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I hope this isn't too narrowly focused. I have a question concerning the inverse of the Bott map as defined in Atiyah's paper, Bott Periodicity and the Index of Elliptic Operators. On page 122 he defines it as the composition

$$K^{-2}(X) \to K(S^2 \times X) \stackrel{\text{index} \bar\partial}{\longrightarrow} K(X).$$

My question is: what is the first map? At first I thought it was $$ K^{-2}(X) = \tilde K(S^2 \wedge X^+) \to \tilde K(S^2 \times X^+) \to \tilde K(S^2\times X) \subset K(S^2 \times X) $$ where the first map is the pullback of the projection and the second is the pullback of the inclusion. But from the 6 term exact sequence, for this first map to be injective (which it needs to be for this to be correct) we need $$ K(S^2 \times X^+, S^2 \times X) \to \tilde K(S^2 \times X^+) $$ to be the zero map which doesn't seem correct since $S^2 \times X^+/S^2 \times X$ is identifiable with $S^2 \sqcup \{pt\}$ and if we pullback a non-trivial bundle over this space to $S^2 \times X^+$ it seems like it won't trivialize.

Thanks!

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2 Answers 2

up vote 2 down vote accepted

Your argument against your definition is not convincing. You say you are writing down an exact sequence for your first map, but in fact you write down one for your second map. The second map is not necessarily injective. We only need it to be injective on the image of the first map.

The definition you write looks fine to me. (A check you could make is whether using it you can establish the three axioms for the composite map that Atiyah gives).

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Ah thanks. I don't know why I made the mistake of thinking the map $\tilde K(S^2\times X^+) \to \tilde K(S^2 \times X)$ had to be injective. –  Eric O. Korman Apr 16 '11 at 20:21

Actually I think I found a better way to look at the map, which shows that it is injective. Consider the pair $(S^2 \times X, \{pt\} \times X)$. Then since $\{pt\} \times X$ is a retract of $S^2 \times X$, the 6-term sequence in $K$-theory splits to give a short exact sequence $$ 0 \to K(S^2 \times X, \{pt\} \times X) \to K(S^2 \times X) \to K(\{pt\} \times X). $$ But $(S^2 \times X) / (pt \times X) \simeq S^2 \wedge X^+$, giving the desired map.

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