2·1. The complex hyperbolic plane

Consider the Hermitian matrix

\begin{equation*}H=\left[\begin{array}{c@{\quad}c@{\quad}c}0 & 0 & 1 \\0 & 1 & 0 \\1 & 0 & 0 \\\end{array}\right]\!.\end{equation*}

The Hermitian form on $\mathbb{C}^3$ associated with H is given by $\langle \mathbf{z}, \mathbf{w} \rangle = \mathbf{w}^* H \mathbf{z}$ . This form has signature (2,1). Then ${\mathbb{C}}^3\setminus{0}$ is the disjoint union of the negative cone $V_{-}$ , the null cone $V_{0}$ , and the positive cone $V_{+}$ , where

\begin{eqnarray*} V_{-} &=& \{ {\bf{z}}\in {\mathbb{C}}^3\;:\; \langle {\bf{z}}, {\bf{z}} \rangle \lt0 \}, \\ V_{0} &=& \{ {\bf{z}}\in {\mathbb{C}}^3\;:\; \langle {\bf{z}}, {\bf{z}} \rangle =0 \}\backslash\{0\} , \\ V_{+} &=& \{ {\bf{z}}\in {\mathbb{C}}^3\;:\; \langle {\bf{z}}, {\bf{z}} \rangle \gt 0 \}.\end{eqnarray*}

Definition 2·1. Let $P\;:\; \mathbb{C}^3\backslash\{0\} \longrightarrow \mathbb{C}P^2$ be the natural projectivisation. The complex hyperbolic plane $\mathbf{H}^2_{\mathbb{C}}$ is defined to be $P(V_{-})$ , and its boundary $\partial\mathbf{H}^2_{\mathbb{C}}$ is $P(V_{0})$ . The Bergman metric on $\mathbf{H}^2_{\mathbb{C}}$ is given by the distance formula

(2·1) \begin{equation} \cosh^2\left(\frac{d(u,v)}{2}\right)=\frac{\langle {\bf{u}}, {\bf{v}} \rangle\langle {\bf{v}}, {\bf{u}} \rangle}{\langle {\bf{u}}, {\bf{u}} \rangle \langle {\bf{v}}, {\bf{v}} \rangle}, \end{equation}

where ${\bf{u}}, {\bf{v}} \in {\mathbb{C}}^3$ are lifts of u,v. Here d(u,v) is the Bergman distance between two points $u,v \in \mathbf{H}^2_{\mathbb{C}}$ .

This model of $\mathbf H^{2}_{\mathbb C}$ is also called the Siegel domain. Let $q_{\infty}$ be the point at infinity, and let $(z_1,z_2)^T$ be a point of $\overline {\mathbf{H}^2_{\mathbb{C}}}=\mathbf{H}^2_{\mathbb{C}} \cup \partial \mathbf{H}^2_{\mathbb{C}}$ . We use $\mathbf{q}_{\infty}=(1,0,0)^T$ to denote the lift of $q_{\infty}$ , and $(z_1,z_2,1)^T$ as the standard lift of $(z_1,z_2)^T$ .

Let $\mathbf{U}(2,1) = \{ A \in \mathbf{GL}(3,\mathbb{C}) \mid A^* H A = H \}$ , and $\mathbf{PU}(2,1) = \mathbf{U}(2,1) / \mathbf{U}(1)$ . Let $\omega$ be a primitive cube root of unity. Then $\mathbf{PU}(2,1)=\mathbf{SU}(2,1)/\langle I,\omega I,\omega^2I\rangle$ , which is the holomorphic isometry group of ${\bf H}^2_{\mathbb C}$ .

Elements of $\mathbf{SU}(2,1)$ fall into three types. An isometry is loxodromic if it has exactly two fixed points on $\partial \mathbf{H}^2_{\mathbb{C}}$ and parabolic if it has exactly one fixed point on $\partial \mathbf{H}^2_{\mathbb{C}}$ . An isometry is called elliptic if it has at least one fixed point in ${\bf H}^2_{\mathbb C}$ . An elliptic element is called regular elliptic if it has three distinct eigenvalues, otherwise, it is called special elliptic.

The trace of a matrix in $\mathbf{SU}(2,1)$ determines the type of the isometry; see Goldman [ Reference Goldman9 , theorem 6·2·4]. Let $A \in \mathbf{SU}(2,1)$ be a non-identity element with real trace. Then A is elliptic if $-1\leq{\textrm{tr}(A)}\lt3$ . In particular, if ${\textrm{tr}(A)}=-1,0,1$ , A is elliptic of order 2, 3 and 4, respectively. Moreover, A is unipotent if ${\textrm{tr}(A)}=3$ .

Let $\iota$ be given on the level of homogeneous coordinates by complex conjugation $(z_1,z_2,z_3)^T\longmapsto(\overline{z_1},\overline{z_2},\overline{z_3})^T$ , which is an anti-holomorphic isometry of $\mathbf{H}^2_{\mathbb{C}}$ . Any other anti-holomorphic isometry may be written as the composition of $\iota$ and an element of $\mathbf{U}(2,1)$ .

2·2. Totally geodesic submanifolds and complex reflections

There are several types of totally geodesic submanifolds in $\mathbf H^{2}_{\mathbb C}$ , classified by their dimension: points (0-dim), real geodesics (1-dim), complex lines (2-dim), Lagrangian planes (2-dim), and the entire space (4-dim). Complex lines are the intersections of projective lines in $\mathbb{C}P^2$ with $\mathbf{H}^2_{\mathbb{C}}$ , which are also called complex geodesics. Lagrangian planes are the images of ${\bf H}^2_{\mathbb R}$ under the action of $\mathbf{PU}(2,1)$ . As the Riemannian sectional curvature of $\mathbf{H}^2_{\mathbb{C}}$ is non-constant, there is no totally geodesic hypersurface in $\mathbf{H}^2_{\mathbb{C}}$ .

The ideal boundary in $\partial\mathbf H^{2}_{\mathbb C}$ of a complex line is called a $\mathbb{C}$ -circle, and that of a Lagrangian plane is called an $\mathbb{R}$ -circle. Let C be a complex line. A polar vector of C is the unique positive vector up to scalar multiplication, perpendicular to C with respect to the Hermitian form. Each positive vector corresponds to a complex line.

There is a special class of elliptic elements of order two in $\mathbf{PU}(2,1)$ related to complex lines.

Definition 2·2. The complex reflection on a complex line C with polar vector ${\bf{n}}$ is a holomorphic isometry of order 2, fixing C pointwise, defined by:

(2·2) \begin{equation} I_{C}({\bf{z}})=-{\bf{z}}+\frac{2\langle {\bf{z}}, {\bf{n}} \rangle}{\langle {\bf{n}}, {\bf{n}} \rangle} {\bf{n}}. \end{equation}

The complex line C is usually called the mirror of $I_{C}$ .

2·3. The Heisenberg group

Let $\mathcal{N}=\mathbb{C}\times \mathbb{R}$ be the Heisenberg group with product

\begin{equation*} (z,t)\cdot(\zeta,\nu)=(z+\zeta,t+\nu+2\Im(z\bar{\zeta})).\end{equation*}

Then $\overline{\mathbf{H}^2_{\mathbb{C}}}$ can be identified with ${\mathcal{N}}\times{\mathbb{R}_{\geq0}}\cup \{q_{\infty}\}$ . Any point $q=(z,t,u)\in{\mathcal{N}}\times{\mathbb{R}_{\geq0}}$ has the standard lift

\begin{equation*}{\bf{q}}=\begin{bmatrix}\frac{-|z|^2-u+it}{2} \\[3pt] z \\[3pt] 1 \\\end{bmatrix}.\end{equation*}

Here (z, t, u) is called the horospherical coordinates of $\overline {\mathbf{H}^2_{\mathbb{C}}}$ .

When $u=0$ , the points are on $\partial\mathbf{H}^2_{\mathbb{C}}$ . Hence, we directly identify $\partial\mathbf{H}^2_{\mathbb{C}}$ with the union $\mathcal{N}\cup \{q_{\infty}\}$ . Fixed $u_0 \gt 0$ , the level set $u=u_0$ is called the horosphere based at $q_{\infty}$ , and the sup-level set $u\ge u_0$ is called the horoball based at $q_{\infty}$ .

The full stabiliser of $\mathbf{q}_{\infty}$ is generated by the isometries of the forms

\begin{equation*}T_{[z,t]}=\left[\begin{array}{c@{\quad}c@{\quad}c}1 & -\overline{z}& \frac{-|z|^{2}+it}{2} \\ 0 & 1 & z \\ 0 & 0 & 1 \end{array}\right],\ R_{\theta}=\left[\begin{array}{c@{\quad}c@{\quad}c}1 & 0& 0 \\ 0 & e^{i\theta} & 0 \\ 0 & 0 & 1 \end{array}\right] \ \text{and} \ D_{\lambda}=\left[\begin{array}{c@{\quad}c@{\quad}c}\lambda & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1/\lambda \end{array}\right],\end{equation*}

where $[z,t]\in\mathcal{N}$ , $\theta,\lambda\in \mathbb{R}$ and $\lambda \neq 0$ . They are called Heisenberg translation, Heisenberg rotation and Heisenberg dilation, respectively. Their actions on $\mathcal{N}$ are as follows:

\begin{equation*} \begin{aligned} T_{(z,t)}\;:\; (\zeta,\nu)&\longmapsto (z+\zeta,t+\nu+2{\textrm{Im}}(z\bar{\zeta}));\\ R_{\theta}\;:\; (\zeta,\nu)&\longmapsto (e^{i\theta}\zeta,\nu);\\ D_{\lambda}\;:\; (\zeta,\nu)&\longmapsto(\lambda \zeta,\lambda^2 \nu). \end{aligned}\end{equation*}

The Cygan metric on $\mathcal{N}$ for $p=(z,t)$ and $q=(w,s)\in\mathcal{N}$ is defined by:

(2·3) \begin{equation} d_{\textrm{Cyg}}(p,q)=|2\langle {\bf{p}}, {\bf{q}} \rangle|^{1/2}= | |z-w|^2-i(t-s+2\Im(z\bar{w})) |^{1/2}. \end{equation}

