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

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.