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Let $S_{s,t}(y)$ be the position at time $t$ of the characteristic curve satisfying $X(s)=y$ and $\dot{X}=u(t,X)$. Smooth uniqueness implies $S_{t,s}=S_{s,t}^{-1}$. Differentiating in the initial point gives the variational equationThe Liouville formula for a fundamental matrix, or the derivative formula for a determinant, therefore givesBecause $u=(-{\varphi }_{x_2},{\varphi }_{x_1})$, the mixed second derivatives cancel and $\mathrm{div}u=0$. The initial Jacobian determinant is one, soThus this Hamiltonian flow preserves planar Lebesgue measure. This calculation assumes smoothness as in the part; a rough-flow conclusion later must be justified by approximation rather than by differentiating a merely continuous velocity.

The Hamiltonian flow of $H$ solvesCompactness and the absence of a boundary make this smooth vector field complete, so the flow exists for all real $t$. The defining equation and Cartan's magic formula giveTherefore the pullback of a differential form differentiation rule yieldsso $({\varphi }_H^t{)}^{\ast }\omega =\omega$. Pullback commutes with the wedge product of differential forms, and consequentlyThus the flow preserves the symplectic volume, in fact the entire symplectic form.

Write $z=(x,v)$ and let $\Phi (t,s,z)$ denote the Hamiltonian flow from time $s$ to time $t$. The characteristic curves satisfy Hamilton's equations:The Hamiltonian Liouville equation then reduces along each curve toThe signs and derivative variables here are those in the PDF.

The global characteristic flow for a Hamiltonian with bounded Hessian follows, for example, from $H\in C^2(\mathbb{R}\times {\mathbb{R}}^{2d})$ and, for every finite $T$,Thus the Hamiltonian vector field $b=({\nabla }_{v}H,-{\nabla }_{x}H)$ is globally Lipschitz continuous in $z$ on each finite time interval and satisfies $|b(t,z)|\le A_T+L_T|z|$. The Picard-Lindelöf theorem gives local existence and uniqueness, while the Gronwall inequality gives, for example,This excludes finite-time escape. There is a unique Hamiltonian flow for all finite forward and backward times, and $\Phi (s,t)$ is the inverse of $\Phi (t,s)$. These sufficient conditions are deliberately stronger than necessary.

For an autonomous Hamiltonian function, the chain rule and Hamilton's equations giveThus the energy is constant along each characteristic:The time-independent hypothesis is implicit in the displayed expression in this subpart. For the time-dependent Hamiltonian function allowed earlier, the correct Hamiltonian energy balance is insteadFor example $H(t,x,v)=t+|v{|}^2/2$ has a unique global Hamiltonian flow, but its value along a curve increases at unit rate. Thus conservation cannot be claimed for arbitrary time-dependent $H$.

For the globally invertible Hamiltonian flow from (a), the general characteristic formula isEvery $\Phi (s,t)$ preserves Lebesgue measure, by (b). Thus composition with it is an isometry of each Lp space. The Minkowski integral inequality givesApplying the Minkowski inequality also in time yields the finite-time Lp bound for Hamiltonian transport:The smoothness assumptions allow the characteristic construction; the norm estimate itself only uses the Lp space data and volume preservation.

For a concrete failure on infinite time, choose $\omega =1$, $f_0=0$, andThis is smooth, time independent and in every finite Lp space; it is invariant under the isotropic harmonic oscillator flow. Therefore the solution grows linearly:This invariant-source secular growth in Hamiltonian transport supplies the counterexample even with zero initial data.

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.

Use the first-order actionVarying $q$ with fixed endpoints and varying $p$ freely givesThe principle of stationary action therefore yields Hamilton's equations

For $H=p^2+q^{-2}$ these become $\dot{q}=2p$ and $\dot{p}=2q^{-3}$. Along their Hamiltonian flow, $H=E$ is conserved andThus $F=pq-ctH$ is an integral of motion exactly when $c=2$.

Write its constant value as $F_0$. The surfaces $pq-2tE=F_0$ in extended $(q,p,t)$-space giveUsing $E{q}^2=p^2{q}^2+1$,The sign of $q$ is fixed. In phase space, each energy orbit is one of the two branchesEach branch approaches $p=-\sqrt{E}$ and $p=+\sqrt{E}$ as $|q|\to \infty$ and turns at $(p,q)=(0,\pm E^{-1/2})$.

In canonical coordinates, Hamilton's equations areA function $F(q,p)$ is an integral of motion if it is constant along every Hamiltonian flow, equivalentlyfor a time-independent $F$. The system is an integrable Hamiltonian system when it has $n$ first integrals $F_1,\dots ,F_n$ whose differentials are linearly independent almost everywhere and which are in mutual Poisson involution: $\{F_i,F_j\}=0$ for all $i,j$.

Hamiltonian Monte Carlo augments the position $x$ by an independent momentum $p\sim N(0,M)$ and uses the Hamiltonian functionFrom the current $x$, draw a fresh $p$, apply a fixed number of leapfrog steps to approximate Hamiltonian flow, and obtain $(x^{\prime },p^{\prime })$. The Metropolis–Hastings acceptance probability isotherwise retain $x$. Momentum negation may be included to make the proposal explicitly reversible. The leapfrog map is volume preserving and reversible, while the acceptance step corrects its discretization error, leaving $\pi$ invariant after the momentum is discarded.

A polarization is an involutive complex rank-half distribution on which the symplectic form vanishes. Polarized prequantum sections satisfy ${\nabla }_{X}s=0$ for $X\in \mathcal{F}$. A real polarization is tangent to a Lagrangian foliation; a complex Kähler polarization uses the antiholomorphic tangent spaces. Only observables whose Hamiltonian flows preserve the chosen distribution act directly on the polarized space. Singular leaves, holonomy and changes of polarization are genuine quantization difficulties.

Differentiating a smooth characteristic flow map in its initial point gives this linear equation, with $Y(s)=I$. The Liouville formula for a fundamental matrix then gives $\det Y(t)=\exp ({\int }_s^t\mathrm{div}b(r,X(r))dr)$. Incompressible Hamiltonian flows preserve Lebesgue measure because this divergence vanishes.