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In cotangent bundle coordinates $(q^1,\dots ,q^n,p_1,\dots ,p_n)$, the Liouville one-form and the canonical symplectic form areThe twisted cotangent symplectic form is closed: pullback of a differential form commutes with the exterior derivative, soTo prove the form is nondegenerate, write a tangent vector in these coordinates as $(u,v)$ and test its pairing with an arbitrary vertical vector $(0,w)$. The base differential two-form has zero pairing with vertical vectors, henceA vector in the kernel of a linear map therefore has $u=0$. Pairing it next with every $(z,0)$ gives $-z^i{v}_i=0$, so $v=0$. Thus the kernel is zero at every point, independently of the size or rank of $\sigma$. Together with closedness, this proves ${\omega }_{\sigma }$ is a symplectic form. The argument is local only for convenience; the nondegenerate property is intrinsic and the coordinate charts cover the whole cotangent bundle.

Put ${\xi }_t=\ker {\alpha }_t$, the contact distribution. If $\dim M=2r+1$, the contact form condition is ${\alpha }_t\wedge (d{\alpha }_t{)}^r\ne 0$. It implies that $d{\alpha }_t{|}_{{\xi }_t}$ is a symplectic form on each contact hyperplane: inserting a vector transverse to ${\xi }_t$ in that volume form leaves the nonzero top exterior power $(d{\alpha }_t{|}_{{\xi }_t}{)}^r$. Hence there is a unique smooth time-dependent vector field $Y_t\in {\xi }_t$ satisfyingIts smoothness follows by inverting the smoothly varying nondegenerate matrix of this contact distribution symplectic form. This is the horizontal generator of contact stability.

Let $R_t$ be the Reeb vector field of ${\alpha }_t$. The one-form ${\dot{\alpha }}_t+{\iota }_{Y_t}d{\alpha }_t$ vanishes on ${\xi }_t$, so it is $h_t{\alpha }_t$. Evaluating on $R_t$, using ${\alpha }_t(R_t)=1$ and ${\iota }_{R_t}d{\alpha }_t=0$, identifies the coefficient:By Cartan's magic formula, the Lie derivative of a differential form isbecause $Y_t$ lies in the contact hyperplane.

Let ${\rho }_t$ solve ${\partial }_t{\rho }_t=Y_t\circ {\rho }_t$, with ${\rho }_0={\mathrm{id}}_M$. Since $M$ is compact without boundary and $Y_t$ is smooth on the closed time interval, its solutions cannot escape and exist for the entire interval. Reversing the time-dependent equation supplies a smooth inverse for each ${\rho }_t$. Thus these flow maps give a smooth isotopy. Differentiating the pullback of a differential form along this isotopy yieldsFor each base point this is a scalar linear equation for a covector, initially ${\alpha }_0$. Its solution isBy finite-time flow completeness on a compact manifold, this construction gives the whole time interval. The contact conformal factor along an isotopy is smooth in $(t,x)$ and nowhere zero, proving Gray stability theorem. The precise domains here are ${\rho }_t:M\to M$ and $\rho :M\times [0,1]\to M$; the extra time factor attached to the already indexed ${\rho }_t$ in the source is a notational slip. The proof produces the required family on its whole specified interval. If a parameter domain of all $\mathbb{R}$ is intended, extend the smooth field slightly beyond $[0,1]$, multiply it by a time cutoff, and use compactness to obtain the extended isotopy $\rho :M\times \mathbb{R}\to M$. Its restriction to $[0,1]$ is unchanged.

We prove the Weinstein neighborhood theorem with the canonical sign convention ${\omega }_0=-d\alpha$. The essential first step is to match the two symplectic forms as bilinear forms on the entire tangent space along $X$, not only after pulling them back to $TX$.

