PropositionStatement: AI-adaptedProof: 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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Scalar curvature is twice the sum of sectional curvatures of orthonormal coordinate planes

Statement

This item assumes ${A C}_{ω}$ , namely countable choice. In the propagated dependency chain, that assumption is required through Scalar curvature, Ricci curvature is symmetric and basis independent, 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.

For every orthonormal basis $(e_{1}, \dots, e_{n})$ of $T_{p} M$ ,

$S (p) = 2 \sum_{1 \leq i < j \leq n} K (span (e_{i}, e_{j})) .$

Facts & Assumptions

[A1]

${A C}_{ω}$ is countable choice and is required here through Scalar curvature, Ricci curvature is symmetric and basis independent, 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]

Scalar curvature is the orthonormal trace $S (p) = \sum_{j} {Ric}_{p} (e_{j}, e_{j})$ . Scalar curvature.

[F4]

In an orthonormal basis, ${Ric}_{p} (X, Y) = \sum_{i} {Rm}_{p} (e_{i}, X, Y, e_{i})$ . Ricci curvature is symmetric and basis independent.

[F2]

For an orthonormal pair $(e_{i}, e_{j})$ , $K (span (e_{i}, e_{j})) = Rm (e_{i}, e_{j}, e_{j}, e_{i})$ . Sectional curvature.

[F3]

The Riemann tensor is skew in its first pair and invariant under interchange of its two pairs. Algebraic symmetries of the Riemann tensor.

Proof

Given: ${A C}_{ω}$ , a point $p$ and an orthonormal basis $(e_{1}, \dots, e_{n})$ of $T_{p} M$ .

1.1

Substituting the Ricci contraction [F4] into the scalar trace [F1] gives $S (p) = \sum_{i, j} Rm (e_{i}, e_{j}, e_{j}, e_{i})$ . The terms with $i = j$ vanish by first-pair skewness in [F3].

A1F1F3F4

2.1

For $i \neq j$ , [F2] identifies the summand with $K (span (e_{i}, e_{j}))$ . Pair interchange in [F3] identifies the summands indexed by $(i, j)$ and $(j, i)$ . Hence the ordered off-diagonal sum in step 1.1 is twice the sum indexed by $i < j$ , which is the asserted formula.

F2F3step 1.1∎

Depends on

Used by

Zero scalar curvature does not imply flatness Counterexample

Dependency tree · two levels

18 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)