Representation Theory, Part 4: The Finite Regular Representation

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 ${\rho_\alpha}$ is equal to the number of conjugacy classes of ${G}$ .
We have ${ \left\lvert G \right\rvert = \sum_\alpha \left( \dim \rho_\alpha \right)^2 }$ .

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

In what follows ${k}$ is an algebraically closed field, ${G}$ is a finite group, and the characteristic of ${k}$ does not divide ${\left\lvert G \right\rvert}$ . 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.

$\displaystyle \begin{array}{|l|lll|} \hline \text{Representation} & \text{Group} & \text{Space} & \text{Action} \\ \hline \rho & V & G & G \rightarrow g \cdot_\rho V \\ \text{Fun}(X) & G & \text{Fun}(X) & (g \cdot f)(x) = f(g^{-1} \cdot x) \\ \text{triv}_G & G & k & g \cdot a = a \\ \rho_1 \oplus \rho_2 & G & V_1 \oplus V_2 & g \cdot (v_1 + v_2) = (g \cdot_{\rho_1} v_1) + (g \cdot_{\rho_2} v_2) \\ \rho_1 \boxtimes \rho_2 & G_1 \times G_2 & V_1 \otimes V_2 & (g_1, g_2) \cdot (v_1 \otimes v_2) \\ &&& = (g_1 \cdot_{\rho_1} v_1) \otimes (g_2 \cdot_{\rho_2} v_2) \\ \text{Res}^G_H(\rho) & H & V & h \cdot v = h \cdot_\rho v\\ \rho_1 \otimes \rho_2 & G & V_1 \otimes V_2 & g \cdot (v_1 \otimes v_2) = (g \cdot_{\rho_1} v_1) \otimes (g \cdot_{\rho_2} v_2) \\ \text{Reg}(G) & G \times G & \text{Fun}(G) & (g_1, g_2) \cdot f(g) = f(g_2 g g_1^{-1}) \\ \rho^\vee & V^\vee & G & (g \cdot \xi)(v) = \xi(g^{-1} \cdot_\rho v) \\ \hline \end{array}$

1. The Regular Representation

Recall that ${\mathrm{Fun}(G)}$ is the vector space of functions from ${G}$ to ${k}$ , with addition being defined canonically. It has a basis of functions ${\delta_g}$ for each ${g \in G}$ , where

$\displaystyle \delta_g(x) = \begin{cases} 1 & x = g \\ 0 & \text{otherwise} \end{cases}$

for every ${x \in G}$ . (Throughout this post, I’ll be trying to use ${x}$ to denote inputs to a function from ${G}$ to ${k}$ .)

Definition Let ${G}$ be a finite group. Then the finite regular representation, ${\mathrm{Reg}(G)}$ is a representation on ${G \times G}$ defined on the vector space ${\mathrm{Fun}(G)}$ , with the following action for each ${f \in \mathrm{Fun}(G)}$ and ${(g_1, g_2) \in G \times G}$ :

$\displaystyle ( g_1, g_2 ) \cdot f(x) \overset{\text{def}}{=} f(g_2 x g_1^{-1}).$

Note that this is a representation of the product ${G \times G}$ , not ${G}$ ! (As an aside, you can also define this representation for infinite groups ${G}$ by replacing ${\mathrm{Fun}(G)}$ with ${\mathrm{Fun}_c(G)}$ , the functions which are nonzero at only finitely many ${g \in G}$ .)

