2.1. Complex hyperbolic space and its isometry group

We will introduce some material on complex hyperbolic geometry in this subsection. More details can be found in Goldman’s book [Reference Goldman6] and Parker’s notes [Reference Parker9].

Let $H$ be a Hermitian matrix given by the following

\begin{equation*} H=\left ( \begin {array}{c@{\quad}c@{\quad}c} 0 & 0 & 1 \\[3pt] 0 & I_{n-1} & 0 \\[3pt] 1 & 0 & 0 \\ \end {array} \right ), \end{equation*}

where $I_{n-1}$ is the identity matrix. $H$ induces a Hermitian form on the complex vector space $\mathbb{C}^{n+1}$ given by

\begin{equation*} \langle z, w \rangle =\bar {w}^t H z=z_1\overline {w_{n+1}}+z_{n+1}\overline {w_1}+z_2\overline {w_2}+\cdots +z_n\overline {w_n}, \end{equation*}

where $z=(z_1,z_2, \cdots, z_n, z_{n+1})^t$ and $w=(w_1, w_2,\cdots, w_n, w_{n+1})^t$ are vectors in $\mathbb{C}^{n+1}$ . Equipped with this Hermitian form, we write $\mathbb{C}^{n,1}$ instead of $\mathbb{C}^{n+1}$ .

Let $V_{+}$ (resp. $V_{0}$ or $V_{-}$ ) be the set of positive (resp. null or negative) vectors in $\mathbb{C}^{n,1}$ , that is the vectors with $\langle z,z \rangle \gt 0$ (resp. $\langle z,z \rangle =0$ or $\langle z,z \rangle \lt 0$ ). Let $P\,:\, \mathbb{C}^{n,1}\rightarrow \mathbb{C}P^n$ be the canonical projection. The complex hyperbolic space is defined to be $\mathbf{H}^n_{\mathbb C}=P(V_{-})$ associated with the Bergman metric, which is the Siegel domain in $\mathbb{C}P^n$ . The boundary of complex hyperbolic space is defined to be $\partial \mathbf{H}^n_{\mathbb C}=P(V_{0})$ , which can be viewed as the one-point compactification of the Heisenberg group.

Let ${\textrm{U}}(n,1)$ be the unitary group preserving the Hermitian form. The holomorphic isometry group of complex hyperbolic space is ${\textrm{PU}}(n,1)=\textrm{U}(n,1)/\textrm{U}(1)$ .

Now, let us focus on the boundary of three-dimensional complex hyperbolic space $\partial \mathbf{H}^3_{\mathbb C}$ . Let $\mathcal{N}=\mathbb{C}^2\times \mathbb{R}$ be the five-dimensional Heisenberg group with the group law

\begin{equation*} (z_1,z_2,t)(\zeta _1,\zeta _2,v)=(z_1+\zeta _1, z_2+\zeta _2,t+v+2\textrm {Im}(\overline {\zeta _1}z_1+\overline {\zeta _2}z_2)). \end{equation*}

Thus, $\partial \mathbf{H}^3_{\mathbb C}$ identifies with $\mathcal{N}\cup \{q_\infty \}$ and the Siegel domain can be identified with $\mathcal{N} \times \mathbb{R}_{+}$ . This identification gives the horospherical coordinates of points in $\mathbf{H}^3_{\mathbb C}$ . In details, the standard lift of $(\zeta _1,\zeta _2,v,u)\in \mathcal{N} \times \mathbb{R}_{+}$ is $\left ( \frac{-|\zeta _1|^2-|\zeta _2|^2-u+iv}{2},\zeta _1,\zeta _2,1 \right )^t$ . For each $u\gt 0$ , the horosphere of height $u$ is the subset of the Siegel domain given by $H_u=\mathcal{N} \times \{u\}$ and horoball of height $u$ is $B_u=\mathcal{N} \times (u,+\infty )$ .

The Cygan metric on $\mathcal{N}$ is defined to be

(2.1) \begin{equation} d_{\textrm{Cyg}}\left ( (z_1,z_2,t), (\zeta _1,\zeta _2,v) \right ) =\left | |\zeta _1 -z_1|^2+|\zeta _2-z_2|^2-it+iv-2i\textrm{Im}\left(\overline{\zeta _1}z_1+\overline{\zeta _2}z_2\right) \right |^{1/2}.\end{equation}

This metric can be extended to a metric on $\overline{\mathbf{H}^3_{\mathbb C}}-\{ q_\infty \}$ as the following

(2.2) \begin{equation} d_{\textrm{Cyg}}\left ( \left(z_1,z_2,t,u_1\right), \left(\zeta _1,\zeta _2,v,u_2\right) \right ) =\left | |\zeta _1 -z_1|^2+|\zeta _2-z_2|^2+|u_2-u_1|-it+iv-2i\textrm{Im}\left(\overline{\zeta _1}z_1+\overline{\zeta _2}z_2\right) \right |^{1/2}.\end{equation}

The Cygan sphere centered at $p_0=(z_1,z_2,t)\in \mathcal{N}$ with radius $r$ is defined by

\begin{equation*} S_{r}(p_0)=\{ p=(\zeta _1,\zeta _2,v,u)\in \mathbf {H}^3_{\mathbb C} \quad | \quad d_{\textrm {Cyg}}(p,p_0)=r \}. \end{equation*}

Let $g=(g_{ij})_{i,j=1}^{4}\in{\textrm{PU}}(3,1)$ such that $g(q_{\infty })\neq q_{\infty }$ , i.e., $g_{41}\neq 0$ . The isometric sphere of $g$ is defined to be

\begin{equation*} \{z\in \mathbf {H}^3_{\mathbb C} \,:\, |\langle z,q_{\infty }\rangle |=|\langle z,g^{-1}(q_{\infty })\rangle |\}. \end{equation*}

The isometric sphere of $g$ is a Cygan sphere centered at $g^{-1}(q_{\infty })$ with radius $r=\sqrt{2/|g_{41}|}$ .

