Alphabeta Math
PropositionStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-13
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Properties of normal coordinates at the center

Statement

Assume ACω. Let (x1,…,xn) be normal coordinates centred at p from an orthonormal ordered basis (e1,…,en) of TpM. Then x(p)=0,∂i∣p=ei,gij(p)=δij,Γkij(p)=0,∂kgij(p)=0. Moreover, if v=∑iviei, then every part of its geodesic lying in the normal neighbourhood has coordinate expression x(γp,v(t))=(tv1,…,tvn).

Facts & Assumptions

Given: An orthonormal supplied basis and its normal coordinate chart on a normal neighbourhood of p.

[F1]

Under The Axiom of Countable Choice (ACω), Normal neighborhood and normal coordinate chart gives x=E−1∘exp⁡p−1 and The exponential map scales geodesic time gives γp,v(t)=exp⁡p(tv) whenever defined in the chart.

[F2]

The differential of exp at zero is the identity gives d(exp⁡p)0p=I, and Coordinate geodesic equation gives the coordinate geodesic equation in both directions.

[F3]

Christoffel formula for the levi civita connection gives both symmetry Γkij=Γkji and the metric-derivative formula after lowering the upper index.

Proof

technique · direct
1.1F1F2given

From [F1], x(p)=E−1(0p)=0. The inverse chart is x−1=exp⁡p∘E, so [F2] gives d(x−1)0(eistd)=d(exp⁡p)0p(ei)=ei; by the definition of coordinate tangent vectors this is ∂i∣p=ei. Orthonormality then gives gij(p)=gp(ei,ej)=δij.

1.2F1

If v=∑iviei and γp,v(t) is in the normal neighbourhood, [F1] yields γp,v(t)=exp⁡p(tv) and therefore x(γp,v(t))=E−1(tv)=(tv1,…,tvn). Thus every radial coordinate line is a geodesic on every connected parameter subinterval for which it remains in the chart.

2.1F2F3step 1.2

Substitute the line from step 1.2 into the coordinate geodesic equation [F2] at t=0. Its second coordinate derivatives vanish, so for every v∈Rn and every k, ∑i,jΓkij(p)vivj=0. Taking v=eistd gives Γkii(p)=0. For i≠j, taking v=eistd+ejstd gives Γkij(p)+Γkji(p)=0; symmetry from [F3] makes both terms zero. Hence every Christoffel symbol vanishes at p.

3.1F3step 2.1

Write Γijℓ=∑mgmℓΓmij. The two equations Γkij(p)=0 and Γkji(p)=0 from step 2.1 and [F3] read respectively ∂kgij+∂igkj−∂jgki=0,∂kgji+∂jgki−∂igkj=0 at p. Adding and using gij=gji gives 2∂kgij(p)=0, so every first metric derivative vanishes.

4.1F1F2F3step 1.1step 1.2step 2.1step 3.1∎

In dimension zero all indexed families and sums are empty, x(p)=0, and every assertion holds. In dimension one step 2.1 uses v=1 and gives the sole symbol and then the sole metric derivative as zero. On an empty manifold there is no centred chart. The zero vector gives the constant radial geodesic; chart-domain endpoints are excluded because the normal source is open, while every included parameter time is covered by step 1.2. The orthonormal basis is supplied, and ACω is inherited only through [F1]--[F2].

Depends on

Dependency tree · 2 levels

Sources