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For a principal bundle $(P,\nabla)\to M$ over a Riemann surface with fiber $G$, the Yang-Mills equation is $\nabla *F=0$, where $*F$ is dual of the curvature with respect to a fixed metric on the surface. Presumably this should imply that $*F$ (which can be thought of as a section of the adjoint bundle) is a fixed element in the center of $\mathfrak g$. I don't know why this is true. This is what I've done so far: in local coordinates writing $\nabla=\partial+A$, we have:

$F_{ij}=\partial_iA_j-\partial_jA_i+[A_i,A_j]$ (curvature in local corrdinates)

$\nabla_i F^{ij}=\partial_iF^{ij}+[A_i,F^{ij}]=0$.

Why should this imply $F^{ij}$ is in the center of Lie algebra? (standard reference for this result is Atiyah-Bott: Y-M equations on Riemann surfaces)

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    $\begingroup$ Apply the operator $\mathrm d_\nabla$ to obtain $F\wedge *F = 0$ (an equation in two-forms valued in the adjoint bundle). Applying the trace on the adjoint representation, we obtain $\lvert F\|^2=0$, where the scalar product on adjoint-valued $2$-forms is defined via the Killing form. If $G$ is compact, the Killing form is nonnegative with maximal isotropic subspace the center of $G$, from which the result readily follows. $\endgroup$
    – Bertram Arnold
    Commented Dec 19, 2020 at 12:26

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In general it is not in the center of $\frak g$. The curvature of a Yang-Mills connection in this dimension is parallel which means it commutes with the image of holonomy representation which need not be the whole of G. This is the statement in Atiyah-Bott.

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