 So Yannir asked me to talk about this theorem that he put on the poster. So if you don't if you don't care about this theorem, it's his fault for putting it on the poster. Anyway, no seriously, I love this theorem. I use it all the time. He takes it so well. Yeah, so the theorem says that if a function phi is strictly plurisubharmonic, then the Hermitian bundle blah blah blah is strictly positive in the sense of knock on all. So I don't I didn't have a sense of how much people knew, how much background people knew. So you either know a lot of complex analysis or differential geometry or both or neither. I'm not gonna say a lot about the the complex analytic side, which is probably an error, but anyway, it's what I'm gonna do. No, that's that's my fault. But but basically what plurisubharmonic means it's if you're in a domain in Cn, then a function is plurisubharmonic if when you restrict it complex one-dimensional sub manifold or say every complex line, then it's subharmonic as a function of the parameter along that line. But we're never gonna really use that so I didn't feel too badly too bad about not getting into the details. But I because the bundles we're gonna be dealing with are in our case always in a dimensional I thought it's a good idea to be a bit precise about connections for vector bundles. So I'm gonna assume that everybody knows what a vector bundle is and then probably everybody knows what a connection is, but for me a connection is going to be a linear map. So X is going to be some manifold pretty soon. It's just going to be a complex manifold. And then I'm gonna have a vector bundle. I could, yeah, probably. I saw it was giving people a lot of trouble. So All right, it's a vector bundle which will soon be holomorphic and then a connection is a linear map from sections of the vector bundle E over X to well sections of say the cotend and bundle of X, tensor E and So it's linear, but it also has one other feature which it makes it like a derivative which is that it satisfies the Leibniz rule. So what that means is if you have a section S and a function F and you apply the connection to that then it's the derivative of the function tensored with S plus the function. Okay, and so what you see from this is if you have a difference of two connections, then it acts like an endomorphism on the fiber because this term doesn't depend on the connection. So the difference of any two connections is basically a linear operator that varies with the base and the other thing about vector bundles is that they're locally trivial and you always have the flat connection for locally trivial bundles which is just the exterior derivative. And so that lets you compute the connection locally from that by looking at the difference and the difference is usually called the connection matrix in that situation. Right, so well this world of connections is kind of well behaved with respect to the kinds of operations you can perform on the spaces. So for example, if you have the dual bundle E, then if E itself is equipped with a connection, then the dual of E has also a connection, nablet check, and that's defined by if you have a section of the dual, so a linear functional and it acts on using S there, but now F is a section, not a function. So this gives you a function on the base by this pairing and then if you apply the connection for the trivial case then you should get a kind of Leibniz rule acting here and then similarly for products, if you have E1 cross E2, then you have a connection that satisfies a Leibniz rule here, F1 times F2, it'll be nabla 1 F1 times F2 plus F1 times nabla 2 F2. So there's a kind of functoriality of the world of connections. But this is not so important for us except in one case, if you take the determinant of a vector bundle, so if it's a vector bundle of rank R, and you take this R skew symmetric product, you get a line bundle and then the connection for that determinant is the trace of the connection for this thing. Okay, maybe so I'm being very very brief and that's fine. So the next thing we want to define for a connection is it's curvature. So well, you want to try to apply a derivative twice in some sense. But you can do that. I mean that space over there, T star tensor E is also a vector bundle. But if you apply, so and it has a, it does have a connection, although you have to choose one for T star. And if you've made that choice, you can apply this new connection again, but that's not what we use to measure the curvature. So what you actually need is an extension of these connection derivatives as exterior derivatives, so meaning you take the skew symmetric part. So I mean, I think the basic idea is you can see in calculus, if you have a flat space and you want to compute some directional derivative along some direction and then directional derivative along another one, and then you want to try to switch the order. Well, if you did that, you'd end up exactly the same place. So you get the same derivative, but if the space is curved, this won't close up. And so a measure of how much it doesn't close up is the difference between the two possible orders of the partial derivatives. And so that's