 This talk will be about proof of the Riemann-Roch theorem. We're just going to do it for non-singular plane curves over complex numbers. This makes the proof a little bit simpler, but the main ideas of the proof remain the same. I'll probably make a few comments about how to extend proof to all curves over any field. But anyway, so we recall the Riemann-Roch theorem. It says 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. L is the dimension of the space of functions with poles only on D. Degree is its D. Degree is the degree. And K is, of course, the canonical divisor. And G is the topological genus of the curve C. Well, the Riemann-Roch theorem is really three different theorems all bundled up. So the three theorems are as follows. First of all, we have what is essentially the theorem proved by Riemann, which says L of D is equal to degree of D plus 1 minus i of 0 plus i of D. This isn't exactly what Riemann proved, but it's pretty close approximation. So what's this number i of D? Well, i of D is equal to the index of speciality of D, whatever that is. By the way, that's a 0, not a D, in case you're wondering why these terms are the same. So what's the index of speciality? Well, if you've done cohomology groups, this is equal to the dimension of the first cohomology group of the line bundle associated to D. However, we haven't done cohomology groups yet, so I'm not going to use this definition. So instead, we're going to use an old-fashioned definition of it, which says it's the dimension of the space of obstructions to solving the Mittag-Leffler problem for the divisor D on the curve C. So what's the Mittag-Leffler problem? Well, the Mittag-Leffler problem is the following. It says, suppose you're given a given pole or given singularity of a function at certain points of C, can you find a function, a meromorphic function on the whole of C with those singularities? So if C is the complex plane, this is very easy if you specify a sort of singular behavior of a function at each point, a finite number of points, it's quite easy to find a function with exactly those singularities. In general, this is not true. For instance, if C is an elliptic curve, there's no meromorphic function with a pole of order one at one point and that's holomorphic elsewhere. So there's some sort of obstruction to solving this problem, and the space of obstructions will turn out to have dimension equal to this number here. There's a key point, which we have to show, which is that I of D is finite. Now, if you treat I of D as the dimension of a certain cohomology group, then this follows from a very general theorem about Seher saying certain cohomology groups of coherent sheaves are always finite dimensional. But what we're going to do is we're going to give an ad hoc proof of this, not using cohomology groups. So that's the first part of the Riemann-Roch theorem. It gives a sort of preliminary version. The second part of the Riemann-Roch theorem is about the genus. And there's a problem with the genus. There are three different ways to define it. So first of all, we can have the topological genus, which is the number of handles of the curve C. So if we draw C something like this, you can think of it as being a sphere with a number of hands attached to it. So this is a sphere with three hands attached. So it has genus three. That depends on the underlying topology of the curve C. Next, we have the geometric genus, which is denoted by P sub G. And I've no idea why the letter P is used, but G stands for geometric. And this is equal to the dimension of the space of polymorphic one-forms on C. In other words, this is just L of K, where K is the canonical divisor, because the canonical divisor is 0's of one-forms. And the third version of the genus is the arithmetic genus. So the arithmetic genus is P sub A, where A stands for arithmetic, just as G stands for geometric. And the arithmetic genus is defined to be I of 0. So it's the space of obstructions to finding a meromorphic function on C with given singular parts at various points. And basic theorem about this is that G is equal to P G is equal to P A. So this is, in some sense, the second part of the Riemann-Roch theorem. Well, first of all, these two turn out to be equal by essentially ser duality, which is turn out to be the third part of the Riemann-Roch theorem. Or you can think of this as being rock's part. And we will show that these two are equal just by doing an explicit calculation. So in a previous lecture, I calculated the topological genus of a plain curve explicitly. And what we're going to do later is calculate the arithmetic genus explicitly. And we'll just see these two are the same. So that's the second part of the Riemann-Roch theorem. It sort of shows that the various forms of the genus are the same. So the third part of the Riemann-Roch theorem is the part due to rock, or possibly ser. So this is rock, which is special case of ser duality in higher dimensions. And this says that I of D is equal to L of K minus D. And the special case D equals L of K minus D. And the special case D equals naught. It says that the arithmetic genus is equal to the geometric genus, because the arithmetic genus is just I of zero, and the geometric genus is just L of K. Now, if we combine Riemann-Roch's theorem with Riemann's theorem, which says that L of D is degree of D plus one minus I of zero plus I of D. And we finally combine this with the part two, which says that I of zero is equal to the topological genus G. So we combine all these three together, and we get Riemann-Roch theorem L of D equals degree of D plus one minus G plus L of K minus D. So the full classical Riemann-Roch theorem follows immediately from these three special cases of it. So what I'm going to do now is explain what is I of D, the index of speciality. So let's recall what the Mittag-Leffler problem is in more detail. So what we're doing is suppose given the singularity of a function, rather