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

$$\rho_1 \boxtimes \rho_2 \coloneqq \left( V_1 \otimes V_2, \cdot \right)$$

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

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

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

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

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

$$\rho_1 \otimes \rho_2 \coloneqq \operatorname{Res}^{G \times G}_G \left( \rho_1 \boxtimes \rho_2 \right)$$

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

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

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.

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

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 $\rho \otimes k \coloneqq \rho \otimes \text{triv}_G$ 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

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

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

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

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

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

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

given by sending every simple tensor via

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

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

$$\rho_1 \boxtimes \rho_2$$

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 $$\rho^1 \coloneqq \operatorname{Res}^{G_1 \times G_2}_{G_1} (\rho).$$ Then by Maschke’s Theorem, we may write $\rho^1$ as a direct sum of the irreducibles $$\bigoplus_\alpha \rho_\alpha^1 \otimes \operatorname{Hom}_{G_1} (\rho_\alpha^1, \rho^1) \simeq \rho^1$$ with the map $v \otimes \xi \mapsto \xi(v)$ being the isomorphism. Now we can put a $G_2$ representation structure on $\operatorname{Hom}_{G_1} (\rho_\alpha^1, \rho^1)$ by $$(g_2 \cdot f)(g) = g_2 \cdot_{\rho} (f(g)).$$ 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 $$\bigoplus_\alpha \rho_\alpha^1 \boxtimes \operatorname{Hom}_{G_1} (\rho_\alpha^1, \rho^1).$$ 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 $$\bigoplus_\alpha \rho_\alpha^1 \boxtimes \operatorname{Hom}_{G_1} (\rho_\alpha^1, \rho^1) \simeq \rho$$ 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. $\Box$

4. Conclusion

In particular, this means that any representation $\rho$ of $G \times G$ decomposes as $$\rho \simeq \bigoplus_{\alpha, \beta} \rho_\alpha \boxtimes \rho_\beta$$ and we even have $$\operatorname{Res}_{G}^{G\times G} \rho \simeq \bigoplus_{\alpha, \beta} \rho_\alpha \otimes \rho_\beta.$$ 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.