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A presentation for the Eisenstein-Picard modular group in three complex dimensions

Published online by Cambridge University Press:  25 July 2023

Jieyan Wang
Affiliation:
School of Mathematics, Hunan University, Changsha 410082, P.R. China
Baohua Xie*
Affiliation:
School of Mathematics, Hunan University, Changsha 410082, P.R. China
*
Corresponding author: Baohua Xie; Email: [email protected]
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Abstract

A. Mark and J. Paupert [Presentations for cusped arithmetic hyperbolic lattices, 2018, arXiv:1709.06691.] presented a method to compute a presentation for any cusped complex hyperbolic lattice. In this note, we will use their method to give a presentation for the Eisenstein-Picard modular group in three complex dimensions.

Type
Research Article
Copyright
© The Author(s), 2023. Published by Cambridge University Press on behalf of Glasgow Mathematical Journal Trust

1. Introduction

To study discrete subgroups of Lie groups, it is a challenging work to determine a presentation of a discrete subgroup. The purpose of this paper is to give a presentation of an arithmetic lattice in ${\textrm{PU}}(3,1)$ , which is the holomorphic isometry group of complex hyperbolic space ${\mathbf{H}}^3_{\mathbb{C}}$ . It is well known that complex hyperbolic space is one of the non-compact rank one symmetric spaces.

As is widely known, the classical modular group $\textrm{PSL}(2,\mathbb{Z})$ is a free product of a cyclic group of order 2 and a cyclic group of order 3. To see this, there are many different ways. One of them is an algebraic method that uses a continued fraction algorithm to compute it directly. The other one is a geometric method that uses the action of $\textrm{PSL}(2,\mathbb{Z})$ on the hyperbolic plane. One can consider $\textrm{PSL}(2,\mathbb{Z})$ as a discrete subgroup of $\textrm{PSL}(2,\mathbb{R})$ that acts isometrically on the hyperbolic plane $\mathbf{H}^2$ . There will be a fundamental domain for this action. The presentation for $\textrm{PSL}(2,\mathbb{Z})$ can be obtained by applying Poincaré’s polygon theorem to the fundamental domain.

There are two kinds of generalizations of $\textrm{PSL}(2,\mathbb{Z})$ in higher dimensional hyperbolic space, (real) hyperbolic space $\mathbf{H}^n_{\mathbb{R}}$ , and complex hyperbolic space $\mathbf{H}^n_{\mathbb{C}}$ . Before we introduce these generalizations in detail, we would like to explain some notations. Let $\mathcal{O}_d$ be the ring of integers in $\mathbb{Q}(\sqrt{-d})$ with $d$ being a positive square-free integer. $\mathcal{O}_d$ is a Euclidean domain when $d=1,2,3,7,11$ .

In the real case, Bianchi groups $\textrm{PSL}(2;\,\mathcal{O}_d)$ are subgroups of $\textrm{PSL}(2, \mathbb{C})$ with entries in $\mathcal{O}_d$ . Here, $\textrm{PSL}(2, \mathbb{C})$ is the orientation-preserving isometry group of $\mathbf{H}^3_{\mathbb{R}}$ . These groups are arithmetic lattices that are finitely presentable. Presentations for Bianchi groups can be obtained geometrically. Considering the actions of the Bianchi groups on hyperbolic space, Swan [Reference Swan13] developed a method to find the Ford domains. Based on these domains and using the generalization theorem of Macbeath [Reference Macbeath7], Swan derived a method to obtain presentations for the Bianchi groups. In particular, finite presentations for the Euclidean Bianchi groups can be derived from a result of Cohn, see [Reference Cohn1] and [Reference Cohn2].