a pole of a function, say C minus one Z minus A minus one plus C minus two Z minus A minus two plus say plus C minus N Z minus A minus N at a point A. So we can ask, is there a function with the same singularity? So we don't care about the holomorphic part of the function. We just wanted to have these singularities. Well, obviously there is. And we can ask more generally given singularities at A1, up to AK, find a meromorphic function with these singularities. Well, if you've got a finite number of points and a finite number of singularities on the complex plane, then it's completely trivial to see you can do this. Mittag-Leffler showed more generally you can do this, even if you've got an infinite number of singularities provided they've got no limit point. And what we want to do is ask about this problem on a compact Riemann surface, C. And now it becomes rather more subtle. As I mentioned earlier, you can't always solve this if C has genes greater than zero. For example, on an elliptic curve, you can't find a function with a first-order pole at one point and no poles anywhere else. So what we want to do is to find what is the space of obstructions to solving this problem? Well, we actually want to solve a slightly more general problem for an arbitrary divisor D, which is sum of n i p i. And what we want to do is we want to pick a singularity at each point p i that appears in D. And we only care about the terms z minus p i for the minus n with n greater than n i. So we're sort of modifying how closely our function has to approximate the singularity. For example, if we've got a point 2 p i, what we do is we specify the coefficient c 3 z minus p i minus 3 plus and so on up to c minus n z minus p i and c minus 1 and c minus 2 can be anything. So if we're specifying the point 2 p i, it becomes easier to find such a function because we're not specifying c minus 2 and c minus 1. Conversely, if the coefficient n i is negative, so if we've got something at minus 2 p i, then we specify the terms c naught, minus p i to the naught plus c 1 z minus p i to the 1. So we're not only specifying the singularity, but we might want to specify the constant term and the first derivative as well. So we have this question. Suppose we've got a divisor D. What is the obstruction to finding singularities, a function with given singularities at each point where we approximate the singularities to an order specified by D? And we're going to let i of D be the dimension of the space of obstructions. So let's write down more explicitly what this space is. So what we do is we write down the space r, which is going to be a restricted product over all points of c of rp. Where rp is the space of Laurent power series at the point p. In other words, rp is the space of all possible meromorphic functions near p. And restricted product means the elements are the form product of fp, where fp is holomorphic for almost all points of c. Points p. So our function is specified to be holomorphic at almost all points p. And as a finite number of points p, we might allow some weird singularity. This space here is called by several names. These days it's usually called a ring of addels, especially if you work over a finite field. It's also called a space of repartitions. An older name for it was valuation vectors. So you can use whichever of these names you like. So r allows us to specify a singularity at every point. And now what we're going to do is we're going to take r, and we're going to quotient out by r of d. What's r of d? Well, this is all products fp, where fp has a pole of order at most np at p, where divisor d is equal to sum of np times p. So this space here is the space of all possible singularities of a point where we measure the order of each singularity by the divisor d. And now we want to also quotient out by k of c, which is going to be the space of meromorphic functions on c. And now this is now going to be the space of obstructions. And you can see that what this is, that this really is a space of obstructions, because here r over rd is the number of ways, is always of specifying a singularity at a point of order given by d. And then we're quotient out by all the possible singularities that are given by a meromorphic function. So i of d is just the dimension of this space here. And now what we can do is we can prove that the Riemann's equality, L of d, is the degree of d plus 1 minus i of 0 plus i of d. And we're going to assume for the moment that i of d is finite. We'll actually check that it's finite after proving this. If it's infinite, then in some sense this is true if we put the i of 0 over here so it doesn't really matter. And first we observe that it's true for d equals 0. And this is completely trivial because i of 0 is then i of d. And this is just 1 and this is just 0. So what we're going to do is we can check that the result for d is equivalent to the result for d plus p, where p is a point of the curve c. And this will prove it for all d because you can get any divisor just by adding a finite number of points to d or maybe subtracting them. So what we want to do is compare these two expressions. So there's the result for d and here's the result for d plus p. And now let's see what all the terms do when we go from d to d plus p. And this term is easy. It increases by 1. So what about this term here? This term increases by either 0 or 1 because if we allow one extra possible place for a pole, there might be an extra function with that pole and there might not be. So we can see some creases by 1 if we can find f. If we can find a function f with f plus d plus p, it's greater than or equal to 0. That means the poles of f are at most d plus p. And also f has pole of order np plus 1 at the point p. So f plus d is not greater than or equal to 0. So it increases by 1 under this condition and by 0 otherwise. And finally, we can see that this term here decreases by 0 or 1 because we can add adding a