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Minimizing along a geodesic is an initial interval property

Statement

Assume countable choice through the declared dependencies. Let (M,g) be a complete connected boundaryless Riemannian manifold, p∈M, v∈TpM a unit vector, and γ(t)=exp⁡p(tv) for t≥0. The set Ap(v)={t≥0:dg(p,γ(t))=t} is an initial interval: if T∈Ap(v) and 0≤s≤T, then s∈Ap(v). Moreover, if the cut time cp(v) is finite, then cp(v)∈Ap(v). Equivalently, once an earlier segment fails to minimize, no later segment along this ray minimizes.

Facts & Assumptions

Given: The complete ray γ and the fixed initial point p.

[A1]

The setup carries ACω through The Axiom of Countable Choice (ACω) and the cut-time supplier. This proof makes no use of full AC.

[F1]

The cut time is the supremum of the positive minimizing times, as defined in Cut time in a unit tangent direction.

[F2]

Riemannian distance is the infimum of lengths of piecewise C1 curves, by Riemannian distance on a connected manifold.

Proof

technique · direct competitor concatenation and an epsilon argument
1.1F2given

Put m(t)=dg(p,γ(t)) and A={t≥0:m(t)=t}. The ray segment from 0 to t has unit speed and length t, so [F2] gives m(t)≤t; also 0∈A. Thus membership in A is exactly the assertion that the segment to γ(t) minimizes.

1.2F2given

For any s,t≥0 and ε>0, [F2] gives one curve from p to γ(s) of length less than m(s)+ε; appending the ray segment gives m(t)≤m(s)+ε+∣t−s∣. Reversing s,t and letting ε↓0 proves ∣m(t)−m(s)∣≤∣t−s∣. This uses one near-minimizer at a time.

2.1F2step 1.1

If T∈A and 0<s<T were not in A, then m(s)<s; by [F2] there is a piecewise C1 curve from p to γ(s) of length strictly less than s. Concatenating it with γ∣[s,T] gives a curve to γ(T) of length less than s+(T−s)=T, contradicting T∈A. The case s=0 is already in [1.1], so A is an initial interval.

2.2F1F2step 1.2

Let c=cp(v) from [F1]. If c<∞, then for every sufficiently small ε>0 the supremum property gives s∈A with c−ε<s≤c; by the Lipschitz estimate [1.2], m(c)≥m(s)−∣c−s∣=2s−c>c−2ε, while the ray segment gives m(c)≤c. Hence m(c)=c, so every finite cut-time endpoint minimizes; for c=+∞ there is no finite endpoint to check.

3.1

At T=0 the segment is constant and minimizing. The empty manifold has no point p; in dimension zero there are no unit directions, and in dimension one the two directions are handled separately. The finite endpoint argument uses one existential near-minimizer for each arbitrary ε, not a selected sequence, so it spends no countable-choice instance beyond the inherited ACω setup [A1]. The contrapositive is the equivalent formulation in the Statement; both implications are established. [A1, F1, step 2.1, step 2.2] □

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