 that it is a great pleasure and also honored to be able to wish you a happy birthday. I had a great, again, privileged and honored to work with Yvonne Chocke-Broyard during the last five years or so about the characteristic coffee problems. So what I would like to do today is to, no, no. So I would like to report on some work we did together and on work which evolved out of our joint collaboration. So the recent results I'm going to talk about are in collaboration with my PhD student team pets. I think this is probably the bottom line of my, or the punch line of my talk. Which has the following. If you have a smooth, light cone without conjugate points on this funny set, I'm going to explain what this means. Then this is the formula for the something called the Troutman-Bondie mass. And so, well, this is the mass. This is pi 3, 1, 14. Well, this is something I'm going to explain to you what it is. I'm going to explain what this is. I'm going to explain to you what this is. I'm going to explain to you what this is. Why is it interesting? Well, you will notice that everything here is positive. So it's in vacuum. So if you want to prove that the Bondie mass is positive, well, it's here. That's your proof. So let me start to tell you what is M. Now, in 58 Troutman and in 62 Bondie published papers where they tell you how to calculate the mass of a null hyper surface. So suppose you have some spacetime. And in this spacetime, you have a surface which is null and goes all the way to infinity. Then they told you what is the right expression for calculating the mass of this object. I'm going to explain you how they invented that and so forth. And I'm going to tell you what this mass is in this setting. I'm going to tell you how it's going to be in the setting of the light cones. And that's going to explain you also these symbols there. But in any case, there is this notion of mass which has been defined by Troutman and Bondie. And it is some kind of limiting integral as you go to infinity. So this is M is the limit as R goes to infinity, 1 over 60 pi of integral over S2 of some expression built out of the metric, which they tell you what it is. And this is this Troutman-Bondie mass. So this is something you associate to any null surface. Well, I'm not going to work with any null surface. I'm going to be interested in null surfaces which are actually light cones. So you take a point in spacetime, and you shoot light rays from this point. And this is the null surface I'm going to look at. So what is known about this, let's say, Troutman-Bondie mass? Well, first it is already known that it is positive. So M is positive if the light cone. So the surface, well, this is a little more general. You don't need to have this light cone here in their settings. So you can just think of this asymptotic region of this characteristic surface. So this mass is positive if the light cone, if n is asymptotic to a well-behaved hypersurface. So in other words, you have a space-like hypersurface. So if you have a space-like hypersurface in here which has the same asymptotics, then this thing is positive. So this is something which can be proved by using the Wheaton equation, the Wheaton type of proof. So this statement is much more general than mine here, because mine is just tied to light cones. So there is no issue that the positivity needs to be established. This simply gives you a very simple and elementary proof of positivity. Well, one other thing which is known is that this Troutman-Bondie mass is decaying. If you move surfaces to the future, so if you take another surface, which is the future to this one, then its mass will be smaller. This also means that if you go backwards, the mass will be larger. So in fact, if you have proved that the mass is positive on this cone, it's positive for everybody before. So you get more than just the proof on the cone. Well, this is a question mark, but it is expected that the limit, as you go with this, so suppose that you can think of this surface as being u is equal to t minus r. And as you go to, you push this surface down, so you go with u to minus infinity. So of this Troutman-Bondie mass is the ADM mass. Well, this hasn't been proved to my knowledge. There is a paper by Ashtekar which proves something like that, assuming a lot of things about the asymptotics of the spacetime, a lot of things which we do not know whether they're true or not. So presumably, and I think it's physically reasonable to expect that this will be true. But then, of course, as I've told you, well, if you go backwards, the mass only increases. So this limit, if it is positive at late time, it's going to be positive always. And therefore, you get an elementary proof of the positivity of the ADM mass as well in this setting for any smooth. In a spacetime where you have a smooth light cone without