 Well, thank you very much for the introduction and thanks to the organizer for organizing this wonderful conference. And as the title suggests, I will be talking about the Einstein equation. So let me say a few words before going on with the slides. So of course, by now I think people are already familiar with the Einstein equations. We'll be talking only about the Einstein vacuum equations, in which case you are talking about a four-money fault with a Lorentzian metric g and verifying reach g of g equal to 0. So this is the shortest equation you can possibly write. And yeah, you can even take away the g. Anyway, so obviously, we have to solve this equation. In the sense, you have to find metrics which verifies the equation. Now, the first thing to be said about it, and of course, people already know, even those who are not familiar with generativity, that these are hyperbolic equations. So they're hyperbolic because of the metric being Lorentzian. And therefore, it leads to an initial value problem. So there is an initial value problem formulation in which you start with a three-money fault, a metric, a Riemannian metric given on sigma 3, and a tensor is symmetric to tensor k bar. And these are like in the case of, you always have to compare, we say, the wave equation, the standard wave equation, where you prescribe phi and the time derivative of phi. So this is prescribing g and k, which will be once you construct the spacetime. This will be essentially the time derivative of the metric. OK, so this is the initial variable. Of course, you have some constraint equations here, which have to be solved. That's a minor detail, though lots of people work on just the constraint equation. And there are lots of interesting results. But we are really interested in the evolution problem. So assume that you have initial conditions and you want to solve. Now, also another thing that I should say, general activities split into problems which are cosmological problems and problems which correspond to isolated physical systems. So I'll talk about the latter. In other words, I assume asymptotic flatness. So what that means is that you are looking for metrics which look Minkowskiian at infinity. Or in terms of the initial data, that means that this metric looks like the Euclidean metric at infinity. And the second fundamental form looks like zero. And of course, you have some precise ways of formulating this. But roughly, that's what it is. Anyway, so this is asymptotic flatness. And then, obviously, we are interested in solving this problem. So there is a local existence result, which we all know. It's due to Yvon Chocke-Brua, 1952, which tells us that at least you can construct, given specific initial data, or some level of differentiability, of course, as usual, you can construct a solution up to a certain time. And then there is also maximum time of existence, which is called the maximum global hyperbolic development. So this is sort of the possible extension that you can make by local existence arguments. Anyway, so keep in mind that that's the maximum global hyperbolic development of the spacetime. And obviously, all of general activity, all of mathematical general activity is about providing information about this maximum global hyperbolic development. So now, the next thing that you do whenever you have a very difficult nonlinear problem, you look for stationary states, right? So stationary states. So these are states which, in some sense, are independent of time. Now, of course, time in general activity is a more complicated notion. But still, you can make sense of it. So it's actually relatively simple. You say that the spacetime is stationary if there is a vector field T, which is space-like, which is time-like at space-like infinity. In other words, if you have a light cone at every point, and as you go towards infinity, the vector field is more and more like the standard d over dT in Minkowski space, right? OK, so a stationary state means there exists such a vector field such that the derivative of the metric with respect to T is equal to 0. So that's how you define stationary T. It's not too difficult. And then, of course, there are examples. So the next thing that we know is that there is this incredible family of solutions of the Einstein equations under this assumption and asymptotically flat assumption, which are the care solutions. So this was a family. So we are typically interested in absolute value of, hey, strictly less than M. A is equal to 0 corresponds to Schwarzschild. A equal to 0. And of course, Minkowski means A equal to 0 equal to M equal to 0. So both these parameters are equal to 0. All right, so these are the stationary states. And then, once you think that you know all the stationary states, which, of course, is a problem, so I'll mention here that there are three obvious problems connected with the care solution, with the care family, then you can conjecture, as people have done, the so-called final state conjecture. Which is that if you take any initial data, then asymptotically, well, it's going to be very complicated. The interactions are going to be very complicated. The nonlinear analysis is going to be incredible, difficult. But the hope is that, eventually, you are going to see in any finite region of state, you are only going to see a care solution, nothing else. So this is like the soliton resolution conjecture, if you want. But I'm interpreting this final state conjecture in the sense that, essentially, things are stable. Because otherwise, I mean, from a physical point of view, it doesn't make any sense, otherwise. So the, yes? That doesn't allow merger of the? Yeah, it does allow merger, also. So this will be part of the final state conjecture. There are all sorts of initial data, very complicated, all sorts of things happen, black holes emerge. Then they will coalesce, maybe some of them, some others will go far away. We need some food for LIGO. But I mean, whatever is done at LIGO is based, essentially, on this picture. So the recent experiments are totally based on the assumption that this picture is correct. Now, in order for the picture to be correct, you have to do a lot of things. So first of all, the simplest thing, maybe, that you could ask is, is care family, so rigidity issue? Rigidity, is this care family rigid? In other words, are there