 OK, so as promised, today I'm going to prove Bernstein's theorem. Start with the one-dimensional base. And it depends. I don't really know how long it's going to take me. So if it takes too long, this is the most interesting case anyway. And all the elements of this case are sort of all you. There's no new idea in the more general case to some extent. So if it happens that way, so be it. So let me first make a small reduction. I'm not going to explain thoroughly. But I'm going to assume that, so remember, we have this metric e to the minus v sub t, depending on a parameter t. And then for a fixed t, I'm going to assume that db bar vt plus the Ricci curvature of the underlying scalar metric is strictly positive. So just to say a word about why you can assume that, remember that our domain is a strictly pseudo convex domain bounded in a Stein manifold. And so one of the characterizations of Stein manifolds, this kind of famous theorem of Grower, is that you have a strictly plurisubharmonic exhaustion. If you like to think about curvatures of line bundles, it means that the trivial line bundle has a metric of strictly positive curvature. And so if you're restricting to some compact subset, you can rescale it to make it negative, for example, or whatever you want. But anyway, so you just add to your metric that strictly plurisubharmonic function. It'll be nice and bounded. And that will make all this positive if it started out being non-negative. And so what our goal is to prove that, so for a given section c, a homomorphic section of the dual, the statement of Berenson's theorem is simply that under the curvature hypotheses I stated last time, which are just, if you take dd bar in x cross omega here, and this is you pull back by the projection map, then that's greater than or equal to 0. So under that assumption, we want to show that the base is one dimensional. We want to show that the Laplacian at a given point, the Laplacian in t of log ct is positive or non-negative. So that's what we'd like to prove. So I'm going to start with, suppose we take a section f, but omega is going to be the unit disc in this case. Now I'm going to take a section of the homomorphic section of the original bundle, not its dual. And what I want to do is to estimate, so the Laplacian of the negative of log norm ft. You'll see why, but this is the main computation that's going to occupy most of the lecture. So let's say we're going to fix one of these guys. And so the first thing I'm going to do is compute the derivative of the norm of this section with respect to t. So I claim that this is equal to the derivative of ft with respect to t minus the Bergman projection of the derivative of phi t with respect to t times ft, inner product with. OK, so if I look at this square norm and I differentiate that. OK, so this quantity here is ft, ft complex conjugate. So and because it's a homomorphic function of t, the d by dt will just kill this quantity. So this is going to be the integral of df dt. And then you're going to have minus d phi dt f sub t times ft bar e to the minus phi t. OK, but the thing is that this section f is a homomorphic. And so if you split up this section here into its homomorphic part and the part that's orthogonal to the homomorphic part, then the orthogonal part is going to disappear. And so you exactly get the Bergman projection, which is what that formula says. Right, so let me just make a remark. So let me say it like this. So here what you have is the inner product of df dt with ft minus the inner product of d phi t dt ft with ft. And the claim is that this part is you're only going to see the homomorphic portion. The part that's perpendicular to the homomorphic is going to be killed by this. That's all I'm saying. OK, so we talked about the churn connection before. So remember the churn connection is the connection whose 0, 1 part is d bar. And then the 1, 0 part, well, and then it's metric compatible. So if we just try to compute the 1, 0 part, if we have some section g, g, then what we're going to get here is this formula d by dt bar of dt. So this is what defines the churn connection. Take this quantity and you subtract this quantity. So that's why they're really easy to compute. But so now you recognize this operator. This is actually the churn connection for this vector bundle. So it's a little bit funny the first time you see it. It's not just the, I mean if you just looked at L2 sections, not homomorphic ones, then the projection wouldn't be here and you'd have the churn connection for the L2 case. But the projection shows up when you're looking at homomorphic section. So this, I'm going to just call this sometimes to shorten the writing, the 1, 0 part of the churn connection. OK, so the next thing I want to compute is the Laplacian of f t squared with respect to t. And it's not as convenient to start from this formula. It's more convenient to start from this one. They're equal, but at least the p doesn't appear that doesn't confuse you. So this is, so when you bring this derivative to this term, this guy's still homomorphic. So it annihilates that term, and this one's also homomorphic. So the first term you get is negative d2 phi dt dt bar squared, and then you get plus, and now this is the