Happy New Year to all! A quick reminder that .
This post will set the stage by examining products of two representations. In particular, I’ll characterize all the irreducibles of in terms of those for and . This will set the stage for our discussion of the finite regular representation in Part 4.
In what follows is an algebraically closed field, is a finite group, and the characteristic of does not divide .
1. Products of Representations
First, I need to tell you how to take the product of two representations.
Definition Let and be groups. Given a representation and a representation , we define
as a representation of on . The action is given by
In the special case , we can also restrict to a representation of . Note that we can interpret itself as a subgroup of by just looking along the diagonal: there’s an obvious isomorphism
So, let me set up the general definition.
Definition Let be a group, and let be a subgroup of . Then for any representation of , we let
denote the representation of on by the same action.
This notation might look intimidating, but it’s not really saying anything, and I include the notation just to be pedantic. All we’re doing is taking a representation and restricting which elements of the group are acting on it.
We now apply this to get out of .
Definition Let and be representations of . Then we define
meaning has vector space and action .
This tensor product obeys some nice properties, for example the following.
Lemma 1 Given representations , , we have
Proof: There’s an obvious isomorphism between the underlying vector spaces, and that isomorphism respects the action of .
To summarize all the above, here is a table of the representations we’ve seen, in the order we met them.
2. Revisiting Schur and Maschke
Defining a tensor product of representations gives us another way to express , as follows. By an abuse of notation, given a vector space we can define an associated -representation on it by the trivial action, i.e. for . A special case of this is using to represent . With this abuse of notation, we have the following lemma.
Lemma 2 Let be an -dimensional vector space over . Then .
Proof: It reduces to checking that is isomorphic to , which is evident. We can then proceed by induction: .
So, we can actually rewrite Maschke’s and Schur’s Theorem as one. Instead of
we now have instead
Now we’re going to explicitly write down the isomorphism between these maps. It suffices to write down the isomorphism , and then take the sum over each of the ‘s. But
So to write the isomorphism , we just have to write down the isomorphism ,
Schur’s Lemma tells us that ; i.e. every just corresponds to multiplying by some constant. So this case is easy: the map
works nicely. And since all we’ve done is break over a bunch of direct sums, the isomorphism propagates all the way up, resulting in the following theorem.
Theorem 3 (Maschke and Schur) For any finite-dimensional , the homomorphism of representations
given by sending every simple tensor via
is an isomorphism.
Note that it’s much easier to write the map from left to right than vice-versa, even though the inverse map does exist (since it’s an isomorphism). (Tip: as a general rule of thumb, always map out of the direct sum.)
3. Characterizing the irreducibles
Now we are in a position to state the main theorem for this post, which shows that the irreducibles we defined above are very well behaved.
Theorem 4 Let and be finite groups. Then a finite-dimensional representation of is irreducible if and only if it is of the form
where and are irreducible representations of and , respectively.
Proof: First, suppose is an irreducible representation of . Set
Then by Maschke’s Theorem, we may write as a direct sum of the irreducibles
with the map being the isomorphism. Now we can put a representation structure on by
It is easy to check that this is indeed a representation. Thus it makes sense to talk about the representation
We claim that the isomorphism for as a representation now lifts to an isomorphism of representations. That is, we claim that
by the same isomorphism as for . To see this, we only have to check that the isomorphism commutes with the action of . But this is obvious, since .
Thus the isomorphism holds. But is irreducible, so there can only be one nontrivial summand. Thus we derive the required decomposition of .
Now for the other direction: take and irreducible. Suppose has a nontrivial subrepresentation of the form . Viewing as representation, we find that is a nontrivial subrepresentation of , and similarly for . But is irreducible, hence . Similarly . So in fact . Hence we conclude is irreducible.
In particular, this means that any representation of decomposes as
and we even have
In the next post I’ll invoke this on the so-called finite regular representation to get the elegant results I promised at the end of Part 2.
Thanks to Dennis Gaitsgory, who taught me this in his course Math 55a. My notes for Math 55a can be found at my website.