Let $\Sigma$ be any Riemann surface, and let $L \rightarrow \Sigma$ be a complex line bundle (which is classified according to its degree). Then the vector bundle $L \oplus L^{-1} \rightarrow \Sigma$ is a rank $2$ complex vector bundle over $\Sigma$, and an easy computation shows that it has trivial Chern class. But, is it true that this vector bundle is isomorphic to the trivial vector bundle over $\Sigma$ of rank $2$?
Thanks in advance.
Let $E$ be a smooth real vector bundle on a compact smooth manifold $X$. Suppose $\operatorname{rank}E > \operatorname{dim}X$, then $E$ admits a nowhere zero section $\sigma$. The span of $\sigma$ defines a trivial real line subbundle $\varepsilon^1$, so we can write $E = E'\oplus\varepsilon^1$ where $\operatorname{rank}E' = \operatorname{rank}E - 1$. If $E$ is endowed with the structure of a complex vector bundle, then the complex span of $\sigma$ defines a trivial complex line subbundle $\varepsilon^1_{\mathbb{C}}$, so we can write $E = E''\oplus\varepsilon^1_{\mathbb{C}}$ where $\operatorname{rank}_{\mathbb{C}}E'' = \operatorname{rank}_{\mathbb{C}}E - 1$.
As $L\oplus L^*$ is a smooth complex vector bundle of real rank four, and $\Sigma$ is a compact Riemann surface or real dimension two, $L\oplus L^* \cong K\oplus\varepsilon^1_{\mathbb{C}}$ where $K$ is a complex line bundle. Note that $$c_1(K) = c_1(K\oplus\varepsilon^1_{\mathbb{C}}) = c_1(L\oplus L^*) = c_1(L) + c_1(L^*) = c_1(L) - c_1(L) = 0.$$
As $c_1 : \operatorname{Vect}^1_{\mathbb{C}}(X) \to H^2(X; \mathbb{Z})$ is an isomorphism (see Proposition $3.10$ of Hatcher's Vector Bundles and K-Theory), $K$ is trivial. Therefore,
$$L\oplus L^* \cong K\oplus\varepsilon^1_{\mathbb{C}} \cong \varepsilon^1_{\mathbb{C}}\oplus\varepsilon^1_{\mathbb{C}} \cong \varepsilon^2_{\mathbb{C}}.$$
That is, $L\oplus L^*$ is trivial.
Note, the exact same proof works if we replace $\Sigma$ by a compact smooth three-dimensional manifold.