Projective Equivalence for the Roots of Unity

TL;DR

This paper classifies pairs of roots of unity that are projectively equivalent under Möbius transformations, establishing a maximum length of 14 for nontrivial such pairs.

Contribution

It provides a complete classification of projectively equivalent roots of unity pairs and determines the maximum length of nontrivial pairs.

Findings

Complete classification of projectively equivalent roots of unity pairs

Maximum length of nontrivial pairs is 14

Implications for symmetry and transformation groups

Abstract

Let $\mu_{\infty}\subseteq\mathbb{C}$ be the collection of roots of unity and $\mathcal{C}_{n}:=\{(s_{1},\cdots,s_{n})\in\mu_{\infty}^{n}:s_{i}\neq s_{j}\text{ for any }1\leq i<j\leq n\}$. Two elements $(s_{1},\cdots,s_{n})$ and $(t_{1},\cdots,t_{n})$ of $\mathcal{C}_{n}$ are said to be projectively equivalent if there exists $\gamma\in\text{PGL}(2,\mathbb{C})$ such that $\gamma(s_{i})=t_{i}$ for any $1\leq i\leq n$. In this article, we will give a complete classification for the projectively equivalent pairs. As a consequence, we will show that the maximal length for the nontrivial projectively equivalent pairs is $14$.

Tables

Table 1. Table 1: The positive-parameter solutions in 𝒮 + / ≈ \mathcal{S}^{+}/\approx

$s_1$	$s_2$	$s_3$	$t_1$	$t_2$	$t_3$
$x$	$y$	$z$	$x$	$y$	$z$
$1/2$	$x$	$-x$	$1/2$	$y$	$-y$
$x$	$1/2+y$	$1/2-y$	$y$	$1/2+x$	$1/2-x$
$x$	$y$	$1/2-x+y$	$2x$	$2y$	$1/2+y$
$x$	$3x$	$1/2+4x$	$4x$	$12x$	$1/2+6x$
$1/6$	$1/3$	$x$	$1/2+x$	$1/3$	$-2x$
$1/6$	$x$	$1/6+x$	$1/3+x$	$3x$	$1/3+2x$
$1/6$	$x$	$1/6+x$	$1/2+x$	$2x$	$1/2+3x$
$1/6$	$x$	$1/6+x$	$1/2+x$	$4x$	$1/2+2x$
$1/6$	$x$	$1/2+x$	$1/2$	$1/6+x$	$1/2+3x$
$1/6$	$x$	$2x$	$1/6+x$	$5/6+2x$	$1/2+3x$
$1/6$	$x$	$2x$	$1/6+x$	$2/3+4x$	$1/3+2x$
$1/6$	$x$	$2x$	$1/3+2x$	$3x$	$6x$
$1/6$	$x$	$1/6+2x$	$1/3+x$	$2x$	$1/3+4x$
$1/6$	$x$	$1/3+2x$	$1/3+x$	$2x$	$1/2+3x$
$1/6$	$x$	$3x$	$1/6+x$	$5/6+3x$	$1/3+2x$
$1/6$	$x$	$3x$	$1/2+x$	$1/6+2x$	$1/3+4x$
$1/6$	$x$	$4x$	$1/3+x$	$5/6+4x$	$1/2+2x$
$1/6$	$x$	$-x$	$1/3+x$	$1/6+x$	$2/3+2x$
$1/6$	$x$	$1/2-x$	$1/6+x$	$x$	$1/2+3x$
$1/6$	$x$	$2/3-x$	$1/6+x$	$5/6$	$1/6-x$
$1/6$	$x$	$2/3-x$	$1/6+2x$	$2x$	$1/3+4x$
$1/6$	$x$	$-2x$	$1/3+x$	$1/6+x$	$1/2+3x$
$1/6$	$x$	$2/3-2x$	$1/2+x$	$5/6$	$1/2-3x$
$1/6$	$1/3+2x$	$1/2+3x$	$x$	$3x$	$1/3+4x$
$1/6$	$1/3+2x$	$1/2+4x$	$x$	$3x$	$1/2+4x$
$1/6$	$1/2+2x$	$2/3+2x$	$x$	$4x$	$1/2+3x$
$1/6$	$1/2+2x$	$1/2+6x$	$x$	$1/6+3x$	$2/3+4x$
$1/3$	$x$	$5/6+2x$	$1/3$	$4x$	$1/2+x$
$1/3$	$x$	$-x$	$1/3$	$1/6+x$	$2/3-2x$
$1/3$	$x$	$5/6-x$	$1/3$	$5/6+2x$	$1/3-4x$
$1/2$	$x$	$1/6-x$	$1/2$	$1/6+x$	$1/2-3x$
$x$	$2x$	$3x$	$1/6+x$	$1/2+3x$	$2/3+2x$
$x$	$1/6+2x$	$1/2+3x$	$5/6+2x$	$1/2+6x$	$1/2+3x$
$x$	$1/6+2x$	$4x$	$5/6+x$	$1/3+4x$	$1/2+2x$
$x$	$1/6+2x$	$2/3+5x$	$5/6+x$	$1/2+6x$	$1/2+2x$
$x$	$1/3+2x$	$3x$	$5/6+x$	$1/6+2x$	$1/3+4x$
$x$	$3x$	$9x$	$1/6+x$	$1/2+4x$	$2/3+3x$
$x$	$1/2+3x$	$4x$	$1/6+2x$	$2/3+4x$	$1/2+6x$
$x$	$4x$	$1/3+5x$	$1/6+x$	$1/2+6x$	$2/3+4x$
$1/12$	$1/4$	$5/6$	$x$	$1/2$	$-x$
$1/12$	$x$	$5/12+x$	$1/6+x$	$3x$	$1/3+2x$
$1/12$	$x$	$5/12+x$	$2/3+x$	$1/3+2x$	$2/3+4x$
$1/12$	$x$	$5/12+x$	$1/3+2x$	$6x$	$1/2+3x$
$1/12$	$1/3+2x$	$3/4+2x$	$x$	$3x$	$1/2+4x$

$w=2,3,5,6$	$R_2$	$R_3$	$R_5$	$\left(R_5:R_3\right)$
$w=7,8$	$\left(R_5:2R_3\right)$	$R_7$	$\left(R_5:3R_3\right)$	$\left(R_7:R_3\right)$
$w=9,10$	$\left(R_5:4R_3\right)$	$\left(R_7:2R_3\right)$	$\left(R_7:3R_3\right)$	$\left(R_7:R_5\right)$
$w=11$	$\left(R_7:4R_3\right)$	$\left(R_7:R_5,R_3\right)$	$(R_7:(R_5:R_3))$	$R_{11}$
$w=12$	$\left(R_7:5R_3\right)$	$\left(R_7:R_5,2R_3\right)$	$(R_7:(R_5:R_3),R_3)$	$(R_7:(R_5:2R_3))$	$\left(R_{11}:R_3\right)$

$O_1$	$\{A_1,A_2,A_3,B_1,B_2,A_1^{\prime }\}$	$192$	$(1,1,1,1,1,1)$
$O_2$	$\{A_1,A_2,A_3,B_1,B_2,A_3^{\prime }\}$	$96$	$(1,1,1,1,1,1)$
$O_3$	$\{A_1,A_2,A_3,B_1,B_2,B_1^{\prime }\}$	$192$	$(1,1,1,1,1,1)$
$O_4$	$\{A_1,A_2,A_3,B_1,B_2,B_3^{\prime }\}$	$32$	$(1,1,1,1,1,1)$
$O_5$	$\{A_1,A_2,A_3,B_1,A_1^{\prime },A_2^{\prime }\}$	$48$	$(1,1,1,1,1,2)$
$O_6$	$\{A_1,A_2,A_3,B_1,A_1^{\prime },B_1^{\prime }\}$	$24$	$(1,1,1,1,1,2)$
$O_7$	$\{A_1,A_2,B_1,B_2,A_1^{\prime },A_2^{\prime }\}$	$24$	$(1,1,1,1,1,2)$
$O_8$	$\{A_1,A_2,B_1,B_2,A_1^{\prime },B_1^{\prime }\}$	$24$	$(1,1,1,1,1,2)$
$O_9$	$\{A_1,A_2,A_3,A_1^{\prime },A_2^{\prime },A_3^{\prime }\}$	$2$	$(1,1,1,1,2,2)$
$O_{10}$	$\{A_1,A_2,A_3,B_1,B_2,B_3\}$	$32$	$(1,1,1,1,1,0)$	$A_1{A}_2{A}_3{B}_1{B}_2{B}_3=-1$
$O_{11}$	$\{A_1,A_2,A_3,B_1,A_2^{\prime },A_3^{\prime }\}$	$24$	$(1,1,1,1,1,0)$	$\left(A_3{B}_1{A}_3^{\prime }{B}_1^{\prime }\right)\left(A_2{B}_1\overline{A_2^{\prime }{B}_1^{\prime }}\right)=1$
$O_{12}$	$\{A_1,A_2,A_3,B_1,A_2^{\prime },B_1^{\prime }\}$	$48$	$(1,1,1,1,1,0)$	$A_2{B}_1\overline{A_2^{\prime }{B}_1^{\prime }}=1$
$O_{13}$	$\{A_1,A_2,B_1,B_2,A_1^{\prime },B_2^{\prime }\}$	$24$	$(1,1,1,1,1,0)$	$A_1{B}_2{A}_1^{\prime }{B}_2^{\prime }=1$
$O_{14}$	$\{A_1,A_2,B_1,B_3,A_1^{\prime },A_2^{\prime }\}$	$48$	$(1,1,1,1,1,0)$	$\left(A_2{B}_3{A}_2^{\prime }{B}_3^{\prime }\right)\left(A_1{B}_3\overline{A_1^{\prime }{B}_3^{\prime }}\right)=1$
$O_{15}$	$\{A_1,A_2,B_1,B_3,A_1^{\prime },B_1^{\prime }\}$	$24$	$(1,1,1,1,1,0)$	$\left(A_2{B}_3{A}_2^{\prime }{B}_3^{\prime }\right)\left(A_1{B}_3\overline{A_1^{\prime }{B}_3^{\prime }}\right)\left(A_2{B}_1\overline{A_2^{\prime }{B}_1^{\prime }}\right)=1$
$O_{16}$	$\{A_1,A_2,B_1,B_3,A_1^{\prime },B_3^{\prime }\}$	$48$	$(1,1,1,1,1,0)$	$A_1{B}_3\overline{A_1^{\prime }{B}_3^{\prime }}=1$
$O_{17}$	$\{A_1,A_2,B_1,B_3,A_2^{\prime },B_3^{\prime }\}$	$24$	$(1,1,1,1,1,0)$	$A_2{B}_3{A}_2^{\prime }{B}_3^{\prime }=1$
$O_{18}$	$\{A_1,A_2,B_1,A_1^{\prime },A_2^{\prime },B_1^{\prime }\}$	$12$	$(1,1,1,1,2,0)$	$A_2{B}_1\overline{A_2^{\prime }{B}_1^{\prime }}=1$
$O_{19}$	$\{A_1,A_2,B_3,A_1^{\prime },A_2^{\prime },B_3^{\prime }\}$	$6$	$(1,1,1,1,0,0)$	$A_2{B}_3{A}_2^{\prime }{B}_3^{\prime }=A_1{B}_3\overline{A_1^{\prime }{B}_3^{\prime }}=1$

$\left(R_{23}:R_3\right)$
$R_{19}+R_5$	$R_{19}+R_3+R_2$	$(R_{19}:R_3)+2R_2$
$(R_{19}:2R_3)+R_3$	$(R_{19}:R_5)+R_2$	$(R_{19}:3R_3)+R_2$
$\left(R_{19}:R_7\right)$	$(R_{19}:(R_5:2R_3))$	${({𝑹}_{\mathrm{𝟏𝟗}}:({𝑹}_{\mathrm{𝟓}}:{𝑹}_{\mathrm{𝟑}}),{𝑹}_{\mathrm{𝟑}})}^{\ast }$
$\left(R_{19}:R_5,2R_3\right)$	$\left(R_{19}:5R_3\right)$
$R_{17}+R_7$	$R_{17}+(R_5:2R_3)$	$R_{17}+R_5+R_2$
$R_{17}+R_3+2R_2$	$(R_{17}:R_3)+(R_5:R_3)$	$(R_{17}:R_3)+2R_3$
$(R_{17}:R_3)+3R_2$	$(R_{17}:2R_3)+R_5$	$(R_{17}:2R_3)+R_3+R_2$
$(R_{17}:R_5)+2R_2$	$(R_{17}:3R_3)+2R_2$	$({𝑹}_{\mathrm{𝟏𝟕}}:({𝑹}_{\mathrm{𝟓}}:{𝑹}_{\mathrm{𝟑}}))+{𝑹}_{\mathrm{𝟑}}^{\ast }$
$({𝑹}_{\mathrm{𝟏𝟕}}:{𝑹}_{\mathrm{𝟓}},{𝑹}_{\mathrm{𝟑}})+{𝑹}_{\mathrm{𝟑}}^{\ast }$	$(R_{17}:4R_3)+R_3$	$(R_{17}:R_7)+R_2$
$(R_{17}:(R_5:2R_3))+R_2$	$({𝑹}_{\mathrm{𝟏𝟕}}:({𝑹}_{\mathrm{𝟓}}:{𝑹}_{\mathrm{𝟑}}),{𝑹}_{\mathrm{𝟑}})+{𝑹}_{\mathrm{𝟐}}^{\ast }$	$(R_{17}:R_5,2R_3)+R_2$
$(R_{17}:5R_3)+R_2$	$(R_{17}:(R_7:2R_3))$	$(R_{17}:(R_5:4R_3))$
${({𝑹}_{\mathrm{𝟏𝟕}}:({𝑹}_{\mathrm{𝟕}}:{𝑹}_{\mathrm{𝟑}}),{𝑹}_{\mathrm{𝟑}})}^{\ast }$	${({𝑹}_{\mathrm{𝟏𝟕}}:({𝑹}_{\mathrm{𝟓}}:\mathrm{𝟑}{𝑹}_{\mathrm{𝟑}}),{𝑹}_{\mathrm{𝟑}})}^{\ast }$	$\left(R_{17}:R_7,2R_3\right)$
$(R_{17}:(R_5:2R_3),2R_3)$	${({𝑹}_{\mathrm{𝟏𝟕}}:({𝑹}_{\mathrm{𝟓}}:{𝑹}_{\mathrm{𝟑}}),{𝑹}_{\mathrm{𝟓}})}^{\ast }$	${({𝑹}_{\mathrm{𝟏𝟕}}:({𝑹}_{\mathrm{𝟓}}:{𝑹}_{\mathrm{𝟑}}),\mathrm{𝟑}{𝑹}_{\mathrm{𝟑}})}^{\ast }$
$\left(R_{17}:2R_5,R_3\right)$	$\left(R_{17}:R_5,4R_3\right)$	$\left(R_{17}:7R_3\right)$

