I put all the definitions I am using at the bottom.
We consider a principal U(1)-bundle $\pi: E \rightarrow M$ with Lie-algebra valued connection $A \in \Omega^1(E, \mathfrak{g}) = \Omega^1(E, \mathfrak{u(1)}) = \Omega^1(E, i\mathbb{R})$ and it's associated curvature form $F_A \in \Omega^2(E,\mathfrak{g})$. As explained in the last point below (in definitions), that form can be identified with a 2-form $\Omega_A \in \Omega^2(M, Ad(E))$. Because U(1) is commutative $Ad(E) = M \times \mathfrak{u(1)}$ is the trivial bundle, and we can identify $\Omega^2(M, Ad(E))= \Omega^2(M,\mathfrak{u(1)}) = \Omega^2(M,i \mathbb{R})$
The second bianchi identity: $$ dF_A = [F_A,A] \Leftrightarrow dF_A+ [A,F_A] = dF_A + A \wedge F_A = 0$$ is equivalent to: $$ d_\nabla \Omega_{A_p} = [\phi, d_A F_{A_\phi}\circ Hor_\phi] = 0 $$ (with $p\in M, \phi \in E_p$) We can thus say, that $\Omega_A$ is closed.
We define the first Chern class as: $$ c_1(E) = [\frac{i}{2\pi} \Omega_A] \in H^2_{dR}(M) $$ Here is my first confusion: Since we only define de Rham cohomology for the ordinary exterior derivative operator and not for the covariant one. But I think we can assume here that we are working with the trivial connection on the trivial bundle $Ad(E) = M \times \mathfrak{g} \rightarrow M$ and thus $\nabla_X = \mathcal{L}_x \Rightarrow d_\nabla = d$. What do you think about that or am I missing something?
$\underline{\textbf{The actual question:}}$
Now we want to show, that for a closed form $\omega \in \Omega^2(M)$: $ [\omega] \in c_1(E)$ implies that $\pi: E\rightarrow M$ admits a principal conection whose curvature 2-form is $\frac{2\pi}{i} \omega $
I don't quite know what I have to show here...
I know that $\omega = \frac{i}{2\pi}\Omega_A+ d\alpha$ for a 1-form $\alpha \in \Omega^1(M)$. That implies that $\frac{2\pi}{i} \omega$ takes values in $i\mathbb{R} = \mathfrak{u(1)}$ and we can thus identify it with $\omega \in \Omega^2(M, Ad(E))$, which determines a unique curvature 2-form $\tilde{F}_A \in \Omega_\rho^2(E, \mathfrak{u(1)})$ via the correspondence explained below in the definitions
Is there anything else to show???
Thank in adance!
Show that for a closed 2-form
$\underline{\textbf{Definitions}}$:
$\underline{\text{Connection 1-form on fiber bundles without G-structure:}} $
The connection map $K \in \Omega^1(E, VE)$ has to satisfy $K \vert VE = Id$ and defines the horizontal subbundle HE $\subset TE$ by $HE = ker (K)$, whose fibers at each point $x \in E $ are complementary to the one of VE ie $T_xE= V_xE \oplus H_xE$.
$\underline{\text{Connection 1-form on fiber bundles with G-structure (Principal bundles):}} $
We require additionally that the resulting parallel transport maps have to respect the G-structure ie they have to be G-equivariant. We showed that this is equivalent to the push-forward of the Right-translation preserving the horizontal subbundle ie $TR_g(HE) = HE $ $\forall g \in G$
$\underline{\text{Lie-Algebra valued connection 1-form on principal bundles:}} $
Since the group G acts by definition freely and transitively on the fibers of our principal bundle E, the vertical subbundle $V_\phi E \subset T_\phi E$ is isomorphic to $\mathfrak{g}$ $\forall \phi \in E$ (via the fundamental vector field $\mathfrak{g} \rightarrow V_\phi E, X \mapsto X^F(\phi)$). That makes it possible to express the connection map $K \in \Omega^1(E, VE)$ as a Lie-Algebra valued 1-form $A \in \Omega^1(E, \mathfrak{g})$.
