 This is part of a series of lectures on elliptic functions. I will start by giving a very quick review of what was in the previous lecture. We were talking about the Weierstrich elliptic function p of z. So this is elliptic, which means it's periodic for lambda in some lattice L, which is a subset of the complex numbers. There's a double pole at z equals zero, and all elliptic functions can be written as polynomials in the elliptic modular function and its derivative. What I'm going to talk about this lecture is variations on the group law associated to it. So we've got this group C modulo of the lattice, and we can think of Weierstrich's function as being a function on this lattice taking values in C union infinity. This is a group. Since L is isomorphic to sum of two copies of z and C is the sum of two copies of r, this is just isomorphic to product of two copies of the circle group. So its structure is very easy to see. And what we want to do is to describe this group structure in terms of the values of p. In other words, we would like to know what is p of z1 plus z2 if we're given p of z1 and p of z2. So this is the sort of analog of the addition formula for sine or cosine. You know sine of z1 plus z2 is sine of z1 times cosine of z2 plus something else and so on. Well, we'd also like to know what it does to the derivative. So we recall that if we've got a map from C modulo L to the projected plane, which takes a point z to p of z, p prime of z, at least of z is not in L. If C is in L, it takes a point of L to the point at infinity. So this is just a point of C squared which is contained in the projected plane. Now I'm just writing the affine coordinates because these are a bit easier to think about. And if we call these points x and y, then we recall the differential equation satisfied by p says that y squared is equal to 4x cubed minus g2x minus g3 for suitable constants g2 and g3. And this is an example of an elliptic curve. So it'll be a curve looking maybe something like this. This is a cubic curve in P2 and cubic curves in P2 are a very famous group law on them where three numbers lie on a line if and only if their sum is zero. So these are the points a, b and c and a plus b plus c is not in the group if a, b and c are collinear. This doesn't mean their sum as vectors. It means their sum in the group law on this curve. And the point at infinity is the zero of the group. So we've got a map between two groups. We've got a map from the group C modulo L to this cubic curve. And obviously we would like this map to be a homomorphism or an isomorphism of groups. So we would like to check that it is indeed. So let's first of all recall what it means for three points to be collinear. So if we've got three points x1, y1, x2, y2, x3, y3, these are collinear if the following determinant vanishes. So we take x1, y1, 1, x2, y2, 1, x3, y3, 1. And if this is, this being zero corresponds to the three points being collinear. And we want xi to be p of xi and yi to be p of, p prime of xi. So what we should do is we should look at the following determinant. We take p of xi1, p prime of xi1, 1, p of xi2, p prime of xi2, 1, p of xi3, p prime of xi3, 1. And we want to know what is this determinant or at least we want to know what are its zeros. Well, this determinant probably vaguely reminds you of the Vandermonde determinant. You remember the Vandermonde determinant is like 1xx squared, 1yy squared, 1zz squared. Of course, you could put in more variables and higher powers if you wanted. You remember this is easy to evaluate because it's divisible by x minus y because it vanishes if x is y because the first two rows are equal. And it's divisible by x minus c and it's divisible by y minus c and both sides are degree three. So this determinant is plus or minus this expression here. And we can do the same trick for this. So we notice that this is zero if z1 equals z2 or z1 equals z3 or z2 equals z3. And we would like to know, does it have any other zeros? It turns out that it does. So if we look at this determinant, we think of this as a function of just z1. And you see we've found two zeros of z1. So as a function of z1, it has a pole of order three at zero. And if you expand it out. And so how many zeros does it have? Well, if you've got an elliptic function, the number of zeros is equal to the number of poles. That's in a fundamental domain. And the reason for this is that if you take a fundamental domain, so here's zero and omega one, omega two, and here's omega one plus omega two. Then you can find the number of zeros of a function f just by integrating the function f prime of z over f of z dz. So this is equal to the number of zeros minus the number of poles for any holomorphic, meromorphic function that you're looking at inside some contour. And if f is elliptic, this is zero. And that's because the integral along the bottom and the integral on the top cancel out with each other by periodicity. And similarly, the integral along the left and the right cancel out by periodicity. So