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For trivial real coefficients, a Lie algebra two-cocycle is an alternating bilinear form satisfying $\kappa ([X,Y],Z)+\kappa ([Y,Z],X)+\kappa ([Z,X],Y)=0$. Adding a form $b([X,Y])$ does not change its Lie algebra cohomology class. The bracket $[(X,a),(Y,b)]=([X,Y],\kappa (X,Y))$ defines a central extension of a Lie algebra by $\mathbb{R}$ precisely when this cocycle identity holds. The constant discrepancy in a moment-map equivariance obstruction is such a cocycle.

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.

Write ${\xi }_P(p)={\frac{d}{dt}|}_0\exp (t\xi )\cdot p$ for the fundamental vector field of a left Lie group action, and use ${\iota }_{X_f}\omega =df$ throughout. Such fundamental fields satisfy $[{\xi }_P,{\eta }_P]=-[\xi ,\eta {]}_P$. A moment map has linear components ${\mu }_{\xi }=⟨\mu ,\xi ⟩$ satisfyingFor the action to be Hamiltonian, these components can be chosen equivariantly, or, infinitesimally,This identifies the span of the component Hamiltonians with the Lie algebra image, rather than identifying the entire infinite-dimensional Poisson algebra $C^{\infty }(P)$ with $\mathfrak{g}$. For an actual isomorphism with $\mathfrak{g}$, injectivity is also needed; a faithful infinitesimal action supplies it.

Preservation of $\omega$ alone gives $d{\iota }_{{\xi }_P}\omega =0$, not necessarily exactness. For example, translation generated by ${\partial }_q$ on the symplectic two-torus has contraction $dp$, whose period is nonzero. That action has no globally defined component Hamiltonian. Once the contractions are exact, choose Hamiltonians linearly in $\xi$. On each connected component of $P$, the discrepancyis constant: apply $[X_f,X_g]=-X_{\{f,g\}}$ and the fundamental-field bracket. The Jacobi identity makes $\kappa$ a Lie algebra two-cocycle. Replacing ${\mu }_{\xi }$ by ${\mu }_{\xi }+b(\xi )$ changes it to $\kappa (\xi ,\eta )-b([\xi ,\eta ])$. Thus the exactness condition and vanishing of this moment-map equivariance obstruction are precisely the required infinitesimal conditions.

For a semisimple Lie algebra, both hold. First Cartan's magic formula givesSemisimplicity implies $[\mathfrak{g},\mathfrak{g}]=\mathfrak{g}$, so every generator is a sum of commutators and has a globally exact contraction. This proves Hamiltonian existence for a semisimple symplectic action.

Next use the nondegenerate Killing form to define $D$ by $B(D\xi ,\eta )=\kappa (\xi ,\eta )$. Antisymmetry makes $D$ skew for $B$. The cocycle identity and invariance of $B$ imply $D[\xi ,\eta ]=[D\xi ,\eta ]+[\xi ,D\eta ]$, so $D$ is a derivation. Choose $Z$ byThen, using cyclicity of trace and $[D,\mathrm{ad}\xi ]=\mathrm{ad}(D\xi )$,Hence $D=\mathrm{ad}Z$ and $\kappa (\xi ,\eta )=B(Z,[\xi ,\eta ])$. Taking $b(\xi )=B(Z,\xi )$ removes the cocycle and proves semisimple moment-map equivariance without merely invoking a cohomology theorem. For a connected group, integrate the infinitesimal equivariance along its one-parameter subgroups to obtain group equivariance. A disconnected group can require an additional discrete-equivariance check.

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.

Use $\omega ={\sum }_{i}d{q}_i\wedge d{p}_i$, ${\iota }_{X_f}\omega =df$ andThen $X_f(g)=\{g,f\}$ and $[X_f,X_g]=-X_{\{f,g\}}$. For a left group action, let ${\xi }_P(p)={\frac{d}{dt}|}_0\exp (t\xi )\cdot p$; its fundamental fields obey $[{\xi }_P,{\eta }_P]=-[\xi ,\eta {]}_P$.

The components of a moment map satisfyThus ${\xi }_P=X_{{\mu }_{\xi }}$. A Hamiltonian moment map additionally has coadjoint equivariance,Here the right side evaluates on $\xi$ as $⟨\mu (p),{\mathrm{Ad}}_{g^{-1}}\xi ⟩$. With that convention its infinitesimal form is $\{{\mu }_{\xi },{\mu }_{\eta }\}={\mu }_{[\xi ,\eta ]}$. Some definitions reserve “moment map” for this equivariant version; choosing Hamiltonians for every generator first gives a weakly Hamiltonian action.

