 So this is a second talk on the chow ring of a variety and will be about churn classes. So we recall in the first lecture for any variety x, or non-singular variety x, we defined a chow ring A of x, which was graded, where A i of x is the ring of, is the rational equivalence classes of cycles of co-dimension i. And what we're going to do now is discuss how given a vector bundle on x, we can get various elements of the chow ring called churn classes C i. So normally in differential geometry, we have churn classes taking values in the cohomology of a complex manifold rather than its chow ring. It turns out the chow ring is kind of closely related to the cohomology ring of a variety x. So here h star stands for the cohomology ring. So what we have that there is a homomorphism from the chow ring to cohomology, which can be defined as follows. If we've got an element of the chow ring of a co-dimension i cycle, then any cycle gives you an element of the homology group. And by Poincaré duality, this sort of corresponds to an element of the co-homology group. I've never been quite sure why you use co-homology rather than homology. I mean, there are some cases when it's slightly technically easier. But my impression is it's using co-homology rather than homology is as much a historical accident as anything else. Anyway, one obvious question is, is this map an isomorphism? The answer is it's not an isomorphism. In fact, it fails quite badly in general. First of all, it's not injective in general. And the reason for this is that a i of x is huge in general. For example, if we take x to be an elliptic curve, then a 1 of x is just x, which could be uncountable. For example, if you're working over complex numbers. However, the second co-homology group of x with integer coefficients is just z. So the ring of churn classes can often be vastly bigger than the co-homology ring. You can also ask, is it surjective? Well, it's obviously not surjective onto h2i plus 1 of x if this is non-empty because there's nothing in the chow ring that can map to odd-dimensional co-homology. You can ask, is the map from a i of x to h2i of x onto... Well, in general, it's not. So more precisely, we should ask, what is the image? And this turns out to be a very difficult and subtle question. So hodge found one obstruction. So there's the hodge structure on the co-homology of x with real coefficients, which writes it as a sum of these groups hpq of x. And what hodge showed is that the image of any cycle is contained in... Sorry, that should be a direct sum, is contained in the group hii of x. So these are also in the integral homology. So you can ask, is the image of a i all of the group hii of x intersected with the homology of x with integer coefficients? And the answer is no. A counter-example of this is found by a tier and here it's a brook who found a torsion element of the co-homology that isn't in the image of a i. And there's a slightly modified version of this conjecture which says that is the map from a i tensed with the rationals to hii of x intersected with h2i of x with q? So we'll ask, is this onto? And this is the famous hodge conjecture, which is a really hard unsolved problem. Incidentally, it's not real clear that hodge ever conjectured this. I have here where hodge is writing about this. And let me just focus it a little bit. He's certainly discussing this problem and all he does is he says beyond a few cases when there's a positive solution, the problem is an unsolved one. And he doesn't actually conjecture it's true. So it should really be called hodge's question rather than hodge's conjecture. Anyway, so the point is you should think of the cowering as being sort of analogous to co-homology. So I'm going to very quickly review characteristic classes. So in algebraic topology, characteristic classes will take a vector bundle over x to co-homology. And in a little while we're going to do an analog of this way we replace co-homology with the cowering. But let's just do co-homology for the moment. And let's look at a simple example. Let's take a line bundle L over and let's take a section f of L. So at each point of x we're choosing an element in the fiber of the line bundle over x. And then we can look at the zeros of f which will be some sort of subset of x. And if we choose f nicely, this will be a reasonably nice subset of x. So it'll be usually be a co-dimension one cycle of x. And this will give us an element of hn-1x with coefficients in z modulo 2z where n is the dimension of x. And as I said people usually switch homology and co-homology. So using Poincare duality this gives us an element of the first co-homology of x with coefficients in z modulo 2z. We need z modulo 2z because we haven't bothered to orient things at all yet. And this actually gives you something called the first stifle Whitney class of the line bundle. And let's just see an example of this. So suppose we take x to be the circle s1. Then there are two obvious line bundles. First of all we can take s1 times the reels which just looks like a cylinder. Or we can take the line bundle to be a sort of mobius strip on the sided surface. So both of these are mapping onto x which is s1. And each of these two line bundles has zero section which looks something like this. And we can take some sort of non-zero section