CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-14

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Stable isomorphism does not imply actual bundle isomorphism

Statement refuted

False claim: if two vector bundles become isomorphic after adding the same trivial summand, then they were already isomorphic.

The real tangent bundle gives a counterexample:

$T S^{2} \oplus ε_{R}^{1} ≅ ε_{R}^{3},$

but $T S^{2} ≇ ε_{R}^{2}$ . This is a real boundary example for the cancellation issue behind Grothendieck completion; it does not assert an equality between complex bundles in $K^{0}$ .

Facts & Assumptions

Given: the unit sphere $S^{2} \subset R^{3}$ .

[F1]

Bundle isomorphisms are fiberwise-linear isomorphisms over the identity, and a section is nowhere zero when it avoids each zero vector (Bundle maps, sections, subbundles, and isomorphisms).

[F2]

Whitney sum has fiber the direct sum of the two bundle fibers (Whitney sum, tensor, dual, Hom, and exterior-power bundles).

[F3]

The even sphere $S^{2}$ has no continuous nowhere-zero tangent vector field (No nowhere zero tangent vector field on an even sphere).

Counterexample

technique · contradiction after an explicit stable isomorphism

1.1

Write $T S^{2} = {(x, v) \in S^{2} \times R^{3} : ⟨ x, v ⟩ = 0}$ . Its normal line is trivialized by $x \mapsto (x, x)$ . Using [F2], define $Φ : T S^{2} \oplus ε_{R}^{1} \to S^{2} \times R^{3}$ by $Φ ((x, v), (x, t)) = (x, v + t x)$ . The continuous inverse sends $(x, w)$ to $((x, w - ⟨ w, x ⟩ x), (x, ⟨ w, x ⟩))$ . Thus [F1] verifies the displayed stable bundle isomorphism fiber by fiber, including at $t = 0$ .

F1F2constructalgebra

1.2

Suppose for contradiction that an actual bundle isomorphism $Ψ : ε_{R}^{2} \to T S^{2}$ exists.

assume-contra

2.1

The constant section $x \mapsto (x, (1, 0))$ of $ε_{R}^{2}$ is nowhere zero. Composing it with $Ψ$ gives a continuous nowhere-zero section of $T S^{2}$ , hence a nowhere-zero tangent vector field in the sense of [F1].

F1step 1.2construct

3.1

This contradicts [F3]. Therefore $T S^{2}$ is not actually trivial, while step 1.1 proves that it becomes trivial after adding one trivial real line. The two bundles in the false claim are $T S^{2}$ and $ε_{R}^{2}$ , with the same summand $ε_{R}^{1}$ added to both.

F3step 1.1step 2.1discharge-contradiction∎

Depends on

Used by

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Dependency tree · two levels

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Sources

Hatcher, Vector Bundles & K-Theory, §1.1 tangent-bundle example (standard reference, not scraped)
Milnor and Stasheff, Characteristic Classes, §2 (standard reference, not scraped)