Let $\mathbb{K}$ be a field (not assumed to be algebraically closed, but we can assume characteristic 0 if necessary), and let $\mathfrak{g}$ be a semisimple Lie algebra.


A Lie subalgebra $\mathfrak{p} \leq \mathfrak{g}$ is said to be **parabolic** if
$$\mathfrak{p}^\perp = \operatorname{nil} \mathfrak{p},$$
where $\mathfrak{p}^\perp$ is the Killing perp, and $\operatorname{nil} \mathfrak{p}$ is the nilradical of $\mathfrak{p}$ (the unique maximal nilpotent ideal in $\mathfrak{p}$).


The nilradical gives us a short exact sequence
$$0 \to \operatorname{nil} \mathfrak{p} \to \mathfrak{p} \to \mathfrak{p}^{\text{red}} \to 0$$
Where $\mathfrak{p}^{\text{red}} = \mathfrak{p} / \operatorname{nil} \mathfrak{p} $ is reductive.
I have seen it mentioned in various places that this short exact admits a splitting on the right in the special case that $\mathfrak{p}$ is parabolic. Just naively, it seems like this should have to do with using the Killing form, since $\mathfrak{p}^\perp = \operatorname{nil} \mathfrak{p}$. But the Killing form is in general degenerate on $\mathfrak{p}$, so I'm not quite sure how to go about this.


Is there a way to construct this splitting (in a manner as choice-free as possible) without extending to an algebraically closed field $\overline{\mathbb{K}}$?



Reductive Levi decomposition of a parabolic subalgebra
There are many such splittings. Any complementary (aka opposite) parabolic subalgebra (i.e. $\mathfrak{q}$ such that $\mathfrak{p} \oplus \mathfrak{q}^\perp = \mathfrak{g}$ and so on) provides a unique splitting $\mathfrak{p} = \mathfrak{p} \cap \mathfrak{q} \oplus \mathfrak{p}^\perp$. The space of complementary parabolic subalgebras is a $\exp \mathfrak{p}^\perp$-torsor or in other words an affine space modelled on $\mathfrak{p}^\perp$.


We can also equate these to a particular adjoint orbit in $\mathfrak{p}$. There is a particular element $\xi^\mathfrak{p}\_\mathfrak{q}$ of $\mathfrak{p}$ called the grading element (aka canonical element) for which $ \mathfrak{q}^\perp \oplus \mathfrak{p} \cap \mathfrak{q} \oplus \mathfrak{p}^\perp$ are the $-1,0,1$ eigenspaces for $\mathrm{ad}\ \xi^\mathfrak{p}\_\mathfrak{q}$. This uniquely identifies the complementary subalgebra and the splitting.


Check out [this paper](https://arxiv.org/abs/1607.00370) on arxiv for an exhaustive treatment of this only assuming the basics of Lie algebras.


Also [this paper](https://scholarworks.umass.edu/cgi/viewcontent.cgi?article=2159&context=math_faculty_pubs), covers the filtrations and gradings we are seeing here quite succintly in the preliminary chapter.


Edit: an additional point is that the span of $\xi^\mathfrak{p}\_\mathfrak{q}$ is exactly the 1-dimensional centre of $\mathfrak{p} \cap \mathfrak{q}$ and its orthocomplement is semisimple.