In fact, it is the restriction to $\mathcal{N}$ of the extended Cygan metric, defined for $p=(z,t,u)$ and $q=(w,s,v)\in\mathcal{N}\times{\mathbb{R}_{\geq 0}}$ by the formula

(2·4) \begin{equation} d_{\textrm{Cyg}}(p,q)= | |z-w|^2+|u-v|-i(t-s+2\Im(z\bar{w})) |^{1/2}. \end{equation}

The Cygan sphere with center $(z_0,t_0)\in\mathcal{N}$ and radius $r\gt 0$ in $\overline {\mathbf{H}^2_{\mathbb{C}}}$ , is defined by

(2·5) \begin{equation} S_{[z_0,t_0]}(r)=\big\{ (z,t,u)\in\overline {\mathbf{H}^2_{\mathbb{C}}}\;:\; \big||z-z_0|^2+u+i(t-t_0+2\Im( z\overline{z_0}))\big|=r^2 \big\}.\end{equation}

Note that $S_{[z_0,t_0]}(r)$ can be obtained by the action of $T_{[z_0,t_0]}$ on $S_{[0,0]}(r)$ . There is another way to define a Cygan sphere:

Definition 2·3. The geographic coordinates $(\alpha,\beta,w)$ of $q=q(\alpha,\beta,w)\in S_{[0,0]}(r)$ is given by the lift

(2·6) \begin{equation} {\bf{q}}={\bf{q}}(\alpha,\beta,w)=\left[ \begin{array}{c} -r^2e^{-i\alpha}/2 \\[2pt] rwe^{i({-}\alpha/2+\beta)} \\[2pt] 1 \end{array} \right], \end{equation}

where $\alpha\in [-\pi/2,\pi/2]$ , $\beta\in [0, \pi)$ and $w\in [-\sqrt{\cos(\alpha)},\sqrt{\cos(\alpha)}]$ . In particular, the ideal boundary of $S_{[0,0]}(r)$ on $\partial\mathbf{H}^2_{\mathbb{C}}$ are the points with $w=\pm\sqrt{\cos(\alpha)}$ .

The following proposition describes the characteristics of the level sets for $\alpha$ and $\beta$ .

Regarding the intersection of Cygan spheres, Goldman, Parker and Will proved the following result.

2·4. Isometric spheres and Ford polyhedron

Let $g=(g_{ij})$ be an element of $\mathbf{PU}(2,1)$ not fixing $\mathbf{q}_{\infty}$ . This implies that $g_{31}\neq 0$ .

Definition 2·7. The isometric sphere of g, denoted by I(g), is the set

(2·7) \begin{equation} I(g)=\left\{ p \in\overline{\mathbf{H}^2_{\mathbb{C}}}\;:\; |\langle {\bf{p}}, {\bf{q}}_{\infty} \rangle | = |\langle {\bf{p}}, g^{-1}({\bf{q}}_{\infty}) \rangle| \right\}. \end{equation}

An isometric sphere is a 3-ball. The spinal sphere of I(g) or g is defined to be $I(g)\cap\partial\mathbf{H}^2_{\mathbb{C}}$ , denoted by $\partial_{\infty}I(g)$ , which is a 2-sphere. Besides, I(g) is exactly the Cygan sphere with center $g^{-1}({\bf{q}}_{\infty})\in\mathcal{N}$ and radius $r_g=\sqrt{2/|g_{31}|}$ . The interior of I(g) is the set

(2·8) \begin{equation}\left\{ p \in \overline{\mathbf{H}^2_{\mathbb{C}}} \;:\; |\langle {\bf{p}}, {\bf{q}}_{\infty} \rangle | \gt |\langle {\bf{p}}, g^{-1}({\bf{q}}_{\infty}) \rangle| \right\}.\end{equation}

The exterior of I(g) is the set

(2·9) \begin{equation}\left\{ p \in \overline{\mathbf{H}^2_{\mathbb{C}}}\;:\; |\langle {\bf{p}}, {\bf{q}}_{\infty} \rangle | \lt |\langle {\bf{p}}, g^{-1}({\bf{q}}_{\infty}) \rangle| \right\}.\end{equation}

We summarise some properties on isometric spheres in the following proposition.

Definition 2·9. The Ford domain $D_{\Gamma}$ for a discrete group $\Gamma \lt \mathbf{PU}(2,1)$ centred at $q_{\infty}$ is the intersection of the closures of the exteriors of all isometric spheres of elements in $\Gamma$ not fixing $\mathbf{q}_{\infty}$ . That is,

\begin{equation*}D_{\Gamma}=\left\{p\in \overline{\mathbf{H}^2_{\mathbb{C}}}\;:\; |\langle \mathbf{p},\mathbf{q}_{\infty}\rangle|\leq|\langle \mathbf{p},g^{-1}(\mathbf{q}_{\infty})\rangle| \ \forall g\in \Gamma \ \text{with} \ g(\mathbf{q}_{\infty})\neq \mathbf{q}_{\infty} \right\}.\end{equation*}

The boundary at infinity of the Ford domain is made up of pieces of isometric spheres. When $q_{\infty}$ is in the domain of discontinuity or is a parabolic fixed point, the Ford domain is preserved by $\Gamma_{\infty}$ , the stabiliser of $q_{\infty}$ in $\Gamma$ . In this case, the fundamental domain for $\Gamma$ is the intersection of $D_{\Gamma}$ with a fundamental domain for $\Gamma_{\infty}$ . Facets of codimensions one, two, three and four in $D_{\Gamma}$ are called sides, ridges, edges and vertices, respectively. Moreover, a bounded ridge is a ridge that does not intersect $\partial {\bf H}^2_{\mathbb C}$ . Otherwise, the ridge is called an infinite ridge.

It is usually very hard to determine the Ford domain $D_{\Gamma}$ because one should check infinitely many inequalities. A common approach is to propose a candidate for the Ford polyhedron and then verify it using the Poincaré polyhedron theorem. The sides of $D_{\Gamma}$ should be paired by isometries, and the images of $D_{\Gamma}$ under these so-called side-pairing maps should give a local tiling of ${\bf H}^2_{\mathbb C}$ . If they do, and if, moreover, the quotient of $D_{\Gamma}$ by the identification given by the side-pairing maps is complete, the Poincaré polyhedron theorem implies that the images of $D_{\Gamma}$ give a global tiling of $\mathbf{H}^2_{\mathbb{C}}$ .

Once a fundamental domain is obtained, one gets a presentation of $\Gamma$ . Its generators are given by the side-pairing maps together with a generating set of $\Gamma_{\infty}$ . Its relations come from ridge cycles, which correspond to the local tilings around each codimension-two face. For more details on the Poincaré polyhedron theorem, see [ Reference Deraux, Parker and Paupert6, Reference Parker and Will20 ].

3·1. The matrix representation of the triangle group $\Delta_{4,\infty,\infty;\infty}$

Suppose $J_1$ and $J_2$ are two complex reflections in $\mathbf{SU}(2,1)$ such that $J_1J_2$ is a unipotent map fixing $q_{\infty}$ . Conjugating by a Heisenberg rotation and a Heisenberg dilation if necessary, we may suppose that $J_1$ and $J_2$ are

(3·1) \begin{equation}J_1=\left[\begin{array}{c@{\quad}c@{\quad}c}-1 & 0& 0 \\ 0 & 1 & 0\\ 0 & 0 & -1 \end{array}\right], \quad J_2=\left[\begin{array}{c@{\quad}c@{\quad}c}-1 & -2 & 2 \\ 0 & 1 & -2 \\ 0 & 0 & -1 \end{array}\right].\end{equation}

We want to find $J_3$ such that $J_2J_3$ is an elliptic element of order 4, and $J_3J_1$ , $J_1J_2$ , $J_1J_3J_2J_3$ are unipotent. Assume that the polar vector of the mirror of $J_3$ is

\begin{eqnarray*} {\bf n}_3 = \left[\begin{matrix} x \\ y+z i \\ 1\end{matrix}\right]\!,\end{eqnarray*}

where $x,y,z \in \mathbb{R}$ . Then $J_3$ has the following form

\begin{equation*}J_3=\left[\begin{array}{c@{\quad}c@{\quad}c}\dfrac{-y^2-z^2}{y^2+z^2+2x}& \dfrac{2x(y- z i)}{y^2+z^2+2x}& \dfrac{2x^2}{y^2+z^2+2x} \\[9pt] \dfrac{2(y+z i)}{y^2+z^2+2x} &\dfrac{y^2+z^2-2x}{y^2+z^2+2x}&\dfrac{2x(y+ z i)}{y^2+z^2+2x} \\[9pt] \dfrac{2}{y^2+z^2+2x} & \dfrac{2(y- z i)}{y^2+z^2+2x} & \dfrac{-y^2-z^2}{y^2+z^2+2x}\end{array}\right]\!.\end{equation*}

Note that $J_3J_1$ is unipotent, which implies that its trace is 3, yielding $x=0$ . Similarly, since $J_1J_2$ and $J_1J_3J_2J_3$ are unipotent, we have $y=1$ and $z= \pm 1$ . The groups corresponding to $z= \pm 1$ are conjugate via an anti-holomorphic isometry. In this paper, we take $z=1$ , then

\begin{equation*}J_3=\left[\begin{array}{c@{\quad}c@{\quad}c}-1& 0& 0 \\ 1+ i& 1 &0 \\1 & 1-i & -1 \end{array}\right]\!.\end{equation*}

Moreover, $\langle J_1, J_2,J_3 \rangle $ is an infinite-index subgroup of the Gauss-Picard modular group $\mathbf{PU}(2,1;\mathbb{Z}[i])$ , proving that $\langle J_1,J_2,J_3\rangle $ is discrete; see [ Reference Falbel, Francsics and Parker7 ]. We thank the referee for pointing out that our results may be derived from the Ford domain of the Gauss-Picard modular group. The discreteness of $\Delta_{4,\infty,\infty;\infty}$ was first noted by Thompson [ Reference Thompson28 ].