conjugate points. So this is my formula, or our formula, for the Troutman-Bondie mass. Well, this one other property which I could mention here that this Troutman-Bondie mass is actually unique when, so unique functional of the field, assume which satisfies two, satisfying two, and invariant under something called super translation. So this is an interesting object by its uniqueness properties. One more last comment here is that, well, since this thing decays and all these things are positive, you get explicit bounds on these integrals on no surfaces in your spacetime. So this is something which could be useful for studies of stability of black holes or other things like that. Well, I have written this formula in the case where your surface starts from a point. But of course, if it started from initial data, you get a boundary term here coming from initial data which you control. So you get some energy type quantity which perhaps might be useful to say something more about the dynamics of your spacetime. It's certainly bounded for all data times. OK, good. So now I've kind of explained to you what M is and 16 pi is. Let's see, so maybe I should start with sigma. And to tell you what sigma is, let me just go back and tell you something about the Cauchy problem in general. And well, let's take the simplest equation possible, Laplacian phi equals 0. Well, the spacelike Cauchy problem, you have to give your initial data on the surface. And its time derivative on the surface, you get a solution. If you are on a light cone, which I'm going to assume from now on for these hyper-surfaces, if you want to solve this equation, so you only provide as initial data phi and C0. OK, so that's how you solve the wave equation. That's the difference between the spacelike Cauchy problem and the light cone. You don't give the function and its derivative. You only give the function and plus some regularity at the vertex, which is well understood. Good, so this is for the wave equation, for Einstein equations. Well, this is something that Yvonne has taught us. If you have a spacelike surface, what we can do is just give the full metric on the surface. All the time derivatives of the metric at the surface. Well, these things are not quite free because there is four relations called the constraints, which especially snow. And if you don't know the area, it's no much point that I write them. But there's four relations between the spacetart of the metric and the time, well, roughly speaking, between these objects. It's not quite true, but that's essentially this. So you give all the components of the metric, all the time derivatives. They have to satisfy four relations called the constraint equations, you get a solution out of this. Now, on the light cone, if you give me g mu nu on this cone and one relation, which I'm going to explain shortly, and regularity at the vertex, then you get a solution. So the claim is that here is your light cone. You get initial data. And there is a subset of, well, there's a neighborhood of a specific subset of the light cone where you get a solution. So this is the picture from a PD point of view. Now, if you want to do the geometric picture here, well, what often thinks that rather than giving these two, one gives gij and something called kij, which is related to these derivatives. And four relations, so plus constraints. So in other words, well, the extra data here, which you're free to prescribe, are just gauge artifacts. So you choose them whatever you want them to be. And for different data, you're going to get the same solution up to isometry. Now here, the geometric data will be something that I'm going to call GAB and a connection coefficient kappa. And so this constraint equation. So let's see. So the G bar ABs, what are those? Well, if I have a null surface, I can induce from this spacetime metric a tensor field on this surface. And so this tensor field I'm going to call G bar. And this is a two covariant symmetric tensor field of signature 0 plus plus. And the 0 is coming from the fact that this is a null surface, so there is a null direction involved. So this is my G bar here. Now, and if I use coordinates RxA, adapt it to this null direction, then J will have no dr square component. It will have no dr dxA component. And it will only have angular components. So this is the G bar. And what is kappa? Well, in this coordinate system, if you want to look at the associated spacetime, this is a connection coefficient here. It's telling you the following. Well, d over dr is a null vector on this null hyper surface. So by standard result, it is auto parallel, right? So if you start moving it along itself, it's going to be parallel to itself. So kappa is this connection coefficient related to here. And the constraint equation is the Ray Chaudhury equation, which says that dr