False statementConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-14

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Christoffel symbols vanishing at one point implies curvature vanishes there

Statement refuted

This item assumes ${A C}_{ω}$ , namely countable choice. In the propagated dependency chain, that assumption is required through Euclidean hypersurface sectional curvature from principal curvatures and Sectional curvature; after those interfaces are fixed, the remaining local or finite argument makes no additional countable-family choice.

False claim: if all Levi–Civita Christoffel symbols vanish at a point, then the Riemann curvature tensor vanishes at that point.

Assume ${A C}_{ω}$ . Normal coordinates make all Christoffel symbols vanish at their centre, but curvature there can be nonzero because the coordinate curvature formula retains first derivatives of those symbols.

Facts & Assumptions

Given: Countable choice.

[A1]

${A C}_{ω}$ is countable choice and is required here through Euclidean hypersurface sectional curvature from principal curvatures and Sectional curvature; after those supplied interfaces are fixed, the remaining local or finite calculation makes no additional countable-family choice.

[F1]

Under countable choice, a supplied ordered orthonormal tangent basis gives normal coordinates, and at their centre $p$ one has $\partial_{i} ∣_{p} = e_{i}$ , $g_{i j} (p) = δ_{i j}$ , and $Γ^{k}_{i j} (p) = 0$ . The Axiom of Countable Choice ( ${A C}_{ω}$ ), Existence of normal neighborhoods, Normal neighborhood and normal coordinate chart, Properties of normal coordinates at the center.

[F2]

In the convention $R (\partial_{i}, \partial_{j}) \partial_{k} = R^{ℓ}_{k i j} \partial_{ℓ}$ , the coordinate curvature formula is $R^{ℓ}_{k i j} = \partial_{i} Γ^{ℓ}_{j k} - \partial_{j} Γ^{ℓ}_{i k} + Γ^{m}_{j k} Γ^{ℓ}_{i m} - Γ^{m}_{i k} Γ^{ℓ}_{j m} .$ Coordinate formula for the curvature tensor.

[F3]

A nonempty regular level set is an embedded submanifold, and its tangent space is the kernel of the defining differential. A regular level set is an embedded submanifold, The tangent space of a regular level set is the kernel.

[F4]

For a Euclidean hypersurface, the shape operator is $S_{ν} X = - ({\overline{\nabla}}_{X} ν)^{⊤}$ ; its eigenvectors are principal directions; and the sectional curvature of the plane spanned by supplied orthonormal principal directions is the product of their principal curvatures. Shape operator, Principal curvatures, Gaussian curvature, and mean curvature of an oriented hypersurface, Euclidean hypersurface sectional curvature from principal curvatures.

[F5]

For an orthonormal pair $(X, Y)$ , $K (span {X, Y}) = Rm (X, Y, Y, X)$ . Sectional curvature.

Refutation

technique · direct counterexample

1.1

Let $S^{2} = {q \in R^{3} : ∥ q ∥^{2} = 1}$ with its induced metric. For $F (q) = ∥ q ∥^{2}$ , one has $d F_{q} (v) = 2 ⟨ q, v ⟩$ , which is surjective at every $q \in S^{2}$ . Thus [F3] makes $S^{2} = F^{- 1} (1)$ an embedded Euclidean hypersurface and gives $T_{q} S^{2} = q^{⊥}$ . The field $ν (q) = q$ is consequently a smooth unit normal.

F3algebra

2.1

Fix $p = (0, 0, 1)$ and the tangent vectors $e_{1} = (1, 0, 0)$ and $e_{2} = (0, 1, 0)$ . They are orthonormal. Because ${\overline{\nabla}}_{X} ν = X$ on the sphere, [F4] gives $S_{ν} X = - X$ ; hence $e_{1}, e_{2}$ are principal directions with curvatures $κ_{1} = κ_{2} = - 1$ . The hypersurface formula in [F4] now gives $K (span {e_{1}, e_{2}}) = (- 1) (- 1) = 1$ .

A1F4step 1.1algebra

3.1

Use [F1] to take the normal coordinates at $p$ associated to the supplied ordered basis $(e_{1}, e_{2})$ . Then every $Γ^{k}_{i j} (p) = 0$ and $\partial_{i} ∣_{p} = e_{i}$ . By [F5] and step 2.1, $R^{1}_{2, 1, 2} (p) = {Rm}_{p} (\partial_{1}, \partial_{2}, \partial_{2}, \partial_{1}) = 1.$

F1F5step 2.1

4.1

At $p$ , the two quadratic Christoffel terms in [F2] vanish, but [F2] and step 3.1 give $1 = R^{1}_{2, 1, 2} (p) = \partial_{1} Γ^{1}_{22} (p) - \partial_{2} Γ^{1}_{12} (p) .$ Thus all Christoffel symbols vanish at $p$ while their first derivatives produce nonzero curvature there, refuting the claim.

F2step 3.1algebra

5.1

The witness is nonempty, boundaryless, two-dimensional, and positive definite. Empty, zero-dimensional, and one-dimensional Riemannian manifolds cannot supply this sectional-curvature witness, but one counterexample suffices to refute the universal claim. Normal-coordinate domains are open, so no chart endpoint is used. Countable choice is used exactly through [F1]; the point, normal, and ordered tangent basis are explicit, so there is no further selection. No biconditional is asserted.

F1step 1.1step 2.1step 3.1step 4.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

44 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

Ved Datar, Lectures on Riemannian Geometry (standard reference, not scraped)
John M. Lee, Riemannian Manifolds: An Introduction to Curvature (standard reference, not scraped)