For technical reasons, we conjugate the matrix representation of $\Delta_{4,\infty,\infty;\infty}$ by

\begin{equation*}T=\left[\begin{array}{c@{\quad}c@{\quad}c} 1& 1& \frac{-1-2 i}{2} \\[2pt] 0 &1&-1 \\ 0 & 0 & 1\end{array}\right]\!.\end{equation*}

That is, let $I_1=T J_1T^{-1}$ , $I_2=T J_2T^{-1}$ , and $I_3=T J_3T^{-1}$ be generators of $\Delta_{4,\infty,\infty;\infty}$ . Specifically:

(3·2) \begin{equation} I_1=\left[\begin{array}{c@{\quad}c@{\quad}c} -1 & 2& 2 \\ 0 & 1 & 2\\ 0 & 0 & -1 \end{array}\right]\!, \quad I_2=\left[\begin{array}{c@{\quad}c@{\quad}c} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{array}\right]\!,\quad I_3=\left[\begin{array}{c@{\quad}c@{\quad}c} -1/2 & -i/2& 1/4 \\[2pt] i & 0 &i/2 \\[2pt] 1 & -i & -1/2\end{array}\right]\!. \end{equation}

3·2. Ford domain of $\Delta_{4,\infty,\infty;\infty}$

Let $\Gamma=\langle I_1, I_2, I_3\rangle$ be the group $\Delta_{4,\infty,\infty;\infty}$ . Let $A=I_1I_2$ , $B=I_2I_3$ . The subgroup generated by A and B, denoted by $\Sigma=\langle A, B\rangle$ , is the even subgroup of $\Gamma$ with index two. By direct calculation, we have the following.

\begin{equation*}A=\left[\begin{array}{c@{\quad}c@{\quad}c}1 & 2& -2 \\ 0 & 1 & -2 \\ 0 & 0 & 1 \end{array}\right]\!, \quad B=\left[\begin{array}{c@{\quad}c@{\quad}c}\frac{1}{2} &\frac{i}{2}&-\frac{1}{4} \\[2pt] i& 0&\frac{i}{2}\\[2pt] -1 &i &\frac{1}{2} \end{array}\right]\!.\end{equation*}

Due to the definition of $\Delta_{4,\infty,\infty;\infty}$ , we have that B is elliptic of order 4, and there are three unipotent elements, A, AB, and $AB^2$ . In addition, $AB^{-1}$ is also unipotent. For future reference, we provide the lifts of their fixed points, as vectors in $\mathbb{C}^{3}$ :

\begin{equation*}\mathbf{p}_{AB}=x_1=\left[\begin{matrix}-\frac{1+2i}{2} \\[2pt] -1 \\ 1 \end{matrix}\right], \ \mathbf{ p}_{AB^2}=x_2=\left[\begin{matrix}-\frac{1}{2} \\[2pt] -1 \\ 1 \end{matrix}\right] \ \text{and} \ \mathbf{ p}_{AB^{-1}}=x_3=\left[\begin{matrix}\frac{-1+2i}{2}\\[2pt] -1\\ 1\end{matrix}\right]\end{equation*}

In this subsection, we will construct a Ford domain $D_{\Sigma}$ whose sides are contained in the isometric spheres of some elements of $\Sigma$ . The reader is referred to Figures 2 and 3 for visualisations of $D_{\Sigma}$ . Figure 2 provides a realistic depiction of the ideal boundary of $D_{\Sigma}$ , centered at $q_{\infty}$ , for motivational purposes. In this subsection, we will verify that the local combinatorial structure matches this figure.

Figure 2.

A realistic view the ideal boundary of the Ford domain of $\Delta(4, \infty,\infty;\infty)$ .

Figure 3.

An abstract picture of the Ford domain of $\Sigma$ . For example, the sphere labelled by B is the spinal sphere $\partial_{\infty}I(B)$ , in the notation of Proposition 3·4, it is $\partial_{\infty}I^{+}_0$ .

Figure 3 Long description

A diagram of the Ford domain of the complex hyperbolic triangle group Delta(4,∞,∞;∞) acting on the complex hyperbolic plane H^2_C. The diagram includes several labeled spheres and points connected by lines. The central sphere is labeled B, representing the spinal sphere. Surrounding B are other spheres labeled A^-1B^-1A, AB^-1A^-1, A^-1BA, B^-1, and ABA^-1. Points such as x1, x2, x3, y1, w1, w2, w3, and w4 are marked with blue dots and connected by lines indicating relationships and transformations. Arrows and lines illustrate the interactions and transformations between these points and spheres, showing the geometric structure and relationships within the Ford domain.

In Figure 2:

1. (i) the green, black and red spheres are the spinal spheres of $A^{-1}BA$ , B and $ABA^{-1}$ , respectively;

2. (ii) the blue, purple and steel-blue spheres are the spinal spheres of $A^{-1}B^{-1}A$ , $B^{-1}$ and $AB^{-1}A^{-1}$ , respectively;

3. (iii) the cyan, yellow and pink spheres are the spinal spheres of $A^{-1}B^{2}A$ , $B^{2}$ and $AB^{2}A^{-1}$ respectively.

Figure 3 is a combinatorial model of Figure 2. Our subsequent computations will show that the combinatorial model is consistent with Figure 2.

Let I(B), $I(B^{-1})$ and $I(B^2)$ be the isometric spheres for B, $B^{-1}$ and $B^2$ , respectively. We give some notation for the isometric spheres, which are the images of I(B), $I(B^{-1})$ , and $I(B^{2})$ by powers of A.

The action of A on $\mathcal{N}$ is given by

(3·3) \begin{equation}(z,t)\longmapsto (z-2, t+4\Im(z)).\end{equation}

In particular, we have the following.

As A is unipotent and fixes $q_{\infty}$ , it is a Cygan isometry and preserves the radii of the isometric spheres. This directly proves

We first consider the intersections of the isometric spheres $I_{k}^{\pm}$ and $I^{\star}_k$ for $k\in \mathbb{Z}$ . By the A-symmetry, $I_{l}^{+} \cap I_{l+k}^{+}$ is non-empty if and only if $I_{0}^{+} \cap I_{k}^{+}$ is non-empty. Similar properties hold for other intersections of pairs of isometric spheres.

Proof. The assertion (i) follows from the fact that the Cygan distance between the centers of $I_{0}^{+}$ and $I_{k}^{+}$ is $|2k|$ , which is greater than $2 \sqrt{2}$ when $|k| \geq 2$ . Similarly, we have the assertion (ii).

For (iii), the Cygan distance between the centers of $I_{0}^{+}$ and $I_{k}^{-}$ is $2 \sqrt{k^2+1}$ , which is greater than $2 \sqrt{2}$ when $|k| \geq 2$ . Hence, $I_{0}^{+}$ is disjoint from $I_{k}^{-}$ if $|k| \geq 2$ . When $k= \pm 1$ , we assume that

\begin{equation*}(x+y i,t) \in \partial_{\infty} I_{0}^{+} \cap \partial_{\infty} I_{ \pm 1}^{-}\end{equation*}

in Heisenberg coordinates with $x,y,t \in \mathbb{R}$ . Then considering the equation of an isometric sphere in the Definition 2·7, it is easy to show that there is no $(x+y i,t)$ lying in $\partial_{\infty} I_{0}^{+} \cap \partial_{\infty} I_{ \pm 1}^{-}$ . This implies that $I_{0}^{+} \cap I_{ \pm 1}^{-} = \emptyset$ .

The assertion (iv) follows from the fact that the Cygan distance between the centers of $I^{\star}_{0}$ and $I^{\star}_{k}$ is $|2k|$ , which is greater than 2 when $|k| \geq 2$ . The assertion (v) follows from the fact that the Cygan distance between the centers of $I^{\star}_{0}$ and $I^{\pm }_{k}$ is $\sqrt{16k^4+24k^2+1}$ , which is greater than $\sqrt{2}+1$ when $|k| \geq 1$ .

For a unipotent element $g\in \mathbf{PU}(2,1)$ , the isometric spheres I(g) and $I(g^{-1})$ are tangent, where the tangent point is exactly the fixed point of g. The assertions (vi), (vii), (viii) follow from the fact that AB, $AB^{-1}$ and $AB^2$ are unipotent.

Definition 3·5. The infinite polyhedron $D_{\Sigma}$ is the intersection of the exteriors of all the isometric spheres in $\{I_{k}^{+}, I_{k}^{-},I^{\star}_{k}\;:\; k\in \mathbb{Z}\}$ . That is,

\begin{equation*}D_{\Sigma}=\{p\in \mathbf{H}^2_{\mathbb{C}}\;:\; |\langle \mathbf{p}, \mathbf{q}_{\infty}\rangle| \leq |\langle \mathbf{p}, A^{k}B^i(\mathbf{q}_{\infty})\rangle|\text{, } \forall k\in \mathbb{Z} \ \text{and} \ i= 1, 2, 3\}.\end{equation*}

We shall use a version of the Poincaré polyhedron theorem for coset decompositions because $D_{\Sigma}$ is stabilised by the cyclic subgroup $\langle A\rangle$ . See [ Reference Deraux, Parker and Paupert6, Reference Parker and Will20 ] for the precise statement of this version of the Poincaré polyhedron theorem. We proceed to show:

Theorem 3·6. The polyhedron $D_{\Sigma}$ is a fundamental domain for the cosets of $\langle A \rangle$ in $\Sigma=\langle A, B \rangle$ . Moreover, the group $\Sigma$ is discrete and has the presentation

\begin{equation*} \langle A, B\;:\; B^4=id\rangle. \end{equation*}

For any $k \in \mathbb{Z}$ , we denote $I_{k}^{+} \cap D_{\Sigma}$ , $I_{k}^{-} \cap D_{\Sigma}$ , and $I^{\star}_{k} \cap D_{\Sigma}$ by $s_{k}^{+}$ , $s_{k}^{-}$ , and $s^{\star}_{k}$ , respectively. We will show that they are 3-faces of $D_{\Sigma}$ . Let

\begin{equation*}w_1=\left[\begin{matrix}{-\frac{1+2\sqrt{2}i}{6}} \\[3pt]{-\frac{\sqrt{2}+i}{3}}\\[2pt]1 \end{matrix}\right],w_2=\left[\begin{matrix}{-\frac{1-2\sqrt{2}i}{6}} \\[3pt]{-\frac{\sqrt{2}-i}{3}}\\[2pt]1\end{matrix}\right],w_3=\left[\begin{matrix}{-\frac{1-2\sqrt{2}i}{6}} \\[3pt]{\frac{\sqrt{2}-i}{3}}\\[2pt]1 \end{matrix}\right]\ \text{and}\, w_4=\left[\begin{matrix}{-\frac{1+2\sqrt{2}i}{6}} \\[3pt]{\frac{\sqrt{2}+i}{3}} \\[2pt]1 \end{matrix}\right]\end{equation*}

be four points. Moreover, one can verify that $B(w_1)=w_4$ , $B(w_4)=w_3$ , $B(w_3)=w_2$ , and $B(w_2)=w_1$ .

