CounterexampleConstruction: AI-adaptedVerification: 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.
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Zero scalar curvature does not imply flatness

False claim

Every Riemannian manifold with identically zero scalar curvature is flat.

Counterexample

This item assumes ${A C}_{ω}$ , namely countable choice. In the propagated dependency chain, that assumption is required through The round sphere has positive constant sectional curvature, Hyperbolic space has negative constant sectional curvature, Curvature of a Riemannian product, and Scalar curvature is twice the sum of sectional curvatures of orthonormal coordinate planes; after those interfaces are fixed, the remaining local or finite argument makes no additional countable-family choice.

Assume ${A C}_{ω}$ and let $r > 0$ . Give

$M = S_{r}^{2} \times H_{r}^{2}$

the Riemannian product metric, with $H_{r}^{2}$ in its upper-half-space model of sectional curvature $- 1 / r^{2}$ . Then $M$ has scalar curvature identically zero, but its Riemann tensor is nonzero at every point. Hence $M$ is not flat and the false claim fails. The countable-choice assumption is inherited through the four curvature and scalar-curvature suppliers named above.

Facts & Assumptions

Given: ${A C}_{ω}$ , $r > 0$ , and the product of the displayed round and hyperbolic surfaces.

[A1]

${A C}_{ω}$ is countable choice and is required here through The round sphere has positive constant sectional curvature, Hyperbolic space has negative constant sectional curvature, Curvature of a Riemannian product, and Scalar curvature is twice the sum of sectional curvatures of orthonormal coordinate planes; after those supplied interfaces are fixed, the remaining local or finite calculation makes no additional countable-family choice.

[F1]

Countable choice permits a choice from every sequence of nonempty sets, and a connection is flat exactly when its curvature tensor vanishes identically. The Axiom of Countable Choice ( ${A C}_{ω}$ ), Curvature of an affine connection.

[F2]

Every finite-dimensional real inner-product space has an orthonormal basis. Every finite-dimensional real or complex inner product space has an orthonormal basis.

[F3]

Under ${A C}_{ω}$ , every tangent two-plane of $S_{r}^{2}$ has sectional curvature $+ 1 / r^{2}$ . The round sphere has positive constant sectional curvature.

[F4]

Under the stated ${A C}_{ω}$ , every tangent two-plane of the upper-half-space $H_{r}^{2}$ has sectional curvature $- 1 / r^{2}$ . Hyperbolic space has negative constant sectional curvature.

[F5]

Product curvature restricts to each factor's curvature, and every mixed plane spanned by one nonzero pure vector from each factor has sectional curvature zero. Curvature of a Riemannian product.

[F6]

For an orthonormal basis $(E_{1}, \dots, E_{n})$ , scalar curvature is $2 \sum_{i < j} K (span {E_{i}, E_{j}})$ . Scalar curvature is twice the sum of sectional curvatures of orthonormal coordinate planes.

Verification

technique · explicit counterexample

1.1

Fix any $(p, q) \in M$ . By [F2], take orthonormal bases $(e_{1}, e_{2})$ of $T_{p} S_{r}^{2}$ and $(f_{1}, f_{2})$ of $T_{q} H_{r}^{2}$ . The product vectors $E_{1} = (e_{1}, 0)$ , $E_{2} = (e_{2}, 0)$ , $E_{3} = (0, f_{1})$ , and $E_{4} = (0, f_{2})$ are orthonormal. By [F3]–[F5], the six coordinate-plane curvatures are $K_{12} = 1 / r^{2}$ , $K_{34} = - 1 / r^{2}$ , and $K_{13} = K_{14} = K_{23} = K_{24} = 0$ .

A1F2F3F4F5algebra

2.1

Substitution of the six values from step 1.1 in [F6] gives ${Scal}_{M} (p, q) = 2 (1 / r^{2} - 1 / r^{2} + 0 + 0 + 0 + 0) = 0$ . Since $(p, q)$ was arbitrary, scalar curvature vanishes identically.

F6step 1.1algebra

3.1

On the pure sphere plane, [F3] and [F5] give ${Rm}^{M} (E_{1}, E_{2}, E_{2}, E_{1}) = K_{12} = 1 / r^{2} \neq 0$ . Thus the Riemann tensor does not vanish at $(p, q)$ ; [F1] says the product is not flat. This is the required failed conclusion despite the zero scalar value in step 2.1.

F1F3F5step 1.1step 2.1algebra

4.1

Both factors and their product are nonempty fixed two- and four-manifolds, so zero- and one-dimensional cases are inapplicable. The hypothesis $r > 0$ makes both metrics nondegenerate and all reciprocal curvature values defined; $r = 0$ is excluded. The round sphere is boundaryless and the hyperbolic upper half-space excludes its height-zero ideal boundary, so no endpoint or manifold-boundary value is asserted. Step 1.1 fixes one arbitrary point before making two finite choices supplied by [F2], so it selects no point-indexed family. The stated ${A C}_{ω}$ is inherited through [F3]–[F6]; [F2] and the remaining calculation make no additional countable-family choice. The item supplies a counterexample to one implication, not a biconditional.

F1F2F3F4F5F6step 1.1step 2.1step 3.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

34 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)