TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedverified 2026-10-15 (gpt-6-luna, gpt-6.1-sol, and delegated Codex review/repair agents; see completed run evidence)

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.

Positive homogeneity forces Legendre degeneracy

Physical scope

Pure conditional result for smooth positively homogeneous Lagrangians; it does not assert every singular system is almost regular.

Statement

Let $L (q, v)$ be $C^{2}$ on a conic open subset of a tangent bundle, and suppose $L (q, a v) = a L (q, v)$ for every $a > 0$ in its domain. At each nonzero $v$ ,

$v^{i} L_{v^{i}} = L, L_{v^{i} v^{j}} v^{i} = 0, E_{L} : = v^{i} L_{v^{i}} - L = 0.$

Consequently the velocity Hessian is singular and the fibre derivative cannot be a local diffeomorphism there; a regular Legendre equivalence theorem cannot be applied. These conclusions do not imply that the motion is static or that the Lagrangian has a specified constraint surface.

Facts & Assumptions

Given: A twice continuously differentiable Lagrangian on a positive-ray conic velocity domain, positively homogeneous of degree one.

[A1]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): Assume countable choice ACω, meaning every sequence of nonempty sets has a choice function; no full Axiom of Choice is assumed.

[F1]

Fibre derivative or Legendre map of a Lagrangian: the fibre derivative has components $p_{i} = L_{v^{i}}$ , and its velocity derivative is the fibre Hessian.

[F2]

Regular, hyperregular and almost-regular Legendre domains: regular inversion requires a locally invertible fibre derivative; almost-regularity is a separate package of image, rank, submersion and fibre hypotheses.

Proof

1.1givenA1algebra

Differentiate $L (q, a v) = a L (q, v)$ with respect to $a$ at $a = 1$ . The chain rule gives $v^{i} L_{v^{i}} = L$ .

2.1givenF1step 1.1algebra

Differentiate the identity in step 1.1 with respect to $v^{j}$ . The product rule gives $L_{v^{j}} + v^{i} L_{v^{i} v^{j}} = L_{v^{j}}$ , hence $L_{v^{i} v^{j}} v^{i} = 0$ . For nonzero $v$ this is a nonzero kernel vector, so the Hessian is singular and the derivative of the fibre map in [F1] is not invertible.

3.1givenF2step 1.1step 2.1∎

The definition $E_{L} = v^{i} L_{v^{i}} - L$ and step 1.1 give $E_{L} = 0$ . The inverse function theorem's necessary invertibility condition fails by step 2.1, so regular Legendre inversion is unavailable; [F2] also makes clear that stronger almost-regular hypotheses do not follow from this degeneracy.

Positive homogeneity forces Legendre degeneracy

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources