Equilateral convex triangulations of $\mathbb R P^2$ with three conical points of equal defect

Fuente: arXiv
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Autores principales: Chernavskikh, Mikhail, Erdnigor, Altan, Kalinin, Nikita, Zakharov, Alexandr
Formato: Preprint
Publicado: 2021
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author Chernavskikh, Mikhail
Erdnigor, Altan
Kalinin, Nikita
Zakharov, Alexandr
author_facet Chernavskikh, Mikhail
Erdnigor, Altan
Kalinin, Nikita
Zakharov, Alexandr
contents Consider triangulations of $\mathbb R P^2$ whose all vertices have valency six except three vertices of valency $4$. In this chapter we prove that the number $f(n)$ of such triangulations with no more than $n$ triangles grows as $C\cdot n^2+ O(n^{3/2})$ where $C = \frac{1}{20} \sqrt{3} \cdot L( \fracπ{3} ) ζ^{-1}(4) ζ(Eis, 2) \approx 0.2087432125056015...$, where $L$ is the Lobachevsky function and $ζ(Eis,2) =\sum\limits_{(a,b)\in\mathbb Z^2\setminus 0}{\frac{1}{|a+bω^2|^4}}$, and $ω^6=1$.
format Preprint
id arxiv_https___arxiv_org_abs_2111_04680
institution arXiv
publishDate 2021
record_format arxiv
spellingShingle Equilateral convex triangulations of $\mathbb R P^2$ with three conical points of equal defect
Chernavskikh, Mikhail
Erdnigor, Altan
Kalinin, Nikita
Zakharov, Alexandr
Combinatorics
Differential Geometry
Metric Geometry
Number Theory
51M09, 57N45, 11P21, 11M36, 11E45
Consider triangulations of $\mathbb R P^2$ whose all vertices have valency six except three vertices of valency $4$. In this chapter we prove that the number $f(n)$ of such triangulations with no more than $n$ triangles grows as $C\cdot n^2+ O(n^{3/2})$ where $C = \frac{1}{20} \sqrt{3} \cdot L( \fracπ{3} ) ζ^{-1}(4) ζ(Eis, 2) \approx 0.2087432125056015...$, where $L$ is the Lobachevsky function and $ζ(Eis,2) =\sum\limits_{(a,b)\in\mathbb Z^2\setminus 0}{\frac{1}{|a+bω^2|^4}}$, and $ω^6=1$.
title Equilateral convex triangulations of $\mathbb R P^2$ with three conical points of equal defect
topic Combinatorics
Differential Geometry
Metric Geometry
Number Theory
51M09, 57N45, 11P21, 11M36, 11E45
url https://arxiv.org/abs/2111.04680