Alphabeta Math
PropositionStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-13
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Injectivity radius at each point is positive

Statement

Assume ACω. Every point p of a boundaryless Riemannian manifold has inj(p)>0. Nevertheless, the global infimum inj(M) can equal zero.

Facts & Assumptions

Given: A boundaryless Riemannian manifold and a point p for the first assertion.

[F1]

Under The Axiom of Countable Choice (ACω), Existence of normal neighborhoods supplies a positive-radius tangent ball on which expp is a diffeomorphism, and Injectivity radius at a point and of a manifold defines inj(p) as the supremum of all such radii.

[F2]

Constant positive metric coefficient is Riemannian, its Levi--Civita symbol vanishes, and the coordinate geodesic equation then has affine solutions (Coordinate criterion for a riemannian metric, Christoffel formula for the levi civita connection, Coordinate geodesic equation, Existence uniqueness and smooth dependence of geodesics).

[F3]

A countable disjoint union of fixed-dimensional smooth manifolds with specified countable bases and atlases is a smooth manifold (Countable disjoint unions of fixed-dimensional smooth manifolds are smooth manifolds).

Proof

technique · direct
1.1

By [F1], some r>0 belongs to the admissible-radius set Rp. Therefore inj(p)=supRpr>0. This also covers inj(p)=+.

F1
1.2

To show that no uniform positive bound follows, for each integer m1 let Cm=R/(2/m)Z. Quotienting intervals of length less than 2/m gives an explicit smooth atlas: overlaps differ by translations by integer multiples of 2/m; rational subintervals give a specified countable basis. Give every such chart the metric ds2. By the transformation rule for translations this is a well-defined Riemannian metric, and [F3] makes M=m1Cm a boundaryless Riemannian one-manifold componentwise.

F2F3construct
2.1

Fix p=[x]Cm and identify TpCm with R by s. The metric coefficient is the constant 1, so [F2] makes the Christoffel symbol zero. Thus the unique maximal geodesic with initial scalar v is t[x+tv], and expp(v)=[x+v]. If 0<r1/m and u,v(r,r) have the same exponential image, then uv=2k/m for an integer k, but uv<2r2/m, forcing k=0. The quotient map is a local translation, so this injective restriction is a diffeomorphism onto its open image. If r>1/m, the two distinct interior vectors 1/m and 1/m have the same image. Hence exactly the radii 0<r1/m are admissible and inj(p)=1/m.

F1F2step 1.2
3.1

It follows that inj(M)=infm11/m=0: zero is a lower bound, and any ε>0 is exceeded downward by 1/m for an integer m>1/ε. Thus the second assertion has an explicit witness. The witness is nonempty and one-dimensional; at dimension zero step 1.1 gives + pointwise, and for an empty manifold the pointwise assertion is vacuous. Zero tangent vectors lie in every test ball. At the critical radius 1/m the colliding vectors are endpoints and therefore excluded, whereas for every larger radius they are included. ACω is propagated through [F1]--[F2]; the witness uses fixed quotient atlases, bases, metrics, and an enumerated disjoint union, so it makes no additional family choice.

F1F2F3step 1.1step 1.2step 2.1

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Sources