 So let me recall where we are. So we're on our way to prove versions of weak, strong, and let's call it the space-like singularity conjecture for the Einstein scalar field system under spherical symmetry. So just to make sure everyone's on the same page. So this is the system. This equation coupled with the scalar field with this energy momentum tensor. So the scalar field then satisfies the wave equation. OK? So under spherical symmetry, and I won't repeat everything that we've sort of said over the last times. But remember, always spherical symmetry is an evolutionary assumption. That's to say, you assume it on initial data and you get that the spacetime that evolves is very symmetric. Then in case the metric finally, it's very symmetric metric can be covered, as we said last time, globally in double null coordinates. So this is the form of the metric. And the Einstein equations coupled to the scalar field take this form. So these are the equations satisfied by the metric. The wave equation it takes, actually. You can write it in this form. And it's useful to point out, again, various things. These equations here are secretly the Reicherduri equation. And you notice immediately this monotonistic property. And the other thing to remember is this quantity here, the so-called Hawking mass. It's a very useful quantity in general in spherical symmetry. Particularly, it satisfies nice evolution equations, which I wrote down in their general form last time. And specialized to this system, the evolution equations look like this. And one thing that we showed last time, given the nature of our initial data, is that this Hawking mass is everywhere non-negative. So we showed that, if you remember. All right. So we ended the last time with a statement of this proposition. So the proposition is the following. Suppose I tell you the following piece of information that I have a first singularity on the center, on gamma. And there is a sequence of trap surfaces that approach this. And of course, a spherical symmetric trap surface, given, so maybe I should also remind you that we also showed, given our assumptions, that we always have this inequality, the u derivative of r, let me also just, that some of you might not, this might not be a completely second nature to everyone in the audience. So the d by du derivative goes this way, d by dv that way. So the d by du derivative of r is negative, we established that last time. So trap surface is equivalent to the statement that the d by dv derivative of r is negative. So suppose I have such a sequence. Then the Penner's diagram of spacetime is this. That's the claim. Moreover, the future null infinity is complete. So r equals 0 is a space-like boundary. So there are no null components of that. And we'll review this immediately. And moreover, the metric is inextendable beyond. Now, I put a star. This is actually a sort of a footnote. So in principle, you should be able to say that you are C0 inextendable beyond. But the most general sense in which you want to say that, strictly speaking, has not been actually shown. So that's another good exercise to try to do, to show that actually you are C0 inextendable. You are manifestly, I claim, C2 inextendable. But you can say certain stronger statements already, but the best statement that you might want to say is actually, strictly speaking, open. So with that footnote understood, then a corollary of this would be, if I can show that generically this happens, or I have the Penrose diagram of Mankowski space, of course, then all the conjectures, weak cosmic censorship, strong cosmic censorship, space-like singularity conjecture would fall. If I could show that generically this holds. Well, so let's show this. So it turns out that given what we already know, there are sort of three pieces of information that I'm telling you in this picture. So the first piece of information is sort of the easiest to understand. So I'm telling you that if you have a first singularity, so prove, so let P be a first singularity, not on gamma, so not this one. So the claim is that in general, what do I know? Since my mother model is tame, I know that the infimum of r in this rectangle is equal to 0. I know that this is preceded by some rectangle completely in the spacetime, and I know that the infimum of r is 0. So without a loss of generality, I claim that I can find the sequence of points which are in the interior of the rectangle, such that r goes to 0, such that the Pi is the sequence drawn, r goes to 0. OK, so now think about the following. Remember, this is a rectangle, which is completely in the spacetime, except for this point. So this is a compact set, which is completely in the spacetime, the lower two edges. So r has an infimum on these lower two edges, which is strictly positive. So here and here, r is bigger than some epsilon, which is bigger than 0. So what this means is that if you go very close here, so which I don't know, let me do this direction, because it's the non-trivial one. So if I go very close here, r was greater than equal to epsilon here, r is less than epsilon, if I'm very close, just on the sequence. So that means that on this no curve, somewhere in between, somewhere here, dv of r had to become negative. So dv of r had to become negative. But then what do we know from good old Raichadoury, from this equation here? We know that, again, remember, to re-prove what we know, you have to remember the covariance of the equation. We know that that's preserved. So to the future now of this point on this null cone, dv of r is less than 0, which means that r here is less than whatever r was there. So first of all, this tells us two things. So the first thing this tells us is the following. So suppose there is a null boundary component