Daj, da zelo v rastroj, puno se je postavljalo naročno reakcije, ki smo pred ahanašili počquet in potem, ki si je pa izgofila res o reakcijov. Tetz del. Zelo, što je začel na zdoroviti, če je začel na reakcij nekud tega. Tako zelo, da se s tudi zelo poč comparing, And you can talk about higher dimensions also, but I will stay in 1 plus 3 in the physical dimension. So four-dimensional manifold metric g, which is Lorentzian, and which verifies the Einstein vacuum equations. In other words, Ricci part of the Riemann-Kovacic tensor of g is equal to zero. And of course, I put myself into the asymptotica de flat regime. There are many other interesting regimes. I mean, there is a cosmological regime, for example, which obviously I will not be talking about at all. But there are many other problems extremely interesting in other contexts. There is also cosmological constant, for example, that you can play with, and so on and so forth. So let's talk about the simplest case, in a sense, which is this one, Ricci equal to zero. So the Einstein equations in the vacuum. So we discussed that these equations are deformorphism invariant. In other words, you have to identify phi star of g, which g verifies any deformorphism of m into itself. You have this huge class of equivalents of solutions of the Einstein equations up to any deformorphism. All right, so once you take that into account, the equations are hyperbolic. So once you stress out the importance, I mean, once you mod out by this gauge group, you can see that the equations are hyperbolic. In other words, the notion of hyperbolicity is really a gauge invariant. It is a gauge dependent, sorry, gauge dependent condition, in a sense. At least in the simplest possible definition, it will be gauge dependent. Then we talked about the initial value problem. So once it's hyperbolic, you talk about the initial value formulation. So, which means that you start with a three-dimensional manifold, a metric, which is Riemannian, and the second fundamental form. So maybe I'll write it like this, sigma jk. And then you study developments of these initial data sets. Our solutions of the Einstein equations, in other words, are four-dimensional. So this is three-dimensional. So you're looking at four-dimensional solutions, which when restricted to sigma, you get back these first and second fundamental forms. And then I talked about stationary solutions. And I said that, obviously, very often whenever you talk about any kind of nonlinear partial differential equations, or even ordinary differential equations, but partial differential equations, the first step in a sense of understanding the general evolution is to really understand the stationary states, or to classify the stationary states. So in the context we talked about, it turns out that people have discovered a large family of such stationary states, which are parametrized by two parameters, a and m. A, typically, we want a to be less or equal to m. Otherwise, there are all sorts of issues that make this the case when a is larger than m, unphysical. And then once you have, once we discuss that, I talked about the final state conjecture. So this is a very general and extremely difficult conjecture and very deep, which contains many other simple kind of problems, which by themselves are huge conjectures. So we discussed the final state conjecture, says that it applies to general initial data sets. So you start with initial data sets, which are asymptotically flat, but eventually very large. And the picture that one has in mind is the final state, is that after asymptotically, after a lot of very complicated dynamics takes place, you are going to see the emergence of only a finite number of black holes, which are care solutions. So in other words, a finite number of stationary states, these care solutions, which diverge from each other. So in other words, in any finite region of space, you are going to see just one black hole. So this is a final state conjecture, which is of course completely out of reach at this time. There are similar such conjecture for other equations, which are much simpler, but for general relativity, there is nothing given remotely close. However, there are simpler questions that one can ask, which as I said by themselves are a huge conjecture. One of them is stability of Minkowski space, stability of Minkowski, which is connected with a finite set conjecture in the sense that if the initial data is efficiently close to the flat initial data, in other words, it's small, small perturbation of Minkowski space, you will get solutions, which look like Minkowski space in the large. So asymptotically, they look like Minkowski space, there are no black holes. So you don't have emergence of black holes. The problem of collapse, which is that if the data is efficiently large, you could form a stationary state, therefore you can form a black hole, and this obviously is very interesting, both mathematical and physical, from physical point of view. Then there is the problem of rigidity, which is to classify, it's a problem of classification of stationary states. How do you know that the finite states are only care solutions, maybe they are asas? The classification of rigidity conjecture is that all stationary states have to be care solutions. There is the problem of stability, the problem of stability, which is that if you make a small perturbation of a care solution, you stay in the care class, you are not going to create something else. This is by itself a huge problem, which has not yet been solved, and about which I will talk. And then finally, there is a cosmic censorship conjecture, which is even beyond all this, something even more complicated, which by itself is a huge conjecture. So this is cosmic censorship conjecture. And then in addition, there are other problems, which have to do with the end body, so-called end body. So there is an analogous of the end body problem in classical mechanics. There is an analogous, of course, in general activity, which is, obviously, extremely important in particular. It's really the two-body problem, the understanding of two-body problem, which is connected with recent LIGO experiments of gravitational waves, right? All right, so anyway. So the problems that I talked, I then said that I'm going to talk about, I'll talk also about Sabitov-Mikowski space, but mainly, as far as black holes are concerned, I want to talk about the problem of collapse, rigidity and stability. So last time we talked, I gave an introduction in the problem of rigidity and also in the problem of collapse, and I didn't have time to introduce the problem of stability, but I'll do it later. And today I thought I'll go in more depths about the rigidity problem and I will start talking about stability. Maybe collapse if we have time, we'll discuss it at the very end. I mean, I will go more in depths in other words, on collapse. OK, so let's start. Now, unfortunately, this is not... I have to make it full. OK, so I'll talk about the problem of rigidity. And maybe I'll remind from last time we had a discussion, first of all, where is the light one? So we had the Schwarzschild solution. So first of all, the Minkowski space. So this was minus... So the metric here is minus dt squared plus... in polar coordinates plus dr squared plus d sigma squared, where this is, say, in one plus three dimensions. Again, this would be the standard metric on the two sphere. And then the Schwarzschild solution, which is a bit more complicated, which has 1 minus 2 M over r dt squared plus 1 minus 2 M over r to the minus 1 dr squared plus r squared, plus this, I should say here, r squared plus r squared d sigma squared, which is, again, this is the standard metric on the unit sphere. And then finally, the Ker's solution, which is even more complicated, but which hopefully soon I'll be able to show you again on the... So let's draw a picture of the Schwarzschild solution here. So here is a way of representing the Schwarzschild solution, which is very useful. So to start with, we see that we have this r equal to 2 M here. And r equal to 2 M here. So this corresponds to that value that we see in the metric. And as I mentioned last time, if you look at the metric, it looks like you are singular at r equal to 2 M, right? So in reality, that singularity is purely a singularity due to the coordinate. So it's very similar to what happens when you write down... Mikowski space, for example, in polar coordinates, you get a singularity at the origin, but that singularity, of course, is just due to your bad choice of coordinates. The same thing here, of course, is complicated. The same thing happened here, and as I mentioned last time, it took people a long time to realize that that singularity was actually a coordinate singularity and not the serious singularity of the spacetime itself. So once you understand that, you realize that you can change coordinates and go across this hyper surface, and therefore this hyper surface is actually completely smooth. And in fact, it turns out that this program is just the boundary between the black hole region. So this is the black hole and this is the domain of outer communication. So exterior, right? So this is... So in fact, you have two such things. You have one on this side and one on this side. And you also have what is called the white hole, but this is of no interest to us and it is not supposed to happen ever in evolution. In evolution you are only going to see either this side or this side. And everything, then at infinity, so you see, you look at the values so far, so you have interesting values so far r equal 0, r equal to 2m, r larger than 2m. There is also here of interest r equal 3m, which I already mentioned last time, and then you have r equal to infinity, and which is represented here by conformal compactification. In as always, by conformally compactifying the metric, you can actually see that r equal to infinity would correspond to what is called a boundary at infinity, which is called scri. There is a scri plus and scri minus. So this corresponds exactly to r equal to infinity, r equal to infinity. And there is this region here of r equal to 3m. So this is important in the discussion that we are going to have, because you can have geodesics or null geodesics, in principle null geodesics in this picture are at 45 degrees, going either this way or this way. The ones which go this way will go to infinity, so they go for infinity time along proper time of the null geodesic, and the ones which go in this direction of course will go into the black hole. Once they go into the black hole, they will hit a singularity at r equal to 0. R equal to 0, unlike r equal to 2m, is a true singularity. In other words, the curvature will blow up. Therefore, the metric cannot be extended in any way. In any reasonable way, you cannot extend the metric beyond r equal to 0. So r equal to 3m, as I said, there are however, in addition to these geodesics, which is easy to visualize, in this picture in RT, you don't see the other coordinates, the theta and phi are not seen in this picture, but if you take that into account, the theta and phi, in other words, you look at the more complicated family of geodesics, you see geodesics that stay tangent, actually, on r equal to 3m. In other words, this will be null geodesics which don't go into the black hole and they don't go to infinity. From the point of view of geometric optics, right? From the point of view of geometric optics, these are good. The ones that go inside, you never see them again. The ones that go to infinity, you also don't see them again. The ones that stay here forever are actually going to influence the dynamics in a major way. OK, so first of all, the care family, which you see, it's a little bit more complicated than to watch it family, but it is impact as a solution. It admits rotations. Of course, the whole point is that it admits rotations. It has two keening vector fields. It has t, which is d over dt, to see m3, and you have a second one, which is d over d phi relative to these coordinates. So it's an axi-symmetric stationary solution. In this class, which depends on the two parameters and m, if you take a to be 0 and then positive, you get the second solution, which is a Schwarzschild solution, which I mentioned earlier. So this is a picture of Schwarzschild, which I started to draw there. So you see these two regions which are external to the black hole.  