# Automorphisms of a split Lie algebra -- strange proof in Bourbaki

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?

• 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 Feb 27, 2017 at 18:30
• @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. Mar 1, 2017 at 3:23