 This lecture will be about the sure indicator of representations. So this is a method for solving the following problems. Suppose that V is an irreducible complex representation of finite group G. So we're going to take G to be finite and V to be finite dimensional throughout. Then we can ask the following questions. First of all, does V have an invariant bilinear form from V tensor V to the complex numbers? And secondly, if so, is it symmetric or anti-symmetric or whatever? And another question we're going to ask is apparently completely unrelated, which is to classify the real irreducible representations of G. So we're going to answer these questions by use of something called the sure indicator. So let's do question one first. We just want to know, does V have a bilinear form? Suppose we have a bilinear form to the complex numbers, which we will assume is invariant and non-zero. Well, if there is a non-zero invariant bilinear form, this is equivalent to saying that V is isomorphic to its dual as a representation of the group G. And this is the same as saying that the character of V is equal to the character of the dual of V for an irreducible representation V, and this just says the character of V is real. In other words, all its entries are real numbers rather than proper complex numbers. So it's easy to tell whether or not an irreducible representation has a bilinear form. What about the dimension of the space of bilinear forms? Well, since bilinear forms correspond to G invariant maps from V to V dual, for V irreducible, the dimension of these must be at most one by Scherzlemmer. You remember Scherzlemmer says that for complex representations, the maps from V to itself is dimension one and from V to anything else is dimension zero. So if the space of bilinear forms is one dimensional, that means bilinear forms can either be symmetric. They're either symmetric. In other words, AB equals BA or anti-symmetric. AB equals minus BA. Of course, in general, a form could be a mixture of both, but if you had a form that was a mixture of both, we could decompose it into its symmetric and anti-symmetric parts, and then the space of forms would be at least two dimensional. So we now have the following question. We know whether an irreducible representation has a bilinear form and we know that if it has a bilinear form, it must be either symmetric or anti-symmetric, and now we have to figure out which case occurs. Well, a bilinear form is going to be symmetric if the trivial representation C is contained in the symmetric square of V and it's anti-symmetric if the trivial representation C occurs in the alternating square of the representation V. And we can figure this out by looking at the characters of the symmetric and alternating square. So you remember the character of the alternating square of V was given by, you take the character of V of G squared and subtract the character of V of G squared and divide it all by two, and the character of the symmetric square of V is similar, but with a sine error. So we take V of G squared minus sine of V of G squared all divided by two. You notice the square is inside the parentheses on one of these and outside on the other. And so at most one has inner product one with the trivial character chi of one, and we want to work out which. So in other words, if we take chi of S squared V and subtract chi of lambda squared V and take its inner product with the trivial character, this is equal to minus one, zero or one. If V has a skew symmetric form, it's zero if V has no invariant bilinear form, and it's one if V has a symmetric bilinear form. So we can work out whether the form on V exists and whether it's symmetric or skew symmetric just by calculating this. And by looking at the expression for chi of S squared V and chi of lambda squared V, we see that this inner product here is given by one over G times sum over G and G of chi of G squared. So this is the famous Schur indicator. So you can work it out for representation and it's one of these numbers and that tells you what possible bilinear forms that are on the representation. We'll see some examples in a moment. I just comment that this case and this case are quite common. This case here for some reason seems to be rather rare. I mean, if you look at a typical group, it quite often has no representations of this sort at all. In fact, quite often all its representations are of this type here. So let's have a look at some examples. First of all, if we look at the, we calculate the representations of various symmetric and alternating groups. And if you look at all the symmetric groups we've done, they all have Schur indicator one. In fact, all complex representations of all symmetric groups always have Schur indicator one, which means they have a symmetric bilinear form and nothing else. And the alternating group A5, they all had Schur indicator one. The alternating group A4, there were a couple of Schur indicator zero. So you remember the character table of A4 looked like this and three minus one, zero, zero. So you see these two characters are real characters. And in fact, their Schur indicator is plus one. These two, the number omega is not real. So the Schur indicator is zero. So these two characters don't have an invariant bilinear form. Well, to see an example with Schur indicator minus one, the simplest example is the quaternion group of order eight. So what I'm going to actually do is I'm going to work out the character table for the dihedral group of order eight and the quaternion group of order eight. So we did the dihedral group of order eight and its character table looks like this. So we've got four one-dimensional characters, and then we found there's one other two-dimensional character. And we can work out the Schur indicators of these. They're all just plus one. So in order to work these out, we need to know what the squares of elements are and the elements of d8 squares follow. So the center squares to one. And then we've got some elements of order two, which square to one. And then we've got an element of order four, which squares to here. So if we work out the Schur indicator of this one here, we find it's two plus two for these two, plus two, plus two, plus two, plus two. Plus two, plus two, plus minus two, plus minus two, which is eight. So we then divide it by the order of the group and we get plus one. If you look at the quaternion group of order eight, you remember the quaternion group has conjugacy classes plus one, minus one, plus or minus i, plus or minus j, plus or minus k. And then just as for d8, there are four one-dimensional characters. Because the benionization is the client four group. And again, by the orthogonality relations, there's a two-dimensional character. And if you look at the character tables of d8 and q8, you'll notice the character tables are actually identical. However, there's a sort of subtle difference in the representations, because if we work out the Schur indicator of this, well, we first have to work out what the squares of everything are. Now the squares are a little bit different, because now everything squares, every element of order, these elements all have order four and square to this element here, not this element. So if you work out the