 Thank you very much. I'm really glad to be here on unfortunately I can't say the whole week, but I'm looking forward to today and tomorrow. I enjoyed today. I'm looking forward to tomorrow very much so I'm going to attempt to sort of Give a talk where I put a few details on the slides, but I'm gonna also draw on the on the board So I think if you if you what I'm drawing on the board should be on the more understandable sides if the slides are too too technical, but feel free to interrupt and so to set the stage I I want to talk about Constantine curvature surfaces, and so I don't hopefully it's since it's such a big deviation from what we had this morning I wanted just to find the basics. So Can you read that? Okay? So the isoprometric problem is sort of a very old geometric problem in various guys's So what we're going to mean by the isoprometric problem is that you say that a set Omega in a man in a Riemannian manifold say is going to be isoprometric if So a set is going to be called isoprometric if it has the least area the boundary area For all regions of the same volume. So I wrote n but really for today You can set n equals 3 so in a 3-manifold you when I say volume an area Then it makes then everything is fine. So you want you want to enclose a fixed amount of volume with the least area Okay, and I'm just going to ignore any issues of issues of regularity of which there are many and quite interesting story but not for today and so what I Wanted to talk about today has to do with critical points for this variational problem So you can ask for the global minimum, but also you can ask for just a further first-order critical point that means that to first order the set minimizes the area for fixed volume and You can also ask that to second order it minimizes area. Okay, and so What I'm going to say is is a surface is constant mean curvature Okay, so a surface is going to be a first-order critical point if it has constant mean curvature the mean curvature is just the trace of the second fundamental form and Stability I won't write it down, but I won't write down the condition Okay, and so stability can be measured along the boundary of the surface I'll always write the boundary of sigma and so it turns out that your second-order stable You're a second-order critical point for the isometric problem if and only if you have the following Inequality for all functions with mean zero. Okay, so here a will be second fundamental form squared and so This condition that that you require that seed have mean zero That's like saying that you preserve the enclosed volume and now you're asking that you don't you can't decrease the area While while enclosing a fixed volume to second order at least Is that okay? Are we good till there? So I just mean that okay. Let me write this over here Yes, this is just Omega if you have any family of surfaces With this being the original one and I I want to fix the volume then Then I just want that the derivative at that time equals zero is Zero okay, and then similarly when I say second order I want a non-negative second derivative at time zero okay, and so The study of coming curvature surfaces is is sort of geometric analysis setting I think and It in the set in the for the isometric problem That's a very old problem and more recently people have studied sort of the critical points and so you have the following uniqueness Results, so if you have it embedded compact constant mean curvature surface in R3 Then it must be a round sphere, so that's proven by alexander of now. There's some other proofs as well If it has genus zero even if it's an immersed then it still it must be around sphere And finally if it's stable then it's also also around here So in particular along with sort of an existence result the symmetric problem This tells us that because in order to find the minima. We found critical points right so in particular The isometric problem in the Euclidean space The solution is the round the round ball right so the round ball enclosing a fixed area has very Enclosing a fixed volume has the least surface area among any such set Okay, so that that statement was known before these do it directly by by a more sort of Sematization techniques or you can sort of classify the critical points and use a existence result for isometric reasons and so If you've taken a sort of a course in in Riemannian geometry where you do you go beyond the definition of of Curvature once you start to look at what does it mean to have certain sorts of curvature? You learn that that Ricci curvature is a thing that controls volume Great, that's sort of the fundamental property of Ricci curvature at least for me if you're a Kaler geometry Maybe you don't agree with me, but so for for comparison geometry really volume is coming with is related to Ricci curvature Okay, so Bishop Gromov. There's also some very I think you don't learn this in your first Riemannian geometry class But it turns out Ricci curvature and the isometric problem have sort of an obvious link You can give some easily bound way the isometric regions behave by knowing something about the Ricci curvature and so What I wanted to talk about instead is the scalar curvature So just in case you don't remember the scalar curvature Will be the trace of the Ricci curvature, which is itself the trace of the sectional curve Or the trace of the correct appropriate sectional curvatures or the curvature tensor and so To start looking at examples of manifolds with various sectional curvature or scalar curvature behavior You start to realize that you don't expect to be have too strong of a control on an area and volume for scalar curvature Okay, so sort of the things that are true for Ricci curvature is definitely not be true for scalar curvature Well, it's it's a much