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A Poisson manifold is a smooth manifold with an antisymmetric contravariant tensor field $\pi =\frac{1}{2}{\pi }^{ij}{\partial }_i\wedge {\partial }_j$ such that the bracketsatisfies the Jacobi identity. Bilinearity, antisymmetry and the Leibniz rule follow directly from this formula. The Poisson bivector is allowed to be degenerate.

Apply the Jacobi identity to coordinate functions $x^i,x^j,x^k$. Necessity gives the coordinate Jacobi condition for a Poisson bivector:For sufficiency, expand $\{f,\{g,h\}\}+\{g,\{h,f\}\}+\{h,\{f,g\}\}$. The terms differentiating $\pi$ are $J^{ijk}{f}_i{g}_j{h}_k$. All terms containing second derivatives cancel. For example, those containing $g_{j\ell }$ have coefficient $({\pi }^{i\ell }{\pi }^{jk}+{\pi }^{k\ell }{\pi }^{ij})f_i{h}_k$, which is antisymmetric under $j\leftrightarrow \ell$ after contraction with $f_i{h}_k$, whereas $g_{j\ell }$ is symmetric. The same argument handles the second derivatives of $f$ and $h$. Thus the displayed condition is both necessary and sufficient.

A symplectic manifold has a smooth nondegenerate closed differential form $\Omega$ of degree two. Write $\Omega =\frac{1}{2}{\Omega }_{ij}d{x}^i\wedge d{x}^j$ and choose the explicit inverse convention ${\pi }^{ik}{\Omega }_{kj}={\delta }_j^i$. This defines a smooth antisymmetric bivector. Differentiating the inverse matrix givesSubstitution into the Jacobi coefficient, followed by relabeling summed indices, givessince the parentheses are the components of $d\Omega$. Hence every symplectic manifold carries a Poisson structure. The opposite overall convention for the inverse bracket gives the same conclusion. Nondegeneracy and closure together are crucial; an arbitrary invertible antisymmetric tensor need not satisfy the Jacobi identity.

For a finite-dimensional Lie algebra $\mathfrak{g}$, its dual space ${\mathfrak{g}}^{\ast }$ carries the Lie-Poisson bracketwhere $d{f}_{\ell },d{g}_{\ell }$ are identified with elements of $\mathfrak{g}$. If $[e_i,e_j]=c_{ij}{}^k{e}_k$ and $x_i(\ell )=\ell (e_i)$, this reads $\{x_i,x_j\}=c_{ij}{}^k{x}_k$. The Jacobi coefficient iswhich vanishes by the Lie algebra Jacobi identity.

In particular, the dual of $\mathfrak{so}(3)$ is ${\mathbb{R}}^3$ with the rotational Lie-Poisson structure on R3,The Poisson bivector has rank two away from the origin and rank zero at the origin. The function $x^2+y^2+z^2$ is a Casimir function, since its bracket with every function vanishes. The centred spheres are its two-dimensional symplectic leaves, and the origin is a point leaf. This is a Poisson manifold but not a symplectic manifold: its bivector is not invertible, and an odd-dimensional manifold cannot have a nondegenerate alternating bilinear form.

A symplectic manifold is a smooth manifold $P$ with a closed, nondegenerate two-form $\omega$. Nondegeneracy forces $\dim P=2n$. Examples are ${\mathbb{R}}^{2n}$ with $\omega ={\sum }_{i}d{q}^i\wedge d{p}_i$, an oriented surface with a nowhere-zero area form, and every cotangent bundle with its canonical form.

A Poisson manifold has a bivector $\Pi$ such that $\{f,g\}=\Pi (df,dg)$ is an antisymmetric bilinear Poisson bracket, is a derivation in each argument, and satisfies the Jacobi identity. Its rank need not be full or constant. The zero bivector gives a Poisson manifold on any smooth manifold, including an odd-dimensional or nonorientable one. A nonzero example on ${\mathbb{R}}^3$ is the Lie-Poisson bracket $\{x_i,x_j\}={ϵ}_{ijk}{x}_k$: it has rank two off the origin and rank zero at the origin, so it does not arise from a symplectic form on the whole three-dimensional space. Even on ${\mathbb{R}}^2$, $\Pi =x{\partial }_x\wedge {\partial }_y$ is Poisson, since its Jacobi condition is a three-vector and hence vanishes, but its degeneracy on $x=0$ prevents it from being the inverse of a global symplectic form. A manifold carrying a degenerate Poisson structure might separately admit some other symplectic structure; these are distinct claims.

At each point of a symplectic manifold, linear nondegeneracy gives a basis with $\omega ={\sum }_i{e}^i\wedge f^i$. Its top wedge product of differential forms isThus ${\omega }^n/n!$ is a global nowhere-zero volume form and selects a consistent symplectic orientation. This proves every symplectic manifold is orientable without choosing a global frame.

