 This talk will be about representations of abelian groups, mostly finite abelian groups acting on complex vector spaces. So the problem we want to look at is, we've got a group G, let's say finite abelian, and we want to know how can it act on a vector space V, which is finite dimensional over the complex numbers. So let's just look at an example. Let's take G to have order three, so it's just three elements, one of G and G squared with G cubed equals one. And then there are three obvious irreducible representations where G can act on the one dimensional complex vector space C, because G can act as one or omega or omega squared. Here, omega is e to the two pi i over three, is a cube root of one. And the question is, can every complex vector space be split up into a sum of these sorts of irreducible? And the answer is yes, because we can take any vector V in a complex vector space and write it like this. We have V plus VG plus, sorry, GV plus G squared V over three, and you can add to that V plus omega GV plus omega squared G squared V over three, and you can add to that V plus omega squared GV plus omega G squared V over three. And you notice that this one G acts as one, this one G acts as omega minus one, unless I've got a sign error, and this one G acts as omega to the minus two. So the vector space splits up as a sum of subspaces where G acts as one or omega to the minus one or omega to the minus two. And of course, each of these subspaces can then be split up into one dimensional pieces. So we've seen that every complex representation of G can be split up into a sum of these three special one dimensional ones. So we can now write down the character table of G. So you remember the character table of G has a column for each entry of G and a row for each character. So if we call these characters chi one, chi two and chi three, then we get one, one, one, one, omega, omega squared and one, omega squared, omega. And it's pretty obvious that something very similar is going to happen if you replace three by some other integer. The character table looked like one G, G squared up to G to the minus one. And the first column will be just ones. The next one will be one zeta and zeta squared up to zeta to the minus one. Next will be one zeta squared, zeta to the four, zeta to the six and so on. The last one will go on until you get zeta to the n minus one and so on, where zeta equals e to the two pi i over n is a primitive n through two of one. You notice by the way that the rows and columns of these tables are all orthogonal in the following sense. Here by orthogonality, you need to use a Hermitian inner product. So we define the inner product of two vectors v and w, the sum of the i times wi complex conjugated, where the i are over the components. And then you can see that the rows and columns of these matrices are orthogonal. This is a very general phenomenon that will also hold for non-Abelian groups. So what makes this work? Well, in order for this to work, we need two things. First of all, we need an nth root of unity. And secondly, we need to be able to divide by n. So first for one over the order of g is in the complex numbers. And secondly, the number zeta equals e to the two pi i over n is in the complex numbers. And of course, if we replace the complex number by any field with these two properties, then the theory is very similar. A little bit later, I'll discuss what happens if one of these two properties fails. So that's more or less done cyclic groups. What about general Abelian groups? Well, that's not very difficult because a general Abelian group g can be written as a product of cyclic groups. So general finite Abelian group. And if we've got a representation of g1 and g2 and so on, so suppose g1 is generated by little g1 and g2 by little g2, then a typical representation of g will be a one-dimensional complex vector space in which g1 is omega one to the n1, g2 is omega two to the n2 and so on, where omega one is a root of unity of order, the order of g1 and so on. You can then see that you can split any vector space into eigenspaces of g1 and these will all be acted on by g2, so you can then split those into eigenspaces of g2 and so on. So any representation of this finite Abelian group of the complex numbers splits as a sum of these one-dimensional representations. You notice, by the way, that the one-dimensional representations of g form a group. So if we've got two representations, chi1, which maps g to the non-zero complex numbers and chi2, which maps g to the non-zero complex numbers, we can simply form the product chi1 and chi2 defined by chi1, chi2 of g is chi1 of g times chi2 of g. And it's easy to check. This makes the set of one-dimensional representations of g into a group. This is called the dual group. And you can easily check that the dual group of a cyclic group is isomorphic to a cyclic group of the same order and the, let's call this dual group g star. And if you've got g1 times g2, then the dual of that is just g1 star times g2 star. So this shows you that if you've got a finite group, then g is actually isomorphic to its dual. But this isomorphism is actually not a very good one to use because there isn't actually a natural isomorphism between these. It depends on the choice of a generator of the