kappa plus tau kappa plus 1 half sigma square plus, is this plus or is this minus? Actually, it's minus here, minus kappa tau plus 1 half sigma square plus sigma square plus 1 half tau square equals 0. So, well, you know already what kappa is. This is this connection coefficient here, which, by the way, you can set to 0 by choosing your r appropriately. But you don't have to. And so this is the equation. Now, what is sigma? Well, I have to define something which was called, I think, kappa chi plus in the last lecture. And it's in my notation, just chi. This is going to be the radial derivative of this, 1 half dr dA B. So then sigma tau is something which was called trace chi plus in the previous lecture. And it's often called theta or theta plus. And this is the trace of this thing with respect to this two-dimensional metric. So as I said, most of the literature uses the symbol tau or tau plus. Well, Yvonne has told me that I have to use tau when he started working on this sign following what I've been told. And so sigma will be just the trace-free part of chi. So sigma AB would be, it's called the shear. It's so chi AB minus 1 half tau GAB. Good. So this is the Rachiduri equation. And this is the only constraint in the theory. And for my purposes, there is a useful observation is that if tau is positive, then I can just get rid of this equation in a, OK, good. This is better. Sorry. Good. If tau is positive, then I can just solve this equation trivially for kappa and get rid of the constraint. So kappa would be. But I do the expanding direction. Don't you need a pass there? kappa times tau. Yeah, I don't care. I don't care here. But I think, let me just check. Maybe you write. No, I think it's all right. kappa is negative? Whatever it is, as long as tau is positive, I can just solve the equation in a stupid way by writing that this is 1 over tau dr tau plus sigma squared. So maybe I have a sign wrong here, right? Let's just. No, I think it's OK. Well, if tau is non-zero, whatever the sign and coefficient there, I can always do this, right? And I have solved my constraint. And then in this sense, this is the same as just giving g. So the initial data, I can freely prescribe this tensor G squared AB, which is a geometric object. And this turns out is good enough to determine the whole solution in this tau larger than 0 case. Now, of course, the assumption tau larger than 0 is restrictive, but when you have a smooth light cone, you necessarily have tau larger than 0. You see, in Minkowski space, tau would be 2 over r, so it has to be dr tau plus 2 over r tau. In Minkowski space, time kappa is 0. And so whether it's plus or minus, it doesn't matter. Tau is 2 over r, but kappa is 0. Kappa is the acceleration. So kappa is this. Tau is 2 over r in Minkowski. And in the standard parameterization, kappa would be 0 in Minkowski. Now, why does this have to be true if you have a smooth light cone? Well, if you take a general space time and start shooting geodesics from some point, these things will start intersecting. So even though the thing was smooth at some stage, I don't think I know how to draw this. But in general light cones, you expect, well, in general. In many situations, you expect light cones to stop being smooth because geodesics will start intersecting. Therefore, any CRM of this kind for any smooth light cone without conjugate points is going to be restricted. In fact, you can ask, are there any space times at all which are reasonably well controlled and which have smooth light cones? And Helmut explained to me once that this is true and follows by a kind of Covino-Shangluan construction with small data space times. So there isn't any issue. But it is a byproduct of what we're doing here that there's a lot of space times here anyway. Just directly can be constructed using these data. So now, what is the story about tau-loader equals 0? Well, this is actually rather easy to see from this equation. If you take once you have, well, suppose when you have a light rays, you can always parametrize them so that kappa equals 0. So this term is a gauge-dependent thing. You can always get rid of it regardless of this story out there. So if I do this, that I see from this equation that tau is a monotonously decreasing function. So this is positive. So the derivative of tau is negative. So tau is decreasing. Now, if you have a light cone in an asymptotically flat spacetime, for large distances, tau will behave like in Minkowski 2 over r will be positive. So whenever you have an asymptotically flat spacetime with a smooth light cone, then necessarily tau will always be positive everywhere. So this hypothesis, that tau is larger than 0, I get it for free if I know that there are smooth light cones in my spacetime. And then I can solve kappa using this equation. And then once I know what kappa is, I