Let $\pi :T^{\ast }M\to M$ and let a point of the cotangent bundle be $(x,p)$ with $p\in T_x^{\ast }M$. Define the canonical Liouville one-form intrinsically by ${\theta }_{(x,p)}(V)=p(d\pi (V))$. In local coordinates it is $\theta =p_{i}d{q}^i$, soThis definition is global, $d\omega =0$ by $d^2=0$, and its coordinate matrix has invertible blocks. Hence $T^{\ast }M$ is symplectic for every smooth $M$, even if $M$ itself is not orientable.

The coordinate Jacobi condition for a Poisson bivector isNecessity follows by applying the Jacobi identity to three coordinate functions. For arbitrary functions, the second-derivative terms in the Jacobiator cancel by antisymmetry of $\Pi$; the terms left are exactly this coefficient multiplying $f_i{g}_j{h}_k$, proving sufficiency.

For a symplectic form, take ${\iota }_{X_f}\omega =df$ and $\{f,g\}=\omega (X_f,X_g)$. With these signs the matrix of $\Pi$ is $-{\omega }^{-1}$. Differentiation gives ${\partial }_{\ell }\Pi =\Pi ({\partial }_{\ell }\omega )\Pi$. The cyclic expression above consequently equalswhich vanishes because $d\omega =0$. This directly proves that a symplectic manifold defines a Poisson manifold.

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 $X_{H}f=\{f,H\}$, agreeing with the canonical bracket $\{q^i,p_j\}={\delta }_j^i$.

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 definesThus $r$ 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 $h\mapsto gh{g}^{-1}$ differentiates at the identity to the Adjoint representation of a Lie group, ${\mathrm{Ad}}_g:\mathfrak{g}\to \mathfrak{g}$. Its infinitesimal version is ${\mathrm{ad}}_{\xi }\eta =[\xi ,\eta ]$. The coadjoint representation on the dual space is defined byThe 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 ${\mathfrak{g}}^{\ast }$:The differentials lie in $({\mathfrak{g}}^{\ast }{)}^{\ast }\simeq \mathfrak{g}$. In a basis with $[e_a,e_b]=c_{ab}{}^c{e}_c$, 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 satisfiesIt 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 ${\mathfrak{g}}^{\ast }$, $X_F(\ell )={\mathrm{ad}}_{d{F}_{\ell }}^{\ast }\ell$ 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: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 $\hat{\mathbf{\xi }}\mathbf{v}=\mathbf{\xi }\times \mathbf{v}$. 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 $\mathbf{\xi }\times \mathbf{M}$. Therefore its nonzero orbits are spheres $|\mathbf{M}|=r$; the origin is a zero-dimensional leaf. In these coordinates,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 isIts 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 $T^{\ast }SO(3)$ by spatial rotations gives the body-frame negative Lie-Poisson bracket and HamiltonianKeeping the sign convention explicit is essential: using the positive bracket with the same body identification would reverse the Euler evolution. The negative bracket giveswith 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 $\mathbf{\Omega }\cdot (\mathbf{M}\times \mathbf{\Omega })=0$ and $\mathbf{M}\cdot (\mathbf{M}\times \mathbf{\Omega })=0$. Trajectories therefore lie on intersections of energy ellipsoids with momentum spheres, the latter being the symplectic leaves.

The body's orientation is reconstructed from $\dot{R}=R\hat{\mathbf{\Omega }}$. Its spatial momentum $\mathbf{J}=R\mathbf{M}$ is constant, since $\dot{\mathbf{J}}=R(\mathbf{\Omega }\times \mathbf{M}+\mathbf{M}\times \mathbf{\Omega })=0$. Thus the body momentum need not be constant even though the spatial conserved momentum is. Poisson reduction retains the dynamics and its momentum leaves while eliminating orientation variables; the unreduced motion is recovered by reconstruction.

A Poisson manifold is a smooth manifold with a smooth antisymmetric bivector $\pi =\frac{1}{2}{\pi }^{ij}{\partial }_i\wedge {\partial }_j$ whose function bracket $\{f,g\}={\pi }^{ij}({\partial }_{i}f)({\partial }_{j}g)$ satisfies the Jacobi identity. The bracket is automatically a derivation in each argument. A symplectic manifold has a nondegenerate differential two-form $\omega$ satisfying $d\omega =0$, that is, a closed differential form; nondegeneracy forces its dimension to be even. It determines a nondegenerate Poisson structure. We use matrix convention $\pi =-{\omega }^{-1}$, so that ${\iota }_{X_H}\omega =dH$ corresponds to $X_H(f)=\{f,H\}$.