a function f of z one has three poles or rather a pole of order three zero zero zero and it's got two zeros at z one and z two. And if you look at this, you see there's a missing zero. There should be three zeros and we've only found two of them. So where's the other one? Well, there's an easy way to find out. We said the number of zeros is equal to the number of poles. There's another nice result which says that the sum of the zeros is equal to the sum of the poles modulo the lattice L. So if the zeros are z one, z two and so on, then z one plus z two plus z n is equal to p one plus p two plus p n plus something in L where the zeros are at these points and the poles are at these points. And to see this, all we do is we take one over two pi i times the integral of z f prime of z over f of z. And again, we integrate over a fundamental domain. And this thing here has a as a pole of order one at zeros with so pole of residue one at the zeros and more generally a pole of residue n is zero order n. So so the sum of the zeros minus the sum of the poles is just this expression here. So this is the sum of the zeros minus the sum of the poles. Well, in this case, the integrals over the top and the bottom don't quite cancel out. This expression here is d by d z of log of f of z. And so log of f of z can change by two pi i n under elements of the lattice L. And if you work this out, it means the top and the bottom integral don't quite cancel out. What happens is you get omega one times some integer n and similarly the left and the right don't quite cancel out. You get omega two times some integer. So this integral turns out to be omega one plus times an integer plus omega two times an integer, which is some element of L. So so that proves this identity here. Now let's apply this to our function f. So so you remember the poles are zero, zero and zero. The zeros were z two, z three and a mystery zero. Well, we've seen that zero plus zero plus zero is equal to z two plus z three plus this mystery zero. So this identifies the mystery zero. It's equal to minus c two minus c three modulo the lattice L. So we've identified the missing zero. So our determinant. Let me just go back since I don't want to write out this determinant again has a third zero. It's also zero if C one plus C two plus C three is equal to zero. And these should all be mod L, of course. So we found the three, three zeros of this as a function of Z. So now if we compare this with the cubic curve, we see that the PZ one, PZ one prime, PZ two, PZ two prime and PZ three, PZ three prime, a collinear if if z i equals z j mod L or z one plus C two plus C three is zero mod modulo the lattice L. So these being collinear is the group law on the elliptic curve and these summing to zero gives you the group law on C modulo modulo L. So the group law on C modulo L corresponds to the group law on the elliptic curve but by this map taking taking Z to P of Z, P prime of Z. I have a sort of exercise you can generalize this so so so here's an exercise. Let's find zeros of the following determinant. Let's take P of Z one, E prime of Z one, E prime of E double prime of Z one, one. And then I'm going to take P of Z two and I'm not going to write all these out. P of Z three, P of Z four and previously we had a map from C modulo L to a cubic curve in the plane. Well if we take Z to P of Z one, P prime of Z one, P double prime of Z one, one. This gives a map from C modulo L to a curve in three dimensional projective space and for the elliptic curve. We saw that three points added up to zero if and only if they lie in a line for this curve. Four points have some zero if and only if they lie on a plane in P three. So you can prove all these by sort of imitating the proof I gave for the curve in the plane. These are actually two special cases of a whole sequence of projective embeddings of our elliptic curve. So we can take a number N to be one or N equals two or N equals three or N equals four and so on. And for N equals one I'm going to map the point Z to the point one in P zero. So this is not terribly exciting, it's just mapping C modulo L to a single point. N equals two I can map Z to the point one P of Z in one dimensional projective space. Of course this is sometimes infinite so you have to sort of renormalize it a bit. So this maps C modulo L to P one and it's almost a double cover. It's a two to one map except at four points. There are four points where P has derivative zero and you get the inverse image of a point and P one is only one point here. So here we get C modulo L maps to a point. P was three, we map Z to one P of Z, P prime of Z. And here we're mapping C over L to a cubic in P two. And the next one we map Z to one P of Z and so on up to P double prime of Z. And this maps C over L to a degree four in P three and it continues like this. So you can map it to a degree five curve in P four and so on. And this gives you a whole sequence of embeddings of C modulo L into various projective spaces. And for all of them you can describe