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Use ${\iota }_{X_f}\omega =df$ and $\{f,g\}=\omega (X_f,X_g)$. For an actual infinitesimal left Lie group action, define ${\xi }_P(p)={\frac{d}{dt}|}_0\exp (t\xi )\cdot p$. Its fundamental vector fields obey $[{\xi }_P,{\eta }_P]=-[\xi ,\eta {]}_P$ with this convention.

Existence and its global obstruction. Preservation of the symplectic form gives ${\mathcal{L}}_{{\xi }_P}\omega =0$. By Cartan's magic formula, ${\iota }_{{\xi }_P}\omega$ is closed. Locally it equals $d{\mu }_{\xi }$; globally this requires its class in $H_{\mathrm{dR}}^1(P)$ to vanish. If it does for every $\xi$, choose potentials linearly in $\xi$ and defineThis gives the component Hamiltonians of a weakly Hamiltonian action. The assumption in the question does not alone guarantee a global moment map: on $T^2$ with $\omega =dx\wedge dy$, translation in $x$ preserves $\omega$, but its contraction is $dy$, whose integral around the $y$ circle is nonzero. It is not an exact one-form. Thus no global real-valued component Hamiltonian exists for that circle action. The intended construction is valid locally, or globally with this exactness condition, for example if $H_{\mathrm{dR}}^1(P)=0$.

Matching the Lie algebra. On each connected component, the moment-map equivariance obstruction is the constantIndeed, the identities $[X_f,X_g]=-X_{\{f,g\}}$ and $[{\xi }_P,{\eta }_P]=-[\xi ,\eta {]}_P$ show that its differential vanishes. The Jacobi identity for the Poisson bracket saysChanging ${\mu }_{\xi }$ by a linear constant $b(\xi )$ changes $c$ to $c-b([\xi ,\eta ])$. Consequently the necessary and sufficient infinitesimal condition is the cocycle must be a coboundary, so that the constants can be chosen withEquivalently its class in $H^2(\mathfrak{g},\mathbb{R})$ vanishes. For a connected group this infinitesimal equivariance integrates to equivariance under its coadjoint action. A disconnected group can impose additional conditions. Here “coincides” means representing the specified Lie brackets; it does not guarantee injectivity when the group action has ineffective generators.

Proof using the Killing form. A nondegenerate Killing form $B$ makes $\mathfrak{g}$ semisimple. First, it is perfect: the $B$-orthogonal complement of $[\mathfrak{g},\mathfrak{g}]$ is the center by invariance, and the center is zero by nondegeneracy. For symplectic fields,by Cartan's magic formula and the antihomomorphism sign. Every commutator therefore has an exact contraction. Perfectness supplies component Hamiltonians for every generator, even when $P$ is not simply connected: this is Hamiltonian existence for a semisimple symplectic action.

It remains to eliminate the cocycle, not merely to name an existence theorem. Define a linear operator $D$ on $\mathfrak{g}$ by $B(D\xi ,\eta )=c(\xi ,\eta )$. Antisymmetry of $c$ makes $D$ skew for $B$. Combining its cocycle identity with invariance of $B$ givesChoose $Z\in \mathfrak{g}$ by $B(Z,\xi )=\mathrm{tr}(D{\mathrm{ad}}_{\xi })$, possible because $B$ is nondegenerate. Since $D$ is a derivation, $[D,{\mathrm{ad}}_{\xi }]={\mathrm{ad}}_{D\xi }$. Cyclicity of the matrix trace then givesThus $D={\mathrm{ad}}_Z$, and $c(\xi ,\eta )=B(Z,[\xi ,\eta ])$. Taking $b(\xi )=B(Z,\xi )$ removes $c$, proving the required condition. This applies to semisimple Lie groups, including $SU(n)$ for $n\ge 2$, $SO(3)$ and $SL(n,\mathbb{R})$ for $n\ge 2$. It is nondegeneracy, not positive definiteness, that matters. Groups with a nonzero central Lie algebra, such as $U(3)$, do not satisfy this Killing-form criterion, although particular actions can still have equivariant moment maps.

