Existence of a non essentially real ideal in a semisimple Lie algebra

Question

Let $\mathfrak{g}$ be the Lie algebra of a semisimple compact Lie group $G$. Denote by $\mathbb{C} \mathfrak g = \mathbb{C} \otimes \mathfrak g$ the complexification of $\mathfrak g$. I am assuming that $\mathfrak g$ is not simple, then there exist an ideal $\mathfrak i \subset\mathfrak g$ that is different from $\{0\}$ and different from the entire Lie algebra $\mathfrak g$.

A subalgebra $\mathfrak h \subset \mathbb C \mathfrak g$ is called essentially real if $\overline {\mathfrak h} = \mathfrak h$. This basically means that there exist $\mathfrak h_{\mathbb R} \subset \mathfrak g$ such that $\mathfrak h = \mathbb{C} \otimes \mathfrak h_{\mathbb R}$.

So, my question is: does it exists an ideal $\mathfrak i \subset \mathbb C \mathfrak g$ that is not essentially real?

In other words, I want an ideal $\mathfrak i$ that is not just the complexification of an ideal from $\mathfrak g$.

Thank you very much for any help.

Answer 1

No such ideal exists.

A simple Lie algebra $\mathfrak{g}$ over a field $k$ is called absolutely simple if for every algebraic extension $K\vert k$ (or equivalently: for an algebraic closure $K\vert k$) , the scalar extension $K\otimes \mathfrak{g}$ is also simple (note that it necessarily is semisimple; for an example where it is not simple, see below).

One can show that if a Lie algebra over a field $k$ is simple but not absolutely simple, it is the scalar restriction of an absolutely simple Lie algebra over some algebraic extension $K \vert k$. Actually, one can compute $K$ as the (associative) subalgebra of $End_k(\mathfrak{g})$ consisting of those elements that commute with all $ad_\mathfrak{g}(x), x \in \mathfrak{g}$. As far as I know, this was first shown by Jacobson in Duke Math. J., Volume 3, Number 3 (1937), 544-548, doi:10.1215/S0012-7094-37-00343-0, and holds for more general kinds of algebras. I wrote a little overview of that in section 4.1 of my thesis, a lot of it is now reproduced in this answer of mine.

Now to your question: Since scalar extension commutes with direct sums, and ideals of semisimple Lie algebras are direct summands, your question is equivalent to asking whether there exists a simple compact real Lie algebra $\mathfrak{g}$ which is not absolutely simple. But by the above theory, and the fact that the only proper algebraic extension of $\Bbb R$ is $\Bbb C$, the only simple but not absolutely simple real Lie algebras are: the simple complex Lie algebras considered as $\Bbb R$-algebras. The first example maybe being $\mathfrak{sl}_2(\Bbb C)$ viewed as a Lie algebra over $\Bbb R$ (six-dimensional); it is simple, but not absolutely simple, as its scalar extension $\Bbb C \otimes_{\Bbb R} \mathfrak{sl}_2(\Bbb C)$ actually is isomorphic to the sum of two copies of $\mathfrak{sl}_2(\Bbb C)$.

However, none of these scalar restrictions of simple complex Lie algebras correspond to compact Lie groups, e.g. because they obviously contain nilpotent elements.