For any given point $(z_1,z_2,t)\in \mathcal{N}$ , let

\begin{equation*} N_{\left ( z_1,z_2, t\right )}=\left [ \begin {array}{c@{\quad}c@{\quad}c@{\quad}c} 1 & -\bar {z}_1 & -\bar {z}_2 & \frac {-|z_1|^2-|z_2|^2+it}{2} \\[4pt] 0 & 1 & 0 & z_1 \\[4pt] 0 & 0 & 1 & z_2 \\[4pt] 0 & 0 & 0 & 1 \\ \end {array} \right ]. \end{equation*}

Then, $N_{(z_1,z_2,t)}\in{\textrm{PU}}(3,1)$ fixes $q_{\infty }$ and translates the origin $(0,0,0)\in \mathcal{N}$ to the given point.

2.2. Picard modular groups

Let $\mathcal{O}_{d}$ be the ring of integers in the imaginary quadratic number field $\mathbb{Q}(\sqrt{-d})$ , where $d$ is a positive square-free integer. When $d\equiv 1, 2$ (mod 4), $\mathcal{O}_{d}=\mathbb{Z}[\sqrt{-d}]$ , and $\mathcal{O}_{d}=\mathbb{Z}\left [\frac{1+\sqrt{-d}}{2}\right ]$ when $d \equiv 3$ (mod 4). The Picard modular group ${\textrm{PU}}(n,1;\,\mathcal{O}_d)$ is the subgroup of ${\textrm{PU}}(n,1)$ with matrix entries in $\mathcal{O}_{d}$ .

In this paper, we will study the Picard modular group in three complex dimensions with $d=3$ . It is well known that $\mathcal{O}_{3}=\mathbb{Z}[\omega ]$ , where $\omega =\frac{-1+\sqrt{3}i}{2}$ is a cubic root of unit.

Theorem 2.1 ([Reference Xie, Wang and Jiang17]). Let $\Gamma ={\textrm{PU}}(3,1;\,\mathbb{Z}[\omega ])$ . Then, $\Gamma$ can be generated by four elements $I_0, M_1, M_2$ and $N_{(1,0, \sqrt{3})}$ , where

\begin{equation*} \begin {array}{c@{\quad}c} I_0=\left [ \begin {array}{c@{\quad}c@{\quad}c@{\quad}c} 0 & 0 & 0 & 1 \\[3pt] 0 & -1 & 0 & 0 \\[3pt] 0 & 0 & -1 & 0 \\[3pt] 1 & 0 & 0 & 0 \\ \end {array} \right ], & N_{\left ( 1,0, \sqrt {3}\right )}=\left [ \begin {array}{c@{\quad}c@{\quad}c@{\quad}c} 1 & -1 & 0 & \omega \\[3pt] 0 & 1 & 0 & 1 \\[3pt] 0 & 0 & 1 & 0 \\[3pt] 0 & 0 & 0 & 1 \\ \end {array} \right ] \end {array}, \end{equation*}

\begin{equation*} \begin {array}{c@{\quad}c} M_1=\left [ \begin {array}{c@{\quad}c@{\quad}c@{\quad}c} 1 & 0&0 & 0\\[3pt] 0 &0&1 & 0 \\[3pt] 0 & 1 & 0&0\\[3pt] 0&0&0&1 \end {array}\right ], & M_2=\left [ \begin {array}{c@{\quad}c@{\quad}c@{\quad}c} 1 & 0&0 & 0 \\[3pt] 0 & -\omega & 0&0 \\[3pt] 0 & 0 & 1&0\\[3pt] 0&0&0&1\end {array}\right ] \end {array}. \end{equation*}

Moreover, the stabilizer subgroup $\Gamma _{\infty }$ that fixes the point at infinity $q_{\infty }$ is generated by $M_1, M_2$ and $N_{(1,0, \sqrt{3})}$ .

Lemma 2.2. The subgroup generated by $M_1$ and $M_2$ has the following presentation

\begin{equation*} \langle M_1, M_2 | M_1^2, M_2^6, (M_1 M_2)^2=(M_2 M_1)^2 \rangle. \end{equation*}

Remark 2.4. For any $M$ in the subgroup generated by $M_1$ and $M_2$ , $M$ can be written as

\begin{equation*}M=M_1^{p}M_2^{j}M_1M_2^{k},\end{equation*}

where $0\leq j \leq 5$ , $0\leq k \leq 5$ and $p=0,1$ .

To obtain a presentation for $\Gamma _{\infty }$ , we define the following notations:

\begin{align*} T_{0}&=N_{(0,0, 2\sqrt{3})},\\ T_{1}&=N_{(0,1, \sqrt{3})}, \\ T_{2}&=N_{(1,0, \sqrt{3})}, \\ T_{3}&=N_{(0,\omega, \sqrt{3})}, \\ T_{4}&=N_{(\omega,0, \sqrt{3})}. \end{align*}

Besides, for simplicity, we define

\begin{align*} T_5&=N_{\left ( -\bar{\omega },-\bar{\omega },-2\sqrt{3} \right )}=T_{0}^{-2}T_1T_2T_3T_4,\\ T_6&=N_{\left ( -\omega,-\omega,2\sqrt{3} \right )}=T_{0}^2T_{3}^{-1}T_{4}^{-1}, \end{align*}

which will be used to describe some relations of ${\textrm{PU}}(3,1;\,\mathbb{Z}[\omega ])$ in Section 4.

Lemma 2.5. Let $T$ be the group generated by $T_{0}$ , $T_{1}$ , $T_{2}$ , $T_{3}$ , and $T_{4}$ . Then, $T$ has the presentation

\begin{equation*} T=\left \langle T_{0}, T_{1}, T_{2}, T_{3}, T_{4} \ \mid \ [T_{1}, T_{2}], [T_{1}, T_{4}], [T_{2}, T_{3}], [T_{3},T_{4}], [T_{1},T_{3}]T_{0}, [T_{2},T_{4}]T_{0} \right \rangle. \end{equation*}

Furthermore, $T$ is a normal subgroup of $\Gamma _{\infty }$ .