any other stationary solutions besides? So these are solutions that you can find explicitly. These are explicit solutions which are found in 62, 63 by Kerr and Schwarzschild even in 1915, very early on. So the rigidity conjecture tells you that there are no other stationary states. The second thing is stability, of course, stability of the care family. So in other words, if I make a small perturbation of the care solution, the expectation is that you are not going to get something else. If you get something else, then it's clearly that the care solutions have no physical reality. They exist in a mathematical universe. They are beautiful. You can study them if you want. But they will have no physical significance. And of course, then this final state conjecture will be false. In fact, if any of these things that I'm writing now will be false, then the final state conjecture is also false. Anyway, so beyond stability, there is also a collapse. The issue of collapse, I mean, can you actually form black holes at all? I mean, of course, we are told that they form when you have a collapse of a star, for example. You will form a black hole. But is it true? I mean, is this picture true? So that's another mathematical question for us. And then if you want to go, so these are three fundamental questions which I like to call them reality, so the test of reality of the black holes. If they are false, then buy black holes from a physical point of view. Now, of course, beyond that, if you want to go to the final state conjecture, there are many, many other things involved. There are many, many steps. Like for example, as you said, take down two black holes. This will be just one black hole. This is the one body problem. But of course, you have to study the two body problem and body problem. Then there are other issues that you have to do is the cosmic censorship, which is another fundamental conjecture in general activity, which says roughly that the only singularities you can have. By the way, singularities, besides stationary states, of course, singularities of non-linear PDs are also fundamental if we want to understand the revolution. Now, it so happens that singularities, the singularities which are known in general activity in this asymptotically fraught regime are the ones of the care family. So the care family, or at least the schwarzschild family, has a singularity in the black hole. So this is a remarkable thing about general activity, that it splits. A black hole splits the spacetime. It's something about the spacetime. It splits the spacetime between a region, which is a black hole region, and the outside region. And the outside region is smooth, is regularity. The only singularities will occur in the black hole region. That's essentially what we believe. Cosmic censorship is a way of formulating that statement, that the only singularity you can have are in black holes. You can't have so-called naked singularities. But you should approve. Of course not. I'm talking about here a program for the next 100 years. I mean, this may take even more than 100 years. But obviously, there are lots of steps, and there are lots of simplifications that you can make. And of course, there is a lot of activity. You can simplify many of these problems. Anyway, the problems that I've been mostly interested in the recent years are these three. So I'm going to talk about the issue of stability in this talk. All right, so let me go to the slides. All right, so how does this go? I've got a button on the side. This one? No, the round disk button on the side of it. The center button is laser, and the side buttons are the advance. OK, very good. So here is a care family. The care, oops, what happened? OK, very good. Now I have to be very careful. Maybe I shouldn't even use this. It's better. So you see, is it OK now? I'm still not fine. Yeah, here in the center, you say, OK, very good. Now I understand. Sorry, I should never touch this button. All right, so here is a care family expressed in so-called boiler-linked coordinates, which are like polar coordinates. And you see the metric looks extremely explicit, even though it's a solution of a very complicated nonlinear problem. You don't have to remember these things. I just want to glance very fast over this formula. The important thing is that all the coefficients that you see here of the metric are independent of t, of the t variable. And that corresponds to the fact that you have stationarity. In other words, this vector field is a kidney vector field, indeed, in the sense of this definition. And you also see that the metric is asymptotically flat. Also, you see that these coefficients are very explicit. And as r goes to infinity, you get the Minkowski space. In addition of stationarity, you also have axisymmetric. So the care family, as you see, these coefficients are also independent of phi, which corresponds to the fact that the vector field is z, which is d over d phi, is indeed a kidney vector field. Now, the Schwarzschild solution is the particular case when a equal to 0, the metric becomes much easier. You see you have this delta over r. But you see, of course, that you have some kind of strange singularities. Of course, these singularities occur here also. But in this case, it's very easy to see that when r is equal to 2M or r is equal to 0, this metric sort of becomes singular. Now, the singularity at r equal to 2M is not a real singularity. It turns out that you can actually extend the solution. And in fact, r equal to 2M is just the boundary between the so-called black hole and the exterior region. Now, you don't have to take. I mean, obviously, if you've never seen these kind of Penrose diagrams, you might be a bit confused. The important thing to look at is maybe, if you're not familiar with this picture, is to look at various values so far. So you see the interesting values are r equal to 2M, which corresponds exactly to the boundary between the black hole and the exterior region of the spacetime. r equal less than 2M, this is the black hole region and maybe the white hole. The white hole is not very physical, so we will throw it out. We are not interested in it. There is a curvature singularity at r equal to 0. Here, the metric actually does become infinite. The curvature of the metric becomes infinite at r equal to 0. Again, I will not be. So