= Solution
A <Poisson bivector> is a smooth antisymmetric contravariant two-tensor
$$
\omega=\frac12\omega^{ij}(x)\partial_i\wedge\partial_j
$$
whose bracket on smooth functions,
$$
\boxed{\{F,G\}=\omega(dF,dG)=\omega^{ij}\partial_iF\partial_jG,}
$$
satisfies the <Jacobi identity>. Bilinearity and antisymmetry are immediate, and the product rule makes the bracket a derivation in each argument. Together these properties define a <Poisson manifold>. Unlike a <symplectic form>, the <Poisson bivector> need not be nondegenerate.
Apply the <Jacobi identity> to the coordinate functions. Since $\{x^i,x^j\}=\omega^{ij}$,
$$
0=\{x^i,\{x^j,x^k\}\}+\{x^j,\{x^k,x^i\}\}+\{x^k,\{x^i,x^j\}\}
=\sum_m\left(\omega^{im}\partial_m\omega^{jk}
+\omega^{jm}\partial_m\omega^{ki}
+\omega^{km}\partial_m\omega^{ij}\right).
$$
Changing $\omega^{im}$ to $-\omega^{mi}$ and rearranging the three summands gives the printed <coordinate Jacobi condition for a Poisson bivector>:
$$
\boxed{\sum_m\left(\omega^{mi}\partial_m\omega^{jk}
+\omega^{mk}\partial_m\omega^{ij}
+\omega^{mj}\partial_m\omega^{ki}\right)=0.}
$$
This is also sufficient: expanding the Jacobiator of three arbitrary smooth functions, all terms involving second derivatives cancel in pairs by antisymmetry. The remaining coefficient of $\partial_iF\partial_jG\partial_kH$ is the coordinate-function Jacobiator displayed above.
For a <Lie algebra>, the natural global space carrying the proposed linear bracket is its <dual space> $\mathfrak g^*$. If $v^i$ is the chosen basis, define its linear coordinate function by $x^i(\ell)=\ell(v^i)$. The <Lie-Poisson bracket> is
$$
\boxed{\{F,G\}(\ell)=\ell([dF_\ell,dG_\ell])
=c^{ij}_kx^k\partial_iF\partial_jG,}
$$
where $dF_\ell,dG_\ell$ are elements of $\mathfrak g=(\mathfrak g^*)^*$. On a general manifold the same coordinate expression gives a local construction; a global one requires compatible transition rules. The use of $\mathfrak g^*$ supplies that compatibility intrinsically.
Here $\omega^{ij}=c^{ij}_rx^r$ and $\partial_m\omega^{jk}=c^{jk}_m$. The left side of the <coordinate Jacobi condition for a Poisson bivector> is therefore
$$
x^r\sum_m\left(c^{mi}_rc^{jk}_m+c^{mk}_rc^{ij}_m+c^{mj}_rc^{ki}_m\right)
=-x^r\sum_m\left(c^{jk}_mc^{im}_r+c^{ij}_mc^{km}_r+c^{ki}_mc^{jm}_r\right)=0.
$$
The final coefficient is the negative of the coefficient of $v^r$ in the <Lie algebra> identity $[v^i,[v^j,v^k]]+[v^j,[v^k,v^i]]+[v^k,[v^i,v^j]]=0$. Thus the <Lie-Poisson bracket> satisfies the <Jacobi identity>.
It remains to find the <Lie algebra structure constants> for the printed rotation fields. Distinguish their original spatial coordinates $(x,y,z)$ from the coordinates $(x^1,x^2,x^3)$ on the <dual space>. Use the conventional <Lie bracket of vector fields>. Their component vectors are $(0,-z,y)$, $(z,0,-x)$ and $(-y,x,0)$, respectively. For example,
$$
[v^1,v^2]=(y,-x,0)=-v^3.
$$
Similarly,
$$
\boxed{[v^2,v^3]=-v^1,\qquad [v^3,v^1]=-v^2,\qquad
c^{ij}_k=-\epsilon_{ijk}.}
$$
The negative sign is essential: these are the fundamental fields of a left rotation action with the stated <Lie bracket of vector fields>, and consequently have the <infinitesimal left-action sign convention> discussed above. The first field here has component $y\partial_z$, as printed in the PDF.
The <Lie-Poisson bracket> on the <dual space> consequently has
$$
\{x^1,x^2\}=-x^3,\qquad
\{x^2,x^3\}=-x^1,\qquad
\{x^3,x^1\}=-x^2.
$$
For the evolution convention $\dot F=\{F,H\}$, the <Hamiltonian function> has derivatives $(2Ax^1,2Bx^2,2Cx^3)$. Substitution gives the <quadratic rotational Lie-Poisson dynamics>
$$
\boxed{\begin{aligned}
\dot x^1&=2(C-B)x^2x^3,\\
\dot x^2&=2(A-C)x^3x^1,\\
\dot x^3&=2(B-A)x^1x^2.
\end{aligned}}
$$
These are Euler-type <Hamilton's equations> on a noncanonical <Poisson manifold>. When $A=1/(2I_1)$, $B=1/(2I_2)$ and $C=1/(2I_3)$ with positive principal inertias, they are the <Euler equations for a torque-free rigid body> in body angular-momentum coordinates. As a check, $H$ is conserved by antisymmetry of the <Poisson bracket>, and $C_0=(x^1)^2+(x^2)^2+(x^3)^2$ is a <Casimir function of a Poisson manifold>. Direct differentiation of $C_0$ in the three equations cancels the terms $4[(C-B)+(A-C)+(B-A)]x^1x^2x^3$. The <Hamiltonian flow> therefore lies on both an energy level and a sphere, a <symplectic leaf> of this signed rotational bracket.
Back to article page
