On the $\text{PGL}_{2}$-invariant quadruples of torsion points of elliptic curves

TL;DR

This paper classifies quadruples of torsion points on elliptic curves whose images under a standard double cover have a cross ratio independent of the curve, revealing special invariant configurations.

Contribution

It provides a complete classification of PGL₂-invariant quadruples of torsion points on elliptic curves, a problem previously not fully resolved.

Findings

Identifies all quadruples with invariant cross ratios across elliptic curves

Establishes conditions for PGL₂-invariance of torsion point configurations

Completes the classification of such invariant quadruples

Abstract

Let $E$ be an elliptic curve and $\pi:E\to\mathbb{P}^{1}$ a standard double cover identifying $\pm P\in E$. It is known that for some torsion points $P_{i}\in E$, $1\leq i\leq4$, the cross ratio of $\{\pi(P_{i})\}_{i=1}^{4}$ is independent of $E$. In this article, we will give a complete classification of such quadruples.

Equations

$$\mathcal{R}=(\{0\}\times [0,1/2]\cup (0,1/2)\times [0,1)\cup \{1/2\}\times [0,1/2])\cap {\mathbb{Q}}^2$$

$${\mathcal{R}}_4=\{\{(r_i,{\theta }_i){\}}_{i=1}^4:(r_i,{\theta }_i)\in \mathcal{R},(r_i,{\theta }_i)\ne (r_j,{\theta }_j)\text{ for }i\ne j\}.$$

$$({\mathbb{P}}^1{)}^4\stackrel{\text{cross ratio}}{\to }(\mathbb{Z}/2\mathbb{Z}{)}^2\backslash ({\mathbb{P}}^1{)}^4/{\text{PGL}}_2$$

$$z$$

$$s_k(q)$$

$$a_4(q)$$

$$a_6(q)$$

$$X(u,q)$$

$$Y(u,q)$$

$$E_q$$

$$\varphi :{\mathbb{C}}^{\ast }/q^{\mathbb{Z}}$$

$$u$$

$$F(r_{\sigma (1)})\le F(r_{\sigma (2)})<F(r_{\sigma (3)})\le F(r_{\sigma (4)})$$

$$s_1(q)=\sum _{n\ge 1}\frac{n{q}^n}{1-q^n}=\sum _{n\ge 1}n\sum _{m\ge 1}(q^n{)}^m=\sum _{n\ge 1}\sum _{m\ge 1}m(q^n{)}^m,$$

$$X(u,q)$$

$$\frac{(X(u_1,q)-X(u_2,q))(X(u_3,q)-X(u_4,q))}{(X(u_1,q)-X(u_4,q))(X(u_3,q)-X(u_2,q))}=c{q}^{r_3-r_2}+\text{higher degree terms of }q\text{ for some }c\ne 0.$$

$$\frac{X(u_3,q)-X(u_4,q)}{X(u_3,q)-X(u_2,q)}=c{q}^{r_3-r_2}+\text{higher degree terms of }q\text{ for some }c\ne 0.$$

$$X(u_i,q)$$

$$S=\{(0,{\theta }_1),(r,{\theta }_2),(r,{\theta }_3),(r_4,{\theta }_4)\}\in {\mathcal{R}}_4,$$

$$\{(0,{\theta }_1),(r,rc+{\theta }_2),(r,rc+{\theta }_3),(r_4,r_{4}c+{\theta }_4)\}.$$

$$A_1=\{c\in \mathbb{Z}:rc+{\theta }_2\equiv 0\text{ mod }1\},$$

$$A_2=\{c\in \mathbb{Z}:rc+{\theta }_3\equiv 0\text{ mod }1\},$$

$$A_3=\{c\in \mathbb{Z}:r_{4}c+{\theta }_4\equiv 0\text{ mod }1\},$$

$$A_4=\{c\in \mathbb{Z}:2rc+{\theta }_2+{\theta }_3\equiv 0\text{ mod }1\},$$

$$A_5=\{c\in \mathbb{Z}:(r_4+r)c+{\theta }_4+{\theta }_2\equiv 0\text{ mod }1\},$$

$$A_6=\{c\in \mathbb{Z}:(r_4+r)c+{\theta }_4+{\theta }_3\equiv 0\text{ mod }1\},$$

$$A_7=\{c\in \mathbb{Z}:(r_4-r)c+{\theta }_4-{\theta }_2\equiv 0\text{ mod }1\},$$

$$A_8=\{c\in \mathbb{Z}:(r_4-r)c+{\theta }_4-{\theta }_3\equiv 0\text{ mod }1\},$$

$$A_1\cup A_2\cup A_4\ne \mathbb{Z},$$

$$A_3\cup A_4\ne \mathbb{Z},$$

$$D_2\backslash D_1=\{(1/9,2/9),(1/9,4/9),(1/6,5/12),(1/5,2/5),(1/5,7/15),(2/9,4/9),(1/4,3/8),(1/3,5/12)\},$$

$$D_1\backslash D_2=\{(1/12,5/12),(1/10,1/2),(1/8,3/8),(1/6,1/2),(3/10,1/2)\},$$

$$C\backslash (D_1\cup D_2)=\{(1/6,1/3),(1/4,1/4),(1/4,1/2),(1/3,1/3),(1/3,1/2),(1/2,1/2)\}.$$

$$\{(r,\theta ),(r,\theta +1/2),(0,1/4),(1/2,1/4)\},$$

$$\{(r,\theta ),(r+1/2,\theta ),(1/4,0),(1/4,1/2)\},$$

$$\{(r,\theta ),(r+1/2,\theta +1/2),(1/4,1/4),(1/4,3/4)\},$$

$$E_{\delta }:y^2=x^4-({\delta }^2+{\delta }^{-2})x^2+1,$$

$${\pi }_{\delta }(P+(\pm {\delta }^{\pm 1},0))=\pm {\pi }_{\delta }(P{)}^{\pm 1}.$$

$${\pi }_{\delta }([2]P)=-\frac{{\delta }^2-2{\pi }_{\delta }(P{)}^2+{\delta }^2{\pi }_{\delta }(P{)}^4}{\delta (1-2{\delta }^2{\pi }_{\delta }(P{)}^2+{\pi }_{\delta }(P{)}^4)}.$$

$$E_{\lambda }:x^3+y^3+z^3=3\lambda xyz,$$

Full text

On the $\textup{PGL}_{2}$-invariant quadruples of torsion points

of elliptic curves

Fedor Bogomolov $\cdot$ Hang Fu

Abstract. Let $E$ be an elliptic curve and $\pi:E\to\mathbb{P}^{1}$ a standard double cover identifying $\pm P\in E$. It is known that for some torsion points $P_{i}\in E$, $1\leq i\leq 4$, the cross ratio of $\{\pi(P_{i})\}_{i=1}^{4}$ is independent of $E$. In this article, we will give a complete classification of such quadruples.

Keywords. Elliptic curves $\cdot$ Torsion points $\cdot$ $q$-series $\cdot$ Congruence subgroups $\cdot$ Modular curves

Mathematics Subject Classification. 14H52 $\cdot$ 11G05 $\cdot$ 11F03 $\cdot$ 20H05 $\cdot$ 40A20

1 Introduction and the Statement of Main Theorem

Let $K$ be an algebraically closed field, $E$ an elliptic curve defined over $K$ with the identity element $O$, $E[n]$ the $n$-torsion subgroup, $E^{*}[n]\subseteq E[n]$ the collection of torsion points of order $n$, $E[\infty]=\cup_{n=1}^{\infty}E[n]$ the collection of all torsion points, and $\pi:E\to\mathbb{P}^{1}(K)$ a standard double cover identifying $\pm P\in E$.