This is again the exterior region. Now I need to emphasize that we are interested whenever we study the black holes from the point of view of these problems, which I mentioned, we are discussing it away from the horizon. So in other words, we are interested in this region all the way to the horizon, maybe a little bit inside the horizon, but certainly we are not interested truly in the black hole region, which terminates in a singularity, and there are many other problems of great interest and we are not going to deduze that alone. So the even horizon r equal to 2m, as I mentioned, magic can be extended past it. As we mentioned, this is not a true singularity. You can have black and white holes, which are in the region r less than 2m. The exterior domains are defined by r larger than 2m. The photon sphere, which I just started to talk about, is this region at r equal to 3m when you can have null geodesics which stay there forever. And that creates problems from the point of view of geometric optics and have an impact on the dynamics of the solutions. Null infinity, which corresponds to these boundaries, here and here, and they are r is equal to infinity, in fact. And, of course, you would have two of them, one to the past and one to the future. Of course, in dynamics we are interested in the future, so, obviously, I would only be interested in scribe plus. OK, so that's a situation of the Schwarzschild. Now, Ker is very similar.  You have these two exterior regions. Again, I'm interested in just one of them. Again, you have the picture here similar to the one of the exterior of Schwarzschild. But in the interior things are slightly different because they don't terminate at r equal to 0, which is singularity. They terminate, in fact, in one of the two roots of this delta. Delta, if you remember, it's exactly the polynomial r squared plus a squared minus 2m over r, which is quadratic, and therefore you have two roots. One of the root corresponds to a boundary of the black hole region, and it's called the Cauchy horizon, and again, it's very interesting. There's a lot of interesting mathematics, but I will not talk about it. I'll talk only about the event horizon, which is now r plus. R plus is one of the roots of r squared plus a squared minus 2m over r. It's actually the bigger root. All right, so, again, external region, event horizon and the black hole region, null infinity, similar to what we had before. This is the external part of the black hole, and again, this is something I said last time, that if you look at the stationary keeling vector field, which is d over dt, I write it here as capital T, this one in the asymptotic region, asymptotic region is, of course, the region where r is very big, and this is what we describe in this picture. So, there, the space time becomes Minkowski and so, if you look at care, you see immediately that for r large, care metric looks like Minkowski metric, and therefore, this one looks like a d over dt in Minkowski space, and obviously, it's time like. It's time like, but as you come close to the horizon, it switches and it becomes space like, and this is a major difficulty that we mentioned last time. This non-empty ergo region, so ergo region exactly is a place where T becomes space like, which is the region near the, near the horizon, and you can have all sorts of problems like non-positive energy and so on and so forth, so this is actually very important in this rigidity issue that I mentioned, and then there is another, there is another thing about of interest, which is the region of trapped null geodesics, so again, you can have trapped null geodesics, and in Schwarzschild, because they don't sit on just one hypersurface, so there are plenty of them in a certain region, which is, let's say, close to r equals 3m, if a is small, but it can become quite large if a is large, so if a becomes like m, the region of trapped null geodesics can go all the way to the horizon, all right? Okay, so anyway, so this region of trapped null geodesics also, of course, plays a fundamental role in the problem of rigidity and I'll mention. Okay, so here are some other properties of care solution, so care is quite remarkable, I mean, it really has absolutely beautiful mathematical structure. To start with, as I mentioned last time, you are interested in null frames, so in particular, you are interested in null pairs e3 and e4, and then associated to null pair, you have a null frame, so the null pair is given by two null vectors e3 and e4, which are normalized such that the metric g of e3 e4 is minus 2, so these are normalized by that condition, but they are both null, and then you look at the space perpendicular to them, so at every point you have such a space, so this is infinitesimally, at every point you pick up vector fields ea, e1, e2, which are perpendicular to these two and which are autonomal among themselves, so these are space-like vectors and these two are null. So, in the particular case when I pick my frame to be exactly this one, this null frame to be exactly, this null pair, I should say, to be exactly this one, then a remarkable thing happens with a curvature, so the curvature, of course, in principle is very complicated, so there are 20 components because of a rich flat, it actually should have 10 components, but it turns out that almost all of them are zero in this frame, so if you calculate components of r relative to this frame, so in other words, if I take ea, e4, eb, e4, so r has four places, so I take r of ea, e4, eb, e4, I get what it's called alpha, so please remember I'll talk again, so I'll talk many times about these components. They play an important role in stability also. Anyway, I have alpha, which is this, alpha bar, which is by symmetry, obtained when you interchange four with the three, so this is alpha bar, and then you have the other components which are beta, beta bar, and rho and rho star. So you see, if you calculate, you get two components here because of the trace of r condition, you get two components here, so it's four, six, eight, and two more, so you get ten. So ten components, so this is called the null decomposition of the curvature relative to this frame, and it has exactly all the degrees of freedom that you need, and the remarkable thing is that for this particular frame everything is zero, plus two. So there is some magical diagonalization that goes on, and as a consequence alpha, beta, beta bar, alpha bar are equal to zero, and the only thing that survives is alpha star, and unfortunately you don't see it here, I don't know why. So in any case you see that this can actually be calculated exactly in this border-linguist coordinates, and they get a very simple form. Yeah, so the raw star you see is defined relative to the whole dual, so I take the whole dual of r. Anyway, so this is some simple algebraic fact. But anyway, the consequence this is what you have. There is a beautiful duality in general activity, especially in four dimensions, which actually plays an important role of course in physics. So now there is in the rigidity problem there is another important property of care that is worthwhile understanding, and that's called the mass simultaneously. I want to tell you a little bit about the mass simultaneously because it will play some role. So you see stationarity, so let me explain it on the blackboard. So stationarity means that there is a killing vector field. So in other words, I'm not talking just about care, now I'm talking about the general stationary solution. So if I have a general stationary solution that means that e t of g is equal to zero. But if you look at what this means, it means that if I take the covariant derivative of the alpha t beta this plus alpha t alpha this is equal to zero. So this is a so-called Keenig equation which follows immediately from this definition. And therefore f alpha beta which you see there, it's the alpha t beta is a two form, because it's anti-symmetric. So it's a two form and I can take it's complexified, I can complexify it by taking as a two form I can define the hodge dual of the two form. And I take f plus i square root of minus one in other words of f. And I do exactly the same thing for the curvature. I take r to be r plus i times the dual of r. So I get a complexified curvature and complexified f. The remarkable thing about f is that d alpha of f beta gamma delta verifies a simple equation and it's a consequence of the fact that t is stationary. So this follows relatively easy that you get this equation I should have t here I made a mistake that should be t and that means that if I take i t, so in other words if I take the contraction of f whose t so f alpha beta t beta that's what I mean here so please read it as t rather than z t rather than z so anyway so it means that d of this one form because now this is a this was a two form if I take the contraction I get a one form d of it is equal to zero it means for simple topological reasons it means that this quantity is given by exterior derivative of a scalar and this is called the Jans potential Yes Just one question Yeah, so r here is a Riemann curvature tensor exactly And the hotch dual How do you define the hotch dual? The Riemann tensor is a fourth tensor Yes, and fourth So the hotch dual is not a stellar Ok, so you define it like this you define it as so if I take this to be mu nu I define