 This talk will be about the Riemann-Roch theorem in algebraic geometry for curves of genus 1. So the Riemann-Roch formula states that L of d is equal to the degree of d plus 1 minus g plus L of k minus d, where as usual d is a divisor, which is a formal sum of points n, i, p, i. The degree of d is just the sum of the n, i. k is the canonical divisor and g is a number called the genus, which we're taking to be 1. And since it's 1, it cancels out with this term 1 here, so we can just cross out these two terms and we get this term here. Now, as usual, you can work out L of the canonical divisor, which is g, which is equal to 1, and the degree of the canonical divisor is 2g minus 2, which is equal to 0. And as usual, L of d is equal to 0 if the degree of d is less than 0. And now we can work out L of d if the degree of d is greater than 0 because this is equal to the degree of d plus L of k minus d. And this term here can be crossed out because the degree of k minus d is now less than 0. There should be a d in there. So this is more or less the Riemann-Rochtheon for genus greater than 1. So we've done it for degree less than 0 and degree greater than 0. What happens for degree equal to 0? Well, L of d is equal to 0 or 1 if the degree of d is equal to 0. And we now have a new phenomenon. So for genus 0, the dimension of L of d was always determined by the degree. And for genus 1 it usually is, but it isn't in the case of degree 0. If L of d is equal to 1 for degree 0, this is equivalent to saying that d is principal. It's the set of zeros and poles of some divisor. And because if L of d is equal to 1, this means there is some function whose poles and zeros are d, which is just the same as saying d is principal. So this is just the same as saying that d is linearly equivalent to the divisor 0. So L of d is equal to 0 for degree 0 divisors corresponds to d being not linearly equivalent to 0. So obviously we would like to see some examples of that which we'll see in a moment. So what we're going to do is just work out everything. We're just going to work out the Riemann-Roch theorem explicitly in using analysis. So we're going to take our curve with the complex numbers modulo lattice where L is a lattice consisting of all things that form m omega 1 plus n omega 2 for m n integers. And functions on the elliptic curve are just periodic functions on c. So f of z plus omega 1 equals f of z and f of z plus omega 2 is equal to f of z. And from this you can deduce several things about the divisor of such a function. First of all, the number of zeros is equal to the number of poles. And you can see that if you take a fundamental region, so every point in the complex plane is conjugate to some point in this region under the group of translations in L. And we want to know how many zeros and poles that are in this region. Well, the number of zeros minus the number of poles is just equal to 1 over 2 pi i times the integral of the logarithmic derivative of f. Where c is just you integrate around the boundary of this region. And if there happen to be zeros actually on the boundary, you need to fiddle things slightly but I won't worry about that. And now we notice this is equal to zero because the interval over this bit counts as out with the integral over this bit because f of z plus omega 1 equals f of z. And similarly the integral over this bit here counts as out with the integral over that bit so we just get zero. So this just says the number of zeros is equal to the number of poles. Or in other words, the degree of f is always equal to zero. So this is a first condition and divisor has to satisfy to be the set of zeros and poles of a function. So the next condition we can have is a slightly more complicated one. What we can do is we can integrate 1 over 2 pi i times the integral of z f prime of z over f of z d z. And this is just equal to sum of n i p i where p i are the zeros of f of order n i. And that follows because f prime over f just has a pole of residue one at zero of f and the pole of residue n i is zero of order n i. So the question is how can we work this out? Well if we integrate it over a fundamental the boundary of a fundamental region then the integral over the left hand side and the integral over the right hand side don't cancel out because z is not periodic. So what happens? Well if you take the integral over this bit minus the plus the integral over this bit, what we end up with is 1 over 2 pi i times omega 1 times the integral from zero to omega 2 of f prime z over f z d z. So the difference between z on here and z on here is just omega 1 which is where that's coming from. So what's this bit equal to? Well this is just equal, well the integral of f prime over f of z is just log of f of z. So we get log of f of omega 2 minus log of f of 0. And f is periodic so you might think these two are the same but they're not quite because logarithm function is multi-valued and only defined up to multiples of 2 pi i. So this bit here is equal to 2 pi i n for some n in z. So the integral over the pink bits becomes n omega 1 for some integer n and obviously the integral over the two green bits becomes m omega 2 for some integer m. So we see that this element here must be in the lattice L. And by the way the notation is a little bit confusing since some of n i p i can either mean the divisor which is a formal linear combination of the p i's or it can mean the complex number where you treat p i's a complex number and just multiply it by n i and add them up and here we're of course treating it as a complex number. So we have two conditions for a function for a divisor to be the divisor of a function. So if some of n i p i is the divisor of a function we know that some of n i is not some of n i p i is in L. So using this we can now investigate the Riemann-Roch theorem. So first of all let's work out what is L of p for p a point. Well this is equal to one. It's obviously at least one because we have all the constant functions. And we want to know it's obviously either one or two. And it's two if there is a function of the pole at p and nowhere else we can ask is there a function f with pole order one at p and no other poles. And the answer is no. Well suppose f is a pole p. Now we know that the sum of the n i must be naught. So f