 Okay, this is the fourth part of lecture one on algebraic geometry, and we will be continuing with some examples of applications of algebraic geometry. So first we will discuss cacaia sets. So cacaia sets are actually something from real analysis. So cacaia set, well there are two slightly different definitions, the difference doesn't really matter that much for our purposes. The first definition says that it's a set such that if you've got a unit line, you can turn the line all the way around. So obviously you can do that if you've got a circle of diameter one, then you can turn around a line of diameter one and a circle. You can do a little bit better by taking a triangle of height one and then if you think about it very quickly you can see you can turn this line around inside the triangle. And cacaia suggested that the minimum possible set was this sort of, I can't remember what it's called, however Besikovic showed that in fact there was no lower limit to the size of a set in which you could turn around a unit interval. So however small a number is you can find a set of areas smaller than that in which you can turn around a unit interval. Slight variation of the definition of a cacaia set says it's just a set containing a unit line in every direction, whether or not you would actually turn the line from one direction to another direction. So cacaia sets turn out to be important in high dimensional harmonic analysis and one harmonic analyst called Tom Wolf suggested a finite field analog of the cacaia set. So this is not quite the same as the two novelists called Thomas Wolf because his name is spelt slightly differently, it's got two Fs at the end. Anyway, he had a sort of finite field cacaia conjecture. So he asked over a finite field the size of a cacaia set over F in F to the n. So this is a finite field is at least Cn F to the n. So this is some constant depending on n. This is the size of the finite field F and a cacaia set means it has to contain a lot line in every direction. So F to the n is a vector space. So it's reasonably clear what is meant by a line and the direction of a line and a cacaia set is just one that contains lines in every direction. And so Wolf's conjecture said that this cacaia sets can't be too small. And this was proved by Dvir in 2008 and he showed you could prove this with Cn being one over n factorial. And his proof was amazingly short for a problem that everybody thought was going to be kind of rather difficult. So we'll describe his proof. It is two steps. Step one says that a cacaia set cannot line a hyper surface of small degree. And step two says that if a set is small, we can find a hyper surface of small degree containing it. So if you put these together, it says that a cacaia set can't be too small. So step one, a cacaia set in F to the n cannot lie in a hyper surface of degree D less than the order of F. And already this is a slightly funny result because cacaia sets seem to have nothing to do with hyper surfaces. They're just a sort of combinatorial subset of vector spaces of a finite field. So it's a little bit surprising that hyper surfaces are appearing in this. Well the proof of this is fairly short. Because F is a polynomial of degree D, which is less than the order of F, defining some hyper surface is a cacaia set. And then we let FD be the highest degree component. So for any V, we can find so that FX plus VT vanishes for all T. And if you think about it, this is just what is meant by the zeros of F being a cacaia set. So this just says that F vanishes on this line. So it's saying for any direction V, we can find a line such that F vanishes. So that's a consequence of it being a cacaia set. So the coefficient FDV of T to the D vanishes. So this is true for any V and FD has degree less than F. So FD must actually be equal to zero. And this is because a polynomial of degree less than the order of F cannot vanish at all points of F. This is just saying that a polynomial of some degree, this number of roots must be at most the degree. So the terms of highest degree of F vanish, so F must be equal to zero. So we assume that our polynomial had degree less than the number of elements of the finite field F. And the places where it vanished form a cacaia set and we find F must be zero. So that shows that a cacaia set cannot vanish on a hypersurface of degree that's too small. That's the first part of the proof. So the second part says we observe that the polynomials of degree at most F minus one form of vector space of dimension n plus F minus one choose n. So this is just a binomial coefficient. So this is an easy result that I will just leave people to do because I'm feeling too late to prove it. So we can find a hypersurface of degree at most F minus one vanishing on any set with less than this number of points. And this follows by linear algebra because the condition that a polynomial vanish at a point is some sort of linear relation between its coefficients. And if we've got a vector space of this dimension and a smaller number of linear conditions on it, then there must be at least one element of this vector, least one non-zero element of this vector space satisfying all these conditions. In other words, there is some polynomial vanishing on all these points, provided the number of points is less than the dimension of this vector space of polynomials. So let's put these two together. The kakaya set cannot lie on a hypersurface of degree less than F, but if it had less than this number of points, then it could. So a kakaya set has at least this number n plus F minus one choose n points. And now we just observe n plus F minus one choose n is just equal to F times F plus one up to F plus n minus one over one times two up to n, which is greater than or equal to F to the n over n factorial, which is exactly the bound we want to a kakaya set must have at least some power of the number of elements of the finite field times some constant. So that's the end of a proof. It's a kind of rather remarkable proof of apparently hard theorem that you can fit on the back of a postcard. Just move that up a bit so you can see this formula I wrote. Final example I wanted to mention to end this lecture with is just a very famous example of 27 lines on a cubic surface. This is sometimes said to be the beginning of high dimensional algebraic geometry. It was almost the first non-trivial result if you don't count things like Bezou's theorem. So I'm not going to prove it. I'm just going to give an example. So this is 27 lines on a cubic surface. So Cayley and Salmon proved that any cubic surface has exactly 27 lines on it as long as it's non-singular. I'm just going to cheat instead of doing it for all cubic surfaces. I'm just going to do it for one cubic surface. I'm going to take the easiest possible one. Let's take w cubed plus x cubed plus y cubed plus c cubed equals nought in three-dimensional projective space. So I'd better explain notation for projective space here because we seem to have four coordinates and I've only got three dimensions. Well projective space is three-dimensional projective space is the set of all quadruples of numbers w, x, y, z except that this is considered to be the same point as lambda w, lambda x, lambda y, lambda z for lambda not equal to zero. So for example if z is non-zero then you can multiply it by lambda to make z equal to one and you find the set of points with z non-zero can just be identified with the set of points w, x, y, one which is isomorphic to three-dimensional space over whatever field you're working with I guess we're working over complex numbers for the moment. So if c is non-zero three-dimensional projective space the points of projective space with z non-zero just form a copy of the three-dimensional affine space and similarly we could take the points with y, x, or w non-zero and those would form three more copies so all together projective space is covered by four copies of affine space. Anyway this is why we seem to have four variables for three-dimensional space anyway so this gives a hypersurface and now we want to write down 27 lines well I can write down one line one line could be the set of points a minus a b minus b obviously a cubed plus minus a cubed plus b cubed plus minus b cubed equals zero so that's a perfectly good line as we vary a and b we seem to be varying two things but remember the projective space this is the same point if you multiply everything by a constant so this is really one-dimensional and now we can permute the coordinates so we can move this minus a to any three positions there and we can multiply by omega where omega is a cube root of one so we can multiply a minus a by cube root of one or minus b by a cube root of one and if you think about it carefully this gives three times three times three possibilities which is 27 possibilities so all together we get 27 obvious lines okay so I guess that's the end of the first lecture the next lecture we will start more systematically studying affine space