This lecture on representation theory will be about the orthogonality relations for a finite group. So in almost always in this lecture G will be finite, finite group, and all representations will be finite dimensional over the complex numbers. So a typical example of such a character table is one we had in the first lecture where you took the character table of S3 and we saw the character table look like this where these rows correspond to the three characters and these correspond to the three conjugacy classes where this conjugacy class is size 3, this is size 1 and this is size 2 and you can see that all the columns are orthogonal and the rows are orthogonal as long as you weight things by the size of the conjugacy class. So the first thing we're going to observe is that any representation of G is unitary. That means it's got a positive definite unitary form on it invariant under group G and this follows because we take any unitary in a product and then take the average under G and we can do this because the complex numbers of characteristics zero so we can divide by the order of G in order to take averages. Over more general fields if you take a non-zero in a product and take the average it might turn out to be zero or something like that. And this is the consequence that any representation is a direct sum of irreducibles and this is true for any unitary representation because given a representation V contained in a representation W you just write W is equal to V plus the orthogonal complement of V under the invariant in a product and this will be invariant under the group G. So as all representations are finite dimensional this shows by induction on the dimension that they're all sums of irreducibles. Next we observe that if V is any representation of G not containing a vector fixed by G other than zero of course then the sum overall G of the character of G is equal to zero. So chi of V of G is the character of V which is just the trace of G on the representation V. So this is a special case of the orthogonality relations just saying that V is orthogonal to the trivial representation with character one everywhere. And the point is if V is any element of the representation V then sum over G of G of V is fixed by G so is zero because we assume that's got no vectors fixed by G. So the sum over G of G V W is equal to naught for any V W. So here we're just taking a unitary in a product on the representation. So in particular the sum over G I of G V I V I is equal to naught here V I is a basis. And this is just sum over G of the trace of G on the representation V which is just sum over G of chi of the character of V at the point W. A special case of this is the number of times any representation V contains the trivial representation with G just acting on C as every element of G fixing it is just the sum over G in G of chi V of G divided by the order of G. That's because this expression is one for the trivial representation and zero for all other irreducible representations. Next we have some basic properties of the character. For chi of V plus W is equal to chi of V plus chi of W. That's sort of obvious because the trace of something on the sum of two vector spaces is the sum of their traces. Similar one is chi of V tensed with W is equal to chi of V tensed with chi of W. And the easy way to see this is just to diagonalize the action of G on V and W. So if we can choose a basis with G V I is lambda I V I and G of W I is equal to mu I W I and then we see the trace that G of V I tensor W J is lambda I times mu J times V I tensor W J. So the trace of this will be the sum of the lambda I's the trace of this will be the sum of the mu J's and the trace of this will be the sum of lambda I times mu J over all I and J. Next we see the character of the dual of V is just the complex conjugate of the character of V. This is this is because you've got a unitary representation and the character of V star. That's G is just trace of G to minus one on V, which is equal to the trace of G on V complex conjugated. Finally, the character of the space of homomorphisms from V to W. So if you've got two vector spaces, then G is going to act on the homomorphism from one to the other. And this will just be the character of W times the character of V complex conjugated. And that's because this is just isomorphic to W tensor with the dual of V because everything is finite dimensional. So it's very easy to work out what the character is for any of these operations on vector space is taking sums or duels or tensor products or whatever just does the obvious thing. Next we have Scherzlemmer, which says that if V and W are irreducible, then the homomorphisms from V to W that commute with the action of G has dimension one if V is equal to W and not if V is not equal to W. And that follows easily because if V is not equal to W, then these are both irreducible. So the image is either north or the whole of W, and the kernel of any map is either north or the whole of W. So both the image and the kernel must be zero. So this is zero. On the other hand, if V is equal to W, notice hom of G V to V is a division algebra. That's because any homomorphism from V to V, if it's none zero, then because V is irreducible, the image must be the whole of V and the kernel must be the whole of V. So it must have an inverse. Well, it must be a division algebra over the complex numbers and the complex numbers are algebraically closed. So it must be just the complex numbers and is one dimensional. And if you're working over fields other than the complex numbers, then this is quite often a non-trivial division algebra. From this, we see the orthogonality relation, which says the sum over G of I, G, J, G is equal to not if I is not equal to J, where these are the characters of irreducible representations and G if I equals J, rather the order of G. So this just says the rows of the character table are orthogonal if you wait them in the right way. And this follows because hom over G of V, J to V, I is either one or zero, depending on whether these are the same or not. On the other hand, by what we said earlier, it's equal to the sum over G of the character of hom V, J to V, I, which is equal to the sum over G of, well, we worked out what the character of this is. So this is the character of I times the complex conjugate of the character of J. Sorry, there should be a complex conjugate in there. So since the number of times the trivial character occurs in something is just given by the sum over the character, we get the orthogonality relations. In particular, if V is equal to a sum of irreducible representations with multiplicity Ni, we see that Ni is just the inner product of the character of V, and the character of the irreducible representation Vi, where this is just short for one over G times the sum over chi, I, G, sorry, chi V of G, chi Vi of G complex conjugated. So it's just the normalized inner product of the character of V with the character of the irreducible representation. An important consequence of this is that a representation V is determined by its character, because since it's completely reducible, it's the sum of irreducibles and the multiplicity of these irreducibles is determined by the character. Overfields of positive characteristic, it's not true that a representation