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I've just been using Bott Periodicity to calculate the K-theory of some simple spaces - spheres, torus, and wedge of spheres. The wedge of spheres is interesting.

Given that $$\tilde{K}(X \vee Y) = \tilde{K}(X) \oplus \tilde{K}(Y)$$ we have that $$\tilde{K}(S^n \vee S^m) = \begin{cases} \mathbb{Z} \oplus \mathbb{Z} & m,n \text{ even} \\ \mathbb{Z} & \text{one of } m,n \text{ even}, \\ 0 &m,n \text{ odd.}\end{cases}$$

The ring structure is trivial in all cases.

Switching to unreduced K-theory we have $$K(S^n \vee S^m) = \begin{cases} \mathbb{Z} \oplus \mathbb{Z}\oplus \mathbb{Z} & m,n \text{ even} \\ \mathbb{Z} \oplus \mathbb{Z} & \text{one of } m,n \text{ even}, \\ \mathbb{Z} &m,n \text{ odd.}\end{cases}$$

If both are odd, or one is odd, I can see what the ring structure is. I wonder what it is when they are both even. $K(S^{2n})$ has ring structure $\mathbb{Z}[H]/(H-1)^2$ and $\tilde{K}(S^{2n})$ is generated by $(H-1)$ and has trivial multiplication.

Analogously I guess the ring structure on the wedge is something like $$\frac{\mathbb{Z}[H,H']}{((H-1)^2,(H'-1)^2)},$$ where $H,H'$ generate $\tilde{K}(S^n),\tilde{K}(S^m)$, but I guess I am not really sure!

(I am tagging ring theory, since it is probably a standard result in algebra, but as usual feel free to re-tag)

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up vote 4 down vote accepted

There is completely general fact: $K(X\vee Y)=K(X)\oplus_{K(pt)}K(Y)$. It means that $K(X\vee Y)=\tilde K(X)\oplus K(pt)\oplus\tilde K(Y)$ and multiplication

  1. between first two summands is the same as in $K(X)$,
  2. between last two summands is the same as in $K(Y)$,
  3. between $\tilde K(X)$ and $\tilde K(Y)$ is trivial.

And this is true for any cohomology theory.

In particular, $K^\bullet(S^n\vee S^m)=H^{\bullet}(S^n\vee S^m)\otimes K^{\bullet}(pt)$ (and one can choose one's favorite generators and relations).


Now, the answer from the last paragraph of OP. Firstly, $\frac{\mathbb{Z}[H,H']}{((H-1)^2,(H'-1)^2)}$ can't be quite right, since additively it is $\mathbb Z^4$ (and not $\mathbb Z^3$) — what is missing, is the relation (3): the right answer is $\frac{\mathbb{Z}[H,H']}{((H-1)^2,(H'-1)^2,(H-1)(H'-1))}$

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thank you for the answer. I am not sure I fully understand it. I think this leads to a ring structure something like I have written originally? Or do I need an additional generator for $K(\text{pt})$? Hmm.... –  Juan S Jun 9 '11 at 7:22
    
got it now, thank you! –  Juan S Jun 9 '11 at 23:26
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