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

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.

## 1 Answer

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.

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.

• Thank you very much for your detailed answer. I was seeking such simple ideals as a way to construct a semisimple subalgebra of $\mathbb C \mathfrak g$ having a specific property. Now it looks that my idea is not going to work. Maybe you can help me with it. If you not mind, please, take a look at this question: math.stackexchange.com/questions/3069692/… – Max Reinhold Jahnke Jan 11 at 10:40