 So, we take a manifold and this is not conformally defilmophic to standard SN. In that case, if we look at the solution set, let's say alpha is any constant. We look at any solution set where u is a smooth function, positive satisfies the equation So, this implies u is equal to less or equal than constant on m. So, this depends only on f, gamma, m, g and that constant. So, what is this? F, Francesco. That's remote control. Ah, thank you. Oh, oops. But it's okay. It's okay. I can go without it. It's all right. I just do that. Yes, sure. Grab the person. No problem, yeah. No problem, that's good enough. So, so I happen to have this. So, so on the manifold, this A g is the Schouten tensor. So, this is the gamma. It's a cone, a cone, something like this. So, and so f is a smooth function, continues up to the boundary, positive inside, there on the boundary, and partial derivative in each lambda i is positive. So, this is ellipticity. So, this g u here, meaning it's a conformal metric. This means we use this conformal metric, insert, so f is equal to 1. So, the eigenvalues of this new Schouten tensor lies on the surface, f equal to 1. So, alpha is any given constant. So, these are all solutions to the problem, to this equation. And it is added that Ricci is bounded from below. So, so, so we have, we are trying to solve this conformal metric equal to 1, this equation. So, this, the last restriction, so if f is sigma k, for k greater or equal to n over 2, so this, so this implies existence. So, this result gives existence for all k in this range. So, because in that case, this condition is naturally satisfied. If this Schouten tensor in the cone, automatically, this will be satisfied, with the right-hand side actually equal to 0. However, for k less than n over 2, this, this is an additional assumption. So, this gives, so that would not give the existence for the problem. So, in this regime, so earlier results, this statement includes earlier results by Alice Tramkowski and Paul Young for k equal to 2 and equal to 4. And also this includes also, so for k bigger than n over 2. So, so let's first, so we describe the proof. So, first, we start from a proposition. It's a known, this is certainly a known statement. So, so this is, let's say this is n, this is n. And this would be the boundary of n. So, you take a point here, you take a point here. Then you go to the boundary. This is the shortest distance. So, if one assumes that the boundary of this n has mean curvature bounded from below by this constant, by a constant, so like a very small bore in Euclidean space is going to have larger mean curvature. So, so we, so this mean curvature is with respect to the inter, inner normal. So, if this is bigger than this alpha, then the distance is controlled by, let's say alpha equal to zero. Let's, let's not just look at that. So, this is 1 over c0. Thanks. This is 1 over c0. So, you see, if alpha is fixed, if c0 is large, then the distance will be small, distance will be small. So, we use this to control things in the proof. The proof, one can prove the is, one can, so let's give a proof. This is a proof of this statement. So, let's say, let's take x. Let's call this, let's make a line like this. This is the shortest geodesic. So, let's say gamma of t goes from zero to a goes to n. So, the n is going to be the distance. Well, we take the shortest geodesics and we parameterize it by arc lengths. And then, want to show that a should be less or equal to that number. So, one can take e1 derivative there, take a vector, take this direction. And one also take the other n minus 1 in the tangent space. So, take orthonormal. Now, transport along gamma. We are going to have an orthonormal frame going that way. So, then, let's say for each i, we take a curve, take another parameter. We, like, move it around. So, we, t is still in the arrow a, and we move around s. So, so that, gamma i, the arrow t is identical to this gamma t. This will lie on the boundary. So, when we, we move this curve around like this and keep that tip on the boundary, and end point is fixed. So, if we take such a curve, because it's the shortest distance, we know that dds, so denoted by that. So, it's at, so this is the first derivative. And because it is minimizing, so we have d square, ds square, s equal to zero, if you calculate the length, you calculate the length, yeah. So, this is the length, then you sum it up, you sum it up. So, this should be, each term should be greater or equal to zero because it's minimizing, so it's a, it's a minimum. Second variation should be like that. Well, then you just make a competition. You will see that it's going to give you this, it's a competition. So, this is a competition, you will see that. And then, you have inequality because you have some. So, use the mean curvature has some bound, and also the Ricci has some bound you inserted. So, then we optimize the right-hand side. We optimize the right-hand side. We can take f of zero is one, because this f a always equal to one zero. So, you make the right-hand side as small as you want, as you can. You will pick up the best f, insert it, and you will get the bound. So, that proves this proposition. So, we want to prove this one. We want to prove this, and we make a contradiction argument. So, proved by contradiction. So, namely, if it's not true, then there exists a sequence of conformal metric. So, it satisfies the equation. And the Ricci is greater or equal than this. And alpha is a fixed