Irreducibility of the $\mathfrak{g}$-module $\mathfrak{o}(k)/ad(\mathfrak{g})$

Question

For a simple lie algebra $\mathfrak{g}$, define $\mathfrak{o}(k)$ to be the orthogonal lie algebra with respect to the Killing form.

In the proof of Theorem 2 in the following paper, https://arxiv.org/pdf/math/0407240.pdf the author mentions that the following is true,

The $\mathfrak{g}$ module $\mathfrak{o}(k)/ad(\mathfrak{g})$ is irreducible if $\mathfrak{g}$ is not of type A while it is a direct sum $W \oplus W^∗ $for some non-self-dual module $W$ if $\mathfrak{g}$ is of type A.

I am unable to see why this is true. Can someone furnish a proof or point to references?

Answer 1

Reposting the answer from my cross-post on MathOverflow.

It's easy to see that $\mathfrak{o}(k) \cong \wedge^2\mathfrak{g}$. Proposition 5.1 in https://doi.org/10.4153/CJM-1997-007-1 is precisely what we need. It says that

Let $\alpha_0$ be the highest root, and let $J$ be the set of simple roots which are not orthogonal to $\alpha_0$. For each $\alpha \in J$, one knows that $\alpha_0-\alpha$ is a root, and it is evident that $e_{\alpha}\wedge e_{\alpha_0-\alpha}$ is a highest weight vector in $\wedge^2\mathfrak{g}$, and therefore generates an irreducible submodule $U_\alpha$. Let $U_2 = \oplus_{\alpha \in J} U_\alpha$ .

By dimension count $ \wedge^2\mathfrak{g} \cong \mathfrak{g} \oplus U_2$. And therefore, $\mathfrak{o}(k)/ad(\mathfrak{g})\cong U_2$ which is claimed (without proof) to be irreducible. However, now that we have an "explicit" form of the quotient, proving irreducibility seems easy.