The Hurwitz problem for abelian differentials
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arXiv
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| Natura: | Preprint |
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2025
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| author | Boulanger, Julien Gutiérrez-Romo, Rodolfo Lanneau, Erwan |
| author_facet | Boulanger, Julien Gutiérrez-Romo, Rodolfo Lanneau, Erwan |
| contents | Fix $g \geq 2$. Let $\mathsf{t}(g)$ be the maximal order of the translation group among all genus-$g$ abelian differentials. By work of Schlage-Puchta and Weitze-Schmithüsen, $\mathsf{t}(g) \leq 4(g - 1)$. They also classify the $g$ attaining this bound. We assume $g$ is outside this class.
We first prove that either $\mathsf{t}(g) = (2(m + 1) / m) (g - 1)$ for some $m \in \mathbb{N} \setminus \{0\}$, when regular genus-$g$ origamis exist, or $\mathsf{t}(g) = 2(g - 1)$, when they do not exist.
In the former case, only some values of $m > 1$ are realizable; $m = 5$ is the smallest. The resulting set of genera, those satisfying $\mathsf{t}(g) = (12/5)(g - 1)$, contains infinitely long arithmetic progressions. The same holds for any odd prime $m$ congruent to $2$ modulo $3$.
In the latter case, "many" strata of the form $\mathcal{H}(g - 1, g - 1)$, $\mathcal{H}(2k^q)$ or $\mathcal{H}(k^{2q})$, where $k \geq 1$ is an integer and $q$ is prime, contain no regular origamis; we derive a complete classification. As an application, we exhibit infinite families of genera $g$ for which $\mathsf{t}(g) = 2(g - 1)$: $g = p + 1$ for prime $p \geq 5$; $g = p^2 + 1$ for prime, but not Sophie Germain prime, $p$; and $g = pq + 1$, for distinct primes $p, q \geq 5$. |
| format | Preprint |
| id |
arxiv_https___arxiv_org_abs_2510_09584 |
| institution | arXiv |
| publishDate | 2025 |
| record_format | arxiv |
| spellingShingle | The Hurwitz problem for abelian differentials Boulanger, Julien Gutiérrez-Romo, Rodolfo Lanneau, Erwan Geometric Topology Number Theory 30F30, 14H37, 32G15, 20D60 Fix $g \geq 2$. Let $\mathsf{t}(g)$ be the maximal order of the translation group among all genus-$g$ abelian differentials. By work of Schlage-Puchta and Weitze-Schmithüsen, $\mathsf{t}(g) \leq 4(g - 1)$. They also classify the $g$ attaining this bound. We assume $g$ is outside this class. We first prove that either $\mathsf{t}(g) = (2(m + 1) / m) (g - 1)$ for some $m \in \mathbb{N} \setminus \{0\}$, when regular genus-$g$ origamis exist, or $\mathsf{t}(g) = 2(g - 1)$, when they do not exist. In the former case, only some values of $m > 1$ are realizable; $m = 5$ is the smallest. The resulting set of genera, those satisfying $\mathsf{t}(g) = (12/5)(g - 1)$, contains infinitely long arithmetic progressions. The same holds for any odd prime $m$ congruent to $2$ modulo $3$. In the latter case, "many" strata of the form $\mathcal{H}(g - 1, g - 1)$, $\mathcal{H}(2k^q)$ or $\mathcal{H}(k^{2q})$, where $k \geq 1$ is an integer and $q$ is prime, contain no regular origamis; we derive a complete classification. As an application, we exhibit infinite families of genera $g$ for which $\mathsf{t}(g) = 2(g - 1)$: $g = p + 1$ for prime $p \geq 5$; $g = p^2 + 1$ for prime, but not Sophie Germain prime, $p$; and $g = pq + 1$, for distinct primes $p, q \geq 5$. |
| title | The Hurwitz problem for abelian differentials |
| topic | Geometric Topology Number Theory 30F30, 14H37, 32G15, 20D60 |
| url | https://arxiv.org/abs/2510.09584 |