[table]capposition=top capbtabboxtable[][]

# Complexification of an infinite volume Coxeter tetrahedron

Jiming Ma  School of Mathematical Sciences, Fudan University, Shanghai,
200433, P. R. China<EMAIL_ADDRESS>
(Date: Jan. 19, 2023)

###### Abstract.

Let $T$ be an infinite volume Coxeter tetrahedron in three dimensional real
hyperbolic space ${\bf H}^{3}_{\mathbb{R}}$ with two opposite right-angles and
the other angles are all zeros. Let $G$ be the Coxeter group of $T$, so

$G=\left\langle\iota_{1},\iota_{2},\iota_{3},\iota_{4}\Bigg{|}\begin{array}[]{c}\iota_{1}^{2}=\iota_{2}^{2}=\iota_{3}^{2}=\iota_{4}^{2}=id,\\\\[3.0pt]
(\iota_{1}\iota_{3})^{2}=(\iota_{2}\iota_{4})^{2}=id\end{array}\right\rangle$

as an abstract group. We study type-preserving representations
$\rho:G\rightarrow\mathbf{PU}(3,1)$, where $\rho(\iota_{i})=I_{i}$ is a
complex reflection fixing a complex hyperbolic plane in three dimensional
complex hyperbolic space ${\bf H}^{3}_{\mathbb{C}}$ for $1\leq i\leq 4$. The
moduli space $\mathcal{M}$ of these representations is parameterized by
$\theta\in[\frac{5\pi}{6},\pi]$. In particular, $\theta=\frac{5\pi}{6}$ and
$\theta=\pi$ degenerate to ${\bf H}^{2}_{\mathbb{C}}$-geometry and ${\bf
H}^{3}_{\mathbb{R}}$-geometry respectively. Via Dirichlet domains, we show
$\rho=\rho_{\theta}$ is a discrete and faithful representation of the group
$G$ for all $\theta\in[\frac{5\pi}{6},\pi]$. This is the first nontrivial
moduli space in three dimensional complex hyperbolic space that has been
studied completely.

###### Key words and phrases:

Complex hyperbolic geomerty, Coxeter polytope, Dirichlet domain, complex
reflection.

###### 2010 Mathematics Subject Classification:

20F55, 20H10, 57M60, 22E40, 51M10.

Jiming Ma was partially supported by NSFC 12171092.

##  1\. Introduction

###  1.1. Motivation

Hyperbolic $n$-space ${\bf H}^{n}_{\mathbb{R}}$ is the unique complete simply
connected Riemannian $n$-manifold with all sectional curvatures $-1$. Complex
hyperbolic $n$-space ${\bf H}^{n}_{\mathbb{C}}$ is the unique complete simply
connected Kähler $n$-manifold with all holomorphic sectional curvatures $-1$.
But the Riemannian sectional curvatures of a complex hyperbolic space are no
longer constant, which are pinched between $-1$ and $-\frac{1}{4}$. This makes
complex hyperbolic geometry much more difficult to study. The holomorphic
isometry group of ${\bf H}^{n}_{\mathbb{C}}$ is $\mathbf{PU}(n,1)$, the
orientation preserving isometry group of ${\bf H}^{n}_{\mathbb{R}}$ is
$\mathbf{PO}(n,1)$. Besides ${\bf H}^{n}_{\mathbb{R}}$ is a totally geodesic
submanifold of ${\bf H}^{n}_{\mathbb{C}}$, $\mathbf{PO}(n,1)$ is a natural
subgroup of $\mathbf{PU}(n,1)$.

Over the last sixty years the theory of Kleinian groups, that is, deformations
of groups into $\mathbf{PO}(3,1)$, has flourished because of its close
connections with low dimensional topology and geometry. More precisely,
pioneered by Ahlfors and Bers in the 1960’s, Thurston formulated a conjectural
classification scheme for all hyperbolic 3-manifolds with finitely generated
fundamental groups in the late 1970’s. The conjecture predicted that an
infinite volume hyperbolic 3-manifold with finitely generated fundamental
group is uniquely determined by its topological type and its end invariants.
Thurston’s conjecture is completed by a series of works of many
mathematicians, which is one of the most great breakthrough in 3-manifolds
theory. See Minsky’s ICM talk [17] for related topics and the reference.

There are also some remarkable works on deformations of groups into
$\mathbf{PU}(2,1)$. Let $\Delta(p,q,r)$ be the abstract $(p,q,r)$ reflection
triangle group with the presentation

$\Delta(p,q,r)=\langle\sigma_{1},\sigma_{2},\sigma_{3}|\sigma^{2}_{1}=\sigma^{2}_{2}=\sigma^{2}_{3}=(\sigma_{2}\sigma_{3})^{p}=(\sigma_{3}\sigma_{1})^{q}=(\sigma_{1}\sigma_{2})^{r}=id\rangle,$

where $p,q,r$ are positive integers or $\infty$ satisfying

$\frac{1}{p}+\frac{1}{q}+\frac{1}{r}<1.$

If $p,q$ or $r$ equals $\infty$, then the corresponding relation does not
appear. The ideal triangle group is the case that $p=q=r=\infty$. A _$(p,q,r)$
complex hyperbolic triangle group_ is a representation $\rho$ of
$\Delta(p,q,r)$ into $\mathbf{PU}(2,1)$ where the generators fix complex
lines, we denote $\rho(\sigma_{i})$ by $I_{i}$. Goldman and Parker initiated
the study of the deformations of ideal triangle group into $\mathbf{PU}(2,1)$
in [13]. They gave an interval in the moduli space of complex hyperbolic ideal
triangle groups, for points in this interval the corresponding representations
are discrete and faithful. They conjectured that a complex hyperbolic ideal
triangle group $\Gamma=\Delta_{\infty,\infty,\infty}=\langle
I_{1},I_{2},I_{3}\rangle$ is discrete and faithful if and only if
$I_{1}I_{2}I_{3}$ is not elliptic. Schwartz proved Goldman-Parker’s conjecture
in [22, 23]. Richard Schwartz has also conjectured the necessary and
sufficient condition for a general complex hyperbolic triangle group
$\Delta_{p,q,r}=\langle I_{1},I_{2},I_{3}\rangle<\mathbf{PU}(2,1)$ to be a
discrete and faithful representation of $\Delta(p,q,r)$. See Schwartz’s ICM
talk [24] for related topics and the reference. Schwartz’s conjecture has been
proved in a few cases [19, 20].

From above we know one way to study discrete subgroups of $\mathbf{PU}(n,1)$
is the deformations of a well-understood representation. From a finitely
presented abstract group $G$, and a discrete faithful representation
$\rho_{0}:G\rightarrow\mathbf{PU}(n,1)$, we may deform $\rho_{0}$ to
$\rho_{1}:G\rightarrow\mathbf{PU}(n,1)$ along a path. We are interested in
whether $\rho_{1}$ is discrete and faithful. Moreover, even when $\rho_{1}$ is
not faithful, but it also has the chance to be discrete. This case is very
interesting, since if we are lucky, we have the opportunity to get a complex
hyperbolic lattice at $\rho_{1}$ [7, 8].

One of the most important questions in complex hyperbolic geometry is the
existence of (infinitely many commensurable classes of) non-arithmetic complex
hyperbolic lattices [16, 10]. This is notorious difficult comparing to its
real hyperbolic counterpart [14]. As results of forty years’ hard work, people
only found 22 commensurable classes of non-arithmetic complex hyperbolic
lattices in $\mathbf{PU}(2,1)$ [3, 7, 8], and 2 commensurable classes of non-
arithmetic complex hyperbolic lattices in $\mathbf{PU}(3,1)$ [3, 5]. Both
$\mathbf{PO}(3,1)$ and $\mathbf{PU}(2,1)$ are subgroups of $\mathbf{PU}(3,1)$.
It is reasonable that deformations of some discrete groups in
$\mathbf{PO}(3,1)$ into the larger group $\mathbf{PU}(3,1)$ may give some
discrete, but not faithful representations, which in turn have the opportunity
to give some new ${\bf H}^{3}_{\mathbb{C}}$-lattices as pioneered by [3, 7,
8]. The author remarks that comparing to results in ${\bf
H}^{2}_{\mathbb{C}}$-geometry [25, 25, 6], any discrete deformations of a
group $G$ into $\mathbf{PU}(3,1)$ with some accidental parabolic or elliptic
element is also highly interesting. For instance, we have no non-trivial
example of a 5-manifold $N$ which admits uniformizable CR-structures now. Here
by nontrivial example, we mean $N$ is neither diffeomorphic to the
$\mathbb{S}^{3}$-bundle over a ${\bf H}^{2}_{\mathbb{R}}$-manifold $F^{2}$,
the trivial $\mathbb{S}^{2}$-bundle over a ${\bf H}^{3}_{\mathbb{R}}$-manifold
$Y^{3}$, nor diffeomorphic to a $\mathbb{S}^{1}$-bundle over a ${\bf
H}^{2}_{\mathbb{C}}$-manifold $X^{4}$.

In this paper, we study deformations of groups into $\mathbf{PU}(3,1)$ via
Dirichlet domains, which is much more difficult and richer than deformations
of groups into $\mathbf{PO}(3,1)$ and $\mathbf{PU}(2,1)$. By this we mean:

  * •

It is well known that the space of discrete and faithful representations of a
group into $\mathbf{PO}(3,1)$ has fractal boundary in general. For example,
the so called Riley slice has a beautiful fractal boundary in $\mathbb{C}$
(see Page VIII of [1]);

  * •

People tend to guess that the space of discrete and faithful representations
of a group into $\mathbf{PU}(2,1)$ has piece-wise smooth boundary (at least
when the deformation space has two dimension). For one of the tractable cases,
the so called complex Riley slice, which is 2-dimensional, see [20];

  * •

Moreover, there are very few results on the space of discrete and faithful
representations of a group into $\widehat{\mathbf{PU}(2,1)}$. Here
$\widehat{\mathbf{PU}(2,1)}$ is the full isometry group of ${\bf
H}^{2}_{\mathbb{C}}$. To the author’s knowledge, the only complete
classification result is in [9]. Where Falbel-Parker completed the study on
the space of discrete and faithful representations of
$\mathbb{Z}_{2}*\mathbb{Z}_{3}$ into $\widehat{\mathbf{PU}(2,1)}$ (with one
additional parabolic element), the moduli space is 1-dimensional.

So deformations of groups into $\mathbf{PU}(3,1)$ may have fractal boundaries
in general (at least when the deformation spaces have large dimensions), but
we are very far from understanding them.

###  1.2. Main result of the paper

In this article, we complexify a Coxeter tetrahedron in the real hyperbolic
space ${\bf H}^{3}_{\mathbb{R}}$ into the complex hyperbolic space ${\bf
H}^{3}_{\mathbb{C}}$.

Using the upper space model of ${\bf H}^{3}_{\mathbb{R}}$, a Coxeter
tetrahedron in ${\bf H}^{3}_{\mathbb{R}}$ is determined by four round circles
$C_{i}$ in $\mathbb{C}$ for $1\leq i\leq 4$. Where each circle $C_{i}$ is the
ideal boundary of a totally geodesic ${\bf
H}^{2}_{\mathbb{R}}\hookrightarrow{\bf H}^{3}_{\mathbb{R}}$, and
$\mathbb{C}\cup\infty$ is the ideal boundary of ${\bf H}^{3}_{\mathbb{R}}$. In
the right subfigure of Figure 1, there is a configuration of four round
circles in $\mathbb{C}$:

  * •

the red, purple, green and blue circles are $C_{1}$, $C_{2}$, $C_{3}$ and
$C_{4}$ respectively;

  * •

$C_{1}$ and $C_{3}$ intersect perpendicularly at two points. $C_{2}$ and
$C_{4}$ also intersect perpendicularly at two points;

  * •

$C_{i}$ is tangent to $C_{i+1}$ for $1\leq i\leq 4$ mod $4$.

This configuration of round circles in $\mathbb{C}$ can be obtained as
follows. We first take $C_{1}$ and $C_{3}$ with the same radius and intersect
perpendicularly at two point, see the red and green circles in right subfigure
of Figure 1. We take original round circles $C_{2}$ and $C_{4}$ which are
coincident, such that they are tangent to both $C_{1}$ and $C_{3}$ (the
original $C_{2}$ and $C_{4}$ are both the dashed circle in Figure 1). We view
$C_{2}$ and $C_{4}$ have angle $\pi$ at this configuration. Then we make
$C_{2}$ with bigger radius, and the center of it to the left of the original
one. We also make $C_{4}$ with bigger but the same radius as $C_{2}$, and the
center of it to the right of the original one. Moreover each of $C_{2}$ and
$C_{4}$ is also tangent to both $C_{1}$ and $C_{3}$. When the radii of $C_{2}$
and $C_{4}$ diverge to the infinity, that is, the limiting case is that
$C_{2}$ and $C_{4}$ are two vertical lines tangent to both $C_{1}$ and
$C_{3}$, the angle between $C_{2}$ and $C_{4}$ converges to zero. So at an
intermediate time, the angle between $C_{2}$ and $C_{4}$ is $\frac{\pi}{2}$.
It is well-known that there is a unique configuration of round circles in
$\mathbb{C}$ satisfies the above angle conditions up to
$\mathbf{PSL}(2,\mathbb{C})$-action (equivalently up to
$\mathbf{PO}(3,1)$-action), see [27].

$\frac{\pi}{2}$$\frac{\pi}{2}$$0$$0$$0$$0$$C_{2}$$C_{4}$$C_{1}$$C_{3}$ Figure
1. The infinite volume Coxeter tetrahedron $T$ (left) in ${\bf
H}^{3}_{\mathbb{R}}$ and the ideal boundary its defining hyperplane
configuration (right).

So let $T$ be the Coxeter tetrahedron in ${\bf H}^{3}_{\mathbb{R}}$, which is
determined by four totally geodesic hyperbolic planes in ${\bf
H}^{3}_{\mathbb{R}}$, such that the ideal boundaries of these hyperbolic
planes is the given configuration of round circles in right subfigure of
Figure 1. Then $T$ is an infinite volume Coxeter tetrahedron with two opposite
angles $\frac{\pi}{2}$ and the other angles are all zeros, see the left
subfigure of Figure 1.

Let $G$ be the Coxeter group of $T$, that is, the reflection group with four
generators the (real) reflections about the defining hyperbolic planes of $T$.
So

$G=\left\langle\iota_{1},\iota_{2},\iota_{3},\iota_{4}\Bigg{|}\begin{array}[]{c}\iota_{1}^{2}=\iota_{2}^{2}=\iota_{3}^{2}=\iota_{4}^{2}=id,\\\\[3.0pt]
(\iota_{1}\iota_{3})^{2}=(\iota_{2}\iota_{4})^{2}=id\end{array}\right\rangle$

as an abstract group. The group $G$ is isomorphic to
$(\mathbb{Z}_{2}\oplus\mathbb{Z}_{2})\ast(\mathbb{Z}_{2}\oplus\mathbb{Z}_{2})$
abstractly. For technical reasons, we also need to consider a subgroup of $G$,
say

$K=\langle\iota_{1}\iota_{3},\iota_{2}\iota_{4},\iota_{1}\iota_{2}\rangle.$

$K$ is an index two subgroup of $G$, and $K$ is isomorphic to
$\mathbb{Z}_{2}\ast\mathbb{Z}_{2}\ast\mathbb{Z}$.

In this paper we study representations $\rho:G\rightarrow\mathbf{PU}(3,1)$,
such that $\rho(\iota_{i})=I_{i}$ is a complex reflection fixing a complex
hyperbolic plane in ${\bf H}^{3}_{\mathbb{C}}$ for $1\leq i\leq 4$, with the
condition $I_{i}I_{i+1}$ is parabolic for $1\leq i\leq 4$ mod $4$. This is a
natural complexification of $T$, since we replace the real reflections by
complex reflections. The moduli space $\mathcal{M}$ is parameterized by
$\theta\in[\frac{5\pi}{6},\pi]$. In particular, $\theta=\frac{5\pi}{6}$ and
$\theta=\pi$ degenerate to ${\bf H}^{2}_{\mathbb{C}}$-geometry and ${\bf
H}^{3}_{\mathbb{R}}$-geometry respectively. By this we mean the group
$\rho_{\frac{5\pi}{6}}(G)$ preserves a totally geodesic ${\bf
H}^{2}_{\mathbb{C}}\hookrightarrow{\bf H}^{3}_{\mathbb{C}}$ invariant, and
$\rho_{\pi}(G)$ preserves a totally geodesic ${\bf
H}^{3}_{\mathbb{R}}\hookrightarrow{\bf H}^{3}_{\mathbb{C}}$ invariant
respectively. See Section 3 for more details.

Using the Dirichlet domains of the $\rho_{\frac{5\pi}{6}}(G)$-action on ${\bf
H}^{2}_{\mathbb{C}}$ and the $\rho_{\pi}(G)$-action on ${\bf
H}^{3}_{\mathbb{R}}$ as guides, the main result of this paper is

######  Theorem 1.1.

$\rho_{\theta}$ is a discrete and faithful representation of the group $G$
into $\mathbf{PU}(3,1)$ for any $\theta\in[\frac{5\pi}{6},\pi]$.

In fact there is a hidden and lucky $\mathbb{Z}_{4}$-symmetry of each
representation $\rho_{\theta}$. More precisely, there is
$J\in\mathbf{PU}(3,1)$ an order-4 regular elliptic element, such that
$I_{i}=JI_{i-1}J^{-1}$ for $i=1,2,3,4$ mod 4. We denote $A_{i}=I_{i}I_{i+1}$
for $1\leq i\leq 4$. Then

$\rho_{\theta}(K)=\left\langle A_{1},A_{2},A_{3},A_{4}\right\rangle,$

such that $A_{i}$ is parabolic and $A_{i}A_{i+1}$ has order two for $1\leq
i\leq 4$. We consider the Dirichlet domain of $\rho_{\theta}(K)$ with the
fixed point of $J$ as the center. The reason that we work on
$\rho_{\theta}(K)$ instead of $\rho_{\theta}(G)$ is that there are subgroups
of $G$ which are infinite dihedral groups. Moreover by Goldman-Parker [12],
the Dirichlet domain of an infinite dihedral group tends to have infinite
facets (depending on the chosen center of the Dirichlet domain). So it seems
the Dirichlet domain of $\rho_{\theta}(G)$ is combinatorically much more
difficult to study than that of $\rho_{\theta}(K)$ for general $\theta$.

Let

$R=\\{A^{\pm 1}_{1},~{}~{}A^{\pm 1}_{2},~{}~{}A^{\pm 1}_{3},~{}~{}A^{\pm
1}_{4},~{}~{}A_{1}A_{2},~{}~{}A_{2}A_{3}\\}$

be a subset of $\rho_{\theta}(K)$ consisting of ten elements. We will show the
partial Dirichlet domain $D_{R}$ is in fact the Dirichlet domain of
$\rho_{\theta}(K)$, then we obtain Theorem 1.1.

Since the proof of Theorem 1.1 is much involved, we prove it in following
steps:

  * •

We consider firstly the case $\theta=\frac{5\pi}{6}$, that is, we consider
${\bf H}^{2}_{\mathbb{C}}$-geometry first. The proof when
$\theta=\frac{5\pi}{6}$ can also be viewed as a model of the proof of Theorem
1.1 for general $\theta$. We remark that for technical reasons the proof for
$\theta\in(\frac{5\pi}{6},\pi]$ in Section 6 does not hold for
$\theta=\frac{5\pi}{6}$, so we must consider $\theta=\frac{5\pi}{6}$
separately. Moreover, even through the discreteness and faithfulness of
$\rho_{\frac{5\pi}{6}}(K)<\mathbf{PU}(2,1)$ are much easier than general
$\rho_{\theta}(K)<\mathbf{PU}(3,1)$, but it is also highly nontrivial;

  * •

We then consider the case $\theta=\pi$, that is, we consider ${\bf
H}^{3}_{\mathbb{R}}$-geometry. We remind the reader that in ${\bf
H}^{2}_{\mathbb{C}}$-geometry and ${\bf H}^{3}_{\mathbb{R}}$-geometry of our
groups, even through the combinatorics of Dirichlet domains for
$\rho_{\frac{5\pi}{6}}(K)$ and $\rho_{\pi}(K)$ are different (and they must be
different since one has dimension four and the other has dimension three), but
they are similar. Moreover, the words which have contributions to Dirichlet
domains both are the set $R$. We also have that the intersection patterns of
Dirichlet domains of $\rho_{\frac{5\pi}{6}}(K)$ and $\rho_{\pi}(K)$ are the
same. This can be seen by comparing Figure 6 and Figure 8 in Sections 4 and 5
respectively. This fact is also very lucky. The part about ${\bf
H}^{3}_{\mathbb{R}}$-geometry (Section 5) can be omitted logically by two
reasons: it is simple, and the proof in Section 6 also covers ${\bf
H}^{3}_{\mathbb{R}}$-geometry of $\rho_{\pi}(K)$. But the reader are
encouraged to read this part before Section 6;

  * •

From above, it is very reasonable to guess that for general
$\theta\in(\frac{5\pi}{6},\pi]$, the Dirichlet domain of $\rho_{\theta}(K)$ is
also given by the set $R$. This is what we do in Section 6, that is, we prove
${\bf H}^{3}_{\mathbb{C}}$-geometry of Theorem 1.1 for general
$\theta\in(\frac{5\pi}{6},\pi]$.

In the procedure of the proof of Theorem 1.1, we also find a method that
parameterizing the intersection of three co-equidistant bisectors in ${\bf
H}^{3}_{\mathbb{C}}$. Moreover, for a group in Theorem 1.1, by the
$\mathbb{Z}_{4}$-symmetry, there are two isometric types of 5-facets, say
$s_{12}$ and $s_{13}$; One isometric type of 4-facets, say $s_{12}\cap
s_{14}$; One isometric type of 3-facets, say $s_{12}\cap s_{13}\cap s_{14}$.
In particular, there is no quadruple intersection of bisectors. So we do not
need the precisely combinatorial structure of these facets. What we really
need is the above mentioned facets are all non-empty. Which is enough for the
Poincaré polyhedron theorem in our (lucky) case. See Section 6 for more
details.

But for more general subgroups of $\mathbf{PU}(3,1)$ in future study,
quadruple intersections of bisectors are unavoidably. We pose the following
question, by which the author has difficult to show. But we believe it is
fundamental in ${\bf H}^{3}_{\mathbb{C}}$-geometry. See Subsection 2.6 for the
notations and background.

######  Question 1.2.

Assume any lifts of four points $q_{0}$, $q_{1}$, $q_{2}$ and $q_{3}$ in ${\bf
H}^{3}_{\mathbb{C}}$ are linearly independent vectors in $\mathbb{C}^{3,1}$,
and the triple intersection

$B(q_{0},q_{1})\cap B(q_{0},q_{2})\cap B(q_{0},q_{3})$

of three bisectors $B(q_{0},q_{i})$ in ${\bf H}^{3}_{\mathbb{C}}$ for
$i=1,2,3$ is nonempty, then the triple intersection is a 3-ball.

See Figure 11 for an example of the boundary of the triple intersection of
bisectors, which seems to be 2-sphere.

To our knowledge, Theorem 1.1 is the first moduli space in ${\bf
H}^{3}_{\mathbb{C}}$-geometry that has been studied completely. There are in
fact infinitely many Coxeter polytopes in ${\bf H}^{3}_{\mathbb{R}}$. In this
paper we only complexify the simplest one. Specially, finite volume Coxeter
polytopes in ${\bf H}^{3}_{\mathbb{R}}$ merit further complexifying/deforming
in ${\bf H}^{3}_{\mathbb{C}}$-geometry to find ${\bf
H}^{3}_{\mathbb{C}}$-lattices. We hope this paper may attract more interest on
this promising direction.