To derive the obstruction, compare the two antihomomorphism identities:On connected $P$, nondegeneracy of $\omega$ makesa constant alternating bilinear form. The two Jacobi identities implyReplacing ${\mu }_{\xi }$ by ${\mu }_{\xi }+b(\xi )$ changes $\kappa$ to $\kappa -b([\xi ,\eta ])$. Thus vanishing of this class in Lie algebra cohomology is the precise moment-map equivariance obstruction. In particular a connected group with $H^2(\mathfrak{g},\mathbb{R})=0$ permits constants to be chosen so that the component Poisson algebra is the Lie algebra. A locally effective action makes this realization faithful; otherwise it realizes the quotient by the infinitesimal kernel. The linear span of the components has the Lie brackets; the entire algebra of polynomial products of them is, of course, larger.

A useful sufficient condition is semisimplicity, which can be proved here without assuming the desired equivariance. Let $B$ be the nondegenerate Killing form and define $D$ by $B(D\xi ,\eta )=\kappa (\xi ,\eta )$. Invariance of $B$ and the cocycle identity giveso $D$ is a derivation. Choose $Z$ uniquely by $B(Z,\xi )=\mathrm{tr}(D\mathrm{ad}\xi )$. Since $[D,\mathrm{ad}\xi ]=\mathrm{ad}(D\xi )$, cyclicity of trace givesHence $D=\mathrm{ad}Z$ and $\kappa (\xi ,\eta )=B(Z,[\xi ,\eta ])$. The shift $b(\xi )=B(Z,\xi )$ removes the obstruction. This proves semisimple moment-map equivariance; for connected $G$ infinitesimal equivariance integrates along its one-parameter subgroups to group equivariance. It does not claim that arbitrary symplectic actions of arbitrary groups have moment maps.

For the Kepler calculation, put $k=Mm$ and work on $r=|\mathbf{r}|>0$, since the Hamiltonian is singular at the origin. The Hamilton equations areThe angular momentum therefore obeysFor the Runge-Lenz vector,The vector triple product identity gives $\mathbf{r}\times \mathbf{L}=\mathbf{r}(\mathbf{r}\cdot \mathbf{p})-r^2\mathbf{p}$. Thereforeand the terms cancel. Since $\dot{f}=\{f,H\}$, this proves

The conserved quantities consequently generate transformations preserving energy. Their brackets give the Kepler dynamical symmetry algebra. On $H<0$, define $\mathbf{A}=\mathbf{K}/\sqrt{-2H}$ as a function on that whole open region. Because $H$ Poisson commutes with both vectors, its brackets areWith ${\mathbf{J}}_{\pm }=(\mathbf{L}\pm \mathbf{A})/2$ one obtains two commuting $\mathfrak{so}(3)$ algebras:Thus the negative-energy algebra is $\mathfrak{so}(4)$. On $H>0$, using $\mathbf{A}=\mathbf{K}/\sqrt{2H}$ instead changes the last bracket to $-{ϵ}_{ijk}{L}_k$, yielding $\mathfrak{so}(3,1)$, with $\mathbf{L}$ rotations and $\mathbf{A}$ boosts. At zero energy the brackets of conserved functions on the characteristic orbit quotient give $\mathfrak{e}(3)=\mathfrak{so}(3)⋉{\mathbb{R}}^3$, with commuting $\mathbf{K}$.

There are two global qualifications. First, an energy hypersurface carries a presymplectic form: its restricted two-form has characteristic direction $X_H$. At $H=0$, the literal vector fields of $\mathbf{K}$ need only commute modulo $X_H$, since differentiating $\{K_i,K_j\}=-2H{ϵ}_{ijk}{L}_k$ still produces an $L_k{X}_H$ term. On nonzero-energy regions the energy-dependent normalization above gives an exact Hamiltonian Lie algebra; one must differentiate that normalization before restricting to a level. Second, completeness is required for a global group action. The collision-excluded Kepler phase space need not have complete hidden-symmetry flows, so the brackets establish local actions (or the appropriate simply connected covers), not an unconditional global action on every unregularized trajectory. Collision regularization supplies the familiar global bound-motion symmetry; this distinction is developed in math.berkeley.edu/~alanw/277papers00/tang.pdf . It does not change the three energy-dependent Lie algebras just derived.