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According to the Wikipedia article on complex Lie algebras , any complex Lie algebra $\mathfrak{g}$ that is isomorphic to its conjugate $\overline{\mathfrak{g}}$ admits a real form.

So as I understand it, this isomorphism exists if and only if there is an antilinear automorphism $\tau: \mathfrak{g} \to \mathfrak{g}$. The wikipedia article then says 'we may assume without loss of generality that $\tau$ is the identity on the underlying real vector space', which seems nonsensical and wrong to me, as this would imply that tau is the identity and hence not antilinear. Anyway, the rest of the argument depends on $\tau$ being an involution; I believe that part is correct, so the question is, why can we assume that $\tau$ is an involution?

My idea was basically that $\tau^2$ is a complex linear automorphism; if $\tau$ is diagonalizable, then $\tau^2=\varphi^2$ for a linear isomorphism $\varphi$. However, I do not know whether this $\varphi$ need be a Lie algebra automorphism; I suspect not. If it is, however, then $\tau \circ \varphi^{-1}$ is the antilinear involution we are looking for. Either way, this depends on $\tau^2$ being diagonalizable, so I do not have much hope for this argument.

Does anyone know a proof (or counterexample) of this fact? For reference, I am currently reading Lie groups, Representation theory and Symmetric spaces due to Ziller, where I have finished the first two chapters.

Callum
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FS123
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  • My answer to https://math.stackexchange.com/q/4398239/96384 might help clear some confusion. – Torsten Schoeneberg Mar 01 '24 at 20:59
  • Thanks for your input! However, I am not sure that you can see this as a representation- for example, $\tau: \mathfrak{gl}(n, \mathbb{C}) \to \overline{\mathfrak{gl}(n, \mathbb{C})}: A \mapsto P^{-1} \overline{A} P$ is a $\mathbb{C}$-linear isomorphism (which corresponds to an antilinear isomorphism $\mathfrak{gl}(n, \mathbb{C}) \to \mathfrak{gl}(n, \mathbb{C})$), but $\tau^2$ is not a multiple of the identity. – FS123 Mar 02 '24 at 15:37
  • I realized some of my previous comments were wrong / misleading, and deleted them. I agree with the questions you raise and don't have answers to them. The article references Knapp's book, which AFAIK is a good one; have you checked the reference, maybe somebody misunderstood something there? – Torsten Schoeneberg Mar 03 '24 at 21:37
  • I checked Knapp's book, and sadly I could not find any reference to this particular result... – FS123 Mar 05 '24 at 20:31
  • I looked through Knapp too and could not find anything resembling that statement. Also, the reference goes to chapter VI which only deals with semisimple LA's anyway. Weird. I wonder if the statement is actually wrong and one needs the extra assumption that the iso between the LA and its dual is an involution (when viewed as a real-linear map say). Now the problem would be to find a counterexample, unfortunately I am not very familiar with LAs without real form (although many exist: https://math.stackexchange.com/a/3903875/96384, https://math.stackexchange.com/q/95540/96384) – Torsten Schoeneberg Mar 06 '24 at 22:56

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