$R_{13}+R_7+2R_2$	$R_{13}+(R_5:2R_3)+2R_2$	$R_{13}+R_5+2R_3$
$R_{13}+R_5+3R_2$	$R_{13}+3R_3+R_2$	${𝑹}_{\mathrm{𝟏𝟑}}+{𝑹}_{\mathrm{𝟑}}+\mathrm{𝟒}{𝑹}_{\mathrm{𝟐}}^{\ast }$
$(R_{13}:R_3)+(R_5:R_3)+2R_2$	$({𝑹}_{\mathrm{𝟏𝟑}}:{𝑹}_{\mathrm{𝟑}})+\mathrm{𝟐}{𝑹}_{\mathrm{𝟓}}^{\ast }$	$({𝑹}_{\mathrm{𝟏𝟑}}:{𝑹}_{\mathrm{𝟑}})+\mathrm{𝟐}{𝑹}_{\mathrm{𝟑}}+\mathrm{𝟐}{𝑹}_{\mathrm{𝟐}}^{\ast }$
$({𝑹}_{\mathrm{𝟏𝟑}}:{𝑹}_{\mathrm{𝟑}})+\mathrm{𝟓}{𝑹}_{\mathrm{𝟐}}^{\ast }$	$(R_{13}:2R_3)+R_5+2R_2$	$({𝑹}_{\mathrm{𝟏𝟑}}:\mathrm{𝟐}{𝑹}_{\mathrm{𝟑}})+\mathrm{𝟑}{𝑹}_{\mathrm{𝟑}}^{\ast }$
$(R_{13}:2R_3)+R_3+3R_2$	$(R_{13}:R_5)+2R_3+R_2$	$({𝑹}_{\mathrm{𝟏𝟑}}:{𝑹}_{\mathrm{𝟓}})+\mathrm{𝟒}{𝑹}_{\mathrm{𝟐}}^{\ast }$
$({𝑹}_{\mathrm{𝟏𝟑}}:\mathrm{𝟑}{𝑹}_{\mathrm{𝟑}})+\mathrm{𝟐}{𝑹}_{\mathrm{𝟑}}+{𝑹}_{\mathrm{𝟐}}^{\ast }$	$({𝑹}_{\mathrm{𝟏𝟑}}:\mathrm{𝟑}{𝑹}_{\mathrm{𝟑}})+\mathrm{𝟒}{𝑹}_{\mathrm{𝟐}}^{\ast }$	$(R_{13}:4R_3)+R_7$
$(R_{13}:4R_3)+(R_5:2R_3)$	$(R_{13}:4R_3)+R_5+R_2$	$(R_{13}:4R_3)+R_3+2R_2$
$(R_{13}:R_7)+2R_3$	$(R_{13}:R_7)+3R_2$	$(R_{13}:(R_5:2R_3))+2R_3$
$(R_{13}:(R_5:2R_3))+3R_2$	$(R_{13}:R_5,2R_3)+2R_3$	$(R_{13}:R_5,2R_3)+3R_2$
$(R_{13}:5R_3)+(R_5:R_3)$	$(R_{13}:5R_3)+2R_3$	$(R_{13}:5R_3)+3R_2$
$(R_{13}:2R_5)+R_5$	$(R_{13}:2R_5)+R_3+R_2$	$(R_{13}:6R_3)+R_5$
$(R_{13}:6R_3)+R_3+R_2$	$(R_{13}:(R_7:2R_3))+2R_2$	$(R_{13}:(R_5:4R_3))+2R_2$
$(R_{13}:R_7,2R_3)+2R_2$	$(R_{13}:(R_5:2R_3),2R_3)+2R_2$	$(R_{13}:2R_5,R_3)+2R_2$
$(R_{13}:R_5,4R_3)+2R_2$	$(R_{13}:7R_3)+2R_2$	$(R_{13}:2(R_5:R_3))+R_3$
$(R_{13}:2R_5,2R_3)+R_3$	$(R_{13}:8R_3)+R_3$	$(R_{13}:(R_7:2R_3),4R_3)$
$(R_{13}:(R_5:4R_3),4R_3)$	$\left(R_{13}:2R_7,R_3\right)$	$(R_{13}:2(R_5:2R_3),R_3)$
$\left(R_{13}:R_7,2R_5\right)$	$(R_{13}:(R_5:2R_3),2R_5)$	$\left(R_{13}:R_7,6R_3\right)$
$(R_{13}:(R_5:2R_3),6R_3)$	$(R_{13}:2(R_5:R_3),R_5)$	$(R_{13}:2(R_5:R_3),3R_3)$
$\left(R_{13}:3R_5,2R_3\right)$	$\left(R_{13}:2R_5,5R_3\right)$	$\left(R_{13}:R_5,8R_3\right)$
$\left(R_{13}:11R_3\right)$
$2R_{11}+R_2$	$R_{11}+(R_7:2R_3)+2R_2$	$R_{11}+(R_5:4R_3)+2R_2$
$R_{11}+(R_5:2R_3)+(R_5:R_3)$	$R_{11}+R_7+2R_3$	$R_{11}+(R_5:2R_3)+2R_3$
$R_{11}+R_7+3R_2$	$R_{11}+(R_5:2R_3)+3R_2$	$R_{11}+(R_5:R_3)+R_3+2R_2$
${𝑹}_{\mathrm{𝟏𝟏}}+\mathrm{𝟐}{𝑹}_{\mathrm{𝟓}}+{𝑹}_{\mathrm{𝟑}}^{\ast }$	$R_{11}+R_5+2R_3+R_2$	${𝑹}_{\mathrm{𝟏𝟏}}+{𝑹}_{\mathrm{𝟓}}+\mathrm{𝟒}{𝑹}_{\mathrm{𝟐}}^{\ast }$
${𝑹}_{\mathrm{𝟏𝟏}}+\mathrm{𝟑}{𝑹}_{\mathrm{𝟑}}+\mathrm{𝟐}{𝑹}_{\mathrm{𝟐}}^{\ast }$	${𝑹}_{\mathrm{𝟏𝟏}}+{𝑹}_{\mathrm{𝟑}}+\mathrm{𝟓}{𝑹}_{\mathrm{𝟐}}^{\ast }$	$2(R_{11}:R_3)$
$(R_{11}:R_3)+(R_7:R_3)+2R_2$	$(R_{11}:R_3)+(R_5:3R_3)+2R_2$	$(R_{11}:R_3)+2(R_5:R_3)$
$(R_{11}:R_3)+(R_5:R_3)+2R_3$	$(R_{11}:R_3)+(R_5:R_3)+3R_2$	$({𝑹}_{\mathrm{𝟏𝟏}}:{𝑹}_{\mathrm{𝟑}})+\mathrm{𝟐}{𝑹}_{\mathrm{𝟓}}+{𝑹}_{\mathrm{𝟐}}^{\ast }$
$(R_{11}:R_3)+R_5+R_3+2R_2$	$({𝑹}_{\mathrm{𝟏𝟏}}:{𝑹}_{\mathrm{𝟑}})+\mathrm{𝟒}{𝑹}_{\mathrm{𝟑}}^{\ast }$	$({𝑹}_{\mathrm{𝟏𝟏}}:{𝑹}_{\mathrm{𝟑}})+\mathrm{𝟐}{𝑹}_{\mathrm{𝟑}}+\mathrm{𝟑}{𝑹}_{\mathrm{𝟐}}^{\ast }$
$({𝑹}_{\mathrm{𝟏𝟏}}:{𝑹}_{\mathrm{𝟑}})+\mathrm{𝟔}{𝑹}_{\mathrm{𝟐}}^{\ast }$	$(R_{11}:2R_3)+R_{11}$	$(R_{11}:2R_3)+R_7+2R_2$
$(R_{11}:2R_3)+(R_5:2R_3)+2R_2$	$({𝑹}_{\mathrm{𝟏𝟏}}:\mathrm{𝟐}{𝑹}_{\mathrm{𝟑}})+{𝑹}_{\mathrm{𝟓}}+\mathrm{𝟐}{𝑹}_{\mathrm{𝟑}}^{\ast }$	$(R_{11}:2R_3)+R_5+3R_2$
$({𝑹}_{\mathrm{𝟏𝟏}}:\mathrm{𝟐}{𝑹}_{\mathrm{𝟑}})+\mathrm{𝟑}{𝑹}_{\mathrm{𝟑}}+{𝑹}_{\mathrm{𝟐}}^{\ast }$	$({𝑹}_{\mathrm{𝟏𝟏}}:\mathrm{𝟐}{𝑹}_{\mathrm{𝟑}})+{𝑹}_{\mathrm{𝟑}}+\mathrm{𝟒}{𝑹}_{\mathrm{𝟐}}^{\ast }$	$(R_{11}:R_5)+(R_5:R_3)+2R_2$
$({𝑹}_{\mathrm{𝟏𝟏}}:{𝑹}_{\mathrm{𝟓}})+\mathrm{𝟐}{𝑹}_{\mathrm{𝟓}}^{\ast }$	$({𝑹}_{\mathrm{𝟏𝟏}}:{𝑹}_{\mathrm{𝟓}})+\mathrm{𝟐}{𝑹}_{\mathrm{𝟑}}+\mathrm{𝟐}{𝑹}_{\mathrm{𝟐}}^{\ast }$	$({𝑹}_{\mathrm{𝟏𝟏}}:{𝑹}_{\mathrm{𝟓}})+\mathrm{𝟓}{𝑹}_{\mathrm{𝟐}}^{\ast }$
$(R_{11}:3R_3)+(R_5:R_3)+2R_2$	$({𝑹}_{\mathrm{𝟏𝟏}}:\mathrm{𝟑}{𝑹}_{\mathrm{𝟑}})+\mathrm{𝟐}{𝑹}_{\mathrm{𝟓}}^{\ast }$	$({𝑹}_{\mathrm{𝟏𝟏}}:\mathrm{𝟑}{𝑹}_{\mathrm{𝟑}})+\mathrm{𝟐}{𝑹}_{\mathrm{𝟑}}+\mathrm{𝟐}{𝑹}_{\mathrm{𝟐}}^{\ast }$
$({𝑹}_{\mathrm{𝟏𝟏}}:\mathrm{𝟑}{𝑹}_{\mathrm{𝟑}})+\mathrm{𝟓}{𝑹}_{\mathrm{𝟐}}^{\ast }$	$(R_{11}:4R_3)+(R_7:2R_3)$	$(R_{11}:4R_3)+(R_5:4R_3)$
$(R_{11}:4R_3)+R_7+R_2$	$(R_{11}:4R_3)+(R_5:2R_3)+R_2$	$(R_{11}:4R_3)+(R_5:R_3)+R_3$
$(R_{11}:4R_3)+R_5+2R_2$	$(R_{11}:4R_3)+3R_3$	$(R_{11}:4R_3)+R_3+3R_2$
$(R_{11}:R_7)+2R_3+R_2$	$({𝑹}_{\mathrm{𝟏𝟏}}:{𝑹}_{\mathrm{𝟕}})+\mathrm{𝟒}{𝑹}_{\mathrm{𝟐}}^{\ast }$	$(R_{11}:(R_5:2R_3))+2R_3+R_2$
$({𝑹}_{\mathrm{𝟏𝟏}}:({𝑹}_{\mathrm{𝟓}}:\mathrm{𝟐}{𝑹}_{\mathrm{𝟑}}))+\mathrm{𝟒}{𝑹}_{\mathrm{𝟐}}^{\ast }$	$({𝑹}_{\mathrm{𝟏𝟏}}:{𝑹}_{\mathrm{𝟓}},\mathrm{𝟐}{𝑹}_{\mathrm{𝟑}})+\mathrm{𝟐}{𝑹}_{\mathrm{𝟑}}+{𝑹}_{\mathrm{𝟐}}^{\ast }$	$({𝑹}_{\mathrm{𝟏𝟏}}:{𝑹}_{\mathrm{𝟓}},\mathrm{𝟐}{𝑹}_{\mathrm{𝟑}})+\mathrm{𝟒}{𝑹}_{\mathrm{𝟐}}^{\ast }$
$(R_{11}:5R_3)+(R_7:R_3)$	$(R_{11}:5R_3)+(R_5:3R_3)$	$(R_{11}:5R_3)+(R_5:R_3)+R_2$
$(R_{11}:5R_3)+R_5+R_3$	$({𝑹}_{\mathrm{𝟏𝟏}}:\mathrm{𝟓}{𝑹}_{\mathrm{𝟑}})+\mathrm{𝟐}{𝑹}_{\mathrm{𝟑}}+{𝑹}_{\mathrm{𝟐}}^{\ast }$	$({𝑹}_{\mathrm{𝟏𝟏}}:\mathrm{𝟓}{𝑹}_{\mathrm{𝟑}})+\mathrm{𝟒}{𝑹}_{\mathrm{𝟐}}^{\ast }$
$(R_{11}:2R_5)+R_3+2R_2$	$(R_{11}:R_5,3R_3)+(R_5:2R_3)$	$(R_{11}:6R_3)+R_7$
$(R_{11}:6R_3)+(R_5:2R_3)$	$(R_{11}:6R_3)+R_5+R_2$	$(R_{11}:6R_3)+R_3+2R_2$
$(R_{11}:(R_7:2R_3))+2R_3$	$(R_{11}:(R_7:2R_3))+3R_2$	$(R_{11}:(R_5:4R_3))+2R_3$
$(R_{11}:(R_5:4R_3))+3R_2$	$(R_{11}:R_7,2R_3)+2R_3$	$(R_{11}:R_7,2R_3)+3R_2$
$({𝑹}_{\mathrm{𝟏𝟏}}:({𝑹}_{\mathrm{𝟓}}:\mathrm{𝟐}{𝑹}_{\mathrm{𝟑}}),\mathrm{𝟐}{𝑹}_{\mathrm{𝟑}})+\mathrm{𝟐}{𝑹}_{\mathrm{𝟑}}^{\ast }$	$(R_{11}:(R_5:2R_3),2R_3)+3R_2$	$(R_{11}:(R_5:R_3),3R_3)+(R_5:R_3)$
$(R_{11}:2R_5,R_3)+2R_3$	$(R_{11}:2R_5,R_3)+3R_2$	$(R_{11}:R_5,4R_3)+(R_5:R_3)$
$({𝑹}_{\mathrm{𝟏𝟏}}:{𝑹}_{\mathrm{𝟓}},\mathrm{𝟒}{𝑹}_{\mathrm{𝟑}})+\mathrm{𝟐}{𝑹}_{\mathrm{𝟑}}^{\ast }$	$(R_{11}:R_5,4R_3)+3R_2$	$(R_{11}:7R_3)+(R_5:R_3)$
$({𝑹}_{\mathrm{𝟏𝟏}}:\mathrm{𝟕}{𝑹}_{\mathrm{𝟑}})+\mathrm{𝟐}{𝑹}_{\mathrm{𝟑}}^{\ast }$	$(R_{11}:7R_3)+3R_2$	$(R_{11}:2R_5,2R_3)+R_5$
$(R_{11}:2R_5,2R_3)+R_3+R_2$	$(R_{11}:8R_3)+R_5$	$({𝑹}_{\mathrm{𝟏𝟏}}:\mathrm{𝟖}{𝑹}_{\mathrm{𝟑}})+{𝑹}_{\mathrm{𝟑}}+{𝑹}_{\mathrm{𝟐}}^{\ast }$
$(R_{11}:(R_7:4R_3))+2R_2$	$(R_{11}:(R_7:2R_3),2R_3)+2R_2$	$(R_{11}:(R_5:4R_3),2R_3)+2R_2$
$(R_{11}:R_7,4R_3)+2R_2$	$({𝑹}_{\mathrm{𝟏𝟏}}:({𝑹}_{\mathrm{𝟓}}:\mathrm{𝟐}{𝑹}_{\mathrm{𝟑}}),\mathrm{𝟒}{𝑹}_{\mathrm{𝟑}})+\mathrm{𝟐}{𝑹}_{\mathrm{𝟐}}^{\ast }$	$(R_{11}:2(R_5:R_3),R_3)+2R_2$
$(R_{11}:3R_5)+2R_2$	$(R_{11}:2R_5,3R_3)+2R_2$	$(R_{11}:R_5,6R_3)+2R_2$
$({𝑹}_{\mathrm{𝟏𝟏}}:\mathrm{𝟗}{𝑹}_{\mathrm{𝟑}})+\mathrm{𝟐}{𝑹}_{\mathrm{𝟐}}^{\ast }$	$(R_{11}:2R_7)+R_3$	$({𝑹}_{\mathrm{𝟏𝟏}}:\mathrm{𝟐}({𝑹}_{\mathrm{𝟓}}:\mathrm{𝟐}{𝑹}_{\mathrm{𝟑}}))+{𝑹}_{\mathrm{𝟑}}^{\ast }$
$(R_{11}:2(R_5:R_3),2R_3)+R_3$	$(R_{11}:2R_5,4R_3)+R_3$	$(R_{11}:10R_3)+R_3$
$(R_{11}:(R_7:4R_3),4R_3)$	$(R_{11}:(R_7:2R_3),6R_3)$	$(R_{11}:(R_5:4R_3),6R_3)$
$\left(R_{11}:2R_7,R_5\right)$	$(R_{11}:2(R_5:2R_3),R_5)$	$\left(R_{11}:2R_7,3R_3\right)$
$(R_{11}:2(R_5:2R_3),3R_3)$	$\left(R_{11}:R_7,2R_5,2R_3\right)$	$(R_{11}:(R_5:2R_3),2R_5,2R_3)$
$\left(R_{11}:R_7,8R_3\right)$	$(R_{11}:(R_5:2R_3),8R_3)$	$(R_{11}:2(R_5:R_3),R_5,2R_3)$
$(R_{11}:2(R_5:R_3),5R_3)$	$\left(R_{11}:4R_5,R_3\right)$	${({𝑹}_{\mathrm{𝟏𝟏}}:\mathrm{𝟑}{𝑹}_{\mathrm{𝟓}},\mathrm{𝟒}{𝑹}_{\mathrm{𝟑}})}^{\ast }$
$\left(R_{11}:2R_5,7R_3\right)$