In the complex case, Picard modular groups ${\textrm{PU}}(n,1;\,\mathcal{O}_d)$ are subgroups of ${\textrm{PU}}(n,1)$ with entries in $\mathcal{O}_d$ . Here ${\textrm{PU}}(n,1)$ is the holomorphic isometry group of $\mathbf{H}^n_{\mathbb{C}}$ . These groups are arithmetic lattices that are finitely presentable. To the best of our knowledge, the presentations for some of them are known. A presentation for the Eisenstein-Picard modular group ${\textrm{PU}}(2,1;\,\mathcal{O}_3)$ was found by Falbel and Parker in [Reference Falbel and Parker5], in which they studied the geometry of the group. A few years later, a presentation for the Gauss-Picard modular group ${\textrm{PU}}(2,1;\,\mathcal{O}_1)$ was given in [Reference Falbel, Francsics and Parker4]. In his paper [Reference Zhao20], Zhao gave a finite system of generators for each one of the Euclidean-Picard modular groups ${\textrm{PU}}(2,1;\,\mathcal{O}_d)$ . We remark that the sister of the Eisenstein-Picard modular group was studied in [Reference Parker10], and a presentation was given in [Reference Zhao19]. In particular, Stover [Reference Stover12] showed that the quotient spaces of $\mathbf{H}^2_{\mathbb{C}}$ by the Eisenstein-Picard modular group and its sister group are the only two arithmetic cusped complex hyperbolic 2-orbifolds with minimal volume. The sister of the other Euclidean-Picard modular groups was studied in [Reference Xie16], in which a finite system of generators of them was constructed. The methods in these papers are geometric, that is, they construct fundamental domains for the Euclidean-Picard modular groups ${\textrm{PU}}(2,1;\,\mathcal{O}_d)$ acting on complex hyperbolic space. On the other hand, one can give a finite system of generators for some of the Picard modular groups using an algebraic method. In [Reference Falbel, Francsics, Lax and Parker3], the authors obtained a system of generators for the Gauss-Picard modular group ${\textrm{PU}}(2,1;\,\mathcal{O}_1)$ by using the continued fraction algorithm. One can also use this method to derive a system of generators for the Eisenstein-Picard modular group ${\textrm{PU}}(2,1;\,\mathcal{O}_3)$ , see [Reference Wang, Xiao and Xie14].

For the Picard modular groups in three complex dimensions ${\textrm{PU}}(3,1;\,\mathcal{O}_d)$ , we know very little. Since it is very difficult to construct a fundamental domain for ${\textrm{PU}}(3,1;\,\mathcal{O}_d)$ acing on $\mathbf{H}^3_{\mathbb{C}}$ , we do not know a presentation of ${\textrm{PU}}(3,1;\,\mathcal{O}_d)$ . In [Reference Xie, Wang and Jiang17], the authors gave a system of finite generators for ${\textrm{PU}}(3,1;\,\mathcal{O}_3)$ by using the continued fraction algorithm. In [Reference Xie, Wang and Jiang18], a finitely generating set of ${\textrm{PU}}(3,1;\,\mathcal{O}_1)$ was given by using a combined geometric and algebraic method. Unfortunately, it seems that it is difficult to give a presentation for these two groups. In a recent work, Mark and Paupert [Reference Mark and Paupert8] present a method to obtain a presentation for Picard modular group ${\textrm{PU}}(2,1;\,\mathcal{O}_d)$ with $d=1,3,7$ , by applying a result of Macbeath [Reference Macbeath7]. Polletta [Reference Polletta11] discussed the application of their method to the Picard modular groups, ${\textrm{PU}}(2,1;\,\mathcal{O}_d)$ , when d = 2, 11, and obtained presentations for these groups. Inspired by their work, in this paper, we obtain a presentation for ${\textrm{PU}}(3,1;\,\mathcal{O}_3)$ (see Theorem 4.4).

To do this, we use the method of Mark and Paupert [Reference Mark and Paupert8]. Let $\Gamma ={\textrm{PU}}(3,1;\,\mathcal{O}_3)$ . Consider the action of $\Gamma$ on $\mathbf{H}^3_{\mathbb{C}}$ . One can construct a suitable horoball $V$ based on the cusp of $\Gamma$ , such that the $\Gamma$ -orbits of $V$ form a covering of $\mathbf{H}^3_{\mathbb{C}}$ . Let $E(V)=\{g\in \Gamma | g(V)\cap V\neq \emptyset \}$ . Then, by Macbeath’s theorem [Reference Macbeath7], $\Gamma$ will be generated by $E(V)$ together with relations $g\cdot h=gh$ for each pair $g,h \in E(V)$ with $V\cap g(V) \cap gh(V) \neq \emptyset$ , where $gh\in E(V)$ .

