Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 140 6 e Solution Created 2026-10-03 Updated 2026-10-05
The level equation is . In particular , so the global affine coordinates on this level are . They identify it withThe diagonal circle action becomes and is free: at least one coordinate is nonzero, so fixing a point forces . Since the circle is compact, the Free proper Lie-group action theorem makes the quotient a smooth two-dimensional smooth manifold.
The map is the Hopf fibration. It is onto the complex projective line, and two unit representatives determine the same complex line precisely when they differ by a scalar of modulus one, hence by a circle orbit. Local sections, for example , show that the induced identification is a diffeomorphism. ThereforeThis identifies the symplectic reduction as a smooth manifold; no computation of its reduced symplectic form is needed.
Past exam of the mathematics course of the University of Cambridge 2021 iii Paper 115 3 c Solution 2026-09-28
The points are ordered orthonormal two-frames, so is the Stiefel manifold . Changing an orthonormal basis of the same plane gives the displayed free right -action. Since is compact, the action is proper, and the quotient is the Grassmannian of unoriented two-planes. The Free proper Lie-group action theorem makesa principal -bundle.