here that emerges from the first singularity. So I claim this immediately tells us that r would have to be equal to 0 on this boundary component. Because you see, I have a sequence now of these segments for which r is going to 0. But of course, du of r is always negative. So this tells me that monotonically r is going to 0 on all null curves that go in this direction. So that's clear. But remember now we know something else, which I advertised at the very last time. So this equation here, we can rewrite it as, let me get the sign of the factor right, we can rewrite it like this. And so in view of this, this also has a sign. So I'm telling you that asymptotically, this r difference is 0. But of course here, dv of r is strictly negative. So this r difference is strictly positive. This minus this is strictly positive. How are you able to say that? So easy exercise, this is now impossible. Because maybe I should say it like this. This minus this is strictly negative. In the limit, this minus that has to be less than equal to this minus this. So it cannot be that asymptotically, this minus this can become 0. So what we've just shown is there can be no component coming out like that. There does not exist. So the same argument allows you to say that first on an ingoing null component, r is equal to 0. In fact, this is easier because dv of r is always less than 0. And then that same inequality allows you to say that this is actually impossible. But note that we really, to show that r has to vanish here and there if there are such null components, that's actually quite general. We only really used Raichadoury. But to show that the components are absent, we use this extra monotonicity, which is special to the Einstein scalar field system. All right, that's great. So the second piece of information that I'm telling you in this proposition is that, so what I'm telling you is that, OK, let's, so where are we, maybe, I should say? That's to say, what is the most general picture that we could have? So of course, under our assumption, I know that the past, the future null infinity, has non-empty complement, because I have at least one trapped surface. I already know that this is complete by what I told you last time. I just told you that the first singularities form a space-like boundary. There are no null components emanating from first singularities. But you can still have null pieces here, and you can still have a null piece there, of which so far I've told you nothing. So let's look near here. So suppose you have a null piece here. Is this possible? So let's just think about it a little bit. So suppose you have a null piece of the boundary here, which is in general allowed by what we've proven so far. And here we're going to use the assumption, namely that we have this sequence. So this is a sequence of trapped surfaces. So dv of r is less than 0 on this sequence, and the sequence converges there. So claim, by the way, without loss of generality, I can put the sequence here. So if the sequence was there, a little exercise using Raishaduri, construct the sequence, which is on this side of that null column, that would be more useful for me. All right, so what did we say? So first of all, what is the r value of these trapped surfaces? I claim to you that r goes to 0 as you go along the sequence. Now you might think that's obvious, because r is 0 here, but of course, no one tells you that r extends in any way continuous to this a priori. So why does r go to 0? And here, we can use our good friend, right. So I want to use our good friend monotonistic, but actually, it's a two-step process. So let's do the two steps. So first of all, if this is the case, you can first go sort of to the past along these, all right, such that you reach the, you know. So you see, OK, this is a trapped surface, but this is an open condition. So there are trapped surfaces earlier. So there are trapped surfaces earlier. So I want to find a marginally trapped surface, so a surface where dv of r is equal to 0. And with the property that if then I go earlier, dv of r is greater than equal to 0. So maybe I'll draw a slightly bigger picture to make this clear. So the claim is the following. At the center, r is equal to 0. And remember, r is greater than 0 everywhere in the bulk. So certainly, as you come off the center, dv of r is greater than 0. So if I know that, let's say, on this null curve, there is a trapped surface, then there is a first surface where dv r is equal to 0. So let me look first at these surfaces. And now I should, I guess, say that this sequence, you don't know if it's on this side or on that side. So it could oscillate. Doesn't matter. So this is a sequence such that dv of r is equal to 0. I don't like this picture. So I claim that actually, I can say something better than this. Namely, I can say that this has to be the picture. So why did this sequence, again, have to be sort of on this side? Well, it's, again, this relation here. So if you are the first, so if you have the property that dv of r is less than equal to 0 here, then, well, certainly, that has to be true from then on by this. And in fact, you get a contradiction view of the fact that I told you that on Rama. So actually, this sequence is, again, here. And it looks like this. That's to say, if I look at a later thing in the sequence and I go backwards, then dv of r is greater than 0. All right. So now, finally, I can tell you that on this sequence r goes to 0, from which it will follow that on my initial sequence of trapped surfaces r goes to 0. Because, of course, if this is the first marginally trapped surface, then dv of r is less than equal to 0 here. And so r here is even less than r there. So to recap, I know that the sequence of, I have a sequence of first