The paper is organized as follows. In Section 2 we give well known background
material on complex hyperbolic geometry. In Section 3, we give the matrix
representations of $G$ into $\mathbf{PU}(3,1)$ with complex reflection
generators. Section 4 is devoted to the Dirichlet domain of
$\rho_{\frac{5\pi}{6}}(K)<\mathbf{PU}(2,1)$ acting on ${\bf
H}^{2}_{\mathbb{C}}$. The Dirichlet domain of $\rho_{\pi}(K)<\mathbf{PO}(3,1)$
acting on ${\bf H}^{3}_{\mathbb{R}}$ is studied in Section 5 (but omitting
some details). With the warming up in Sections 4 and 5, we prove Theorem 1.1
in Section 6.

Acknowledgement: The author would like to thank his co-author Baohua Xie [15],
the author learned a lot from Baohua on complex hyperbolic geometry.

##  2\. Background

The purpose of this section is to introduce briefly complex hyperbolic
geometry. One can refer to Goldman’s book [11] for more details.

###  2.1. Complex hyperbolic space

Let ${\mathbb{C}}^{n,1}$ denote the vector space ${\mathbb{C}}^{n+1}$ equipped
with the Hermitian form of signature $(n,1)$:

$\langle{\bf{z}},{\bf{w}}\rangle=\bf{w}^{*}\cdot H\cdot\bf{z},$

where $\bf{w}^{*}$ is the Hermitian transpose of $\bf{w}$,

$H=\begin{pmatrix}Id_{n}&0\\\ 0&-1\\\ \end{pmatrix},$

and $Id_{n}$ is the $n\times n$ identity matrix. Then the Hermitian form
divides ${\mathbb{C}}^{n,1}$ into three parts $V_{-},V_{0}$ and $V_{+}$. Which
are

$\displaystyle V_{-}$ $\displaystyle=$ $\displaystyle\\{{\bf
z}\in{\mathbb{C}}^{n+1}-\\{0\\}:\langle{\bf z},{\bf z}\rangle<0\\},$
$\displaystyle V_{0}$ $\displaystyle=$ $\displaystyle\\{{\bf
z}\in{\mathbb{C}}^{n+1}-\\{0\\}:\langle{\bf z},{\bf z}\rangle=0\\},$
$\displaystyle V_{+}$ $\displaystyle=$ $\displaystyle\\{{\bf
z}\in{\mathbb{C}}^{n+1}-\\{0\\}:\langle{\bf z},{\bf z}\rangle>0\\}.$

Let

$[~{}~{}]:{\mathbb{C}}^{n+1}-\\{0\\}\longrightarrow{\mathbb{C}}{\mathbf{P}}^{n}$

be the canonical projection onto the complex projective space. Then the
complex hyperbolic space ${\bf H}^{n}_{\mathbb{C}}$ is the image of $V_{-}$ in
${\mathbb{C}}{\mathbf{P}}^{n}$ by the map $[~{}~{}]$. The ideal boundary of
${\bf H}^{n}_{\mathbb{C}}$, or boundary at infinity, is the image of $V_{0}$
in ${\mathbb{C}}{\mathbf{P}}^{n}$, we denote it by $\partial{\bf
H}^{n}_{\mathbb{C}}$. In this paper, we will denote by

$\mathbf{q}=(z_{1},z_{2},\cdots,z_{n+1})^{t}$

a vector in ${\mathbb{C}}^{n,1}$ (note that we use the boldface $\mathbf{q}$),
and by

$q=[z_{1},z_{2},\cdots,z_{n+1}]^{t}$

the corresponding point in ${\mathbb{C}}{\mathbf{P}}^{n}$. Here the
superscript $``t"$ means the transpose of a vector. Consider the natural
biholomorphic embedding of ${\mathbb{C}}^{n}$ onto the affine patch of
${\mathbb{C}}{\mathbf{P}}^{n}$, that is

(2.1) $\left(\begin{matrix}z_{1}\\\ z_{2}\\\ \vdots\\\
z_{n}\end{matrix}\right)\longmapsto\left[\begin{matrix}z_{1}\\\ z_{2}\\\
\vdots\\\ z_{n}\\\ 1\end{matrix}\right].$

Then ${\bf H}^{n}_{\mathbb{C}}$ is identified with the unit ball
$\mathbb{B}^{n}$ in ${\mathbb{C}}^{n}$ by this embedding, and $\partial{\bf
H}^{n}_{\mathbb{C}}$ is identified with the unit sphere
$\mathbb{S}^{2n-1}=\partial\mathbb{B}^{n}$ in ${\mathbb{C}}^{n}$.

There is a typical anti-holomorphic isometry $\iota$ of ${\bf
H}^{n}_{\mathbb{C}}$. $\iota$ is given on the level of homogeneous coordinates
by complex conjugate

(2.2) $\iota:\left[\begin{matrix}z_{1}\\\ z_{2}\\\ \vdots\\\
z_{n+1}\end{matrix}\right]\longmapsto\left[\begin{matrix}\overline{z_{1}}\\\
\overline{z_{2}}\\\ \vdots\\\ \overline{z_{n+1}}\end{matrix}\right].$

###  2.2. Totally geodesic submanifolds and complex reflections

There are two kinds of totally geodesic submanifolds in ${\bf
H}^{n}_{\mathbb{C}}$:

  * •

Given any point $x\in{\bf H}^{n}_{\mathbb{C}}$, and a complex linear subspace
$F$ of dimension $k$ in the tangent space $T_{x}{\bf H}^{n}_{\mathbb{C}}$,
there is a unique complete holomorphic totally geodesic submanifold contains
$x$ and is tangent to $F$. Such a holomorphic submanifold is called a
_$\mathbb{C}^{k}$ -plane_. A $\mathbb{C}^{k}$-plane is the intersection of a
complex $k$-dimensional projective subspace in $\mathbb{C}{\bf P}^{n}$ with
${\bf H}^{n}_{\mathbb{C}}$, and it is holomorphic isometric to ${\bf
H}^{k}_{\mathbb{C}}$. A $\mathbb{C}^{1}$-plane is also called a _complex
geodesic_. The intersection of a $\mathbb{C}^{k}$-plane with $\partial{\bf
H}^{n}_{\mathbb{C}}=\mathbb{S}^{2n-1}$ is a smoothly embedded sphere
$\mathbb{S}^{2k-1}$, which is called a _$\mathbb{C}^{k}$ -chain_.

  * •

Corresponding to the compatible real structures on $\mathbb{C}^{n,1}$ are the
real forms of ${\bf H}^{n}_{\mathbb{C}}$. That is, the maximal totally real
totally geodesic sub-spaces of ${\bf H}^{n}_{\mathbb{C}}$, which have real
dimension $n$. A maximal totally real totally geodesic subspace of ${\bf
H}^{n}_{\mathbb{C}}$ is the fixed-point set of an anti-holomorphic isometry of
${\bf H}^{n}_{\mathbb{C}}$. We have gave an example of anti-holomorphic
isometry $\iota$ in (2.2) of Subsection 2.1. For the usual real structure,
this submanifold is the real hyperbolic $n$-space ${\bf H}^{n}_{\mathbb{R}}$
with curvature $-\frac{1}{4}$. Any totally geodesic subspace of a maximal
totally real totally geodesic subspace is a totally real totally geodesic
subspace, which is isometric to the real hyperbolic $k$-space ${\bf
H}^{k}_{\mathbb{R}}$ for some $k$.

Since the Riemannian sectional curvatures of the complex hyperbolic space are
non-constant, there are no totally geodesic hyperplanes in ${\bf
H}^{n}_{\mathbb{C}}$ when $n\geq 2$.

Let $L$ be a $(n-1)$-dimensional complex plane in ${\bf H}^{n}_{\mathbb{C}}$,
a polar vector of $L$ is the unique vector (up to scaling) in
${\mathbb{C}}^{n,1}$ perpendicular to this complex plane with respect to the
Hermitian form. A polar vector of a $(n-1)$-dimensional complex plane belongs
to $V_{+}$ and each vector in $V_{+}$ corresponds to a $(n-1)$-dimensional
complex plane. Moreover, let $L$ be a $(n-1)$-dimensional complex plane with
polar vector ${\bf n}\in V_{+}$, then the complex reflection fixing $L$ with
rotation angle $\theta$ is given by

(2.3) $I_{\bf n,\theta}({\bf z})=-{\bf
z}+(1-\mathrm{e}^{\theta\mathrm{i}})\frac{\langle{\bf z},{\bf
n}\rangle}{\langle{\bf n},{\bf n}\rangle}{\bf n}.$

The complex plane $L$ is also called the _mirror_ of $I_{\bf n,\theta}$. In
this paper, we only consider the case $\theta=\pi$, then the complex
reflection has order 2.

###  2.3. The isometries

The complex hyperbolic space is a Kähler manifold of constant holomorphic
sectional curvature $-1$. We denote by $\mathbf{U}(n,1)$ the Lie group of the
Hermitian form $\langle\cdot,\cdot\rangle$ preserving complex linear
transformations and by $\mathbf{PU}(n,1)$ the group modulo scalar matrices.
The group of holomorphic isometries of ${\bf H}^{n}_{\mathbb{C}}$ is exactly
$\mathbf{PU}(n,1)$. It is sometimes convenient to work with
$\mathbf{SU}(n,1)$. The full isometry group of ${\bf H}^{n}_{\mathbb{C}}$ is

$\widehat{\mathbf{PU}(n,1)}=\langle\mathbf{PU}(n,1),\iota\rangle,$

where $\iota$ is the anti-holomorphic isometry in Subsection 2.1.

Elements of $\mathbf{SU}(n,1)$ fall into three types, according to the number
and types of fixed points of the corresponding isometry. Namely,

  * •

an isometry is loxodromic if it has exactly two fixed points on $\partial{\bf
H}^{n}_{\mathbb{C}}$;

  * •

an isometry is parabolic if it has exactly one fixed point on $\partial{\bf
H}^{n}_{\mathbb{C}}$;

  * •

an isometry is elliptic when it has (at least) one fixed point inside ${\bf
H}^{n}_{\mathbb{C}}$.

An element $A\in\mathbf{SU}(n,1)$ is called regular whenever it has distinct
eigenvalues, an elliptic $A\in\mathbf{SU}(n,1)$ is called special elliptic if
it has a repeated eigenvalue. Complex reflection about a totally geodesic
${\bf H}^{n-1}_{\mathbb{C}}\hookrightarrow{\bf H}^{n}_{\mathbb{C}}$ is an
example of special elliptic element in $\mathbf{SU}(n,1)$.

###  2.4. Bisectors, spinal spheres and Dirichlet domain

Dirichlet domain (or Dirichlet polyhedron) is a fundamental tool to study a
discrete subgroup in $\mathbf{SU}(n,1)$, which is defined in terms of
(infinitely many) bisectors.

######  Definition 2.1.

Given two distinct points $q_{0}$ and $q_{1}$ in ${\bf H}^{n}_{\mathbb{C}}$
with the same norm (e.g. one could take lifts $\mathbf{q_{0}},\mathbf{q_{1}}$
of them such that
$\langle\mathbf{q_{0}},\mathbf{q_{0}}\rangle=\langle\mathbf{q_{1}},\mathbf{q_{1}}\rangle=-1$),
the _bisector_ $B(q_{0},q_{1})$ is the projectivization of the set of negative
vector $\mathbf{x}$ in ${\mathbb{C}}^{n,1}$ with

$|\langle\mathbf{x},\mathbf{q_{0}}\rangle|=|\langle\mathbf{x},\mathbf{q_{1}}\rangle|.$

The spinal sphere of the bisector $B(q_{0},q_{1})$ is the intersection of
$\partial{\bf H}^{n}_{\mathbb{C}}$ with the closure of $B(q_{0},q_{1})$ in
$\overline{{\bf H}^{n}_{\mathbb{C}}}={\bf H}^{n}_{\mathbb{C}}\cup\partial{{\bf
H}^{n}_{\mathbb{C}}}$. The bisector $B(q_{0},q_{1})$ is a topological
$(2n-1)$-ball, and its spinal sphere is a $(2n-2)$-sphere.

A bisector $B(q_{0},q_{1})$ separates ${\bf H}^{n}_{\mathbb{C}}$ into two
half-spaces, one of them containing $q_{0}$.

######  Definition 2.2.

The _Dirichlet domain_ $D_{\Gamma}$ for a discrete group
$\Gamma<\mathbf{PU}(n,1)$ centered at $q_{0}$ is the intersection of the
(closures of the) half-spaces containing $q_{0}$ of all bisectors corresponds
to elements in $\Gamma$ not fixing $q_{0}$. That is,

$D_{\Gamma}=\\{p\in{\bf H}^{n}_{\mathbb{C}}\cup\partial{\bf
H}^{n}_{\mathbb{C}}:|\langle\mathbf{p},\mathbf{q_{0}}\rangle|\leq|\langle\mathbf{p},g(\mathbf{q_{0}})\rangle|,\
\forall g\in\Gamma\ \mbox{with}\ g(q_{0})\neq q_{0}\\}.$

######  Definition 2.3.

For a subset $R\subset\Gamma$, the _partial Dirichlet domain_ $D_{R}$ for a
discrete group $\Gamma<\mathbf{PU}(n,1)$ centered at $q_{0}$ is

$D_{R}=\\{p\in{\bf H}^{n}_{\mathbb{C}}\cup\partial{\bf
H}^{n}_{\mathbb{C}}:|\langle\mathbf{p},\mathbf{q_{0}}\rangle|\leq|\langle\mathbf{p},g(\mathbf{q_{0}})\rangle|,\
\forall g\in R\ \mbox{with}\ g(q_{0})\neq q_{0}\\}.$

From the definition, one can see that parts of bisectors form the boundary of
a Dirichlet domain. For $g\in\Gamma$, when the center $q_{0}$ is clear from
the context, we also denote $B(q_{0},g(q_{0}))$ by $B(g)$, and we will denote
by $s(g)=B(g)\cap D_{\Gamma}$, which is a $(2n-1)$-facet of $D_{\Gamma}$ in
general. Facets of codimension one in $D_{\Gamma}$ will also be called sides.
In general, facets of codimension two in $D_{\Gamma}$ will be called ridges.
Facets of dimension one and zero in $D_{\Gamma}$ will be called edges and
vertices respectively. Moreover, a bounded ridge is a ridge which does not
intersect $\partial{\bf H}^{n}_{\mathbb{C}}$, and if the intersection of a
ridge $r$ and $\partial{\bf H}^{n}_{\mathbb{C}}$ is non-empty, then $r$ is an
infinite ridge.

It is usually very hard to determine $D_{\Gamma}$ because one should check
infinitely many inequalities. Therefore a general method will be to guess that
a partial Dirichlet domain $D_{R}$ is in fact the Dirichlet domain
$D_{\Gamma}$, and then check it using the Poincaré polyhedron theorem. The
basic idea is that 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 ${\bf H}^{n}_{\mathbb{C}}$. If they do (and if the
quotient of $D_{\Gamma}$ by the identification given by the side-pairing maps
is complete), then the Poincaré polyhedron theorem implies that the images of
$D_{\Gamma}$ actually give a global tiling of ${\bf H}^{n}_{\mathbb{C}}$. So
$D_{\Gamma}$ is a fundamental domain of $\Gamma$-action on ${\bf
H}^{n}_{\mathbb{C}}$.

Once a fundamental domain is obtained, one gets an explicit presentation of
$\Gamma$ in terms of the generators given by the side-pairing maps together
with a generating set for the stabilizer $p_{0}$. Where the relations
correspond to so-called ridge cycles, which correspond to the local tilings
bear each co-dimension two facet. For more on the Poincaré polyhedron theorem,
see [7, 20].

###  2.5. Hermitian cross product in ${\bf H}^{2}_{\mathbb{C}}$ and double
intersection of bisectors

The intersections of bisectors in ${\bf H}^{2}_{\mathbb{C}}$ are a little
easier to describe than in higher dimensions. We show this in this subsection.

For complex hyperbolic plane ${\bf H}^{2}_{\mathbb{C}}$ with Hermitian form
given by

$H=\begin{pmatrix}Id_{2}&0\\\ 0&-1\\\ \end{pmatrix}.$

If

$p=\left[\begin{matrix}p_{1}\\\ p_{2}\\\ p_{3}\end{matrix}\right],\quad
q=\left[\begin{matrix}q_{1}\\\ q_{2}\\\ q_{3}\end{matrix}\right]$

are two points in $\mathbf{H}^{2}_{\mathbb{C}}$, then the Hermitian cross
product of $p$ and $q$ is a point in ${\mathbb{C}}{\mathbf{P}}^{2}$ defined by

$p\boxtimes
q=\left[\begin{array}[]{c}\overline{p}_{3}\overline{q}_{2}-\overline{p}_{2}\overline{q}_{3}\\\\[2.0pt]
\overline{p}_{1}\overline{q}_{3}-\overline{p}_{3}\overline{q}_{1}\\\\[2.0pt]
\overline{p}_{1}\overline{q}_{2}-\overline{p}_{2}\overline{q}_{1}\end{array}\right],$

see Page 45 of [11]. Any lift of $p\boxtimes q$ is orthogonal to lifts of both
$p$ and $q$ with respect to the Hermitian form $\langle\cdot,\cdot\rangle$. It
is a Hermitian version of the Euclidean cross product in ${\mathbb{R}}^{3}$.

In order to analyze 2-faces of a Dirichlet polyhedron in ${\bf
H}^{2}_{\mathbb{C}}$, we must study the intersections of bisectors. From the
detailed analysis in [11], we know that the intersection of two bisectors is
usually not totally geodesic and it can be somewhat complicated. In this
paper, we shall only consider the intersections of _co-equidistant bisectors_
, i.e. bisectors equidistant from a common point. When $p,q$ and $r$ are not
in a common complex line, that is, their lifts are linearly independent in
$\mathbb{C}^{2,1}$, then the locus

$B(p,q,r)=B(p,q)\cap B(p,r)$

of points in ${\bf H}^{2}_{\mathbb{C}}$ equidistant to $p$, $q$ and $r$ is a
smooth disk that is not totally geodesic, and is often called a _Giraud disk_.
The following property is crucial when studying fundamental domain.

######  Proposition 2.4 (Giraud).

If $p$, $q$ and $r$ in ${\bf H}^{2}_{\mathbb{C}}$ are not in a common complex
line, then the Giraud disk $B(p,q,r)$ is contained in precisely three
bisectors, namely $B(p,q)$, $B(q,r)$ and $B(p,r)$.

Note that checking whether an isometry maps a Giraud disk to another is
equivalent to checking that corresponding triples of points are mapped to each
other.

In order to study Giraud disks, we will use spinal coordinates. The complex
spine of the bisector $B(p,q)$ is the complex line through the two points $p$
and $q$. The real spine of $B(p,q)$ is the intersection of the complex spine
with the bisector itself, which is a (real) geodesic. The real spine is the
locus of points inside the complex spine which are equidistant from $p$ and
$q$. Bisectors are not totally geodesic, but they have a very nice foliation
by two different families of totally geodesic submanifolds. Mostow [18] showed
that a bisector is the preimage of the real spine under the orthogonal
projection onto the complex spine. The fibres of this projection are complex
lines ${\bf H}^{1}_{\mathbb{C}}\hookrightarrow{\bf H}^{2}_{\mathbb{C}}$ called
the complex slices of the bisector. Goldman [11] showed that a bisector is the
union of all totally real totally geodesic planes containing the real spine.
Such Lagrangian planes are called the real slices or meridians of the
bisector. The complex slices of $B(p,q)$ are given explicitly by choosing a
lift $\mathbf{p}$ (resp. $\mathbf{q}$) of $p$ (resp. $q$). When $p,q\in{\bf
H}^{2}_{\mathbb{C}}$, we simply choose lifts such that
$\langle\mathbf{p},\mathbf{p}\rangle=\langle\mathbf{q},\mathbf{q}\rangle$. The
complex slices of $B(p,q)$ are obtained as the set of negative lines
$(\overline{z}\mathbf{p}-\mathbf{q})^{\bot}$ in ${\bf H}^{2}_{\mathbb{C}}$ for
some arc of values of $z\in\mathbb{S}^{1}$, which is determined by requiring
that
$\langle\overline{z}\mathbf{p}-\mathbf{q},\overline{z}\mathbf{p}-\mathbf{q}\rangle>0$.

Since a point of the bisector is on precisely one complex slice, we can
parameterize the Giraud torus $\hat{B}(p,q,r)$ in ${\bf P}^{2}_{\mathbb{C}}$
by
$(z_{1},z_{2})=(e^{it_{1}},e^{it_{2}})\in\mathbb{S}^{1}\times\mathbb{S}^{1}$
via

(2.4)
$V(z_{1},z_{2})=(\overline{z}_{1}\mathbf{p}-\mathbf{q})\boxtimes(\overline{z}_{2}\mathbf{p}-\mathbf{r})=\mathbf{q}\boxtimes\mathbf{r}+z_{1}\mathbf{r}\boxtimes\mathbf{p}+z_{2}\mathbf{p}\boxtimes\mathbf{q}.$

The Giraud disk $B(p,q,r)$ corresponds to
$(z_{1},z_{2})\in\mathbb{S}^{1}\times\mathbb{S}^{1}$ with

$\langle V(z_{1},z_{2}),V(z_{1},z_{2})\rangle<0.$

It is well known that this region is a topological disk if it is non empty
[11]. The boundary at infinity $\partial B(p,q,r)$ is a circle, given in
spinal coordinates by the equation

$\langle V(z_{1},z_{2}),V(z_{1},z_{2})\rangle=0.$

Note that the choices of different lifts of $p$, $q$ and $r$ affect the spinal
coordinates by rotation on each of the $\mathbb{S}^{1}$-factors.

A defining equation for the trace of another bisector $B(u,v)$ on the Giraud
disk $B(p,q,r)$ can be written in the form

$|\langle V(z_{1},z_{2}),\mathbf{u}\rangle|=|\langle
V(z_{1},z_{2}),\mathbf{v}\rangle|,$

provided that $\mathbf{u}$ and $\mathbf{v}$ are suitably chosen lifts. The
expressions $\langle V(z_{1},z_{2}),\mathbf{u}\rangle$ and $\langle
V(z_{1},z_{2}),\mathbf{v}\rangle$ are affine in $z_{1}$ and $z_{2}$.

This triple bisector intersection can be parameterized fairly explicitly,
because one can solve the equation

$|\langle V(z_{1},z_{2}),\mathbf{u}\rangle|^{2}=|\langle
V(z_{1},z_{2}),\mathbf{v}\rangle|^{2}$

for one of the variables $z_{1}$ or $z_{2}$ simply by solving a quadratic
equation. A detailed explanation of how this works can be found in [4, 6, 7].

###  2.6. Triple Hermitian cross product in ${\bf H}^{3}_{\mathbb{C}}$ and
triple intersection of bisectors

Now we consider triple intersections of bisectors in ${\bf
H}^{3}_{\mathbb{C}}$.

