 This algebraic geometry lecture will describe two examples of projective varieties, the verinace surface in P5 and the variety of all lines in space. We start with the verinace surface. This consists of all points in five-dimensional projective space that form x squared, xy, xz, y squared, yz, z squared. So you can think of this as being a map from the projective plane P2 to P5, where you take a point x, y, z of the projective plane to this point here. Well, obviously this map isn't on two because P5 is dimension five and this only is dimension two. So we would like to describe this subset of P5 by giving explicit equations for it. So let's write a point of P5 in the following form, w00, w01, w02, w11, w12, w22. So for example, w12 corresponds to the fact that there's a y and a z where x corresponds to the number 0, y to 1 and z to 2. So we want to find relations between these numbers w that are automatically satisfied for all these points. Well, it's pretty obvious. We see that wij, wkl, is equal to wik, wjl for all ijkl, where we put wij is equal to wji corresponding to the fact that xy is equal to yx and so on. So this is a reasonably large family of quadratic equations satisfied by all these points here. And you can check the map from P2 to P2, the set of points satisfying all these equations is onto. The proof is sort of similar to the way we showed the map from P1 times P1 to something was onto. And it's similar to the example we're going to give next of lines in space. I won't actually give the details of this. So this map can be regarded as a map from P2 to P5. We can also think of it as being a map from affine space A3 to A6 in some sense, where A6 we think was being the symmetric square of the space A3. So symmetric square is just a way of saying we can sort of all degree two monomials in x, y and z where x, y and z are bases of A3. So S squared of An is dimension n times n plus 1 over 2, which for n equals 3 gives us 3 by 4 over 2, which is 6. Of course, there's no reason why you should stick to symmetric squares. You could also map A3 to the symmetric cube of A3, which would have dimension 3 times 4 times 5 over 1 times 2 times 3. Which is equal to 10 and so on. And there's no reason why you should stop at A3. You can also have maps from Am to the symmetric nth power of Am. And this will give a map from Pm minus 1 to P of something fairly large that I'm feeling too lazy. So the images of these maps from projective space to projective space are all called veronets of varieties. And you can define them by a similar but more complicated set of equations. So the next example is going to be the variety of all lines in P3. So the idea is we want to find a variety whose points correspond to the straight lines in P3. So let's stop and just think about what the dimension of it should be. Well, if you've got a line in P3 and you take a random plane and let's say this plane here, then the line will generically pass through a point in the plane. So that gives you two degrees of freedom. And then it can point in a direction in... Well, in ordinary Euclidean space, the direction will be a point on the sphere S2. So that gives you another two dimensions. So we expect this variety to have dimension 4. Well, a point, so a line in P3 corresponds to a two-dimensional subspace, four-dimensional vector space. That's right, that is a vector space rather than affine space because we're picking the origin. So we're picking a two-dimensional subspace of the vector space k4. This is a special case of a grass manian. A general grass manian gmn can be thought of as the set of m-dimensional subspaces of the vector space k to the m plus n. So we can think about what these are. So g0, m and gm0 are just points. They're all the trivial. g1m is the set of one-dimensional subspaces of an m plus one-dimensional vector space. So that's just projective space Pm. gmn is more or less the same as gnm. Because here this is an m-dimensional subspace of k to the m plus n. And we can look at its dual. In other words, a set of all linear transformations vanishing on this. So if we take the dual of this in k to the m plus n dual, which is just isomorphic to k to the m plus n, we get an n-dimensional subspace. So flipping m and n doesn't really affect this grass manian very much. So we see in particular gm1 is also isomorphic to m-dimensional projective space. So the first non-trivial case that isn't a point of projective space is g22, which is equal to the two-dimensional subspaces of k4. So in general, grass manians are special cases of Hilbert schemes. So a Hilbert scheme is a scheme, which is a generalization of a variety we will discuss later, whose points correspond to certain configurations such as algebraic sets or more generally sub-schemes of projective space satisfying certain conditions. The simplest condition is that it should be a linear subspace of projective space such as a line or a plane or something like that. So grass manians are the simplest cases of Hilbert schemes and g22 is the simplest non-trivial example of a grass manian. So what we want to do is to try and make this grass manian g22 into a projective variety. When we will do that, we will embed g22 into five-dimensional projective space. So how do we do this? We pick a line in p3, which corresponds to a point of the grass manian. And now we can pick two points on the line. So the first point will be a0, a1, a2, a3, and the second will be b1, b0, b1, b2, b3. These should be colognes between them. So these are only defined up to multiplication by scalars. And from these, we want to somehow produce a point of p5. So we need to give six numbers. And what I'm going to do is I'm going to put these two things together. a0, a1, a2, a3, b0, b1, b2, b3. And here I get a little 2 by 4 matrix. And now I'm going to put sij to be the determinant of columns ij. So, for instance, s0, 1 is the determinant of a0, a1, b0, b1, which is a0, a1, minus a1. So it's a0, b1, minus a1, b0. And similarly for the others. So altogether we get six coordinates, s0, 1, s0, 2, s0, 3, s1, 2, s1, 3, s2, 3. Of course, something like s3, 0 is going to be the same as s0, 3 up to sign. So there's no point in putting it in. And something like s1, 1 is automatically going to be 0. So we can miss out those as well. Now notice this point of p to the 5 only depends on the line that we chose. So first of all, if we multiply all the a's by