Going live, yeah, I think there's a little delay. Okay, perfect. Welcome everyone. Thank you for joining us for today's low physics webinar. My name is Alejandra, I'll be your host today. Today we're presenting on the stability of care of black holes by Rita Teixeira da Costa. She got her bachelor at the Institute of Superior Technical in Portugal, and then she moved to Cambridge where she did her master's and PhD and she got recently her PhD at the University of Cambridge under the supervision of Michalis da Ferros. She's now an NSF post-doctorate fellow at Princeton and a junior research fellow at Trinity College in Cambridge. Rita's research focused on differential equations arising like with connection with general relativity and her works really represents an important regular mathematical progress on several like open questions related to black hole stability. She recently won the dissertation prize given by the Association for Women in Mathematics. And according to one of the experts letter, this is something I was talking to Rita, says that her thesis is an outstanding piece of work that resolves a major problem in the field of general relativity. It is a pleasure to have Rita today. Remember that you can ask questions over email through our YouTube channel or Twitter and then the questions will be read at the end of the talk. So now without further ado, we'll turn in time to over to Rita. So thanks for joining us and Rita, the floor is yours. Thank you so much. And it's a bit embarrassing that you talked me up so much because now I feel the expectation is very high. But thank you, it's really a pleasure to be part of this. And it's great to meet, to have met all of the organizers and hopefully to talk virtually with other people in Latin America who are interested in this. So I want to talk about the stability of Kerr black holes. And I know that not everyone in the audience will be coming from mathematics background. So I want to start with a kind of more gentle introduction. And then at the end, things will get a little more serious. And this is basically what Alejandro said, I should start gently, but at the end everyone should be dying for me to shut up. So let's start with the gentle introduction. So as I said, ooh, okay. As I said, I'm going to talk about the black holes ability problem. And this is a 65 year old problem. And I want to start by convincing you that this is still an open problem. So a disclaimer here is because this audience is, I know composed of people who come from many different backgrounds in physics and probably mathematics as well. This talk is really about general relativity. And in particular, it's about the vacuum Einstein equation. So Richie tensor of G is equal to zero where G is the Lorenzian metric on a Lorenzian manifold before the mention. So space time. Okay. And this is all that I'll do. So very classical general relativity. And the black holes stability problem, as far as I know, it really goes back to this work of Regio and Wheeler and in the fifties. What they do in this, in this paper is that they know that the Schwarzschild solution is a stationary black hole. And so it is in some sense an equilibrium state for the Einstein equations. And so they ask, well, if this is an equilibrium state, this is a stable equilibrium or an unstable equilibrium. Okay. And I've written here the Schwarzschild metric. Okay. At this point, as you can see, obviously from the title and the abstract, people were still not completely sure what the Schwarzschild solution meant. And a few years later, it became clear that this question that they were trying to answer, the question of stability for Schwarzschild was false. And it was false for trivial reasons.  And so you'll end up with a Kerr black hole, which at the time weren't known yet. So really the Schwarzschild solution, the exterior, the black hole exterior, this can only be stable within the larger class of Kerr black holes. Okay. And this, as you probably all know, this is a more general stationary solution to the Einstein vacuum equations. This is the metric. It's a little more ugly. This is now not strictly symmetric. It's axisymmetric. And we still have a mass here, a mass M, but now we have a parameter A, which represents the rotation of the black hole. And I'll use this notation. I'll say that if it's smaller than M, then we have a sub-extreme of black hole. If it's equal to M, we have an extreme of black hole. The case where you don't satisfy the inequality A smaller equal to M is not a case where you have a black hole. Okay. And I want to point out, I just want to point out here that this stability within the larger class of black holes, this actually can be made precise because this has been done in recent work of the pharmaceutical Rodney Askin-Taylor from 2021. And they give a very precise meaning to the statement stability of Schwarzschild within the larger class of black holes.  Okay. And there's also earlier work of Klanerman-Chevdell, which also proves this statement, but for preservations which lie in a symmetry class. That is not the case in this other work. This is without assumptions of symmetry. Okay. So really these works on Schwarzschild, the formalism for investigating the question of stability in the sixties, it was really gearing up, I suppose. So there was after Reggie and Wheeler, there were works by Bardeen Press, someone whose name starts with V, which I can never pronounce Visevara, Visevara. And other people who were writing down many equations, which were relevant for the investigation of stability of Kerr. But all of these equations were very much tied to the spherically symmetric aspect of Schwarzschild. And so it wasn't clear exactly how to adapt this formalism to understand Kerr black holes. And really a big breakthrough in understanding the stability of Kerr black holes of rotating black holes came from Tkolsky only in the seventies. So Tkolsky took the Newman-Penrose formalism, which was developed in the sixties for understanding linearized, for understanding perturbations of the Einstein equations, really, and he understood something very spectacular. He understood that if you look at the Newman-Penrose formalism, and you focus on the so-called extreme curvature components, then they satisfy an equation, which completely decouples from the rest of the system of linearized Einstein equations. So if you look at linearized gravity in this Newman-Penrose formalism, you will find that you have a million equations, but there's two particular components which satisfy equations which decouple from the rest of the system. And those two particular components happen to be gauging variant components at the linear level. And so if you want to understand the stability of Kerr, you have to start by understanding the linear stability of Kerr. The Einstein equations are nonlinear, so you first have to understand what happens when you do linearized gravity, when you assume that perturbations of Kerr are so small that the quadratic terms don't matter, just the linear terms matter. And it turns out, right, something completely non-obvious that when you write these linearized Einstein equations, suddenly you have a huge system, but there's two special ones which decouple from the rest are completely independent. And the variables that these equations relate to are variables which are gauging variants, so extremely physically relevant. Okay, and this is true in the case of gravitational perturbations, but it's also true in the case of electromagnetic perturbations and others. And this is why this equation came to be known as the Tchaikovsky master equation. There's a way of putting together different of writing in a single equation by introducing this parameter S. There's a way of writing in a single equation, the equations for many types of perturbations of black holes, again at the linear level. Okay, so what after doing this, Tchaikovsky, I think with his PhD advisor, I think Pres was his PhD advisor, they then looked at separable solutions of this equation, of this master equation. And I should point out that, the existence of these separable solutions is not obvious. It's something which follows from very non-trivial work of Carter in the sixties who found that Kerr had this additional conserved quantity which allowed the wave equation to be separated. But anyway, so they knew by then that you could have separable solutions to this Tchaikovsky equation. And so they looked at solutions of this form and looked at possible unstable modes. So unstable modes, now what I mean by this is just modes which have positive imaginary part of omega. So if you take positive imaginary part of omega here, you'll see that this is a solution which is exponentially growing in time. So you have here an instability, okay? You want to, maybe I should have pointed this out but if you look at Kerr, the exact Kerr solution, this alpha, this extreme curvature component that I mentioned, it