Angular momentum as the moment map for rotations of a cotangent bundle

For $\xi =({\xi }_1,{\xi }_2,{\xi }_3)$ let $\widehat{\xi }=\left(\begin{array}{ccc}0& -{\xi }_3& {\xi }_2\\ {\xi }_3& 0& -{\xi }_1\\ -{\xi }_2& {\xi }_1& 0\end{array}\right)\mathrm{.}$ Then $\widehat{\xi }u=\xi \times u$, and $\xi \mapsto \widehat{\xi }$ is a linear bijection ${\mathbb{R}}^3\to \mathfrak{s}\mathfrak{o}(3)$. Direct expansion of the coordinate cross product gives $[\widehat{\xi },\widehat{\eta }]=\widehat{\xi \times \eta }$ and $-\frac{1}{2}\mathrm{tr}(\widehat{\xi }\widehat{\eta })=\xi \cdot \eta$.

Expanding [F4] gives $p\cdot (\xi \times q)=p_1({\xi }_2{q}_3-{\xi }_3{q}_2)+p_2({\xi }_3{q}_1-{\xi }_1{q}_3)+p_3({\xi }_1{q}_2-{\xi }_2{q}_1)\mathrm{.}$ Grouping these six terms by ${\xi }_i$ gives $\xi \cdot (q\times p)$. The same six-term expansion identifies $w\cdot (u\times v)$ with the determinant whose columns are $w,u,v$. Thus the scalar triple-product identity is proved from the coordinate definition.

For $R\in SO(3)$, for every $w$ the determinant identity in step 1.2 gives $(Rw)\cdot ((R\xi )\times (Ru))=\det (R)\det (w,\xi ,u)=w\cdot (\xi \times u)$. Orthogonality also makes this last expression $(Rw)\cdot R(\xi \times u)$. As $Rw$ ranges over all vectors, nondegeneracy of the dot product gives $R(\xi \times u)=(R\xi )\times (Ru)$, hence ${\mathrm{Ad}}_R\widehat{\xi }=R\widehat{\xi }{R}^{-1}=\widehat{R\xi }$. The trace pairing of step 1.1 identifies $\mathfrak{s}\mathfrak{o}(3{)}^{\ast }$ with ${\mathbb{R}}^3$; under that identification the definition of the coadjoint action gives ${\mathrm{Ad}}_R^{\ast }v=Rv$, since $v\cdot R^{-1}\xi =(Rv)\cdot \xi$.

By [F3], the fundamental field of $\widehat{\xi }$ on ${\mathbb{R}}^3$ is ${\xi }_Q(q)={\frac{d}{dt}\mid }_0\exp (-t\widehat{\xi })q=-\widehat{\xi }q=-\xi \times q\mathrm{.}$ Substituting into [F2] gives ${\mu }^{\xi }(q,p)=-p(-\xi \times q)=p\cdot (\xi \times q)\mathrm{.}$

The scalar triple product identity proved in step 1.2, $p\cdot (\xi \times q)=\xi \cdot (q\times p)$ identifies this with the linear functional $\xi \mapsto \xi \cdot (q\times p)$; under the trace-pairing identification of step 2.1, the covector $\mu (q,p)\in \mathfrak{s}\mathfrak{o}(3{)}^{\ast }$ is therefore the vector $q\times p$.

The resulting map $\mu (q,p)=q\times p$ satisfies the component moment equations by [F2]. The cotangent lift of $q\mapsto Rq$ sends the covector represented by $p$ to the one represented by $Rp$, since $p\cdot R^{-1}v=(Rp)\cdot v$. Thus under the lifted action, $(Rq)\times (Rp)=R(q\times p)$, which is exactly the coadjoint action computed in step 2.1. Together with the component equations this proves equivariance and the moment-map assertion. If $q=0$, $p=0$ or the two vectors are parallel, the formula gives zero without any division or freeness assumption. The countable-choice assumption is exactly [A1].