ON THE LIMIT SET OF A COMPLEX HYPERBOLIC TRIANGLE GROUP

Abstract

Let $\Gamma =\langle I_{1}, I_{2}, I_{3}\rangle $ be the complex hyperbolic $(4,4,\infty )$ triangle group with $I_1I_3I_2I_3$ being unipotent. We show that the limit set of $\Gamma $ is connected and the closure of a countable union of $\mathbb {R}$-circles.

1 Introduction

The purpose of this paper is to study the limit set of a discrete complex hyperbolic triangle group.

Recall that a complex hyperbolic $(p, q, r)$ triangle group is a representation $\rho $ of the abstract $(p, q, r)$ reflection triangle group

$$ \begin{align*} \langle \sigma_1,\sigma_2,\sigma_3\ |\ \sigma_1^2=\sigma_2^2=\sigma_3^2=(\sigma_1\sigma_2)^p=(\sigma_2\sigma_3)^q=(\sigma_3\sigma_1)^r =\text{Id}\rangle \end{align*} $$

into ${\mathrm {PU}}(2,1)$ such that $I_j=\rho (\sigma _j)$ are complex involutions, where $2\leq p\leq q\leq r\leq \infty $ and $1/p+1/q+1/r<1$ .

For a given triple $(p,q,r)$ with $p,q,r\geq 3$ , it is a classical fact that there is a $1$ -parameter family $\{\rho _t: t\in [0,\infty )\}$ of nonconjugate complex hyperbolic $(p, q, r)$ triangle groups (see for example [8]). Here $\rho _0$ is the embedding of the hyperbolic reflection triangle group, that is, an $\mathbb {R}$ -Fuchsian representation (preserving a Lagrangian plane of ${\mathbf {H}^{2}_{\mathbb {C}}}$ ) and so the limit set is an $\mathbb {R}$ -circle. In [9], Schwartz conjectured that $\rho _t$ is discrete and faithful if and only if neither $w_A=I_1I_3I_2I_3$ nor $w_B=I_1I_2I_3$ is elliptic. Moreover, $\rho _t$ is discrete and faithful if and only if $w_A$ is nonelliptic when $p<10$ , or $w_B$ is nonelliptic when $p>13$ . For a discrete complex hyperbolic triangle group, it would be interesting to know its limit set.

In [10], Schwartz studied the limit set of the complex hyperbolic $(4,4,4)$ triangle group with $(I_1I_2I_1I_3)^7$ = Id.

Theorem 1.1 [10].

Let $\langle I_1, I_2, I_3 \rangle $ be the complex hyperbolic $(4,4,4)$ triangle group with $I_1I_2I_1I_3$ being elliptic of order $7$ . Let $\Lambda $ be its limit set and $\Omega $ its complement. Then $\Lambda $ is connected and the closure of a countable union of $\mathbb {R}$ -circles in $\partial {\mathbf {H}^{2}_{\mathbb {C}}}$ . The quotient $\Omega /\langle I_1I_2, I_2I_3 \rangle $ is a closed hyperbolic $3$ -manifold.

Recently, in [1], Acosta studied the limit set of the complex hyperbolic $(3,3,6)$ triangle group with $I_1I_3I_2I_3$ being unipotent.

Theorem 1.2 [1].

Let $\langle I_1, I_2, I_3 \rangle $ be the complex hyperbolic $(3,3,6)$ triangle group with $I_1I_3I_2I_3$ being unipotent. Let $\Lambda $ be its limit set and $\Omega $ its complement. Then $\Lambda $ is connected and the closure of a countable union of $\mathbb {R}$ -circles in $\partial {\mathbf {H}^{2}_{\mathbb {C}}}$ , and contains a Hopf link with three components. The quotient $\Omega /\langle I_1I_2, I_2I_3 \rangle $ is the one-cusped hyperbolic $3$ -manifold m $023$ in the SnapPy census.