Here is the needed symplectic splitting along a Lagrangian submanifold. Write $E=TM{|}_X$ and $L=TX$. Choose a smooth vector bundle complement $N$ of $L$, for example using a Riemannian metric. Since $L$ is a Lagrangian subspace in every fiber, the pairing mapis an isomorphism: its kernel is $N\cap L^{\omega }=N\cap L=0$, and both bundles have rank $n$. Let $s:L^{\ast }\to N$ be its inverse and put $b(\eta ,\zeta )=\omega (s\eta ,s\zeta )$. Define $a:L^{\ast }\to L$ byand set $\tilde{s}=s+a$. Nondegeneracy of the dual pairing gives a unique smooth $a$. Since $b$ is alternating,Thus $K=\tilde{s}(L^{\ast })$ is a smooth Lagrangian complement to $L$. The mapis a vector bundle isomorphism, is the identity on $TX$, and satisfiesThe last equality uses the canonical splitting of the tangent bundle of $T^{\ast }X$ along its zero section.

Choose a Riemannian metric near $X$ for which $K$ is orthogonal to $TX$. The tubular neighborhood construction using its Riemannian exponential map gives a diffeomorphismafter shrinking around the zero section. It restricts to the given inclusion of $X$ and has differential $A$ along it. Consequently $\beta ={\psi }^{\ast }\omega$ and ${\omega }_0$ agree pointwise as ambient bilinear forms at every point of the zero section.

We use the following precise relative Moser theorem. If two closed symplectic forms ${\gamma }_0,{\gamma }_1$ near a compact embedded submanifold $S$ agree as ambient bilinear forms along $S$, then, on smaller neighborhoods, there is a diffeomorphism $\kappa$ fixing $S$ pointwise with ${\kappa }^{\ast }{\gamma }_1={\gamma }_0$. One sufficient version allows a symplectic path ${\gamma }_t$ with ${\partial }_t{\gamma }_t=d{\eta }_t$ and ${\eta }_t=0$ as ambient covectors along $S$: solve ${\iota }_{Z_t}{\gamma }_t=-{\eta }_t$ and use its flow. The vector field vanishes on $S$, so the flow fixes $S$, and compactness permits a common smaller neighborhood for the full time interval.

For completeness, all hypotheses can be checked directly here. Put $\delta =\beta -{\omega }_0$ and ${\gamma }_t={\omega }_0+t\delta$. At the zero section, every ${\gamma }_t$ equals ${\omega }_0$. The nondegenerate property is open, and compactness of $X\times [0,1]$ allows a single smaller neighborhood on which all ${\gamma }_t$ are symplectic forms. Choose it fiberwise star-shaped. Write $H_s(q,p)=(q,sp)$ and let $E$ be the vertical radial vector field. The radial homotopy primitive near a zero sectionis smooth, is zero as an ambient covector along the zero section, and satisfiesThe radial homotopy operator gives this identity, using $d\delta =0$ and $H_0^{\ast }\delta ={\pi }^{\ast }{i}_0^{\ast }\delta =0$. Smoothness at $s=0$ follows because contraction with $E$ supplies a factor of $s$ after evaluation at $(q,sp)$; the vanishing of $\delta$ along $X$ supplies additional vanishing. Therefore solve ${\iota }_{Z_t}{\gamma }_t=-\eta$. By Cartan's magic formula, its flow maps ${\kappa }_t$ satisfyAfter shrinking the starting neighborhood, these flows exist for $0\le t\le 1$ and fix $X$. This is a relative local construction, not an assertion of completeness throughout the noncompact cotangent bundle.

Finally set $\phi =\psi \circ {\kappa }_1$, let $U_0$ be its sufficiently small domain, and let $U=\phi (U_0)$. ThenThus the map is a symplectomorphism of neighborhoods and agrees with the specified inclusion at every point of $X$.

The Poisson bracket in the original coordinates isFor $z=(q,p)$, setThen ${\omega }_0(u,v)=u^{T}Jv$ and $\{F,G\}=\nabla F^{T}J\nabla G$. The pullback definition of a canonical transformation gives the pointwise matrix conditionSince $A$ is invertible, this is equivalent to $AJ{A}^T=J$. Indeed, invert $A^{T}JA=J$ and use $J^{-1}=-J$ to obtain $A^{-1}J{A}^{-T}=J$, then multiply by $A$ and $A^T$. Applying the same calculation to the inverse implication gives equivalence. These are the two equivalent symplectic matrix identities that connect the symplectic form with Poisson brackets.

