 So this talk is one of several talks on representation theory and it will be an introduction to representations of groups. So what is a representation of a group and why should we wish to find them? Well, if you're really ambitious, what you might want to do is classify all symmetries of all mathematical objects. Obviously, that's a little bit over ambitious, but let's see what happens. Well, first of all, you notice that the set of symmetries of a mathematical object is a group. So we can identify symmetries. This kind of corresponds to groups. So our problem splits into two problems. First, we could classify all groups and secondly, given a group G, find all things that G acts on. Well, the first problem is obviously hopelessly impossible. There's no way we can classify all groups. We can't even classify all finite groups. The second problem is essentially representation theory. So a thing that G acts on is called a representation of the group. So our problem is, suppose someone hands us a group, we want to be able to find all possible representations that's things the group acts on. So what sort of things can G act on? Well, it could act on any mathematical object, topological spaces or the algebras or whatever, but in representation theory, we're mostly interested in the following things. So G can act on, well, first of all, it can act on a set. And an action of G on a set is called a permutation representation. In the early days of the group theory, permutation representations were essentially the same as groups. A group was basically defined as a set of permutations of some set, usually finite. Another thing a group can act on is modules over a ring. And this ring would quite often be a field. And these are called linear representations for fairly obvious reasons, their study is more or less linear algebra. So the ring could be the integers or it could be a field. And if it's a field, it could be the complex numbers or it could be a field of characteristic P greater than zero. And representations over the integers and representations over fields of characteristics greater than zero, turn out to be rather complicated for reasons we'll discuss in a moment, but representations over the complex numbers of finite groups turn out to be rather well behaved. So let's have a look at some examples of representations. So let's pick a group G and we will take G to be rotations of an icosahedron. So here is an icosahedron and we are going to look at all its rotations and you see it's got 20 faces and each face we can rotate three ways. So it has 60 symmetries and let's find some representations of it. Well, first of all, we notice that it has 20 faces and it has 30 edges and it has 12 vertices. So these give us three permutation representations and in general, if a group acts on a mathematical object you can get lots of permutation representations by just looking at points in the object it's acting on. There are some more subtle ones. For example, the vertices all form pairs of opposite vertices, so we have six pairs of opposite vertices which gives six diagonals. So as well as acting on 12 vertices it could act on six diagonals. So there are sometimes some less obvious permutation representations. Well, G also acts on three-dimensional real vector space because the icosahedron lives in our three three-dimensional space and any rotation of the icosahedron is just a linear transformation of our three. So we get a linear representation on three-dimensional real vector space. As we'll see in a few lectures time there are plenty of other actions of this group on real vector spaces. So how do we classify all representations? Well, we can first look at decomposable representations. So these are representations that can be split as two smaller representations. So if we've got a set S acted on by G we might be able to write S as a union of S1 union S2 with S1 S2 acted on by G. If we've got a vector space V we might be able to write V as about some of V1 plus V2 with V1 V2 acted on by G. And if the set S or the vector space can be split up like this we say S or V is decomposable. Of course there should be a non-trivial splitting. So we want S1 S2 not empty and we want V1 and V2 not zero vector space. So any set acted on by G is a union of in decomposable permutation representations. So when you're talking about in decomposable permutation representations we usually call these transitive actions which means that for any two points there's an element of G taking one to the other. And these sub-representations of S are often usually called the orbits of G on S where two elements are in the same orbit if there's an element of G taking one to the other. So for permutation representation it's pretty obvious how to split things up. We just split them up into transitive actions and moreover transitive actions are sort of easy to classify that if G acts transitively on S then S is more or less the same as G over H for a subgroup H of G. So this is just the set of cosets of H and you can easily see that if you've got a subgroup you get a transitive action and conversely if you've got a transitive action then there's a subgroup or more precisely a conjugacy class of subgroups corresponding to it. So that sort of classifies all permutation representations provided you can classify all subgroups of G. However, classifying all subgroups of a group turns out to be usually a hopelessly complicated problem except in a few easy cases. So we can decompose permutation representations in decomposable ones. Can we do the same with linear representations? Well, if the dimension of V is finite we can decompose V as a direct sum V1 plus V2 and so on of in decomposable representations. That follows rather easily by induction on the dimension of vector space. If the dimension of V is infinite things get really rather hairy. First of all, if your vector space is infinite you probably ought to be thinking about a topology on it and you want to have these to be closed sub-spaces otherwise you get into a real mess. Then you can get weird representations that you can just keep chopping on smaller and smaller pieces and this process never comes to an end. So for infinite dimensional representations you can't always decompose into in decomposable ones. We're going to avoid this problem just by sticking to finite dimensional representations and we'll simplify further by usually talking about finite groups. So we would like to classify all in decomposable representations and let's do of a finite group on a finite dimensional vector space. Well, we still run into a problem. This is the problem of the difference