can change my r parameter so that this kappa becomes 0. So after doing this, change r and get a equation, Reicher-Dury equation with tau equals 0. And so get Reicher-Dury with kappa equals 0. Good, well, so at this state, at least you know what sigma square is. Now here I'm assuming that r is, I am in this kappa equals 0 gauge. So I started with any G squared AB. I am assuming that this derivative, so that this function here, the G squared AB, dr G squared AB is positive. I can solve for kappa and go to a geodesic parameterization where kappa is 0. So then this field is manifestly geodesic. Now let's see. So I can tell you now what psi is. I should have written before. So psi is just an explicit integral. Did I? OK, good, well. So you know what psi is now, right? So this is your tau. This is 2r, which is coming from light cones in Minkowski spacetime. And so this is this expression here. Now what is this phi? The phi is the only thing I cannot, I don't know, right offhand how to give you an explicit expression for, but I can tell you how I can construct this. What you have to do is you solve the equation dr square phi minus sigma square phi. Is that efficient here? I think maybe there's a 2 here or something like that. Equal 0, with the condition phi equal 0 at 0. And d phi over dr at r equals 0 is 1. And then this big phi would be lim as r goes to infinity of phi overall. So when you play a little with this equation, it's rather simple. You can see that this limit exists and is larger equals 0. So the only thing you need to worry about is phi equals 0, because if capital phi is 0, I have a problem there. So what I'm saying is to obtain this formula, there are two restrictive conditions I need to impose. First, that tau is positive, and second, that phi which could perhaps vanish in some direction is positive. So if I take any initial data like that, which are asymptotically flat in the usual sense, so g bar AB, this is the angular part of the metric, so it has to decay to Minkowski spacetime allowed infinity in a suitable sense, then as soon as I have these two properties, I have this formula for the Schrodinger-Bundemanns. Now, for the experts, just a comment on this. Wait, wait, wait, wait, OK. It's actually psi. It should be psi, but yeah. So if you think in terms of your conformal completion, so you have your Scriplus here, and you have your light cone emanating from a point, and you're assuming that this light cone is globally smooth in spacetime, so there are no conjugate points and no self-crossing and anything like that, then when you're getting to Scriplus, the thing which could perhaps happen is that you get a conjugate point sitting exactly on Scriplus. So if you do all your scaling, it's not an infinite light cone anymore, but it's a light cone at finite distance, and you could have a conjugate point which sits precisely here, and this would correspond to phi, capital phi equals 0. So I have to exclude this possibility. So I have to assume that my light cone is smooth and there isn't a perverse thing which happens precisely at Scriplus. OK, so we don't know what psi is. Well, it looks like zeta, but it's actually psi. Now psi is whatever you need more to have a complete connection in the d over dr direction. So if you write dA d over dr, well, there's a fine normalization which we use. So kappa together with psi are just the connection on the space on the bundle spanned by d over dr. So that's psi here. And now this psi, you can calculate it out of the initial data. There's an equation for it which I'm not going to write down. Or actually, well, I can just write it for you, but it looks like something like that. dr plus tau psi A, well, modular coefficients, you get something like that plus dA tau plus dA kappa equals 0. So in vacuum, you get an equation like that, which if you're given kappa, which we have, and if you're given tau, which we have, because this is this trace of the chi, and if you're given sigma, which we have, because it's the trace free part of psi, we can just integrate the equation starting from the tip of the latecon, and we get our xi out of this. So now you know all the ingredients. And well, let me show you how you prove this formula, because it's really elementary. Where do I get more blackboard? Probably just pulling this down. Say it again? Right, yeah. So the definition of M, well, you take any textbook which has a bondy expansions, and you get a formula which I told you, but I'm going to give you the formula in when you translate it to this formula is. And so, you know, I mean, this is so simple, I can't believe that this isn't already somewhere in the literature. So if someone has seen this formula, I'd be glad to see it, because how could people have overlooked it? But OK, so well, there's a calculation which you take, as I said, take the