PropositionStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-13

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Affine reparametrization of a geodesic is a geodesic

Statement

If $γ : I \to M$ is an affinely parametrized geodesic and $ℓ (t) = a t + b$ maps an interval $J$ with nonempty interior into $I$ , then $γ \circ ℓ$ is a geodesic. For any supplied Riemannian metric, its speed at $t$ is $∣ a ∣$ times the speed of $γ$ at $a t + b$ .

Facts & Assumptions

Given: The geodesic $γ$ , constants $a, b$ , and intervals in the statement.

[F1]

Geodesic of an affine connection defines the geodesic equation by $D_{s} γ^{'} = 0$ .

[F2]

Riemannian speed and length defines speed as the Riemannian norm of the velocity.

Proof

1.1F1givenalgebra

Put $\tilde{γ} = γ \circ ℓ$ . The ordinary and covariant chain rules give ${\tilde{γ}}^{'} (t) = a γ^{'} (a t + b)$ and $D_{t} {\tilde{γ}}^{'} = a^{2} (D_{s} γ^{'}) (a t + b) = 0$ by [F1]. Hence $\tilde{γ}$ is geodesic.

2.1F2step 1.1∎

By homogeneity of the norm in [F2], $∣ {\tilde{γ}}^{'} (t) ∣ = ∣ a ∣ ∣ γ^{'} (a t + b) ∣$ . If $a = 0$ , the reparametrized curve is constant and both formulas give zero; negative $a$ reverses the parameter and uses the absolute value. Dimensions zero and one require no change, empty manifolds have no curves, and included endpoints use the corresponding one-sided chain rule. The constants and curve are supplied, so no choice principle is used.

Depends on

Dependency tree · 2 levels

Sources

Ved Datar, Lectures on Riemannian Geometry, Remark 15.1.2, pp.113–114