Vanishing Chern classes on vector bundle of $S^2$

Question

Suppose $E\to S^2$ be a complex vector bundle. If $c_1(E)=0$, does it imply that $E$ is a trivial bundle? And why if so? This question is motivated from Audin, Damian: Morse Theory and Floer Homology

Answer 1

Yes, complex bundles over $S^2$ are completely characterized by their first Chern class. This follows from the fact that $U(1) \rightarrow U(n)$ given by extending by the identity matrix induces an isomorphism on $\pi_1 = \mathbb{Z}$.

To see this implication, $[X,U(n)]$ classifies rank n complex bundles over $SX$, the suspension of $X$. When $X=S^1$ we have from above that $[S^1, U(n)]=\mathbb{Z}$, and it is not hard to see that this is exactly given by evaluating the first Chern class of the associated bundle over $S^2$, evaluated on the fundamental class.

Answer 2

This is basically the same as Connor's answer, but maybe exploiting the fact that $S^2$ is low-dimensional instead of the fact that $S^2$ is a suspension. Interpret the first Chern class $c_1 \in H^2(BU(n); \mathbb{Z})$ as a map $BU(n) \to K(\mathbb{Z},2)$. This map fits into a fiber sequence $$BSU(n) \to BU(n) \xrightarrow{c_1} K(\mathbb{Z},2).$$ So, if a vector bundle $E$ on $X$ has $c_1(E) = 0$, then its structure group can be reduced from $U(n)$ to $SU(n)$. But $BSU(n)$ is $3$-connected and $S^2$ is $2$-dimensional, so every map $S^2 \to BSU(n)$ is nullhomotopic, i.e., every $SU(n)$-bundle $E$ is trivial.