When is $SO^0(n,1)$ isomorphic to a complex Lie group?

Question

The group $SO^0(3,1)$ is isomorphic to a complex Lie group, namely $PSL_2(\mathbb{C})$. Are there further examplex when $SO^0(n,1)$ isomorphic to a complex Lie group? An obvious necessary condition is that the dimension is even, i.e. $n$ mod $4$ is either $0$ or $3$.

Answer 1

No. Let me work entirely on the level of Lie algebras. If a real Lie algebra $\mathfrak{g}$ has a complex structure, then its complexification $\mathfrak{g} \otimes_{\mathbb{R}} \mathbb{C}$ splits up, as a complex Lie algebra, as the direct sum of $\mathfrak{g}$ and its "conjugate" $\overline{\mathfrak{g}}$ (this means $\mathfrak{g}$ with the conjugate complex structure): in particular, it cannot be simple.

But the complexification of $\mathfrak{so}_{n, 1}(\mathbb{R})$ is $\mathfrak{so}_{n+1}(\mathbb{C})$, which is simple unless $n = 3$. When $n = 3$, there is an exceptional isomorphism

$$\mathfrak{so}_4(\mathbb{C}) \cong \mathfrak{so}_3(\mathbb{C}) \times \mathfrak{so}_3(\mathbb{C})$$

but this doesn't generalize.

What does generalize, however, is the following: the exceptional isomorphism you describe fits into a family of exceptional isomorphisms

$$\mathfrak{sl}_2(\mathbb{R}) \cong \mathfrak{so}(2, 1)$$ $$\mathfrak{sl}_2(\mathbb{C}) \cong \mathfrak{so}(3, 1)$$ $$\mathfrak{sl}_2(\mathbb{H}) \cong \mathfrak{so}(5, 1)$$ $$\mathfrak{sl}_2(\mathbb{O}) \cong \mathfrak{so}(9, 1).$$

See the references here for details.