How statement and proof provenance work
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A nilpotent-by-nilpotent extension need not be nilpotent
Counterexample
The assertion that an extension of a nilpotent Lie algebra by a nilpotent Lie algebra must be nilpotent is false. For the affine algebra with , the sequence
has nilpotent kernel and quotient, but is not nilpotent.
Facts & Assumptions
Given: The displayed affine Lie algebra over a field , with the first map the inclusion and the second the quotient map.
Its lower central series satisfies for every (The two-dimensional affine Lie algebra is solvable, not nilpotent).
The quotient bracket is for an ideal (Quotient Lie algebras).
Every nonzero abelian Lie algebra is nilpotent of class one (Abelian Lie algebras are nilpotent of class one).
Refutation
The line is an ideal because and . The displayed inclusion is injective, the quotient map is surjective, and its kernel is exactly , so the sequence is short exact.
Nevertheless, [L1] says that every lower-central term of from onward is the nonzero line . Thus the ambient algebra is not nilpotent.
The kernel is one-dimensional abelian. The quotient is spanned by , and [L2] gives , so it too is one-dimensional abelian. Hence both kernel and quotient are nilpotent by [L3].
Steps 1.1, 1.2, and 2.1 satisfy the hypotheses of a nilpotent-by-nilpotent extension and explicitly fail its proposed nilpotence conclusion. The construction is finite and uses no choice principle.
Depends on
Used by
Nothing in the library uses this result yet.
Dependency tree · two levels
7 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.
Sources
- Milne, Lie Algebras, nilpotent extensions (standard reference, not scraped)