of the line bundle. So I'm going to take some non-zero section that might look something like this. Just take some sort of function. And it will have some zeros and you can ask what the zeros look like. So over the circle the zeros of the section might look something like this. And if you take a section of the mobius strip over s1 zeros might look something like this. And what you notice is that here there are always an even number of zeros. And here the number of zeros is always odd. I'm assuming you've chosen a nice generic function so there aren't an infinite number of zeros or anything like that. The zeros should all be single zeros so I'm not allowed to have double zeros or anything like that. And this gives us an element of the first cohomology group of s1 with z modulo 2z. Which is of course just z modulo 2z. And here it's the zero element and here it's always the non-zero element. So we can distinguish these two line bundles over s1 by counting the number of zeros of a generic section of the line bundle. And that's always either even or odd. And the whole theory of characteristic classes is really just a generalization of this fundamental example. If you've got a line bundle or a vector bundle you can look at zeros of sections of it and these will give you homology or cohomology classes. So now let's look at churn classes. So previously I was just discussing real vector bundles over a topological space. Now I'm going to look at complex vector bundles. So the fibre is now a complex vector space. And these are going to take values either in the chow ring if you're an algebraic geometre or the cohomology ring if you're doing differential geometry. And as I said these two are pretty similar so it doesn't really matter which you use. So let's take a complex line bundle L over x which is either a non-singular algebraic variety or a manifold or something. And now it's before we take a section of L, f and then we look at the zeros of f which will be a subset of x and it will have real co-dimension equal to 2 in general or complex co-dimension equal to 1. Now if there's real co-dimension 2 and you're a differential geometre this gives you an element of the 2n-2 homology group which by Poincare duality you can think of as being more or less the second co-homology group of x with integer coefficients. If you're an algebraic geometre then this is giving you an element of a1 of x the chow group of rational equivalence classes of cycles of co-dimension 1. So we get the first churn class given by the cycle of zeros of a section. So this gives us the first churn class which is denoted usually by c sub 1 of x. Well I've actually been cheating a little bit because there's a slight problem. If you're doing differential geometry there's no problem but an algebraic geometry we have the following problem. What if L has no non-zero sections? Well this happens quite often for example we might take L to be the dual of the bundle O1 over P1 and as we saw earlier there are no non-zero sections of this bundle. So how do we define churn classes if there aren't any sections? Well that's not too difficult because you can check that churn classes have the property that if you tense a two line bundles this is just c1 of L plus c1 of M. Now if you've got any line bundle L possibly with no sections then you can tensor it with some high power of cells bundle. So we take N large and positive and this will have lots and lots of sections if N is large enough. And now we can just define c1 of L to be c1 of L tense with ON minus c1 of O of N. And the additivity property implies that this is well defined and gives you a perfectly good first churn class of any line, any complex line bundle. And by the way this is by no means the only way to define churn classes. There are many other ways of defining them. I'll just give one example. If you're fond of sheath theory and cohomology then you can look at the following collection of sheaves. So here this is the exponential map so this doesn't work in algebraic geometry. This is only something you can do if you're doing complex analytic geometry. Then the long exact sequence of cohomology goes from h1x with coefficients in the sheath O star to h2 of x with coefficients in z. And this more or less classifies line bundles and this is of course just the second cohomology class. And this map from here to here is more or less just the first churn class. Well, so that explains what the first churn class is. What about higher churn classes? So what happens if v is a bundle of rank greater than 1? What we want to do is to define some churn classes for it. And as usual there are many different ways of doing this. The way I'm going to describe was introduced by Grotendick. And what you do is you sort of start from v and you derive a line bundle from it as follows. What we do is suppose v is a line bundle over x. Then what we do is we take p of v to be the projected bundle of x. So we've got a map from p of v mapping to x and the fibre at x is just a projective space p of dimension r minus 1 where r is the rank of v. And all you do is you take the fibre of v over x which is a vector space and take the associated projective space which has dimension 1 less. And p of v has a line bundle O of 1 which I'm not going to try and define because I always get it confused