= Solution
A <Poisson manifold> is a smooth manifold $M$ with a bilinear bracket on smooth functions that is antisymmetric, obeys the Leibniz rule and satisfies the <Jacobi identity>. Equivalently, a <Poisson tensor> $\pi$ gives $\{f,g\}=\pi(df,dg)$. Its rank need not be constant or maximal. Here define Hamiltonian evolution by $X_Hf=\{f,H\}$, agreeing with the canonical bracket $\{q^i,p_j\}=\delta^i_j$.
A smooth <Lie group action> sends each $x$ along its orbit $Gx$, with <stabilizer> $G_x$. If the action is free and proper, the orbit space is a smooth quotient manifold and the projection is a submersion. Without these hypotheses the quotient can have singular orbit-type strata, or even fail to be Hausdorff. For an action preserving the <Poisson bracket>, invariant functions form a Poisson subalgebra. On a smooth quotient $M/G$, their identification with smooth quotient functions defines
$$
\{\bar f,\bar g\}_{M/G}\circ r=\{\bar f\circ r,\bar g\circ r\}_M.
$$
Thus $r$ is a <Poisson map>. The quotient is generally Poisson, not necessarily symplectic: different reduced <momentum> sectors can be different leaves. An arbitrary smooth <group action> alone does not guarantee this bracket descent.
For a <Lie group> $G$, conjugation $h\mapsto ghg^{-1}$ differentiates at the identity to the <Adjoint representation of a Lie group>, $\operatorname{Ad}_g:\mathfrak g\to\mathfrak g$. Its infinitesimal version is $\operatorname{ad}_\xi\eta=[\xi,\eta]$. The <coadjoint representation> on the dual space is defined by
$$
\langle\operatorname{Ad}^*_g\ell,\xi\rangle=\langle\ell,\operatorname{Ad}_{g^{-1}}\xi\rangle,\qquad\langle\operatorname{ad}^*_\xi\ell,\eta\rangle=-\langle\ell,[\xi,\eta]\rangle.
$$
The inverse in the first formula makes this a left action. These definitions distinguish the representation on infinitesimal generators from its dual action on momenta.
There is a natural positive <Lie-Poisson bracket> on $\mathfrak g^*$:
$$
\boxed{\{F,G\}_+(\ell)=\langle\ell,[dF_\ell,dG_\ell]\rangle.}
$$
The differentials lie in $(\mathfrak g^*)^*\simeq\mathfrak g$. In a basis with $[e_a,e_b]=c_{ab}{}^ce_c$, linear coordinates satisfy $\{\ell_a,\ell_b\}_+=c_{ab}{}^c\ell_c$. The bracket extends by the chain rule and Leibniz rule to smooth functions. Its coordinate Jacobi condition reduces precisely to the <Jacobi identity> for $c_{ab}{}^c$, proving that it is Poisson. Reversing its overall sign also gives a <Poisson bracket> and is useful for body-frame conventions below.
On a general <Poisson manifold>, the vectors $X_f$ span a distribution. The <Jacobi identity> closes this family under commutators, and its orbits under Hamiltonian flows form the <symplectic foliation>. On each leaf the induced two-form satisfies
$$
\omega(X_f,X_g)=\{f,g\}.
$$
It is well-defined after quotienting out differential covectors giving zero <Hamiltonian vector field>, is nondegenerate on the leaf, and is closed by the <Jacobi identity>. Leaf dimensions can vary, so ordinary constant-rank foliation terminology must be qualified. A Hamiltonian trajectory remains in one leaf; a <Casimir function of a Poisson manifold> is constant along all such trajectories, but its level set can contain more than one leaf.
For $\mathfrak g^*$, $X_F(\ell)=\operatorname{ad}^*_{dF_\ell}\ell$ under the convention above. Hence the connected <coadjoint orbits> are exactly its <symplectic leaves> for connected $G$; for disconnected $G$, take connected orbit components. The orbit two-form is the <Kirillov–Kostant–Souriau symplectic form>:
$$
\boxed{\omega_\ell(\operatorname{ad}^*_\xi\ell,\operatorname{ad}^*_\eta\ell)=\langle\ell,[\xi,\eta]\rangle.}
$$
If a generator is in the <stabilizer>, its pairing with every bracket vanishes, proving independence of the choice of tangent generator. Conversely, vanishing pairing with all tangent generators puts it in that <stabilizer>, proving nondegeneracy. The cyclic derivative relation giving $d\omega=0$ is the <Lie algebra> <Jacobi identity>. This explains the symplectic structure of the orbit rather than just naming it.
Identify the <SO(3) Lie algebra> with $\mathbb R^3$ by the hat map $\widehat{\boldsymbol\xi}\mathbf v=\boldsymbol\xi\times\mathbf v$. Its Lie bracket becomes the vector cross product, and the invariant Euclidean pairing identifies the dual with angular-momentum vectors $\mathbf M$. The <coadjoint action> is ordinary rotation, with infinitesimal action $\boldsymbol\xi\times\mathbf M$. Therefore its nonzero orbits are spheres $|\mathbf M|=r$; the origin is a zero-dimensional leaf. In these coordinates,
$$
\{F,G\}_+=\mathbf M\cdot(\nabla F\times\nabla G),\qquad C(\mathbf M)=|\mathbf M|^2
$$
is a <Casimir function of a Poisson manifold>, since its gradient is parallel to $\mathbf M$. For tangent vectors $\mathbf v,\mathbf w$ to a nonzero sphere, the positive orbit form is
$$
\omega_{\mathbf M}(\mathbf v,\mathbf w)=\frac{\mathbf M\cdot(\mathbf v\times\mathbf w)}{r^2}.
$$
Its area integral with the corresponding orientation is $4\pi r$, and it reverses sign with the negative bracket.
For a torque-free <rigid body>, choose body <angular momentum> $\mathbf M$ and principal moments $I_1,I_2,I_3>0$. Reduction of $T^*SO(3)$ by spatial rotations gives the body-frame negative <Lie-Poisson bracket> and Hamiltonian
$$
\{F,G\}_{\mathrm{body}}=-\mathbf M\cdot(\nabla F\times\nabla G),\qquad H=\frac12\sum_{i=1}^3\frac{M_i^2}{I_i}.
$$
Keeping the sign convention explicit is essential: using the positive bracket with the same body identification would reverse the Euler evolution. The negative bracket gives
$$
\boxed{\dot{\mathbf M}=\mathbf M\times\boldsymbol\Omega,\qquad\Omega_i=M_i/I_i,\qquad\dot M_1=(I_3^{-1}-I_2^{-1})M_2M_3,}
$$
with cyclic expressions for the other two components. These are the <Euler equations for a torque-free rigid body>. The Hamiltonian and $C$ are conserved because $\boldsymbol\Omega\cdot(\mathbf M\times\boldsymbol\Omega)=0$ and $\mathbf M\cdot(\mathbf M\times\boldsymbol\Omega)=0$. Trajectories therefore lie on intersections of energy ellipsoids with <momentum> spheres, the latter being the <symplectic leaves>.
The body's orientation is reconstructed from $\dot R=R\widehat{\boldsymbol\Omega}$. Its spatial <momentum> $\mathbf J=R\mathbf M$ is constant, since $\dot{\mathbf J}=R(\boldsymbol\Omega\times\mathbf M+\mathbf M\times\boldsymbol\Omega)=0$. Thus the body <momentum> need not be constant even though the spatial conserved <momentum> is. \b[Poisson reduction retains the dynamics and its <momentum> leaves while eliminating orientation variables; the unreduced motion is recovered by reconstruction.]
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