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

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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.

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Schur's lemma for pointwise constant sectional curvature

Statement

This item assumes ${A C}_{ω}$ , namely countable choice. In the propagated dependency chain, that assumption is required through Sectional curvature, Ricci curvature is symmetric and basis independent, Scalar curvature, and Contracted second Bianchi identity; the tensor-determination step is choice-free.

Let $(M, g)$ be connected of dimension $n \geq 3$ . If, at each point $p$ , $K (σ)$ has the same value for every two-plane $σ \subseteq T_{p} M$ , then

$k (p) : = \frac{S (p)}{n (n - 1)}$

is smooth, equals that common sectional curvature, and is constant on $M$ . No smoothness of the pointwise common value is assumed in the hypothesis.

Facts & Assumptions

[A1]

${A C}_{ω}$ is countable choice and is required here through Sectional curvature, Ricci curvature is symmetric and basis independent, Scalar curvature, and Contracted second Bianchi identity; the pointwise tensor argument makes no additional countable-family choice.

[F1]

Algebraic curvature tensors are determined by their sectional curvatures. Sectional curvatures determine the Riemann tensor.

[F2]

Sectional curvature uses the positive Gram determinant and, on an orthonormal pair, is $Rm (X, Y, Y, X)$ . Sectional curvature.

[F3]

Ricci curvature is the orthonormal contraction of $Rm$ . Ricci curvature is symmetric and basis independent.

[F4]

Scalar curvature is the metric trace of Ricci. Scalar curvature.

[F5]

The contracted Bianchi identity is $div Ric = (1 / 2) d S$ . Contracted second Bianchi identity.

[F6]

The Levi–Civita connection preserves $g$ . Levi civita connection.

[F7]

A smooth function whose differential vanishes is constant on every connected component. A smooth function with zero differential is constant on each connected component.

Proof

Given: ${A C}_{ω}$ , the connected Riemannian manifold in the statement.

1.1

Since $S$ is smooth by [F4] and $n (n - 1) \neq 0$ , the displayed function $k$ is smooth. Fix $p$ and one two-plane $σ \subseteq T_{p} M$ , and put $κ = K (σ)$ . By hypothesis every two-plane at $p$ has curvature $κ$ . The metric model $A_{κ} (X, Y, Z, T) = κ (g (Y, Z) g (X, T) - g (X, Z) g (Y, T))$ has the algebraic curvature symmetries by direct expansion and, by [F2], the same sectional quotient. Thus [F1] gives ${Rm}_{p} = A_{κ}$ .

A1F1F2F4algebra

2.1

Contracting the model in an orthonormal basis using [F3] gives ${Ric}_{p} = (n - 1) κ g_{p}$ ; taking its trace using [F4] gives $S (p) = n (n - 1) κ$ . Consequently $κ = k (p)$ . Since $p$ and $σ$ were arbitrary, every sectional curvature at $p$ equals the smooth function $k (p)$ , and globally $Ric = (n - 1) k g$ and $S = n (n - 1) k$ .

F3F4step 1.1algebra

3.1

Metric compatibility [F6] gives $div (k g) = d k$ . Substitute the two identities from step 2.1 into [F5]: $(n - 1) d k = (n (n - 1) / 2) d k$ . Because $n \geq 3$ , the coefficient $(n - 1) (n - 2) / 2$ is nonzero, so $d k = 0$ .

F5F6step 2.1algebra

4.1

If $M$ is nonempty, connectedness makes it one connected component, so [F7] and step 3.1 make $k$ constant on $M$ . If $M$ is empty under the library's connected-empty convention, the unique empty function agrees vacuously with every constant, so the conclusion still holds.

F7step 3.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

27 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

Will J. Merry, Differential Geometry (2021) (standard reference, not scraped)
Ved Datar, Lectures on Riemannian Geometry (standard reference, not scraped)
John M. Lee, Riemannian Manifolds: An Introduction to Curvature (standard reference, not scraped)