is determined by its character, and essentially this breaks down because in positive characteristic representations are not completely reducible. So we've done orthogonality of rows, we would like to do orthogonality of columns, but there's a little bit of a problem. We first got to prove that there are enough characters. So how do we construct lots of non trivial characters. So the regular representation of G. So the regular representation is given by the action of G on the group algebra and C of G. So this is this is just a vector space with basis. It's G and C, and the multiplication is defined in the obvious way the multiplication this algebra is the same as the multiplication in the group. So this is a vector space of dimension is just the order of G. So the vector of the regular representation is pretty easy to work out. This is equal to the order of G for G equals one obviously, and not if G is not equal to to one because if she is not equal to one it, it permutes the basis without fixing any of them so just trace zero. So we can now work out the regular representation as a sum of irreducible representations. What's N I will N I is the inner product of the regular representation of G with the character of V I. And if you work it out this is very easy because this is zero except the identity element and the identity element. This is just the dimension of V I so this is the dimension of V I. So this is equal to the sum over all representations, each occurring with number of times equal to its dimension. By the way, this fails for infinite groups in general. In particular this says that for finite groups this says every representation occurs in the regular representation which makes it easy to find for infinite groups, irreducible representations quite often don't occur in the regular representation and they're much harder to find. One consequence of this if we just take the dimension of both sides we see the order of G, which is dimension of this is equal to the sum over all irreducible representations, the dimension of V I squared because dimension of this is just the sum of the squares of the dimension of the irreducible representations. So, let's carry on a bit and actually show that there are plenty of irreducible representations. You notice the center of the group algebra. As a basis, consisting of the element sum over G in C I of G was C I is a conjugacy class of G. So the dimension of the center is just the number of conjugacy classes. And the center acts faithfully on the representation C of G. That means if, if an element of the center is non zero attacks non trivially on this. Moreover, you can also see the center acts as a scalar on any irreducible representation V I and from the fact that it acts faithfully on C C of G which is a sum of irreducibles and it acts as a scalar on any irreducible. This implies the number of irreducibles must be greater than or equal to the dimension of the center of C of G, which equals the number of conjugacy classes. On the other hand, we also know the number of irreducibles must be less than or equal to the number of conjugacy classes by the orthogonality relations. Because the character of irreducibles are functions on the conjugacy classes and they're all orthogonal so the number of irreducibles is at most the number of conjugacy classes. And if we put these two together, you see that the number of irreducibles is equal to the number of conjugacy classes, which is very useful because it tells us when we found all the irreducibles. As soon as we found enough, the number of irreducibles we found is the number of conjugacy classes, we know we found them all. By the way, this thing fails in a characteristic greater than zero. If you're looking at representations over fields of characteristic P, then the number of irreducibles can be strictly less than the number of conjugacy classes. And the reason this fails is that in this proof, we use the fact that the regular representations are some of irreducible representations and that fails in characteristic P and this can also fail in characteristic P. Well, having done that, we can now get the column orthogonality relations. So we first write down the row orthogonality relations. So the row orthogonality relations say that X times C times X bar transpose is equal to G times the identity matrix. So I'll just say what everything is. So this is the character table. This is a matrix C1, C2, and so on with zeros up here, where Ci is the size of the ith conjugacy class. And this is the conjugate transpose of the character table. This is the order of G, and this is just the matrix 1111 and so on. And if you write this out, you'll see the row orthogonality that we proved earlier is just the same as this relation here. And now we see that X is an invertible matrix, so we can just move it to the other side. And we find that X bar transpose times X is equal to G times C to the minus one. And what this says is the columns are orthogonal. So this says if we've got two characters, sorry, if we've got two elements, then the sum over all irreducible characters of Chi i of G1 times Chi i of G2 is equal to not if Chi G1 and G2 not conjugate. And it's equal to the order of G over the size of the conjugacy class if G1 and G2 are conjugate, where this is the size of the conjugacy class of G1. So that covers the basic orthogonality properties of the character table. So let's just have a quick summary.  The character table is square. In other words, the number of irreducible characters is the number of conjugacy classes and the rows and columns are orthogonal, where orthogonal means you've got to weight them and you're using a Hermitian inner product and so on. And the norms of the rows are just the order of the group G and the norms of the columns of the order of G divided by the size of the conjugacy class. Finally, the sum of the squares of the irreducible representations is the order of G. So these are the basic facts about the character table. And they're very powerful. We're just finished with an example. Let's take G to be the alternating group A5. So G has ordered 60 and the number of conjugacy classes is five. And from this information, we can work out the dimensions of the irreducible representations. This sort of illustrates the bizarre fact about character theory because there just doesn't really seem to be enough information to work out the degrees of the characters here. I mean, all we know is the order of the group and the number of conjugacy classes. You know, it sort of sounds like there might be lots of groups like that. Well, all we need to know is the sum of the squares of the irreducible representations is the order of G. And we know there's one irreducible representation of dimension one. So we want to solve one squared plus A squared plus B squared plus C squared plus D squared equals 60. And if you do some trial and error, it's not very difficult to see. There's only one solution of this equation with these being positive integers, which is one squared plus three squared plus three squared plus four squared plus five squared equals 60. So the dimension of the irreducible representations are one, three, three, four and five. Okay, next lecture, we will use these properties to calculate a few character tables of finite groups.