 So this lecture will be a second talk on the vacant gestures. And what we will be discussing is how the Riemann-Roch theorem implies the functional equation for the zeta function. So let's just quickly recall what the functional equation for the zeta function is. So in the previous lecture, we defined the zeta function of a curve over a finite field to be sum over all positive divisors of 1 over norm of the divisor to the power of s. And this had an Euler product expansion where this is a product over prime divisors. And it was also related to the number of points of the curve over finite field. So this is a sum over m where m, m is the number of points of the curve over the field with q to the m elements. And we sometimes write the zeta function in terms of a variable t where t equals q to minus s. And what we want to show in this lecture is, first of all, the zeta function is the following form. So sometimes you say the zeta function is rational. Well, this isn't actually a rational function of s, but it is a rational function of t. Here, this is a polynomial of degree 2g, where g is, of course, the genus of the curve. So that's the rationality. We also want to show the functional equation which says that q to the g minus 1 times s times zeta of s is invariant under s goes to 1 minus s. So we should also just quickly recall what the Riemann-Roch theorem says. So the Riemann-Roch theorem says that l of d is equal to the degree of d plus 1 minus g plus l of k minus d where this is the canonical divisor. And that's the degree of divisor and so on. Well, the first problem is the Riemann-Roch theorem tells us about l of d. So we need to rephrase the definition of the zeta functions that involves l of d. So we have zeta of s is sum over norm of d to the s. And we can write this as a sum over n greater than 0. And then we sum over the degree of d equal to an integer n, 1 over q to the ns. And here d is of course a positive divisor. And this is because the norm of a divisor d is just equal to q to its degree. So at least this is now linked up with the zeta function. So in order to use this we need to ask how many divisors are there of degree n. In this we recall that we can divide divisors up into classes. So we say d1 and d2 are in the same class if and only if d1 minus d2 is principal. This means of the form given by the zeros and poles of some function f. And there's a key point we're going to use, which is that there are only a finite number of classes of given degree. This is finiteness of the class number of an imaginary quadratic field. This class number is usually denoted by h. And we're just going to assume it. So here we can rewrite the number of divisors of degree n as follows. So the number of divisors of degree n that are greater than zero. I mean the divisor isn't greater than zero, not that n is greater than zero. There's now a sum over a of q to the l of a minus one over q minus one. Over the h classes of degree n. The number of classes of any given degree is the same as the number of classes of any other degree because you can just multiply by the divisor of some given degree to go from one to the other. And this is really rather good because this is now linked to the Riemann-Roch theorem. And let's explain why this is true. Well, the number of positive divisors of degree n in a class a is q to the l of a minus one. Sorry, that minus one is a bit ambiguous. It shouldn't be in the exponent. It should be down there divided by q minus one. The reason for this is this is the number of non-zero functions f with f plus a is greater than or equal to zero. So this gives the number of positive divisors an equivalent to a except that you've got to remember that f is equal to lambda f. Whenever lambda is in the finite field and there's none zero. So we have to divide by the number of elements of this finite field, which is this factor q minus one here. So we can think of this as being the number of lines in a vector space of dimension l of a, which correspond to divisors equivalent to a that are positive. So that's where this factor comes from. So if we put this together, we now get the following form of the zeta function. We find the zeta function is sum over n. And then we sum over all divisors with degree of D equals n of one over one, one over q to the NS. And this is now equal to a sum over n, one over q to the NS times the number of divisors, which we saw was a sum over all classes a degree n. So that's really a finite sum q, the LA minus one divided by q minus one. So now we can we can see that the zeta function is rational. So let's prove rationality. So first of all, let's look at the terms with n less than or equal to two g minus two. Well, this is obviously rational, because there are only a finite number of such terms and each term is obviously rational in q to the minus s. Now, if n is greater than two g minus two, then L of a is equal to the degree of a plus one minus G by by Riemann's part of the Riemann rock theorem. So so this is credit to Riemann. So if n is less than two g minus two, then we get this extra contribution from L of L of k minus a and we don't quite know what this is. So our sum here the terms for degree n greater than two g minus two can now be written as follows we get a sum of n greater than two g minus two of one over q to the NS. And then we have to multiply this by q to the n plus one minus g minus one divided by q minus one and we must remember to put in a factor of h because this sort of corresponds to the sum over the h classes. And now we can we can work out what this is this is let's take out we can take out the factor of h over q minus one which isn't terribly interesting. And this bit becomes this is just a geometric series or rather two geometric series so we can sum them explicitly and we get q to the one minus g plus two g minus one times one minus s divided by one minus q to the one minus s. And then we get another geometric series coming from this minus one terms will get h over q minus one times q to the two g minus one s over one minus q to minus s. And now we can see this is rational is