weaker addition and so However, it turns out that for small volumes. So if you're interested in very small regions Scalar curvature does play a certain role. Okay, so let me draw a try to draw a picture and so So what my picture is supposed to be is you supposed to imagine a Compact manifold where the scalar curvature has up as a maximum at this point. Okay, so Okay, and so what it turns out is that then the point where the scalar curvature is the largest is where the small isoparametric regions turn out to concentrate. Okay, so There is a blue so if I ask on this in this manifold, how do I enclose? point zero zero zero one Volume with the least area. I'm going to do something like like this and so Heat so you don't know anything about where these regions are for large volumes I can't say anything. Well, I may be in a rotationally symmetric example You could say something but for very small volumes that answer is actually very simple is that ice for regions and actually more generally stable CMC regions like to concentrate around points of maximal scalar curvature or locally maximum It's stable the stable CMC problem okay, and I'll just point out that there's there's a big difference in the Stable CMC case and the isoparametric case because let's say this was the global maximum But this was a local maximum if this is the local maximum. You could find Stable CMC surfaces which were not isoparametric Okay, so stable CMC surfaces just feel the effect nearby But to be isoparametric you have to really beat every other surface Because that's like being a global minimum versus a local minimum And so the the real key to this sort of result is that on small scales Remania manifold look flat Okay, and so it turns out that when you start to study this problem the scalar curvature is telling you What's the deviation from flatness in a certain sense? That's very vague explanation, but the The reason that scalar curvature can control the area and volume is because of the additional input of flatness in some sense That's sort of a big explanation Okay, so what I want to talk about today is You take this problem as an idea you say, okay, I sort of understand the behavior of these sorts of questions for manifolds which are flat on small scales and I that that lets me understand small regions So then the question is what about manifolds that are flat on large scales and Let me understand look at large regions. Okay, and so These sorts of examples have sort of a completely different story Coming from general relativity, but I'm not going to start I'm not going to tell that story all and I'll just tell it from the geometric point of view okay, and so For today the key example of a manifold that's flat in some sense on large scales is For me it's going to be the Schwarzschild metric and So the exact form of the metric is not so important, but what you should know is That it forms a family of rotationally symmetric scalar flat manifolds and If you look at the metric and so for me bars are always going to be Euclidean so The metric is going to be some conformal factor to the Euclidean metric and for it the distance from the origin very large The conformal factor is going to go to one right so this metric the Schwarzschild metric as you go towards infinity You go toward G bar Okay, and there's some parameter here M Which has some some physical meaning it's called the mass but for us again We can just think of it as just a parameter describing these metrics Okay, when M is zero you get Euclidean space, and so it turns out There's also there's a long story relating scalar curvature in this number M But it turns out that M will always be positive otherwise the metric is not going to be well behaved in the compact and so it's going to turn out that Not only does scalar curvature have an effect, but also the mass M is going to have an effect on the positioning of large isoparametric and CMC regions and so What I wanted to talk about today now that we're sort of to the real topic is so what happens to large Stable CMC surfaces if you're asymptotic to the Schwarzschild metric so that means that I'm going to consider like manifolds with metrics, which are this Schwarzschild metric which is like Euclidean plus some correction and then I'm going to allow some either even a further correction to perturb things a bit So you're not the the manifold the the metrics will not be exactly rotationally symmetric and in the compact part of the manifold You won't have any real control so you sort of imagine them as Out here They're looking roughly like this, but then there's something quite confusing going on and You could have some some boundary for me for technical reasons if you're an expert I'm going to assume that boundary is minimal and there's no other closed minimal surfaces and so So for me, I cut the Schwarzschild at the I cut it off here and I don't worry anymore and so the the first result in This direction proved by his skin now and His getting out proved an existence result and I'll later talk about a uniqueness result that they prove And so what they prove is that so in this manifold. It's not clear That you can find any CMC large CMC regions or large stable CMC or isoparametric regions Okay, so it turns out even the existence of the isoparametric region You can try to fix the volume and min this area Even the existence for that problem is a bit difficult because you could have some loss of mass at infinity Right, there's there's some non-compactness to the problem So you could have like this usual picture where a minimizer would rather just run off to infinity in the limit There's nothing that's something you're you're a priori worried about. Okay, so it's not even clear that There exists any CMC surfaces Nor clear that there's just any that you you know