 This lecture is part of an online algebraic geometry course covering schemes and will be about the cotangent chief of a scheme. So the problem is as follows. Given a space X, which might be a scheme. Well, in differential geometry, we can construct the tangent space, or other tangent bundle Tx of X mapping to X where the points of the tangent bundle of X are roughly speaking tangent vectors at points of X. And we want to do an analog of this in algebraic geometry. Well, there are two analogs of the tangent bundle, we could either construct the tangent bundle as a scheme mapping to X, which we can do, or we can construct a tangent sheath. Over X, where the sections of the sheath local sections are the same as tangent vector fields. And we'll actually construct the sheath rather than the bundle, although you, you can construct the bundle from the sheath without too much difficulty. Well, it turns out that instead of constructing the tangent sheath, it's easier to construct the cotangent sheath, which is essentially just a sort of dual of the tangent sheath, at least if X is non singular. And so here the sections are just one forms of the form. So if F is a smooth function of differential manifold, we know that there's you can take a one form D F corresponding to it, which is a one form of the manifold, we want to do the same in algebraic geometry. Well, what you notice in differential geometry that this map D is an order one differential operator, and it's also normalized. And we discussed last lecture, how to construct normalized order one differential operator. So let's just recall very quickly what we did. So here we, if B is an A algebra, we defined the concept of nth order differential operators on B, and we found a universal nth order differential operators follow. So suppose we set I to be the kernel of the map from B times over AB to B, just taking B tends to C to BC. Then we found there's a map from B to I to the nought over I to the n plus one taking B to one tensor B. And this is a universal order n differential operator. So I to the nought is of course just make it clear it's just B tensor over AB. Well, that's not quite what we want. What we want is the map from B to I to the one over I to the n plus one, which is a universal normalized order n operator. So normalized for differential operator just means D of one equals zero. And this map here takes B to B times one minus one times B. What we want? Well, the operator D, little D we're trying to construct has order one, which suggests we should just take this map with n equals one here. So we take the map I one over I two, which has a map from B to it taking beta, B tensor one minus one tensor B, and we're going to call this map D. And remember this was the, we defined this to be the module of differential forms of B over A. And we're going to take this to be the cotangent sheaf at least for affine schemes and then we'll glue these together to get a cotangent sheaf arbitrary schemes. So for instance, let's just check this gives a reasonable result. If B is a polynomial algebra in n variables. Then we saw that omega B over A is a free B module with basis DX one up to DXN. So it's elements are just things the form sum of fi times DXI and these do indeed look like one forms. Well, they've got polynomial coefficients rather than smooth coefficients but that's close enough. And the map D from B to omega B over A just takes an element F to sum of delta fi over delta xi times DXI just the same as in differential geometry. So what about general schemes X that aren't affine so that works fine for affine schemes. What about none affine schemes. X. Well, method one we could cover X by open affine's UI and glue things together. So we could construct a module of one forms over each UI and just check that glue together. We're not actually going to do that because it's a minor headache trying to work out how to glue things together. Instead, we're going to do it in one step. So, so first of all, let's let's let's first look at the affine case where X is the spectrum of some ring B and why is the spectrum of some ring a. Then we've got these maps from B tensor over AB to be a goes like that. And then there's this map taking this to be which just takes B times C to BC. If we convert this into schemes, what we discover is we've got scheme X times over Y X mapping to X mapping to X mapping to Y. And this map here is just the diagonal morphism mapping X the diagonal of X times over Y X. And now if we think about what the ideal is. So the ideal I is the kernel of this map here. We see the corresponding sheaf. So what we do is we get a sheaf of ideals I on X times over Y X, which is just the sheaf of ideals to find the closed sub scheme X. So I over I squared is now sheaf on X times X. And if we pull it back to X by applying delta star of I over I squared. We now get a sheaf on X, which corresponds to the sheaf omega B over a on B. So we define the cotangent sheaf would call this omega X over Y on X to be pulled back by the diagonal of I over I squared where I is is the is the sheaf of ideals of the diagonal embedding of X in X times over Y X. And you can check that this is this is omega X over Y is a quasi coherent. X. Oh, X module. And it's coherent if various nice conditions on X are satisfied. I think if X and Vex is not hearing a finite type over why and some then this is coherent. Let's see some examples of it. Well, for affine rings we've already done several examples so let's do the simplest non affine example and just take X to be P projective space over field and why to be a point just the spectrum of the field. And we can ask what is omega X over why. Well, instead of calculating if I'm going to figure it out by doing some inspired guessing. First of all, it should be the cotangent space. We have P one of X and P one is one dimensional so it should be an invertible sheaf because P one of X is non singular so expect it's tangent and cotangent bundles to be to be to be vector bundles which should correspond to locally free sheaves. And we know the locally free sheaves on P one local. So we know the invertible sheaves on P one, they're just to the form on for N in Z. So the cotangent sheaf is going to be isomorphic to on for some value of N, and we want to know what is N. Well, we can figure it out as follows so sections of O N have N zeros on P one. So we can figure out what N is by figuring out how many zeros a differential form has on P one so so if we choose coordinates XY for P one, we can take an a one in P one with coordinates. X and just look at the differential form DX on a one and try and figure