In practice, our process is as follows.

  1. Step 1: give a presentation for the stabilizer subgroup $\Gamma _{\infty }$ of $\Gamma$ , which fixing the point at infinity $q_{\infty }$ , see Corollary 2.6.

  2. Step 2: give a suitable coarse fundamental domain $D_{\infty }$ of $\Gamma _{\infty }$ , see Definition 3.1.

  3. Step 3: determine the covering depth of $\Gamma$ , which is at most 4, see Section 3.2.

  4. Step 4: find all the $\mathcal{O}_3$ -rational points of depth at most 4 in $D_{\infty }$ and a system of representatives $\{p_{\alpha }\,:\, \alpha \in \{0,31,32,41,42,43\}\}$ of $\Gamma _{\infty }$ -orbits of these $\mathcal{O}_3$ -rational points, see Section 3.3.

  5. Step 5: give a map $A_{\alpha }\in \Gamma$ sending $q_{\infty }$ to $p_{\alpha }$ for each $\alpha \in \{0,31,32,41,42,43\}$ , see Section 4.1.

  6. Step 6: determine $\gamma,W,W^{\prime}\in \Gamma _{\infty }$ such that $A_{c}^{-1}W^{-1}A_{a}\gamma A_{b}=W^{\prime}$ for each triple $(A_a,p_b,p_c)$ , where $A_{a}\in \{I_0,A_{31},A_{32},A_{41},A_{42},A_{43}\}$ , $p_b,p_c$ be two $\mathcal{O}_3$ -rational points of depth at most 4 in $D_{\infty }$ and $A_{b},A_{c}$ be two maps sending $q_{\infty }$ to $p_b,p_c$ , respectively. See Section 4.2.

  7. Step 7: obtain a presentation $\Gamma =\langle S | R \rangle$ , where $S$ consists of the generators of $\Gamma _{\infty }$ and $A_{\alpha }$ with $\alpha \in \{0,31,32,41,42,43\}$ , and $R$ consists of relations of $\Gamma _{\infty }$ and the forms $A_{c}^{-1}W^{-1}A_{a}\gamma A_{b}=W^{\prime}$ given in Step 6.

2. Background

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 2.3. The subgroup generated by $M_1$ and $M_2$ is a finite group with order $72$ , which is isomorphic to the semidirect product of $\mathbb{Z}_6 \times \mathbb{Z}_6$ with $ \mathbb{Z}_2$ .

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. The action of PU(3, 1; $\mathbb{Z}$ [ $\omega$ ])

Let $\Gamma ={\textrm{PU}}(3,1;\,\mathbb{Z}[\omega ])$ , and $\Gamma _{\infty }$ be the stabilizer subgroup of ${\textrm{PU}}(3,1;\,\mathbb{Z}[\omega ])$ fixing the point at infinity.

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 $\mathbb{Z}[\omega ]\subset \mathbb{C}$ .

Lemma 3.2. The $\Gamma _\infty$ -translates of $D_\infty$ cover $\partial{\mathbf{H}}_{\mathbb{C}}^3$ .

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}$ .

Definition 3.3. The depth of a $\mathbb{Z}[\omega ]$ -rational point $x$ is given by $|\langle v, \mathbf{q}_{\infty }\rangle |^2=|v_4|^2$ , where $v=(v_1,v_2,v_3,v_4)^t$ is any primitive integral lift of $x$ .

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 $B\left ((0,1,\sqrt{3}),\sqrt{2}\right )$ .

Remark 3.5. This proof of Proposition 3.4 was given by the referee. We would like to thank him for allowing us to write it here.

Corollary 3.6. The covering depth of $\Gamma$ is at most 4.

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. Generators and relations of PU (3, 1; $\mathbb{Z}$ [ $\omega$ ])

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 $A_{a}(\gamma p_{b})=W(p_{c})$ .

Lemma 4.1. $M_1$ commutes with $I_0, A_{31}, A_{32}, A_{41}, A_{42}, A_{43}$ , and $M_2$ commutes with $I_0$ , $A_{31}$ , $A_{32}$ , $A_{41}$ .

Proof. All of these can be computed directly.

