Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-14
How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Engel triangularization theorem

Statement

Let V be finite-dimensional over any field, and let ρ:ggl(V) be nil. Then there is a flag

0=V0V1Vn=V

with dimVi=i and ρ(g)ViVi1. Equivalently, there is a basis in which every ρ(x) is strictly upper triangular.

Facts & Assumptions

Given: A nil representation of a Lie algebra g on a finite-dimensional vector space V.

[L1]

On a nonzero finite-dimensional nil module there is a nonzero vector annihilated by all of g (Engel's common-zero-vector lemma).

[L2]

An invariant subspace and its quotient carry the restricted and induced representations (Subrepresentations, quotient representations, and intertwiners).

[L3]

The canonical projection onto a vector-space quotient is linear and surjective (The quotient vector space V/W and its canonical projection).

Proof

technique · induction on $\dim V$
1.1

If V=0, the empty flag has the required property and the empty basis gives the matrix assertion.

basegiven
1.2

Assume V0 and the theorem for all nil modules of dimension smaller than dimV.

ihgiven
1.3

By [L1], choose 0vV with gv=0, and set V1=kv. This is an invariant one-dimensional subspace annihilated by g.

L1L2
2.1

Every induced operator on V/V1 is nilpotent, since a power of ρ(x) that vanishes on V also vanishes on the quotient. By [L2] and step 1.2, the quotient has a flag 0=V1Vn=V/V1 with gViVi1. Taking inverse images under the projection [L3] and adjoining 0V1 gives the required flag of V.

L2L3step 1.2step 1.3
3.1

Choose vectors adapted to the flag, in the finite sequential sense. The containment ρ(g)ViVi1 says every matrix sends the ith basis vector into the span of earlier vectors, hence is strictly upper triangular in this ordering. Conversely a common strictly upper-triangular basis supplies exactly this flag. The zero case was covered in step 1.1.

step 1.1step 2.1discharge-induction

Depends on

Used by

Dependency tree · two levels

9 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