Let $\mathfrak{g}$ be a simple, complex Lie algebra and let $\mathfrak{h}$ be a Cartan subalgebra of $\mathfrak{g}$ (defined to be the maximum abelian subalgebra of $\mathfrak{g}$ containing only ad-diagonalisable elements). Then we have:

If $X \in \mathfrak{g}$ and $[X,H]=0$ for all $H\in \mathfrak{h}$, then $X \in \mathfrak{h}$.

Can someone please help me out? Either by giving a proof or directing me to one, because I can't seem to find one anywhere. Thanks:)

(p.s. is the simple, complex part necessary for the result?)


1 Answer 1


To answer your last question first: Neither "complex" nor "simple" are really necessary. One can relax the conditions and has the following statement:

If $\mathfrak{g}$ is a (finite-dim.) semisimple Lie algebra over a field of characteristic $0$, and $\mathfrak{h}$ is a subalgebra which is maximal among those consisting of semisimple elements, then $\mathfrak{h} = \mathfrak{z}_{\mathfrak{g}}(\mathfrak{h})$.

Actually, one can show more, namely that $\mathfrak{h}$ is self-normalising and nilpotent -- which is another common definition for Cartan subalgebras. If one works over $\mathbb C$ or any algebraically closed field, one can replace "semisimple elements" in the above statement by "ad-diagonalisable", but in general one has to be careful about this, cf. my answer to linked question.

That those two definitions of Cartan subalgebras are equivalent is proposition 3.1.5 in my thesis, for which I combined ideas from Humphreys' book on Lie algebras with some from Seligman's "Rational Methods in Lie Algebras" and "A note on the centralizer ..."by Xiaoxi Xue. It is also exercise 3 to vol VII, §2 of Bourbaki's books on Lie groups and Lie algebras. The proof for the relevant part here goes like this:

Call $Z:=\mathfrak{z}_{\mathfrak{g}}(\mathfrak{h})$ that centraliser of $\mathfrak{h}$. First show that $Z$ is self-normalising, and that for each element $x$ it contains, it also contains its semisimple and nilpotent parts $x_s$ and $x_n$. (For these two facts, one only needs that $\mathfrak{h}$ consists of semisimple elements, not that it's maximal). Now by maximality it follows that every semisimple element of $Z$ must already be in $\mathfrak{h}$.

From this it follows that $Z$ is a nilpotent Lie algebra. Namely, if $x \in Z$, then we just noted that $x_s \in \mathfrak{h}$, hence the restriction of $ad(x)$ to $Z$ acts via the restriction of $ad(x_n)$, which is nilpotent, so all elements of $Z$ are $ad_Z$-nilpotent.

So we have shown that $Z$ is nilpotent and self-normalising, i.e. a Cartan subalgebra in the classical definition. But for those it's well-known (e.g. Bourbaki loc.cit. VII.2.4 Thm. 2) that all its elements are semisimple. But then the above implies that actually $Z=\mathfrak{h}$ all along, and we're done!

I admit that proof seems very indirect. I would be happy to see a more direct proof.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .