So here's a picture. So for example, this could be our compact set, sorry, no, this could be, this part down here under this plane could be our compact set c. And then the remainder of the manifold up here is isotropic diffeomorphic to r3 or yn minus a ball. And the boundary is just this. And because this c set c is compact, the boundary has to be compact as well, okay? Because it's compact, it can still consist of multiple components, yeah? But finally, many compact components and we don't make any assumptions about their topology. But when we talk about these sigmas that I introduced last class, the black hole horizons, what I will, the language I will use is that each connected component is one horizon. So if the boundary is not connected, we will just have multiple black holes if all of them are black holes, okay? Not necessarily connected. Static horizons, sigma n minus one are connected. So a picture like this shows you what I mean. So down here we would have our compact set c and the boundary of Mn then has two components and each of them could or could not be a static black hole horizon depending on whether the function n vanishes and the mean curvature vanishes, okay? And then there was a third question that was already asked in the auditorium and I didn't get it correctly and then it was again asked to me and maybe you could come up later and give me your name so I can thank you in the lecture notes and the question was whether a set that has n equals zero in the Romanian manifold is necessarily the cross section of a killing horizon. And the answer is no and I was given an example that I wanted to share because it's very similar to the metric that we discussed yesterday but different. So n equals zero, t equals zero does not have to be a cross section of a killing horizon and the example is the metric that looks like this and it doesn't even matter in which dimension I do this where n is one minus two m one over r and f looks almost the same but with a different mass. So if those were the same masses this would again be one of the metrics we studied carefully last class but now these have different mass parameters and we assume that zero is less than m one is less than m two then n equals zero is a hyper surface yeah and we cross it with t equals zero so it has such a set and that set here in this example coincides with when r is two m one and t zero as usual but now we have to be careful to know which mass but upstairs n equals zero is not a killing horizon because it's not a null hyper surface and horizons should be null hyper surfaces and how you can see that it's not a null hyper surfaces by computing it's normal and checking the causal character of the normal and if you compute the normal to the n equals zero set looking at this it will be f times dr is this function at two m one is not zero so this doesn't this makes complete sense and this normal is space like so it's a time like hyper surface and not a null hyper surface so much for the bad news the good news is that this doesn't create any problems for us because if you compute the mean curvature of this surface here all the sigma if you compute the mean curvature of the surface sigma and try to compute it because it's at the boundary again in the three dimensional set Romanian manifold or n dimensional Romanian manifold then it degenerates to infinity so you cannot extend the Romanian metric in t equals zero to this boundary so Romanian part g f squared dr squared does r squared d omega squared does not extend to r equals two m one and in particular the mean curvature is r approaches two m one goes off to infinity so in the Romanian picture this is an inner boundary and this inner boundary has n equals zero so n extends nicely to it but the Romanian metric does not extend nicely to it and indeed its mean curvature even blows up in this setting so this wouldn't be one of the surfaces we would be allowed to consider in our setup because we asked that the Romanian metric also extends to the inner boundary so this is not going to cause us any problems besides it actually doesn't solve vacuum Einstein equations anyway but even if it did it wouldn't shock us now okay great so now to black hole uniqueness so what do I even mean by black hole uniqueness by black hole uniqueness I mean the only static vacuum black hole in three plus one dimensions which is isolated to be schwarz yield and this is still very imprecise right so we can do it differently and we can ask a question we can ask a question question one can there exist single isolated black hole and I'll use bh as the abbreviation for black hole from that one in vacuum which is in static equilibrium with itself and not schwarz yield the answer to this question will be no which is why it's called black hole uniqueness and this is in the single black hole case and the intuitive reasoning why you may be expecting such a thing goes like this imagine you have a single isolated black hole which sits in vacuum and is in static equilibrium so it doesn't move it's isolated from all other influences so there are no external forces coming from very far away on it it's in vacuum so there are no external forces coming from any matter around it it's a single black hole so there are no other black holes exerting any gravitational forces on it so no external forces full stop but also it's only a surface right it's only a horizon of a black hole so there's no internal structure which could be in any way inhomogeneous so there's also no intrinsic reason why it could be anything but spherically symmetric so we can expect and it's static so they can all be dynamical forces pulling on it and making it into say an ellipsoid like the earth that rotates right so there's no pull on it at all except its own gravitation and there's no reason to anticipate it shouldn't be spherically symmetric but if it's spherically symmetric in addition to having all these other properties then we're exactly in the scenario where Schwartz looked for solutions and