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Edit: The reasoning on which this question is based is wrong. See my own answer for a counterexample.

Let $L$ be a (finite dimensional) semisimple Lie algebra over a field $k$ with $char(k) = 0$.

Let us call a subalgebra of $L$ toral if it is abelian and consists of semisimple elements. It can be shown with Bourbaki that the Cartan subalgebras of $L$ are precisely the maximal toral ones: See exercise 3) to ch. VII §2 of Bourbaki's book on Lie Groups and Lie Algebras; this is also noted in the first point of the "resumé" to this volume.

Now Humphreys proves in ch. 8.1 of his book Introduction to Lie algebras and Representation Theory -- under the standing assumption that $k$ is algebraically closed -- that a subalgebra of $L$ consisting of semisimple elements is automatically abelian, i.e., the assumption "abelian" in the above definition of "toral" is redundant. (And accordingly is left out of his definition of "toral".)

Now I wondered: Does it not follow that the same is true for general $k$? Namely, let $K|k$ be an algebraic closure, $A$ a subalgebra of $L$ consisting of semisimple elements, then $A_K$ is a subalgebra of $L_K$ consisting of semisimple elements, so by Humphreys, $A_K$ is abelian and a fortiori so is $A$. But then:

Cartan subalgebras of $L$ = subalgebras of $L$ maximal w.r.t. the property "consisting of semisimple elements" (which, automatically, are abelian)

Questions: Is there a direct reference for this? Or is my reasoning wrong?

Edit: To put it in other words, I claim

Let $L$ be a semisimple Lie algebra over a field $k$ with $char(k) = 0$ (but not necessarily algebraically closed). Let $A$ be a subalgebra consisting of semisimple elements. Then $A$ is abelian.

and would like to have a reference for this (or a counter-example, which would surprise me). From the above claim it follows that in the French edition of ch. VII/VIII of Bourbaki's Groupes et algèbres de Lie, p. 259, 1), in

c'est aussi l'ensemble des sous-algèbres commutatives de $\mathfrak{g}$ dont tous les éléments sont semi-simples.

the word "commutatives" is redundant.

Dietrich Burde
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Torsten Schoeneberg
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1 Answers1

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I have surprised and embarassed myself, my reasoning and claim is wrong. The point is that $A$ consisting of semisimple elements does not imply that $A_K$ consists of semisimple elements, because there are sums intervening. Example: $k = \mathbb{R}, K = \mathbb{C}$; $L = \mathbb{R} \pmatrix{i & 0 \\0 & -i} + \mathbb{R} \pmatrix{0 & 1 \\-1 & 0} + \mathbb{R} \pmatrix{0 & i \\i & 0}$, the compact real form of $\mathfrak{sl}_2$. One can check that this whole Lie algebra consists of semisimple elements, but of course its complexification $\mathfrak{sl}_2$ does not, and none of them are abelian.

I will edit the post and leave this as a counterexample for others.

Torsten Schoeneberg
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  • This is a good example. I was thinking that any semisimple $x\in L$ remains semisimple over the algebraic closure (because by definition it is diagonalisable over the algebraic closure), but of course, this is not enough. – Dietrich Burde Jan 04 '14 at 12:04