this is a black hole by itself. It's, of course, a fascinating thing. That's where the singularities will occur. But I will be interested. When we talk about stability, we are interested in the external part of the black hole all the way up to the horizon. So this is the part that I do here. There are some other interesting numbers. r equal 3M corresponds to the existence of so-called trap-nalgeodesics, which are lead to lots of problems, but especially mathematically, I would say. And finally, there is r equal to infinity, which corresponds in this picture to these boundaries. So these boundaries are really at infinity. Now, the light here propagates at 45 degrees. So as you see that if you are inside, you cannot escape. You will eventually hit the singularity. While, of course, if you are outside, if you are unlucky and you go in this direction, you will get into the black hole and you hit the singularity, but you can also escape. You can escape to infinity. So that's the difference between the external, very roughly between the external and the black hole region. So now, sorry, I'm not doing well here. So the k-space time is very similar, at least in the exterior region, except that the horizon now is at r equal r plus, which is the root of this polynomial. So again, it's very explicit. The black hole is slightly different in the sense that you don't have that singularity at r equal to 0, but I'm not going to get into this region. I'll be interested in just this. And as you see, the picture is very similar. This is a space like hypersurface, which would play the role that I want to give initial data later on. I want to perturb the care solution, and I want to give initial data on something like this. All right, so there are, again, the external region, r largest and r plus, event horizon, which is exactly the boundary, the black hole region, which is r less than r plus. Here is, again, the same picture, but this is the one that you should look at. That's external region. This is the event horizon, where r was equal to r plus in care. Now here, you also see the vector field capital T, which is the stationary kinetic vector field. You see it's time-like. It means it goes in this direction as you approach infinity. So remember, this is the boundary at infinity. While now, the remarkable thing that happens is that T becomes actually space-like near the horizon. So even though it started by being time-like near the horizon, it's space-like, and that leads to lots of issues. I mean, it leads a lot of interesting physical issues, but also mathematically to lots of difficulties, because, for example, energy is always tied to a kinetic vector field, in particular time-like kinetic vector field. So coercivity is, of course, tied to the fact that it's time-like. In places where it will be space-like, coercivity will disappear. So you have, anyway, this is something that I just want to draw your attention at this point. So this is now the conjecture of the stability conjecture. So stability of external care. So in other words, I'm interested only in this region. I take this initial data. I take the initial data here exactly of the care solution. So care solution has, of course, associated initial data on any space-like hypersurface. I take that data and I perturb it a little bit. And the conjecture says that small perturbations of a given external care solution have maximum future development. In other words, I develop solutions as far as I can, which converge to another care solution. So in particular, it will be complete in this direction. And it will look asymptotically. It will look like a care solution, but maybe a different one. In other words, I can start with A and M, and the final states will be different from this, which leads, again, to many other difficulties. We know. I mean, whatever you have these kind of situations, you have to do careful analysis. All right, so what are the difficulties of the black hole stability problem? Well, the first one, which is common to any geometric problem, in particular, general relativity, is a so-called gauge covariance. I mean, you write down the equations, to write it like this, it's useless. You want to write it in some kind of coordinates, let's say. But then, as you pick coordinates, you maybe will get a system of nonlinear wave equation if you are careful. But it may be that instabilities in that system of coordinates will mean nothing else that you picked up the wrong coordinates. So you have to be very careful what you mean. You have to work as much as you can in an invariant fashion, which is not easy, given the nature of this equation. So in any case, this issue of gauge covariance is connected with proper linearization. If you change variables in nonlinear equations, the linear equations will look very different. So then you can try to linearize at the level of metric coefficients in a given coordinate system. That's not a very good idea, as it turns out. You can use connection coefficients, or you can use curvature. So you can linearize at the level of the curvature. This is actually the way to go. And I'll explain maybe a little bit more later on. And then another problem, of course, is that these equations are quasi-linear. So in other words, the characteristics depend heavily on the solution itself. And the characteristics, of course, have to play a fundamental role in the story, because this is where gravitational waves, or light, or whatever you have metafields, they will propagate along null characteristics. So there is a non-trivial null structure. I mean, we heard about the null condition. These equations can be viewed at some level as some system of non-linear wave equations. So you expect the null condition to play an important role. However, if you try to use a null condition in the coordinates used by von Schoke-Bruau, which are called wave coordinates, it's not there. There is no null condition. There is a so-called weak null condition, but that's very different. Anyway, so to find the correct null condition is also a highly non-trivial issue. To detect, in other words, and of course, this also has to do with the problem of coordinates and gauges. Now, even if you look at the linear problem in care, in other words, you fix your background. So this is much simpler. You fix your background. This is sort of a baby