what they look like, but it turns out that okay There do exist stable CMC surfaces and you know exactly what they look like and I'll draw that Here, okay, okay So it turns out that you can find a Foliation near infinity by stable CMC surfaces Okay, and so notice in Euclidean space if this was exactly Euclidean space. This is easy I just take very large balls and take bigger and smaller ones But in Euclidean space there's no uniqueness for those balls So I could take a large sphere and move it a little bit And it's also CMC with the same mean curvature and same area And what's gonna be the question here is is it true for these spheres? So what happens if you move these spheres a little bit? Can you can they remain CMC or? What happens if you move them a lot can you find another one which is CMC and it turns out the answer is going to be no So that's what I'll come to next So but for the existence result we do know that there are candidates for CMC Surfaces and so and in some sense they're centered Okay, so the if you look at sort of what is the center of these spheres? Obviously you're in a manifold so it doesn't make sense But you're nearly Euclidean in the region where the spheres are and so you can define some sort of Euclidean Center and the Euclidean center is going to be roughly the center of the manifold whatever that means so these spheres are not very far off Okay, so they enclose the center and their center is roughly the center and What I just said is going to be important for my description of the uniqueness results Okay, so the theorem that I wanted to talk about today is the following so if You're asymptotically short shield in the sense that you're a perturbation of short shield a lower order perturbation with six derivatives Who cares and you will assume that you have positive mass because obviously if you're going to prove a uniqueness theorem You want to exclude Euclidean space so Euclidean space has zero mass. That's just there's no extra terms So I want to add an extra term. So once you're not Euclidean space Then if your scalar curvature vanishes identically Then a large stable embedded CMC surface is here Okay, so in particular there's no So there's no such surface for example over here. That's large stable CMC Okay Is that there are any questions about? Yeah, so so these are the only large stable CMC surfaces in the in the in the manifold okay, and so You you can't do anything you can't have any hope besides saying things for large surfaces And this is sort of like what I was talking about before with the small surfaces You don't expect to understand the medium surfaces in such a manifold Because they could live entirely in the piece which is just compact is doesn't have anything to do with Euclidean and okay, and so What's that so If you in short-shield if you if you look at the doubled short-shield for example Then there are definitely Extra CMC surfaces that you sort of don't expect and so I don't know if there are large CMC surfaces you don't expect But you sort of you don't imagine that Surfaces that go are allowed to go through the the boundary They could behave very very poorly. I mean, I guess that's not totally understood This is sort of this is sort of the smallest manifold in which to prove it you name this result in some sense You could try to you could try to enlarge your manifold improve a different a stronger you think this result. Okay, so again for The experts You you might people you might have assumed that this has this theorem is true If you say non-negative scalar curvature instead of zero scalar curvature and that turns out to be false So You can I can construct an example with non-negative scalar curvature which violates this theorem I could I could change the hypothesis on the scalar curvature to a longer sentence And then it would still be true But it there's no nice way of saying the exact hypothesis which what we can prove And so again if you try to weaken asymptotically short-shield to asymptotically flat So here we're sort of assuming a sort of asymptotics if you try to do something slightly weaker It's also false by these examples of curl auto and chain And so what I wanted to talk about for my remaining time Is what goes into the uniqueness result in particular? I should say what other people have done before me because this builds on several a list of several results by other people so um the Essentially the way the story goes Is that people proved uniqueness results for these surfaces Under various conditions on the surface So you assume that you have a large stable cmc surface and it satisfies x Additional assumption Then it must lie in the in the canonical foliation or then it does not exist Right so and then what we did was finish the last the last list of of conditions and put you put everything together And you got a uniqueness result okay, so what Husky and yaw prove in their their original paper is that if You're centered in a certain sense If you're weakly centered then you're you lie in this foliation Okay, so let me try to draw a picture of this so what Husky and yaw were able to prove is that if You contain a ball of a certain size related to the mean curvature Then this picture is wrong and actually you're in the in the canonical foliation Okay, and so this sort of assumption Is essentially assuming something about the inner radius of the manifold compared to the mean curvature Okay, and that that will come back when I talk about what we proved um Oh, I'm sorry, and then this Condition was subsequently weakened weakened by ching and tian to just assume To say that there exists a set Which is very large say but fixed and once you contain that set Then you're in the in the foliation. Okay, and so I'll I'll say a word about how you might prove such a result