Schur indicator, we get two plus two, plus two, plus two, plus minus two. We're going to get two twos there, plus minus two, plus minus two, plus minus two, plus minus two, plus minus two, plus minus two, which is minus eight. And then when you divide that by the order of the group, we get minus one. So this representation here has a skew symmetric invariant bilinear form, whereas this one has a symmetric invariant bilinear form. So I'll give an application of the Schur indicator to groups of odd order. So suppose the order of G is odd. Well, what's the Schur indicator? It's sum over G in G of chi of G squared. What's that? Well, that's easy to work out because if G is odd, then the map from G to G squared is a bijection from G to itself. So this is equal to the sum over G in G of chi of G, which is just the inner product of chi with the one-dimensional irreducible character chi one. So this is equal to one if chi is trivial and not if chi is non-trivial and irreducible. So the only irreducible character of a group of odd order with an invariant bilinear form is the trivial character. All other representations have no invariant bilinear form and come in complex conjugate pairs. There's a sort of funny application of this. We know the sum of di squared is equal to the order of G, where the di is the dimension of the ith character, which is I guess it's chi by one. And since the characters come in complex conjugate pairs, we can write this as let's say the sum over the non-trivial characters where we're going to take one of each complex conjugate pair is equal to G minus one over two. Now, in fact, di is always odd because it divides the order of G. This is a theorem that we'll prove a little bit later. But assuming it for the moment, we see that di squared is congruent to one modulo eight because every odd number squared is congruent to one modulo eight. So we see that n minus one over two is congruent to G minus one over two mod eight where n is equal to the number of irreducible characters of chi, which is the same as the number of conjugacy classes. And if we double this, we find that chi is equal to the number of conjugacy classes mod 16. So in particular, we see that any group of order of order less than 19 must actually be a billion because the number of conjugacy classes must be equal to the order of the group mod 16 so it must be equal to the order of the group. So all conjugacy classes of size one, in fact, even if there's order 19, it must still be a billion because 19 is prime. But when you get to order 21, there is actually none a billion group of order 21. And you can see from this that it must have exactly five conjugacy classes, which is of course not hard to check directly. Another application of the shirt indicator is to classify the real representations or the real irreducible representations of group G. Here there are three types because the G invariant maps from V to V must be a division algebra over the reals. And there are three possible division algebras, could be the reals or the complex numbers or the quaternions. And for each of these, you can turn this into a complex irreducible representation and the way you do that is slightly different in these three cases. So, so let's look at the three possible types. So we've got here's the representation V and here I'm going to let its dimension be N or 2N or 4N. And here the HOMs from V to G from V to itself can either be the reals or the complex numbers or the quaternions. So this side here are real representations, V is a real vector space. Now from each of these I'm going to construct a complex representation of G. So for the first one, I just take V tensed over R with C. And this turns out to be an irreducible complex representation and its dimension as a complex representation is N. And furthermore, this has a symmetric bilinear form. So the bilinear form on this is going to be symmetric. And you can see that because this space here always has a symmetric bilinear form. You can take a positive definite quadratic form and just make it invariant under the group G. And this is also invariant under norm one elements of R, C or H. At least if you, again, you can make it invariant under these by just averaging over the norm one elements of this. So the invariant symmetric form on V just becomes bilinearity an invariant symmetric form on V tensed over with C. Well, if V is, if the hom from V to V is the complex numbers, then instead of taking V tensed with C, we just take V is an irreducible complex representation. Because now that the, if the linear maps from V to V are isomorphic to the ring C, then that means C is really a vector space over C. We also get a second one because we can take a sort of complex conjugate of V. So we actually get two different representations. And on these representations, there's no invariant complex bilinear form. I mean, there's a real one, but it's not complex. And finally on the quaternions, we again take V and we can now put a complex structure on it by using I in the quaternions to give complex structure. So it's automatically a complex vector space. Now it has an anti-symmetric bilinear form. And this anti-symmetric bilinear form can be defined by setting AB equals AB times J. So I'm going to write the quaternions acting on the right plus I times ABK. And you can check that this formula makes this into a complex bilinear map that is anti-symmetric. So it gives an anti-symmetric form on this. And here V has complex dimension N and V bar has complex dimension N. And here V has complex dimension 2N. So this gives the, if you know the real irreducible representations, you can work out the complex irreducible representations from it. And conversely, you can work backwards. I'm not actually going to do this because it's not too dissimilar, but it's slightly trickier than going from the real representations to the complex representations. And to go from the complex representations to the real representations, you notice the sure indicator in these three cases is plus 1, 0, or minus 1. So if you know the complex representations, you can find the real representations by just working out the sure indicator and then you just get the real representations by reversing this process. For example, suppose you want to know the real representations of A4. Well, we look at the character table, which looks like this, and we look at the sure indicator, which is plus 1, plus 1, 0, 0. Well, plus 1 gives us a real representation of the same dimension. So we get real representations of dimension 1 and 3. For sure indicator 0, what happens is these two complex representations both become the same real representation of dimension 2. So here are the dimensions of the real irreducible representations of A4. Let's just do the quaternion group as an example with a quaternionic representation. So we write down the character table, which should be minus 1. And here the sure indicators of plus 1, plus 1, plus 1, plus 1, and minus 1. So here we get real representations of dimension 1 by taking a real form of these complex numbers. And here this complex representation of dimension 2 becomes a real representation of dimension 4. So here are the dimensions of the four real irreducible representations of the quaternions.