I am reading Jost's Riemannian Geometry and Geometric Analysis (7-Ed) and having a question about the Yang-Mills functional (page 183).
Let $M$ be a compact manifold and $E$ be a metric bundle with dimension $n = 2$ whose structure group is $SO(2)$. For a metric connection $D$, its curvature $F_D = D\circ D$ can be viewed as a $Ad(E)$-valued 2-form where $Ad(E)$ means that each fiber is skew-symmetric. Since the Lie algebra $o(2)$ is trivial and the structure group is $SO(2)$, $Ad(E)$ is a trivial line bundle: $Ad(E) \simeq M \times \mathbb{R}$, meaning that $F_D$ is just an ordinary $2$-form (am I right?).
Jost showed that $F_D$ is harmonic if and only if $D$ is a Yang-Mills connection. Then he concluded that the existence and uniqueness of the curvature of a Yang–Mills connection are consequences of Hodge theory.
I don't understand this conclusion. Based on my understanding, the Hodge theory says that given a cohomology class $\mu \in H^p_{dR}(M,\mathbb{R})$, there exists a unique harmonic $p$-form $f$ such that $\mu = [f]$. Therefore, I would say that if $D_1$ and $D_2$ are two Yang-Mills connections such that the closed forms $F_{D_1}$ and $F_{D_2}$ are cohomologous, then $F_{D_1} = F_{D_2}$. Can we say or is there a theorem stating that the curvature is independent of the choice of Yang-Mills connections?
Thanks in advance.
As you have noticed, $\mathrm{Ad}(E)\cong M \times \mathbb{R}$. We also have $d_A=d$ in this case. Given any connection $1$-form $A$, you have $$ F_A=d_AA(=dA+[A,A])=dA $$ and the Yang-Mills equations become $$ d_A^*F_A=d^*F_A=0 $$ and the Bianchi-Identity (which is always true) $$ d_AF_A=dF_A=0. $$ This is a partial answer to your last question: Regardless of the choice of connection $A$, you have $d_A=d$, but not necessarily $d^*F_A=0$. Now, for a the curvature of a connection to solve the Yang-Mills equation is equivalent to the curvature, as a "normal" 2-form, being harmonic.
But now Hodge theory tells you that each cohomology class contains exactly(!) 1 harmonic form. This is existence and uniqueness. If the curvature satisfies the YM-equations, it has to be the harmonic representative of this cohomology class. On thing which you are maybe(?) not aware of is that $$ \Delta F_A=(d^*+d)^2F_A=0 \iff d^*F_A=0 \text{ and }dF_A=0. $$ You can prove this by identity by manipulating the integral $\int_M \langle F_A, \Delta F_A \rangle dvol_g$.