In this paper, we are interested in describing the limit set of the complex hyperbolic $(4,4,\infty )$ triangle group with $I_1I_3I_2I_3$ being unipotent. The main result is the following theorem.

Theorem 1.3. Let $\Lambda $ be the limit set of the complex hyperbolic $(4,4,\infty )$ triangle group $\langle I_1, I_2, I_3 \rangle $ with $I_1I_3I_2I_3$ being unipotent. Then:

1. (1) $\Lambda $ contains two linked $\mathbb {R}$ -circles;

2. (2) $\Lambda $ is the closure of a countable union of $\mathbb {R}$ -circles;

3. (3) $\Lambda $ is connected.

However, the quotient of the complement of the limit set has been described as follows.

Theorem 1.4 [6].

Let $\Omega $ be the discontinuity set of the complex hyperbolic $(4,4,\infty )$ triangle group with $I_1I_3I_2I_3$ being unipotent. Then the quotient $\Omega /\langle I_1I_2, I_2I_3 \rangle $ is the two-cusped hyperbolic $3$ -manifold s $782$ in the SnapPy census.

2 Preliminaries

In this section, we briefly recall some basic facts and notation about the complex hyperbolic plane ${\mathbf {H}^{2}_{\mathbb {C}}}$ . We refer to Goldman’s book [5] and Parker’s notes [7] for more details.

2.1 The space ${\mathbf {H}^{2}_{\mathbb {C}}}$ and its isometries

Let $\mathbb {C}^{2,1}$ denote the three-dimensional complex vector space endowed with a Hermitian form H of signature $(2,1)$ . We take H to be the matrix

$$ \begin{align*}H=\begin{bmatrix} 0 & 0 & 1\\ 0 & 1 & 0\\1 & 0 & 0 \end{bmatrix}.\end{align*} $$

The corresponding Hermitian form is given by

$$ \begin{align*} {\langle {\mathbf{z},\mathbf{w}}\rangle }=z_1\bar{w}_3+z_2\bar{w}_2+z_3\bar{w}_1.\end{align*} $$

Here $\mathbf {z}=[ z_1,z_2,z_3]^t$ and $\mathbf {w}=[w_1,w_2,w_3 ]^t$ are column vectors in $\mathbb {C}^{2,1}\setminus \{0\}$ . Let $\mathbb {P}:\mathbb {C}^{2,1}\setminus \{0\}\rightarrow \mathbb {CP}^2$ be the natural projection map onto complex projective space. Define

$$ \begin{align*} V_0 &=\{\mathbf{z}\in \mathbb{C}^{2,1}\setminus\{0\}:\langle\mathbf{z},\mathbf{z}\rangle=0\},\\ V_- &=\{\mathbf{z}\in \mathbb{C}^{2,1}: \langle\mathbf{z},\mathbf{z}\rangle<0\},\\ V_+ &=\{\mathbf{z}\in \mathbb{C}^{2,1}: \langle\mathbf{z},\mathbf{z}\rangle>0\}. \end{align*} $$

The complex hyperbolic plane ${\mathbf {H}^{2}_{\mathbb {C}}}$ is defined as $\mathbb {P}V_-$ and its boundary $\partial {\mathbf {H}^{2}_{\mathbb {C}}}$ is defined as $\mathbb {P}V_0$ . We will denote the point at infinity by $q_{\infty }$ . Note that a standard lift of $q_{\infty }$ is $[1,0,0]^t$ .

