Moment maps are unique without normalization

Given: ${\mathrm{A}\mathrm{C}}_{\omega }$, the cotangent bundle $T^{\ast }\mathbb{R}={\mathbb{R}}^2$ with canonical coordinates $(q,p)$, the translation action of $G=\mathbb{R}$ lifted to the cotangent bundle, and the tautological moment map.

${\mathrm{A}\mathrm{C}}_{\omega }$ is countable choice; it is used only through the fundamental-field and cotangent suppliers.

For the lifted action of a group acting on $Q$, the tautological map has components ${\mu }^{\xi }(q,p)=-p({\xi }_Q(q))$ and satisfies the component moment equations; the companion lemma proves its coadjoint equivariance, so it is an equivariant moment map. The cotangent lift of an action is Hamiltonian with the tautological moment map, The tautological cotangent moment map is equivariant.

For $Q=\mathbb{R}$ with the translation action, the fundamental field of $\xi =1$ is the constant field ${\xi }_Q=-{\mathrm{\partial }}_q$, the lifted action is $t\cdot (q,p)=(q+t,p)$, and the tautological moment map is $\mu (q,p)=p$. Fundamental vector fields for a left action, The cotangent lift of an action is Hamiltonian with the tautological moment map.

The coadjoint action of an abelian group is trivial, and for connected $M$ every translate $\mu +\delta$ of an equivariant moment map by a coadjoint-fixed covector is again an equivariant moment map. The coadjoint representation, action and orbits, Moment maps for one action form an affine space over coadjoint-fixed covectors.