The radicals of Lie algebras

DEFINITION 5.9   A radical of a Lie algebra $L$ is a maximal solvable ideal of $L$ .

LEMMA 5.10   Let $\mathfrak{a}$ be an ideal of a Lie algebra $L$ . If $L/\mathfrak{a}$ and $\mathfrak{a}$ are both solvable Lie algebras, then $L$ is also solvable.

PROOF.. Since $L/\mathfrak{a}$ is solvable, there exists a positive integer $N_1$ such that

$$\operatorname{Comm}^{N_1}(L/\mathfrak{a})=0.$$

Then we obviously have

$$\operatorname{Comm}^{N_1}(L)\subset \mathfrak{a}.$$

On the other hand, since $\mathfrak{a}$ is solvable, there exists a positive integer $N_2$ such that

$$\operatorname{Comm}^{N_2}(\mathfrak{a})=0.$$

We thus have

$$\operatorname{Comm}^{N_1+N_2}(L)=\operatorname{Comm}^{N_2}(\operatorname{Comm}^{N_1}(L))\subset \operatorname{Comm}^{N_2}(\mathfrak{a})=0.$$

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LEMMA 5.11   Every Lie subalgebras and quotients of solvable Lie algebras are solvable.

PROOF.. Obvious. $\qedsymbol$

LEMMA 5.12   Let $\mathfrak{a},\mathfrak{b}$ be ideals of a Lie algebra $L$ . If $\mathfrak{a},\mathfrak{b}$ are both solvable (as Lie algebras), then $\mathfrak{a}+\mathfrak{b}$ is also solvable.

PROOF..

$$\mathfrak{a}+\mathfrak{b}/\mathfrak{b}\cong \mathfrak{a}/\mathfrak{a}\cap \mathfrak{b}$$

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PROPOSITION 5.13   For a finite dimensional Lie algebra $L$ over a field $k$ , there exists a unique maximal solvable ideal of $L$ . So we may call it the radical of $L$ .

PROOF.. Let $\mathfrak{a}_0$ be a solvable ideal of $L$ which has the maximal dimension among solvable ideals. Then for any solvable ideal $\mathfrak{b}$ of $L$ , $\mathfrak{a}_0+\mathfrak{b}$ is also solvable. Thus by the choice of $\mathfrak{a}_0$ we see that

$$\mathfrak{a}_0+\mathfrak{b}=\mathfrak{a}_0. \quad($$That is, $$\mathfrak{a}_0\supset \mathfrak{b}.)$$

Thus we see that $\mathfrak{a}_0$ is the largest solvable ideal of $L$ .

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COROLLARY 5.14   Let $L$ be a finite dimensional Lie algebra over a field $k$ . Let $\mathfrak{r}$ be its radical. Then:

1. $L/\mathfrak{r}$ is semisimple.
2. $L$ is semisimple if and only if $\mathfrak{r}=0$ .
3. A quotient $L/\mathfrak{a}$ is semisimple if and only if $\mathfrak{r} \subset \mathfrak{a}$ .

PROOF.. (1) follows immediately from the definition and Lemma 5.8. (2) is also easy.

(3): $L/\mathfrak{a}$ contains

$$(\mathfrak{r}+\mathfrak{a})/\mathfrak{a}(\cong \mathfrak{r}/(\mathfrak{r}\cap \mathfrak{a}))$$

as a solvable ideal. $\qedsymbol$