why we want to take the skew symmetric part. You look at composition of two derivatives minus the composition in the opposite order. So anyway, assuming that you know that, then the curvature is just a second derivative. And the miraculous, well, in the flat case, we saw that that's zero. The miraculous thing that happens for vector bundles is that this composition is no longer differentiates. So all the second derivatives appear in the symmetric portion. So anyway, this is some endomorphism with values in two forms. So you feed it to two directions and then you get an endomorphism of the vector bundle. So that's just a quick recap of what connections and curvature are. And they don't really tell you very much. And somehow the idea is to try to tie. So there are many, I guess they don't tell you very much because there are many, many connections for a given vector bundle. As I said, if you have a global endomorphism and you add it to some connection, you get another one. That's actually how you get all of them. But the hope is that once you have some additional symmetry on the space, you can isolate some special connections. So for example, if you have a way of measuring the angles between vectors in the vector bundle, and that's a metric, then you can ask for connections that are compatible with this metric. So what does that mean? That means if you have two sections, say F1 and F2, and so over each fiber, you look at the something like the angle between them and you try to differentiate this function, then you're supposed to get the derivative, well, this equality anyway. So that will narrow down the connections somewhat, but usually not enough. So there's still too many of these to really have some kind of canonical connection associated to the geometry. But there are some special cases. So maybe I try to talk about one of them now. So one interesting case where there's an additional symmetry happens when the vector bundle E is the cotangent bundle of your manifold X. And so then we know that NABLA maps sections of T star X to sections of the tensor square of T star X. I think these are, the value is a two tensor. And so any two tensor can be decomposed into a product, a symmetric two tensor, and then an anti-symmetric two tensor. Now, a section of this vector bundle is just a differential form. And so we have already an operation that assigns to a differential one form, a differential two form, which is the exterior derivative, that for example. So then we'll say that a connection is symmetric, lambda is symmetric, if not nabla, symmetric, if the kind of two form part. So any map into, any map into the text values in here will split as a delta superscript S for symmetric and delta nabla superscript lambda for anti-symmetric. And so you say it's symmetric if the anti-symmetric part is just this operator D. So in other words, there's no, none of the differentiation goes into the symmetric part. Now if, if our manifold is also a Riemannian manifold is Riemannian, say with metric G, then you have that situation. You have a metric and, well, this G is a metric for X, then there's an associated metric for the dual bundle. And so you also have the possibility of finding a metric compatible connection. And the theorem is that the intersection of the set of symmetric connections and metric compatible connections consist of exactly one connection. So that's called the Levy-Chiwida theorem. This is a unique symmetric connection compatible with the metric associated to the dual, the cotangent bundle. That's not that hard to prove once you phrase it this way. So that's a nice geometric theorem. It tells you that if you want to exploit the geometry of a Riemannian manifold, well, of differential one forms on a Riemannian manifold, by doing calculus on it, there is a special way to differentiate that will give you a lot of information about the geometry. So another interesting situation for us occurs when the manifold is complex and the vector bundle is holomorphic. So the metric induces an isomorphism to the dual space by the Ries representation theorem. And then you define the dual metric from that isomorphism. Ries representation theorem is an isometric theorem. So now suppose that X is a complex manifold and the vector bundle E is holomorphic. So that means that the transition functions are holomorphic functions. And so if you apply d bar to them, then you get zero locally. And so what that allows you to do is to define a d bar operator. So you can either start with the holomorphic vector bundle and then define d bar by picking a frame. Or you could say that you have a complex manifold and some operator that takes a section of this vector bundle to something that looks locally like a section tensored with a zero one form. And it satisfies d bar squared equals zero. So you can do it either way. But it's sort of, in some sense, this thing is given to you by the data one way or another. So for this discussion for a finite dimensional vector bundle, I'll just show you how to define it. But later we're going to define it in a different way because our vector bundle is going to be more canonically given. It won't be so easy to find frames. Okay, so if you have, so if E has finite rank R, and you take a frame, say E1 up to ER, then you can write any section F as FiEI. And there's a summation implied here. I'm going to forget it, so I might as well introduce the summation convention. If you see an upper index and a lower index that are the same, you sum. And so how do you define d bar of this? You just take d bar of the Fi and tensor with EI. So that's the d bar operator. It's really analogous to the exterior operator that we talked about here. But there is no round d operator unless your vector bundle is flat. Now you could also have a metric. Okay, so now we're in the complex category. So I really want this metric to be her mission, not symmetric. But anyway, given a metric for E, let's say call it H, you can, well, there are many metric compatible connections, but maybe I jumped the gun a little bit here. So if we take sort of a general connection, then it maps sections of E to sections of T star tensor E. But then this T star because of the complex structure also splits into two parts. There's the 1, 0 part, and then there's the 0, 1 part. So just like in the Levy-Chivita theorem, the connection also splits as a 1, 0 and a 0, 1 part. And so now this map here maps sections of E to sections of E, well, 0, 1 forms with values in E. But so does d bar. So we can impose a condition just like in the Levy-Chivita theorem of asking that the 0, 1 part is equal to d bar. So the definition of such connections, well, I'm going to call those connections. So by definition, this is called a complex connection. I don't know if there's a standard name or not. And so now there's an analogous theorem to the Levy-Chivita theorem. So given a holomorphic vector bundle E with a Hermitian metric H, there exists a unique H-compatible complex connection. And that's even easier to prove than that one depending on what you call easy or hard. And I suppose you try to combine these two cases. Well, so you've got a complex manifold, X. And then, of course, it has a complex structure, almost complex structure, which is a multiplication. It's an endomorphism of the tangent or cotangent bundle, whichever you prefer, that squares to minus the identity. And then if you've got a Riemannian metric G, you can ask for it to be invariant under this operation, meaning if you take two vectors and you apply this J to them and you measure the angle between them, it should be the same as the angle for the original two vectors. So such metrics are called Hermitian, even though this is not exactly, not yet exactly the same word as this Hermitian. These are actually metrics for the tangent or cotangent bundle, depending which one you're using. But if you, well, there's this little construction, if you take t star of X and then embed it into its complexification, then this J here is diagonalizable on this complexification. And so you can split it into its eigenvalues, but also because it's a real operator, it's invariant with respect to complex conjugation, and the eigenvalues are plus or minus the square root of minus one. So you decide which one is plus and which one is minus, and then you, the one zero space to be the eigenvectors of J with eigenvalue plus square root of minus one, and then the other eigenvalue is called the zero one space. And so essentially what you have is you have this, the real cotangent bundle here, then you complexify, and then when you do this splitting, it sort of lives here. And so if you take a vector here, think of it as a vector in there, and then project it onto this space, you see that you get an isomorphism. And so that means that these two spaces here are isomorphic, but this one will automatically be a holomorphic vector bundle. And so that gives this guy the structure of a holomorphic vector bundle. And so now you've got a Hermitian metric, which is also a Riemannian metric for this holomorphic. Well, you have to check that it's a Hermitian metric, but it is. And so now you've got a Hermitian metric, which is also a Riemannian metric for a vector bundle that's holomorphic. And so now you've got these two different connections from these two theorems. There's the Levy-Civita connection, and then there's the complex connection, this is, I didn't say it, but this is called the churn connection. So if you're given a manifold, you could ask, is there a situation in which these two connections are the same? And it's not always the case. Those are special manifolds. And so you just define them, they're called Kailer manifolds. X is Kailer if this Levy-Civita connection is equal to the churn connection. Kailer manifolds will be important for us a little bit. There's a bunch of different ways to know what a Kailer manifold is, or whether or not a given manifold is Kailer. So if you've got a complex manifold and you've got a Hermitian metric, Riemannian metric, then you can make a differential form. So you can check from this Hermitian symmetry connection that this is a two-form. And then the theorem is, so following our equivalent, 1x is Kailer, 2, this form is closed. So this is a symplectic formulation of Kailer manifolds. 