Representation on the dual space of a simple representation

Question

Let $V$ be a finite-dimensional representation of the simple Lie algebra $\mathfrak{g}$. We also give to the dual space $V^*$ of $V$ a structure of representation of $\mathfrak{g}$ by setting $(g \cdot f)(v) = -f(g \cdot v)$. How to prove that if $V$ is simple (resp. semi-simple) then $V^*$ is simple (resp. semi-simple) ?

Answer 1

Let $\langle -, - \rangle: V \times V^* \to \mathbb{C}$ represent the natural pairing between the original space and its dual, so that $\langle v, f \rangle = f(v)$. Note that by the definition of the action of $\mathfrak{g}$ on the dual space, we have the "skew-adjoint" property $$ \langle v, x \cdot f \rangle = (x \cdot f)(v) = - f(x \cdot v) = - \langle x \cdot v, f \rangle.$$

For any subspace $U \subseteq V$, we can define the annihilating subspace $$ U^\perp = \{ f \in V^* \mid \langle u, f \rangle = 0 \text{ for all } u \in U \} \subseteq V^*.$$ Now suppose that $U \subseteq V$ is a subrepresentation, so $x \cdot u \in U$ for all $x \in \mathfrak{g}$. Then we check that the complement space $U^\perp$ is also subrepresentation of $V^*$: for any $f \in U^\perp$ we have $$ 0 = \langle x \cdot u, f \rangle = - \langle u, x \cdot f \rangle$$ by the skew-adjoint property, showing that $x \cdot f \in U^\perp$. Hence $U^\perp$ is a subrepresentation.

Nondegeneracy of $\langle -, - \rangle$ implies that $\dim U^\perp = \dim V - \dim U$, and so we know that a nontrivial subrepresentation $U$ of $V$ gives a nontrivial subrepresentation $U^\perp$ of $V^*$. The converse also works (the proof is exactly the same), and hence $V$ is simple if and only if $V^*$ is. I'll leave it up to you to figure out how to prove that $V$ is semisimple if and only if $V^*$ is.