the only ones he found and he made a construction that is not losing anything on the way is Schwartz's shield so once we recovered spherical symmetry intuitively at least that for now we recovered that this is Schwartz's shield okay, this is the intuitive argument of course this is not a proof and we will see that the proof goes entirely differently and I am not aware of any proof idea that uses any of these notions explicitly and the main reason for this I believe is that we don't have a notion of force in relativity and neither do we have a notion of acceleration which would put us in a position to use a notion of force okay, maybe we don't have definitions of these notions so we need to do without them and you will see that the proofs work entirely differently than this intuitive argument but why you would even study such a problem arises from this intuitive argument okay, so now here to question two what about configurations of multiple black holes multiple such black holes and static equilibrium and the answer again is cannot occur so depending on whether you're a physicist or a mathematician you read this you read multiple to include one or not if you read multiple as to include one then cannot occur is not the right answer then the right answer should be then you have to be Schwartz's as well we will see that in the theorem next class okay, so but here let's read multiple as an everyday person or as I believe physicists do and exclude one okay, so why is that intuitively well if we had multiple say at least two black holes that were in static equilibrium they would gravitationally attract each other and that would make them accelerate and that would make them be non-static which is not the scenario we're looking at so no so then that means there's only one and if there's only one we already know that the intuitive answer should be the it should be Schwartz's yield pardon oh just repeat the argument okay, so if you had two black holes or more in static equilibrium they would exert gravitational pull on each other which should make them accelerate towards each other which makes them stop being static and there's only starting static configurations right and then that we're back in case one so what we'll do today is we'll study carefully a proof of this by Israel that was the first proof that was ever given of any kind of static back in black hole uniqueness in 1967 and tomorrow we'll we'll study the multiple black hole case with the proof by Banting and Masoud Al-Alam from the 80s that's a very good question we're gonna translate these questions to math statements in a second the asymptotics are hidden in the word isolated so let's translate it to mathematics so what's a single black hole in static equilibrium will of course be a static space time or let's write static system in the terminology we introduced yesterday m3gn it's easier to handle with closed in particular the single saying connected in a boundary which is a static horizon as defined last class we call this means that both g and n extend smoothly to the surface and mean curvature zero and the lapse n is also zero isolated will be translated to mathematics by saying that m3 gn is asymptotically short shielding with some mass m and r vacuum will be of course translated to the vacuum equations that I told you that come from the Einstein equations in vacuum and I'll rewrite them for convenience and I introduced the name static vacuum equations for them and then instead of single black hole we have multiple of black holes we will have partial m the boundary consisting of a union of finitely many this joint union as we discussed yesterday of finitely many surfaces sigma 2i and sigma 2i is a static horizon connected so in the single black hole case the boundary is connected and in the multiple black hole case the boundary is disconnected in each component is a single black hole so we can take now let's focus on the one black hole case so the boundary of all the manifolds we're looking at will be connected yes by inner oh there is typically if you say you have an annulus you have an inner boundary and an outer boundary okay if we have an asymptotically flat asymptotically short shielding manifold there's several different ways of drawing them and we draw them like this now so here's our black hole horizon here's where s goes to infinity and in some sense we call this the inner boundary because this is the infinitely far away outer boundary very good question and actually i wanted to draw exactly this picture anyway to tell you that i'm also using a very funny convention that is used in this field on the orientation of the normal which is going to confuse you also so we're calling a normal in this direction nu is always going to denote a normal this is going to be the inner normal sorry the outer normal because it points out to infinity so this is the outward uninormal pointing to infinity if you're a geometer and you don't typically work with asymptotically flat things your outward would point exactly the opposite direction because it would point out of the manifold here out always means towards the infinity and if you read papers in asymptotically flat or asymptotically short shielding manifolds some of them are like some of them in the archive at least are wrong because they cite theorems and use the wrong orientation of the normal so this is a really delicate issue so I'll say it again for me an outward uninormal to any surface in an asymptotically short shielding manifold will always point towards the infinity not this way then automatically by the gauss-godotzi equations the static vacuum equations for over there imply that several things are constant so first they imply that h is zero on sigma two and this is in fact also true for multiple black holes so I'll put a little i in brackets there to indicate that if there several black holes it's true on each individual one individually what is this this is the second fundamental form of this boundary surface inside the three-manifold