linearization. You assume that you are in care, and you look at some linear problem, like, for example, just the standard wave equation in care. This is already a highly non-trivial problem, which was only understood after 30 or 40 years of efforts. It was only understood in the last 10 years essentially by mathematicians by introducing all sorts of new techniques, which take into account the fact that care has limited symmetry. So the standard vector fields are not enough. The vector fields which are tied to symmetries of care, the fact that there is an ergo region in which you have issues even with the energy, and the fact that you have presence of trapped null geodesics, which is actually, it turns out, to be technically the most difficult of all. So what are these trapped null geodesics? These are geodesics that don't escape either in the black hole or at infinity, and stay, in fact, in spacetime at finite values so far. And as we know from geometric optics, they will create lots of problems. So finally, but not least, you do expect no matter what kind of linearization you do to have some kind of linear instability connected with gauges. And this is due to the change in A and M. So the final, as I said, the final A and M will be different from the ones you perturb. And that has to be read in some kind of instability of the linearized equations. And also, the position of the final black hole also will change. So you have to take that into account also. That will also create instability. All right, so OK. So now, here is a little bit of technical stuff, but it's not very difficult. Let me say it here. So you see, typically, when you have a spacetime, the best way to analyze it is not using coordinates but using frames. So why are frames important? So if frames are simply vector fields in spacetime, which are null, so another word, G is 3, is 3, is G4E4, equal to 0. And therefore, they correspond somehow to propagation. So that's why they are very, very important. In coordinates, it's very hard to write things like this. So that's why frames are very important. And then once you have, you also normalize this so that E3E4 is equal to minus 2. And then once you have that, you can also look at the spacetime perpendicular to these two. And you write down E1, E2, two other vector fields, which are orthonormal relative to themselves. And of course, perpendicular to this one. So that's what it's called an R frame, which plays a fundamental role in just about anything you want to do in GR nowadays. So the first thing to take into account is this null frame. Once you have the null frame, you can construct connection coefficients, which simply means, which are typically denoted by gamma, which simply means that you take derivative of relative to the frame, let's say alpha e beta e gamma. So you take covariant derivative of the frame relative to the frame. And then you project it again in the direction of the frame. So this gives you the so-called gammas. And then next thing you do is you look at the curvature. So this is the curvature tensor. And you also decompose the curvature tensor. So the curvature tensor is a 4 tensor. You also decompose the relative to the frame. And you get various coefficients. For example, if you look at RE4, E1, or E2, so let's call it A, capital A refers to E1 or E2, E4, EB. This is called, say, alpha AB. And it's a certain components of the curvature. And what I wrote there are various combinations of different components of R relative to the relative to the null frame. And very often, in general activity, you will see these kind of notations, right? So anyway, so again, what you have to keep in mind, however, maybe I don't want to scare you with too many notations. What you should keep in mind is that the gamma, the gamma's verifies equations which relate it to curvature. So for example, this is a very important equation. E4 of gamma times gamma is equal to curvature. Of course, I'm writing it in a symbolic manner. Here you should think about various components of gamma, various components of gamma, and various components of curvature. So you can write. You'll have a lot of, you can fill the blackboard with many equations. Oh, sorry. Yeah, so you can fill the blackboard with various equations. And finally, the other equation. So there is another one relative to E3. So there is something similar. And then there is something similar about curvature. So there is E3 of r plus gamma times r is a derivatives of curvature, but derivatives which will be different from the E3 derivative, which you put it here. And E4, so very roughly times r is also. Of course, again, as I said, I mean, if you write down all the components here, you can fill out the blackboard. The important thing to realize is that these are the most fundamental of all. And these are, in fact, nothing else but Bianchi identities. Bianchi identities. And it's not too difficult to see in a first approximation that they are actually, well, first of all, they can be derived from the standard Bianchi identities, which are beta, gamma, delta, sigma. Plus, you take cyclic permutations of these indices. So let me write it just like this, equal to 0. And the fact that you also have r alpha beta, you have this kind of divergence. So this is a divergence of gamma, delta, equal to 0. So this actually, this equation follows from this one. They are, in fact, equivalent if the Ricci of G is equal to 0. In other words, if we are in the Einstein-Vacuum equations. Anyway, so this is what I want to say is that you can view this as glorified Maxwell equations. And somehow, this is extremely important because it tells you that somehow the hyperbolic character of the equation is most explicit in terms of the curvature. So the curvature being also invariant, of course, this is obviously extremely important. So this is the hyperbolic character of the equation. And therefore, the way to think about all this is that somehow you first estimate the curvature using energy estimates. Energy, because since they are glorified Maxwell equations, in order for Maxwell equations, they have energy estimates and so on. So you can do something very similar here. It's a bit more complicated, but it's very, very similar. Once you have the curvature, then you can analyze the gammas by thinking of these as transport equations where the curvature is