The general reduced dimension is dim M minus two dim G

For $v\in {\mathbb{R}}^3$ put $A_v(u)=v\times u$. The coordinate formula for the cross product shows that $v\mapsto A_v$ is a vector-space isomorphism ${\mathbb{R}}^3\to \mathfrak{s}\mathfrak{o}(3)$, and the vector triple-product identity gives $[A_v,A_w]=A_{v\times w}$. Moreover $R{A}_v{R}^{-1}=A_{Rv}$ for $R\in SO(3)$. After identifying the dual by the Euclidean inner product, the definition in [F1] therefore gives ${\mathrm{Ad}}_R^{\ast }a=Ra$. In particular, the coadjoint stabilizer $G_{\alpha }$ of $\alpha =e_3$ is the circle of rotations about the $e_3$-axis, so $\dim G_{\alpha }=1$.

Let $(q,p)\in {\mu }^{-1}(\alpha )$. By [F2], $q\times p=e_3\mathrm{\ne }0$, so $q$ and $p$ are linearly independent. A rotation fixing the cotangent point $(q,p)$ fixes both vectors and hence is the identity. Thus the $SO(3)$-stabilizer of every point of the level is trivial. By [F5], $\alpha$ is a regular value, and the $G_{\alpha }$-action on the level is free. It is proper by [F3], and the level is nonempty because $e_1\times e_2=e_3$.

By step 1.1 the coadjoint stabilizer is one-dimensional, so by [F4] the reduced space has dimension $\dim M_{\alpha }=\dim M-\dim G-\dim G_{\alpha }=6-3-1=2.$

The claimed general formula would give $\dim M-2\dim G=6-6=0$, which contradicts the computed dimension $2$ of the reduced space for this regular value.

Since the correct general formula subtracts $\dim G+\dim G_{\alpha }$ and this nonzero coadjoint value has a proper stabilizer, the false statement fails; the zero-level formula $\dim M-2\dim G$ is a special case in which $G_{\alpha }=G$.