2024-01-31--notes-RT.md

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title: Introducing the character table of a finite group
date: 2024-01-31

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\newcommand{\trivial}{\mathbf{1}}

Let $G$ a finite group; we consider finite dimensional representations of $G$ on $\mathbb{C}$-vector spaces.

Previews!

Let $L_1,\cdots,L_m$ be a complete set of non-isomorphic irreducible representations for the finite group $G$, and let $\chi_i$ be the character of $L_i$.

Recall that any $G$-representation $V$ can be written as a direct sum of irreducible subrepresentations: $$G = \bigoplus_{j=1}^N S_j$$ where each $S_j$ is an irreducible representation. For each $1 \le i \le m$, we say that the multiplicity of $L_i$ in $V$ -- written $[V:L_i]$ -- is the natural number given by $$[V:L_i] = \#\{j \mid S_j \simeq L_i\}$$

Thus we have $$V \simeq \bigoplus_{i=1}^m L_i^{\oplus [V:L_i]}$$ where the notation $W^{\oplus d}$ means $$W^{\oplus d} = W \oplus W \oplus \cdots \oplus W \quad \text{($dParseError: KaTeX parse error: Expected 'EOF', got '}' at position 9: copies)}̲.$

Next week, we are going to sketch a proof of the following facts

Theorem: : a. The number $m$ of irreducible representations of $G$ is equal to the number of conjugacy classes in $G$.

 b. $\chi_1,\cdots,\chi_m$ are an *orthonormal basis* for the
    space $\operatorname{Cl}(G)$ of $\mathbb{C}$-value class
    functions on $G$.

 c. For any $G$-representation $V$, let $\chi$ be the character of $V$.
    Then the multiplicity $[V:L_i]$ is given by
	$$[V:L_i] = \langle \chi,\chi_i\rangle.$$
	
	In particular, write $k_i = [V:L_i]$; then
	$$\chi = k_1 \chi_1 + k_2 \chi_2 + \cdots + k_m \chi_m.$$
	
 d. The multiplicity with which $L_i$ appears in the *regular representation*
    $F[G]$ is given by
	$$[F[G]:L_i] = \dim_F L_i.$$
	
	In particular, if $d_i = \dim_F L_i$, then
	$$|G| = \sum_{i=1}^m d_i^2.$$

The Hermitian inner product on class functions, again

Enumerate the conjugacy classes of $G$, say $C_1,\cdots,C_m$ and choose a representative $g_i \in C_i$ for each $i$.

Write $c_i = |C_G(g_i)|$ for the number of elements in the centralizer of $g_i$, and notice that $$|\operatorname{Cl}(g_i)| = |G|/|C_G(g_i)| = |G|/c_i.$$

Recall that for two class functions $f_1,f_2 \in \operatorname{Cl}(G)$ we have defined $$\langle f_1,f_2 \rangle = \dfrac{1}{|G|} \sum_{g \in G} f_1(g) \overline{f_2(g)}.$$

Lemma: : We have $$\langle f_1,f_2 \rangle = \dfrac{1}{c_i} \sum_{i = 1}^m f_1(g) \overline{f_2(g)}.$$

The character table.

The matrix is known as the character table of the group $G$. Consider the $m \times m$ matrix whose rows are indexed by the irreducible characters $\chi_1,\cdots,\chi_m$ and whose columns are indexed by the conjugacy class representatives $g_1,\dots,g_m$, and whose entry in the $i$-th row and $j$-th column is given by $\chi_i(g_j)$. $$\begin{array}{l|llll} & g_1 & g_2 & \cdots & g_m \\ & c_1 & c_2 & \cdots & c_m \\ \hline \chi_1 & \chi_1(g_1) & \chi_1(g_2) & \cdots & \chi_1(g_m) \\ \chi_2 & \chi_2(g_1) & \chi_2(g_2) & \cdots & \chi_2(g_m) \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ \chi_m & \chi_m(g_1) & \chi_m(g_2) & \cdots & \chi_m(g_m) \end{array}$$

Remark: : a. The fact that the $\chi_i$ form an orthonormal basis for the space $\operatorname{Cl}(G)$ is equivalent to the statement that the above matrix is unitary, in the sense that for $1 \le i,j \le m$ we have

   $$\sum_{k=1}^m c_k \chi_i(g_k) \overline{\chi_j(g_k)} = \delta_{i,j}.$$

b. Since the transpose of unitary matrix is also unitary, we find
   for $1 \le i,j \le m$ that

   $$\sum_{k=1}^m c_i \chi_k(g_i) \overline{\chi_k(g_j)} = \delta_{i,j}.$$

We are now going to compute the character table for a few finite groups

Cyclic groups

Let $G = \mathbb{Z}/N\mathbb{Z}$ for a natural number $N$. If $\zeta$ is a primitive $N$-th root of unity, we have for $0 \le i < N$ a homomorphism $$\rho_i:G \to F^\times$$ determined by the equation $$\rho_i(1 + N\mathbb{Z}) = \zeta^i.$$

Since $G$ is abelian, its conjugacy classes are singletons; thus the number of irreducible representations is $N = |G|$.