3, x has an atlas of charts in which the metric g is Euclidean to second order. And then the fourth I'm not really going to use here is that if you take the Levy-Civita connection, so it's a connection for say Tx or T star x. And so j is an endomorphism of T star x, which is T star x tensor Tx. So there's an induced connection, and you ask that this j be flat with respect, or parallel with respect to this connection. So the Levy-Civita connection applied to j is zero. So these are four different ways of defining Kailer manifolds. Okay, so let's get back to holomorphic vector bundles, because we still need to know what positive in the sense of Nakano means. Okay, I'll just state this as a fact. Maybe I'll try to explain it in words, and then write it down. So if you look at this complex connection, well any connection for a holomorphic vector bundle splits as a 1, 0, and a 0, 1 part. And then if it's a complex connection, then the 0, 1 part is d bar. That means that if you use a connection matrix, for example, the connection matrix will have to consist of 1, 0 forms. It'll be a matrix of 1, 0 forms, because the 0, 1 part is just the exterior differentiation. And so this implies that the curvature is actually a 1, 1 form. And then it's also a real object because it's a curvature of a connection, so it's Hermitian in some sense, and we use that to try to define its positivity. Okay, so let's let omega, omega be the curvature-turn connection for some holomorphic Hermitian vector bundle, EH. So then this curvature form induces a quadratic form TX tensor E. And the way it's defined is if you've got two indecomposable vectors, then you pair them by, well as follows. So while these C and A are tangent vectors, so you can stick them into omega. And now you've got a 1, 1 endomorphism. And now you've got an endomorphism, sorry. And this should be the conjugate. And then you can act on F, and now you can take the inner product with respect to G. I guess there's a lot of letters that look like they should mean the same thing. But so that gives you, and then you extend this by linearity to this full vector bundle. And then you want to use this to say what it means to be positive definition. We say that this EH is positive in the sense of Griffith's, or Griffith's positive. If there exists some constant C such that when you act on these indecomposable vectors, there's one missing, missing, well okay, it's not really missing. So this should be greater than or equal to norm of C squared times C times the metric H, and this is the same metric of F. So this norm is computed with respect to any choice of Riemannian metric for the underlying manifold X. And we're not saying that this number C is any measure of precise positivity because if we change the metric we might have to change C. But because of that it's independent of the choice of this metric. Yes, this is sort of the least restrictive, but it's really the one we're going to focus on. Now the most, well it depends on what you mean, but the most restrictive notion of positivity is is called Nakano positivity. H is, so I should have said this before, but N is going to be the dimension, the complex dimension of the underlying manifold X. So if you take C1 up to CN, and then F1 up to FN, then CI, so this should be greater than or equal to some constant times the sum of CI CK, CK times H of FJ. And then there's a corresponding notion of negative. So you make C negative and change the inequality on both of them. So this one is clearly stronger than this one, so why not always use this one? Well, I mean there's that sort of a dumb question, I guess sometimes the first one holds and the second one doesn't, so you have no choice. But actually this first one is in some sense more natural. So for example if you have a Griffith's positive vector bundle, then its dual is automatically this negative. But that's not the case for Nakano positive. So it's not a great notion, but it's really useful for getting estimates on eigenvalues of Laplacians. Maybe it's worth pointing out that when the, there's sort of two situations in which Nakano and Griffith's positivity are exactly the same. The first one is when the base manifold is of dimension 1, and it's sort of trivially, look at it, it's trivially the same. But the other situation is when the rank of the vector bundle is 1, because then all the FIs are proportional point-wise. And so you get the same condition. Okay, so unfortunately these conditions as stated are a little bit hard to work with from the analytic point of view, particularly for the theorem that we want to prove. I mean that that thing is sort of hard to confirm directly. So it's maybe simplified a little bit to have a couple of propositions that give a criterion for positivity that's computed slightly differently. So this is a proposition. So EH is positive in the sense of Griffith's and only if. So we've got this E and this H. So now I'm going to look at the dual bundle. Dual bundle has a dual metric, which I'll write as H star. And then I'm going to take a section which is holomorphic of the dual bundle. And then I can compute the square norm of this section of the dual bundle and take the logarithm of this. And the proposition