# Differing definitions of Cartan subalgebras

I'm quite new to Lie algebras in general and I have recently come across two differing definitions of a Cartan subalgebra. The first is from J-P. Serre's book 'Complex Semisimple Lie Algebras' wherein a Cartan subalgebras is defined as follows:

Serre: The Cartan subalgebra, $$\mathfrak{h}$$, of a finite-dimensional Lie algebra $$\mathfrak{g}$$ over the base field $$\mathbb{C}$$ is a subalgebra which satisfies the following two conditions:

1) $$\mathfrak{h}$$ is nilpotent.

2) $$\mathfrak{h} = N_{\mathfrak{g}}(\mathfrak{h})$$, i.e. $$\mathfrak{h}$$ is self-normalising.

However in Knapp's 'Lie Groups Beyond an Introduction' one finds the following, differing definition.

Knapp: Let $$\mathfrak{h} \subseteq \mathfrak{g}$$ be a nilpotent subalgebra of a finite-dimensional complex semi-simple Lie algebra. We can decompose $$\mathfrak{g}$$ into its generalised weight-spaces $$\mathfrak{g}_{\alpha}$$ relative to the adjoint representation $$\text{ad}(\mathfrak{h})$$, whereupon a Cartan subalgebra is defined as a nilpotent subalgebra $$\mathfrak{h}$$ such that $$\mathfrak{h} = \mathfrak{g}_0$$.

Given these two differing definitions, it seems natural that I should be able to prove that for a finite-dimensional complex semi-simple Lie algebra, $$\mathfrak{g}_0$$ is nilpotent for $$\mathfrak{h}$$ nilpotent and self-normalising. Is this true, or have I misunderstood something. Thank you!

This equivalence is proposition 4 in book VII, §2 of Bourbaki's treatise on Lie Groups and Lie Algebras. The direction you're interested in is proved via a straightforward application of Engel's Theorem:

(Proof for "$$\mathfrak h$$ nilpotent and self-normalising" $$\Rightarrow$$ "$$\mathfrak{h}=\mathfrak g_0$$".) There is a natural Lie algebra action of $$\mathfrak{h}$$ on the quotient space $$V:=\mathfrak{g}_0/\mathfrak{h}$$ (via adjoint). If $$V \neq 0$$, since $$\mathfrak h$$ is nilpotent, Engel's theorem guarantees that there is a $$0 \neq v \in V$$ such that $$\mathfrak{h} \cdot v =0$$. Unravelling the definitions, this means any representative $$x \in \mathfrak{g}_0 \setminus \mathfrak h$$ of $$v$$ would be in the normaliser of $$\mathfrak h$$, but we had assumed $$\mathfrak h$$ to be self-normalising, contradiction. So $$V = 0$$ which proves the claim, which obviously is stronger than what you seek.

As pointed out in the other answer, if $$\mathfrak{g}$$ is semisimple, one can even prove that $$\mathfrak h = \mathfrak g_0$$ is abelian; the argument above however works in any (finite-dimensional) Lie algebra $$\mathfrak g$$, and obviously here in general $$\mathfrak h = \mathfrak g_0$$ need not be abelian.

For completeness, the other direction (which in Bourbaki is hidden in a rabbit hole of references, in loc. cit. §1 proposition 10) goes like this:

(Proof for "$$\mathfrak h$$ nilpotent and $$\mathfrak h = \mathfrak g_0$$" $$\Rightarrow$$ "$$\mathfrak{h}$$ is self-normalising".) Let $$x$$ be in the normaliser of $$\mathfrak g_0$$. For any $$y \in \mathfrak h$$, we have $$z:=ad_y(x) \in \mathfrak h$$ (by definition of normaliser and $$\mathfrak h = \mathfrak g_0$$). But then because $$\mathfrak h$$ is nilpotent (or by definition of $$\mathfrak g_0$$), there is $$n \in \mathbb N$$ (depending on $$y$$) such that $$0=(ad_y)^n(z) = (ad_y)^{n+1}(x)$$. But the existence of such $$n$$ for each $$y \in \mathfrak h$$ implies by definition that $$x \in \mathfrak{g}_0$$. We have thus shown that $$\mathfrak{g}_0$$ is self-normalising.

Yes, $$\mathfrak g_0$$ is nilpotent. More than is true, since actually $$\mathfrak g_0$$ is abelian.

• How would one go about seeing this from its definition as the 0 weight-space? Commented Apr 1, 2020 at 16:25
• It's not trivial at all. Commented Apr 1, 2020 at 16:26