marginally trapped surfaces. First in the sense I said, such that dv of r equals 0. And I want to show that r goes to 0. So I claim that this is clear, simply from the following statement. So let's look at this r difference. And compare it to this r difference. So over here, dv of r is greater than equal to 0, from here to here. On the other hand, du dv of r is less than equal to 0. So that tells you that this r difference is less than equal to this r difference. All right, but now if you think about the geometry of this, this point is too big. So this point is converging there. So if I draw these rectangles for later, later points, they look like this. So this r difference is always less than equal to that r difference. But that r difference goes to 0, because the dimension of the rectangle here is going to 0, and the data here is regular. So that's actually immediate from this nice one. So to recap, what I've just shown you is that on this sequence of trapped surfaces, which again are on this side of that backwards null cone coming from this point, r is going to 0. Well now, OK, I claim that immediately I know that if there is a null piece coming out from there, r better be 0 identically here. Well, why is that the case? Well, again, I know that r here is very small. And if I go on the outgoing null cone coming from there, dv of r is less than 0. So r is even smaller than what it is here. So I have now these null curves. And on this whole null curve, r is smaller than something which is tending to 0, as I put the point there. On the other hand, again, r is monotonically decreasing in this direction. So indeed, r has to be 0. But now the exact same argument that we used to say that you could not have these null segments tells you that, so in fact, this is a contradiction. Again, you're contradicting always this. So that's great. So finally, what this is telling you is that if I have this sequence, then the r equals 0, I have to have a curve of first singularities that goes into this point. OK, so the last issue, and this is in some sense the one that we actually have to use some more information that I haven't told you yet, is how do I know that I do not have a null piece coming from there? And this will be interesting exactly because this property is the property that won't be true in greater generality. So what can I say here? So here's the point. So remember, I've already told you a few extra things about what happens in this situation. So let's sort of assume for sake of contradiction that near this point, the Penrose diagram looks like this. So what have I already told you? I've already told you, yes. Say again, sorry? The sequence of first singularities is somewhere over there. I haven't depicted it. So the claim is the following, that the existence of the sequence is just telling me I'm just using it here. So for this part, if you want, all I need, all I am using is the existence of one trapped surface, just one, because that's what's telling me that the future of null infinity indeed has a non-trivial. It's passed as a non-trivial complement. So I now want to look at near this point and exclude this possibility. This one? This one? Oh, yeah, I don't know what this was supposed to be. Nothing. This was the sequence. There's no claim that this sequence has anything to do with this point. In fact, I'm not really, you know, basically you should think of what I'm going to tell you now. It's a completely, quote, local statement near this point. I'm going to say, you know, you cannot have this Penrose diagram near this point. So let's do this. So what other information do I know? Well, I know this Penrose inequality. That's to say, I know that r goes to some r plus along here, which is less than infinity. OK? Let's see what I can get just from knowing that. So I claim to you that, so this is sort of a funny little exercise, so I claim to you that you can already show the following. So of course, all right. So I can talk about the limit of r on this boundary, where by limit I just mean the limit in this direction. So why can I talk about that? Remember, r is almost monotonic in this direction. Its derivative can change sign exactly once, which means that eventually it is monotonic. So I can certainly talk about the limit of r here. So I can talk about r there. So let me affirm the following statement that actually r, whatever it is, has to tend to r plus as you go this way. So this is sort of a nice little exercise. So suppose it doesn't. Well, it certainly cannot tend to something bigger than r plus. I claim that you can easily see contradicts the fact that the u of r is less than 0. That's a very easy argument. But actually, if r were to tend to something less than r plus, strictly less than r plus, so suppose r tended to r plus minus epsilon, sorry. Then you could find a sequence of points like this where, I don't know, r equals r plus minus epsilon over 2, and r equals r plus minus, I don't know. Maybe let me make this minus, I need a number of epsilon over 4, that's fine. You like that? So in particular, so that's sort of an easy exercise. And the sequence of points have the, I claim you can choose them, so that this point is lying on this null curve, backwards null curve coming from there. And now you can sort of throw away a bunch of these points, so you can always wait in the sequence such that the next one of these points has a lower u value than this point. So you can find the sequence of points of if you want intervals like this, and these intervals go there. All right, this is my favorite way of saying this, just because, so what is the r difference here? r difference is epsilon over 4, right? But now I want to use this relation, but I'm going to use it in the other direction, trivial fact analysis. So