In any case, we now can make ${\mathrm{Reg}(G)}$ into a representation of ${G}$ by this restriction, giving ${\mathrm{Res}_G^{G \times G} \left( \mathrm{Reg}(G) \right)}$ , which I will abbreviate as just ${\mathrm{Reg}^\ast(G)}$ through out this post (this is not a standard notation). The action for this is

$\displaystyle (g \cdot_{\mathrm{Reg}^\ast(G)} f)(x) \overset{\text{def}}{=} \left( (g, g) \cdot_{\mathrm{Reg}(G)} f \right)(x) = f\left( g^{-1} x g \right).$

Exercise Consider the invariant subspace of ${\mathrm{Reg}^\ast(G)}$ , which is

$\displaystyle \left( \mathrm{Reg}^\ast(G) \right)^G = \left\{ f : G \rightarrow V \mid f(g^{-1} x g) = f(x) \; \forall x,g \in G \right\}.$

Prove that the dimension of this space is equal to the number of conjugacy classes of ${G}$ . (Look at the ${\delta_g}$ basis.)

Recall that in general, the invariant subspace ${\rho^G}$ is defined as

$\displaystyle \rho^G \overset{\text{def}}{=} \left\{ v \in V \mid g \cdot_\rho v = v \; \forall g \in G \right\}.$

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 ${V}$ , we define the \textbf} by

$\displaystyle V^\vee \overset{\text{def}}{=} \mathrm{Hom}(V,k)$

i.e. it is the set of maps from ${V}$ to ${k}$ . If ${V}$ is finite-dimensional, we can think of this as follows: if ${V}$ consists of the column vectors of length ${m}$ , then ${V^\vee}$ is the row vectors of length ${m}$ , which can be multiplied onto elements of ${V}$ . (This analogy breaks down for ${V}$ infinite dimensional.) Recall that if ${V}$ is finite-dimensional then there is a canonical isomorphism ${V \simeq (V^\vee)^\vee}$ by the map ${v \mapsto \mathrm{ev}_v}$ , where ${\mathrm{ev}_v : V^\vee \rightarrow k}$ sends ${\xi \mapsto \xi(v)}$ .

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

Definition Let ${\rho = (V, \cdot_\rho)}$ be a ${G}$ -representation. Then we define the dual representation ${\rho^\vee}$ by

$\displaystyle \rho^\vee = \left( V^\vee, \cdot_{\rho^\vee} \right) \quad\text{where}\quad \left( g \cdot_{\rho^\vee} \xi \right)(v) = \xi \left( g^{-1} \cdot_\rho v \right).$

Lemma 1 If ${\rho}$ is finite-dimensional then ${(\rho^\vee)^\vee \simeq \rho}$ by the same isomorphism.

Proof: We want to check that the isomorphism ${V = (V^\vee)^\vee}$ by ${v \mapsto \mathrm{ev}_v}$ respects the action of ${G}$ . That’s equivalent to checking

$\displaystyle \mathrm{ev}_{g \cdot_\rho v} = g \cdot_{(\rho^\vee)^\vee} \mathrm{ev}_v.$

But

$\displaystyle \mathrm{ev}_{g \cdot v}(\xi) = \xi(g \cdot_\rho v)$

and

$\displaystyle \left( g \cdot_{(\rho^\vee)^\vee} \mathrm{ev}_v \right)(\xi) = \mathrm{ev}_v(g^{-1} \cdot_{\rho^\vee} \xi) = \left( g^{-1} \cdot_{\rho^\vee} \xi \right)(v) = \xi(g \cdot_\rho v).$

So the functions are indeed equal. $\Box$

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

Lemma 2 For any finite-dimensional ${\rho_1}$ , ${\rho_2}$ we have ${\mathrm{Hom}_G(\rho_1, \rho_2) \simeq \mathrm{Hom}_G(\rho_1 \otimes \rho_2^\vee, \text{triv}_G)}$ .

Proof: Let ${\rho_1 = (V_1, \cdot_{\rho_1})}$ and ${\rho_2 = (V_2, \cdot_{\rho_2})}$ . We already know that we have an isomorphism of vector homomorphisms

$\displaystyle \mathrm{Hom}_{\textbf{Vect}}(V_1, V_2) \simeq \mathrm{Hom}_{\textbf{Vect}} (V_1 \otimes V_2^\vee, k)$

by sending each ${T \in \mathrm{Hom}_{\textbf{Vect}}(V_1, V_2)}$ to the map ${T' \in \mathrm{Hom}_{\textbf{Vect}} (V_1 \otimes V_2^\vee, k)}$ which has ${T'(v \otimes \xi) = \xi(T(v))}$ . So the point is to check that ${T}$ respects the ${G}$ -action if and only if ${T'}$ does. This is just a computation. $\Box$

You can deduce as a corollary the following.

