 So this talk will be about the Riemann-Roch theorem for genus one curves. More precisely, we'll be showing how to use the genus, the Riemann-Roch theorem to classify genus one curves. So previously, we showed that if we take a copy of the projective line, then this satisfies the Riemann-Roch theorem. In the case of genus nought and conversely, we showed that if we've got a curve satisfying the Riemann-Roch theorem for genus zero, then it must be a copy of the projective line. We also showed that if we've got an elliptic curve performed c modulo elatis, then this satisfies the Riemann-Roch theorem for genus one. And what we're going to do today is to show that if we've got a curve satisfying the Riemann-Roch theorem for genus one, then you can write it as the complex numbers modulo elatis. And so suppose c is a curve satisfying the Riemann-Roch theorem for genus one. Now the Riemann-Roch theorem in the case of genus one says that L of d is equal to degree of d plus L of k minus d, where as usual k is the canonical divisor and L is the dimension of the space of functions with poles only on d. As usual, we find that degree of k is equal to zero and L of k is equal to one from this. So the Riemann-Roch theorem becomes L of d is equal to degree of d whenever degree of d is greater than zero, because in that case the degree of this will be less than zero, so it just vanishes. So what we want to do is to try and classify curves satisfying this condition. So how do we do that? Well we're going to pick a point P on the curve, and then we're going to look at the dimension of n of P for n equals zero, one, two, three, and so on, where you recall this is the dimension of the space of functions which have a pole of order at most n at the point P and no poles elsewhere. So let's just calculate this using the Riemann-Roch theorem. Well this is pretty easy. So we just take n to be zero, one, two, three, four, five, or six, and then what L of nP? Well L of zero is always just one, and then by the Riemann-Roch theorem these numbers are one, two, three, four, five, and six. So it's almost linear except there's a sort of little glitch when you go from zero to one. Let's try and figure out what functions you get. Well here we get the space of constants, which is obviously spanned by one. Here well the dimension of L of P is the same as the dimension of L of zero so we get nothing new, so there are no new functions. Here the dimension goes up by one so we get a new function, let's call it x, which has a pole of order two. And similarly when we go up to here we get a new function, let's call it y, which has a pole of order three. And here you may think we get a new function but we don't really because the new function we get has a pole of order four but it's pretty obvious what that is. We can just take this function of the pole of order two and square it so we just get x squared. When we go to a pole of order five well we can get that by multiplying x and y together so we get x, y, and when we go to six well we can get a pole of order six by multiplying y by itself or we can multiply x by itself three times. So well we can now stop here because if we count up we've got one, two, three, four, five, six, seven functions living inside a six-dimensional vector space. That means there must be a linear relation between them. So we get a linear relation of the following form. It says a y squared plus b y plus c x y equals d x cube plus e x squared plus f times x plus g for some a, b, c, d, e, f, g in the complex numbers. So this is giving us a curve in p three, sorry, in p two I guess. And we can simplify it as follows. So first of all, we can just change y to y plus a constant. And this will eliminate the term b times y. At least if the characteristic is not equal to two because if we do that we can't always get rid of that. Similarly we can change y to y plus constant times x. And this will eliminate the term c times x times y. Again, that characteristic isn't two. Then we can change x to x plus a constant. And this will eliminate the term in x squared. Then we can change y to y times a constant. So the term in y squared has coefficient one. And then we can change x to x times a constant to make the coefficient of x cubed equal to, well for historical reasons, we make it equal to four not one. So our equation now looks like y squared equals four x cubed minus g two x minus g three for some g two and g three in the complex numbers. Again, this notation is done mostly for historical reasons. So this is an affine curve and we can look at the corresponding projective curve if we like it, z, z, z y squared is four x cubed minus g two x z squared minus g three z cubed. So this is a curve in p two. And for the curve in p two the point p that we started with corresponds to the point at infinity, which is just the point where x, in projective space is the point where x, y, z is equal to zero, one, zero. So we've started with an arbitrary curve, I guess complete or compact, satisfying the Riemann-Roch theorem for genus zero and shown that we can write it in this form, assuming it's non-simple and so on. Well, now the next question is how do we show that this is C modulo elatis? Well, the first thing to do is to try and identify what it looks like topologically over the complex numbers. So suppose for any curve which is of a form y squared equals x minus a x minus b x minus c or y squared equals x minus a x minus b x minus c x minus d. So we're doing this case because we can get it for free. Then what we do is we get a map from C to the projective line, just taking a point x, y to the point x. And you see this is almost a double cover except at the points a, b, c and either zero and either d or infinity. So at the points a or b or c or either d in this case or