Good luck to everyone taking the January TST for the IMO 2015 tomorrow!

Now that we have products of irreducibles under our belt, I’ll talk about the finite regular representation and use it to derive the following two results about irreducibles.

- The number of (isomorphsim classes) of irreducibles is equal to the number of conjugacy classes of .
- We have .

These will actually follow as corollaries from the complete decomposition of the finite regular representation.

In what follows is an algebraically closed field, is a finite group, and the characteristic of does not divide . As a reminder, here are the representations we’ve already seen in the order we met them, plus two new ones we’ll introduce properly below.

**1. The Regular Representation **

Recall that is the vector space of functions from to , with addition being defined canonically. It has a basis of functions for each , where

for every . (Throughout this post, I’ll be trying to use to denote inputs to a function from to .)

DefinitionLet be a finite group. Then thefinite regular representation, is a representation on defined on the vector space , with the following action for each and :

Note that this is a representation of the product , not ! (As an aside, you can also define this representation for infinite groups by replacing with , the functions which are nonzero at only finitely many .)

In any case, we now can make into a representation of by this restriction, giving , which I will abbreviate as just through out this post (this is not a standard notation). The action for this is

ExerciseConsider the invariant subspace of , which is

Prove that the dimension of this space is equal to the number of conjugacy classes of . (Look at the basis.)

Recall that in general, the invariant subspace is defined as

**2. Dual Representations **

Before I can state the main theorem of this post, I need to define the dual representation.

Recall that given a vector space , we define the \textbf} by

i.e. it is the set of maps from to . If is finite-dimensional, we can think of this as follows: if consists of the column vectors of length , then is the row vectors of length , which can be multiplied onto elements of . (This analogy breaks down for infinite dimensional.) Recall that if is finite-dimensional then there is a canonical isomorphism by the map , where sends .

Now we can define the dual representation in a similar way.

DefinitionLet be a -representation. Then we define thedual representationby

Lemma 1If is finite-dimensional then by the same isomorphism.

*Proof:* We want to check that the isomorphism by respects the action of . That’s equivalent to checking

But

and

So the functions are indeed equal.

Along with that lemma, we also have the following property.

Lemma 2For any finite-dimensional , we have .

*Proof:* Let and . We already know that we have an isomorphism of vector homomorphisms

by sending each to the map which has . So the point is to check that respects the -action if and only if does. This is just a computation.

You can deduce as a corollary the following.

ExerciseUse the lemma to show .

Finally, we want to talk about when being irreducible. The main result is the following.

Lemma 3Consider a representation , not necessarily finite-dimensional. If is irreducible then so is .

When is finite dimensional we have , and so it is true for *finite-dimensional* irreducible that is also irreducible. Interestingly, this result fails for infinite-dimensional spaces as this math.SE thread shows.

*Proof:* Let . Let be a -invariant subspace of . Then consider

This is a -invariant subspace of , so since is irreducible, either or . You can check that these imply and , respectively.

**3. Main Result **

Now that we know about the product of representations and dual modules, we can state the main result of this post: the complete decomposition of .

Theorem 4We have an isomorphism

Before we can begin the proof of the theorem we need one more lemma.

Lemma 5Let be a representation of . Then there is an isomorphism

*Proof:* Let . Given a map which respects the action, we send it to the map with . Conversely, given a map which respects the action, we send it to the map so that .

Some very boring calculations show that the two maps are mutually inverse and respect the action. We’ll just do one of them here: let us show that respects the action given that respects the action. We want to prove

Using the definition of

The remaining computations are left to a very diligent reader.

Now let’s prove the main theorem!

*Proof:* We have that is the sum of finite-dimensional irreducibles , meaning

But using our lemmas, we have that

We know that is also irreducible, since is (and we’re in a finite-dimensional situation). So

Thus we deduce

and we’re done.

**4. Corollaries **

Recall that , the space underlying , has a basis with size . Hence by comparing the dimensions of the isomorphsims, we obtain the following corollary.

Theorem 6We have .

Moreover, by restriction to we can obtain the corollary

Now let us look at the -invariant spaces in this decomposition. We claim that

Indeed, {Proposition 1} in {Part 1} tells us that we have a bijection of vector spaces

Then we can write

by the lemma, where we have also used Schur’s Lemma at the last step. So that means the dimension of the invariant space is just the number of irreducibles.

But we already showed that the invariant space of has dimension equal to the conjugacy classes of . Thus we conclude the second result.

Theorem 7The number of conjugacy classes of equals the number of irreducible representations of .

Hooray!

Time permitting I might talk about the irreducibles of in subsequent posts. No promises here though.

*Thanks to Dennis Gaitsgory, who taught me this in his course Math 55a. My notes for Math 55a can be found at my website.*