between in decomposable and irreducible representations. So V is called decomposable if V is equal to V1 plus V2 with V1 and V2 not zero. V is called reducible if V has a subspace V1 acted on by G and of course we take V1 not equal to zero and V1 not equal to the whole space V otherwise it's kind of trivial. And the problem is that it's obvious that decomposable spaces are reducible so irreducible spaces which are not reducible are in decomposable. So irreducible implies in decomposable and the trouble is the converse isn't always true. We can sometimes get in decomposable vector spaces that aren't reducible and the standard example is the following. We can take the group of reals acting on a two-dimensional space over R by the following matrices. We take the matrix 11x0 and this is reducible. It's got an obvious subspace consisting of the vectors nor something which is a map to itself by these but it's in decomposable which is fairly easy to check. If we call this subspace V1 and V2 then we see we have a rather bizarre phenomenon that G acts trivially on V1 and V2 over V1 meaning every element of G acts as the identity but not on V2 itself which is a bit bizarre. I mean you thought that if a group acts trivially on a subspace and the quotient space it should act trivially on the whole space but here's a counter example. You're going to say this group is infinite and I said I was going to be talking only about finite groups so this example is cheating a bit. Let's do an example for finite groups. This time I'm going to take G to be the cyclic group of order P and I'm going to act on a two-dimensional vector space over the field with P elements and it's going to act like this where you can think of G as being the additive group of this finite field and again the same thing happens. Well this is a vector space over a field of characteristic greater than P sorry greater than zero. So indeed composable and irreducible representations may be different for infinite groups and they may be different if you're even for finite groups of the field as characteristic P over the key point is that for complex representations of a finite group irreducible is the same as indeed composable and when this happens representation theory becomes very much easier. What it means is that all representations are what is known as completely reducible meaning there are direct some of irreducible representations and the rule of thumb is that if your representations are irreducible then life is easy so this happens for complex representations of finite groups it also happens for complex representations of compact groups and for unitary representations of more general objects in cases when you get in decomposable representations that are not irreducible you're doing things like modular representation theory which turns out to be much more complicated so we now have the problem classify all irreducible representations of G where we're going to take this to be finite for the moment and these to be complex representations let's start by looking at a very simple example just to see what happens let's take G to be a group of order two generated by one element G G squared equals one and let's suppose G acts on V which is a complex vector space and it's really easy to figure out what G does to this V just splits as V plus plus V minus where here G acts as one and here G acts as minus one and that's because we can write any vector V as V plus G V over two plus V minus G V over two and this is in V minus and this is in V plus so furthermore V plus is with G just acting as one can be written as a sum of one-dimensional spaces you can write as one-dimensional spaces any way you like and this will be they will all be invariant under G and similarly V minus is also a sum of one-dimensional spaces so every complex representation can be split up into a sum of one-dimensional spaces and there are only two of them and we draw this by doing the character table of G so the character table is a way of listing all representations of the group and for the group of order two of the character table will look like this first of all we write out well you think we're writing out the elements these aren't the elements these are the conjugacy classes but since G is a billion we've only got two conjugacy classes each with one element next we write out the first representation by writing down its character and this is a Greek letter chi which is often used for a character and the first character is going to be denoted by this so this is a character of a one-dimensional representation and what is this character well each entry here is the trace of the element on the representation V well here V is just going to be dimension of V is going to be one and G is going to be one on V so both one and G have trace one so the character is just one and one the other representation let's call it chi two again one dimension so the element one is trace one but now the element G acts as minus one so it is trace minus one so here's the character table of this little group of order two so the rows are characters one for each irreducible representation and the columns correspond to conjugacy classes and the entries just give the trace of some element of G on the corresponding representation and you notice the trace only depends on the conjugacy class because trace of A, G, A to the minus one is equal to the trace of G of course for this group of order two that doesn't matter because it's a B in the end but in general the character table will look like this so why do we use the trace in order to represent the representation? That I've never really quite understood it's a sort of mysterious fact discovered by the early pioneers of representation theory like Frobenius that giving the trace of every element is a really efficient and useful way of describing a representation and we'll see a little bit later that you can usually reconstruct the representation just knowing the trace of every element so this is a really compact way of describing a representation so we now have the problem given a group G how do we calculate its character table? Well we've done this for a group of order two let's do something slightly more complicated so let's take G to be the group S3 the symmetric group on three letters 1, 2 and 3 which is order factorial equals 6 and let's write down its character table and for this we're going to use a combination of inspired guesswork and just plain cheating so let's first of all write down the conjugacy classes so there's the conjugacy class well first of all there's the identity element next there's a conjugacy class consisting of