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

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Geodesics have constant speed for a metric-compatible connection

Statement

Let $γ : I \to M$ be a geodesic for a metric-compatible affine connection on a Riemannian manifold. Then $g (γ^{'}, γ^{'})$ and the speed $∣ γ^{'} ∣$ are constant on $I$ .

Facts & Assumptions

Given: The geodesic, metric, and compatible connection in the statement.

[F1]

Geodesic of an affine connection gives $D_{t} γ^{'} = 0$ .

[F2]

Metric compatible connection on a riemannian vector bundle gives the product rule for differentiating the metric pairing along a curve.

[F3]

A function continuous on an interval $I$ whose derivative vanishes at every interior point of $I$ is constant on $I$ ; consequently two such functions with the same derivative differ by a constant makes a differentiable real function with zero derivative on an interval constant.

Proof

1.1F1F2given

Metric compatibility and the symmetry of $g$ give $\frac{d}{d t} g (γ^{'}, γ^{'}) = g (D_{t} γ^{'}, γ^{'}) + g (γ^{'}, D_{t} γ^{'}) = 2 g (D_{t} γ^{'}, γ^{'}) = 0$ by [F1] and [F2].

2.1F3step 1.1∎

By [F3], $g (γ^{'}, γ^{'})$ is constant on the interval. It is nonnegative, so its nonnegative square root $∣ γ^{'} ∣$ is constant as well. This includes the zero-speed constant geodesics and shows that a nonconstant geodesic never has zero velocity. In dimension zero the constant is zero; empty manifolds give no curves. Included parameter endpoints follow by continuity from the interior, and no choices are made.

Depends on

Dependency tree · 2 levels

Sources

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