what this is telling you is that if you go at a later time, I just don't have so much, then the r difference is at least the sum of these. But there are infinitely many, so you'd get an infinite r difference, which is a contradiction. So I let you think about this picture, but the claim is that this is immediately contradicted by this monotonous. So you actually have this. So that's great, but we haven't yet disproved anything. So you could have this, right? So I have to tell you some more information, which is actually more difficult to show, but you can show it. So it turns out that by really using now the wave equation and its decay properties in the exterior, one can show two things about the exterior that will be important. So the first thing is that the event horizon is complete. I've never yet told you that the event horizon is complete. I've told you that null infinity is complete, but I've never told you that the event horizon is complete. What that just means is that this on the event horizon, let's call the event horizon u equals u0, is equal to infinity, so fact. Again, this fact depends on the analysis of waves in black hole exteriors, something that's been a subject of much research for some time. So this is one thing. But I'll give you another fact. So fact two is that if I look at the integral of dv phi in absolute value along the event horizon, dv, this is finite. Now, can you see the statement? Maybe you can. So yeah, it's a little bit funny. What's nice, so this is some sort of, this is a statement of decay for phi along the event horizon. And I claim this statement is invariant to the normalization of null coordinates. So if I rescale the coordinate v because I'm taking dv here, this is actually, so this is an invariant quantity. So let me tell you one third fact. So these are, if you want, difficult facts. But the third fact is actually immediate from the sort of monotonistic properties of m. Because I know that there is at least one trapped surface, then I know that m is strictly positive near here. In fact, I know that m is greater than some epsilon in some region like this. And the reason you know that is that on a trapped surface, 2m has to be bigger than r. And then you can follow the monotonicity of m like this. So I know that there's at least one trapped surface. That's all I need to be able to give you that. So all right, well, how can we argue from this point on? OK, so because I don't want to spend infinite time on this, I'll leave the various statements as exercises. But they're quite easy to do. I'll say it in words. And you can try to fill in the details. So the first thing you immediately notice is the following. I can look at that shaded region, that purported characteristic rectangle that supposedly exists here. Look at the shaded region. I can integrate this equation twice because of the finiteness of r. The left-hand side of this is bounded. It's just the r difference of two endpoints of the rectangle. So it's bounded, whatever it is. r itself is bounded below, away from 0, m is finite. r is bounded above, maybe that's more pertinent, by r plus in that region, again, by the d ur less than 0 monotonicity. So it follows that the double integral of omega squared is finite. So similarly, if you want the double integral of r squared omega squared is finite because r is bounded. So that's actually the spacetime volume. So just a remark. So if I'm in the situation, then the spacetime volume of this is bounded. So now let's look at this equation here. And let's integrate it twice. So it's maybe most convenient to, at this point, choose a particular normalization of the v-coordinate. So I'm going to use, this is not necessary, but it's sort of makes me feel better. I'm going to use a normalization of the v-coordinate such that omega squared is equal to 1 on the event horizon. And of course, because I just told you that the integral of omega squared is infinite, that means that this will not be a bounded v-coordinate. But nonetheless, I'll draw it like this. So this is the event horizon. And of course, this will now be v equals infinity. And omega squared is equal to 1 here. And again, imagine I have this sort of rectangle. I can translate the coordinates so that the edge of the rectangle is, let's say, v equals 0. It doesn't matter. So let me try to integrate the equation store in this region. So here's the point. I'm also going to really consider a very, very small region. So that whatever omega is here, it's very near 1. In fact, if you want, for convenience, let me rescale the u-coordinate here such that omega squared is equal to 1. So if I integrate this equation twice, as far as the left hand side is concerned, I have no initial data terms because the log of omega is initially 1. So I'm just going to get the log of omega, let's say here, and some double integral here. So what's the double integral? It's the double integral of this, of this, and of this. So I just told you that this double integral is bounded. It's finite. It's something. So it's uniformly bounded, no matter which point I take. So what about this double integral? I claim that it's also bounded. Maybe the easiest way of seeing that is putting this term here on the left hand side and integrating this again twice over u and v. So you get then that this double integral is bounded. And OK, they only differ by the different weights of r. But r is bounded above and below. So the last thing that you have to understand is this double integral, this double integral. And it's a very nice exercise, I think. It's not a difficult exercise to show the following. So given this assumption here, I know that if I put my v equals 0, very, very late, the integral of dv of v is finite. It's fact I can even make it small