Simple, semisimple, solvable, and nilpotent Lie algebras: definition

DEFINITION 5.4   For a Lie algebra $L$ , let us define the following ideals of $L$ .

1. $\operatorname{Comm}(L)=[L,L]$ , and inductively,
   $$\operatorname{Comm}^j(L)=\operatorname{Comm}(\operatorname{Comm}^{j-1}(L)).$$

2. $\operatorname{ad}(L)(L)=[L,L]$ , and inductively,
   $$\operatorname{ad}^j(L)(L)=\operatorname{ad}(\operatorname{ad}^{j-1}(L)).$$

LEMMA 5.5   For any Lie algebra $L$ and for any positive integer $j$ , we have

$$\operatorname{Comm}^j(L) \subset \operatorname{ad}^j(L).$$

PROOF.. Inductively, we have

$$\operatorname{Comm}^j(L) =[\operatorname{Comm}^{j-1}(L),\operator... ...{Comm}^{j-1}(L)]\subset [L,\operatorname{ad}^{j-1}(L)]=\operatorname{ad}^j(L).$$

$\qedsymbol$

DEFINITION 5.6   A Lie algebra $L$ over a field $k$ is said to be

1. semisimple if it has no abelian ideals.
2. simple if it has no non trivial ideals and $\dim(L)>1$ .
3. solvable if $\operatorname{Comm}^N (L)=0$ for some $N\in \mathbb{Z}_{>0}$ .
4. nilpotent if $\operatorname{ad}(L)^N(L)=0$ for some $N\in \mathbb{Z}_{>0}$ .

PROPOSITION 5.7   We have the following implications.

1. Simple Lie algebras are semisimple.
2. Nilpotent Lie algebras are solvable.

PROOF.. (1)is Easy. (2) follows from Lemma 5.5. $\qedsymbol$

Semisimple algebras and solvable ones are ``orthogonal''. For now we only mention the following

LEMMA 5.8   Non zero solvable algebra $L$ cannot be semisimple.

PROOF.. Let $N_0$ be a positive integer such that

$$\operatorname{Comm}^N_0 (L)\neq 0 , \qquad \operatorname{Comm}^{N_0+1} (L)=0.$$

Then $\operatorname{Comm}^{N_0}(L)$ is a non-zero abelian ideal of $L$ . $\qedsymbol$