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A positive even sphere has no nowhere-zero tangent field

Statement

For every integer $m \geq 1$ , the tangent bundle of the unit sphere $S^{2 m} \subset R^{2 m + 1}$ has no continuous nowhere-zero section. In particular $T S^{2}$ is not trivial.

Facts & Assumptions

Given: A positive integer $m$ and the unit sphere $S^{2 m}$ .

[F1]

Homotopic sphere self-maps have equal degree (Degree is homotopy invariant and multiplicative under composition).

[F2]

The identity on $S^{2 m}$ has degree $1$ , and its antipodal map has degree $(- 1)^{2 m + 1} = - 1$ (Degree of identity constant reflection and antipodal sphere maps).

Proof

1.1givenconstructalgebra

Suppose a continuous nowhere-zero tangent field $V$ exists, and set $u (x) = V (x) / ∥ V (x) ∥$ . Tangency means $⟨ x, u (x) ⟩ = 0$ , while both vectors have norm one. Consequently $H (x, t) = \cos (π t) x + \sin (π t) u (x)$ has norm one for every $(x, t) \in S^{2 m} \times [0, 1]$ , is continuous, and has endpoints $H (x, 0) = x$ and $H (x, 1) = - x$ . It is a homotopy from the identity to the antipodal map.

2.1F1F2step 1.1∎

By [F1] these endpoints have equal degree, contradicting their degrees $1$ and $- 1$ in [F2]. Thus no such field exists. A trivial positive-rank tangent bundle has a nowhere-zero section given by a constant nonzero vector in its trivialization, so $T S^{2}$ cannot be trivial.

Depends on

Used by

The standard sphere immersion and its normal line Example

Dependency tree · two levels

8 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

Allen Hatcher, Algebraic Topology, degree properties and vector fields on spheres, section 2.2 (standard reference, not scraped)