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I apologize that this is perhaps not adequate for mathoverflow but I have struggled with this for days now and become desperate...

The reduced K-group $\tilde{K}(S^0)$ of the zero sphere is the ring $\mathbb{Z}$ as being the kernel of the ring morphism $K(S^0)\to K(x_0)$. The ring structure on $K(S^0)$ and $K(x_0)$ comes from the tensor product $\otimes$ of vector bundles.

If $H$ is the canonical line bundle over $S^2$ then $(H-1)^2=0$ where the product comes from $\otimes$. The Bott periodicity theorem states that the induced map $\mathbb{Z}\left[H\right]/(H-1)^2\to K(S^2)$ is an isomorphism of rings. So $\tilde{K}(S^2)\cong \mathbb{Z}\left[H-1\right]/(H-1)^2$, I think, and every square in $\tilde{K}(S^2)$ is zero.

The reduced external product gives rise to a map $\tilde{K}(S^0)\to \tilde{K}(S^2)$ which is a ring (?) isomorphism (see e.g. Hatcher Vector Bundles and K-Theory, Theorem 2.11.) but not every square in $\tilde{K}(S^2)$ is zero then. How can this be?

Aside from that I do not understand the relation of $\otimes:K(X)\otimes K(X)\to K(X)$ and the composition of the external product with map induced from the diagonal map $K(X)\otimes K(X)\to K(X\times X)\to K(X)$.

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    $\begingroup$ You forgot to ask your question? $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Jan 28, 2010 at 20:38
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    $\begingroup$ Oh, the question is 'how can this be'? :) $\endgroup$
    – roger123
    Commented Jan 28, 2010 at 20:46

2 Answers 2

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Reduced $K$-groups are ideals of the standard $K$-groups. $\tilde K(X) \subset K(X)$ is the ideal of virtual-dimension-zero elements.

In particular, the reduced K-theory $\tilde K(S^2)$ is not $\mathbb{Z}[H]/(H-1)^2$, but rather the ideal of this generated by $(H-1)$. In particular, any element in this group does square to zero.

Additionally, the "exterior product" isomorphism $f: \tilde K(X) \to \tilde K(X \wedge S^2)$, which is an isomorphism, is not a ring map: it takes an element $x$ to the exterior product $x \wedge (H-1)$. Instead, it satisfies $f(x) f(y) = (H-1) f(xy) = 0$. This is because the suspension is covered by two contractible open subsets, and so all products must vanish.

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  • $\begingroup$ The ideal of $\mathbb Z[H]/(H-1)^2$ generated by $H-1$, no? $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Jan 28, 2010 at 20:52
  • $\begingroup$ err, yes. I'm not sure why I wrote it like that. edited. $\endgroup$
    – Tyler Lawson
    Commented Jan 28, 2010 at 20:55
  • $\begingroup$ ah, so $S^0$ is really an exception! $\tilde{K}(S^{2n})\cong \mathbb{Z}$ for all $n\geq 0$ as a group and as a ring for $n=0$ but for $n>1$ the ring structure is trivial. This helps me much. Thank you. $\endgroup$
    – roger123
    Commented Jan 28, 2010 at 22:10
  • $\begingroup$ Hatcher says in 2.3. page 60 that $\tilde{K}(S^2)$ is $\mathbb{Z}[\alpha]/\alpha^2$. Because he pulls alpha back it has to be a vector bundle and not only a virtual one but what is $\alpha$ then? Isn't $H-1$ just a virtual bundle? $\endgroup$
    – roger123
    Commented Jan 29, 2010 at 14:44
  • $\begingroup$ On that page he actually says $K(S^2)$ has that form, not the reduced K-theory $\tilde K(S^2)$. In this case $\alpha$ is $H-1$, and in particular it's a virtual bundle rather than coming from an actual bundle. $H$ is the one that comes from an actual bundle. $\endgroup$
    – Tyler Lawson
    Commented Jan 29, 2010 at 18:23
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The map $K(X)\otimes K(X)\to K(X)$ induced by $\otimes$, and the composition $K(X)\otimes K(X)\to K(X\times X)\to K(X)$ of the external product and the diagonal map coicinde. Just seeing what both do to a pair of vector bundles shows this.

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