 This lecture is part of an online algebraic geometry course on schemes and will be about invertible sheaves on one-dimensional projective space over a field k. So one-dimensional projective space, the projective line over a field is in algebraic geometry, is something like the hydrogen atom in quantum mechanics. It's a very simple objects, which is complicated enough to illicit, to exhibit some non-trivial phenomena and yet so easy that you can calculate everything explicitly. So we're going to do the sort of simplest non-trivial examples of sheaves on the simplest non-affine space. So, so far we're looking at quasi-coherent sheaves on on affine schemes and these are essentially the same as modules over rings, except they're modules over rings talked about in a sort of funny geometric language, but nothing essentially new is turning up. So this is the first non-affine scheme on which we're going to look at quasi-coherent sheaves. And the problem we want to solve is find all invertible sheaves and invertible sheaves are more or less the same as line bundles, unless you're being picky about line bundles being fiber-spaces or whatever. And you remember if you've got two invertible sheaves, you can take their tents of product and get another invertible sheaves. They form a group. So what we really want to do is to find the group, the Picard group of P1 over a field. And let's start by just sort of guessing the answer. There are some obvious line bundles over P1. There's one obvious line bundle and you can get several other obvious ones from it. So P, well, why do P1? Let's do Pn is just the set of lines in Kn plus 1. So there's a map from the points of Pn to one-dimensional vector spaces, which takes any point x naught up to xn to the line of all multiples of x naught lambda x1 up to lambda xn for lambda in k. In other words, given a point of projective space, there's a one-dimensional vector space associated with it. So this is informally a line bundle. So you remember I said a line bundle is roughly something that assigns a one-dimensional vector space to every point of your manifold. And if this line bundle is called L, we can do various things to L. We can take L tensor L, which is L squared. And similarly, we can take Ln for n greater than or equal to zero. And we can also take the dual of L. Just by taking the dual of each vector space. So we can think of this as being L for minus one. So we get a line bundle Ln for n in z. And what we will do is we will show that these are all distinct, which isn't immediately obvious, and that they form all line bundles over P1. Well, we'll do Pn later as well. But so this lecture, we're going to show all line bundles over the one-dimensional projective line. I sum off it to one of these. So how do we do it? Well, we recall that projective space is the union of two copies of affine space. And you can picture it like this. So here's 1a1 and here's the other a1. And you sort of glue them together along the open subset, which is a1 minus a point. And an inversible sheaf of P1 is going to be obtained by taking an inversible sheaf on one of these a1s, which looks, you can think of it as being assigned a line bundle to each of the one-dimensional vector space to each point of a1. And you take another inversible sheaf on the other a1 and you kind of glue them together over the intersection. So we must take this bit here and this bit here and glue these together. Well, obviously to do this, we first need to know what are the invertible sheaves on a1. Well, that's easy because a1 is affine. So sheaves correspond to modules over the coordinate ring, which is just k of x. Now, invertible sheaves over ring can be quite complicated. So we saw an example last time where you remember we looked at ring Z minus 5 and found there were non-trivial invertible sheaves over it. However, kx is a principal ideal domain. And over a principal ideal domain, all invertible sheaves are trivial. They're just isomorphic to the obvious one-dimensional module. And to see this is quite easy. First of all, it's easy to check that an invertible sheaf corresponds to a finitely generated module. And finitely generated modules over principal ideal domains are just of the form kx over some f. And if f is not zero, it's easy to see this is not invertible. Sorry, that direct sums of modules like this. So the only possibility is to take a sum of several copies of kx. And it's pretty obvious if we take more than one copy of kx, that's not invertible. So kx is the only invertible module. So now what we've got to do is to take these two invertible modules over the affine line that we've identified and figure out how to glue them together. So what we've got is we've got modules m1, which is equal to kx over the ring kx. And we've got a module m2, which has to be ky because that's the only invertible module over ky. So here are the coordinate rings of the two copies of the affine line. And so here we've got the spectrum of kx. And then we've got the spectrum of ky. And the intersection is the spectrum of ky over xy minus one. So we're setting y equals x minus one. And m1 and m2, if you restrict them to this space here, become isomorphic to the sheaf associated to this module here. So we've got that the picture we've got is we've got this module m1, which is kx. And it maps to the module kxy. Remember, y is x minus one over this space here. And then we've got another copy of it, ky, which again maps to kxy. So this blue thing here is a sort of sheaf over this. And this pink thing here is a sheaf over this. And if we restrict them both to this, we get slightly smaller copies here. So if I call this blue thing, it's going to be here. And this pink thing corresponds to this space here. Now we've got to glue them by finding some sort of isomorphism from between these two things here. In other words, we've got to find isomorphism between these two modules. So and we need to know what are the isomorphisms from kxy to kxy, where these are thought of as modules over the ring kxy. And of course, and y is actually minus one here. This isn't the polynomial ring in two variables. Well, the isomorphisms obviously just correspond to units. So they're just units of kxy. And the units of kxy are just a form a times x to the n for n and integer and a in the field k. And the fact that a doesn't really matter because