TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07

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.

No nowhere zero tangent vector field on an even sphere

Statement

For every positive even integer $n$ , no continuous map $v : S^{n} \to R^{n + 1}$ can satisfy both $⟨ v (x), x ⟩ = 0$ and $v (x) \neq 0$ for all $x$ . Thus an even-dimensional sphere has no continuous nowhere-zero tangent vector field.

Facts & Assumptions

Given: The objects and hypotheses in the statement above.

[F1]

For $n \geq 1$ and continuous sphere self-maps $f, g$ , homotopic maps have the same degree and $\deg (g \circ f) = \deg (g) \deg (f)$ . Every homotopy equivalence $S^{n} \to S^{n}$ has degree $1$ or $- 1$ . (Degree is homotopy invariant and multiplicative under composition)

[F2]

On $S^{n} \subset R^{n + 1}$ for $n \geq 1$ , the identity, a constant map, any single coordinate reflection, and the antipodal map have degrees $1$ , $0$ , $- 1$ , and $(- 1)^{n + 1}$ respectively. (Degree of identity constant reflection and antipodal sphere maps)

Proof

1.1

If such $v$ existed, $w (x) = v (x) / ∣ v (x) ∣$ would be continuous, orthogonal to $x$ , and of norm one. Hence $H (x, t) = \cos (π t) x + \sin (π t) w (x)$ has squared norm $\cos^{2} (π t) + \sin^{2} (π t) = 1$ for all $0 \leq t \leq 1$ .

givenalgebra

2.1

The homotopy has endpoints $H (x, 0) = x$ and $H (x, 1) = - x$ . By F1 their degrees agree, but by F2 these degrees are $1$ and $(- 1)^{n + 1} = - 1$ because $n$ is even. The unequal integers contradict the existence of $v$ .

F1F2step 1.1∎

Depends on

Used by

A sphere has a nowhere zero tangent vector field iff its dimension is odd Theorem

Dependency tree · two levels

6 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

Hatcher, Algebraic Topology, Theorem 2.28, p.135 (standard reference, not scraped)