it as epsilon gamma delta mu nu ok, this is our dual gamma delta alpha beta gamma delta excuse me, alpha beta gamma delta so alpha beta gamma delta is defined like this where this is just the volume form in four dimensions right, so in other words I act only on two of the indices not on all four ok so ok, so this is just I need it all this so I need this sigma as you see this is called the Jans potential and plays a very important role so now I use this complex curvature plus a combination of a quantity which is defined with respect to f it's a quadratic quantity in f and you can simply define it like this q alpha beta mu nu is alpha is f times f a complex tensor which plays somehow the role again of sort of a complexified volume form we see it here, it has this part which is exactly the volume form and then the real part is given by this anyway so you can create in other words an interesting tensor which is formed by these two and the remarkable thing is that first of all in care you can calculate these quantities you can calculate f squared and you can calculate this one so sigma is exactly 1 minus 2m divided by this and the remarkable thing is that s is zero in care that's just a calculation it's not too difficult so s vanishes in care but remarkably and this is a theorem of Mars in 1999 the condition that s is equal to zero makes sense of course for any stationary solution that can construct this Mars-Simon tensor so Mars result says that s is equal to zero characterizes care locally so in other words this Mars-Simon tensor plays the same role as the Riemann curvature tensor plays in Riemann geometry it characterizes the care solution so that's an important concept so any solution which has a killing vector and to describe s is equal to zero locally has to be isometric to the care solution it has to be like asymptotically time-like so obviously it has to be time-like but not necessarily time-like you only need it to be asymptotically time-like but why asymptotically if it's a local if it's a local theorem because it characterizes care everywhere even in the Ergo region you always expect an Ergo region so we always expect a place where t will have to become space-like so it's always time-like where you are far away but it could become space-like in the Ergo region and in the Ergo region of course you have to you also this s is equal to zero characterizes the entire care not just the asymptotic part I understand the entire care but locally yeah but locally at every point locally at every point in particular in points in the Ergo region then it's not local I don't understand this theorem can you state if you want to have it global you need additional conditions at infinity that's all you need I mean the way without giving additional conditions at infinity the only thing I can say is that in a neighborhood of any point I can show that if the mass amount is zero the space time is care but it can be cared in whatever different morphies so to get exactly the care solution the way we know it you need also some kind of global conditions but anyway for the moment I think the local characterization is good enough so go back to the rigidity conjecture which I mentioned last time so the rigidity conjecture says that care family exhaust all stationary asymptotically for the regular vacuum black hole so of course you need a notion a general notion of stationary solution which we discussed last time and then you want to show that with the definition of a black hole solution in other words a stationary solution which has certain properties under these conditions you show that the space time has to be care of course you always talk about stationary solution of the Einstein equations of the Einstein vacuum equations so this is what we were two days ago we started to I started to tell you what is known so there is a known result in the static case which I'm not going to mention anymore I mentioned it last time this is the state of this result is very very good in other words we really understand the static case in full generality so the actually symmetric case is actually much more complicated the general case is much more complicated general means again that T so this stationary kinetic vector field is said to be static if in addition to the fact that it's a kinetic vector field it satisfies another integrability condition which we mentioned last time but I'm not going to talk about it now and so now we look at the general case the beautiful result due to Cart and Robinson which is at least 40 years old which says that if in addition to stationarity you also know that the spacetime is actually symmetric then you are in care so that's a result which is based on elliptic theory it's a reduction to harmonic maps beautiful result highly nontrivial and well known and then in addition to that this result it's an observation due to Hawking which was that if you also assume analyticity then the result is true in general in other words analyticity and stationarity implies axial symmetry and I mentioned a little bit about this result last time and I also mentioned that this result is not very good in that this assumption of analyticity can almost never be doesn't come naturally from the equations and the reason very simply was that if t is time like in region where t is time like the equations reduce the ancient equations in those regions reduce to an elliptic system and therefore elliptic system naturally have analytic solutions real analytic solutions but if you are in the Ergo region where t becomes space like you are there is no way the equations become hyperbolic even more complicated there is a transition between elliptic and hyperbolic these are mixed problems which are extremely complicated in general there is no reason to expect analyticity R equal to m is a limit between space like and time like 14 so I'm really talking now about the journal case I cannot talk about Schwarzschild is R equal to 2m so now it's I just have a horizon it carries a little bit more complicated it's not R equal to 2m it's another value it's this root but you do have an Ergo region you have a non-tripe Ergo region in Schwarzschild you don't have an Ergo region it's just in care in any case this hocking result is only interesting in so far is that it tells you that all explicit solutions of ancient equations have to be care but if you look beyond explicit then there is no reason to think that this tells you anything does it follow from hocking result that at least in the region where t is time like if that region exists the solution might occur no, of course not because you can't yeah you need a whole thing global, you need a whole global picture in order to conclude ok, so now why is it clear that you need perhaps you don't need to know the whole solution you can show that if it's outside certain sphere no, but you can't everything starts exactly, everything starts from I'll mention about this in a second everything starts from the horizon the whole construction is based on something that happens in the horizon as I mentioned last time let me repeat that argument of hocking so this is a journal stationary solution it's not too complicated under the assumptions we have made to show that a horizon exists it's a soft argument so there exists a horizon and it's also not difficult to show that the keeling vector field the stationary keeling vector field becomes space like in other words it's tangent on the horizon and therefore it rotates along the horizon ok, so it does something like this and from that fact it's not hard to say that there is another keeling vector field exactly on the horizon a second keeling vector field which this time is in this direction in other words it's exactly tangent to the generators of the horizon the horizon that you remember is a null hypersurface so it has null generators so the second keeling vector field is going to be tangent to these null generators ok, so in other words you get a second keeling vector field but this is a keeling vector field only on the horizon and the next thing you have to extend in the interior and that's where analyticity is used in other words analyticity is used exactly near the horizon in order to extend and once you have analyticity everywhere the result, the observation of hocking extend the second keeling vector field everywhere and therefore you reduce yourself you reduce yourself to this case the actually symmetric case but it's a very complicated argument and in fact the depth of the argument is in here this one, what he did is rather soft and obviously it's not at all conclusive in general ok, so so this is exactly the observation of hocking that there exist a second keeling vector field along the horizon now you see, once you have a second keeling vector field you could do the following thing so you here is why things are not going to work so you have you have the horizon and you have a second keeling vector field so you have now let's call it say Z ok, it's actually called it's actually called K but if you know how to extend K from the horizon everywhere inside and you have T, K and T you can show by simple argument that K and T generate together they generate a rotation so they generate a second keeling vector field another keeling vector field which is actually anaxial keeling vector field ok, so the question is given K here can I extend it, well K has to be extended as a keeling vector field so I have to have the alpha K beta so it has to satisfy the keeling equation equal to zero if I take a second derivative here and I commute derivatives and use the fact that the space is rich in flat I get so this will be zero it's easy to see that this is going to be zero and you'll be left with essentially a wave equation for K is equal to zero so in other words you want to extend it in such a way that the line version of K is equal to zero so you have K given here and here from Hawking's argument and you think maybe I should extend it like this the line version of K is equal to zero because this is at least consistent with the fact that K should be keeling so what is the problem but the problem you can see already in much simpler situations imagine that I have an alkan so imagine that I have sort of an alkan like this in Minkowski space so I'm simplifying the situation to Minkowski space so I have a sphere here and I have light cones going this way and suppose I want to solve K is equal to zero with data here and here so this is the initial data is now a characteristic initial data so I'm prescribing data here and here and I want to solve this one so you say well it should be able to do it but it turns out that it's an ill post problem so it's an ill post problem you don't have solutions so these equations don't exist in general so they exist of course if the data here is analytic then you can extend of course but in general you cannot and that's sort of a major obstacle so this is the reason why hocking argument will not work you will not be able to extend it naturally unless you have an electricity so this is a typical to ill post problem so ill post problems play a major role in mathematical physics partial differential equations there are many situations where these things occur and of course they are very interesting and there are methods developed by mathematicians to deal with them which I will talk about it a little bit later so our approach is in fact based on these new methods which were developed by mathematicians to deal with ill post problems so it happens that you can see it from here actually already that though if I give you an arbitrary phi zero I will not be able to solve this equation if I I can do something nevertheless which is interesting which is that if I have two solutions I can show that if they coincide here they will have to coincide everywhere so uniqueness still holds even though you don't have existence uniqueness holds so there are in other words ill post problems for which uniqueness holds but existence doesn't so the whole idea of our approach to the rigidity problem is to use this geometric continuation arguments based on uniqueness and which also require this analytic tool which are called Karleman estimates so let me mention first of all the main obstruction so what is a main obstruction maybe let me actually talk about this a little bit later so let me go on and mention some interesting results which are a little bit simpler than the general rigidity problem so again the general rigidity problem is to show that in the absence of analyticity I still have uniqueness of the Karle solution I want to show you some results which are a little bit easier than that but which are still interesting because they show the importance of some concepts which play a role of course in the big problem so here is a first kind of problem suppose I have a money fault which is Richy flood so in fact actually in this kind of result I don't even need G to be Lorentzian it can be any pseudo-Romanian metric but anyway in particular it's Lorentzian plus and also that dimension here doesn't matter I can take any dimension suppose I have a keeling vector field Z so Z is keeling in this domain O so I have a domain here and I have the boundary of the domain so this is the boundary and so I know the Z is keeling and I want to extend it I want to extend it outside in a sense what Hawking was trying to do he had a keeling vector field exactly on the horizon and wanted to extend it so the issue of extension is what I'm going to address here so if everything is analytic you don't even need this condition so actually you can replace the Einstein equation in fact that's exactly what Hawking did his result did not require the Einstein equations at all, analyticity is like replacing the Einstein equation by the Koshiriman equations so in the case of analyticity you don't need essentially any condition you just need the Z is keeling and you need some topological conditions that show immediately that Z can be extended everywhere so in the analytic case there is no issue whatsoever simple topological condition give you extensions of Z on the whole manifold but of course if the manifold is smooth only then zeta remanes be keeling in the extension ja ja ja, of course I want Z to stay keeling, ja, I want to extend it so that it's still keeling, obviously otherwise of course you can have lots of extensions but I want a keeling extension so the question is so what can you do now in the smooth regime in the non-alytic regime so of course the results are going to be restrictive but interesting in the following sense so the result which I mentioned here is that if I look at the neighborhood of a point P and I assume that the boundary verifies a certain condition which I call null convexity condition so if the null convexity condition it's called pseudo convexity for more general PDE but in this case it's much nicer to call it null convexity so the null convexity condition can be expressed geometrically in this simple form which is that if you take the Hessian of a function here which defining function of the boundary in other words I take a function f which is zero of the boundary the boundary is equal to zero and is negative inside the domain and it's sufficiently smooth so if I take the Hessian in the direction of any null vector field which is tangent to the boundary so I take at all the tangent space at the boundary in particular just at the point P is good enough so has to be null so I'm looking only at null things so g of x is equal to zero then I want the Hessian in the direction of x to be negative at point P so that's a very simple condition which is called a null convexity condition so if this null convexity condition is satisfied then the result says that z can be extended past P in a full neighborhood of the point P and of course z will be killing in that neighborhood, so that's the result in addition if x is perpendicular to T then you are interested there is a companion result maybe I'll talk about this a little bit later but I will concentrate on what's not read here so that's a result here is it again so z is killing the boundary is strong in null convex strong in null convex means exactly this then z extends as a to a neighborhood of the point P so this is so this is the first result that I want to talk a little bit about it today now here is a second result which is in a sense slava, in a sense this answers to some of your questions so here is a second result which says the following, so let me describe the picture here, so again you have the care solution, think of it as a care solution take any point on the boundary, on the horizon but not here, take any point here and the claim is that I can create in a neighborhood of the point I can create a new solution of the answer equation which is exactly equal to care on this side but which is different from care on the other side it's stationary, verifies the answer equation but it doesn't have any additional kidney vector fields it's an obvious contradiction to what Hawking wanted to do obviously it holds in the nonality case in the nonality case of course it's always true that you can extend but in the nonality case you can see that if I take any point on the boundary on the horizon I will not be able to extend because I can always, in other words I'm not going to be extend in the sense of creating something which has a second kidney vector field so again I start to scare here in a small neighborhood of a point B on the horizon I can construct a solution in a small neighborhood of the point which is exactly care on this side it's verifies the answer and vacuum equations everywhere it's stationary on the other side also but it doesn't have any extra kidney vector fields yes my basic confusion with the whole where you present things is as follows so you started by saying that outside where it is time like things are analytic for this you have no problem then you said, ok let's look inside there this argument does not it does not apply so things get complicated and now you are showing how they are complicated but what about the outside what if I sit outside and start moving inside outside things were analytic then at some point you expect that they are going to become nonanalytic starting from sky starting from r equal 10m but but I have no information there I need an extra you know that it's analytic there but that's not good enough, lots of analytic solutions but supposedly I start moving inside are you sure that things will become nonanalytic where they are going to become nonalytic they will probably blow up somewhere I mean they will probably blow up even before their origin I mean they will be local there is no way I can extend them in any reasonable way but will the solution stay analytic perhaps it will but they will stay nonanalytic until they blow up what will blow up? the solution, I mean the solution exists only locally so the only thing I can say with this solution of what? you have a nonlinear problem for each e zero I have a kidney vector field which is say time black so I have a point in my manifold where t is time black we assume that we have a black hole we assume that we have a black hole we assume that we have a black hole no, but I am talking about the result which I want to apply I want to apply result about analytic about I can't say anything I mean from the information that I have at this point which is away from the horizon I can't say anything else as they approach the horizon there is no no way to control the solution that starts here control it here yeah, the solution exists you tell me but so what I don't have any way of connecting what's here with what's here it's a nonlinear problem there is no way I can do that where is the surface where things can go wrong it's the surface the information that I have from my analytic solution starting from a point will not even go all the way to here the information is purely local the information that I can extract is purely local I don't understand what you are so the solution is analytic there but so what there are lots of analytic solutions I can't tell anything you see what I am saying you are going for the most pessimistic possibility the most pessimistic possibility is that the solution will actually be analytic you mean here? no, absolutely not because again analytic solutions of a nonlinear complicated nonlinear PDE exist only locally they are defined only locally they will blow up typically and the information that I get from points here in this neighborhood of this point have nothing to do with the only information which I have is here here I know that there is a second kinevector field there is no other information I want to use this information you have to go from here to here I don't know how to do that my first impulse would be to try to cross this problematic surface you know that on the right I can't I can't in fact there is absolutely no way you can do it so you can do something else I think that you can try to do you can come all the way from infinity because you can hope that also at infinity at the boundary at infinity you get some additional information that's true but that also has never worked so there are lots of difficulties also at infinity there are difficulties which have to do with somehow the metric at infinity is pretty irregular itself at infinity so you can't do it there are lots of things that people have tried they never worked anyway so this is the second result so so let's with these two results I might give you some conclusions already so there exists no as explicit stationary so this is the result of Hawking there exists no as a stationary solutions close to an extremal care or care Newman and this is exactly the result that I mentioned which is the result of Yonescu Alexakis and myself right it also can be generalized but it