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For the Arnold-Liouville theorem, work on a symplectic manifold of dimension $2d$. Let $F_1=H,F_2,\dots ,F_d$ be smooth functions with pairwise vanishing Poisson brackets, and supposeon a connected component $N$ of a common level. If $N$ is compact, then $N$ is a $d$-dimensional torus. Its commuting Hamiltonian vector fields generate translations, and the Hamiltonian motion on it is linear in angular coordinates. Compactness ensures completeness of these restricted vector fields.

Moreover a neighbourhood of $N$ admits action-angle variables $(I_1,\dots ,I_d,{\theta }_1,\dots ,{\theta }_d)$, with each angle defined modulo $2\pi$, for whichOne compatible symplectic form convention is $\omega ={\sum }_{i}d{\theta }_i\wedge d{I}_i$. The equations becomeThe resulting motion is periodic when the frequencies are commensurable and otherwise quasiperiodic on a subtorus. This is Liouville integrability with $d$ independent first integrals in involution. Compactness and regularity cannot be discarded from the torus conclusion. On a general Poisson manifold, apply this theorem on a regular symplectic leaf of dimension $2d$; the ambient dimension alone does not specify the required number of integrals.

Take the separate oscillator energiesThey depend on disjoint canonical coordinate pairs, soThus they are first integrals in involution. Their differentialsare nonzero precisely when $(x_i,y_i)\ne (0,0)$ and have disjoint supports. Hence they are functionally independent on the specified open set.

To understand all further global integrals of motion, use the actionsDirect differentiation gives $d{x}_i\wedge d{y}_i=d{\theta }_i\wedge d{R}_i$, so these are action-angle variables withWhen ${\alpha }_2/{\alpha }_1$ is irrational, every orbit is dense in its fixed-action two-dimensional flat torus. One elementary proof samples the orbit at times $2\pi j/{\alpha }_1$: the first angle returns, and the second undergoes an irrational rotation. These samples are dense in the second circle. Allowing a fixed additional time supplies any desired first angle, proving density in the whole flat torus.

A continuous, globally defined integral of motion must have the same value on an orbit and its closure. Therefore it is constant on every fixed-action flat torus. A $C^1$ such integral has the form $F=\varphi (R_1,R_2)$, with $\varphi$ $C^1$, by evaluating $F$ at a fixed choice of angles. HenceIt cannot be independent of $E_1,E_2$. This proves there are at most two globally independent differentiable integrals. If a third global integral were independent at a point on an excluded coordinate plane, independence would persist in a neighbourhood and hence at nearby points of the regular open set, again a contradiction.

The word global matters: on an angular chart, ${\alpha }_2{\theta }_1-{\alpha }_1{\theta }_2$ is a local integral of motion independent of the actions. It is not single-valued on the flat torus when the frequency ratio is irrational. Thus the assertion would be false for arbitrary local, rather than global, integrals.

On a $2n$-dimensional symplectic manifold, a completely integrable Hamiltonian system has $n$ first integrals in involution $F_1=H,F_2,\dots ,F_n$ whose differentials are independent almost everywhere, hence on a dense open regular set:An equivalent formulation allows $H$ to be a function of the $n$ commuting independent integrals. The independence requirement is on the regular set, permitting exceptional critical points and singular fibres.

For example, on ${\mathbb{R}}^{2n}$ take $H={\sum }_{j=1}^n{E}_j$ with $E_j=(p_j^2+{\alpha }_j^2{q}_j^2)/2$ and ${\alpha }_j>0$. The disjoint coordinate pairs make all Poisson brackets $\{E_i,E_j\}$ zero, and the differentials are independent whenever no pair $(q_j,p_j)$ vanishes. Replacing $E_1$ by $H$ is an invertible linear map of this list. Thus $F_1=H,F_j=E_j$ for $j\ge 2$ give the required example. Irrational frequency ratios affect orbit density, but not complete integrability.

Define the canonical momentaWhen the velocity Hessian is nonsingular, these equations can be inverted for ${\dot{q}}_i(q,p,t)$, and the Hamiltonian is the Legendre transformIts differential isbecause the $p_{i}d{\dot{q}}_i$ terms cancel. The Euler-Lagrange equation identifies $\partial \mathcal{L}/\partial q_i={\dot{p}}_i$, so coefficient comparison gives Hamilton's equations

For the symmetric top, the momenta areThusThe Legendre transform gives

Both $\varphi$ and $\psi$ are cyclic coordinates, so $p_{\varphi }$ and $p_{\psi }$ are conserved. The Hamiltonian has no explicit time dependence and is also conserved. Since $H$ is independent of $\varphi ,\psi$,under the Poisson bracket. Hence three independent first integrals in involution are

Functions $F_i$ for which $\{F_i,F_j\}=0$ for all pairs. Their Hamiltonian vector fields commute, and each is constant along the flows of the others. When they are integrals of motion of a Hamiltonian, they are first integrals in involution.