bondy definition, right? So bond is definition, and you're going to get that the mass is equal 1 over 16 pi. And there is a new symbol which now is zeta coming in. So this is, you can just believe me, there's a calculation you have to do. You're still don't know what zeta is. It obviously can't be wrong, but I'm going to tell you what zeta is. And what is tau 2? Well, if you are asymptotically flat, then tau is equal, for large distances, behaves like the expansion of Lydkon-Dinminkowski, which is 2 plus over r, plus a correction term, which decays like r square, plus higher order terms. So this correction term tau 2 is the thing which appears here. And what is zeta? Well, zeta, it's again one of these, there's an equation for zeta, which is important and well, which you see the positivity coming out of it right away, which says that dr plus tau zeta plus the curvature of these two-dimensional matrix GAB minus 1 half my psi square. And this is exactly this term, 1 half over 2 coming there. And plus a divergence of this psi is 0. So this is one of the equations you get out of the Einstein equations, which determines my function zeta. And by the way, I should say that zeta is actually the following. Well, once you have your Lydkon, and you have the cross-section of this Lydkon, then I've told you that tau is the divergence of the Lydkon in this direction, zeta is actually minus the divergence in this direction. And don't ask me why. I mean, this notation psi and zeta is coming from our joint work with Yvonne. And so I'm keeping it for historical reason. Maybe if I had to rewrite everything, I would use those notations. But in any case, this is one of the Einstein equations that you're going to get in vacuum. So you can integrate this equation with the right boundary value, set the origin. And in fact, for small r, you know that zeta is equal to minus 2 over r plus o of 1. And for large r, just by analyzing this equation, see, once you know zeta, you can calculate this. You can calculate this. You know the two-dimensional matrix you can calculate this. You know what tau is. You just integrate radially, and you get zeta. And so for large r, zeta is equal to minus 2 over r plus a square, a term over r2 plus o of r minus 3. So now zeta 2 is a coefficient which appears in the expansion of zeta for large r. So let's start again. You give me a G squared AB and a kappa so that tau is positive. And this is a necessary condition for a smooth light cone, anyway. Then I can calculate kappa from an equation which I probably erased, and then go to a gate where kappa equals 0. Then I can solve the equation for zeta, which have I erased it or it's already somewhere. Are you erased it? For zeta, not for psi, sorry, for psi. Thank you. So I can integrate psi. Once I know psi, I can integrate zeta. I can calculate the asymptotic expansion of these things. So I can find this coefficient here. I know the coefficient of expansion of tau here. I plug it in this integral, and I'm going to. This is my mass. So by the way, this is obtained by transforming Bondi's expression into expressing Bondi's expression in this framework. So Bondi's assuming various coordinate conditions, so you have to build them in. And if they're satisfied, you're going to get this expression. And now this expression involves only things which are intrinsic to this hypersurface because everything has been defined in terms of G bar AB. Everything I did here starting from G bar I can obtain. So I don't care what's happening with coordinates of the surface or anything like that. And I could just say, well, this is a definition of the mass. Which would be a silly thing to do. But I could do this without. But in fact, it is exactly the same as Bondi's mass when Bondi's mass is defined. So for example, in spacetime, this must go. Good, so now we want to know how to prove this. That should be, let's see how I'm doing this time. OK. Yeah, well, I don't need to. Yeah, so I probably just need 10 minutes or less to finish the proof. I'll stop there. So I need the right to do the equation. I don't want to erase it. Well, you see, this equation produces positivity when you integrate it over spheres for a very simple reason. The divergence goes away. Because if you integrate the divergence, it goes away. And you can use Gauss-Bonnais to integrate R bar and to get something simple. And this is our positive term. Now, this term tau zeta here is exactly what you need when you're differentiating an integral of zeta d with respect to the physical measure. Because you're going to get a term which is the derivative of zeta. And the derivative of the measure produces exactly this tau squared, tau tau. So if you use the equation, you're going to find that this is minus integral of R bar, which is 2 pi, 4 pi, 8 pi, 8 pi, 8 pi. On this here, 8. Because R is not the