$\left(R_7:4R_3\right)$	$\left(R_7:5R_3\right)$	$\left(R_7:R_5,2R_3\right)$
$(R_7:(R_5:2R_3))$	$\left(R_7:6R_3\right)$	$\left(R_7:2R_5\right)$
$\left(R_7:R_5,4R_3\right)$	$\left(R_7:2R_5,R_3\right)$	$\left(R_7:(R_5,2R_3),2R_3\right)$
$\left(R_7:(R_5,4R_3)\right)$	$\left(R_7:2R_5,3R_3\right)$	$\left(R_7:3R_5\right)$
$\left(R_7:2(R_5,R_3),R_3\right)$	$\left(R_7:(R_5,2R_3),4R_3\right)$	$\left(R_7:(R_5,4R_3),2R_3\right)$
$\pm \left({\zeta }_5+{\zeta }_5^2+{\zeta }_5^3+{\zeta }_5^4+{\sum }_{i=1}^6\left({\zeta }_3+{\zeta }_3^2\right){\zeta }_7^i\right)$
$\pm \left({\zeta }_3+{\zeta }_3^2+{\zeta }_{10}+{\zeta }_{10}^9+{\sum }_{i=1}^6\left({\zeta }_5+{\zeta }_5^4\right){\zeta }_7^i\right)$
$\pm \left(\left({\zeta }_{10}+{\zeta }_{10}^9-1\right)\left({\zeta }_7+{\zeta }_7^6\right)+{\sum }_{i\ne 1,6}\left({\zeta }_5+{\zeta }_5^4\right){\zeta }_7^i\right)$

$R_5+R_3+8R_2$	$R_5+3R_3+5R_2$	$R_5+5R_3+2R_2$
$2R_5+7R_2$	$2R_5+2R_3+4R_2$	$2R_5+4R_3+R_2$
$3R_5+R_3+3R_2$	$3R_5+3R_3$	$4R_5+2R_2$
$(R_5:R_3)+9R_2$	$(R_5:R_3)+2R_3+6R_2$	$(R_5:R_3)+4R_3+3R_2$
$(R_5:R_3)+6R_3$	$(R_5:R_3)+R_5+R_3+5R_2$	$(R_5:R_3)+R_5+3R_3+2R_2$
$(R_5:R_3)+2R_5+4R_2$	$(R_5:R_3)+2R_5+2R_3+R_2$	$(R_5:R_3)+3R_5+R_3$
$2(R_5:R_3)+6R_2$	$2(R_5:R_3)+2R_3+3R_2$	$2(R_5:R_3)+4R_3$
$2(R_5:R_3)+R_5+R_3+2R_2$	$2(R_5:R_3)+2R_5+R_2$	$3(R_5:R_3)+3R_2$
$3(R_5:R_3)+2R_3$	$4(R_5:R_3)$	$(R_5:2R_3)+R_3+7R_2$
$(R_5:2R_3)+3R_3+4R_2$	$(R_5:2R_3)+5R_3+R_2$	$(R_5:2R_3)+R_5+6R_2$
$(R_5:2R_3)+R_5+2R_3+3R_2$	$(R_5:2R_3)+R_5+4R_3$	$(R_5:2R_3)+2R_5+R_3+2R_2$
$(R_5:2R_3)+3R_5+R_2$	$(R_5:2R_3)+(R_5:R_3)+R_3+4R_2$	$(R_5:2R_3)+2(R_5:R_3)+R_3+R_2$
$(R_5:2R_3)+2(R_5:R_3)+R_5$	$2(R_5:2R_3)+5R_2$	$2(R_5:2R_3)+2R_3+2R_2$
$2(R_5:2R_3)+R_5+R_3+R_2$	$2(R_5:2R_3)+2R_5$	$2(R_5:2R_3)+(R_5:R_3)+2R_2$
$3(R_5:2R_3)+R_3$	$(R_5:3R_3)+8R_2$	$(R_5:3R_3)+2R_3+5R_2$
$(R_5:3R_3)+4R_3+2R_2$	$(R_5:3R_3)+R_5+R_3+4R_2$	$(R_5:3R_3)+2R_5+3R_2$
$(R_5:3R_3)+2R_5+2R_3$	$(R_5:3R_3)+(R_5:R_3)+5R_2$	$(R_5:3R_3)+(R_5:R_3)+2R_3+2R_2$
$(R_5:3R_3)+(R_5:R_3)+2R_5$	$(R_5:3R_3)+2(R_5:R_3)+2R_2$	$(R_5:3R_3)+2(R_5:2R_3)+R_2$
$2(R_5:3R_3)+4R_2$	$2(R_5:3R_3)+2R_3+R_2$	$2(R_5:3R_3)+R_5+R_3$
$2(R_5:3R_3)+(R_5:R_3)+R_2$	$3(R_5:3R_3)$	$(R_5:4R_3)+R_3+6R_2$
$(R_5:4R_3)+3R_3+3R_2$	$(R_5:4R_3)+5R_3$	$(R_5:4R_3)+R_5+5R_2$
$(R_5:4R_3)+R_5+2R_3+2R_2$	$(R_5:4R_3)+2R_5+R_3+R_2$	$(R_5:4R_3)+3R_5$
$(R_5:4R_3)+2(R_5:R_3)+R_3$	$(R_5:4R_3)+(R_5:2R_3)+4R_2$	$2(R_5:4R_3)+3R_2$
$2(R_5:4R_3)+2R_3$	$2(R_5:4R_3)+(R_5:R_3)$
$R_7+R_3+7R_2$	$R_7+3R_3+4R_2$	$R_7+5R_3+R_2$
$R_7+R_5+6R_2$	$R_7+R_5+2R_3+3R_2$	$R_7+R_5+4R_3$
$R_7+2R_5+R_3+2R_2$	$R_7+3R_5+R_2$	$R_7+(R_5:R_3)+R_3+4R_2$
$R_7+2(R_5:R_3)+R_3+R_2$	$R_7+2(R_5:R_3)+R_5$	$R_7+(R_5:2R_3)+5R_2$
$2R_7+5R_2$	$R_7+(R_5:2R_3)+2R_3+2R_2$	$2R_7+2R_3+2R_2$
$2R_7+R_5+R_3+R_2$	$R_7+(R_5:2R_3)+2R_5$	$2R_7+2R_5$
$2R_7+(R_5:R_3)+2R_2$	$R_7+2(R_5:2R_3)+R_3$	$2R_7+(R_5:2R_3)+R_3$
$3R_7+R_3$	$(R_7:R_3)+8R_2$	$(R_7:R_3)+2R_3+5R_2$
$(R_7:R_3)+4R_3+2R_2$	$(R_7:R_3)+R_5+R_3+4R_2$	$(R_7:R_3)+2R_5+3R_2$
$(R_7:R_3)+2R_5+2R_3$	$(R_7:R_3)+(R_5:R_3)+5R_2$	$(R_7:R_3)+(R_5:R_3)+2R_3+2R_2$
$(R_7:R_3)+(R_5:R_3)+2R_5$	$(R_7:R_3)+2(R_5:R_3)+2R_2$	$(R_5:3R_3)+2R_7+R_2$
$(R_7:R_3)+2(R_5:2R_3)+R_2$	$(R_7:R_3)+2R_7+R_2$	$(R_7:R_3)+(R_5:3R_3)+4R_2$
$2(R_7:R_3)+4R_2$	$2(R_7:R_3)+2R_3+R_2$	$2(R_7:R_3)+R_5+R_3$
$2(R_7:R_3)+(R_5:R_3)+R_2$	$(R_7:R_3)+2(R_5:3R_3)$	$2(R_7:R_3)+(R_5:3R_3)$
$3(R_7:R_3)$	$(R_7:2R_3)+R_3+6R_2$	$(R_7:2R_3)+3R_3+3R_2$
$(R_7:2R_3)+5R_3$	$(R_7:2R_3)+R_5+5R_2$	$(R_7:2R_3)+R_5+2R_3+2R_2$
$(R_7:2R_3)+2R_5+R_3+R_2$	$(R_7:2R_3)+3R_5$	$(R_7:2R_3)+2(R_5:R_3)+R_3$
$(R_5:4R_3)+R_7+4R_2$	$(R_7:2R_3)+(R_5:2R_3)+4R_2$	$(R_7:2R_3)+R_7+4R_2$
$2(R_7:2R_3)+3R_2$	$2(R_7:2R_3)+2R_3$	$2(R_7:2R_3)+(R_5:R_3)$
$(R_7:3R_3)+7R_2$	$(R_7:R_5)+7R_2$	$(R_7:3R_3)+2R_3+4R_2$
$(R_7:R_5)+2R_3+4R_2$	$(R_7:3R_3)+4R_3+R_2$	$(R_7:R_5)+4R_3+R_2$
$(R_7:3R_3)+2R_5+2R_2$	$(R_7:R_5)+2R_5+2R_2$	$(R_7:3R_3)+(R_5:R_3)+4R_2$
$(R_7:R_5)+(R_5:R_3)+4R_2$	$(R_7:3R_3)+2(R_5:R_3)+R_2$	$(R_7:R_5)+2(R_5:R_3)+R_2$
$(R_7:3R_3)+2(R_5:2R_3)$	$(R_7:3R_3)+2R_7$	$(R_7:R_5)+2(R_5:2R_3)$
$(R_7:R_5)+2R_7$	$2(R_7:3R_3)+2R_2$	$2(R_7:R_5)+2R_2$
$2(R_7:4R_3)+R_2$	$2(R_7:R_5,R_3)+R_2$	$2(R_7:(R_5:R_3))+R_2$
$(R_7:5R_3)+2(R_5:R_3)$	$(R_7:R_5,2R_3)+2(R_5:R_3)$	$(R_7:(R_5,R_3),R_3)+2(R_5:R_3)$
$(R_7:(R_5,2R_3))+2(R_5:R_3)$	$2(R_7:5R_3)$	$2(R_7:R_5,2R_3)$
$2(R_7:(R_5,R_3),R_3)$	$2(R_7:(R_5,2R_3))$

$x$	$y$	$z_1$	$x$	$x-z_1$	$x-y$
$x$	$y$	$z_2$	$y-z_2$	$y$	$y-x$
$x$	$y$	$z_3$	$z_3-y$	$z_3-x$	$z_3$
$x$	$y$	$z_4$	$-x$	$z_4-x$	$y-x$
$x$	$y$	$z_5$	$z_5-y$	$-y$	$x-y$
$x$	$y$	$z_6$	$y-z_6$	$x-z_6$	$-z_6$

Table 9. Table 2: The zero-parameter solutions in 𝒮 + / ≈ \mathcal{S}^{+}/\approx