Let $q_{0}$, $q_{1}$, $q_{2}$ and $q_{3}$ be four points in
$\mathbf{H}^{3}_{\mathbb{C}}$. We take lifts $\mathbf{q_{i}}$ of $q_{i}$ such
that

$\langle\mathbf{q_{0}},\mathbf{q_{0}}\rangle=\langle\mathbf{q_{1}},\mathbf{q_{1}}\rangle=\langle\mathbf{q_{2}},\mathbf{q_{2}}\rangle=\langle\mathbf{q_{3}},\mathbf{q_{3}}\rangle.$

We also assume
$\\{\mathbf{q_{0}},\mathbf{q_{1}},\mathbf{q_{2}},\mathbf{q_{3}}\\}$ are
linearly independent as vectors in $\mathbb{C}^{3,1}$. We have three bisectors
$B(q_{0},q_{1})$, $B(q_{0},q_{2})$ and $B(q_{0},q_{3})$ in ${\bf
H}^{3}_{\mathbb{C}}$. We now parameterize the triple intersection

$B(q_{0},q_{1})\cap B(q_{0},q_{2})\cap B(q_{0},q_{3}).$

The bisector $B(q_{0},q_{1})$ has a decomposition into a set of complex slices
(each of them is a totally geodesic
$\mathbf{H}^{2}_{\mathbb{C}}\hookrightarrow\mathbf{H}^{3}_{\mathbb{C}}$),
these complex slices are obtained as the set of negative lines in
$(w\mathbf{q_{0}}-\mathbf{q_{1}})^{\perp}$ for some arc of values
$w\in\mathbb{S}^{1}$, see [11, 4]. Similarly, $B(q_{0},q_{2})$ and
$B(q_{0},q_{3})$ also have decompositions into a set of complex slices which
are parameterized by $(w\mathbf{q_{0}}-\mathbf{q_{2}})^{\perp}$ and
$(w\mathbf{q_{0}}-\mathbf{q_{3}})^{\perp}$ for some arc of values
$w\in\mathbb{S}^{1}$.

Consider triple Hermitian cross product with respect to the Hermitian form
$H$. Recall that for three linearly independent vectors

(2.5)
$a=\left(\begin{matrix}a_{1},a_{2},a_{3},a_{4}\end{matrix}\right)^{t},~{}~{}b=\left(\begin{matrix}b_{1},b_{2},b_{3},b_{4}\end{matrix}\right)^{t},~{}~{}c=\left(\begin{matrix}c_{1},c_{2},c_{3},c_{4}\end{matrix}\right)^{t}$

in $\mathbb{R}^{4}$, the _generalized cross product $a\times b\times c$_ of
$a,b,c$ is a vector

$d=\left(\begin{matrix}d_{1},d_{2},d_{3},d_{4}\end{matrix}\right)^{t}.$

Where $d_{i}$ is the coefficient of $e_{i}$ in the determinant of the matrix

$\begin{pmatrix}a_{1}&b_{1}&c_{1}&e_{1}\\\ a_{2}&b_{2}&c_{2}&e_{2}\\\
a_{3}&b_{3}&c_{3}&e_{3}\\\ a_{4}&b_{4}&c_{4}&e_{4}\\\ \end{pmatrix}.$

The vector $d=a\times b\times c$ is perpendicular to $a$, $b$ and $c$ with
respect to the standard Euclidean metric in $\mathbb{R}^{4}$. Now for three
linearly independent vectors $a$, $b$, $c$ as in (2.5) in $\mathbb{C}^{3,1}$
with $a_{i},b_{i},c_{i}\in\mathbb{C}$, the _triple Hermitian cross product
$a\boxtimes b\boxtimes c$_ of $a$, $b$, $c$ is a vector

$d=\left(\begin{matrix}d_{1},d_{2},d_{3},d_{4}\end{matrix}\right)^{t}.$

Where by definition

$d=\begin{pmatrix}-1&0&0&0\\\ 0&-1&0&0\\\ 0&0&-1&0\\\ 0&0&0&1\\\
\end{pmatrix}\cdot(\bar{a}\times\bar{b}\times\bar{c}),$

here for example

$\bar{a}=\left(\begin{matrix}\overline{a_{1}},\overline{a_{2}},\overline{a_{3}},\overline{a_{4}}\end{matrix}\right)^{t}$

with $\overline{a_{i}}$ the complex conjugate of $a_{i}$. Then $d=a\boxtimes
b\boxtimes c$ is the vector

$\left(\begin{matrix}\det\begin{pmatrix}\overline{a_{2}}&\overline{b_{2}}&\overline{c_{2}}\\\
\overline{a_{3}}&\overline{b_{3}}&\overline{c_{3}}\\\
\overline{a_{4}}&\overline{b_{4}}&\overline{c_{4}}\\\
\end{pmatrix},-\det\begin{pmatrix}\overline{a_{1}}&\overline{b_{1}}&\overline{c_{1}}\\\
\overline{a_{3}}&\overline{b_{3}}&\overline{c_{3}}\\\
\overline{a_{4}}&\overline{b_{4}}&\overline{c_{4}}\\\
\end{pmatrix},\det\begin{pmatrix}\overline{a_{1}}&\overline{b_{1}}&\overline{c_{1}}\\\
\overline{a_{2}}&\overline{b_{2}}&\overline{c_{2}}\\\
\overline{a_{4}}&\overline{b_{4}}&\overline{c_{4}}\\\
\end{pmatrix},\det\begin{pmatrix}\overline{a_{1}}&\overline{b_{1}}&\overline{c_{1}}\\\
\overline{a_{2}}&\overline{b_{2}}&\overline{c_{2}}\\\
\overline{a_{3}}&\overline{b_{3}}&\overline{c_{3}}\\\
\end{pmatrix}\end{matrix}\right)^{t}.$

By direct calculation, we have

$\langle a,a\boxtimes b\boxtimes c\rangle=\langle b,a\boxtimes b\boxtimes
c\rangle=\langle c,a\boxtimes b\boxtimes c\rangle=0$

with respect to the Hermitioan form $H$ on $\mathbb{C}^{3,1}$.

Since a point in a bisector $B$ lies in precisely one complex slice of $B$, we
can parameterize $B(q_{0},q_{1})\cap B(q_{0},q_{2})\cap B(q_{0},q_{3})$ by

$V(w_{1},w_{2},w_{3})=(\overline{w_{1}}\mathbf{q_{0}}-\mathbf{q_{1}})\boxtimes(\overline{w_{2}}\mathbf{q_{0}}-\mathbf{q_{2})}\boxtimes(\overline{w_{3}}\mathbf{q_{0}}-\mathbf{q_{3}})$

with
$(w_{1},w_{2},w_{3})\in\mathbb{S}^{1}\times\mathbb{S}^{1}\times\mathbb{S}^{1}$.
Up to sign and rewriting, we can parameterize $B(q_{0},q_{1})\cap
B(q_{0},q_{2})\cap B(q_{0},q_{3})$ by

(2.6)
$V(z_{1},z_{2},z_{3})=\mathbf{q_{1}}\boxtimes\mathbf{q_{2}}\boxtimes\mathbf{q_{3}}+z_{1}\cdot\mathbf{q_{0}}\boxtimes\mathbf{q_{2}}\boxtimes\mathbf{q_{3}}+z_{2}\cdot\mathbf{q_{0}}\boxtimes\mathbf{q_{1}}\boxtimes\mathbf{q_{3}}+z_{3}\cdot\mathbf{q_{0}}\boxtimes\mathbf{q_{1}}\boxtimes\mathbf{q_{2}}$

with
$(z_{1},z_{2},z_{3})\in\mathbb{S}^{1}\times\mathbb{S}^{1}\times\mathbb{S}^{1}$
such that $\langle V(z_{1},z_{2},z_{3}),V(z_{1},z_{2},z_{3})\rangle$ is
negative.

We remark that it is reasonable to guess that for fixed
$\\{q_{0},q_{1},q_{2},q_{3}\\}$ in $\mathbf{H}^{3}_{\mathbb{C}}$, if there is
$(z_{1},z_{2},z_{3})\in\mathbb{S}^{1}\times\mathbb{S}^{1}\times\mathbb{S}^{1}$
such that $V=V(z_{1},z_{2},z_{3})$ is negative in (2.6), then all
$(z_{1},z_{2},z_{3})\in\mathbb{S}^{1}\times\mathbb{S}^{1}\times\mathbb{S}^{1}$
satisfying this condition should be a 3-ball in
$\mathbb{S}^{1}\times\mathbb{S}^{1}\times\mathbb{S}^{1}$. But the author has
difficult to show this. See Question 1.2.

##  3\. The moduli space of representations of $G$ into $\mathbf{PU}(3,1)$

In this section, we give the matrix representations of $G$ into
$\mathbf{PU}(3,1)$ with complex reflection generators and
$\iota_{i}\iota_{i+1}$ is mapped to a parabolic element for $i=1,2,3,4$ mod
$4$.

Let $G$ be the abstract group with the presentation

$G=\left\langle\iota_{1},\iota_{2},\iota_{3},\iota_{4}\Bigg{|}\begin{array}[]{c}\iota_{1}^{2}=\iota_{2}^{2}=\iota_{3}^{2}=\iota_{4}^{2}=id,\\\\[3.0pt]
(\iota_{1}\iota_{3})^{2}=(\iota_{2}\iota_{4})^{2}=id\end{array}\right\rangle.$

$G$ is isomorphic to
$(\mathbb{Z}_{2}\oplus\mathbb{Z}_{2})\ast(\mathbb{Z}_{2}\oplus\mathbb{Z}_{2})$
abstractly. Then
$K=\langle\iota_{1}\iota_{3},\iota_{2}\iota_{4},\iota_{1}\iota_{2}\rangle$ is
an index two subgroup of $G$, which is isomorphic to
$\mathbb{Z}_{2}\ast\mathbb{Z}_{2}\ast\mathbb{Z}$.

###  3.1. The Gram matrices of four complex hyperbolic planes in
$\mathbf{H}^{3}_{\mathbb{C}}$ for the group $G$

Recall that for two $\mathbb{C}$-planes $\mathcal{P}$ and
$\mathcal{P}^{\prime}$ in $\mathbf{H}^{3}_{\mathbb{C}}$ with polar vectors $n$
and $n^{\prime}$ such that $\langle n,n\rangle=\langle
n^{\prime},n^{\prime}\rangle=1$:

  * •

If $\mathcal{P}$ and $\mathcal{P}^{\prime}$ intersect in a $\mathbb{C}$-line
in $\mathbf{H}^{3}_{\mathbb{C}}$, then the angle $\alpha$ between them has
$|\langle n,n^{\prime}\rangle|=\cos(\alpha)$;

  * •

If $\mathcal{P}$ and $\mathcal{P}^{\prime}$ are hyper-parallel in
$\mathbf{H}^{3}_{\mathbb{C}}$, then the distance $d$ between them has
$|\langle n,n^{\prime}\rangle|=\cosh\frac{d}{2}$;

  * •

If $\mathcal{P}$ and $\mathcal{P}^{\prime}$ are asymptotic in
$\mathbf{H}^{3}_{\mathbb{C}}$, then $|\langle n,n^{\prime}\rangle|=1$.

We consider $\mathbb{C}$-planes $\mathcal{P}_{i}$ for $i=1,2,3,4$, so each
$\mathcal{P}_{i}$ is a totally geodesic ${\bf
H}^{2}_{\mathbb{C}}\hookrightarrow{\bf H}^{3}_{\mathbb{C}}$. Let
$n^{\prime}_{i}$ be the polar vector of $\mathcal{P}_{i}$ in
$\mathbb{C}{\mathbf{P}}^{3}-\overline{\mathbf{H}^{3}_{\mathbb{C}}}$. We assume

  * •

the angle between $\mathcal{P}_{1}$ and $\mathcal{P}_{3}$ is $\frac{\pi}{2}$;

  * •

the angle between $\mathcal{P}_{2}$ and $\mathcal{P}_{4}$ is $\frac{\pi}{2}$;

  * •

the planes $\mathcal{P}_{i}$ and $\mathcal{P}_{i+1}$ are asymptotic for
$i=1,2,3,4$ mod $4$.

Then we can normalize the Gram matrix of $\\{n^{\prime}_{i}\\}^{4}_{i=1}$ into
the following form

$\mathcal{G}^{\prime}=(\langle n^{\prime}_{i},n^{\prime}_{j}\rangle)_{1\leq
i,j\leq 4}=\begin{pmatrix}1&1&0&\mathrm{e}^{-4\theta\mathrm{i}}\\\ 1&1&1&0\\\
0&1&1&1\\\ \mathrm{e}^{4\theta\mathrm{i}}&0&1&1\\\ \end{pmatrix}.$

Where up to anti-holomorphic isometry of ${\bf H}^{3}_{\mathbb{C}}$, we may
assume $4\theta\in[a,a+\pi]$ for any $a\in\mathbb{R}$, that is
$\theta\in[\frac{a}{4},\frac{a+\pi}{4}]$. Moreover $4\theta=2k\pi$ for
$k\in\mathbb{Z}$ corresponds to the case of an infinite volume 3-dimensional
real hyperbolic Coxeter tetrahedron, see [27]. By [2], for a Gram matrix
above, there is a unique configuration of four $\mathbb{C}$-planes
$\mathcal{P}_{i}$ in ${\bf H}^{3}_{\mathbb{C}}$ for $i=1,2,3,4$ up to
$\mathbf{PU}(3,1)$ realizing the Gram matrix.

In fact, we can re-normalize $n^{\prime}_{i}$ above into

$n_{1}=n^{\prime}_{1},~{}~{}~{}~{}~{}n_{2}=\mathrm{e}^{-\theta\mathrm{i}}\cdot
n^{\prime}_{2},~{}~{}~{}~{}~{}n_{3}=\mathrm{e}^{-2\theta\mathrm{i}}\cdot
n^{\prime}_{3},~{}~{}~{}~{}~{}n_{4}=\mathrm{e}^{-3\theta\mathrm{i}}\cdot
n^{\prime}_{4},$

Then we can re-normalize the Gram matrix to the following form

(3.1) $\mathcal{G}=(\langle n_{i},n_{j}\rangle)_{1\leq i,j\leq
4}=\begin{pmatrix}1&\mathrm{e}^{\theta\mathrm{i}}&0&\mathrm{e}^{-\theta\mathrm{i}}\\\
\mathrm{e}^{-\theta\mathrm{i}}&1&\mathrm{e}^{\theta\mathrm{i}}&0\\\
0&\mathrm{e}^{-\theta\mathrm{i}}&1&\mathrm{e}^{\theta\mathrm{i}}\\\
\mathrm{e}^{\theta\mathrm{i}}&0&\mathrm{e}^{-\theta\mathrm{i}}&1\\\
\end{pmatrix}.$

From now on we may assume that $\theta\in[\frac{3\pi}{4},\pi]$. It is easy to
see

$\det(\mathcal{G})=-1-2\cos(4\theta),$

and the eigenvalues of $\mathcal{G}$ are

$1\pm 2\cos(\theta),~{}~{}~{}1\pm 2\sin(\theta).$

We have the followings:

  * •

when $\theta=\pi$, all entries of $\mathcal{G}$ are reals, so it degenerates
to ${\bf H}^{3}_{\mathbb{R}}$-geometry;

  * •

when $\theta=\frac{5\pi}{6}$, $\mathcal{G}$ has eigenvalues

$0,~{}~{}~{}2,~{}~{}~{}1+\sqrt{3},~{}~{}~{}1-\sqrt{3},$

so it degenerates to ${\bf H}^{2}_{\mathbb{C}}$-geometry;

  * •

when $\theta\in(\frac{5\pi}{6},\pi]$, $\mathcal{G}$ has signature $(3,1)$, we
have ${\bf H}^{3}_{\mathbb{C}}$-geometry;

  * •

when $\theta\in[\frac{3\pi}{4},\frac{5\pi}{6})$, $\mathcal{G}$ has signature
$(2,2)$. We will not study them in this paper.

So our moduli space is

(3.2) $\mathscr{M}=\left[\frac{5\pi}{6},\pi\right].$

From the Gram matrix (3.1), there is a $\mathbb{Z}_{4}$-symmetry of the
configurations of $\mathbb{C}$-planes above. Take

$J=\begin{pmatrix}-1&0&0&0\\\ 0&\rm{i}&0&0\\\ 0&0&-\rm{i}&0\\\ 0&0&0&1\\\
\end{pmatrix}.$

Then $J^{*}HJ=H$, $J$ has order 4 with fixed point

(3.3) $p_{0}=\left[\begin{matrix}0,0,0,1\end{matrix}\right]^{t}$

in ${\bf H}^{3}_{\mathbb{C}}$. Note that $\det(J)=-1$, so
$J\in\mathbf{U}(3,1)$, but $J\notin\mathbf{PU}(3,1)$, and
$\rm{e}^{\frac{\pi\rm{i}}{4}}\cdot J\in\mathbf{PU}(3,1)$. But we can also use
$J$ to study the $\mathbb{Z}_{4}$-symmetry for simplicity of notations.

We take

(3.4)
${n_{1}}=\left[\begin{array}[]{c}\displaystyle{\frac{\sqrt{1-2\cos(\theta)}}{2}}\\\\[12.91663pt]
\displaystyle{\frac{\sqrt{1-2\sin(\theta)}}{2}}\\\\[12.91663pt]
\displaystyle{\frac{\sqrt{1+2\sin(\theta)}}{2}}\\\\[12.91663pt]
\displaystyle{\frac{\sqrt{-1-2\cos(\theta)}}{2}}\\\ \end{array}\right]$

in ${\mathbb{C}}{\mathbf{P}}^{3}-\overline{{\bf H}^{3}_{\mathbb{C}}}$. When
$\theta\in[\frac{5\pi}{6},\pi]$, all entries of $n_{1}$ are non-negative
reals. Let $n_{i}$ be $J^{i-1}(n_{1})$ for $i=2,3,4$. Then

$(\langle n_{i},n_{j}\rangle)_{1\leq i,j\leq 4}$

is $\mathcal{G}$ in (3.1). By [2], there is a one-to-one correspondence
between the Gram matrix and configurations $\mathbb{C}$-planes, so there is a
$\mathbb{Z}_{4}$-symmetry of the configurations $\mathbb{C}$-planes above.

######  Remark 3.1.

The author notes that the $\mathbb{Z}_{4}$-symmetry above is lucky since it
will simplify the calculations dramatically. The $\mathbb{Z}_{4}$-symmetry is
also a little mysterious since the author can not explain it from geometric
point of view.

Now for any $\theta\in\mathcal{M}$, let
$\rho_{\theta}:G\rightarrow\mathbf{PU}(3,1)$ be the representation with
$\rho_{\theta}(\iota_{i})=I_{i}$ the order two $\mathbb{C}$-reflection about
$\mathcal{P}_{i}$. We also denote by

$\Gamma=\Gamma_{\theta}=\langle I_{1},I_{2},I_{3},I_{4}\rangle.$

  * •

When $\theta=\frac{5\pi}{6}$, $\Gamma$ preserves a totally geodesic ${\bf
H}^{2}_{\mathbb{C}}\hookrightarrow{\bf H}^{3}_{\mathbb{C}}$ invariant, so we
will view the representation as degenerating to a representation into
$\mathbf{PU}(2,1)$. But the discreteness and faithfulness of this
representation are still non-trivial, see Section 4 for more details.

  * •

When $\theta=\pi$, $\Gamma$ preserves a totally geodesic ${\bf
H}^{3}_{\mathbb{R}}\hookrightarrow{\bf H}^{3}_{\mathbb{C}}$ invariant. We have
a 3-dimensional real hyperbolic Coxeter tetrahedron, so both the discreteness
and faithfulness of this representation are trivial. The Dirichlet domain of
this group representation in ${\bf H}^{3}_{\mathbb{R}}$ is not difficult, see
Section 5 for more details.

  * •

When we decrease $\theta$ from $\pi$ to $\frac{5\pi}{6}$, we still have a
representation of $G$ into $\mathbf{PU}(3,1)$, but the discreteness of the
representation is highly non-trivial. This is what we will do in the paper.

###  3.2. Matrices representations of the group $G$ into $\mathbf{PU}(3,1)$

For each $\theta\in\mathcal{M}$, we now give the matrix presentation of
$I_{i}$, the order-two complex reflection with mirror $\mathcal{P}_{i}$ for
$i=1,2,3,4$.

From (2.3) and the vector $n_{1}$ in (3.4), it is easy to see

$I_{1}=\left(\begin{array}[]{cccc}-\displaystyle{\frac{1+2\cos(\theta)}{2}}&a_{12}&a_{13}&a_{14}\\\\[5.0pt]
a_{12}&-\displaystyle{\frac{1+2\sin(\theta)}{2}}&a_{23}&a_{24}\\\\[5.0pt]
a_{13}&a_{23}&\displaystyle{\frac{-1+2\sin(\theta)}{2}}&a_{34}\\\\[5.0pt]
-a_{14}&-a_{24}&-a_{34}&\displaystyle{\frac{-1+2\cos(\theta)}{2}}\\\
\end{array}\right),$

where

$a_{12}=\frac{\sqrt{(1-2\sin(\theta))\cdot(1-2\cos(\theta))}}{2},$
$a_{13}=\frac{\sqrt{(1+2\sin(\theta))\cdot(1-2\cos(\theta))}}{2},$
$a_{14}=-\frac{\sqrt{(2\cos(\theta)-1)\cdot(1+2\cos(\theta))}}{2},$
$a_{23}=\frac{\sqrt{(1+2\sin(\theta))\cdot(1-2\sin(\theta))}}{2},$
$a_{24}=-\frac{\sqrt{(2\sin(\theta)-1)\cdot(1+2\cos(\theta))}}{2},$

and

$a_{34}=-\frac{\sqrt{(1+2\sin(\theta))\cdot(-1-2\cos(\theta))}}{2}.$

Since $\theta\in[\frac{5\pi}{6},\pi]$, all the terms $a_{ij}$ above are real.

Then $I_{1}$ is the order two $\mathbb{C}$-reflection with mirror
$\mathcal{P}_{1}$. Let $I_{i}=JI_{i-1}J^{-1}$, then $I_{i}$ is the order two
$\mathbb{C}$-reflection with mirror $\mathcal{P}_{i}$ for $i=2,3,4$.

By direct calculation we have

$\det(I_{1})=\det(I_{2})=\det(I_{3})=\det(I_{4})=-1,$

and

$I^{*}_{i}HI_{i}=H$

for $i=1,2,3,4$, here $I^{*}_{i}$ is the Hermitian transpose of $I_{i}$.
Moreover

$(I_{1}I_{3})^{2}=(I_{2}I_{4})^{2}=id,$

and $I_{1}I_{2}$ is parabolic, so we get a representation $\rho$ of $G$ into
$\mathbf{PU}(3,1)$ (with additional condition that $I_{i}I_{i+1}$ is parabolic
for $i=1,2,3,4$ mod 4). Note that $\det(I_{i})=-1$, so
$I_{i}\in\mathbf{U}(3,1)$, but $I_{i}\notin\mathbf{PU}(3,1)$, and ${\rm
e}^{\frac{\pi\rm{i}}{4}}\cdot I_{i}\in\mathbf{PU}(3,1)$. But we can also use
the matrices $I_{i}$ to study these representations for simplicity of
notations.

In the following, we also denote $A_{i}=I_{i}I_{i+1}$ for $i=1,2,3,4$ mod $4$.
Then $A_{i}$ is parabolic, $A_{1}A_{2}=A_{3}A_{4}$ and $A_{2}A_{3}=A_{4}A_{1}$
have order 2. Then

$\rho_{\theta}(K)=\langle A_{1},A_{2},A_{3},A_{4}\rangle$

is an index two subgroup of $\Gamma=\rho_{\theta}(G)$. We have
$A_{1}A_{2}=I_{1}I_{3}$ is

$\left(\begin{array}[]{cccc}2\cos(\theta)&0&0&-\sqrt{4\cos^{2}(\theta)-1}\\\\[5.0pt]
0&2\sin(\theta)&-\sqrt{1-4\sin^{2}(\theta)}&0\\\\[5.0pt]
0&-\sqrt{1-4\sin^{2}(\theta)}&-2\sin(\theta)&0\\\\[5.0pt]
-\sqrt{4\cos^{2}(\theta)-1}&0&0&-2\cos(\theta)\\\ \end{array}\right).$

The matrix $A_{1}$ is a little complicated, we omit it, but it is easy to get
it from the matrices of $I_{1}$ and $J$.

