 This talk will be about how to use the representations of finite abelian groups in order to prove Dirichlet's theorem on primes in arithmetic progression, which says that if you've got an arithmetic progression of the form a, n plus b, there are infinitely many primes of this form. Here a and b are co-prime integers. In order to show how Dirichlet's proof works, I'll start by recalling Euler's proof that there are infinitely many primes. So Euler started with the Riemann zeta function, which is 1 over 1 to the s plus 1 over 2 to the s and so on, and he showed that it could be written as an infinite product over all primes of 1 over 1 minus p to the minus s. Euler's easily just by writing each of these as a geometric series and expanding out. So we have 1 plus 1 over 2 to the s plus 1 over 4 to the s and so on, times 1 plus 1 over 3 to the s plus 1 over 9 to the s and so on. The same thing for 1 plus 1 over 5 to the s. And if you just multiply these all together, you find the formula for zeta of s. Now you notice in particular that zeta of 1 is infinite because this is just the harmonic series. Now what we do is we take logarithms of both sides here. So we have log of zeta of s is sum over n greater than or equal to 1 and sum over p prime. I should have said p was sum over primes here of p to minus ns over n. So you see that this term is infinite at s equals 1 because it's log of infinity, which is infinite. Now you notice the terms for n greater than 1 have a finite sum. Because this sum is just sum over n greater than 1 and the sum overall p of p to minus ns over n and you can easily estimate this. This is at most sum over n greater than 1 and m greater than 1 of m to minus n, which is equal to sum over m greater than 1 of 1 over m times m minus 1, which is just equal to 1. So the sum of the terms for n greater than 1 is finite. So the sum of the terms for n equals 1, which is the sum over terms of p prime, 1 over p is infinite. So this is Euler's form of the theorem that there are infinitely many primes. I'll mark that although this sum is infinite, it's only just infinite. It grows very, very slowly. In fact, so slowly that if you get a computer and try adding up the reciprocals of all primes, it doesn't look as if it's tending to infinity at all. If you add up the first few billion or something, it looks as if it's tending to a limit slightly bigger than 3. In fact, the sum of the reciprocals of all primes for primes less than n is about log of log of n, which is pretty close to being a constant function. So now we get on to Dirichlet's theorem. Dirichlet's theorem uses Dirichlet characters, which are denoted by Kai. And to Dirichlet character is just an irreducible representation of the group of integers modulo n that are co-prime to n under multiplication. Instead of explaining this in detail, I'll just write out a first few examples. So let's take n equals 3, for example, then z over 3z, the elements co-prime to 3 are just one or two mod 3. So we have two elements of this group. So we have two characters. The first character is the trivial character, and the other character takes value as 1 on their identity and minus 1 on the element of order 2. And to each Dirichlet character, I'm going to associate an L series, and the L series is going to be sum over n of Kai of n over n to the s, where we put Kai of n equals 0 if n is not co-prime to n. So the Dirichlet series in these two cases is going to be 1 over 1 to the s, plus 1 over 2 to the s, plus 1 over 4 to the s, plus 1 over 5 to the s, plus 1 over 7 to the s, and so on. And this will be 1 over 1 to the s, minus 1 over 2 to the s, plus 1 over 4 to the s, minus 1 over 5 to the s, plus 1 over 7 to the s, and so on. I'll do another example. Let's just do any equals five. See what's going on. This time there are four elements in Z over NZ star. It can be one, two, three or four, mod five. And there are going to be four characters. And the first character is the identity character. And now this group, the element two has order four. So you can choose the value of the character at two to be any fourth root of one. So it can be like this. Once you've chosen this, you can fill in everything else because the character is multiplicative like this. So there are four characters and we get four L series. So the first one is one over one to the S, plus one over two to the S, plus one over three to the S, plus one over four to the S, plus one over six to the S and so on. And the next one will be one over one to the S, plus I over two to the S, minus I over three to the S, minus one over four to the S, plus one over six to the S, plus I over seven to the S and so on. So you can see what's going on. And I won't bother writing up the other two because those are both fairly obvious. So these are the famous Dirichlet L series. So the Dirichlet L series, L S of chi in general is sum over all N of chi of N over N to the S. That's N greater than or equal to one. And just as the Riemann zeta function can be written as a product, this can be written as a product. It's a product over all primes of one over one minus chi of P times P to minus S. And you can prove this just as you prove it for the Riemann zeta function, you just expand this out as a product of geometric series. And we can take logarithms of both sides and we find log of L of S of chi is equal to, well, now you can just expand the logarithm of this, which would be the sum of the logarithms of all these. So it would be sum over all primes and sum over N greater than or equal to one of chi of P to the N over N times P to the N S. That's just because log of one plus X is equal to X minus X squared over two plus X cubed over three and so on. So now we're going to use that to prove Dirichlet's theorem. Well, actually what I'm going to do is I'm just going to prove Dirichlet's theorem in a special case. We'll prove that there are infinitely many primes which are three mod eight. And you'll be able to see that most of this proof goes through for any other arithmetic progression. Well, what we do is we take N to be eight and we write out the Dirichlet character's modulo eight. So we have four non-zero elements of Z modulo eight Z star, which are going to be one, three, five and seven. And there are going to be four characters, chi one, chi two, chi three, chi four. And the first one has values like that. And this is not a cyclic group. It's a product of two groups of order two. And we worked out the character table of that group. So it looks something like this. So you can see all the rows and all the columns are orthogonal as they ought to be. And the various Dirichlet L series look like one over one to the S plus one over three to the S plus one over five to the S plus one over seven to the S and so on. And here we get one over one to the S minus one over three to the S plus one over five to the S minus one over seven to the S. And here we get one over one to the S plus one over three to the S minus one over five to the S minus one over seven to the S. And here we get one over one to the S minus one over three to the S minus one over five to the S plus one over seven to the S and so on. And now what you notice is that this term here is infinite at S