Topologically, the complex hyperbolic plane ${\mathbf {H}^{2}_{\mathbb {C}}}$ is homeomorphic to the unit ball of $\mathbb {C}^2$ and its boundary $\partial {\mathbf {H}^{2}_{\mathbb {C}}}$ is homeomorphic to the unit $3$ -sphere $S^3$ . Note that any point $q\neq q_{\infty }$ of ${\mathbf {H}^{2}_{\mathbb {C}}}$ admits a standard lift $\mathbf {q}$ given by

$$ \begin{align*}\mathbf{q}=\begin{bmatrix} (-{\lvert{z}\rvert}^2-u+it)/2 \\z \\1 \end{bmatrix},\end{align*} $$

where $z\in \mathbb {C},t\in \mathbb {R}$ and $u>0$ . Let $\mathbb {R}_{\geq 0}=\{x\in \mathbb {R}:x\geq 0\}$ . Then the triple $(z,t,u)\in \mathbb {C}\times \mathbb {R}\times \mathbb {R}_{\geq 0}$ is called the horospherical coordinates of q. Let $\mathcal {N}=\mathbb {C}\times \mathbb {R}$ be the Heisenberg group with group law given by

$$ \begin{align*} [z_1,t_1]\cdot[z_2,t_2]=[z_1+z_2,t_1+t_2-2\,\text{Im}(\overline{z_1}z_2)]. \end{align*} $$

Then $\partial {\mathbf {H}^{2}_{\mathbb {C}}}=\mathcal {N}\cup \{q_{\infty }\}$ .

Let $\mathrm {U}(2,1)$ be the subgroup of ${\mathrm {GL}}(3,\mathbb {C})$ preserving the Hermitian form H. Let ${\mathrm {SU}}(2,1)$ be the subgroup of $\mathrm {U}(2,1)$ consisting of unimodular matrices. The full group of holomorphic isometries of ${\mathbf {H}^{2}_{\mathbb {C}}}$ is ${\mathrm {PU}}(2,1)={\mathrm {SU}}(2,1)/\{\omega I: \omega ^3=1\}$ , which acts transitively on points of ${\mathbf {H}^{2}_{\mathbb {C}}}$ and pairs of distinct points of $\partial {\mathbf {H}^{2}_{\mathbb {C}}}$ .

An element of ${\mathrm {PU}}(2,1)$ is called elliptic if it has a fixed point in ${\mathbf {H}^{2}_{\mathbb {C}}}$ . If an element is not elliptic, then it is called parabolic or loxodromic if it has exactly one fixed point in $\partial {\mathbf {H}^{2}_{\mathbb {C}}}$ or exactly two fixed points in $\partial {\mathbf {H}^{2}_{\mathbb {C}}}$ , respectively. A parabolic element of ${\mathrm {PU}}(2,1)$ is called unipotent if it admits a lift to ${\mathrm {SU}}(2,1)$ that is unipotent. These terms will also be used for elements of ${\mathrm {SU}}(2,1)$ .

2.2 Totally geodesic subspaces and related isometries

There is no totally geodesic subspace of real dimension three of ${\mathbf {H}^{2}_{\mathbb {C}}}$ . Except for the points, geodesics and ${\mathbf {H}^{2}_{\mathbb {C}}}$ (they are obviously totally geodesic), there are two kinds of totally geodesic subspaces of real dimension two: complex lines and Lagrangian planes. A complex line is the intersection of a projective line in $\mathbb {C}\mathbb {P}^2$ with ${\mathbf {H}^{2}_{\mathbb {C}}}$ . The boundary of a complex line is called a $\mathbb {C}$ -circle. A Lagrangian plane is the intersection of a totally real subspace in $\mathbb {C}\mathbb {P}^2$ with ${\mathbf {H}^{2}_{\mathbb {C}}}$ . The boundary of a Lagrangian plane is called an $\mathbb {R}$ -circle. In particular, if an $\mathbb {R}$ -circle contains $q_{\infty }=[1,0,0]^t$ , it is called an infinite $\mathbb {R}$ -circle.

An elliptic isometry whose fixed point set is a complex line is called a complex reflection. The complex reflections we will use in this paper have order $2$ and we call them complex involutions.

Similarly, every Lagrangian plane is the set of fixed points of an antiholomorphic isometry of order $2$ , which is called a real reflection on the Lagrangian plane.