Proof. Let $\Pi \,:\,\mathcal{N} \to \mathbb{C}^2$ be defined by $\Pi (z_1,z_2,t)=(z_1,z_2)$ , for all $(z_1,z_2,t)\in \mathcal{N}$ . Then, we have the sequence

\begin{equation*} 0 \rightarrow \mathbb {R} \rightarrow \mathcal {N} \xrightarrow {\Pi } \mathbb {C}^2 \rightarrow 0. \end{equation*}

This induces the following exact sequence

\begin{equation*} 0 \rightarrow \mathbb {R} \rightarrow {Isom}(\mathcal {N}) \xrightarrow {\Pi _{\ast }} {Isom}(\mathbb {C}^2) \rightarrow 0. \end{equation*}

Since $T_{0}$ , $T_{1}$ , $T_{2}$ , $T_{3}$ , and $T_{4}$ are Heisenberg translations, the group $T$ is a subgroup of ${Isom}(\mathcal{N})$ . When we restrict to the subgroup $T$ , we obtain the exact sequence

\begin{equation*} 0 \rightarrow 2\sqrt {3}\mathbb {Z} \rightarrow T \xrightarrow {\Pi _{\ast }} \mathbb {Z}^4 \rightarrow 0. \end{equation*}

Observe that the kernel of $\Pi _{\ast }$ is generated by $T_{0}$ . The generators of $\Pi _{\ast }(T)\cong \mathbb{Z}^4$ are $\Pi _{\ast }(T_{1})$ , $\Pi _{\ast }(T_{2})$ , $\Pi _{\ast }(T_{3})$ , and $\Pi _{\ast }(T_{4})$ , of which each pair is commutative. Hence, the relations of $T$ are generated by the commutators

\begin{equation*} [T_{1}, T_{2}], [T_{1}, T_{4}], [T_{2}, T_{3}], [T_{3},T_{4}], [T_{1},T_{3}]=T_{0}^{-1}, [T_{2},T_{4}]=T_{0}^{-1}. \end{equation*}

It is obvious that $T_0$ commutes with $T_1,T_2,T_3,T_4$ .

Note that $\Gamma _{\infty }$ is generated by $T_{2},M_1$ and $M_2$ . Since

\begin{equation*} M_1 T_{1} M^{-1}_1=T_{2}, M_1 T_{2} M^{-1}_1=T_{1}, M_1 T_{3} M^{-1}_1=T_{4}, M_1 T_{4} M^{-1}_1=T_{3}, \end{equation*}

and

\begin{equation*} M_2 T_{1} M^{-1}_2=T_{1}, M_2 T_{2} M^{-1}_2=T_{0} T_{4}^{-1}, M_2 T_{3} M^{-1}_2=T_{3}, M_2 T_{4} M^{-1}_2=T_{2} T_{4}, \end{equation*}

we have

\begin{equation*} M_1 T_{0} M^{-1}_1=T_{0}, M_2 T_{0} M^{-1}_2=T_{0}. \end{equation*}

Thus, the group $T$ is normal in $\Gamma _{\infty }$ .

We can obtain the presentation of the stabilizer subgroup $\Gamma _{\infty }$ by using the procedure introduced in Lemma 15 of [Reference Mark and Paupert8]. In short, we know that $\Gamma _{\infty }$ has two subgroups, $T$ and $\langle M_1, M_2 \rangle$ , where $T$ is normal in $\Gamma _{\infty }$ . Then, $\Gamma _{\infty }$ admits a presentation $\langle{\bf gen} |{\bf rel} \rangle$ , where the set $\bf gen$ consists of $M_1, M_2$ and the generators of $T$ , the set $\bf rel$ consists of relations of those two subgroups and of the form $M_i T_{j} M_{i}^{-1}=G\in T$ with $i=1,2$ and $j=0,1,2,3,4$ .

Corollary 2.6. The stabilizer subgroup $\Gamma _{\infty }$ has a presentation as the following:

\begin{equation*} \Gamma _{\infty }=\left \langle \begin {array}{c} {T_0, T_1, T_2, T_3, T_4,} \\[4pt] {M_1, M_2 } \end {array} \ \Bigg \vert \ \begin {array}{c} {[T_{1}, T_{2}], [T_{1}, T_{4}], [T_{2}, T_{3}], [T_{3},T_{4}], [T_{1},T_{3}]T_{0}, [T_{2},T_{4}]T_{0}, } \\[4pt] {M_1 T_{1} M^{-1}_1=T_{2}, M_2 T_{1} M^{-1}_2=T_{1}, M_2 T_{2} M^{-1}_2=T_{0} T_{4}^{-1},}\\[4pt] {M_1 T_{3} M^{-1}_1=T_{4}, M_2 T_{3} M^{-1}_2=T_{3}, M_2 T_{4} M^{-1}_2=T_{2} T_{4},}\\[4pt] {M^2_1, M^6_2, (M_1 M_2)^{2}=(M_2 M_1)^{2}} \end {array} \right \rangle. \end{equation*}

3.1. A coarse fundamental domain for $\Gamma _{\infty }$

Definition 3.1. The coarse fundamental domain $D_\infty \subset \partial \mathbf{H}^3_{\mathbb{C}}$ for the cusp stabilizer $\Gamma _\infty$ is defined to be

\begin{equation*}D_\infty =\{(z_{1},z_{2},t) \in \partial \mathbf {H}^3_{\mathbb {C}} \vert z_{1}\in \Delta, z_{2}\in \Delta,0\leq t \leq 2\sqrt {3}\},\end{equation*}

where $\Delta$ is the isosceles triangle in $\mathbb{C}$ with vertices $0,\frac{1}{2}-\frac{i \sqrt{3}}{6},\frac{1}{2}+\frac{i \sqrt{3}}{6}$ , see Figure 1.

Figure 1.

A fundamental domain $\triangle$ for the group of orientation-preserving symmetries of $\mathbb{Z}[\omega ]\subset \mathbb{C}$ .

Proof. The restriction of $\Gamma _\infty$ on the subsets ${0}\times \mathbb{C} \times \{0\}$ or $\mathbb{C} \times{0} \times \{0\}$ is the group of orientation-preserving symmetries of $\mathbb{Z}[\omega ]\subset \mathbb{C}$ , which is the $(2,3,6)$ -triangle group. A fundamental domain for this group acting on $\mathbb{C}$ is the isosceles triangle $\Delta$ . By the same argument of [Reference Falbel and Parker5, Reference Xie, Wang and Jiang18], we conclude that the fundamental domain for $\Gamma _\infty$ on $\partial{\mathbf{H}}_{\mathbb{C}}^3$ is contained in $D_\infty$ .