constant. And the maximum of Ui goes to infinity. Without loss of generality, Xi will goes to a point on L. So, we want to rule out this picture. Rule out such a picture. So, usually one tries to know, first know things about blow-up points and so on. So, we will first prove, actually, there can only be one blow-up point. So, step one is that inequality. So, step one says, the Ui, this sequence of function, is actually less than a constant. You measure the distance with respect to the original metric to Xi. So, you go minus M minus 2 over 2, or X belongs to M minus Xi. This certainly implies there can only be one blow-up point, because away from this Xi is bounded from above. And we know much more than that statement. So, the lemma for that is the following. So, suppose I have a situation that I have a sequence of point Yi. For the time being, one can just think of this Yi as this maximizing point. So, suppose I have a sequence of Yi, and within radius, so let's say Ki goes to infinity, there's a sequence of number. So, within this radius, a shrinking radius, which is dictated by the height. So, this is determined by the standard bubble blow-up somehow. So, this Ki goes to infinity. The speed that goes to infinity is much slower than the speed Ui of Yi goes to infinity. So, let's say Ui of Yi goes to infinity. And we know that the supremum, the Ui here, Ui here in the ball, is bounded by a constant Ui Yi. So, C1 is a positive constant fixed. So, even though it's not a maximum point, but inside this ball, it's almost a maximum point, modular fixed constant. So, we have, if we know there's a situation like this, then the conclusion is, we know the volume we measure with Yi, and in this ball, the ball we draw there, and smaller. So, there exists Ki prime, less than Ki, but still goes to infinity. So then, for Ki, this will go to 1. So, this ball, if looking at the GI, this ball already taking almost all the volume, all the volume, the remaining volume will be going to the arrow. And in fact, it will, in the proof, it will show, see that the top will actually goes to the volume of the standard ball, and also the, of course, the bottom. And actually, the diameter of the whole ball also goes to the diameter of SN. So this, from the proof, when we'll see that. So, we rescale. This is, this piece is M. This is Yi. Now, we take the tangent space like that. So, our coordinate is there. So, we know that this is, we know that this is less than C1 for X less than Ki. And this Ki goes to infinity. So, that's the assumption here. So, inside, we have that. This is the good scaling. So, the equation will be maintained, except the metric will be stretched. This will be dilated. So, this metric is the rescaling. So, this will go to G flat, because we are taking from a point and in very high norm locally, locally everywhere. So, this will go there. So, therefore, so, then once we have upper bound, we can have gradient bound. We only describe in, in our N, but that proof actually goes to remaining manifold. So, we have green bound, and then we have secondary bound. And here, our F is concave. So, we have C2 alpha bound by Evans Krilov. And then we have shoulder estimate. So, we have all the bound. So, Ui tilde is bounded on any compact subset. So, therefore, this Ui tilde is going to go to some U star tilde. And this will be positive, and this will be very small. C3, C4, C5, infinity actually. So, we have all the, we have a convergence. So, so, therefore, this is the limiting profile, and then this will satisfy the equation in our N. And we know this is actually, there's a Liouville type theorem. This can be classified. So, we actually describe not only our N solution, if it can be approximated in large balls, it's also classified. So, anyway, so, by the Liouville type theorem, so, one can write down this limit profile. And because of the normalization, you will see that the A star is bounded above and below by constant, and X star at the center will not move too far. These are universal constants. So, that says, so, this is a standard way of doing a blow-up. So, somehow, if you have a, say, you have a maximum point here, say, everywhere around, so, we just take this ball. So, then, we can always do this process if one is able to do the rescaling, having limit, and have a Liouville type theorem. So, whenever you know that, you know that in this very, very small ball, you have very, very good control of your blow-up solution. The ball is shrinking. So, now the question is, what happens outside? Well, so, here, we know we have a ball, we have a ball. This is like K i prime, U i, Y i to this. This is a shrinking ball. So, that corresponds to when we take a standard sphere, because this corresponds to standard sphere. You take this, raise to this power, and G flat. This is standard sphere. So, this is a standard sphere. Here, we cut off a little ball. We cut off a little ball. So, this ball is shrinking. This little sphere is shrinking, and outside here is there. So, this is your G i, G i. So, it's like that. So, then the outside, minus this outside, will be something, you know, coming out from here. So, in principle, this piece, one can get very large. One can, in principle, it's possible like that. But, we have that proposition, that this very strong control tells us that this little piece will be small. This will be large, because, let's say this, because this little piece, the metric, everything, is like this one, very, very small. So, we have