Exercise Use the lemma to show ${\mathrm{Hom}_G(\rho, \text{triv}_G) \simeq \mathrm{Hom}_G(\text{triv}_G, \rho^\vee)}$ .

Finally, we want to talk about when ${\rho^\vee}$ being irreducible. The main result is the following.

Lemma 3 Consider a representation ${\rho}$ , not necessarily finite-dimensional. If ${\rho^\vee}$ is irreducible then so is ${\rho}$ .

When ${\rho}$ is finite dimensional we have ${(\rho^\vee)^\vee \simeq \rho}$ , and so it is true for finite-dimensional irreducible ${\rho}$ that ${\rho^\vee}$ is also irreducible. Interestingly, this result fails for infinite-dimensional spaces as this math.SE thread shows.

Proof: Let ${\rho = (V, \cdot_\rho)}$ . Let ${W}$ be a ${\rho}$ -invariant subspace of ${V}$ . Then consider

$\displaystyle W^\perp = \left\{ \xi \in V^\vee : \xi(w) = 0 \right\}.$

This is a ${\rho^\vee}$ -invariant subspace of ${V^\vee}$ , so since ${\rho^\vee}$ is irreducible, either ${W^\perp = V^\vee}$ or ${W^\perp = \{0\}}$ . You can check that these imply ${W=0}$ and ${W=V}$ , respectively. $\Box$

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 ${\mathrm{Reg}(G)}$ .

Theorem 4 We have an isomorphism

$\displaystyle \mathrm{Reg}(G) \simeq \bigoplus_{\alpha} \rho_\alpha \boxtimes \rho_\alpha^\vee.$

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

Lemma 5 Let ${\pi}$ be a representation of ${G \times G}$ . Then there is an isomorphism

$\displaystyle \mathrm{Hom}_{G \times G}(\pi, \mathrm{Reg}(G)) \simeq \mathrm{Hom}_G(\mathrm{Res}^{G \times G}_G(\pi), \text{triv}_G).$

Proof: Let ${\pi = (V, \cdot_\pi)}$ . Given a map ${T : V \rightarrow \mathrm{Fun}(G)}$ which respects the ${G \times G}$ action, we send it to the map ${\xi_T : V \rightarrow k}$ with ${\xi_T(v) = T(v)(1)}$ . Conversely, given a map ${\xi : V \rightarrow k}$ which respects the ${G}$ action, we send it to the map ${T_\xi : V \rightarrow \mathrm{Fun}(G)}$ so that ${T_\xi(v)(x) = \xi\left( (x,x^{-1}) \cdot v \right)}$ .

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 ${\xi_T(v)}$ respects the ${G}$ action given that ${T}$ respects the ${G \times G}$ action. We want to prove

$\displaystyle \xi_T\left( (g,g) \cdot_\pi v \right) = g \cdot_\text{triv} \xi_T(v) = \xi_T(v).$

Using the definition of ${\xi_T}$

$\displaystyle \begin{aligned} \xi_T\left( (g,g) \cdot_\pi (v) \right) &= T\left( (g,g) \cdot_\pi v \right)(1) \\ &= \left( (g,g) \cdot_{\mathrm{Fun}(G)} T(v) \right)(1) \\ &= T(v)\left( g 1 g^{-1} \right) = T(v)(1) = \xi_T(v). \end{aligned}$

The remaining computations are left to a very diligent reader. $\Box$

Now let’s prove the main theorem!