complex conjugate, you're going to get this same kind of formula on the other side. So this will be the integral of dft dt minus, now I'm going to bring back the p. OK, so now this is some vector u. So there's a little bit of a thing I'm doing here which is not completely kosher, but I'm going to ignore it, but I'll admit it first. So it's not at all clear that if you have a homomorphic section of this thing, so something that's L2 in these fibers, that then when you multiply it by some function, it stays L2. Of course it's not homomorphic anymore, but why should it be L2? So that's OK if you assume that you're, as we have, that this function extends to some neighborhood, because then you're multiplying by a bounded function. So there is always that sitting in the background. But you could do this kind of thing on a not approximate domain, and then you would only get densely defined operators. So if you wanted to do this in some more singular situations, you would have to deal with densely defined operators. Yeah, no, it just extends to some larger domain. So therefore it's bounded because it's smooth, it's bounded. Right, so anyway, we've got this vector u here, and now I just want to write it in terms of its orthogonal decomposition as a homomorphic part and an anti-holomorphic part. And so this will be, you're going to apply the projection operator, but this guy is still homomorphic. And so it stays the same, and then the projection acts here. OK, and then what you want to do is take, so this is just Pythagoras' theorem. You just write this as u, and then you have pu, and then you have u minus pu, and Pythagoras' theorem just tells you that these guys are equal. And so this is our u, but when we apply p, this term doesn't change. And so we just get sub t of d phi t dt f sub t minus d phi t dt f sub t squared. OK, so now let's take a look at this term here, and just this whole expression. So I actually discussed more abstractly integrals of this form as a consequence of Hormander's theorem. So let me just remind you. So if you take this expression here, so let's say we call this guy v, then v solves the equation d bar u equals, v is a solution, u equals v, of d bar u equals d bar of this thing, but just on x. So maybe I should emphasize that now because there's a t floating around too. Because f is holomorphic on the fibers, it comes out of this. So this v is a solution of this equation, but it's not only a solution. It's the minimal solution, solution of minimal norm because any two solutions differ by something holomorphic, and this guy is perpendicular to the holomorphic thing. And so that implies that this term over here is less than or equal to the integral over x of whatever estimate we get for this from Hartford. Well, whatever it's got norm less than the norm of the Hormander solution. And the norm of the Hormander solution is less than or equal to the norm of this vector, which is this. So that's just that application of Hormander's theorem to the solution of minimal. OK, so I just erased it, but what I wanted to do is calculate not the t Laplacian of this guy, but rather the negative of its logarithm. If you look at negative d2 dt dt bar log norm ft squared. Let me just multiply this by norm ft squared to get an expression where I can use these two derivatives that I wrote down. So this is equal to negative d2 dt dt bar norm ft squared t plus 1 over norm ft squared t times the absolute value of d by dt of norm ft squared. Let's just differentiating this, is that OK? I sort of wrote that quickly, but just applying the derivatives twice, and this is what you get. OK, so let me first look at this term, 1 over norm ft squared t d ft squared dt squared. This is equal to, with my notation of nabla 1, 0. And then let me just rewrite what I had here, negative d2 dt bar of norm ft t. This is going to be greater than or equal to, because I'm using this as the minus sign, integral over x, d2 dt dt dt bar minus d bar along the x directions, dft dt squared with respect to theta ft squared e to the minus ft dv minus the norm of nabla 1, 0 ft. So altogether, I don't see any log yet. What are you asking me about this formula? There's no log yet, so it's just from the formula here. I'm just using this formula here. This is the nabla 1, 0. I haven't done the logs yet. I wrote down that formula, and now I'm going to take these two things and put it in here. So d2 dt dt bar of log 1 over norm ft squared t, this we've now proved is greater than or equal to, 1 over norm ft squared t times the integral of this quantity here and plus 1 over norm ft squared t times the absolute value of the inner product nabla 1, 0 d by dt of ft with ft. OK, so there's two terms here. There's this term here, which we can't yet tell if it's positive or not, but if you're really good at Kramer's rule, you already know that it's non-negative. And then there's this term, which is non-positive, because the Cauchy-Schwarz inequality goes the wrong way. And we're trying to prove, ultimately, as you'll see, we're trying to prove that it's positive. There's a little more I have to do. So anyway, let me just remind you what this theta was. So remember in Hormander's theorem, we looked at, now I'm going to write it in terms of this