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:

1.  $\mathfrak{g}/ \mathfrak{p}$ is irreducible as a $\mathfrak{p}$-module;

2.  $\mathfrak{u}_-$ is irreducible as an $\mathfrak{l}$-module;
  
3.  $\mathfrak{u}_-$ is an abelian Lie algebra;
  
4.  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.