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

Mikhail Chernavskikh, Altan Erdnigor, Nikita Kalinin, Alexandr Zakharov

TL;DR

This work counts convex equilateral triangulations of $\mathbb{R}P^2$ with three equal-defect conical points by translating the problem into centrally symmetric equilateral triangulations of $S^2$ and then into lattice-point problems in the Eisenstein and Epstein frameworks. The moduli space of flat metrics on $S^2$ with six centrally symmetric conical points is parametrized by $(a,b,c,d)$ via a quadratic form $Q(a,b,c,d)=ab+ac+ad+bc+bd+cd$, yielding quadratic lattice growth and explicit volume constants. A key parametrization uses the shifted Eisenstein lattice $\widetilde{\mathit{Eis}}$, establishing a bijection between labelled triangulations and 4-tuples $(\vec{a},\vec{b},\vec{c},\vec{d})$ modulo a $\mathbb{Z}_6$ action, which, together with zeta-function techniques, produces the main asymptotic $f(n)=\frac{1}{20}\sqrt{3}\,\mathrm{Л}(\frac{\pi}{3})\zeta^{-1}(4)\zeta(\mathrm{Eis},2)\,n^2+O(n^{3/2})$. The constant $C$ is numerically $0.2087\ldots$, and extensive computational data for small $n$ align with the asymptotic growth, validating the method and the predicted quadratic rate.

Abstract

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 $ω^6=1$.

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

TL;DR

This work counts convex equilateral triangulations of with three equal-defect conical points by translating the problem into centrally symmetric equilateral triangulations of and then into lattice-point problems in the Eisenstein and Epstein frameworks. The moduli space of flat metrics on with six centrally symmetric conical points is parametrized by via a quadratic form , yielding quadratic lattice growth and explicit volume constants. A key parametrization uses the shifted Eisenstein lattice , establishing a bijection between labelled triangulations and 4-tuples modulo a action, which, together with zeta-function techniques, produces the main asymptotic . The constant is numerically , and extensive computational data for small align with the asymptotic growth, validating the method and the predicted quadratic rate.

Abstract

Consider triangulations of whose all vertices have valency six except three vertices of valency . In this chapter we prove that the number of such triangulations with no more than triangles grows as where , where is the Lobachevsky function and , and .

Paper Structure

This paper contains 6 sections, 8 theorems, 50 equations, 8 figures.

Table of Contents

  1. Introduction
  2. Triangulations of $\mathbb{R} P^2$ with three marked points with defects $2\pi/3$
  3. Moduli space of flat metrics on $S^2$ with six pair-wise centrally symmetric conical points of equal defect
  4. A parametrization of equilateral triangulations of $S^2$ with six centrally-symmetric points with defects $2\pi/3$
  5. Examples and computer computations
  6. Acknowledgments

Key Result

Lemma 2.1

There exist a bijection between labelled triangulations $(T,A,B,C,\Delta)$ of $\mathbb{R} P^2$ with $n$ triangles and labelled triangulations $(\tilde{T},A,B,C)$ with $2n$ triangles.

Figures (8)

  • Figure 1: An octahedron with vertices $AA'BB'CC'$, two Fermat--Torricelli points in the faces $ABC, ACB'$, and the corresponding parallelogram with sides $a,b$.
  • Figure 2: A developing of the octahedron $AA'BB'CC'$ on $\mathbb{R}^2$ is presented, $A=0$, vertices $A,B,C,A',B',C'$ go to the lattice $\mathit{Eis}$. Note that a triangulation of $\mathbb{R} P^2$ can be obtained from the grey area by gluing $AB$ to $AB$, then $BC$ to $B'C'$ and then $BC'$ to $B'C$. Note that the Torricelli centers of the faces do not belong to $\mathit{Eis}$ but belong to $\widetilde{Eis}$, e.g. see the Torricelli center of $ABC$.
  • Figure 3: In this case $z=\frac{1+\omega}{3}=\frac{1}{1-\omega^2}, (a,b,c,d)=(1,1,1,1)$.
  • Figure 4: In this case $z=1,(a,b,c,d)=(1,1,1,0)$, and we count this triangulation four times as $(a,b,c,d)=(1,1,1,0),(1,1,0,1),(1,0,1,1),(0,1,1,1)$. Two of them are isometric while another two differ by relabelling $B\to C$.
  • Figure 5: The leftmost picture: $z=\frac{1+\omega}{3},(a,b,c,d)=(4,1,1,1)$ (counted four times). The central pictures and the rightmost picture are representatives for the tuple $(2,1,1,0)$ (counted $12=8+4$ times). Namely, $z=1,(a,b,c,d)=(2,1,1,0)$ in the central picture (counted eight times). The rightmost picture: $z=1,(a,b,c,d)=(1,2,1,0)$ (counted four times).
  • ...and 3 more figures

Theorems & Definitions (15)

  • Lemma 2.1
  • Lemma 3.1
  • proof
  • Theorem 3.1
  • proof
  • Theorem 3.2
  • Lemma 4.1
  • proof
  • Definition 1
  • Lemma 4.2
  • ...and 5 more