Edit: In the exercise I tried to solve it wasn't stated explicitly that we were dealing with line bundles (though the usage of the symbol $L$ for the vector bundle was a strong suggestion that this was the case), hence why my ponderings in the question post are about vector bundles of arbitrary rank.
Let $M$ be a smooth manifold, $E$ a complex rank $r$ vector bundle on $M$. My definition of the first Chern class $c_1(E)$ of $E$ is
$$
c_1(E)=\left[\text{tr}\left(\frac{i}{2\pi}F^\nabla\right)\right] \in H_{\text{dR}}^2(M)
$$
where $F^\nabla$ is the curvature tensor corresponding to any connection $\nabla$ on $E$.
If $E$ admits a flat connection $\nabla$, then $F^\nabla=0$, and hence $c_1(E)=0$. I want to prove that the converse is true too: if $c_1(E)=0$, then $E$ admits a flat connection.
I know the identity $c_1(E)=c_1(\Lambda^r E)$. Since $\Lambda^r E$ is a line bundle, $c_1(\Lambda^r E)$ is simply the cohomology class of the curvature $F^{\nabla^{\Lambda^r E}}$, where $\nabla^E$ is some connection on $E$ and $\nabla^{\Lambda^r E}$ is the induced connection on $\Lambda^r E$. So $c_1(\Lambda^r E)=0$ means that $F^{\nabla^{\Lambda^r E}}$ is an exact $2$-form.
Maybe I should first show the case that if a complex line bundle $L$ has vanishing first Chern class, then $L$ admits a flat connection, and maybe I can then show that a flat connection on $\Lambda^r E$ is induced by a flat connection on $E$?
Any hints are welcome.
Context: all I know about characteristic classes is what's treated in chapter 5 of Tu's Differential Geometry book, which I've learned is not the modern way of doing things. And see below the exercise that I'm trying to solve.
Best Answer
If $L$ admits a flat connection then obviously $c_1(L)=0$. Conversely, suppose $c_1(L)=0$. Pick a connection $\nabla$ on $L$. Then, by definition, $F_\nabla=d\eta$ for some $1$-form $\eta\in\Omega^1(X)$. As you know from the course, the space of connections is an affine space, and we can define the connection $\nabla'=\nabla-\eta$. Say that the local connection $1$-forms are $A$ and $A'$ respectively. Then the local curvature $2$-form is $$F_{\nabla'}=dA'=d(A-\eta)=dA-d\eta=F_{\nabla}-d\eta=0$$ Hence $\nabla'$ is a flat connection on $L$.