A History of the Hurwitz Problem Concerning Branched Coverings

Author

James Alexander Byars, Boise State UniversityFollow

Publication Date

8-2023

Date of Final Oral Examination (Defense)

June 2023

Type of Culminating Activity

Thesis

Degree Title

Master of Science in Mathematics

Department

Mathematics

Major Advisor

Jens Harlander, Ph.D.

Advisor

Uwe Kaiser, Ph.D.

Advisor

John Clemens, Ph.D.

Abstract

From a d-sheeted branched covering f : M → N, where M and N are surfaces, one can read off the branch datum
D(f) = {M, N, d, (A₁, . . . , A_r)},
where A_i = [e_i1, . . . , e_ini] is a partition of d. Furthermore, a relationship between the Euler characteristics between M and N is known, called the Riemann-Hurwitz formula
𝒳(M) = d𝒳(N) − v(D(f)),
where v(D(f)) = ∑(e_ij − 1). The Hurwitz problem asks for a characterization of realizable abstract branching data. More precisely, given an abstract branch datum
D = {M, N, d, (A₁, . . . , A_r)},
where
𝒳(M) = d𝒳(N) − v(D),
how do we know that there is a branched covering f : M → N such that D = D(f)?

Hurwitz himself found an answer (at least for N = S²) in terms of permutations α₁, . . . , α_r ∈ S_d, whose cycle structures realize the given partitions: the cycle structure of α_i is A_i.

Eighty years later Gersten solved the Hurwitz problem, at least in the case M = N = S², using cancellation diagrams. He showed that the datum D = {S², S², d, (A₁, . . . , A_r)}, where 𝒳(S²) = d𝒳(S²) − v(D), that is 2 = 2d − v(D), is realizable if a cancellation diagram exists for a family of words in the group
Γ = < x₁, . . . , x_r | x₁ . . . x_r = 1 >
that is determined by the given partitions. The appeal of Gersten's solution lies in the fact that the cancellation diagram depicts the actual branched covering. Thus, his solution is more concrete than the solution of Hurwitz.