We review some terminology from [Reference Mark and Paupert8]. A vector $v\in \mathcal{O}_{3}^{4}$ is called primitive if for every $0\neq \lambda \in \mathcal{O}_{3}$ , $\frac{1}{\lambda }v \in \mathcal{O}_{3}^{4}$ implies that $\lambda$ is a unit. The units in $ \mathcal{O}_{3}$ are $1,\omega, \omega ^2$ . Moreover, every $\mathbb{Z}[\omega ]$ -rational point in $\mathbb{C}P^3$ has a primitive representation, which is unique up to multiplication by a unit in $ \mathcal{O}_{3}$ .

3.2. Covering depth of $\Gamma$

Recall that $u(n)=\frac{2}{\sqrt{n}}$ is the height at which balls of depth $n$ appear, in the sense of Corollary 1 of [Reference Mark and Paupert8]. Let $B\left ((z_1,z_2,t),r\right )$ denote the open Cygan ball centered at $p=(z_1,z_2,t) \in \partial{ \mathbf{H}}_{\mathbb{C}}^3$ with radius $r$ . It is easy to see that the isometric sphere of $g\in{\textrm{PU}}(3,1;\,\mathbb{Z}[\omega ])$ has radius $r\leq \sqrt{2}$ . In particular, the isometric sphere of $I_0$ is a Cygan sphere centered at $(0,0,0)$ with radius $r=\sqrt{2}$ . We will show that four Cygan balls of radius $\sqrt{2}$ will cover the set $D_\infty \times \{ u(5) \}$ .

Proposition 3.4. Let $u_0=2/\sqrt{5}$ and $H_{u_0}$ be the horosphere of height $u_0$ based at $\infty$ . Then, the prism $D_\infty \times \{ u_0 \}$ is covered by the intersections with $H_{u_0}$ of the isometric spheres

\begin{equation*}B\left ((0,0,0),\sqrt {2}\right ), B\left ((0,0,2\sqrt {3}),\sqrt {2}\right ),B\left ((0,1,\sqrt {3}),\sqrt {2}\right ),B\left ((1,0,\sqrt {3}),\sqrt {2}\right ).\end{equation*}

Proof. The method of the proof is similar to the proof of the main result in [Reference Xie, Wang and Jiang18]. We proceed with the following steps.

• If $z_1,z_2\in \Delta$ and $t\in [0, 2/\sqrt{3}]$ , then

  \begin{equation*}(|z_1|^2+|z_2|^2+u_0)^2+t^2\leq (2/3+u_0)^2+4/3=3.77\ldots \lt 4.\end{equation*}
  So the point is in $B\left ((0,0,0),\sqrt{2}\right )$ .

• If $z_1,z_2\in \Delta$ and $t\in [4/\sqrt{3}, 2\sqrt{3}]$ , then

  \begin{equation*}(|z_1|^2+|z_2|^2+u_0)^2+(2\sqrt {3}-t)^2\leq (2/3+u_0)^2+4/3=3.77\ldots \lt 4.\end{equation*}
  So the point is in $B\left ((0,0,2\sqrt{3}),\sqrt{2}\right )$ .

• If $z_1,z_2\in \Delta$ with $\textrm{Re}(z_2)\leq \textrm{Re}(z_1)$ , and $t\in [2/\sqrt{3}, 4/\sqrt{3}]$ then consider

  \begin{equation*}(|z_1|^2+|z_2|^2+u_0)^2+\left (t-\sqrt {3}+2\textrm {Im}(z_1)\right )^2.\end{equation*}
  Note that $ -\textrm{Re}(z_1)/\sqrt{3}\leq \textrm{Im}(z_1)\leq \textrm{Re}(z_1)/\sqrt{3}$ . For a given $\textrm{Re}(z_1)=x$ , this expression is bounded above by
  \begin{equation*}f(x)=(1-2x+4x^2/3+4x^2/3+u_0)^2+(1/\sqrt {3}+2x/\sqrt {3})^2.\end{equation*}
  Elementary calculus shows that this function attains its maximum values at one of the endpoints. (To see this, note that $f^{\prime}(0)\lt 0\lt f^{\prime}(1/2)$ and $f^{\prime\prime}(x)\gt 0$ .) We have
  \begin{eqnarray*} f(0) &=&(1+u_0)^2+1/3=3.922\ldots \lt 4, \\ f(1/2) &=&(2/3+u_0)^2+4/3=3.77\ldots \lt 4. \end{eqnarray*}
  Thus, the point is in $B\left ((1,0,\sqrt{3}),\sqrt{2}\right )$ .

• If $z_1,z_2\in \Delta$ with $\textrm{Re}(z_1)\leq \textrm{Re}(z_2)$ , and $t\in [2/\sqrt{3}, 4/\sqrt{3}]$ then consider

  \begin{equation*}(|z_1|^2+|z_2|^2+u_0)^2+\left (t-\sqrt {3}+2\textrm {Im}(z_2)\right )^2.\end{equation*}
  Note that $ -\textrm{Re}(z_2)/\sqrt{3}\leq \textrm{Im}(z_2)\leq \textrm{Re}(z_2)/\sqrt{3}$ . For a given $\textrm{Re}(z_2)=x$ , this expression is bounded above by the same function $f(x)$ . As above, this function is less than $4$ and so the point is in $B\left ((0,1,\sqrt{3}),\sqrt{2}\right )$ .