We will need the following lemma, which is [4, Proposition 3.1].

Proposition 2.1 [4].

If $I_1$ and $I_2$ are reflections on the $\mathbb {R}$ -circles $\mathcal {R}_1$ and $\mathcal {R}_2$ :

1. (i) $I_1\circ I_2$ is parabolic if and only if $\mathcal {R}_1$ and $\mathcal {R}_2$ intersect at one point;

2. (ii) $I_1\circ I_2$ is loxodromic if and only if $\mathcal {R}_1$ and $\mathcal {R}_2$ do not intersect and are not linked;

3. (iii) $I_1\circ I_2$ is elliptic if and only if $\mathcal {R}_1$ and $\mathcal {R}_2$ are linked or intersect at two points.

2.3 Limit set

Let $\Gamma $ be a discrete subgroup of $PU(2,1)$ . The limit set of $\Gamma $ is defined as the set of accumulation points of any orbit in ${\mathbf {H}^{2}_{\mathbb {C}}}$ under the action of $\Gamma $ . It is the smallest closed nonempty $\Gamma $ -invariant subset of $\partial {\mathbf {H}^{2}_{\mathbb {C}}}$ . The complement of the limit set of $\Gamma $ in $\partial {\mathbf {H}^{2}_{\mathbb {C}}}$ is called the discontinuity set of $\Gamma $ .

3 The group

Let $\omega =-1/2+i\sqrt {3}/2$ be the primitive cube root of unity. The complex involutions $I_1$ , $I_2$ and $I_3$ are given by

$$ \begin{align*} I_1=\left[ \begin{array}{@{}ccc@{}} -1 & 2(1+\omega) & 2 \\ 0 & 1 & -2\omega \\ 0 & 0 & -1 \\ \end{array} \right], \quad I_2=\left[ \begin{array}{@{}ccc@{}} -1 & 2\omega & 2 \\ 0 & 1 & -2(1+\omega) \\ 0 & 0 & -1 \\ \end{array} \right], \quad I_3=\left[ \begin{array}{@{}ccc@{}} 0 & 0 & 1 \\ 0 & -1 & 0 \\ 1 & 0 & 0 \\ \end{array} \right]. \end{align*} $$

The products $I_2I_3$ and $I_3I_1$ are elliptic elements of order 4 and $I_2I_1$ is unipotent. In fact, $\langle I_1, I_2, I_3 \rangle $ is a discrete complex hyperbolic $(4,4,\infty )$ triangle group. Moreover, the element $I_1I_3I_2I_3$ is unipotent.

From Theorem 1.4, one can see that the group $\langle I_1,I_2, I_3 \rangle $ is a subgroup of the Eisenstein–Picard modular group $\mathrm {PU}(2,1;\mathbb {Z}[\omega ])$ of infinite index and has no fixed point. In [3], Falbel and Parker studied the geometry of the Eisenstein–Picard modular group $\mathrm {PU}(2,1;\mathbb {Z}[\omega ])$ . Moreover, they obtained a presentation of $\mathrm {PU}(2,1;\mathbb {Z}[\omega ])$ .

Theorem 3.1 [3].

The Eisenstein–Picard modular group $\mathrm {PU}(2,1;\mathbb {Z}[\omega ])$ has a presentation

$$ \begin{align*} \langle P,Q,R \mid R^2=(QP^{-1})^6=PQ^{-1}RQP^{-1}R=P^3Q^{-2}=(RP)^3=1 \rangle, \end{align*} $$

where

$$ \begin{align*} P=\left[ \begin{array}{@{}ccc@{}} 1 & 1 & \omega \\ 0 & \omega & -\omega \\ 0 & 0 & 1 \\ \end{array} \right],\quad Q=\left[ \begin{array}{@{}ccc@{}} 1 & 1 & \omega \\ 0 & -1 & 1 \\ 0 & 0 & 1 \\ \end{array} \right], \quad R=\left[ \begin{array}{@{}ccc@{}} 0 & 0 & 1 \\ 0 & -1 & 0 \\ 1 & 0 & 0 \\ \end{array} \right]. \end{align*} $$

By using this presentation, the complex involutions $I_1$ , $I_2$ and $I_3$ can be expressed as follows.