in a minute, but let me just let me go through the the um Results and then so we were able to shrink this set down to a point. So if you contain a point then That's enough to force you into the into the foliation Any given point if you contain a point and you're very large So but but very large depends on the point so and so somehow What what we prove is that if you contain a point Then you eventually miss this compact like we prove as you eventually miss any given compact set and so what the these three results Tell you is that if you have a counter example to the uniqueness theorem Then you don't look like this but you look like this Okay, so The sphere and its center are very far to one side the manifold Okay So i'll come back and talk about these these guys Okay, and so there's been several works related to this and so um What I wanted to talk about first is how do you sort of how does such You prove sort of these theorems. I'll be a big brief here But the general idea Is the following quite simple identity Which is remember for me bars mean euclidean and no bars mean this other metric, okay and so Assume that you have a constant mean curvature surface sigma in your manifold Then you what you It is for so for the husk and yaw or the ching and tian arguments You're going to the surface of the the surface will lie very far out in the manifold So you can say what's the difference between the euclidean and the g mean curvature Okay, so it's quite a nice exercise In terms of how does mean curvature could change other conformal change And then because there's extra terms you can add those in So it turns out you can compute that the difference of the euclidean and the G mean curvatures has something to do with this quantity m And then there's this r to the minus three times and x dot the normal vector, okay and um Okay, so but on the other hand we have a way of getting zero Okay, so Both of these terms are going to give me zero So what i'm going to do is i imagine the first term This is like saying what happens to the the euclidean area when i move In a fixed direction a so i'm right. This is the first derivative area if i just move the surface in a direction a So for euclidean space, that's a isometry so that obviously doesn't change area And if you work it out the mean curvature times the speed Along the normal speed which is that is is going to be that first derivative. So that's the first term The second term is just the divergence theorem Right, so if i integrate New dot a over the boundary i can use the euclidean divergence theorem to bring it inside and the divergence of a constant vector field is zero Okay, so putting that together i get That the integral of h bar minus h times this new dot a it has to be zero precisely Okay, so that's coming from and there i've used the fact that you're c mc Right to pull in the h Few paying attention so that's one of the only places you use it in an obvious way in some other places You'll use it in a more subtle way and so If you assume that you have a sphere which has radius r And center r times a fixed vector so that that that's sort of like saying that if you rescale the picture So it has radius one the center is a Okay, so variant notion of center And so what using this this estimate for how the The mean curvature changes between the two metrics What you can do is you you have this thing which you know is zero I don't know if you can see that this thing, you know is zero, but then you can compute it Okay And now here if you're really bored you can figure out how to compute this integral and so You're using here implicitly that sigma is quite round and that's a sort of technical Note, but it turns out that what pops out is a Okay, so in some what we find The zero in order for you to be a quite round c mc surface Okay Is that clear so the idea is to use this fun Argument that I described before to have an integral which on one hand is zero The other hand is the the size of a in some sense Okay, that tells you that you have to be centered and then from there you can use the implicit function theorem to tell you that you actually Delay in the filiation You're you're near in the filiation. So then you can you can use something much sort of more basic in some sense So, all right, there we go And so just one It turns out that the the step from replacing sigma by the sphere is quite Because some of these terms are a bit singular so there's like R to the minus three term and that that could play So if you if you don't Investments on how spherical you are this argument might not work Okay, so and ching and tian what their contribution was is they were able to improve this approximation Technique and that's that and in this can you have this this Additional assumption that the inner radius was somehow under control and ching and ching and tian were able to handle The the argument even if the inner radius was not under control in a certain sense Okay, and so then i'll be very brief here, but if the inner radius is not very large Then you need a sort of a different different argument and just i'll just say use an argument related to the positive mass there Okay, so I won't I won't explain that so now to get sort of to the stage of what I want to talk about today is What we're left with so i'll just to recap we use this sort of flux type argument You compare euclidean to g mean curvature and you use this funny argument to show that if a The center a was sort of you're sort of centered Then you had to actually be in the in the foliation And so thus the thing that's remaining is that you're really quite far out Okay, so you look like this And so it turns out that you could try to run the same argument you say oh i'll just check the same integral But it turns out that These sort of flux integrals