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PropositionStatement: Literature-sourcedProof: AI-generatedPipeline-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)
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Conservation of quadratic geodesic integrals

Physical scope

Conditional mathematical statement about affine geodesics of a metric-selected connection; no physical trajectory or observation is asserted.

Statement

Let (M,g) be a smooth Lorentzian four-manifold, with no global time orientation assumed, and let ∇ be its metric-selected connection, namely the unique torsion-free affine connection with ∇g=0. Let K be a smooth symmetric covariant 2-tensor satisfying ∇(aKbc):=(∇aKbc+∇bKca+∇cKab)/3=0. Here a rank-two Killing tensor means precisely such a K; this extends the time-oriented definition by the same tensor equation. If γ:I→M is an affinely parametrized geodesic on a connected open interval I, with tangent v=γ˙, then

λ⟼Kab(γ(λ))va(λ)vb(λ)

is constant on I.

Facts & Assumptions

Given: The metric-selected connection, a smooth rank-two Killing tensor K, and a smooth affinely parametrized geodesic γ:I→M.

[F1]

Killing tensors and quadratic geodesic integrals defines a Killing tensor on a time-oriented Lorentzian four-manifold by symmetry and the cyclic equation ∇(aKbc)=0. In the Statement we explicitly use that same equation without a time-orientation assumption; symmetry and the equation are hypotheses, not an application of that definition outside its scope.

[F2]

Geodesic of an affine connection: affine parameter means Dλv=0 along γ.

[F3]

Existence and uniqueness of the metric-selected connection of a nondegenerate metric, and its extension to tensor fields supplies the unique metric-selected connection and the contraction product rule for every smooth nondegenerate symmetric metric on a smooth four-manifold, without time orientation; Covariant derivative along a curve defines its induced derivative along γ.

[A1]

The metric-selected connection and tangent-bundle convention use only countable choice ACω (The Axiom of Countable Choice (ACω)).

Proof

1.1givenA1F3algebra

Define IK(λ):=Kab(γ(λ))va(λ)vb(λ). The product rule in [F3] gives IK′=(∇vK)(v,v)+2K(Dλv,v).

2.1F1F2step 1.1algebra

By [F2], Dλv=0. Also vavbvc is symmetric in its indices, so (∇vK)(v,v)=vavbvc∇aKbc=vavbvc∇(aKbc)=0 by [F1]. Thus IK′=0 on I.

3.1F4step 2.1∎

Apply [F4] to IK on the connected interval I. Therefore Kabvavb is constant along γ.

Depends on

Dependency tree · 2 levels

Sources