Order	$s_1$	$s_2$	$s_3$	$t_1$	$t_2$	$t_3$
26	1/26	3/26	9/26	2/13	11/26	8/13
30	1/30	1/15	1/10	1/30	1/10	4/15
30	1/30	1/15	1/10	1/30	1/6	11/15
30	1/30	1/15	1/10	1/15	1/6	3/10
30	1/30	1/15	1/10	1/15	7/30	8/15
30	1/30	1/15	1/10	1/10	7/30	11/30
30	1/30	1/15	1/10	1/10	11/30	19/30
30	1/30	1/15	1/10	2/15	3/10	13/30
30	1/30	1/15	2/15	1/30	1/6	13/15
30	1/30	1/15	2/15	1/15	1/6	13/30
30	1/30	1/15	2/15	1/15	7/30	7/10
30	1/30	1/15	2/15	1/10	7/30	7/15
30	1/30	1/15	2/15	1/10	4/15	17/30
30	1/30	1/15	2/15	1/6	7/15	7/10
30	1/30	1/15	1/6	1/30	1/10	11/15
30	1/30	1/15	1/6	1/30	2/15	13/15
30	1/30	1/15	1/6	1/15	2/15	3/10
30	1/30	1/15	1/6	1/15	1/6	8/15
30	1/30	1/15	1/6	1/15	7/30	23/30
30	1/30	1/15	1/6	1/10	1/6	4/15
30	1/30	1/15	1/6	1/10	7/30	8/15
30	1/30	1/15	1/6	1/6	3/10	7/15
30	1/30	1/15	1/6	1/6	13/30	7/10
30	1/30	1/15	1/5	1/10	3/10	11/15
30	1/30	1/15	7/30	1/10	1/6	3/10
30	1/30	1/15	7/30	1/10	4/15	7/10
30	1/30	1/15	3/10	1/30	1/10	13/15
30	1/30	1/15	3/10	1/15	1/10	1/6
30	1/30	1/15	3/10	1/15	2/15	13/30
30	1/30	1/15	3/10	1/15	1/6	7/10
30	1/30	1/15	3/10	1/10	4/15	11/15
30	1/30	1/15	3/10	1/6	3/10	8/15
30	1/30	1/15	1/3	1/10	1/6	1/3
30	1/30	1/15	1/3	1/10	7/30	2/3
30	1/30	1/15	11/30	1/15	1/6	11/15
30	1/30	1/15	7/15	1/15	1/6	23/30
30	1/30	1/15	1/2	2/15	7/30	1/2
30	1/30	1/10	2/15	1/30	1/6	3/10
30	1/30	1/10	2/15	1/30	4/15	17/30
30	1/30	1/10	2/15	1/30	1/3	2/3
30	1/30	1/10	2/15	1/15	3/10	13/30
30	1/30	1/10	2/15	1/15	11/30	8/15
30	1/30	1/10	2/15	1/10	11/30	7/15
30	1/30	1/10	1/6	1/30	2/15	3/10
30	1/30	1/10	1/6	1/30	1/6	7/15
30	1/30	1/10	1/6	1/15	1/6	7/30
30	1/30	1/10	1/6	1/10	4/15	11/30
30	1/30	1/10	1/6	1/10	11/30	8/15
30	1/30	1/10	1/5	2/15	13/30	3/5
30	1/30	1/10	4/15	1/15	1/6	3/10
30	1/30	1/10	4/15	1/10	7/30	11/30
30	1/30	1/10	4/15	2/15	3/10	13/30
30	1/30	1/10	4/15	1/6	7/15	19/30
30	1/30	1/10	1/3	1/30	2/15	2/3
30	1/30	1/10	1/3	1/15	1/6	1/3
30	1/30	1/10	1/3	1/10	11/30	2/3
30	1/30	1/10	1/3	2/15	13/30	2/3
30	1/30	1/10	7/15	1/30	2/15	23/30
30	1/30	1/10	7/15	1/15	2/15	7/30
30	1/30	1/10	7/15	1/10	4/15	8/15
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120	1/120	1/15	17/120	3/40	29/60	7/12
120	1/120	1/15	37/120	1/40	23/120	7/15
120	1/120	1/15	11/20	1/10	11/24	71/120
120	1/120	1/15	67/120	1/12	23/60	8/15
120	1/120	1/15	67/120	1/10	9/20	7/12
120	1/120	1/15	37/60	11/120	47/120	8/15
120	1/120	3/40	17/60	1/24	4/15	17/40
120	1/120	3/40	7/12	1/60	2/15	29/60
120	1/120	3/40	7/12	1/12	23/60	31/60
120	1/120	1/12	23/40	1/20	13/60	19/60
120	1/120	1/12	23/40	1/15	3/10	5/12
120	1/120	11/120	1/2	7/120	43/120	1/2
120	1/120	11/120	3/5	1/20	4/15	2/5
120	1/120	1/10	71/120	1/20	19/60	7/15
120	1/120	5/24	43/60	1/20	19/60	2/5
120	1/120	5/24	43/60	1/15	7/15	11/20
120	1/120	7/30	29/40	1/20	5/12	31/60
120	1/60	1/24	19/40	1/12	11/60	7/15
120	1/60	1/24	19/40	2/15	3/10	7/12
120	1/60	1/24	19/40	3/20	23/60	41/60
120	1/60	1/15	3/4	1/10	9/40	3/8
120	1/60	3/20	7/15	1/20	9/40	13/40
120	1/40	1/15	7/24	3/40	5/24	8/15
120	1/40	1/15	7/24	1/10	11/40	23/40
126	1/126	1/21	23/63	1/14	22/63	71/126
126	1/126	1/18	13/21	5/126	20/63	43/63
132	1/132	5/132	6/11	1/22	10/33	8/11
132	1/132	3/44	19/33	1/22	10/33	35/66
132	1/132	4/33	27/44	1/22	23/66	16/33
132	1/66	2/33	9/22	1/33	5/44	5/12
140	1/140	1/28	19/35	2/35	3/10	9/14
140	1/140	3/20	23/35	2/35	33/70	4/7
140	1/70	1/20	13/28	1/14	9/35	43/70
150	1/150	2/75	41/150	1/15	19/75	83/150
150	1/150	2/75	7/10	1/10	11/50	17/50
150	1/150	2/75	7/10	1/10	19/50	33/50
150	1/150	2/75	121/150	23/150	19/50	83/150
150	1/150	1/15	37/50	2/25	19/50	13/25
150	1/150	7/50	7/15	1/25	19/50	12/25
156	1/156	5/52	47/78	2/39	11/26	22/39
168	1/168	1/28	61/168	1/14	19/56	31/56
168	1/168	3/56	17/21	1/14	55/168	83/168
168	1/168	3/56	17/21	13/168	8/21	31/56
168	1/168	2/21	43/56	1/28	19/56	31/56
168	1/168	5/24	5/7	1/21	10/21	47/84
180	1/180	1/60	1/12	29/180	17/45	11/20
180	1/180	2/45	97/180	1/10	43/90	11/18
180	1/180	1/18	11/20	1/45	17/90	22/45
180	1/180	1/18	11/20	1/30	16/45	59/90
180	1/90	1/36	13/20	7/90	7/36	11/18
180	1/90	1/18	38/45	1/20	11/45	25/36
180	1/90	1/15	5/18	2/45	7/36	7/20
210	1/210	2/35	1/14	1/70	59/210	13/35
210	1/210	2/35	19/210	1/30	3/10	13/35
210	1/210	2/35	71/210	1/15	1/3	89/210
210	1/210	2/35	5/14	1/14	71/210	89/210
210	1/210	2/35	3/7	1/14	71/210	3/7
210	1/210	2/35	7/10	1/15	59/210	13/35
210	1/210	2/35	151/210	1/14	2/7	13/35
210	1/210	1/14	68/105	11/210	4/15	37/105
210	1/210	1/14	11/15	2/35	19/70	5/14
210	1/210	29/210	5/7	11/210	101/210	17/30
210	1/210	1/7	169/210	2/35	2/5	17/35
210	1/105	9/70	31/42	8/105	17/42	37/70
210	1/105	1/5	11/21	2/35	2/5	33/70
210	1/70	1/14	23/35	1/15	9/35	11/21
210	1/70	3/35	32/35	2/35	4/21	8/15
252	1/126	1/28	17/36	11/126	5/18	10/21
300	1/150	2/75	7/10	1/20	11/50	67/100
420	1/70	1/20	13/28	2/35	4/21	8/15

Equations

$${\mathcal{C}}_n:=\{(s_1,\cdots ,s_n)\in {\mu }_{\infty }^n:s_i\ne s_j\text{ for any }1\le i<j\le n\}.$$

$$(0,1/30,1/15,1/10,2/15,1/6,1/5,3/10,11/30,7/15,3/5,11/15,5/6,9/10)$$

$$(0,1/60,1/30,1/15,1/10,2/15,1/5,11/30,31/60,2/3,5/6,9/10,14/15,29/30)$$

$$(s_1,\cdots ,s_n)\sim (t_1,\cdots ,t_n)$$

$$(1,s_1,s_2,s_3)\sim (1,t_1,t_2,t_3)\text{ in }{\mathcal{C}}_4.$$

$$\mathcal{S}:=\{[s_1,s_2,s_3,t_1,t_2,t_3]:(1,s_1,s_2,s_3)\sim (1,t_1,t_2,t_3)\text{ in }{\mathcal{C}}_4\}$$

$${\mathfrak{G}}_s:=\{g_s\text{ fixes }(s_1,s_2,s_3)\text{ and maps }(t_1,t_2,t_3)\text{ to some element of}$$

$$\{(t_1,t_2,t_3{)}^{\pm 1},(t_1,t_1/t_3,t_1/t_2{)}^{\pm 1},(t_2/t_3,t_2,t_2/t_1{)}^{\pm 1},(t_3/t_2,t_3/t_1,t_3{)}^{\pm 1}\}\}\simeq C_2^3,$$

$${\mathfrak{G}}_t:=\{g_t\text{ fixes }(t_1,t_2,t_3)\text{ and maps }(s_1,s_2,s_3)\text{ to some element of}$$

$$\{(s_1,s_2,s_3{)}^{\pm 1},(s_1,s_1/s_3,s_1/s_2{)}^{\pm 1},(s_2/s_3,s_2,s_2/s_1{)}^{\pm 1},(s_3/s_2,s_3/s_1,s_3{)}^{\pm 1}\}\}\simeq C_2^3$$

$$(1,s_1,s_2,s_3)\sim (1,t_1,t_2,t_3)\sim (1/t_1,1,t_2/t_1,t_3/t_1)\sim (1,1/t_1,t_3/t_1,t_2/t_1),$$

$$[s_1,s_2,s_3,t_1,t_2,t_3]\approx [s_1^{\prime },s_2^{\prime },s_3^{\prime },t_1^{\prime },t_2^{\prime },t_3^{\prime }]$$

$$[s_1,s_2,s_3,t_1,t_2,t_3]=\sigma \cdot g\cdot [s_1^{\prime },s_2^{\prime },s_3^{\prime },t_1^{\prime },t_2^{\prime },t_3^{\prime }]$$

$$\{\begin{array}{l}\begin{array}{rlrlrlrlrlrl}& S_1=\overline{s_1}{s}_2{s}_3,& & S_2=s_1\overline{s_2}{s}_3,& & S_3=s_1{s}_2\overline{s_3},& & T_1=\overline{t_1}{t}_2{t}_3,& & T_2=t_1\overline{t_2}{t}_3,& & T_3=t_1{t}_2\overline{t_3},\\ & A_1=S_3{T}_2,& & A_2=S_1{T}_3,& & A_3=S_2{T}_1,& & B_1=-\overline{S_2{T}_3},& & B_2=-\overline{S_3{T}_1},& & B_3=-\overline{S_1{T}_2},\\ & A_1^{\prime }=S_3\overline{T_2},& & A_2^{\prime }=S_1\overline{T_3},& & A_3^{\prime }=S_2\overline{T_1},& & B_1^{\prime }=-\overline{S_2}{T}_3,& & B_2^{\prime }=-\overline{S_3}{T}_1,& & B_3^{\prime }=-\overline{S_1}{T}_2,\end{array}\end{array}$$

$$0$$

$$\{\begin{array}{l}{A}_1{A}_2{A}_3{B}_1{B}_2{B}_3{A}_1^{\prime }{A}_2^{\prime }{A}_3^{\prime }{B}_1^{\prime }{B}_2^{\prime }{B}_3^{\prime }=1,\\ A_1{A}_2{A}_3{B}_1{B}_2{B}_3=A_1^{\prime }{A}_2^{\prime }{A}_3^{\prime }{B}_1^{\prime }{B}_2^{\prime }{B}_3^{\prime }=-1,\\ A_2{B}_3{A}_2^{\prime }{B}_3^{\prime }=A_3{B}_1{A}_3^{\prime }{B}_1^{\prime }=A_1{B}_2{A}_1^{\prime }{B}_2^{\prime }=1,\\ A_3{B}_2\overline{A_3^{\prime }{B}_2^{\prime }}=A_1{B}_3\overline{A_1^{\prime }{B}_3^{\prime }}=A_2{B}_1\overline{A_2^{\prime }{B}_1^{\prime }}=1.\end{array}$$

$$[s_1^2,s_2^2,s_3^2,t_1^2,t_2^2,t_3^2]=[-A_3\overline{B_2^{\prime }},-A_1\overline{B_3^{\prime }},-A_2\overline{B_1^{\prime }},-A_2{B}_3^{\prime },-A_3{B}_1^{\prime },-A_1{B}_2^{\prime }]$$

$$G=\sum _{i=1}^m{\xi }_i=0,$$

$$1+{\zeta }_p+\cdots +{\zeta }_p^{p-1}=0.$$

$${\zeta }_6+{\zeta }_6^5+{\zeta }_5+{\zeta }_5^2+{\zeta }_5^3+{\zeta }_5^4=0,$$

$$\{\begin{array}{l}{A}_1{A}_2{A}_3{B}_1{B}_2{B}_3=A_1^{\prime }{A}_2^{\prime }{A}_3^{\prime }{B}_1^{\prime }{B}_2^{\prime }{B}_3^{\prime }=-1,\\ A_2{B}_3{A}_2^{\prime }{B}_3^{\prime }=A_3{B}_1{A}_3^{\prime }{B}_1^{\prime }=1,\\ A_3{B}_2\overline{A_3^{\prime }{B}_2^{\prime }}=A_1{B}_3\overline{A_1^{\prime }{B}_3^{\prime }}=1,\end{array}$$

$$\{\begin{array}{l}{a}_1+a_2+a_3+b_1+b_2+b_3=a_1^{\prime }+a_2^{\prime }+a_3^{\prime }+b_1^{\prime }+b_2^{\prime }+b_3^{\prime }=1/2,\\ a_2+b_3+a_2^{\prime }+b_3^{\prime }=a_3+b_1+a_3^{\prime }+b_1^{\prime }=0,\\ a_3+b_2-a_3^{\prime }-b_2^{\prime }=a_1+b_3-a_1^{\prime }-b_3^{\prime }=0,\end{array}$$

$$\frac{(s_3/s_1-1)(s_2-1)}{(s_2/s_1-1)(s_3-1)}=\frac{(t_3/t_1-1)(t_2-1)}{(t_2/t_1-1)(t_3-1)}\in \mathbb{R}.$$

$$\{e_k=\left|{\zeta }_n^k-1\right|:1\le k<n\}.$$

$$e_k=e_{-k}$$

$$e_k=\prod _{pi=k}{e}_i\text{ for prime }p\mid n.$$

$$p\in \{2,3,5,7,11,13,17,19,23\}.$$

$$\fixTABwidth{T}\parenMatrixstack{1&11111000000\\ 000000111111\\ 010001010001\\ 001100001100\\ 00101000-10-10\\ 100001-10000-1}$$

$$\{A_1^{\pm 1},A_2^{\pm 1},A_3^{\pm 1},-B_1^{\pm 1},-B_2^{\pm 1},-B_3^{\pm 1},A_1^{\prime \pm 1},A_2^{\prime \pm 1},A_3^{\prime \pm 1},-B_1^{\prime \pm 1},-B_2^{\prime \pm 1},-B_3^{\prime \pm 1}\}.$$

$$U_{24}:=\{A_1^{\pm 1},A_2^{\pm 1},A_3^{\pm 1},B_1^{\pm 1},B_2^{\pm 1},B_3^{\pm 1},A_1^{\prime \pm 1},A_2^{\prime \pm 1},A_3^{\prime \pm 1},B_1^{\prime \pm 1},B_2^{\prime \pm 1},B_3^{\prime \pm 1}\}.$$

$$U_{12}:=\{A_1,A_2,A_3,B_1,B_2,B_3,A_1^{\prime },A_2^{\prime },A_3^{\prime },B_1^{\prime },B_2^{\prime },B_3^{\prime }\}.$$

Full text

Projective Equivalence for the Roots of Unity

Hang Fu

Abstract. Let $\mu_{\infty}\subseteq\mathbb{C}$ be the collection of roots of unity and $\mathcal{C}_{n}:=\{(s_{1},\cdots,s_{n})\in\mu_{\infty}^{n}:s_{i}\neq s_{j}\text{ for any }1\leq i<j\leq n\}$. Two elements $(s_{1},\cdots,s_{n})$ and $(t_{1},\cdots,t_{n})$ of $\mathcal{C}_{n}$ are said to be projectively equivalent if there exists $\gamma\in\textup{PGL}(2,\mathbb{C})$ such that $\gamma(s_{i})=t_{i}$ for any $1\leq i\leq n$. In this article, we will give a complete classification for the projectively equivalent pairs. As a consequence, we will show that the maximal length for the nontrivial projectively equivalent pairs is $14$.