so second fundamental form extrinsic curvature whatever name you prefer sigma two equals in this single case the boundary of m3 in m3 vanishes we already assume that the mean curvature vanishes this is telling you that the trace-free part of the mean curvature is also vanishing so the trace-free part is defined as h minus in two dimensions one-half trace h times sigma and the trace is also with respect to sigma where sigma is the induced metric sigma two so okay so we have our surface sigma two sitting inside our space m3 which has a metric and also of course a left function and then it has a second fundamental form with respect to the normal pointing outward to infinity and the mean curvature was assumed to be zero and now automatically from the static vacuum equations we also get that the trace-free part as defined there is also zero which means if you're not familiar with this it's called conformal also uh... so sorry umbilic which means if you stand at any point on the surface in normal direction like you stand on the earth and you rotate in your tangents and look at your tangent space at your felt point the curvature of what you're looking at is the same in each direction right now the mean curvature is also zero so then the curvature is flat the extrinsic curvature vanishes this vanishing finish the sentence this vanishing of the trace-free part of the second fundamental form alone just tells you the curvature is the same in each direction and doesn't depend on the angle and I will maybe not be able to stop myself to call surfaces with this condition umbilic this is the name for this condition okay so this just follows from the equation another thing that just follows from the equation number two is that the normal derivative of n how would you write this nabla nu n or the inner product of the gradient of n with nu or dn nu it's all the same thing I'll write it this way is constant on sigma two the level of the normal the left's function n is zero on the surface and it's normal derivative is a constant so it's independent of the fault point on the surface that's what this is saying to those who know what that means this is equivalent to the surface gravity being a constant and again if the surface had multiple components this would be true on each component but it would be a different constant potentially on each component and then the third thing that holds automatically is that scalar curvature of the metric sigma so the scalar curvature of this two-dimensional surfaces or in other words the Gauss curvature is constant as well so we know very very much about the intrinsic and extrinsic geometry of a static horizon in a static vacuum and space time we assume the mean curvature to be zero capital H so this is this is the capital H and we assume to be zero okay so we're assuming the mean curvature to be zero in order to for it to be a horizon so minimal surface condition and then we get for free that it's also ambilic from the static equations by Gauss-Gaudazi this is R within an index sigma yeah okay we can let me write it more pre-leaf more questions and this is for those of you who are interested in doing homework to verify this by explicit computations can be your first homework exercise for today exercise six so here comes the theorem and in the original version let me first write that it's by Ishael 1967 and the precise references in the references I gave you yesterday so Israel when he first wrote this down he put enormous amounts of conditions that I'm not going to write because today partially because we know these things are priority we don't need to assume them anymore so some of me he formed one thing he assumed for example was this but today we know it follows anyway okay so I'm not going to write it in the original version I'm just going to write it in a way that it makes sense today so let m3 bn the ecstatic system vacuum asymptotic to Schwarzschild some mass m bigger than zero with a single black hole foundry human attention this is a new condition that we haven't seen yet that the uh... differential n doesn't vanish anywhere in m3 the corresponding space time is isometric globally tool for the to the schwarzschild man space time of mass m this assumption is essential for the proof that israel gave as you will see and it really is meant as there is no point where it vanishes not there's some point where it doesn't vanish yeah it's nowhere vanishing and before I show you anything about the proof and let me show what this condition allows us to do when that of course will be then useful in the proof but let me first not mention anything from the proof explicitly anyway so the condition that dn is non-zero in m3 on sigma two of course is zero because it's a static horizon tells us some sense automatically that this is connected why is that this says actually does something which is called fully aching the manifold the condition that the gradient doesn't vanish anywhere by the implicit function theorem tells you that each point in the manifold lies exactly in one level set of the lab's function yeah so and these are my names for the level sets this is the level set of value n level sets 48 m3 and because the horizon has n equals zero it's one of the level sets of this fully aching horizon we called before sigma two but in this node then we could also call it sigma two zero as a level set in this fully aching the most inner one and asymptotically n goes like one minus m over r s asymptotically by assumption it's like one minus m over s plus lower order terms and this implies that n goes to one and for n equals one minus a small epsilon we have that sigma one minus epsilon is approximately a large sphere of some radius okay so the asymptotics of this n are almost such that if this if this wasn't there then n being constant would mean s being constant okay but s is the radius in our coordinate system so that would be a large sphere with radius s that you can compute this is one minus epsilon so epsilon would be m over s so