Vanishing of the first Chern class of a complex vector bundle Suppose that $E\to M$ is a $\mathbb{C}^n$-bundle with a metric. This is equivalent to saying that there exists a chart $\{U_\alpha\}$ of $M$ and $\phi_{\alpha,\beta} \colon U_\alpha\cap U_\beta\to  U(n)$ such that $\{\phi_{\alpha,\beta}\}$ satisfies a cocycle condition. (In this case, we say the structure group of $E$ is reduced to $U(n)$.)
Is the following true? If so, can you give a sketch of proof or give me a reference?
Question: The first Chern class $c_1(E)=0$ if and only if the structure group of $E$ can be reduced to $SU(n)$?
I know that if $L$ is a line bundle over $M$, then $c_1(L)=0$ if and only if $L$ is trivial. The above question seems to be a natural generalization. 
 A: This is an interesting question! The 'if' part is true and I think the 'only if' part is also (but this would need verification).
Suppose that $E$ is a $SU(n)$-bundle. Given any $SU(n)$-connection form on $E$, its curvature form $\Omega$ is a basic $\mathfrak{su}(n)$-valued 2-form on the principal $SU(n)$-bundle associated to $E$. As such, it can also be understood as a $\mathfrak{su}(n)$-valued closed 2-form on $M$. According to Chern-Weil theory, the first Chern class is given by the cohomology class of $\frac{i}{2\pi} \mathrm{tr}\,\Omega$, which vanishes since the matrices in $\mathfrak{su}(n)$ are traceless.
Suppose that $E$ is  $U(n)$-bundle with vanishing first Chern class. The structure group of $E$ can be reduced to $SU(n)$ if and only if there exists a $SU(n)$-principal subbundle in the $U(n)$-principal bundle $P$ associated to $E$, which is equivalent to the existence of an 'equivariant' function $f : P \to U(n)/SU(n)$ (with the canonical left $U(n)$-action on the quotient). In turn, this is equivalent to the existence of a section of the associated fiber bundle $P(U(n)/SU(n))$. Since $SU(n)$ is a normal subgroup of $U(n)$, we deduce that $Q = P(U(n)/SU(n))$ is a principal bundle, so it admits a section if and only if it is trivial. In fact, $U(n)/SU(n)$ being isomorphic to $U(1)$, it is a $U(1)$-principal bundle. Letting $U(1)$ act on $\mathbb{C}$ in the standard way, we can form the complex line bundle $F = Q(\mathbb{C})$. Thus, the reduction of the structure group of $E$ from $U(n)$ to $SU(n)$ is equivalent to the line bundle $F$ being trivial, that is, as you pointed out, equivalent to $F$ having a vanishing first Chern class.
I think, but I am not quite sure, that looking at how a connection form on $P$ induces a connection form on $Q$, we can infer that the vanishing of the first Chern class of $E$ implies the vanishing of the first Chern class of $F$. If true, the reduction to $SU(n)$ would be guaranteed.
