= Solution
Use $\iota_{X_f}\omega=df$ and $\{f,g\}=\omega(X_f,X_g)$. For an actual infinitesimal left <Lie group action>, define $\xi_P(p)=\left.\frac d{dt}\right|_0\exp(t\xi)\cdot p$. Its <fundamental vector fields> obey $[\xi_P,\eta_P]=-[\xi,\eta]_P$ with this convention.
\b[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 $H^1_{\rm dR}(P)$ to vanish. If it does for every $\xi$, choose potentials linearly in $\xi$ and define
$$
\langle\mu(p),\xi\rangle=\mu_\xi(p),\qquad d\mu_\xi=\iota_{\xi_P}\omega.
$$
This 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 $H^1_{\rm dR}(P)=0$.
\b[Matching the Lie algebra.] On each connected component, the <moment-map equivariance obstruction> is the constant
$$
c(\xi,\eta)=\{\mu_\xi,\mu_\eta\}-\mu_{[\xi,\eta]}.
$$
Indeed, 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> says
$$
c([\xi,\eta],\zeta)+c([\eta,\zeta],\xi)+c([\zeta,\xi],\eta)=0.
$$
Changing $\mu_\xi$ by a linear constant $b(\xi)$ changes $c$ to $c-b([\xi,\eta])$. Consequently the necessary and sufficient infinitesimal condition is \b[the cocycle must be a coboundary], so that the constants can be chosen with
$$
\boxed{\{\mu_\xi,\mu_\eta\}=\mu_{[\xi,\eta]}.}
$$
Equivalently 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.
\b[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,
$$
\iota_{[\xi,\eta]_P}\omega=d[\omega(\xi_P,\eta_P)]
$$
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 $B$ gives
$$
D[\xi,\eta]=[D\xi,\eta]+[\xi,D\eta].
$$
Choose $Z\in\mathfrak g$ by $B(Z,\xi)=\operatorname{tr}(D\operatorname{ad}_\xi)$, possible because $B$ is nondegenerate. Since $D$ is a derivation, $[D,\operatorname{ad}_\xi]=\operatorname{ad}_{D\xi}$. Cyclicity of the <matrix trace> then gives
$$
\begin{aligned}
B([Z,\xi],\eta)&=B(Z,[\xi,\eta])=\operatorname{tr}(D[\operatorname{ad}_\xi,\operatorname{ad}_\eta])\\
&=\operatorname{tr}([D,\operatorname{ad}_\xi]\operatorname{ad}_\eta)=B(D\xi,\eta).
\end{aligned}
$$
Thus $D=\operatorname{ad}_Z$, 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 $n\ge2$, $SO(3)$ and $SL(n,\mathbb R)$ for $n\ge2$. 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>.
\b[The isotropic oscillator.] On $T^*\mathbb R^3$, put
$$
z_j=\frac{x_j+ip_j}{\sqrt2},\qquad\omega=i\sum_j dz_j\wedge d\bar z_j,\qquad H=z^\dagger z.
$$
The 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> is
$$
\mu_A=z^\dagger Az,\qquad X_{\mu_A}z=-iAz.
$$
Since $\{z_j,\bar z_k\}=-i\delta_{jk}$, direct differentiation yields
$$
\boxed{\{\mu_A,\mu_B\}=-iz^\dagger[A,B]z=\mu_{-i[A,B]}.}
$$
The 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 $zz^\dagger$. The central generator $A=I$ gives \b[$\mu_I=H$], 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 $\mu_A=-x^TKp$. For $Kv=\Omega\times v$, this is $\Omega\cdot(x\times p)$, so its three components are the usual <angular momentum> $L_i=\epsilon_{ijk}x_jp_k$.
The remaining six independent Hermitian directions can be taken as real symmetric matrices. Their moment functions span
$$
Q_{ij}=\frac12(x_ix_j+p_ip_j),\qquad Q_{ij}=Q_{ji}.
$$
The 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>.
Back to article page
