 Today, I want to get to give some kind of a more sketch less complete, more than a sketch less than a complete proof of the L2 extension theorem. Sometimes it's called the Osawa Takegoshi theorem. So I guess that it's more general than the original theorem they had, but less general than it could be. But this will be, you actually can't prove the full L2 extension theorem with this technique that I'm, or at least I don't know how to prove it with this technique. I'll give you some case which is maybe the easiest to explain. So we have a Stein-Keylor manifold, and then we have a holomorphic function that's bounded, so it maps into the unit disk, holomorphic. And then I'm going to let z be just the set of all zeros of this t. And I want to assume that it's smooth in some strong sense. So dt restricted to z is never 0. And then aside from this, we have a holomorphic line bundle. And then we have a metric, Hermitian metric, e to the minus phi with a familiar condition, db bar phi plus the Ricci curvature of omega is non-negative. Then the conclusion is that for all holomorphic sections of this line bundle L over z, such that the integral of f squared e to the minus phi a of omega finite, I need a little bit more. This holomorphic section over x, such that interpolates this initial data on z. And then it has a bound on its L2 norm. And I guess there's some constant here, which I'm going to write down in a second. The important part of this theorem was that the constant was universal. So it just sort of only depends on the universe, so it doesn't depend on anything else. But what the sharp constant is was not known until relatively recently. Anyway, with these kinds of normalizations, the sharp constant is pi, and that's going to be important in the last lecture. So the universality of this thing really allows you to do a lot of great things, like study sort of line bundles asymptotically in some sense, so a very, very large curvature. You can take twists of them, and you still always have this universal thing. For compact Kaler, if you have a smooth, well, okay. The answer is yes, but it isn't pi, so it depends a little bit. See, what's maybe hidden here a little bit is that this sub-manifold is cut out by a bounded holomorphic function, which of course you can't do on any compact manifold. But even if you're given any hypersurface in a Stein manifold, in principle, in general, you cannot find a holomorphic function that cuts it out. You'll need, right, and then the curvature of the line bundle, of which that is a section of, will enter into this. Well, it will depend on more than that, so it will depend on what kind of metrics you can put on that, and it's not completely clear in general. So you have to be given all that data, otherwise. But then once you know that, you can write down the sharp constant. It hasn't actually been done in all cases, only in a case where you can find a metric of non-negative curvature for this funny line bundle, correct. So this theorem wouldn't apply there, but there is an analogous theorem. So the analogous theorem is that you have, okay, so I'll just do it quickly, because I don't want to spend a lot of time on it, no, no, it's fine. So you have some z inside x, and then you take the line bundle associated to z. And then, of course, there's also a section t of this line bundle that exactly cuts out this z and generates the ideal blah, blah, blah. So what you need, and this is what I was telling Tomash, is you need this extra metric for this line bundle, and the curvature of this metric matters for this condition. So then the condition becomes something like d, d bar phi plus the Ricci of omega has to be greater than or equal to d, d bar lambda, and also greater than or equal to 1 plus delta d, d bar lambda for some positive number delta. So for example, if you have non-negative thing, then you can forget about this condition. It's just this one, but in general, we don't know. The other thing is that this metric has to restrict to z as something reasonable. Okay, and then, and then the L2 norm will change a little bit, and the denominator, you have to make sense of it, this will be the denominator instead. And then if this thing is positive, then the constant is I think it's 1 over delta plus 1 times pi. So if you really can find a flat thing and let this delta go to infinity, you'll recover the pi case, but anyway, that's what happens in the more general one. Okay, so well I made this claim that this is a sharp constant, so you at least have to see that it's true in one case, but it's easy to see. If you take x to be the disc, omega to be the Euclidean metric, and phi to be 0, so your line bundle is trivial. And then take t of z just equal to z, so that your capital Z is just the origin. So you're trying to extend something from the origin to the disc with a given value there, and the claim is that the constant will give you this. If you have a constant at the center, you just take that constant to be the function. And that extension will satisfy the sharp estimate. So, and somehow this looks like maybe a silly example, but it's the example we're gonna try to reduce to. So what we're gonna try to do is actually shrink x onto z in some nice way. So that the weight essentially becomes a constant. And then the thing you're trying to extend is already extended because you have some kind of infinitesimal neighborhood. And so it's gonna satisfy the sharp estimate. And then what we wanna prove is that as you grow this domain along this deformation, the constant only gets better. So that's