will be zero. And so you want to be as close, if you want stability of Kerr to hold, you want to be as close to zero as you can. You don't want to be exponentially growing in time. So they looked at this kind of separable solution and they tried to rule out the existence of unstable modes, okay? And why can you expect that there might be unstable modes? Well, in principle, these are possible because on Kerr there is this funny feature which is called super radiance or sometimes, and sometimes this is referred to as the Penrose process. So in Kerr, you have this region of space time, which is the Ergo region where the vector field which is stationary ceases to be time-like, okay? And starts becoming space-like, okay? So what happens is just that the direction of time is somehow not aligned with the direction of the stationary killing field, which is kind of funny. And it leads to this Penrose process which is a theoretical experiment where you send in a particle to a Kerr black hole and once it's in the Ergo region, it divides into two and you buy this process because of the energy being measured with respect to this stationary killing field which now is space-like, you can extract, you can have one of the particles come back to infinity with more energy than the particle which you shoot it and had, okay? So that's the Penrose process and the similar thing is possible for waves. This is called super radiance. So if you have this condition that omega times omega minus this factor which depends only on the azimuthal number of your separable solution, little m. And on the Kerr black hole parameters, if you have this inequality, then you cease to have control over the energy of your waves. They might be, their energy might grow similar to the Penrose process where the energy of the particle that comes out is larger than the energy of the particle that goes in. And actually these super radiance instabilities, they are present, for instance, in massive Klein Gordon. This was already conjectured by President Tchaikovsky and their famous black hole bomb paper in nature. And then it was shown in 2015 by Slapendoc Rothman. So super radiance instabilities do happen. The fact that you have this, the Ergo region and the possibility of the Penrose process, this can lead to having unstable mode solutions to your equation, okay? It is possible. And so they were worried about this and they looked numerically at the possibility of this happening. So they write, no instabilities, so they looked by numerical integration of the separable to perturbation equations for the Kerr metric. So exactly what I wrote here, no instabilities are found in any of the dozen or so lowest angular modes tested for any value of specific angular momentum. These results add credibility to the use of the Kerr metric in detail that's your physical models. Now, since this paper of President Tchaikovsky, I think many people have credited them with solving the black hole stability problem because they write, they've looked for unstable solutions and they couldn't find any. But for us mathematicians, this is still a little unsatisfying. And I think President Tchaikovsky were not satisfied themselves. Their abstract is quite toned down. So they do not in any way claim to have solved the problem stability. They simply say, we haven't found any unstable modes and this somehow adds credibility to the use of Kerr, although the jury is somehow still out, okay? So for us mathematicians, now the task is clear. You want to show that solutions to the Tchaikovsky equation remain bounded and in fact decay in time. So you want to show that these unstable modes don't exist and you want to even show that the decay in time so that as time goes to infinity, you are approaching more and more the value of alpha on the particular on the Kerr background, which is zero. Okay, hopefully this makes sense. Okay, so how does a mathematician now approach this? Well, informally, okay, this is not rigorous. The solution to the Tchaikovsky equation we expect is sort of like an infinite superposition of the separable solutions. And maybe now I can go a bit slower and explain what are the things going on here. So omega is a time frequency. M is an azimuthal number. And the fact that you can separate these out, this just follows from somehow from the fact that you have a stationary killing field, the d by dt and then you have axis symmetry of Kerr. And then the non-trivial part aspect that I mentioned is the separation of the r and theta variables.  And the separation constant is this capital lambda here. Okay, that's what that's a notation that I'm using. There's a separation constant. So for instance, if you were on Schwarzschild black holes, you would have a spherical symmetric space time. The separation would be pretty obvious and the lambda would just be L times L plus one, like in the spherical harmonics, if that makes sense. So you should think of lambda as a generalization of that because now you no longer have C or you have sort of ellipses. And because you have ellipses, the lambda is not going to be so easy to write down. In fact, it will depend on A times omega. Okay, so in this infinite superposition, again, I'm going more slowly over this. So U here and S, they satisfy ODEs. The S satisfies an angular OD, which mostly I don't care about, but it's kind of an OD that constrains the value of the separation constant lambda. And U satisfies a radial OD, which I very much care about.  So the OD looks like this. It's like a one D Schrodinger equation.  So really I'm writing this in Schrodinger form, right? So I have second derivative, I have here an energy level and I have some potential. Okay, and the potential is not super nice looking. So here's the potential. You can write it explicitly in terms of the curve parameters, M, A, so delta, if you remember, was also a function of R, which depends on the curve parameters. And you can write it explicitly in terms of the frequency parameters as well. So you see here, omega, the time frequency, you see the azimuthal number M, and you see this separation constant capital M, okay? And okay, here the prime is a derivative with respect to the so-called tortoise coordinate, which is a coordinate, which is a rescaled radial coordinate. So we're working in the in the curve series. So we're going between, for R between R plus, the value at the event horizon and infinity. But now here we are rescaling that so that infinity stays at infinity, but R plus the value of the event horizon gets mapped to minus infinity. Okay, and so this is what the potential looks like. Now we can hide it again. So what does a mathematician want to do? Well, we don't actually want to just exclude non-zero solutions, which have positive imaginary parts. Remember, those are the ones that the cost can press for trying to exclude because they corresponded to exponentially growing in time solutions. So it makes sense to want to rule that out. But actually for us, we actually want to also be able to rule out solutions which have omega on the real axis. Why? Because if you think about it, if omega is on the real axis, these are just solutions which are signs or cosines. So they are bounded, but they are not decaying in time, okay? And we would like to have, we would like at the end that our solution to the Koski equation is something which is decaying in time. And in fact, as mathematicians, have a way of just being able of making do with this last statement. So if you can show that you can rule out solutions which have omega on the real axis, then you're done, okay? We can have a proof which only goes through, which only needs the information on the real axis. And this is actually quite fortunate because this separation makes perfect sense if you are on the real axis. So these angular functions will form a complete basis if you're on the real axis. But if you're on the upper half plane and you go away from the real axis enough, suddenly there will be problems with your angular ODE, branch points, et cetera, et cetera, which makes things very nasty. And it's not clear that when you look at the complex plane, that the angular functions form a complete basis. So it's not clear that you could make sense of this statement here, this informal statement that a solution is this infinite superposition. You can make sense of it in the real axis. And that's why it's very fortunate that for us mathematicians, we have a way of just using the real axis. And it's really not surprising that we have a way of just using the real axis. It's just a Fourier transform, right? So that's all that we're using here. And you don't need to understand Fourier transforms, you don't really need to go to complex values of the Fourier parameters, okay? So omega is on the real axis. And now let me tell you more concretely what we want to show. What we want to show is, there's many words here, a quantitative uniform infrequency, non-existence of non-trivial