s would be m over epsilon okay so now they're lower order terms so this is only an approximate identity so then of course from this condition we don't only know, bless you that the level sets foliate the manifold we also know that they all have the same topology and of course large spheres have spherical topology so which means that all of them have spherical topology and in particular they are connected so this argument shows us that our horizon sigma two zero is the connected surface and in particular is a sphere so this already allowed us to fix the topology of this boundary and it shows us that the assumption that we have a single black hole is actually superfluous and we could have dropped it because it follows from this together with the asymptotics okay let me say this again in practice we wouldn't need to write this assumption here because it follows from this non-degeneracy condition of the of the differential of n as I explained in particular also it follows that it's a sphere now if you look better at this condition our sigma the scalar curvature being constant if we know it's a sphere and the scalar curvature is constant then this constant has to be positive by Gauss-Bohne theorem so in our three we have that this is a constant and indeed this must be a positive constant by Gauss-Bohne so maybe it's now a good time for a picture so I'll draw it here so now I'm looking from the top onto the Romanian manifold I'm not drawing it curved I'm just drawing it flat and then we have our boundary of m3 which is sigma 2 which is also sigma 2 0 in our foliation being part of a foliation here where we have sigma 2 say n equals one half and here we have a leaf sigma 2 n equals three quarters and then maybe a little bit further out again we have a leaf sigma n is seven over eight or and so on we never reach one because one is at the infinity of the manifold okay but this is like a set of concentric spheres only we don't know they're round we don't know they're concentric and the background is curved but intuitively we have this foliation by these spheres topologically we know their spheres and we know the inner one is a round sphere because we know it has constant scalar curvature but the others don't need to be round and there could be anything but topologically they're spheres also we don't know if they're extrinsically round so what is their second fundamental form except for the horizon where we know it's extrinsically flat another thing that we can learn from the asymptotics the following and we said last time that the static vacuum equations imply that the function n is harmonic and as a consequence of this by the divergence theorem in fact the integral over sigma two n and now it's n one of the normal derivative of n dA where this is the area measure with respect to sigma sigma so let's go in this picture now on each of these surfaces we have a normal nu pointing towards infinity and an induced metric sigma we're going to give them all the same names and the second fundamental form h and the mean curvature capital h anything else I don't think so so we can take the normal derivative of n on any of these level sets and integrate with respect to the area measure of this over this surface and because the function is harmonic we will get and this is the same as integrating the exact same thing of a different level set so the integral of the normal derivative of n over this level set is the same as on this level set is the same as on this level set is the same as the integral over the horizon for any in two in the range zero to one that n takes okay by the divergence theorem then plugging in the asymptotics and end of G we can compute that this integral which is independent of any of these level sets is equal to four pi times the mass and this is another homework exercise so in particular on the horizon the integral over nu of n dA is four pi n and we assume m to be positive this tells us that nu of n is positive so far we only knew it's a constant or vice versa if we assume the surface gravity nu of n to be positive then we know that the mass is positive okay as we like so now this constant is positive and this constant is positive and i can erase this now what else can we learn from this we know n is zero and i already implicitly used it in here n is zero on this and n goes out to one here so n is increasing n is increasing to infinity from from zero to one as s goes to infinity but if n is increasing wait a second then the normal derivative of n must be positive so we didn't need to assume that anyway so this implies also that nu and n is positive and this implies that m is positive so Israel wouldn't have needed to assume this either and there were some other conditions as i said that i didn't write this is just the conventional way to write it so let me say again we know from the Gauss-Kodatsu version of the static equations that nu of n is a constant on the horizon from the fact that n is growing or increasing we know that nu of n is positive so it's a positive constant on the horizon okay so now let's begin with the proof and the basic idea is already hidden in some of the things i've said the basic idea is to use n as sort of a radial coordinate in the space it's a radial so in order to make it coordinate we still need two other functions because we're in a three-dimensional manifold so what Israel did now is he says okay we know this is a round sphere so let's just take theta and phi as coordinates on this and n is a coordinate and this is a function in this direction which is increasing and it has no vanishing, never a vanishing gradient so we can use the gradient of n to flow these coordinates out to infinity so we can take theta and phi on the surface the angular coordinates and flow them out and have coordinates theta and phi everywhere they might not mean the thing we usually know, but they're nice coordinate functions I'll write that down let's call them y1 and y2 or in other words y capital i, now my capital indices are 1 and 2 