a constant or all the b's by a constant, obviously all these determinants get multiplied by the square of that constant so it doesn't change that point in p5. Secondly, we had a lot of choice about which two points were picked on the line. So we can pick other points by replacing b by a multiple of a. Well, this corresponds to adding a multiple of the first row to the second row in all these determinants, but that doesn't change the value of the determinant. So changing which points of the line we pick as long as these are distinct points isn't going to change this point of p5. So we've got a defined map from lines in p3 to points of p5. Well, is this map onto? Well, no, because the lines in p3 form a four dimensional space and p5 obviously has dimension 5. So not all points of p5 can be the image of a point of the form sij given by all these determinants. So there must be some relation between these numbers sij and what is this relation? Well, the relation between them is the famous Plukka relation which looks like this. So s01, s23 minus s02, s13 plus s03, s12 equals zero. And this is easy to prove. Each term of a form a something times b something times a something times b something occurs twice with opposite sign. So everything just cancels out. There's a pretty good chance I got a sign error wrong in this somewhere and that they won't cancel out because the signs of the Plukka relation are a bit tricky, but what do you know? Stupid errors I've made. This Plukka relation holds between these. Next we can ask are there any other relations satisfied by these points? And the answer is no. That if we've got a set of numbers sij satisfying the Plukka relation in p5 then there is a line in g22 mapping to that point. So we want to show the map g22 mapping to the solutions of the Plukka relation is on to. Well, some sij must be nonzero so can assume it is one. So we may as well assume that s01 is equal to one. But then we know that s23 times s01 is equal to some combination of the others two by the Plukka relation. So s23, sorry, s23 is determined by s02, s03, s12, s13. And now we can just pick the two points, 1, 0, s12, s13 and 0, 1, s02, s03. So here we can pick, here's a point and here's a point. And so this determines a line in g22 with image the given point of p5. So any point satisfying the Plukka relations is indeed the image of a point of the Grassmannian. So we found a description of all lines in three-dimensional space. It's more or less the same as the set of points of this quadric in five-dimensional space. Remember a quadric just means the solutions of a degree two equation and here we've got a degree two equation. So we can use this to do things like find the cohomology of a quadric in p5 over, say, the complex numbers. It doesn't really matter if you don't know what cohomology is because we're not going to make any serious use of it. What we notice is the Grassmannian is a union of the following subsets. Well, first of all, we can take all points of the following form. So this is a point of the Grassmannian corresponding to the line going along this point and along this point where star means any number we don't really care about what it is. Well, that doesn't cover all of them because this coordinate might be zero. So we should also include one, something, naught, something, naught, naught, one, something. Notice, by the way, that if we choose this thing to be one, we may as well choose this coordinate to be zero because we can subtract the multiple of the first row from the second to make this zero. And similarly, if this is none zero, we may as well make it one and can then kill off this coordinate to make this zero. And similarly, if both of these are zero and this is none zero, we can push them to this form. And there are several other forms the point can take as follows. Try to get this right. And then the first coordinate might be zero, so we might get zero, one, star, star, and zero, zero. That should be zero, one, star, zero, one, star, zero, zero, zero, zero, one. And finally we might get zero, zero, one, zero, zero, zero, zero, one. So we've divided the Grassmanian into six pieces. And you can see that these form six copies of various little affine spaces. So here we've got a copy of affine space A4, here we've got A3, here we've got A2, here we've got A2, here we've got A1, and here we've got A0. And this gives us a very nice cellular decomposition of the Grassmanian. So over the complex numbers, if you go to an algebraic topology course, you can see the dimensions of the cohomology groups coming right off from this. So the zeroth cohomology group is dimension one, the second is dimension one, the fourth is dimension two, because there are two of these. The sixth is dimension one, and the eighth is dimension one. If you don't like cohomology groups, then instead you can use this to answer an alternative question, how many points, so how many points does the Grassmanian G22 have over a finite field K? Well that's quite easy to answer because over a finite field K, An has Q to the n points with Q elements of the finite field. So since we've divided up the Grassmanian into disjoint copies of affine space, we see that the Grassmanian G22, or equivalent to the Quadric in five-dimensional projective space, has Q to the nought plus Q to the one plus two Q to the two plus Q to the three plus Q to the four points. So you've got two copies of Q squared because there are two copies of two-dimensional affine space. If you want to compare this with four-dimensional projective space, this has Q to the nought plus Q to the one plus Q squared plus Q cube plus Q to the four points. So we see that the Grassmanian G22 is definitely different from four-dimensional projective space. It's either got a different number of points over a finite field or it's got different cohomology groups if you work over the complex numbers. Incidentally, this relation between this show, this strongly suggests there's some relation between the cohomology groups of a variety over the complex numbers and the number of points of the variety over a finite field. This is the theme of the vague conjectures over finite fields which give, extend this to all varieties, basically saying the number of points of a variety over a finite field is closer related to its cohomology over the complex numbers. For the next lecture, we will look at more complicated Grassmanians.