$K$-theoretical interpretation of Bott periodicity.

Question

We consider complex Bott periodicity $\pi_{i-1}(U) \simeq \pi_{i+1}(U)$. Why we can say that the $K$-theoretical interpretation of this assertion is $\tilde{K}(X) \cong \tilde{K}(\Sigma^2(X))$? Where I denote with $\Sigma$ the reduced suspention and with $\tilde{K}(X)= \langle X, BU \times \mathbb{Z} \rangle$ the homotopical classes of preserving basepoint maps. I'm agree that if we say that $\tilde{K}(X) \cong \tilde{K}(\Sigma^2(X))$ than follows that $\Omega(U) \cong BU \times \mathbb{Z}$... Thanks.

Answer 1

Firstly, you can find all that I am going to write about in a pretty little book by Allen Hatcher on K-theory.

You can describe all rank-$k$ vector bundles (not just the stable ones) on the sphere $S^n$ with the so-called clutching functions (Hatcher, p. 22). In this approach, you view the sphere as being assembled from two copies of $D^n$ glued together at the ‘equator’ $S^{n-1}$. Since $D^n$ is contractible, vector bundles over it are trivial, so the only non-trivial information about the bundles on $S^n$ comes from the clutching function $S^{n-1} \to GL(k, \mathbb C) \cong U(k)$. This describes how to relate the two trivial bundles by giving the linear transformation at every point of the equator. It’s not hard to show that homotopy classes of these functions then correspond bijectively with equivalence classes of vector bundles.

When you pass to the stable vector bundles (i.e. K-theory), you can let $k$ in the above go to infinity -- this loses a lot of information but makes everything simpler and computable. So this is a direct connection between classes in K-theory and homotopy groups of $U$.

The Bott periodicity in K-theory then takes the following form: (Hatcher p. 41):

There is an isomorphism $$\mu : K(X) \otimes K(S^2) \to K(X \times S^2).$$

Proving this theorem is the hard part of proving the periodicity. One works with generalized clutching functions on $X \times S^2$ shows that's enough to work with linear functions and after that, it's a simple matter of checking that the mapping is bijective. In any case, for the reduced theory, one gets (Hatcher, p. 54) $$ \widetilde K(X) \otimes \widetilde K(S^2) \to\widetilde K(X \wedge S^2) \cong \widetilde K(S^2 X) $$ and since $\widetilde K(S^2)$ is just $\mathbb Z$, we obtain your claim (note that reduced vs. unreduced suspension shouldn't matter here).