Proof: We have that ${\mathrm{Reg}(G)}$ is the sum of finite-dimensional irreducibles ${\rho_\alpha \boxtimes \rho_\beta}$ , meaning

$\displaystyle \mathrm{Reg}(G) = \bigoplus_{\alpha, \beta} \left( \rho_\alpha \boxtimes \rho_\beta \right) \otimes \mathrm{Hom}_{G \times G}\left( \rho_\alpha \boxtimes \rho_\beta, \mathrm{Reg}(G) \right).$

But using our lemmas, we have that

$\displaystyle \mathrm{Hom}_{G \times G}\left( \rho_\alpha \boxtimes \rho_\beta, \mathrm{Reg}(G) \right) \simeq \mathrm{Hom}_G(\rho_\alpha \otimes \rho_\beta, \text{triv}_G) \simeq \mathrm{Hom}_G(\rho_\alpha, \rho_\beta^\vee).$

We know that ${\rho_\beta^\vee}$ is also irreducible, since ${\rho_\beta}$ is (and we’re in a finite-dimensional situation). So

$\displaystyle \mathrm{Hom}_G\left( \rho_\alpha, \rho_\beta^\vee \right) \simeq \begin{cases} k & \rho_\beta^\vee = \rho_\alpha \\ \{0\} & \text{otherwise}. \end{cases}$

Thus we deduce

$\displaystyle \mathrm{Reg}(G) \simeq \bigoplus_{\alpha} \left( \rho_\alpha \boxtimes \rho_\alpha^\vee \right) \otimes k \simeq \bigoplus_{\alpha} \left( \rho_\alpha \boxtimes \rho_\alpha^\vee \right)$

and we’re done. $\Box$

4. Corollaries

Recall that ${\mathrm{Fun}(G)}$ , the space underlying ${\mathrm{Reg}(G)}$ , has a basis with size ${\left\lvert G \right\rvert}$ . Hence by comparing the dimensions of the isomorphsims, we obtain the following corollary.

Theorem 6 We have ${\left\lvert G \right\rvert = \sum_\alpha \left( \dim \rho_\alpha \right)^2}$ .

Moreover, by restriction to ${G}$ we can obtain the corollary

$\displaystyle \mathrm{Reg}^\ast(G) \simeq \bigoplus_\alpha \mathrm{Res}_{G}^{G \times G} \left( \rho_\alpha \otimes \rho_\alpha^\vee \right) = \bigoplus_\alpha \rho_\alpha \otimes \rho_\alpha^\vee.$

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

$\displaystyle \left( \rho_\alpha \otimes \rho_\alpha^\vee \right)^G \simeq k.$

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

$\displaystyle \left( \rho_\alpha \otimes \rho_\alpha^\vee \right)^G \simeq \mathrm{Hom}_G(\text{triv}_G, \rho_\alpha \otimes \rho_\alpha^\vee).$

Then we can write

$\displaystyle \begin{aligned} \mathrm{Hom}_G(\text{triv}_G, \rho_\alpha \otimes \rho_\alpha^\vee) &\simeq \mathrm{Hom}_G\left(\text{triv}_G, \left( \rho_\alpha^\vee \otimes \rho_\alpha \right)^\vee \right) \\ &\simeq \mathrm{Hom}_G\left(\rho_\alpha^\vee \otimes \rho_\alpha, \text{triv}_G \right) \\ &\simeq \mathrm{Hom}_G\left(\rho_\alpha \otimes \rho_\alpha^\vee, \text{triv}_G \right) \\ &\simeq \mathrm{Hom}_G\left(\rho_\alpha, \rho_\alpha \right) \\ &\simeq k \end{aligned}$

by the lemma, where we have also used Schur’s Lemma at the last step. So that means the dimension of the invariant space ${(\mathrm{Reg}^\ast (G))^G}$ is just the number of irreducibles.

But we already showed that the invariant space of ${(\mathrm{Reg}^\ast (G))^G}$ has dimension equal to the conjugacy classes of ${G}$ . Thus we conclude the second result.