Proposition 3.2. Let $M=PQ^{-1}$ and $T=QM^3$ , then:

• • $I_2=-TQ^4T(PM^2)^{-2}M^3$ ;

• • $I_1=I_2T^2Q^2$ ;

• • $I_3=R$ .

4 The limit set

Lemma 4.1. Let $G_0=\langle I_1,I_2,I_3I_2I_3 \rangle $ , as a subgroup of $\langle I_1,I_2, I_3 \rangle $ . Let $\mathcal {L}_0$ be the Lagrangian plane, whose boundary at infinity is the infinite $\mathbb {R}$ -circle given by $\mathcal {R}_0=\{[x+i\sqrt {3}/2,\sqrt {3}x-\sqrt {3}/2]\in \mathcal {N}: x\in \mathbb {R}\}\cup \{q_\infty \}$ . Then $G_0$ preserves $\mathcal {L}_0$ .

Proof. Using horospherical coordinates,

$$ \begin{align*}\mathcal{L}_0=\{(x+i\sqrt{3}/2,\sqrt{3}x-\sqrt{3}/2,u)\in{\mathbf{H}^{2}_{\mathbb{C}}}: x\in \mathbb{R},u>0\},\end{align*} $$

and one can compute

$$ \begin{align*}I_1(x+i\sqrt{3}/2,\sqrt{3}x-\sqrt{3}/2,u)&=(-x-1+i\sqrt{3}/2,\sqrt{3}(-x-1)-\sqrt{3}/2,u),\\I_2(x+i\sqrt{3}/2,\sqrt{3}x-\sqrt{3}/2,u)&=(-x+1+i\sqrt{3}/2,\sqrt{3}(-x+1)-\sqrt{3}/2,u), \end{align*} $$

$$ \begin{align*} &I_3I_2I_3(x+i\sqrt{3}/2,\sqrt{3}x-\sqrt{3}/2,u)\\ &\quad =\bigg( \frac{(x+1/2)^2-1+u}{2(x-1/2)^2+2u}+i\frac{\sqrt{3}}{2},\sqrt{3} \frac{(x+1/2)^2-1+u}{2(x-1/2)^2+2u} -\frac{\sqrt{3}}{2},\frac{u}{((x-1/2)^2+u)^2}\bigg). \end{align*} $$

Thus, $I_1\mathcal {L}_0=I_2\mathcal {L}_0=I_3I_2I_3\mathcal {L}_0=\mathcal {L}_0$ . Therefore, the group $G_0$ preserves the Lagrangian plane $\mathcal {L}_0$ with boundary $\mathcal {R}_0$ at infinity.

In the same way, we can prove the following result.

Lemma 4.2. Let $G_1=\langle I_1, I_2, I_3I_1I_3 \rangle $ , as a subgroup of $\langle I_1,I_2, I_3 \rangle $ . Let $\mathcal {L}_1$ be the Lagrangian plane, whose boundary at infinity is the infinite $\mathbb {R}$ -circle given by $\mathcal {R}_1=\{[x+i\sqrt {3}/2,\sqrt {3}x+\sqrt {3}/2]\in \mathcal {N}: x\in \mathbb {R}\}\cup \{q_\infty \}$ . Then $G_1$ preserves $\mathcal {L}_1$ .

Proposition 4.3. The limit set of $\langle I_1,I_2, I_3 \rangle $ contains an $\mathbb {R}$ -circle.

