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Nilpotent Lie algebras are solvable

Statement

For every Lie algebra $g$ and every $r \geq 0$ ,

$g^{(r)} \subseteq γ_{2^{r}} (g) .$

Consequently every nilpotent Lie algebra is solvable.

Facts & Assumptions

Given: A Lie algebra $g$ .

[L1]

The derived series satisfies $g^{(r + 1)} = [g^{(r)}, g^{(r)}]$ (Derived series and solvable Lie algebras).

[L2]

The lower central series satisfies $γ_{q + 1} = [g, γ_{q}]$ , and nilpotence means that a lower term vanishes (Lower central series and nilpotent Lie algebras).

Proof

technique · direct

1.1L2algebra

We first prove $[γ_{p}, γ_{q}] \subseteq γ_{p + q}$ for all $p, q \geq 1$ . For $p = 1$ this is [L2]. If it holds for $p - 1$ and every second index, Jacobi gives $[[g, γ_{p - 1}], γ_{q}] \subseteq [g, [γ_{p - 1}, γ_{q}]] + [γ_{p - 1}, [g, γ_{q}]] \subseteq [g, γ_{p + q - 1}] + [γ_{p - 1}, γ_{q + 1}] \subseteq γ_{p + q}$ .

2.1L1L2step 1.1algebra

At $r = 0$ , $g^{(0)} = g = γ_{1}$ . If $g^{(r)} \subseteq γ_{2^{r}}$ , then [L1] and step 1.1 imply $g^{(r + 1)} \subseteq [γ_{2^{r}}, γ_{2^{r}}] \subseteq γ_{2^{r + 1}}$ . Thus the displayed containment holds for every $r$ .

3.1L1L2step 2.1∎

If $g$ is nilpotent, choose a given bound $c$ with $γ_{c + 1} = 0$ as in [L2]. Taking $r = c + 1$ gives $2^{r} \geq c + 1$ , so descent of the lower series and step 2.1 yield $g^{(r)} \subseteq γ_{2^{r}} \subseteq γ_{c + 1} = 0$ . Hence $g$ is solvable by [L1]. This includes $g = 0$ and uses only the supplied finite bound.

Depends on

Dependency tree · 2 levels

Sources

Kirillov, An Introduction to Lie Groups and Lie Algebras, §5.4