4
$\begingroup$

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 $\overline{\mathbb{K}}$?

$\endgroup$
1
  • 2
    $\begingroup$ A Cartan $\mathfrak h$ of $\mathfrak g$ which is contained in $\mathfrak p$ maps isomorphically to a Cartan of $\mathfrak p^{\mathrm{red}}$. If we have a Cartan which splits in $\mathfrak g$ then the quotient map to $\mathfrak p^{\mathrm{red}}$ is an isomorphism on root spaces $\mathfrak p_\alpha$ such that $\mathfrak p_{\alpha},\mathfrak p_{-\alpha}\neq 0$, and is the $0$ map otherwise. So we get a splitting $\mathfrak p^{\mathrm{red}}\to \mathfrak p$ by reversing the isos on each $\mathfrak p_{\alpha}$ and $\mathfrak h$. I can't speak to the non-split case, though. $\endgroup$
    – Grant B.
    Nov 24 at 6:29
3
$\begingroup$

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 on arxiv for an exhaustive treatment of this only assuming the basics of Lie algebras.

Also this paper, 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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.