Proof. By Lemma 4.1, the subgroup $G_0=\langle I_1,I_2,I_3I_2I_3 \rangle $ is an $\mathbb {R}$ -Fuchsian subgroup of $\langle I_1,I_2, I_3 \rangle $ . Since $I_1I_2$ and $I_1I_3I_2I_3$ are unipotent and $I_2I_3I_2I_3$ is elliptic of order 2, the restriction $G_0|_{\mathcal {L}_0}$ is a $(2,\infty ,\infty )$ -reflection triangle group. Thus, the limit set of $G_0$ is $\partial \mathcal {L}_0=\mathcal {R}_0$ . Therefore, the limit set of $\langle I_1,I_2, I_3 \rangle $ contains the $\mathbb {R}$ -circle $\mathcal {R}_0$ .

Remark 4.4. The $(2,\infty ,\infty )$ -reflection triangle group is a noncompact arithmetic triangle group [11].

Now, let us consider the images of $\mathcal {R}_0$ and $\mathcal {R}_1$ by the group $\langle I_1, I_2, I_3 \rangle $ . Since $\mathcal {R}_0$ is the limit set of $G_0$ , the image $I_j \mathcal {R}_0$ , with $j=1,2$ , is the limit set of the group $I_j G_0 I_j$ . One can see that $I_j G_0 I_j=G_0$ . Thus, $\mathcal {R}_0$ is stabilised by both $I_1$ and $I_2$ . Similarly, $\mathcal {R}_1$ is stabilised by both $I_1$ and $I_2$ .

Lemma 4.5. The limit sets $I_3 \mathcal {R}_0$ and $\mathcal {R}_0$ are linked and the limit sets $I_3 \mathcal {R}_0$ and $\mathcal {R}_1$ intersect at one point.

Proof. Since $I_3 \mathcal {R}_0$ is the limit set of $I_3G_0I_3=\langle I_3I_1I_3, I_3I_2I_3, I_2 \rangle $ , it contains the parabolic fixed point $P_{I_2I_3I_1I_3}$ . Therefore, $I_3\mathcal {R}_0\cap \mathcal {R}_1=\{P_{I_2I_3I_1I_3}\}$ .

Since both $I_3 \mathcal {L}_0$ and $\mathcal {L}_0$ contain the elliptic fixed point $P_{I_2I_3I_2I_3}\in I_3\mathcal {L}_0\cap \mathcal {L}_0$ , the product of reflections on the Lagrangian planes $I_3\mathcal {L}_0$ and $\mathcal {L}_0$ is elliptic. Therefore, by Proposition 2.1, the two $\mathbb {R}$ -circles $I_3 \mathcal {R}_0$ and $\mathcal {R}_0$ must be linked or intersect at two points.

We claim that $I_3 \mathcal {R}_0$ and $\mathcal {R}_0$ do not intersect. One can compute that the points of $I_3 \mathcal {R}_0$ are given by

$$ \begin{align*}\bigg[ \frac{8(4x^3-3x+3)}{16x^4+72x^2-48x+21}+i\frac{4\sqrt{3}(12x^2-4x+3)}{16x^4+72x^2-48x+21},\frac{-32\sqrt{3}(2x-1)}{16x^4+72x^2-48x+21}\bigg].\end{align*} $$

Suppose that $I_3 \mathcal {R}_0\cap \mathcal {R}_0\neq \emptyset $ , then

$$ \begin{align*} \frac{8(4x^3-3x+3)}{16x^4+72x^2-48x+21}+i\frac{4\sqrt{3}(12x^2-4x+3)}{16x^4+72x^2-48x+21}=x+i\sqrt{3}/2 \end{align*} $$

should have solutions for x. However, this is impossible by a simple computation. Thus, $I_3 \mathcal {R}_0\cap \mathcal {R}_0=\emptyset $ . Therefore, $I_3 \mathcal {R}_0$ and $\mathcal {R}_0$ are linked.

Similarly, we have the following result.

Lemma 4.6. The limit sets $I_3 \mathcal {R}_1$ and $\mathcal {R}_1$ are linked and the limit sets $I_3 \mathcal {R}_1$ and $\mathcal {R}_0$ intersect at one point.

