 This lecture is part of Berkeley Math 115, an introductory undergraduate course on number theory and will be more about Dirichlet characters as part of the proof of Dirichlet's theorem. So we just quickly recall that a Dirichlet character is a homomorphism from the multiplicative group z modulo nz star to the non-zero complex numbers and it's usually denoted by chi. So an alternative way of saying this is that chi is a function from integers to complex numbers such that chi of m plus n is equal to chi of m and chi of m equals zero if m and n are co-prime and chi is multiplicative so chi of mn is equal to chi of m times chi of n and we should also make sure it's non-trivial by saying that chi of one is equal to one. And last lecture we saw several examples of this and in particular we saw that the number of characters of z modulo nz star is equal to the order of z over nz star. And what we're going to do now is to just discuss some properties of characters in more detail. And before we go on just notice that we can actually define a character for any finite abelian group. A character is just going to be a homomorphism from chi to the non-zero complex numbers and again we can check that the number of characters of a finite abelian group is equal to the order of the group chi and the proof of this is more or less the same as what we did last lecture for in the special case of z over nz star. So you remember what we did is we pretty much decomposed z over nz star into a product of cyclic groups and proved the theorem for cyclic groups and then used the Chinese remainder theorem to to deduce it for general n. And any finite abelian group can also be written as a product of cyclic groups so we can more or less copy the previous proof. In fact the set of characters of g is itself a group because we can define the product of two characters if we've got two characters chi1 and chi2 we can define a new character chi1 chi2 of g to be chi1 of g times chi2 of g. So for any finite abelian group we get a new group called the character group. In fact we get a sort of pairing from g times its character group to the complex numbers or non-zero complex numbers so this just takes g and a character to the value of the character on on g and if you want to be really clever you can think of an element of g as being a character of the character group because if we fix g then this gives a homomorphism from characters to complex numbers you know chi1 chi2 of g is equal to chi1 of g times chi2 of g. So g is really a homomorphism from the character group to the non-zero complex numbers. So you have this rather confusing thing that the characters of the group of characters of g is g itself. If you've done linear algebra you remember there's something similar for vector spaces that any vector spaces are dual and the dual of the dual of a vector space is the vector space you first thought of so characters of finite abelian groups are kind of similar. Another way in which characters turn up is is that characters are very closely related to Fourier theory. So you remember if we've got a periodic function f such that f of x plus 2 pi is equal to f of x on the real numbers then that then you can generally write f as a sum of functions sine 2 pi n and cosine 2 pi n with 2 pi n x and cosine 2 pi n x with various Fourier coefficients of these. Now instead of using sines and cosines it's often easier to use 2 pi shouldn't be there it's often easier to use e to the i n x for all integers n and these functions e to the i n x are just the characters of the group the re if you take the reels modulo 2 pi z so periodic functions are just functions on the reels modulo 2 pi z and this group has a collection of characters and the characters are just these exponentials e to the pi n x which are more or less the same as sines and cosines if you don't worry about linear combinations. So you should think of the characters of a finite abelian group as being analogous to these functions e to the i n x for periodic functions on the reels and now just as you can expand any function as a linear combination of these functions e to the i n x you can expand any function on g as a linear combination of its characters and we're going to do this for z over n z star so the first key um so what we want to do is to show that any function on z over n z star is a linear combination of these Dirichlet characters we'll be using this in the proof of Dirichlet's theorem and what you notice is that the if you take the space of functions on here these just form a vector space of dimension equal to phi of n because it just has a basis of elements that are one on some element of this group in zero elsewhere and we know the number of characters is also phi of n so to show everything is a linear combination of characters all we need to do is to show the characters are linearly independent as vectors in a vector space and to do this we're going to show that they are in fact orthogonal so in order to make them orthogonal we we should quickly explain what we mean by orthogonality so if we've got two functions on z modulo n z star we just define the inner product to be sum of n in z modulo n z star of f of n times chi of n and then we put a complex conjugate sign on there this makes sure that the inner product of f with itself is always greater than or equal to zero because it's a sum of f n times f n bar so so um these form a complex vector space with a nice positive definite emission in a product on it and now we want to show that any two characters are orthogonal if chi one is not equal to chi of two so this is just sum over chi one of n times chi two of n complex conjugated um and we notice that chi two of n bar times chi two of n is always equal to one because um it's a it's a root of unity so is absolute value one so this is just sum over n of chi one of n chi two of n minus one and this is just the sum over n of chi three of n where chi three is the character i one chi two to minus one so all we want to do is to show that this is zero if i three is not the the unit character i mean if a character is one everywhere then obviously the sum is none zero so we want to show if chi three is not one then this sum is none zero so suppose chi of m is none zero for sum m in z modulo n z star but what you do is we notice that sum over chi three of n that's chi three is equal to the sum over i sorry the sum over chi three of n times m because if you multiply by the um number m that's a that's a bijection from the integer's mod n to itself and this is also equal to chi three of m times the sum of chi three of n because chi three is multiplicative so we have the following identity chi three of m minus one times um sum over n of chi three of n is equal to zero and if i three of m is not equal to one then this is none zero so this implies the sum