solutions of the Tukolsky equation with good boundary conditions. So let me break that down for you, starting with the end. Good boundary conditions, what do I mean by this? Well, we want to be studying solutions of the Tukolsky ODE, which arise from physically reasonable solutions of the Tukolsky PDE. So physically reasonable solutions of the Tukolsky PDE should be sort of in going at infinity, sorry, out, sorry, in going, outgoing at infinity, outgoing, sorry, at infinity and in going at the event horizon. And we just want to carry so that they have finite energy. So that this, just this idea is something that we want to carry over to the Tukolsky ODE. So we have a second order ODE in order to study it, we need to specify boundary conditions. And there's an obvious definition of what should be the good boundary conditions, which comes from looking at the current geometry and figuring out what are the physically reasonable solutions. Okay, so that explains the good boundary conditions. Now, why we want to show non-existence of non-trivial solutions that I've already said, right? So we want to rule out solutions which have omega real. But now why do I want to do this in a uniform and frequency way? I want to do this in a uniform and frequency way because you have to imagine that I have here an infinite superposition. So it's not enough to rule this out for, it's not enough to say fix omega, then there is no, then the only solution is zero. I actually have to be able to say that this still holds when you reach arbitrarily large omegas because my infinite superposition is going to involve arbitrarily large omegas, ones which you cannot simulate on your computer. Okay, so I want to show this in a uniform and frequency kind of way. And I want to do quantitative statements. This is a bit more, this a little harder to explain for non-mathematicians, but basically the idea is that I want at the end of my day not to just have the statement that they don't exist, but actually to have an estimate that tells me by how much they don't exist.  So that at the end of the day when I prove stability of current, which I doubt that it will be done by me and my lifetime, but once this is proven, you want the stability to say, consider that you're sufficiently close to Kerr and that sufficiently close being measured explicitly in terms of the Kerr parameters, then you will converge to a Kerr solution. Okay, okay, so let me start, so let me start trying to now explain how we go about doing this, how we go about showing the quantitative uniform and frequency non-existence of non-zero solutions. And I want to start by kind of throwing away that word uniform, which maybe you didn't like and actually be more in the setting of the fixed frequency case, which I said we would need to go beyond. But let's start with that case. That case is basically what Tukolsky and Press are doing.  We cannot go all the way to infinity. Okay, so let's start with the fixed frequency. So we fixed some omega, we fixed some time frequency omega, we fixed an azimuthal number M and we fixed a separation constant lambda. And now we asked, are there non-trivial solutions of Tukolsky-OB, which have good boundary conditions? And there's actually a very systematic way of answering this question in general. And that systematic way is to find the quantity for our ODE, which is both conserved and coercive. And on the conserved side, we have very good news. So this is something that you've seen already at the beginning of my talk and now I'm talking in more detail about. So on the conserved side, we have very good news because we have a stationary killing field. And that stationary killing field induces an energy, which is associated to the stationary killing field and which you can write in terms of my ODE notation like this. So this is a conservation where you have zero equals something. And that's something is boundary condition at infinity multiplied by some factor and plus boundary condition at minus infinity, meaning the horizon, multiplied by some factor, okay? And now this factor is written in terms of omega and the azimuthal number M explicitly and we have here this definition of omega plus explicitly in terms of the black hole parameters. Okay, so this is, we have a conserved quantity. Now, is this quantity coercive? Coercive means does this quantity control the solution? And we see that actually it's not. It's not because it's not in general, okay? Because if we have, if this factor here, omega times omega minus M omega plus happens to be positive, then yes, we do have coercivity. We have coercivity in the following sense. So if this factor is positive, we have boundary condition at infinity squared times something positive plus something positive times boundary condition at the horizon squared. And the only way that this can be equal to zero is if both of the boundary terms vanish. So the only way that this can be equal to zero is if you is zero. So in that case, we have answered this question and in the negative, there are no non-trivial solutions. And moreover, we see that this conservation law, it really controls our solution, okay? So it tells us, it gives us this very, very, suddenly you just know that this guy is positive and suddenly you're able to say by this conservation law, you has to be zero, okay? So it's really controlling our solution. But if this guy is not positive, if it's negative, then what you have here is the boundary term at infinity competing with the boundary term at the event horizon. So they could be both very, very large as long as they cancel each other out. So it's really not controlling your solution at all. You are not able to say anything about your solution, but if you have this condition. In this condition, you've seen it before, it was called super radians. So if you want, every time that I say coercive, you can replace it by non-super radians, okay? And you see that this is obviously very problematic in the case of rotating black holes. So if you don't have a rotating black hole, then this guy will be zero. So you'll just end up with omega squared here. And so you'll always be in that first case where you have positive times boundary term plus positive times boundary per term. You'll always be in the case where you do have that control over your solution. So it's really just in the rotating case that you have this problem. Super radians is just the property of the rotating black holes. You don't have an ergo region in Churchill. You just have an ergo region in Kerr. Okay, and as I've told you before, in general, ODEs are going to be modally unstable. So I've talked about the case where you have a Klein-Gordon mass, which was shown by Schlappen-Bach-Rothmann with many, many people before doing heuristics and numeric arguments. You can also show this if you, instead of just adding a mass, you just add a potential which is compactly supported away from the event horizon and away from infinity. And this was shown by Moschides in 2017. But for the Tkolsky equation, there's a kind of algebraic miracle. Even though you expect that generically, you'll have modes instability. For the Tkolsky equation, there's an algebraic cancellation which happens, which somehow magically tells you that there will be mode stability. So it tells you that the answer to this question here is no. Whenever you are looking at solutions of the Tkolsky OD which have good boundary conditions, they're going to have to be zero, okay? So fix omega and lambda, then the only solution with good boundary conditions is the zero solutions. And this is the case for sub-extremal Kerr and it's also the case for extremal Kerr if you additionally eliminate this funny frequency, okay? And there's just a little note, I said that it was important not to just have a non-existent, a soft non-existent, but to actually say by how much do these things not exist. And that's exactly something which is done by Schlapp and Dr. Rothman for Waves and which I did for the general spins in 2019. Okay, so how this is obviously a very important statement because this is the proof of what Tkolsky and Press were grasping at in that paper. So if you take nothing else from this talk, please take away the idea that it wasn't Tkolsky and Press that proved stability of Kerr, it was whoever proved mode stability. And these are, I'll tell you in a second who these people are. So how does the classical proof of mode stability go? So this is by now a classical result and the classical proof is as follows. So I said that there was this systematic way, you find a conserved non-super radiant quantity, okay? There is a way of sort of cheating that system if you want, which is to say, well, I can't find that for my original equation, but I'm going to find another equation which has is an isospectral to my equation. So it has the same if you want quasi normal modes, has the same mode solutions, has mode solutions for the same frequencies. I'm going to find something which is isospectral. And for this isospectral thing, this T0 tilde, I'm going to show that there is a conserved coercive quantity. So the way that this will work is I'm going to say, okay, for T0 tilde, since I have a conserved coercive quantity, I will have the no mode solutions on the real axis. We'll have no modes on the real axis. And by the isospectrality, that implies that I'll have no modes for the Tkolsky equation. So the classical proof gives you basically an idea of how to construct this isospectral equation.  