Corollary 4.7. The union of $\mathcal {R}_i$ and $I_3\mathcal {R}_i$ ( $i=0,1$ ) is connected.

Proof. Since $\mathcal {R}_0$ and $\mathcal {R}_1$ are infinite $\mathbb {R}$ -circles, we obtain $\mathcal {R}_0\cap \mathcal {R}_1=\{q_{\infty }\}$ . From Lemmas 4.5 and 4.6, $I_3\mathcal {R}_0\cap \mathcal {R}_1=\{P_{I_2I_3I_1I_3}\}$ and $I_3 \mathcal {R}_1 \cap \mathcal {R}_0=\{P_{I_1I_3I_2I_3}\}$ . It is obvious that $I_3\mathcal {R}_0\cap I_3\mathcal {R}_1=\{I_3q_{\infty }\}=[0,0]\in \mathcal {N}$ . Now, there is a path in $\mathcal {R}_0\cup \mathcal {R}_1\cup I_3\mathcal {R}_0\cup I_3\mathcal {R}_1$ between any two points in it. Thus, the union is connected. See Figure 1.

Figure 1 A schematic view of the four $\mathbb {R}$ -circles. Here $\mathcal {R}_0$ and $\mathcal {R}_1$ are two lines intersecting at infinity.

Proof of Theorem 1.3.

(1) This is a consequence of Lemmas 4.5 or 4.6.

(2) From Proposition 4.3, the limit set $\Lambda $ contains an $\mathbb {R}$ -circle. Then the $\Gamma $ -orbit of the $\mathbb {R}$ -circle is contained in $\Lambda $ . Since $\Lambda $ is the smallest closed nonempty invariant subset of $\partial {\mathbf {H}^{2}_{\mathbb {C}}}$ under the action of $\Gamma $ , it is the closure of the $\Gamma $ -orbit of the $\mathbb {R}$ -circle. Thus, $\Lambda $ is the closure of a countable union of $\mathbb {R}$ -circles.

(3) Let n be a positive integer and $\gamma =\gamma _1\gamma _2\cdots \gamma _n\in \Gamma $ , where $\gamma _i\in \{I_1,I_2,I_3\}$ for $i=1,\ldots ,n$ . Let $\mathcal {U}_0=\mathcal {R}_0\cup \mathcal {R}_1$ and $\mathcal {U}_i=\gamma _1\cdots \gamma _i\mathcal {U}_0$ . Since $\mathcal {R}_0\cap \mathcal {R}_1=\{q_{\infty }\}$ , the subset $\mathcal {U}_i$ of $\Lambda $ is connected for $i=0,1,\ldots ,n$ . For $i\in \{0,1,\ldots ,n-1\}$ , we see that

$$ \begin{align*}\mathcal{U}_i\cap\mathcal{U}_{i+1}=\gamma_1\cdots\gamma_i\mathcal{U}_0\cap\gamma_1\cdots\gamma_{i+1}\mathcal{U}_0=\gamma_1\cdots\gamma_i(\mathcal{U}_0\cap\gamma_{i+1}\mathcal{U}_0).\end{align*} $$

By Lemmas 4.5 and 4.6, $\mathcal {U}_0\cap \gamma _{i+1}\mathcal {U}_0\neq \emptyset $ , so $\mathcal {U}_i\cap \mathcal {U}_{i+1}\neq \emptyset $ . Thus, there is a path in $\Lambda $ from $q_{\infty }$ to $\gamma q_{\infty }$ . From item $(2)$ , $\Lambda $ is the closure of the $\Gamma $ -orbit of an $\mathbb {R}$ -circle. Hence, $\Lambda $ is connected.

Remark 4.8. We note that $\Lambda $ is not slim (see [2] for the definition). In other words, there are three distinct points of $\Lambda $ lying in the same $\mathbb {C}$ -circle.