The isotropic oscillator. On $T^{\ast }{\mathbb{R}}^3$, putThe action $z\mapsto Uz$, $U\in U(3)$, preserves the Hermitian norm and the symplectic form, and therefore preserves $H$. A Hermitian matrix $A$ parametrizes the Lie algebra generator $\xi =-iA$. Its unitary moment map of an isotropic oscillator isSince $\{z_j,{\bar{z}}_k\}=-i{\delta }_{jk}$, direct differentiation yieldsThe matrix $-i[A,B]$ is Hermitian and corresponds precisely to $[{\xi }_A,{\xi }_B]$; no central defect remains. The full moment map can be identified, using the pairing with Hermitian matrices, with the rank-one matrix $z{z}^{†}$. The central generator $A=I$ gives ${\mu }_I=H$, whose flow is $z(t)=e^{-it}z(0)$. Globally $U(3)=(SU(3)\times U(1))/{\mathbb{Z}}_3$, so the “$U(1)$ factor” refers to this central circle, rather than a literal direct-product decomposition.

The real orthogonal subgroup acts geometrically by $(x,p)\mapsto (Rx,Rp)$, the cotangent lift of a diffeomorphism of ${\mathbb{R}}^3$. If $K$ is real antisymmetric and $A=iK$, then ${\mu }_A=-x^{T}Kp$. For $Kv=\Omega \times v$, this is $\Omega \cdot (x\times p)$, so its three components are the usual angular momentum $L_i={ϵ}_{ijk}{x}_j{p}_k$.

The remaining six independent Hermitian directions can be taken as real symmetric matrices. Their moment functions spanThe diagonal ones are the individual oscillator energies; their sum is $H$. The off-diagonal ones correlate two oscillator modes. Their flows mix positions and momenta and are not cotangent lifts of the obvious spatial rotation action: for a real symmetric $A$, $\dot{x}=Ap$ and $\dot{p}=-Ax$, rather than $\dot{x}=Kx$, $\dot{p}=Kp$. All nine quadratic functions commute with $H$, since the entire unitary group action preserves the oscillator Hamiltonian.

For a covector $p\in T_x^{\ast }X$, define the cotangent lift of a diffeomorphism byThe inverse derivative here maps $T_{f(x)}X$ to $T_{x}X$. This is a smooth diffeomorphism, its inverse is $(f^{-1}{)}_{\#}$, and it satisfies $\pi \circ f_{\#}=f\circ \pi$.

For $v\in T_{(x,p)}{T}^{\ast }X$, the pullback of a differential form and the defining formula for the Liouville one-form giveHence $f_{\#}^{\ast }\alpha =\alpha$, and applying the exterior derivative also proves $f_{\#}^{\ast }\omega =\omega$. The cotangent lift of a diffeomorphism thus preserves both canonical forms, with no extra choice of metric or connection.

Fix a smaller angular strip and a momentum ball $B^{\prime }⋐B$; the canonical transformation will be defined there and map into the original domain for small $ϵ$. This domain restriction is natural for a near-identity change of coordinates with a nonzero momentum shift.

First solve the scalar cohomological equation on a Diophantine torusThe preceding result gives a real-analytic periodic $u$ on every smaller strip. We need a near-identity angular diffeomorphism $\chi$ and a constant vector $a$ such thatHere is a complete analytic construction of this translated conjugacy of a Diophantine vector field.

Write $w=ϵ\nabla u$ and, for a current approximation, putAssume $A$ is close to the identity. For a correction $\Delta \chi =Av$, the linearized defect isThe identity follows by differentiating the definition of $e$, so the term $(De)v$ already contains the current error. ChooseThe matrix average is invertible near the identity, and the right-hand side has zero mean. The preceding torus small-divisor estimate therefore solves this vector equation componentwise. Update ${\chi }_{+}=\chi +Av$, $a_{+}=a+\Delta a$. The exact new error isIt is quadratic in the current defect.