##  4\. Dirichlet domain of $\rho_{\frac{5\pi}{6}}(K)<\mathbf{PU}(2,1)$ in
${\bf H}^{2}_{\mathbb{C}}$

In this section, we will prove Theorem 1.1 for $\theta=\frac{5\pi}{6}$ via the
Dirichlet domain of $\rho_{\frac{5\pi}{6}}(K)$ with center $p_{0}$. We note
that the proof of Theorem 1.1 when $\theta\in(\frac{5\pi}{6},\pi]$ in Section
6 does not work for $\theta=\frac{5\pi}{6}$, since (the lifts of) four points

$\left\\{p_{0},~{}~{}~{}I_{1}I_{2}(p_{0}),~{}~{}I_{1}I_{3}(p_{0}),~{}~{}I_{1}I_{4}(p_{0})\right\\}$

are linearly dependent when $\theta=\frac{5\pi}{6}$ (see Lemma 6.2). In spite
of this, the proof in this section is a model for the proof of Theorem 1.1 for
general $\theta$ in Section 6.

###  4.1. Matrices in $\mathbf{PU}(2,1)$

From (3.4) we have

(4.1)
$n_{1}=\left[\begin{matrix}\displaystyle{\frac{\sqrt{1+\sqrt{3}}}{2}},~{}~{}~{}~{}0,~{}~{}~{}~{}\frac{\sqrt{2}}{2},~{}~{}~{}~{}\frac{\sqrt{\sqrt{3}-1}}{2}\end{matrix}\right]^{t}$

when $\theta=\frac{5\pi}{6}$.

So the second entry of each $n_{i}$ is zero for $i=1,2,3,4$. We take

(4.2) $n=\left[\begin{matrix}0,1,0,0\end{matrix}\right]^{t},$

in $\mathbb{C}{\bf P}^{3}-\overline{{\bf H}^{3}_{\mathbb{C}}}$, then $\langle
n,n_{i}\rangle=0$ for $i=1,2,3,4$. The intersection of ${\bf
H}^{3}_{\mathbb{C}}$ and the dual of $n$ in $\mathbb{C}{\bf P}^{3}$ with
respect to the Hermitian form is a copy of ${\bf H}^{2}_{\mathbb{C}}$, which
is

$\mathcal{P}=\left\\{\left[\begin{array}[]{c}z_{1}\\\ 0\\\ z_{3}\\\ 1\\\
\end{array}\right]\in{\bf H}^{3}_{\mathbb{C}}\right\\}.$

Each $I_{i}$ preserves this ${\bf H}^{2}_{\mathbb{C}}\hookrightarrow{\bf
H}^{3}_{\mathbb{C}}$ invariant. So we delete the second column and the second
row of the matrix $I_{i}$ in Subsection 3.2, we get new matrices in
$\mathbf{U}(2,1)$. But we still denote them by $I_{i}$ for the simplification
of notations. We have

$I_{1}=\left(\begin{array}[]{ccc}\displaystyle{\frac{\sqrt{3}-1}{2}}&\displaystyle{\frac{1}{\sqrt{\sqrt{3}-1}}}&-\displaystyle{\frac{\sqrt{2}}{2}}\\\\[12.0pt]
\displaystyle{\frac{1}{\sqrt{\sqrt{3}-1}}}&0&-\displaystyle{\frac{1}{\sqrt{\sqrt{3}+1}}}\\\\[16.0pt]
\displaystyle{\frac{\sqrt{2}}{2}}&\displaystyle{\frac{1}{\sqrt{\sqrt{3}+1}}}&-\displaystyle{\frac{\sqrt{3}+1}{2}}\\\
\end{array}\right),$

$J=\left(\begin{matrix}-1&0&0\\\ 0&-\rm{i}&0\\\ 0&0&1\\\ \end{matrix}\right),$

and

$A_{1}=I_{1}I_{2}=\left(\begin{array}[]{ccc}\displaystyle{\frac{3+\rm{i}+\sqrt{3}\rm{i}-\sqrt{3}}{2}}&-\rm{i}\displaystyle{\sqrt{\sqrt{3}-1}}&\displaystyle{\frac{\sqrt{2}(\sqrt{3}+\rm{i})}{2}}\\\\[10.0pt]
\displaystyle{\sqrt{\sqrt{3}-1}}&-\sqrt{3}\rm{i}&\displaystyle{\sqrt{\sqrt{3}+1}}\\\\[10.0pt]
\displaystyle{\frac{\sqrt{2}(\sqrt{3}+\rm{i})}{2}}&-\rm{i}\displaystyle{\sqrt{\sqrt{3}+1}}&\displaystyle{\frac{3-\rm{i}+\sqrt{3}\rm{i}+\sqrt{3}}{2}}\\\
\end{array}\right),$

$A_{i}=JA_{i-1}J^{-1}$ for $i=2,3,4$ and

$A_{1}A_{2}=I_{1}I_{3}=\left(\begin{array}[]{ccc}-\sqrt{3}&0&\sqrt{2}\\\\[2.0pt]
0&-1&0\\\\[2.0pt] -\sqrt{2}&0&\sqrt{3}\\\ \end{array}\right).$

In fact the above $I_{i}$ for $i=1,2,3,4$ and $J$ are in $\mathbf{U}(2,1)$,
but not in $\mathbf{PU}(2,1)$ since the determinants of them are not 1. But we
can also use these matrices to study the group action on ${\bf
H}^{2}_{\mathbb{C}}$ for simplicity of notations.

###  4.2. A partial Dirichlet domain $D_{R}$ of
$\rho_{\frac{5\pi}{6}}(K)<\mathbf{PU}(2,1)$ in ${\bf H}^{2}_{\mathbb{C}}$

In this subsection, we will define a subset
$R\subset\rho_{\frac{5\pi}{6}}(K)$, from which we have the partial the partial
Dirichlet domain $D_{R}$ in ${\bf H}^{2}_{\mathbb{C}}$.

First note that $I_{1}I_{2}$ is parabolic with fixed point

$q_{12}=[\displaystyle{\frac{-1+\rm{i}}{\sqrt{2}\sqrt{\sqrt{3}+1}}},~{}~{}1,~{}~{}\frac{1+\rm{i}}{\sqrt{2}\sqrt{\sqrt{3}-1}}]^{t}$

in $\partial{\bf H}^{2}_{\mathbb{C}}$. We denote by
$p_{ij}=I_{i}I_{j}(p_{0})\in{\bf H}^{2}_{\mathbb{C}}$ and
$B_{ij}=B(p_{0},p_{ij})$ the bisector with respect to the two points $p_{0}$
and $p_{ij}$ for certain $i,j\in\\{1,2,3,4\\}$. Now

  * •

$p_{0}=[0,0,1]^{t}$;

  * •

$p_{12}=[\displaystyle{\frac{\sqrt{2}(\sqrt{3}+\rm{i})}{2}},~{}~{}~{}\sqrt{\sqrt{3}+1},~{}~{}~{}\frac{3-\rm{i}+\sqrt{3}+\sqrt{3}\rm{i}}{2}]^{t}$;

  * •

$p_{21}=[-\displaystyle{\frac{\sqrt{2}(\sqrt{3}-\rm{i})}{2}},~{}~{}~{}-\rm{i}\sqrt{\sqrt{3}+1},~{}~{}~{}\frac{3+\rm{i}+\sqrt{3}-\sqrt{3}\rm{i}}{2}]^{t}$;

  * •

$p_{23}=[-\displaystyle{\frac{\sqrt{2}(\sqrt{3}+\rm{i})}{2}},~{}~{}~{}-\rm{i}\sqrt{\sqrt{3}+1},~{}~{}~{}\frac{3-\rm{i}+\sqrt{3}+\sqrt{3}\rm{i}}{2}]^{t}$;

  * •

$p_{32}=[\displaystyle{\frac{\sqrt{2}(\sqrt{3}-\rm{i})}{2}},~{}~{}~{}-\sqrt{\sqrt{3}+1},~{}~{}~{}\frac{3+\rm{i}+\sqrt{3}-\sqrt{3}\rm{i}}{2}]^{t}$;

  * •

$p_{34}=[\displaystyle{\frac{\sqrt{2}(\sqrt{3}+\rm{i})}{2}},~{}~{}~{}-\sqrt{\sqrt{3}+1},~{}~{}~{}\frac{3-\rm{i}+\sqrt{3}+\sqrt{3}\rm{i}}{2}]^{t}$;

  * •

$p_{43}=[-\displaystyle{\frac{\sqrt{2}(\sqrt{3}-\rm{i})}{2}},~{}~{}~{}\rm{i}\sqrt{\sqrt{3}+1},~{}~{}~{}\frac{3+\rm{i}+\sqrt{3}-\sqrt{3}\rm{i}}{2}]^{t}$;

  * •

$p_{41}=[-\displaystyle{\frac{\sqrt{2}(\sqrt{3}+\rm{i})}{2}},~{}~{}~{}\rm{i}\sqrt{\sqrt{3}+1},~{}~{}~{}\frac{3-\rm{i}+\sqrt{3}+\sqrt{3}\rm{i}}{2}]^{t}$;

  * •

$p_{14}=[\displaystyle{\frac{\sqrt{2}(\sqrt{3}-\rm{i})}{2}},~{}~{}~{}\sqrt{\sqrt{3}+1},~{}~{}~{}\frac{3+\rm{i}+\sqrt{3}-\sqrt{3}\rm{i}}{2}]^{t}$;

  * •

$p_{13}=[\sqrt{2},0,\sqrt{3}]^{t}$;

  * •

$p_{24}=[-\sqrt{2},0,\sqrt{3}]^{t}$.

We fix lifts $\bf{p_{0}}$ and $\bf{p_{ij}}$ of $p_{0}$ and $p_{ij}$ as vectors
in $\mathbb{C}^{2,1}$, such that the entries of $\bf{p_{0}}$ and $\bf{p_{ij}}$
are just the same as entries of $p_{0}$ and $p_{ij}$ above. We have

$\langle{\bf p_{0}},{\bf p_{0}}\rangle=\langle{\bf p_{ij}},{\bf
p_{ij}}\rangle=-1$

for ${\bf p_{ij}}$ above. All of these lifts have the same norm will be very
convenient later.

Recall that $A_{i}=I_{i}I_{i+1}$ for $i=1,2,3,4$ mod 4. Let
$R\subset\rho_{\frac{5\pi}{6}}(K)$ be the set of ten words in $\Gamma$:

$\left\\{(I_{1}I_{2})^{\pm 1},~{}~{}~{}(I_{2}I_{3})^{\pm
1},~{}~{}~{}(I_{3}I_{4})^{\pm 1},~{}~{}~{}(I_{4}I_{1})^{\pm
1},~{}~{}~{}I_{1}I_{3},~{}~{}~{}I_{2}I_{4}\right\\}.$

We will show the partial Dirichlet domain $D_{R}$ centered at the fixed point
$p_{0}$ of $J$ is in fact the Dirichlet domain of $\rho_{\frac{5\pi}{6}}(K)$.
The main tool for our study is the Poincaré polyhedron theorem, which gives
sufficient conditions for $D_{R}$ to be a fundamental domain of the group. We
refer to [20] for the precise statement of this version of Poincaré polyhedron
theorem we need. The main technical result in this section is

######  Theorem 4.1.

$D_{R}$ is the Dirichlet domain of $\rho_{\frac{5\pi}{6}}(K)$ acting on ${\bf
H}^{2}_{\mathbb{C}}$ with center $p_{0}$. Moreover, the group
$\rho_{\frac{5\pi}{6}}(K)=\langle A_{1},A_{2},A_{3},A_{4}\rangle$ has a
presentation

$\langle
A_{1},A_{2},A_{3},A_{4}:A_{1}A_{2}A_{3}A_{4}=id,(A_{1}A_{2})^{2}=(A_{2}A_{3})^{2}=(A_{3}A_{4})^{2}=(A_{4}A_{1})^{2}=id\rangle.$

So $\rho_{\frac{5\pi}{6}}:K\rightarrow\mathbf{PU}(2,1)$ is a discrete and
faithful presentation of $K$.

###  4.3. Intersection patterns of the bisectors for $D_{R}$

In this subsection, we will study the information on intersection patterns of
the bisectors for $D_{R}$. We summarize the intersections of them in Table 1
and we will show this carefully. Moreover, Table 1 should be compared with
Figures 5 and 6. Where Figure 5 is a realistic view of the boundary of the
partial Dirichlet domain $D_{R}$ of
$\rho_{\frac{5\pi}{6}}(K)<\mathbf{PU}(2,1)$. For example, the sphere labeled
by $B_{41}$ is in fact the spinal sphere $B_{41}\cap\partial{\bf
H}^{2}_{\mathbb{C}}$. Unfortunately, since the union of these spinal spheres
are “twisted” in $\partial{\bf H}^{2}_{\mathbb{C}}$, it seems impossible to
see all the spinal spheres from only one point of view. Figure 6 is an
abstract picture of the boundary of the partial Dirichlet domain $D_{R}$,
which is much more transparent.

$B_{12}\cap B_{21}$, tangent | $B_{12}\cap B_{41}=\emptyset$
---|---
$B_{12}\cap B_{23}=\emptyset$ | $B_{12}\cap B_{14}\neq\emptyset$
$B_{12}\cap B_{32}=\emptyset$ | $B_{12}\cap B_{13}\neq\emptyset$
$B_{12}\cap B_{34}\neq\emptyset$ | $B_{12}\cap B_{24}=\emptyset$
$B_{12}\cap B_{43}=\emptyset$ | $B_{13}\cap B_{24}=\emptyset$
Table 1. The intersections of bisectors we should be concerned with up to
$J$-action.

We first consider the intersections of the bisector $B_{12}$ with other
bisectors.

######  Proposition 4.2.

For the bisector $B_{12}$ of $I_{1}I_{2}$, we have

  1. (1)

$B_{12}$ is tangent to $B_{21}$;

  2. (2)

$B_{12}$ does not intersect $B_{23}$, $B_{32}$, $B_{43}$, $B_{41}$ and
$B_{24}$;

  3. (3)

$B_{12}$ intersects $B_{34}$ in a non-empty Giraud disk $B_{12}\cap B_{34}$.
Moreover, the disk $B_{12}\cap B_{34}$ lies in the component of ${\bf
H}^{2}_{\mathbb{C}}-B_{13}$ which does not contain the point $p_{0}$. In
particular, $B_{12}\cap B_{34}$ does not lie in the partial Dirichlet domain
$D_{R}$.

###### Proof.

The trace of $I_{1}I_{2}$ is $3$ (when we normalize them such that
$\det(I_{1}I_{1})=1$), so $I_{1}I_{2}$ is unipotent. By Theorem 6.1 of [21],
the Dirichlet domain (with center $p_{0}$) of the infinite cyclic group
$\langle I_{1}I_{2}\rangle$ has two sides $B_{12}$ and $B_{21}$. These two
bisectors intersect exactly in the fixed point $q_{12}$ of $I_{1}I_{2}$. This
proves (1) of Proposition 4.2.

We now consider $B_{12}\cap B_{23}$. It is easy to see $\bf{p_{12}}$,
$\bf{p_{23}}$ and $\bf{p_{0}}$ are linearly independent in $\mathbb{C}^{2,1}$.
In Equation (2.4), we take ${\bf q}=\bf{p_{12}}$, ${\bf r}=\bf{p_{23}}$ and
${\bf p}=\bf{p_{0}}$, then we can parameterize the intersection $B_{12}\cap
B_{23}$ of the bisectors $B_{12}$ and $B_{23}$ by $V=V(z_{1},z_{2})$ with
$\langle V,V\rangle<0$. Where

(4.3)
$V=\left(\begin{array}[]{c}-\displaystyle{\frac{(\rm{i}\rm{e}^{r\rm{i}}-\rm{e}^{s\rm{i}}+2-2\rm{i})\sqrt{3}+\rm{i}\rm{e}^{r\rm{i}}-\rm{i}\rm{e}^{s\rm{i}}+2-4\rm{i}}{\sqrt{\sqrt{3}+1}}}\\\\[17.22217pt]
-\displaystyle{\frac{(\rm{e}^{r\rm{i}}+\rm{e}^{s\rm{i}}-2)\sqrt{3}-\rm{i}\rm{e}^{r\rm{i}}-\rm{i}\rm{e}^{s\rm{i}}-4+6\rm{i}}{\sqrt{2}}}\\\\[17.22217pt]
\displaystyle{\frac{1-\rm{i}+(1+\rm{i})\sqrt{3}}{\sqrt{\sqrt{3}-1}}}\\\
\end{array}\right),$

and
$(z_{1},z_{2})=(\rm{e}^{r\rm{i}},\rm{e}^{s\rm{i}})\in\mathbb{S}^{1}\times\mathbb{S}^{1}$.
Now $\langle V,V\rangle=V^{*}\cdot H\cdot V$ is

(4.4)
$\displaystyle(8\sqrt{3}+4)\sin(s)-(8\sqrt{3}+20)\cos(r)-(8\sqrt{3}+16)\cos(s)$
$\displaystyle+(2\sqrt{3}+2)\sin(r-s)+4\cos(r-s)-8\sin(r)+16\sqrt{3}+48.$

With Maple, the minimum of (4.4) is 12.752 numerically. In particular, any $V$
in (4.3) is a positive vector. So $B_{12}\cap B_{23}=\emptyset$.

We then consider $B_{12}\cap B_{32}$. In Equation (2.4), we take ${\bf
q}=\bf{p_{12}}$, ${\bf r}=\bf{p_{32}}$ and ${\bf p}=\bf{p_{0}}$, then we can
parameterize the intersection $B_{12}\cap B_{32}$ of the bisectors $B_{12}$
and $B_{32}$ by $V=V(z_{1},z_{2})$ with $\langle V,V\rangle<0$. Where

(4.5)
$V=\left(\begin{array}[]{c}\displaystyle{\frac{(\rm{e}^{r\rm{i}}+\rm{e}^{s\rm{i}}-4)\sqrt{3}+\rm{e}^{r\rm{i}}+\rm{e}^{s\rm{i}}-6}{\sqrt{\sqrt{3}+1}}}\\\\[17.22217pt]
\displaystyle{\frac{(\rm{e}^{r\rm{i}}-\rm{e}^{s\rm{i}}-2\rm{i})\sqrt{3}+\rm{i}\rm{e}^{r\rm{i}}+\rm{e}^{s\rm{i}}}{\sqrt{2}}}\\\\[17.22217pt]
-\sqrt{6}{\sqrt{\sqrt{3}+1}}\\\ \end{array}\right),$

and
$(z_{1},z_{2})=(\rm{e}^{r\rm{i}},\rm{e}^{s\rm{i}})\in\mathbb{S}^{1}\times\mathbb{S}^{1}$.
Now $\langle V,V\rangle=V^{*}\cdot H\cdot V$ is

(4.6)
$\displaystyle-(10\sqrt{3}+12)\cdot(\cos(r)+\cos(s))+2\sqrt{3}\cdot(\sin(r-s)+\cos(r-s))$
$\displaystyle+14\sqrt{3}-6\sin(r)+6\sin(s)+36.$

With Maple, the minimum of (4.6) is 4.78 numerically. In particular, any $V$
in (4.5) is a positive vector. So $B_{12}\cap B_{32}=\emptyset$.

For $B_{12}\cap B_{43}$, in Equation (2.4), we take ${\bf q}=\bf{p_{12}}$,
${\bf r}=\bf{p_{43}}$ and ${\bf p}=\bf{p_{0}}$. Then we can parameterize the
intersection $B_{12}\cap B_{43}$ of the bisectors $B_{12}$ and $B_{43}$ by
$V=V(z_{1},z_{2})$ with $\langle V,V\rangle<0$. Where

(4.7)
$V=\left(\begin{array}[]{c}\displaystyle{\frac{(\rm{e}^{s\rm{i}}+\rm{i}\rm{e}^{r\rm{i}}-2-2\rm{i})\sqrt{3}+\rm{i}\rm{e}^{r\rm{i}}+\rm{e}^{s\rm{i}}-4-4\rm{i}}{\sqrt{\sqrt{3}+1}}}\\\\[17.22217pt]
-\displaystyle{\frac{(\rm{e}^{r\rm{i}}+\rm{e}^{s\rm{i}}-4)\sqrt{3}+\rm{i}\rm{e}^{r\rm{i}}-\rm{i}\rm{e}^{s\rm{i}}-2}{\sqrt{2}}}\\\\[17.22217pt]
(1-\rm{i}){\sqrt{\sqrt{3}-1}}\\\ \end{array}\right),$

and
$(z_{1},z_{2})=(\rm{e}^{r\rm{i}},\rm{e}^{s\rm{i}})\in\mathbb{S}^{1}\times\mathbb{S}^{1}$.
Now $\langle V,V\rangle=V^{*}\cdot H\cdot V$ is

(4.8) $\displaystyle
2\cdot(\cos(r-s)-\sin(r-s))-(6\sqrt{3}+20)\cdot(\cos(r)+\cos(s))$
$\displaystyle-4\sqrt{3}\sin(r-s)+(8\sqrt{3}+10)\cdot(\sin(r)-\sin(s))+20\sqrt{3}+54.$

With Maple, the minimum of (4.8) is 20.51 numerically. In particular, any $V$
in (4.7) is a positive vector. So $B_{12}\cap B_{43}=\emptyset$.

Since $J(B_{41}\cap B_{12})=B_{12}\cap B_{23}$, and we have proved that
$B_{12}\cap B_{23}$ is empty, then $B_{41}\cap B_{12}$ is empty.

For $B_{12}\cap B_{24}$, in Equation (2.4), we take ${\bf q}=\bf{p_{12}}$,
${\bf r}=\bf{p_{24}}$ and ${\bf p}=\bf{p_{0}}$. Then we can parameterize the
intersection $B_{12}\cap B_{24}$ of the bisectors $B_{12}$ and $B_{24}$ by
$V=V(z_{1},z_{2})$ with $\langle V,V\rangle<0$. Where

(4.9)
$V=\left(\begin{array}[]{c}\displaystyle{\frac{(\rm{e}^{s\rm{i}}-1)\sqrt{3}+\rm{e}^{s\rm{i}}-3}{\sqrt{\sqrt{3}+1}}}\\\\[17.22217pt]
\displaystyle{-\frac{(\rm{e}^{s\rm{i}}-1+2\rm{i})\sqrt{3}-\rm{i}\rm{e}^{s\rm{i}}+2\rm{e}^{r\rm{i}}-6-\rm{i}}{\sqrt{2}}}\\\\[17.22217pt]
\sqrt{2}{\sqrt{\sqrt{3}+1}}\\\ \end{array}\right),$

and
$(z_{1},z_{2})=(\rm{e}^{r\rm{i}},\rm{e}^{s\rm{i}})\in\mathbb{S}^{1}\times\mathbb{S}^{1}$.
Now $\langle V,V\rangle=V^{*}\cdot H\cdot V$ is

(4.10) $\displaystyle
2\sqrt{3}\cdot(\cos(r-s)-\sin(s))-2\sin(r-s)-(10\sqrt{3}+8)\cos(s)$
$\displaystyle-(12+2\sqrt{3})\cos(r)+(-2+4\sqrt{3})\sin(r)+6\sqrt{3}+32.$

With Maple, the minimum of (4.10) is 4.58 numerically. In particular, any $V$
in (4.9) is a positive vector. So $B_{12}\cap B_{24}=\emptyset$. We end the
proof of (2) of Proposition 4.2.

