Representations of finite groups; a matrix view

Posted: February 8, 2011 in Representations of Finite Groups
Tags: ,

Throughout G is a finite group and \rho : G \longrightarrow \text{GL}(V) is a representation of G.

We defined an irreducible representation in here. The definition can also be given in terms of matrices.

Theorem 1. Let \deg \rho = \dim_{\mathbb{C}} V=n. Then \rho is reducible if and only if there exist a \mathbb{C}-basis \mathcal{B} of V and an integer 1 \leq m < n such that for all g \in G

[\rho(g)]_{\mathcal{B}} = \begin{pmatrix} A & B \\ 0 & C \end{pmatrix},

where 0 here means an m \times m zero matrix. The matrices A,B and C depend on g.

Proof. Well, by definition, \rho is reducible if and only if V is not simple as a \mathbb{C}[G] – module. So \rho is reducible if and only if V has a non-zero \mathbb{C}[G]-submodule W \neq V. Since \mathbb{C}[G] is a semisimple ring (see the corollary in this post), V is a semisimple \mathbb{C}[G]-module. Thus V=W \oplus U, for some \mathbb{C}[G]-submodule U of V. Now, choose a \mathbb{C}-basis \{v_1, \cdots , w_m \} for W and extend it to a basis \mathcal{B} for V. Note that 1 \leq m < n, because W \neq (0), V. Let g \in G. Since W is a \mathbb{C}[G]-module, we have \rho(g)(v_i) \in W, for all 1 \leq i \leq m. So \rho(g)(v_i) is a \mathbb{C}-linear combinations v_1, \cdots , v_m. It is now clear that the matrix [\rho(g)]_{\mathcal{B}} looks like the matrix given in the theorem. Conversely, if for some basis \mathcal{B}=\{v_1, \cdots , v_n \} of V, the matrix [\rho(g)]_{\mathcal{B}} looks like the one given in the theorem, then we let W = \text{span} \{v_1, \cdots , v_m \}. It is clear from the form of [\rho(g)]_{\mathcal{B}} that \rho(g)(v_i) \in W, for all 1 \leq i \leq m and hence W is a \mathbb{C}[G]-submodule of V. Thus V cannot be irreducible. \Box

Theorem 2. There exists a basis \mathcal{B} of V and square matrices A_1, \cdots , A_k such that for every g \in G

[\rho(g)]_{\mathcal{B}} = \begin{pmatrix} A_1 & 0 & \cdots & 0 \\ 0 & A_2 & \cdots & 0 \\ . & . & \cdots & . \\ . & . & \cdots & . \\ 0 & 0 & \cdots & A_k \end{pmatrix}.

Proof. Since V is a semisimple \mathbb{C}[G]-module, V = \bigoplus_{i=1}^k V_i, for some integer k and some simple \mathbb{C}[G]-submodules V_i of V. In fact, from the theory of semisimple modules, it is clear that V_1, \cdots , V_k are all simple \mathbb{C}[G]-submodules of V. Now, choose a \mathbb{C}– basis \mathcal{B}_i for each V_i and let A_i =[\rho(g)|_{V_i}]_{\mathcal{B}_i}. Let \mathcal{B} = \bigcup_{i=1}^k \mathcal{B}_i. Obviously \mathcal{B} is a basis for V and [\rho(g)]_{\mathcal{B}} is the matrix given in the theorem. \Box

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