In this article, we will continue to study the image of $E[\infty]$ under $\pi$. See [1, 2, 3, 4] for the background and prior results. In particular, [3, Conjecture 2], the guiding problem of our project, predicts that if $K=\mathbb{C}$ and $\pi_{1}(E_{1}[2])\neq\pi_{2}(E_{2}[2])$, then $\left|\pi_{1}(E_{1}[\infty])\cap\pi_{2}(E_{2}[\infty])\right|$ is bounded by some universal constant. A partial result [5, Theorem 1.3] has recently been claimed under the assumption $\left|\pi_{1}(E_{1}[2])\cap\pi_{2}(E_{2}[2])\right|=3$. On the other hand, in [2, Theorem 1.1], we are able to find $E_{1}$ and $E_{2}$ such that $\left|\pi_{1}(E_{1}[\infty])\cap\pi_{2}(E_{2}[\infty])\right|\geq 22$. A crucial fact utilized in our construction is that, up to an automorphism of $\mathbb{P}^{1}(K)$, $\pi(E^{*}[4])$ is independent of $E$. The same statement is also correct if $\pi(E^{*}[4])$ is replaced with $\pi(E^{*}[3])$. These two interesting examples raise the question of when such phenomena may happen. In this article, we will establish a modified version of [3, Conjecture 3] as Theorem 1. We note that the original statement of [3, Conjecture 3] is inaccurate due to the infinitude of cases (1) and (2).

Let

[TABLE]

be a fundamental domain of $\pm\backslash\mathbb{Q}^{2}/\mathbb{Z}^{2}$. There is a natural left $\textup{SL}_{2}(\mathbb{Z})$-action on the fourth unordered configuration space of $\mathcal{R}$,

[TABLE]

For $\gamma\in\textup{SL}_{2}(\mathbb{Z})$ and $\{(r_{i},\theta_{i})\}_{i=1}^{4}\in\mathcal{R}_{4}$, we define $\gamma\cdot\{(r_{i},\theta_{i})\}_{i=1}^{4}=\{F((r_{i},\theta_{i})\cdot\gamma^{T})\}_{i=1}^{4}$, where $F(r,\theta)\in\mathcal{R}$ is the unique element such that $F(r,\theta)\equiv\pm(r,\theta)\textup{ mod }\mathbb{Z}^{2}$.

Theorem 1**.**

Let $S=\{(r_{i},\theta_{i})\}_{i=1}^{4}\in\mathcal{R}_{4}$, $n$ their common order, $\{w_{1},w_{2}\}$ a basis of $E[n]$. Write $E[r_{i},\theta_{i}]=[nr_{i}]w_{1}+[n\theta_{i}]w_{2}$. Then the image of $\{E[r_{i},\theta_{i}]\}_{i=1}^{4}$ under $\pi$ inside $\mathfrak{S}_{4}\backslash(\mathbb{P}^{1})^{4}/\textup{PGL}_{2}$ is a constant (independent of $K$, $E$, $\pi$, and $\{w_{1},w_{2}\}$) if and only if $S$ is $\textup{SL}_{2}(\mathbb{Z})$-equivalent to exactly one of the following:

1. (1)

$\{(0,a),(0,1/4),(0,1/2-a),(1/2,1/4)\}$, where $a\in\{0\}\cup\{1/(2r):r\in\mathbb{Z},r\geq 3\}$, 2. (2)

$\{(0,b),(1/4,0),(1/4,1/2),(1/2,b)\}$, where $b\in\{1/(2r):r\in\mathbb{Z},r\geq 2\}$, 3. (3)

$\{(0,1/3),(1/3,0),(1/3,1/3),(1/3,2/3)\}$, 4. (4)

$\{(0,1/4),(1/4,0),(1/4,1/4),(1/4,1/2)\}$, 5. (5)

$\{(0,1/4),(1/2,0),(1/2,1/4),(1/2,1/2)\}$, 6. (6)

$\{(0,1/6),(1/6,0),(1/6,1/6),(1/3,2/3)\}$, 7. (7)

$\{(0,1/6),(1/6,0),(1/6,1/3),(1/3,5/6)\}$, 8. (8)

$\{(0,1/6),(1/3,0),(1/3,1/6),(1/3,1/3)\}$, 9. (9)

$\{(0,1/6),(1/3,1/6),(1/3,1/2),(1/3,5/6)\}$, 10. (10)

$\{(0,1/8),(1/4,1/8),(1/4,3/8),(1/2,1/8)\}$, 11. (11)

$\{(0,1/12),(1/3,1/12),(1/3,11/12),(1/2,1/4)\}$.

Moreover, if we fix the quotient map

[TABLE]

then the constant is:

• •

$27/4$* for the cases (1), (2), and (5),*

• •

[math]* for the cases (3), (6), (9), and (11),*

• •

$1/2$* for the cases (4) and (10),*

• •

$8/3$* for the cases (7) and (8).*

We will prove Theorem 1 in three steps: (I) we obtain a necessary condition for $S$ to give a constant image; (II) we find all $S$ that satisfy the necessary condition found in (I); and (III) we prove that all $S$ found in (II) indeed give a constant image.

2 Proof of Theorem 1: Step (I)

Since we simply attempt to find a necessary condition in this section, we can temporarily assume that $K=\mathbb{C}$, so that the analytic uniformization can be applied.

Theorem 2**.**

[6, Page 410, Theorem 1.1]* For $u,q\in\mathbb{C}$ with $\left|q\right|<1$, define quantities*

[TABLE]

• •

$E_{q}$* is an elliptic curve, and $X$ and $Y$ define a complex analytic isomorphism*

[TABLE]

• •

For every elliptic curve $E/\mathbb{C}$ there is a $q\in\mathbb{C}^{*}$ with $\left|q\right|<1$ such that $E$ is isomorphic to $E_{q}$.

Definition 3**.**

• •

For any $r\in\mathbb{Q}$, define $F(r)\in[0,1/2]$ to be the unique element such that $F(r)\equiv\pm r\textup{ mod }\mathbb{Z}$. A set $\{r_{1},r_{2},r_{3},r_{4}\}\subseteq\mathbb{Q}$ is said to be “good” if

[TABLE]

for some permutation $\sigma\in\mathfrak{S}_{4}$.

• •

For any $(r,\theta)\in\mathbb{Q}^{2}$, define $F(r,\theta)\in\mathcal{R}$ to be the unique element such that $F(r,\theta)\equiv\pm(r,\theta)\textup{ mod }\mathbb{Z}^{2}$. A set $\{(r_{i},\theta_{i})\}_{i=1}^{4}\subseteq\mathbb{Q}^{2}$ is said to be “good” if there exist $a,b\in\mathbb{Z}$ such that $\gcd(a,b)=1$ and $\{ar_{i}+b\theta_{i}\}_{i=1}^{4}$ is “good”.

Lemma 4**.**

Let $u_{i}=\exp(2\pi\sqrt{-1}\theta_{i})q^{r_{i}}$, $1\leq i\leq 4$, be four distinct torsion points on $\pm\backslash\mathbb{C}^{*}/q^{\mathbb{Z}}$. If $\{r_{1},r_{2},r_{3},r_{4}\}$ is “good”, then the cross ratio of $\{X(u_{i},q)\}_{i=1}^{4}$ is not a constant function of $q$.

Proof.

Assume that $(r_{i},\theta_{i})\in\mathcal{R}$ and they are ordered lexicographically. First note that

[TABLE]

then

[TABLE]

In particular, we find that if $u_{i}\neq u_{j}$, then the lowest degree terms of $X(u_{i},q)$ and $X(u_{j},q)$ are different. If $u_{1}\neq 1$, then the cross ratio of $\{X(u_{i},q)\}_{i=1}^{4}$ is

[TABLE]

If $u_{1}=1$, then the cross ratio of $\{X(u_{i},q)\}_{i=1}^{4}$ is

[TABLE]

Since $r_{2}<r_{3}$ by the assumption, the cross ratio is nonconstant. ∎

Corollary 5**.**

Let $u_{i}=\exp(2\pi\sqrt{-1}\theta_{i})q^{r_{i}}$, $1\leq i\leq 4$, be four distinct torsion points on $\pm\backslash\mathbb{C}^{*}/q^{\mathbb{Z}}$. If $\{(r_{i},\theta_{i})\}_{i=1}^{4}$ is “good”, then the cross ratio of $\{X(u_{i},q)\}_{i=1}^{4}$ is not a constant function of $q$.

Proof.