has zero at some point q not equal to p. Well then the sum of n i p i for f is just equal to p minus q or I guess minus p plus q which is not an element of the lattice L. So f cannot exist. So we can now show that if the degree of d is greater than zero then L of d is less than or equal to the degree of d. So we know that in any case L of d plus p is always less than or equal to one plus L d. So it's enough to check for degree of d equals one. And if the degree of d is equal to one then it's easier to check. So we're trying to show L of d is equal to one less than or equal to one. If the degree of d is equal to one we can put d equals p plus e where e degree of e is equal to zero. And if L of e equals naught then L of d is less than or equal to one plus L of e equals one so we're okay. On the other hand if L of e is equal to one which is the other possibility then e the divisor e is equivalent to zero. That means L of d equals L of p plus e equals L of p because if a divisor is equivalent to zero then adding it to some divisor doesn't change the dimension of the space of functions and we've just seen this is equal to one. So in any case we'll show that if the degree of d is equal to one then L of d is also equal to one. Now in order to prove the Riemann-Roch theorem we now have to show the following problem. I want to show that L of d is at least the degree of d or the degree of d is greater than zero. And so we've already shown that the left-hand side of this is at most equal to the right-hand side and that was fairly straightforward showing it's at least equal to the right-hand side is somehow fundamentally harder because we've actually got to construct functions. I mean we've got to construct enough functions with divisor at most d to show that this is at least something. So somehow proving that functions exist is harder than showing they don't exist because we actually have to find them. And to do this we can use the Weistreis-P function. So we just recall that we define a Weistreis function P of z to be some of lambda in L of one over z minus lambda squared. Except if you check carefully this doesn't converge. So that doesn't quite work and we have to kind of twiddle with this a little bit. So it's actually one over z squared so that's the term for lambda zero. And we sum over all the terms for lambda and L lambda not equal to zero of one over z minus lambda squared. And then we add a constant to this to make everything converge. So this is just a sort of renormalization constant to get rid of an infinite constant we get in this definition. And then the P function is periodic. So here for lambda in L. And it's got a pole of order two at all points of L which is pretty obvious you can just see these poles here. And now we instead of having a pole of order two lattice points we want to have a zero of order one because this is much easier to deal with. We can get that by integrating this twice and then exponentiating them. So we define the zeta function device tri-zeta function nothing to do with the Riemann zeta function to be minus the integral of P of z. And then you can see that zeta has a pole of order one and residue one at z equals zero. And this is periodic but zeta isn't quite periodic because there's a constant of integration that comes in. So we find zeta of z plus omega i is going to be zeta of z plus some constant which is often denoted by eta of i. So now we want to get from this pole of order one and residue one to a zero of order one. So we define the sigma function which is informally e to the integral of zeta of z dz. Or in other words d by dz of log of sigma of z is equal to zeta of z. And since we have to exponentiate, well the integral involves a logarithm term which is only defined by the multiples of two pi i. But since we exponentiate we get rid of these indeterminacies and sigma is actually a single valued function. So sigma has zero of order one at all points of L. However it's not periodic. We see as before we get constants of integration we find zeta of z sigma of zeta plus omega j is equal to e minus e to the eta j times z plus omega j over two if you work everything out. So what's really going on here is that sigma is a section of a line bundle. Back in the 19th century when people invented these sigma functions they didn't know what a line bundle was because line bundles hadn't been invented but they were sort of working with line bundles except not saying so explicitly. So whenever you see a sort of fudge factor in a periodicity like this. One way of thinking of it is that these functions are really sections of line bundles. And now we can find functions with with given zero so suppose D is a divisor on our elliptic curve, given by some of NiPi. Now if it was on a genus zero curve we could find the function of these zeros just by taking product of z minus Pi to the Ni because this function here is a zero of order one at Pi. Well we can do the same thing on an elliptic curve by taking fc is equal to product of sigma of z minus Pi to the Ni. So this differs from this expression here in that we have this extra function sigma going on. Well the problem is f might not be periodic so we can ask is f periodic. And we can see that f of zeta plus omega j is equal to f of z times this big product over j equals one and two and i of. i equals one and two product over i of f of z times product of minus x of eta j minus Pi times Ni. So we want to know when this factor is one well it's one if first of all the sum of the Ni is equal to zero. And now you notice that these we earlier found that these were the necessary conditions to have a function f with divisor D and we've now seen that these are sufficient conditions. So we've found that these conditions were equivalent to saying there exists f with f equals sum of Ni Pi equals D. So we found necessary and sufficient conditions for a function to exist and we can now use this to prove the Riemann-Roch theorem. What we want to do first of all for degree of D equals one we want to show L of D equals one and do this we can pick P to be sum of Ni Pi and we're now thinking of this as being an element of C not as a divisor. And then L of D minus P is equal to one because this has degree zero. So there's a function f with with this as a function f with f