Proof. First note that the fixed point of B in ${\bf H}^2_{\mathbb C}$ is

\begin{equation*}\mathbf{p}_{B}=\left[\begin{matrix}-1/2 \\ 0 \\ 1 \end{matrix}\right]\!.\end{equation*}

A direct computation shows that $\mathbf{p}_{B}$ lies in $I^{+}_0$ , $I^{-}_0$ , and $I^{\star}_0$ . Moreover, $q_{\infty}$ , $B(q_{\infty})$ and $B^2(q_{\infty})$ are linearly independent. By Proposition 2·5, $I_{0}^{-} \cap I^{\star}_{0}$ is a non-empty Giraud disk. Similarly, each of $I_{0}^{+} \cap I^{-}_{0}$ and $I_{0}^{+} \cap I^{\star}_{0}$ is a non-empty Giraud disk.

We now consider the triple intersection $I_{0}^{+}\cap I^{-}_{0}\cap I^{\star}_{0}$ . From (2.6), the geographic coordinates $(\alpha,\beta,w)$ of points in $I^{\star}_0$ are given by

(3·4) \begin{equation} {\bf{q}}={\bf q}(\alpha,\beta,w)=\left[ \begin{array}{c} -e^{-i\alpha}/2 \\[3pt] we^{i({-}\alpha/2+\beta)} \\ 1 \end{array} \right]\!, \end{equation}

where $\alpha\in [-\pi/2,\pi/2]$ , $\beta\in [0, \pi)$ and $w\in [-\sqrt{\cos(\alpha)},\sqrt{\cos(\alpha)}]$ . Since ${\bf{q}}\in I^{\star}_0$ , we have

\begin{equation*}|\langle {\bf{q}}, (q_{\infty}) \rangle |^2 =|\langle {\bf{q}}, B^{-2} (q_{\infty}) \rangle |^2=1.\end{equation*}

We calculate $|\langle {\bf{q}}, B^{-1} (q_{\infty}) \rangle |^2 $ to be

(3·5) \begin{align} \begin{aligned} w^2+\frac{1+\cos(\alpha)}{2}+w\left(\sin\left(\frac{\alpha-2 \beta}{2}\right)-\sin\left(\frac{\alpha+2 \beta}{2}\right)\right) \end{aligned} \end{align}

and $|\langle {\bf{q}}, B (q_{\infty}) \rangle |^2 $ to be

(3·6) \begin{align} \begin{aligned} w^2+\frac{1+\cos(\alpha)}{2}-w\left(\sin\left(\frac{\alpha-2 \beta}{2}\right)-\sin\left(\frac{\alpha+2 \beta}{2}\right)\right). \end{aligned} \end{align}

Then the triple intersection $I_{0}^{+} \cap I^{-}_{0} \cap I^{\star}_{0}$ is given by

(3·7) \begin{align} \left\{ \begin{aligned} w^2+\frac{1+\cos(\alpha)}{2}=1; \qquad & \\[4pt] w\left(\sin\left(\frac{\alpha-2 \beta}{2}\right)-\sin\left(\frac{\alpha+2 \beta}{2}\right)\right)=0. \qquad& \end{aligned} \right. \end{align}

The solutions are

\begin{equation*}\alpha=\pi, \quad \text{or}\quad w=0, \quad \text{or}\quad \beta=0.\end{equation*}

Since $\alpha\in [-\pi/2,\pi/2]$ , we know that $\alpha=\pi$ is not a solution of our equation. If $w=0$ , then $\alpha= 0$ , which corresponds to the point $\mathbf{p}_{B}$ . If $\beta=0$ , the solutions lie in a meridian of $I^{\star}_0$ . In this case, the set ${\bf{q}}(\alpha,0,w)$ is a union of two geodesics which intersect at $\mathbf{p}_{B}$ , where $w = \pm \sqrt{{(1-\cos(\alpha))}/{2}}$ .

Moreover, in the case of $\beta=0$ , we consider the intersection of $I_{0}^{+}\cap I^{-}_{0}\cap I^{\star}_{0}$ with $\partial{\bf H}^2_{\mathbb C}$ . Note that $|\langle {\bf{q}}, {\bf{q}} \rangle |^2 =w^2-\cos(\alpha)$ . If $\bf{q}\in\partial{\bf H}^2_{\mathbb C}$ , we have $w=\pm \sqrt{\cos(\alpha)}$ . Hence the equation $\pm \sqrt{\cos(\alpha)}= \pm \sqrt{{(1-\cos(\alpha))}/{2}}$ gives the intersection point. Specifically,

\begin{equation*}\alpha= \pm \arccos\left(\frac{1}{3}\right)\!,\;\;\beta=0,\;\;\; w= \pm \frac{1}{3^{1/2}}.\end{equation*}

The coordinates of the points $w_i$ for $i=1, 2, 3, 4$ are as given above.

From Proposition 3·7, we have the following three propositions on the combinatorics of 3-faces $s_{0}^{+}$ , $s_{0}^{-}$ and $s_{0}^{\star}$ . The readers are referred to Figure 4 for illustrations.

Figure 4.

3-sides of $D_{\Sigma}$ . The left one is $s^\star_0$ , the two shaded disks are $s^\star_0 \cap \partial {\bf H}^2_{\mathbb C}$ . The central one is $s^{+}_0$ , the $B^{-1}$ -labelled sectors are $s^{+}_0 \cap s^{-}_0$ , the $B^{2}$ -labelled sectors are $s^{+}_0 \cap s^{\star}_0$ . The right one is $s^{-}_0$ , the labelled sectors have similar meanings. In each sub-figure, the fixed point of B is indicated in cyan (for the online edition) or grey (for the print edition).

The proof of Proposition 3·9 is similar and is omitted.

The side pairing maps on $D_{\Sigma}$ are defined by:

\begin{equation*}A^kBA^{-k}\;:\;s_{k}^{+}\longrightarrow s_{k}^{-},\quad A^kB^{-1}A^{-k}\;:\;s_{k}^{-}\longrightarrow s_{k}^{+}, \quad A^kB^{2}A^{-k}\;:\;s_{k}^{\star}\longrightarrow s_{k}^{\star}.\end{equation*}

By the A-symmetry, we may only consider the following properties.

Proof. The ridge $s_{0}^{+}\cap s_{0}^{-}$ is defined by the triple equality

\begin{equation*}|\langle z, \mathbf{ q}_{\infty}\rangle|=|\langle z, B^{-1}(\mathbf{ q}_{\infty})\rangle|=|\langle z, B(\mathbf{ q}_{\infty})\rangle|.\end{equation*}

The map B sends

\begin{equation*}\{\mathbf{ q}_{\infty}, B^{-1}(\mathbf{ q}_{\infty}), B(\mathbf{ q}_{\infty}) \}\end{equation*}

to

\begin{equation*}\{B(\mathbf{q}_{\infty}), \mathbf{ q}_{\infty}, B^2(\mathbf{ q}_{\infty})\}.\end{equation*}

So B sends the ridge $s_{0}^{+} \cap s_{0}^{-}$ to $s_{0}^{-} \cap s_{0}^{\star}$ . Another claim in Proposition 3·11 can be proved similarly.

Proof of Theorem 3·6. After Propositions 3·8, 3·9, 3·10 and 3·11.

Local tessellation. We prove the tessellations around the sides and ridges of $D_{\Sigma}$ .

1. (1) Since $A^kB^{\pm}A^{-k}$ sends the exterior of $I_k^{\pm}$ to the interior of $I_k^{\mp}$ , we see that $D_{\Sigma}$ and $A^kB^{\pm}A^{-k}(D_{\Sigma})$ have disjoint interiors and cover a neighbourhood of each point in the interior of the 3-side $s_k^{\mp}$ .

2. (2) For the ridge $s_{0}^{+} \cap s_{0}^{\star}$ , the ridge circle is

   \begin{align*} s_{0}^{+} \cap s_{0}^{\star} \xrightarrow{B} s_{0}^{+} \cap s_{0}^{-} \xrightarrow{B} s_{0}^{-} \cap s_{0}^{\star} \xrightarrow{B^2} s_{0}^{+} \cap s_{0}^{\star}.\end{align*}
   This gives the relation $B^4=id$ ; By a standard argument as in [ Reference Parker and Will20 ], we have that $D_{\Sigma} \cup B^2(D_{\Sigma}) \cup B^{-1}(D_{\Sigma})$ covers a small neighborhood of $s_{0}^{+} \cap s_{0}^{\star}$ .

Completeness. We must construct a system of consistent horoballs at the parabolic fixed points. First, we consider the side pairing maps on the parabolic fixed points $x_1=p_{AB}$ , $x_2=p_{AB^2}$ and $x_3=p_{AB^{-1}}$ . We have

\begin{eqnarray*} B &:& p_{AB} \longrightarrow p_{BA},\\ A&:& p_{BA} \longrightarrow p_{AB},\\ B^2 &:& p_{AB^{2}} \longrightarrow p_{B^2A} ,\\ A &:& p_{B^{2}A} \longrightarrow p_{AB^{2}},\\B^{-1} &:& p_{AB^{-1}} \longrightarrow p_{B^{-1}A},\\A &:& p_{B^{-1}A} \longrightarrow p_{AB^{-1}}.\end{eqnarray*}

Up to powers of A, the cycles for the parabolic fixed points are the following

\begin{eqnarray*} &&p_{AB} \xrightarrow{B} p_{BA}\xrightarrow{A} p_{AB},\\ &&p_{AB^2} \xrightarrow{B^2} p_{B^2A}\xrightarrow{A} p_{AB^2},\\&& p_{AB^{-1}} \xrightarrow{B^{-1}} p_{BA^{-1}}\xrightarrow{A} p_{AB^{-1}}.\end{eqnarray*}

That is, $p_{AB}$ , $p_{AB^2}$ and $p_{AB^{-1}}$ are fixed by AB, $AB^2$ and $AB^{-1}$ respectively. The elements AB, $AB^2$ and $AB^{-1}$ are unipotent and preserve all horoballs at $p_{AB}$ , $p_{AB^2}$ and $p_{AB^{-1}}$ respectively.

This completes the proof of Theorem 3·6.

4·1. Two disks $E_B$ and $E_{B^{-1}}$

We will construct a set of disks

\begin{equation*}\big\{A^{k}(E_B \cup E_{B^{-1}})\big\}_{k \in \mathbb{Z}},\end{equation*}

which cap off all the “holes” in $\partial_{\infty} D_{\Sigma}$ . By cutting $\partial_{\infty} D_{\Sigma}$ along these disks, we obtain an infinite cylinder, which is topologically a solid torus.

Recall the points $x_{i}$ for $i=1,2,3,4$ in Section 3, and consider two additional points

\begin{equation*}y_1=\left[\begin{matrix}-\frac{1}{2} \\[4pt] -\frac{\sqrt{3}-i}{2} \\[4pt] 1 \end{matrix}\right] \in \partial_{\infty}I^{+}_0\cap \partial_{\infty}I^{\star}_0, \quad\quad y_2=\left[\begin{matrix}-\frac{1}{2} \\[4pt] -\frac{\sqrt{3}+i}{2} \\[4pt] 1\end{matrix}\right] \in \partial_{\infty}I^{-}_0\cap \partial_{\infty}I^{\star}_0.\end{equation*}

Now, consider the topological circle $C_{B}$ consisting of the following four arcs:

1. (i) let $C_5$ be the arc in $s^{+}_0 \cap \partial \mathbf{H}^2_{\mathbb C}$ joining $x_1$ to $y_1$ given by

   \begin{equation*} C_5=\left\{\begin{bmatrix}-\frac{1}{2}-e^{2si}-i\sqrt{2\cos{(2s)}}e^{\frac{2s-\pi}{4}i} \\[4pt]-\sqrt{2\cos{(2s)}}e^{\frac{2s-\pi}{4}i}+i \\[4pt]1\end{bmatrix}\;:\; s\in \left[\frac{5\pi}{6},\pi\right]\right\}\!;\end{equation*}

2. (ii) let $C_1$ be the $\mathbb{C}$ -arc in $s^{\star}_0 \cap \partial \mathbf{H}^2_{\mathbb C}$ joining $y_1$ to $x_2$ given by

   \begin{equation*} C_1=\left\{\begin{bmatrix}-\frac{1}{2} \\[4pt]e^{is} \\1\end{bmatrix}\;: \ s\in \left[\frac{5\pi}{6},\pi\right]\right\}\!;\end{equation*}

3. (iii) let $C_6$ be the arc in $s^{-}_1 \cap \partial \mathbf{H}^2_{\mathbb C}$ joining $AB^2(y_1)$ and $x_1$ given by

   \begin{equation*} C_6=\left\{\begin{bmatrix}-\frac{5}{2}-2i-e^{2si}+(2-i)\sqrt{2\cos{(2s)}}e^{\frac{2s-\pi}{4}i} \\[4pt]-2-i+\sqrt{2\cos{(2s)}}e^{\frac{2s-\pi}{4}i} \\[4pt]1\end{bmatrix}\;: \ s\in \left[\frac{5\pi}{6},\pi\right]\right\}\!;\end{equation*}

4. (iv) let $C_2$ be the $\mathbb{C}$ -arc in $s^{\star}_1 \cap \partial \mathbf{H}^2_{\mathbb C}$ joining $x_2$ and $AB^2(y_1)$ given by

   \begin{equation*} C_2=\left\{\begin{bmatrix}-\frac{5}{2}-2e^{is} \\[4pt] -2-e^{is} \\[1pt] 1 \end{bmatrix}\;: \ s\in \left[\frac{5\pi}{6},\pi\right]\right\}\!.\end{equation*}

We will show later that the union of $C_5$ , $C_1$ , $C_6$ , and $C_2$ forms a simple closed curve, denoted $C_B$ . We use ruled surfaces to construct a disk $E_B$ with boundary $C_{B}$ . One ruled surface, with directrices $C_5$ and $C_1$ , is denoted by $E_B^l$ ; another, with directrices $C_6$ and $C_2$ , is denoted by $E_B^r$ . Let $c_i$ represents the coordinates of $C_i$ in $\mathcal{N}$ for $i=1,2,5,6$ . The ruled surfaces can be described as

\begin{equation*} E_B^l=(1-a)\cdot c_5+a\cdot c_1,\quad E_B^r=(1-a)\cdot c_6+a\cdot c_2\end{equation*}

for $a\in[0,1]$ . In Figure 5, $E_B^l$ is shown in blue and $E_B^r$ in green. We define $E_B = E_B^l \cup E_B^r$ . It is easy to verify that $E_B^l\cap E_B^r$ is the straight-line segment from $x_1$ to $x_2$ in $\mathcal{N}$ . In Figure 5, $E_{B}$ patches up the “hole” of $\partial_{\infty} D_{\Sigma}$ enclosed by the spinal spheres $\partial_{\infty}I_0^{+}$ , $\partial_{\infty}I_1^{-}$ , $\partial_{\infty}I^{\star}_0$ and $\partial_{\infty}I^{\star}_1$ .

Figure 5.

The disks EB (the union of the blue triangle [ $x_1$ , $y_1$ , $x_2$ ] and green triangle [ $x_1$ , $x_2$ , $AB^2(y_{1})$ ]) and EB-1 (the union of the cyan triangle [ $x_2$ , $y_2$ , $x_3$ ] and yellow triangle [ $x_2$ , $x_3$ , $AB^2(y_2)$ ]).

Figure 6.

The disk $E_{B}$ caps off the “hole” of $\partial_{\infty} D_{\Sigma}$ enclosed by the spinal spheres $\partial_{\infty}I_0^{+}$ (black, remaining black in print), $\partial_{\infty}I_1^{-}$ (steelblue, appearing grey in print), $\partial_{\infty}I^{\star}_0$ (yellow, appearing light grey in print) and $\partial_{\infty}I^{\star}_1$ (pink, appearing light grey in print).

As an example, we give the Heisenberg coordinates of $E_B^l$ :

(4·1) \begin{equation} \begin{cases} x=\cos{(s)}a -\sqrt{2\cos{(2s)}}\sin{\frac{2s+\pi}{4}}\cdot(1-a)\\[4pt] y=\sin{(s)}a+\Big(1+\sqrt{2\cos{(2s)}}\cos{\frac{2s+\pi}{4}}\Big)\cdot(1-a)\\[4pt] t=-2\Big(\sin{(2s)}+ \sqrt{2\cos{(2s)}}\sin{\frac{2s+\pi}{4}}\Big)\cdot(1-a),\end{cases}\end{equation}

where $a\in[0,1]$ and $s\in[{5\pi}/{6},\pi]$ . We use projection mapping to give results on whether $E_B^l$ intersects the spinal spheres.

Proof. According to the equations of $C_5$ and $C_1$ , it is easy to know

\begin{equation*}\Pi(C_5)\cap \Pi(C_1)=\Pi(y_1) \cup \Pi(x_1).\end{equation*}

So $\Pi(C_5) \cup \Pi(C_1)$ is a simple closed curve. It is also easy to check that $\Pi(C_5)$ and $\Pi(C_1)$ are all contained in $\Pi(\partial_{\infty}I_1^{+})$ and $\Pi(\partial_{\infty}I_0^{+})$ . We have completed the first half of the proof.

For (i), according to

\begin{equation*}\Pi(x_1)=\Pi(x_2)=\Pi(\partial_{\infty}I_0^{+})\cap\Pi(\partial_{\infty}I_1^{-})=\Pi\big(\partial_{\infty}I_0^{-}\big)\cap\Pi(\partial_{\infty}I_1^{+}),\end{equation*}

and the affine convexity of $\Pi(\partial_{\infty}I_0^{+})$ and $\Pi(\partial_{\infty}I_1^{+})$ , we know that

\begin{equation*}\Pi\big(E_B^l\big)\cap\Pi\big(\partial_{\infty}I_0^{-}\big)=\Pi\big(E_B^l\big)\cap\Pi\big( \partial_{\infty}I_1^{-}\big)=\Pi(x_1).\end{equation*}

For (ii), we prove that for a given $s\in[{5\pi}/{6},\pi]$ , the projection of each straight line on $E_B^l$ , with parameterisation (4.1), intersects $\Pi(C_1)$ at one point. Write $d_u=\big(\cos{(s)},\sin{(s)}\big)$ , which is the coordinate of some point in $\Pi(C_1)$ . It is also the vector from the origin O in $\mathbb{C}$ to this point. Write $d_l=\big({\partial x}/{\partial a},{\partial y}/{\partial a}\big)\big |_{a=1}$ , which is the direction of the projection of the straight line on $E_B^l$ . Note that when $a=0$ , the corresponding point of $E_B^l$ is in $C_5$ , and when $a=1$ , the corresponding point is in $C_1$ . By calculating the inner product of $-d_l$ and $d_u$ , we can determine whether the point in the straight line of $\Pi(E_B^l)$ is far away from $\Pi(C_1)$ when a decreases from 1 to 0. In fact, the inner product of $-d_l$ and $d_u$ is

\begin{equation*} \begin{aligned} &\left(\sin{\left(\frac{s}{2}\right)}-\cos{\left(\frac{s}{2}\right)}\right)\sqrt{\cos{(2s)}}+ \sin{(s)}-1\\[4pt] &=-\sqrt{2}\sin{\left(\frac{\pi}{4}-\frac{s}{2}\right)} \cdot \left(\sqrt{2}\sin{\left(\frac{\pi}{4}-\frac{s}{2}\right)}+\sqrt{\cos{(2s)}}\right), \end{aligned} \end{equation*}

which is non-negative when $s\in[{5\pi}/{6},\pi]$ . By the affine convexity of $\Pi( \partial_{\infty}I^{\star}_0)$ , we have $\Pi(E_B^l)\cap \Pi(\partial_{\infty}I^{\star}_0)=\Pi(C_1)$ .