rational. Well, it's not rational and it's rational in q to minus s. And furthermore, we can see the denominator is one minus q to the one minus s times one minus q to minus s. Or maybe something dividing that. So this proves rationality of the zeta function and with a little bit of care you can see the numerator is a polynomial degree at most to G but we weren't bothered doing that because that's pretty trivial. And now we want to talk about the functional equation. So what we should do is multiply by q to the G minus one s and ask if it if it if this is invariant under s goes to one minus s. Well, so is this stuff up here invariant under s goes to one minus s. And the answer is no it isn't, but it's very nearly invariant. It turns out you can make it invariant if we just priddle this thing a little bit so we got this bit by summing over n greater than two g minus two. And if we add all the terms for this for all n, but just keep the terms for this that are where n is greater than two g minus two then it will be invariant so so let's see what happens when we do this. So, let's take this term and write it out so we're getting, let's put in the factor of q to the g minus one times s. And then we've got this h over q minus one which is the boring term. And then we've got q to the one minus g plus two g minus one times one minus s, all divided by one minus q to one minus s. And then we're comparing it with is minus q to the g minus one s times h over q minus one times one over one minus q to minus s. And now these two terms here are exchanged by s goes to one minus s, which is, which is what we want for the functional equation. The problem is, this doesn't prove the functional equation for the zeta function because we only included some of the terms so we had zeta of s was sum over a and n, which is the degree of a of one over q to the n s times q to the L of a minus one over q minus one. Break this up into several terms. First of all, we've got the sum over all n of one over q to the n s times minus one over q minus one. So here, this minus one is this minus one here, and that's giving us this term here. Secondly, we've got a sum over n greater than two g minus two of one over q to the n s times q to the n plus one minus g over q minus one. So I guess we should sum over all a as well. And that's essentially this term here. And what's important to notice is that here we're summing over n greater than two g minus two. But here we're summing over all n. So that's accounted for the function equation for some of the terms but we've got some terms left over because we've got the terms sum over n less than or equal to two g minus two of one over q to the n s times q to the L of a minus one. I guess we should sum over that as well. So we've still got these terms to deal with. So what do we do about them? Well, so far, we haven't used the full Riemann-Roch theorem. All we've used is Riemann's part of the Riemann-Roch theorem, which just tells us about divisors of degree greater than two g minus two. In order to handle these leftover terms, we need to use the full power of the Riemann-Roch theorem. So let's look and see what happens. So what have we got? Well, we want to multiply by q to the g minus one times s because it's only when you do that that it becomes invariant under s goes to s minus one. And then we've got this sum over n less than or equal to n less than or equal to two g minus two of one over q to the n s times q to the L of a. And then we've got this extra factor of, well, there's a factor one over q minus one. I'm going to put it outside here so that it doesn't confuse us because it's just a rather boring constant. And let's recall what the Riemann-Roch theorem says it says that L of a is equal to the degree of a plus one minus g plus L of k minus a where a is any divisor. I guess I should write D for divisor rather than divisor class, but whatever. So this should be a sum over divisors D. And what we're going to do is we're going to pair the divisor D with the divisor k minus D. So here the Riemann-Roch theorem is relating a divisor to k minus the divisor. So this is going to correspond to changing s to one minus s in the functional equation. And what we do is we do this for each individual divisor. So the divisor D gives us a factor of q to the g minus one s times one over q to the n s times q to the L of a. And the divisor k minus D gives us a factor of q to the g minus one s times one over q to the two g minus two minus n s times q to the L of k minus a. And now we can use the Riemann-Roch theorem to identify this bit here. And this is just q to the L of a plus g minus one minus n. So what we're doing is we're comparing these two expressions. And now you can see that they're related to each other just by changing s to one minus s. So we've divided the Riemann-Zeta function into a sum of lots of terms. And each of these terms satisfies the functional equation. So we see that q to the g minus one s times Zeta of s is invariant under s goes to one minus s. So that sort of completes the proof of the functional equation. I should confess I've been cheating a little bit during this proof. I will tacitly assume several things which really ought to have been proved. First of all, I assume finiteness of the class number. There's another more subtle thing I just very quietly assumed, which I assumed there are divisors of degree, any given positive integer. And that's actually something that isn't trivial and needs proving, but whatever. I'm just going to ignore it. So to summarize, we've proved the functional equation and rationality. So rationality follows some Riemann's part of the Riemann-Rock, where you just know L of a divisor for a divisor having large degree. And Rock's part of the Riemann-Rock gives you the functional equation because it pairs off any divisor of degree at most 2g minus 2 with another divisor k minus d. Okay, so next lecture may have a proof of the Riemann hypothesis for the Riemann's zeta function of a curve, following the proof by Bomberian Stepanov, who showed that it is actually possible to deduce this from the Riemann-Rock theorem.