Moment maps for one action form an affine space over coadjoint-fixed covectors

Fix $\xi \in \mathfrak{g}$. By [F1] and [F2] the two components ${\mu }_1^{\xi }$ and ${\mu }_2^{\xi }$ are Hamiltonian functions for the same vector field $-{\xi }_M$, namely the unique $X_{{\mu }_i^{\xi }}=-{\xi }_M$; hence ${\mu }_1^{\xi }-{\mu }_2^{\xi }$ is locally constant, and constant because $M$ is connected.

Conversely let $\mu$ be an equivariant moment map and $\delta \in ({\mathfrak{g}}^{\ast }{)}^G$. The components of $\mu +\delta$ are ${\mu }^{\xi }+\delta (\xi )$; adding the constant $\delta (\xi )$ changes no differential, so the component equations hold for $\mu +\delta$ by [F1]. Moreover $(\mu +\delta )(g\cdot p)=g\cdot \mu (p)+\delta =g\cdot \mu (p)+g\cdot \delta =g\cdot (\mu +\delta )(p)$ for all $g,p$, so $\mu +\delta$ is equivariant.

Define $\delta (\xi ):={\mu }_1^{\xi }-{\mu }_2^{\xi }$, a real number. Since $\xi \mapsto {\mu }_1^{\xi }-{\mu }_2^{\xi }$ is linear by [F1], the assignment $\delta :\mathfrak{g}\to \mathbb{R}$ is a linear functional, so $\delta \in {\mathfrak{g}}^{\ast }$ and ${\mu }_1(p)-{\mu }_2(p)=\delta$ for every $p\in M$.

Both maps are equivariant, so for all $g\in G$ and $p\in M$ $\delta ={\mu }_1(g\cdot p)-{\mu }_2(g\cdot p)=g\cdot {\mu }_1(p)-g\cdot {\mu }_2(p)=g\cdot \delta ,$ the last step by linearity of the coadjoint action. Hence $\delta \in ({\mathfrak{g}}^{\ast }{)}^G$.

Steps 1.1--3.1 show that any two equivariant moment maps differ by an element of $({\mathfrak{g}}^{\ast }{)}^G$, and step 1.2 shows that every translate by an element of $({\mathfrak{g}}^{\ast }{)}^G$ is again an equivariant moment map; hence the solution set is either empty or an affine space under $({\mathfrak{g}}^{\ast }{)}^G$.