By Definition 3, there exist $a,b\in\mathbb{Z}$ such that $\gcd(a,b)=1$ and $\{ar_{i}+b\theta_{i}\}_{i=1}^{4}$ is “good”. Since $\gcd(a,b)=1$, there exist $c,d\in\mathbb{Z}$ such that $\gamma=\begin{pmatrix}a&b\\ c&d\end{pmatrix}\in\textup{SL}_{2}(\mathbb{Z})$. Let $(r^{\prime}_{i},\theta^{\prime}_{i})=(r{}_{i},\theta{}_{i})\cdot\gamma^{T}$, $\tau=\dfrac{\log(q)}{2\pi\sqrt{-1}}$, $\tau^{\prime}=\gamma^{-T}\cdot\tau=\dfrac{d\tau-c}{-b\tau+a}$, $q^{\prime}=\exp\left(2\pi\sqrt{-1}\tau^{\prime}\right)$, and $u^{\prime}_{i}=\exp(2\pi\sqrt{-1}\theta^{\prime}_{i})q^{\prime r^{\prime}_{i}}$. Then by [6, Page 50, Theorem 6.2],

[TABLE]

Since $\{r^{\prime}_{1},r^{\prime}_{2},r^{\prime}_{3},r^{\prime}_{4}\}$ is “good” by the assumption, the cross ratio of $\{X(u^{\prime}_{i},q^{\prime})\}_{i=1}^{4}$ is nonconstant, and the cross ratio of $\{X(u_{i},q)\}_{i=1}^{4}$ is also nonconstant. ∎

3 Proof of Theorem 1: Step (II)

Lemma 6**.**

$S\in\mathcal{R}_{4}$* is not “good” if and only if $S$ is $\textup{SL}_{2}(\mathbb{Z})$-equivalent to exactly one of the quadruples listed in Theorem 1.*

Proof.

Suppose that $S$ is not “good”. Without loss of generality, we assume that $S$ is the minimal element within its $\textup{SL}_{2}(\mathbb{Z})$-orbit with respect to the lexicographical order. It must be of the form

[TABLE]

where $0\leq r\leq r_{4}\leq 1/2$ and $\theta_{2}\neq\theta_{3}$. By the minimality of $S$, if $\theta_{1}=0$, then $r=0$; if $\theta_{1}\neq 0$, then $1/\theta_{1}=\textup{ord}(\theta_{1})\geq\textup{ord}((r,\theta_{2})),\textup{ord}((r,\theta_{3})),\textup{ord}((r_{4},\theta_{4}))$. Let the lower triangular matrix $\begin{pmatrix}1&0\\ c&1\end{pmatrix}\in\textup{SL}_{2}(\mathbb{Z})$ act on $S$, we get

[TABLE]

Consider the following eight sets

[TABLE]

For any $1\leq i\leq 8$, $A_{i}$ is an empty set or an arithmetic progression. Whenever $A_{i}\neq\varnothing$, $d_{i}$ is the difference of its consecutive terms. We note that if $d_{1}$ is even, then $d_{1}=2d_{4}$; if $d_{1}$ is odd, then $d_{1}=d_{4}$.

We claim that $\cup_{i=1}^{8}A_{i}=\mathbb{Z}$ and $\sum_{i=1}^{8}1/d_{i}\geq 1$.

Take any $c\in\mathbb{Z}$, we know that $\{\theta_{1},rc+\theta_{2},rc+\theta_{3},r_{4}c+\theta_{4}\}$ is not “good”. If $c\notin\cup_{i=1}^{3}A_{i}$, then by the minimality of $S$, $\min\{F(\theta_{1}),F(rc+\theta_{2}),F(rc+\theta_{3}),F(r_{4}c+\theta_{4})\}=F(\theta_{1})$. Therefore, two of $F(rc+\theta_{2})$, $F(rc+\theta_{3})$, and $F(r_{4}c+\theta_{4})$ must be equal, which exactly means $c\in\cup_{i=4}^{8}A_{i}$. If $A_{i}\neq\varnothing$, then $A_{i}$ covers $1/d_{i}$ of $\mathbb{Z}$. Since $\cup_{i=1}^{8}A_{i}=\mathbb{Z}$, we have $\sum_{i=1}^{8}1/d_{i}\geq 1$.

Now we use this criterion to find all possible $S$ with the help of computer programs [8].

• •

Assume that $\min\{d_{i}\}_{i=1}^{8}\geq 4$.

Let $B_{1}=\{(r,r_{4})\in\mathbb{Q}^{2}:0\leq r\leq r_{4}\leq 1/2,\sum_{i=1}^{8}1/d_{i}\geq 1,\min\{d_{i}\}_{i=1}^{8}\geq 4\}$.

• $\bullet$

If $d_{5},d_{7}\geq 27$, then $\sum_{i=1}^{8}1/d_{i}\leq 1/4+3/5+4/27=539/540<1$.

• $\bullet$

Let $B_{11}=\{(r,r_{4})\in B_{1}:d_{5},d_{7}\leq 26\}$.

• $\bullet$

If $4\leq d_{5}\leq 26$, $d_{7}\geq 27$, and $d_{1}\geq 23$, then $\sum_{i=1}^{8}1/d_{i}\leq 3/4+4/23+2/27=2479/2484<1$.

• $\bullet$

Let $B_{12}=\{(r,r_{4})\in B_{1}:d_{1}\leq 22,d_{5}\leq 26\}$.

• $\bullet$

If $4\leq d_{7}\leq 26$, $d_{5}\geq 27$, and $d_{1}\geq 23$, then $\sum_{i=1}^{8}1/d_{i}\leq 3/4+4/23+2/27=2479/2484<1$.

• $\bullet$

Let $B_{13}=\{(r,r_{4})\in B_{1}:d_{1}\leq 22,d_{7}\leq 26\}$.

• •

Assume that $\min\{d_{i}\}_{i=1}^{8}=3$ and $A_{1}\cup A_{2}\cup A_{4}\neq\mathbb{Z}$.

Let $B_{2}=\{(r,r_{4})\in\mathbb{Q}^{2}:0\leq r\leq r_{4}\leq 1/2,\sum_{i=1}^{8}1/d_{i}\geq 1,\min\{d_{i}\}_{i=1}^{8}\geq 3\}$.

• $\bullet$

If $d_{5}=3$ and $d_{7}=n$, then $d_{1},d_{3},d_{4}\geq n/3$. If $n>42$, then $\sum_{i=1}^{8}1/d_{i}\leq 2/3+14/n<1$.

• $\bullet$

Let $B_{21}=\{(r,r_{4})\in B_{2}:d_{5}=3,d_{7}\leq 42\}$.

• $\bullet$

If $d_{7}=3$ and $d_{5}=n$, then $d_{1},d_{3},d_{4}\geq n/3$. If $n>42$, then $\sum_{i=1}^{8}1/d_{i}\leq 2/3+14/n<1$.

• $\bullet$

Let $B_{22}=\{(r,r_{4})\in B_{2}:d_{7}=3,d_{5}\leq 42\}$.

• $\bullet$

If $r_{4}=1/3$ and $d_{1}=n$, then $d_{5},d_{7}\geq n/3$. If $n>24$, then $\sum_{i=1}^{8}1/d_{i}\leq 1/3+16/n<1$.

• $\bullet$

Let $B_{23}=\{(r,r_{4})\in B_{2}:r_{4}=1/3,d_{1}\leq 24\}$.

• $\bullet$

If $r=1/6$ and $d_{3}=n$, then $d_{5},d_{7}\geq n/6$. If $n>75$, then $\sum_{i=1}^{8}1/d_{i}\leq 2/3+25/n<1$.

• $\bullet$

Let $B_{24}=\{(r,r_{4})\in B_{2}:r=1/6,d_{3}\leq 75\}$.

• $\bullet$

If $r=1/3$ and $d_{3}=n$, then $d_{5},d_{7}\geq n/3$. Since we assumed $A_{1}\cup A_{2}\cup A_{4}\neq\mathbb{Z}$, we only need $\sum_{i\neq 1}1/d_{i}\geq 1$. If $n>39$, then $\sum_{i\neq 1}1/d_{i}\leq 2/3+13/n<1$.

• $\bullet$

Let $B_{25}=\{(r,r_{4})\in B_{2}:r=1/3,d_{3}\leq 39,\sum_{i\neq 1}1/d_{i}\geq 1\}$.