initially by choosing this time to be very late. And again, by choosing this sort of rectangle to be very short, I can certainly just by regularity, ensure that this integral, the integral of du phi in norm, is less than epsilon. So then you look at the wave equation. This is the wave equation. And I've written it in this form. And when you write it in this form, it's very nice because you don't see omega at all. You just see r. So somehow, here is my exercise for you, integrating in a smart way the wave equation, using the fact that the r difference in this rectangle, so what I'll call r minus r, r of x minus r of y, for any two points, is less than epsilon if I make the rectangle small enough. You can show that this double integral is also bounded. And now I claim this is a contradiction because what you've shown is you've shown that log of omega in absolute value is uniformly bounded in this whole region everywhere. But that means that omega is like 1. Log of omega is uniformly bounded. It means omega is like 1. It's bounded away from 0, which means that the integral of omega squared is like the integral of 1 du dv. But the range of the v coordinates is infinite. So it's a contradiction. OK. So anyway, that's the proof of that. Exercise, fill in the details, or maybe find a simpler arc. So now, we've completed this proposition. So now let me tell you the actual meat of Chris D'Aulou's work for this model, which is precisely the statement that, generically, we are in the domain of that proposition. So this is the theorem. Actually, it's good that there's some musical accompaniment. All right. So the statement is for generic initial data, and I'll get back to this in just a second, then if there exists a first singularity on the center. So remember, if there does not exist a first singularity in the center by what we said previously, the Penrose diagram is that of Minkowski's space. And again, everything is true. So if there exists a first singularity in the center, then it's accompanied by such a sequence, then there exists a sequence. So there exists a sequence of trapped surfaces, as in the proposition. So corollary, weak cosmic censorship, strong cosmic censorship, well, certainly the C2 version. But as I said, that's not the right version. In principle, the C0 version. Maybe it's relatively easy to show a slightly weaker notion of the C0 version. Namely, you cannot extend C0 as a spherically symmetric space time. But in any case, I just put the asterisk that there's a nice little exercise to complete here. And the space-like singularity conjecture, true for Einstein's scalar field in spherical symmetry. So that's his steer. So let me just discuss a little bit the proof. Because somehow the proof is much more subtle than anything we've done on the board so far. And so I really can't give it justice, but I can tell you the mechanism, which is very pretty and relatively easy to describe. So here is the mechanism. So the first, if you want, thing that he shows is the following. If you have this first singularity on the center, then something has to go wrong at this first singularity. But what has to go wrong? Remember, we got a load of mileage for first singularities outside of the center saying that R had to go to 0. In fact, we're using this all the time. Of course, here that R goes to 0, you know it. Because you're on the center, and so R is 0 here. So how can you characterize what goes wrong here? So first of all, he is able to show the following that 2m over R, so this is in some sense a dimensionless quantity, this has to be bigger than some sort of universal constant in this model, which is strictly positive. So the limit, and let me not say precisely what I mean in what sense the limit, but particular there has to be a sequence of points, such that this quantity here is strictly positive in the limit, in fact, bigger than some particular constant that you can actually compute. So this is sort of claim 1. And this is already actually quite difficult to prove. We're very nice to have sort of analogous statements for other mother models. There really is not anything available like this. So that's the first statement. So now I'm going to rewrite this equation here in yet another way. I'm going to write it, so sort of looking at this expression and recalling the definition of m. I can rewrite this expression. Let me write it correctly. With the right sign, I'll get the sign wrong. I'll get the 2m over R. So all I've done, you can see it from this form here. All I've done is I've solved this equation for omega squared and plugged it in there. OK? So again, I'm going to, what am I going to do now? I'm going to think of this as an evolution equation for dv of R on the backwards null cone emanating from this first singularity. So this equation is telling me du dvR is equal to something and I think of it as an evolution equation for this. And I try to integrate this equation as a linear equation in dv of R. So here's the point. Because I know that this is true in the limit, I can rewrite this as 1 over R times something bounded below away from 0. And then this, if you want, without loss of, I mean, by the argument that we already saw actually, I know that this is greater than, I mean, this expression here is less than 1. So this is a factor that only makes things bigger. And I can integrate this duR. So what does that tell you? It tells you the following. Since R is going to 0, the integral of this diverges. So that's telling you that the quantity dvR is becoming 0 because you have to also remember the sign. So dvR is becoming 0 as you go over there. So what does this mean? So I claim that this is a blue shift effect. Because this means that if you have two observers here and here, and the observer normalizes