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

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.

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.