x operator, so this is our theta. Well, we required it to be greater than or equal to theta. But for example, if we set this equal to theta, our hypothesis is that this is positive, so that'll be fine, so we can use this guy. And now, if you look at the full dd bar of phi, it's equal to d2 phi dt dt bar. The board was shaking. I thought I was getting dizzy. dt wedge dt bar plus d bar x phi t, sorry, d bar x d phi t dt. There's a minus sign here. Again, there's a minus sign. And then plus dx d phi t dt bar wedge dt bar plus dx d bar x. And so now, if you try to compute the determinant of this thing, that's taking this dd bar phi to the power n plus 1, n is the dimension of x. Then, if you calculate this, use some appropriate determinant formula, which comes from Kramer's rule. You get d2 phi dt dt bar minus d bar x d phi t dt squared theta. And then theta to the n over n factorial wedge dt wedge dt bar. There should be square roots of minus 1. I don't like to put in. But anyway, so it's this positive thing. But anyway, this is a determinant formula for our matrix of the form, so a v v theta. The determinant of that in this guy is invertible. This is what you'll get. So that gives us the non-negativity here, but then we're still in a little bit of trouble with that thing. And moreover, this isn't what I'm supposed to be doing. I'm supposed to be computing dd bar of log of the square norm of a dual section of the dual bundle. Let's move to that. So the first thing I'm going to do, actually, let me see. How should I say this best? So we want to show the following. So fix some t. I guess it's arbitrary, but fix it anyway. We want to show that this quantity, this is a little bit pedantic way to say it, but we want to show that the Laplacian at the point tau equals t is non-negative. If I show that for every t, that's exactly what I'm asking for. This is what we need to show. This is a statement of Bernstein's theorem. So that's what I'm going to do now. First, so if we take ft, the homomorphic section that we were using here, if this thing satisfies nabla 1, 0 d by dt, d by d tau equals t, if this is 0, then from this formula here, we see that v2 dt, so this means that tau at tau equals t of log 1 over norm ft squared, then this will be greater than or equal to 0, because both of these terms involve this derivative. So I'm going to choose a section ft well adapted to this particular c for which this is going to be true. So how do I do that? So we have our fixed t there. So I claim that, well, it's the re-representation theorem tells you that there exists some, let's call it f0. This is not a section. I mean, it's not a section of the bundle h. It's just an element in the vector space living over t. So it is a section of l, but not a section of this bundle. So there exists such a thing, such that this dual vector ct, its norm is exactly the norm of this f0. The way you compute and ct, the way it acts on any g in this hvt, is just the inner product of, I guess, g with f. So this is f0. That's just a re-representative of the dual vector. So in particular, the other thing, yeah, I said that there. So this is f0 squared t. But we can write this again as ct f0 when you plug in g equals f0. This is what you get. But of course, this is non-negative, so I can put absolute value there. And then, because these guys are equal, I can square it and divide by this thing. So that's my expression for that norm. So remember, this norm, as I said last time, is actually the supremum over all possible g. And all I'm saying is that I've chosen f0 to realize the maximum. And so also greater than or equal to ctg. OK, so we're almost done. This f0 depends on little t, but little t is not moving. It's fixed. So I'm just going to prove this for every little t, but I'm going to fix it. Now let's define section. So let me, I don't know, call it capital F. So how do I define this thing? This is where I use the triviality of the bundle. So this f at a point tau is going to be this little f0 minus tau minus t times nabla 1 0 d by d tau of this f0 at tau equals t. So I just do this linear shift. And so then you can see that this ft is equal to little f0 at nabla 1 0 d by d tau of f at tau equals t is 0. So this thing is independent of tau. So you're only differentiating this and differentiating this expression. When you set tau equals t, the other derivative disappears. So this therefore implies by what I was saying here, if you choose a section whose derivative is 0, you're going to get that this quantity is non-negative. d tau, d tau bar of log norm equals t. This is, you know, it's greater than or equal to, all I need is it's greater than or equal to 0 tau, OK? t is not moving. I computed that derivative and it took me a very long time. And then this was the estimate for tau equals t. It doesn't. It's just a fixed vector associated to t. The t parameter is living there. But this doesn't depend on tau. It only depends on t. It's the dual vector of ct. Yeah, OK, I can fix it. If there's an error there, I'm happy to fix it. Right. But on the other hand, what we're really interested in is expressions like this, because those will give us the norm somehow. But the numerator, so if you look at c tau, f tau squared and take the logarithm of that, that's already subharmonic, because the definition of a holomorphic section is that this quantity depends holomorphically on tau. So this is already subharmonic. It's Laplacian is greater than or equal to 0. And then when you add the negative of the logarithm of this guy, you also get something subharmonic. Well, it's not subharmonic, but it's sort of subharmonic at tau equals t. So what we've shown now is that the Laplacian with respect to tau of this quantity at tau equals t is non-negative. This is still not what we want, but we're very close now. Yeah. There's a minus 2, because I got lazy to erase it and write it as 1 over. Yeah, the exponent is minus 2. Sorry, I should have not been lazy. It should be 1 over. And then it has the property that it kills off this term. Essentially, this is the whole point. The rest is a kind of soft argument. So now let's look at log norm c tau squared tau star minus log of c tau in this bold face f tau squared over the norm f tau squared tau. So what do we know about this function? Let's call it phi of tau. OK, so if you look at its definition, at tau equals t, this function vanishes, because the norm at tau equals t is exactly reliant. But then everywhere else, this norm, by definition, is the supremum over all possible sections. So therefore, the tau equals t is a minimum. So the Laplacian is non-negative. So since we already know that the second Laplacian is non-negative, we get that. So a minimum at tau equals t. Therefore, putting together that fact as well, the Laplacian tau log equals t. So that's the proof of Barenson's theorem. It's kind of analysts proof, right? It's not very illuminating, maybe. I don't know. Yeah, yes? It's not flat, but it's right. So then you can compute. That's right. This is exactly what I'm doing. So I was going to say that. Right. And that's what this term is here. This is a Griffith curvature formula. And there's a second fundamental form. And what I basically did was compute it. That's right. That's right, because your orthogonal projection to the sub-bundle is exactly the Bergman projection. So you're getting the second fundamental form. It's just a question of what your, this is a proof of a Griffith formula in this case, I guess, if you want. Before I use the estimate. So this is after the, so let me say it in your words. So what we have is this bigger vector bundle of just L2 sections. So usually, I mean, the reason I avoided this language was because initially I had in mind to prove another theorem, which I'm not going to have time to do. But in general, that's not a holomorphic vector bundle. But in this case, because it's trivial, it's a holomorphic vector bundle. But then it has this metric coming from the weight e to the minus vt. And so it has non-zero curvature. In fact, the curvature is given by this, this quantity here, for just the L2 bundle. And then if you have a holomorphic sub-bundle, then you look at your orthogonal projection down to it. And you can compare the curvature of the sub-bundle with the curvature of the bundle. And the comparison, by definition, is the second fundamental form. And so that was what I had when I had equality before I had this inequality. And then what we want to do is estimate the second fundamental form. And the statement is that because you're looking at the projection to the holomorphic subspace, Hormander's theorem will allow you to estimate the second fundamental form. And this is the estimate you get. A big bundle, I never discussed here. But it's just the space of L2 sections of X, sections of L over X, that are not necessarily holomorphic. So it's a much larger Hilbert space. What's that? Right, just measurable L2 things. But because of the way we, well, there are issues that I'm confused about. Like, what happens with all these issues, where here we had a section that extends up to the boundary and so on? Like, what happens when you're looking at L2 things that are not necessarily smooth? You have to be a lot more careful about really proving that that's a holomorphic vector bundle in some sense, because you end up with a lot of densely defined operators. This way, I just worked directly on H. But what you say is completely correct. So I think it's not going to be particularly illuminating to prove the general theorem, because the computations are really very similar to these. It's just that you use this other test with the test form that I constructed last time. And you compute dd bar of this test form, and you get some formula. And really, the only input for positivity, again, is this SEMS kind of Kramer's rule and Hormander's theorem. But it's, I guess Yanir and I were talking about this over lunch. This test form, I don't really know what it means. It's not something I'd ever seen outside of Barenson's actual paper. Just a very, very convenient way to prove not going to know positivity. But anyway, so once you have done these calculations, all the other ones are more or less the same. So I'll skip that and take the last 10 minutes to get started on the next portion, which is the L2 extension theorem, for which I'm only going to use this Barenson theorem. OK, so just before I state the theorem, I mentioned these before, but let me just say something about