the strategy, it's a kind of model case. Okay, so now let's start into the proof. So first, we have to do a little bit of reduction in order to apply the technique that we're gonna do. One of the reductions we're gonna use is to shrink to some relatively compact pseudo convex subdomain, which I'll keep calling x. And then, and therefore the section that we started with is actually gonna be defined on some neighborhood of this domain. So we can, in other words, we're gonna assume that f is defined up to the boundary as is all the data as r omega e to the minus phi t. I think that's all the data. So how do you reduce to this case? Well, the claim is that once you know how to do this on some relatively compact subset, shrinking the z into its intersection with this compact subset, it's only gonna be smaller norms than this one. So this will be a universal upper bound. And then you just use some weak star compactness to pass to the limit. So it's just basic real analysis to make this reduction. So that's what we're gonna assume from now on. And then the next step, so this is the reduction. The next step is that, so the claim, which I'm not gonna prove, is that there exists some capital F, maybe capital F tilde, homomorphic section on the closure of x such that the restriction of f tilde to z is this little f. So, oh, and another reduction that we need to, we need to also assume that this singular Hermitian metric is also smooth. And that was the spiel that I gave last, at the end of last lecture about how on Stein manifolds you can always smooth the sections and still keep their curvature and non-negative, or in this case, this Ricci twisted curvature and non-negative. Okay, so how do you do something like this? Well, this is really not part of the yoga I'm espousing here. This is a more standard thing. If you have this submanifold z, well, you can sort of make local extensions of f here and then you can take some other extension or some other any function in the rest. And then you try to patch them together. And then it's really a problem about some Schief-Kohmology. And if you use this Dolbo isomorphism thing then it's just about solving the D-bar equation. Hermander's theorem will show you on Stein manifolds there are no obstructions. If it's a domain in Cn, for example, then you can really just, because this thing is extending past the boundary, you can really just extend constant. I mean, you choose some holomorphic transverse section which you can do on Stein manifolds. In Cn, I guess. And then you're just going to extend holomorphically to some neighborhood, extend to f constant, then you cut it off, and then you have to somehow use the D-bar equation to correct the smooth extension to a holomorphic one. There's a little bit involved. You have to insert some singularity here because when you solve the D-bar theorem, you get some solution, but you don't know if it's the original thing you want. You want whatever you add to this to correct it to vanish on the initial data set. And so you need to introduce some singularity. And this is where you lose the estimate. So if you introduce some singularity, it'll be something like the metric 1 over t squared. Anything that's L2 with respect to this metric will have to vanish along the zero set of t. And then, but, well, it's not global in this, in general. I mean, in this case, I suppose it is. Anyway, you're going to lose some curvature when you introduce this singularity. But because you're on a Stein manifold, remember yesterday I said that the trivial bundle has a metric of positive curvature. And you can twist it, it's still a trivial bundle. So you can get back as much curvature as you want. And then it's another way of saying the vanishing theorem, I guess, more directly. So okay, so this you can do. And so if you do it on a slightly larger domain than the x you're interested in, then its restriction to this subdomain that we've reduced to is automatically L2 because everything is smooth. So it means that, so this guy, this f tilde is automatically in L2. So you've got some extension and therefore there is some constant so that that extension is less than or equal to this number. But this point, the constant depends on the data too much. You can't take limits or anything like that. So you need to make a much better L2 extension than this one because you don't know much about this one. So the thing to do is to take, so let f be the extension of minimal L2 norm. So the whole rest of the talk is about trying to get estimates for the extension of minimal norm of this extension. All right, so in order to estimate it, I need a manageable expression for it. So I'm going to formulate this, the norm in some dual way. So, okay, I'm going to introduce some notation. So let's let A for analytic be the set of all polymorphic sections with a finite L2 norm. X bar means the closure of X, sorry. Yeah, so up to the boundary. That's what makes it L2. Otherwise it couldn't be blowing up. So let me define this L2 ideal of Z, call it I phi of Z, is going to be the set of all G in this A phi such that the restriction of G to Z is identically zero. And so the annihilator of this guy, that's just going to be the set of all bounded linear functionals on this A which kill anything in this ideal. And so then this is some trivial linear algebra proposition. So for all f, polymorphic on Z, say smooth up to the boundary, maybe just write it like this, up to the boundary so that we have some L2 extension to begin with. Minimal extension f0 of f satisfies its squared L2 norm is equal to the