3.3. $\mathbb{Z}$ $[\omega]$ -rational points in $D_\infty$

We will give the points of depth at most 4 in $D_\infty$ . The standard lift of a $\mathbb{Z}[\omega ]$ -rational point $q=(z_1,z_2,t)$ in $D_\infty$ and the primitive lift of $q_{\infty }$ are given by

\begin{equation*}\label {eq:lift} {\bf {q}_{\infty }}=\left [ \begin {array}{c} 1 \\[3pt] 0 \\[3pt] 0 \\[3pt] 0 \\ \end {array} \right ],\quad \quad {\bf {q}}=\left [ \begin {array}{c} \dfrac {-|z_1|^2-|z_2|^2+it}{2} \\[6pt] z_1 \\[3pt] z_2 \\[3pt] 1 \\ \end {array} \right ], \end{equation*}

where $z_1,z_2\in \mathbb{C}$ and $t\in \mathbb{R}$ . Note that the standard lift of $q$ may not be integral. But we can choose some $\lambda \in \mathbb{Z}[\omega ]$ such that $\lambda \bf{q}$ is a primitive integral lift. The depth of $q$ is

\begin{equation*}|\langle \bf {q}_{\infty }, \lambda \bf {q}\rangle |^2=|\lambda |^2.\end{equation*}

In order to find the $\mathbb{Z}[\omega ]$ -rational points of depth $h$ for $1\leq h\leq 4$ , we need to find $\lambda \in \mathbb{Z}[\omega ]$ such that $|\lambda |^2=h$ up to multiplication by a unit. Let $\lambda =a+b\omega$ with $a, b\in \mathbb{Z}$ . Then, $|\lambda |^2=a^2-ab+b^2$ , and so we need to find $a, b\in \mathbb{Z}$ such that $a^2-ab+b^2=h$ for $1\leq h\leq 4$ . Note that there are no $\mathbb{Z}[\omega ]$ -rational points of depth 2. By a simple calculation, we have the following:

Now we can find the point $q=(z_1,z_2,t)\in D_\infty$ such that $\lambda \bf{q}$ is a primitive integral lift for a fixed $\lambda$ , that is, $\lambda z_1\in \mathbb{Z}[\omega ]$ , $\lambda z_2\in \mathbb{Z}[\omega ]$ , and $\lambda \frac{-|z_1|^2-|z_2|^2+it}{2}\in \mathbb{Z}[\omega ]$ . We will list depths that contain $\mathbb{Z}[\omega ]$ -rational points in $D_\infty$ . Furthermore, we need only consider one representative from each $\Gamma _\infty$ -orbit. We list the set of $\Gamma _\infty$ -orbit up to depth 4 in the following:

• Depth 1: $(0,0,0)$ and $(0,0,2\sqrt{3})$ , both of them are in the same $\Gamma _\infty$ -orbit;

• Depth 3: $\left(0,0,\frac{2}{3}\sqrt{3}\right)$ in one $\Gamma _\infty$ -orbit, and $\left(0,0,\frac{4}{3}\sqrt{3}\right)$ in the other $\Gamma _\infty$ -orbit;

• Depth 4: $(0,0,\sqrt{3})$ in one $\Gamma _\infty$ -orbit; $\left(\frac{1}{2},\frac{1}{2},\frac{1}{2}\sqrt{3}\right)$ and $\left(\frac{1}{2},\frac{1}{2},\frac{3}{2}\sqrt{3}\right)$ in the other two different orbits.

Integral lifts of the representatives of $\Gamma _\infty$ -orbits of these points are as follows:

\begin{equation*} \begin {array}{c@{\quad}c@{\quad}c@{\quad}c} p_0=\left [ \begin {array}{c} 0 \\[3pt] 0 \\[3pt]0 \\[3pt] 1 \end {array}\right ], & p_{31}=\left [ \begin {array}{c} -1 \\[3pt] 0 \\[3pt]0 \\[3pt] i\sqrt {3} \end {array}\right ], & p_{32}=\left [ \begin {array}{c} -2 \\[3pt] 0 \\[3pt]0 \\[3pt] i\sqrt {3} \end {array}\right ], \end {array} \end{equation*}

\begin{equation*} \begin {array}{c@{\quad}c@{\quad}c} p_{41}=\left [ \begin {array}{c} i\sqrt {3} \\[3pt] 0 \\[3pt] 0 \\[3pt]2 \end {array}\right ],& p_{42}=\left [ \begin {array}{c} \frac {-1+\sqrt {3}i}{2} \\[3pt] 1 \\[3pt]1 \\[3pt] 2 \end {array}\right ], & p_{43}=\left [ \begin {array}{c} \frac {-1+3\sqrt {3}i}{2} \\[3pt] 1 \\[3pt] 1 \\[3pt]2 \end {array}\right ] \end {array}. \end{equation*}

4.1. The generators

Let $A_{\alpha }\in \Gamma$ be a map sending $q_{\infty }$ to $p_\alpha$ , where $\alpha$ is in the set $\{0,31,32,41,42,43\}$ . The matrix form of $A_{\alpha }$ is given by

\begin{equation*} A_0=I_0=\left [ \begin {array}{c@{\quad}c@{\quad}c@{\quad}c} 0 & 0 &0& 1 \\[3pt] 0 & -1 & 0&0\\[3pt] 0 &0& -1 & 0 \\[3pt] 1 & 0 & 0&0 \end {array}\right ], A_{31}=\left [ \begin {array}{c@{\quad}c@{\quad}c@{\quad}c} -1 & 0&0 & i\sqrt {3} \\[3pt] 0 & 1 & 0&0\\[3pt] 0 & 0 & 1&0 \\[3pt] i\sqrt {3} & 0 &0 & 2 \end {array}\right ],A_{32}=A_{31}^{-1}, \end{equation*}

\begin{equation*}A_{41}=\left [ \begin {array}{c@{\quad}c@{\quad}c@{\quad}c} -i\sqrt {3} & 0 &0 & -2\\[3pt] 0 & -1 & 0& 0\\[3pt] 0 & 0 & -1& 0 \\[3pt] -2 & 0 &0 & i\sqrt {3} \end {array}\right ], A_{42}=\left [ \begin {array}{c@{\quad}c@{\quad}c@{\quad}c} 1 & 0 &0 & 0\\[9pt] \dfrac {-1-\sqrt {3}i}{2} & 1 & 0& 0\\[9pt] \dfrac {-1-\sqrt {3}i}{2}& 0 & 1& 0 \\[9pt] -1-\sqrt {3}i & \dfrac {1-\sqrt {3}i}{2} &\dfrac {1-\sqrt {3}i}{2}& 1 \end {array}\right ], \end{equation*}

\begin{equation*} A_{43}=\left [ \begin {array}{cccc} -2-i\sqrt {3} & \dfrac {-3+\sqrt {3}i}{2} & \dfrac {-3+\sqrt {3}i}{2} & i\sqrt {3}\\[9pt] \dfrac {-1+\sqrt {3}i}{2} & 1 & 0 & 0\\[9pt] \dfrac {-1+\sqrt {3}i}{2}& 0 & 1& 0 \\[9pt] -1+\sqrt {3}i & \dfrac {1+\sqrt {3}i}{2} & \dfrac {1+\sqrt {3}i}{2} & 1 \end {array}\right ]. \end{equation*}

