 This talk on mathematical representation theory will be about induced representations mostly of finite groups. So mainly be defining what an induced representation is and then giving an example of one. So, suppose we have a homomorphism of groups from h to g, then if we've got a representation of g, in other words a vector space acted on by g, it's pretty obvious how to get a representation of h from it because you can just restrict the action of g to the action of h, even if the map from h to g isn't injective. And we've got the problem. Can we find a sort of opposite map? Can we go from a representation of h to a representation of g in some natural way? Now, it's not going to be an inverse of this map here. Inverses don't really quite make sense. Instead, it's something called an adjoint. So the answer is yes. We take an adjoint map of this restriction map. So we can call this this map here restriction and there are actually two different sorts of adjoints. There can be a left adjoint or a right adjoint. So I'll explain what these are. And these actually give us in general two slightly different concepts of induced modules. And easy way to see this is actually to sort of forget about the fact we're working with groups for the moment and look at a homomorphism of the corresponding rings. So we're going to take the the group ring of h. We may as well take it over the complex numbers and the group ring of g. So we've got a homomorphism from the group ring of h to the group ring of g. And if v is a module over s, we can restrict it to a module over r just by defining the r action using the homomorphism from r to s. So we've got a restriction map here. Now if we've got a module of w over r, there are two natural ways to turn it into a module over s. First of all, we can take hom over r of s to w and we can make this into an s module using the action of s on itself by right multiplication. On the other hand, we can take w and we can map it to s tensor over r with w. So this will now be an s module because s can act on this bit. Now this map here is a left adjoint. So what does left adjoint mean? Well, it means that if you've got a map of r modules from w to v, it's more or less the same as a map of s modules from s tensor over r w to v. That's essentially the the meaning of an adjoint, although the exact definition is a slightly more fussy. On the other hand, this is a right, this one here is a right adjoint. What this means is that maps of r modules from v to w are the same as maps of s modules from w to this thing here. And in fact, these properties define these objects here up to canonical isomorphism, they turn out to be given by by the these expressions I've written down. So if we take, if you remember, r is really a group algebra and so is s, we see that we have two ways to turn an h module into a g module. We can take c of g tensor over c of h of w, or we can take h over c of h c of g to w. And these are called induced representations. The problem is they're not actually the same and the term induced representation is in general a little bit ambiguous because there are two possible different things it can mean. And for this reason, some people will call this one a produced representation. So I think produced representation usually refers to this induced representation in turns out to refer to either of these. Well, the ambiguity doesn't really matter if you're doing representations of finite groups such that h is a subgroup of g. So if h is a subgroup of g and if g is finite, or more generally if g over h is finite, then c of g is a finitely generated free c of h module. And this makes everything particularly easy. In fact, if c of g is a finitely generated c of h module, then these two modules are actually chronically isomorphic. So for finitely generated so for finite groups where h is a subgroup of g, there's only one concept of induced representation, which can be defined in either of these two ways. Furthermore, it behaves particularly well because left adjoints are always right exact. I'll explain, remind you what this is in a moment and right adjoints always left exact. So if the right adjoint is equal to the left adjoint, then it's right and left exact and so it's exact. So in this particular case, not only are these the same, but they're both exact, meaning they preserve exact sequences. So for finite groups, we don't need to worry about this ambiguity. So for most of the time, I'm going to be just talking about representation of finite groups, but before doing that, I'll just show you what can go wrong if h is not a subgroup of g or not isomorphic to a subgroup of g. Well, in that case, things get rather more complicated. Let's look at the simplest possible example. We just take the group of order 2 back into the group of order 1. So this is going to be h and this is going to be g. And then we've got two ways of mapping modules over h to modules over g. We can either take w to cg tensor over c of hw, which essentially takes w to w modulo 1 minus sigma of w. Here sigma is going to be one of the two elements of z modulo 2z. So we're going to take sigma as a generator of order 2 of this group here. On the other hand, we have the map taking w to home over c of h to c of g to w. And this takes w to the fixed points under sigma. And these are different and they're not necessarily exact. And you can see this fairly easily if you just look at the following exact sequence of h modules. Here I'm going to let sigma act as minus 1 on these two modules here. And this is going to be multiplication by 2. And now if I look at the modules, w goes to cg tensor over c of hw. I get the following sequence. I get z over 2z goes to z over 2z goes to z over 2z goes to 0. And this is multiplication by 2 and it's not injective. So this form of induction doesn't preserve exactness. On the other hand, if I take w to the fixed points, this sequence now becomes nought goes to nought goes to nought goes to z over 2z. And you see this is not on 2. So if h is not a subgroup of g, things get a lot more complicated. The two sorts of induction become different and they need not be exact. What you get instead is homology and cohomology of groups. If we take g to be the trivial group, you can take derived functions of this and these are called homology of h and you can