Past exam of the mathematics course of the University of Cambridge 2018 ii Paper 1 24F Solution Created 2026-09-24 Updated 2026-10-03
A function element is a pair consisting of a domain and a holomorphic function belonging to the given complete analytic function. Its germ of a holomorphic function at is the equivalence class , where two elements are equivalent when they agree on a neighbourhood of .
For every function element and sufficiently small open , setThese sets form a basis for the space of germs of holomorphic functions. The projectionrestricts to a homeomorphism . The inverse projections are compatible complex charts, making a Riemann surface and a covering map and local biholomorphism. In such a chart, the evaluation map on a space of germs has coordinate expressionIt is therefore holomorphic on every chart, hence analytic on each component of .
Now let be a nonconstant analytic map of compact connected Riemann surfaces, and let be its finite branch-value set. Each point of is a branch value of a holomorphic map. At every point of the local degree is one, so the holomorphic inverse-function theorem supplies a neighbourhood on which is biholomorphic. For , the fibre is finite. Choose disjoint inverse neighbourhoods around all its points and shrink their common image; compactness, or equivalently properness of , ensures that no additional preimages enter. Thusis an ordinary unramified covering map.
Fix . Given a loop based at , lift it from every and send to the endpoint of its lift. The path lifting theorem makes this a permutation; concatenation of loops composes the permutations and reversal gives inverses. Its image is the monodromy group of a covering. The punctured surface is path connected. Given two points of the fibre, join them there by a path; its projection is a loop whose lift joins those points. Hence the monodromy action is transitive.
Finally considerIt has degree four. Its finite critical points satisfywith branch values and ; the double poles at zero and infinity give the branch value infinity. For a regular value, the fibre isThe two commuting involutions and preserve and act transitively on this fibre. They generate all four deck transformations, so the covering is normal and its monodromy group isWith the displayed ordering of the fibre, its elements areThis is the monodromy group of z squared plus z to the minus two.
Past exam of the mathematics course of the University of Cambridge 2023 ii Paper 1 24F Solution Created 2026-09-23 Updated 2026-09-25
A germ of a holomorphic function at is an equivalence class of pairs consisting of a neighbourhood of and a holomorphic function . Two pairs represent the same germ if their functions agree on some neighbourhood of .
The space of germs of holomorphic functions isFor every holomorphic on an open , the setis declared open; these sets form a basis. The forgetful mapmaps each homeomorphically onto . The inverses of these restrictions give the charts defining the complex structure, so is a local biholomorphism.
The evaluation map isIt is well-defined by the germ equivalence relation. On the chart , its coordinate expression iswhich is holomorphic. Hence is analytic, as stated in evaluation map on a space of germs.
Now putThis gives the germ surface of the square root of z to the eighth minus one. An explicit gluing description is obtained by pairing the eight roots into four adjacent pairs and cutting the plane along four disjoint arcs joining the members of each pair. On the complement of the cuts choose one branch of . Take two copies, labelled by and , and glue the upper bank of each cut in one copy to the lower bank in the other, and conversely. The cut interiors are restored by the gluing, while the eight endpoints remain absent because they are not in .
On this surface defineFor each , the holomorphic function is nonzero near , so it has a unique local square root with . DefineThis is well-defined, injective, and analytic in the displayed local sheets; its inverse on its image is . Thus is the required analytic embedding, and it intertwines both the forgetful and evaluation maps.
To compactify, first add one point above each of the eight roots of unity. These are simple branch points of the double cover. Since the polynomial has even degree eight, the even-degree hyperelliptic model has two distinct, unbranched points above infinity. Consequently
Finally apply the Riemann-Hurwitz formula to the degree-two meromorphic map . Its only ramification consists of the eight simple finite branch points, soThereforeThis is the compactification of y squared equals x to the eighth minus one.