we can multiply by units of kx and ky without changing the isomorphism class of our invertible module, because that just corresponds to doing an isomorphism on, say this big blue module or this pink module. So we may as well take a equals one. That's because the units of kx are just nonzero elements of k. I guess k should be nonzero, a should be nonzero there. And the units of ky are also nonzero elements of k, so we can just eliminate these. So we've got a preliminary classification of a module. If we go back here, we're taking our blue module kx, mapping it to kxy, and we're taking our pink module ky and mapping it to kxy. And now we have to identify these and we can identify these by mapping one to y to the n, or we could go back the other way and map one to x to the n. It makes no difference because x is y to the minus one. So this is the gluing of two modules, we're identifying one with the other. And this gives us a sheaf over p1 called, it's called the twisting sheaf or says twisting sheaf, where you put a one here for reasons that will become clear a little bit late. No, we don't. We put an n there, sorry. One is the sales twisting sheaf. So we've got a preliminary classification. Every invertible sheaf over p1 is of the form o, n to sum n in z. And it's not immediately clear that these are all different. So let's show that these are all different. Well, what we can do is we can look at the global sections, gamma of o, n. So you remember this is just sections of o, n over the whole of projected space. And let's think what a global section is. Well, to choose a global section of this sheaf, we've got to choose a global section over the first a1, which is an element of kx. And then we've got to choose a global section over the second copy of the affine line, which is an element of k of y. And then we've got to glue them together. So let's think what this means. Well, an element of kx is just a polynomial p of x. And then we've got to choose another polynomial q of y in here. And then we've got to glue them together. So we've got to have that the gluing is given by y to the n times p of x must be equal to q of y because we were multiplying by y to the n to get from one to the other. And you remember x is equal to y to the minus one, of course, otherwise, this wouldn't make sense. So y to the n times p y to minus one must be equal to q of y where p and q are polynomials. And this just is equivalent to saying that p is a polynomial of degree at most n. And the dimension of these is n plus one if n is greater than or equal to zero and zero if n is less than zero. So we've worked out the dimension of the global sections of O of n. And it's given as follows for n equals minus three minus two minus one naught one two three. The dimension is zero zero zero one two three four and so on. So this shows that these are all distinct. And what about the sheaves O n for n less than zero? How do we show they're distinct? Well, there are a couple of ways of doing this. First, we can observe that O of m tensor with O of n is isomorphic to O of m plus n. And this is because if you glue two modules by multiplying by y to the m and glue two other modules by multiplying by y to the n and take the tensor product, it means you end up gluing by multiplying by y to the m plus n. And by using this, you can see that immediately if all these are different, then everything must be different as well because you can just, for instance, you can show these two are different by tensioning them with with O of three and then you will get two distinct modules. Alternatively, you can see the dual of O m is O of minus m. And again, that follows because if you glue two things by multiplying by y to the m, then you glue their duals by multiplying by y to the minus m. So we see that they're all distinct. So all the O n are distinct. So the Picard group of P one over a field k is isomorphic to the integers because we found all the line bundles and found out what the group law is. And it's obviously just isomorphic to the integers. One thing you should note, by the way, is that these are all finite dimensional over the field k. So the dimension means the dimension of the field k. In other words, the global sections of these modules are actually quite small. Notice, for example, that if we took the module that the ordinary sheaf over, say, the affine line k of x, the space of sections is the whole of k of x, which is infinite dimensional over k. So global sections of sheaves over projective spaces of a tendency to be much smaller than over affine spaces. This is a very special case of a very general theorem saying cohomology of coherent sheaves over projective varieties is finite dimensional that we probably won't get to this course. But anyway, now we should give some examples of this. The first was that I mentioned last lecture that if we try to define tensor products of sheaves by saying f tensor g of u is equal to f of u tensor with g of u, then this just fails completely. And I sort of promised a counter example to it, and we can now have a look at a counter example because let's look at o minus one, tensor o of o of one. And the global sections of this is just as dimension one. On the other hand, the global sections of o of minus one has dimension zero. So if we tense it with the global sections of o of one, well, this is dimension two. Well, that doesn't really help because if you take a zero dimensional space and tense it with a two dimensional space, this still has dimension zero. So these two spaces are not equal. And this means here if we take f equals o of minus one and g equals o of one and u to be the whole of projective space, we find an example where these two spaces are not equal. This is why we cannot use this naive formula as the definition of a tensor product of sheaves. It just completely fails. And it's not failing for some weird pathological non-notary example. It's failing for the simplest and easiest examples of invertible line bundles over projective line. You can't get any nicer than this example. There are some other things that go wrong. So first of all, let's just have a quick look at what are the morphisms from the sheaf o of m to o of n. Well, this is easy to work out. It's just isomorphic to the morphisms from o of zero to o of n minus m because we can just tense everything with o of minus n. So