• •

Assume that $\min\{d_{i}\}_{i=1}^{8}=2$, $A_{1}\cup A_{2}\cup A_{4}\neq\mathbb{Z}$, and $A_{5}\cup A_{6}\neq\mathbb{Z}$.

Let $B_{3}=\{(r,r_{4})\in\mathbb{Q}^{2}:0\leq r\leq r_{4}\leq 1/2,\sum_{i=1}^{8}1/d_{i}\geq 1,\min\{d_{i}\}_{i=1}^{8}\geq 2\}$.

• $\bullet$

If $d_{5}=2$ and $d_{7}=n$, then $d_{1},d_{3},d_{4}\geq n/2$. Since we assumed $A_{5}\cup A_{6}\neq\mathbb{Z}$, we only need $\sum_{i\neq 5}1/d_{i}\geq 1$. If $n>20$, then $\sum_{i\neq 5}1/d_{i}\leq 1/2+10/n<1$.

• $\bullet$

Let $B_{31}=\{(r,r_{4})\in B_{3}:d_{5}=2,d_{7}\leq 20,\sum_{i\neq 5}1/d_{i}\geq 1\}$.

• $\bullet$

If $r_{4}=1/2$ and $d_{1}=n$, then $d_{5},d_{7}\geq n/2$. If $n>24$, then $\sum_{i=1}^{8}1/d_{i}\leq 1/2+12/n<1$.

• $\bullet$

Let $B_{32}=\{(r,r_{4})\in B_{3}:r_{4}=1/2,d_{1}\leq 24\}$.

• $\bullet$

If $r=1/4$ and $d_{3}=n$, then $d_{5},d_{7}\geq n/4$. Since we assumed $A_{1}\cup A_{2}\cup A_{4}\neq\mathbb{Z}$, we only need $\sum_{i\neq 1}1/d_{i}\geq 1$. If $n>68$, then $\sum_{i\neq 1}1/d_{i}\leq 3/4+17/n<1$.

• $\bullet$

Let $B_{33}=\{(r,r_{4})\in B_{3}:r=1/4,d_{3}\leq 68,\sum_{i\neq 1}1/d_{i}\geq 1\}$.

• •

Assume that $\min\{d_{i}\}_{i=1}^{8}=1$ and $r\neq 0$.

• $\bullet$

Since we assumed $r\neq 0$, we have $r=r_{4}$ and $A_{7},A_{8}=\varnothing$. If $d_{1}$ is even, we need $\sum_{i=1}^{6}1/d_{i}=9/d_{1}\geq 1$; if $d_{1}$ is odd, we need $\sum_{i=1}^{6}1/d_{i}=6/d_{1}\geq 1$.

• $\bullet$

Let $B_{41}=\{(r,r_{4})\in\mathbb{Q}^{2}:0\leq r=r_{4}\leq 1/2,d_{1}=2,3,4,5,6,8\}$.

Let $B=B_{11}\cup B_{12}\cup B_{13}\cup B_{21}\cup B_{22}\cup B_{23}\cup B_{24}\cup B_{25}\cup B_{31}\cup B_{32}\cup B_{33}\cup B_{41}$, which is a finite set. If $S$ is not “good”, then we have several possibilities: $(r,r_{4})\in B$, $A_{1}\cup A_{2}\cup A_{4}=\mathbb{Z}$, $A_{5}\cup A_{6}=\mathbb{Z}$, or $r=0$. In the following, whenever we consider $(r,r_{4})\in B$, we assume that $A_{1}\cup A_{2}\cup A_{4}\neq\mathbb{Z}$ and $A_{5}\cup A_{6}\neq\mathbb{Z}$.

We claim that

• •

if $\gcd(d_{1},d_{3})=1$, then ($A_{1}=\varnothing$ or $A_{3}=\varnothing$) and ($A_{2}=\varnothing$ or $A_{3}=\varnothing$);

• •

if $\gcd(d_{1},d_{5})=1$, then ($A_{1}=\varnothing$ or $A_{5}=\varnothing$) and ($A_{2}=\varnothing$ or $A_{6}=\varnothing$);

• •

if $\gcd(d_{1},d_{7})=1$, then ($A_{1}=\varnothing$ or $A_{7}=\varnothing$) and ($A_{2}=\varnothing$ or $A_{8}=\varnothing$);

• •

if $\gcd(d_{3},d_{5})=1$, then ($A_{3}=\varnothing$ or $A_{5}=\varnothing$) and ($A_{3}=\varnothing$ or $A_{6}=\varnothing$);

• •

if $\gcd(d_{3},d_{7})=1$, then ($A_{3}=\varnothing$ or $A_{7}=\varnothing$) and ($A_{3}=\varnothing$ or $A_{8}=\varnothing$);

• •

if $\gcd(d_{5},d_{7})=1$, then ($A_{5}=\varnothing$ or $A_{7}=\varnothing$) and ($A_{6}=\varnothing$ or $A_{8}=\varnothing$).

Let us prove the first one, others can be proved similarly. If $\gcd(d_{1},d_{3})=1$ and $A_{1},A_{3}\neq\varnothing$, then $A_{1}\cap A_{3}\neq\varnothing$, i.e., there exists $c\in\mathbb{Z}$ such that $F(rc+\theta_{2})=F(r_{4}c+\theta_{4})=0$. Since $\{\theta_{1},rc+\theta_{2},rc+\theta_{3},r_{4}c+\theta_{4}\}$ is not “good”, if $\theta_{1}\neq 0$, then $F(rc+\theta_{3})=0$, which implies $\theta_{2}=\theta_{3}$, contradiction; if $\theta_{1}=0$, then $r=0$, which implies $(0,\theta_{1})=(r,\theta_{2})=(0,0)$, contradiction.

Based on this observation, we define

• •

$C_{1}=\{(r,r_{4})\in B:\gcd(d_{1},d_{3})=1,\sum_{i\neq 3}1/d_{i}<1,\sum_{i\neq 1,2}1/d_{i}<1\}$;

• •

$C_{2}=\{(r,r_{4})\in B:\gcd(d_{1},d_{5})=1,\sum_{i\neq 1,2}1/d_{i}<1,\sum_{i\neq 5,6}1/d_{i}<1,\sum_{i\neq 1,5}1/d_{i}<1\}$;

• •

$C_{3}=\{(r,r_{4})\in B:\gcd(d_{1},d_{7})=1,\sum_{i\neq 1,2}1/d_{i}<1,\sum_{i\neq 7,8}1/d_{i}<1,\sum_{i\neq 1,7}1/d_{i}<1\}$;

• •

$C_{4}=\{(r,r_{4})\in B:\gcd(d_{3},d_{5})=1,\sum_{i\neq 3}1/d_{i}<1,\sum_{i\neq 5,6}1/d_{i}<1\}$;

• •

$C_{5}=\{(r,r_{4})\in B:\gcd(d_{3},d_{7})=1,\sum_{i\neq 3}1/d_{i}<1,\sum_{i\neq 7,8}1/d_{i}<1\}$;

• •

$C_{6}=\{(r,r_{4})\in B:\gcd(d_{5},d_{7})=1,\sum_{i\neq 5,6}1/d_{i}<1,\sum_{i\neq 7,8}1/d_{i}<1,\sum_{i\neq 5,7}1/d_{i}<1\}$.

Let $C=B\backslash(C_{1}\cup C_{2}\cup C_{3}\cup C_{4}\cup C_{5}\cup C_{6})$. If $S$ is not “good”, then we have several possibilities: $(r,r_{4})\in C$, $A_{1}\cup A_{2}\cup A_{4}=\mathbb{Z}$, $A_{5}\cup A_{6}=\mathbb{Z}$, or $r=0$.

Assume that $(r,r_{4})\in C$ and $r\neq r_{4}$. If $\cup_{i=4}^{8}A_{i}\neq\mathbb{Z}$, then $\cup_{i=1}^{3}A_{i}\neq\varnothing$. Since we assumed $\min\{d_{i}\}_{i=1}^{8}\geq 2$ and $A_{5}\cup A_{6}\neq\mathbb{Z}$, any two of $\{A_{i}\}_{i=4}^{8}$ cannot cover $\mathbb{Z}$. If at least three of $\{A_{i}\}_{i=4}^{8}$ cover $\mathbb{Z}$, then $\theta_{2},\theta_{3},\theta_{4}\in(1/(2n))\mathbb{Z}/\mathbb{Z}$, where $n=\textup{lcm}(d_{4},d_{5},d_{7})$. If $d_{1}=2n$ or $d_{3}=2n$, then $\cup_{i=1}^{3}A_{i}\neq\varnothing$.

