Given a complex Lie algebra $\mathfrak{g}$, with dual space $\mathfrak{g}^*$, fix $\operatorname{ad}: \mathfrak{g} \rightarrow \mathfrak{gl}(\mathfrak{g})$, $\operatorname{ad}(X) = [X,\_]$ the adjoint representation of $\mathfrak{g}$, and fix $\operatorname{ad}^*: \mathfrak{g} \rightarrow \mathfrak{gl}(\mathfrak{g^*})$, $\operatorname{ad}^*(X) = -\operatorname{ad}(X)^\top$ the coadjoint representation. I am trying to prove that the adjoint and coadjoint representations are isomorphic.
I have been told that the Killing form $K: \mathfrak{g} \times \mathfrak{g} \rightarrow \mathbb{C}$ induces such isomorphism. To do so, I've defined a map $K^\flat: \mathfrak{g} \rightarrow \mathfrak{g}^*$ given by
\begin{align}
K^\flat(X): \mathfrak{g} & \rightarrow \mathbb{C} \\ Y & \mapsto K(X,Y).
\end{align}
Questions
-
How do I prove that $K^\flat$ is a representation homomorphism? It feels very easy, but somehow I cannot seal the deal on this.
-
Is there a better approach to this problem?
Any help would be greatly appreciated.
Best Answer
I'll denote the adjoint operator by $\operatorname{ad}_x \colon \mathfrak{g} \to \mathfrak{g}$, so that we have $\operatorname{ad}_x^* = - \operatorname{ad}_x^t$, which sends a linear functional $\varphi \colon \mathfrak{g} \to \mathbb{C}$ to $\operatorname{ad}_x^* \varphi = - \varphi \circ \operatorname{ad}_x$. Let $K \colon \mathfrak{g} \times \mathfrak{g} \to \mathbb{C}$ be any symmetric bilinear form, then we get a map $K^\flat \colon \mathfrak{g} \to \mathfrak{g}^*$ by defining $K^\flat(x) = K(x, -)$. Recall that for $K^\flat$ to give an isomorphism of representations between $\operatorname{ad}$ and $\operatorname{ad}^*$, we need the equation $K^\flat(\operatorname{ad}_x y) = \operatorname{ad}^*_x K^\flat(y)$ for all $x, y \in \mathfrak{g}$. Since both sides of this equation are elements of $\mathfrak{g}^*$, we can "check" both sides for all $z \in \mathfrak{g}$.
A key property of the killing form in particular is that it is invariant, meaning that $K([x, y], z) = K(z, [y, z])$ for all $x, y, z \in \mathfrak{g}$. Now we work out that $$ \begin{aligned} (\operatorname{ad}^*_x K^\flat(y))(z) &= -(K^\flat(y) \circ \operatorname{ad}_x)(z) \\ &= -K(y, [x, z]) \\ &= K([x, y], z) \\ &= K(\operatorname{ad}_x y, z) \\ &= (K^\flat(\operatorname{ad}_x y))(z). \end{aligned}$$
Notice that the invariance property is crucial to proving that $K^\flat$ is a homomorphism of representations. In order for $K$ to be an isomorphism as well, you need that the Killing form is nondegenerate (equivalently, $\mathfrak{g}$ is a semisimple Lie algebra).