4.2. The relations

By applying generators to points of depth at most 4, we obtain the cycles. The general method to get these can be found in [Reference Mark and Paupert8]. Here, we just describe it briefly. Let $A_{a}$ be a map in the set of generators $\{I_0,A_{31},A_{32},A_{41},A_{42},A_{43}\}$ . Let $p_{b},p_{c}$ be two $\mathbb{Z}[\omega ]$ -rational points of depth at most 4 which are listed in Section 3.3. Let $\gamma \in \Gamma _{\infty }$ . If $A_{a}(\gamma p_b)$ and $p_{c}$ have the same depth, and there exists a map $W\in \Gamma _{\infty }$ such that $A_{a}(\gamma p_{b})=W(p_{c})$ , then one obtains $A_{c}^{-1}W^{-1}A_{a}\gamma A_{b}=W^{\prime}\in \Gamma _{\infty }$ , here $A_{b}$ and $ A_{c}$ be two maps sending $q_{\infty }$ to $p_{b}$ and $ p_{c}$ , respectively. Therefore, one obtains a relation $A_{c}^{-1}W^{-1}A_{a}\gamma A_{b}W^{\prime -1}=Id$ .

In practice, by using the following observations, we can simplify the process of solving the equations $A_{a}(\gamma p_{b})=W(p_{c})$ .

Proof. All of these can be computed directly.

Proof. The first item is obvious. To see the second item, it is suffice to show that $M(p_b)=p_b$ . Since $p_b=A_b(q_{\infty })$ and $M\in \Gamma _{\infty }$ ,

\begin{equation*} M(p_b)=MA_b(q_{\infty })=A_b M (q_{\infty })=A_b(q_{\infty })=p_b. \end{equation*}

Proof. (1). If $M$ commutes with $A_a, A_b$ and $A_c$ , then $M(p_b)=p_b$ and $M(p_c)=p_c$ . Therefore,

\begin{equation*} A_a \gamma _2 p_b=A_a M\gamma _1 M^{-1}p_b=M A_a\gamma _1 p_b=MW_1M^{-1}p_c, \end{equation*}

which means that $W_2=MW_1M^{-1}$ . It implies that

\begin{equation*} A_{c}^{-1}W_{2}^{-1}A_{a}\gamma _{2} A_{b}=A_c^{-1}MW_1^{-1}M^{-1}A_a M\gamma _1 M^{-1}A_b=MA_c^{-1}W_1^{-1}A_a\gamma _1A_bM^{-1}=MW^{\prime}_1M^{-1} \end{equation*}

and so $W^{\prime}_2=MW^{\prime}_1M^{-1}$ . Thus,

\begin{equation*} A_{c}^{-1}W_{2}^{-1}A_{a}\gamma _{2} A_{b}{W^{\prime}_2}^{-1}=M\cdot A_{c}^{-1}W_{1}^{-1}A_{a}\gamma _{1} A_{b}{W^{\prime}_1}^{-1} \cdot M^{-1} =M \cdot Id \cdot M^{-1}=Id. \end{equation*}

(2). If $M$ commutes with $A_a,A_b$ , but not $A_c$ , then

\begin{equation*} A_a \gamma _2 p_b=A_a M\gamma _1 M^{-1}p_b=M A_a\gamma _1 p_b=MW_1p_c, \end{equation*}

and so $W_2=MW_1$ . Thus,

\begin{equation*} A_{c}^{-1}W_{2}^{-1}A_{a}\gamma _{2} A_{b}=A_c^{-1}W_1^{-1}M^{-1}A_a M\gamma _1 M^{-1}A_b=A_c^{-1}W_1^{-1}A_a\gamma _1A_bM^{-1}=W^{\prime}_1M^{-1}, \end{equation*}

and so $W^{\prime}_2=W^{\prime}_1M^{-1}$ . Thus,

\begin{equation*} A_{c}^{-1}W_{2}^{-1}A_{a}\gamma _{2} A_{b}{W^{\prime}_2}^{-1}= A_{c}^{-1}W_{1}^{-1}A_{a}\gamma _{1} A_{b}{W^{\prime}}_1^{-1} =Id. \end{equation*}

Our main result is the following.

Theorem 4.4. Let $\Gamma ={\textrm{PU}}(3,1;\,\mathbb{Z}[\omega ])$ . Then, $\Gamma$ has a presentation $\langle S | R \rangle$ , where $S$ consists of $I_0, A_{31}, A_{32}, A_{41}, A_{42}, A_{43}$ and the generators of $\Gamma _{\infty }$ , $R$ consist of the relations of $\Gamma _{\infty }$ , the relations given in Lemma 4.1 and the relations from Tables 1 to 16.

Table 1.

Let $\gamma =N_{(z_1,z_2,t)}$ be a Heisenberg translation. If $A_a(\gamma p_b)=W(p_c)$ , then $A_c^{-1}W^{-1}A_a \gamma A_b=W^{\prime}$ . The case $A_a=I_0$ , $p_b=p_0, p_{31},p_{32},p_{41}$

Table 2.

Let $\gamma =N_{(z_1,z_2,t)}$ be a Heisenberg translation. If $A_a(\gamma p_b)=W(p_c)$ , then $A_c^{-1}W^{-1}A_a \gamma A_b=W^{\prime}$ . The case $A_a=I_0$ , $p_b=p_{42}$

Table 3.

Let $\gamma =N_{(z_1,z_2,t)}$ be a Heisenberg translation. If $A_a(\gamma p_b)=W(p_c)$ , then $A_c^{-1}W^{-1}A_a \gamma A_b=W^{\prime}$ . The case $A_a=I_0$ , $p_b=p_{43}$

Table 4.

