 Okay, this is algebraic geometry lecture 11 in which we will discuss quotients of algebraic varieties by groups. So first of all, we just recall for any algebraic variety or any algebraic set, we can go to its coordinate ring. And so the algebraic set will be given by some subset of a to the n. And the coordinate ring will be given by the polynomial ring x1 up to xn, quotient out by the ideal of this algebraic set. The algebraic set is y, i of y is functions vanishing on y. So this can be thought of as the ring of all polynomial functions on the algebraic set y. And this coordinate ring has three properties. First of all, it's an algebra over K. Secondly, it's finitely generated. It's obvious. And third slightly more subtle property is it is no nil potent elements. On the other hand, if you've got an algebra with these three properties, it's finitely generated over K and there's no nil potent elements, then by the strong nil still and that's it corresponds to an algebraic set. The algebraic set it corresponds to isn't quite unique because it depends on your choice of generators. If you choose seven generators, it will be contained in seven dimensional affine space. But somehow if you choose a different collection of generators, the corresponding algebraic sets are essentially isomorphic in some sense in the sense that there are polynomial maps between them that are inverses of each other. Informally, we can think of algebraic sets up to some sort of isomorphism as being more or less equivalent to algebras with these three properties. Technical term is a category of algebraic sets is equivalent to the opposite of the category of coordinate rings. So now we have an application of this. Suppose we have an algebraic set Y acted on by a group G. And the question is, can we form a quotient Y over G where we identify points in Y if they are equivalent under G? Well, actually, this doesn't really work. We can't form a quotient by identifying points of Y under elements of G as we'll see a little bit later in some of the examples. But we would still like to form something that we can count as being a quotient. Well, if you try forming equivalence classes of points of Y under G, you immediately run into problems. For instance, how do we embed this inside some affine space? We said an algebraic set was contained in an affine space. And if you start identifying points inside affine space, it's not all clear how you're going to embed that quotient in affine space. However, if you look at it from the point of view of coordinate rings, this is much easier to do because the functions on Y modulo G should correspond to functions on Y invariant under G. So what we can do is if we've got a group G acting on this algebraic set Y, then G will also act on this ring here, the coordinate ring of Y. And we can look at the ring of invariants where we take the coordinate ring kx1 up to xn, quotient it out by I of Y. And then you quite often denote the invariance of something by putting a superscript G. So this is the invariant ring. And if the invariant ring satisfies these three properties, then it's the coordinate ring of some algebraic set, which it would be reasonable to call the quotient Y over G. Well, let's check these properties. First of all, it's obviously an algebra over k. Secondly, it obviously has no nil potent element. So the problem is, is it finitely generated? And the answer is sometimes it is finitely generated and sometimes it isn't. And in general, it can be quite tricky to work at which. So Hilbert proved that in many cases it was finitely generated and seems to have thought that it was quite likely that it would always be finitely generated. In the 1950s, Nagata actually found an example where this ring of invariants was not finitely generated. So it's actually a fairly tricky problem. Anyway, let's look at a few examples of this. So one example we've all seen. Let's take an to be affine space and G to be the group S permutations. The light has gone off. Anyway, never mind. GN is the symmetric group of permutations of coordinates. So the coordinate ring, we take space of polynomials in n variables, take the invariants under the symmetric group. So this is just the ring of invariant functions. And you know what the ring of invariant functions is. It's generated by the elementary invariant polynomials where e1 is x1 plus plus xn e2 is x1 x2 plus all other things. And so I'm up to en, which is x1 times x2 times xn. Well, this is the coordinate ring of affine space. So we see that if we take affine space and quotient it out by the symmetric group on acting on its coordinates, this is actually isomorphic to affine space again. This example is actually really misleading because the answer is particularly easy. In general, quotients of even affine space by groups are in general very, very complicated. It's only if a group happens to be a reflection group or something very close to a reflection group that you get a really nice answer in general and SN happens to be a reflection group. You have to be a little bit careful about this. For instance, if we take the real line and have the group z modulo 2z acting on it just by taking x goes to minus x. So it sort of identifies points like that. Now if you take the real line and quotient out like that, you might think the quotient is going to be a half closed interval. But in fact, the quotient. So that would be the quotient if you identify points under the group. But you notice this isn't the same as what you get from this construction here. So the quotient under this construction is just the whole real line again. And what's happening is the points here are what you get by identifying 2 and minus 2 say. Whereas you also get some points here which you get by identifying say 2i and minus 2i. So you might think these are complex points, but in fact they count as real points of the quotient for slightly complicated reasons. So the points of the quotient might not be what you expect. Next example, let's look at a special linear group in n variables over k acting on k to the n. Let's take the general linear group. Then the orbits of points here are either the point zero or everything else. So you might think that the quotient is going to have two points. In fact, the quotient only has one point because the only invariant functions, the only invariant polynomials acted on by gln are constants. So here we just get the field k and this is the coordinate ring of a point. So although you might guess that the quotient of k n is two points because you can either take nought or everything else. In fact, if you calculate the quotient by taking invariant polynomials, you only get one point. So the third example is classical invariant theory. Here we take the group G to be SL2 over say the complex numbers and it acts on things called binary context. So if you look at a nx to the n plus a n minus one x to the n minus one y, so nought y to the n. Well, you can act on x and y by SL2 acting on x and y is a x plus b y c x plus d y. And this will transform this quantic into some other quantic and will act on the coefficients a n a n minus one and so on in some rather complicated way. So a nought up to a n is a point in n plus one dimensional affine space. So we can ask what is the quotient of n plus one dimensional affine space modulo SL2c. And the coordinate ring of this will be all polynomials in a n a n minus one that invariant under SL2c. And these are called invariants and finding them as part of classical invariant theory. A simple example if we take a two x squared plus a one x y plus a nought y squared and a typical invariant is the discriminant. Which is a one squared minus four a nought a two or b squared minus four a c in usual notation for quadratic equations. So you have this question is the ring of invariants finitely generated. And this was proved by Paul Gordon in a very complicated proof. So Gordon was known as the king of invariant theory because he was really good at proving finiteness of invariants. Well, Gordon's proof of this was so complicated that people kind of despaired of pushing it any further. In late 19th century, Hilbert found an astonishing way of proving in very great generality that rings of invariants are finitely generated. And we'll describe Hilbert's proof because it's actually remarkably short. By the way, there's a rumor that Gordon is supposed to be very negative about Hilbert's work and said this isn't mathematics, it's theology. It's not at all clear that Gordon really said this and the story that he'll that Gordon didn't like Hilbert's worker, mostly nonsense. Hilbert Gordon thought very highly of Hilbert and even used Hilbert's work and results himself. What he did do was when Hilbert's papers first submitted Gordon was the referee and Gordon quite correctly pointed out there was some gaps in Hilbert's paper. So this might be where this rumor about Gordon comes from. Anyway, in the next lecture we'll be giving Hilbert's proof.