And here I'm just going to add a weight here, w, so that TR is self-adjoint in L2w, which just means that once I put TR here, I'm going to be able to swap it. So suppose that you are given, so this is just the definition of some u tilde. Now suppose that you are given a differential operator, T tilde, which satisfies this property. So TR here is just the operator associated to the Tkolsky ODE. So and suppose you're given this T tilde, which satisfies T tilde minus TR applied to that kernel equal to zero. Then you can show that the equation for your u tilde will be T tilde, u tilde equals zero. So it's easy to see why. So this is all back of the envelope computation. You put T tilde here, and now you put it on the inside. And when you put it on the inside, you add and subtract TR, okay? So T tilde minus TR, and here TR. Now this term disappears because of the condition star that we impose. And the first term here, it's going to, you can use the self-adjointness of the weight to put the place the TR, not on K, but on u, okay? So this is equal to the same thing with TR, now on u. And the equation that you satisfies is TR, u equals zero. So now you see that you have T tilde, u tilde equals zero. So you have this way of producing an equation, which, so you have this way of taking your equation and producing a new equation. And now you want to show two things about this method. You want to be able to this method to work for your setting.  Because if you have injectivity, then they will have the same, then they will be isospectral. They will have the same quasi-normal modes, okay? And you also need to guess, so I said that you have to be given a T tilde. So you also need to guess this T tilde. You also need to basically guess what will be the new equation. And you want that guess to be very good because you want that this new equation has a conserved course of quantity. So you want it to not have any super radiance, okay? And this is really, really done by guessing. And the first guess of how this was done was by this absolutely incredible work of whiting in 1989. So again, if you don't take anything else from this talk, please take away that it was whiting, you know? If you want to claim that someone has solved the black house ability problem, please give this honor to whiting because he came up with some amazing guesses of what should be the kernel, what should be the T tilde, which allowed him to show that the thing that Tekowski and Press were grasping at is true. If you fix omega M and lambda, so if you fix a separable set of frequencies, then the only non-trivial solution with good boundary conditions is, the only solution with good boundary conditions is the trivial one. It's the zero solution, okay? So whiting did this for sub-extreme occur and then in 2019 I did the same thing. So I had to come up with new guesses for what the K and the T tilde should be in the extremal case because whiting's transformation, it really breaks down in the extremal limits. So when you look at the extremal case, you have to make completely new guesses. And it turns out, you know, when you look at these proofs, it turns out that the correct choices are to take a kernel, which is kind of an exponential and to take T tilde to be something called a confluent Hoin operator. I won't go into this very much, but you know, it turns out from these two papers that this is the right thing to do in Kerr spacetime. And the reason why this is very hard is because these choices are really fine tuned to the spacetime. So for instance, this XE here, it's normalized with respect to the Kerr parameters and it's discontinuous when A goes to M. So in whiting's case, it blew up in the extremal limit and that's why you needed a new set of guesses in the extremal case. You need a new set of, you need a new kernel and you need kind of a new form of this confluent Hoin operator. Okay. And moreover, this is really, really annoying to implement if you are looking at real omega, which as I said, as mathematicians, this is what we're interested in. So whiting did this in the upper half plane. For the unstable modes of Tkosky and Press, we're also considering. But mathematicians, we want to do it on the real axis because we have a way of using real axis to infer results. And if you want to apply this method on the real axis, then you need a regularization argument. And the reason is very simple. If you take an exponential kernel like this and if you imagine that C is like omega, if omega is on the real axis, then is on the upper half plane, then this will give you exponential decay. So here, all of these back of the envelope competitions that I did, they will be easy to justify because the kernel has exponential decay. But if you omega is on the real axis, you won't have that exponential decay. So all of this will be harder to justify. And that's why you need a regularization argument, which is quite annoying. Okay. So I, let's see. I don't have that much time already, but half of, I'm through with half of this list. So, okay. So, you see, there are reasons to sort of hope that you could have a different proof of this very classical result of modestability. And here is a new approach, which is joint work with Mark Kozalsk, was at Central Brazil at Pesquisas Fizikis and University College Dublin. So the approach is really to go back to the classics in every sense of the word. So this is a paper, which if you're familiar with black hole perturbation, you've seen many, many times. This is Lever's paper for quasi-normal modes on curved black holes. And what he does is he gives you a condition for the existence of a mode solution, a non-trivial and non-trivial separable solution with good boundary conditions, okay? So he says that a non-trivial mode solution exists if the frequency omega is a root of a certain continued fraction equation. And so how is this a continued fraction equation? First of all, well, it's a continued fraction equation because you have to imagine that this guy is after the minus and this guy is after that minus. So if you're not familiar with that notation, that's how it works. And the way and these coefficients, alpha, beta and gamma, they are also explicit in the black hole parameters and they're explicit in the frequency parameters. And here they are. So alpha has this horrible expression depending on the index N, which appears here in the continued fraction equation and on some intermediate constants, which he gives here. And maybe you'll notice that capital M, the mass of the black hole is absence. That's because he uses a very unfortunate notation that M is equal to one half, but really it's hiding here. So every, and this ALM is my lambda, my separation constant basically. So you have this very nasty expression and you can ask, well, how does he get this? Well, he gets this basically by doing the compositions of the solutions of the ODE with respect to bases of special functions and trying to enforce the boundary condition. So in two minutes, that's what he does. So when he plugs in these expansions for the solutions, he gets a three-term recursion relation. These are the recursion coefficients. And when you solve that recursion relation, that you will get this continued fraction equation, okay? Nice, okay, so one way that you can kind of think about proving most stability, if you have this is, well, this is a condition for existence of a non-trivial mode. So if you can just show that this condition, this continued fraction equation has those solutions on the real axis, for instance, you're done. Or on the upper half plane, you know, you're done. You have managed to show this theorem, which I said was so important. There's a problem with doing that, of course. And the problem is you look at this page and you don't wanna do it because the intermediate constants here, in theory, you can do it, but in practice, how are you going to deal with the fact that you have all of this complexity here going on? It's not easy to show that there are no solutions to this continued fraction on the real axis, especially when the dependence on the omegas is so complicated. The dependence of alpha, beta and gamma on the omegas is so complicated. Okay, so here's a different, so here's an idea. Maybe the reason why this looks horrible is because Lever chose the wrong kind of intermediate constants you consider.  