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.


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.

  • $\begingroup$ 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. $\endgroup$ – Dietrich Burde Jan 4 '14 at 12:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.