singular Hermitian metrics. I'm going to say and unsay something about them. I'm going to say and sort of allow myself not to use singular Hermitian metrics. But I want to state the theorem for singular Hermitian metrics because it's a good theorem. So what is a singular Hermitian metric? Well, so far, we don't really know how to define them for vector bundles unless the rank is 1. But for holomorphic line bundles, we do know how to define these, and they're very, very useful. So what's a singular Hermitian metric? Singular Hermitian metric is, let me put it in quotes, metric e to the minus phi. So this is not even a point-wise defined object. So you really should just think of it as a measurable section of L star, tensor L star bar. Just like a smooth metric as a section of that with certain obvious positivity properties at the points where it's actually defined. OK? And so now you want to say what this singular part means. So if c is a local holomorphic section of L, then of course, you can compute the length of this c. Let's say the square length. So that's the square length. It's almost everywhere defined. And then you take the logarithm of this thing. And let's give this thing a name. This is now a function phi that depends on this frame c. And we'll just ask for this one, this function, to be in L1 loke. So this implies that you can define the curvature current, dd bar phi c. And it's independent of c. Because if you were to switch to another frame, then the new function would differ from the old function by log squared norm of the lower vanishing holomorphic function, the frame change. So that's a harmonic thing. So dd bar of that will be 0. It's pluriharmonic even. So we just call this dd bar phi. Hence the notation for a metric. OK, so now this is a 1-1 current. And we can ask for it to be positive. It may or may not be, but if it is, that's equivalent to saying that this local function is a plurisubharmonic function, locally given by plurisubharmonic function. So that's a theorem in complex analysis that one of the several equivalent ways of defining a plurisubharmonic function is it's a upper semi-continuous function whose dd bar is a non-negative 1-1 current. So these guys, you can try to imagine working with them in this Bachner-Kedaira identity that I wrote down before when I gave the sketch of proof of Hormander's theorem. I've never quite been able to use them correctly in that way. But I sort of wonder if you really could. So you want to get some positive lower bound on eigenvalues of Laplacian. And this current really should be enough to give you that. But even writing down the integration by parts formula, you sort of need smooth functions and smooth forms with compact support and so on. So it might actually be possible to do it with currents. I don't really know. When you're worried about issues of regularity, though, you really do run into trouble if you can't approximate these things by smooth things. So plurisubharmonic functions in CN can be approximated by smooth things. It's just a convolution with a radial function kind of trick. And it's the sub mean value property, the other characterization of plurisubharmonic functions that tells you that you get the right kind of approximation. But when you have, for example, a complex manifold, then doing this approximation is only local. And then you have trouble patching together the approximations. And so there are some spaces where you can do it. And there are some spaces where we don't really know much about the regularization. So for example, a large class of important spaces in which we don't really know how to regularize these singular Hermitian metrics appropriately is compact scalar manifolds. That's a pretty significant class. But a class where we do know how to deal with them is Stein manifolds. So if you can embed it in CN, then you can use some trick from CN to deal with it. Yeah, so Ritzberg is exactly the Stein theorem for continuous ones, not smooth. And then de Maïve was the one who really opened this whole thing up. But still, his approximation techniques work well on Stein manifolds. And if you're not on a Stein manifold, then you lose a little bit of positivity. And these theorems are really, in algebraic geometry, the critical cases that you can't handle if you lose positivity. Anyway, what I want to say is for projective manifolds, a projective manifold is a submanifold of projective space. And if you remove a hyperplane from projective space, you have CN. And so if you remove the intersection of this hyperplane with your projective manifold, you get a closed submanifold of the remaining thing, which is CN. So that's Stein. It's even better. It's cut out by polynomials, so it's affine. And so on projective manifolds, because we're doing things like integration, we can throw away a set of measure 0, we can do the approximation as though we were on a Stein manifold. So these things are very useful. Singular metrics on projective manifolds are on Stein manifolds, but they do have some issues. And so I guess I'll just kind of use that as a stopping point. Next time I will state the L2 extension theorem with these singular Hermitian metrics and objects, and I'll give a proof using Bernstein's theorems. Thanks.