Why curvature is equivariant as a moment map?

In Atiyah and Bott's famous paper "The Yang-Mills Equations over Riemann Surfaces", they treated curvature as a moment map of the gauge group acts on the connection space of a principal bundle $P$ over a compact Riemann surface $M$.

I know how to see the moment map is a weakly Hamiltonian. But I don't how they check the equivariant condition of it. I don't even know what is the coadjoint action of gauge group.

Is there some good understandings? Thanks!

• If $A$ is a connection, $F_A$ is its curvature, $g$ a gauge transformation then $F_{g\cdot A}= gF_Ag^{-1}$. This is seen immediately by working locally. Have a look at sec 8.1.1. of www3.nd.edu/~lnicolae/Lectures.pdf – Liviu Nicolaescu Jan 16 '16 at 11:46

A gauge transformation can be viewed as a section of $P \times_G G$, where $G$ acts via conjugation. Hence the Lie algebra $\mathfrak{gau}$ of infinitesimal gauge transformation is the space of sections of the adjoint bundle $P \times_G \mathfrak{g}$. Via integration the dual space $\mathfrak{gau}^*$ can be identified with the space of sections of the dual bundle; the coadjoint bundle $P \times_G \mathfrak{g}^*$. As usual, an $Ad$-invariant scalar product on $\mathfrak{g}$ yields an isomorphism between the adjoint and the coadjoint bundle. Hence $\mathfrak{gau}$ is self-dual and under this identification the coadjoint action of the gauge group corresponds to the adjoint action. In other words, you just lift the isomorphism $(\mathfrak{g}^*, Ad^*_G) = (\mathfrak{g}, Ad_G)$ to the corresponding bundles and then to the section spaces.
Now equivariance is rather obvious: the curvature (as a $2$-form with values in $Ad\,P$) transforms via the adjoint action under gauge transformations.