equals D minus P because this has degree zero and also has the property that the sum of all points in it in C is zero. And now we see that L of D is greater than or equal to L of D minus P is greater than or equal to one and we already know that L of D is at most one for D having degree one. So degree D equals one implies L of D equals one. Now now we do the case where the degree of D is greater than one. So suppose the degree of D plus P is greater than one and we want to show that L so we want to show that L of D plus P is greater than L of D. This will show by induction on the degree that L of D plus P is equal to the degree D plus P. Well what we need to do is to find a function f with f plus D plus P greater than or equal to zero but f plus D is not greater than or equal to zero. I can't say less than zero because this isn't a total order because that will show that there are more functions with these poles than with these poles which is what we want. Well what we do is we can choose those Q i's so that D plus P minus the sum of the Q i's but we're going to choose a divisor like this. And here we're going to choose a Q i so Q i is not equal to P and if D is sum of N i P i we want sum of N i P i plus P minus sum of the Q i is equal to zero in C. And we want the number of Q i is equal to the degree of D plus P. And then this divisor here has the property that it has degree zero and the sum of all the points in it is zero in C. So we can find the function f whose divisor of zeros is equal to this. And now we see that f has a pole of maximal order at the point P. So f plus D is not greater than or equal to zero because this is equal to minus P plus something or other. So this shows that L of D is equal to the degree of D whenever the degree of D is greater than or equal to one. Now we can write this out explicitly. So here's the possible degree of D. It can go minus three, minus two, minus one, naught, one, two, three and so on. And L of D is zero, zero, zero, one, two, three. And here it can be naught or one because as we said it's vaginas zero. We actually get an ambiguity here depending on whether D is linearly equivalent to zero or not. And L of K minus D is just L of minus D. So this is easy to work out. It goes three, two, one, zero or one, zero, zero, zero. And zero here if and only if it's zero here because we notice that L of D equals L of minus D because you can just change a function f to one over f. This is if degree of D is equal to zero of course. And if we subtract these we find L of D minus L of K minus D is now equal to minus three, minus two, minus one, naught, one, two, three. And now you notice that this thing here is a polynomial in a degree of D. And polynomial is of course just the degree of D. But so L of D is a bit badly behaved because it's got a sort of kink in it at zero and similarly L of K minus D has a kink in it near zero. But if you take their difference these kinks cancel out and we get a nice polynomial. So one other thing you can do with this is we can check to see whether we get a unique factorization domain. So you remember genus zero meant that we got lots of unique factorization domains. So if you look at the functions with poles at a single point, then that's polynomial ring. For genus one, we don't generally get unique factorization domains. So let's just pick a point P because it was zero in our curve C over L. And that would be the ring of functions, theomorphic functions with poles only at P equals zero. So you can think of this as being the coordinate ring of the affine curve. And what we're going to do is to check that R is not a unique factorization domain. In fact, we'll come up with an explicit example of an unique factorization. So let's just look at the following points. The points zero, omega one over two, omega two over two, and omega one plus omega two over two. And we're going to find a function with a zero of order minus three at naught. In other words, a pole of order three and zeros of order one at these two points. And you can do this because this has degree zero and the sum of all these with multiplicities is in the lattice L. So let's call this function Y. And then I'm going to three more functions, A, B, and C with the following poles and zeros. I'm going to take pole of order two here and zero of order two at one of these three points. And then we notice that Y squared is now a unit times A, B, C. And this gives two completely different factorizations of Y squared. So these are none unique factorizations. Well, you may wonder, maybe we can, maybe Y and A, B, C aren't irreducible. Well, they are irreducible. For instance, you can't write A as a product of two simpler functions because then they would both have to have zeros of order one at omega one over two and we can't have a zero of order one and no other zeros on an elliptic curve. So the factorization of this function just isn't unique. We can actually write out these functions explicitly in terms of the Weierstrass function. So Y, for example, is just the derivative of the Weierstrass function and the function A is the Weierstrass function minus the Weierstrass function at omega one over two. And these are similar except you take omega two over two and that point and so on. And so what's going on is that there is a group called the Jacobian given as follows. It's the degree zero divisors form a group. So these are the things sum of nipi with sum of ni equal zero. And then you quotient it out by the principal divisors of the form f. So this is you can think of this as being the obstruction to being things being unique factorization domain. And it's also the obstruction to finding a function with a given principal divisor. So for genus zero, we saw that Jacobian was just zero because every degree zero divisor corresponded to a function. And this sort of meant that everything was unique factorization domain. For g equals one, you can see we've more or less shown that divisor is actually the same as the elliptic curve. And for g greater than one, j is more complicated. And we'll be looking at an example when g is equal to two or three a little bit later. And so next lecture I'll be saying a little bit more about elliptic curves and show how to classify genus one curves using the Riemann-Roch theorem.