Assume that $(r,r_{4})\in C$ and $r=r_{4}$. If $d_{1}=5,8$, then $\cup_{i=4}^{6}A_{i}\neq\mathbb{Z}$, so $\cup_{i=1}^{3}A_{i}\neq\varnothing$. If $d_{1}=6$, then any two of $\{A_{i}\}_{i=4}^{6}$ cannot cover $\mathbb{Z}$ and $d_{1}=2d_{4}$, so $\cup_{i=1}^{3}A_{i}\neq\varnothing$.

• •

Let $D_{1}=\{(r,r_{4})\in C:r\neq r_{4},\sum_{i=4}^{8}1/d_{i}<1$ or $(d_{5}=2,\sum_{i=4,6,7,8}1/d_{i}<1)$ or $d_{1}=2\textup{lcm}(d_{4},d_{5},d_{7})$ or $d_{3}=2\textup{lcm}(d_{4},d_{5},d_{7})\}\cup\{(r,r_{4})\in C:r=r_{4},d_{1}=5,6,8\}$.

Assume that $(r,r_{4})\in C$ and $\cup_{i=1}^{3}A_{i}\neq\varnothing$. If

[TABLE]

then $\theta_{2},\theta_{3},\theta_{4}\in(1/n)\mathbb{Z}/\mathbb{Z}$, where $n=\textup{lcm}(d_{1},d_{3})$. Since $\{\theta_{1},rc+\theta_{2},rc+\theta_{3},r_{4}c+\theta_{4}\}$ is not “good”, if, for example, $F(rc+\theta_{2})=0$, then by the minimality of $S$, $F(\theta_{1})=F(rc+\theta_{3})$ or $F(\theta_{1})=F(r_{4}c+\theta_{4})$, which implies $\theta_{1}\in(1/n)\mathbb{Z}/\mathbb{Z}$ as well.

• •

Let $D_{2}=\{(r,r_{4})\in C:\sum_{i=1,3,5,7}1/d_{i}<1\textup{ or }(r=r_{4},\sum_{i=1,3,5}1/d_{i}<1),\sum_{i=3,4}1/d_{i}<1\}$.

We want to check that for any $(r,r_{4})\in D_{1}\cap D_{2}$, any $\theta_{1},\theta_{2},\theta_{3},\theta_{4}\in(1/n)\mathbb{Z}/\mathbb{Z}$, and any $a,b\in\mathbb{Z}$ such that $\gcd(a,b,n)=1$, where $n=\textup{lcm}(d_{1},d_{3})$, whether $\{b\theta_{1},ar+b\theta_{2},ar+b\theta_{3},ar_{4}+b\theta_{4}\}$ is “good” or not. This is a finite calculation, so we can use computer programs to do that. The answer is: if $S$ is not “good”, then $(r,r_{4})\notin D_{1}\cap D_{2}$.

By the computer programs,

[TABLE]

If $S$ is not “good”, then we have several possibilities: $(r,r_{4})\in C\backslash(D_{1}\cap D_{2})$, $A_{1}\cup A_{2}\cup A_{4}=\mathbb{Z}$, $A_{5}\cup A_{6}=\mathbb{Z}$, or $r=0$.

Assume that $(r,r_{4})\notin D_{1}$, $r\neq r_{4}$, and $\cup_{i=1}^{3}A_{i}=\varnothing$. Then $\cup_{i=4}^{8}A_{i}=\mathbb{Z}$ and $\theta_{2},\theta_{3},\theta_{4}\in(1/(2n))\mathbb{Z}/\mathbb{Z}$, where $n=\textup{lcm}(d_{4},d_{5},d_{7})$. Also, for any $c\in\mathbb{Z}$, $\min\{F(\theta_{1}),F(rc+\theta_{2}),F(rc+\theta_{3}),F(r_{4}c+\theta_{4})\}=F(\theta_{1})$, so the smallest two of $F(rc+\theta_{2})$, $F(rc+\theta_{3})$, and $F(r_{4}c+\theta_{4})$ must be equal. We can use computer programs to find all $(\theta_{2},\theta_{3},\theta_{4})$ that satisfy these conditions. The answer is:

• •

If $(r,r_{4})=(1/5,2/5)$, then $(\theta_{2},\theta_{3},\theta_{4})=(1/10,3/10,9/10)$. Let $(a,b)=(-1,2)$, then $\{2\theta_{1},0,2/5,2/5\}$ is not “good”, which implies $2/5\leq 2\theta_{1}\leq 3/5$, contradiction.

• •

If $(r,r_{4})=(1/4,1/2)$, then $(\theta_{2},\theta_{3},\theta_{4})=(1/8,3/8,1/8)$. Let $(a,b)=(1,2)$, then $\{2\theta_{1},1/2,0,1/4\}$ is not “good”, which implies $\theta_{1}=1/8$. This is the case (10) of Theorem 1. It is easy to check that it is indeed not “good” and minimal.

• •

If $(r,r_{4})=(1/3,1/2)$, then $(\theta_{2},\theta_{3},\theta_{4})=(1/12,11/12,1/4)$. Let $(a,b)=(1,4)$, then $\{4\theta_{1},1/3,0,1/2\}$ is not “good”, which implies $\theta_{1}=1/12$. This is the case (11) of Theorem 1. It is easy to check that it is indeed not “good” and minimal.

• •

Otherwise, such $(\theta_{2},\theta_{3},\theta_{4})$ does not exist.

Assume that $(r,r_{4})\notin D_{2}$, $r\neq r_{4}$, and $\cup_{i=1}^{3}A_{i}\neq\varnothing$. We note that $A_{1}\cup A_{2}\cup A_{4}\neq\mathbb{Z}$, $A_{2}\cup A_{5}\cup A_{7}\neq\mathbb{Z}$, and $A_{1}\cup A_{6}\cup A_{8}\neq\mathbb{Z}$ are automatic for each case. If $A_{3}\cup A_{4}=\mathbb{Z}$, then $(r,r_{4})=(1/4,1/2)$ and $(\theta_{2},\theta_{3},\theta_{4})\equiv(\theta_{2},1/2-\theta_{2},0)\textup{ mod }1$, which implies $\theta_{1}\equiv\pm\theta_{2}\textup{ mod }1$ or $\theta_{1}\equiv 1/2\pm\theta_{2}\textup{ mod }1$. Without loss of generality, we can assume that $\theta_{1}=\theta_{2}$. Let $(a,b)=(1,2)$, then $\{2\theta_{1},1/4+2\theta_{1},1/4-2\theta_{1},1/2\}$ is not “good”, which implies $4\theta_{1}\equiv 0,1/4,3/4\textup{ mod }1$. By the minimality of $S$, $\theta_{1}=1/16$. However, when $\theta_{1}=1/16$, $\{1/8,3/8,1/8,1/2\}$ is “good”, so $A_{3}\cup A_{4}=\mathbb{Z}$ cannot happen. If, for example, $A_{1}\cup A_{3}\cup A_{5}\cup A_{7}=\mathbb{Z}$, then $\theta_{2},\theta_{4}\in(1/n)\mathbb{Z}/\mathbb{Z}$, where $n=\textup{lcm}(d_{1},d_{3})$. We can use computer programs to find all $(\theta_{2},\theta_{4})$ that satisfy this condition. The answer is:

• •

If $(r,r_{4})=(1/8,3/8)$, then $(\theta_{2},\theta_{4})=(0,1/2)$. Let $(a,b)=(0,1)$, then $\{\theta_{1},0,\theta_{3},1/2\}$ is not “good”, which implies $\theta_{1}\equiv\pm\theta_{3}\textup{ mod }1$. Let $(a,b)=(4,1)$, then $\{\theta_{1},1/2,1/2+\theta_{3},0\}$ is not “good”, which implies $\theta_{1}\equiv 1/2\pm\theta_{3}\textup{ mod }1$. If, for example, $\theta_{1}\equiv\theta_{3}\equiv 1/2-\theta_{3}\textup{ mod }1$, then $\theta_{1}=1/4$, contradiction.