already given to. And then somehow these other things here are viewed as relations. I mean, the true hyperbolic character is included in here. This, of course, follows. So the equations here follow from this one. But somehow these are typically used in combinations in this one in order to get all sorts of relations. These are solutions. These care solutions, they are obtained at the level of the curvature or what? So I'm going to say something right now. OK, so this was just preparation exactly for this. All right, so why are these important? You can see it immediately when you talk about, so these are the non-structured equations that I mentioned. You can see immediately when you apply all these to care, because in care, it turns out, so one remarkable fact about care, is that they exist a principal null frame. In other words, E3, E4, which you can write down explicitly in terms of coordinates. They exist and such that all curvature components, these are alpha, beta, beta, bar, alpha, bar, all these curvature components are zero. They vanish in that frame. And only the so-called rho and i, rho, star survive. So these are the sorts of things. There is this R, E4, E3, E4, and there is another one, which is just like that. But let me not get into these details. OK, I just want to give you some kind of idea. So the very important thing is that this principal null frame diagonalizes the curvature of the care metric. Otherwise, it would be incredibly complicated. I mean, anybody who has worked a little bit with care, you see that if you write it in coordinates, you have a never-ending series of calculations that leads to nowhere. So it's very, very important to have conceptual understanding of care. Anyway, so this is a principal null frame is essential. In Schwarzschild, yeah, in addition, of course, you have connection coefficients. Some connection coefficients are also zero. Never mind exactly which ones. But some are also zero, which is very important. In Schwarzschild, in addition, you have that rho star, right? So you have that rho star there. Rho star will be actually zero in Schwarzschild. And also, further connection coefficients are also going to be zero. So many things are zero. In fact, the only non-varnishing components of gamma will be, well, whatever. I mean, I don't necessarily want to write down now, again, what they are. But they are fewer and fewer. Of course, if you are in Minkowski's space, in addition, you also have the rho is zero. And then these ones are zero, these two. And you are left just to these two, which are just geometric. These are just two over R in terms of polar coordinates. So these are very simple to calculate. So that's kind of the story. And as you see, these frames are very important. Now, OK, so now you want to think about perturbations of care. So what are you going to think about perturbation of care? Of course, you can try again to look at the level of the metric. But that's a mess. You don't get anywhere. So this is infinitely better. You imagine now a space time, a general space time, which is the solution of the Einstein equations, where you have a frame which is close to a principal null frame. So in other words, you have a null frame E3, E4, E1, E2, such that all connection coefficients and the curvature components that you have here, they are all of epsilon, relative to a small parameter epsilon that you have in the problem. So this will be all of epsilon. And also, the part of the curvature, which is not 0 in care, you want to say that it deviates very little from the position of the old care. So this will be this all of epsilon here. And then, of course, you linearized. You linearized the equations that we had before, exactly these kind of equations. You linearized by saying, well, gamma is going to be gamma care plus a gamma tilde, where all gamma tilde is all for the epsilon. R is R care, exactly the curvature of care, plus something which is all of epsilon. And of course, you get a system of equation now at the linearized level, which looks like this. So these are the linear quantities. And you see, it's very complicated to a large extent because of the presence of these things, because this leads to a coupling which is very, very hard to understand. In Minkowski's space, this will be essentially all 0s. That's why the stability of Minkowski's space could be done. It's much simpler. But in this case, it's far more complicated. So let me tell you a little bit of what we still know. Now, one important thing is that you see, in this definition, you can say, well, but this frame is arbitrary. I mean, which frame I'm talking about. It's an arbitrary frame, which is good up to all the epsilon. Well, one thing that you can try to do is to see how to change, if you want to go from one frame to another frame, by a frame transformation. And here, in this story, everything which is all of epsilon squared is good, because all of epsilon squared terms are the terms that we love. We like in all linear equations. Of course, they are still very complicated, obviously. But at least you seem to be reducing the problem to something that you have some control on. So the quadratic terms, I will neglect. And then look at only terms like this. These are all of epsilon. So this is very simple. I mean, just write down what the general transformation which preserved an alpha frame. So I want to go from one alpha frame to another alpha frame. Now, the remarkable fact, which I think plays a fundamental role in everything that is done with linearized gravity, is that the extreme null components, alpha, alpha bar. So remember, alpha, alpha bar were, for example, this one. This was alpha. And if I interchange E4 with E3, I'll get the so-called alpha bar. So these components are invariant up to all of epsilon squared terms with respect to these kind of transformations. So in other words, they somehow don't care about the gauge I pick. They are OK, no matter what they do. So this is very important, because this is something that I have to start with. As always, I have no idea what to do. All other linear curvature components of the curvature. So all other components of the curvature are not invariant. So that's also important. And now, based on these observations of Tawkowski, was able to show in 1973 that these curvature components which are invariant actually