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.

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

Statement

For every integer $n \geq 1$ , $S^{n}$ admits a continuous nowhere-zero tangent vector field if and only if $n$ is odd.

Facts & Assumptions

Given: The objects and hypotheses in the statement above.

[F1]

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. (No nowhere zero tangent vector field on an even sphere)

[F2]

For $n = 2 m - 1 \geq 1$ , identify $R^{2 m}$ with $C^{m}$ . The map $v : S^{n} \to R^{2 m}$ given by $v (x) = i x$ is a continuous unit tangent vector field. (An odd sphere admits a nowhere zero tangent vector field)

Proof

1.1

If a field exists, $n$ cannot be positive even by F1. Every positive integer is even or odd, so $n$ must be odd.

F1given

2.1

Conversely, for odd $n \geq 1$ write $n = 2 m - 1$ with $m \geq 1$ . F2 constructs the continuous unit tangent field $x \mapsto i x$ . This proves the reverse implication as well.

F2∎

Depends on

Used by

Degree of the antipodal map in low dimensions Example

Dependency tree · two levels

4 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)