 This algebraic geometry video will be about automorphisms of affine and projective space. So we've defined a category of algebraic varieties or algebraic closed algebraic sets. And an obvious question is what is the group of automorphisms of an algebraic set? In general, the group of automorphisms of some algebraic set is almost always the trivial group that most algebraic varieties don't have any interesting automorphisms at all. They're just very ugly and complicated. However, affine space and projective space are very symmetrical and have lots of automorphisms. So let's first of all look at automorphisms of affine space, A, N. And of course, we should first do the case of one-dimensional affine space. Well, one-dimensional affine space is really easy because we know its coordinate ring is just K of X. So morphisms from A1 to A1 just correspond to homomorphisms of algebras from K of X to K of X. And it's obvious what these are. These just take X to a polynomial P of X. So morphisms from A1 to itself are just the same as polynomials with coefficients in K. And we want to ask which of these are invertible. Well, again, that's pretty obvious. It's invertible if P of X is equal to AX plus B with A none zero. So completely easy to check that these things have inverses and nothing else does. So the group of automorphisms of one-dimensional affine space is just the so-called... Well, this group is sometimes called the AX plus B group for fairly obvious reasons. You can identify it with the set of two by two matrices of the form AB01. And it's fairly easy structure as a group. It's non-commutative as you can easily check. It's got a normal subgroup consisting of the elements of the form 11B0, which is isomorphic to the additive group of K. And the quotient is isomorphic to the multiplicative group of K. This group is a counter-example to quite a lot of things. For example, over the real numbers, this is the simplest example of a group where the left harm measure is not the same as a right harm measure. But anyway, there's really not much else to say about it. So what about A2? So here we have the automorphisms of polynomials and two variables. And by analogy, there are some pretty obvious automorphisms. You can map KX1, sorry, KXY2, where we can map X to AX plus BY plus C and Y to DX plus EY plus F, where ABDE has determinant non-zero. So this would be the analog of what happens for affine space. And this certainly does give you a group of automorphisms of affine space. However, the full group is much bigger. So what other automorphisms can you get? Well, for example, we can have the following automorphism. We map X to X, not terribly exciting. And we map Y to Y plus any polynomial in X. And this obviously has an inverse where we map X to X and Y to Y minus this polynomial in X. So we have an infinite-dimensional Abelian subgroup. So it's a very large infinite-dimensional group of automorphisms. And of course, you can do lots of other things. Instead of fixing X, we could fix Y or some other linear polynomial in X. And then we could start composing these. So the automorphism group is getting pretty hairy. Well, what else can you say about it? Suppose you've got an automorphism of affine space where we map all the XI to FI. I'm taking affine space to FI of X1 up to Xn. So in general, an endomorphism of affine space will be given by n polynomials in n variables. And the question is, when is this an automorphism? So when is F1 up to Fn an automorphism of An? Well, there's an obvious necessary condition. We can look at the Jacobian, which remembers this matrix of elements delta FI over delta Xj. And we can take the determinant of this. Now, the Jacobian of Fg is equal to the Jacobian of F times the Jacobian of g. So if Fg equals 1, then the Jacobian of F is invertible as a matrix with entries in polynomials. And so it's reasonably natural to conjecture that this is an automorphism of affine space if and only if the Jacobian is invertible. So it's a one implication. It's pretty trivial. The other one is the Jacobian conjecture, which the Jacobian conjecture says that if the Jacobian is invertible, this implies F is an automorphism. It's still an unsolved problem, as Frasier notes. One of the most notorious unsolved problems in mathematics is because of the large number of incorrect published solutions of it. In fact, I think there's one mathematician who managed to publish three different solutions of this, all of which were wrong. So it's really notorious for having subtle pitfalls that people fall into. OK, well, that's done morphisms of affine space, morphisms of the affine line that easy and morphisms of affine space in high dimensions are a real headache to deal with. Now let's look at morphisms of projected space. As before, we should start off by looking at the case of endomorphisms of P1, or automorphisms of P1. Well, if we've got a map from P1 to P1, then we can look at the inverse image. Sorry, let me try again. We get a map. We can