that you used before vanish So you can show they vanish just by this funny like first variation and divergence theorem argument But then you get that it's equal to something in terms of a But then you get that that's equal to zero. So you haven't proved anything So the the previous argument doesn't work for in that in that case is what i'm trying to say Yeah, you you get f of a you know that f of a is equal to zero you'd like to say that implies But if they is zero That's that's not E zero for all a bigger than one in some sense. So you don't get any information and so It turns out that that's not so i mean After the fact that's not so surprising because there exist counter examples So it the problem is now in for these guys the scalar curvature and the mass Play a sort of a similar role Okay, in the previous results the mass was this quantity m was everything But here now the scalar curvature starts clawing its way back into the problem And you turns out that if you just assume non-negative scalar curvature you can construct You can construct those guys The double piece has a different problem, which is that they go through the compact part in a certain way Is that the answer? Okay, so um it turns out that in some sense these sorts of integrals can't work out but if you do sort of a very Careful what's called liapunov-schmidt analysis In some way you can you can rule you can show that vanishing scalar curvature is enough to rule these out So i also don't want to talk about that because It's it's a bit hard to describe okay, but What it turns out Is what you need is that Analysis to work You need to know something about how the inner radius and the mean curvature behave So mean curvature is usually like one over the radius so for a sphere in euclidean space Mean curvature is one over r two over r depending on how you measure it, okay, and So this sort of theorem needs some sort of a prior input that the mean curvature and the inner radius are of the same size so that's like saying that this distance And this distance whatever that exactly means Which i'll write say one over h that's another sort of way of measuring the radius of a sphere c mc sphere This this theorem needs it there of the same size So what's missing now is proving that they must be of the same size And so that's what that's the theorem that i wanted to finish by by talking about, okay, so Under in the assumptions that we've been talking about actually this theorem itself only needs not negative scalar curvature or whatever We don't really talk about that but You have this a priori estimate that says the the inner radius the distance to the sphere And the mean curvature Are roughly of the same size You're not so worried about this being very large. You're worried about it being very small So what you're worried about Is that this sphere is getting very large But drifting off very slowly, so it's like getting very large But only drifting to infinity very very slowly And that that turns out to to really hurt your most of the analytic techniques that you have because the error terms are just going to dominate Right because if you integrate something like one over r Maybe you bound that one one over r cubed You might end up trying to bound that by the inner radius, right one over r is is Smaller than one over r not right because it can only it only gets better far out But then if that if that term you don't understand you have you have a big problem and so um Yeah, so where's that you is it used? so so this theorem is false if you don't assume asymptotic to short shield, um, so Which asymptotically short shield that's a stronger So I assume that i'm nearly euclidean, but then the next term is m over r And then I allow any any terms and it's actually this isn't this is not true if you if you just assume yeah and so um sort of the key Input in the following theorem, which I think is one of my favorite It's one of my favorite theorems Is that so if you have a stable cmc surface in a three manifold Then it turns out that you have the following relation Between this integral of the mean curvature squared And some other terms and these terms You can think of as just being non negative. So we were gonna we're interested in scalar non negative This is a trace free second final form. Just throw it away. It's not it's a non negative term Okay, and that tells us that integral of h squared is smaller than 16 pi okay, and if you're If you're good at computing things in euclidean space say so for the realm sphere in euclidean space The integral of h squared if you've normalized it like I have is exactly 16 pi okay, and so The proof Some sort of conformal balancing argument, which appears in many related related problems nowadays But what I just said is essentially this so if If you have any closed genus zero surface Then 16 pi minus integral h bar squared. So the euclidean version of this term is negative right so This is this is related to the the wilmore inequality not the most not the recently proven wilmore inequality But sort of the classical wilmore inequality, which says L2 norm of the second of h like the integral of h squared in euclidean space is always at least 16 pi okay And so the idea of the the sort of The estimate that I Let me bring the estimate back The idea of this estimate to get Somehow this this estimate you want to play these two things off each other. So obviously This is not a contradiction because the things which are positive and are negative are different right one is euclidean One is g but we've seen that it's quite quite a good idea to try to change The barred to the unbarred and pick up this m term. Okay, and so, um, I'll I'm nearly finished. I'll just sort of Tell you briefly the idea and so I'll put some I'll put some formulas on the board. So