Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 57 / 5

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.

Let $\mu :P\to \mathbb{R}$ be the moment map for the circle group, assume $c$ is a regular value, and assume the circle action on $N={\mu }^{-1}(c)$ is free. The action is proper because the circle is compact. Then $N$ has dimension $2n-1$ and $P_c=N/SO(2)$ is a smooth manifold of dimension $2n-2$.

If ${\xi }_P$ generates the action, $\omega ({\xi }_P,v)=d\mu (v)=0$ for every $v\in TN$. In fact $(TN{)}^{\omega }$ is exactly the line spanned by ${\xi }_P$, since $TN=\ker d\mu$ is a hyperplane and $\omega$ is nondegenerate. Therefore the kernel of the restricted two-form is precisely the orbit direction. The restriction is invariant and horizontal, so it descends to a unique nondegenerate closed two-form ${\omega }_c$ withThis proves the Marsden-Weinstein theorem in the circle case. If the Hamiltonian $H$ is invariant, then $\{\mu ,H\}=0$: its Hamiltonian flow stays on $N$, projects to the quotient and is generated there by the descended function $H_c$. This is a reduced Hamiltonian system in $2n-2$ local variables. At a critical level or a nonfree orbit the quotient can be singular, so the smooth dimension assertion needs the stated hypotheses.

For a mass-$m$ particle in the plane with central force potential $V(r)$, work on $r>0$ and writeExterior differentiation gives $\omega =dr\wedge d{p}_r+d\theta \wedge d\ell$. Rotation translates $\theta$ and has moment map $\ell$. Fix $\ell ={\ell }_0$ and quotient out $\theta$ to obtain the planar rotational symplectic reductionThe reduced equations are $\dot{r}=p_r/m$ and ${\dot{p}}_r={\ell }_0^2/(m{r}^3)-V^{\prime }(r)$. The centrifugal term arises from the conserved angular momentum; it is not an extra force imposed by hand. Reconstruct the angle from $\dot{\theta }={\ell }_0/(m{r}^2)$. For ${\ell }_0\ne 0$ the origin cannot lie on the level; for ${\ell }_0=0$ the same local reduction holds away from the origin, while the full phase-space origin is a singular fixed orbit.