next up previous
Next: Theorem of Engel Up: generalities in finite dimensional Previous: Simple, semisimple, solvable, and

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

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

Then we obviously have

$\displaystyle \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

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

We thus have

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

$ \qedsymbol$

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..

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

$ \qedsymbol$

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

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

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

$ \qedsymbol$

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

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

as a solvable ideal. $ \qedsymbol$


next up previous
Next: Theorem of Engel Up: generalities in finite dimensional Previous: Simple, semisimple, solvable, and
2007-12-19