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

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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Sectional curvature depends on an ordered basis of the plane

Statement refuted

This item assumes ${A C}_{ω}$ , namely countable choice. In the propagated dependency chain, that assumption is required through Sectional curvature is independent of the basis of the plane; after those interfaces are fixed, the remaining local or finite argument makes no additional countable-family choice.

False claim: the sectional curvature assigned to a tangent two-plane depends on the choice or ordering of a basis of that plane.

In fact it depends only on the unoriented two-plane.

Facts & Assumptions

Given: ${A C}_{ω}$ , a Riemannian manifold, a point $p$ , a tangent two-plane $σ \subseteq T_{p} M$ , and two supplied ordered bases $(X, Y)$ and $(X^{'}, Y^{'})$ of $σ$ .

[A1]

${A C}_{ω}$ is countable choice and is required here through Sectional curvature is independent of the basis of the plane; after those supplied interfaces are fixed, the remaining local or finite calculation makes no additional countable-family choice.

[F1]

If $(X, Y)$ and $(X^{'}, Y^{'})$ are two ordered bases of the same tangent two-plane, their sectional-curvature quotients are equal. Sectional curvature is independent of the basis of the plane.

Refutation

technique · direct

1.1

The two supplied ordered pairs are bases of the same plane $σ$ . Therefore [F1] directly gives $K (X^{'}, Y^{'}) = K (X, Y)$ .

A1F1

2.1

In particular, $(Y, X)$ is another ordered basis of $σ$ , so [F1] gives $K (Y, X) = K (X, Y)$ . Reversing orientation therefore carries no extra curvature datum.

F1step 1.1

3.1

Steps 1.1–2.1 apply to every tangent two-plane, so they refute both choice-of-basis and orientation dependence. On an empty, zero-dimensional, or one-dimensional manifold there are no tangent two-planes, making the false claim vacuous rather than producing an exception. Positive definiteness makes the Gram determinant nonzero for each basis; degenerate bilinear forms are outside the Riemannian hypothesis. The argument is pointwise, uses no parameter endpoint, and the plane and both bases are supplied, so it makes no further family choice beyond the stated inherited assumption. No biconditional is asserted.

F1step 1.1step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

9 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)