For (iii), because

\begin{equation*}\Pi\big(E_B^l\big)\subset\Pi\big(\partial_{\infty}I_0^{-}\big)\cap\Pi\big( \partial_{\infty}I_1^{-}\big),\end{equation*}

which is outside all $\Pi(\partial_{\infty}I_k^{*}\cup \partial_{\infty}I_k^{+}\cup \partial_{\infty}I_k^{-})_{k\in \mathbb{Z}}$ for $k\neq 0,1$ , we get the conclusion.

Figure 7.

The projections of $\partial_{\infty} I_0^{\pm}$ , $\partial_{\infty}I_1^{\pm}$ , $\partial_{\infty}I_{0}^{\star}$ , $\partial_{\infty}I_1^{\star}$ , $E_{B}$ , $E_{B^{-1}}$ and F.

Instead of depicting the entire projections of $E_B^l$ and $E_B^r$ , we show the projections of several representative straight lines from these surfaces in Figure 7, represented by blue and green segments, respectively. The projections of $E_{B}$ and $E_{B^{-1}}$ are the union of blue and green crescents and the union of yellow and cyan crescents respectively. The projection of $\partial_{\infty} I_0^{+}$ , $\partial_{\infty} I_0^{-}$ , $\partial_{\infty} I_1^{+}$ and $\partial_{\infty} I_1^{-}$ are the round disk with radius $\sqrt{2}$ and centered at i, $-i$ , $-2+i$ and $-2-i$ , respectively. The projections of $\partial_{\infty} I_{0}^{\star}$ and $\partial_{\infty} I_1^{\star}$ are round disks with radius 1 and centered at 0 and $-2$ respectively. The projection of F is the union of magenta, coral and navy curves. (Colours online. In print: blue and navy appear as dark grey; green, yellow, cyan and coral appear as light grey; magenta appears as grey.) Figure 8 gives an enlarged view of $\Pi(E_B^l)$ .

Figure 8.

An enlarged view of $\Pi(E_B^l)$ and part of $\Pi(\partial_{\infty}I^{\star}_0)$ .

For the relationship between $E_B^l$ and the remaining spinal spheres, we have:

Proof. Let $p(s,a)=(x(s,a),y(s,a),t(s,a))\in E_B^l$ be the point given by Equation (4.1). For (i), consider the Cygan distance between $B^{-1}(q_{\infty})$ and p(s, a). Let

\begin{equation*}g(s,a)=d_{\textrm{Cyg}}^4\big(B^{-1}(q_{\infty}), p\big)=\big(x^2+(y-1)^2\big)^2+(t-2x)^2\end{equation*}

with $a\in[0,1]$ and $s\in[{5\pi}/{6},\pi]$ . Then

\begin{equation*} \begin{aligned} \frac{\partial g(s,a)}{\partial a}\Big |_{a=0}&=16\cos^{3/2}{(2s)}\left(\sin{\frac{s}{2}}-\cos{\frac{s}{2}}\right) +4(\sin{(3s)}-\cos{(4s)}+\sin{(s)})-12\\ &\ge 0, \end{aligned}\end{equation*}

for $s\in[{5\pi}/{6},\pi]$ by Lemma 4·3 later. By the affine convexity of the isometric spheres, we know that for $a\in[0,1]$ and $s\in[{5\pi}/{6},\pi]$ ,

\begin{equation*}d_{\textrm{Cyg}}\big(B^{-1}(q_{\infty}), p(s,a) \big)\ge d_{\textrm{Cyg}}\big(B^{-1}(q_{\infty}), p(s,0) \big),\end{equation*}

and the equality holds if and only if $a=0$ , which are points on $C_5$ .

For (ii), let

\begin{equation*}h(s,a)=d_{\textrm{Cyg}}^4\big(AB^2A^{-1}(q_{\infty}), p(s,a) \big)=((x+2)^2+y^2)^2+(t-2y)^2,\end{equation*}

where $a\in[0,1]$ and $s\in[{5\pi}/{6},\pi]$ . By Lemma 4·3 later, we have

\begin{equation*} \begin{aligned} \frac{\partial h(s,a)}{\partial a}\Big |_{a=1}=&\left(96\cos^3{\frac{s}{2}}+12\cos{\frac{s}{2}}+ 4\sin{\frac{s}{2}}\right)\sqrt{\cos{(2s)}}+ 52\cos{(s)}-28\sin{(s)} \\ &\qquad+ 12\cos{(2s)}+4\cos{(3s)}- 8\sin{(2s)}+ 40 \le 0, \end{aligned}\end{equation*}

for $s\in[{5\pi}/{6},\pi]$ . By the affine convexity of the isometric spheres, we know that for $a\in[0,1]$ and $s\in[{5\pi}/{6},\pi]$ ,

\begin{equation*}d_{\textrm{Cyg}}\big(AB^2A^{-1}(q_{\infty}), p(s,a) \big)\ge d_{\textrm{Cyg}}\big(AB^2A^{-1}(q_{\infty}), p(s,1) \big)\ge1.\end{equation*}

The first equality holds if and only if $a=1$ , and the second equality holds if and only if $s=\pi$ , corresponding to the point $x_2$ .

For (iii), consider the plane

\begin{equation*} -3 (X + 1 + Y)+T=0\end{equation*}

in the Heisenberg group. On the one hand, it is easy to check that $t-3(x + 1 + y)\le0$ , and the equality holds if and only if $p(s,a)=x_2$ . On the other hand,

\begin{equation*}\partial_{\infty}I_1^{+}=\left\{\begin{bmatrix}2i-e^{-\alpha i}\pm(2+i)\sqrt{2\cos{\alpha}}\cdot e^{({-}\frac{\alpha}{2}+\beta)i}-\frac{5}{2} \\[4pt]-2+i\pm\sqrt{2\cos{\alpha}}\cdot e^{({-}\frac{\alpha}{2}+\beta)i} \\[1pt]1\end{bmatrix}\; : \ \alpha\in \left[-\frac{\pi}{2},\frac{\pi}{2}\right],\ \beta\in [0,\pi)\right\}\!.\end{equation*}

Let $p_0\in \partial_{\infty}I_1^{+}$ be a point with Heisenberg coordinates $(x_0,y_0,t_0)$ . Then

\begin{equation*} \begin{aligned} -3(x_0 + 1 + y_0)+t_0&=\mp\sqrt{2\cos{\alpha}}\left(\cos{\left(\frac{\alpha}{2}-\beta\right)}+\sin{\left(\frac{\alpha}{2}-\beta\right)}\right)+2\sin{\alpha}+4\\ &\geq -2\sqrt{\cos{\alpha}}+2\sin{\alpha}+4\gt 0. \end{aligned}\end{equation*}

Hence, $E_B^l$ and $\partial_{\infty}I_1^{+}$ are separated by the above plane.

Lemma 4·3. Let

\begin{equation*} \begin{aligned} f_1(s)&=16\cos^{3/2}{(2s)}\left(\sin{\frac{s}{2}}-\cos{\frac{s}{2}}\right) +4(\sin{(3s)}-\cos{(4s)}+\sin{(s)})-12\\ \text{and}\quad f_2(s)&=\left(96\cos^3{\frac{s}{2}}+12\cos{\frac{s}{2}}+ 4\sin{\frac{s}{2}}\right)\sqrt{\cos{(2s)}}+ 52\cos{(s)}-28\sin{(s)} \\ &\qquad\qquad\qquad\qquad\qquad\qquad+ 12\cos{(2s)}+4\cos{(3s)}- 8\sin{(2s)}+ 40 \end{aligned}\end{equation*}

be two functions. For $s\in[{5\pi}/{6},\pi]$ , we have:

1. (i) $f_1(s)\ge0$ , and the equality holds if and only if $s={5\pi}/{6}$ or $s=\pi$ ;

2. (ii) $f_2(s)\le0$ , and the equality holds if and only if $s={5\pi}/{6}$ or $s=\pi$ .

Proof. For (i), it is easy to see that $4\cos^{3/2}{(2s)}(\sin{\frac{s}{2}}-\cos{\frac{s}{2}})\gt 0$ and

\begin{equation*}-\sin{(3s)}+\cos{(4s)}-\sin{(s)}+3\gt 0\end{equation*}

when $s\in[{5\pi}/{6},\pi]$ . Then

\begin{equation*} \begin{aligned} &\frac{f_1(s)}{2}\cdot\left(4\cos^{3/2}{(2s)}\left(\sin{\frac{s}{2}}-\cos{\frac{s}{2}}\right)+({-}\sin{(3s)}+\cos{(4s)}-\sin{(s)}+3)\right)\\ &=-21 + 9\cos{(6s)} - 10\cos{(4s)} + 6\sin{(5s)} - 2\sin{(3s)} + 22\sin{(s)} \\ &\qquad\qquad\qquad\qquad+ 23\cos{(2s)} - \cos{(8s)} - 2\sin{(7s)}. \end{aligned} \end{equation*}

Let $x=\sin{(s)}$ , the above equation is written as:

\begin{equation*}F_1(x)=-32x(2x-1)(2x^6-x^5+2x^3-2x^2-x+1),\end{equation*}

where $x\in[0,{1}/{2}]$ . Observe that $2x^6-x^5+2x^3=x^3(2x^3+2-x^2)\gt 0$ and $1-2x^2-x\gt 0$ for $x\in (0,1/2)$ . The roots of $F_1(x)$ in $[0,{1}/{2}]$ are only 0 and ${1}/{2}$ . Thus we have that $f_1(s)\ge0$ , and the equality holds exactly when $s={5\pi}/{6}$ or $s=\pi$ .