in this case says that this E is positive in the sense of Griffith's, if and only if this function here is plural sub-harmonic. So meaning if you study this function on some coordinate chart in X, you look at a complex line and you look at this function restricted to that complex line, that should be a sub-harmonic function. So you compute the Laplacian and the Euclidean coordinates on that line and it should be greater than or equal to zero. And if it's, sorry, yeah, so you could say this E star or you could say EH is Griffith's negative. Yeah, just if you just, you remove the stars and change the word positive to negative, then right, yes. But I'm going to be trying to check Griffith's positivity and the point is that it's a condition checked on the dual bundle. And Nakano positivity, as I said, doesn't follow from Nakano negativity of the dual, but there is a similar proposition to this one in the Nakano case and we are going to use it to prove Berenstin's theorem. In fact, we're going to have two versions of Berenstin's theorem. One, when the base is one dimensional and then I'm going to use this criterion and then one for a general base and I'm going to use the next criterion that I write down for you. Okay, so, okay, to do that I need to do a little bit of setup. No, no, but this is a local, local condition. Sorry, so all these things, these conditions up here are point wise and then this condition you confirm locally and you check that it's independent of the choice of holomorphic sections, as long as you span. Yeah, if you like, it's a good point. I just don't have time to go into all the details, but I really should. I want to point out one other thing about this particular definition. So if you are a fan of complex analysis, then you know that things like plurisubharmonic functions or even subharmonic functions, although they originally started out as being smooth functions, they pretty quickly got, the definitions got generalized in view of this elliptic regularity that Rafe mentioned earlier, to much weaker conditions. In fact, if this function is just locally integrable then you can define dd bar of this function as a current and still talk about what, that thing being positive. And so this condition here would only require that the logarithm of the square norm with respect to the metric doesn't have to be a smooth function, just should be locally integrable. And then you can define dd bar of this. So if this e is actually a line bundle, then you get an object called a singular Hermitian metric. And then it has a curvature current, which is a one-one current. But if it's a vector bundle, this thing will not be capturing the whole curvature. And in fact, you can't. You can't really define the curvature of a singular, any kind of reasonable class of singular metrics for vector bundles. But strangely, you can say what, what it means for a, to be a vector bundle with a metric whose curvature is positive in the sense of griffits. But there is no curvature there. It's just a single statement. There's positive curvature without being curvature. And that, that has many applications I won't so much get into here. It's particularly important in algebraic geometry. Okay. Right. So I want to do this, the Nakano version of this. Okay. So again, all these things are local. And so let me just look at my holomorphic vector bundle, e, sitting over, say, the unit ball in Cn. And then on this unit ball, I'm going to let theta jk, or maybe jk bar, these will be n minus one, n minus one forms. So what you do is you take the local coordinates t on this b. And this will be all the dt i, except for i equal to j. And now the dt bar i's, except for i equal to k bar, wedge together in such a way that theta jk bar wedge dt j wedge dt bar k is just the, the big measure in the unit ball. Okay. And now the next thing you do is you take an n-tuple u. I'm just defining something and then I'm going to state a thing that I'm not going to prove. Okay. And now what I want from these is that if I apply the churn connection to uj, say, I'm just going to check Nakano positivity at the origin, say, I want these guys to vanish there. So it's not difficult to show that given any n-tuple of non-zero vectors, you can find local sections that give you those non-zero vectors at the origin. And then their derivative is zero. We'll see that in action later on. And then we, we make this test form, t, t for test. It's a quadratic form, but it's an n minus one, n minus one form. So this is going to be negative h ui uj bar and then theta raise these indices. This is jk bar, jk bar. So now I'm using my, oops, ij bar. I'm using my summation convention. So this is a, and two sums over the indices from one to n. So this is an n minus one, n minus one form. Okay. And then the statement is that, so EH is positive, Nakano positive, if and only if you compute square root of minus one dd bar of this t u at the origin, then it's exactly the, the big volume form, I guess, at, at the origin. Chalk's getting shorter. Okay. So this doesn't have the same advantage as this one because it's quite complicated to see what sorts of conditions you need from the metric in order