Loop space and stable homotopy theory

Question

The Bott periodicity theorem for unitary group $U(n)$ says that $$ \pi_{i-1}(U) \simeq \pi_{i+1}(U) $$ How can I prove, using this theorem, that $$ \Omega (U) \simeq BU \times \mathbb{Z} ?$$ What is precisely the loop space $\Omega(U)$?

Answer 1

My memory of this is that the K-theory functor on the category of compact Hausdorff spaces is represented by $BU\times \mathbb{Z}$, that is, $K(X)= [X,BU\times \mathbb{Z}]$ for all compact Hausdorff spaces $X$. If we're dealing with reduced K-theory, then we're looking at basepoint preserving homotopy classes of maps: $\tilde{K}(X)=\langle X,BU\times \mathbb{Z} \rangle$.

The K-theoretic form of Bott periodicity is that $\tilde{K}(X)\cong \tilde{K}(\Sigma^2 X)$, i.e., $\langle X, BU\times \mathbb{Z} \rangle \cong \langle \Sigma^2 X, BU\times \mathbb{Z} \rangle$. The (reduced) suspension/loopspace adjunction implies that $\langle \Sigma^2 X, BU\times \mathbb{Z}\rangle \cong \langle X,\Omega^{2} (BU\times \mathbb{Z}) \rangle$. Therefore, $\langle X, BU\times \mathbb{Z} \rangle \cong \langle X, \Omega^{2} (BU\times \mathbb{Z}) \rangle$ for all compact Hausdorff spaces $X$. Yoneda's lemma implies that $\Omega^{2} (BU\times \mathbb{Z})\cong BU\times \mathbb{Z}$ in the homotopy category of pointed topological spaces, i.e., that $\Omega^{2} (BU\times \mathbb{Z})$ is homotopy equivalent to $BU\times \mathbb{Z}$.

However, $\Omega (BU\times \mathbb{Z})\cong U$; indeed, it's basic that taking the loop spaces of the classifying space of a topological group always gives you back the topological group. Therefore, $\Omega U\cong BU\times \mathbb{Z}$ (and vice-versa!).

I hope this helps!