Lemma 4.2. Suppose that $A_{a}(\gamma p_{b})=W(p_{c})$ and $A_{c}^{-1}W^{-1}A_{a}\gamma A_{b}=W^{\prime}$ . Let $M\in \langle M_1, M_2 \rangle$ .

  1. 1. If $A_a M=M A_a$ , then $A_{a}(M\gamma p_{b})=MW(p_{c})$ and $A_{c}^{-1}W^{-1}M^{-1}A_{a}M\gamma A_{b}=W^{\prime}$ .

  2. 2. If $A_b M=M A_b$ , then $A_{a}(\gamma M p_{b})=W(p_{c})$ and $A_{c}^{-1}W^{-1}A_{a}\gamma M A_{b}=W^{\prime}M$ .

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*}

Lemma 4.3. Suppose that $A_{a}(\gamma _{i} p_{b})=W_{i}(p_{c})$ and $A_{c}^{-1}W_{i}^{-1}A_{a}\gamma _{i} A_{b}=W^{\prime}_{i}$ with $i=1,2$ . Let $\gamma _{2}=M\gamma _1 M^{-1}$ with $M\in \langle M_1,M_2 \rangle$ .

  1. 1. If $M$ commutes with $A_a,A_b$ and $A_c$ , then $W_2=MW_1M^{-1}$ and $W^{\prime}_2=MW^{\prime}_1M^{-1}$ .

  2. 2. If $M$ commutes with $A_a,A_b$ , but not $A_c$ , then $W_2=MW_1$ and $W^{\prime}_2=W^{\prime}_1M^{-1}$ .

Moreover, in both of the two cases, the relation coming from $\gamma _2$ can be derived from the relation coming from $\gamma _1$ .

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 116.

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 $p_b=p_{43}$

Remark 4.5. One can see that $A_{31}=(T_0I_0)^{-2}(M_1M_2^3)^{2}$ , $A_{32}=(T_0I_0)^{2}(M_1M_2^3)^{2}$ , and $A_{41}=(T_0I_0)^3$ from Table 1. From Tables 7 to 8, one can obtain that $A_{42}=I_0T_5I_0$ and $A_{43}=T_0I_0T_6I_0$ , here $T_5=T_{0}^{-2}T_1T_2T_3T_4$ , $T_6=T_{0}^2T_{3}^{-1}T_{4}^{-1}$ .

Remark 4.6. In fact, $\Gamma$ has a presentation with 15 generators and 591 relations. One can simplify the presentation in GAP(via the command IsomorphismSimplifiedFpGroup ) to get a new presentation with 4 generators and 511 relations. The details of this presentation are available there [Reference Wang and Xie15].

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

Corollary 4.7. The abelianization $\Gamma ^{\prime}$ of $\Gamma$ is $\mathbb{Z}/2\mathbb{Z}$ .

Acknowledgement

The authors are grateful to Prof. John Parker for his valuable comments and suggestions. The authors would like to thank the anonymous referee, whose constructive suggestions helped improving earlier versions of the manuscript. This work was supported by NSFC (No.11701165, No.11871202, No.12271148).

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Figure 0

Figure 1. A fundamental domain $\triangle$ for the group of orientation-preserving symmetries of $\mathbb{Z}[\omega ]\subset \mathbb{C}$.

Figure 1

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 $A_a=I_0$, $p_b=p_0, p_{31},p_{32},p_{41}$

Figure 2

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 $A_a=I_0$, $p_b=p_{42}$

Figure 3

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 $A_a=I_0$, $p_b=p_{43}$

Figure 4

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 $A_a=A_{31}$

Figure 5

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 $A_a=A_{32}$

Figure 6

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 $A_a=A_{41}$

Figure 7

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 $A_a=A_{42}$

Figure 8

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 $A_a=A_{43}$

Figure 9

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 $p_b=p_{42}$

Figure 10

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 $p_b=p_{42}$

Figure 11

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 $p_b=p_{43}$

Figure 12

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 $p_b=p_{43}$

Figure 13

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 $p_b=p_{42}$

Figure 14

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 $p_b=p_{42}$

Figure 15

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 $p_b=p_{43}$

Figure 16

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 $p_b=p_{43}$