for this to be a well-defined current. I'm still thinking about whether or not you could, you could do something with this to define a notion of singular occurrence for Nakano positive situation. But that, that's an open question on how to define positive, Nakano positively curved vector bundles. Again, it is known that you can't define the curvature in any reasonable sense of currents. Okay. Yes. Wait a second. So I want to be a little bit careful how I answer that because it's a little bit confusing. So there's something called ample in the sense of hard charm. Is that what you mean? Or are you really talking about the curvature of a metric for the tautological? Yeah. So the curvature of the metric, yeah. So if that metric is, has positive curvature, that's the same as Griffith's positivity. But ample in the sense of hard charm talks about some sections, whether or not the sections embed something or other. And that, that's different. Yeah. Yeah. So the answer is yes. But it is an open question whether or not this other notion ample in the sense of hard charm is the same as positive in the sense of Griffith's. So your question, the answer to your question is yes. Another question. I mean, in some sense, but it's not, I mean, this quadratic form is defined on some associated bundle, the squiggly bracket quadratic form. I guess that's sort of related to that. But whether or not it's any kind of tautological bundle that I don't think, but I don't know maybe. I don't know if it buys you a lot, but in any case, yeah, I don't know. I don't know for sure if there's some other way of expressing it. But this way is kind of cute and we will use this computation. Okay. So I'd like to finish by just stating a theorem that I'm going to use later and hope maybe next, I'll start next lecture by giving a sketch of the proof of this theorem, but it's in the notes that I circulated. So really, this will seem a little bit disconnected from everything that I said till now, but it's not really. It involves a lot of the things I mentioned. For example, Kaler manifolds. So this is the theorem. And it's in one form due to Hormander. And sometimes it's just called Hormander's theorem, but this version was probably proved by Skoda and written down by Demae. Anyway, that's what I was told by some people in the field that knew. Not that Demae has proved a lot of theorems, but this one I think he just wrote down. Okay. So here, L will be holomorphic line bundle. So you can do this for a vector bundle, but I'm going to do it for a line bundle because that's the only case we'll use. So on Kaler manifold XG. So this X is a Kaler manifold. G is a Kaler metric, but this X is also a complete Kaler manifold, but not necessarily with the metric G. Okay. So you have some complete Kaler metric, but this isn't it. Maybe. And then this line bundle has a metric which, so the usual notation, the notation I use, I don't know how usual it is, but to me that's a notation for a metric for L. And it's not, usually this notation is used locally, but so let me tell you what it means. If you have a vector in L and you have a metric for L, then you can compute the square norm of that vector. Now if you have a non-vanishing section of L, then L is trivial, at least in the neighborhood of this non-vanishing point. And so you can compute the norm of that thing, and then any other vector is a functional multiple of that one, and its norm is the square norm times the norm of the initial vector. This is the norm of the initial vector in local notation. But anyway, the point is that the curvature is just dd bar of phi. So you can think of this globally or locally. And now we need a curvature assumption. So I assume that you look at dd bar phi plus the Ricci curvature, so the trace of the curvature of the Levy-Civita connection for G or the Kaler connection, you assume that this is greater than or equal to some form theta, which is a non-negative 1, 1. Then for any d bar closed, L-valued section, or 0, 1 form, alpha on X. There exists a section u of L such that d bar u equals alpha, and you compute the L2 norm of this u with respect to the metric for L and the volume form from the metric G. This will be less than or equal to the integral of, and now you compute the square norm of alpha. So alpha is an L-valued 0, 1 form, so you need to use both the metric for the manifold and the metric for the line bundle. That's not what I meant, sorry. So instead of using a metric for the manifold, I actually need to use this Hermitian 1, 1 form, and this way I get rid of any constant. So the constant here is really 1, so this is provided, the right-hand side is finite. So this is a really useful theorem in one and several complex variables for constructing holomorphic sections or holomorphic functions, in fact. You have a tremendous amount of control on them if your weight or your metric here has sufficiently positive curvature. I will use this theorem to estimate the some second fundamental form of some vector sub-bundle, of a vector bundle, and in that way prove Bernstein's theorem about the Nakano positivity. So I'll give more explanations of that next time. Sorry for going over time.