Gauss curvature, but the scalar curvature, which is 2. So this produces 8 pi. The minus psi squared term becomes plus 1 half integral of psi squared. And the divergence goes away. So I have this very clever idea to write this as the integral of S2 of psi plus 8 pi R is equal 1 half integral of psi squared. Good. Now, this thing turns out to extract this coefficient of this integral when you integrate, go to infinity. So integrate at 0 by the boundary condition of the vertex. You get 0. At infinity, there's something I'm using which happens, which says, I don't need this equation, that if I integrate lemus R goes to infinity of integral of zeta plus 8 pi R, is while you think that this is integral over this sphere of z2, actually not quite. You get minus 2 tau 2. And this is easy to see. Well, or roughly, how does this go? For large R, zeta behaves like minus 2 over R. So in this integral, when you look at the leading order behavior, we're interacting minus 2 over R over a sphere. You get 4 pi from the area. The 2 gives you 8 pi. With a minus. And from the area, you get 4 pi R squared, which cancels out 1 R. So the leading order term is minus 8 pi R. So this cancels this one. Now, there is a sub-leading order term, which comes from the fact that you're integrating 1 over R against not the space, the flat spacetime measure, but the physical measure. You do the calculation, this produces this. So in other words, if I call this 1, then my mass is 1 over 16 pi, 1, which is positive because of this. This is my first integral over there. Plus and now, well, minus because I had in this formula for m, which vanished somewhere. OK, well, here, right? So I have integral of zeta 2, which is here. I have minus 2 tau 2 and plus tau 2. And I get minus integral of tau 2. So now, it's a question of, I think I have five minutes. I could do it in three minutes, but you can just believe me that if you take the equation here with kappa equals 0 and you integrate it in a clever way, then this coefficient tau 2, which is coming from this integral, is exactly this thing. So that's all there is. And this is how you prove that the Troutman-Bondy mass is positive, using a lot of ideas which Ivan helped me understand and which, which, OK. But we developed together this formula to understand the light counts. Near the vertex under the behavior of the initial data. Congratulations. When I suggested this to you, you said, this is, I don't want to do this. OK, thank you very much. Question, please, what happens if you try to do this with out vertex? Do they just get boundary terms from the? Yeah, so this is, I've tried to, because obviously, well, first, if you have just a null surface coming out of initial data, then you get a boundary term, which you control by the initial data. And you get this formula with boundary terms. The term here is obvious, because if you're integrating this, then you're going to get this at the initial value. The other one is a little more messy, but you can work it out. And I wish I understood it, because if I did, there could be perhaps some kind of Pender's inequality floating around. So that's something I'm thinking about. And I certainly like to understand. There is this work of, I have to go by exactly, Charles. How does he compare? How does he compare? Idea. Does it compare? Do they look at the boundary mass? Do they prove positivity? OK, good. I didn't know this. Do they assume small data, or do they assume? Well, they have to make certain assumptions of the life on, like you have to make proof of it, because if you don't check up. Good. As I said, I'm so astonished that this has been overlooked. I've never seen it before. So I'm just proving the boundary mass. Positivity of mass is usually a big bill, and you do complicated things, and things like that. No, you don't have to do it. In this setting. Right, of course it could be absolutely clear. It's also obviously proof. Yes, so I certainly have a look at this reference. So a recent paper by Schell and Arautakis. Alexakis. Alexakis. Asperos. OK, good. Please. Sorry, so the sigma and the x-sine here, they are the square of connection coefficients, right? So, but they are defined with respect to the foliation of the line. Right, so the foliation is fixed rather rigidly here. By, well, first the parameter is a defined parameter along the light cone. And you're fixing the asymptotic behavior of the metric. And so, yes. So there is some freedom left, but, well, that's what it is in this case. So you're saying for this, there's a specific choice made of infinity. Yes, I'm assuming that the metric goes to, well, it's this G squared AB is asymptotically meaning Gaussian in usual sense that people assume. So this determines a lot of normalizations here and some irrelevant one, which obviously should be not relevant for those integrals, but.