equals one. However, the other three terms of finite and non-zero at S equals one. The reason is, first of all, you can see there, the sum is finite, for instance, this one is an alternating series, so it converges. This one is an alternating series if you group the terms in pairs, so you take these two, minus these two, and so on. And this one, you have to be a little bit more careful about what you can again see the sum is finite. And you can see the sum is non-zero because if you take these four terms in each of them, the sum is positive, and the next four terms, you can see the sum is positive, and the next four, the sum is positive, and so on. So it's quite easy to check that these three, all sum to something that's finite and non-zero. Now what we do is we take the function f of n, which is one if n is three mod eight, and naught if n is not congruent to three mod eight. And we write f as a linear combination of these four characters by expanding it out as a sort of Fourier series. So we know f is the inner product of f with xi one bar times chi one plus f, chi two bar, chi two plus f, chi three bar, chi three plus f, chi four bar times chi four, except I guess we should divide this all by four, which is the order of the group G. So we see that this is chi one minus chi two, plus chi three minus chi four all over four. So you can see that if we take this linear combination of these four characters, we get something that's one when something is three mod eight and naught otherwise. And now we look at the corresponding logarithms of L series. So we take log of L of s chi one minus log L s chi two plus log L s chi three minus log L s chi four. So these coefficients here one minus one one minus one are just these four coefficients one minus one one minus one. And we see that this is equal to sum of f p to the n over n p to the s n, which is just sum over p and n, where p to the n is congruent to three mod eight, one over n times p to the n s. And now we can prove Dirichlet's theorem because we see it this term is infinite at s equals one. Whereas these terms here are finite at s equals one. So that implies this is infinite at s equals one because it's something infinite plus three finite terms. And the terms with n greater than one of finite by exactly the same argument that we used for Euler's proof that there are infinitely many primes. So the terms that then equals one at s equals one must be infinite, which means there are infinitely many primes that are three modulo eight. So that's Dirichlet's very ingenious proof. Well, I said that this worked for arbitrary values of n, except there's one thing that doesn't work all that, obviously for arbitrary values of n, which is where we said that these L series were none zero at s equals one. We sort of checked it for n equals eight by more or less doing an explicit calculation in each case. And you can sort of see that the sum is going to be positive. So we have this problem. Show L s of chi is not equal to zero at s equals one. So this is actually a sort of very special case of the Riemann hypothesis. So there's a Riemann hypothesis for these functions, which says the only zeros apart from some trivial zeros that negative integers have real part equal to a half. So you can think of this as being a tiny special case of the Riemann hypothesis. And in order to prove this, I'm going to quote a couple of results from complex analysis. So first of all, we're going to assume the result that L of s of chi can be extended to a meromorphic function of s. The only poll is when s equals one and chi is equal to chi one. So this is the character that is one everywhere. So chi one of n is one for all n except when it's zero. The second result we're going to use that if f s equals sum of a n over n to the s for the real part of s large. And if a n is greater than or equal to zero for all n, so if it has non-negative coefficients, if f has no singularities for s greater than or equal for s greater than s naught then the series converges for s greater than s naught. So this is a sort of, actually, it's a slight generalization of the theorem that says if you've got a power series with positive coefficients, then with radius of convergence rho then it actually has a singularity at the point rho. So that's a standard theorem from complex analysis. And this is actually a mild generalization of it. Anyway, we're going to use these two theorems to show that the Dirichlet L functions don't vanish at s equals one. And the key point here is to look at the following variation of a zeta function. I'm going to define zeta of s to be a product of all characters of L s of chi. If you want to know where this comes from, it's actually something called the decained zeta function of a cyclotomic field. I won't explain this in more detail because we don't need to use this. Anyway, what we do is we look at the logarithm of this decained zeta function and we know what the logarithms of all characters are. So it's just a sum over all chi of sum over all p and n of chi of p to the n over n times p to the n s. So this is just the logarithm of the corresponding L series. And now we can use the orthogonality relations for these numbers chi again in order to work out what sum over all characters is. And this is just a sum over all p and n with p to the n common to one modulo n of n over n times p to the n s. This is essentially because if you sum over all characters then by the orthogonality relations, that's going to be zero unless p to the n is the identity element of the group z modulo n. In particular, we notice that all coefficients are greater than or equal to zero. So if we exponentiate all coefficients, are greater than or equal to zero. So this is a key fact we need. We need to have positive coefficients in order to apply this result I mentioned earlier which needs positive coefficients. So now we notice that zeta k of nought does not converge. You can see this because all its coefficients are integers, non-negative integers when s is nought. So it's got an infinite number of terms so it obviously doesn't converge. So zeta k of s has a singularity somewhere. Well, all the factors have no singularities except for one of them which has a singularity at s equals one. So the only possibility is at s equals one. Well, at s equals one, l of s chi one has a pole of order one. And this means that none of the other l functions can have a zero there because if they did it would cancel out this pole and the zeta kine zeta function wouldn't have a singularity. So ls of chi is none zero for chi not equal to chi one. So that gives a proof that the Dirichlet L series don't vanish at s equals one. Incidentally, there's another proof you can give using the fact that this is the zeta kine zeta function because there's something called a class number formula which shows you can actually evaluate the residue of this function at s equals one in terms of the class number of a cyclotomic field and the class number is none zero which shows that this must actually have a pole at s equals one, but it's actually rather easier to do it like this. Okay, so the next lecture will be talking a bit more about representation theory of non-Abelian finite groups.