Solvable Lie algebra

A Lie algebra $𝔤$ is solvable iff its derived series

𝔤 (0) = 𝔤, 𝔤 (𝑛 + 1) = [𝔤 (𝑛), 𝔤 (𝑛)]

terminates in the zero subalgebra, #m/def/lie i.e. $𝔤 (𝑛) = 0$ for some $𝑛 \in ℕ$ .¹ Clearly this is a special case of a Nilpotent Lie algebra.

Properties

If $𝔤$ is solvable, then so too are all subalgebras and homomorphic images.
If $𝔞 ⊴ 𝔤$ is a solvable ideal such that the quotient $𝔤 / 𝔞$ is solvable, then $𝔤$ is solvable.
If $𝔞, 𝔟 ⊴ 𝔤$ are solvable ideals, then so to is $𝔞 + 𝔟$ .

Proof of 1–3

Clearly if $𝔞 \leq 𝔤$ , then $𝔞 (𝑛) \leq 𝔤 (𝑛)$ for $𝑛 \in ℕ 0$ , so if the latter terminates so to does the former. Similarly given a epimorphism $𝜑 : 𝔤 ↠ 𝔥$ we have $𝜑 (𝔤 (0)) = 𝔥 (0)$ , and given $𝜑 (𝔤 (𝑛)) = 𝔥 (𝑛)$

𝜑 (𝔤 (𝑛 + 1)) = 𝜑 ([𝔤 (𝑛), 𝔤 (𝑛)]) = [𝜑 (𝔤 (𝑛)), 𝜑 (𝔤 (𝑛))] = [𝔥 (𝑛), 𝔥 (𝑛)] = 𝔥 (𝑛 + 1)

proving ^P1 by induction.

Let $𝜋 : 𝔤 ↠ 𝔤 / 𝔞$ be the projection, and say $(𝔤 / 𝔞) (𝑛) = 0$ . Then $𝜋 (𝔤 (𝑛)) = 0$ so $𝔤 (𝑛) \leq k e r 𝜋 = 𝔞$ . But then applying ^P1 the derived series of $𝔤 (𝑛)$ must terminate, and thus the derived series of $𝔤$ terminates, proving ^P2.

By the second isomorphism theorem we have the isomorphism

𝔞 + 𝔟 𝔟 ≅ 𝔞 𝔞 \cap 𝔟

Since the latter is the homomorphic image of $𝔞$ , by ^P1 it is solvable, and thus $𝔞 + 𝔟$ is solvable by ^P2, proving ^P3.

Solvable Lie algebra

Properties

See also

Footnotes