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$\DeclareMathOperator{\Inn}{\operatorname{Inn}}\DeclareMathOperator{\Aut}{\operatorname{Aut}}\DeclareMathOperator{\Out}{\operatorname{Out}}\DeclareMathOperator{\g}{\mathfrak{g}}$According to Wikipedia, outer automorphisms as symmetries of a Dynkin diagram follows from the general fact, that for a complex or real simple Lie algebra $\g$, the automorphism group $\Aut(\g)$ is a semidirect product of $$\Inn(\g) \rtimes \Out(\g)$$

i.e., the short exact sequence splits:

 $${{1 ⟶ \Inn(\g) ⟶ \Aut(\g) ⟶ \Out(\g) ⟶ 1}}$$

 

 > Question: Why the complex simple Lie algebras, we already know $\Aut(\g)=\Inn(\g) \rtimes \Out(\g)$, but it took a late recent work in 2010 to settle that the real simple Lie algebra also has $\Aut(\g)=\Inn(\g) \rtimes \Out(\g)$? (Especially given the following facts that the complex and real simple Lie algebras seem not be so different?)   Facts:

  •  Complex Lie algebra is a Lie algebra over the field $F$ as a complex $\mathbb{C}$.

  • Real Lie algebra is a Lie algebra over the field $F$ as a real $\mathbb{R}$.

  • Given a complex Lie algebra $\mathfrak g$, a real Lie algebra $\mathfrak{g}_0$ is said to be a  real form of $\mathfrak g$ if the complexification $$\mathfrak{g}_0 \otimes_{\mathbb{R}} \mathbb{C} \simeq \mathfrak{g}$$ is isomorphic to $\mathfrak{g}$.

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    $\begingroup$ The real case is always harder than the complex case; over the complex numbers the existence of eigenvalues is a huge help and you don't have that over the reals. $\endgroup$
    – Qiaochu Yuan
    Commented Oct 11, 2020 at 4:28
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    $\begingroup$ You didn't define $\mathrm{Inn}(\mathfrak{g})$. There's a possible ambiguity. It might mean $\mathrm{Aut}(\mathfrak{g})^0(\mathbf{R})$, the group of real points of the algebraic connected component of the identity. Or its identity component as a Lie group, which is smaller (and usually not Zariski-closed). It is the latter group that is considered in Gündogan's paper. (An amusing case is when $\mathfrak{g}=\mathfrak{so}(4,4)$, in which case as a Lie group, $\mathrm{Aut}(\mathfrak{g})$ has 24 connected components.) $\endgroup$
    – YCor
    Commented Oct 11, 2020 at 7:22
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    $\begingroup$ When $K$ is a more general field of characteristic zero and $\mathfrak{g}$ has no $K$-anisotropic factor there's a reasonable substitute for $\mathrm{Inn}(\mathfrak{g})$: the subgroup $I$ of $\mathrm{Aut}(\mathfrak{g})$ generated by exponentials of nilpotent derivations of $\mathfrak{g}$: in the real case this yields the identity component in the Lie topology (this fails in the presence of anisotropic factors: if $\mathfrak{g}$ is anisotropic this group $I$ is reduced to the identity). Then $I$ is normal in the automorphism group, but I don't know if this splits in general. $\endgroup$
    – YCor
    Commented Oct 11, 2020 at 8:30
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    $\begingroup$ @YCor 24 connected components: known to Cartan, it seems? See (1927a, p. 225): “Je signalerai simplement ce résultat assez curieux, c’est que, dans la Géométrie cayleyenne à 7 dimensions dont l’absolu est une quadrique de la forme $$x_1^2+x_2^2+x_3^2+x_4^2-x_5^2-x_6^2-x_7^2-x_8^2=0$$ le groupe des déplacements proprement dits se complète par 23 autres familles de transformations.” Also (1927b, p. 420). $\endgroup$
    – Francois Ziegler
    Commented Nov 14 at 4:39

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