Does equality of the complexifications imply equality of the algebras?

Question

Let $\mathfrak{g}$ and $\mathfrak{h}$ be two real Lie algebras. Now suppose their complexifications are isomorphic, that is, $$\mathfrak{g}_{\mathbb{C}}\simeq\mathfrak{h}_{\mathbb{C}}.$$.

Can I say anything about the "isomorphicness" of $\mathfrak{g}$ and $\mathfrak{h}$? i.e. do I necessarily have $$\mathfrak{g}\simeq\mathfrak{h}$$

Specifically I am curious about whether $$\mathfrak{su}(4)_{\mathbb{C}}=\mathfrak{sl}(4,\mathbb{C})\simeq \mathfrak{so}(6,\mathbb{C})=\mathfrak{so}(6,\mathbb{R})_\mathbb{C}$$ implies $$\mathfrak{su}(4)\simeq \mathfrak{so}(6,\mathbb{R})$$

Answer 1

I've constructed a counter example.

$$so(3,1)_\mathbb{C}\simeq so(4,\mathbb{C})\simeq sl(2,\mathbb{C})\oplus sl(2,\mathbb{C})=su(2)_\mathbb{C}\oplus su(2)_\mathbb{C}=(su(2)\oplus su(2))_\mathbb{C}$$

but it is not true that $so(3,1)\simeq su(2)\oplus su(2)$. In particular one is compact and the other is non-compact.

However, $so(4)\simeq su(2)\oplus su(2)$, so perhaps $g_\mathbb{C}\simeq h_\mathbb{C}$ implies $g\simeq h$ only if $g$ and $h$ are both compact, but I have not proven this.

Answer 2

A simpler counterexample in dimension $3$ is $\mathfrak{g}=\mathfrak{so}_3(\mathbb{R})$ and $\mathfrak{h}=\mathfrak{sl}_2(\mathbb{R})$. We have $$\mathfrak{g}_{\mathbb{C}}\cong\mathfrak{h}_{\mathbb{C}}\cong \mathfrak{sl}_2(\mathbb{C}),$$ but $\mathfrak{g}$ and $\mathfrak{h}$ are not isomorphic as real Lie algebras, because one algebra has a $2$-dimensional subalgebra, the other one does not. If we furthermore assume that the real Lie algebras are simple, then we can use the machinery of the real classification, i.e., that we have one compact real form and one split form etc. The classification result in dimension $3$ for the simple real Lie algebras then is exactly $\mathfrak{so}_3(\mathbb{R})$ and $\mathfrak{sl}_2(\mathbb{R})$.