with O of minus 1. You either take lines in the vector space or elements of the dual and one of them gives you O of 1 and the other gives you O of minus 1. So we've got a line bundle on p of v and this gives us a first churn class C1 of O of 1. Well the trouble is that's not in the chowering of x. It's in the chowering of the projective bundle of v. So in order to get elements of the chowering of x we've got to show how p of v is related to the chowering of x. Well normally the chowering of x is probably hopelessly complicated to describe and the chowering of p of v is also hopelessly complicated to describe but it turns out that we can describe p of v in terms of the chowering of x. So to simplify notation I'm going to take this element here and call it xi and we can now describe what the chowering of p of x looks like. So the chowering of p of v is a module over the chowering of x and in fact it's a free module of rank r where r is the rank of the vector bundle and it has a basis consisting of the elements 1, xi, xi squared and so on up to xi to the r minus 1. So the reason why this is a free module of rank r is the argument is kind of similar to the proof that the chowering of a r minus 1 dimensional vector space is free over z of rank r so this would be what you get if x was a point and a relative version of that just says that the chowering of a projected bundle is a free module over the chowering of the underlying space. Well now we can look at the element xi to the r and this will be a linear combination of these basis elements here so we can write xi to the r minus c1 xi to the r minus 1 plus c2 xi to the r minus 2 and so on plus or minus cr equals 0 for some ci in ai of x. By the way there are some sign conventions about churn classes so these signs here may vary a little bit from author to author. Anyway these values here are of course now the churn classes of the vector bundle v. By the way the churn class cr has a very simple geometric interpretation. This is just represented by the cycle of a section of v. So just as for line bundles you can sometimes take a section of a vector bundle and the zeros of that section will generally have co-dimension r and will be a cycle representing the churn class cr. So that particular churn class is easier to describe. So that's a sort of sketch of how you define churn classes of a vector bundle. In the rest of this talk what I'll try and do is very briefly sketch an application of churn classes to the Riemann-Roch theorem in higher dimensions. So the Riemann-Roch problem is to compute the dimension of the global sections of a vector bundle or a sheaf or whatever. And usually there are no nice formulas for this. Instead what we can get a nice formulas for h0 of the sheaf minus the dimension of h1 of the sheaf plus the dimension of h2 and so on. So alternating sums of co-homology groups are much easier to handle than individual co-homology groups. So h0 of s is of course just the same as gamma of s. And if you're lucky then you might be able to find a vanishing theorem that makes all these zero in which case you can calculate the dimension of h0 of s. But in general we can only calculate the alternating sum of the dimensions which I'll call chi of s where this stands for the Euler characteristic. So the Riemann-Roch theorem for curves says that this is equal to the degree of s plus 1 minus g for curves. And the problem is to generalise this to high dimensional varieties. And this was done in general by Hirtzer-Brook. The case for surfaces and free folds had already been done before Hirtzer-Brook came along but Hirtzer-Brook found the way to do it in all dimensions. And Hirtzer-Brook formula says that dimension of chi of s is equal to the degree of the churn character of s that I'll explain in a moment times the Todd class of x and then you take the nth degree and part of this. Now what I'm going to do is explain what all the various bits of this formula mean and then we'll show that this really is the same as the usual Riemann-Roch theorem for curves. So first of all this is called the churn character and is defined in terms of churn classes in a way I'll explain in a bit and is in the churn ring of x. This is called, this is the Todd class of the tangent bundle of x. And again I'll explain what Todd classes are in a moment and this is again an element of A of x. So the product is an element of the chow ring. This means you take the degree n piece, so this is the piece in A n of x and elements of A n x are represented by points and so you can count the number of points and the number of points is just the degrees. So this is the number of points in the zero cycle that you get by multiplying these two together. So now what I've got to do is explain what the churn character and what the Todd class are. So these are both defined in terms of the churn classes of various bundles. So what you can do is you can form the churn classes into a churn polynomial just by taking the c i's to be their coefficients. And I'm going to formally write this as a product of 1 plus A i t. So the A i are more or less minus the inverses of the roots of this polynomial, ignoring the fact that its coefficients are a bit funny. Now the churn character of S is just given by the sum of e to the A i, where the A i are these numbers here. Now this