I am looking for a sequence of topological spaces $X\_n$, $n\in\mathbb N$, with the following property. Let $\tilde{K}^0(X\_n)$ be the complex reduced $K$-theory group of $X\_n$ (with respect to some choice of base point). I would like, for each $n$, for there to be a class $\xi\_n\in\tilde{K}^0(X\_n)$ such that $\xi\_n$ cannot be represented as the difference of two vector bundles of rank at most $n$.


**Question:** How could one choose/construct $X\_n$ and $\xi\_n$ for all $n$?


*Comments:* I thought one might be able to take $X\_n=S^n$ (say for $n$ even), which has a Bott vector bundle $\beta\_n$ whose dimension increases with $n$. The class of $\beta\_n-1$ generates $\tilde{K}^0(S^n)$, and I suspect it cannot be represented by the difference of two lower-rank bundles, but I'm not sure if this is true or how to show it.



"High-dimensional" classes in topological $K$-theory
Let $X\_n = S^{2n+2}$.


Since $\operatorname{ch} : K(S^{2n+2})\otimes\_{\mathbb{Z}}\mathbb{Q} \to H^{\text{even}}(S^{2n+2}; \mathbb{Q})$ is an isomorphism, there is a complex vector bundle $E \to S^{2n+2}$ of rank $n + 1$ with $\operatorname{ch}\_{n+1}(E) \neq 0$, i.e. $c\_{n+1}(E) \neq 0$. Let


$$\xi\_n = E - \varepsilon^{n+1} \in \widetilde{K}(S^{2n+2}).$$


Note that $c(\xi\_n) = c(E)c(\varepsilon^{n+1})^{-1} = c(E) = 1 + c\_{n+1}(E)$. On the other hand, if $F \to S^{2n+2}$ and $G \to S^{2n+2}$ are vector bundles of rank at most $n$, then $c(F - G) = c(F)c(G)^{-1} = 1$. Since $c\_{n+1}(E) \neq 0$, we see that $c(\xi\_n) \neq 1$, so $\xi\_n$ cannot be represented as the difference of two vector bundles of rank at most $n$.