For $B_{12}\cap B_{34}$, in Equation (2.4), we take ${\bf q}=\bf{p_{12}}$,
${\bf r}=\bf{p_{34}}$ and ${\bf p}=\bf{p_{0}}$. Then we can parameterize the
intersection $B_{12}\cap B_{34}$ of the bisectors $B_{12}$ and $B_{34}$ by
$V=V(z_{1},z_{2})$ with $\langle V,V\rangle<0$. Where

(4.11)
$V=\left(\begin{array}[]{c}\displaystyle{\frac{(\rm{e}^{r\rm{i}}+\rm{e}^{s\rm{i}}-4)\sqrt{3}+\rm{e}^{r\rm{i}}+\rm{e}^{s\rm{i}}-6+2\rm{i}}{\sqrt{\sqrt{3}+1}}}\\\\[17.22217pt]
\displaystyle{\frac{(\rm{i}-\sqrt{3})(\rm{e}^{s\rm{i}}-\rm{e}^{r\rm{i}})}{\sqrt{2}}}\\\\[17.22217pt]
(\rm{i}-\sqrt{3})\sqrt{2}{\sqrt{\sqrt{3}+1}}\\\ \end{array}\right),$

and
$(z_{1},z_{2})=(\rm{e}^{r\rm{i}},\rm{e}^{s\rm{i}})\in\mathbb{S}^{1}\times\mathbb{S}^{1}$.
Now $\langle V,V\rangle=V^{*}\cdot H\cdot V$ is

(4.12)
$\displaystyle-(8\sqrt{3}+12)\cdot(\cos(r)+\cos(s))+(2\sqrt{3}-2)\cos(r-s)$
$\displaystyle+4\cdot(\sin(r)+\sin(s))+14\sqrt{3}+26.$

There is $(r,s)$ such that (4.12) is negative, so
$V=V(\rm{e}^{r\rm{i}},\rm{e}^{s\rm{i}})$ is a negative vector. A sample point
is $(r,s)=(0,-\frac{\pi}{16})$.

Figure 2. $B_{12}\cap B_{34}$ lies in the half space of ${\bf
H}^{2}_{\mathbb{C}}-B_{13}$ which does not contain the fixed point $p_{0}$ of
$J$. The red circle is $B_{12}\cap B_{34}\cap\partial{\bf H}^{2}_{\mathbb{C}}$
in coordinates of $(r,s)\in[-\pi,\pi]^{2}$. So the small disk bounded by the
red circle is $B_{12}\cap B_{34}\cap{\bf H}^{2}_{\mathbb{C}}$. The cyan region
consists of points such that whose distance to $p_{13}$ is smaller than whose
distance to $p_{0}$. This figure illustrates that the disk $B_{12}\cap B_{34}$
lies in the component of ${\bf H}^{2}_{\mathbb{C}}-B_{13}$ which does not
contain the point $p_{0}$.

But for $V$ in (4.11), $|V^{*}\cdot H\cdot{\bf p_{0}}|^{2}$ is the constant
$8+8\sqrt{3}$, which is 21.85 numerically. For $V$ in (4.11), $|V^{*}\cdot
H\cdot{\bf p_{13}}|^{2}$ is

(4.13) $\displaystyle
4(1+\sqrt{3})\cdot(\cos(r-s)-(\cos(r)+\cos(s))\sqrt{3}-\sin(r)-\sin(s)+3).$

With Maple, the maximum of (4.13) with the condition $V^{*}\cdot H\cdot V=0$
is 14.23 numerically, which is smaller than $|V^{*}\cdot H\cdot{\bf
p_{0}}|^{2}$. Moreover, the minimum of (4.13) with the condition $V^{*}\cdot
H\cdot V=0$ is 5.85. In particular, the ideal boundary of $B_{12}\cap B_{34}$,
say $B_{12}\cap B_{34}\cap\partial{\bf H}^{2}_{\mathbb{C}}$, is disjoint from
$B_{13}$. Then we have $B_{12}\cap B_{34}$ is disjoint from $B_{13}$ by well-
known properties of bisectors in ${\bf H}^{2}_{\mathbb{C}}$, see [11]. For the
sample point $V=V(\rm{e}^{0\cdot\rm{i}},\rm{e}^{-\frac{\pi}{16}\rm{i}})$, it
is easy to see its distance to $p_{13}$ is smaller than its distance to
$p_{0}$. We have $B_{12}\cap B_{34}$ lies in the half space of ${\bf
H}^{2}_{\mathbb{C}}-B_{13}$ which does not contain the fixed point $p_{0}$ of
$J$. In particular, $B_{12}\cap B_{34}$ does not lie in the partial Dirichlet
domain $D_{R}$. See also Figure 2 for an illustration of this fact. This ends
the proof of (3) of Proposition 4.2.

∎

######  Proposition 4.3.

For the bisector $B_{13}$ of $I_{1}I_{3}$, we have $B_{13}$ does not intersect
$B_{21}$, $B_{23}$, $B_{43}$, $B_{41}$ and $B_{24}$.

###### Proof.

We have proved $B_{12}\cap B_{24}=\emptyset$. By the $\langle
J\rangle=\mathbb{Z}_{4}$ symmetry, we have

$B_{13}\cap B_{23}=\emptyset,~{}~{}~{}~{}B_{13}\cap B_{41}=\emptyset.$

The fact that $B_{13}\cap B_{21}=\emptyset$ and $B_{13}\cap B_{43}=\emptyset$
can be proved similarly to $B_{12}\cap B_{24}=\emptyset$ in Proposition 4.2.

For $B_{13}\cap B_{24}$. It is easy to see $p_{0}$, $p_{13}$ and $p_{24}$ lie
in the $\mathbb{C}$-line

$l=\left\\{\left[z_{1},~{}~{}0,~{}~{}1\right]^{t}\in{\bf
H}^{2}_{\mathbb{C}}\right\\}.$

Now it is easy to see $B_{13}\cap B_{24}\cap l=\emptyset$. Then from the
projection of ${\bf H}^{2}_{\mathbb{C}}$ to $l$, we get $B_{13}\cap
B_{24}=\emptyset$.

∎

######  Proposition 4.4.

For the bisectors $B_{12}$, $B_{13}$ and $B_{14}$, we have:

  1. (1)

Each of the intersections $B_{12}\cap B_{13}$, $B_{12}\cap B_{14}$ and
$B_{13}\cap B_{14}$ is a non-empty Giraud disk;

  2. (2)

The triple intersection $B_{12}\cap B_{13}\cap B_{14}$ is an arc in each of
the Giraud disks $B_{12}\cap B_{13}$, $B_{12}\cap B_{14}$ and $B_{13}\cap
B_{14}$.

###### Proof.

For $B_{12}\cap B_{14}$, in Equation (2.4), we take ${\bf q}=\bf{p_{12}}$,
${\bf r}=\bf{p_{14}}$ and ${\bf p}=\bf{p_{0}}$. It is easy to see these three
vectors are linearly independent. Then we can parameterize the intersection
$B_{12}\cap B_{14}$ of the bisectors $B_{12}$ and $B_{14}$ by
$V=V(z_{1},z_{2})$ with $\langle V,V\rangle<0$. Where

(4.14)
$V=\left(\begin{array}[]{c}\displaystyle{\frac{(\rm{e}^{s\rm{i}}-\rm{e}^{r\rm{i}})\sqrt{3}-\rm{e}^{r\rm{i}}+\rm{e}^{s\rm{i}}-2\rm{i}}{\sqrt{\sqrt{3}+1}}}\\\\[17.22217pt]
\displaystyle{\frac{(\rm{e}^{r\rm{i}}-\rm{e}^{s\rm{i}}-2\rm{i})\sqrt{3}+\rm{i}\rm{e}^{r\rm{i}}+\rm{i}\rm{e}^{s\rm{i}}}{\sqrt{2}}}\\\\[17.22217pt]
-\rm{i}\sqrt{2}{\sqrt{\sqrt{3}+1}}\\\ \end{array}\right)$

and
$(z_{1},z_{2})=(\rm{e}^{r\rm{i}},\rm{e}^{s\rm{i}})\in\mathbb{S}^{1}\times\mathbb{S}^{1}$.
Now $\langle V,V\rangle=V^{*}\cdot H\cdot V$ is

(4.15) $\displaystyle 2\sqrt{3}\cdot(\sin(r-s)-\cos(r)-\cos(s)-\cos(r-s)+1)$
$\displaystyle-4\cos(r-s)-2\sin(r)+2\sin(s)+8.$

There are $(r,s)$ such that (4.15) is negative, that is
$V=V(\rm{e}^{r\rm{i}},\rm{e}^{s\rm{i}})$ is a negative vector. For example
when $(r,s)=(0,0)$. Then $B_{12}\cap B_{14}$ is non empty, it is a Giraud
disk.

Now for $V$ in (4.14), $|V^{*}\cdot H\cdot{\bf p_{0}}|^{2}$ is $2+2\sqrt{3}$,
and $|V^{*}\cdot H\cdot{\bf p_{13}}|^{2}$ is

(4.16) $\displaystyle
4\sqrt{3}(\sin(s)-\sin(r)-\cos(r-s))+4(\sin(s)-\sin(r)-\cos(r-s))$
$\displaystyle+6\sqrt{3}+6.$

The solutions of $V$ in (4.14) with the condition

$|V^{*}\cdot H\cdot{\bf p_{13}}|^{2}=2+2\sqrt{3}$

are

$\\{s=-\frac{\pi}{2}\\},~{}~{}~{}\\{r=\frac{\pi}{2}\\},~{}~{}~{}\\{r=s\\}.$

It is easy to see when $s=-\frac{\pi}{2}$ or $r=\frac{\pi}{2}$ then $V$ is
positive. But when $r=s$ with

$r\in\left(-\arccos(\frac{\sqrt{3}}{3}),-\arccos(\frac{\sqrt{3}}{3})\right),$

$V$ is a negative point in $B_{12}\cap B_{14}$. So $B_{12}\cap B_{14}\cap
B_{13}$ is an arc in $B_{12}\cap B_{14}$. See Figure 3 for $B_{12}\cap B_{14}$
in coordinates $(r,s)\in[-\pi,\pi]^{2}$.

Figure 3. The small disk bounded by the red circle is the Giraud disk
$B_{12}\cap B_{14}$. The three purple lines are points (in the Giraud torus
$\hat{B}(p_{0},p_{12},p_{14}$) such that its distances to $p_{13}$ and $p_{0}$
are the same. So the red half disk is the part of $B_{12}\cap B_{14}$ which
lies in the half space of ${\bf H}^{2}_{\mathbb{C}}-B_{13}$ containing the
fixed point of $J$, and then it is the ridge $s_{12}\cap s_{14}$.  Figure 4.
The small disk bounded by the purple circle is the Giraud disk $B_{12}\cap
B_{13}$. The three red lines are points (in the Giraud torus
$\hat{B}(p_{0},p_{12},p_{13}$) such that its distances to $p_{14}$ and $p_{0}$
are the same. So the part of $B_{12}\cap B_{13}$ which lies in the half space
of ${\bf H}^{2}_{\mathbb{C}}-B_{14}$ containing the fixed point of $J$ is the
purple half disk, and then it is the ridge $s_{12}\cap s_{13}$.

For $B_{12}\cap B_{13}$, we have proved that $B_{12}\cap B_{13}$ is non-empty,
so it is a Giraud disk. In Equation (2.4), we take ${\bf q}=\bf{p_{12}}$,
${\bf r}=\bf{p_{13}}$ and ${\bf p}=\bf{p_{0}}$. It is easy to see these three
points are linearly independent. Then we can parameterize the intersection
$B_{12}\cap B_{13}$ of the bisectors $B_{12}$ and $B_{13}$ by
$V=V(z_{1},z_{2})$ with $\langle V,V\rangle<0$. Where

(4.17)
$V=\left(\begin{array}[]{c}\displaystyle{\frac{(\rm{e}^{s\rm{i}}-1)\sqrt{3}+\rm{e}^{s\rm{i}}-3}{\sqrt{\sqrt{3}+1}}}\\\\[17.22217pt]
\displaystyle{\frac{(-\rm{e}^{s\rm{i}}-1)\sqrt{3}+\rm{i}\rm{e}^{s\rm{i}}+2\rm{e}^{r\rm{i}}-\rm{i}}{\sqrt{2}}}\\\\[17.22217pt]
-\sqrt{2}{\sqrt{\sqrt{3}+1}}\\\ \end{array}\right),$

and
$(z_{1},z_{2})=(\rm{e}^{r\rm{i}},\rm{e}^{s\rm{i}})\in\mathbb{S}^{1}\times\mathbb{S}^{1}$.
Now $\langle V,V\rangle=V^{*}\cdot H\cdot V$ is

(4.18) $\displaystyle 2\sqrt{3}\cdot(\sin(s)-\cos(r)-\cos(s)-\cos(r-s)+1)$
$\displaystyle+2\sin(r-s)-2\sin(r)-4\cos(s)+8.$

Now for $V$ in (4.17), $|V^{*}\cdot H\cdot{\bf p_{0}}|^{2}$ is $2+2\sqrt{3}$,
and $|V^{*}\cdot H\cdot{\bf p_{14}}|^{2}$ is

(4.19) $\displaystyle
4\sqrt{3}\cdot(\sin(r-s)-\sin(r)-\cos(s))+4\cdot(\sin(r-s)-\sin(r)-\cos(s))$
$\displaystyle+6\sqrt{3}+6.$

The solutions of $V$ in (4.17) with the condition

$|V^{*}\cdot H\cdot{\bf p_{14}}|^{2}=2+2\sqrt{3}$

are

$\\{s=0\\},~{}~{}~{}\\{r=\frac{\pi}{2}\\},~{}~{}~{}\\{r=s-\frac{\pi}{2}\\}.$

It is easy to see when $r=\frac{\pi}{2}$ and $r=s-\frac{\pi}{2}$ then $V$ is
positive. But when $s=0$ with

$r\in\left(-\arccos(\frac{\sqrt{3}}{3}),-\arccos(\frac{\sqrt{3}}{3})\right),$

$V$ is a negative point in $B_{12}\cap B_{13}$. See Figure 4 for $B_{12}\cap
B_{13}$ in coordinates $(r,s)\in[-\pi,\pi]^{2}$.

Similarly, $B_{13}\cap B_{14}$ is a non-empty Giraud disk. We omit it. ∎

From Propositions 4.2, 4.3 and 4.4, we have the following two propositions,
which study the combinatorics of 3-sides and ridges of $D_{R}$. They are
crucial for the application of the Poincaré polyhedron theorem. For each
$I_{i}I_{j}\in R$, the side $s_{ij}$ by definition is $B_{ij}\cap D_{R}$,
which is a 3-dimensional object. The reader may compare to Figure 6, which
illustrates the ideal boundary behaviors of these 3-sides $s_{ij}$.

######  Proposition 4.5.

The side $s_{12}=B_{12}\cap D_{R}$ is 3-ball in ${\bf
H}^{2}_{\mathbb{C}}\cup\partial{\bf H}^{2}_{\mathbb{C}}$:

  * •

$s_{12}\cap\partial{\bf H}^{2}_{\mathbb{C}}$ is a disk with the point $q_{12}$
in its interior;

  * •

The frontier of $s_{12}\cap{\bf H}^{2}_{\mathbb{C}}$ is a disk consisting of
two half disks $s_{12}\cap s_{14}$ and $s_{12}\cap s_{13}$ glued along the arc

$s_{12}\cap s_{13}\cap s_{14}=B_{12}\cap B_{13}\cap B_{14}.$

######  Proposition 4.6.

The side $s_{13}=B_{13}\cap D_{R}$ is 3-ball in ${\bf
H}^{2}_{\mathbb{C}}\cup\partial{\bf H}^{2}_{\mathbb{C}}$:

  * •

$s_{13}\cap\partial{\bf H}^{2}_{\mathbb{C}}$ is an annulus;

  * •

The frontier of $s_{13}\cap{\bf H}^{2}_{\mathbb{C}}$ are two disks, each of
them consists of two half disks: one component is the union of $s_{13}\cap
s_{12}$ and $s_{13}\cap s_{14}$ glued along the arc

$s_{12}\cap s_{13}\cap s_{14}=B_{12}\cap B_{13}\cap B_{14},$

and the other component is union of $s_{13}\cap s_{32}$ and $s_{13}\cap
s_{34}$ glued along the arc

$s_{13}\cap s_{32}\cap s_{34}=B_{13}\cap B_{32}\cap B_{34}.$

We have similar properties for the 3-side $s_{21}$. Since there is a $\langle
J\rangle=\mathbb{Z}_{4}$ symmetry, we omits the statements of the
combinatorics of $s_{23}$, $s_{32}$, $s_{34}$, $s_{43}$, $s_{41}$, $s_{14}$
and $s_{24}$.

$B_{23}$$B_{32}$$B_{34}$$B_{14}$$B_{43}$$B_{41}$$B_{13}$ Figure 5. A realistic
view of the boundary of Dirichlet domain of
$\rho_{\frac{5\pi}{6}}(K)<\mathbf{PU}(2,1)$. For example, the sphere labeled
by $B_{41}$ is in fact the spinal sphere $B_{41}\cap\partial{\bf
H}^{2}_{\mathbb{C}}$. The other labels have similar meanings. In this figure,
we can not see $B_{21}$ and $B_{12}$, and the black one is $B_{24}$. The brown
abd green spheres labeled by $B_{43}$ and $B_{34}$ are tangent at a point
$p_{34}$, which is disjoint from the purple sphere labeled by $B_{13}$.

######  Remark 4.7.

From Figures 5 and 6, we strongly believe that $\partial{\bf
H}^{2}_{\mathbb{C}}\cap D_{R}$ is a genus three handlebody. In other words, if
we denote by $\mathcal{H}_{ij}$ the half space of ${\bf
H}^{2}_{\mathbb{C}}-B_{ij}$ which does not contain the point $p_{0}$ for
$I_{i}I_{j}\in R$ (so $\mathcal{H}_{ij}$ contains $p_{ij}$).
$\overline{\mathcal{H}_{ij}}$ is the closure of $\mathcal{H}_{ij}$ in
$\overline{{\bf H}^{2}_{\mathbb{C}}}$. Then

$(\cup_{I_{ij}\in R}\overline{\mathcal{H}_{ij}})\cap\partial{\bf
H}^{2}_{\mathbb{C}}$

is a union of 3-balls in $\partial{\bf H}^{2}_{\mathbb{C}}$ (some of them are
tangent), so it is a handlebody. This handlebody is the union of all the
3-balls bounded by spheres in Figure 5. We have $\partial{\bf
H}^{2}_{\mathbb{C}}\cap D_{R}$ is the complement of $(\cup_{I_{ij}\in
R}\overline{\mathcal{H}_{ij}})\cap\partial{\bf H}^{2}_{\mathbb{C}}$ in
$\partial{\bf H}^{2}_{\mathbb{C}}$, which is the region outside all of the
spheres in Figure 5. We guess that $(\cup_{I_{ij}\in
R}\overline{\mathcal{H}_{ij}})\cap\partial{\bf H}^{2}_{\mathbb{C}}$ is an
unknotted genus three handlebody. In this paper we do not care it, and we do
not show this rigorously.

$B_{21}$$B_{23}$$B_{32}$$B_{34}$$B_{14}$$B_{12}$$B_{43}$$B_{41}$$B_{24}$$B_{13}$
Figure 6. The abstract picture of the boundary of Dirichlet domain of
$\rho_{\frac{5\pi}{6}}(K)<\mathbf{PU}(2,1)$. What we draw is in fact
$D_{R}\cap\partial{\bf H}^{2}_{\mathbb{C}}$. Here the once-punctured disk
labeled by $B_{ij}$ for $j=i\pm 1$ mod 4 is in fact $B_{ij}\cap
D_{R}\cap\partial{\bf H}^{2}_{\mathbb{C}}$. The annulus labeled by $B_{13}$
(and $B_{24}$ respectively) is in fact $B_{13}\cap D_{R}\cap\partial{\bf
H}^{2}_{\mathbb{C}}$ (and $B_{24}\cap D_{R}\cap\partial{\bf
H}^{2}_{\mathbb{C}}$ respectively).

###  4.4. Using the Poincaré polyhedron theorem

With the preliminaries in Subsection 4.3, we have the side pairing maps of
$D_{R}$ as follows:

  * •

$I_{2}I_{1}:s_{12}\rightarrow s_{21}$;

  * •

$I_{3}I_{2}:s_{23}\rightarrow s_{32}$;

  * •

$I_{4}I_{3}:s_{34}\rightarrow s_{43}$;

  * •

$I_{1}I_{4}:s_{41}\rightarrow s_{14}$;

  * •

$I_{1}I_{3}:s_{13}\rightarrow s_{13}$;

  * •

$I_{2}I_{4}:s_{24}\rightarrow s_{24}$.

We have

######  Proposition 4.8.

$I_{2}I_{1}$ is a homeomorphism from $s_{12}$ to $s_{21}$:

  1. (1)

$I_{2}I_{1}$ sends the ridge $s_{12}\cap s_{14}$ to the ridge $s_{21}\cap
s_{24}$;

  2. (2)

$I_{2}I_{1}$ sends the ridge $s_{12}\cap s_{13}$ to the ridge $s_{21}\cap
s_{23}$.

###### Proof.

The ridge $s_{12}\cap s_{14}$ is defined by the triple equality

$|\langle{\bf z},{\bf p_{0}}\rangle|=|\langle{\bf
z},p_{12}\rangle|=|\langle{\bf z},{\bf p_{14}}\rangle|$

with ${\bf z}\in\mathbb{C}^{3,1}$. From $I_{2}I_{1}$’s action on the set

$\\{p_{0},p_{12},p_{14}\\},$

we get the set

$\\{p_{21},p_{0},p_{24}\\}.$

So $I_{2}I_{1}$ maps $s_{12}\cap s_{14}$ to $s_{21}\cap s_{24}$.

The proof of (2) of Proposition 4.8 is similar. ∎

Similarly, we have

######  Proposition 4.9.

The side pairing map $I_{1}I_{3}$ is a self-homeomorphism of $s_{13}$:

  1. (1)

$I_{1}I_{3}$ exchanges the ridges $s_{13}\cap s_{12}$ and $s_{13}\cap s_{32}$;

  2. (2)

$I_{1}I_{3}$ exchanges the ridges $s_{13}\cap s_{14}$ and $s_{13}\cap s_{34}$.

######  Proposition 4.10.

$I_{3}I_{2}$ is a homeomorphism from $s_{23}$ to $s_{32}$:

  1. (1)

$I_{3}I_{2}$ sends the ridge $s_{23}\cap s_{24}$ to the ridge $s_{32}\cap
s_{34}$;

  2. (2)

$I_{3}I_{2}$ sends the ridge $s_{21}\cap s_{23}$ to the ridge $s_{31}\cap
s_{32}$.

######  Proposition 4.11.

$I_{4}I_{3}$ is a homeomorphism from $s_{34}$ to $s_{43}$:

  1. (1)

$I_{4}I_{3}$ sends the ridge $s_{34}\cap s_{32}$ to the ridge $s_{24}\cap
s_{43}$;

  2. (2)

$I_{4}I_{3}$ sends the ridge $s_{34}\cap s_{13}$ to the ridge $s_{43}\cap
s_{41}$.

######  Proposition 4.12.

$I_{1}I_{4}$ is a homeomorphism from $s_{41}$ to $s_{14}$:

  1. (1)

$I_{1}I_{4}$ sends the ridge $s_{41}\cap s_{24}$ to the ridge $s_{14}\cap
s_{12}$;

  2. (2)

$I_{1}I_{4}$ sends the ridge $s_{41}\cap s_{43}$ to the ridge $s_{14}\cap
s_{13}$.

######  Proposition 4.13.

The side pairing map $I_{2}I_{4}$ is a self-homeomorphism of $s_{24}$:

  1. (1)

$I_{2}I_{4}$ exchanges the ridges $s_{24}\cap s_{21}$ and $s_{24}\cap s_{41}$;

  2. (2)

$I_{2}I_{4}$ exchanges the ridges $s_{24}\cap s_{23}$ and $s_{24}\cap s_{43}$.

Proof of Theorem 4.1. After above propositions, we prove the tessellation
around the sides and ridges of the partial Dirichlet domain $D_{R}$.