The chain rule yieldsIf $f$ is a canonical transformation, the symplectic matrix identity $AJ{A}^T=J$ makes this expression $\{F,G{\}}_{Q,P}\circ f$. Thus preservation of the symplectic form implies preservation of every Poisson bracket, for all $C^1$ functions.

Conversely, assume this Poisson bracket identity for every $F,G$. Take $F(Z)=u\cdot Z$ and $G(Z)=v\cdot Z$, for arbitrary constant vectors $u,v\in {\mathbb{R}}^{2n}$. It gives $u^{T}AJ{A}^{T}v=u^{T}Jv$. Since this holds for every $u,v$, $AJ{A}^T=J$, hence $A^{T}JA=J$. Therefore preservation of all Poisson brackets is equivalent to canonicity. Only first derivatives of $f,F,G$ enter this argument.

Under the given rank hypothesis, the necessary and sufficient local condition isVanishing only at the single point would not suffice: an extension preserves the coordinate Poisson brackets throughout its domain. Necessity follows from the coordinate criterion just proved.

For sufficiency, put $Q_i=a_i$ and $Y_i=X_{Q_i}$. Because the symplectic form is a nondegenerate bilinear form at every point, the independent differentials $d{Q}_i$ give independent fields $Y_1,\dots ,Y_n$. They commute, sinceMoreover $d{Q}_j(Y_i)=\{Q_j,Q_i\}=0$, so they span the tangent spaces of the $n$-dimensional fibres of the submersion $Q$. Choose a local section $s(Q)$ transverse to those fibres. Its joint local flow defines coordinates $(Q,t)$ byThe inverse function theorem makes $\Psi$ a local diffeomorphism, and commuting flows give ${\Psi }_{\ast }{\partial }_{t_i}=Y_i$.

In these coordinates ${\iota }_{{\partial }_{t_i}}{\Psi }^{\ast }{\omega }_0=d{Q}_i$. Each flow preserves the symplectic form, by Cartan's magic formula. Thus its coefficients are independent of $t$, andwhere $\beta =s^{\ast }{\omega }_0$ is a closed differential two-form on the base. On a small ball, the Poincare lemma gives $\beta =d\alpha$, with $\alpha ={\sum }_i{\alpha }_i(Q)d{Q}_i$. DefineThenConsequently the original coordinates are extended to the required canonical transformation $(Q,P)$, with the prescribed first half $Q=a$. Its Jacobian matrix is invertible by construction. This is the canonical completion of independent commuting functions. For the printed $C^2$ data, the flows and resulting canonical transformation can be taken $C^1$, which is sufficient to preserve the symplectic form; smooth data give smooth coordinates.

Use the standard symplectic form ${\omega }_0={\sum }_{j=1}^{n}d{q}_j\wedge d{p}_j$. A canonical transformation is a diffeomorphism $f$ preserving this symplectic form:Thus, writing $f(q,p)=(Q,P)$, the defining identity is ${\sum }_{j}d{Q}_j\wedge d{P}_j={\sum }_{j}d{q}_j\wedge d{p}_j$. For a local canonical transformation, require this on its open domain. In particular preservation of the symplectic form, rather than just of phase-space volume, is the condition in dimensions greater than two.

Throughout these solutions choose the Hamiltonian vector field convention ${\iota }_{X_F}\omega =dF$. It gives $\dot{q}=F_p$, $\dot{p}=-F_q$ and agrees with the positive coordinate Poisson bracket used below. The opposite convention for Hamiltonian vector fields changes signs in the Lie bracket of vector fields identity but none of the commuting or integrability conclusions.

The Hamiltonian perturbation preserving one regular level is explicit:Linearity of the Hamiltonian vector field in the differential of a smooth map givesIt vanishes on ${\Sigma }_0$. At a point of ${\Sigma }_c$, its contraction with the symplectic form isSince the differentials are independent there, this vanishes exactly when $c=0$. Thus on every nonempty nonzero regular fibre the two Hamiltonian vector fields differ at every point, not merely as functions somewhere:Mutual Poisson commutation of the $G_j$ is unnecessary; the given first-integral property does not by itself make the perturbed flow preserve every common level, nor is that requested.