• •

If $(r,r_{4})=(1/4,1/2)$, then $(\theta_{2},\theta_{4})=(0,1/2)$. Let $(a,b)=(2,1)$, then $\{\theta_{1},1/2,1/2+\theta_{3},1/2\}$ is not “good”, which implies $\theta_{1}=1/2$ or $\theta_{3}=0$, contradiction.

• •

Otherwise, such $(\theta_{2},\theta_{4})$ does not exist.

As the case $(r,r_{4})\in D_{1}\cap D_{2}$ before, we can use computer programs to check that for any $(r,r_{4})\notin D_{1}\cap D_{2}$ with $r\neq r_{4}$, any $\theta_{1},\theta_{2},\theta_{3},\theta_{4}\in(1/n)\mathbb{Z}/\mathbb{Z}$, and any $a,b\in\mathbb{Z}$ such that $\gcd(a,b,n)=1$, where $n=\textup{lcm}(d_{1},d_{3})$, whether $\{b\theta_{1},ar+b\theta_{2},ar+b\theta_{3},ar_{4}+b\theta_{4}\}$ is “good” or not. The answer is: if $S$ is not “good”, then $S$ is the case (6) or case (7) of Theorem 1.

Assume that $r=r_{4}=1/4$ or $1/3$. Without loss of generality, we also assume that $A_{1}\cup A_{3}\cup A_{5}\neq\mathbb{Z}$, $A_{2}\cup A_{3}\cup A_{6}\neq\mathbb{Z}$, $A_{4}\cup A_{5}\neq\mathbb{Z}$, and $A_{4}\cup A_{6}\neq\mathbb{Z}$. If $\cup_{i=1}^{3}A_{i}=\varnothing$, we must have $r=r_{4}=1/3$ and $\{\theta_{2}+\theta_{3},\theta_{2}+\theta_{4},\theta_{3}+\theta_{4}\}\equiv\{0,1/3,2/3\}\textup{ mod }1$, which implies $(\theta_{2},\theta_{3},\theta_{4})=(1/6,1/2,5/6)$. Let $(a,b)=(1,2)$, then $\{2\theta_{1},1/3,1/3,0\}$ is not “good”, which implies $\theta_{1}=1/6$. This is the case (9) of Theorem 1. It is easy to check that it is indeed not “good” and minimal. We note that $A_{3}\cup A_{4}\neq\mathbb{Z}$, $A_{2}\cup A_{5}\neq\mathbb{Z}$, and $A_{1}\cup A_{6}\neq\mathbb{Z}$ are automatic for each case, so if $\cup_{i=1}^{3}A_{i}\neq\varnothing$, then $\theta_{1},\theta_{2},\theta_{3},\theta_{4}\in(1/d_{1})\mathbb{Z}/\mathbb{Z}$, which implies $S$ is the case (3) or case (4) of Theorem 1. It is easy to check that both of them are indeed not “good” and minimal.

Assume that $r=r_{4}=1/2$ and $0\leq\theta_{2}<\theta_{3}<\theta_{4}\leq 1/2$. Let $(a,b)=(0,1)$, then $\{\theta_{1},\theta_{2},\theta_{3},\theta_{4}\}$ is not “good”, which implies $\theta_{1}=\theta_{3}$. Let $(a,b)=(1,1)$, then $\{\theta_{1},1/2+\theta_{2},1/2+\theta_{1},1/2+\theta_{4}\}$ is not “good”, which implies $\theta_{1}=1/4$. By the minimality of $S$, $\textup{ord}((1/2,\theta_{2})),\textup{ord}((1/2,\theta_{4}))\leq 4$, so $\theta_{2}=0$ and $\theta_{4}=1/2$. This is the case (5) of Theorem 1. It is easy to check that it is indeed not “good” and minimal.

Assume that $A_{1}\cup A_{2}\cup A_{4}=\mathbb{Z}$ and $r=1/4$. Then $(\theta_{2},\theta_{3})=(0,1/2)$. Let $(a,b)=(2c,1)$, then $\{\theta_{1},0,1/2,2cr_{4}+\theta_{4}\}$ is not “good”, which implies $\theta_{1}\equiv\pm(2cr_{4}+\theta_{4})\textup{ mod }1$ for any $c\in\mathbb{Z}$. Now $\theta_{1}\equiv 2cr_{4}+\theta_{4}\equiv 2c^{\prime}r_{4}+\theta_{4}\textup{ mod }1$ or $\theta_{1}\equiv-2cr_{4}-\theta_{4}\equiv-2c^{\prime}r_{4}-\theta_{4}\textup{ mod }1$ for some $0\leq c<c^{\prime}\leq 2$, which implies $r_{4}=1/4$ or $1/2$. If $r_{4}=1/4$, then $\theta_{4}\equiv-2r_{4}-\theta_{4}\textup{ mod }1$, which implies $\theta_{4}=1/4$ or $3/4$, and therefore $\theta_{1}=1/4$. These two cases are clearly equivalent and have been found previously. If $r_{4}=1/2$, then $\theta_{1}=\theta_{4}$. By the minimality of $S$, $\theta_{1}\in\{1/(2R):R\in\mathbb{Z},R\geq 2\}$. This is the case (2) of Theorem 1. Any element within its $\textup{SL}_{2}(\mathbb{Z})$-orbit must be of the form

[TABLE]

so it is indeed not “good”. If $S$ is not minimal, then the minimal element must be $\{(0,\theta_{1}),(0,1/4),(0,1/2-\theta_{1}),(1/2,1/4)\}$. Therefore, there exist $a,b\in\mathbb{Z}$ such that $\{a/4,a/4+b/2\}\equiv\{0,1/2\}\textup{ mod }1$, which implies $b$ is odd, and $\{b\theta_{1},a/2+b\theta_{1}\}\equiv\{0,0\}\textup{ mod }1$, which implies $b$ is even, contradiction.

Assume that $A_{1}\cup A_{2}\cup A_{4}=\mathbb{Z}$ and $r=1/3$. Then $(\theta_{2},\theta_{3})=(0,1/3)$ and $\theta_{1}\leq 1/3$. If $\theta_{1}=1/3$, then by the minimality of $S$, $(r_{4},\theta_{4})=(1/3,2/3)$, $(1/2,0)$, or $(1/2,1/2)$. The first case has been found previously. Let $(a,b)=(0,1)$ for the second and $(a,b)=(-1,1)$ for the third, contradiction. Now we assume that $\theta_{1}<1/3$. Let $(a,b)=(3c,1)$, then $\{\theta_{1},0,1/3,3cr_{4}+\theta_{4}\}$ is not “good”, which implies $\theta_{1}\equiv\pm(3cr_{4}+\theta_{4})\textup{ mod }1$ for any $c\in\mathbb{Z}$. Now $\theta_{1}\equiv 3cr_{4}+\theta_{4}\equiv 3c^{\prime}r_{4}+\theta_{4}\textup{ mod }1$ or $\theta_{1}\equiv-3cr_{4}-\theta_{4}\equiv-3c^{\prime}r_{4}-\theta_{4}\textup{ mod }1$ for some $0\leq c<c^{\prime}\leq 2$, which implies $r_{4}=1/3$ or $1/2$. If $r_{4}=1/2$, then $\theta_{4}\equiv-3r_{4}-\theta_{4}\textup{ mod }1$, which implies $\theta_{4}=1/4$ and also $\theta_{1}=1/4$. Let $(a,b)=(1,2)$, contradiction. If $r_{4}=1/3$, then $\theta_{1}\equiv\pm\theta_{4}\textup{ mod }1$. Let $(a,b)=(-1,1)$, then $\{\theta_{1},1/3,0,\theta_{4}-1/3\}$ is not “good”, which implies $\theta_{1}\equiv\pm(\theta_{4}-1/3)\textup{ mod }1$. If $\theta_{1}\equiv-\theta_{4}\equiv\theta_{4}-1/3\textup{ mod }1$, then $(\theta_{1},\theta_{4})=(1/3,2/3)$, which has been found previously. If $\theta_{1}\equiv\theta_{4}\equiv-\theta_{4}+1/3\textup{ mod }1$, then $(\theta_{1},\theta_{4})=(1/6,1/6)$. This is the case (8) of Theorem 1. It is easy to check that it is indeed not “good” and minimal.