And so with Casals, we introduced this idea for the intermediate constants. So we call the masses M1, M2 and M3, they're again completely explicit, okay? And S, M, omega, little M, A, et cetera. And it turns out that if you plug in this intermediate constants, then suddenly, this, the coefficients in the continued fraction equation do turn out to look much, much simpler. So Lever wrote alpha and gamma independently since in the continued fraction equation, which is what I'm interested in, alpha appears always multiplying gamma. I'm just going to put the product here. And then I'm going to look at this and I'm going to look at beta. And now I'm going to write it in terms of my intermediate constants and this is what I get. So what I find is that the dependence on these masses which I introduced is only through the so-called symmetric polynomials. So sigma one, sigma two, sigma three. And sigma I means symmetric polynomial of degree I. So sigma one will be M1 plus M2 plus M3. Sigma three will be M1 times M2 times M3, okay? Third degree, okay? And so this already looks much, much nicer. So it's shorter than this for sure. And so on the surface, this might seem like it's just a cosmetic improvement, but in fact, it's much, much more because since I'm writing here, since it turns out that the dependence on the recursion coefficients in their dependence on the intermediate constants is only through symmetric polynomials, what that means is that the continued fraction equation is invariant under swapping any of these masses. Swapping M1 and M2, swapping M2 and M3, swapping M1 and M3, all of this will lead you to exactly the same continued fraction, okay? And let me tell you why this is interesting. So let me just write again that to Kolsky ODE, which I showed you before, that Schrodinger type ODE with a very explicit potential. This looks horrible with the potential written like this, but now I've introduced these intermediate constants. So let me take advantage of that to write my potential in a somewhat neater form with these intermediate constants, okay? So what do I see here? Well, if I change M1 and M2, nothing changes. The potential is the same because you see the dependence on M1 and M2 is very simple. But if I change M1 and M3, or if I change M2 and M3, suddenly something different is going on, right? And what I said before is that, so the condition for existence of a mode does not change, but the equation does change. So that gives you now a new idea of how to come up with how to produce from your equation and new equation, okay? And in this new equation that you produce when you apply the symmetries will be isospectral because we just showed that the continued fraction condition, the condition that tells you if there's a mode is invariant. So I've shown that the condition, the quasino mode spectrum, if you want, has symmetries corresponding to exchanging M1 and M3 and corresponding to changing M2 and M3. And changing M1 and M3, actually you can see very easily here, is trivial, it's just swapping the sign of S and that's that symmetry we know, this is the Tukolsky-Sarbensky identities, which are very familiar in the physics literature, they go back to the 70s. But exchanging M2 and M3, this is more interesting. This really drastically changes the equation, okay? And it turns out that what this does is that it recovers Whiting's OD. So without having to do an integral transformation, we have arrived at exactly the same ODE as Whiting did in his paper. And that is an ODE, which has a conserved coercive quantity. And since that OD has a conserved coercive quantity, it has the property of mode stability. And by the isospectrality, we conclude that we have mode stability. And all of this without having to go through Whiting's integral transformation. So this is different way of generating that isospectral equation to reach mode stability. And I should point out that this theorem here, this was conjectured by Aminor Brasian Hatsuga, who really came up with this idea of writing the Tukolsky equation in terms of these masses. And they came up with this idea by comparing the Tukolsky equation with some equations arising in supersymmetric quantum chromodynamics. So maybe the high energy physicists will be happy to hear these words now. Okay, so this is one advantage of this method is that we don't have to do the Whiting's integral transformation. Another advantage is that it turns out that it's more flexible. So as I said before, the Whiting, this method, the classical method of guessing the kernel and guessing a T tilde, this is very rigid. You see, I said that the choices that we have to make are always very fine tuned to the space time. And so it's very, very, every time that you encounter new space time have to start from scratch. In contrast, this method of finding these intermediate constants, it's more flexible. And to showcase this, we are able to look at sub-extreme occur the center for which no analog of Whiting's result exists, no transformation has been proposed. And we, which works, okay, has been proposed. And so we look at the sub-extreme occur the center. And now we select some masses, some intermediate constants, which are not three, but four. They are again explicit in the black hole parameters and in the frequency parameters. So kappa here is a surface gravity at some horizon in occur the center, omega plus is the angular velocity at that horizon. And again, we can show that the Tekolsky ODE has the spectral symmetries of exchanging the masses. And exchanging, so the exchanging the masses is a symmetry of the spectrum, but it's not a symmetry of the equation. And since it's not a symmetry of the equation, we actually can show that motizability holds for some sub-radiant frequencies. If the match is not perfect as in occur, so in occur, we are able to then show motizability for all sub-radiant frequencies. Here we are only able to rule out the sub-radiant frequencies which lie in this orange region. Okay, so it's still an open problem to understand what happens in the blue region. It's still an open problem to decide whether occur the center is modally stable or not. Okay, so we are progressively getting more hardcore. And now I think we are getting to the most hardcore, which I've titled painful but necessary. So I said at the beginning that we need to do, we need to prove non-existence of the non-zero solutions in a uniform and omega way. And the reason why is because we have that superposition, that infinite superposition. And we need to make sure that when we look at infinite omegas and ems and lambdas, we need to make sure that we still have no modes. So that's just a text here again. And through motizability, and I just want to emphasize again, why the motizability doesn't give us this? Because motizability, what it says is fix an omega, then there are no modes. So basically, you can pick a very large box of omegas, but you can't go all the way to infinity. So motizability tells us, fix a very large box, whatever box, you can fix any box, but it will be a box. Then there will be no modes in this blue region. And of course, you have to exclude the zero frequency. I don't know if I mentioned this properly, but in the motizability statement, you have to exclude the zero frequency. The proof does not work at the zero frequency. So you have very close to zero, this doesn't work. And I did mention that for extreme occur, very close to this other funny frequency, it did not work. Okay, so through motizability, we know that we have this huge place where there's no modes, but we still don't know. Okay, and we still have to prove that there can be no modes coming in from infinity, right? This is the uniform and omega part. And we also need to make sure that there are no modes coming in from that to zero that we excluded from the zero frequency that we excluded. And if you're interested in extreme occur, then you have to make sure that there are no modes coming out from this cut right here. Okay, that's what we need to do in order to show stability of occur. Okay, it's not enough to do motizability. However, this in some sense is much easier in the following sense, okay? So you look at this equation again, it can be this. So this one, if I tell you that omega is very large, then you know, for instance, at this term here, S squared, this S squared term is completely irrelevant. If omega is very large, you know that you don't care about the entire potential. You just care about some part of that potential.  