LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-14

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Sectional curvature is independent of the basis of the plane

Statement

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

The quotient defining $K (σ)$ is unchanged under every change of ordered basis of the two-plane $σ$ .

Facts & Assumptions

[A1]

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

[F1]

Sectional curvature is the quotient of $Rm (X, Y, Y, X)$ by the Gram determinant of the ordered basis $(X, Y)$ . Sectional curvature.

[F2]

The Riemann tensor is alternating in each pair. Algebraic symmetries of the Riemann tensor.

Proof

Given: ${A C}_{ω}$ , two ordered bases $(X, Y)$ and $(X^{'}, Y^{'})$ of the same two-plane, with $X^{'} = a X + b Y$ , $Y^{'} = c X + d Y$ , and $Δ = a d - b c \neq 0$ .

1.1

Multilinearity and first-pair alternation in [F2] give $Rm (X^{'}, Y^{'}, Y^{'}, X^{'}) = Δ Rm (X, Y, Y^{'}, X^{'})$ . Applying last-pair alternation to $(Y^{'}, X^{'})$ gives a second factor $Δ$ , so the numerator is $Δ^{2} Rm (X, Y, Y, X)$ .

A1F2algebra

2.1

If $G$ is the Gram matrix of $(X, Y)$ , the Gram matrix of $(X^{'}, Y^{'})$ is $A G A^{T}$ for $A = (\begin{array}{cc} a & b \\ c & d \end{array})$ , so its determinant is $Δ^{2} \det G$ . The nonzero common factor $Δ^{2}$ cancels from the quotient in [F1], proving basis independence.

F1step 1.1algebra∎

Depends on

Used by

Sectional curvature depends on an ordered basis of the plane False statement

Cited to discharge well-definedness by Sectional curvature.

Dependency tree · two levels

12 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)