Let $\gamma =N_{(z_1,z_2,t)}$ be a Heisenberg translation. If $A_a(\gamma p_b)=W(p_c)$ , then $A_c^{-1}W^{-1}A_a \gamma A_b=W^{\prime}$ . The case $A_a=A_{31}$

Table 5.

Let $\gamma =N_{(z_1,z_2,t)}$ be a Heisenberg translation. If $A_a(\gamma p_b)=W(p_c)$ , then $A_c^{-1}W^{-1}A_a \gamma A_b=W^{\prime}$ . The case $A_a=A_{32}$

Proof. Let $\gamma =N_{(z_1,z_2,t)}M\in \Gamma _{\infty }$ , where $N_{(z_1,z_2,t)}$ be a Heisenberg translation and $M\in \langle M_1, M_2 \rangle$ . We need to solve $A_{a}(\gamma p_{b})=A_{a}N_{(z_1,z_2,t)}M(p_{b})=W(p_{c})$ for every pair $(A_{a},p_{b})$ .

Table 6.

Let $\gamma =N_{(z_1,z_2,t)}$ be a Heisenberg translation. If $A_a(\gamma p_b)=W(p_c)$ , then $A_c^{-1}W^{-1}A_a \gamma A_b=W^{\prime}$ . The case $A_a=A_{41}$

If $MA_{a}=A_{a}M$ , then by Lemma 4.2,

\begin{equation*}A_{a}N_{(z_1,z_2,t)}M(p_{b})=A_{a}M\cdot M^{-1}N_{(z_1,z_2,t)}M(p_b)=MA_{a}N_{(\tilde {z}_1,\tilde {z}_2,\tilde {t})}(p_b)=M\widetilde {W}({p_{c}}),\end{equation*}

where $N_{(\tilde{z}_1,\tilde{z}_2,\tilde{t})}=M^{-1}N_{(z_1,z_2,t)}M$ . Moreover, suppose that $A_{c}^{-1}\widetilde{W}^{-1}A_{a}N_{(\tilde{z}_1,\tilde{z}_2,\tilde{t})}A_{b}=\widetilde{W^{\prime}}\in \Gamma _{\infty }$ , then

\begin{equation*} A_{c}^{-1}\widetilde {W}^{-1}M^{-1}A_{a}N_{({z}_1,{z}_2,{t})}MA_{b}=A_{c}^{-1}\widetilde {W}^{-1}A_{a}N_{(\tilde {z}_1,\tilde {z}_2,\tilde {t})}A_{b}=\widetilde {W^{\prime}}. \end{equation*}

This means that the relation coming from $N_{(z_1,z_2,t)}M$ is the same as the relation from $N_{(\tilde{z}_1,\tilde{z}_2,\tilde{t})}$ . Thus, it is equivalent to solve $A_{a}N_{(\tilde{z}_1,\tilde{z}_2,\tilde{t})}(p_{b})=\widetilde{W}({p_{c}})$ . If $MA_{b}=A_{b}M$ , then by Lemma 4.2,

\begin{equation*}A_{a}N_{(z_1,z_2,t)}M(p_{b})=A_{a}N_{(z_1,z_2,t)}(p_{b})=W(p_{c}).\end{equation*}

Similarly, the relation coming from $N_{(z_1,z_2,t)}M$ is the same as relation from $N_{(z_1,z_2,t)}$ . Therefore, in both of the two cases, it suffices to solve $A_{a}N_{(z_1,z_2,t)}(p_{b})=W(p_{c})$ .

By Lemma 4.1, $M$ commutes with $I_0$ , $A_{31}$ , $A_{32}$ and $A_{41}$ . Thus, we have to solve $A_{a}N_{(z_1,z_2,t)} (p_{b})=W(p_{c})$ for every pair $(A_a,p_b)$ .

Besides, we should solve $A_{a}N_{(z_1,z_2,t)}M (p_{b})=W(p_{c})$ for the pairs

\begin{equation*}(A_a,p_b)\in \{(A_{42},p_{42}),(A_{42},p_{43}),(A_{43},p_{42}),(A_{43},p_{43})\},\end{equation*}

with $M\neq Id$ , since $M$ does not commute with $A_{42}$ and $A_{43}$ . For these four cases, it suffices to solve $A_{a}N_{(z_1,z_2,t)}M p_{b}=Wp_{c}$ with $M=M_2^{j}M_1M_2^{k}$ and $M=M_2^{j}$ ( $j,k=1,2,3,4,5$ ) by Lemma 4.2, since $M=M_1^{p}M_2^{j}M_1M_2^{k}$ (see remark 2.4) and $M_1$ commutes with $A_{42}$ and $A_{43}$ .

In practice, there should be several solutions $(\gamma,W)$ of $A_{a}(\gamma p_{b})=W(p_{c})$ for fixed triple $(A_a,p_b,p_c)$ . However, we do not need to consider all of the solutions. Suppose that $(\gamma _1,W_1)$ and $(\gamma _2,W_2)$ be two different solutions. If $\gamma _2=M\gamma _1 M^{-1}$ for some $M\in \langle M_1, M_2 \rangle$ commuting with $A_a$ and $A_b$ , then by Lemma 4.3, the relation coming from $\gamma _2$ can be derived from the relation coming from $\gamma _1$ . Thus, the relation coming from $\gamma _2$ can be omitted.

The solutions of $A_{a}(\gamma p_{b})=W(p_{c})$ and $A_c^{-1}W^{-1}A_a \gamma A_b=W^{\prime}$ are given in Tables 1–16.

Table 7.

Let $\gamma =N_{(z_1,z_2,t)}$ be a Heisenberg translation. If $A_a(\gamma p_b)=W(p_c)$ , then $A_c^{-1}W^{-1}A_a \gamma A_b=W^{\prime}$ . The case $A_a=A_{42}$

Table 8.

Let $\gamma =N_{(z_1,z_2,t)}$ be a Heisenberg translation. If $A_a(\gamma p_b)=W(p_c)$ , then $A_c^{-1}W^{-1}A_a \gamma A_b=W^{\prime}$ . The case $A_a=A_{43}$

As follows, we shall explain the computations for the case $A_a=I_0$ and $p_b=p_0$ in Table 1. The others are similar.

