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 $\Omega(U)$?

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    $\begingroup$ $\Omega U\cong BU\times\mathbb Z$ implies $\pi_{i-1}(U)\cong\pi_{i+1}(U)$ but I don't think there is an implication in the opposite direction. $\endgroup$
    – Grigory M
    Jul 1, 2013 at 14:03
  • $\begingroup$ @ArthurStuart I've provided an answer to your question below; let me know if it answers your question ... $\endgroup$ Jul 7, 2013 at 12:35

1 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!

  • $\begingroup$ I don't know how can I prove that are equivalent $\pi_{i-1}(U) \simeq \pi_{i+1}(U)$ and $\mathbb{Z} \times BU \to \Omega(U)$ ia a weak homotopy equivalence. $\endgroup$ Jul 9, 2013 at 12:19
  • $\begingroup$ @ArthurStuart The condition $\pi_{i-1}(U)\cong \pi_{i+1}(U)$ for all positive integers $i$ is equivalent to the statement that $\mathbb{Z}\times BU$ and $\Omega(U)$ have isomorphic homotopy groups. However, two spaces (or even two CW complexes) may have isomorphic homotopy groups without being homotopy equivalent. On the other hand, Whitehead's theorem states that a map $f:X\to Y$ between CW complexes that induces isomorphisms on all homotopy groups must be a homotopy equivalence. So, in general, I don't think there's a way of proving the assertion that you wish to prove. $\endgroup$ Jul 9, 2013 at 14:03
  • $\begingroup$ @ArthurStuart We aren't explicitly given the isomorphisms $\pi_{i-1}(U)\cong \pi_{i+1}(U)$. For example, $\mathbb{RP}^2$ and $\mathbb{S}^2\times \mathbb{RP}^{\infty}$ have isomorphic homotopy groups (their universal covers, $\mathbb{S}^2$ and $\mathbb{S}^2\times \mathbb{S}^{\infty}$, are homotopy equivalent). However, $\mathbb{RP}^2$ and $\mathbb{S}^2\times \mathbb{RP}^{\infty}$ cannot be homotopy equivalent because their homology groups aren't isomorphic: $\mathbb{RP}^{\infty}$ has non-trivial homology in arbitrarily large degrees whereas $\mathbb{RP}^2$ does not! $\endgroup$ Jul 9, 2013 at 14:13
  • $\begingroup$ What is the difference among $[\cdot, \cdot]$ and $\langle \cdot, \cdot \rangle$? Why $\Omega(BU \times \mathbb{Z}) \cong U$? $\endgroup$ Jul 12, 2013 at 12:19
  • $\begingroup$ I'm sorry, how can I prove that $\Omega(U) \simeq BU \times \mathbb{Z}$? Where $BU:= colim_n BU(n)$ and $\mathbb{Z}$ has the diescrete topology?... I know that $K(X) \simeq \tilde{K}(X) \otimes \mathbb{Z} \simeq K(pt)$. $\endgroup$ Jul 12, 2013 at 12:29

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