Keywords. Roots of unity $\cdot$ Diophantine equations $\cdot$ Unlikely intersections

Mathematics Subject Classification. 11Y50 $\cdot$ 11L03 $\cdot$ 11D72

1 Introduction and the statements of main results

Let $\mu_{\infty}\subseteq\mathbb{C}$ be the collection of roots of unity and $\textup{Arg}/(2\pi):\mu_{\infty}\to\mathbb{Q}/\mathbb{Z}$ the canonical isomorphism.

Definition 1**.**

Let

[TABLE]

Two elements $(s_{1},\cdots,s_{n})$ and $(t_{1},\cdots,t_{n})$ of $\mathcal{C}_{n}$ are said to be projectively equivalent if there exists $\gamma\in\textup{PGL}(2,\mathbb{C})$ such that $\gamma(s_{i})=t_{i}$ for any $1\leq i\leq n$. We write $(s_{1},\cdots,s_{n})\sim(t_{1},\cdots,t_{n})$ to denote that they are projectively equivalent. Let $\mathcal{C}_{n}^{+}$ be the isomorphic image of $\mathcal{C}_{n}$ under $(\textup{Arg}/(2\pi))^{n}$. The corresponding equivalence relation in $\mathcal{C}_{n}^{+}$ is also denoted by $\sim$.

The following examples are immediate.

Example 2**.**

• •

Rotation: $(s_{1},\cdots,s_{n})\sim(ss_{1},\cdots,ss_{n})$ for any $(s_{1},\cdots,s_{n})\in\mathcal{C}_{n}$ and $s\in\mu_{\infty}$.

• •

Inversion: $(s_{1},\cdots,s_{n})\sim(1/s_{1},\cdots,1/s_{n})$ for any $(s_{1},\cdots,s_{n})\in\mathcal{C}_{n}$.

• •

If $n\leq 3$, then $(s_{1},\cdots,s_{n})\sim(t_{1},\cdots,t_{n})$ for any $(s_{1},\cdots,s_{n}),(t_{1},\cdots,t_{n})\in\mathcal{C}_{n}$.

Our objective in this article is to describe $\mathcal{C}_{n}/\sim$ for general $n$. In particular, it is reasonable to expect that the nontrivial projectively equivalent pairs cannot be arbitrarily long. While the existence of such an upper bound can be implied by a result of Schlickewei [5], our next result shows that the maximal length is $14$.

Theorem 3**.**

Up to rotations, inversions, permutations, and Galois conjugations, there are two nontrivial projectively equivalent pairs in $\mathcal{C}_{14}^{+}$:

[TABLE]

and

[TABLE]

There are no nontrivial projectively equivalent pairs in $\mathcal{C}_{n}^{+}$ for $n\geq 15$.

Since for $n\geq 4$,

[TABLE]

$\mathcal{C}_{n}/\sim$ is determined by $\mathcal{C}_{4}/\sim$. Now let us collect the solutions to the Diophantine equation

[TABLE]

Definition 4**.**

Let

[TABLE]

and $\mathcal{S}^{+}$ the isomorphic image of $\mathcal{S}$ under $(\textup{Arg}/(2\pi))^{6}$.

We notice that one solution in $\mathcal{S}$ can actually produce many more solutions.

Proposition 5**.**

The groups $\textup{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$, $\mathfrak{G}_{0}:=\left\langle(123)(456),(12)(45)\right\rangle\times\left\langle(14)(25)(36)\right\rangle\simeq\mathfrak{S}_{3}\times C_{2}$,

[TABLE]

and

[TABLE]

act on $\mathcal{S}$ naturally.

Proof.

If $[s_{1},s_{2},s_{3},t_{1},t_{2},t_{3}]\in\mathcal{S}$, then

[TABLE]

so $[s_{1},s_{2},s_{3},1/t_{1},t_{3}/t_{1},t_{2}/t_{1}]\in\mathcal{S}$. The rest are straightforward. ∎

Definition 6**.**

Let $\mathfrak{G}:=\left\langle\mathfrak{G}_{0},\mathfrak{G}_{s},\mathfrak{G}_{t}\right\rangle$, a group of order $768$. In $\mathcal{S}$, we write

[TABLE]

to denote that

[TABLE]

for some $\sigma\in\textup{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ and $g\in\mathfrak{G}$. The corresponding equivalence relation in $\mathcal{S}^{+}$ is also denoted by $\approx$.

As we will see in a minute, the size of $\mathcal{S}$ is quite large, so this equivalence relation is really helpful. With this simplification, we are able to describe the structure of $\mathcal{S}^{+}/\approx$ now. We use the additive $\mathcal{S}^{+}$ rather than the multiplicative $\mathcal{S}$ because $\theta$ is typographically simpler than $\exp(2\pi i\theta)$.

Theorem 7**.**

$\mathcal{S}^{+}/\approx$* can be classified as follows:*

• •

$1$* three-parameter solution,*

• •

$3$* two-parameter solutions,*

• •

$41$* one-parameter solutions,*

• •

$730$* zero-parameter solutions.*

The positive-parameter solutions are listed in Table LABEL:table1, where $x,y,z\in\mathbb{Q}/\mathbb{Z}$ can be taken arbitrarily as long as $[s_{1},s_{2},s_{3},t_{1},t_{2},t_{3}]\in\mathcal{S}^{+}$. The zero-parameter solutions are listed in Table LABEL:table2. (Since it is too long, we put it at the end of this article to keep the text uninterrupted.)

After introducing some necessary preliminaries in Section 2, we will give the proofs of Theorem 7 and 3 in Section 3 and 4, respectively. As the final results suggest, the proofs consist mostly of algorithms and computations, which are realizable with the help of Mathematica 11.0 [7]. The total running time is about seven hours on the author’s personal laptop. The source code will be attached with this article.

Whenever we perform the computations on the computer, we will always apply the most efficient algorithms we have. But when we describe the algorithms in this article, we will simply outline their key ideas in order to avoid too many cumbersome details. As a compensation, we will provide some concrete examples for better illustration.

2 Preliminaries

2.1 Vanishing sums of roots of unity

Without loss of generality (since the group $\mu_{\infty}$ is divisible), suppose that $[s_{1}^{2},s_{2}^{2},s_{3}^{2},t_{1}^{2},t_{2}^{2},t_{3}^{2}]\in\mathcal{S}$. Define

[TABLE]

where $\overline{\xi}=1/\xi$ represents the complex conjugate of $\xi\in\mu_{\infty}$, then

[TABLE]

and

[TABLE]

Conversely, given $(A_{1},A_{2},A_{3},B_{1},B_{2},B_{3},A_{1}^{\prime},A_{2}^{\prime},A_{3}^{\prime},B_{1}^{\prime},B_{2}^{\prime},B_{3}^{\prime})$ satisfying (2.1) and (2), then

[TABLE]

is almost a solution (which may be degenerate sometimes) in $\mathcal{S}$.

Thus we are led to study how $24$ roots of unity can sum to zero. More generally, we consider the cyclotomic relation

[TABLE]

where $\xi_{i}\in\mu_{\infty}$. This topic has been studied by many people. Among them, the article of Poonen and Rubinstein [4] enlightens us quite a lot. In their work, in order to count the number of interior intersection points and the number of regions made by the diagonals inside a regular polygon, they classify the cyclotomic relations up to $m=12$.

We will use their method substantially in the sequel. Let us first collect some of their notations. We call $w(G)=m$ the weight of $G$. $G$ is said to be minimal if $\sum_{i\in I}\xi_{i}\neq 0$ for any nonempty and proper subset $I\subseteq\{1,\cdots,m\}$. Any relation can be decomposed as a sum of minimal relations (which may not be unique). Write $\zeta_{n}=\exp(2\pi i/n)$ for any positive integer $n$. For each prime $p$, let $R_{p}$ be the relation

[TABLE]

Its minimality follows from the irreducibility of the cyclotomic polynomial. Also we can “rotate” any relation by multiplying an arbitrary root of unity to obtain a new relation. If we subtract $R_{3}$ from $R_{5}$, cancel the $1$’s and incorporate the minus signs into the roots of unity, we obtain a new minimal relation

[TABLE]

which we will denote $(R_{5}:R_{3})$. In general, if $G$ and $H_{1},\cdots,H_{j}$ are relations, we will use the notation $(G:H_{1},\cdots,H_{j})$ to denote any relation obtained by rotating the $H_{i}$ so that each shares exactly one root of unity with $G$ which is different for each $i$, subtracting them from $G$, and incorporating the minus signs into the roots of unity. For notational convenience, we will write $(R_{5}:4R_{3})$ for $(R_{5}:R_{3},R_{3},R_{3},R_{3})$, for example. Note that in general there will be many relations of type $(G:H_{1},\cdots,H_{j})$ up to rotational equivalence.

The following results will be needed in our arguments.

Lemma 8**.**

[3, Theorem 1; 4, Lemma 1]* If the relation $G=\sum_{i=1}^{m}\xi_{i}=0$ is minimal, then there are distinct primes $p_{1}<\cdots<p_{k}\leq m$ so that each $\xi_{i}$ is a $p_{1}\cdots p_{k}$-th root of unity, after the relation has been suitably rotated.*

Lemma 9**.**

[4, Lemma 2]* The only minimal relations (up to rotation) involving only the $2p$-th roots of unity, for $p$ prime, are $R_{2}$ and $R_{p}$.*

Lemma 10**.**

[4, Lemma 3]* Suppose $G$ is a minimal relation, and $p_{1}<\cdots<p_{k}$ are picked as in Lemma 8 with $p_{1}=2$ and $p_{k}$ minimal. If $w(G)<2p_{k}$, then $G$ (or a rotation) is of the form $(R_{p_{k}}:H_{1},\cdots,H_{j})$ where the $H_{i}$ are minimal relations not equal to $R_{2}$ and involving only $p_{1}\cdots p_{k-1}$-th roots of unity, such that $j<p_{k}$ and $\sum_{i=1}^{j}[w(H_{i})-2]=w(G)-p_{k}$.*

Lemma 11**.**

(Modified from [4, Lemma 4])* Suppose the relation $G$ is stable under complex conjugation. If $R_{p}$, for $p$ prime, occurs in $G$, then it is itself stable under complex conjugation, or can be paired with another $R_{p}$ which is its complex conjugate.*

Lemma 12**.**

[4, Lemma 5]* Let $G$ be a minimal relation of type $(R_{p}:H_{1},\cdots,H_{j})$, $p\geq 5$, where the $H_{i}$ involve roots of unity of order prime to $p$, and $j<p$. If $G$ is stable under complex conjugation, then the particular rotation of $R_{p}$ from which the $H_{i}$ were “subtracted” is also stable (and hence so is the collection of the relations subtracted).*

Lemma 13**.**

[4, Theorem 3]* The minimal relations of weight up to $12$ are as follows:*

The reason why we take $[s_{1}^{2},s_{2}^{2},s_{3}^{2},t_{1}^{2},t_{2}^{2},t_{3}^{2}]\in\mathcal{S}$ rather than the more natural $[s_{1},s_{2},s_{3},t_{1},t_{2},t_{3}]\in\mathcal{S}$ at the beginning of this section is to make the relation (2.1) stable under complex conjugation, so that Lemma 11 and 12 can be applied.

2.2 Smith normal form

The relations (2) can be reduced to

[TABLE]

which are equivalent to the linear equations

[TABLE]

where the lower case unknowns are the images of corresponding upper case unknowns under $\textup{Arg}/(2\pi)$. Note that the numbers on the right-hand side of (2*′*) belong to $\mathbb{Q}/\mathbb{Z}$. An important tool to deal with such equations is the following well-known result in elementary algebra.

Theorem 14**.**

Let $M$ be an $m\times n$ matrix of rank $r$ over $\mathbb{Z}$. There exist invertible $m\times m$ and $n\times n$ matrices $S$ and $T$ such that the product $SMT$ is of the form $D=\begin{pmatrix}D_{0}&0\\ 0&0\end{pmatrix}$, where $D_{0}$ is an $r\times r$ diagonal matrix with positive entries $d_{i}$ such that $d_{i}\mid d_{i+1}$ for any $1\leq i<r$. $D$ is called the Smith normal form of $M$.

2.3 Cyclotomic units

We should note that although the method developed by Poonen and Rubinstein works perfectly for their own problem, it is not enough for us. The main reason is, briefly speaking, our weight $24$ is much larger than their weight $12$. More details will be explained in Section 3.7.

If $[s_{1},s_{2},s_{3},t_{1},t_{2},t_{3}]\in\mathcal{S}$, then

[TABLE]

The numbers of the form $\xi-1$ with $\xi\in\mu_{\infty}$ are related to the so-called cyclotomic units (whose definition is irrelevant to us). The following fundamental result regarding the cyclotomic units will be needed in Algorithm 37.

Theorem 15**.**

[1; 2; 6, Theorem 8.9]* Let $n$ be a positive integer and $E_{n}$ the multiplicative abelian group generated by*

[TABLE]

Then for any relation $R$ in $E_{n}$, $R^{2}$ is a $\mathbb{Z}$-linear combination of the relations

[TABLE]

and

[TABLE]

In fact, what we will use is a weaker form of Theorem 15: any relation in $E_{n}\otimes_{\mathbb{Z}}\mathbb{Q}$ is a $\mathbb{Q}$-linear combination of the relations (5) and (6).

3 Proof of Theorem 7

We decompose the relation (2.1) into a sum of minimal relations, and then consider the largest prime $p$ involved. Since (2.1) has weight $24$, by Lemma 8, we have

[TABLE]

We divide the proof into four parts: $p=17,19,23$, $p=11,13$, $p=2,3$, and finally the most tough $p=5,7$. In Section 3.7, we will explain why different strategies are required for different $p$.

3.1 Symmetry of the relations (2.1)

and (2)

Since the associated matrix of (2*′*)

[TABLE]

has rank $6$, we can fix six unknowns and then try to solve the remaining six. However, there are $\binom{12}{6}=924$ ways to do that, so we need to reduce the number of cases first. We do that by considering the following natural group actions of $\mathfrak{G}$.

• •

$\mathfrak{G}$ permutes the $24$-element set

[TABLE]

This action is transitive and faithful.

• •

Modulo negation, $\mathfrak{G}$ permutes the $24$-element set

[TABLE]

This action is equivalent to the previous one.

• •

Modulo negation and inversion, $\mathfrak{G}$ permutes the $12$-element set

[TABLE]

This action is transitive but not faithful. The kernel is

[TABLE]

• •

$\mathfrak{G}$ permutes the $\binom{12}{n}$-element set

[TABLE]

The following two lemmas will be essentially used in Algorithm 27.