very, very strong convergence. So, this will be very, very large. And that, we know the Ricci has a lower bound. We know the Ricci has a lower bound. So, then, that proposition says, the diameter of the remaining will go to zero. So, the diameter of the remaining will go to zero. So, and the diameter of this remaining, so you have something outside, out here, okay? So, that piece, the diameter is shrinking to zero. And also, we have B-sharp theorem using, again, the Ricci. So, it tells that this volume will also go to zero. It's less than constant, optional i's, and will go to zero. So, outside this piece will have diameter and volume, or go to zero, and that lemma follows. That lemma follows, that this lemma follows. So, actually, you show that both the top goes here, bottom goes here. Then, why we have such an inequality? So, if that inequality is violated, then we are able to produce, we can take another point. So, this is the maximum of that quantity. Xi is the maximum of Ui. So, then, we are going to have Xi sitting here, and we are going to have Xi tilde somewhere else. And you take a maximum point, you have this. So, this will, this distance, this power. So, then, it will give you a scenario, you will be able to draw a board like this, draw a board like that. So, this will conserve as your Yi, and that can also serve as your Yi. You will be getting a small board around it, which will take a whole volume. So, you have two volume of M, that's a contradiction, because the total volume should be Sn. So, in each small board you get an M, so that's a contradiction. So, what I'm saying is, then, this statement follows from this level. So, now, the picture is your Ui on the manifold will have one maximum point, Xi, Xi is here. And we have a control that this Uix is bounded by distance x to xi and minus 2 over 2. So, that's the control. So, it's a way, it's bounded. Once we have zero derivative control, because whenever we have an upper bound, we are going to have gradient estimates and then second derivative and so on. So, therefore, here, we can, these estimates can be, can go to log, let's put log. K goes from 1 to 2. This is our gradient and second derivative estimates in the scale. We have it in the scale. So, we have these estimates. So, in particular, this gives Hanok, because for K equal to 1 here, a gradient log of that is bounded. So, anything away, you take a piece Ui as Hanok. So, it's maximum mean and the proportional. We show that this Ui, so this xi will go to x infinity. Let's say x infinity is sitting here. It goes there. So, anything away from x infinity, we are going to see that the limit will be zero. It will be zero. So, this is actually just because we know that, we already know that if you circle a fixed distance outside, the volume should go to zero. The volume means Ui for the 2n over n minus 2dVg. That's the volume. And in this range, so including here, we go to zero. But we have Hanok. So, Ui maximum mean are comparable. So, integral goes to zero means Ui itself goes to zero. So, that shows this step. So, the picture now is anything away from here. So, Ui will be going to zero. So, we know that this Ui goes to infinity here. Away will go to zero and there's Hanok everywhere. So, the maximum mean are comparable. So, now look at, so let's pick up a fixed point here. Let's take p not equal to x infinity. So, then look at Vi of x is Ui of x divided by Ui of p. So, because of the Hanok, so this is bounded above and below by positive constants away from x infinity. And also, we have derivative estimates. So, we have derivative estimates. So, away from x infinity, these derivatives are bounded for Vi. So, therefore, this says that the Vi, after passing through subsequence, will converges to the infinity, which is positive. And this is in C11, look, M minus x infinity. And this convergence is C1 alpha or background alpha for every alpha less than one. Because we have after two derivatives control. So, this function go to zero, but because of Hanok, they are all comparable rate. I just normalize it. I get Vi, it will converges to V infinity. So, this V infinity will satisfy the equation. We looked at before, and it will be the boundary. So, the eigenvalues of the Schultz tensor will belongs to the boundary. So, of course, along the way, there is an issue that V infinity is not C2. So, whatever we talk about eigenvalues belongs here. We should interpret as viscosity sense. If we talk of Ricci curvature of GV infinity, it's greater than zero. Apply some sort, something like Bishop Gromov later on. So, we mean that this V infinity can be approximated from Vi. And those theorems applied to Vi can be passed to that. So, we are not really saying, so all those theorems hold for C11 metric. But that passage is not difficult. Not bother with the smoothness of V infinity, I'm trying to say. Just apply everything on that. So, then this shows this step. So, we know that the limit, when we normalize the limit, will end up with here. So, solutions of such things are relevant in studying this problem. And step four is to show, to give some description of this V infinity. There exists a constant, which is now negative, but finite, such that the limit exists. This function V infinity, when it goes to near this X infinity. So, when multiplying this like fundamental solution is, it has a limit. It has a limit. The limit at the moment can be zero. So that's actually an important point to address. When it's a show, eventually it