supremum of Cf squared over norm C squared star for all C in this annihilator. Absolutely. I'm already assuming after this reduction that I'm working in a domain inside a Stein manifold where Z sticks out like this. And then I'm saying suppose I take a holomorphic section which is defined on this piece, say, which is therefore is L2. As I've argued, there's some L2 extension. And now I want to write down its norm for you by some dual formulation. And I'm claiming that this is the formulation. If you just take any extension for any f, which is in A2, A of phi, such that f restricted to Z is little f. Is it OK? I didn't let you finish your question, but did I answer it? No, I think so. This restriction to this? Yeah, yeah. It's nice. Sure. Yeah, I mean, or it could stick out a little bit. OK, so if you think about this, it doesn't matter what extension you put in here. If you have another extension, then the difference must vanish on Z. It's going to be killed by this expression. So you expect this somehow only to depend on little f. And then it's not hard to confirm that it is the norm of the minimal extension. OK. So essentially what you want to do is try to estimate this quantity. But the trouble is that these annihilators, they're not so manageable directly. I mean, how do you estimate something like this? Well, you'd like to try to estimate by some duality, and you could in principle do that with a re-representation theorem, but the trouble is here that you're in the annihilator, not in the whole space. So you have to be a little careful, but it's not difficult what to do. You choose a good dense subspace. So for example, if I take, say, H, which is, let me write it like this. So this is just going to mean sections of L that are smooth with compact support in Z. So this time it's sort of slightly relevant. When I mean compact support in Z, the support is a compact subset of this subdomain X. So it doesn't go up to the boundary. And then for each one of these guys, you can define this C sub H, which takes, say, some F in here to the integral over Z of F H bar, e to the minus phi dA over dt squared. So these guys, they generate a dense subspace of all the elements in the annihilator. And to check that, you just have to check that if all these guys kill some F, then F actually must vanish on Z. Then from Hanbanak or something, you see that it's a dense subspace. But if these are just these little bumps along Z, and so if this is zero for all little bumps along Z, then this F must vanish on Z. So this collection of all such things is dense. So therefore, in order to compute this L2 norm, we can just, in fact, we have to estimate this just for the C sub H. The have to is not the important part. The fact that we can is the important part. Okay, so let's take a look at... So therefore, enough to estimate... So you have your fixed guy, and then you're going to look at C H star in this annihilator. Yeah, right? Because this norm is the soup over these, and if you have a dense subset, you just have to estimate these guys. Okay, so let's take a look at the numerator of this expression. So let me write L2Z for the Hilbert space, which is L2 on the space Z with the weight e to the minus phi over dt squared omega. So just the L2 norm that I've been using all along there. But now I'm not asking for a holomorphic section, just L2. And then in here we'll have a subspace, which I'll call E because I'm running out of letters. This is just this L2 intersect with holomorphic sections. Those are kind of the ones I'm interested in. And this is very similar to the story we told when we talked about the Bergman kernel. This is a closed subspace by this Bergman's inequality again. And so you have, again, an orthogonal projection to this bounded projection, which is it also called Bergman kernel. I think maybe there's one person here that actually met Bergman. Did you meet him? Nope. He wasn't there? Oh, you did? Yeah, there is one person. Okay. I think he would have been really happy that I'm saying Bergman. Okay. Apparently you had to really not get cornered by him in the hallways because if he cornered you, he would try to tell you how important this Bergman kernel is. Anyway, it's a nice idea. Bergman's inequalities was his main contribution, I think. Okay. So anyway, in terms of this information, let me write down some proposition, which is not difficult. It's a lot like what we've already done before. So for each H, one of these compactly supported guys, C, H, F. Or maybe I'll say F. Yeah, it doesn't matter which extension I use here. And then remember that our F restricted to Z is little f. So if I want to estimate this denominator, this is less than or equal to the integral over Z of little f times the integral over Z of not H, but rather the Bergman projection of H. Okay. So this is almost the Cauchy-Schwarz inequality. The only thing is that if you look at this formula here, because this F is holomorphic and L2 on Z, if you split H into its orthogonal component in the direction of the holomorphic things and the orthogonal component, then the orthogonal component is killed. So you can replace H by pH, and then you apply Cauchy-Schwarz. So, all right, good. So let's see, all together then what we have now is that this C H F squared over norm C H squared is less than or equal to the integral over Z of pH squared e to the minus phi Da over dt squared divided by the norm C H squared times this L2 norm right here. So basically, we want to prove that this thing is less than or equal to pi. We prove that, then we're good. Okay. H is not