infinity in this case, there are two points of the curve c that map to the same point. So what has happened is we get C is got by gluing together two copies of P1, except you have to glue them a little bit carefully. So let's draw the two copies of P1. And we've got these points a, b, c and either d or infinity on both copies. Or rather, these are the points that are going to map to d or infinity. And we've got to kind of glue them together. We sort of glue them together like this. What we do is we cut these open with a, you know, you imagine these being made out of paper or something. You get your pair of scissors and just cut them open so there's a little hole here. And then you glue them together. So, and we're going to glue this piece to this piece. And we're going to glue this piece to this piece and this piece to this piece and that piece to that piece. So let's draw what we're getting again. Here I'll draw them slightly differently so you can see them being joined together. So what I'm going to do is I'm going to distort them slightly. So now here are the four bits. I'm going to have got a green bit and a dark blue bit. And I think this was a pink bit and a light blue bit or whatever. And I glue it to a second copy of. So you see, this is just a sphere where I've made a couple of slits in it and then I've deformed it a bit. And I'm doing that blue bit to that blue bit and this green bit to that green bit. And maybe I should have done it the other way around to make it a bit less confusing. And if I glue them up like this, you can see that what you're getting is just a torus. So it ends up looking something like this. So in other words, the genus of this complex curve in the sense of algebraic topology, where the genus is the number of handles on a sphere, is the same as the genus G that appears in the Riemann-Roch theorem. Well, so far, we've got our curve as a cubic equation in X and Y, but we haven't quite written it as C modulo alattis. So we've got two ways of representing it as an elliptic curve. We can either write it as the complex numbers modulo alattis, or we can write it as this plane curve Y squared equals 4X cubed minus G2X minus G3. And how do we get from one to the other? Well, we saw we can get from C modulo alattis to the plane curve by taking Y to be the bias-strice function, differentiated, and X to be the bias-strice function. To go back the other way, what we're going to do is to use this magic differential dx over Y. So this is a one form on the curve. And now you may think this one form has got a singularity at Y equals 0. And the answer is no, it doesn't actually have a singularity at Y equals 0. And let's see why. The reason is if we write Y squared equals X minus A times X minus B times X minus C. Let's forget about this factor of 4 for the moment. Then you see 2Y dy is equal to X minus A X minus B plus X minus B X minus C plus X minus C X minus A times dx. So dx over Y is equal to 2Y dy divided by all this junk here. We should also divide by Y. So here we have X minus A X minus B plus all the rest of it. I can't be bothered to write out all over again. And now you see the Y's cancel out. And the key point is this thing is none 0 at Y equals 0 because if Y is equal to 0, then X is equal to A or B or C. And if X is equal to A or B or C, then 2 of these terms are 0. And the third one is very definitely none 0 because of course A is not equal to B. B is not equal to C and C is not equal to A. Otherwise, we would have a covert singularity. So this side is now holomorphic at Y equals 0. So this is also really holomorphic at Y equals 0. It just happens that dx and Y both have a 0 at these points. Now what we've got is a rather nice one form that's actually holomorphic everywhere on the curve. I guess I should have checked the point of infinity as well, but I'm feeling too lazy to do that. And now what we can do is we can look at the integral from point P to any other point Z. And we're going to integrate dx over Y. So this is going to be some fixed point. And this is going to be some point varying on our curve. And the problem is this isn't really quite well defined because if we draw a curve like this, we might have a point P here and a point Z here, but then there are lots of different paths from P to Z. So we could go along here or we could go along here. And these would give different values of this integral. So Cauchy's theorem says that we get the same value if two paths are homotopic. But this pink path and this green path are certainly not homotopic. So this integral is not actually well defined. It depends on the path. Well, so what's the ambiguity? Well, there are two ambiguities. First of all, if we just integrate round a loop like this, we can get something that's nonzero. And secondly, if we integrate round a loop like this, we also might get something nonzero. But every loop you can do on the torus is homotopic to an integral multiple of some multiple of the pink loop and some multiple of the green loop. So if we put omega 1 is the integral over the pink loop of dx over y and omega 2 is the integral over the green loop of dx over y, then the ambiguity of the integral from P to Z dx over y is equal to m times omega 1 plus n times omega 2. And this is just a point of a certain lattice L, which is just the lattice generated by omega 1 and omega 2. So we've now found the lattice L corresponding to our elliptic curve. It's given by the ambiguity in this integral here. So what we get is a well-defined map from the curve y equals 4x cubed minus g2x minus g3 to the complex numbers modulo this lattice L. So that's how you get from an algebraic curve to the complex numbers modulo a lattice. Okay, that's all for the Riemann Rock in the genus 1. And next we'll be looking at the Riemann Rock for genus 2.