The dual representation

Posted: February 11, 2011 in Representations of Finite Groups
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We introduced the dual representation \rho^* of a representation \rho of a finite group G in the third construction in this post. We will keep notations we used in that post. I’m now going to show that \rho^* is indeed a representation and it is irreducible if and only if \rho is irreducible.

Remark 1\rho^* is a representation of G.

Proof. First we need to show that \rho^* is well-defined, i.e. \rho^*(g) \in \text{GL}(V^*) for all g \in G. The fact that \rho^*(g) is \mathbb{C}-linear follows trivially from the fact that the elements of V^* are linear. Also, if \rho^*(g)(f)=0, for some g \in G and f \in V^*, then f \rho(g^{-1})(v) = 0, for all v \in V. But, since \rho(g^{-1}) is an isomorphism, \rho(g^{-1}) is onto and hence f = 0. Finally, \rho^* is a group homomorphism because if g_1,g_2 \in G and f \in V^*, then

\rho^*(g_1g_2)(f) = f \rho(g_2^{-1}g_1^{-1})=f \rho(g_2^{-1}) \rho(g_1^{-1})=\rho^*(g_1)(f \rho(g_2^{-1}))= \rho^*(g_1) \rho^*(g_2)(f). \Box

Remark 2. Suppose that U is a \mathbb{C}[G]-submodule of V^* and let W = \{v \in V: \ f(v)=0, \ \forall f \in U \}.

1) W is a \mathbb{C}[G]-submodule of V.

2) If \rho is irreducible, then \rho^* is irreducible.

Proof. 1) f \rho(g)=\rho^*(g^{-1})(f) =0 over W for all f \in U, because \rho^*(g^{-1})(U) \subseteq U. So \rho(g)(W) \subseteq W.

2) Since \rho is irreducible, either W=(0) or W=V. Clearly if W=V, then U=(0). Now, supppose that W=(0). Let \dim_{\mathbb{C}}V = \dim_{\mathbb{C}}V^*=n and choose a basis \{f_1, \cdots , f_m \} for U. Consider the natural linear map \varphi : V \longrightarrow \bigoplus_{i=1}^m (V/\ker f_i), which is injective because W=(0). But \dim_{\mathbb{C}} (V/\ker f_i) \leq 1, because V/\ker f_i are all \mathbb{C}-subspaces of \mathbb{C}. Hence

n=\dim_{\mathbb{C}}V \leq \sum_{i=1}^m \dim_{\mathbb{C}} (V/\ker f_i) \leq m,

which implies that m=n and so U=V^*. \Box

Remark 3.  Suppose that W is a \mathbb{C}[G]-submodule of V and let U=\{f \in V^* : \ f(W)=(0) \}.

1) U is a \mathbb{C}[G]-submodule of V^*.

2) If \rho^* is irreducible, then \rho is irreducible.

Proof.  1) Since \rho(g^{-1})(W) \subseteq W, we have \rho^*(g)(f)(W)=f \rho(g^{-1})(W)=(0), for all f \in U.

2) Since \rho^* is irreducible, either U=(0) or U=V^*. Clearly if U=V^*, then W=(0). So suppose that U=(0). If W \neq V, then there exists a \mathbb{C}-subspace W_1 \neq (0) of V such that V=W \oplus W_1. Choose 0 \neq v_1 \in W_1 and extend it to a basis \{v_1, \cdots , v_n \} for V. Now define f \in V^* by f(\sum_{i=1}^n c_i w_i) = c_1w_1. Clearly 0 \neq f \in U. Contradiction! \Box

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Comments
  1. hoormh says:

    how can i prove this when G is lie group ??differentiable of f:v_______c?

  2. Line says:

    How do you know that V/ ker f_i is a subspace of C?

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