# Equivalent fomulations of Bott periodicity

Is there an easy way to see the equivalence of the two statements of Bott periodicity. $$BU \times \mathbb{Z} \simeq \Omega^2BU$$ and $$K(X)\otimes K(S^2) \cong K(X\times S^2)$$

The easy way for me to think about this things is via the Yoneda lemma. This works better with reduced $K$-theory. This is defined for a pointed space $X$ as $$\tilde K^0 (X) := ker(K^0(X)\to K^0(*))$$ It is easy'' to see that your second statement reduces to the fact that multiplication by the generator of $\tilde K^0(S^2)$ induces an isomorphism $\tilde K^0(X) \cong \tilde K^0(X\wedge S^2)$ (this can be seen in many ways, one is noting that the projections of a product to two factors have canonical pointed sections and so induce direct summands in $K$-theory).

The key fact about reduced $K$-theory is that it is representable. That is there is a natural isomorphism $$\tilde K^0(X) \cong [X,BU\times \mathbb{Z}]_*$$ where $[,]_*$ are the homotopy classes of maps. So your second statement says that there is a natural isomorphism $$[X,BU\times \mathbb{Z}]_* \cong [\Sigma^2 X,BU\times\mathbb{Z}]_*$$ By using the suspension-loopspace adjunction you get a natural isomorphism $$[X, BU\times \mathbb{Z}]_* \cong [X,\Omega^2(BU\times\mathbb{Z})]_*$$ Finally the Yoneda lemma tells us that this must correspond to an homotopy equivelence $BU\times \mathbb{Z}\sim \Omega^2(BU\times\mathbb{Z})$, which is your first statement.

If you look closely all steps are reversible, so the two statements are in fact equivalent.

EDIT: For future reference I am putting the steps to recover the unreduced statement from the reduced statement

I am adding the steps on how to prove the reduced formulas and the unreduced formulas to be equivalent. First note that the basepoint inclusion $*\to X$ has a retraction $X\to *$, so $$K(X) = \tilde K(X)\oplus \mathbb{Z}$$ Moreover, if $X,Y$ are pointed spaces the inclusion $X=X\times * \to X\times Y$ has a retraction, so the induced map on $K$-theory $$\tilde K^0(X\times Y) \to \tilde K^0(X)$$ has a section. Since the same is true for $Y$ from the additivity of $K$-theory it follows that the map induced by the inclusion $X\vee Y\to X\times Y$ $$\tilde K^0(X\times Y)\to \tilde K^0(X\vee Y) = \tilde K^0(X)\oplus \tilde K^0(Y)$$ splits. Now consider the cofiber sequence $$X\vee Y \to X\times Y \to X\wedge Y$$ This induces an exact sequence $$\tilde K^0(\Sigma(X\times Y)) \to \tilde K^0(\Sigma X\vee \Sigma Y) \to \tilde K^0(X\wedge Y) \to \tilde K^0(X\times Y) \to \tilde K^0(X\vee Y)$$ We now that the rightmost arrow splits and the rightmost arrow splits too by a similar argument (in fact it splits even at the space level). So we have the desired isomorphism $$K(X\times Y) = \mathbb{Z}\oplus\tilde K^0(X\times Y) = \mathbb{Z}\oplus\tilde K^0(X) \oplus \tilde K^0(Y) \oplus \tilde K^0(X\wedge Y)$$

• This works directly only if one has $K(X)\otimes K(S^2) \cong K(X\times S^2)$ for all spaces $X$ and not only for compact Hausdorff spaces. In the latter case one gets a priori only a weak homotopy equivalence. Depending on the model for BU (CW-complex or not) one might get really a homotopy equivalence. I guess, the standard Grassmannian model for $BU$ has at least the homotopy type of a CW-complex? – Lennart Meier Apr 3 '14 at 22:58
• Yeah $BU$ has the homotopy type of a CW-complex ($BU_n$ has an explicit cell decomposition and I think that it passes to the limit, even if I never really checked). Anyway in my experience trying too hard to distinguish between homotopy equivalences and weak equivalences is not very useful. – Denis Nardin Apr 3 '14 at 23:02
• Thanks for answering. The problem I am have with this is that $[-,BU\times\mathbb{Z}]$ represents the unreduced $K$ functor. Also since $K(S^2)=\mathbb{Z}\oplus\mathbb{Z}$, I think $K(X)\cong K(X\times S^2)$ is false. – Rene Schipperus Apr 3 '14 at 23:03
• $BU\times \mathbb{Z}$ represents the reduced $K$-theory in pointed spaces. Also I wrorte $\tilde K^0(X)\cong \tilde K^0(X\wedge S^2)$, where I used the smash product not the cartesian product. – Denis Nardin Apr 3 '14 at 23:04
• I cannot read very well your last formula, if you want I'll edit in the details on how to get the unreduced result from the reduced result. It essentially comes from $K(X\times S^2) = \tilde K(X\wedge S^2) \oplus \tilde K(X) \oplus \tilde K(S^2) \oplus \mathbb{Z}$ – Denis Nardin Apr 4 '14 at 0:36