# Definition of Cartan subalgebra in Erdmann-Wildon

Sorry in advance if this is too basic but I am just beginning to learn about lie algebras. I am reading Introdution to Lie Algebras by Karin Erdmann and Mark J. Wildon and I don't understand the definition of Cartan subalgebras (p. 94):

Definition 10.2: A Lie subalgebra $$H$$ of a Lie algebra $$L$$ is said to be a Cartan subalgebra (or CSA) if $$H$$ is abelian and every element $$h\in H$$ is semisimple, and moreover $$H$$ is maximal with these properties.

I understand what a "$$h\in H$$ is semisimple" means when $$H$$ is a semisimple Lie algebra ($$n=0$$ where $$h=d+n$$ is the Abstract Jordan decomposition, p. 87). However in our case $$H$$ isn't semisimple if $$\dim H>0$$, so I don't understand why this makes sense. I thought maybe it meant that $$h$$, seen as an element of $$L$$, is semisimple, but the authors emphasis after the definition that "we do not assume $$L$$ is semisimple in this definition".

What is the meaning of "$$h\in H$$ is semisimple"?

I'm thinking that this could mean $$\mathsf{ad}(h)\in \mathsf{gl}(V)$$ is diagonalisable (I'm just interested in the complex case), but I'm sure the authors would have stated explicitely that we can extend the definition to the non-semisimple case (or maybe I missed it).

• This definition of Cartan subalgebra seems only adapted to semisimple Lie algebras. The usual definition for a general (finite-dimensional) Lie algebra is a nilpotent subalgebra equal to its normalizer.
– YCor
Commented May 15, 2020 at 13:11

Definition: A semi-simple element of an abstract Lie algebra $$L$$ is an element $$x\in L$$ for which the adjoint linear transformation $$\operatorname{ad}(x)$$ is a semi-simple endomorphism of the vector space $$L$$".
• Right, and maybe one should clarify that in the source, it means that every $h\in H$ is semisimple as element of $L$; i.e. the endomorphism $ad_L(h) \in End(L)$ is semisimple (not just it's restriction $ad_H(h)\in End(H)$, which by $H$ being abelian is identically $0$ anyway). Commented May 15, 2020 at 16:29
You are right: it does mean $$\mathrm{ad}(h)$$ is diagonalisable.