three transpositions and finally there's a conjugacy class consisting of the two three cycles so our character table is going to have three columns and let's find out its first representation well that's really easy because we can just take the dimension of our vector space to be 1 and let G act trivially and then every element as trace 1 so this is the trivial character which obviously every group has can we think of any other characters? Well you remember every symmetric group has this notion of having even cycles and odd cycles so there's a homomorphism from s3 to z modulo 2z which takes even cycles to 1 or 0 and odd cycles to element 1 and z2 and this gives us an action on the complex numbers where even cycles go to 1 and odd ones go to minus 1 so we've got a one dimensional representation and even cycles which are that one and that one of trace 1 and odd cycles of trace minus 1 can we think of any other representations? Well yes there's another one because G acts on c3 you see it acts on 3 points we can just take a basis consisting of 3 elements x1, x2 and x3 with G permuting these 3 basis elements or if you want to write in coordinates it's just permuting the coordinates and that gives us a representation and what is its character? Well we need to work out the trace of each element on this 3 I'm going to write it in green for a reason you'll see in a moment and the trace of any other element if you think about it is just the number of points fixed by that element because for example a transposition will have a matrix that looks like this and a 3 cycle will have a matrix that looks like this and that has trace 0 because it's got no fixed points so I get 3, 1, 0 as my character and next we ask is this irreducible? and the answer is no because this splits as a sum of 2 representations so first of all you can have a subrepresentation with x1 equals x2 equals x3 and you can see that's just the same as the trivial one dimensional representation that's got another representation given by x1 plus x2 plus x3 equals 0 so this actually splits as the sum of something 2-dimensional and something 1-dimensional and we can work out the trace of the 2-dimensional bit by subtracting the trivial representation from this so we get our representation chi3 which looks like 2, 0, minus 1 so here is our character table except we should cross out this green bit because it was just sort of scaffolding that we were using to work out this representation and this character table has a lot of properties first of all we notice the columns are orthogonal as you can easily check next you notice the rows are orthogonal well you may be a bit suspicious about this because if you look carefully at the rows you can see that strictly speaking they are not orthogonal I mean 1 times 1 plus 1 times minus 1 plus 1 times 1 is certainly not 0 so what on earth do I mean well that's because you've got to weight them by the sum by the number of elements in each conjugacy class so we weight by size of the conjugacy class if you like you can think when you work out whether these two rows are orthogonal you shouldn't be summing over conjugacy classes you should be summing over all elements of G so you have to sum 3 times this now we find 1 times 1 minus 3 times these 2 plus 2 times that 2 is indeed 0 so the rows are orthogonal and you see also all the rows have norm equal to the order of G where the norm means the inner product of a row with itself again weighting by the conjugacy class you notice the columns also have a norm and the norm of the column is the order of a group divided by the number of elements in each conjugacy class so here's a conjugacy class of 3 elements so the order of G over 3 is 2 which is 1 squared plus minus 1 squared so there are lots and lots of orthogonality relations you also notice that the character table is square so I'm going to kind of cheat by just announcing that this is all the characters there are and so in general for a group you can describe all the representations by just writing down the character table so this is what the character table of a rather easy small group looks like what does the character table of a really big group look like well I've got an example here we can look at the character table of the monster group which is order about 10 to 54 and if you look in the let's see if I switch to not present change so here if I hold this up you can see a small piece of the character table of the monster actually takes 8 pages like this not just 2 so it's got about 40,000 entries but that's a lot more efficient than trying to write out the multiplication table of the monster which has so many entries that it wouldn't fit into the visible universe if you tried writing it out so I can show you a small piece of the monster character table in more detail so if you look at this you can see the top left hand corner of the monster character table if I zoom in a bit you can see the first column there's the order of the monster and these are the first entries which are the trace of the identity element on the representation so you can see that the first non-trivial representation is dimension 196883 and the next one is even worse and so on so that's what a really big character table looks like so I just finish by answering the question what is the point of finding representations of a group well the answer is that using representation theory you can give quite easy proofs of things that would otherwise be extraordinarily difficult to prove for example we have the Burnside p to the aq to the b theorem so what does this say it says that a group p to the aq to the b is solvable in particular it can't be simple here p and q are of course primes and this was originally proved by Burnside using representation theory and the representation theory proof is amazingly short and it looks like a piece of black magic you do these slightly bizarre calculations with characters of groups using a little bit of modular arithmetic and the fact that a group of order p to the aq to the b just sort of drops out for some mysterious reason there are proofs of this without using representation theory but they are incredibly difficult and no proofs were found for many decades after Burnside found his proof using representation theory incidentally the smallest order of a non solvable group is 60 which is 2 squared times 3 times 5 which you see is divisible by 3 different primes so Burnside's theorem only works for 2 primes and definitely can't be extended to 3 primes ok that's the end of the introductory lecture the next lecture will be on representations of a billion groups which is the easiest case