can be generalized anyway I will not say any words about this so this says that if I am sufficiently close to care I have a stationary solution which is sufficiently close to care then I am in care because probably I won't be able to say much more about the problem maybe I should mention at least that much so how do you say that you are close to care so I have a stationary solution of the initial equation I have a Killing vector field T which is time like in outside the ergo region so this is at infinity so in this situation we have introduced this additional tensor which is mass-simon tensor so this is a complex tensor which has a property that if it vanishes then you are in care and therefore the result so this is Alexakis Yonescu and myself says that if s, this mass-simon tensor is sufficiently small so in other words I have a stationary solution such that the mass-simon is sufficiently small then you are in care so this is the only result that we have there are many versions of this result but they all need smallness the full problem is far from being solved because we don't know how to do the large deviations from the care solution but we do have a conjecture which again I mentioned last time which is this conjecture was Yonescu and Alexakis which is the rigidity conjecture holds through provided that there are no T-trap null geodesics so this is you see the presence of trap null geodesics is easily seen that it has something to do with lack of null convexity so the null convexity fails so if I start if I start from the horizon and I try to extend this second keening vector field so let's call it K so if you remember Hawking produced a keening vector field on the horizon and the point here was to extend the importance of Hawking's observation is that this keening vector field was defined everywhere on the horizon so I have a keening vector field defined everywhere on the horizon I need to extend it and it turns out that in the spirit of this result that I mentioned here the null convexity condition is actually verified in a neighborhood in a very very small neighborhood of this of the horizon and as a consequence because of this I can extend so this is the result that I just mentioned or rather maybe a variant of this result this result was quite local here I need something which is global but in any case relatively simple variant of the result will allow you to extend the keening vector field slightly away from the horizon and then if you want to keep going so you want to continue to extend at some point you hit trap null jodesics trap region and as you hit the trapping region this null convexity will just evaporate in fact the null convexity instead of being strict null convexity it becomes zero so in other words instead of having the second the Hessian instead of having the condition dfx is negative there it becomes actually zero and I am dead I cannot extend it any longer so the trapping is a serious to extension so so what do you do well it turns out that since you are only interested in stationary solution so I am not interested in extending all possible solutions I am interested in extension only the solutions which verify leti of g is equal to zero and this is sort of a global thing I mean this is verified everywhere by the definition of stationarity and as a consequence and it is not very hard to see as a consequence the only obstruction that you see are not also here if you remember these are null vector fields right but in reality because of that condition I am only interested in those vector fields which are perpendicular to t right and these are the only obstructions for the problem of stationarity when I know that I have a stationary vector field are in those directions which are null and for which g of xt is equal to zero so that's the notion of t-trap null geodesic null geodesic is said to be trapped if it stays in a finite region it says to be t-trapped if it also stays perpendicular to t and these are the only obstructions so as long as those don't exist you are fine and in fact in care they don't exist so it's easy to see, this is a calculation in care you can show that there are no t-trap null geodesic so there are plenty of trap null geodesic but none of them is t-trapped and that's fundamental because it plays a major role also in stability in the stability problem so it's because of this that in a neighborhood of care you can still do it so if the mass simultaneously is sufficiently small it means you are in a neighborhood of care then this condition is still verified in a neighborhood of care and you can do it and finally and I'll finish with this rigidity there is this conjecture of Alex Akiks and Esken myself which is a rigidity conjecture holds through provided there are no t-trap null geodesics so I don't need any smallness condition I don't need mass simultaneously to be small just the conjecture is that so it's a purely geometric condition as long as there are no t-trap null geodesics then we are in care and then as I mentioned last time it's not at all clear I would think that the conjecture is true but I don't believe that you can rule out t-trap null geodesics from abstract principles so t-trap null geodesics probably do exist and therefore there exist probably exceptional cases to the rigidity conjecture in as always probably there are some very isolated space times which are stationary which verify all the other conditions that you need and yet they are not they are not cared and well I wanted but unfortunately I lost a lot of time at the beginning I wanted to give you some sense of the proofs of these things I would not be able to do it I just want to say that these t-trap null convexity conditions so these kind of results here which are highly nontrivial in their own right require require a lot of geometric analysis based on Karleman's estimates so the geometric versions of Karleman's estimates which are needed but I will not I think I should say anything more because I won't have the time maybe I'll go to the conclusions again so there exist no other stationary solution there exist no other stationary solutions close to an external care external not extremal sorry, apology is not extremal it's external then the rigidity conjecture is satisfied and also finally there is also the dimension which I think it's actually very interesting there is that locally you cannot do extensions typically if I start from some point on the horizon and I look only locally in other words I'm not using the full global information along the horizon then I cannot do the extension so there isn't this is important because it shows you that what allows you to have this non convexity condition is the intersection of the two horizons so it's here that this non convexity condition is verified but it fails here and certainly it fails here so locally I will never be able to extend unlike what Hawkin did and certainly the only way I can do it is if I start from the intersection of the two horizons ok, so this maybe I don't know it's already 25 so so you take maybe few minutes and then I'll start talking about stability instead of talking about rigidity I'll start talking I mean instead of saying more about rigidity which of course interesting thing will be to prove these results that I mentioned in particular this one this I think has a very nice and very pretty proof which exemplify the importance of geometric ideas but let me not go into it because I want to talk about the most complicated problem in connection with this with this what I call the tests of reality which is the problem of stability and this is the most important problem I mean it's by far the most important problem in connection with with this tests of reality and of course obviously physically it's immensely interesting because if it so happens that the solutions are unstable then obviously the very concept of a black hole which is based on the care solutions will go out the window it will have to find other explanations for what the astrophysicists see so but hopefully they are stable and let's see why alright so first of all condition a smaller than that means that it doesn't rotate too fast it doesn't rotate too fast exactly ok so here is a conjecture so again here is a picture of the care solution right we are interested only in the external part and in order to formulate this stability I have to take a space like hypersurface I take any I take an arbitrary space like hypersurface which looks roughly like this it goes from here to here but again I'm interested only in this part or maybe to go a little bit beyond but not much and I look at the restriction of the care solution to this space like hypersurface it will give me an initial data set right so the initial data set is a trivial initial data set of the care solution and then I want to make now a small perturbation of it so I change the care solution somewhere by very little right and of course I want to change it in such a way that the constrain equations are still satisfied remember that the constrain equations are very important in general activity but anyway this is not a big deal you can always do that so you can always find perturbations of the care solutions which verify the constrain equations you look at those initial data and you start you look at the maximal global hyperbolic development which is a concept that we discussed last time which means you extend the solution as far as you can the danger of course will be that it stops in finite time in other words observers which live on the solution will hit some singularity at some point and they will be destroyed so they exist only for a short time of course the stability will have to mean in other words this maximal global hyperbolic development will have to be complete so this is what it says small perturbation of a given exterior care solution have maximal future developments which in fact converge to a care solution but not necessarily the one you started with so this is very important you don't necessarily go back to the same solution it may be another solution that's what makes the problem hard I mean but this is one of the reasons why it's hard because you don't know a priori you don't know where you would converge so let me talk a little bit about this is a quotation from Chandra Sehar which was I don't know he has this beautiful book on mathematical theory of black holes and this is what he said the treatment of perturbations of care spacetime has been prolixious in its complexity perhaps at a later time the complexity will be unraveled by deeper insights but meantime the analysis has led into a realm of the Rococo splendorous, joyful and immense ornate so of course Chandra Sehar was sort of a great admirer of of art and beauty and he thought that this subject is very beautiful but very Rococo unfortunately and but he thought that later time it will be the beauty would be unraveled and things will