take derived functions of this and that's called cohomology of h. So cohomology and homology of groups and induced representations are really both special cases of the same construction. Anyway, what we're going to do for the rest of the talk is forget about the case when h is not a subgroup of g and just look at the case when h is a subgroup of g. So we're now going to suppose h is a subgroup of g and we're going to suppose g is finite. If g isn't finite, then again things become rather complicated. So in this case, we have an induced representation from h to g of w can be identified as either z of c of g tensor over c of h of w or as hom over c of h from c of g to w. So we don't need to worry about the different sorts of induced representation and these are both exact so everything's very nice. And the first question we want to ask is suppose w is a representation of h with character chi. So you remember the character of a representation is given by chi of g is just the trace of g on w and we saw earlier that if you've got a representation at least a complex representation is determined by its character. So we can ask what is the character of the induced representation of g and this is fairly easy to work out. Since h is a subgroup of g, we just have to look at c of g tensor over c of h of w and this is just equal to sum of chi of v where chi are the elements of g modulo h. So I think these are left coset representatives. Although I must admit I always get left cosets and right cosets muddled up so they might maybe they're right coset representatives I don't know. And we want to work out the trace of g on this. Well what will happen is that in general an element g might permute a few of these and if it maps one of these g i of v's to a different g j of v then it gives no contribution to the trace. So the contributions to the trace are nonzero if g g i of v is equal to g i of v and then we get the trace of g i to minus one g g i on v as a contribution to the trace of g on that. So the trace of g on sum of g i of v is just equal to sum over all the g i's of the trace of g i to minus one g g i on w. I guess I seem to have changed w to v at some point so let's just change everything to a v. So this is over the sum of the g i such that g i to the minus one g g i is in the group h. If it isn't then we sort of think we just take this term to be zero. So you can draw a picture of the character as follows. Suppose we draw the group g think of g as being a group like this and inside g are going to be um we're going to have a subgroup h and then we might have various conjugates of h so this might be g i to the minus one h g i and then the character of h you can think of is going to be some sort of function on h and what we do to get the character of of the induced representation is we just take this function and copy it to all the conjugates of h and then we add those ops so you can think of the the character of induced representations pictorially as being just taking a sum of all conjugates of the character of the original representation. This will probably become clear if I do an actual example so let's just take g to be the group s3 so let's draw a picture of the group of um s3 of order six so it's got six elements one two three one three two one two two three three one and one and now we can take a subgroup h like this so here's my subgroup h and i'm going to take a character of h and i'm going to take my character of h to be one on the element one and minus one on the element three one and now what i do is i take all the conjugates of h under g so it's got three conjugates which which look like this and like this and i just take the character on h and just sort of conjugate it to all these three conjugates so get minus one one there and minus one one there and i should of course add these all up so the character of the induced representation is three on the identity element and minus one on these three elements so let's recall the character table of s3 well it's got three conjugacy classes one one two and one two three and you remember we worked out the characters of this earlier it's got three characters um who's given like this so here's the character table and now we've got an induced character which looks like three on the identity element and it's minus one of these elements and it's not here and you can see this is two zero minus one plus um one minus one one so the induced representation is a sum of uh these two representations of the of the of the symmetric group s3 um generally um it's if you try doing calculations with representations algebraically using this formula um it's it's rather confusing and hard to see what's going on um what you should do is think of the character of induced representation geometrically it's just taking a character of h and just conjugating it and adding these up and if you do that it's usually much easier to work at what the induced representation is there's one um trivial but very important example of an induced representation here we just take h to be the subgroup of g consisting of the um trivial subgroup with just one element and now if we take v to be the one dimensional representation of h we can ask what is the induced representation from h to g of of v well that's easy to work out if you do it pictorially you just take h to be the um subgroup with just one element and the representation just as character one on h and now we take the sum over all conjugates of h over all um coset representatives g over h well there um there's one of these for each element of g and each conjugate is again going to be this trivial one-dimensional representation of h so what we do is we get a sum of um g copies of the trivial representation of h so the induced representation its character is equal um the character of an element of g is equal to the order of g if g is equal to one and not if g is not equal to one and you see this is just the character of the regular representation of g which is just the action of g on the group ring c of g so next lecture i'll be showing how to use induced representations to prove a famous theorem of Frobenius which says that if you've got a Frobenius group then it has a kernel you remember a Frobenius group is a permutation group such that um no non-trivial element fixes more than one element