over n of chi three of n is equal to zero so um if we've got any character which isn't one everywhere then the sum over all that then the sum of its elements over z modulo n is always zero and this implies orthogonality of characters well since the characters are orthogonal this obviously implies they're linearly independent which implies they span the space of all functions on z modulo n z star and the way we're going to use this is we're going to take a function f of um n which is equal to one if n is congruent to sum number b modulo n and not if n is not congruent to b modulo n and we're going to use this to show that infinitely many primes which are congruent to b modulo big n by by expressing this function as a linear combination of characters um now we should discuss um convergence of um um Dirichlet series so we've got the series sum over um n of chi of n over n to the s which is going to be chi of one over one to the s plus chi of two over two to the s and so on and we want to know where it converges the first thing to notice is that it converges absolutely for the real part of s greater than one well we're just using s to be real so let's just say s greater than one and this follows because one over one to the s plus one over two to the s plus one over three to the s and so on converges for s greater than one and chi of chi of n has absolute value at most one so by comparison with this series it converges um nice thing is it also converges or s greater than zero if chi is not always one if chi is always one it diverges for s equals one but for all the other characters it actually converges um which is kind of useful to know and um for these you just use Dirichlet's convergence theorem it says that if some numbers a one if you've got some numbers a i and b i such that a one plus a two and so on plus a n is always bounded and um b i is decreasing and tends to zero then the sum of a i b i converges and to see this we just look at an expression like a one b one plus a two b two plus a three b three plus a four b four say and we can rewrite this as follows it's a one plus a two plus a three plus a four times um b one plus a one plus a plus sorry a two plus a three plus a four times b two minus b one plus a three plus a four times b three minus b two plus a four times um b four minus b three so you notice these um things are all um bounded and all these sums are bounded so suppose the sum is bounded by say some number m then this is going to be bounded by um at most m times um say two b one because the sum of all these numbers will be sort of at most two b one so so this sum here is going to be at most um sort of say two m times b one um similarly if we take the partial sums um a i b i plus plus a n b n this will be at most two m times b i and this tends to zero because b i tends to zero and so um Cauchy's um principle as a convergence says that if all these partial sums um sort of tend to zero as i tends to zero then our original series converges so sum of a i b i converges now we're going to apply this by taking a i to be chi of i for some character i and b i equals one over n to the s then the sum of the a i b i is is just the Dirichlet series l of s which is chi one over one to the s plus chi two over two to the s and so on and these numbers here are obviously um decreasing and they tend to zero so all we've got to do is to show that um the numbers a one plus a two and so on plus a n have bounded and this follows because um um the sum over if we take i mod n of chi i then this is bounded that that that that sorry this is not bounded this is equal to zero so if we take the sum of the first n values of chi i this will be zero and the sum of the next n will be zero and so on so um the sum of any number of them will be at most the sum of the first n of them which is going to be bounded it would be bounded by that most n in fact if chi has period n um you you can give better bounds than that but that's uh turns out to be a kind of slightly subtle problem um but final topic we i just want to mention is the notion of primitive characters so um you remember that some characters modulo n aren't really new for instance if we had n equals four for example we had one character um which gave us an l series one over one to the s plus one over three to the s plus one over five to the s and another one which gave us one over one to the s minus one over three to the s plus one over five to the s and so on and this one was not new and this one really was new um it was a nothing to do with any previous characters we had and and the characters that are new are called primitive so a character mod n is called primitive if it's not equal on on z modulo nz star to a character for some smaller um n so so here um this is equal to the character which is just one everywhere um on z modulo four z star which is why this this isn't really a new character and we can count how many new characters there are suppose n is a prime power p to the n well then we get um um that there are altogether psi of p to the n characters but we can get psi of p to the n minus one um old characters because we can take any character modulo p to the n minus one and that will give us a non primitive character modulo p to the n so the number of primitive characters is going to be psi of p to the n minus psi of p to the n minus one which is p minus one times p to the n minus one minus p to the p minus one times p to the n minus two that's provided n is greater than or equal to two so that will be um um so that will be p to the n minus two times p minus one squared for n greater than two but for n um equal one we've got to be a little bit careful because um that that no longer quite makes sense that the phi of p to the p to the zero has is given by slightly different formulas so so we get p minus two characters if n equals one that's primitive characters and in general if n is a product of prime powers the number of primitive characters is the product of the primitive characters for each of these prime powers notice by the way this is actually zero if p equals two so for numbers of the form two times something odd there are no primitive characters um we saw this earlier when we found there were no primitive characters um modulo two or modulo six um and for something like so if you want the number of primitive characters for 12 for example well 12 is equal to two squared times three and there's one primitive character for two squared and there's one for three so all together we get one times one equals one primitive character modulo 12 which is the one we found last lecture um okay so next lecture what we'll be doing is explaining how to deduce Dirichlet's theorem if you know that all the l series don't vanish at s equals one and after that the next lecture we will show that they don't vanish at s equals one