holomorphic. H is compactly supported. Smooth. Yeah, I'm sorry. It's not the best. It should really be... That's really okay. It's infinity zero. Either way, I should write it. I'm a little bit allergic to sheaf theory, so... Okay. So this is what we're going to try to prove. And now is where this degeneration strategy comes in, and we're going to use this Berenson theorem I spent the last couple of days discussing to get some information about what happens along degenerations. So, okay, I'm going to be more degeneration. So this is going to be the first, and it'll be slightly off, but it's important to try it this way. First, see why it fails. Okay, so I'm going to use a notation L for the left-half plane. It's only convenient to use the left-half plane rather than the unit disk because of... I want to... Ultimately, I'm going to want to say that some function is increasing, and if I use the unit disk, I'd have to do something funnier. Convenient choice, but not a crucial one. And now let's let xt be our degeneration. So remember, what I said before is we're going to try to collapse x onto the initial data set. And if we do this correctly, then we will get some information about how... I mean, I can sort of repeat this process for every xt, and we'll get some information about how this constant grows when we do that. Okay, so xt is going to be the set of all x in capital X, such that log t of x squared is less than t. And then this t is just going to be some negative number. So eventually, this t will be the real part of tau. So I'm looking for something symmetric and you'll see why. So this is certainly a degeneration in the sense that x0 is x and then xt is approximately z for t approximately minus infinity. So now let's define some Hilbert spaces over this left half plane. So let me define h of phi sub tau to be the set of all f which are holomorphic on say x sub real part of tau, sections of l, such that and now I'm going to make a modification of the just normalized constants. I'm going to put e to the minus real part of tau. I want x real part of tau. And then the rest is the same as before. I want this to be finite. These are Hilbert spaces and then we get kind of a vibration of the left half plane by Hilbert spaces. And of course, there's a vertical symmetry, but so what? Okay, so now for each such Hilbert space I can repeat this procedure. And then now you see that the dependence on the real part of tau or on tau in general, doesn't appear in the numerator because that's integrating over the sub manifolds. It only appears in the denominator. And so if we could apply Berenson's theorem to this family of Hilbert spaces then we'd be in really good shape because we would know that the logarithm of this denominator is sub harmonic but it only depends on the real part of tau when you look at the definition. So the sub harmonic function of tau that only depends on the real part of tau is a convex function of the real part of tau. So then suppose you knew that this thing was also bounded. Okay, well if you have on the negative axis a bounded convex function it can never decrease. So it has to increase. That means that this constant will just get better as t goes to zero. If it goes down, then by convexity it has to go to plus infinity backwards. And so that's why I take the negative, the left half plane. Okay. Right, so if you could also show that this thing, that the logarithm was bounded or that this thing is bounded then you win because at time t equals zero you have the case you're interested in and you don't know the constant but at time t equals minus infinity you're in that infinitesimal neighborhood where you know the constant is pi just to take the thing you started with as the extension. So that's a little bit cheating but it's essentially easy to see. Well, if you have a really, really skinny, so here's your z and you have this sort of tiny neighborhood. So take any extension and try to estimate the L2 norm on this thing then it's essentially going to be the value at the center times the area of the transversal disk which is pi times the thickness. But we have the square of the thickness normalizing the measure here. So the constant will be pi for t very, very large negative. So that's the rigorous version. Okay, but the problem here is that we can't actually apply Berenson's theorem because we really don't know these are different spaces that we're looking at holomorphic functions of. So we can't compare different spaces for different t and so we might not get a vector bundle. Anyway, it's at least not clear. It's not a serious problem. I'm going to fix it in a second. So one way to do the integral over all of x is to just take the weight to be I guess identically plus infinity outside x t. It's kind of potential well. And then you'll get zero outside and then you can think about holomorphic things on all of x so by some density argument you see it's the same. But that's a problem too because we need smooth metrics. So what I'm going to do essentially is smooth that in a good way so I can apply. So I need to smooth it in a way that's still plurisubharmonic because the hypothesis in Berenson's theorem is that the metrics are plurisubharmonic. I mean after you add the Ricci curvature. So, okay. So, no I didn't say it yet because I haven't written the but you're right. Basically, once I write this thing down if you've been paying attention it's done. Yeah, I will. I wanted to say it before I write it. So that's my strategy for telling you things twice. Okay, so if you wanted just a quick repetition. You're