should be positive. But first, one should show that it has a limit. So, we want to know what happens to this V infinity. V infinity outside are just positive function. When going to here, what's the behavior near that possible singularity? Well, the proof of that is, so we know that this is the conformal equation, applying to that should be less or equal to zero on M. So, this is the structure of the equation. So, therefore, let's call it LG. LG acting on that is less or equal to zero, and V infinity is positive. We have this. Once we have this, so this is just a linear statement, then we know we can define the limit of, this is the distance from the origin from X infinity, and minimum V infinity on the boundary of this ball. This is finite. And this is a linear statement. We are not using anything nonlinear. We have a nice electric operator here. So, the limit if along this going to the center is finite. But step two, so we have this estimate. We have gradient log V infinity is bounded by distance to X infinity minus 1. This give you Hanak. So, that give you a spherical Hanak. So, you take here, you take a circle, then on this circle, the maximum will be comparable. So, it's a consequence of that. So, then that means the lim soup is also finite. So, the lim soup will be finite because on every sphere, around it, the max is controlled by the mean. And we already know that lim if is always finite. That's a general result. But for this statement, we need to show that this A and capital A are the same. Namely, if I look at this quantity, X infinity is at the center, I take balls going in on each sphere, the max and mean, I want them to go to the same thing. I shouldn't have oscillation. I don't want oscillation. So, this is ruled out by an argument I showed, already showed before. So, need to prove A equal to A. So, if A bigger than A, I'm running out of time. So, then we are going to have balls shrinking and we have two points oscillate. So, then we normalize it to a unit size picture. And by passing limit, we end up with a solution which is in, we end up with a solution, something in here. So, lambda of this A, this A, we will end up with something like this and G flat will be here, ending Rn minus 0. And then, we have a symmetry result saying this has to be a radialistic metric, but this carried oscillation on the unit ball boundary contradicts with that. So, we saw this argument actually before. So, anyway, we use that argument and we will see that we actually have a limit. So, this V infinity near this singularity will have this nice limit exist. So, after that, it is very important to prove that this A is actually equal to, not equal to 0. So, to prove that, we divided into two cases. One case is the case where sort of corresponds to K sigma K, case elementary symmetric function for K greater or equal to n minus 2. So, in that case, to prove this A positive is through a construction of a barrier function. And for K less than n over 2, that requires a different construction of barrier function plus a generalization of power levels as parametric inequality. So, anyway, so that will give this A positive. This step is a crucial step in the proof. It's more technical, so I'll skip it. So, next one is to then we are having a contradiction now. So, here, the contradiction, again, we use this Ricci bound. So, the manifold is like this. This is the M manifold. So, this is closed here. So, here, there's a point X infinity. So, point X infinity. So, the metric now we are going to have GV infinity is equal to V infinity 4 over n minus 2 in the background metric. So, V infinity away is just positive C11. Everything is C11. And near here, V infinity is indeed like a constant. A is positive. And the distance is truly like a constant. So, when we have this power, we have this, you put it here. That means this point is being opened as a plane. This asymptotics at infinity is a flat plane. So, near here, so the metric is sort of open it. Open that. And outside is truly an RM because of this. If you put a flat, you will see RM. So, this is why it's very important to have a precise control of this V infinity near singularity. So, now we know that the reach of this V infinity is greater or equal than 0. So, that means the volume of GV infinity with boar. This is boar of V infinity boar. And divided by VRN, R will be non-increasing. Which is greater or equal to 0. This is Bishop Gromov. You can compare to standard model. This is RN. So, the volume will have this. However, limit, well, R goes to 0. This quantity is clearly equal to what? Because near the origin, everything looks like Euclidean. So, the important thing is actually when R goes to infinity, this quantity is also 1. Because, so here inside is, you take any point here. You measure large boar. This part is compact. So, it has finite volume. So, when we take larger and larger boar, the only thing you see will be a large boar of Euclidean space. So, you are going to see 1. So, both ends is 1. So, that means this is just identically equal to 1. But the Bishop Gromov theorem has rigidity part. So, this is monotone if it's just identically equal to 1. So, the rigidity part says that M minus this with respect to the metric. This is going to be isometric RN and G-flat. So, that means the original metric MG is actually conformally equivalent to SN and G canonical. And that ends the proof. Because, start from the beginning, we are saying that the manifold is not SN. Thank you for your patience.