Assume that $A_{5}\cup A_{6}=\mathbb{Z}$, and $r\neq 0$ or $1/2$. Then $r+r_{4}=1/2$ and $(\theta_{3},\theta_{4})\equiv(\theta_{2}+1/2,-\theta_{2})\textup{ mod }1$. Let $(a,b)=(2c,1)$, then $\{\theta_{1},2cr+\theta_{2},2cr+\theta_{2}+1/2,2cr+\theta_{2}\}$ is not “good”. If $2cr+\theta_{2}\equiv 0$ or $1/2\textup{ mod }1$, then $\theta_{1}=0$ or $1/2$, contradiction. By the minimality of $S$, $\min\{F(\theta_{1}),F(2cr+\theta_{2}),F(2cr+\theta_{2}+1/2),F(2cr+\theta_{2})\}=F(\theta_{1})$, which implies $0<F(2cr+\theta_{2})\leq 1/4$ for any $c\in\mathbb{Z}$. Since $0<2r\leq 1/2$, we must have $F(\theta_{2})=1/4$ and also $r=r_{4}=1/4$, then $(r,\theta_{3})=(r_{4},\theta_{4})$, contradiction.

Assume that $r=0$ and $0\leq\theta_{1}<\theta_{2}<\theta_{3}\leq 1/2$. Clearly, $r_{4}\neq 0$. Let $(a,b)=(c,1)$, then $\{\theta_{1},\theta_{2},\theta_{3},cr_{4}+\theta_{4}\}$ is not “good”, which implies $\theta_{2}\equiv\pm(cr_{4}+\theta_{4})\textup{ mod }1$ for any $c\in\mathbb{Z}$. Now $\theta_{2}\equiv cr_{4}+\theta_{4}\equiv c^{\prime}r_{4}+\theta_{4}\textup{ mod }1$ or $\theta_{2}\equiv-cr_{4}-\theta_{4}\equiv-c^{\prime}r_{4}-\theta_{4}\textup{ mod }1$ for some $0\leq c<c^{\prime}\leq 2$. Since $r_{4}\neq 0$, we have $c=0$, $c^{\prime}=2$, and $r_{4}=1/2$. Since $\theta_{4}\equiv-r_{4}-\theta_{4}\textup{ mod }1$, we have $\theta_{4}=1/4$ and also $\theta_{2}=1/4$. Let $(a,b)=(1,2)$, then $\{2\theta_{1},1/2,2\theta_{3},0\}$ is not “good”, which implies $\theta_{1}+\theta_{3}=1/2$. By the minimality of $S$, $\theta_{1}\in\{0\}\cup\{1/(2R):R\in\mathbb{Z},R\geq 3\}$. This is the case (1) of Theorem 1. As the case $A_{1}\cup A_{2}\cup A_{4}=\mathbb{Z}$ and $r=1/4$ before, it is indeed not “good” and minimal. ∎

4 Proof of Theorem 1: Step (III)

Theorem 7**.**

Let $\{(r_{i},\theta_{i})\}_{i=1}^{4}\in\mathcal{R}_{4}$ be one of the quadruples listed in Theorem 1, $n$ their common order, $\{w_{1},w_{2}\}$ a basis of $E[n]$. Write $E[r_{i},\theta_{i}]=[nr_{i}]w_{1}+[n\theta_{i}]w_{2}$. Then whenever $\{E[r_{i},\theta_{i}]\}_{i=1}^{4}$ is well-defined, i.e.,

• •

$\textup{char}(K)\neq 2$* for the cases (1), (2), and (5),*

• •

$\textup{char}(K)\nmid n$* for the cases (3), (4), (6), (7), (10), and (11),*

• •

$\textup{char}(K)\neq 3$* and there exists a point of order $2$ (i.e., $E$ is ordinary if $\textup{char}(K)=2$) for the cases (8) and (9),*

the image of it under $\pi$ inside $\mathfrak{S}_{4}\backslash(\mathbb{P}^{1})^{4}/\textup{PGL}_{2}$ is a constant.

Proof.

Let us first collect some basic facts that we will need for the Jacobian form and Hessian form.

If $\textup{char}(K)\neq 2$, then any $E/K$ can be transformed to the Jacobian form

[TABLE]

where $\delta^{4}\neq 0,1$. We take the origin to be $O_{\delta}=(\delta,0)$ and $\pi_{\delta}:E_{\delta}\rightarrow\mathbb{P}^{1},(x,y)\mapsto x$. Its $2$-torsion points are $(\pm\delta^{\pm 1},0)$, which induce

[TABLE]

The fixed points of those nontrivial ones $\{0,\infty\}$, $\{\pm 1\},$ and $\{\pm i\}$ constitute $\pi_{\delta}(E_{\delta}^{*}[4])$, where $i$ is a primitive fourth root of $K$. Moreover,

[TABLE]

If $\textup{char}(K)\neq 3$, then any $E/K$ can be transformed to the Hessian form

[TABLE]

where $\lambda^{3}\neq 1$. We take the origin to be $O_{\lambda}=(1:-1:0)$ and $\pi_{\lambda}:E_{\lambda}\rightarrow\mathbb{P}^{1},(x:y:z)\mapsto-(x+y)/z$. Its $3$-torsion points are

[TABLE]

where $\omega$ is a primitive cube root of $K$. The addition formula is

[TABLE]

The doubling formula is

[TABLE]

We use the Jacobian form for the cases (1), (2), and (5).

Let $Q\in E_{\delta}^{*}[2]$, $Q_{1},Q_{2}\in E_{\delta}^{*}[4]$ such that $[2]Q_{1}=[2]Q_{2}=Q$ and $Q_{1}+Q_{2}\neq O_{\delta}$, and $P\in E_{\delta}[\infty]$ such that $P\neq Q_{1},Q_{2}$. Then $\pi_{\delta}$ maps $(P,P+Q,Q_{1},Q_{2})$ to $(\pi_{\delta}(P),-\pi_{\delta}(P),0,\infty)$, whose cross ratio is a constant.

We use the Jacobian form for the cases (4) and (10).

The case (4) is done. Now we deal with the case (10). Fix a basis $\{w_{1},w_{2}\}$ of $E_{\delta}[8]$ such that $[4]w_{2}=(-\delta,0)$ and $[4]w_{1}=(\delta^{-1},0)$. Let $a=\pi_{\delta}(w_{2})$ and $b=\pi_{\delta}([2]w_{1}+w_{2})$, then $\pi_{\delta}([2]w_{1}+[3]w_{2})=-b^{-1}$ and $\pi_{\delta}([4]w_{1}+w_{2})=a^{-1}$. The cross ratio of $(a,b,-b^{-1},a^{-1})$ is

[TABLE]

Let $\pi_{\delta}([2]w_{2})=0$ and $\pi_{\delta}([4]w_{1}+[2]w_{2})=\infty$, then we know that

[TABLE]

If, for example, taking “$+$” for both, then the cross ratio

[TABLE]

is a constant.

We use the Hessian form for the cases (3), (8), and (9).

The case (3) is done. Now we deal with the cases (8) and (9). Let $Q=(a:a:1)\in E_{\lambda}^{*}[2]$, $P_{0}=(1:-\omega:0)$, $P_{1}=(-1:0:1)$, $P_{2}=(-\omega:0:1)$, and $P_{3}=(-\omega^{2}:0:1)$. Then $\pi_{\lambda}$ maps $(P_{0},P_{1},P_{0}+Q,P_{1}+Q,P_{2}+Q,P_{3}+Q)$ to $(\infty,1,a,-1-a^{-1},-\omega-\omega^{2}a^{-1},-\omega^{2}-\omega a^{-1})$. The cross ratio of $(\infty,1,-\omega-\omega^{2}a^{-1},-\omega^{2}-\omega a^{-1})$ is

[TABLE]

The cross ratio of $(a,-1-a^{-1},-\omega-\omega^{2}a^{-1},-\omega^{2}-\omega a^{-1})$ is

[TABLE]

Both of them are constants.

