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For a symplectic left Lie group action, use actual fundamental fields ${\xi }_P$ and ${\iota }_{X_f}\omega =df$. They satisfy $[{\xi }_P,{\eta }_P]=-[\xi ,\eta {]}_P$. Cartan's magic formula then gives ${\iota }_{[\xi ,\eta {]}_P}\omega =d[\omega ({\xi }_P,{\eta }_P)]$. Thus every commutator generator has a globally exact contraction. A semisimple Lie algebra is perfect, so every generator has a global Hamiltonian. The separate semisimple moment-map equivariance result removes the constant two-cocycle obstruction and gives an equivariant moment map for a connected group.

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 $\omega ={\sum }_{i}d{q}_i\wedge d{p}_i$, ${\iota }_{X_f}\omega =df$ 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.