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LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passaudited 2026-09-14
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Negative Laurent powers are cleared by Hopf-line stabilization

Statement

Assume AC. If

f(z)=∑j=−rsajzj

is Laurent-polynomial clutching data for a bundle E over X×S2, then zrf(z) is polynomial. In the fixed clutching convention this multiplication tensors the glued bundle by pr⁡S2∗γr. The original K-class is recovered by multiplying by the inverse unit [γ]−r.

Facts & Assumptions

Given: AC, r,s≥0, and normalized Laurent clutching data f supplied by Uniform Laurent approximation through bundle automorphisms.

[F1]

Under the fixed convention, transition maps multiply under tensor product (Clutching construction for bundles over a suspension).

[F2]

The external product pulls the Hopf line from S2 to X×S2 (External product in complex K-theory).

[F3]

For β=[γ]−1, one has β2=0 and hence [γ]−1=1−β (Hopf-line calculation of K⁰(S²)).

[A1]

AC is inherited from [F2] for the reduced product convention and from [F3]; the exponent-clearing calculation itself is finite algebra.

Proof

technique · direct
1.1algebra

Multiplication gives zrf(z)=∑j=−rsajzj+r, whose exponents range from 0 to r+s. On ∣z∣=1 the scalar zr is nonzero, so zrf(z) remains an automorphism.

2.1F1F2A1step 1.1

The line γr has transition zr. By [F1] and [F2], tensoring the bundle [E,f] with pr⁡S2∗γr multiplies its transition by zr. Therefore [E,zrf]=[E,f] [γ]r in K0(X×S2).

3.1F3A1step 2.1algebra∎

By [F3], [γ] is a unit with [γ]−r=(1−β)r=1−rβ. Multiplying the equality in step 2.1 by this unit gives [E,f]=[E,zrf] [γ]−r, so clearing the negative powers loses no class information. This includes r=0, when no change occurs.

Depends on

Dependency tree · 2 levels

Sources