verify decoupled but non-conservative linear wave equations. So in other words, there will be some linear wave equations with lower the terms which are non-conservative. They are not Lagrangian, they are non-conservative. Now, however, they were extremely excited. Physicists were very excited. So they immediately declared victory. Tawkowski maybe other people were more reasonable. But you read a lot in the physical literature that somehow this solves the problem of stability completely. I mean, even though this is just at the level of linear stability. Well, actually, Whiting proved something using this, proved that the Tawkowski equations, so these wave equations which are not conservative, have no exponentially growing modes. So you mentioned that the location of the black hole is also. Right. So this has to do with the frame, for example, the change of frame. Is this reflected in any way in these analysis, or how do I see? Well, I mean, the whole point is that these are invariant. There are certain components of the curvature. This is quite the remarkable fact. There are components of the curvature which do not depend on this fact. Otherwise, you wouldn't be able to do anything at this level. Anyway, so using this, Whiting was able to show that in the linearized gravity equation is exactly what I wrote earlier. There are no exponentially growing modes. So and then in principle, one can show, now using gauges, that all other curvature and connection components can be controlled once you have this. I mean, once you have control of alpha bar, which are this one, then in principle, you can control all the other components. Even this is highly non-trivial. But the question is, does this solve the problem? And of course, we all know the answer. If lack of exponentially growing modes for the linearized equations was enough to deduce non-linear stability, then the presence of shockwaves, extreme sensitivity to data, and turbulence in fluids will be ruled out. Because in all those cases, we have more than there are no exponentially growing modes. You can show that there are no growing modes at all in linear theory. So obviously, this doesn't solve anything. But of course, it was a very important circle of observations and ideas. Anyway, so what are the main results we know on stability? Of course, we know the global stability of Minkowski space. And here I should say that what was important was maybe not so much just the result, but the fact that it uses geometric methods which turn out to be important in many other problems. In particular, it's also important in this problem. The classical vector field method, non-trivial use of characteristics, the non-trivial use of characteristics is fundamentally important. So in other words, you use foliations based by null characteristics, generalized null conditions, which also played an important role, I mean a fundamental role actually, and some kind of very complicated bootstrap argument, which of course by now we all know we can do it in many difficult problems. All right, so now, of course, the more stability of the care family, which I mentioned, poor man's stability. So this is a stability that I mentioned earlier that if you fix a care solution and you look at the wave equation, just a standard wave equation on the care solution, then you can show sufficient information about the solutions, which are consistent with what you might need in the non-linear problem. Of course, the non-linear problem is much more difficult because you don't even know the linear equations. But this is extremely important, at least to study the standard wave equation. If you don't understand the standard wave equation, you don't understand anything. And so this was enormously important. All these techniques which were developed by these people were very important. There was a new vector field method, which developed as a consequence of this, which is not just based on symmetry, but it's based on many other geometric features of black holes. For example, the horizon, there is a redshift. The horizon, there is something that has to do with the ergo region, the degeneracy of the ergo region. There is something that has to do with the trapping, which I told you the trapping means that you have this non-geodesic truth sitting and lead to all sorts of technical difficulties. The far region, which has to do with something that, in a sense, you can say was already understood in the stability of Migosky's base. Anyway, more. How much time do I have? 8 minutes, OK. So let's make it 15 minutes, OK? 15 minutes, OK. It's very generous. Very generous. Great. So what are the results we know? There is a scattering construction of the dynamical black holes by Daphermas, Holtzegel, and Rodniansky, which means that you solve the problem from infinity, rather than from initial data. This is much, turns out to be much easier, but still quite remarkable that it was done. Then there is a result of Fionnesco and myself, which I call, in the axially symmetric case, so I'll mention axially symmetry in a second, which is based on some partial linearization. So you don't solve the full problem, but you solve something which is still nonlinear. So in a sense, it's a better approximation than just a wave equation on the care solution. It's something which is maybe more relevant. Then there is a remarkable result of Daphermas, Holtzegel, and Rodniansky, which just appeared at the beginning of the year, which is linear stability near Schwarzschild. And finally, work in progress of Seftel and myself on the nonlinear stability of Schwarzschild, but nonlinear. I mean, we are trying our luck with a full nonlinear stability. For axially symmetric polarized perturbation, so this is actually in terms of linear stability. Of course, linear stability also plays a role here, but the linear stability in our case will be sort of a subset of this. But we are trying our luck with a full nonlinear problem. Now, Daphermas, Holtzegel, and Rodniansky, so I think the central point of their approach, which is the result I just mentioned, is a remarkable discovery of a new tensor, which is an S tangent to tensor, at the level of two derivative of curvature. So remember, alpha, alpha, bar are just components of the curvature. And they come in the Teogolski equation. So here you take something which involves two