Lie algebras MOC

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, lie i.e. 𝔤(𝑛) =0 for some 𝑛 ∈ℕ.¹ Clearly this is a special case of a Nilpotent Lie algebra.

Properties

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

Proof of 1–3

Clearly if 𝔞 ≤𝔤, then 𝔞(𝑛) ≤𝔤(𝑛) for 𝑛 ∈ℕ0, 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) ≤ker⁡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+𝔟𝔟≅𝔞𝔞∩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.

See also

• Radical of a Lie algebra

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Footnotes

1. 1972. Introduction to Lie Algebras and Representation Theory, §3,1, p. 10 ↩