Lemma 16**.**

The action of $\mathfrak{G}$ on $U_{12,6}$ has the following orbits:

Let us take $O_{18}$ as an example to explain each column of the table. The second and third columns say that the orbit containing $\{a_{1},a_{2},b_{1},a_{1}^{\prime},a_{2}^{\prime},b_{1}^{\prime}\}$ has $12$ elements. If we consider $\{a_{1},a_{2},b_{1},a_{1}^{\prime},a_{2}^{\prime},b_{1}^{\prime}\}$ as constants and $\{a_{3},b_{2},b_{3},a_{3}^{\prime},b_{2}^{\prime},b_{3}^{\prime}\}$ as unknowns, then (2*′*) becomes

[TABLE]

By Theorem 14,

[TABLE]

The fourth column says that the diagonal of the Smith normal form is $(1,1,1,1,2,0)$. Multiplying the inverse of the first matrix on both sides, we get

[TABLE]

The fifth column says that (2*′*) has a solution if and only if $a_{2}+b_{1}-a_{2}^{\prime}-b_{1}^{\prime}=0$. Assuming this condition is satisfied, then

[TABLE]

where $x$ is a free variable. Multiplying the inverse of the first matrix on both sides, we get

[TABLE]

Lemma 17**.**

**

• •

Every element of $U_{12,7}$ has a $6$-element subset that belongs to $O_{1}$, $O_{3}$, $O_{5}$, $O_{7}$, $O_{11}$, or $O_{14}$.

• •

Every element of $U_{12,8}$ has a $6$-element subset that belongs to $O_{1}$, $O_{2}$, $O_{6}$, $O_{7}$, or $O_{15}$.

• •

Every element of $U_{12,9}$ has a $6$-element subset that belongs to $O_{1}$.

3.2 $\boldsymbol{p=17,19,23}$

Let $\mu_{n}$ be the collection of $n$-th roots of unity. If $n=\prod_{i=1}^{k}p_{i}^{n_{i}}$ is the prime factorization of $n$, then every $\xi\in\mu_{n}$ can be uniquely written as a product $\xi=\prod_{i=1}^{k}\xi_{i}$, where $\xi_{i}\in\mu_{p_{i}^{n_{i}}}$ for any $1\leq i\leq k$. We call $\xi_{i}$ the $p_{i}$-component of $\xi$.

Lemma 18**.**

The relations $G=0$ with $w(G)=24$ and $p=17,19,23$ are as follows:

Proof.

Since $w(G)<2p$, by Lemma 10, $G$ must be of the form

[TABLE]

where $\sum_{i=1}^{j}[w(H_{i})-2]+\sum_{i=1}^{k}w(G_{i})=w(G)-p\leq 7$. Since the minimal relations of weight up to $9$ are given in Lemma 13, we are done. ∎

Now we want to find all the solutions of (2.1) and (2) by assuming that (2.1) is of the form $G$, where $G$ is one of the entries listed in Lemma 18.

By Lemma 12, the marked entries cannot be stable under complex conjugation, so we can ignore them. By Lemma 11, if $G\neq(R_{17}:R_{3})+(R_{5}:R_{3})$, then

Condition 19**.**

$G$ can be decomposed as a sum of minimal relations such that each minimal relation is itself stable under complex conjugation, or can be paired with another minimal relation which is its complex conjugate.

The following two lemmas show that this is also true for $G=(R_{17}:R_{3})+(R_{5}:R_{3})$.

Lemma 20**.**

Let $G=\xi_{1}G_{1}+\xi_{2}G_{2}$ be a relation which is stable under complex conjugation, where $\xi_{i}\in\mu_{\infty}$ and $G_{i}$ are minimal relations. Assume that all the terms of $G_{i}$ belong to $\mu_{n_{i}}$. If $\xi_{1}G_{1}$ and $\xi_{2}G_{2}$ are not conjugate to each other, then $\xi_{i}\in\mu_{2n_{i}}$.

Proof.

By the assumptions, two terms of $G_{i}$ must be conjugate to each other, so $\xi_{i}^{2}\in\mu_{n_{i}}$. ∎

Lemma 21**.**

If $G=(R_{17}:R_{3})+(R_{5}:R_{3})$ is stable under complex conjugation, then both $(R_{17}:R_{3})$ and $(R_{5}:R_{3})$ are stable under complex conjugation.

Proof.

Assume that

[TABLE]

is stable under complex conjugation. By Lemma 20, $\xi_{1}\in\mu_{204}$ and $\xi_{2}\in\mu_{60}$. Since the terms of $G$ with nontrivial $17$-components are stable under complex conjugation, $\xi_{1}$ must have a trivial $17$-component and also $\xi_{1}^{2}\in\mu_{17}$. Therefore, $\xi_{1}=\pm 1$, and consequently, $\xi_{2}=\pm 1$. ∎

The longest minimal relation occurring in $G$ has weight at least $17$, so at least $9$ elements of $U_{12}$ are known. It is easy to see (directly or by Lemma 17) that if at least $9$ elements of $U_{12}$ are given, then $U_{12}$ can be completely determined. In particular, if at least $9$ elements of $U_{12}$ belong to $\mu_{n}$, then all elements of $U_{12}$ belong to $\mu_{n}$. Therefore, we are led to consider the $q$-components of $G$ for prime $q$.

Definition 22**.**

Let $G=\sum_{i=1}^{m}\xi_{i}=0$ be a relation such that all the $q$-components of $\xi_{i}$ belong to $\mu_{q}$ (i.e., higher powers of $q$ are not involved). Let $n_{k}$ be the number of $\zeta_{q}^{k}$ in the $q$-components of $\xi_{i}$. We call $(n_{k})_{k=0}^{q-1}$ the $q$-type of $G$. We say that $(n_{j+k})_{k=0}^{q-1}$ is a rotation of $(n_{k})_{k=0}^{q-1}$ for any $j\in\mathbb{Z}/q\mathbb{Z}$, and $(n_{jk})_{k=0}^{q-1}$ is a Galois conjugate of $(n_{k})_{k=0}^{q-1}$ for any $j\in(\mathbb{Z}/q\mathbb{Z})^{\times}$.

Consider the $q$-component of (2*′*):

[TABLE]

Define the finite sets

[TABLE]

In practice, we only need to calculate $\mathcal{P}_{q}$ for $q\leq 13$. For $q=2$ and $3$, we have

[TABLE]

and

[TABLE]

Algorithm 23**.**

For a given relation of the form $G$, we check that whether there are any possible $q$-types of $G$ that belong to $\mathcal{P}_{q}$. If not, then we conclude that the relation (2.1) cannot be of the form $G$.

Let us give some examples to illustrate Algorithm 23.

Suppose that $G=(R_{23}:R_{3})$. By Lemma 12,

[TABLE]

The $2$-type of $G$ is a rotation of $(22,2)$, which does not belong to $\mathcal{P}_{2}$. Therefore, the relation (2.1) cannot be of the form $G$.

Suppose that $G=(R_{19}:R_{3})+2R_{2}$. By Lemma 11 and 12,

[TABLE]

where $\xi\in\mu_{114}$. The $2$-type of $(R_{19}:R_{3})$ is a rotation of $(18,2)$, and the $2$-type of $R_{2}$ is $(1,1)$, so the $2$-type of $G$ must be a rotation of $(20,4)$, which belongs to $\mathcal{P}_{2}$. Therefore, Algorithm 23 fails for $q=2$. The $3$-type of $(R_{19}:R_{3})$ is $(18,1,1)$, and the $3$-type of $R_{2}$ is a rotation of $(2,0,0)$, so the $3$-type of $G$ must be $(22,1,1)$ or $(18,3,3)$, which does not belong to $\mathcal{P}_{3}$. Therefore, the relation (2.1) cannot be of the form $G$.

Suppose that $G=(R_{17}:R_{5},4R_{3})$. By Lemma 12,

[TABLE]

for some $1\leq j<k\leq 8$. The $2$-type of $G$ is $(12,12)$, which belongs to $\mathcal{P}_{2}$. The $3$-type of $G$ is $(16,4,4)$, which belongs to $\mathcal{P}_{3}$. Therefore, Algorithm 23 fails for $q=2$ and $3$. The $5$-type of $G$ is $(20,1,1,1,1)$, which does not belong to $\mathcal{P}_{5}$. Therefore, the relation (2.1) cannot be of the form $G$.

It turns out that, by taking $q=2$, $3$, or $5$, Algorithm 23 works for any $G$ listed in Lemma 18. Therefore, there are no solutions of (2.1) and (2) such that (2.1) is of the form $G$.

The careful readers may have noticed that in Table LABEL:table2, there are one solution of order $34$ and four solutions of order $102$. Although the prime $17$ is involved, their associated relations (2.1) can only be decomposed as $n_{2}R_{2}+n_{3}R_{3}$ for some $n_{2}$ and $n_{3}$, which will be discussed in Section 3.4.

3.3 $\boldsymbol{p=11,13}$

The number of relations $G=0$ with $w(G)=24$ and $p=11,13$ is much larger than the number of relations $G=0$ with $w(G)=24$ and $p=17,19,23$. Since our purpose is finding the solutions of (2.1) and (2) rather than classifying the vanishing relations, we can first exclude some cases with certain shapes.

Lemma 24**.**

**

• •

If at least $7$ terms of $(n_{k})\in\mathcal{P}_{11}$ are $1$, then $(n_{k})$ must be a Galois conjugate of $(4,6,1,\cdots,1,6)$.

• •

If at least $9$ terms of $(n_{k})\in\mathcal{P}_{13}$ are $1$, then $(n_{k})$ must be a Galois conjugate of $(6,4,1,\cdots,1,4)$, $(4,5,1,\cdots,1,5)$, or $(2,6,1,\cdots,1,6)$.

By Lemma 10, if $p=13$, or $p=11$ and the longest minimal relation occurring in $G$ has weight at most $21$, then $G$ must be of the form (7). If $G$ is not of this form, then $G=G_{0}$ or $G_{0}+R_{2}$, where $G_{0}$ is minimal and stable under complex conjugation. By Lemma 8, $G_{0}$ can be written as $\xi\sum_{i=0}^{10}f_{i}\zeta_{11}^{i}$, where $\xi\in\mu_{\infty}$ and each $f_{i}$ is a sum of at least two elements of $\mu_{210}$. Since $[\mathbb{Q}(\zeta_{2310}):\mathbb{Q}(\zeta_{210})]=10$, we know that $f_{i}-f_{j}=0$ for any $0\leq i<j\leq 10$.

If $w(G_{0})=22$, then $w(f_{i}-f_{j})=4$ for any $0\leq i<j\leq 10$. But since the only relation of weight $4$ is $2R_{2}$, all $f_{i}$ must be the same, $G_{0}$ cannot be minimal. In general, if $f_{i}$ is a sum of two elements of $\mu_{\infty}$, then $\xi f_{i}=\xi f_{11-i}=\overline{\xi f_{i}}$.

If $w(G_{0})=24$ and the $11$-type of $G_{0}$ is $(4,2,\cdots,2)$, then $w(f_{0}-f_{1})=6$. If $f_{0}-f_{1}=3R_{2}$ or $2R_{3}$, then $G_{0}$ cannot be minimal. If $f_{0}-f_{1}=(R_{5}:R_{3})$, then since both $\xi f_{0}$ and $\xi f_{1}$ are stable under complex conjugation, by Lemma 12, $G$ must be

[TABLE]

or a Galois conjugate of

[TABLE]

The $2$-type of $G$ is a rotation of $(24,0)$ or $(22,2)$, which does not belong to $\mathcal{P}_{2}$. Therefore, the relation (2.1) cannot be of the form $G$.

If $w(G_{0})=24$ and the $11$-type of $G_{0}$ is a Galois conjugate of $(2,3,2,\cdots,2,3)$, then $w(f_{0}-f_{1})=5$. If $f_{0}-f_{1}=R_{2}+R_{3}$, then $f_{0}-f_{10}=\overline{f_{0}-f_{1}}=R_{2}+R_{3}$, $G_{0}$ cannot be minimal. If $f_{0}-f_{1}=R_{5}$, since $\xi f_{0}$ is stable under complex conjugation, $\xi f_{1}$ is forced to be stable under complex conjugation, $G$ must be a Galois conjugate of

[TABLE]

The $2$-type of $G$ is a rotation of $(18,6)$, which does not belong to $\mathcal{P}_{2}$. Therefore, the relation (2.1) cannot be of the form $G$.

Suppose that (2.1) is of the form (7), and some of the $H_{i}$ and $G_{i}$ has weight greater than $12$, then $G=(R_{13}:H_{1})$, $(R_{11}:H_{1})$, $(R_{11}:H_{1})+R_{2}$, or $R_{11}+G_{1}$. For each case except for $R_{11}+(R_{11}:R_{3})$, the $p$-type of $G$ is $(25-p,1,\cdots,1)$, which does not belong to $\mathcal{P}_{p}$ by Lemma 24. Since $R_{11}+(R_{11}:R_{3})=(R_{11}:R_{3})+R_{11}$, we can assume that none of the $H_{i}$ and $G_{i}$ has weight greater than $12$, so that Lemma 13 is enough for us.

(2.1) cannot be of the form $(R_{p}:H_{1},\cdots,H_{j})+R_{2}$ because the $2$-components of $(R_{p}:H_{1},\cdots,H_{j})$ belong to $\mu_{2}$, but the $2$-components of $R_{2}=\zeta_{4}+\zeta_{4}^{3}$ belong to $\mu_{4}$.

If (2.1) is of the form $(R_{p}:H_{1},\cdots,H_{j})+\sum_{q<p}n_{q}R_{q}$, then by Lemma 11, $(R_{p}:H_{1},\cdots,H_{j})$ is stable under complex conjugation. By Lemma 12, among $H_{i}$, all but possibly one entries of Lemma 13 must occur even times, and only $R_{3}$, $R_{5}$, $(R_{5}:2R_{3})$, $(R_{5}:4R_{3})$, $R_{7}$, $(R_{7}:2R_{3})$, $(R_{7}:4R_{3})$, or $R_{11}$ can be the exception.

If (2.1) is of the form $(R_{13}:H_{1},\cdots,H_{j})+\sum_{q<13}n_{q}R_{q}+G_{1}$, where $n_{q}\in\{0,1\}$ for any $q$, then by Lemma 20, the $13$-components of $\sum_{q<13}n_{q}R_{q}+G_{1}$ must be trivial. By Lemma 24, if $j\leq 3$, then $\{H_{1},\cdots,H_{j}\}$ must contain $2R_{5}$, $2(R_{5}:R_{3})$, $2(R_{5}:2R_{3})$, or $2R_{7}$.

If (2.1) is of the form $(R_{11}:H_{1},\cdots,H_{j})+\sum_{q<11}n_{q}R_{q}+G_{1}$, where $n_{q}\in\{0,1\}$ for any $q$, and $G_{1}\neq R_{11}$ or $(R_{11}:R_{3})$, then by Lemma 20, the $11$-components of $\sum_{q<11}n_{q}R_{q}+G_{1}$ must be trivial. By Lemma 24, if $j\leq 3$, then $\{H_{1},\cdots,H_{j}\}$ must contain $2(R_{5}:2R_{3})$ or $2R_{7}$.

Lemma 25**.**

After these eliminations, the relations $G=0$ with $w(G)=24$ and $p=11,13$ are as follows:

By Lemma 20, $(R_{11}:R_{5},3R_{3})+(R_{5}:2R_{3})$ and $(R_{11}:(R_{5}:R_{3}),3R_{3})+(R_{5}:R_{3})$ cannot be stable under complex conjugation, so we can ignore them. For $R_{11}+(R_{5}:2R_{3})+(R_{5}:R_{3})$, we can apply Algorithm 23 by taking $q=3$ without knowing whether $(R_{5}:2R_{3})$ and $(R_{5}:R_{3})$ are stable under complex conjugation.