Examples of genuinely nonsymplectic Poisson geometries include the zero bracket on ${\mathbb{R}}^3$ and the Lie-Poisson bracket $\{x_i,x_j\}={ϵ}_{ijk}{x}_k$ on ${\mathbb{R}}^3$. Their three-dimensional underlying manifold cannot carry a nondegenerate two-form. In the latter case the Poisson rank is two away from the origin and zero at the origin, with spheres and the origin as symplectic leaves. Even in even dimension a given Poisson structure need not come from a symplectic form: $\pi =x{\partial }_x\wedge {\partial }_y$ on ${\mathbb{R}}^2$ satisfies Jacobi but vanishes along $x=0$. The statement concerns this degenerate structure, not a claim that ${\mathbb{R}}^2$ admits no other symplectic structure.

For any candidate bivector, applying the Jacobi identity to coordinate functions gives the necessary conditionIt is sufficient as well: expanding $\{f,\{g,h\}\}$ and its cyclic companions cancels all terms with second derivatives by antisymmetry of $\pi$. What remains is $J^{ijk}({\partial }_{i}f)({\partial }_{j}g)({\partial }_{k}h)$. This is the coordinate Jacobi condition for a Poisson bivector.

For a symplectic form, differentiate $\omega \pi =-I$ to obtain ${\partial }_{\ell }\pi =\pi ({\partial }_{\ell }\omega )\pi$. Substituting this into the cyclic expression givesbecause $d\omega =0$. Thus the inverse symplectic structure always satisfies the Jacobi identity. Conversely, for a nondegenerate Poisson structure, multiplying by three inverse matrices proves closure of its associated two-form.

In momentum space, choose Hamiltonian $H(\mathbf{p})=E(\mathbf{p})$ and the constant Poisson bivectorIts Poisson bracket is $\{f,g\}=e\mathbf{B}\cdot (\nabla f\times \nabla g)$. The Hamiltonian vector field therefore givesexactly the prescribed motion. The sign here uses the signed charge $e$ as printed; an electron's physical charge may be negative. Since the magnetic field is uniform, every derivative of ${\pi }^{ij}$ vanishes, so every term in $J^{ijk}$ is zero. Jacobi holds independently of the dispersion function $E$. This is the momentum-space Poisson structure for a uniform magnetic field.

For $e\mathbf{B}\ne 0$ this bivector has rank two and $C(\mathbf{p})=\mathbf{B}\cdot \mathbf{p}$ is a Casimir function of a Poisson manifold: $\{C,f\}=0$ for every $f$. Its symplectic leaves are the planes $C=c$. For axes with $\mathbf{B}=B_0{\mathbf{e}}_3$, the leaf form is $\omega =(e{B}_0{)}^{-1}d{p}_1\wedge d{p}_2$ in our inverse-sign convention. Also $\dot{E}=\{E,E\}=0$, so trajectories on a Fermi surface lie on its intersections with these planes, typically curves. The energy surface is invariant for this Hamiltonian, but need not be a Poisson submanifold; the bracket has been defined on the whole momentum space. If $e\mathbf{B}=0$, the Poisson structure and motion vanish.

A Poisson structure is a smooth antisymmetric matrix field ${\omega }^{ab}$ defining a bracketthat obeys the Jacobi identity. Bilinearity, antisymmetry and the product rule follow from this formula; Jacobi is the remaining condition. Applying Jacobi to the coordinate functions givesThese are also sufficient: expanding Jacobi for arbitrary functions cancels all second-derivative terms by antisymmetry, leaving these coefficients multiplying their first derivatives.

For ${\omega }^{ab}={ϵ}^{abc}{x}^c$, the nonconstant function $f={\sum }_a(x^a{)}^2$ is a Casimir function of a Poisson manifold, becauseby symmetry of $x^b{x}^c$. This is the usual three-dimensional Lie-Poisson bracket.

Since the matrix $M$ is symmetric, ${\partial }_{d}H={\sum }_b{M}^{db}{x}^b$. The Hamilton equations are ${\dot{x}}^a=\{x^a,H\}$, and henceOne valid choice of coefficients is $Q^{abc}={\sum }_d{ϵ}^{adc}{M}^{db}$. If symmetric coefficients in $b,c$ are preferred, the same equations useIn vector notation the evolution is $(Mx)\times x$, consistent with the bracket convention and with preservation of $|x{|}^2$.

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.

With the displayed signed Lie-Poisson bracket and evolution $\dot{F}=\{F,H\}$, the equations are ${\dot{x}}_1=2(C-B)x_2{x}_3$, ${\dot{x}}_2=2(A-C)x_3{x}_1$, and ${\dot{x}}_3=2(B-A)x_1{x}_2$. The energy and the Casimir function of a Poisson manifold $x_1^2+x_2^2+x_3^2$ are conserved. For $A=1/(2I_1)$, $B=1/(2I_2)$, $C=1/(2I_3)$ with positive inertias, these are the Euler equations for a torque-free rigid body. Reversing the bracket while fixing $H$ reverses time orientation.