So if omega is very large or very small, we as mathematicians or physicists, we have arguments to say, well, the only part of the potential which will matter will be this one. And now we can just do analysis. We can just analyze what can happen with that kind of potential. If the potential has a simple form, you know, with our first course in quantum mechanics where we learn to deal with these equations, we might be able to figure out what's going on. If the potential is simple. So in that sense, it is easier to deal with. It is easier to deal with the limits where omega is very large or omega is very small. In some sense, what is hard is to deal with, in just a sense that we don't know what to do. It is harder to look at the case, the multibilty case where omega is neither small nor large and we can't throw out any part of this potential, okay? So I want to tell you that, you know, this is in some sense easier in the sense that we have strategies to deal with it. We can throw away parts of the potential. We only care about the leading order terms and we have strategies to deal with the leading order terms. It's for the mode stability range, which we really have no good concrete strategies. As I hopefully I illustrated by, you know, the big question marks that corresponded to that part of my talk. Okay, so it's easier, but at the same time, it's very, very painful. And it takes many, many pages to do because if you want to rigorously justify that intuition that we all have that you only care about the leading order part of the potential and that leading order part is going to behave in a certain way to justify this, it's still going to take you a lot of work, okay? And this big, big task of going through this, this boring stuff, this was done, this was something that I did with Slap and Talk Rothman in 2020. So what we show is that there are no modes coming in from the high frequencies or in no modes coming in from the omega equals zero frequency in the force of extremal range as a smaller than M, okay? And so combining that with the mode stability result, you can show the solutions to the Tekolsky equation are bound and in fact, decay in time. And of course, you know, the way that I'm presenting this, it sounds like we were the first to come up with this, but that's not true at all. So this is, we're really standing on the shoulders of the shoulders of giants, so to speak. This, our work comes after decades of work on the Tekolsky equation and on showing this kind of boundness and decay in time. And this is a picture illustrating just that, okay? So of course, people looked at this very much for the Schwarzschild case. And, you know, this was fundamental to us, especially this work of the Fermes-Holzer-Rodnianski. And it was, it is also fundamental that people have studied this problem in the case where you're looking at scalar waves, S equals zero and the Tekolsky equation. So that was a much simpler form of the Tekolsky equation. And that study in the full sub-extremal range was done by the Fermes-Rodnianski Slapintok Rothman back in 14. So somehow it is by drawing very heavily on both of these works that then we're able to show this result, okay? And so these works I've mentioned here, the Fermes-Holzer-Rodnianski and Taylor, but I want to also mention that looking at the Tekolsky equation, I just want to bring back that point that looking, studying the Tekolsky equation, this is the first step to understanding stability of Kerr. And in fact, these works on the Tekolsky equation have paved the way for people to show, for instance, the K of the linear Eisenstein-Sanbackim equations when the Kerr-Backeau is very, very slowly rotating. This was done by Andersenback Dalbouma and Hafner Hinson-Vaagé in 2019 to study, you know, coupled electromagnetic and gravitational perturbations. This is work done by Georgie. Et cetera, et cetera, okay? So this really is the first step is understanding the Tekolsky equation and improving this statement, okay? And this statement, as I said, it not only entails more stability, it entails something which is painful but necessary. And I want to tell you a little bit about that painful but necessary part in the five minutes that I have left. Okay, so blazing through it. So let me start with the limit where omega goes to zero. So in the limit where omega goes to zero, you actually have something quite bad in your potential. If you look at the potential again, you will see that if omega is very small, but it's non-zero, then this term here does not vanish. And this term is a long range, is a long range part of your potential. And it's a long range part of it's a, it's part of the imagine, your potential has an imaginary component, which is long range. Both things are awful. And I think they have been identified as being awful by people who do numerics as well. You don't want long range potentials in general. So there's many works in the literature about how to get rid of this long range potential by kind of considering, again, transformations from the Tkolsky equation to some other equation, which does not have this longer range potential. A little bit like in the mode of stability, but typically people consider different types of transformations. So the transformation that we will use, it's something which was done by the thermos, which was also used by the pharmaceutical Rodniowski, is something called the Shander-Sekar transformation. So basically you take new variables, which are null derivatives of your Tkolsky variable in the right null direction. Okay. And you find that if you do this S times, you know, S being your spin weight, so plus or minus two, if you're looking at gravitational perturbation. So if you do this twice in the case of gravitational perturbations, the new variable, the Shander-Sekar variable satisfies a new equation, which has a real potential. The imaginary part is gone. And also it's short range. So it goes like one over R squared. Okay. So that's really important in the low frequencies. This is an idea, which is really important. Now let's look at the high frequencies. Well, the high frequencies, you see something appearing in Kerr, which I haven't mentioned at all before in this huge talk about Kerr. So I mentioned super radians, but I never mentioned trapping. I never mentioned any photon spheres. Now is the time when photon spheres appear. If you're looking at the very large limits of frequencies, then you can have trapping. And this new potential that comes with the Shander-Sekar transform variable, this really is like the scalar wave. So you have a kind of trapping at high frequency. You have this kind of photon sphere. Now in Kerr, this is not really a photon sphere. It's more like a photon region. It's thick in R, unlike in Schwarzschild where it really is a sphere. And it comes in addition to the problem of super radians. So this photon region intersects the Erger region. So you could be worried that you have trapped waves which are undergoing that Penrose process that I called super radians. So somehow they are trapped without being able to go into the event horizon, go into the back hole or escape to infinity. And they're always interacting with the Erger region, maybe gaining more and more energy. So that can really worry you. And because when you look at the geometry, it does look like these two things are coupled. The trapping set intersects the Erger region. However, when you look in the ODE world, suddenly you see that, thank God, this doesn't happen. In the ODE world, you recover a photon sphere. So you don't have that thick range. So there's a single R value for each omega M lambda for which you have trapping. And moreover, if omega M lambda is super radiant, then it cannot be trapped. So you can show that if you are undergoing the Penrose process, you cannot be trapped, which is very nice, okay? So it prevents that situation that I said, where you have accumulation of energy. Okay, and this observation was first done by the Fermi's Rodney Asking in 2009 for slowly rotating black holes. And then it was shown in the full sub-external range by the Fermi's Rodney Asking shop in Dr. Rothman in 2014. So this is another place where we really draw on this earlier work. Okay, let me just finish with this. So in the high frequency, maybe I didn't mention this enough, but when you do this Chandrasekhar transformation, there are no free lunches. So when you do this Chandrasekhar transformation, yes, you got rid of your imaginary long range part of the potential, but you will get a right hand side. So you will no longer have a decoupled equation. You will have here some coupling to what? Well, to all of the lower order Chandrasekhar variables. So to the Shkolsky equation and some of its derivatives. Okay, of course, notice that as usual, there will be a free lunch if you're looking at virtual. You know, not just in the issue of not having super radians, but also you won't have this right hand side if you're looking at virtual. So it's really a career issue. It's really a rotation issue that you have this, you have to pay this price. Okay, but we are interested in rotating black holes. We know that rotating black holes are out there. And so we are willing to pay this price. This price, what it tells us is that of course, the analogy between this new Chandrasekhar variable and the scalar wave setting, it will not be perfect. It will not be perfect because we have something here, not zero. Okay, and one way that you can see the imperfectness