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Let $\mathfrak g$ be a complex Lie algebra. If $\mathfrak g$ is semisimple then its real forms are completely understood. In particular there is a compact real form of $\mathfrak g$, which is unique up to inner automorphism.

Suppose $\mathfrak g$ is not semisimple and write $\mathfrak g = \mathfrak r \ltimes \mathfrak s$ using the Levi-Malcev theorem, where $\mathfrak r$ is the radical of $\mathfrak g$ and $\mathfrak s$ is a maximal semisimple subalgebra. Assume that $\mathfrak g$ has a real form. My question is:

Can a real form $\mathfrak g_0$ of $\mathfrak g$ be taken so that $\mathfrak g_0 = \mathfrak r_0 \ltimes \mathfrak s_0$, $\mathfrak r_0$ the radical of $\mathfrak g_0$ and $\mathfrak s_0$ a maximal semisimple subalgebra of $\mathfrak g_0$ which is a compact real form of $\mathfrak s$?

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  • $\begingroup$ Thanks Qiaochu Yuan, I have edited the question. $\endgroup$
    – Claudio Gorodski
    Commented Oct 24, 2022 at 20:41
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    $\begingroup$ To be clear, you mean "does there exist a real form of $\mathfrak{g}$ such that..."? $\endgroup$
    – Qiaochu Yuan
    Commented Oct 24, 2022 at 20:43
  • $\begingroup$ Yes, that is what I meant. $\endgroup$
    – Claudio Gorodski
    Commented Oct 24, 2022 at 20:44
  • $\begingroup$ I am not sure I understand. Levi Malcev theorem implies that the real form ${\mathfrak g}_0$ has a Levi decomposition whose complexification gives a Levi decomposition of $\mathfrak g$ and by the uniqueness of the Levi decomposition, must be the original Levi decomposition up to automorphism. $\endgroup$
    – Venkataramana
    Commented Oct 25, 2022 at 10:05
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    $\begingroup$ I think $\mathfrak g= \mathbb C^2 \rtimes \mathfrak{sl}_2\mathbb C$ is a counter example. The only real form is $\mathfrak g_0=\mathbb R^2 \rtimes \mathfrak{sl}_2\mathbb R$. $\endgroup$
    – Claudio Gorodski
    Commented Oct 25, 2022 at 14:38

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