On a simple real Lie algebra whose complexification is not simple

Question

I wish to prove the following result:

Theorem: Let $\mathfrak{h}$ be a real simple Lie algebra and $\mathfrak{h}_{\mathbb{C}}$ be its complexification, which is not simple. Then, there is a complex simple Lie algebra such that $\mathfrak{h} \equiv \mathfrak{g}_{\mathbb{R}}$, where $\mathfrak{g}_{\mathbb{R}}$ is $\mathfrak{g}$ viewed as a real Lie algebra with twice the dimension.

The only hint I have at this moment is to decompose $\mathfrak{h}_{\mathbb{C}}$ into a direct sum of simple Lie algebras. However, that would require me to know that $\mathfrak{h}$ comes from a compact Lie group and $\mathfrak{h}_{\mathbb{C}}$ has a trivial center. With the latter seems easy to deal with, the first one need not be necessarily true. So, is there any way I can get around this?

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EDIT:

The initial hint was not completely true! I have tried directly using the fact that $\mathfrak{h}_{\mathbb{C}}$ can be decomposed into a direct sum of simple Lie algebras, but it is not really useful. Instead, I came up with another way:

Since $\mathfrak{h}_{\mathbb{C}}$ is semi-simple, it has a compact real form $\mathfrak{k}$. That is, $\left( \mathfrak{h}_{\mathbb{C}} \right)_{\mathbb{R}} \equiv \mathfrak{k} \oplus \iota \mathfrak{k}$. We consider $\mathfrak{g} = \mathfrak{k}_{\mathbb{C}}$.

Is there some way we can argue that $\mathfrak{k}$ has to be simple so that $\mathfrak{g}$ becomes simple and then we have $\mathfrak{h} \equiv \mathfrak{g}_{\mathbb{R}}$?

Answer 1

There is a far more general approach to dealing with simple Lie algebras over a characteristic $0$ field $K$ which do not stay simple under scalar extension to some field $L \vert K$, due to Jacobson. I outlined it in my thesis and largely quoted it in https://math.stackexchange.com/a/3932095/96384.

For the case at hand, I would distill it to the following:

Assume $\mathfrak h$ is a real simple Lie algebra such that $\mathfrak h_\mathbb C$ is not simple. Let $V$ be the adjoint representation of $\mathfrak h$, i.e. $V$ "is" $\mathfrak h$ as a vector space, with $\mathfrak h$ acting on it via $h.v := [h,v] = \mathrm{ad}(h)(v)$.

Because $\mathfrak h$ is simple, the representation $V$ is simple, so its endomorphism ring

$$D:= \mathrm{End}_{\mathfrak h} (V) := \{ f \in \mathrm{End}_{\mathbb R}(V) : \mathrm{ad}(h) \circ f = f \circ \mathrm{ad}(h) \text{ for all } h \in \mathfrak h\}$$

is a skew field by Schur's Lemma. Actually one can easily show (see above link) that in this case it is commutative, so here it is a finite-dimensional field extension of $\mathbb R$, so it's either $\mathbb R$ itself or $\mathbb C$.

But now take note of the plain linear algebra fact that for any finite-dimensional real vector space $V$, we have $\mathbb C \otimes \mathrm{End}_{\mathbb R}(V) \simeq \mathrm{End}_{\mathbb C}(V_\mathbb C)$, and this identification is compatible with the respective adjoint actions of $\mathfrak h$ and $\mathfrak h_{\mathbb C}$. Hence

$$\mathbb C \otimes_\mathbb R D \simeq \mathrm{End}_{\mathfrak h_\mathbb C}(V_\mathbb C).$$

But since by assumption $\mathfrak h_\mathbb C$ is not simple (rather semisimple with $\ge 2$ simple summands), the right hand side, again via Schur's Lemma and semisimplicity, is a $\mathbb C$-vector space of dimension $\ge 2$.

This cannot be if $D=\mathbb R$. So $D= \mathbb C$ and has a natural action on $\mathfrak h$ (compatible with the scalar action of $\mathbb R$), so we can now view $\mathfrak h$ as the scalar restriction to $\mathbb R$ of a complex simple Lie algebra.

(And we can infer more precisely that $\mathfrak h_\mathbb C$ has exactly $2$ simple summands. With a little more work one can show these two summands are isomorphic to each other.)