Each $\rho_i$ determines a 1-dimensional irreducible representation, and $i \ne j \implies \rho_i \not \simeq \rho_j$.

Orthonormality of characters implies that $$\langle \rho_i, \rho_j \rangle = \delta_{i,j}$$

This equality can actually be deduce in a more elementary way. For example, write ParseError: KaTeX parse error: Undefined control sequence: \trivial at position 1: \̲t̲r̲i̲v̲i̲a̲l̲ ̲= \rho_0.

Note that ParseError: KaTeX parse error: Undefined control sequence: \trivial at position 16: \langle \rho_1,\̲t̲r̲i̲v̲i̲a̲l̲ ̲\rangle = \dfra… since $\zeta$ is a root of $\dfrac{T^N-1}{T-1} = T^N + T^{N-1} + \cdots + T + 1$.

Consider a function $f \in F[\mathbb{Z}/N\mathbb{Z}]$. Using orthonormality of characters we deduce $$f = \sum_{i=0}^{N-1} \langle f, \rho_i \rangle \rho_i.$$

The assigment $i \mapsto \langle f,\rho_i \rangle$ is usual known as the discrete Fourier transform of $f$.

The group $D_6 = S_3$.

Let $G = S_3$ be the symmetric group on 3 letters, so that $|G| = 6$.

Note that the subgroup $H=\langle (123) \rangle \subset G$ has index 2 and is thus a normal subgroup. (In fact, $H$ is the alternating group $A_3$).

Notice that the centralizer of $(123)$ coincides with $H$, so that $(123)$ has exactly $[G:H] = 2$ conjugates (namely, $(123)$ and $(132)$). On the other hand, the centralizer of $(12)$ is $\langle (12) \rangle$ so that $(12)$ has $3$ conjugates; namely $(12)$, $(13)$ and $(23)$.

Thus $1$, $(12)$, and $(123)$ is a full set of representatives for the conjugacy classes of $G$.

In particular, we expect to find 3 irreducible representations of $G$.

There are exactly two homomorphisms $G \to F^\times$ which contain $H$ in their kernel; one is the trivial mapping ParseError: KaTeX parse error: Undefined control sequence: \trivial at position 1: \̲t̲r̲i̲v̲i̲a̲l̲, and the other is the sign homomorphism $\operatorname{sgn}:G \to F^\times$; it is the unique mapping for which $H \subset \ker \operatorname{sgn}$ and $\operatorname{sgn}((12)) = -1$.

Thus ParseError: KaTeX parse error: Undefined control sequence: \trivial at position 1: \̲t̲r̲i̲v̲i̲a̲l̲ and $\operatorname{sgn}$ are 1-dimensional irreducible representations of $G$.

It remains to find one more irreducible representation.

Let $\Omega = \{1,2,3\}$ and consider the standard action of $G = S_3$ on $\Omega$. Write $\chi = \chi_\Omega$ for the character of this representation.

You will see for homework that $\chi(\sigma)$ is equal to the number of fixed points of $\sigma$ on $\Omega$.

We have seen that the trivial representation ParseError: KaTeX parse error: Undefined control sequence: \trivial at position 1: \̲t̲r̲i̲v̲i̲a̲l̲ appears in the representation $F[\Omega]$; thus ParseError: KaTeX parse error: Undefined control sequence: \trivial at position 22: …ga] = W \oplus \̲t̲r̲i̲v̲i̲a̲l̲.

You will argue in homework that the character of $W$ is given by ParseError: KaTeX parse error: Undefined control sequence: \trivial at position 15: \psi = \chi - \̲t̲r̲i̲v̲i̲a̲l̲.

Thus $\chi$ and ParseError: KaTeX parse error: Undefined control sequence: \trivial at position 8: \chi - \̲t̲r̲i̲v̲i̲a̲l̲ are given by:

$$\begin{array}{l|ll} & \chi(\sigma) & \psi(\sigma) \\ \hline 1 & 3 & 2\\ (12) & 1 & 0 \\ (123) & 0 & -1 \end{array}$$

Now, we compute $$\langle \psi, \psi \rangle = 1/6 \cdot 2 \cdot 2 + 1/2 \cdot 0 \cdot 0 + 1/3 \cdot -1 \cdot -1 = 2/3 + 1/3 = 1$$ This shows that $\psi$ is an irreducible representation. Thus the character table of $G$ is given by

ParseError: KaTeX parse error: Undefined control sequence: \trivial at position 69: …& 3 \\ \hline \̲t̲r̲i̲v̲i̲a̲l̲ ̲& 1 & 1 & 1 \\ …