Stably isomorphic complex vector bundles which are not isomorphic

Solution 1:

Here's an example of a stably trivial rank two complex vector bundle $E \to S^5$ which is not trivial. In particular, this provides an example with $E_1 = E$, $E_2 = \varepsilon_{\mathbb{C}}^2$ (the trivial rank two complex vector bundle), and $n = 1$.

Viewing $S^5 \subset \mathbb{C}^3$, consider the bundle $T\mathbb{C}^3|_{S^5}$. The Euler vector field is a nowhere-zero section $\sigma$ of $T\mathbb{C}^3|_{S^5}$, so it spans a trivial complex line subbundle of $T\mathbb{C}^3|_{S^5}$. Therefore, there is a rank two complex subbundle $E$ of $T\mathbb{C}^3|_{S^5}$ such that $T\mathbb{C}^3|_{S^5} \cong E\oplus\varepsilon_{\mathbb{C}}^1$. As $T\mathbb{C}^3|_{S^5}$ is trivial, $E$ is stably trivial. On the other hand, as a real bundle, $E$ is a rank four subbundle of $TS^5$ and we have $TS^5 \cong E\oplus\varepsilon_{\mathbb{R}}^1$ where the trivial line bundle is spanned by $i\sigma$; an alternative description of $E$ is $TS^5\cap i(TS^5)$ where we view $TS^5$ as a subbundle of the complex vector bundle $T\mathbb{C}^3|_{S^5}$. If $E$ were trivial, then $TS^5$ would also be trivial, but $S^n$ is parallelisable if and only if $n \in \{0, 1, 3, 7\}$, so $E$ is non-trivial.

More generally, for every $n \geq 2$, we can construct a stably trivial rank $n$ complex vector bundle $E \to S^{2n+1}$ which is not trivial. Note however that the above argument for non-triviality fails when $n = 3$, so a separate argument is needed.