PropositionStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-10

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Length is additive under concatenation and invariant under reversal

Statement

Length adds under finite concatenation and is unchanged by reversal.

Facts & Assumptions

Given: Piecewise $C^{1}$ curves, with matching endpoints for concatenation.

[F1]

Riemannian length is independent of piecewise c one subdivision: Riemannian length is independent of admissible finite subdivision and of corner derivative conventions.

[F2]

Line integrals under reversal and concatenation: Let $γ$ be a piecewise- $C^{1}$ path, and let $f$ be a continuous scalar field and $F$ a continuous vector field on a set containing its trace. Then $\int_{γ^{-}} f d s = \int_{γ} f d s, \int_{γ^{-}} F \cdot d r = - \int_{γ} F \cdot d r .$ If piecewise- $C^{1}$ paths $α, β : [0, 1] \to R^{n}$ satisfy $α (1) = β (0)$ , and $f$ and $F$ are continuous on a set containing both traces, then $\int_{α * β} f d s = \int_{α} f d s + \int_{β} f d s,$ $\int_{α * β} F \cdot d r = \int_{α} F \cdot d r + \int_{β} F \cdot d r .$

Proof

technique · direct

1.1F1F2given

For curves $α, β$ on $[0, 1]$ , their concatenation has derivatives $2 \dot{α} (2 t)$ on the first half and $2 \dot{β} (2 t - 1)$ on the second. Substitution gives the two contributions $L_{g} (α)$ and $L_{g} (β)$ , hence their sum. The same finite integral calculation as for scalar line integrals applies to these scalar speeds.

2.1F1F2step 1.1∎

The reversed curve $γ^{-} (t) = γ (a + b - t)$ has velocity $- \dot{γ} (a + b - t)$ and therefore the same norm at the reversed time. Substitution reverses the integration limits and cancels the minus sign, giving $L_{g} (γ^{-}) = L_{g} (γ)$ . Repeating the first calculation proves finite concatenation; constant pieces and zero-length intervals contribute zero.

Source locator

Lee, Chapter 13, pp.337–340, Proposition 13.25, Lemma 13.28 and Theorem 13.29; finite piecewise $C^{1}$ refinements and pauses are treated explicitly here.

Depends on

Used by

Dependency tree · two levels

8 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

John M. Lee, Introduction to Smooth Manifolds, second edition (standard reference, not scraped)