asking about the deformation part? No. Okay, so let's let so the function that we want to add to our initial weight phi on x is the function psi t it's going to be a map from x cross L sub tau, but the tau is living in the left half plane. So it maps maybe I should write it like this. So it maps x tau to well so you want to sort of get a good plurisubharmonic approximation of this phi restricted to x sub real part of tau. So what you do is you take a little more space in the log of mod t of x squared minus the real part of tau and that's good as long as it's well that's good as long as it's positive but it can become negative when you get close to z. And so what you do is you take the max of this and zero. And so anyway for t of x less part of tau squared less than real part of tau this thing is zero and so you get this part but then when it gets bigger than when you get outside this region you get this contribution but it's not necessarily large. So what we do is define our metric e phi tau to be e to the minus phi plus multiplied by some number p times c tau and p is going to be just very very large but fixed. And so in the proof what we're ultimately going to do is just take it sufficiently large to prove certain estimates which I'm not going to do here but it's really just basic calculus to estimate those things but this is the right kind of approximation to get the kind of space that we're looking for for application of Bernstein's theorem. So let me just say it again so the limit as p goes to infinity of the integral over x of let me just try it. I might be off by adding some real part of tau here but I have blackboard amnesia so I can't really so this depends on p not putting the p dependence in the notation but as you go to infinity you get this integral so I think I must have to add something like plus real part of tau so you can take my word for it or read in the notes how you handle these technicalities but this is not we're going to work from here we're not actually going to take this limit I don't know what you said this in order to get this anyway that's not so important so I may have a typo but I can't tell here right so now you know all the data here extends up to the boundary so actually this is the same Hilbert space as if you leave out this psi sub t so now you do have an actual trivial vector bundle with a non-trivial metric given by this so now you really can't apply Bernstein's theorem and so we have to do what we said we have to do so I didn't name these things but these are going to live in the space h phi sub let's just use t now just assume t is real so we want to show that this is less than or equal to pi and I'm not going to as I said do all the details so I mean you have to you have to do some estimates that involve some basic complex analysis and it's not really worth doing here it's better to just do it the way I said before just think about if you know that you can go down to zero then you do get the sharp constant and so all you have to know I guess I'm done all you have to know is that this thing is a convex increasing function so you need to show that this norm is bounded and then Bernstein's theorem implies that it's convex strictly speaking the logarithm is convex but then the exponential of the logarithm right and the boundedness is essentially the same computation as this kind of thing so what is that ch acting on some section f it's not exactly that but it's the x analog well no it is so this is the integral of say f h bar e to the minus phi over dt squared omega over z yeah so there is a little bit to do here sorry I'm saying this a little sloppy but let me just explain how you get that this is bounded by the L2 norm of f so there is a step here you have to use this Bergman's inequality to say that the value of f at a point of z is controlled by the L2 norm on some neighborhood of z right and so once you know that everything else is a smooth compactly supported thing so that's really it there's some constant that depends on h times the norm of f okay so that's the end of the proof as far as I'm going to give it like I said there's some details to fill in but they're not the kind of interesting details of the Berenson theorem part of the story more some basic analysis but so let me give you a slight preview of what what I want to do next lecture and maybe end a couple minutes short nobody will hate me for that okay so what what have we what have we done today well we've proved this theorem right so this thing looks like a cylinder around z and the radius is t which is a real part of tau and so the integral of any extension if that thing is really skinny extensions more or less constant in the perpendicular direction so the integral is more or less the integral of little f over the disk and then integrated over z but if little f is constant it's the central value times the area so the area is pi times t squared but I have or maybe it's not t it's log t so the way I've defined it it's log t so this is one over the t squared as I'm defining it maybe that was not the right thing sorry it's defined by log t so mod t squared is less than e to the t so that's not right sorry e to the t very very small number because t is negative and you have the area as e to the t so the radius is e to the t over 2 sorry so that's is that the confusion many things many confusions okay right so just a couple words so we use this this Berenson theorem to prove this theorem here by sort of degenerating something and then we got this family of Hilbert spaces and if we had tried to use this approach