# Representation theory of finite groups (3/4)

In the previous post we introduced characters and looked at the orthogonality relations. In this post we continue our study of representation by looking at the regular representation – a gem of representation theory. Someone at Discord told me that

It is the whole reason group theory can be transported to operator algebras

which says a lot about its importance already. Maybe another blog post will look more deeply at this representation in the distant future. In this post we will really just touch the surface of the theory behind it.

The (left) regular representation is simply $$G$$ acting on itself by left translation. We make this more precise. Let $$\mathbb{C}G = \left \{ \sum_{g \in G} \alpha_g g \right \}$$ so in other words $$\mathbb{C}G$$ is all formal linear combinations of elements of $$G$$. We put a vector space structure on $$\mathbb{C}G$$ the obvious way, i.e. $$\sum_{g \in G} \alpha_g g + \sum_{g \in G} \beta_g g := \sum_{g \in G} (\alpha_g + \beta_g)g$$ and $$c\sum_{g \in G} \alpha_g g = \sum_{g \in G} (c\alpha_g) g$$. Two elements $$\sum_{g \in G} \alpha_g g, \sum_{g \in G} \beta_g g$$ are equal exactly when $$\alpha_g = \beta_g \enspace \forall g \in G$$. This shows that $$G$$ is indeed a basis for $$\mathbb{C}G$$. Furthermore we make $$\mathbb{C}G$$ into an inner product space with: $$\left \langle \sum_{g \in G} \alpha_g g, \sum_{g \in G} \beta_g g \right \rangle = \sum_{g \in G} \alpha_g\overline{\beta_g}$$ The (left) regular representation $$L : G \to GL(\mathbb{C}G)$$ of $$G$$ is defined by $$L_g \sum_{h \in G} \alpha_h h = \sum_{h \in G} \alpha_h gh$$ We will start with working towards showing that the regular representation decompose very nicely. From now on, whenever we say regular representation, we mean the left regular representation.

Lemma 3.1: The regular representation of $$G$$ is a unitary representation.

Proof. We first show that the regular representation is, well, a representation. Firstly, $$L_g$$ is a linear transformation: $$L_g(\alpha g_1 + g_2) = (\alpha g_1 + g_2)h = \alpha g_1h + g_2h = \alpha L_g(g_1)+L_g(g_2)$$ and hence \begin{alignat*}{2} L_{g_1g_2}\sum_{h \in G} \alpha_h h &= \sum_{h \in G} L_{g_1g_2} \alpha_h h \\ &= \sum_{h \in G} g_1g_2 \alpha_h h \\ &= g_1 \sum_{h \in G} g_2\alpha_h h \\ &= L_{g_1}L_{g_2} \end{alignat*} For unitaryness we compute: \begin{alignat*}{2} \left \langle L_g\sum_{h \in G} \alpha_h h, L_g\sum_{h \in G} \beta_h h \right \rangle &= \left \langle \sum_{h \in G} \alpha_h gh, \sum_{h \in G} \beta_h gh \right \rangle \\ &= \left \langle \sum_{x \in G} \alpha_{xg^{-1}} x, \sum_{x \in G} \beta_{xg^{-1}} x \right \rangle \\ &= \sum_{x \in G} \alpha_{xg^{-1}}\overline{\beta_{xg^{-1}}}\end{alignat*} where we have used the “substitution”-trick mentioned earlier with $$x= gh$$. Now $$\sum_{x \in G} \alpha_{xg^{-1}}\overline{\beta_{xg^{-1}}} = \sum_{h \in G} \alpha_{h}\overline{\beta_h} = \langle \sum_{h \in G} \alpha_h h , \sum_{h \in G} \beta_h h \rangle$$ which finishes the proof. $$\blacksquare$$

We continue our investigation by the regular representation by calculating its character. To do this we start by fixing an ordered basis of $$\mathbb{C}G$$: $$\mathcal{B} = \{g_1,\dots,g_n\}$$. Then $$L_{g}(g_j) = gg_j = g_i$$ for some $$i$$. Hence if $$[L_g]$$ is the matrix of $$L_g$$ with respect to the basis $$\mathcal{B}$$, we see that $$[L_g]_{ij} = \left\{\begin{matrix} 1 & \text{ if } gg_j = g_i \\ 0 & \text{ otherwise } \end{matrix}\right.$$ Now $$gg_i = g_i$$ if and only if $$g=e$$ and so $$\chi_{L}(g)= \left\{\begin{matrix} |G| & \text{ if } g=e \\ 0 & \text{ otherwise } \end{matrix}\right.$$ As the calculation over shows, the character of $$L$$ is very nice to work with. The fact that $$\chi_L(e) = |G|$$ is no surprise for we choosed $$G$$ as a basis to start with. This nevertheless connects $$L$$ to the group in a very specific way: the order of the regular representation is exactly the order of the group. This somewhat trivial insight will be proven very fruitful towards our goal of classifying all groups of order $$p^2$$. This comes down to what is known as the dimension theorem, which we will prove in the next text, after a little detour into the land of algebraic number theory.