their time to R, which is the only canonical thing to do, then the observer here will see radiation coming from here as infinitely blue shifted. So that's something you can think about very easily. Now in this model, you have radiation sitting around because you have phi. And what you want to show now is that generically, that blue shift tends to concentrate energy between, where is a good picture, between here and any close enough point like this to make this trapped. So that's basically what he shows. In fact, what he shows is that if you happen to have a solution such that there were no trapped surfaces that went there, then by adding arbitrarily small perturbation, in some sense, which lives here, then this blue shift effect forces the existence of those trapped surfaces. So that's essentially what he shows. Now one interesting thing about the argument is that in order to exploit this, it's very useful to allow the initial data to be slightly irregular. That's to say to always consider initial data that are not smooth, see infinity, but slightly irregular. Slightly irregular but still in the class of a well-posed theory. So part of the difficulty is identifying such a class that gives you the flexibility to be able to carry out this argument. So he also sort of discovered, so class refers to the function space defining the initial data. And the second interesting thing to point out in this context is the generosity is necessary. And so this is necessary. I want to stress this was not thought to be the case before. I already mentioned that in the context of the dust model, weak cosmic censorship is actually violated generically within the context of spherical symmetry. But in that case, it really is because of a failure of the model. The Einstein scalar field system is as good as it gets as far as models are concerned. The Mather equation is completely linear, so there's nothing wrong about the Mather. The Mather certainly does not form singularities on its own accord. So the original hope was that there would never be naked singularities in this model. And so in parallel with this proof, he actually constructed examples where the Penrose diagram looks like this or like this. So there are examples, OK, spherical symmetric examples that look like this, but they are not generic. So the theorem tells you that they are not generic. So in particular, these examples were you to perturb them, you would have the Penrose diagram that is drawn over there. OK. So that's Kriste Zoulou's work. So this is great. I have 20 minutes. And in 20 minutes, I will give an abridged version of lecture four, but in some sense, I think that's exactly the amount of time that we need for it. OK. So let me write here, maybe. I warned you at the beginning, of course, that I'm not going to keep to my schedule. OK. So lecture four, the C0 stability of She-Heroisen without symmetry. Without symmetry. OK. So before I say anything about Kr, let me say something far more elementary, OK? Let me say something about Reisen and Nordstrom, OK? So I've already drawn in a previous lecture the Penrose diagram of Reisen and Nordstrom solution. So let me draw it again. So remember, Reisen and Nordstrom is a solution of the Einstein Maxwell equation, so Timini is, OK. So exercise, adorn this with indices so that that is the correct energy momentum tensor for a Maxwell field, OK? And then the Maxwell equations, of course, are just that without any mother, which is charged, are just this, right? So Reisen and Nordstrom is a spherical symmetric solution of that, OK? It has two ends, and its Penrose diagram as defined in the sort of notations of this course looks like this, OK? And this boundary is actually completely regular, so I can extend the solution here, move the extents, and of course, any such extension is manifestly non-unique, OK? So let's just focus on this rectangle. Let's just focus on this rectangle, OK? So if I focus on this rectangle, then, OK, the global aspects, the fact that I have two ends, et cetera, it's really irrelevant, OK? What is going on in this rectangle that's different from before? Because before I gave you an argument in the Einstein-Scaler field model that you could not have this behavior. So what's the difference? Well, OK, I'm going to, OK. Sorry, but we like graffiti a lot in Greece, so I'm going to do some graffiti. So let's understand the extra terms that you would have in the Maxwell case. So I claim that actually, even though this is a lot of equations to write down, in spherical symmetry, the Maxwell part is not dynamic. So it just gives you a constant times something depending on the metric, which appears in the equations. So we call that constant 2 squared. So I have an extra term here, which looks like this. And I have an extra term here also that looks like this. So if you won't forget about the scalar field, there's no scalar field now, bye-bye scalar field, but you have q, these are the equations that Reisen and Nordstrom satisfies. So actually, exercise, understand completely the Penrose diagram of Reisen and Nordstrom just from the equations without writing down the metric explicitly. It's actually an illuminating exercise. So in particular, I claim to you that the reason that this can now happen is because of this term. So in particular, you lose the very nice monotonicity that DUDV has assigned. And if you look at the equation here, then it turns out that inside the black hole near the Cauchy horizon, this term here dominates over that term. And it's this term here, let's say in this equation, that is driving the phenol. So the first thing you might ask is whether in spherical symmetry, having Cauchy horizon is stable. And somehow the easiest model to look at that is the model where I just couple both