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This is about ch. VIII § 5 no. 3 Proposition 5 in Bourbaki's book on Groupes et algèbres de Lie (unchanged on p. 109 in the Hermann 1975 or Springer 2006 edition). To clarify, I am sure the statement is true, but I am not convinced by the proof given there.

The assertion is that for a "splittable" semisimple Lie algebra $\mathfrak{g}$, the group $Aut_0(\mathfrak{g})$ acts simply transitively on the set of épinglages of $\mathfrak{g}$. Here, $Aut_0(\mathfrak{g})$ are those automorphisms of $\mathfrak{g}$ which become elementary automorphisms (= products of exp(ad n) for nilpotent n) after base extension to an algebraic closure. (This is defined two pages earlier, alas, there seems to be a misprint: $Aut_0$ and $Aut_e$ are flipped in lines 14 and 16.) Epinglages are defined in §4 no. 1.

Now I understand the proof of "simple" in the first two lines, and then I also understand the single statements in the following, until the very last sentence. There they say that $\mathfrak{h}_i$ and $X_\alpha^i$ generate $\mathfrak{g}_i$ for $i =1,2$; but what is $\mathfrak{g}_i$? It has not occured before. Is $\mathfrak{g} = \mathfrak{g}_1 = \mathfrak{g}_2$? (This seems implausible, because an épinglage contains $X_\alpha^i$ for $\alpha$ in a basis of the root system, so with $\mathfrak{h}_i$ they should generate Borel subalgebras.) In any case, I do not see how to finish the argument, where they somehow conclude that the automorphism $\varphi$, defined over the algebraic closure, descends to the given base field.

Question 1: How to understand the final step in Bourbaki's proof?

Now I also think that I can prove the assertion of transitivity with just a small alteration, as follows: By loc.cit., ch. VIII § 3 no. 3 Corollaire to Prop. 10, there is an element $f$ of $Aut_e(\mathfrak{g})$ that transforms $\mathfrak{h}_1$ to $\mathfrak{h}_2$, and consequently the épinglage $e_1$ to an épinglage $f_1(e_1) = (\mathfrak{g}, \mathfrak{h}_2, B', X'_\alpha)$. By prop. 4 and cor. 2 to prop. 2, there further is an element $f_2$ of $Aut_0(\mathfrak{g}, \mathfrak{h}_2) ( \subset Aut_0(\mathfrak{g}))$ that transforms the épinglage $f_1(e_1)$ to $e_2$. Set $\psi := f_2 \circ f_1$.

Question 2: Is this proof correct?

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  • $\begingroup$ Did you figure it out by now? This was asked a long time ago. I am reading Bourbaki at the moment, so I will eventually stumble on this $\endgroup$ Feb 27, 2017 at 18:30
  • $\begingroup$ @PatrickDaSilva: It has been a long time, but I think I eventually was confident that my proposed alternative works. I added it in a footnote to the paper where I needed and quoted the result, and the referee never brought it up. Let me know if you have thoughts on this when you arrive there in your reading. $\endgroup$ Mar 1, 2017 at 3:23

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