Theorem 3.2: Let $$L$$ be the regular representation of $$G$$. Then the decomposition $$L \sim \deg(\varphi^{(1)})\varphi^{(1)} \oplus \cdots \oplus \deg(\varphi^{(s)})\varphi^{(s)}$$ where $$\{\varphi^{(1)}, \dots, \varphi^{(s)} \}$$ is a complete set of representatives of the pairwise inequivalent irreducible representations of $$G$$.

Proof. $$\langle \chi_{L}, \chi_{\varphi^{(i)}} \rangle = \frac{1}{|G|} \sum_{g \in G} \chi_L(g)\overline{\chi_{\varphi{(i)}}(g)} = \frac{1}{|G|}|G| \deg(\varphi^{(i)})$$ and so the theorem follows from theorem 2.7 from the previous article. $$\blacksquare$$

Corollary 3.3: If $$\varphi^{(1)}, \dots, \varphi^{(s)}$$ is a complete list of all inequivalent irreducible representations of $$G$$, then $$|G| = (\deg(\varphi^{(1)}))^2+\dots+(\deg(\varphi^{(s)}))^2$$.

Proof. Evaluating the character of $$L$$ in $$e$$ we get that: \begin{alignat*}{2} |G|&=\deg(\varphi^{(1)})\chi_{\varphi^{(1)}}(e) + \dots + \deg(\varphi^{(s)})\chi_{\varphi^{(s)}}(e) \\ &= (\deg(\varphi^{(1)}))^2 + \dots + (\deg(\varphi^{(s)}))^2 \end{alignat*} which is the desired conclusion. $$\blacksquare$$

We will end this post by proving a very nice requirement for when a group is abelian, extending on a result from the first post. Let us fix a complete list of all inequivalent irreducible representations of $$G$$:  $$\varphi^{(1)}, \dots, \varphi^{(s)}$$. In this article, from now on $$\chi_{i}$$ will denote $$\chi_{\varphi^{(i)}}$$  . Furthermore we let $$d_i = \deg(\varphi^{(i)})$$. A function $$f: G \to \mathbb{C}$$ that is constant on conjugacy classes is called a class function. It is clear that the class functions form a subspace of the group algebra, and it can be proven that this is actually the center of the group algebra, and hence we denote them by $$Z(L(G))$$. We will not use this fact, so we only borrow the notation. As we proved earlier, all characters are class functions.  Let $$\text{Cl}(G)$$ be the set of conjugacy classes of $$G$$. Then for $$C \in \text{Cl}(G)$$ we define the function $$\delta_{C}: G \to \mathbb{C}$$ by $$\delta_{C}(g) = \left\{\begin{matrix} 1 & \text{ if } g \in C \\ 0 & \text{ otherwise } \end{matrix}\right.$$ Then as class functions are constant on conjugacy classes we can represent $$f \in Z(L(G))$$ like $$f = \sum_{C \in \text{Cl}(G)} f(C)\delta_C$$ Furthermore the orthogonality $$\langle \delta_C,\delta_{C’} \rangle = \frac{1}{|G|} \sum_{g \in G} \delta_C(g)\overline{\delta_{C’}(g)} = 0$$ whenever $$C \neq C’$$ shows us that $$\{\delta_C \}_{C \in \text{Cl}(G)}$$ is a linearly independent set and so we conclude that $$\{\delta_C\}_{C \in \text{Cl}(G)}$$ is a  basis for $$Z(L(G))$$. This shows that $$\dim(Z(L(G)) = |\text{Cl}(G)|$$.

We know from the first post that if $$G$$ is abelian then all its irreducible representations have degree $$1$$. Conversely if all irreducible representations have degree $$1$$, then Corollary 3.3 implies there are exactly $$|G|$$ irreducible representations. Since the characters of inequivivalent representations are linearly independent by the orthogonality relations, it follows that the space spanned by all the characters of inequivalent representations have at least dimension $$|G|$$. But! All the characters are living inside $$Z(L(G))$$ and we have just seen that $$\dim(Z(L(G))) = |\text{Cl}(G)|$$, so at one side $$\dim(Z(L(G))) \geq |G|$$ but $$|\text{Cl}(G)|\leq|G|\leq \dim(Z(L(G))) = \text{Cl}(G)$$. So we deduce the equality $$\text{Cl}(G)=|G|$$, or in other words – $$G$$ is abelian. Let us state that properly:

Theorem 3.4: Let $$G$$ be a finite group. $$G$$ is abelian if and only if all irreducible representations have degree $$1$$.