PropositionStatement: 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.

Curvature tensor of constant sectional curvature

Statement

This item assumes ${A C}_{ω}$ , namely countable choice. In the propagated dependency chain, that assumption is required through Constant sectional curvature and space form; the tensor-determination argument is choice-free.

Assume ${A C}_{ω}$ as inherited through Constant sectional curvature and space form; the algebraic proof below uses no additional choice. A Riemannian manifold has constant sectional curvature $K$ if and only if

$R (X, Y) Z = K (g (Y, Z) X - g (X, Z) Y) .$

Equivalently,

$Rm (X, Y, Z, W) = K (g (Y, Z) g (X, W) - g (X, Z) g (Y, W)) .$

Facts & Assumptions

[A1]

${A C}_{ω}$ is countable choice and is required here through Constant sectional curvature and space form; the pointwise tensor argument makes no additional countable-family choice.

[F1]

Constant sectional curvature $K$ means that every tangent two-plane has sectional curvature $K$ , with the stated low-dimensional convention and inherited ${A C}_{ω}$ . Constant sectional curvature and space form.

[F2]

Algebraic curvature tensors with equal sectional curvatures on every two-plane are equal. Sectional curvatures determine the Riemann tensor.

Proof

Given: ${A C}_{ω}$ , a real number $K$ and the Riemannian metric $g$ .

1.1

Define $A_{K} (X, Y, Z, W) = K (g (Y, Z) g (X, W) - g (X, Z) g (Y, W))$ . Directly exchanging arguments shows that $A_{K}$ is skew in each pair and invariant under pair interchange; its three cyclic terms cancel pairwise, so it has all algebraic curvature symmetries. Moreover $A_{K} (X, Y, Y, X) = K (g (X, X) g (Y, Y) - g (X, Y)^{2})$ .

A1F2algebra

2.1

If the manifold has constant sectional curvature $K$ , step 1.1 shows that $A_{K}$ and $Rm$ give the same quotient on every two-plane. By [F2], $Rm = A_{K}$ . Conversely, if $Rm = A_{K}$ , division of the last identity in step 1.1 by the positive Gram determinant gives sectional curvature $K$ on every two-plane, which is [F1].

F1F2step 1.1

3.1

The four-tensor identity says for every $W$ that $g (R (X, Y) Z, W) = g (K (g (Y, Z) X - g (X, Z) Y), W)$ . Nondegeneracy of $g$ yields the vector-valued formula, and pairing that formula with $W$ gives the converse equivalence.

step 2.1algebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

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