derivatives of curvature, which turns out to actually verify a conservative equation. In other words, it's an equation which has just a wave equation in Schwarzschild, applied to this p plus a potential times c. And this potential has good properties in some sense. It's called the regular potential. So this has actually good energy estimates and more of it and so on and so forth. So you can start applying all the technology developed in the last 15 years, which I alluded to before, in order to control the decay properties, boundaries and decay properties of this p. Once you have that, then you have to reconstruct all the other quantities. I mean, p, of course, is just the level of two derivative of one component of a curvature. So you have to go back and go down and get all the other quantities. In particular, those quantities which are gauge dependent. And therefore, you have to make gauge choices. And of course, that's a highly non-trivial part. In fact, remarkably, this sort of thing, they refer to a computation of Chandrasekar. And remarkably, they don't even prove this, which to me, it's like the most important thing. They say it's just a calculation. But then the rest of the paper is about 170 pages just to do this stuff. So it's a very complicated thing. Anyway, now I said I want to talk about axi asymmetries. I want to simplify. So the problem in axi asymmetry turns out to be a lot simpler. And let me explain why. So axi asymmetry means that you assume that the metrics you are looking at, which are perturbations of care or Schwarzschild, have an additional symmetry, which corresponds to a keying vector field z. Of course, care itself has that symmetry. So it's totally consistent. So it turns out that using this z, you can construct the so-called Nernst potential, which is a complex scalar. So it's x plus iy, where x is just g of zz. So it's just something simple to calculate. And it turns out that phi verifies a wave equation. So in fact, it's actually a non-linear wave equation. In fact, for people who recognize immunity, that it's a wave map with values in hyperbolic space. So this is part of the symmetry reduction. Then it turns out that you can work a little bit harder and do a decomposition also of the metric. So the metric decomposes itself into a 2 plus 1 metric. And the scalar, this complex scalar. And now you have, again, the wave map, which is this one. But now, of course, you have to relate the wave map also to the curvature. So you have that the Ricci part of this 2 plus 1 metric, which is here, is tied to this wave map through this relation. I should say, for people who love wave maps, of course, they usually do wave maps in Minkowski space. And there are lots of beautiful results which were approved in recent years. This would be a great challenge to apply their methods to the wave map, the true wave map, which is coupled, actually, with a metric. So this is a quasi-linear system. It's much more complicated. Anyway, but I'm interested in something else here. OK, so this is a wave map. Now, here are two problems that I've been working on in recent years. So one is what I call modern problem number one. What you do, again, I cannot do the stability in general. It's still immensely complicated in this case. But you can try to simplify things a little bit by keeping the phi to satisfy this equation and simplifying the metric to be the care metric. You take G to be the care metric. So then, you see, you get now a system. G is actually fixed. And you have the N's potential of care itself, which exists. The care itself has an N's potential corresponding to the vector field Z, the axial symmetry. And you look now for solutions of this equation. So this is the equation returning components for phi. This is how it looks around the solution that you already know that you have, which is exactly the one of the N's potential of the care metric. So in other words, you have this system of non-year wave maps. But you have a solution, a specific solution. When you take X to be A and Y to be B, you get a specific solution. And now, you want to perturb around it. So that's problem number one. Mother problem number two is that you assume in addition, you assume polarization. So if you go back to what we have here, I'm going to assume that Y is equal to 0. And it's easy to see that this system is consistent. In other words, Y equal to 0 is a true solution. So if I start up with Y equals 0, it would stay Y equal to 0. So this is actually, when you do axial symmetry polarization, you really, it's an honest result. Because you actually solved the true Einstein equations. So the problem now becomes, when Y is equal to 0, so this simplifies a lot. So the wave equation, instead of being a coupled system, is now just one wave equation, which is d'Ambition of phi is dI phi dI phi. This actually, you can solve explicitly by exponentiating. And you get just a solution of the standard wave equation. And of course, you tie it to the Ricci curvature of this spacetime metric. So remember, this is a 2 plus 1 metric. You tie it by this. So this is a system that you have to solve, these two equations. And it's easy to see that in this context, the final state has to be Schwarzschild. So it's really a problem of the stability of Schwarzschild solution in axial asymmetric polarized. So the conjecture number one, which is more of a problem than number one, the stationary wave map is stable under general axial asymmetric perturbations. So again, given small initial data, close to the else potential of care, or a space like hyper surface, there exists a global solution such that phi is equal to phi 0 and phi converges to phi 0. Now unfortunately, I don't have time to talk about this. I just want to say that on this, we have made a lot of progress. We proved all the important linear stability results with UNESCO. So the conjecture is certainly consistent at the linear level. And my student, John Stoggin, is now doing the full nonlinear result. So this is something that I think we understand. So let me go to, I won't have time to talk about this. There are some interesting things about that, but I won't have time. So let me go back to modern problem number two, which is work in progress with Jeremy Seftel. So this is again, this is a system that you want to one. So you are looking for metrics of this form. So