of this correspondence is to compare definitions of energy that you can take. So one definition of energy is a definition that you take for scalar waves. So scalar waves, this is a definition of energy which has appeared before. It's the energy associated to the d by dt killing field. There's a very concrete way of writing that. If you now translate the formula that I gave you before for you to see, this is what it looks like. So this is exactly the same expression that you saw before, but now because you have a right-hand side, you won't have a conservation. You will have this right-hand side here. Okay, so you can compare this with another notion of energy which is the notion of energy introduced by Tukolsky and Press to look at specifically spin weighted fields. So the solutions of the Tukolsky equation. So this is actually, so this will be of course, since it was an energy designed for the Tukolsky equation, it is an energy which is conserved. But if you notice, so you have equals to zero, but if you notice on the left-hand side, what do we have that is equal to zero, you will see that somehow it doesn't look like what I showed you before. Now there's these additional terms, these strange additional terms CS over DS. And these are some constants which are again explicit in the black hole parameters and the frequency parameters. Everything is explicit here. However, explicit though they may be, they're still very bad. And they're bad because it can be degenerating in some parts of the frequency space. So for very large omegas, some region of the very large omegas space, some region of infinity, they are going to zero. So you're losing control. When this goes to zero, you lose control over the boundary term at infinity, right? You, this identity loses control and it only gives you control of the boundary term at minus infinity. So that's kind of annoying.  So you have to kind of balance, when you do all of this analysis, you have to kind of balance. Sometimes you're going to have to, sometimes you can deal with these frequency weights which behave weirdly. And so you'll want to use this energy, but sometimes you can't deal with those frequency weights. And so you'll have to use this other energy. And so you'll have to deal with the fact that it's not conserved. So these kinds of things, they really complicate the analysis. But at the end of the day, it is possible to use all of these ingredients to get a proof. And, you know, I said that I would increase in horribleness as we reach the end of the talk. So this is just a simplified picture of what that analysis actually entails of how we divide up the space of very large frequencies and very small frequencies to do our analysis. And I think that's a very good place to stop. Thank you. Thank you, Rita. Thank you for this very nice talk. Let me see if we have some questions. If someone here in the Zoom talk, in the Zoom call wants to ask a question, well, I'll go to you too. So we have one, one says, hi, Rita, would the gravitational waves emitted by this type of black holes carry relevant information about this stability? And if so, could we hope to measure it in an experiment setup? Yes, I think so. And, you know, all of these, the gravitational waves, they're all, well, this, you know, in the Tukovsky equation, all of the things that LIGO is picking up, if I understand correctly, and if I don't, then Alejandro will correct me as he has before, you know, when you look at the, what LIGO is picking up is the alpha minus two component of this Tukovsky master equation. So, yes, definitely. But, you know, I should point out that it is definitely an expectation that Kerr is stable. I mean, one of the reasons why we use it as an astrophysical, I mean, there's two reasons why we use it as an astrophysical model. One is that we think that it's the only black hole which is stationary and four dimensions. So if you want to have a stationary state, which is a black hole, it has to be Kerr. And there are some uniqueness theorems that support this claim. Although for mathematicians, you know, there's still some work to be done on the assumptions of those theorems. So that's one reason is it's the only black hole which is stationary in four dimensions. And the other reason is we think it's stable. Because, you know, if we didn't think it's stable, then we wouldn't use it for astrophysical models. So yeah, as far as I know, no astrophysicist doubts that black holes are stable. Kerr black holes, at least in the sub-extreme range. I don't think that anyone doubts that they're stable. And I'm sure that there's lots of numerical work which is much more convincing than this paper of Tkolsky and Press, which I showed you. This was really the early tests of stability. Okay, yeah, I think related to that, there was another question that maybe you sort of address it here. It is computer might not be able to provide a proof, but maybe a counter example or something like that. What does the numerics tell us about the stability of Kerr black holes? Yeah, I think the numerics tells us that they are stable. But again, I think you, Alejandro, are more up to answer this question than I am. Certainly since this paper of Tkolsky, many people have computed numerically the quazinormal modes. And I think, okay, quazinormal modes means solutions like this, when omega is in the lower half plane for imaginary part negative, but whenever they do these searches, whenever they can do these computations, as far as I know, they also keep in mind, they also check just to be safe that there's nothing in the upper half plane, nothing on the real axis. So there's definitely, since the seventies, there's been big advances in computing. So I do think that the state of the art in terms of numerical tests is much better. And not just from the point of view of the Tkolsky equation, but probably full Einstein equations, right? Yeah, cool. Thank you. Someone here in the Zoom call wants to ask a question. Yes, I have a question for Rita. It's very nice to talk, Rita. Thank you. My question is regarding with this photon region that you were mentioning at the end of this talk in the sense because of the nature of the care black hole, is it possible to in the in the observation of these photons coming from the photon region, not the photons here, to have a special signature with respect with other type of black hole like in astrophysical environment? Yeah, remember the imaging of the black hole that they observed with the radio telescopes that there were photons that were trapped and then you can have this ray tracing and you have a special shape. If it is in the case of the care black hole and this region to have a special, how to distinguish the two different black holes? Or see if it is rotating a lot or not so much? I don't know exactly because this is not the kind of thing that I work on. I think Alejandro again will be able to answer this question better. I do think that for the question of rotating more, rotating less, there probably is some connection. You can probably see this from the ray tracing. Is that wrong Alejandro? Yes, yes, although like we don't have enough of the resolution to. Sure, I mean in practice now, but simulation wise definitely tell the difference. I think to distinguish curve from a different type of black hole, that seems like it would be much more difficult because you're not getting a full image of the curve black hole by doing the ray tracing. So I don't see why there should be some uniqueness. You see why the image that you get by ray tracing should be uniquely characterized curve, why you shouldn't be able to construct some other geometry which gives you the same. I think that there are some people who do like work on this type of problem, like trying to come up with geometries that mimic have the same ray tracing, have the same image when ray traced than curve. We have a low physics webinar like in last December, I think like someone talk about this type of mimicry. For the audience that wants to, they might want to revisit that video. Okay. So another question that is very small, I mean also related with this part of, what is kind of the implication in the case of Hawking radiation? If it in the case of, is the form region makes, is evaporate faster or slower with respect with the non-rotating black hole? This is a question which I definitely don't know how to answer. I don't know. You see, well, I did part three in Cambridge as we were discussing before the seminar and obviously we do talk about Hawking radiation, but it's very conspicuous that whenever you talk about it and whenever you sort of actually do the computation, suddenly you're in Schwarzschild. You're in a kind of Oppenheimer-Sneider-like space time, which is like Schwarzschild. And actually there's another student of my advisor who was PhD student, same time as me, who was working on the problem of scattering on Oppenheimer-Sneider space time, precisely with the goal