First, $I_{1}I_{3}$ is a self-homeomorphism of $s_{13}$, and $I_{1}I_{3}$
exchanges the two components of ${\bf H}^{2}_{\mathbb{C}}-B_{13}$. Then
$D_{R}$ and $I_{1}I_{3}(D_{R})$ have disjoint interiors, and they together
cover a neighborhood of each point in the interior of the side $s_{13}$. The
cases of the other 3-sides are similar.

Secondly, we consider tessellations about ridges. Recall that
$A_{1}=I_{1}I_{2}$, $A_{2}=I_{2}I_{3}$, $A_{3}=I_{3}I_{4}$ and
$A_{4}=I_{4}I_{1}$.

(1). For the ridge $s_{14}\cap s_{12}$, the ridge circle is

$s_{14}\cap s_{12}\xrightarrow{A^{-1}_{1}}s_{21}\cap
s_{24}\xrightarrow{(A_{2}A_{3})^{-1}}s_{24}\cap
s_{41}\xrightarrow{A^{-1}_{4}}s_{14}\cap s_{12}.$

Which gives the relation $A^{-1}_{4}\cdot A^{-1}_{3}A^{-1}_{2}\cdot
A^{-1}_{1}=id$. By a standard argument as in [20], we have $D_{R}\cup
A_{1}(D_{R})\cup A^{-1}_{4}(D_{R})$ covers a small neighborhood of $s_{14}\cap
s_{12}$.

(2). For the ridge $s_{13}\cap s_{14}$, the ridge circle is

$s_{13}\cap s_{14}\xrightarrow{A_{4}}s_{41}\cap
s_{43}\xrightarrow{A_{3}}s_{34}\cap s_{31}\xrightarrow{A_{1}A_{2}}s_{13}\cap
s_{14}.$

Which gives the relation $A_{1}A_{2}\cdot A_{3}\cdot A_{4}=id$. By a standard
argument as above $D_{R}\cup A^{-1}_{4}(D_{R})\cup(A_{1}A_{2})^{-1}(D_{R})$
covers a small neighborhood of $s_{13}\cap s_{14}$.

(3). For the ridge $s_{13}\cap s_{12}$, the ridge circle is

$s_{13}\cap s_{12}\xrightarrow{A^{-1}_{1}}s_{21}\cap
s_{23}\xrightarrow{A^{-1}_{2}}s_{32}\cap
s_{31}\xrightarrow{A_{1}A_{2}}s_{13}\cap s_{12}.$

Which gives the relation $A_{1}A_{2}\cdot A^{-1}_{2}\cdot A^{-1}_{1}=id$. We
have $D_{R}\cup A_{1}(D_{R})\cup A_{1}A_{2}(D_{R})$ covers a small
neighborhood of $s_{13}\cap s_{12}$.

(4). For the ridge $s_{24}\cap s_{23}$, the ridge circle is

$s_{24}\cap s_{23}\xrightarrow{A^{-1}_{2}}s_{32}\cap
s_{34}\xrightarrow{A^{-1}_{3}}s_{43}\cap
s_{42}\xrightarrow{A_{2}A_{3}}s_{24}\cap s_{23}.$

Which gives the relation $A_{2}A_{3}\cdot A^{-1}_{3}\cdot A^{-1}_{2}=id$. Then
$D_{R}\cup A_{2}(D_{R})\cup A_{2}A_{3}(D_{R})$ covers a small neighborhood of
$s_{13}\cap s_{12}$.

In Figure 6, we take the three ridges in the same class as in $s_{14}\cap
s_{12}$ by cyan colors (more precisely, the intersections of these ridges with
$\partial{\bf H}^{2}_{\mathbb{C}}$, so we have a set of arcs). Similarly, we
color other ridge classes by orange, red and green colors.

Since the identification around $q_{12}\in\partial{\bf H}^{2}_{\mathbb{C}}$ is
given by $A_{1}$, which is unipotent. So the identification space given by the
side-pairing maps is complete at the ideal point $q_{12}$. By $\langle
J\rangle$-symmetry, the identification space of $D_{R}$ by these side pairing
maps is complete.

By Poincaré polyhedron theorem, the partial Dirichlet domain $D_{R}$ is in
fact the Dirichlet domain of $\rho_{\frac{5\pi}{6}}(K)<\mathbf{PU}(2,1)$. Now
we have the presentation

$\rho_{\frac{5\pi}{6}}(K)=\left\langle
A_{1},A_{2},A_{3},A_{4}\Bigg{|}\begin{matrix}(A_{1}A_{2})^{2}=(A_{2}A_{3})^{2}=A_{1}A_{2}A_{3}A_{4}=id\end{matrix}\right\rangle.$

Note that with the relations

$(A_{1}A_{2})^{2}=id,~{}~{}~{}(A_{2}A_{3})^{2}=id,~{}~{}~{}A_{1}A_{2}A_{3}A_{4}=id,$

it is trivial to get

$(A_{3}A_{4})^{2}=id,~{}~{}~{}(A_{4}A_{1})^{2}=id.$

So the relations above are also $\mathbb{Z}_{4}$-invariant, we have a discrete
and faithful representation of $K$ into $\mathbf{PU}(2,1)$. This ends the
proof of Theorem 4.1, and so the proof of Theorem 1.1 when
$\theta=\frac{5\pi}{6}$.

##  5\. Dirichlet domain of $\rho_{\pi}(K)<\mathbf{PO}(3,1)$ in ${\bf
H}^{3}_{\mathbb{R}}$

In this section, we let $\theta=\pi$. Then $\rho_{\pi}(K)$ preserves a totally
geodesic ${\bf H}^{3}_{\mathbb{R}}\hookrightarrow{\bf H}^{3}_{\mathbb{C}}$
invariant. Even through the discreteness of $\rho_{\pi}(K)$ is trivial, but we
consider the Dirichlet domain of $\rho_{\pi}(K)<\mathbf{PO}(3,1)$ in ${\bf
H}^{3}_{\mathbb{R}}$, which is the baby case for Section 6.

###  5.1. $\rho_{\pi}(K)$-invariant totally geodesic ${\bf
H}^{3}_{\mathbb{R}}\hookrightarrow{\bf H}^{3}_{\mathbb{C}}$

Since the $\rho_{\pi}(K)$-invariant totally geodesic ${\bf
H}^{3}_{\mathbb{R}}\hookrightarrow{\bf H}^{3}_{\mathbb{C}}$ is not the obvious
one in ${\bf H}^{3}_{\mathbb{C}}$, we first describe it. When $\theta=\pi$, as
the notations in Subsection 3.1, we have

(5.1)
$n_{1}=\left[\begin{matrix}\frac{\sqrt{3}}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2}\end{matrix}\right]^{t},~{}~{}n_{2}=\left[\begin{matrix}-\frac{\sqrt{3}}{2},\frac{\rm{i}}{2},-\frac{\rm{i}}{2},\frac{1}{2}\end{matrix}\right]^{t},$

and

(5.2)
$n_{3}=\left[\begin{matrix}\frac{\sqrt{3}}{2},-\frac{1}{2},-\frac{1}{2},\frac{1}{2}\end{matrix}\right]^{t},~{}~{}n_{4}=\left[\begin{matrix}-\frac{\sqrt{3}}{2},-\frac{\rm{i}}{2},\frac{\rm{i}}{2},\frac{1}{2}\end{matrix}\right]^{t}.$

We fix lifts $\bf{n_{i}}$ of $n_{i}$ as vectors in $\mathbb{C}^{3,1}$, such
that the entries of $\bf{n_{i}}$ are just the same as entries of $n_{i}$
above. Let $\mathcal{L}$ be the intersection of the real span of
$\bf{n_{1}},\bf{n_{2}},\bf{n_{3}},\bf{n_{4}}$ and ${\bf H}^{3}_{\mathbb{C}}$,
that is $\mathcal{L}$ is the set

$\left\\{[{\bf x}=x_{1}{\bf n_{1}}+x_{2}{\bf n_{2}}+x_{3}{\bf n_{3}}+x_{4}{\bf
n_{4}}]\in{\bf H}^{3}_{\mathbb{C}}~{}~{}|~{}~{}x_{i}\in\mathbb{R}\right\\}.$

Since $J(n_{i})=n_{i+1}$ mod $4$, for

$x=x_{1}n_{1}+x_{2}n_{2}+x_{3}n_{3}+x_{4}n_{4}\in\mathcal{L},$
$J(x)=x_{4}n_{1}+x_{1}n_{2}+x_{2}n_{3}+x_{3}n_{4}\in\mathcal{L}.$

So

(5.3) $J_{\mathcal{L}}=\begin{pmatrix}0&0&0&1\\\ 1&0&0&0\\\ 0&1&0&0\\\
0&0&1&0\\\ \end{pmatrix}$

is the matrix representation of $J$ on $\mathcal{L}$ with basis
$\\{n_{1},n_{2},n_{3},n_{4}\\}$. The fix point of $J$ in $\mathcal{L}$ is

$x_{0}=n_{1}+n_{2}+n_{3}+n_{4},$

that is, we may take $x_{1}=x_{2}=x_{3}=x_{4}=\frac{1}{2}$.

Now

(5.4)
$I_{1}=\left(\begin{array}[]{cccc}\frac{1}{2}&\frac{\sqrt{3}}{2}&\frac{\sqrt{3}}{2}&-\frac{\sqrt{3}}{2}\\\\[3.0pt]
\frac{\sqrt{3}}{2}&-\frac{1}{2}&\frac{1}{2}&-\frac{1}{2}\\\\[3.0pt]
\frac{\sqrt{3}}{2}&\frac{1}{2}&-\frac{1}{2}&-\frac{1}{2}\\\\[3.0pt]
\frac{\sqrt{3}}{2}&\frac{1}{2}&\frac{1}{2}&-\frac{3}{2}\\\ \end{array}\right)$

as given in Subsection 3.1. Consider

$I_{1}(x_{1}n_{1}+x_{2}n_{2}+x_{3}n_{3}+x_{4}n_{4})=y_{1}n_{1}+y_{2}n_{2}+y_{3}n_{3}+y_{4}n_{4},$

we have the matrix representation of $I_{1}$-action on $\mathcal{L}$ with
basis $\\{n_{1},n_{2},n_{3},n_{4}\\}$ is

(5.5) $I_{\mathcal{L},1}=\begin{pmatrix}1&-2&0&-2\\\ 0&-1&0&0\\\ 0&0&-1&0\\\
0&0&0&-1\\\ \end{pmatrix}.$

Then $I_{\mathcal{L},i}=J_{\mathcal{L}}\cdot I_{\mathcal{L},i}\cdot
J^{-1}_{\mathcal{L}}$ for $i=2,3,4$ is the matrix representation of
$I_{i}$-action on $\mathcal{L}$ with base $\\{n_{1},n_{2},n_{3},n_{4}\\}$.

We transform $\mathcal{L}$ into the Klein model of ${\bf H}^{3}_{\mathbb{R}}$,
we also transform $J_{\mathcal{L}}$ and $I_{\mathcal{L},i}$ into matrices in
the Lie group $\mathbf{PO}(3,1)$. Now the quadratic form with respect to the
basis $\\{{\bf n_{1}},{\bf n_{2}},{\bf n_{3}},{\bf n_{4}}\\}$ is given by

(5.6) $H_{\mathcal{L}}=(\langle{\bf n_{i}},{\bf n_{j}}\rangle)_{1\leq i,j\leq
4}=\begin{pmatrix}1&-1&0&-1\\\ -1&1&-1&0\\\ 0&-1&1&-1\\\ -1&0&-1&1\\\
\end{pmatrix}.$

The quadratic form of the Klein model of ${\bf H}^{3}_{\mathbb{R}}$ is given
by $H_{\mathcal{R}}=H=diag\\{1,1,1,-1\\}$. Consider the matrix

(5.7) $C=\begin{pmatrix}1&0&0&0\\\ 0&0&1&0\\\
-\frac{1}{\sqrt{3}}&-\frac{2}{\sqrt{3}}&-\frac{1}{\sqrt{3}}&\frac{1}{\sqrt{3}}\\\
1&1&1&0\\\ \end{pmatrix}$

with real entries. Then $\det(H_{\mathcal{L}})=-3$,
$\det(C)=\frac{1}{\sqrt{3}}$, and $C\cdot H_{\mathcal{L}}\cdot C^{t}=H$.

We denote by

$J_{k}=(C^{t})^{-1}\cdot J_{\mathcal{L}}\cdot C^{t},$

and

$I_{k,i}=(C^{t})^{-1}\cdot I_{\mathcal{L},i}\cdot C^{t}$

for $i=1,2,3,4$. Then

$J_{k}\cdot H\cdot J^{*}_{k}=H$

and

$I_{k,i}\cdot H\cdot I^{*}_{k,i}=H$

for $i=1,2,3,4$. So now $J_{k}$ and $I_{k,i}$ for $i=1,2,3,4$ are the matrix
presentations in the Klein model of ${\bf H}^{3}_{\mathbb{R}}$ for $J$ and
$I_{i}$ for $i=1,2,3,4$ acting on $\mathcal{L}$. It is simple to get the
matrices of $J_{k}$ and $I_{k,i}$ for $i=1,2,3,4$ from the matrix $C$,
$J_{\mathcal{L}}$ and $I_{\mathcal{L},i}$, we do not write down them
explicitly here.

###  5.2. The partial Dirichlet domain $D_{R}$ in ${\bf H}^{3}_{\mathbb{R}}$

From the transformation in Subsection 5.1, in this subsection, we always
consider the Klein model of ${\bf H}^{3}_{\mathbb{R}}$. Consider the partial
Dirichlet domain $D_{R}$ with center point

$p_{0}=(C^{t})^{-1}([\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2}]^{t})=[-\frac{1}{2},-\frac{1}{2},\frac{\sqrt{3}}{2},\frac{3}{2}]^{t}$

in ${\bf H}^{3}_{\mathbb{R}}$, here $p_{0}$ is the fixed point of $J_{k}$ in
the Klein model of ${\bf H}^{3}_{\mathbb{R}}$. Where

$R=\left\\{(I_{k,1}I_{k,2})^{\pm 1},(I_{k,2}I_{k,3})^{\pm
1},(I_{k,3}I_{k,4})^{\pm 1},(I_{k,4}I_{k,1})^{\pm
1},I_{k,1}I_{k,3},I_{k,2}I_{k,4}\right\\}$

is a subset of the group

$\langle I_{k,1}I_{k,2},I_{k,2}I_{k,3},I_{k,3}I_{k,4},I_{k,4}I_{k,1}\rangle.$

As usual, we denote by $p_{ij}=I_{k,i}I_{k,j}(p_{0})\in{\bf
H}^{3}_{\mathbb{R}}$ and $B_{ij}=B(p_{0},p_{ij})$ the bisector in ${\bf
H}^{3}_{\mathbb{R}}$ with respect to the two points $p_{0}$ and $p_{ij}$ for
certain $i,j\in\\{1,2,3,4\\}$. Then

$p_{12}=I_{k,1}I_{k,2}(p_{0})=[\frac{3}{2},-\frac{3}{2},\frac{\sqrt{3}}{2},\frac{5}{2}]^{t},$
$p_{21}=I_{k,2}I_{k,1}(p_{0})=[-\frac{5}{2},-\frac{7}{2},\frac{\sqrt{3}}{2},\frac{9}{2}]^{t},$
$p_{23}=I_{k,2}I_{k,3}(p_{0})=[-\frac{7}{2},-\frac{5}{2},\frac{\sqrt{3}}{2},\frac{9}{2}]^{t},$
$p_{32}=I_{k,3}I_{k,2}(p_{0})=[-\frac{3}{2},\frac{3}{2},\frac{\sqrt{3}}{2},\frac{5}{2}]^{t},$
$p_{34}=I_{k,3}I_{k,4}(p_{0})=[-\frac{3}{2},\frac{3}{2},\frac{3\sqrt{3}}{2},\frac{7}{2}]^{t},$
$p_{43}=I_{k,4}I_{k,3}(p_{0})=[-\frac{7}{2},-\frac{5}{2},\frac{7\sqrt{3}}{2},\frac{15}{2}]^{t},$
$p_{41}=I_{k,4}I_{k,1}(p_{0})=[-\frac{5}{2},-\frac{7}{2},\frac{7\sqrt{3}}{2},\frac{15}{2}]^{t},$
$p_{14}=I_{k,1}I_{k,4}(p_{0})=[\frac{3}{2},-\frac{3}{2},\frac{3\sqrt{3}}{2},\frac{7}{2}]^{t},$
$p_{13}=I_{k,1}I_{k,3}(p_{0})=[\frac{1}{2},\frac{1}{2},\frac{\sqrt{3}}{2},\frac{3}{2}]^{t},$
$p_{24}=I_{k,2}I_{k,4}(p_{0})=[-\frac{5}{2},-\frac{5}{2},\frac{3\sqrt{3}}{2},\frac{9}{2}]^{t}.$

Figure 7. Part of a realistic view of Dirichlet domain of
$\rho_{\pi}(K)<\mathbf{PO}(3,1)$ from a point view inside $D_{R}$. The
steelblue sphere is the ideal boundary of ${\bf H}^{3}_{\mathbb{R}}$.

We fix lifts $\bf{p_{0}}$ and $\bf{p_{ij}}$ of $p_{0}$ and $p_{ij}$ as vectors
in $\mathbb{R}^{3,1}$, such that the entries of $\bf{p_{0}}$ and $\bf{p_{ij}}$
are just the same as entries of $p_{0}$ and $p_{ij}$ above. We have

$\langle{\bf p_{0}},{\bf p_{0}}\rangle=\langle{\bf p_{ij}},{\bf
p_{ij}}\rangle=-1$

for ${\bf p_{ij}}$ above.

Now for $p=[x,y,z,1]^{t}\in{\bf H}^{3}_{\mathbb{R}}$, if $p$ lies in the
bisector $B_{12}$ of $I_{k,1}I_{k,2}$, that is, the distances of $p$ to
$p_{0}$ and $p_{12}$ equal, then

$|{\bf p_{0}}^{*}\cdot H\cdot{\bf p}|=|{\bf p_{12}}^{*}\cdot H\cdot{\bf p}|.$

So the bisector $B_{12}$ is given by all $[x,y,z,1]^{t}$ with the conditions

$|-\frac{1}{2}\cdot x-\frac{1}{2}\cdot y+\frac{\sqrt{3}}{2}\cdot
z-\frac{3}{2}\cdot 1|=|\frac{3}{2}\cdot x-\frac{3}{2}\cdot
y+\frac{\sqrt{3}}{2}\cdot z-\frac{5}{2}\cdot 1|$

and

$x^{2}+y^{2}+z^{2}<1.$

Which is a round disk in the Klein model of ${\bf H}^{3}_{\mathbb{R}}$.
Similarly, we can give all the bisectors $B_{21}$, $B_{23}$, $B_{32}$,
$B_{34}$, $B_{43}$, $B_{41}$, $B_{14}$, $B_{13}$ and $B_{24}$.

Let $A_{k,i}=I_{k,i}I_{k,i+1}$ for $i=1,2,3,4$ mod 4. It is not difficult to
show the following directly (without using the over-killed methods in Section
6), we omit the details.

######  Proposition 5.1.

$D_{R}$ is the Dirichlet domain for $\rho_{\pi}(K)<\mathbf{PO}(3,1)$ acting on
${\bf H}^{3}_{\mathbb{R}}$ with center $p_{0}$. Moreover, the group
$\rho_{\pi}(K)=\langle A_{k,1},A_{k,2},A_{k,3},A_{k,4}\rangle$ is discrete and
has a presentation

$\rho_{\pi}(K)=\left\langle
A_{k,1},A_{k,2},A_{k,3},A_{k,4}\Bigg{|}\begin{array}[]{c}A_{k,1}A_{k,2}A_{k,3}A_{k,4}=id,\\\\[3.0pt]
(A_{k,1}A_{k,2})^{2}=(A_{k,2}A_{k,3})^{2}=id,\\\\[3.0pt]
(A_{k,3}A_{k,4})^{2}=(A_{k,4}A_{k,1})^{2}=id\end{array}\right\rangle$

$\partial{\bf H}^{3}_{\mathbb{R}}$$\partial{\bf
H}^{3}_{\mathbb{R}}$$\partial{\bf H}^{3}_{\mathbb{R}}$$\partial{\bf
H}^{3}_{\mathbb{R}}$$B_{12}$$B_{14}$$B_{41}$$B_{43}$$B_{23}$$B_{21}$$B_{34}$$B_{32}$$B_{24}$$B_{13}$
Figure 8. The abstract picture of the boundary of Dirichlet domain of
$\rho_{\pi}(K)<\mathbf{PO}(3,1)$ look out side from $p_{0}$, which is a
2-sphere with labeled disks. For example, the disk labeled by $B_{12}$ means
$B_{12}\cap D_{R}$. The out disk is $B_{13}\cap D_{R}$. Each of the four disks
labeled by $\partial{\bf H}^{3}_{\mathbb{R}}$ means a disk in $\partial{\bf
H}^{3}_{\mathbb{R}}$.

See Figure 7 for the partial Dirichlet domain of $D_{R}<\rho_{\pi}(K)$ acting
on ${\bf H}^{3}_{\mathbb{R}}$. The color rule in Figure 5 is:

  * •

$B_{12}$, gray;

  * •

$B_{21}$, cyan;

  * •

$B_{23}$, blue;

  * •

$B_{32}$, yellow;

  * •

$B_{34}$, green;

  * •

$B_{43}$, brown;

  * •

$B_{41}$, aquamarine;

  * •

$B_{14}$, red;

  * •

$B_{13}$, purple;

  * •

$B_{24}$, black.

For example, the brown rectangle in Figure 7 is $B_{43}\cap D_{R}$, which is a
part of the frontier of $D_{R}$. The aquamarine and red rectangles are
$B_{41}\cap D_{R}$ and $B_{14}\cap D_{R}$ respectively. The black hexagon is
$B_{24}\cap D_{R}$. But the other $B_{ij}\cap D_{R}$ is not so clear from this
realistic figure. See also Figure 8 for an abstract picture of the Dirichlet
domain of $\rho_{\pi}(K)<\mathbf{PO}(3,1)$.

Figures 7 and 8 in ${\bf H}^{3}_{\mathbb{R}}$-geometry should be compared with
Figures 5 and 6 in ${\bf H}^{2}_{\mathbb{C}}$-geometry of
$\rho_{\frac{5\pi}{6}}(K)$. In particular, both the Dirichlet domains are
given by the same set $R$. Moreover, the intersection patterns of the sides of
$D_{R}$ are also the same. Which give the hint that the Dirichlet domain of
$\rho_{\theta}(K)$ for general $\theta\in(\frac{5\pi}{6},\pi]$ is also given
by the set $R$. This is what we do in Section 6.

##  6\. Dirichlet domain of $\rho_{\theta}(K)$ in ${\bf H}^{3}_{\mathbb{C}}$

In this section, we prove Theorem 1.1 for all $\theta\in(\frac{5}{6},\pi]$.

###  6.1. Some preliminaries

For any $\theta\in(\frac{5}{6},\pi]$ in the moduli space $\mathcal{M}$, we
denote by $p_{ij}=I_{i}I_{j}(p_{0})\in{\bf H}^{3}_{\mathbb{C}}$, and
$B_{ij}=B(p_{0},p_{ij})$ the bisector in ${\bf H}^{3}_{\mathbb{C}}$ with
respect to the points $p_{0}$ and $p_{ij}$. We also consider the partial
Dirichlet domain $D_{R}$ for the same set $R$ as in Sections 4 and 5. Recall
$R\subset\rho_{\frac{5\pi}{6}}(K)$ be the set of ten words in
$\rho_{\theta}(K)$:

$\left\\{(I_{1}I_{2})^{\pm 1},~{}~{}~{}(I_{2}I_{3})^{\pm
1},~{}~{}~{}(I_{3}I_{4})^{\pm 1},~{}~{}~{}(I_{4}I_{1})^{\pm
1},~{}~{}~{}I_{1}I_{3},~{}~{}~{}I_{2}I_{4}\right\\}.$

######  Remark 6.1.

The main technical difference between the partial Dirichlet domains $D_{R}$
when $\theta\in(\frac{5}{6},\pi]$ and $\theta=\frac{5\pi}{6}$ is:

  * •

$B_{12}\cap B_{34}=\emptyset$ when $\theta$ is near to $\pi$;

  * •

$B_{12}\cap B_{34}\neq\emptyset$ when $\theta$ is near to $\frac{5\pi}{6}$ (in
particular, when $\theta=\frac{5\pi}{6}$, $B_{12}\cap B_{34}$ is non empty as
shown in Proposition 4.2).