become simpler despite 50 years of theoretical progress and the worth of numerical and indirect astrophysical observations this conjecture is still open and I I'll try in the rest of this lecture I'll try to tell you what's known alright so first of all there were I'll go very fast about these results and then I'll mention again later on but what's known is only linear stability so linear result so what happened I'll discuss more about this you can actually linearize the Einstein equations but of course you have to be careful because when you linearize the Einstein equations you have to take into account the gauge so you have to remember that the solution is actually a class of equivalence of solutions so the gauges are very important you have to linearize in such a way that what you get is gauge invariant and that's not easy but anyway it can be done and this was understood by Fils and this is reasonable well there were some results of Schwarzschild sorry, the results in Schwarzschild by starting with Wheeler in 1957 Reggie Wheeler the very influential paper that they wrote just on the linear stability of Schwarzschild there was some results of Dinoveshara and Zerilii in 1970 and then for care the most important results are due to Tawkowski pres Tawkowski in 1973 all these results are on linear stability but linear stability done the way Fils is do which is they expand, they do some kind of expansions in psychological harmonics or in the case of care it's a little bit more complicated and they in other words they do use the solutions in modes and they are interested only in the mode stability so they want to see that the modes don't explode somehow so the lack of exponentially growing modes is what they are interested these results gave some indication that they are not exponentially growing modes but the true result on this mode stability was due to Whiting in 1989 where he gave a sort of a complete picture almost complete picture actually of the fact that there are no exponentially growing modes so that was in 1989 by Whiting now Fils is declare victory of course typically if they show result then they immediately claim that they understand the whole thing often I just here I want to point out that if lack of exponentially growing modes for the linearized equations was enough to deduce nonlinear stability then the presence of shock waves extreme sensitivity to do with data and turbulence in fluids will be ruled out so if you think that lack of exponentially growing modes is enough to understand stability of the nonlinear problem then you have a problem because in all these cases when you linearize for all these phenomena when you linearize you see immediately there are no exponentially growing modes for turbulence you see that for higher areas numbers there are exponentially growing modes for higher areas numbers you see but you have many other instabilities which there are no exponentially growing modes so ok, the typical example here is a typical example which is typical to shock waves so u t plus u u x is equal to zero when you linearize you just get u t equal to zero and of course no, ok, but that's not the point but when I linearize I linearize around care or I linearize around Minkowski or I linearize around something which is given to me so in this case I linearize around the trivial solution and nevertheless so it's always linearized around one solution, not more so I just take u equal constant for example or I take u equal to zero and there are no solutions that you can come up with no, that's not that's not how the linear stability is done that's just no way no, it's done relative to care so I fix the care so I fix one solution and I linearize around it I don't linearize around all of them that's much more complicated I mean nobody can do that so the linearization is always done relative to a given solution anyway, we can argue more about this later I think your turbulence remark is out of place because the various numbers okay, let's discuss it later okay so I claim that this is true also in turbulence but we'll discuss it later so the weak linear instability is to be expected in view of final care difference from the one sweeper term in this covariance of the Einstein vacuum equations stability requires quantitative decay of the final state and leading to lack of decay for the linearized field so this is something about what we'll need to have in order to prove stability I'll talk about this later on so let me actually go to sort of a more concrete discussion about the kind of stability and instability that you see in linearized for linearized gravity okay, so first of all here is a care solution again so this you have seen so I'm not going to say much more okay, so let me talk about let's talk about so I'm going to address exactly the issue raised by Slava so let's discuss it in full generality so let's imagine that I have only a problem that I have to solve n of u is equal to zero so this can be an ODE or partial differential equation that can be something very general and let's assume that I have a solution so this is equal to zero and I want to talk about various notions of stability so so let's for example you can think of phi zero to be not just any solution but say a stationary solution so something which doesn't depend on time so the simplest notion of stability which of course we know already for ordinary differential equation is that of orbital stability which is that if I take a perturbation so I take a psi I look for solutions of the nonlinear equations which have a small psi perturbation let's say initially the perturbation is small and I want to know what happens for a long time so let's say a time t equals zero I know that psi is small and the issue of orbital stability is whether psi stays bounded for all time asymptotic stability is the question of whether not only psi stays bounded for all time but the perturbation tends to zero in other words phi zero plus psi tends to back to the original solution phi zero right so the second type of kind of stability questions that you want or I might say if you really want to understand this the first thing that typically you do is you linearize so you look at the linearized equation and you want to understand what kind of behavior for the linearized equations you have which you later on will transfer it to the nonlinear equation so first of all there is an issue of mode stability and in other words I look at the linearized equation and I decompose it into modes so I look at some kind of I look at the eigenvalues of the linearized equations and I show that there are no growing modes in fact mode stability in that case will mean that I have to prove that there are no growing modes the lack of growing modes by itself is however and this is already I am addressing your issue lack of even if I am able to show that there are no growing modes even imply boundedness of the solutions and there are counter examples where you can show that there are no growing modes and yet the solutions are not for PDs not for ordinary differential equations so for PDs and then the other thing that you might want beyond boundedness and this is something that is absolutely essential if you want to study the full nonlinear problem so if you are just interested in linearization if you are interested in the nonlinear problem boundedness is not good enough it is just too weak an information because in the nonlinear problem at some point you will have to integrate the corrections to the linearized equations will have to be integrated it will have to be integrated by all time and boundedness will just simply not be enough because you don't have any integrability excuse me, for ordinary differential equations you can have polynomial curve even for ordinary differential equations you can have job and log in polynomial curve yes ok, fine so that is a very good point so even for ODEs this is not sufficient there is another question you said there is no exponential mode but are there polynomial growing modes? so this question here should be for polynomial or exponential mode now having no exponentially growing modes does not exclude the possibility that there are growing modes but even if you don't have growing modes if you have boundedness of all the modes it doesn't mean that you have boundedness and certainly boundedness will have no that's the point I'm trying to make even boundedness will be very far away from nonlinear stability so again of course mathematics is speaking your right but it's not like in physics we don't know that you have to be careful and that doesn't always imply that you should be careful there are three fine, but that's not the statement the statement about the problem of stability of the care solution was very clear so if the linearized equations around a fixed care has no exponentially growing modes, that's it so that's wrong and what I meant about for example what I meant about about the Euler equation is that if you look at u dot grad u is equal to zero so you have the Euler equations and I start again, I'm just starting that u equals zero there's no turbulence around u equals zero of course they you have turbulence if you have but you can have turbulence too because this could blow up even if I make a very small perturbation of u equals zero this equation will blow up and presumably they blow up is connected with turbulence but turbulence has to do with this one but in the limit as nu is equal to zero this equation is very relevant I agree with you that turbulence turbulence is something much more complicated which has to do with the interaction between the Euler equation and the fact that nu here turns to zero but nevertheless the fact that this actually around u equal to zero there's no exponentially growing modes and nevertheless it's not stable we can talk more about this later but in the case of the black hole stability the statement was very clear if you linearize around care then you have stability of the black hole and that statement is clearly wrong let me go on to the next statement so the stationary case we still want to look at the linearized problem and I want to point out what can go wrong even beyond so again I take a stationary solution I look at the linearized problem which is n prime of i zero i is equal to zero and one thing that can happen which is the worst if I have an exponentially growing mode so if I have an exponentially growing mode so I am done there is nothing I can do but there are other things that can happen namely if I have a family of stationary solutions so phi zero is just one of my family so I have phi lambda and lambda equal to zero I have my phi zero but I have in reality a continuum family which is the case of course of the care solution the care solution has a two parameter family it's a and m so if I am looking in this general case I am looking at the family so I am looking at the fact that n of