the scalar field and the Maxwell, but I don't even bother to make the scalar field charged. So this is a model. It admits Reisen and Nordstrom as an explicit solution. Unfortunately, Reisen and Nordstrom is the only spherically symmetric electro-vacuum solution. So if you want dynamics in spherical symmetry, you have to add matter. So let's just add the scalar field. But let's not even make it charged. So this is the energy momentum tensor. So you still have the wave equation. You have these equations. And just that you still also have this term. So I should note that there are extra terms on these equations here. These equations are deduced from these. So a while ago, I showed that when you consider general solutions of these equations. By the way, in these equations, if you want q to be non-zero, so if you don't want these to reduce to Christodoulou's model, initial data has to have two n's. It has to have two n's. That's another nice, easy exercise. So initial data has to have two n's. And another very, very easy exercise is the following. Any time you have a tamed matter model, and this model is, again, tamed in the sense that we said, and you have two-ended initial data, then weak cosmic censorship is true, sort of trivially. And that just has to do with the fact that there's no center for first singularities to come from. So all the first singularities have to have r equals 0. And if you think about Richard Dury, you have that statement. So in this case, you always have weak cosmic censorship. You always have complete, future non-infinity. And the statement is the following. So if q is not equal to 0, there will always be null pieces of the boundary that come from here, which are non-empty. So this is something that I proved a while back. It also uses a study I did jointly with Rodnianski about the decay of waves in the exterior, because you really need to know rather precise information about the stability of the exterior regions here, and particularly the behavior of fields on the event horizon in order to say that this is the case. So in general, you may also have a space-like singularity here, although actually strictly speaking, in this model, the argument that says that the r equals 0 first singularities form a space-like singularity does not hold because of the failure of that monotonistic, duv r less than equal to 0. But in any case, the rest of the boundary, you can show that r is equal to 0. But actually, what I showed moreover is that if your initial data was close to Reissner-Nordstrom initial data, then there will be no such piece. And these will close off the space-like singularity. What I showed is that you can still extend this C0. So the metric extends. So in particular, in this model in spherical symmetry, the C0 extendability, inextendability formulation strong cosmic censorship is false in this model in spherical symmetry. Now, even though, so from the point of view of C0, this is a Cauchy horizon. This is regular. There's no problem whatsoever. But from the point of view of higher regularity, generically, there is a problem. And I had some partial result on that, but I couldn't quite get the full result. And actually, much more recently, Luke and Oh have shown that for generic initial data, the boundary is entirely singular. So they have a condition on initial data, which is a generosity condition. So for somehow, you should think with probability one, they have a precise statement in their paper. Then the boundary is entirely singular. And OK, they can say in particular that you are C2 inextendable. And again, the proof morally suggests that you have actually a stronger statement, maybe even a statement at the level of H1 in extendability, which is a formulation of strong cosmic censorship due to Stodulu. But there's still a little thing to complete in order to be able to say that. But in any case, you have a weaker formulation of strong cosmic censorship is true in this model in spherical symmetry. Now, in this spherical symmetric world, there's still nice things you might want to consider. So the reason that you need to have two-ended data here is because the scalar field itself is uncharged. So this Q, to be known as 0, has to sit on some topology. So this is the topology, right? You can consider one-ended collapse in this model by making, for instance, the scalar field charged. And actually, this model has been considered recently by Maxime van der Mortel. And he's been able to show a version of the stability of a piece of the Cauchy horizon. But under assumptions on the behavior of the event horizon, and actually, those assumptions are thought to be true always. But it's still much more difficult, even in that model, to retrieve those assumptions from assumptions of the level of initial data. So even in spherical symmetry, there are nice things to do. OK. In the last five minutes, I can state this result, which, in some sense, is the conclusion of the story of the C0 formulation of strong quantum censorship. So you might say, this is all very well. But at the end of the day, this is a spherical symmetric model. And what is generic within spherical symmetry may be completely un-generic outside the spherical symmetry, because spherical symmetry itself is measure 0 in the sort of probability 0, I'll have to say it, for general initial dot. OK. So is this really relevant more generally? So here's our theorem with, so I'll erase these equations now, with Jonathan Luke. This is a theorem released a few years ago now on the archive. So let me say it as follows. So theorem, so there are two statements maybe I'll make. So I'll make, in some sense, what's in some sense first the more general statement. So the more general statement is the following. So let's forget about symmetry