For (ii), it is easy to see that

\begin{equation*}\left(24\cos^3{\frac{s}{2}}+3\cos{\frac{s}{2}}+ \sin{\frac{s}{2}}\right)\sqrt{\cos{(2s)}}\gt 0\end{equation*}

and

\begin{equation*}13\cos{(s)}-7\sin{(s)}+4\cos{(2s)}+\cos{(3s)}- 2\sin{(2s)}+10\lt0\end{equation*}

when $s\in[{5\pi}/{6},\pi]$ . Then

\begin{equation*} \begin{aligned} &\frac{f_2(s)}{2}\cdot\Big(\left(24\cos^3{\frac{s}{2}}+3\cos{\frac{s}{2}}+ \sin{\frac{s}{2}}\right)\sqrt{\cos{(2s)}}\\ &\qquad-\big(13\cos{(s)}-7\sin{(s)}+4\cos{(2s)}+\cos{(3s)}- 2\sin{(2s)}+10\big)\Big)\\ &=-306 + 32\sin{(4s)} - \cos{(6s)} + 248\sin{(2s)} +95\cos{(4s)} \\ &\qquad +4\sin{(5s)}+109\sin{(3s)}-268\cos{(s)} + 271\sin{(s)}\\ &\qquad +212\cos{(2s)} + 256\cos{(3s)} + 12\cos{(5s)}. \end{aligned} \end{equation*}

Let $x=\sin{(s)}$ , the above equation is written as:

\begin{equation*}F_2(x)=32\left(x-\frac{1}{2}\right)x\left((39-6x^2+5x)\sqrt{1-x^2} + x^4 + \frac{5x^3}{2}+\frac{47x^2}{2}-\frac{35x}{8}-\frac{309}{8}\right),\end{equation*}

where $x\in[0,{1}/{2}]$ . Define

\begin{equation*}G(x)=(39-6x^2+5x)\sqrt{1-x^2} + x^4 + \frac{5x^3}{2}+\frac{47x^2}{2}-\frac{35x}{8}-\frac{309}{8}.\end{equation*}

We claim that $G(x)\gt 0$ for $x\in [0,{1}/{2}]$ . Note that $\sqrt{1-x^2}\geq 1-{x^2}/{2}-{x^4}/{6}$ and $39-6x^2+5x\gt 0$ for $x\in [0,{1}/{2}]$ . Then

\begin{equation*} \begin{aligned} G(x)&\geq (39-6x^2+5x)\left(1-\frac{x^2}{2}-\frac{x^4}{6}\right)+x^4 + \frac{5x^3}{2}+\frac{47x^2}{2}-\frac{35x}{8}-\frac{309}{8}\\ &=x^6- \frac{5}{6}x^5 -\frac{5}{2}x^4 - 2x^2 + \frac{5}{8}x + \frac{3}{8}\\ &=\left(x-\frac{1}{2}\right)^6+\frac{13}{6}x\left(x-\frac{1}{2}\right)^4 -\frac{23}{12}x^4 -\frac{3}{4}x^3-\frac{89}{48}x^2 + \frac{65}{96}x + \frac{23}{64}. \end{aligned}\end{equation*}

It is easy to verify that ${45}/{96}x -{23}/{12}x^4 -{3}/{4}x^3\geq 0$ and $-{89}/{48}x^2 +{20}/{96}x+{23}/{64}\geq 0$ for $x\in [0,{1}/{2}]$ . Then we have $-{23}/{12}x^4 -{3}/{4}x^3-{89}/{48}x^2 + {65}/{96}x + {23}/{64}\geq 0$ for $x\in [0,{1}/{2}]$ . Now we have that $F_2(x)\le0$ and the roots of $F_2(x)$ in $[0,{1}/{2}]$ are only 0 and ${1}/{2}$ . Hence $f_2(s)\le0$ , and the equality holds exactly when $s={5\pi}/{6}$ or $s=\pi$ .

Now, we have completed the proofs of (ii) and (iii) of Proposition 4·4 below.

Proof. For (i), by Proposition 3·4, we know that $I^{\star}_{0}\cap I^{-}_{1}=\emptyset$ , $I^{\star}_{0}\cap I^{\star}_{1}=x_2$ , $I_{0}^{+}\cap I_{1}^{-}=x_1$ , and $I_{0}^{+}\cap I^{\star}_{1}=\emptyset$ . The curves $C_5$ , $C_1$ , $C_6$ and $C_2$ are contained in $I_{0}^{+}$ , $I^{\star}_{0}$ , $I^{\star}_{1}$ and $I^{-}_{1}$ , respectively. Therefore, we need only to check that $C_5\cap C_1=y_1$ . By the equations of $C_5$ and $C_1$ , we have $C_5 \cap C_1$ is given by

\begin{equation*} \begin{bmatrix}-\frac{1}{2}-e^{2si}-i\sqrt{2\cos{(2s)}}e^{\frac{2s-\pi}{4}i} \\[3pt]-\sqrt{2\cos{(2s)}}e^{\frac{2s-\pi}{4}i}+i \\[3pt]1\end{bmatrix}=\begin{bmatrix}-\frac{1}{2} \\[3pt]e^{s'} \\1\end{bmatrix}\end{equation*}

for some $s, s'\in [{5\pi}/{6},\pi]$ . This holds if and only if $s=s'={5\pi}/{6}$ , which corresponds to the point $y_1$ .

Another topological circle $C_{B^{-1}}$ consisting of four arcs:

1. (i) let $C_7$ be the arc in $s^{-}_0 \cap \partial \mathbf{H}^2_{\mathbb C}$ joining $y_2$ and $x_3$ given by

   \begin{equation*} C_7=\left\{\begin{bmatrix} -\frac{1}{2}-e^{2si}+\sqrt{2\cos{(2s)}}e^{\frac{\pi+2s}{4}i} \\[3pt] -i+i\sqrt{2\cos{(2s)}}e^{\frac{\pi+2s}{4}i} \\[3pt] 1\end{bmatrix} \;: \ s\in \left[0,\frac{\pi}{6}\right]\right\}; \end{equation*}

2. (ii) let $C_3$ be the $\mathbb{C}$ -arc in $s^{\star}_0 \cap \partial \mathbf{H}^2_{\mathbb C}$ joining $x_2$ and $y_2$ given by

   \begin{equation*} C_3=\left\{\begin{bmatrix} -\frac{1}{2} \\[3pt] -e^{si} \\ 1\end{bmatrix} \;: \ s\in \left[0,\frac{\pi}{6}\right]\right\}\!; \end{equation*}

3. (iii) let $C_8$ be the arc in $s^{+}_1 \cap \partial \mathbf{H}^2_{\mathbb C}$ joining $x_3$ and $AB^2(y_2)$ given by

   \begin{equation*} C_8=\left\{\begin{bmatrix} -\frac{5}{2}+2i-e^{2si}+(1-2i)\sqrt{2\cos{(2s)}}e^{\frac{\pi+2s}{4}i} \\[3pt] -2+i-i\sqrt{2\cos{(2s)}}e^{\frac{\pi+2s}{4}i} \\[3pt] 1\end{bmatrix} \;: \ s\in \left[0,\frac{\pi}{6}\right]\right\}; \end{equation*}

4. (iv) let $C_4$ be the $\mathbb{C}$ -arc in $s^{\star}_1 \cap \partial \mathbf{H}^2_{\mathbb C}$ joining $AB^2(y_2)$ and $x_2$ given by

   \begin{equation*} C_4=\left\{\begin{bmatrix} -\frac{5}{2} + 2e^{si} \\[3pt] -2+e^{si} \\[3pt] 1 \end{bmatrix} \;: \ s\in \left[0,\frac{\pi}{6}\right]\right\}. \end{equation*}

Similarly, let $c_i$ represent the coordinates of $C_i$ in $\mathcal{N}$ for $i=3,4,7,8$ . We then construct two more ruled surfaces:

\begin{equation*} E_{B^{-1}}^l=(1-a)\cdot c_8+a\cdot c_4,\quad E_{B^{-1}}^r=(1-a)\cdot c_7+a\cdot c_3 \end{equation*}

with $a\in[0,1]$ .

Define $E_{B^{-1}}=E_{B^{-1}}^l\cup E_{B^{-1}}^r$ . We also have $E_{B^{-1}}^l\cap E_{B^{-1}}^r$ is the straight-line segment from $x_3$ to $x_2$ in $\mathcal{N}$ . In Figure 5, $E_{B^{-1}}^l$ is the cyan part, and $E_{B^{-1}}^r$ is the yellow part. The proof of Proposition 4·5 is similar to the proof of Proposition 4·4, and we omit it.

From the equations of $E_B$ and $E_{B^{-1}}$ , we have:

Proof. Let $\mathcal{L}$ be the line

\begin{equation*} \{[x,0]\;:\; x\in \mathbb{R}\}\end{equation*}

in $\mathcal{N}$ . Then item (i) can be checked directly.

Now, the complement of $\mathcal{S} \cup_{k \in \mathbb{Z}}A^{k}(E_B \cup E_{B^{-1}})$ in $\partial \mathbf{H}^2_{\mathbb C}$ is isotopic to the boundary of a small neighborhood of $\mathcal{L} \cup_{k \in \mathbb{Z}}A^{k}(E_B \cup E_{B^{-1}})$ . Moreover, this isotopy can be performed in an A-equivariant fashion. Therefore, the complement is topologically an A-invariant $(\mathbb{R}^2-\mathbb{D}^2) \times \mathbb{R}$ .

4·2. Three more disks and seventeen more arcs

More disks are used to cut out a 3-ball in $\partial_{\infty} D_{\Sigma}$ . The quotient space of this 3-ball by suitable side-parings is the 3-manifold at infinity in Theorem 1·2.