I'm sorry, but I'll get a point out which term How the terms work and there's one there's one final conclusion about the structure that I wanted to make Okay, so If if we're going to use this this this formula, which I had before so that says the The h mean curvature is the collinear curvature plus some term Which is like x dot new with a certain scaling factor and here m is as usual this this all important quantity m mass and so Okay here. You just square this great and so now Okay, I didn't do anything except So I've squared h Then this quantity I said is negative and in particularly you can write it exactly It turns out to be the using gauss-bonnet and the gauss equations It's negative l2 norm of the trace-free second fundamental form Okay, so that and that term will turn out to be important in a second And so This term Is going to tell us something about the inner radius Okay, so this term has a sign. It's everything is squared Okay, and so that means we can bound it from if we put it on the other side We can bound it from below so it doesn't matter exactly you should think of that as a good term but The argument seems quite hopeless because if you compute the size of this term it seems really big okay but you notice that This term would vanish if you knew the euclidean mean curvature was constant And that has something and that has to do with it, right if you delete Then this is that you can use the divergence theorem to the inside and this is exactly the thing which has Laplacian zero Right, this is the great the gradient of the green's function But the is not inside of the surface So there's no there's no delta. You don't pick up a delta. That's where you're far out. Okay And so now what's funny is that in order to to deal with the fact that This euclidean mean curvature is not constant you have to use this sort of Sure type inequality so the shirt like sure's theorem says for example that if you have something who's Trace whose second form has is pure as a Zero trace free part so it's pure trace Then the mean curvature is actually constant Okay, so that's that's this this says something like this in l2 And they are these two terms play off of each other and you can you can complete them there. Yeah Here Yeah in in this This is for the manifold and so yeah, I've suppressed some amount of like changing back and between euclidean So I'll I'll stop there. Sorry if that end was a little technical so The thus the in some sense what happens is that So you you want to So the problem is to a stable cmc condition only feels things along the boundary, right? That that says the mean curvature is constant and some integral Expression along the only I'll go back to the big now everyone's going to get sick So yeah, so these two things cmc and stable cmc only feel the boundary So if you don't assume something like Asymptotically short shield it turns out that there's these examples by carlotto and chain Where in some piece of the manifold? It's exactly flat sort of in a half space So then you can just just look over here and you can be stable cmc In this flat space and you you don't feel the rest of the manifold at all and so What assuming what the the stronger assumption of asymptotically short shall does is it sort of smears out this The the thing that's going to help you so that the mass which is what helps you the m over r Is sort of everywhere rather than rather than just potentially in the half So so so then if there's no m over r term then the manifold is flat if you have non-negative scalar curvature That's the positive mass there So m over squared of r. We don't know so the m over r to the The one half plus point one Whatever that is, right? So so those I don't I don't know exactly what happens here um So we can say something Which is that if you're interested in the iso parametric problem So you assume you impose a stronger condition on your surface in exchange for a weaker condition on the metric So if I ask not just what are the second order critical points, but what are the minimizers? okay for the minimizers then You can prove that going off like this no matter what the metric is this is bad For the minimum for the for what's the least Area in closing that volume and it turns out that the canonical foliation is indeed the minimizer No matter what the metric does But there could exist other critical points That that that answer your question Okay, so so that So for the result that I talked about at the very end you don't you don't use it but it turns out that in In this sort of result. So once you know that the Inner radius and the mean curvature are comparable Then you can do sort of a Lyapunov-Schmidt. I don't know if you know it a Lyapunov-Schmidt redot You can sort of perturb off of euclidean space in some sense You know that this sphere is round in this year a cmc with respect to euclidean And then you say put this new metric which is nearly euclidean and how does that change the mean curvature? Let me do the best I can to account for that by wiggling the surface slightly And then it turns out you can What the effect of that is and then you can say Is it possible to pick a sphere? Which then becomes exactly constant in curvature Okay, and so it turns out that when you say is it possible that has to do with a certain function that you can compute And that function turns out to have have a from both the scalar curvature and the mass And so by by looking at those two things together It turns out that you can see there's no real geometric I don't think there's like a geometric way of seeing this but Just by computing this function you can see that the curvature could beat the mass term in a certain sense If it was allowed to sort of bump do some bumps it could like go like this and that would allow large cmc spheres