Past exam of the mathematics course of the University of Cambridge 2003 iii Paper 57 5 b Solution Created 2026-10-03 Updated 2026-10-14
Let be the moment map for the circle group, assume is a regular value, and assume the circle action on is free. The action is proper because the circle is compact. Then has dimension and is a smooth manifold of dimension .
If generates the action, for every . In fact is exactly the line spanned by , since is a hyperplane and 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 withThis proves the Marsden-Weinstein theorem in the circle case. If the Hamiltonian is invariant, then : its Hamiltonian flow stays on , projects to the quotient and is generated there by the descended function . This is a reduced Hamiltonian system in 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- particle in the plane with central force potential , work on and writeExterior differentiation gives . Rotation translates and has moment map . Fix and quotient out to obtain the planar rotational symplectic reductionThe reduced equations are and . The centrifugal term arises from the conserved angular momentum; it is not an extra force imposed by hand. Reconstruct the angle from . For the origin cannot lie on the level; for the same local reduction holds away from the origin, while the full phase-space origin is a singular fixed orbit.
Past exam of the mathematics course of the University of Cambridge 2006 iii Paper 64 6 Solution Created 2026-10-03 Updated 2026-10-14
The Marsden-Weinstein theorem needs a Hamiltonian action, not merely an action by symplectomorphisms. Assume an equivariant moment map exists, choose a coadjoint-fixed regular value (zero is the usual choice), and assume the action on is free and proper. Regularity makes a submanifold of codimension . Equivariance and invariance of make invariant under , and freeness and properness give a smooth quotient . Thus
Let and . For and ,Moreover the conormal of is spanned by the independent , so nondegeneracy of gives . The coadjoint-fixed value makes these orbit directions tangent to . Hence the kernel of is exactly the orbit tangent space. The restricted form is invariant and horizontal, so descends uniquely to a two-form withIt is closed because is closed. If a quotient tangent vector annihilates , any lift lies in the kernel just identified and is vertical; the quotient vector is therefore zero. This proves nondegeneracy and completes symplectic reduction.
If the action is not free, singular strata or orbifold phenomena may occur. At a general noncentral regular value one quotients by its coadjoint stabilizer , not by all of ; the regular dimension is . These are genuine hypotheses behind the stated dimension formula.
Both proposed examples can be given explicitly. For symplectic reduction of an isotropic oscillator, take two modes withIts Hamiltonian flow is , a circle action with moment map . For , is the sphere of radius , and the action is free. The Hopf fibration identifies its quotient with . On the chart , use and choose a section with real positive. Pulling back to this section givesWriting gives , whose integral is . This is the Fubini-Study form scaled by in the convention of area at unit scale. The reduced dimension is . At , the circle fixes the origin and the smooth regular-level hypotheses fail.
For a unit-mass free particle in the plane, use the rotation circle action with angular-momentum moment map . Choose so the entire level avoids and the action is free. Polar canonical coordinates haveFixing and quotienting the rotation angle leavesThis planar rotational symplectic reduction replaces a planar free trajectory by radial motion with the centrifugal potential. A nonzero moment value is allowed because the circle is abelian and every value is coadjoint-fixed.