Everything here is over $\mathbb{C}$. Let $\mathfrak{g}$ be a finite-dimensional simple Lie algebra and let $\mathfrak{p}$ be a parabolic subalgebra (relative to some fixed Borel subalgebra that is unimportant for this question). Then $\mathfrak{p}$ has a decomposition $$ \mathfrak{p} = \mathfrak{l} \oplus \mathfrak{u_+},$$ where $\mathfrak{l}$ is a reductive subalgebra (the Levi factor of $\mathfrak{p}$) and $\mathfrak{u}_+$ is a nilpotent ideal (the nilradical of $\mathfrak{p}$). Finally, we can decompose $\mathfrak{g}$ as
$$\mathfrak{g} = \mathfrak{u}_- \oplus \mathfrak{l} \oplus \mathfrak{u}_+$$ (as $\mathfrak{l}$-modules), where $\mathfrak{u}_-$ and $\mathfrak{u}_+$ are dual to each other via the Killing form of $\mathfrak{g}$.
Assume further that the following (equivalent) conditions hold:
$\mathfrak{g}/ \mathfrak{p}$ is irreducible as a $\mathfrak{p}$-module;
$\mathfrak{u}_-$ is irreducible as an $\mathfrak{l}$-module;
$\mathfrak{u}_-$ is an abelian Lie algebra;
2 and 3 with $\mathfrak{u}_-$ replaced by $\mathfrak{u}_+$.
Buzzwords here are "Hermitian symmetric space" and "generalized flag variety." There is a classification of these in terms of root systems but I don't want to use that.
I need to understand the decomposition of ${\bigwedge}^2 \mathfrak{u}_- $ into irreducible modules for $\mathfrak{l}$. Using the classification of these parabolics, you can just see explicitly what the highest weight of $\mathfrak{u}_-$ is, and then it's not too hard to compute what ${\bigwedge }^2 \mathfrak{u}_-$ is, but I would like a more elegant way to see what's going on here.
I have been informed that there is some version of the BGG resolution that will be helpful for this - this evidently gives the highest weights of ${\bigwedge }^2 \mathfrak{u}_-$ in terms of the affine action of some elements of the Weyl group on the highest weight of $\mathfrak{u}_-$, but at this point I'm stuck. I don't know enough (ok, anything really) about the BGG resolution to know where to look for this stuff. Either an explanation or a reference would be much appreciated.