Equivalence of Two Cartan Subalgebra Definitions in Semi-Simple Lie Algebra

Question

I'm currently working out the equivalence of the following definitions, when the ambient space $\mathfrak{g}$ is a complex, semi-simple Lie algebra.

(1) $\mathfrak{h}$ is called a Cartan-1 subalgebra when $\mathfrak{h}$ is maximally abelian (not contained in a larger abelian subalgebra) and $ad$-diagonalizable, that is, $ad_X$ is diagonalizable for any $X \in \mathfrak{h}$.

(2) $\mathfrak{h}$ is called Cartan-2 subalgebra when $\mathfrak{h}$ is its own normalizer, $N(\mathfrak{h}) = \mathfrak{h}$ and $\mathfrak{h}$ is nilpotent.

Here, the normalizer of $\mathfrak{h}$ in $\mathfrak{g}$ is $N(\mathfrak{h}) := \{ X \in \mathfrak{g} \; | \; ad_X(\mathfrak{h}) \subset \mathfrak{h} \}$. For clarity, I'm trying to prove

Theorem - Let $\mathfrak{g}$ be a complex semi-simple Lie algebra. Then a sub-algebra $\mathfrak{h}$ is Cartan 1 iff it is Cartan 2.

So far, I've been able to deal with one direction. I showed Cartan 1 implies Cartan 2. For the converse Cartan 2 implies Cartan 1, I would like to proceed as follows:

pf: C2 -> C1: First, we can show any $\mathfrak{h}$ that is Cartan 2 will be maximally nilpotent. So, it suffices to show that $\mathfrak{h}$ is abelian and diagonalizable, as we will then get maximally abelian for free. On the other hand, we actually need only show diagonalizability of $\mathfrak{h}$. For then take any $X \in \mathfrak{h}$ and $ad_X$ is diagonalizable, acts nilpotently on $\mathfrak{h}$ by Engel's Theorem. We conclude that $ad_X|_{\mathfrak{h}} \equiv 0$ so that $\mathfrak{h}$ is abelian. Now, on to proving diagonalizability. Here, I want to use the Jordan-Chevalley decomposition to write $ad_X = S +N$ for some $S,N \in \mathfrak{gl}(\mathfrak{g})$. Doing some clever work, we can show that $S, N$ are actually derivations on $\mathfrak{g}$. Since $\mathfrak{g}$ is semi-simple, we see that $S,N$ are inner derivations. That is, $S = ad_Y, N = ad_Z$ for some $Y, Z \in \mathfrak{g}$. But then since $S, N$ are polynomials in $ad_X$ by Jordan-Chevalley, we see that they preserve the subalgebra $\mathfrak{h}$. Hence, $Y, Z \in N(\mathfrak{h}) = \mathfrak{h}$, by definition. Ok, my suspicion is that we can finish the proof by using the non-degeneracy of the killing form to show that $Z =0$ and hence $ad_X = S$ is diagonalizable. I have been unable to put the pieces together yet, though.

I'd very much appreciate someone who can help me finish this proof or offer guidance on a simpler way to prove Cartan-2 implies Cartan 1.

Answer 1

What you want is proven in greater generality in Bourbaki's Lie Groups and Lie Algebras, chapter VII §2 no.4 (your statement is a special case of Theorem 2). As usual, Bourbaki's proofs go by reference to earlier proofs; unravelling those, it looks like they followed a strategy very similar to yours. The missing link at the end is what constitutes lemma 2 and Proposition 11 in loc.cit. §1 no.3, and it goes like this:

Non-degeneracy of the Killing form $\kappa(\cdot, \cdot)$ implies that $\mathfrak{h}$ is reductive (this is a much earlier result from ch. I of loc.cit., and not hard to prove). (I am not sure if at this point in your argument we are allowed to assume $\mathfrak{h}$ abelian without making an earlier argument circular -- if yes, then of course this is redundant anyway and in the following replace $\mathfrak{c}$ by $\mathfrak{h}$.) Now let $\mathfrak{c}$ be the centre of $\mathfrak{h}$ and let $x\in \mathfrak{c}$ be $ad$-nilpotent in $\mathfrak{g}$. Then for all $y \in \mathfrak{h}$, $ad(x)$ and $ad(y)$ commute, hence $\kappa(x,y)=0$. But it is another fact that

the restriction of the Killing form to $\mathfrak{h} \times \mathfrak{h}$ is non-degenerate, $(*)$

so this implies that actually $x=0$. Using that the nilpotent part of $ad$ of any element $x' \in \mathfrak{c}$ is in $\mathfrak{c}$ itself (by being a polynomial in $ad(x')$) one can conclude that every element of $\mathfrak{c}$ is actually $ad$-semisimple, which is enough to conclude.

Now to prove $(*)$, one has to go down another rabbit hole of propositions which might simplify in your algebraically closed case. The crucial part is that if one has a finite-dimensional representation of any nilpotent Lie algebra $\mathfrak{h}$, and for weights $\lambda$ of $\mathfrak{h}$ looks at generalised eigenspaces $V^\lambda$, and one has an $\mathfrak{h}$-invariant bilinear form on $V$, then $V^\lambda \perp V^\mu$ unless $\lambda+\mu=0$; meaning that if the bilinear form is non-degenerate, so must be its restriction to $V^\lambda \times V^{-\lambda}$ for all $\lambda$. (And in our case, $V=\mathfrak{g}, \lambda=0$ and $\mathfrak{h}= \mathfrak{g}^0$ by being self-normalising.) For more precise arguments, this is Proposition 9(v) and 10(iii) in loc.cit. ch. 7 §1.