1. (i) let $C_{17}$ be the $\mathbb{C}$ -arc in $s^{+}_0 \cap \partial \mathbf{H}^2_{\mathbb C}$ joining $x_1$ and $B^{-1}(x_3)$ given by

   \begin{equation*} C_{17}=\left\{\begin{bmatrix}-\frac{3}{2}-i\sqrt{2}e^{is} \\[3pt]i-\sqrt{2}e^{is} \\[3pt]1\end{bmatrix}\;: \ s\in \left[\frac{3\pi}{4},\frac{9\pi}{4}\right]\right\}.\end{equation*}

2. (ii) let $C_{19}$ be the $\mathbb{C}$ -arc in $\partial \mathbf{H}^2_{\mathbb C}$ joining $q_{\infty}$ and $B^{-1}(x_3)$ given by

   \begin{equation*} C_{19}=\left\{\begin{bmatrix}-\frac{1}{2}+si \\[3pt]1 \\1\end{bmatrix}\;: \ s\in [2,\infty]\right\}.\end{equation*}

3. (iii) let $C_{20}$ be the $\mathbb{C}$ -arc in $\partial \mathbf{H}^2_{\mathbb C}$ joining $x_1$ and $q_{\infty}$ given by

   \begin{equation*} C_{20}=\left\{\begin{bmatrix}-\frac{1}{2}-si \\[3pt]-1 \\1\end{bmatrix}\;: \ s\in [2,\infty]\right\}.\end{equation*}

Now, we construct a disk F. First, we take vertical upward rays (part of $\mathbb{C}$ -arcs) along the first half of $C_{17}$ , with paramatrization

\begin{equation*}[-\sqrt{2}\cos{(s)}, 1-\sqrt{2}\sin(s), -2\sqrt{2}\cos{(s)}+a]\end{equation*}

in Heisenberg coordinates with $s\in [{3\pi}/{4},{3\pi}/{2}]$ and $ a\in [0,+\infty]$ . Denote this part by $F_1$ . Next, we take vertical downward rays (part of the $\mathbb{C}$ -arcs) along the second half of $C_{17}$ , with paramatrization

\begin{equation*}[-\sqrt{2}\cos{(s)}, 1-\sqrt{2}\sin(s), -2\sqrt{2}\cos{(s)}+a]\end{equation*}

in Heisenberg coordinates with $s\in [{3\pi}/{2},{9\pi}/{4}]$ and $ a\in [0,-\infty]$ . Denote this part by $F_2$ . Finally, we take a vertical half-plane $F_3$ , with paramatrization

\begin{equation*}[0, (1+\sqrt{2})s, t]\end{equation*}

in Heisenberg coordinates with $s\in[1+\infty]$ and $t\in [-\infty,+\infty]$ . Let

\begin{equation*}C_F=C_{17}\cup C_{19}\cup C_{20} \quad \text{and} \quad F=F_1\cup F_2\cup F_3.\end{equation*}

They are shown in Figure 10. We draw a part of F together with several spinal spheres in Figure 11. Using the previous method similarly, we give the following results.

Figure 9.

The topology of the Ford domain of $\Delta_{4,\infty,\infty;\infty}$ . After capping off the holes in Figure 2 by red disks (the quadrilaterals [ $A^{-1}$ ( $x_3$ ), $A^{-1}$ ( $y_2$ ), $A^{-1}$ ( $x_2$ ), $B^2$ ( $y_2$ )] and [ $x_2$ , $y_2$ , $x_3$ , $AB^2$ ( $y_2$ )]) and purple disks (the quadrilaterals [ $A^{-1}$ ( $x_2$ ), $A^{-1}$ ( $y_1$ ), $A^{-1}$ ( $x_1$ ), $B^2$ ( $y_1$ )] and [ $x_1$ , $y_1$ , $x_2$ , $AB^2$ ( $y_1$ )]), we need two once-punctured disks $D_{-}$ and $D_{+}$ , and a triangle F. The boundary of $D_{-}$ (green) is the curve [ $A^{-1}$ ( $x_3$ ), $B^2$ ( $y_2$ ), $w_4$ , $B^2$ ( $y_1$ ), $A^{-1}$ ( $x_1$ ), $B^{2}$ ( $y_1$ ), $w_3$ , $B^2$ ( $y_2$ ), $A^{-1}$ ( $x_3$ )], the boundary of $D_{+}$ (blue) is the curve [ $x_1$ , $AB^2$ ( $y_1$ ), $A(w_3)$ , $AB^2$ ( $y_2$ ), $x_3$ , $AB^2$ ( $y_2$ ), $A(w_4)$ , $AB^2$ ( $y_1$ ), $x_1$ ], and the triangle F has an orange boundary (online; light grey in print).

Figure 10.

Disks $F_1$ (coral), $F_2$ (magenta) and $F_3$ (blue), curves $C_{17}$ (green), $C_{19}$ (red) and $C_{20}$ (tan).

Figure 11.

Add part of the disk F to Figure 2.

Proof. For item (i), by the equations of the arcs of $C_F$ , it is easy to verify that any pair intersect only at one end point.

For item (ii), according to the structure of F, we naturally have $\partial F=C_F$ . First, the projection $\Pi(F_3)$ is a ray in $\mathbb{C}$ , satisfying

\begin{equation*}\Pi\big(F_3\cap(\partial_{\infty}I_k^{+}\cup \partial_{\infty}I_k^{-}\cup\partial_{\infty}I_k^{*})\big)=(0,1+\sqrt{2})\in\Pi(\partial_{\infty}I_0^{+}).\end{equation*}

Second,

\begin{equation*}\Pi(F_2)=\big({-}\sqrt{2}\cos{(s)},1-\sqrt{2}\sin{(s)}\big)\subset \partial\Pi(\partial_{\infty}I_0^{+}),\end{equation*}

where $s\in [{3\pi}/{4},{3\pi}/{2}]$ . Therefore, the interior of $F_2$ is disjoint from $\partial_{\infty}I^{+}_k(k\neq 1)$ , $\partial_{\infty}I^{-}_k$ , and $\partial_{\infty}I^{\star}_k(k\neq 1)$ for $k \in \mathbb{Z}$ . We need only to check that if $F_2\cap(\partial_{\infty}I^{+}_1\cup \partial_{\infty}I^{\star}_1)$ is an empty set.

On the one hand, consider the plane

\begin{equation*}3(X + 1 + Y) - 2-T=0\end{equation*}

in the Heisenberg group. For any point $(x_0,y_0,t)\in F_2$ , where $(x_0,y_0,t_0)\in C_{17}^1$ and $t\leq t_0$ , we have

\begin{equation*} \begin{aligned} 3(x_0 + 1 + y_0) - 2-t&\ge 3(x_0 + 1 + y_0) - 2-t_0\\ &=3({-}\sqrt{2}\cos{(s)}+1+1-\sqrt{2}\sin(s))-2+2\sqrt{2}\cos{(s)}\\ &=4-\sqrt{2}\cos{(s)}-3\sqrt{2}\sin(s)\ge 0. \end{aligned}\end{equation*}

This means that $F_2$ is on one side of this plane.

On the other hand, for any point $(x_0,y_0,t_0)\in s^{+}_1$ ,

\begin{equation*} \begin{aligned} 3(x_0 + 1 + y_0)-2-t_0&=\mp\sqrt{2\cos{\alpha}}\left(\cos{\left(\frac{\alpha}{2}-\beta\right)}+\sin{\left(\frac{\alpha}{2}-\beta\right)}\right)-2\sin{\alpha}-6\\ &\le 2\sqrt{\cos{\alpha}}-2\sin{\alpha}-6\lt0. \end{aligned}\end{equation*}

Moreover,

\begin{equation*}\partial_{\infty}I_1^{*}=\left\{\begin{bmatrix}-\frac{e^{-\alpha i}}{2}\pm2\sqrt{\cos{\alpha}}\cdot e^{({-}\frac{\alpha}{2}+\beta)i}-2 \\[3pt]-2\pm\sqrt{\cos{\alpha}}\cdot e^{({-}\frac{\alpha}{2}+\beta)i} \\[3pt]1\end{bmatrix}\;: \ \alpha\in \left[-\frac{\pi}{2},\frac{\pi}{2}\right],\ \beta\in [0,\pi)\right\}.\end{equation*}

So for any point $(x_0,y_0,t_0)\in \partial_{\infty}I^{*}_1$ ,

\begin{equation*} \begin{aligned} 3(x_0 + 1 + y_0)-2-t_0 &=-5\pm\sqrt{\cos{\alpha}}\left(3\cos{\left(\frac{\alpha}{2}-\beta\right)}+\sin{\left(\frac{\alpha}{2}\right)}-\beta\right)-\sin{\alpha}\lt0. \end{aligned}\end{equation*}

Therefore, $F_2$ and $(\partial_{\infty}I^{+}_1\cup \partial_{\infty}I^{\star}_1)$ are separated by the above plane.

Table I gives the other curves we used.

Table I.

Some curve information

Consider a curve $C_{+}$ in $\partial_{\infty} D_{\Sigma}-(\cup_{k \in \mathbb{Z}} A^{k}(E_B \cup E_{B^{-1}}))$ which consisting of eight edges:

1. (i) glue two copies of $C_6$ together along $x_1$ ;

2. (ii) take a copy $C_{21}$ , $C_{22}$ , $C_{23}$ , and $C_{24}$ ;

3. (iii) glue two copies of $C_8$ together along $x_3$ .

These edges were then glued together along $AB^2(y_1)$ , $A(w_3)$ , $AB^2(y_2)$ , and $A(w_4)$ to obtain the simple closed curve $C_{+}$ (the union of blue edges in Figure 9). Similarly, there is a curve $C_{-}=A^{-1}(C_{+})$ that is the union of green edges in Figure 9.

6·1. 3-manifold at infinity of $\Delta_{p,\infty,\infty;\infty}$ for $3 \leq p \leq 9$

In [ Reference Ma and Xie14 ], Ma and Xie proved the following

The magic 3-manifold in the SnapPy Census [ Reference Culler, Dunfield, Goerner and Weeks2 ] is the complement of the chain link $6_1^3$ in the 3-sphere, which is hyperbolic with three cusps and volume 5.3334895669 numerically. Recall that we have proved that the 3-manifold at infinity of the even subgroup of $\Delta_{4,\infty,\infty;\infty}$ is the complement of the chain link $8^4_1$ in the 3-sphere. Compared with the above results, we denote by $M_p$ the 3-manifold at infinity of the even subgroup of $\Delta_{p,\infty,\infty;\infty}$ for $3 \leq p \leq 9$ . Recall that Schwartz’s conjecture in [ Reference Schwartz26 ] predicts that this group is discrete, so $M_p$ is well-defined. For $p \leq 9$ , the complex hyperbolic group $\Delta_{p,\infty,\infty}$ is of type A in the sense of Schwartz [ Reference Schwartz26 ]. For more details, see [ Reference Schwartz26 ]. We note that in the group $\Delta_{p,\infty,\infty;\infty}= \langle I_1,I_2,I_3 \rangle $ for $3 \leq p \leq 9$ , $I_1I_2$ , $I_1I_3$ and $I_1I_3I_2I_3$ are parabolic by definition. Moreover, one can verify that $I_1(I_3I_2)^kI_3$ is parabolic for any $2 \leq k \leq p-2$ . Therefore, there are (at least) p conjugacy classes of parabolic elements in $\Delta_{p,\infty,\infty;\infty}$ . In particular, there are (at least) $p-3$ hidden conjugacy classes of parabolic elements in $\Delta_{p,\infty,\infty;\infty}$ which are unexpected. We propose the following conjecture.