There is a genuine missing nonemptiness hypothesis in the literal statement. Independent differentials on an empty set is a vacuous condition, and any two vector-field restrictions to an empty set are equal. For example, on ${\mathbb{R}}^2$ let $r=1$, $G_1=H=e^q$, and $\rho =1$. Every nonempty ${\Sigma }_c$, $0<c<1$, is regular, while ${\Sigma }_c=\varnothing$ for $-1<c\le 0$. In particular, the restrictions at $c=-1/2$ can never be unequal for any $K$. Therefore the printed universal inequality requires either all the compared fibres to be nonempty, or the pointwise formulation displayed above. The explicit $K$ fully proves either intended qualification.

A function $F$ is an integral of motion of $H$ when it is constant along every trajectory of the Hamiltonian flow. Infinitesimally,The antisymmetry of a Lie bracket of the Poisson bracket gives $X_{F}H=\{H,F\}=0$. Hence $H$ is also an integral of motion of $F$.

For the commuting assertion, ${\mathcal{L}}_{X_F}\omega =d({\iota }_{X_F}\omega )=d^{2}F=0$, because $\omega$ is closed. The contraction identity for a Lie bracket of vector fields then givesSince the symplectic form is a nondegenerate bilinear form at every point, the Hamiltonian vector fields commute:With the chosen convention the general identity is $[X_F,X_H]=X_{\{H,F\}}$. Equivalently, it follows by applying the Jacobi identity for the Poisson bracket to an arbitrary test function. Local flows commute whenever both compositions are defined; no global completeness assumption is needed.

The Arnold-Liouville theorem concerns a connected compact component $L$ of a regular common level of $n$ commuting independent smooth functions $F=(F_1,\dots ,F_n)$ on a $2n$-dimensional symplectic manifold. It states that $L$ is a Lagrangian torus, and that a neighbourhood of $L$ has action-angle variablesin which every $F_i$ depends only on $I$. If $H=F_1$ or, more generally, $H=h(F)$, its Hamiltonian flow satisfiesAngles have period $2\pi$ here. The theorem is local near this regular compact component; it does not assert globally defined action-angle variables across singular levels or over an entire base with monodromy.

First, $L$ has dimension of a manifold $n$ by the submersion theorem. The Hamiltonian vector fields $X_{F_i}$ are independent, tangent to $L$, and commute. They span $TL$. Moreoverso $L$ is a Lagrangian submanifold. Compactness makes these restricted vector fields complete. Their joint flow is an ${\mathbb{R}}^n$-action on $L$. Its orbits are open because the fields span $TL$, so connectedness gives one orbit. The stabilizer $\Lambda$ of a point is discrete by the inverse function theorem, andCompactness forces $\Lambda$ to be a full-rank Euclidean lattice: otherwise an unbounded linear coordinate transverse to its span would descend to the quotient. Thus $L$ is an $n$-torus.

Next choose a small ball of regular values near $F(L)$ and a neighbourhood $N$ of this component which is a product family of compact tori $L_c$. This local trivialization follows directly by choosing transverse vector fields $Z_j$ with $d{F}_i(Z_j)={\delta }_{ij}$ and lifting short radial paths in the base; compactness of $L$ gives a uniform neighbourhood where their flows exist. Hence $N$ deformation retracts onto $L$. The commuting joint flows on nearby fibres have smoothly varying full-rank Hamiltonian period lattices. A basis $t^{(j)}(c)\in {\mathbb{R}}^n$ of their periods can be chosen smoothly on this small ball: continue the return equations from a basis at the central fibre, using their nonsingular vertical flow derivatives and the implicit function theorem. No global choice of lattice basis is needed.

Because $\omega {|}_L=0$ and $N$ retracts onto $L$, its closed symplectic form is exact on $N$. Choose a one-form $\lambda$ with $d\lambda =-\omega$. Let ${\gamma }_j(c)$ be the smoothly continued cycles represented by the period basis and define the action integralsFor a transverse variation $v$ of the fibre, differentiating a cycle integral and using Cartan's magic formula eliminates the integral of the exact term. With the cycle parametrized by the joint-flow time $t^{(j)}$, this givesThusThe period matrix is nonsingular, so $I$ gives coordinates on the base. The corresponding Hamiltonian vector fields satisfyTheir time-$2\pi$ flows return along the basis cycles. They commute because each $I_j$ is a function of the commuting $F_i$. Consequently they give a free torus action on $N$.

