"High-dimensional" classes in topological $K$-theory

Question

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.

Answer 1

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 $n$.