Represenation Theory, Part 3: Products of Representations

Happy New Year to all! A quick reminder that ${2015 = 5 \cdot 13 \cdot 31}$ .

This post will set the stage by examining products of two representations. In particular, I’ll characterize all the irreducibles of ${G_1 \times G_2}$ in terms of those for ${G_1}$ and ${G_2}$ . This will set the stage for our discussion of the finite regular representation in Part 4.

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

1. Products of Representations

First, I need to tell you how to take the product of two representations.

Definition Let ${G_1}$ and ${G_2}$ be groups. Given a ${G_1}$ representation ${\rho_1 = (V_1, \cdot_{\rho_1})}$ and a ${G_2}$ representation ${\rho_2 = (V_2, \cdot_{\rho_2})}$ , we define

$\displaystyle \rho_1 \boxtimes \rho_2 \overset{\text{def}}{=} \left( V_1 \otimes V_2, \cdot \right)$

as a representation of ${G_1 \times G_2}$ on ${V_1 \otimes V_2}$ . The action is given by

$\displaystyle (g_1, g_2) \cdot (v_1 \otimes v_2) = \left( g_1 \cdot_{\rho_1} v_1 \right) \otimes (g_2 \cdot_{\rho_2} v_2).$

In the special case ${G_1 = G_2 = G}$ , we can also restrict ${\rho_1 \boxtimes \rho_2}$ to a representation of ${G}$ . Note that we can interpret ${G}$ itself as a subgroup of ${G \times G}$ by just looking along the diagonal: there’s an obvious isomorphism

$\displaystyle G \sim \left\{ (g,g) \mid g \in G \right\}.$

So, let me set up the general definition.

Definition Let ${\mathcal G}$ be a group, and let ${\mathcal H}$ be a subgroup of ${\mathcal G}$ . Then for any representation ${\rho = (V, \cdot_\rho)}$ of ${\mathcal G}$ , we let

$\displaystyle \mathrm{Res}^{\mathcal G}_{\mathcal H} (\rho)$

denote the representation of ${\mathcal H}$ on ${V}$ 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 ${\rho_1 \otimes \rho_2}$ out of ${\rho_1 \boxtimes \rho_2}$ .

Definition Let ${\rho_1 = (V_1, \cdot_{\rho_1})}$ and ${\rho_2 = (V_2, \cdot_{\rho_2})}$ be representations of ${G}$ . Then we define

$\displaystyle \rho_1 \otimes \rho_2 \overset{\text{def}}{=} \mathrm{Res}^{G \times G}_G \left( \rho_1 \boxtimes \rho_2 \right)$

meaning ${\rho_1 \otimes \rho_2}$ has vector space ${V_1 \otimes V_2}$ and action ${g \cdot (v_1 \otimes v_2) = (g \cdot_{\rho_1} v_1) \otimes (g \cdot_{\rho_2} v_2)}$ .

This tensor product obeys some nice properties, for example the following.

Lemma 1 Given representations ${\rho}$ , ${\rho_1}$ , ${\rho_2}$ we have

$\displaystyle \rho \otimes \left( \rho_1 \oplus \rho_2 \right) \simeq \left( \rho \otimes \rho_1 \right) \oplus \left( \rho \otimes \rho_2 \right).$

Proof: There’s an obvious isomorphism between the underlying vector spaces, and that isomorphism respects the action of ${G}$ . $\Box$

To summarize all the above, here is a table of the representations we’ve seen, in the order we met them.

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

2. Revisiting Schur and Maschke

Defining a tensor product of representations gives us another way to express ${\rho^{\oplus n}}$ , as follows. By an abuse of notation, given a vector space ${k^m}$ we can define an associated ${G}$ -representation ${k^m = (k^m, \cdot_{k^m})}$ on it by the trivial action, i.e. ${g \cdot_{k^m} v = v}$ for ${v \in k^m}$ . A special case of this is using ${k}$ to represent ${\text{triv}_G}$ . With this abuse of notation, we have the following lemma.

Lemma 2 Let ${M}$ be an ${m}$ -dimensional vector space over ${k}$ . Then ${\rho^{\oplus m} \simeq \rho \otimes M}$ .

Proof: It reduces to checking that ${\rho \otimes k \overset{\text{def}}{=} \rho \otimes \text{triv}_G}$ is isomorphic to ${\rho}$ , which is evident. We can then proceed by induction: ${\rho \otimes (k \oplus k^{t-1}) \simeq (\rho \otimes k) \oplus (\rho \otimes k^{t-1})}$ . $\Box$

So, we can actually rewrite Maschke’s and Schur’s Theorem as one. Instead of

$\displaystyle \rho \simeq \bigoplus_\alpha \rho_\alpha^{\oplus n_\alpha} \quad\text{where}\quad n_\alpha = \dim \mathrm{Hom}_G(\rho, \rho_\alpha)$

we now have instead

$\displaystyle \bigoplus_\alpha \rho_\alpha \otimes \mathrm{Hom}_G(\rho, \rho_\alpha) \simeq \rho.$

