Past exam of the mathematics course of the University of Cambridge 2015 ii Paper 2 20F i Solution Created 2026-09-24 Updated 2026-10-06
Take and the complete analytic function generated by a logarithm germ. Its continuation surface identifies with via with and logarithm value . Thus . It is a universal cover; the deck transformations act transitively on each fibre, so it is also a regular covering map in the deck-transitive sense. It is not bijective, since every nonzero has infinitely many logarithm values. It also meets the alternative convention in which “regular” merely requires an evenly covered neighbourhood at each base point.
Past exam of the mathematics course of the University of Cambridge 2015 ii Paper 2 20F Solution 2026-10-06
Two holomorphic function elements represent the same germ of a holomorphic function at if they agree on some neighbourhood of . For a holomorphic on an open , set . These sets form a basis of a topology on the space of germs of holomorphic functions: every germ lies in one, and if a germ lies in two, equality of the representatives on a smaller neighbourhood supplies a basis of a topology set in their intersection. Declaring arbitrary unions of basis of a topology sets open therefore gives a topology.
Different base points are separated by disjoint small discs in . If two different germs have the same base point, choose representatives on one connected small disc. Their sheets are disjoint: any common germ elsewhere on that disc would force equality throughout the disc by the identity theorem. This proves the Hausdorff property.
The projection maps each sheet homeomorphically onto . Use these maps as complex charts. On overlaps their transition functions are the identity, hence holomorphic; thus each connected component is a Riemann surface, and is a local biholomorphism, an analytic germ projection. For completeness, second countability of a component follows by following continuation chains through a countable basis of a topology of rational discs: starting at one germ, finite chains of such discs give only countably many continuation sheets, by uniqueness on overlaps, and they cover every germ in that component.
A complete analytic function consists of all germs reachable from one initial germ by analytic continuation. The supplied path-lifting equivalence identifies these with its path component. Since the germ space is locally path connected, connected components and path components agree. Thus its continuation surface is exactly one component .
A local biholomorphism need not evenly cover the whole base; the germ projection is an analytic covering in this local sense. The examples below distinguish a genuine regular covering map from a surjective nonregular projection, with the meaning of regular stated explicitly.
Past exam of the mathematics course of the University of Cambridge 2026 ii Paper 1 24G a Solution Created 2026-09-24 Updated 2026-09-25
A covering evenly covers every base point. It is regular covering map when deck transformations act transitively on fibres. For a concrete nonregular example, restrict toIt is a surjective three-sheeted unbranched cover with monodromy , so its deck group is not transitive.