only on boundary from 3 sigma 2 sigma 2 0 and flow them flow m3 to get coordinates sorry along right gradient of n with respect to the three metric to get coordinates y1 y2 on m3 now if you're worried this won't work globally in this condition that the gradient doesn't vanish tells you that it works globally in n direction but of course on the sphere we don't have global coordinates so in reality you will need to do this on a patch on the sphere and then you'll get coordinates here and then on a different patch on the sphere but I'll completely gloss over that detail so now that we have these coordinates they are super useful to study the problem and let's introduce one more name let's introduce a function row to be a function on m3 given as 1 over the normal derivative of n it's a little bit stupid I drew the picture right over there so I have to walk back and forth all the time so now for any point this function row is defined as follows take a point in the manifold it will sit exactly on one of those lever sets so it will have a value of n and that will be its n coordinate and will also have a y1 and a y2 coordinate of course but it also will have a normal derivative of n to this surface evaluated at it this is the number and this is how the function row works and because n is growing this is a positive function okay so this is really a function on m3 and if you do the computation we write the metric g of our space by writing it as row squared dn squared plus sigma where sigma is the metric on the surface sigma 2n to which the point belongs so if we have a point it belongs to one of the sigma 2n's this is the induced metric on this surface this is dn and this point and this row is evaluated as I just explained at this point so this is how you can rewrite the metric this looks familiar to Schwarzschild that's because it works for Schwarzschild as well okay so this we have and how to compute this is by the chain rule okay now that we have this we still have the static vacuum equations that we know to hold and we can plug in this form of the metric into the static vacuum equations and if you do that we can take exactly the same approach as you do when you split the Einstein equations into constraints and evolution equations and now split the static vacuum equations for into constraints and evolution equations for the function row and the metric sigma quotes on evolution because it's not really evolution in time it's evolution in n and then you get tons of equations that isn't a good idea I leave the theorem you get tons of equations so what we're gonna do is we're gonna forget some of them but keep the following three I'm gonna write them and then say something about them they look awful the last one is a little shorter okay so row we introduced h is the mean curvature n is the coordinate n, h comma n is the partial derivative with respect to the coordinate n of the mean curvature this is the Laplacian with respect to sigma and all the remaining things I believe should be clear so these are this is an evolution equation for row and these two seem to look like evolution equations for h on each sigma 2 n where n is in 0 and 1 and of course we're dividing by n so we need to be a little bit more careful but we gloss over those details here all of these can be made sense of also in the limit n going to 0 and now here comes the second cool idea by Israel we introduce a number or a function tau as the determinant of the matrix sigma ij this is the matrix of sigma with respect to these angular coordinates that we chose and we take its determinant think of this as the the quantity you want to take the square root out if you want to compute the area element okay and this is how we're going to use it that's one thing and then the other thing is then I want you to notice that the bracket is positive, is not negative okay there's a square and there's a square so what he does now is he takes the first two and transforms them into inequalities by dropping the bracket and he uses this quantity introduces this quantity and writes down two very nice inequalities between all the things in here including the tau okay so drop the square brackets then we get two nice density inequalities the partial n derivative of the following thing square root of tau h root of rho n is less than or equal to minus two square root of tau divided by n the plus in sigma of square root of rho and the other one is partial root n square root of tau over rho times h n plus or over rho is less than or equal to minus n square root of tau the plus in sigma of l n rho plus r sigma okay this is a matter of computation it's a little bit lengthy but there's nothing ingenious going on at all and all of these equations you can find in slightly different language in Heuser's book or I'll give you a reference at the end of the last class to a paper of mine where you can find them in this notation yes that's a very good question the initial the row is like one on nu of n right and nu of n was a constant on the horizon and that constant gives you the initial condition and same for h, h starts like zero, yeah okay so now as she already indicated this is very nice to look at as kind of also an evolution in the quality for the term oops in the braces and so what we're going to do is we're going to integrate it over n and then integrate also over sigma 2n so integrate from zero to one integrate oops sigma 2n all of this and now of course the integral from zero to one is the n and now we have to take a brief moment to discuss how we integrate over sigma 2n we already have what will be the functional determinant included in our inequalities as a square root in each term on each side so we're not going to integrate square root of tau dy1 dy2s but we're instead going to integrate just dy1 dy2 because the functional determinant is already in here dy1 dy2 both inequalities so if this happens let's first look at the right-hand side both right-hand sides are proportional to this functional determinant so we