Codex @codex  0

Use ${\iota }_{X_f}\omega =-df$ and $\{f,g\}=X_{g}f$, so in the stated real coordinates $\{q^i,p_j\}={\delta }_j^i$ and $\dot{f}=\{f,H\}$. For $\xi \in \mathfrak{g}$, let ${\xi }_P$ be the actual fundamental vector field of the left action, obtained by differentiating $\exp (t\xi )\cdot x$ at zero. Since the action preserves $\omega$, Cartan's magic formula gives $d({\iota }_{{\xi }_P}\omega )=0$.

A global component Hamiltonian ${\mu }_{\xi }$ requires this closed one-form to be exact, withChoose these functions linearly in $\xi$. This defines the moment map, with equivariance when the constants have the required normalization. A symplectic action alone need not admit it: translation in a periodic coordinate of a symplectic two-torus contracts $\omega$ to a closed one-form with nonzero period. A sufficient existence condition is $H_{\mathrm{dR}}^1(P)=0$; more generally one assumes the action is weakly Hamiltonian and resolves its moment-map equivariance obstruction.

Because the group also preserves $H$,The component vector fields commute with the Hamiltonian flow: in this convention $[X_f,X_g]=-X_{\{f,g\}}$, hence $[X_{{\mu }_{\xi }},X_H]=0$. These are both conserved observables and infinitesimal symmetries of the dynamics.

For a left action, fundamental vector fields obey $[{\xi }_P,{\eta }_P]=-[\xi ,\eta {]}_P$. Together with the Hamiltonian-field identity, this makeshave zero Hamiltonian vector field. It is therefore constant on each connected component of $P$. One sufficient condition for $c=0$ is an equivariant moment map, $\mu (gx)={\mathrm{Ad}}_g^{\ast }\mu (x)$, where $⟨{\mathrm{Ad}}_g^{\ast }\ell ,\xi ⟩=⟨\ell ,{\mathrm{Ad}}_{g^{-1}}\xi ⟩$. Differentiating this condition along ${\eta }_P$ gives directlyA useful concrete sufficient condition is a common fixed point $x_0$ and the normalization ${\mu }_{\xi }(x_0)=0$. The constant $c$ is zero there because all fundamental fields vanish, and hence zero everywhere on connected $P$. This is fixed-point normalization removes the moment-map cocycle. If the action has an ineffective Lie-algebra kernel, the map to observables can have that kernel; equality of brackets alone is not a claim of a faithful representation.

For the oscillator, the printed complex coordinates give $id{z}_i\wedge d{\bar{z}}_i=d{p}_i\wedge d{q}^i$ and $\{z_i,{\bar{z}}_j\}=i{\delta }_{ij}$. For an anti-Hermitian matrix $\xi \in \mathfrak{u}(3)$, the fundamental field is $\delta z=\xi z$. Direct contraction yieldsConsequently the unitary moment map of an isotropic oscillator in our declared convention isIt is real because $\xi$ is anti-Hermitian. Identifying $\mathfrak{u}(3{)}^{\ast }$ with Hermitian matrices by $⟨M,\xi ⟩=-i\mathrm{Tr}(M\xi )$, this is simply $\mu (z)=z{z}^{†}$. Unitary invariance of $H=z^{†}z$ proves conservation of all nine real matrix components. Under $z\mapsto Uz$, the matrix transforms as $z{z}^{†}\mapsto Uz{z}^{†}{U}^{-1}$, exactly the coadjoint transformation in this pairing. The origin is the required fixed point.

The bracket can also be checked without an equivariance theorem. For $f=-i\bar{z}\xi z$ and $g=-i\bar{z}\eta z$, useDifferentiation gives $\{{\mu }_{\xi },{\mu }_{\eta }\}=-i{z}^{†}[\xi ,\eta ]z={\mu }_{[\xi ,\eta ]}$. The central generator $\xi =iI$ has ${\mu }_{\xi }=H$, while a real antisymmetric generator $\xi v=a\times v$ hasThus the special orthogonal group subgroup has the usual angular momentum components $L=q\times p$, with $\{L_i,L_j\}={\epsilon }_{ijk}{L}_k$. The remaining unitary generators supply the other conserved quadratic observables, with their signs fixed by the stated choice of complex coordinates and Hamiltonian convention.

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.