condition about what happens at p is going to change the space of obstructions by at most 1. And if you go and think about the definition of i of d a little bit, you will see that this decreases by 0 exactly when you can find the function f with this property. So now what happens in going from here to here, this term increases by 1 and either this term increases by 1 or this term decreases by 1 and exactly one of these two cases happens. So this equation is equivalent to this equation and therefore this equation is true for all devices d. Well, one thing we haven't yet shown is that i of d is finite. So we need to check this. Well, it's enough to show that i of 0 is finite because we saw that i increases or decreases by at most 1 whenever you add a point to d. And in fact, we're not just going to show i of 0 is finite. We're going to evaluate it as well, which we'll need in the next talk. And for simplicity, we're going to take c to be a non-singular plane curve over the complex numbers. I'll briefly say what happens if c isn't non-singular. So it might be given by an equation f x, y, z equals 0 in the projective plane where the degree of f might be equal to d. And the problem is we want to show, given certain poles at points p i of c, can we find functions with these poles? In other words, we've sort of got a metagletal problem. And the first step is we can move all poles to the points at infinity. So by the points at infinity, we mean the points where z is equal to 0. So they don't lie in the affine plane. And we can just do this by finding a function with those poles and with no poles elsewhere except at infinity, which is not very difficult to do. So what we can do is we can reduce to the case when all the singularities that we specified are at infinity. And now what we're going to do is we look at the dimension of the space of possible poles at infinity of order less than or equal to n. Well, this has dimension d times n, because this is the number of points at infinity. And this is the order of the pole. So at each point at infinity, we can sort of specify a pole of like z minus something to the minus 1 plus something times z minus that to minus 2. And so on up to z minus whatever it is minus n. And you see there are n coefficients for each point at infinity. On the other hand, we can also calculate the dimension of the space of functions on the curve with whose only poles are at infinity. And for this, we look at the space of polynomials in x, y of degree less than or equal to n. So these will have a pole of order at most n at infinity and will be holomorphic elsewhere. And this space has dimension n plus 1, n plus 2 over 2. However, that doesn't give us the space of function of the curve c, because some polynomials in x and y are actually 0. So we see the space of functions on c, given by polynomials of degree at most n in x and y, has dimension n plus 1 times n plus 2 over 2. And then we've got to subtract n minus d plus 1, n minus d plus 2 over 2. And this term comes from the fact that if you take functions of the form f times something, then this is going to be 0, so you need to subtract it. You remember f is the function such that f, y, z equals 0 is the curve c. And this thing here will be a polynomial of degree n minus d. So the space of such polynomials is this dimension. This applies if n is at least d. If n is too small, then the space of polynomials of degree n minus d isn't given by this expression, but by 0. And this term here is equal to dn plus 3 over 2d minus d squared over 2. Well, now we can work out the dimension of the space of obstructions rather easily. So the space of poles or functions on c of polynomials of degree less than or equal to n is now dn plus 3 over 2d minus d squared over 2 minus 1 here, as usual. This is for n greater than or equal to d. And what's this minus 1 coming from? Well, this is coming from constant functions, which we don't contribute to singular parts of infinity. And now we compare with the space dn of all possible singularities at infinity. So this is the space of all conceivable singularities of degree at most with poles of order at most n at infinity. And this is the dimension of the functions we actually get with those singularities. And if you compare these, we see the difference. We cross off these two. And the difference is constant, at least if n is bigger than d. And this is equal to the dimension of the space of obstructions to the misaglethala problem. So it's i of naught. And you can now see this is just equal to d minus 1, d minus 2 over 2. And this is equal to minus that is equal to i of d. So this is equal to the arithmetic genus. And in particular, we see that it's finite, which is the main thing we wanted to prove. Later on in the next talk, we will actually have to use this exact value of the arithmetic genus in order to prove rock's part of the real rock theorem. So we should keep this in mind. I'm going to finish by just briefly saying what happens if c has n double points. Well, all that happens in this case is we find that this space here increases by n. At least if n is large because we get an extra n-dimensional space of functions because for each double point, we can sort of split it apart and then we can get functions which differ here and here. So the result of this is that the arithmetic genus, which is minus this bit here, decreases by n. So the arithmetic genus is d minus 1, d minus 2 over 2 minus n. So this is for a degree d curve with n double points. OK, so so far what we have done is we have verified Riemann's part of the Riemann rock theorem and we have checked that the index of speciality is finite and we have shown that the arithmetic genus is actually equal to the topological genus because we remember we worked out the topological genus and it was also given by this number. So the next talk, I will finish off the proof of the Riemann rock theorem by proving rock's part of the Riemann rock theorem.