out what it does at infinity. And so we can figure out how many zeros it we need to work out how many zeros DX has we notice DX is none zero on affine space. So let's work out what happens at infinity so we put Y equals one over X, and we find D Y equals minus one over X squared DX DX is equal to minus one over Y squared D Y. So we see there is an order to pole at infinity. So the sheaf of one forms on P one is isomorphic to O of minus two because there are minus two zeros as there's a double pole of any one form. Actually, we may as well take a quick look at the tangent bundle of the tangent sheaf, which can be defined as the dual of the cotangent sheaf, which is the dual of O of minus two, which is O of two. So so tangent vector fields just correspond to sections of O of two. So we can figure out what the dimension of the space of global tangent vector fields is is the dimension is just three because O of two has a three dimensional space of sections. In fact, if we like, we can write down what these tangent vector fields are, they're just D by DX, X D by DX and X squared D by DX. And you may take a look at then say why do you stop at X squared by D by DX, what on earth is wrong with X cubed D by DX. Well, it turns out that X cubed D by DX has a pole at infinity. So if we look at infinity, as before we put Y equals one over X and take a look at what happens when Y equals zero. Well, D by DX is equal to D by D Y times minus one over X squared. So D by DX becomes minus Y squared D by D Y and this becomes minus Y D by D Y and this becomes D by D Y and this becomes minus one over Y D by D Y. So, so this one here is not acceptable because it's got a pole at infinity. We see that there are, there's a three dimensional space of holomorphic vector fields on on the projective line. And these form a basis for the Lie algebra of the automorphism group of the projective line. And the automorphism group of the projective line is just the projective general linear group PGL over two of K, which has dimension three. So it's Lie algebra is indeed three dimensional. So that was a sort of calculation of the cotangent bundle of the projective line where we sort of cheated a bit by doing some bit of wild guessing. Now let's work out the cotangent bundle of n dimensional projective space. So what we get is an exact sequence nought goes to omega of P to the n goes to O of minus one to the n plus one goes to O of zero goes to nought. And we have a slightly more honest calculation of the cotangent bundle of P one because for n equals one, this just corresponds to the sequence nought goes to O of minus two, just O of minus one plus O of minus one, just zero, which we mentioned a few times when we were when we were discussing. And that achieves on the projective line. And so, how do we prove this? Well, the easy way is to look at that graded modules over K x nought up to x n. So we have the following sequence of graded modules. We have a new piece of paper because this is going to take a very amount of space. And we get nought goes to things of form sum of f i d x i, the sum of f i x i zero. And this goes to a free module with basis d x one up to d x n, so d x nought up to d x n. And this is just omega of K x nought up to x n. And then this maps to K x nought up to x n. This maps to K and this maps to zero. And here just takes d x i to x i, so it doesn't have image one, which is why we get a K there. So, we've got this sequence of graded modules over the polynomial ring in n plus one variables. You remember that if we've got a graded module over the polynomial ring, this gives us a family of sheaves over projective space. This family of sheaves is going to look like this. We get omega of P n, as we will show shortly, and a free module with basis d x nought up to d x n. Well, we have x i has degree one. This is a sum of n plus one copies of something that is either O one or O minus one. And it's really hard to remember which I think it's O minus one. And here we've got a free module on one generator, so we just get O of zero. And here we get K. And you remember finite dimensional graded modules, when you turn them into sheaves, they just become zero. So, we've got the slightly odd phenomenon. This map of graded modules is not surjective, but the corresponding map of sheaves is surjective. So, that correspondence and that correspondence are easy to see. And what we've got to do to finish off is to show that the sheaf omega of P to the n corresponds to this graded module over the polynomial ring. And to see this, we remember that P n is covered by open sets. U i where x i is not zero, roughly speaking. And the coordinate ring of U i can be taken to be K x nought over x i up to x n over x i. The one forms over this, so omega of this ring is just a free module, spanned by D x nought over x i up to D x n over x i. So there are only n of these because of course D x i over x i is equal to zero. On the other hand, if we look at this graded module here, the sheaf corresponding to it. So, if we take the graded module of things, sum of f i D x i with sum of f i x i zero. The corresponding, to get the corresponding sheaf over each open set U i, you take the module of degree zero elements. Of the module you get by inverting x i, so U i is where x i is none zero. So we're just looking at the elements, sum of f i D x i such that sum of f i x i zero. And the degree of f i D x i is equal to zero. And f i is in K x nought up to x n and then we allow x nought to be inverted. And now you can check that if you ignore for the moment the condition about the degrees and just ask for the things satisfying this condition. You can see these are spanned by as a module by x i D x sorry x j D x i minus x i D x j. You can see that this now satisfies the condition that sum of f i x i was not. And conversely, if this satisfies this condition you can easily check that it's a linear combination of these. Now we make this degree zero. And we see that we're just getting the elements x j D x i minus x i D x j and now we can make it degree zero by dividing by the square of D x i. And well if we go back that omega was the free module satisfied by these things here will D x j over x i is just equal to D x j times x i minus x j times D x i over x i squared. So, in other words, if we take the. Sheaf correspondent this graded module. It's canonically, we can canonically identify it with the sheaf of one forms over each of these open sets you I and therefore over the whole of projective space. So that identifies the cotangent space of projective space. Next lecture we'll be doing some more examples of the cotangent sheaf.