Corollary of the Bott Periodicity Theorem, $\widetilde{K}(S^{2n+1}) = 0$

Question

In Hatcher's Vector Bundles and K-Theory, he states the Bott Periodicity Theorem:

Theorem 2.11: The homomorphism $\beta : \widetilde{K}(X) \longrightarrow \widetilde{K}(S^2X)$, $\beta(a) = (H-1) \ast a$ is an isomorphism for all compact Hausdorff spaces $X$.

As a corollary of this however, he states the following:

Corollary 2.12: $\widetilde{K}(S^{2n+1}) = 0$ and $\widetilde{K}(S^{2n}) \cong \mathbb{Z}$, generated by the $n$-fold reduced external product $(H-1) \ast \cdots \ast (H-1)$.

I don't understand how it follows that $\widetilde{K}(S^{2n+1}) =0$.

Answer 1

Note that $S^{2n+1} = S^2(S^{2n-1})$, so $\widetilde{K}(S^{2n+1}) \cong \widetilde{K}(S^2(S^{2n-1})) \cong \widetilde{K}(S^{2n-1})$ and hence

$$ \widetilde{K}(S^{2n+1}) \cong \widetilde{K}(S^{2n-1}) \cong \dots \cong \widetilde{K}(S^3) \cong \widetilde{K}(S^1).$$

Rank $n$ complex vector bundles on $S^1$ are classified by $\pi_0(GL(n, \mathbb{C})) = \pi_0(U(n))$ by the clutching construction. But $U(n)$ is path-connected for every $n$, so every complex vector bundle on $S^1$ is trivial. Therefore $K(S^1) \cong \mathbb{Z}$ and hence $\widetilde{K}(S^1) = 0$.