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!


2 Answers 2


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.

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

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