For convergence, reserve a fixed outer strip on which $w$ is analytic and work on shrinking inner strips whose total loss is less than half the starting width. Let $s=n+\tau$ and choose loss parameters ${\delta }_j=d{2}^{-j}$ with $0<d<1$ small enough that the fixed multiples of these losses used at each stage fit in the reserved total width. The torus small-divisor estimate and Cauchy estimates give, with constants uniform while $A$ stays close to the identity and $\chi$ remains in the reserved strip,andIndeed, the first remainder is bounded using $\Vert D{e}_j\Vert \le C{\delta }_j^{-1}\Vert e_j\Vert$, and the second using the uniformly bounded second derivatives of $w$ on the reserved strip. Taking a slightly larger exponent $\mu$ covers all these losses. Start with ${\chi }_0=\mathrm{id}$ and $a_0=0$, so $e_0=-w=O(ϵ)$. If a small constant $K$ satisfies $C{d}^{-\mu }K{2}^{2\mu }\le 1$ and $\Vert e_0\Vert \le K$, induction gives $\Vert e_j\Vert \le K{2}^{-2\mu j}$. The estimates on $\Delta {\chi }_j$ and $\Delta a_j$ are summable. Choosing $ϵ$ still smaller keeps the derivatives close to the identity and the images inside the reserved strip, closing the induction. The real periodic analytic limits $\chi ,a$ solve the conjugacy equation. Since $⟨w⟩=0$, the very first constant correction is zero; the first new defect is $O({ϵ}^2)$, and the remaining summed constant corrections are $O({ϵ}^2)$. Also the summed change of $\chi$ is $O(ϵ)$. This proves the claimed estimates as well as existence, without leaving an unsolved linear remainder.

Now set $\eta (x)=ϵ\nabla u(x)+a$ and define the symplectic cotangent lift with a closed momentum shiftThe cotangent lift of a diffeomorphism is symplectic. Translation by $\eta$ is also symplectic because the one-form $\eta \cdot dx=ϵdu+a\cdot dx$ is closed. The constant term need not be exact on the flat torus; closedness is sufficient. Thus their composition is the required near-identity canonical transformation.

Expanding the quadratic kinetic term gives a transformed linear coefficient $D{\chi }^{-1}(\omega +\eta \circ \chi )$, which is exactly $\omega$ by the conjugacy equation. The constant-in-$y^{\prime }$ part issince ${\mathcal{D}}_{\omega }u=⟨V⟩-V$. Consequently defineand, for $ϵ\ne 0$,In the exact identities, squared Euclidean norms mean the real polynomial ${\sum }_j{v}_j^2$, continued bilinearly on complex strips, without complex conjugation. The latter is uniformly bounded and analytic on the retained strip because $a=O({ϵ}^2)$. At $ϵ=0$ use the identity map, $E_0=0$, and $Q_0=|y^{\prime }{|}^2$. We have the exact requested normal formHere $E_{ϵ}=O(ϵ)$, $Q_{ϵ}$ is genuinely homogeneous quadratic in $y^{\prime }$, and its coefficient matrix differs from the identity by $O(ϵ)$. Thus $|Q_{ϵ}-|y^{\prime }{|}^2|\le C|ϵ||y^{\prime }{|}^2$. The remainder depends only on the angular coordinate. If the printed final $\tilde{V}(x)$ uses the old angle $x$, the same term is $\frac{1}{2}|\nabla u(x)+a/ϵ{|}^2$ evaluated at $x=\chi (x^{\prime })$; this is just the corresponding coordinate convention.

Define the cotangent lift of a diffeomorphismIf $\alpha$ is the canonical one-form and $W\in T_{(x,\xi )}{T}^{\ast }X$, thenHence $f_{\#}^{\ast }\alpha =\alpha$.

A diffeomorphism of the entire cotangent bundle preserving the Liouville one-form is the cotangent lift of a diffeomorphism of the base. It preserves the zero section and the Liouville vector field, hence commutes with the cotangent fiber-dilation flow. The limit as the fiber contracts forces it to cover its restriction to the zero section. Evaluation of the preserved one-form then forces the inverse-transpose derivative action on each covector.

If $x=\chi (x^{\prime })$ is an angular diffeomorphism and $\eta \cdot dx$ is a closed differential one-form, the map $y=\eta (\chi (x^{\prime }))+D\chi (x^{\prime }{)}^{-T}{y}^{\prime }$ is a symplectomorphism. It combines a cotangent lift of a diffeomorphism with translation by a closed differential one-form. In particular $\eta =du+a$ is allowed on a torus even when the constant one-form $a\cdot dx$ is not exact.