phi lambda is equal to zero so this is a non-linear problem and I differentiate with respect to lambda what I get is that d over d lambda of phi lambda at lambda equal to zero is in the kernel of n prime of phi zero so in other words I have an eigen function if you want of the linearized problem with zero eigen value and this is clearly non-trivial so it's a non-trivial eigen function so this is one thing that can happen whenever I have a continuum family in my non-linear PDE as a continuous family of stationary solutions I have that kind of difficulty here so of course the eigen values can lead to instabilities the second thing that I can have is that it could be that my equation has some admit some class of the homomorphism which takes solutions into solutions so this is of course a case of general activity where for every given metric you have a whole family you have a four parameter class of the homomorphism so in this case I can do exactly the same thing I can take phi zero of psi lambda so phi zero of psi lambda is still a solution because of the fact that this is invariant the equations are invariant relative to the homomorphism and now if I differentiate with respect to lambda and I take lambda equal to zero I get the d over d lambda of this thing here is also an eigen value a non-trivial eigen value eigen function and only eigen functions of my system so all these things can lead to degeneracy in principle but in reality will show that somehow these are not so dangerous in a sense of which I'll explain it in a second now finally we have the interesting instability of phi zero which means if you have negative eigen values then you are dead there is nothing you can do, I mean in that case you don't expect any stability so here is in the linearized case here is what can happen again the presence of a continuous family of stationary states will have to imply because of that instability will have to imply that the final states may differ from the initial states phi zero in other words you make a perturbation of phi zero but in reality you don't converge back to phi zero you converge back to something entirely different and that you'll have to find in other words the final state has to be found the presence of a continuous family of invariant morphism require us to truck dynamically the gauge condition so now you have another difficulty which is that to solve the actual nonlinear problem you have to truck the different morphism which is sort of a problem of center of mass frame you have to find the correct center of mass frame in which the solutions do actually converge towards the final state so that's the second one and that somehow this kind of thing here is called modulation theory so this modulation theory is understood relatively well in simple very simplified much simple problem but in general activity of course is certainly much less understood now the way to think therefore in terms of linear stability somehow what you want to connect with these things here is that quantitative linear stability in order to prove nonlinear stability you want some concept of quantitative linear stability in other words I want that my linear theory should be such that after you account for this one and two in other words after you account for this eigenfunctions corresponding to zero eigenvalues all solutions of this equation decay sufficiently fast so it's not enough to have non-expansionary growing modes but you actually have in fact decay and to decay has to be sufficiently fast in order to have stability so modulation as I said is just a method of producing solution of the nonlinear problem accounting of these two things in other words you produce solution of the nonlinear problem by dynamically tracking down the final states and dynamically tracking down this this gauge condition the final gauge condition in which the problem makes sense I wanted to say so when this linear stability physics result that you refer to it was more than just absence of growing modes because there was a very detailed classification of quasi normal modes things between infinity so you cannot say that physicists stopped or nothing grows fine there was detailed understanding achieved which actually there is a role for the result about gravitational waves that you like to cite the ring down phase the ring down phase of that Slava, let me apologize a little bit let me apologize I didn't mean at all to say that physicists have done nothing on the country the fact that there are non-expansionary growing modes is not just a growth not just absence of growth there is also exponential decay for many perturbations which also this linearized analysis reveals but there is nothing to do it has nothing to do I will discuss it but there is something else what they see what they see but it doesn't but the proof look there is a difference that's what I'm trying to say it's very nice you get very pretty pictures but it's not at all rigorous maybe it's a truth maybe it's a truth maybe it's a truth but the truth is not expressed is not explained in any theoretical way in terms of understanding the nonlinear problem listen, Slava, let's talk about this later because I love to argue with you about it ok, so once this is discussed to prove nonlinear stability you have to, as I said find this gauge condition which you have to find dynamically you need a robust mechanism for deriving sufficient decay because otherwise you don't have you don't have any chance of producing either the final state or the correct gauge condition you need a version of the null condition with respect to the gauge what this means is that somehow the structure of the nonlinear problem has to be such that you get you get certain cancellations so maybe since this has something to do with this argument that I have with Slava let me mention something connected with it so, you see the Einstein equations in a sense can be viewed at a very, very simplified level is being not too different from equations like this down version of phi, so this is 3 plus 1 down version and here you have something which is say a quadratic nonlinearity and I take the simplest possible one so I take dt phi squared ok and I do the most trivial thing which is what I mentioned earlier I want to take, I look at the stability of phi equal to 0 so I look at the initial data phi equal to 0, dt phi equal to 0 and I perturb it in other words I take epsilon here something small epsilon something small and of course there are no exponentially growing modes obviously I mean this is a linear level is just down version of phi is equal to 0 right and yet this problem blows up in finite time and in fact if you look at the quasi linear equivalent of this so for example if you look at something like dt phi which is quasi linear the corresponding solution is actually a shock wave so it typically shock waves of course occur in reality and they have nothing to do with exponentially growing modes they are in instability which is far deeper than exponentially growing mode the level of exponentially growing mode this actually not only is bounded this decays, so phi actually decays like a power of t say Is this a good model for what is going on in the black hole because what is this dt phi squared but give me a good model it's part of what a mathematician has to do for example physics would say that perturbations of black hole either escape to infinity in the form of gravitational waves or fall inside the horizon where is this intuition in your discussion but this is even more trivial than that there is no black hole it's even easier everything goes to infinity it's as simple as possible I don't understand what you are saying I mean this is the easiest possible thing it's much easier than a black hole problem a black hole problem has many other issues you see the black hole problem has this trapped non-geodesics for example which are much more complicated than this case so I don't really understand I don't really understand what you are saying I mean this is by far the simplest possible problem that you can think of in terms of black hole stability as a bad model it's not a good model because obviously what happens for the Einstein equations if you take into account the gauge condition you can show that these kind of interactions don't exist is it the model for outgoing or for outgoing part of the radiation this is for outgoing of course you are in the whole space so it's for outgoing going to infinity there is nothing incoming here there is part of your perturbation there is back how do you call it the back scattering which exist because of the non-linear problem everything moves out to infinity so it's infinity more trivial than the black hole stability problem again in the black hole stability problem there are many other issues I'll talk about it but one of them for example is just this strapping which you don't have in Minkowski space there is no strapping here and the fact that the linear level look at this I mean what can you do you can't do anything better not only you get that there are no expansionary growth modes 5 decays like t to the minus 1 these things form shock waves right strapping is not at infinity because at infinity I would expect that given that you already proved the theorem of stability of Minkowski space by the time things skip to infinity they will remain small in the case of stability of Minkowski space there are no expansionary growth modes and yet it's a hard problem no but it's just I don't understand so there is a certain amount of intuition physicist is not great look we are not talking about physicists are great the only point I have the highest opinion I have the highest opinion of physicists all I'm saying is that often they declare victory when it's not the case that's all I'm saying did they make very important contribution absolutely I'm an observer here there are some students for mathematicians you basically telling them you have to reinvent the wheel start from scratch physicist didn't do anything you should just like renalize from scratch intuition about things going there falling into the horizon and so on this is nowhere present you say that intuition is not important what is important is this example but this intuition is trivial the fact that things going to the horizon everybody knows you look at the solution you see that things going to the horizon I don't understand your point the event horizon is already the event horizon obviously things going to the horizon what is this great intuition what question normal modes have no player role in this absolutely no role at this stage they play a role in the observations I agree but that's a totally different story that's beyond once you have the stability I mean once for example in this case once I have stability I can go on and talk about about more refined features of the solutions quasi normal modes refined features of the solutions are not