of understanding Hawking radiation from our mathematical perspective. But you see, when you look at rotating black holes, there's no Oppenheimer-Sneider. So I have no idea how I would do that computation. And I don't know what people, how people think about doing that computation for rotating black holes. I'm sure that there are back of the envelope arguments involving black hole thermodynamics that tell you something, but I don't know. I have no idea. And also black holes. Because I was wondering if this kind of radiation that the black hole may impact the stability as well. I mean, maybe down in the smoke or something like that. You know, if you leave the world of classical general relativity and you add quantization, the statement will be different. The black holes ability problem will be different. I am only interested, you know, as I said at the beginning with the disclaimer, just pure general relativity. We just want to see what does Einstein's theory predict before adding on anything else. And it's not clear how to add on these other things. I mean, it's not clear how to, I don't know, I don't think that Hawking radiation is very understood, is properly understood for mathematicians yet. So for instance. Thanks. Sure, thank you. Thank you very much. Let me see. I don't see, there was another question, but I think you addressed that, which is like, yeah, I'm not sure if black holes in our universe are described by the curve solution, but what are the implications for them? I think you touched on that. So that's... Well, I mean, black holes are not described by the curve space time in the sense that, you know, we expect that they are dynamical, right? And the black hole solution is stationary. So in that sense, it is true. But, you know, this is a bit like, it still might be a good model, basically. Even if, you know, these other effects, you can still think of them as lower order effects. Well, another reason why you would think that the curse black hole is not good is because when the curve solution is an empty, it's just the curve solution. There's nothing else. There's, you don't have nothing around it. You know, in the universe, you have things around black holes, you have stars, but it still might be a good model for the local geometry near black hole. Thank you. Okay, we have another question on YouTube. I think this will be our last one. So they're asking, this might be out of topic, but have you considered how the presence of charge in the black hole would affect stability? So this is something that I mentioned very briefly in the Shoulders of Giants part. So this is something which one of our colleagues here at Princeton, Eleanor Georgie is working on. And it, well, it does affect the stability problem in the sense that, you know, mathematically, it makes it much, much harder. Because so if you consider, so there's a generalization of curve which has charge, which is Curr Newman black holes. And Curr Newman black holes, just their generalization with charge. And if you look at their stability, when you do linearize stability, you need to consider at the same time, gravitational and electromagnetic perturbations and they couple together, which means that then you cannot have, you know, I showed you at the beginning that you have this nice Tekolsky master equation which describes the electromagnetic perturbations and it describes the gravitational perturbations, but you know, it describes them decoupled. This is for Curr. If you look at Curr Newman, they will be coupled. So there's not a single Tekolsky equation. So things will be much, much harder. And from what I understand, from also from listening to her talks, that is especially, that will probably be especially relevant in the case where you have a large rotation. So if I understand correctly, some of other colleagues here at Princeton have done some simulations where they show that if you have sort of, so there's this idea in the physics literature I think that charge should not matter too much. And they kind of wanted to investigate this claim that charge doesn't matter. And I think they put a charge scalar field or something in around the Curr or Curr Newman black hole. And if you have spherical symmetry, so, you know, if you're Schwarzschilder, Reisenorsturm or whatever it is, then that's maybe it's a Reisenorsturm charge. Well, it dissipates. The charge dissipates very quickly. But when you have rotation, the charge stays on for much longer. So this will probably be more impactful in the more rotating case, in the really where the black hole is rotating. And yeah, it does impact things. And it's not clear at all how it will impact. It's still, we have to wait for Elena to show us. Cool. I think there was another question in this lag in our communication, but so I'm gonna rephrase it because I don't forget exactly what was, but then the question is there a way to know the numbers of symmetries and those masses. And if you could please comment why you call them masses. Okay, sure. So I can comment on why you call them masses. And then we have to repeat it. So I call them masses because in supersymmetric quantum chromodynamics in this, so I said that this theorem is really was conjectured before by Amino, Krasian hats to them. And it's something very clever. And it's based on the same kind of ideas that we then did the Curd-Desider case. And their idea was there's this paper about supersymmetric quantum chromodynamics. I know nothing about this. So my explanation will be very, very basic. But from what I understand, there's something called cyber within theory. And they look at hypermultiplits in this theory. I can't tell you exactly what that means now, but the point is they have masses, okay? This is, and they have, you know, if you look at three hypermultiplits, they have masses M1, M2 and M3, okay? And that's why I call them masses. And actually their idea for putting out this conjecture is based on two things. So one thing is that it turns out in the quantum chromodynamics world and the cyber within theory, exchanging the masses, it's, this is invariant, the theory is invariant, the predictions will be invariant. And so by analogy with that, Amino, Krasian hats to the, this is my interpretation, the conjecture that you should also have invariance here. And then they did some numerics, which showed that they seem to have, the equations seem to have the same positive normal modes when you change the masses versus when it didn't, right? Okay, so what was the other question? How many symmetries or masses like, do you know, because here I think you show three and then in the other one for the ADS one was four. Is there a way that they can tell how many to specter function? I don't know exactly, but you know, I can tell you, I can tell you this, which is there's a, there's some reason to think that here it should be three and in the other it should be four. And it's related to how many singularities there are in the ODE. So there's a, so in ODE theory, so you have the Tukowski ODE, right? And there's books from the nineties and earlier about ODE theory, which tells you that there can be singularities in your ODE and there's a characterization of those singularities. And for the ODE for Kerr, you have a singularity, you have singularities where? Well, you have them at the event horizon, r equals r plus, you have them at the Koshy horizon. So the two roots of that delta and you have a singularity at infinity, okay? In Kerr-Desider, where do you have roots? Again, you have roots at the roots of delta and you have a root, where do you have singularities? You have singularities at the roots of delta, which now will be four and you have a singularity at infinity. But there's some work in the 98 by Suzuki and I forgot the other people. So the group of three Japanese co-authors who show that somehow one of those singularities can be removed so that you end up with just four singularities. So Kerr, you have three singularities, you have three masses. And then Kerr-Desider, you have four singularities, you have four masses, okay? That's kind of the idea, but really this is all by a very hand-wavy, this is by drawing an analogy with super symmetric quantum chromodynamics that we come up with these masses and these definitions of masses. And yeah, I can't tell you exactly, it's just somehow you see from counting number of parameters that in Kerr-Desider, you will want to have four masses from counting parameters in Kerr, you will see that you want three. But it's a bit like this, it's counting, you know, counting degrees of freedom if you want. Okay, thank you. I think we're way past the hour. Thank you very much for contributing to Love Physics and for everyone, please stay tuned for our next webinars. Thank you very much. Thank you so much. Bye-bye. Bye-bye. I'm sorry to have taken so long. No worries. No worries. Okay, I think... You haven't beaten a record, our record has been of an hour and a half. Okay.