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.

Ricci curvature is symmetric and basis independent

Statement

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

Ricci curvature is a smooth symmetric covariant two-tensor. For every orthonormal basis $(e_{1}, \dots, e_{n})$ of $T_{p} M$ ,

${Ric}_{p} (X, Y) = \sum_{i = 1}^{n} {Rm}_{p} (e_{i}, X, Y, e_{i}),$

and the value is independent of the basis.

Facts & Assumptions

[A1]

${A C}_{ω}$ is countable choice and is required here through 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]

Ricci curvature is the trace of $Z \mapsto R (Z, X) Y$ . Ricci curvature.

[F2]

The Riemann tensor has first- and last-pair skewness and pair-interchange symmetry. Algebraic symmetries of the Riemann tensor.

[F3]

Contraction written using a basis and its dual is independent of that basis. Contraction is independent of the basis formula.

Proof

Given: ${A C}_{ω}$ , a point $p$ , tangent vectors $X, Y \in T_{p} M$ , and a local frame near $p$ .

1.1

In a basis $(b_{i})$ with dual basis $(b^{i})$ , [F1] is $\sum_{i} b^{i} (R (b_{i}, X) Y)$ . This is precisely a tensor contraction, so [F3] proves basis independence. In a smooth local frame, the same finite sum has smooth curvature and dual-frame coefficients; it is bilinear in $X, Y$ and smooth in $p$ , hence defines a smooth covariant two-tensor.

F1F3

2.1

If $(e_{i})$ is orthonormal, its metric dual is $g (e_{i}, \cdot)$ , so step 1.1 becomes the displayed $Rm$ sum. For every $i$ , pair interchange gives $Rm (e_{i}, X, Y, e_{i}) = Rm (Y, e_{i}, e_{i}, X)$ ; applying first- and last-pair skewness gives $Rm (Y, e_{i}, e_{i}, X) = Rm (e_{i}, Y, X, e_{i})$ . Summing proves $Ric (X, Y) = Ric (Y, X)$ .

A1F2step 1.1algebra∎

Depends on

Used by

Scalar curvature Definition
Ricci decomposition of the Riemann tensor in dimension at least three Proposition
Scalar curvature is twice the sum of sectional curvatures of orthonormal coordinate planes Proposition
Contracted second Bianchi identity Theorem
Schur's lemma for pointwise constant sectional curvature Theorem

Dependency tree · two levels

17 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)