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

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The round sphere has positive 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; after those interfaces are fixed, the remaining local or finite argument makes no additional countable-family choice.

Assume ${A C}_{ω}$ . Let $r > 0$ . For $n \geq 2$ , the round sphere

$S_{r}^{n} = {x \in R^{n + 1} : ∣ x ∣ = r}$

with its metric induced from Euclidean space has constant sectional curvature $1 / r^{2}$ . For $n = 0$ or $n = 1$ , its sectional-curvature domain is empty.

The choice assumption is inherited through both the smooth orthogonal projection constructions used by the Gauss–Weingarten suppliers and the sectional-curvature interface.

Facts & Assumptions

Given: Countable choice, integers $n \geq 0$ , a real number $r > 0$ , and, when $n \geq 2$ , a point $x \in S_{r}^{n}$ and a tangent two-plane $σ \subseteq T_{x} S_{r}^{n}$ .

[A1]

${A C}_{ω}$ is countable choice and is required here through Sectional curvature; after those supplied interfaces are fixed, the remaining local or finite calculation makes no additional countable-family choice.

[F1]

Countable choice supplies a choice function for every countable family of nonempty sets. The Axiom of Countable Choice ( ${A C}_{ω}$ ).

[F2]

A nonempty regular level set is an embedded submanifold, and its tangent space is the kernel of the differential. A regular level set is an embedded submanifold, The tangent space of a regular level set is the kernel.

[F3]

Constant Euclidean metric coefficients give zero Levi–Civita symbols, and a connection is function-linear in its direction and satisfies the Leibniz rule in the differentiated field. Christoffel formula for the levi civita connection, Connection laws in directional form.

[F4]

For a unit normal $ν$ , the page's sign convention is $({\overline{\nabla}}_{u} ν)^{⊤} = - S_{ν} u$ and $g (S_{ν} u, v) = ⟨ I I (u, v), ν ⟩$ . Weingarten equation and adjointness of the shape operator.

[F5]

The Gauss equation is ${Rm}^{S} (u, v, z, w) = {Rm}^{R^{n + 1}} (u, v, z, w) + ⟨ I I (u, w), I I (v, z) ⟩ - ⟨ I I (u, z), I I (v, w) ⟩$ . Gauss equation for a Riemannian submanifold.

[F6]

Euclidean space has zero Riemann curvature. Euclidean space has zero curvature.

[F7]

Sectional curvature is the Riemann numerator divided by the positive Gram determinant of a basis of the two-plane. Sectional curvature.

Verification

technique · direct calculation

1.1

Put $Q (y) = ∣ y ∣^{2}$ . On $Q^{- 1} (r^{2})$ , one has $d Q_{y} (w) = 2 ⟨ y, w ⟩$ and $d Q_{y} (y) = 2 r^{2} \neq 0$ , so $r^{2}$ is a regular value; the level is nonempty because $(r, 0, \dots, 0)$ lies in it. By [F2], $S_{r}^{n}$ is an embedded hypersurface and $T_{y} S_{r}^{n} = \ker d Q_{y} = y^{⊥}$ .

F2algebraconstruct

2.1

The field $ν (y) = y / r$ has unit length and is normal by step 1.1. In Cartesian coordinates the metric coefficients are $δ_{a b}$ , so [F3] gives ${\overline{\nabla}}_{\partial_{a}} \partial_{b} = 0$ ; applying the connection laws to $ν = \sum_{a} (y^{a} / r) \partial_{a}$ gives ${\overline{\nabla}}_{u} ν = u / r$ for every tangent $u$ . This derivative is tangent by step 1.1, and [F4] therefore gives $S_{ν} u = - u / r$ . Since the normal space is spanned by $ν$ , [F4] then gives $I I (u, v) = - (1 / r) g (u, v) ν$ .

F3F4step 1.1algebra

3.1

Let $(u, v)$ be any supplied basis of $σ$ . Substitute $X = u$ , $Y = Z = v$ , $W = u$ and the formula from step 2.1 into [F5]. The ambient term is zero by [F6], while the two quadratic terms give ${Rm}^{S_{r}^{n}} (u, v, v, u) = (1 / r^{2}) (g (u, u) g (v, v) - g (u, v)^{2})$ . The denominator is positive by [F7], so division yields $K (σ) = 1 / r^{2}$ , independently of $x$ and $σ$ .

A1F5F6F7step 2.1algebra

4.1

The explicit point in step 1.1 proves that every $S_{r}^{n}$ here is nonempty. When $n = 0$ or $n = 1$ there is no tangent two-plane, so the curvature function has empty domain rather than a numerical exception; for $n \geq 2$ , [F7] excludes a degenerate Gram denominator. The required endpoint condition is $r > 0$ : at $r = 0$ the level is not regular and $1 / r^{2}$ is undefined. Countable choice [F1] is assumed because [F4]–[F5] inherit it from their smooth orthogonal projection construction and [F7] inherits it through the Riemann-tensor symmetries; the point, normal, and the basis used above are explicit or supplied, so the calculation makes no additional choice. The result is a direct equality and asserts no biconditional.

F1F2F4F5F7step 1.1step 2.1step 3.1∎

Depends on

Used by

Zero scalar curvature does not imply flatness Counterexample

Dependency tree · two levels

39 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)