group. So this isomorphism is a bit iffy. It's not canonical or anything. However, there is a canonical isomorphism g to the dual of its dual. So this is a bit like vector spaces. You know finite-dimensional vector space is isomorphic to its dual, but not in a canonical way because you have to choose a base. Whereas it's canonically isomorphic to the dual of its dual. And you can see this because if you've got a character acting on an element of g, this is a non-zero complex number. And if you fix the character, this is a homomorphism from g to the complex numbers. But if you fix g, it's a homomorphism from characters to the complex numbers. You can think of an element g as being an element of the dual of its dual group. This is a special case of Pontriagin duality for finite groups. So Pontriagin duality for finite groups is actually kind of trivial as we've seen. Pontriagin duality really becomes rather more interesting when g is allowed to be an infinite locally compact group. Anyway, so let's see an example of a representation of a non-cyclic group. Here we're going to take the simplest possible example. Let's just take the group z modulo 2z times z modulo 2z and look at its character table. Well, its character table looks like this. It's got four elements, nought nought, nought one, one nought and one one. And it's got four characters. And the first character is just one one one one. And the second character looks like one one, minus one, minus one. And third like one minus one one minus one. And the fourth like one minus one minus one one. And now you can see the characters of these two groups inside this as follows. So if I call this chi one, chi two, chi three, then this is chi two times chi three. Then you can see the character table of the first z modulo two z in sitting in here like this. So you remember the character table of z modulo two z just looks like one, one, one minus one. And the character table of the second one is sort of sitting inside like this. And in some sense, you just sort of multiply these two character tables together to get the character table of the product group. Again, you can see the rows and the columns are orthogonal. You can show this either by just sort of brute force checking since we have explicit formulas for everything, or you can just prove it directly. So what we want to do for an Abelian group is to prove the sum over g of chi i g times chi j g complex conjugate is equal to naught for i not equal to j. And we may as well multiply chi i by chi, by the chi j to the minus one. So this is just equal to sum over g of chi i chi j to the minus one, g times one. So we just want to show that the sum over g of chi of g is equal to naught if chi is not equal to the identity character. And if chi is not equal to the identity character, we can just choose some element h with chi of h not equal to one. And then we can change g to g h. So sum over g of chi of g is just sum over g of chi of g h. That's just a change of variable. And this is the sum over g of chi of g times chi of h because chi is multiplicative. So you see this expression is equal to this expression times chi of h. So sum over g, chi of g is equal to naught as the chi of h is not equal to one. So this shows orthogonality of characters for any Abellion group. In particular, you see the characters, you can think of a character chi as being a function from g to the complex numbers and these form a base for the vector space of functions from g to the complex numbers. That means you can expand any function of g to the complex numbers as in terms of the characters as sum over f i i bar. I guess I better divide by the order of g of chi i g. So this is called the Fourier series of the function f. And the reason for that is that it's more or less the same as the usual Fourier series for a periodic function. So suppose I take g to be the circle group which is the reals quotient out by two pi times z. So okay, this isn't actually a finite group but it's compact which isn't too far from being finite. And it's got characters, the characters are chi i which just take a number x to e to the i n x. I guess if I'm using i for a complex number I better call this J instead of i. Sorry, let's call that n. So for each integer n and z, I've got a character taking x in s one to this periodic function. And these are orthogonal because the integral from zero to one of e to the i m x e to the i n x bar the x is equal to naught for m not equal to n. So that's just like the orthogonality relation we had here except I'm instead of summing I'm integrating over the group. And we have a Fourier series which says that if f is a periodic function on the group G then f is equal to the sum over or n of the integral of f of x e to the minus n i x dx divided by two pi times e to the i n x. And if you look at this expression here it's almost identical to this expression here except we've replaced the summation by an integral we've replaced the order of G by the measure of G and so on. So the Fourier series expansion for finite abelian groups is and the Fourier series for a periodic function of both really special cases of a Fourier series for abelian groups. In particular, we notice I guess I better just remark the dual of the circle group s one is the group of integers z. And