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I asked this also here, but maybe it's also appropriate to ask it here.

Bruno Harris' proof (or I guess also Bott's original proof) of Bott periodicity (see here for instance) shows that there is a homotopy equivalence $h\colon\mathbb{Z}\times BU \rightarrow \Omega^2 (\mathbb{Z}\times BU)$, hence there is a natural isomorphism of set-valued functors $KU\rightarrow KU^{-2}$. Karoubi calls this "weak Bott periodicity" in his book.

Now, is it possible to deduce "strong Bott periodicity" from that? I.e. that external multiplication by the Bott element is an isomorphism?

It would suffice to show that the maps $KU(X)\rightarrow KU^{-2}(X)$ are $KU(X)$-module homomorphisms, i.e. I think one has to show that a diagram like

$\begin{matrix} \mathbb{Z}\times BU\wedge\mathbb{Z}\times BU &\stackrel{\otimes}{\rightarrow}&\mathbb{Z}\times BU\\\ \downarrow{h\wedge id}&&\downarrow{h}\\\ \Omega^2(\mathbb{Z}\times BU)\wedge \mathbb{Z}\times BU&\stackrel{\otimes}{\rightarrow}&\Omega^2(\mathbb{Z}\times BU) \end{matrix} $

commutes up to homotopy (the lower $\otimes$ should be something like pointwise multiplication of maps $S^2\rightarrow \mathbb{Z}\times BU$ by elements of $\mathbb{Z}\times BU$). Of course, the problem is that the map $h$ is not very explicit.

Is it at all possible to do that? Many thanks.

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up vote 11 down vote accepted

As a matter of history, the original Bott maps are very explicit, and in fact they are $E_{\infty}$ maps with respect to the actions of the linear isometries operad (as shown in the first chapter of $E_{\infty}$ ring spaces and $E_{\infty}$ ring spectra). Bott himself, in his paper Raoul Bott. Quelques remarques sur les théorèmes de périodicité. Bull. Soc. Math. France 87 1959 293–310, showed how to derive the strong form from the weak form by showing that his original maps are homotopic to the adjoints of the evident maps obtained by tensoring with the Bott classes. He does this for the real and quaternionic cases as well.

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Thank you for the reference to Bott's paper. I was not aware of this one. –  ruediger Mar 30 '12 at 8:24

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