## 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! $\Box$

Because $f_i \in V^*$ and so $f_i: V \to \mathbb{C}$ is a $\mathbb{C}$-linear map. Hence $V/\ker f_i$ is isomorphic to a subspace of $\mathbb{C}$.