I started reading Brian C.Hall's book on matrix Lie group and Lie algebra representations and I come across the following:
Dual representation is irreducible if and only if the representation is irreducible.
I read several explanations to the above statement, but I am wondering if the following sloppy way of proof also works.
Let's say we have a vector space $V = a_1 v_1 + a_2 v_2 +…+a_n v_n$ written in its bases and the dual space $V^* = b_1 w_1 + b_2 w_2 +…+b_n w_n$ written in the dual bases so that $w_i(v_j) = \delta_{ij}$, then
$$V^*(V) = a_1b_1 + a_2 b_2 +…+a_nb_n$$
but on the other hand, if G is a matrix Lie group and $A$ is any element of $G$; $\Pi(A)$ and $\Pi^*(A)$ are representations of G acting on a finite-dimensional vector space V and its dual space $V^*$ respectively, then the following holds
$$V^*(V) =\Pi(A)^*V^*(\Pi(A)V) = a_1b_1 + a_2 b_2 +…+a_nb_n$$
Then isn't it obvious that if $\Pi(A)$ is irreducible then $\Pi(A)^*$ must be irreducible since $\Pi(A)V = V$ or vice-versa?
I feel like I am misunderstanding something badly.
Best Answer
I can't say much about your attempted proof, since I don't know what $\Pi$ is (perhaps include the definition in your question?). However, I think an extremely clear way to see this statement is at the level of characters: $V$ is irreducible iff $\langle \chi_V, \chi_V\rangle_G=1$, where $\langle\cdot,\cdot\rangle_G$ is the inner product on class functions on $G$. Then $\chi_{V^\ast} = \overline{\chi_V}$ (the complex conjugate) so we have that $$ \langle \chi_{V^\ast},\chi_{V^\ast}\rangle_G = \frac{1}{|G|} \sum_{g\in G} \chi_{V^\ast}(g) \overline{\chi_{V^\ast}(g)} = \frac{1}{|G|} \sum_{g\in G} \chi_{V}(g) \overline{\chi_{V}(g)} = \langle \chi_V, \chi_V\rangle. $$ (This is for a complex representation of a finite group, but the idea can be made precise for other groups, e.g. Swap the sum for an integral for locally compact $G$).