Choose a section over the action-coordinate ball and use this torus action to define angles $\theta$. Since $X_{I_j}={\partial }_{{\theta }_j}$, the symplectic form takes the formThe closed base two-form $\beta$ is exact on the ball: write $\beta =d\alpha$, $\alpha ={\sum }_j{a}_j(I)d{I}_j$, by the Poincare lemma. Changing the angular origins by ${\theta }_j^{\prime }={\theta }_j+a_j(I)$ removes this term, since ${\sum }_{j}d{\theta }_j^{\prime }\wedge d{I}_j=\omega$. This completes the construction of action-angle variables. The functions $F_i$ are constant on the fibres, so are functions of $I$ alone, and Hamilton's equations give the claimed straight-line motion. This proves both the topological and symplectic parts of the Arnold-Liouville theorem.

Set $s=|z{|}^2$. Contraction of the symplectic form with the rotation vector field givesWith the convention fixed in the unheaded solution, ${\iota }_K\omega =-dH$, so the Hamiltonian function isIts derivative is $dH=ds/(1+s{)}^2$. In the chart $w=1/z$, $H=1/(1+|w{|}^2)+C$, so it extends smoothly across infinity; the additive constant is the kernel freedom already identified. Thus $K=X_H$ globally, proving that the action is a Hamiltonian group action. With $C=0$, the moment map for rotation of the complex projective line takes values in $[0,1]$, reaching its endpoints at the two fixed poles.

One can check the normalization in polar coordinates: $\omega =2r(1+r^2{)}^{-2}dr\wedge d\varphi$ has total symplectic area $2\pi$, not $4\pi$, and $H=r^2/(1+r^2)$. Equivalently, in a polar angle with $r=\tan (\vartheta /2)$, $H=(1-\cos \vartheta )/2$. This explains the half-height factor that would be lost by replacing the printed form with the unscaled round-sphere area form. Under the opposite convention ${\iota }_{X_H}\omega =dH$, the Hamiltonian is the negative of this function, up to a constant.

Fix the convention ${\iota }_{X_f}\omega =-df$ for a Hamiltonian vector field. Nondegeneracy of the symplectic form gives one unique smooth $X_f$ for each smooth real-valued function $f$. Define the Poisson bracket byIn coordinates with $\omega =\sum d{q}^i\wedge d{p}_i$, this is $\sum (f_{q^i}{g}_{p_i}-f_{p_i}{g}_{q^i})$. Thus the sign convention is explicit and agrees with the usual coordinate Poisson bracket. By Cartan's magic formula, ${\mathcal{L}}_{X_f}\omega =d{\iota }_{X_f}\omega +{\iota }_{X_f}d\omega =0$. The contraction–Lie derivative commutator identity now givesHence the Hamiltonian Lie algebra homomorphism is $\boxed{\Phi (f)=X_f,[X_f,X_g]=X_{\{f,g\}}}$. It is linear and onto the space of Hamiltonian vector fields by definition. For completeness, the Jacobi identity for the bracket on functions follows from closure of $\omega$: evaluating $d\omega =0$ on $X_f,X_g,X_h$ and using the displayed commutator relation gives the cyclic Jacobi sum zero. Thus this really is a map of Lie algebras, not just a bracket-preserving notation.

Its kernel is determined by nondegeneracy:On a connected manifold these are precisely the real constants; on a disconnected manifold the constant may differ on each component. Therefore the quotient by these functions is isomorphic to the Lie algebra of Hamiltonian vector fields. The printed “homeomorphis” is interpreted as homomorphism: the map before taking this quotient is not an isomorphism because of its nontrivial kernel. The alternative convention ${\iota }_{X_f}\omega =df$ requires a corresponding bracket sign change to retain this homomorphism.