immediately recover the area element square dy1 dy2 is da so fine great so we left something that depends only on n which is constant on sigma 2n so it can be taken out of the interior integral and something which is a Laplacian now an integral over Laplacian over closed surface by the divergence theorem vanishes okay so we use divergence theorem the exact same argument applies to this Laplacian of the logarithm here we get an integral over the scalar curvature over the surface by Gauss-Bounet we can compute this explicitly we already know it's a sphere so we get a number divergence theorem so we've treated the right-hand sides on the left-hand side we're going to do a Foubini theorem to interchange the order of integration which allows us to use the fundamental theorem of calculus to get rid of this derivative Foubini and fundamental theorem calculus and then we can complete it completely explicitly compute everything and because by the fundamental theorem we just need to evaluate these expressions without the square root of tau because it disappears into the area element for n equals one and for n equals zero at the boundaries of the integral okay for n equals one this means we need to exploit the asymptotics that we've prescribed and for n equals zero it means we need to use the initial conditions that we prescribed on the horizon okay so asymptotics plus conditions on horizon and all of this together leads us to certain numerical inequalities between the only things that we have and so quantities at infinity is the only quantity at infinity we know is m so all the things that we compute by exploiting asymptotics will depend on m that'll be m or m squared or a square root of m or something and the conditions on the horizon all the numbers are zero except the scalar curvature which is a positive constant and the normal derivative of the labs which is a positive constant these two constants we still have and so if we express everything in terms of this what we get is one on four nu of n at n equals zero and also let's call this 12 and also the area of sigma two divided by four pi is less as bigger than or equal to one on four nu of n at n equals zero squared 13 but we also have which was called five this integral identity that the area of any of these guys divided by four pi equals times nu of n and n equals zero recall we said the integral over nu of n divided by four pi is m now i'm because nu of n is a constant i can take it out of the integral and just get the area okay so now we have this involves three quantities the mass the normal derivative which we know is a positive constant and the area of the horizon the area is a good replacement for the scalar curvature by Gauss-Poulet so these are the three numbers we have and we have two inequalities and one identity and the magic is that the the directions of these inequalities are exactly opposite so all of this can only hold true if both of these have to be identities okay so this implies by a computation that equality holds in 12 and 13 so here comes the full magic of this we derived lots of equations from those we dropped positive or non-negative terms to get inequalities then we integrated these inequalities simplified everything plugged in everything we knew and got inequalities these inequalities are identities so everything we ever integrated also had to be identities which means that this the content of this square bracket had to be zero to begin with because they're both positive this means that each now each level set is ambilic and on each level set row so nu of n is a constant okay i don't have room to write this anywhere i'll write it here in a color so this is from above the square bracket zero so h circle is zero and row is constant and that's nu of n is constant on each sigma two n we started knowing that this is true on the horizon now we know it on each leaf of the foliation in this picture okay and now if you plug this information back into seven eight and nine row being independent of the angles tells you row n is also independent of the angle so h is constant as well on each of them so each of them is now a constant mean curvature surface which is ambilic and has constant nu of n we already started by knowing they have constant n because that's how we define them then let's look at this now all of this is independent of the angles this is independent of the angles this is zero so the scalar curvature is also constant okay so all of these things are constant or zero even on each leaf so now each leaf is intrinsically around sphere and extrinsically around sphere in the sense that it's ambilic and it has constant mean curvature and in fact these equations if we use them quantitatively and not just qualitatively as i just did we can compute all these constants in terms of n okay and then this allows us to explicitly compute that everything is schwarz yield okay so this is really a fun idea to to split everything as a problem on surfaces and of course we made tremendous use of this condition which looks like a little bit of a technical assumption because how many critical points will a harmonic function even have yeah so now it's the question that someone you already asked last time like why do we make so much regularity assumptions that that you can interpret as a regularity assumption as well and there's a generalization of this i need to look up the names so i don't say anything bad by by miller some hagin robinson and someone called cyford from 1973 the references in the heusler book who to take this argument and use more advanced techniques and a lot of knowledge about harmonic functions to to reduce the amount of the number of points where this can happen so much that they can actually go through them and get rid of this condition okay get rid of and at the same time they also get rid of a lot of the other regularity assumptions that we made as you asked yes last class also much lower regularity at the horizon regularity the horizon and let's end here for today thanks