Let $\gamma =N_{(z_1,z_2,t)}\in \Gamma _{\infty }$ be a Heisenberg translation. Recall that $z_1,z_2\in \mathbb{Z}[\omega ], t/\sqrt{3}\in \mathbb{Z}$ , and the integers $|z_1|^2+|z_2|^2$ and $t/\sqrt{3}$ have the same parity. One can compute that the depth of $I_0(\gamma p_0)$ is

\begin{equation*} \textrm {dep}=\frac {(|z_1|^2+|z_2|^2)^2+t^2}{4}. \end{equation*}

Obviously, we have the following.

• If $\textrm{dep}=0$ , then $z_1=z_2=t=0$ .

• If $\textrm{dep}=1$ , then $|z_1|=|z_2|=1, t=0$ or $|z_1|^2+|z_2|^2=1,t=\pm \sqrt{3}$ .

• If $\textrm{dep}=3$ , then $z_1=z_2=0, t=\pm 2\sqrt{3}$ or $|z_1|^2+|z_2|^2=3, t=\pm \sqrt{3}$ .

• If $\textrm{dep}=4$ , then $|z_1|^2+|z_2|^2=4, t=0$ or $|z_1|=|z_2|=1, t=\pm 2\sqrt{3}$ .

It is obvious that there are many solutions of $N_{(z_1,z_2,t)}$ . Note that $M_1N_{(z_1,z_2,t)}M_1^{-1}=N_{(z_2,z_1,t)}$ and $M_2N_{(z_1,z_2,t)}M_2^{-1}=N_{(-\omega z_1,z_2,t)}$ . Thus according to Lemma 4.1 and Lemma 4.3, it suffices to consider the following 12 cases.

• $\textrm{dep}=0\ :$ $N_{(0,0,0)}=Id$ .

• $\textrm{dep}=1\ :$ $N_{(1,1,0)}=T_0^{-1}T_1T_2$ , $N_{(1,0,\sqrt{3})}=T_2$ , $N_{(1,0,-\sqrt{3})}=T_0^{-1}T_2$ .

• $\textrm{dep}=3\ :$ $N_{(0,0,2\sqrt{3})}=T_0$ , $N_{(0,0,-2\sqrt{3})}=T_0^{-1}$ , $N_{(0,1-\omega,\sqrt{3})}=T_1T_3^{-1}$ , $N_{(0,1-\omega,-\sqrt{3})}=T_0^{-1}T_1T_3^{-1}$ .

• $\textrm{dep}=4\ :$ $N_{(0,2\omega,0)}=T_0^{-1}T_3^2$ , $N_{(1,1+2\omega,0)}=T_1T_2T_0^{-1}T_3^2$ , $N_{(1,1,2\sqrt{3})}=T_1T_2$ , $N_{(1,1,-2\sqrt{3})}=T_0^{-2}T_1T_2$ .

Finally, for each one of the above cases, we find $W(p_c)$ and $W^{\prime}$ such that $A_c^{-1}W^{-1}I_0 \gamma I_0=W^{\prime}$ , where $W,W^{\prime}\in \Gamma _{\infty }$ .

Table 9.

Let $\gamma =N_{(z_1,z_2,t)}$ be a Heisenberg translation and $M\in \langle M_1, M_2 \rangle$ . If $A_a(\gamma M p_b)=W(p_c)$ , then $A_c^{-1}W^{-1}A_a \gamma M A_b=W^{\prime}$ . The case $A_a=A_{42}$ and $p_b=p_{42}$

Table 10.

Let $\gamma =N_{(z_1,z_2,t)}$ be a Heisenberg translation and $M\in \langle M_1, M_2 \rangle$ . If $A_a(\gamma M p_b)=W(p_c)$ , then $A_c^{-1}W^{-1}A_a \gamma M A_b=W^{\prime}$ . The case $A_a=A_{42}$ and $p_b=p_{42}$

Table 11.

Let $\gamma =N_{(z_1,z_2,t)}$ be a Heisenberg translation and $M\in \langle M_1, M_2 \rangle$ . If $A_a(\gamma M p_b)=W(p_c)$ , then $A_c^{-1}W^{-1}A_a \gamma M A_b=W^{\prime}$ . The case $A_a=A_{42}$ and $p_b=p_{43}$

Table 12.

Let $\gamma =N_{(z_1,z_2,t)}$ be a Heisenberg translation and $M\in \langle M_1, M_2 \rangle$ . If $A_a(\gamma M p_b)=W(p_c)$ , then $A_c^{-1}W^{-1}A_a \gamma M A_b=W^{\prime}$ . The case $A_a=A_{42}$ and $p_b=p_{43}$

Table 13.

Let $\gamma =N_{(z_1,z_2,t)}$ be a Heisenberg translation and $M\in \langle M_1, M_2 \rangle$ . If $A_a(\gamma M p_b)=W(p_c)$ , then $A_c^{-1}W^{-1}A_a \gamma M A_b=W^{\prime}$ . The case $A_a=A_{43}$ and $p_b=p_{42}$

Table 14.

Let $\gamma =N_{(z_1,z_2,t)}$ be a Heisenberg translation and $M\in \langle M_1, M_2 \rangle$ . If $A_a(\gamma M p_b)=W(p_c)$ , then $A_c^{-1}W^{-1}A_a \gamma M A_b=W^{\prime}$ . The case $A_a=A_{43}$ and $p_b=p_{42}$

Table 15.

Let $\gamma =N_{(z_1,z_2,t)}$ be a Heisenberg translation and $M\in \langle M_1, M_2 \rangle$ . If $A_a(\gamma M p_b)=W(p_c)$ , then $A_c^{-1}W^{-1}A_a \gamma M A_b=W^{\prime}$ . The case $A_a=A_{43}$ and $p_b=p_{43}$

Table 16.

Let $\gamma =N_{(z_1,z_2,t)}$ be a Heisenberg translation and $M\in \langle M_1, M_2 \rangle$ . If $A_a(\gamma M p_b)=W(p_c)$ , then $A_c^{-1}W^{-1}A_a \gamma M A_b=W^{\prime}$ . The case $A_a=A_{43}$ and $p_b=p_{43}$

From the presentation of $\Gamma$ , we get the following useful information by using GAP.