the group law by saying all the points lie on some hyperplane. If you've done algebraic geometry you may recognize these as being projective embeddings associated with various powers of a certain line bundle. There's something called a line bundle on an elliptic curve and its first, second, third and fourth powers give you these four embeddings. The reason why we usually use the case N equals three is N equals one and two are too small because the curve isn't isomorphic to its image. And N equals four is just more complicated than the case N equals three because, you know, dealing with a curve in P two is obviously a lot easier than dealing with a more complicated curve in P three. So that's why everybody uses this embedding of C over L as a cubic in P three. It's the simplest case that gives you an embedding. Well, we've got a bit of a problem. We found an addition formula for which relates P of Z one plus C two to P of Z one and P of Z two. But it doesn't quite because if you look at the determinant we had, it's also got the derivative of Z one and Z two. And we can eliminate this because you remember there's a differential equation relating this. So what we really want to do is to write, have an explicit expression for just P of Z one and Z two in terms of P of Z one, P of Z two and their derivatives. And I'll show you how to extract that from the previous result. The problem is really the following. It's really a geometric problem that purely algebraic problem. Suppose X one, Y one, X two, Y two and X three, Y three lie on a line and on a cubic. Y squared equals four X cubed minus G two X minus G three. And the problem is find X three in terms of X one and X two. In other words, we want to describe the group law on the cubic explicitly. Are we going to do this as follows? We notice that X one, X two and X three are roots of a certain cubic equation. I'm going to write down a bit of this equation. Well, it's going to be four X cubed minus Y two minus Y one over X two minus X one squared times X squared plus something times X that I don't care about plus something that I don't care about. So where does this factor come from here? Well, what you do is use the fact that they lie on a line which is given by one X one, Y one, one X two, Y two, one X three, Y three equals zero. And if you write this out, it says X one, Y two minus X two, Y one minus Y two minus Y one X three plus X two minus X one and Y three equals zero. And this says that Y three is equal to Y two minus Y one over X two minus X one, X three plus some constant that I don't care about. Now the cubic equation says that four X cubed minus Y squared minus something or other in X and a constant term is equal to zero. So if you notice Y is given by this expression here. So Y squared is really that this expression here. And I don't care about the linear in the constant term. All I need to know is that if you've got a cubic equation, then the sum of the roots of the equation is that is the term coefficient of X squared divided by the coefficient of X cubed. So we find X one plus X two plus X three is equal to Y two minus Y one over X two minus X one squared divided by four. And this gives us the answer. So this tells us what X three is in terms of Y two Y one X one and X X two and X one. And if we recall that X I is equal to P of Z I and Y I is equal to P dash of Z I and expand this out. We just find that P of Z one plus Z two is equal to quarter of P prime of Z two minus P prime of Z one over P and Z two minus P and Z one. All squared minus P of Z one minus P of Z two. So this gives us the addition formula or the bias stress P function. As you see, it's really just the group law for a cubic curve in a slightly disguised form. We can also put Z one equals Z two and get the duplication formula for the bias stress P function. So P of Z times two is equal to a quarter of P double prime Z squared over P prime of Z squared minus two P of Z. You see, if C one is equal to Z two, then this vanishes and this vanishes. So you have to use Droppertal's rule to figure out what this quotient is, which means you take another derivative of P. So this is the duplication formula for the bias stress P function, which is the sort of analog of the duplication formula for sine and cosine. Now I'm going to give an example of how to use this. So an elliptic curve is isomorphic as a group to S one times S one. So it is n squared points Z with n Z equals zero. So these are the points of order n and they form a little group. And number theorists are very interested in these points because they give Galois representations. So the idea is if you've got a cubic curve with coefficients in the rationals, say Y squared equals four X cubed minus G two X minus G three. If G two and G three are rational, then if you look at points of order n, these are algebraic coefficients. And these coefficients generate extensions of the rationals and you can using using the elliptic curve gives you very good control over these extensions. They give you nice