phi, this derivative with respect to phi, it's exactly the Keeling vector field. So x is independent of phi, and the metric g is independent of phi. And you have this nice reduction. And the equations really look like this. So these are the 2 plus 1 dimensional equations. Now, in principle, you could think that this equation is so simple. I mean, there should be no problem to solve this, which is what we thought, but it turned out to be totally wrong. I mean, there's nothing you can do with this equation. If you try to linearize, you get a mess, and you never get anywhere. So we really had to go back to the standard picture using curvature and connection, and so on and so forth, which is what we did. So again, we introduce a frame very similar to before. It was a null frame E3, E4, Etheta. But now, just because of the reduced picture, you can do this just in the 2 plus 1 picture, which seems simplified. So a lot of these components of connection and curvature are simpler. In particular, they are all scalars now. So this is a scalar, this is a scalar. And anyway, I don't want to bore you with all these things. They still are sort of this frame that you can write down. These are the equations written in full details, but I don't want to scare you. They are not that complicated, actually. It takes just a little bit to get adjusted. But in any case, I want to go back to this business of the Okolsky equation. So there are these Okolsky equations, which now are scalars. So in my case, alpha-alpha-bar are scalars. And these are the equations that actually they satisfy. Alpha-alpha-bar, again, just like in the general case, they will be invariant up to of epsilon squared. So there are two invariants of the problem up to terms of all of epsilon squared. And they verify this kind of equation. So this is the standard wave equation. Actually, you can take note the metric to be the Schwarzschild metric, in fact. So you can think of being Schwarzschild. And these are some derivatives of alpha-bar, derivative of alpha-bar, and similarly here. And you can very easily see that this equation is not conservative. So there is not much you can do at this level. So the idea is to find a quantity, to find a good quantity, which linearizes the only component of the curvature, which is, say, the one which is close to raw care, is non-trivial. And I cannot, I mean, saying that it's close to raw care is kind of misleading, because raw care only in a system of coordinates, I can talk about the true value of raw and care. But I don't have a system of coordinates to start with. So I cannot even talk about this. So the issue is how to linearize without doing this, how to find a quantity which is invariant, and which is tight to raw, and which is small, which is zero in Schwarzschild. So this was sort of a major thing for us. So we found that there is such a quantity, which is at the level of this x, which was e to z 2 phi in the previous slide. And you see, it's just derivatives. So this is clearly a covariant expression. It's a tensorial expression, minus this raw, which is the one that you want to linearize, minus this. This quantity tends to be zero in Schwarzschild, tends to be an invariant of order of epsilon squared. And it satisfies a good wave equation, unfortunately, but one which is still a wave equation which looks like this. And unfortunately, it's still tied to the Teukovsky quantities, alpha, alpha bar. So we're hoping to get something which doesn't have anything on the right-hand side. It will be just all epsilon squared. But it turns out to be much more complicated. However, there were some very simple relations that are satisfied between this p-frag, this quantity which we have introduced, and alpha and alpha bar, which are some kind of simple oddies like this. And inspired by the fact that we knew that Aferman's Holtzege and Rodiansky already had a good quantity, which was at the level of the second derivative, I'll finish right now, which was at the level of second derivatives. We realized that what we had to do is just to differentiate this with respect to E3, then differentiate with respect to E4, commute to this, get rid of these terms because of these relations, and you get finally a wave equation which looks very simple and it verifies it. Obviously, it's a wave equation with the potential. The potential is very nice. It has good energy estimates, and therefore, you can do more of it, and you can do all sorts of things. And once you have that, then you have to work a little bit harder and get the full story. But unfortunately, given the time, I'll have to stop. So I'll stop here. Ah, here. Let's take a seat. Thank you. No questions? Yes? I have a question about this final state conjecture. Yes. Did you say that Kerr is unique and state? No, I didn't say. I said that I want to prove that it's unique. No, no, I mean the conjecture, so. Sorry? The conjecture is the single Kerr black hole. You stir and stir. The weight, this is different. I mean, we're talking about different things. Any other stationary states besides Kerr? That's a rigidity conjecture, all right? And so far, we don't know. We have only partial results. I was thinking, suppose you take initial data, which have the form of two approximate Kerr black holes far apart, and you hit them velocities in opposite directions. You have to ask Thibault about this. I mean, really, we are, unfortunately, we are still very far from that kind of results, unfortunately. So we are really at the one-body problem. I mean, from a mathematical theory of GR is just at the level of the one-body problem at this stage. The two-body problem is something that maybe if we go to Tuesday talk, maybe we'll learn. I mean, I think it would be very, very nice if we start getting into the two-body problem also. But so far, I haven't seen any serious attempt to do that in the mathematics. So any more questions or comments? Of course, needless to say, this experiment, this recent remarkable experiment at LIGO, it pre-assumes, based, of course, on numerical experiments, pre-assumes that the two-body problem is stable. Therefore, in particular, the one-body problem is also stable. So the stability is not under question from a physical point of view at this stage. But we don't have any serious mathematical results at this point. OK, so let's thank the speaker again.