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 \[W^{\oplus d} = W \oplus W \oplus \cdots \oplus W \quad \text{($d$ copies)}.\]

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

Theorem:

1. The number \(m\) of irreducible representations of \(G\) is equal to the number of conjugacy classes in \(G\).

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

3. 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 \[\chi = k_1 \chi_1 + k_2 \chi_2 + \cdots + k_m \chi_m.\]

4. 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:

1. 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

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

2. 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 \(\mathbf{1}= \rho_0\).

Note that \[\langle \rho_1,\mathbf{1}\rangle = \dfrac{1}{N}\sum_{i=0}^N \zeta^i = 0\] 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 \(\mathbf{1}\), 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 \(\mathbf{1}\) 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 \(\mathbf{1}\) appears in the representation \(F[\Omega]\); thus \[F[\Omega] = W \oplus \mathbf{1}.\]

You will argue in homework that the character of \(W\) is given by \(\psi = \chi - \mathbf{1}\).

Thus \(\chi\) and \(\chi - \mathbf{1}\) 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

\[\begin{array}{l|llll} & 1 & (12) & (123) \\ & 6 & 2 & 3 \\ \hline \mathbf{1}& 1 & 1 & 1 \\ \operatorname{sgn} & 1 & -1 & 1 \\ \psi & 2 & 0 & 1 \end{array}\]