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A nondegenerate Poisson bivector inverts to a closed differential form of degree two. Differentiating $S=P^{-1}$ gives ${\partial }_{d}S=-S({\partial }_{d}P)S$. Contracting the coordinate Jacobi condition for a Poisson bivector with three copies of $S$ yields ${\partial }_i{S}_{jk}+{\partial }_j{S}_{ki}+{\partial }_k{S}_{ij}=0$, which is precisely closure of the resulting symplectic form. An overall sign in the inverse convention does not change closure.

The Jacobi identity for a Poisson bivector reduces to the displayed cyclic identity in every local chart. Necessity follows by applying the bracket to coordinate functions. Sufficiency follows because the second-derivative terms in an arbitrary Jacobiator cancel by antisymmetry, leaving this coefficient multiplying first derivatives of the three functions.

Let $P=({\omega }^{ab})$ be a smooth antisymmetric matrix field. It defines a Poisson bivector throughThis Poisson bracket is bilinear, antisymmetric and a derivation in each argument. It defines a Poisson manifold when it also satisfies the Jacobi identity. Apply that identity in the form $\{\{f,g\},h\}+\{\{g,h\},f\}+\{\{h,f\},g\}=0$ to the coordinate functions. Since $\{x^a,x^b\}={\omega }^{ab}$, it givesThis is the coordinate Jacobi condition for a Poisson bivector. It is also sufficient: in the Jacobi identity for arbitrary functions the terms containing second derivatives cancel by antisymmetry, leaving $J^{abc}({\partial }_{a}f)({\partial }_{b}g)({\partial }_{c}h)$. Thus the coordinate condition captures the whole obstruction.

Now suppose the Poisson bivector is nondegenerate. Write $S=P^{-1}$, so $S_{ab}{\omega }^{bc}={\delta }_a^c$, and define the 2-formAn overall minus sign in identifying the symplectic form depends on the convention for Hamiltonian vector fields; it does not affect the closure argument. Differentiating the inverse matrix givesContract the coordinate Jacobi identity with $S_{ia}{S}_{jb}{S}_{kc}$. Antisymmetry gives ${\omega }^{dc}{S}_{kc}=-{\delta }_k^d$, and similarly for the other terms, henceThe cyclic expression is precisely the coefficient of the exterior derivative $d\Omega$. ThereforeSince $S$ is antisymmetric and nondegenerate, $\Omega$ is a symplectic form. This proves the closure of the inverse of a nondegenerate Poisson bivector. The matrix entries ${\omega }^{ab}$ are scalar functions; the codomain in the source's matrix description should be read as the space of antisymmetric matrices, rather than a vector-valued individual entry.

A Poisson bivector is a smooth antisymmetric contravariant two-tensorwhose bracket on smooth functions,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 $\{x^i,x^j\}={\omega }^{ij}$,Changing ${\omega }^{im}$ to $-{\omega }^{mi}$ and rearranging the three summands gives the printed coordinate Jacobi condition for a Poisson bivector: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 ${\partial }_{i}F{\partial }_{j}G{\partial }_{k}H$ is the coordinate-function Jacobiator displayed above.

For a Lie algebra, the natural global space carrying the proposed linear bracket is its dual space ${\mathfrak{g}}^{\ast }$. If $v^i$ is the chosen basis, define its linear coordinate function by $x^i(\ell )=\ell (v^i)$. The Lie-Poisson bracket iswhere $d{F}_{\ell },d{G}_{\ell }$ are elements of $\mathfrak{g}=({\mathfrak{g}}^{\ast }{)}^{\ast }$. On a general manifold the same coordinate expression gives a local construction; a global one requires compatible transition rules. The use of ${\mathfrak{g}}^{\ast }$ supplies that compatibility intrinsically.

Here ${\omega }^{ij}=c_r^{ij}{x}^r$ and ${\partial }_m{\omega }^{jk}=c_m^{jk}$. The left side of the coordinate Jacobi condition for a Poisson bivector is thereforeThe 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,Similarly,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 hasFor the evolution convention $\dot{F}=\{F,H\}$, the Hamiltonian function has derivatives $(2A{x}^1,2B{x}^2,2C{x}^3)$. Substitution gives the quadratic rotational Lie-Poisson dynamicsThese 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 $4[(C-B)+(A-C)+(B-A)]x^1{x}^2{x}^3$. The Hamiltonian flow therefore lies on both an energy level and a sphere, a symplectic leaf of this signed rotational bracket.