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The spaces $O$ and $O/U$ that appear in Bott periodicity represent the functors $KO^7(X)$ and $KO^6(X)$ respectively. Is there an interpretation of the map $KO^7(X) \to KO^6(X)$ induced by the quotient map $O \to O/U$? Or is it always zero?

What about $KO^3(X) \to KO^2(X)$ induced by $Sp \to Sp/U$?

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    $\begingroup$ The quotient map $O\to O/U$ induces a nontrivial map on $\pi_0$. $\endgroup$ Jul 21, 2015 at 0:07
  • $\begingroup$ Thanks. Maybe I should have asked the question like this: what is KO^6(O)? What is KO^2(Sp)? $\endgroup$
    – Bott asker
    Jul 21, 2015 at 2:14
  • $\begingroup$ Bott periodicity provides homotopy equivalences between successive spaces for KO or K spectra, e.g. a homotopy equivalence $SO/U\to\Omega(Spin)$. So, how it can be that the map you quote $KO^7\to KO^6$ is induced by the projection?!? A good collection on these maps and spaces is a series of papers by Cartan, as well as a paper by Douady for the complex case. $\endgroup$
    – user51223
    Jul 21, 2015 at 10:54
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    $\begingroup$ I think, in the stable world you're asking: If we identify the cofiber of the "realification map" $r\colon KU \to KO$ with $\Sigma^{-1}KO$, what is the map $KO \to \Sigma^{-1}KO$? As $r$ is $KO$-linear, $KO$-linear maps $KO \to \Sigma^{-1}KO$ are in 1-1 correspondence with $\pi_1KO$ and the map is non-zero (as $KU$ does not split in copies of $KO$), it must be $\eta$. So your transformation $KO^7(X) \to KO^6(X)$ seems to be multiplication with $\eta \in KO^{-1}(pt)$. $\endgroup$ Jul 22, 2015 at 9:50
  • $\begingroup$ May I add that in general, there is a transformation $KO^{*+1}(X)→KO^*(X)$ induced by multiplying with $e$ where $KO_*\simeq\mathbb{Z}[e,\alpha,\beta^{\pm 1}]/\langle 2e,e^3,e\alpha,\alpha^2−4\beta\rangle$ which fits into Bott's exact sequence $$\cdots\to KO^{*+1}(X)\stackrel{\cdot e}{\to}KO^*(X)\stackrel{\chi}{\to}K^{*+2}(X)\stackrel{r}{\to}KO^{*+2}(X)\to\cdots$$ where $\chi$ is induced by complexification followed by multiplication with $z^{-1}$ and $r$ is the realification. Maybe this might be more useful if you're just looking for a transformation. $\endgroup$
    – user51223
    Jul 22, 2015 at 13:04

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