But in any case $B_{12}\cap B_{34}$ does not lie in the partial Dirichlet
domain $D_{R}$, see Proposition 6.4.

Now the fixed point of $J$ is $p_{0}=[0,0,0,1]^{t}$, and

(6.1)
$p_{12}=\left[\begin{array}[]{c}-\rm{e}^{-\theta\rm{i}}\sqrt{2\cos(2\theta)+1}\\\\[8.61108pt]
\displaystyle{\frac{(-2\rm{e}^{-\theta\rm{i}}+1+\rm{i})\sqrt{-1-2\cos(\theta)+2\sin(\theta)+2\sin(2\theta)}}{2}}\\\\[17.22217pt]
\displaystyle{\frac{(-2\rm{e}^{-\theta\rm{i}}+1-\rm{i})\sqrt{-2\sin(2\theta)-2\cos(\theta)-1-2\sin(\theta)}}{2}}\\\\[17.22217pt]
\rm{e}^{-2\theta\rm{i}}-\rm{e}^{\theta\rm{i}}+1\\\ \end{array}\right],$ (6.2)
$p_{21}=\left[\begin{array}[]{c}\rm{e}^{\theta\rm{i}}\sqrt{2\cos(2\theta)+1}\\\\[8.61108pt]
\displaystyle{\frac{(-2\rm{i}\rm{e}^{\theta\rm{i}}+1+\rm{i})\sqrt{-1-2\cos(\theta)+2\sin(\theta)+2\sin(2\theta)}}{2}}\\\\[17.22217pt]
\displaystyle{\frac{(2\rm{i}\rm{e}^{\theta\rm{i}}+1-\rm{i})\sqrt{-2\sin(2\theta)-2\cos(\theta)-1-2\sin(\theta)}}{2}}\\\\[17.22217pt]
\rm{e}^{2\theta\rm{i}}-\rm{e}^{-\theta\rm{i}}+1\\\ \end{array}\right],$

(6.3) $p_{13}=\left[\begin{array}[]{c}\sqrt{2\cos(2\theta)+1}\\\\[4.30554pt]
0\\\ 0\\\ -2\cos(\theta)\\\ \end{array}\right].$

From the coordinates of these points in ${\bf H}^{3}_{\mathbb{C}}$, it is easy
to get all coordinates of $p_{i,i\pm 1}$ for $i=1,2,3,4$ mod 4 and $p_{24}$ by
the $J$-action. We fix lifts $\bf{p_{0}}$ and $\bf{p_{ij}}$ of $p_{0}$ and
$p_{ij}$ as vectors in $\mathbb{C}^{3,1}$, such that the entries of
$\bf{p_{0}}$ and $\bf{p_{ij}}$ are just the same as entries of $p_{0}$ and
$p_{ij}$ above. We have

$\langle{\bf p_{0}},{\bf p_{0}}\rangle=\langle{\bf p_{ij}},{\bf
p_{ij}}\rangle=-1$

for ${\bf p_{ij}}$ above.

First we have

######  Lemma 6.2.

For any $\theta\in(\frac{5\pi}{6},\pi]$ in the moduli space $\mathcal{M}$, the
four vectors $\bf{p_{0}}$, $\bf{p_{12}}$, $\bf{p_{13}}$ and $\bf{p_{14}}$ are
linearly independent in $\mathbb{C}^{3,1}$.

###### Proof.

This is proved by a direct calculation. Consider the matrix $S$ such that its
four columns are the vectors $p_{0}$, $p_{12}$, $p_{13}$ and $p_{14}$ given by
(6.1), (6.3) and $J$-action respectively. Then

$\det(S)=(-2\rm{i}\cos(\theta)-2\rm{i}\cos(3\theta)-2\sin(3\theta)+\rm{i})\sqrt{4\cos^{2}(2\theta)-1}.$

Now it is easy to see $\det(S)=0$ when $\theta=\frac{5\pi}{6}$ and $\det(S)$
is non-zero when $\theta\in(\frac{5\pi}{6},\pi]$. ∎

Lemma 6.2 also explains why we should prove Theorem 1.1 when
$\theta=\frac{5\pi}{6}$ separately in Section 4, since the proof in this
section does not hold when these four vectors $\bf{p_{0}}$, $\bf{p_{12}}$,
$\bf{p_{13}}$ and $\bf{p_{14}}$ span a three dimensional subspace of
$\mathbb{C}^{3,1}$.

######  Lemma 6.3.

For any $\theta\in(\frac{5\pi}{6},\pi]$ in the moduli space $\mathcal{M}$, the
four vectors $\bf{p_{0}}$, $\bf{p_{12}}$, $\bf{p_{34}}$ and $\bf{p_{13}}$ are
co-planar in $\mathbb{C}^{3,1}$.

###### Proof.

This is also proved by a direct calculation. In fact we have

$2\rm{e}^{\theta\rm{i}}\cdot{\bf p_{0}}+{\bf p_{12}}+{\bf
p_{34}}+2\rm{e}^{-\theta\rm{i}}\cdot{\bf p_{13}}$

is the zero vector in $\mathbb{C}^{3,1}$. ∎

###  6.2. Intersection patterns of the bisectors of the partial Dirichlet
domain $D_{R}$

In this subsection, we will study intersection patterns of the bisectors for
$D_{R}$. The reader may compare to Table 1 in Section 4.2 (there is only one
difference on the intersection $B_{12}\cap B_{34}$ as noted in Remark 6.1).

We first consider the intersections of the bisector $B_{12}$ with other
bisectors, comparing to Proposition 4.2 in Subsection 4.3, the proof here is
much more involved.

######  Proposition 6.4.

For any $\theta\in(\frac{5\pi}{6},\pi]$, the bisector $B_{12}$ of $I_{1}I_{2}$
have the following properties:

  1. (1)

$B_{12}$ is tangent to $B_{21}$;

  2. (2)

$B_{12}$ does not intersect $B_{23}$;

  3. (3)

$B_{12}$ does not intersect $B_{32}$;

  4. (4)

$B_{12}$ does not intersect $B_{43}$;

  5. (5)

$B_{12}$ does not intersect $B_{41}$;

  6. (6)

$B_{12}$ does not intersect $B_{24}$;

  7. (7)

$B_{12}$ intersects $B_{34}$ in a non-empty Giraud disk $B_{12}\cap B_{34}$ in
${\bf H}^{3}_{\mathbb{C}}$ when $\theta\in[\frac{5\pi}{6},\pi]$ but near to
$\frac{5\pi}{6}$. $B_{12}\cap B_{34}=\emptyset$ when
$\theta\in[\frac{5\pi}{6},\pi]$ but near to $\pi$. When $B_{12}\cap B_{34}$ is
non-empty, $B_{12}\cap B_{34}$ lies in the component of ${\bf
H}^{2}_{\mathbb{C}}-B_{13}$ which does not contain the point $p_{0}$. In
particular, in any case $B_{12}\cap B_{34}$ does not lie in the partial
Dirichlet domain $D_{R}$.

###### Proof.

The proof of (1) of Proposition 6.4 runs the same line as the proof of (1) of
Proposition 4.2.

For the proof of (2). We now consider the intersection $B_{12}\cap B_{23}$. It
is easy to see that for any $\theta\in(\frac{5\pi}{6},\pi]$, the span of
$\bf{p_{0}}$, $\bf{p_{12}}$, $\bf{p_{23}}$ is a 3-dimensional subspace of
$\mathbb{C}^{3,1}$. The intersection of ${\bf H}^{3}_{\mathbb{C}}\subset{\bf
P}^{3}_{\mathbb{C}}$ with the projection of this 3-dimensional subspace into
${\bf P}^{3}_{\mathbb{C}}$ is denoted by $L$, then $L$ is a totally geodesic
${\bf H}^{2}_{\mathbb{C}}\hookrightarrow{\bf H}^{3}_{\mathbb{C}}$.

We re-denote $\bf{p_{0}}$, $\bf{p_{12}}$ and $\bf{p_{23}}$ by $\bf{e_{1}}$,
$\bf{e_{2}}$ and $\bf{e_{3}}$. Then we denote
$H_{L}=H_{L}(p_{0},p_{12},p_{23})$ by the matrix $({\bf e_{i}}^{*}\cdot
H\cdot{\bf e_{j}})_{1\leq i,j\leq 3}$, we have

$H_{L}=\begin{pmatrix}-1&-\rm{e}^{-2\theta\rm{i}}+\rm{e}^{\theta\rm{i}}-1&-\rm{e}^{-2\theta\rm{i}}+\rm{e}^{\theta\rm{i}}-1\\\
-\rm{e}^{2\theta\rm{i}}+\rm{e}^{-\theta\rm{i}}-1&-1&D\\\
-\rm{e}^{2\theta\rm{i}}+\rm{e}^{-\theta\rm{i}}-1&\bar{D}&-1\\\ \end{pmatrix}.$

Here

$D=2\rm{e}^{-3\theta\rm{i}}-2\rm{e}^{-2\theta\rm{i}}+2\rm{e}^{\theta\rm{i}}-2\cos(2\theta)-4,$

and $\bar{D}$ is the complex conjugate of $D$. Now $\det(H_{L})$ is

$(-4\cos(4\theta)-22\cos(2\theta)-1+16\cos(3\theta)+12\cos(\theta))\cdot(1+2\cos(\theta))^{2}.$

So $H_{L}$ is the Hermitian form with signature $(2,1)$ on the subspace with
the basis $\\{{\bf e_{1}},{\bf e_{2}},{\bf e_{3}}\\}$ when
$\theta\in(\frac{5\pi}{6},\pi]$.

The vector $x_{1}{\bf e_{1}}+x_{2}{\bf e_{2}}+x_{3}{\bf e_{3}}$ is denoted by
the vector ${\bf x}$ in $\mathbb{C}^{3}$, and the vector $y_{1}{\bf
e_{1}}+y_{2}{\bf e_{2}}+y_{3}{\bf e_{3}}$ is denoted by the vector ${\bf y}$
in $\mathbb{C}^{3}$, here

${\bf x}=\left(\begin{array}[]{c}x_{1}\\\ x_{2}\\\ x_{3}\\\
\end{array}\right),\quad{\bf y}=\left(\begin{array}[]{c}y_{1}\\\ y_{2}\\\
y_{3}\\\ \end{array}\right)$

with $x_{i},y_{i}\in\mathbb{C}$. So

${\bf E_{1}}=\left(\begin{array}[]{c}1\\\ 0\\\ 0\\\
\end{array}\right),\quad{\bf E_{2}}=\left(\begin{array}[]{c}0\\\ 1\\\ 0\\\
\end{array}\right),\quad{\bf E_{3}}=\left(\begin{array}[]{c}0\\\ 0\\\ 1\\\
\end{array}\right)$

in $\mathbb{C}^{3}$ present the vectors ${\bf e_{1}}$, ${\bf e_{2}}$ and ${\bf
e_{3}}$ in $\mathbb{C}^{3,1}$.

Comparing to Subsection 2.5, we define the Hermitian cross-product
$\boxtimes_{L}$ on the space $H_{L}$ (which is isometric to ${\bf
H}^{2}_{\mathbb{C}}$) with respect to the basis
$\\{\bf{e_{1}},\bf{e_{2}},\bf{e_{3}}\\}$ by

$\mathbf{x}\boxtimes_{L}\mathbf{y}=\left[\begin{matrix}\mathbf{x}^{*}H_{L}(1,2)\cdot\mathbf{y}^{*}H_{L}(1,3)-\mathbf{y}^{*}H_{L}(1,2)\cdot\mathbf{x}^{*}H_{L}(1,3)\\\
\mathbf{x}^{*}H_{L}(1,3)\cdot\mathbf{y}^{*}H_{L}(1,1)-\mathbf{y}^{*}H_{L}(1,3)\cdot\mathbf{x}^{*}H_{L}(1,1)\\\
\mathbf{x}^{*}H_{L}(1,1)\cdot\mathbf{y}^{*}H_{L}(1,2)-\mathbf{y}^{*}H_{L}(1,1)\cdot\mathbf{x}^{*}H_{L}(1,2)\end{matrix}\right].$

Here $\mathbf{x}^{*}H_{L}$ is a one-by-three matrix, and
$\mathbf{x}^{*}H_{L}(1,2)$ is the second entry of $\mathbf{x}^{*}H_{L}$. The
other terms have similar meanings.

Then the intersection $B_{12}\cap B_{23}\cap L$ is parameterized by
$V=V(z_{1},z_{2})\in\mathbb{C}^{3}$ with $\langle V,V\rangle<0$ with respect
to the Hermitian form $H_{L}$. Where

$V={\bf E_{2}}\boxtimes_{L}{\bf E_{3}}+z_{1}\cdot{\bf E_{1}}\boxtimes_{L}{\bf
E_{3}}+z_{2}\cdot{\bf E_{1}}\boxtimes_{L}{\bf E_{2}}$

and
$(z_{1},z_{2})=(\rm{e}^{r\rm{i}},\rm{e}^{s\rm{i}})\in\mathbb{S}^{1}\times\mathbb{S}^{1}$.

We note that

$E_{1}\boxtimes_{L}E_{2}=\left[\begin{array}[]{c}D_{1}\\\\[4.30554pt]
2\rm{i}\sin(\theta)-2\rm{i}\sin(3\theta)+2\cos(2\theta)+1+2\rm{i}\sin(2\theta)\\\\[4.30554pt]
2\cos(3\theta)+2\cos(\theta)-2\cos(2\theta)-2\\\ \end{array}\right],$

where

$\displaystyle
D_{1}=-5\rm{i}\sin(2\theta)+\rm{i}\sin(3\theta)-3\rm{i}\sin(4\theta)+2\rm{i}\sin(5\theta)+4+11\cos(2\theta)$
$\displaystyle-3\cos(3\theta)-2\cos(5\theta)-10\cos(\theta)+3\cos(4\theta).$

We also have

$E_{1}\boxtimes_{L}E_{3}=\left[\begin{array}[]{c}D_{2}\\\\[4.30554pt]
-2\cos(3\theta)-2\cos(\theta)+2\cos(2\theta)+2\\\\[4.30554pt]
2\rm{i}\sin(\theta)-2\rm{i}\sin(3\theta)+2\cos(2\theta)+1+2\rm{i}\sin(2\theta)\\\
\end{array}\right],$

where

$\displaystyle
D_{2}=\rm{i}\sin(2\theta)+3\rm{i}\sin(3\theta)-\rm{i}\sin(4\theta)+2\rm{i}\sin(\theta)-8-7\cos(2\theta)+8\cos(\theta)$
$\displaystyle+7\cos(3\theta)-3\cos(4\theta),$

and

$E_{2}\boxtimes_{L}E_{3}=\left[\begin{array}[]{c}4\cos(5\theta)+28\cos(3\theta)+32\cos(\theta)-14\cos(4\theta)-32\cos(2\theta)-33\\\\[4.30554pt]
D_{3}\\\\[4.30554pt] D_{4}\\\ \end{array}\right],$

here

$\displaystyle
D_{3}=\rm{i}\sin(2\theta)+3\rm{i}\sin(3\theta)-\rm{i}\sin(4\theta)+2\rm{i}\sin(\theta)+8+7\cos(2\theta)-8\cos(\theta)$
$\displaystyle-7\cos(3\theta)+3\cos(4\theta)$

and

$\displaystyle
D_{4}=5\rm{i}\sin(2\theta)-\rm{i}\sin(3\theta)+3\rm{i}\sin(4\theta)-2\rm{i}\sin(5\theta)+4+11\cos(2\theta)-3\cos(3\theta)$
$\displaystyle-2\cos(5\theta)-10\cos(\theta)+3\cos(4\theta).$

The vector $V\in\mathbb{C}^{3}$ is a three-by-one matrix, we denote by
$V(i,1)$ the entry in the $i$-th row of $V$. Then

$V(1,1)\cdot{\bf e_{1}}+V(2,1)\cdot{\bf e_{2}}+V(3,1)\cdot{\bf e_{3}}$

is a vector in $\mathbb{C}^{3,1}$, we denote it by $W$. The projection of $W$
into ${\mathbb{C}}{\mathbf{P}}^{3}$ is a point in $B_{12}\cap B_{23}\cap L$ if
$W$ is a negative vector. Now $W^{*}\cdot H\cdot W$ is a very complicated
term. But with Maple, $W^{*}\cdot H\cdot W$ has minimum 734.88 numerically
when $r,s\in[-\pi,\pi]$ and $\theta\in[\frac{5\pi}{6},\pi]$ at a point when
$\theta=\frac{5\pi}{6}$. In particular, any $W$ above is a positive vector in
$\mathbb{C}^{3,1}$, so $B_{12}\cap B_{23}\cap L=\emptyset$. Then by the
projection from ${\bf H}^{3}_{\mathbb{C}}$ to $L$, we have $B_{12}\cap B_{23}$
in ${\bf H}^{3}_{\mathbb{C}}$ is the empty set.

We remark here the minimum of $W^{*}\cdot H\cdot W$ when
$\theta=\frac{5\pi}{6}$ is 734.88, which is different from the minimum 12.752
in Proposition 4.2. The reason is that even through the Hermitian from $H_{L}$
is related to $H$, but it is not the same as $H$. This end the proof of (2).

For the proof of (3), so we consider $B_{12}\cap B_{32}$. For any
$\theta\in(\frac{5\pi}{6},\pi]$, the span of ${\bf p_{0}}$, ${\bf p_{12}}$,
${\bf p_{32}}$ is a 3-dimensional subspace of $\mathbb{C}^{3,1}$. The
intersection of ${\bf H}^{3}_{\mathbb{C}}\subset{\bf P}^{3}_{\mathbb{C}}$ with
the projection of this 3-dimensional subspace into ${\bf P}^{3}_{\mathbb{C}}$
is denoted by $L$, then $L$ is a totally geodesic ${\bf
H}^{2}_{\mathbb{C}}\hookrightarrow{\bf H}^{3}_{\mathbb{C}}$.

We re-denote ${\bf p_{0}}$, ${\bf p_{12}}$ and ${\bf p_{32}}$ by ${\bf
e_{1}}$, ${\bf e_{2}}$ and ${\bf e_{3}}$. We denote
$H_{L}=H_{L}(p_{0},p_{12},p_{32})$ by the matrix $({\bf e_{i}}^{*}\cdot
H\cdot{\bf e_{j}})_{1\leq i,j\leq 3}$, then

$H_{L}=\begin{pmatrix}-1&-\rm{e}^{-2\theta\rm{i}}+\rm{e}^{\theta\rm{i}}-1&-\rm{e}^{2\theta\rm{i}}+\rm{e}^{-\theta\rm{i}}-1\\\
-\rm{e}^{2\theta\rm{i}}+\rm{e}^{-\theta\rm{i}}-1&-1&6\cos(\theta)-4\cos(2\theta)\\\
-\rm{e}^{-2\theta\rm{i}}+\rm{e}^{\theta\rm{i}}-1&6\cos(\theta)-4\cos(2\theta)&-1\\\
\end{pmatrix}.$

Now $\det(H_{L})$ is

$-(4\cos(\theta)-3)\cdot(-1+2\cos(\theta))\cdot(4\cos(\theta)^{2}-2\cos(\theta)-3)\cdot(1+2\cos(\theta))^{2}.$

So $H_{L}$ is the Hermitian form with signature $(2,1)$ on the subspace with
the basis $\\{{\bf e_{1}},{\bf e_{2}},{\bf e_{3}}\\}$ when
$\theta\in(\frac{5\pi}{6},\pi]$.

The vector $x_{1}{\bf e_{1}}+x_{2}{\bf e_{2}}+x_{3}{\bf e_{3}}$ is denoted by
the vector ${\bf x}$ in $\mathbb{C}^{3}$, and the vector $y_{1}{\bf
e_{1}}+y_{2}{\bf e_{2}}+y_{3}{\bf e_{3}}$ is denoted by the vector ${\bf y}$
in $\mathbb{C}^{3}$, here

${\bf x}=\left(\begin{array}[]{c}x_{1}\\\ x_{2}\\\ x_{3}\\\
\end{array}\right),\quad{\bf y}=\left(\begin{array}[]{c}y_{1}\\\ y_{2}\\\
y_{3}\\\ \end{array}\right)$

with $x_{i},y_{i}\in\mathbb{C}$. So

${\bf E_{1}}=\left(\begin{array}[]{c}1\\\ 0\\\ 0\\\
\end{array}\right),\quad{\bf E_{2}}=\left(\begin{array}[]{c}0\\\ 1\\\ 0\\\
\end{array}\right),\quad{\bf E_{3}}=\left(\begin{array}[]{c}0\\\ 0\\\ 1\\\
\end{array}\right)$

in $\mathbb{C}^{3}$ present the vectors ${\bf e_{1}}$, ${\bf e_{2}}$ and ${\bf
e_{3}}$ in $\mathbb{C}^{3,1}$.

Comparing to Subsection 2.5, we define the Hermitian cross-product
$\boxtimes_{L}$ on the space $H_{L}={\bf H}^{2}_{\mathbb{C}}$ with respect to
the basis $\\{{\bf e_{1}},{\bf e_{2}},{\bf e_{3}}\\}$ by

$\mathbf{x}\boxtimes_{L}\mathbf{y}=\left[\begin{matrix}\mathbf{x}^{*}H_{L}(1,2)\cdot\mathbf{y}^{*}H_{L}(1,3)-\mathbf{y}^{*}H_{L}(1,2)\cdot\mathbf{x}^{*}H_{L}(1,3)\\\
\mathbf{x}^{*}H_{L}(1,3)\cdot\mathbf{y}^{*}H_{L}(1,1)-\mathbf{y}^{*}H_{L}(1,3)\cdot\mathbf{x}^{*}H_{L}(1,1)\\\
\mathbf{x}^{*}H_{L}(1,1)\cdot\mathbf{y}^{*}H_{L}(1,2)-\mathbf{y}^{*}H_{L}(1,1)\cdot\mathbf{x}^{*}H_{L}(1,2)\end{matrix}\right].$

Then the intersection $B_{12}\cap B_{32}\cap L$ is parameterized by
$V=V(z_{1},z_{2})\in\mathbb{C}^{3}$ with $\langle V,V\rangle<0$ with respect
to the Hermitian form $H_{L}$. Where

$V={\bf E_{2}}\boxtimes_{L}{\bf E_{3}}+z_{1}\cdot{\bf E_{1}}\boxtimes_{L}{\bf
E_{3}}+z_{2}\cdot{\bf E_{1}}\boxtimes_{L}{\bf E_{2}}$

and
$(z_{1},z_{2})=(\rm{e}^{r\rm{i}},\rm{e}^{s\rm{i}})\in\mathbb{S}^{1}\times\mathbb{S}^{1}$.