Galois extension with nice Galois groups. So number theorists wish to know what are the points of order n. So we now have a problem. Let's find the points of order n on this curve. Another way of doing this is to find an elliptic function vanishing on the non zero points of order n. We can't have it vanishing on all the points of order n and nowhere else because the number of poles has to be equal to the number of zeros. So what we're going to do is to put a big pole at zero and zeros at all the other points of order n. So let's do this for n equals two and n equals three. Well, n equals two is easy. All we do is we look at the derivative of P and we notice this vanishes at the points of order two. We notice there are three points of order two. If I draw a fundamental domain has zero omega one, omega two there and the points of order two are just these three points here. So these three points of order two and so these points will be omega one over two, omega two over two and omega one plus omega two over two. By the way, these points of order two turn up so often that a common convention is to call the periods two omega one and two omega two. So these points of order two are sometimes written as omega one, omega two and omega one plus omega two. But what we notice is that P of z plus omega one over two is still even. So it's derivative vanishes at zero. So what we're doing is using the fact that if you've got an even periodic function with period omega, so I suppose I've got some sort of periodic function. I don't know what it might be. It might look something like this. You notice the derivative is always going to be zero at these half periods. And that's just what's going on here. The P is an even periodic function. So it's derivative automatically vanishes at all the half periods. Well, P prime of z has order three because it's got a pole of order three. So it has a pole of order three at zero and zeros at the points of order two. I'm using the word order in two different ways. It's got zeros of order one at the points of the elliptic curve of order two. So n equals two is easy. Now let's try n equals three. And this time we've got eight non-zero points of order three. So if you draw a picture, here's the fundamental domain zero, omega one, omega two. And there will be nine points of order one or three sort of living on a little lattice like that. And we're trying to find elliptic function that is zeros of order one here. But here it has a pole order eight because there are eight zeros. Now you notice that if three z is in the lattice L, then P of z is equal to P of two z. And because two z is congruent to minus c and P is even so P of z is equal to P of minus c. So now we can recall that we had a duplication formula. So this is just a quarter of P double prime of z squared over P prime of z squared minus two times P of z. So this gives us the points of order three. We just want three P of z minus a quarter of P prime of z squared. So P double prime of z squared over P prime of z squared. So this should be zero. Well, this isn't quite the right answer because P prime of z also vanishes points of order two. So this is giving us a few extra poles that we don't really want. What we can do is we can now just multiply it by the square of this. And we may as well get rid of the four as well. So we get 12 P of z times P prime of z squared minus P double prime of z squared vanishes at the points of order three. And if you look at this, you see this as a pole of order eight at the zero. So we found all the zeros because we found eight zeros and it only has eight zeros. So this is the function we want that vanishes at the points of order three. Obviously, you can continue this for other values of n. So for n greater than three, we can do something similar. And we can find a formula for P of nz and find functions vanishing at the n squared points of order n. However, it's pretty obvious that for n bigger than three, the algebra is going to get more and more complicated. And you're probably going to want a computer algebra system to go much further than this. So this sort of illustrates that there are enormous numbers of algebraic identities involving the viastrice function. In fact, books will sometimes contain pages and pages of them. I've got a few examples here from Whitaker and Watson. You see, here are some complicated looking identities between the viastrice function. These go on for several pages and some of them are pretty scary. I mean, I was looking at this one here. Let me just magnify it so you can see it. Okay, so this one here, I honestly have no idea how to prove it or what it's good for. The really scary thing about this identity is this apparently was an undergraduate mathematics exam question back in 1897. So back in 1897, undergraduates were expected to do things that I have no idea how to do. Actually, I suspect most of the undergraduates at that time couldn't do it either, but anyway. Okay, so next lecture will probably be on the Jacobi elliptic functions.