tension with line bundles is a really nice operation that preserves nearly everything you can think of. And this is just the global sections of o of n minus m, which we've just worked out. It has dimension equal to n minus m plus one for n minus m. It's greater than zero. And it has dimension zero if n is less than n. So we've got very good control over all the morphisms between these line bundles. There's actually a slightly easier way to describe morphisms that we'll see a bit later when you describe these bundles using in terms of graded modules over graded rings. We're just doing things very crudely by hand, gluing things together. So our descriptions of these are still a little bit clumsy. And now we can look at the following exact sequence. So suppose we take the sheaf o of zero and take the sum of two copies of it, then we can map this to the sheaf o of one in two ways. You remember the space of morphisms from o of zero to o of one is two dimensional. So we can map this to say one plus y. You remember the morphisms from this to that correspond to a polynomial degree at most one and y. So we can take the polynomial one and the polynomial y. And we notice this is surjective because the first morphism from o of zero to o of one is surjective on the first copy of a one. And the second morphism is surjective on the second copy of a one, meaning it's surjective on all the fibers. So this is a surjective morphism. And you can work out the kernel. We don't actually need the kernel very much, but it's not difficult to check. The kernel can be identified with o minus one, if you like. So we really just want the fact that this is surjective. Now let's tensor it with o minus one. And we get nought goes to o minus two goes to o minus one plus o minus one goes to o of zero goes to zero. Now you may object that tension with things doesn't necessarily preserve exactness and you're absolutely right. However, for invertible sheaves it does preserve exactness because invertible sheaves look locally like a free module or a free sheaf. And tensioning with free modules or free sheaves does preserve exactness and exactness is a local property. So we've got an exact sheaf here. And now this is a bit weird because there are no morphisms from o of zero to o of minus one. So we can't get any, so no morphisms other than zero. So we can't split this. There's no morphism going that way. This is non-split. Well, what's the big deal about that? You've come across non-split sequences before. Well, what is rather surprising about this is that this is essentially just the ring of coordinate. This is just the sheaf of coordinate functions on P1. So if we're working over an affine scheme, you know any module nought goes to a goes to b, goes to r, goes to nought splits. This is over the ring r because this is just a one-dimensional free module and you can always split this by just taking a generator of this and mapping it to b. And so over spectrum, over an affine scheme spec of r, nought goes to the sheaf m goes to the sheaf n, goes to o, goes to nought. If this is surjective, it always splits. And so here we have something that always holds for affine schemes that the sheaf of coordinate functions is always a projective module. So free sheaves are projective over affine schemes. But here we have a sort of free sheaf that isn't projective over projective space. The word projective for modules and sheaves and the word projective for projective space don't have all that much to do with each other. It's just an unfortunate coincidence of terminology. So if you're working with, so an emphasise again that for sheaves over say general schemes, free sheaves need not be projective. This always confuses differential geometries because in differential geometry, if you've got a vector bundle corresponding to a free sheaf, then it is projective. So in differential geometry, if you've got a free sheaf there, you can always split it. So differential geometry and algebraic geometry are different. Free sheaves just aren't projective in algebraic geometry. Another very important thing that goes a bit wrong is, let's take this exact sequence again, nought goes to o minus two, just o minus one plus o minus one goes to o zero, goes to zero. And look at what happens if we take global sections. So we might try getting nought goes to the global sections of o minus two, goes to the global sections of o minus one plus global sections of o minus one goes to global sections of o zero. And if you've taken a course on sheaf theory, you know that if you've got an exact sequence of sheaves, then taking global sections this far is exact. And you can check it's exact very easily. This is just zero, this is just zero, this is just zero, and this is just one dimensional. So it's completely trivial, this is exact. However, this map here is not on two. So if you've got a surjective map of sheaves, the corresponding map of global sections need not be on two. And again, if you're talking about quasi-coherent sheaves over an affine scheme, then the global sections from, then taking global sections does preserve maps being surjective. So this is a new phenomenon for non-affine schemes. The fact that this isn't onto is a really major problem, and it's in the entire cause of cohomology. So you later see that if you take an exact sequence of say quasi-coherent sheaves and take global sections, we get an exact sequence like this, but we can't put a zero here. What we can put here is a first cohomology group with coefficients in A, then we get a first cohomology group with coefficients in B, then we get a first cohomology group with coefficients in C, and so on. So first cohomology groups are a way of fixing the fact that taking global sections isn't exact. Well, this isn't exact either. So obviously what we do is we put in a second cohomology group and we sort of carry on like this and so on. So I'm emphasising three things that go wrong, well, they not go wrong. They're different for schemes over non-affine spaces. First of all, tens of products don't behave in quite the way you might expect. Secondly, free sheaves aren't necessarily projective and these sequences don't split. And thirdly, if you've got a surjective map of sheaves, the global sections need not be surjective. Okay, that's all about P1 and we'll have more about quasi-coherent sheaves next lecture.