Suppose $G=2R_{11}+R_{2}$. If $R_{2}=1+(-1)$, then $G$ can be decomposed as $12R_{2}$, which will be discussed in Section 3.4. Otherwise, $G=\xi R_{11}+\overline{\xi}R_{11}+\zeta_{4}+\zeta_{4}^{3}$. From the relations (2), we see that either $\xi\in\mu_{44}$ or $\xi^{3}\in\mu_{44}$. If $\xi\in\mu_{44}$, then again $G$ can be decomposed as $12R_{2}$. Otherwise, we can apply Algorithm 23 by taking $q=3$.

Suppose $G=\xi_{1}(R_{11}:R_{3})+\xi_{2}(R_{11}:R_{3})$. If $\xi_{1}(R_{11}:R_{3})$ and $\xi_{2}(R_{11}:R_{3})$ are not conjugate to each other, then by Lemma 20, $\xi_{1},\xi_{2}\in\mu_{132}$. Otherwise, $\xi_{1}=\overline{\xi_{2}}$. Without loss of generality, let $A_{1}=\xi_{1}\zeta_{6}$, then from the relations (2), either $A_{1}B_{2}A_{1}^{\prime}B_{2}^{\prime}=1$ or $A_{1}B_{3}\overline{A_{1}^{\prime}B_{3}^{\prime}}=1$ does not contain $\xi_{1}\zeta_{6}^{5}$. Therefore, the $3$-components cannot be canceled, which implies that either $\xi_{1}^{2}\in\mu_{66}$ or $\xi_{1}^{4}\in\mu_{66}$. In any case, we can apply Algorithm 23 by taking $q=3$.

Suppose $G=\xi_{1}(R_{11}:R_{3})+\xi_{2}(R_{5}:R_{3})+\xi_{3}(R_{5}:R_{3})$. If $\xi_{2}(R_{5}:R_{3})$ and $\xi_{3}(R_{5}:R_{3})$ are not conjugate to each other, then by the following lemma, $\xi_{1},\xi_{2},\xi_{3}\in\mu_{660}$. Otherwise, $\xi_{1}=\pm 1$ and $\xi_{2}=\overline{\xi_{3}}$. From the relations (2), either $\xi_{2}^{k}\in\mu_{330}$ for some $1\leq k\leq 6$, or $\xi_{2}$ can be canceled, which means that $\xi_{2}$ can be taken arbitrarily. In any case, we can apply Algorithm 23 by taking $q=11$.

Lemma 26**.**

Let $G=\xi_{1}G_{1}+\xi_{2}G_{2}+\xi_{3}G_{3}$ be a relation which is stable under complex conjugation, where $\xi_{i}\in\mu_{\infty}$ and $G_{i}$ are minimal relations. Assume that all the terms of $G_{i}$ belong to $\mu_{n}$. If any two of $\xi_{i}G_{i}$ are not conjugate to each other, then $\xi_{i}\in\mu_{2n}$.

Proof.

If $\xi_{1}\notin\mu_{2n}$, then any two terms of $\xi_{1}G_{1}$ cannot be conjugate to each other. Since $\xi_{2}G_{2}$ is minimal, and $\xi_{1}G_{1}$ is not conjugate to $\xi_{2}G_{2}$, some term of $\xi_{1}G_{1}$ must be conjugate to some term of $\xi_{3}G_{3}$. Therefore, $\xi_{1}\xi_{3}\in\mu_{n}$. By the same reasoning, $\xi_{1}\xi_{2}\in\mu_{n}$. If $\xi_{2}\in\mu_{2n}$ or $\xi_{3}\in\mu_{2n}$, then we are done. Otherwise, we must have $\xi_{2}\xi_{3}\in\mu_{n}$, then $\xi_{1}^{2}\xi_{2}^{2}\xi_{3}^{2}\in\mu_{n}$, $\xi_{1}\xi_{2}\xi_{3}\in\mu_{2n}$, and finally $\xi_{1},\xi_{2},\xi_{3}\in\mu_{2n}$. ∎

By Lemma 11 and the same reasoning of Lemma 21, any other $G$ listed in Lemma 25 satisfy Condition 19. It turns out that, for those unmarked entries, we can apply Algorithm 23 by taking $q=2$, $3$, $5$, $7$, or $11$. However, for the remaining cases, we need a new algorithm.

Algorithm 27**.**

Given a relation $G=G_{0}+n_{2}R_{2}+n_{3}R_{3}+n_{5}R_{5}$ of weight $24$, where $G_{0}$ is stable under complex conjugation but not necessarily minimal. Let $X=\{X_{j}\}_{j=1}^{n}$ be the partition of the terms of $G_{0}$ into conjugate pairs. $X$ is allowed to contain at most one $\{1\}$ and at most one $\{-1\}$. For any $1\leq i\leq 19$, let $Y_{i}$ be a fixed element (for example, the $(i,2)$-entry of Lemma 16) in the orbit $O_{i}$.

• •

Suppose $n=6$. For any $1\leq i\leq 19$, we assign each element of $Y_{i}$ a term of $G_{0}$ such that any two of them do not come from the same $X_{j}$. Then we solve the equations (2′**) as we have demonstrated after Lemma 16 to get a full set of $U_{24}$. There will be zero, one, or two free variables depending on whether $1\leq i\leq 9$, $10\leq i\leq 18$, or $i=19$. Then we check whether $G_{0}\subseteq U_{24}$ and $U_{24}\backslash G_{0}$ is possible to be of the form $n_{2}R_{2}+n_{3}R_{3}+n_{5}R_{5}$. If yes, then we use (3) to get the solutions in $\mathcal{S}$.

• •

Suppose $n=7$. It suffices to take $i=1$, $3$, $5$, $7$, $11$, and $14$ (see Lemma 17).

• •

Suppose $n=8$. It suffices to take $i=1$, $2$, $6$, $7$, and $15$.

• •

Suppose $n\geq 9$. It suffices to take $i=1$ and select the elements of $Y_{1}$ from the subset $\cup_{j=1}^{9}X_{j}\subseteq G_{0}$.

Applying Algorithm 27 to those marked entries listed in Lemma 25, we will get $21$ zero-parameter solutions: ten have order $66$, seven have order $78$, one has order $110$, and three have order $132$.

To illustrate Algorithm 27, let us give an example to see how to get the solution

[TABLE]

from the relation

[TABLE]

Now $n=7$, so we can take $i=14$. Let

[TABLE]

By Lemma 16, (2*′*) has a solution if and only if $a_{1}+a_{2}+2b_{3}-a_{1}^{\prime}+a_{2}^{\prime}=0$. Now this condition is satisfied, so (2*′*) has the solution

[TABLE]

where $x$ is a free variable. Now $G_{0}\subseteq U_{24}$ and the image of $U_{24}\backslash G_{0}$ under $\textup{Arg}/(2\pi)$ is

[TABLE]

It is not hard to see that $U_{24}\backslash G_{0}$ can be decomposed as $R_{3}+4R_{2}$ if and only if $x=1/44$ or $23/44$. If we take $x=23/44$, then by (3), we get

[TABLE]

If we take $x=1/44$, we will get the same solution.

If $n=6$ and $i=19$, then $U_{24}\backslash G_{0}$ may have two free variables, which will bring additional difficulties. But fortunately, this situation can always be avoided in practice.

3.4 $\boldsymbol{p=2,3}$

By Lemma 9, the relations $G=0$ with $w(G)=24$ and $p=2,3$ are $12R_{2}$, $9R_{2}+2R_{3}$, $6R_{2}+4R_{3}$, $3R_{2}+6R_{3}$, and $8R_{3}$.

Algorithm 28**.**

Given a relation $G$ of weight $24$ that satisfies Condition 19, then $G$ can be decomposed as

[TABLE]

where $G_{0}$ is allowed to be empty and not necessarily minimal, $G_{i}$ are minimal for $i\neq 0$, and $\xi_{i}$ are unknowns to be determined. Let $X=\{X_{j}\}_{j=1}^{12}$ be the partition of the terms of $G$ into conjugate pairs. We assign each element of $U_{12}$ a term of $G$ such that any two of them do not come from the same $X_{j}$. Then (2*′*) becomes a system of six linear equations in $k\leq 6$ unknowns. Let $M$ be the associated matrix of (2*′*), $D$ the Smith normal form of $M$, and

[TABLE]

Assume that all the terms of $G_{i}$ belong to $\mu_{n}$ for some even $n$. Depending on the value of $n$, we choose another nonnegative integer $m$.

• •

If $d>m$, then we solve the equations (2′**) and use (3) to get the solutions in $\mathcal{S}$.

• •

If $d\leq m$, then all the terms of $G$ belong to $\mu_{nd}$. This situation will be discussed in Section 3.6.

Applying Algorithm 28 to $G=n_{2}R_{2}+n_{3}R_{3}$, where $2n_{2}+3n_{3}=24$, by choosing $m=0$, we will get all the positive-parameter solutions and $133$ zero-parameter solutions. We note that none of the positive-parameter solutions comes from $G=8R_{3}$.

To illustrate Algorithm 28, let us give an example to see how to get the solution

[TABLE]

from the relation $G=12R_{2}$. Let

[TABLE]

then (2*′*) becomes

[TABLE]

Although we have seen how to solve such equations after Lemma 16, let us do it again for the completeness of this example. By Theorem 14,

[TABLE]

Multiplying the inverse of the first matrix on both sides, we get

[TABLE]

Note that the last entry of the matrix on the right-hand side is [math], so

[TABLE]

where $x$ is a free variable. Multiplying the inverse of the first matrix on both sides, we get

[TABLE]

Substituting them into $U_{12}$, we get

[TABLE]

By (3), we get

[TABLE]

Clearly, the second solution can be obtained by replacing $x$ with $1/2+x$ in the first solution.

3.5 $\boldsymbol{p=5,7}$

The purpose of this section to show that

Lemma 29**.**

If the associated relation (2.1) of $[s_{1},s_{2},s_{3},t_{1},t_{2},t_{3}]\in\mathcal{S}$ has $p=5,7$, then $s_{i},t_{i}\in\mu_{n}$ for some $n\in N:=\{240,252,280,300,336,360,420\}$.

As before, we first try to classify the relations $G=0$ with $w(G)=24$ and $p=5,7$. The following lemma shows that all the minimal relations with $p=5$ have been given in Lemma 13.

Lemma 30**.**

Suppose $G$ is a minimal relation with $p=5$. Then $G=R_{5}$ or $(R_{5}:jR_{3})$ for some $j<5$.

Proof.

By Lemma 8, $G$ can be written as a rotation of $\sum_{i=0}^{4}f_{i}\zeta_{5}^{i}$, where each $f_{i}$ is a sum of elements of $\mu_{6}$. Since $[\mathbb{Q}(\zeta_{30}):\mathbb{Q}(\zeta_{6})]=4$, we know that $f_{0}-f_{i}=0$ for any $0\leq i\leq 4$. By another rotation, we can assume that $f_{0}$ contains $1$, which belongs to some $R_{2}$ or $R_{3}$ by Lemma 9.

If $f_{0}$ contains $1$ but not $\zeta_{3}$ or $\zeta_{3}^{2}$, then $f_{i}$ contains either $1$ or $\zeta_{6}+\zeta_{6}^{5}$. By the minimality of $G$, $G=R_{5}$ or $(R_{5}:jR_{3})$, where $j$ is the number of $f_{i}$ that contains $\zeta_{6}+\zeta_{6}^{5}$.

If $f_{0}$ contains $1+\zeta_{3}$, then $f_{i}$ contains either $1+\zeta_{3}$ or $\zeta_{6}$. By the minimality of $G$, not all $f_{i}$ contain $1+\zeta_{3}$, so $G=(R_{5}:jR_{3})$, where $j$ is the number of $f_{i}$ that contains $1+\zeta_{3}$. ∎

Therefore, the relations $G=0$ with $w(G)=24$ and $p=5$ are one-to-one corresponding to the partitions of $24$ such that every summand belongs to $\{2,3,5,6,7,8,9\}$ and some summand belongs to $\{5,6,7,8,9\}$. However, the situation becomes much more complicated when $p=7$.

Definition 31**.**

Let $G=G_{1}+\cdots+G_{k}+2H_{1}+\cdots+2H_{j}$, where $G_{i}$ and $H_{i}$ are minimal relations. We call $2H_{1}+\cdots+2H_{j}$ the even part of $G$.

Lemma 32**.**

Suppose $G=0$ is a relation with $w(G)=24$ and $p=5,7$. If the even part of $G$ is empty, $2R_{2}$, or $2R_{3}$, and $G$ is stable under complex conjugation, then all the terms of $G$ belong to $\mu_{420}$.

Proof.

Suppose $G=G_{1}+\cdots+G_{k}+\sum_{q\leq 7}n_{q}R_{q}$, where $G_{i}\neq R_{q}$ are minimal. By the assumptions, $k\leq 3$ and any two of $G_{i}$ are not conjugate to each other. By Lemma 8, 20, and 26, all the terms of $G_{1}+\cdots+G_{k}$ belong to $\mu_{420}$, and consequently, all the terms of $G$ belong to $\mu_{420}$. ∎

Lemma 33**.**

Suppose $G=0$ is a relation with $w(G)=24$ and $p=5,7$. If the even part of $G$ is not empty, $2R_{2}$, or $2R_{3}$, and $G$ can be decomposed as $G=G_{0}+n_{2}R_{2}+n_{3}R_{3}+n_{5}R_{5}$, where $G_{0}$ is minimal, of weight at least $11$, and stable under complex conjugation, then up to Galois conjugation, $G_{0}$ must be one of the following:

Proof.

If $w(G_{0})=15$ or $w(G_{0})\geq 17$, then the even part of $G$ must be empty, $2R_{2}$, or $2R_{3}$. If $G$ is of the form (7), we can apply Lemma 10, 12, and 13. The last three entries can be obtained by the same reasoning of (8), (9), and (10). ∎

Applying Algorithm 27 to these entries, the results confirm Lemma 29.

Lemma 34**.**

Suppose $G=0$ is a relation with $w(G)=24$ and $p=5,7$. If the even part of $G$ is not empty, $2R_{2}$, or $2R_{3}$, and $G$ cannot be decomposed as $G=G_{0}+n_{2}R_{2}+n_{3}R_{3}+n_{5}R_{5}$, where $G_{0}$ is minimal and of weight at least $11$, then $G$ must be one of the following:

Let $G$ be one of these entries. Suppose $G=G_{1}+\cdots+G_{k}+\sum_{q\leq 7}n_{q}R_{q}$, where $G_{i}\neq R_{q}$ are minimal, is stable under complex conjugation. By Lemma 11, $G_{1}+\cdots+G_{k}$ is also stable under complex conjugation.

Suppose $k=2$. If $w(G_{1})+w(G_{2})\geq 18$, and $G_{1}$ and $G_{2}$ are not conjugate to each other, then by the same reasoning of Lemma 32 and 33, all the terms of $G$ belong to $\mu_{420}$. Otherwise,

Lemma 35**.**

If $G_{0}=G_{1}+G_{2}$, where $G_{1},G_{2}\neq R_{q}$ are minimal, and $w(G_{0})=12$, $13$, $14$, or $16$, is stable under complex conjugation, then $G_{0}$ satisfies Condition 19.