Now we’re going to explicitly write down the isomorphism between these maps. It suffices to write down the isomorphism ${\rho_\alpha \otimes \mathrm{Hom}_G(\rho, \rho_\alpha) \rightarrow \rho_\alpha^{\oplus n_\alpha}}$ , and then take the sum over each of the ${\alpha}$ ‘s. But

$\displaystyle \mathrm{Hom}_G(\rho, \rho_\alpha) \simeq \mathrm{Hom}_G(\rho_\alpha^{\oplus n_\alpha}, \rho_\alpha) \simeq \mathrm{Hom}_G(\rho_\alpha, \rho_\alpha)^{\oplus n_\alpha}.$

So to write the isomorphism ${\rho_\alpha \otimes \mathrm{Hom}_G(\rho_\alpha, \rho_\alpha)^{\oplus n_\alpha} \rightarrow \rho_\alpha^{\oplus n_\alpha}}$ , we just have to write down the isomorphism ${\rho_\alpha \otimes \mathrm{Hom}_G(\rho_\alpha, \rho_\alpha) \rightarrow \rho_\alpha}$ ,

Schur’s Lemma tells us that ${\mathrm{Hom}_G(\rho_\alpha, \rho_\alpha) \simeq k}$ ; i.e. every ${\xi \in \mathrm{Hom}_G(\rho_\alpha, \rho_\alpha)}$ just corresponds to multiplying ${v}$ by some constant. So this case is easy: the map

$\displaystyle v \otimes \xi \mapsto \xi(v)$

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 ${\rho}$ , the homomorphism of ${G}$ representations

$\displaystyle \bigoplus_\alpha \rho_\alpha \otimes \mathrm{Hom}_G(\rho, \rho_\alpha) \rightarrow \rho$

given by sending every simple tensor via

$\displaystyle v \otimes \xi \mapsto \xi(v)$

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 ${G_1 \times G_2}$ 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 ${G_1}$ and ${G_2}$ be finite groups. Then a finite-dimensional representation ${\rho}$ of ${G_1 \times G_2}$ is irreducible if and only if it is of the form

$\displaystyle \rho_1 \boxtimes \rho_2$

where ${\rho_1}$ and ${\rho_2}$ are irreducible representations of ${G_1}$ and ${G_2}$ , respectively.

Proof: First, suppose ${\rho = (V, \cdot_\rho)}$ is an irreducible representation of ${G_1 \times G_2}$ . Set

$\displaystyle \rho^1 \overset{\text{def}}{=} \mathrm{Res}^{G_1 \times G_2}_{G_1} (\rho).$

Then by Maschke’s Theorem, we may write ${\rho^1}$ as a direct sum of the irreducibles

$\displaystyle \bigoplus_\alpha \rho_\alpha^1 \otimes \mathrm{Hom}_{G_1} (\rho_\alpha^1, \rho^1) \simeq \rho^1$

with the map ${v \otimes \xi \mapsto \xi(v)}$ being the isomorphism. Now we can put a ${G_2}$ representation structure on ${\mathrm{Hom}_{G_1} (\rho_\alpha^1, \rho^1)}$ by

$\displaystyle (g_2 \cdot f)(g) = g_2 \cdot_{\rho} (f(g)).$

It is easy to check that this is indeed a ${G_2}$ representation. Thus it makes sense to talk about the ${G_1 \times G_2}$ representation

$\displaystyle \bigoplus_\alpha \rho_\alpha^1 \boxtimes \mathrm{Hom}_{G_1} (\rho_\alpha^1, \rho^1).$

We claim that the isomorphism for ${\rho^1}$ as a ${G_1}$ representation now lifts to an isomorphism of ${G_1 \times G_2}$ representations. That is, we claim that

$\displaystyle \bigoplus_\alpha \rho_\alpha^1 \boxtimes \mathrm{Hom}_{G_1} (\rho_\alpha^1, \rho^1) \simeq \rho$

by the same isomorphism as for ${\rho^1}$ . To see this, we only have to check that the isomorphism ${v \otimes \xi \mapsto \xi(v)}$ commutes with the action of ${g_2 \in G_2}$ . But this is obvious, since ${g_2 \cdot (v \otimes \xi) = v \otimes (g_2 \cdot \xi) \mapsto (g_2 \cdot \xi)(v)}$ .

Thus the isomorphism holds. But ${\rho}$ is irreducible, so there can only be one nontrivial summand. Thus we derive the required decomposition of ${\rho}$ .

Now for the other direction: take ${\rho_1}$ and ${\rho_2}$ irreducible. Suppose ${\rho_1 \boxtimes \rho_2}$ has a nontrivial subrepresentation of the form ${\rho_1' \boxtimes \rho_2'}$ . Viewing as ${G_1}$ representation, we find that ${\rho_1'}$ is a nontrivial subrepresentation of ${\rho_1}$ , and similarly for ${\rho_2}$ . But ${\rho_1}$ is irreducible, hence ${\rho_1' \simeq \rho_1}$ . Similarly ${\rho_2' \simeq \rho_2}$ . So in fact ${\rho_1' \boxtimes \rho_2' \simeq \rho_1 \boxtimes \rho_2}$ . Hence we conclude ${\rho_1 \boxtimes \rho_2}$ is irreducible. $\Box$