restrict this to a map from A1 to P1. And we get a map from an open subset of A1 to A1, which is contained in P1, because P1 is just equal to A1. Let's make it A1 together with a point of infinity. So we've got a map from an open subset of A1 to itself. This open subset might possibly be empty if A1 is just mapped to the point of infinity, but let's not worry about that. Now, this means we've just got a regular map on an open subset of A1, and this is just given by a rational function, Px over Q of x. So conversely, if you've given a rational function, giving a map from an open subset of A1 to P1, then this can be extended to a map from P1 to P1. So this gives us all the morphisms from P1 to itself. And now we want to figure out which of these have inverses. And this isn't too difficult to figure out. They have inverses if they're of the form Ax plus B over Cx plus D, where A, B, C, D has determined this non-zero. In fact, if you represent this rational map here by a little 2 by 2 matrix, you can see that composition of rational maps actually corresponds to multiplication of matrices by a fairly easy calculation that I will omit. So this almost gives us the group of automorphisms of the projective line. And we have the group GL2 of the field K, which are the matrices A, B, C, D with determinant not equal to 0. However, this isn't quite the full group of automorphisms because the diagonal matrices of the form A, A all act as the trivial automorphism by taking x to Ax over A, which is obviously just equal to x. So we should quotient out by this. The quotient is called the projective general linear group because it's the group of automorphisms of the projective space. And it's just the general linear group of a field quotient out by the diagonal matrices of the form A0, A. So this group will have dimension 4. And this group here is dimension 1. So altogether, we get a three-dimensional group of automorphisms of the projective space. We notice that this contains the Ax plus B group as a subgroup. If we just take all the matrices with C equals 0 and D equals 1, then we get the group of automorphisms of the affine line being a subgroup of a group of automorphisms of the projective space. So what we see is that this is actually the same as what you get in complex analysis. If you remember, the Riemann sphere in complex analysis is a group of automorphisms which are maybe fractional linear transformations. And we find the group of automorphisms of the projective line is the same as the group of automorphisms of the Riemann sphere. However, in the affine case, the group of, let me just try again. So what I meant to say is the endomorphisms of the projective line are the same as endomorphisms of the Riemann sphere. That's a complex analytic map from the Riemann sphere to itself. However, for the affine line, the group of endomorphisms from the affine line are fairly restricted. They just consist of polynomials. Whereas the group of endomorphisms of the complex plane in complex analysis is much bigger. That consists of all entire functions, which are that there are lots of entire functions that aren't polynomials. So in other words, we see that for the projective line, algebraic geometry is very similar to doing things analytically. Whereas for the affine line, there are far more endomorphisms if you allow analytic functions than if you allow algebraic functions. So this is a very special case of a paper by Seher called Analytic Geometry and Algebraic Geometry, except the title was in French. It says roughly speaking that if your projective manifolds are, if you're working with projective manifolds, then analytic maps tend to be algebraic. Whereas if you're working with affine manifolds, then analytic things are much more common than algebraic things and quite often not algebraic. Should have one more example involving morphisms of the affine plane. Here we have an endomorphism of the affine plane to itself, which takes x, y to x, x, y. And we can ask, what is the image? Well, this isn't very difficult to work out. The image just consists of everything with x none 0. So we get everything here and everything here. If x is 0, then this corner is 0, but this corner also has to be 0. So we also get the origin, but we don't get the rest of the y axis. And the point of this example is the image can be quite complicated. The image is not open. It's not closed. It's not even locally closed. So there are no particularly easy statements you can make about what the image of a map of a closed set is under a morphism. There's a theory of Chevalet saying that it always has to be a constructable set, which means it can be constructed from open and closed sets by taking unions and intersections and complements. But in general, there's no easier sort of type of set that the image of a closed set is. On the other hand, we will see that for projected manifolds, the image of a closed set is always closed. OK, so the next lecture will be on the ax-grovendic theorem.