We use the Jacobian form for the cases (6) and (7).

Let $P_{0},P_{1},P_{2},P_{3}\in E_{\delta}^{*}[3]$ such that $P_{i}+P_{j}\neq O_{\delta}$, and $a,b,c,d$ their images under $\pi_{\delta}$. We have already known that the cross ratio of $(a,b,-c,-d)$ is a constant, so the cross ratio of $(a^{-1},b^{-1},-c^{-1},-d^{-1})$ is the same constant. The cross ratio of $(a,-b,c^{-1},-d^{-1})$ is

[TABLE]

We know that (see, for example, [1, Proof of Corollary 3.6 (C)])

[TABLE]

which implies

[TABLE]

so $abcd=-1$ and $\dfrac{ac+bd}{ad+bc}=\omega$. Therefore,

[TABLE]

is a constant.

We use the Hessian form for the case (11).

Let $P_{1}=(-1:0:1)$, $P_{2}=(-\omega:0:1)$, $P_{3}=(-\omega^{2}:0:1)$, $\{v_{1},v_{2}\}$ a basis of $E_{\lambda}[4]$, $v_{2}=(a_{1}:b_{1}:1)$, $[2]v_{1}+v_{2}=(a_{2}:b_{2}:1)$, and $[2]v_{2}=(c:c:1)$. Then $\pi_{\lambda}$ maps $(P_{1}+v_{2},P_{2}+v_{2},P_{3}+v_{2},[2]v_{1}+v_{2})$ to $(-a_{1}^{-1}(b_{1}+1),-a_{1}^{-1}(\omega b_{1}+\omega^{2}),-a_{1}^{-1}(\omega^{2}b_{1}+\omega),-(a_{2}+b_{2}))$, whose cross ratio is

[TABLE]

if we have $a_{1}b_{1}(a_{2}+b_{2})=-1$. Now let us check this is true. From the doubling formula, we know that $(a_{1},b_{1})$, $(b_{1},a_{1})$, $(a_{2},b_{2})$, and $(b_{2},a_{2})$ are the solutions of $\left(\dfrac{y(x^{3}-1)}{y^{3}-x^{3}},\dfrac{x(1-y^{3})}{y^{3}-x^{3}}\right)=(c,c)$. From the second coordinate, we get $y^{3}=\dfrac{cx^{3}+x}{x+c}$. Substituting it into the first coordinate, we get $y=-\dfrac{cx}{x+c}$, i.e., $x^{-1}+y^{-1}=-c^{-1}$. Thus we have $\dfrac{cx^{3}+x}{x+c}=y^{3}=\left(-\dfrac{cx}{x+c}\right)^{3}$, which can be simplified to $cx^{4}+2c^{2}x^{3}+(2c^{3}+1)x^{2}+2cx+c^{2}=0,$ so $a_{1}b_{1}a_{2}b_{2}=c$. Therefore, $a_{1}b_{1}(a_{2}+b_{2})=a_{1}b_{1}a_{2}b_{2}(a_{2}^{-1}+b_{2}^{-1})=-1$. ∎

5 A Corollary of Theorem 1

Theorem 8**.**

[6, Page 412, Proposition 1.3]* Define a normalized theta function $\Theta(u,q)$ by the formula*

[TABLE]

• •

$\Theta(u,q)$* converges for all $u,q\in\mathbb{C}^{*}$ with $\left|q\right|<1$ and satisfies the functional equation*

[TABLE]

• •

$\Theta(u,q)$* is related to the function $X(u,q)$ described in Theorem 2 by the formula*

[TABLE]

Corollary 9**.**

Suppose that $\{(r_{i},\theta_{i})\}_{i=1}^{4}\in\mathcal{R}_{4}$ gives a constant image. Write $u_{i}=\exp(2\pi\sqrt{-1}\theta_{i})q^{r_{i}}$, then for $\left|q\right|<1$, we have the infinite product identity

[TABLE]

Proof.

By Theorem 8, the cross ratio of $\{X(u_{i},q)\}_{i=1}^{4}$ can be written as

[TABLE]

∎

6 Further Discussion

Let $S=\{(r_{i},\theta_{i})\}_{i=1}^{4}\in\mathcal{R}_{4}$ and $n$ their common order. Let

[TABLE]

be the extended upper half plane,

[TABLE]

the principal congruence subgroup of level $n$, and $X(n)=\Gamma(n)\backslash\mathcal{H}^{*}$. Let

[TABLE]

and $\mu_{S}$ the composition of

[TABLE]

and

[TABLE]

Then $\mu_{S}$ is a meromorphic function from $X(n)$ to $\mathbb{P}^{1}$. Let

[TABLE]

and

[TABLE]

By the proof of Corollary 5, $\Gamma(n)\subseteq\Gamma_{S}\subseteq\Delta_{S}\subseteq\textup{SL}_{2}(\mathbb{Z})$. The map $\mu_{S}$ factors through $X(\Gamma_{S})=\Gamma_{S}\backslash\mathcal{H}^{*}$ and $X(\Delta_{S})=\Delta_{S}\backslash\mathcal{H}^{*}$. In summary, we have the following commutative diagram:

[TABLE]

Theorem 1 classifies all $S$ such that $\mu_{S}$ is a constant map, thus gives a complete answer to [3, Conjecture 3]. The next natural question is whether

Conjecture 10**.**

[3, Conjecture 4]* $\Gamma_{S}=\Delta_{S}$ for all but finitely many $S$, up to the $\textup{SL}_{2}(\mathbb{Z})$-equivalence in $\mathcal{R}_{4}$.*

Now we give an example to show that sometimes $\Gamma_{S}\neq\Delta_{S}$ can indeed happen. Consider

[TABLE]

By the second part of Theorem 8,

[TABLE]

where $\zeta_{5}=\exp(2\pi\sqrt{-1}/5)$. By the first part of Theorem 8, these three expressions are equal for any $\left|q\right|<1$. Actually, for this example, we have $\Gamma_{S}/\pm\Gamma(5)\cong(\mathbb{Z}/2\mathbb{Z})^{2}$ and $\Delta_{S}/\pm\Gamma(5)\cong A_{4}$. Moreover, we note that $S$, $\begin{pmatrix}1&2\\ 1&3\end{pmatrix}\cdot S$, and $\begin{pmatrix}1&1\\ 2&3\end{pmatrix}\cdot S$ give a partition of the collection of all projective torsion points of order $5$.

If $n=p\geq 5$ is a prime, then by Lemma 6, $S$ is “good”. The calculations in the proof of Lemma 4 imply that at some cusp of $X(\Delta_{S})$, the $q$-expansion is not inside $\mathbb{C}((q))$. Therefore, $\Delta_{S}\neq\textup{SL}_{2}(\mathbb{Z})$ and

[TABLE]

By [7, Page 412, Theorem 6.25], any subgroup $H\subsetneq\textup{PSL}_{2}(\mathbb{Z}/p\mathbb{Z})$ contains at most one copy of $\mathbb{Z}/p\mathbb{Z}$. If $\mathbb{Z}/p\mathbb{Z}\subseteq H$, then the normalizer $N_{\textup{PSL}_{2}(\mathbb{Z}/p\mathbb{Z})}(\mathbb{Z}/p\mathbb{Z})\supseteq H$ is a subgroup of order $p(p-1)/2$.

• •

If $p\nmid[\Delta_{S}:\pm\Gamma(p)]$, then the quotient map $X(p)\to X(\Delta_{S})$ is unramified at all cusps of $X(\Delta_{S})$.

• •

If $p\mid[\Delta_{S}:\pm\Gamma(p)]$, then for some $\tau\in\mathbb{P}^{1}(\mathbb{Q})$, the isotropy subgroup

[TABLE]

satisfies

[TABLE]

The ramified cusps of $X(\Delta_{S})$ are corresponding to the left cosets in

[TABLE]

whose size is

[TABLE]

Acknowledgments. The first author was partially supported by the HSE University Basic Research Program, Russian Academic Excellence Project ‘5-100’, and EPSRC programme grant EP/M024830. The second author would like to express his gratitude for a pleasant stay at Laboratory of Algebraic Geometry, HSE, where a substantial part of this article was accomplished.