Proof.

Let $G_{0}=\xi_{1}(R_{p_{1}}:\cdots)+\xi_{2}(R_{p_{2}}:\cdots)$. If $p_{1}\neq p_{2}$, then the conclusion follows from the same reasoning of Lemma 21. Otherwise, $G_{0}$ must be one of

[TABLE]

We can get the conclusion by verifying every $\xi_{1},\xi_{2}\in\mu_{60}$ or $\mu_{84}$. ∎

If $k=3$, then $w(G_{1})+w(G_{2})+w(G_{3})\geq 18$. If any two of $G_{i}$ are not conjugate to each other, then by the same reasoning of Lemma 32 and 33, all the terms of $G$ belong to $\mu_{420}$.

For $k=4$, we have the following lemma.

Lemma 36**.**

If $G=\xi_{1}(G_{5}:R_{3})+\xi_{2}(G_{5}:R_{3})+\xi_{3}(G_{5}:R_{3})+\xi_{4}(G_{5}:R_{3})$ is stable under complex conjugation, then either $G$ satisfies Condition 19 or $\xi_{i}\in\mu_{60}$.

Proof.

Consider the graph $\mathfrak{g}$ (in the sense of graph theory) with vertices $\xi_{i}$. If some term of $\xi_{i}(G_{5}:R_{3})$ and some term of $\xi_{j}(G_{5}:R_{3})$ are conjugate to each other, we draw an edge $\{\xi_{i},\xi_{j}\}$ between $\xi_{i}$ and $\xi_{j}$, where $i=j$ is allowed. If $\{\xi_{i},\xi_{j}\}\in\mathfrak{g}$, then $\xi_{i}\xi_{j}\in\mu_{30}$.

If there is only one edge at $\xi_{i}$, then either $\xi_{i}(G_{5}:R_{3})$ is itself stable under complex conjugation, or can be paired with another $\xi_{j}(G_{5}:R_{3})$ which is its complex conjugate. The conclusion then follows from Lemma 26 and 35.

Now we assume that there are at least two edges at each vertex. Suppose the conclusion is false by assuming $\xi_{1}\notin\mu_{60}$, then $\{\xi_{1},\xi_{1}\}\notin\mathfrak{g}$. Without loss of generality, assume $\{\xi_{1},\xi_{2}\},\{\xi_{1},\xi_{3}\}\in\mathfrak{g}$. If $\{\xi_{2},\xi_{2}\}$, $\{\xi_{2},\xi_{3}\}$, or $\{\xi_{3},\xi_{3}\}\in\mathfrak{g}$, then by the same reasoning of Lemma 26, we have $\xi_{1}\in\mu_{60}$, which is a contradiction. Therefore, $\{\xi_{2},\xi_{4}\},\{\xi_{3},\xi_{4}\}\in\mathfrak{g}$, and by the same reasoning, $\{\xi_{1},\xi_{4}\},\{\xi_{4},\xi_{4}\}\notin\mathfrak{g}$. In summary,

[TABLE]

Thus $\xi_{2}(G_{5}:R_{3})+\xi_{3}(G_{5}:R_{3})$ can also be decomposed as $\overline{\xi_{1}}(G_{5}:R_{3})+\overline{\xi_{4}}(G_{5}:R_{3})$. ∎

We have shown that all the entries listed in Lemma 34 either satisfy Condition 19 or confirm Lemma 29. Now suppose $G=G_{0}+2G_{1}+\cdots+2G_{k}$ is of the form (11), and define $D$, $d$, $n$, and $m$ as in Algorithm 28. For a general $G$, we can apply Algorithm 28 by choosing $m$ such that for any $d\leq m$, $nd$ divides some element of $N$. However, for some cases, it will be much more efficient to apply Algorithm 23 and 27. Here are some examples.

If the even part of $G$ is $4R_{2}$, $2R_{2}+2R_{3}$, $2R_{5}$, $6R_{2}$, or $4R_{3}$, we can always apply Algorithm 27, provided $U_{24}\backslash G_{0}$ has at most one free variable.

If $G=G_{0}+2G_{1}$, then $d\leq 4$ or $d=\infty$. We can always try to

• •

Apply Algorithm 23 by taking $q=5$ or $7$ to conclude that (2.1) cannot be of the form $G$,

• •

Apply Algorithm 23 by taking $q=5$ or $7$ to conclude that $d=\infty$ cannot happen,

• •

Apply Algorithm 27 to conclude that $d=\infty$ cannot happen.

If $G=2G_{1}+2R_{3}$ or $2G_{1}+2R_{2}$, then it turns out that the diagonal of $D$ must be $(1,0)$, $(1,2)$, $(1,4)$, or $(1,6)$. The last three cases confirm Lemma 29. We can apply Algorithm 27 to conclude that the first case cannot happen.

After applying a combination of Algorithm 23, 27, and 28 to all the entries listed in Lemma 34, Lemma 29 can be fully confirmed.

3.6 The application of Theorem 15

Suppose $[s_{1},s_{2},s_{3},t_{1},t_{2},t_{3}]\in\mathcal{S}$ such that $s_{i},t_{i}\in\mu_{n}$. Let

[TABLE]

then the identity (4) implies

[TABLE]

Algorithm 37**.**

Given a positive integer $n$, let $\widehat{E}_{n}$ be the $\mathbb{Q}$-vector space generated by the free symbols $\{\widehat{e}_{k}:1\leq k<n\}$, $\widehat{F}_{n}$ the subspace generated by the relations (5) and (6) (where $e_{k}$ are replaced with $\widehat{e}_{k}$), $\widehat{F}_{n}^{\perp}$ the orthogonal complement of $\widehat{F}_{n}$, $\{\widehat{f}_{1},\cdots,\widehat{f}_{j}\}$ a basis of $\widehat{F}_{n}^{\perp}$. For any $1\leq\sigma_{1},\sigma_{2},\sigma_{3}<n$ such that any two of them are not equal, we calculate the inner products

[TABLE]

By Theorem 15,

[TABLE]

• •

We first find all $(\sigma_{1},\sigma_{2},\sigma_{3})$ and $(\tau_{1},\tau_{2},\tau_{3})$ satisfying (4′**),

• •

and then among them, find all $(s_{1},s_{2},s_{3})$ and $(t_{1},t_{2},t_{3})$ satisfying (4),

• •

and finally among them, find all solutions such that their associated relations (2.1) cannot be decomposed as $n_{2}R_{2}+n_{3}R_{3}$ for some $n_{2}$ and $n_{3}$.

Applying Algorithm 37 to all $n\in N$ given in Lemma 29, we will get $576$ zero-parameter solutions.

3.7 Comparisons of Algorithm 23,

27, 28, and 37

In summary, we obtain the complete list of $\mathcal{S}$ by finding all the solutions of (2.1) and (2) such that (2.1) is of the form $G$. Depending on the largest prime $p$ involved in $G$, we use the following strategies:

• •

$p=17,19,23$: We first give a classification for all such $G$, and then apply Algorithm 23 to prove that (2.1) cannot be any of them.

• •

$p=11,13$: We first apply Algorithm 23 to give some necessary conditions that (2.1) must satisfy. We then eliminate those unqualified $G$ and give a classification for the rest. For those survived $G$, we first try to apply Algorithm 23 to prove that (2.1) cannot be of that form. If Algorithm 23 fails, we apply Algorithm 27 to find the solutions.

• •

$p=2,3$: We first give a classification for all such $G$, and then apply Algorithm 28 to find the solutions.

• •

$p=5,7$: We first prove that Lemma 29 holds unless each minimal relation occurring in $G$ has a moderate weight, for which we are able to give a classification. We then apply Algorithm 23, 27, and 28 together to show that Lemma 29 holds in general. For each element given in Lemma 29, we apply Algorithm 37 to find the solutions.

We see that the first step for each case is classification. On the one hand, there are only a few minimal relations with $p\leq 5$, so it is easy to classify all $G$ with $p\leq 5$. On the other hand, when $p\geq 11$, $w(G)$ is smaller or slightly larger than $2p$, so that we can apply Lemma 10 to classify all $G$ with $p\geq 11$. However, for $p=7$, we are unable to give a full description for those $G$ containing a long minimal relation. To tackle this issue, we first prove Lemma 29 and then apply Algorithm 37.

Although Algorithm 37 is powerful, it fails to give those positive-parameter solutions since their orders can be arbitrarily large. This is why we need Algorithm 28.

Now suppose $G=G_{0}+2G_{1}+\cdots+2G_{k}$ is of the form (11), and define $d$ and $n$ as in Algorithm 28. If $w(G_{i})$ is large for some $i$, then Algorithm 28 will be very inefficient. If $d=\infty$, then Algorithm 37 fails. If $nd\neq\infty$ is large, then Algorithm 37 will be very inefficient. This is why we need Algorithm 23 and 27.

4 Proof of Theorem 3

Recall that

[TABLE]

Now we will find the longest nontrivial projectively equivalent pairs in four steps. For $k=3$, $2$, $1$, and [math], we assume that all $Z_{i}$ are at least type-$k$ and determine how large $n$ can be, where an element of $\mathcal{S}^{+}$ is said to be type-$k$ if it is equivalent to some $k$-parameter solution.

The $k=3$ case, which can be easily done by hand, serves as a prototype for the following two algorithms.

Lemma 38**.**

If all $Z_{i}$ are type-three, then $n\leq 5$.

Proof.

We choose $Z_{3}$ and $Z_{4}$ from the rows of

and assume that the resulting pair $(0,s_{1},s_{2},s_{3},s_{4})\sim(0,t_{1},t_{2},t_{3},t_{4})$ is nontrivial. This is only possible when we take the $j$-th and $(j+3)$-th rows for some $1\leq j\leq 3$.

• •

If $j=1$, we have $x=-x$ and $x-z_{1}=z_{4}-x$, then $x=1/2$ and $z_{4}=-z_{1}$. The resulting pair is

[TABLE]

• •

If $j=2$, we have $y-z_{2}=z_{5}-y$ and $y=-y$, then $y=1/2$ and $z_{5}=-z_{2}$. The resulting pair is

[TABLE]

• •

If $j=3$, we have $z_{3}-y=y-z_{6}$ and $z_{3}-x=x-z_{6}$, then $y=1/2+x$ and $z_{6}=2x-z_{3}$. The resulting pair is

[TABLE]

Since $\{1,4\}$, $\{2,5\}$, and $\{3,6\}$ are mutually disjoint, we have $n\leq 5$. ∎

Algorithm 39**.**

Suppose all $Z_{i}$ are at least type-two, and at least one $Z_{i}$ is type-two. We will find the largest $n$ inductively. Let $[\widehat{s}_{1},\widehat{s}_{2},\widehat{s}_{3},\widehat{t}_{1},\widehat{t}_{2},\widehat{t}_{3}]$ be a general type-two solution (with two free variables). For any

[TABLE]

that we already have, we solve the equation

[TABLE]

• •

If the resulting $s_{1},s_{2},t_{1},t_{2}$ are constants, we add their lowest common denominator $d$ into a set $D$.

• •

If the resulting $s_{1},s_{2},t_{1},t_{2}$ have free variables, we extend the previous pair to

[TABLE]

When this process terminates, we get a set $D$ and some long projectively equivalent pairs with free variables. Finally, in order to get those long projectively equivalent pairs without free variables, we find all $Z_{i}$ such that $s_{1},s_{2},t_{1},t_{2}\in(1/d)\mathbb{Z}/\mathbb{Z}$ for some $d\in D$. The conclusion is: if all $Z_{i}$ are at least type-two, then $n\leq 10$.

Algorithm 40**.**

Suppose all $Z_{i}$ are at least type-one, and at least one $Z_{i}$ is type-one. Let $\mathcal{S}_{k}^{+}$ be the collection of general type-$k$ solutions (with $k$ free variables). For any

[TABLE]

we try to solve the equation (12). However, due to the large size of $\mathcal{S}_{1}^{+}$, it is better to solve (12) only partially. (We have used the same trick in Algorithm 28.)

• •

If we know that the resulting $s_{1},s_{2},t_{1},t_{2}\in(1/d)\mathbb{Z}/\mathbb{Z}$ are constants, we add $d$, which may not be their lowest common denominator, into a set $D_{1}$.

• •

If the resulting $[s_{1},s_{2},s_{3},t_{1},t_{2},t_{3}]$ has one free variable, we add it into a set $\mathcal{S}_{1,new}^{+}$.

When this process terminates, we get two sets $D_{1}$ and $\mathcal{S}_{1,new}^{+}$. For any

[TABLE]

we try to solve the equation (12) again. Since both of them have exactly one free variable, (12) becomes

[TABLE]

where $V_{0},\widehat{V}_{0}\in(\mathbb{Q}/\mathbb{Z})^{4}$ and $V_{1},\widehat{V}_{1}\in\mathbb{Z}^{4}$.

• •

If we know that the resulting $s_{1},s_{2},t_{1},t_{2}\in(1/d)\mathbb{Z}/\mathbb{Z}$ are constants, we add $d$, which may not be their lowest common denominator, into a set $D_{2}$.

• •

Long projectively equivalent pairs with free variables possibly exist only when $V_{1}$ and $\widehat{V}_{1}$ are linearly dependent. It turns out that the greatest common divisor of $V_{1}$ must be $1$ or $2$. Moreover,

[TABLE]

Therefore, for each element of $\mathcal{S}_{1}^{+}\cup\mathcal{S}_{1,new}^{+}$, we can substitute its free variable $x$ with some $c_{0}+c_{1}x$, where $c_{0}\in(1/12)\mathbb{Z}/\mathbb{Z}$ and $c_{1}\in\{\pm 1,\pm 2\}$.

When this process terminates, we get a set $D_{2}$ and some long projectively equivalent pairs with free variables. Finally, in order to get those long projectively equivalent pairs without free variables, we find all $Z_{i}$ such that $s_{1},s_{2},t_{1},t_{2}\in(1/d)\mathbb{Z}/\mathbb{Z}$ for some $d\in D_{1}\cup D_{2}$. The conclusion is: if all $Z_{i}$ are at least type-one, then $n\leq 14$. Moreover, if $n\geq 11$, then $s_{1},s_{2},t_{1},t_{2}\in(1/30)\mathbb{Z}/\mathbb{Z}$.

Therefore, if $n$ is the maximal length, then either $s_{1},s_{2},t_{1},t_{2}\in(1/30)\mathbb{Z}/\mathbb{Z}$ or at least four $Z_{i}$ are type-zero. It turns out that the latter condition implies $s_{1},s_{2},t_{1},t_{2}\in(1/84)\mathbb{Z}/\mathbb{Z}$ or $(1/120)\mathbb{Z}/\mathbb{Z}$. Once we find all $Z_{i}$ such that $s_{1},s_{2},t_{1},t_{2}\in(1/84)\mathbb{Z}/\mathbb{Z}$ or $(1/120)\mathbb{Z}/\mathbb{Z}$, Theorem 3 will be proved.

Acknowledgments. The author is deeply grateful to Fedor Bogomolov for suggesting this problem and indicating the crucial Theorem 15.