# 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!

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 $$\mathrm d^*\omega = \mathrm d \omega = 0$$.