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

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Complex K-theory of spheres

Statement

Assume AC. For $m \geq 0$ ,

${\tilde{K}}^{0} (S^{2 m}) ≅ Z, {\tilde{K}}^{0} (S^{2 m + 1}) = 0.$

The even generator is the $m$ -fold reduced external product of the Bott class $β$ , with $S^{0}$ interpreted as a based two-point space and the empty product as its rank-difference generator. Equivalently, ${\tilde{K}}^{q} (S^{n})$ is $Z$ when $q - n$ is even and is zero when $q - n$ is odd.

Facts & Assumptions

Given: AC, based spheres, and the Hopf Bott class $β$ .

[F1]

Multiplication by $β$ is the natural twofold-suspension isomorphism in every degree (Complex Bott periodicity).

[F2]

Complex bundles on $S^{1}$ are classified by clutching data on $S^{0}$ and ${GL}_{n} (C)$ is path-connected in the complex case (Clutching classifies vector bundles over spheres in the stable range).

[F3]

${\tilde{K}}^{0} (S^{2}) = Z β$ (Hopf-line calculation of K⁰(S²)).

[A1]

AC is propagated from [F1] and [F3]; the $S^{0}$ and $S^{1}$ base calculations themselves are finite and choice-free.

Proof

technique · direct

1.1F2algebra

A bundle on the based two-point space $S^{0}$ is a pair of finite-dimensional complex vector spaces. The reduced kernel records the dimension at the nonbasepoint minus the dimension at the basepoint, so ${\tilde{K}}^{0} (S^{0}) ≅ Z$ . On $S^{1}$ , [F2] reduces a rank- $n$ bundle to two clutching values in the same path component of ${GL}_{n} (C)$ , so it is trivial. Thus $K^{0} (S^{1}) ≅ Z$ by rank and ${\tilde{K}}^{0} (S^{1}) = 0$ .

2.1F1F3A1step 1.1induction

Apply [F1] repeatedly to the two base groups in step 1.1. It gives ${\tilde{K}}^{0} (S^{2 m}) ≅ {\tilde{K}}^{0} (S^{0}) = Z$ and ${\tilde{K}}^{0} (S^{2 m + 1}) ≅ {\tilde{K}}^{0} (S^{1}) = 0$ for all $m \geq 0$ . At each even step the isomorphism is external product with $β$ , so the generator is the stated $m$ -fold product; for $m = 1$ it agrees with [F3].

3.1F1step 2.1algebra∎

By definition, suspension shifts the reduced degree and [F1] makes it two-periodic. Hence ${\tilde{K}}^{q} (S^{n})$ depends only on the parity of $q - n$ ; step 2.1 gives $Z$ in even parity and zero in odd parity. This includes $n = 0$ , $m = 0$ , and the zero group without a hidden exception.

Complex K-theory of spheres

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources