Solutions of Yang-Mills equation in the case of $G=U_1$ If $P\to M$ is a principal $U_1$-bundle, and $A$ is a connection on $P$, then it's curvature $F_A$ is a $2$-form with coeficient in $P\times_G\mathfrak{u}_1$, where $\mathfrak{u}_1$ is the Lie algebras of $U_1$.
In what I have learnt, I can write Yang-Mills equations as
$$
\begin{cases}
\mathrm{d}_AF_A=0\\
\mathrm{d}_A^*F_A=0
\end{cases}
$$
An exercise ask me to show that the curvature of $A$ can be identified as an element of $\Omega_M^2$, that is, without coefficient. Does that mean what I need to show is that $P\times_G\mathfrak{u}_1$ is a trivial bundle $M\times\Bbb{C}$? But how can I show this?
Furthermore, it let me show that $A$ is a Yang-Mills connection if and only if $F_A$ is a harmonic form, it's clear from Yang-Mills equation we have
$$
\Delta F_A=\mathrm{d}\mathrm{d}^*F_A+\mathrm{d}^*\mathrm{d}F_A=\mathrm{d}0+\mathrm{d}^*0=0
$$
But how does converse hold? Thanks in advance for anyone's help!
 A: The curvature form $F_A$ takes values in the bundle $\text{Ad}(P)= P\times_{G,\text{Ad}}\mathfrak g$.
But here $G=\text U(1)$ is abelian, so the Adjoint action on $\mathfrak u(1)$ is trivial, and hence $\text{Ad}(P)$ is trivial because all the transition functions $\text{Ad}(g_{ij})$ are trivial.
$$
\text{Ad}(P) \cong M\times \mathfrak u(1) = M \times i\mathbb R \quad \text{(or just $M\times \mathbb R$)}
$$
The Bianchi identity ($\mathrm d_A F_A = 0$, which is always true) and the Yang-Mills equation both involve the operator
$$\mathrm d_A: \Omega^*(M,\text{Ad}(P))\to\Omega^{*+1}(M,\text{Ad}(P))$$
which satisfies a formula which can be succinctly written down as:
$$
\mathrm d_A = \mathrm d + \text{ad}_A = \mathrm d + [A\wedge \cdot \, ]
$$
but again the $\text{ad}$ action is trivial. So $\mathrm d_A$ just reduces to $\mathrm d$ (on $M$) here.
(Some details are swept under the rug here, but it does work out!)
The second thing is from Hodge theory.
For a harmonic form $\omega\in \Omega^*(M)$, $~\Delta \omega = 0$, we have
$$
\begin{align}
0 &= \int_M \langle \Delta \omega, \omega \rangle ~\mathrm dx \\
&= \int_M \langle \mathrm d \mathrm d^* \omega,\omega\rangle 
+ \langle \mathrm d^* \mathrm d \omega, \omega \rangle ~\mathrm dx \\
&= \int_M |\mathrm d^* \omega|^2 
+ |\mathrm d \omega|^2 ~\mathrm dx \\
\end{align}
$$
implies that $\mathrm d^*\omega = \mathrm d \omega = 0$.