should be considered as a formal expression and if you expand it out formally you find that this is a polynomial in the c i. In fact, it's easy to work out the first few terms. The degree zero term is the rank of S, the degree one term is just c one, the degree two term is c one squared minus c two. And that's divided by two and so on. I may wonder why you define this rather funny churn character, but the reason is the churn character is a ring homomorphism from vector bundles to the, well actually I guess you should really take the chow ring and tense it with the rationals because we've had to define a few rationals. Now you make vector bundles into a ring by using addition and tensor product as the addition, direct sum and tensor product as the sum and product of the rings and you have to fiddle about to make sure it's actually a ring but we won't worry about this. So in other words the churn character of S plus T is the churn character of S plus the churn character of T and the churn character of S tensed with T is the churn character of S tensed with the churn character of T, sort of times the churn character of T. So that's the churn character, it's a natural ring homomorphism. The Todd class of a vector bundle which will usually be the tangent bundle is given by the sum of a i over one minus e to the minus a i and again this is a formal polynomial in the churn classes. So the first two are one plus c1 over two plus c1 squared plus c2 over 12 and if you were looking at the lecture, some of the lectures I give on surfaces, you may notice this as being part of Nourta's formula for complex surfaces. You may wonder where this funny expression here comes from. Well, the Hitzl-Bruck-Riemann-Roch theorem and the Grothendrick-Riemann-Roch theorem are basically proved by brute force. You basically calculate both sides in all possible cases and check they're the same and in the calculations for projective space, this function appears because the function one over one minus e to the minus c has the funny property that if you raise it to the power of N and multiply it by dz, this has residue one at nought for N equals one, two, three and so on. So this is a fairly easy exercise in complex analysis to check that this is essentially the only function with the property that's residue is one for these numbers here and zero for other values of N. So that's where this funny expression comes from for the Todd class. So now what we can do is we can just check that Hitzl-Bruck's formula really is the usual Riemann-Roch theorem for curves. So let's just unravel it. We'll take x to be a curve and s to be a line bundle and just work out everything. So the churn classes of s, well the zeroth churn class is one and the first churn class is just d where this is a divisor. So the line bundle is the line bundle associated with the divisor. You remember line bundles and divisors on curves more or less correspond to each other as long as you take divisors to linear equivalents. So the churn character of s is just one plus d. If you take t to be the tangent bundle, then we need to work out the Todd class. Well c nought of t is just one and c one of t is now minus the canonical divisor. The cotangent bundle corresponds to the canonical divisor so the tangent bundle corresponds to minus this. And if you work out the Todd class using the funny formulas I wrote down earlier, you find it's one minus k over two. So Hitzl-Bruck's formula in this case says the dimension of chi of the line bundle is the degree of the one plus d, one minus k over two. Except you should take the degree one piece of this. So it's just the degree of d minus k over two because that's the degree one part of it. Which is the degree of d minus, now the degree of k is just two g minus two over two. So we get degree of d plus one minus g which is the usual Riemann-Roch formula. So now just very briefly discuss growth index version of the Riemann-Roch formula which again uses churn classes. So growth index version says that if you've got a map between varieties x and y, suppose you've got a sheaf on x, then you can push it forward to a sheaf on y. And of course as we've seen it's a good idea to take an alternating sum of derived functions of this. So we take minus R one f star s plus R two f star of s. So this is the relative analog of taking a sort of Euler characteristic and alternating sum of co homology groups. And we can also take the churn character of s which gives us something in A of x. And we can take the churn character of y, sorry the churn character of all this rubbish here. And we get something in A of y. And there's a push forward from cycles on x to cycles of y which we can call f star. And we can ask does this diagram commute? And the answer is no, it doesn't commute. So if we go around this way we get f star of the churn class of s. And if we go around this way we get the churn class of R f star of s. And these are not the same. They differ by... One of them is equal to the other times a certain Todd class of a sort of relative tangent sheaf. Now hits a Brooks version of the Riemann rock formula you just take y to be a point. And the relative tangent sheaf is then just the tangent sheaf of x. And this turns into the hits of Brooks Riemann rock formula. OK, that'll be all about churn classes for the moment.