English

The Hurwitz problem for abelian differentials

Geometric Topology 2025-10-13 v1 Number Theory

Abstract

Fix g2g \geq 2. Let t(g)\mathsf{t}(g) be the maximal order of the translation group among all genus-gg abelian differentials. By work of Schlage-Puchta and Weitze-Schmith\"usen, t(g)4(g1)\mathsf{t}(g) \leq 4(g - 1). They also classify the gg attaining this bound. We assume gg is outside this class. We first prove that either t(g)=(2(m+1)/m)(g1)\mathsf{t}(g) = (2(m + 1) / m) (g - 1) for some mN{0}m \in \mathbb{N} \setminus \{0\}, when regular genus-gg origamis exist, or t(g)=2(g1)\mathsf{t}(g) = 2(g - 1), when they do not exist. In the former case, only some values of m>1m > 1 are realizable; m=5m = 5 is the smallest. The resulting set of genera, those satisfying t(g)=(12/5)(g1)\mathsf{t}(g) = (12/5)(g - 1), contains infinitely long arithmetic progressions. The same holds for any odd prime mm congruent to 22 modulo 33. In the latter case, "many" strata of the form H(g1,g1)\mathcal{H}(g - 1, g - 1), H(2kq)\mathcal{H}(2k^q) or H(k2q)\mathcal{H}(k^{2q}), where k1k \geq 1 is an integer and qq is prime, contain no regular origamis; we derive a complete classification. As an application, we exhibit infinite families of genera gg for which t(g)=2(g1)\mathsf{t}(g) = 2(g - 1): g=p+1g = p + 1 for prime p5p \geq 5; g=p2+1g = p^2 + 1 for prime, but not Sophie Germain prime, pp; and g=pq+1g = pq + 1, for distinct primes p,q5p, q \geq 5.

Keywords

Cite

@article{arxiv.2510.09584,
  title  = {The Hurwitz problem for abelian differentials},
  author = {Julien Boulanger and Rodolfo Gutiérrez-Romo and Erwan Lanneau},
  journal= {arXiv preprint arXiv:2510.09584},
  year   = {2025}
}

Comments

46 pages, 1 figure, comments welcome!

R2 v1 2026-07-01T06:29:48.795Z