We note that $E_{1}\boxtimes_{L}E_{2}$ is

$\left[\begin{array}[]{c}(1+2\cos(\theta))(8\cos^{2}(\theta)\rm{e}^{-\theta\rm{i}}+3\rm{i}\sin(2\theta)+3\rm{i}\sin(\theta)-14\cos(\theta)^{2}+5\cos(\theta))\\\\[4.30554pt]
3+8\cos^{3}(\theta)\rm{e}^{\theta\rm{i}}-10\cos^{2}(\theta)-\rm{i}\sin(2\theta)+2\cos(\theta)\\\\[4.30554pt]
4\cos(\theta)(\cos(\theta)-1)(1+2\cos(\theta))\\\ \end{array}\right],$

$E_{1}\boxtimes_{L}E_{3}$ is

$\left[\begin{array}[]{c}(1+2\cos(\theta))(-8\cos^{2}(\theta)\rm{e}^{\theta\rm{i}}+3\rm{i}\sin(2\theta)+3\rm{i}\sin(\theta)+14\cos(\theta)^{2}-5\cos(\theta))\\\\[4.30554pt]
4\cos(\theta)(1-\cos(\theta))(1+2\cos(\theta))\\\\[4.30554pt]
-3-8\cos^{3}(\theta)\rm{e}^{-\theta\rm{i}}+10\cos^{2}(\theta)-\rm{i}\sin(2\theta)-2\cos(\theta)\\\
\end{array}\right].$

and $E_{2}\boxtimes_{L}E_{3}$ is

$=\left[\begin{array}[]{c}(5-4\cos(\theta))(1+2\cos(\theta))(8\cos^{2}(\theta)-6\cos(\theta)-3)\\\\[4.30554pt]
(1+2\cos(\theta))(8\cos^{2}(\theta)\rm{e}^{-\theta\rm{i}}+3\rm{i}\sin(2\theta)+3\rm{i}\sin(\theta)-14\cos(\theta)^{2}+5\cos(\theta))\\\\[4.30554pt]
(1+2\cos(\theta))(8\cos^{2}(\theta)\rm{e}^{\theta\rm{i}}-3\rm{i}\sin(2\theta)-3\rm{i}\sin(\theta)-14\cos(\theta)^{2}+5\cos(\theta))\\\
\end{array}\right].$

The vector $V\in\mathbb{C}^{3}$ is a three-by-one matrix, and

$V(1,1)\cdot{\bf e_{1}}+V(2,1)\cdot{\bf e_{2}}+V(3,1)\cdot{\bf e_{3}}$

is a vector in $\mathbb{C}^{3,1}$, we denote it by $W$. The projection of $W$
into ${\mathbb{C}}{\mathbf{P}}^{3}$ is a point in $B_{12}\cap B_{32}\cap L$ if
$W$ is a negative vector. With Maple, $W^{*}\cdot H\cdot W$ is

(6.4)
$\displaystyle\left(-256\cdot(\frac{1}{2}+\cos(\theta))^{3}\cdot(\cos(\theta)-\frac{3}{4})\cdot(-\frac{1}{2}+\cos(\theta))\cdot(\cos^{2}(\theta)-\frac{\cos(\theta)}{2}-\frac{3}{4})\right)\cdot
h,$

where $h$ is

$\displaystyle
21-8\cos(3\theta)+36\cos(2\theta)+4\cos(r-2\theta)+10\cos(r+2\theta)-50\cos(\theta)+14\cos(r)$
$\displaystyle-14\cos(s)+2\cos(-2\theta+r-s)+2\cos(\theta+r-s)-4\cos(3\theta+r)+10\cos(-\theta+s)$
$\displaystyle+12\cos(\theta+s)-10\cos(s-2\theta)-4\cos(s+2\theta)-2\cos(-3\theta+r-s)$
$\displaystyle-4\cos(r-s)-12\cos(-\theta+r)-10\cos(\theta+r)+4\cos(-3\theta+s).$

Figure 9. The positivity of the term $h$ in the proof of (3) of Proposition
6.4. In this figure, we take $h=h(r,s)$. The blue surface is the graph of the
function $h$ when $\theta=\frac{5\pi}{6}$. The red plane is $h=0$. This
elucidates (but does not prove) $h$ is positive when $\theta=\frac{5\pi}{6}$.
For general $\theta\in(\frac{5\pi}{6},\pi]$, the graph of $h=h(r,s)$ is a
surface which is similar to the surface in the figure and it is also disjoint
from the plane $h=0$.

Note that the first term of (6.4) is positive when
$\theta\in[\frac{5\pi}{6},\pi]$. By Maple, $h$ has minimum 6.5305 numerically
when $r,s\in[-\pi,\pi]$ and $\theta\in[\frac{5\pi}{6},\pi]$ at a point when
$\theta=\frac{5\pi}{6}$. In particular, any $W$ above is a positive vector in
$\mathbb{C}^{3,1}$, so $B_{12}\cap B_{32}\cap L=\emptyset$. Then by the
projection from ${\bf H}^{3}_{\mathbb{C}}$ to $L$, we have $B_{12}\cap B_{32}$
in ${\bf H}^{3}_{\mathbb{C}}$ is the empty set. See Figure 9 for an
illustration this fact when $\theta=\frac{5\pi}{6}$. This ends the proof of
(3).

For the proof of (4), now we consider $B_{12}\cap B_{43}$. So for any
$\theta\in(\frac{5\pi}{6},\pi]$, the span of ${\bf p_{0}}$, ${\bf p_{12}}$,
${\bf p_{43}}$ is a 3-dimensional subspace of $\mathbb{C}^{3,1}$. The
intersection of ${\bf H}^{3}_{\mathbb{C}}\subset{\bf P}^{3}_{\mathbb{C}}$ with
the projection of this 3-dimensional subspace into ${\bf P}^{3}_{\mathbb{C}}$
is denoted by $L$, then $L$ is a totally geodesic ${\bf
H}^{2}_{\mathbb{C}}\hookrightarrow{\bf H}^{3}_{\mathbb{C}}$.

We re-denote ${\bf p_{0}}$, ${\bf p_{12}}$ and ${\bf p_{43}}$ by ${\bf
e_{1}}$, ${\bf e_{2}}$ and ${\bf e_{3}}$. We denote
$H_{L}=H(p_{0},p_{12},p_{43})$ by the matrix $({\bf e_{i}}^{*}\cdot H\cdot{\bf
e_{j}})_{1\leq i,j\leq 3}$, then

$H_{L}=\begin{pmatrix}-1&-\rm{e}^{-2\theta\rm{i}}+\rm{e}^{\theta\rm{i}}-1&-\rm{e}^{2\theta\rm{i}}+\rm{e}^{-\theta\rm{i}}-1\\\
-\rm{e}^{2\theta\rm{i}}+\rm{e}^{-\theta\rm{i}}-1&-1&D_{5}\\\
-\rm{e}^{-2\theta\rm{i}}+\rm{e}^{\theta\rm{i}}-1&\overline{D}_{5}&-1\\\
\end{pmatrix},$

where

$\displaystyle
D_{5}=-32\cos^{3}(\theta)\rm{e}^{\theta\rm{i}}+6\rm{i}\sin(2\theta)+2\rm{i}\sin(\theta)+24\cos^{2}(\theta)+4\cos(\theta)-3.$

Now $\det(H_{L})$ is

$-64\cos^{6}(\theta)+160\cos^{5}(\theta)+144\cos^{4}(\theta)-168\cos^{3}(\theta)-96\cos^{2}(\theta)+40\cos(\theta)+20.$

So $H_{L}$ is the Hermitian form with signature $(2,1)$ on the subspace with
the basis $\\{{\bf e_{1}},{\bf e_{2}},{\bf e_{3}}\\}$ when
$\theta\in(\frac{5\pi}{6},\pi]$.

The vector $x_{1}{\bf e_{1}}+x_{2}{\bf e_{2}}+x_{3}{\bf e_{3}}$ is denoted by
the vector ${\bf x}$ in $\mathbb{C}^{3}$, and the vector $y_{1}{\bf
e_{1}}+y_{2}{\bf e_{2}}+y_{3}{\bf e_{3}}$ is denoted by the vector ${\bf y}$
in $\mathbb{C}^{3}$, here

${\bf x}=\left(\begin{array}[]{c}x_{1}\\\ x_{2}\\\ x_{3}\\\
\end{array}\right),\quad{\bf y}=\left(\begin{array}[]{c}y_{1}\\\ y_{2}\\\
y_{3}\\\ \end{array}\right)$

with $x_{i},y_{i}\in\mathbb{C}$. So

${\bf E_{1}}=\left(\begin{array}[]{c}1\\\ 0\\\ 0\\\
\end{array}\right),\quad{\bf E_{2}}=\left(\begin{array}[]{c}0\\\ 1\\\ 0\\\
\end{array}\right),\quad{\bf E_{3}}=\left(\begin{array}[]{c}0\\\ 0\\\ 1\\\
\end{array}\right)$

in $\mathbb{C}^{3}$ present the vectors ${\bf e_{1}}$, ${\bf e_{2}}$ and ${\bf
e_{3}}$ in $\mathbb{C}^{3,1}$.

Comparing to Subsection 2.5, we define the Hermitian cross-product
$\boxtimes_{L}$ on the subspace $H_{L}$ (isometric to ${\bf
H}^{2}_{\mathbb{C}}$) with respect to the basis
$\\{\bf{e_{1}},\bf{e_{2}},\bf{e_{3}}\\}$ by

$\mathbf{x}\boxtimes_{L}\mathbf{y}=\left[\begin{matrix}\mathbf{x}^{*}H_{L}(1,2)\cdot\mathbf{y}^{*}H_{L}(1,3)-\mathbf{y}^{*}H_{L}(1,2)\cdot\mathbf{x}^{*}H_{L}(1,3)\\\
\mathbf{x}^{*}H_{L}(1,3)\cdot\mathbf{y}^{*}H_{L}(1,1)-\mathbf{y}^{*}H_{L}(1,3)\cdot\mathbf{x}^{*}H_{L}(1,1)\\\
\mathbf{x}^{*}H_{L}(1,1)\cdot\mathbf{y}^{*}H_{L}(1,2)-\mathbf{y}^{*}H_{L}(1,1)\cdot\mathbf{x}^{*}H_{L}(1,2)\end{matrix}\right].$

Then the intersection $B_{12}\cap B_{43}\cap L$ is parameterized by
$V=V(z_{1},z_{2})\in\mathbb{C}^{3}$ with $\langle V,V\rangle<0$ with respect
to the Hermitian form $H_{L}$. Where

$V={\bf E_{2}}\boxtimes_{L}{\bf E_{3}}+z_{1}\cdot{\bf E_{1}}\boxtimes_{L}{\bf
E_{3}}+z_{2}\cdot{\bf E_{1}}\boxtimes_{L}{\bf E_{2}}$

and
$(z_{1},z_{2})=(\rm{e}^{r\rm{i}},\rm{e}^{s\rm{i}})\in\mathbb{S}^{1}\times\mathbb{S}^{1}$.
The vector $V\in\mathbb{C}^{3}$ is a three-by-one matrix, and

$V(1,1)\cdot{\bf e_{1}}+V(2,1)\cdot{\bf e_{2}}+V(3,1)\cdot{\bf e_{3}}$

is a vector in in $\mathbb{C}^{3,1}$, we denote it by $W$. The projection of
$W$ into ${\mathbb{C}}{\mathbf{P}}^{3}$ is a point in $B_{12}\cap B_{43}\cap
L$ if $W$ is a negative vector. Now $W^{*}\cdot H\cdot W$ is a very
complicated term. But with Maple,

$W^{*}\cdot H\cdot W=-2(1+2\cos(\theta))^{3}\cdot h,$

where $h$ is

$\displaystyle 2213-1992\cos(3\theta)+718\cos(r)+3041\cos(2\theta)+...$
$\displaystyle-3\cos(s-7\theta)-16\cos(s+7\theta)+\cos(r-s-6\theta).$

(We omit many terms of $h$). Note that the first term of $W^{*}\cdot H\cdot W$
is positive when $\theta\in[\frac{5\pi}{6},\pi]$. By Maple, $h$ has minimum
1521.583 numerically when $r,s\in[-\pi,\pi]$ and
$\theta\in[\frac{5\pi}{6},\pi]$ at a point when $\theta=\frac{5\pi}{6}$. In
particular, $W$ above is a positive vector in $\mathbb{C}^{3,1}$, so
$B_{12}\cap B_{43}\cap L=\emptyset$. Then by the projection from ${\bf
H}^{3}_{\mathbb{C}}$ to $L$, we have $B_{12}\cap B_{32}$ in ${\bf
H}^{3}_{\mathbb{C}}$ is the empty set. This ends the proof of (4).

Note that $B_{12}\cap B_{41}=J^{-1}(B_{12}\cap B_{23})$. Since we have proved
that $B_{12}\cap B_{23}=\emptyset$, we have $B_{12}\cap B_{41}=\emptyset$.
This ends the proof of (5).

For the proof of (6), so we consider $B_{12}\cap B_{24}$. For any
$\theta\in(\frac{5\pi}{6},\pi]$, the span of ${\bf p_{0}}$, ${\bf p_{12}}$,
${\bf p_{24}}$ is a 3-dimensional subspace of $\mathbb{C}^{3,1}$. The
intersection of ${\bf H}^{3}_{\mathbb{C}}\subset{\bf P}^{3}_{\mathbb{C}}$ with
the projection of this 3-dimensional subspace into ${\bf P}^{3}_{\mathbb{C}}$
is denoted by $L$, then $L$ is a totally geodesic ${\bf
H}^{2}_{\mathbb{C}}\hookrightarrow{\bf H}^{3}_{\mathbb{C}}$.

We re-denote $\bf{p_{0}}$, $\bf{p_{12}}$ and $\bf{p_{24}}$ by $\bf{e_{1}}$,
$\bf{e_{2}}$ and $\bf{e_{3}}$. We denote $H_{L}=H(p_{0},p_{12},p_{24})$ by the
matrix $({\bf e_{i}}^{*}\cdot H\cdot{\bf e_{j}})_{1\leq i,j\leq 3}$, then

$H_{L}=\begin{pmatrix}-1&-\rm{e}^{-2\theta\rm{i}}+\rm{e}^{\theta\rm{i}}-1&2\cos(\theta)\\\
-\rm{e}^{2\theta\rm{i}}+\rm{e}^{-\theta\rm{i}}-1&-1&D_{6}\\\
2\cos(\theta)&\overline{D}_{6}&-1\\\ \end{pmatrix},$

where

$D_{6}=-8\cos^{2}(\theta)\rm{e}^{\theta\rm{i}}-\rm{e}^{\theta\rm{i}}-\rm{e}^{-2\theta\rm{i}}-1.$

Now $\det(H_{L})$ is

$32\cos^{5}(\theta)-24\cos^{3}(\theta)-4\cos^{2}(\theta)+4\cos(\theta)+1.$

So $H_{L}$ is the Hermitian form with signature $(2,1)$ on the subspace with
the basis $\\{{\bf e_{1}},{\bf e_{2}},{\bf e_{3}}\\}$ when
$\theta\in(\frac{5\pi}{6},\pi]$.

The vector $x_{1}{\bf e_{1}}+x_{2}{\bf e_{2}}+x_{3}{\bf e_{3}}$ is denoted by
the vector ${\bf x}\in\mathbb{C}^{3}$, and the vector $y_{1}{\bf
e_{1}}+y_{2}{\bf e_{2}}+y_{3}{\bf e_{3}}$ is denoted by the vector ${\bf
y}\in\mathbb{C}^{3}$, here

${\bf x}=\left(\begin{array}[]{c}x_{1}\\\ x_{2}\\\ x_{3}\\\
\end{array}\right),\quad{\bf y}=\left(\begin{array}[]{c}y_{1}\\\ y_{2}\\\
y_{3}\\\ \end{array}\right)$

with $x_{i},y_{i}\in\mathbb{C}$. So

${\bf E_{1}}=\left(\begin{array}[]{c}1\\\ 0\\\ 0\\\
\end{array}\right),\quad{\bf E_{2}}=\left(\begin{array}[]{c}0\\\ 1\\\ 0\\\
\end{array}\right),\quad{\bf E_{3}}=\left(\begin{array}[]{c}0\\\ 0\\\ 1\\\
\end{array}\right)$

in $\mathbb{C}^{3}$ present the vectors ${\bf e_{1}}$, ${\bf e_{2}}$ and ${\bf
e_{3}}$ in $\mathbb{C}^{3,1}$.

Comparing to Subsection 2.5, we define the Hermitian cross-product
$\boxtimes_{L}$ on the subspace $H_{L}$ (which is isometric to ${\bf
H}^{2}_{\mathbb{C}}$) with respect to the basis
$\\{\bf{e_{1}},\bf{e_{2}},\bf{e_{3}}\\}$ by

$\mathbf{x}\boxtimes_{L}\mathbf{y}=\left[\begin{matrix}\mathbf{x}^{*}H_{L}(1,2)\cdot\mathbf{y}^{*}H_{L}(1,3)-\mathbf{y}^{*}H_{L}(1,2)\cdot\mathbf{x}^{*}H_{L}(1,3)\\\
\mathbf{x}^{*}H_{L}(1,3)\cdot\mathbf{y}^{*}H_{L}(1,1)-\mathbf{y}^{*}H_{L}(1,3)\cdot\mathbf{x}^{*}H_{L}(1,1)\\\
\mathbf{x}^{*}H_{L}(1,1)\cdot\mathbf{y}^{*}H_{L}(1,2)-\mathbf{y}^{*}H_{L}(1,1)\cdot\mathbf{x}^{*}H_{L}(1,2)\end{matrix}\right].$

Then the intersection $B_{12}\cap B_{24}\cap L$ is parameterized by
$V=V(z_{1},z_{2})\in\mathbb{C}^{3}$ with $\langle V,V\rangle<0$ with respect
to the Hermitian form $H_{L}$. Where

$V={\bf E_{2}}\boxtimes_{L}{\bf E_{3}}+z_{1}\cdot{\bf E_{1}}\boxtimes_{L}{\bf
E_{3}}+z_{2}\cdot{\bf E_{1}}\boxtimes_{L}{\bf E_{2}}$

and
$(z_{1},z_{2})=(\rm{e}^{r\rm{i}},\rm{e}^{s\rm{i}})\in\mathbb{S}^{1}\times\mathbb{S}^{1}$.
The vector $V\in\mathbb{C}^{3}$ is a three-by-one matrix, we denote by
$V(i,1)$ the entry in the $i$-th row of $V$. Then

$V(1,1)\cdot{\bf e_{1}}+V(2,1)\cdot{\bf e_{2}}+V(3,1)\cdot{\bf e_{3}}$

is a vector in $\mathbb{C}^{3,1}$, we also denote it by $W$. The projection of
$W$ into ${\mathbb{C}}{\mathbf{P}}^{3}$ is a point in $B_{12}\cap B_{24}\cap
L$ if $W$ is a negative vector. Now $W^{*}\cdot H\cdot W$ is a very
complicated term. But with Maple,

$W^{*}\cdot H\cdot W=-(1+2\cos(\theta))^{3}\cdot h,$

where $h$ is

$\displaystyle
799-798\cos(3\theta)+1180\cos(2\theta)-222\cos(5\theta)-24\cos(7\theta)-86\cos(s-4\theta)...$
$\displaystyle 236\cos(r-2\theta)+110\cos(r+2\theta).$

(We omit many terms of $h$). Note that the first term of $W^{*}\cdot H\cdot W$
is positive when $\theta\in[\frac{5\pi}{6},\pi]$. By Maple, $h$ has minimum
275.152 numerically when $r,s\in[-\pi,\pi]$ and
$\theta\in[\frac{5\pi}{6},\pi]$ at a point when $\theta=\frac{5\pi}{6}$. In
particular, any $W$ above is a positive vector in $\mathbb{C}^{3,1}$, so
$B_{12}\cap B_{24}\cap L=\emptyset$. Then by the projection from ${\bf
H}^{3}_{\mathbb{C}}$ to $L$, we have $B_{12}\cap B_{24}$ in ${\bf
H}^{3}_{\mathbb{C}}$ is the empty set. This ends the proof of (6).

Now we consider (7). Recall in Lemma 6.3, we have proved that ${\bf p_{0}}$,
${\bf p_{12}}$, ${\bf p_{34}}$ and ${\bf p_{13}}$ are co-planar. So for any
$\theta\in(\frac{5\pi}{6},\pi]$, the span of ${\bf p_{0}}$, ${\bf p_{12}}$,
${\bf p_{34}}$ and ${\bf p_{13}}$ is a 3-dimensional subspace of
$\mathbb{C}^{3,1}$. The intersection of ${\bf H}^{3}_{\mathbb{C}}\subset{\bf
P}^{3}_{\mathbb{C}}$ with the projection of this 3-dimensional subspace into
${\bf P}^{3}_{\mathbb{C}}$ is denoted by $L$, then $L$ is a totally geodesic
${\bf H}^{2}_{\mathbb{C}}\hookrightarrow{\bf H}^{3}_{\mathbb{C}}$.

By definition $p_{0}\in L$. We will show if the intersection $B_{12}\cap
B_{34}\cap L$ is non-empty, then it is a Giraud disk in $L$. Moreover, in this
case, $B_{12}\cap B_{34}\cap L$ lies in the half space of $L-B_{13}$ which
does not contain the fixed point $p_{0}$ of $J$. In particular, $B_{12}\cap
B_{34}$ does not lie in the partial Dirichlet domain $D_{R}$.

We re-denote ${\bf p_{0}}$, ${\bf p_{12}}$ and ${\bf p_{34}}$ by ${\bf
e_{1}}$, ${\bf e_{2}}$ and ${\bf e_{3}}$. We denote
$H_{L}=H(p_{0},p_{12},p_{34})$ by the matrix $({\bf e_{i}}^{*}\cdot H\cdot{\bf
e_{j}})_{1\leq i,j\leq 3}$, then

$H_{L}=\begin{pmatrix}-1&-\rm{e}^{-2\theta\rm{i}}+\rm{e}^{\theta\rm{i}}-1&-\rm{e}^{-2\theta\rm{i}}+\rm{e}^{\theta\rm{i}}-1\\\
-\rm{e}^{2\theta\rm{i}}+\rm{e}^{-\theta\rm{i}}-1&-1&16\cos^{3}(\theta)-8\cos(\theta)-3\\\
-\rm{e}^{2\theta\rm{i}}+\rm{e}^{-\theta\rm{i}}-1&16\cos^{3}(\theta)-8\cos(\theta)-3&-1\\\
\end{pmatrix}.$

Now $\det(H_{L})$ is

$128\cos^{5}(\theta)-96\cos^{3}(\theta)-16\cos^{2}(\theta)+16\cos(\theta)+4.$

So $H_{L}$ is the Hermitian form with signature $(2,1)$ on the subspace with
the basis $\\{\bf{e_{1}},\bf{e_{2}},\bf{e_{3}}\\}$ when
$\theta\in(\frac{5\pi}{6},\pi]$.

The vector $x_{1}{\bf e_{1}}+x_{2}{\bf e_{2}}+x_{3}{\bf e_{3}}$ is denoted by
the vector ${\bf x}$, and the vector $y_{1}{\bf e_{1}}+y_{2}{\bf
e_{2}}+y_{3}{\bf e_{3}}$ is denoted by the vector ${\bf y}$, here

${\bf x}=\left(\begin{array}[]{c}x_{1}\\\ x_{2}\\\ x_{3}\\\
\end{array}\right),\quad{\bf y}=\left(\begin{array}[]{c}y_{1}\\\ y_{2}\\\
y_{3}\\\ \end{array}\right)$

with $x_{i},y_{i}\in\mathbb{C}$. So

${\bf E_{1}}=\left(\begin{array}[]{c}1\\\ 0\\\ 0\\\
\end{array}\right),\quad{\bf E_{2}}=\left(\begin{array}[]{c}0\\\ 1\\\ 0\\\
\end{array}\right),\quad{\bf E_{3}}=\left(\begin{array}[]{c}0\\\ 0\\\ 1\\\
\end{array}\right)$