more generally, as I mentioned there's a dual for any locally compact abelian group and an analog of this. Now I'm going to give a few examples of what happens over other fields and also what happens for infinite groups. So you remember in order for this to work well we needed to use the fact that the complex numbers had lots of roots of unity and also a characteristic zero. So first let's look at what happens if there are no roots of unity or it's not enough roots of unity but let's assume the characteristic is still zero. Well, what happens then is things get slightly more complicated but they're still fairly easy to handle. So for example, for G being of order that's suppose G is cyclic of order three then we find there are now two representations over the reals. One representation is dimension one and if this is generated by G we have the element G acts as one. And secondly, there's an irreducible representation of dimension two where G acts as rotation by pi over three radians. So it's just a rotation by third of a revolution. And in some sense this representation dimension two has got by gluing together the two one dimensional complex representations. So if you take this representation let's call this V and to complexify it it turns out to be the sum of V one and V two wherein V one G is equal to omega and V two G is equal to omega squared. So it's not too difficult to deal with the case when the field doesn't have roots of unity. The only difference is you find some representations some irreducible representations might have dimension greater than one. However, the second case is if the characteristic is greater than naught can get a lot more complicated. Suppose the characteristic P is greater than naught if P doesn't divide the order of the group then you're fine you can just continue as before. So if P does not divide the order of the group G this is easy. If P divides the order of the group G things get complicated. First of all, there are lots of indeed composable representations but need not be irreducible. So you can't divide everything up into a sum of irreducible representations. And suppose for example that G is a generation of G and G is ordered P to the N. Well, we notice that G minus one to the P to the N is equal to zero because we're working in characteristic P the binomial coefficients P to the N choose I for all equal to zero for not less than I less than P to the N. So this is nil potent. Well, if you've got a nil potent matrix over a field you know it has a sort of block form. It can be written as a lot of blocks each of which has zeros down the diagonal ones just above it and so on. So G minus one might look something like this. And you see each of these blocks corresponds to an indeed composable representation but it's reducible because it's got a one dimensional invariant subspace. Well, that's for cyclic groups. Cyclic groups aren't too bad because we can sort of classify nil potent endomorphisms and pushing them in Jordan normal form. However, if G is non-cyclic this gets very complicated. This runs into the problem of trying to classify two commuting nil potent matrices over a field. And this turns out to be a complete nightmare if the field has reasonably large dimension. As far as I know, nobody really knows a good way to classify all indeed composable representations of even an abelian group in characteristic greater than zero. Finally, I'll just say what happens? Suppose G is infinite and acts on V with the dimension V being infinite. Well, here you've got to be rather careful about your definitions otherwise you run into some rather horrible examples. So you might say, if V is infinite dimensional obviously we should give it a topology and the action should be continuous. So let's take V to be say a Banach space and let's take G to be the integers. And the question is can we find an irreducible representation of G which is the integers on a Banach space of dimension greater than one. And this representation should be continuous. So it should be by bounded operators. And this is the notorious invariant subspace problem. So it asks, can you fight if you've got an operator on a Banach, bounded operator on a Banach space in infinite dimension doesn't necessarily have an invariant subspace. And the answer is it doesn't necessarily have an invariant subspace. So the answer to this question is yes. And the first example was found by Enflow and his example was nightmarishly complicated. It took about 10 or 12 years for people to actually verify that his example was correct. It was so difficult. I mean, several people tried to verify it and simply gave up. So incidentally, this is still open for Hilbert spaces. So what this example shows is not that the integers has an irreducible representation in infinite dimensions, but you've got to be rather careful how you define irreducible representations in infinite dimensions because you really, really do not want to count this thing as a representation of Z. So you have to define infinite dimensional representations in ways it excludes these rather pathological examples. Okay, so next talk, I'll be giving some more examples of representations of Abelian groups and showing how to use them to prove Dirichlet's theorem about primes in arithmetic progression.