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Nilpotent-by-nilpotent extensions are always nilpotent

Statement

If an ideal and the corresponding quotient Lie algebra are nilpotent, then the ambient Lie algebra is nilpotent.

Facts & Assumptions

Given: A field $k$ and the two-dimensional Lie algebra $a = k x \oplus k y$ with $[x, y] = y$ .

[L1]

Nilpotence is termination of the lower central series (Lower central series and nilpotent Lie algebras).

[L2]

The bracket in a quotient by an ideal is computed on coset representatives (Quotient Lie algebras).

Refutation

technique · direct

1.1

The line $I = k y$ is an ideal because $[x, y] = y \in I$ and $[y, y] = 0$ . Its bracket is zero, so $γ_{2} (I) = 0$ and $I$ is nilpotent by [L1].

givenL1algebra

2.1

The quotient $a / I$ is spanned by $x + I$ and is abelian: [L2] gives $[x + I, x + I] = I$ . Hence its second lower-central term is zero, so it too is nilpotent. Thus $0 \to I \to a \to a / I \to 0$ has nilpotent kernel and quotient.

L1L2step 1.1algebra

3.1

Nevertheless, $γ_{2} (a) = k y$ and $γ_{r + 1} (a) = [a, k y] = k y$ for every $r \geq 2$ , since $[x, y] = y \neq 0$ . The lower central series never vanishes, so $a$ is not nilpotent by [L1]. This exact extension is therefore a counterexample over every field; it is finite and uses no choice.

L1step 2.1algebra∎

Depends on

Used by

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Dependency tree · two levels

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Sources

Milne, Lie Algebras, Aside 2.3 (standard reference, not scraped)