really nothing to do the fact that the solutions don't blow up before it could very well be that if I make up a version of the solution I get solutions which actually form singularities how do you know that they don't form singularities for me it's an interesting thing to do and that you think it's a prime feature for me it looks like it goes hand in hand with stability but perhaps I am wrong go hand in hand with stability the existence of quasi normal modes and the evaluation is something I mean it's it's an extra feature of the equations which is very interesting and relevant but it has nothing to do with the actual issue of whether solutions of these equations are global or not stability of Minkowski space I don't have to know anything about quasi normal modes In term of what means quasi normal modes Well, look I apologize I should not be your friend No, but I should also apologize because I made myself misunderstood I don't mean to say at all on the country you'll see later on if you had a little bit more patience I would have talked about what the physicists have done in more details and you see that it's extremely important what they have done it's just that the only statement I made is that the physicists often declare victory when the victory is not there that's all I'm saying what they have done is not interesting in fact not only interesting but fundamental in fact sorry sorry to everybody so maybe I should finish and continue next time all I wanted to do is maybe I'll show this slide and then I'll finish so we talked about the general problem I talked about the problem of stability in general so I had n of phi is equal to 0 and I looked at I looked at stability around phi 0 and we discussed various types of stability and linear stability linear instabilities so to the actual Einstein equation so for the Einstein equation the role of this phi 0 is of course a care solution so I have the care solution and the ratio of it is equal to 0 so we have indeed two parameter families so m and a are parameters which are continuous parameters so I can do exactly what we did in that case I can do it here so I can linearize the Einstein equations so I find that this derivative of g with respect to m is actually an eigenfunction corresponding to 0 eigenvalue so in other words I have a whole family I have a one parameter family of solutions non-trivial kernel in other words non-trivial kernel here this is two dimension and in addition I have this gauge gauge invariance so gauge invariance means now that if I look at one parameter group of deformation which is generated by a vector field so I look at a vector field on my manifold and I look at deformation the one parameter group generated by x I find out because of the fact that phi t star of g and g are in fact the same so the solutions are invariant with respect to this deformation I immediately conclude that if I take the derivative with respect to this x of the metric g I get again that that this is a 0 this provides an eigenfunction so 0 eigenvalue and of course the whole group of deformation is 4 dimensional so which means that if I look now at the dimension of the kernel of this of this delta of Ritchie of this linearized Einstein equation the dimension of the kernel is 2 which comes from this and then I have 4 times infinity which comes from this deformation in other words this is a highly infinite dimensional kernel and you can imagine that because of this a problem is going to be very difficult so I should stop here ok so I'll stop here I don't know if there are any other questions I apologize but what to be the final mass you mean no absolutely no I mean you have to track it down I mean it's like in in modulation sorry for simple and only a problem you only find the final state it's close to the initial one but you don't know a priority what it is you cannot determine it from the data in other words you have to really do the whole dynamics in order to determine so this is one major problem but of course the one who is a kernel in other words the one who is a gauge is even more complicated because if you don't put yourself into the correct gauge and the correct gauge also has to be dynamically constructed if you are not in the correct gauge you don't have decay if I don't have decay I cannot control the nonlinear because you aim a syncopics to be very immediate but don't you try to do orbit stability which you see so orbit stability you can never do orbit stability for a nonlinear quasi linear problem so it's very rare that you can do directly orbit stability for complicated quasi I mean for example sorry I think I know an example you can do it but conditionally for the waterways there is a thing maybe some stationary solutions but it's conditional to global existence but I mean you can but if I don't have global existence it's another problem I mean no I cannot, I don't know how to separate the two I mean in this kind you have to do both existence in other words when you prove stability you show that there are no other singularities which is you can say it's cosmic censorship baby cosmic censorship mother because there are no singularities close to a care solution we have a question existence problem is more important than the stability in your no you they are totally connected you cannot do one without the others we don't know no but I'm talking about this problem I don't know there are examples I totally agree there are all the examples which are due to Arnold has a whole set of examples where you can prove you can prove orbital stability directly by using variational methods but in this case they don't work so whenever I have something even something like this I wouldn't know how to do it but in particular if I have something quasi-linear there is absolutely I believe there is no way of doing one without the other but of course in a certain sense you do both and orbital because your final state is not the original one you started with so it's a combination of orbital and asymptotic stability orbital because the final state is different asymptotic because you actually converging to the final state it's not that you stay close to the final state you converge to the final state yeah can I follow up on the question that can you predict the final mass but presumably to first term perturbation you can derive some rigorous estimate first out of perturbation if you initial perturbation is expanded in quasi normal modes then you could predict what is going to escape to infinity and what will fall into the black hole assume that this perturbative analysis perturbative analysis should make a rigorous estimate which is of the same type you cannot so perturbative the perturbative analysis so for example there is the work of your colleague which has done a lot of beautiful work on asymptotic analysis none of this can be turned into proofs and I'll show you why it's very easy to see why it will not turn into proof so you have to develop completely different methods to do that completely out of reach totally out of reach I have to convince you obviously because you don't seem convinced at all but I'll try to go we have another week if you are here next week I'll try to convince you it's very very far so it's a little bit maybe from that point of view it might be of interest to you because it did say in quarterfistor which is even more complicated you have asymptotic analysis but it's very hard to turn the asymptotic analysis into some kind of rigorous proof and probably you require you need different methods which are not perturbative so in a sense what we are doing is the development of such methods but for this classical problem but absolutely asymptotic analysis you'll never be able to close this so you'll be here next week but we can are you coming back in July? I'll be here on Friday alright


Hello friends, I am Sanjay Gupta. In this video, I am going to demonstrate you how you can print cubes of natural numbers using recursion in C. Before starting, you can subscribe my YouTube channel by clicking on the link which is available at the bottom right corner of this video. You can download my app TechMS that is available on Google Play. You can connect with me by following these details which are available in front of you. Now coming on to the code, you can see the first statement which is a comment. It is showing print cubes of natural numbers using recursion in C. Then a header file stdo.h is included. Then a function is declared which is display. Its return type is void. It will receive two arguments as input of integer type. Then main function is defined which is having two variables and an i. i is initialized with one because natural number starts from one. Then printf and scanf are returned. Printf will print enter value of n message on console. The entered value will be received by scanf inside n variable. Then I have called display function. It is passing two arguments i and n. So when we call a function, control automatically transfers to its definition. Then return statement is available because return type of main is int. So return zero is available here. So whenever we call display function, value of i and n are passed to this definition. Here in this definition, value of i is received inside i and n is received inside n variable. Then inside this definition, you can see display function is called. That's why it is converted into recursive function because whenever a function is called inside its definition, that function is known as recursive function. To terminate recursive function, a base condition is required. Here base condition is i equals to n. So if this condition is true, then i cube will be displayed and return will terminate execution of this recursive process. Otherwise else will be executed. It will display cube of i and then it will call display function again for i plus one and n. So every time value of i is incremented here in calling of recursive function and n will remain same. So at the end, this base condition will become true and last cube will be displayed with the help of this print and return will terminate the execution of recursive function call. Now I am going to execute this code. You can see the output. Cubes of all natural numbers starting from 1, 2, 3 up to 10 are displayed here. You can pass any particular number here till that number you want to print cubes of natural numbers. So I hope you have understood this demonstration so that you can print cubes of all the natural numbers starting from 1 to n. Remember that this base condition will be executed only once when it is true. Otherwise all time this else will be executed because it is calling a display function again and again. So implement this code so that you can understand how you can print cubes of natural numbers using recursive. If you want to watch more programming related videos, you can subscribe my YouTube channel. Thank you for watching this video.