completely. Since we're forgetting about symmetry, let's also forget about matter. Because in some sense, the motivation for the matter that we've been considering is exactly, we had to add matter to evade Perkov's theorem to have non-trivial dynamics in spherical symmetry. But somehow the conceptual question of cosmic censorship, after all, is first and foremost a question already about the vacuum equations. So vacuum, but now no symmetry whatsoever. So suppose I have a black hole horizon. So suppose I have a black hole horizon. And I assume that, OK, the horizon, who knows what happens at early times, how it didn't exist. I mean, whatever, how it formed in the spacetime, however you want to say that. But suppose that at late times it settles down to Kerr. And it settles down at the expected inverse polynomial rates that it is believed that perturbations on event horizons decay in the very late time regime where they're governed by tails. So let's assume that your geometry settles down on the event horizon to Kerr. So it is actually believed that generic vacuum spacetimes will eventually either disperse or settle down to a finite number of Kerr's moving far away from each other. So we are now talking about, let's say, the neighborhood of one of them. OK. So suppose this is what you know about the event horizon. Then the statement is settles down to Kerr asymptotically. So let me write here this is sort of with Jonathan Luke. So the statement is then, well, of course, this is a spacetime without any symmetries. So I cannot define Penrose diagrams in the way I did. But actually, the claim is that the future of the event horizon and, let's say, an ingoing null cone can actually be covered by double null coordinates. You are not strictly symmetric, but you are still covered with double null coordinates. So I can, again, draw the range of those coordinates. And the statement is that the spacetime exists in a region that contains a characteristic rectangle in double null coordinates that looks like this. And in particular, this is a Cauchy horizon beyond which the spacetime is extendable. So what this tells you is if you believe sort of the completely standard picture of gravitational collapse for the vacuum, then the generic spacetime either disperses or will have a piece of Cauchy horizon across which the solution is extendable C0. And remember, extendable C0 means, morally speaking, that the observers, they are not destroyed. So they reach the Cauchy horizon without being destroyed. So an amplification, but also specialization of this statement is the following. So suppose, in particular, that you start here with initial data, which is globally near two-ended Kerr initial data for the Einstein vacuum equations without any symmetry. So it is believed that now, OK, this is more than a belief, sort of all the ideas necessary to prove rigorously, the fact that the exterior regions of the Kerr spacetime are stable and the solution will settle down to two nearby Kerr solutions. So this is what's sometimes called the nonlinear stability of Kerr problem, all right? So this is something that now can be shown, sort of putting together a lot of things. I mean, can be shown it's difficult to write something which is self-contained and readable, but in principle, this is now mostly a technical problem. So in particular, in this case, you will be everywhere close to Kerr, and you will settle down inverse polynomially to two nearby Kerrs. So in this case, while the theorem already tells you that, OK, near these event horizons from these points in the Penrose diagram, whatever this means now in this context, you'll have a little piece of Cauchy horizon. And actually, in this case, you can say something more. The entire bifurcate Cauchy horizon of Kerr is stable, and you can, in fact, extend the spacetime across the entire bifurcate horizon so that all observers that were incomplete in the original spacetime pass safely to the extension. C0. So somehow, not only in exact Kerr, but in small perturbations of Kerr initial data, it's exactly the same phenomenon that the spacetime, its Penrose diagram, is bounded by a globally null surface across which the metric is extended. So in any case, corollary of this already, so this would be an open set in the modular space of initial data for which the predicate of the inextendability statement that strong cosmic censorship wanted in its C0 formulation is false. So corollary of this is already C0 formulation of strong cosmic censorship is false. And this is a definitive statement. It is false for the vacuum Einstein equations without any symmetry. Now, again, just like with the spherical symmetric case and the work of Luke and O, it is thought that, generically, this boundary still will be singular in a weaker sense. And there are now many methods that have been developed to be able to show that, at least in a neighborhood of Kerr. So in particular, what that would mean is that, in a neighborhood of Kerr, some weaker formulations of strong cosmic censorship should be true and should be provable. Now, just remember one thing. To disprove a version of strong cosmic censorship, it's sufficient to find one open set in the modular space of solutions without symmetry of what you understand, one truly open set for which the statement is false. To prove a version of strong cosmic censorship, weak cosmic, whatever your favorite thing, then, OK, this is something about the whole modular space or generic in the modular space. So a complete proof of some formulation of strong cosmic censorship that may be true still looks like it is very, very far off. But a complete resolution of the picture in a neighborhood of Kerr is now very much within reach. All right, let me end there. Thanks a lot. Thank you.