The conditions on A to define a principal connections are
$(i)$ $A(X^F(\phi)) = X$
$(ii)$ $R^*_g(A) = Ad_{g^{-1}} \circ A $
$\underline{\text{Lie algebra valued curvature 2-form}} $
The corresponding curvature 2-forms $F_K \in \Omega^2(E,VE)$ and $F_A \in \Omega^2(E,\mathfrak{g})$ are defined as: $F_K(X,Y) = -K[H(X),H(Y)]$ and $F_A(X,Y) = -A[H(X),H(Y)]$ with $X,Y \in TE$ where H denotes the complementary projection to K that gives the horizontal part of a vector in TE. (We showed that these forms vanish iff the connection on the bundle admits flat sections around the neighborhood of every point in the bundle.)
$\underline{\text{Associated bundles:}} $
Given a principal bundle $ \pi: E \rightarrow M$, another Manifold F and a smooth Group action $\rho : G \times F \rightarrow F, (g,x) \mapsto gx$, we can define the associated bundle $\pi^\rho: E^\rho = E \times_{\rho} F :=(E \times F)/G \rightarrow M$ where elements in $E^\rho$ are identified by $[\phi, x] = [\phi g^{-1}, gx] \in (E\times F)/G = E^\rho$ (thats how we define the group action on the product). This construction ensures that $\pi^\rho: E^\rho \rightarrow M$ has the same transition functions as $\pi: E \rightarrow M$ and has standard fiber F.
$\underline{\text{Connection and curvature on the adjoint bundle:}} $
Given any linear representation $\rho : G \rightarrow GL(V)$ of our group on a vector space V, we can "project" horizontal and $\rho$ -equivariant forms $\hat{\omega} \in \Omega_\rho ^k(E, V) \subset \Omega^k(E, V) $ down to the tangent space ie it is isomorphic to uniquely determined forms $\omega \in \Omega^k(M, E^\rho)$. They are related by: $\omega_p(X_1,...,X_k) = [\phi, \hat{\omega}_\phi(Hor_\phi(X_1), ... , Hor_\phi(X_1))]$ with $p \in M, \phi \in E_p$.
The curvature 2-form $F_A \in \Omega^1(E,\mathfrak{g})$ is by definition horizontal and ad-invariant. We can use the above correspondence to express it as a form on the tangent space of M: $\Omega_A \in \Omega^2(M, E^\rho = Ad(E)) $ where $Ad(E) = E \times_{Ad} \mathfrak{g}$
In the case of a commutative structure group, the adjoint representation becomes trivial. Elements in the adjoint bundle are thus identified by $[\phi, x] = [\phi g^{-1}, gx] = [\phi g^{-1}, x] \in Ad(E) = E \times_{Ad} \mathfrak{g}$ with $\phi \in E, x \in \mathfrak{g}$. That means that $\pi^{Ad}: Ad(E) = M \times \mathfrak{g} \rightarrow M$ is the trivial bundle.
$\underline{\text{Exterior covariant derivative and covariant derivative on associated bundle:}}$
The covariant exterior derivative is defined on bundle-valued forms. It is analogous to the ordinary exterior derivative taking into account that the bundle-valued output is not canonically constant, but we have to chose a connection. It can be shown, that via the correspondence described above: $$ d_\nabla \omega_p = [\phi, (d\circ H) \hat{\omega}_\phi] $$ for $p\in M, \phi \in E_p$ and $\omega \in \Omega^k(M, E^\rho), \hat{\omega} \in \Omega^k_\rho (E,V)$.
It turns out that: $d\circ H = d +A\wedge () := d_A$
Just in case that you feel the urge to accept an answer and have the feeling that my comments helped you, here is one:
You question boils down into the problem of showing a formula of the kind $$ \Omega^{\nabla+\omega}~"="~\Omega^\nabla+\text{d}\omega $$ or alike, for a covariant derivative $\nabla$ and an $\text{End}(E)$-valued 1-form $\omega$ (for which $\nabla+\omega$ is again a covariant derivative). Here I put the $=$ in quotation marks since I'm not entirely sure if it really holds like that, however, a quick computation evaluating the LHS gives me the RHS plus something which I'd expect to vanish, but I'm not completely sure.
However, once you have shown this formula, it is to be interpreted as $\Omega^{(\cdot)}$ is a map of the affine spaces of all covariant derivatives and the Chern class aka. de Rham class of $\Omega^\nabla$.