The shifting trick for a nonzero coadjoint orbit

Let $a_r$ denote the outward Euclidean area form on $S_r^2$: $a_{r,\beta }(u,v)=(\beta \mathrm{/}r)\cdot (u\times v)$ for tangent vectors $u,v$. Since $d{D}_r$ multiplies both tangent vectors by $r$, $D_r^{\ast }{a}_r=r^2{\omega }_{S^2}$. Comparing with [F1] gives ${\omega }_r=r^{-1}{a}_r$. Thus the product uses the negative of this KKS form, not negative $r{a}_r$.

The stabilizer of $\alpha$ is the rotation group of its perpendicular plane, hence isomorphic to $SO(2)$, compact and of dimension one. The level is nonempty: choose a unit $q\perp \alpha$ and put $p=\alpha \times q$, giving $q\times p=\alpha$ by the vector triple-product identity. At every point of the level, $q,p$ are independent. The differential is $d{\mu }_{(q,p)}(u,v)=u\times p+q\times v$. If $\xi$ annihilates its image, the scalar triple-product identity gives $p\times \xi =0$ and $\xi \times q=0$, so $\xi =0$; finite-dimensional duality proves surjectivity. A rotation fixing $q,p$ also fixes $q\times p$, so fixes a basis and is identity. Thus the stabilizer action is free. For any compact group $C$ acting on a Hausdorff manifold $L$, the inverse image of a compact set $D\subseteq L\times L$ under $(g,x)\mapsto (gx,x)$ is closed in $C\times {\mathrm{pr}}_{2}D$, hence compact. This proves properness here. All hypotheses of [F4] hold, giving $\dim M_{\alpha }=6-3-1=2$.

By [F2] the product moment map is $\mathrm{\Psi }=\mu -\beta$; its zero level consists of the pairs $(q,p,\beta )$ with $q\times p=\beta \in S_r^2$, and the diagonal action makes this zero level the equivariant image of the saturated level ${\mu }^{-1}(S_r^2)$.

At any shifted zero-level point, $q\times p=\beta \mathrm{\ne }0$, so the same derivative calculation as step 1.2 shows that $d\mathrm{\Psi }(u,v,0)=d\mu (u,v)$ is surjective. Thus zero is regular. A diagonal stabilizer fixes $q,p$ and is identity by the same basis argument. The action is proper by the compact-group argument in step 1.2, since $SO(3)$ is compact (it is closed and bounded in matrix space). The shifted level is nonempty by the point constructed in step 1.2 together with $\beta =\alpha$. Since the product dimension is $6+2=8$ and the group dimension is three, [F5] gives reduced dimension $8-2\cdot 3=2$.

Consequently the two-dimensional reduced manifold $M_{\alpha }$ is exhibited as the zero reduction of $M\times S_r^2{}^{-}$. On the level ${\mu }^{-1}(\alpha )$ the slice map is $(q,p)\mapsto (q,p,\alpha )=(q,p,q\times p)$, and [F2] identifies its quotient by $G_{\alpha }$ with the shifted zero quotient by $SO(3)$; all hypotheses for the symplectomorphism in [F2] have been verified in steps 1.2 and 3.1. The two reduced forms agree because their pullbacks to this slice are both the restriction of the canonical form on $T^{\ast }{\mathbb{R}}^3$: the orbit coordinate is constant on the slice, so its form pulls back to zero. The excluded value $\alpha =0$ has a point orbit and does not meet the regular/free argument used here.