Free groups generated by two parabolic maps

Abstract

In this paper we consider a group generated by two unipotent parabolic elements of $\textrm{SU}(2,1)$ with distinct fixed points. We give several conditions that guarantee the group is discrete and free. We also give a result on the diameter of a finite ${\mathbb R}$-circle in the Heisenberg group.

1 Introduction

The study of free, discrete groups has a long history dating back to Schottky and Klein in the nineteenth century. We will be particularly interested in groups generated by two unipotent parabolic maps in $\textrm{SU}(2,1)$ and their action on complex hyperbolic space and its boundary. The conditions we give could be thought of as complex hyperbolic analogues of the results proved by Lyndon and Ullman [7] and by Ignatov [4] giving conditions under which two parabolic elements of $\textrm{PSL}(2,{\mathbb C})$ generate a free Kleinian group. Our work is very closely related to the well-known Riley slice of Schottky space, where Riley considered the space of conjugacy classes of subgroups of $\textrm{PSL}(2,\mathbb {C})$ generated by two non-commuting parabolic maps, see in [6]. In [11] Parker and Will considered a related problem, namely they also studied groups with two unipotent generators, but they made the additional assumption that the product of these maps is also unipotent. We will comment on the relationship between our results and those in [11] below.

The main theme of the paper concerns groups generated by two Heisenberg translations with distinct fixed points. We normalise so that the fixed points are $\infty$ and o, the origin in the Heisenberg group. Specifically, we consider the group generated by

$$\begin{aligned} A=\left( \begin{matrix} 1 &{}\quad -\sqrt{2}s_1e^{-i\theta _1} &{}\quad -s_1^2+it_1 \\ 0 &{}\quad 1 &{}\quad \sqrt{2}s_1e^{i\theta _1} \\ 0 &{}\quad 0 &{}\quad 1 \end{matrix}\right) , \quad B=\left( \begin{matrix} 1 &{}\quad 0 &{}\quad 0 \\ \sqrt{2}s_2e^{i\theta _2} &{}\quad 1 &{}\quad 0 \\ - s_2^2+it_2 &{}\quad -\sqrt{2}s_2e^{-i\theta _2} &{}\quad 1 \end{matrix}\right) . \end{aligned}$$

(1)

From this, it might appear that the space of such pairs of transformations has dimension six, and is parameterised by $s_j,\,t_j,\,\theta _j$ for $j=1,\,2$. In fact, it has dimension four. There is a further normalisation we can do using the stabiliser of the pair $\{o,\infty \}$. This depends on $(k,\psi )\in {\mathbb R}_+\times [0,2\pi )$ which act as follows:

$$\begin{aligned} (s_1,t_1,\theta _1;s_2,t_2,\theta _2)\longmapsto (s_1k, t_1k^2, \theta _1+\psi ; s_2/k, t_2/k^2,\theta _2+\psi ). \end{aligned}$$

(2)

In order to show the symmetry in the parameters, we choose not to make this normalisation in the statement of the results. But in some of the proofs we use it to simplify calculations, for example by choosing $\theta _2=0$.

We want to find conditions on $s_j$, $t_j$, $\theta _j$ that ensures $\langle A,B\rangle$ is discrete and freely generated by A and B. To do so, we will use Klein’s combination theorem (sometimes called the ping-pong theorem) or variants of it on the boundary of complex hyperbolic space. This is given in the following proposition.

Proposition 1.1

Let A and B be the Heisenberg translations fixing $\infty$ and o respectively, given by (1). If the fundamental domains $D_A\subset \partial \textbf{H}^2_{\mathbb C}$ for $\langle A\rangle$ and $D_B\subset \partial \textbf{H}^2_{\mathbb C}$ for $\langle B\rangle$ satisfy $D_A^\circ \cap D_B^\circ \ne \emptyset$ and $\overline{D}_A\cup \overline{D}_B=\partial \textbf{H}^2_{\mathbb C}$, then $\langle A,B \rangle$ is free and discrete.

Our results are also related to proofs of discreteness of complex hyperbolic isometry groups using other variations on Klein’s combination theorem. For example, see Goldman and Parker [2], Wyss-Gallifent [12], Monaghan, Parker and Pratoussevitch [8] or Jiang and Xie [13].

Our main theorem is:

Theorem 1.2

Let $(s_1e^{i\theta _1},t_1)$ and $(s_2e^{i\theta _2},t_2)$ be non-trivial elements of the Heisenberg group. Here, $\theta _1$ and $\theta _2$ are only defined when $s_1\ne 0$ and $s_2\ne 0$. Let A and B given by (1) be the associated Heisenberg translations fixing $\infty$ and o respectively. Replacing one of these by its inverse if necessary, we suppose $-\pi /2\le (\theta _1-\theta _2)\le \pi /2$. If one of the following four conditions is satisfied then $\langle A,\,B\rangle$ is discrete and freely generated by A and B. The conditions are

• (1)
  $$\begin{aligned} |s_1^2+it_1|^{1/2}|s_2^2+it_2|^{1/2}\ge & {} 2^{1/2}\left( \left( 1-\frac{t_2}{|s_2^2+it_2|}\right) ^{1/3}+\left( 1+\frac{t_2}{|s_2^2+it_2|}\right) ^{1/3}\right) ^{3/4}\\{} & {} \quad \times \left( \left( 1-\frac{t_1}{|s_1^2+it_1|}\right) ^{1/3}+\left( 1+\frac{t_1}{|s_1^2+it_1|}\right) ^{1/3}\right) ^{3/4}, \end{aligned}$$

• (2) if $s_1\ne 0$ then

  $$\begin{aligned} s_1\,|s_2^2+it_2|^{1/2} \ge {\left\{ \begin{array}{ll} \frac{2~{s_2}^3}{|s_2^2+it_2|^{3/2}}\,\cos (\theta _1-\theta _2)+2 &{} \hbox {if } s_2\ne 0, \\ 2 &{} \hbox {if } s_2=0. \end{array}\right. } \end{aligned}$$

• (3) if $s_2\ne 0$ then

  $$\begin{aligned} |s_1^2+it_1|^{1/2}\,s_2 \ge {\left\{ \begin{array}{ll} \frac{2~{s_1}^3}{|s_1^2+it_1|^{3/2}}\,\cos (\theta _1-\theta _2)+2 &{} \hbox {if } s_1\ne 0, \\ 2 &{} \hbox {if } s_1=0. \end{array}\right. } \end{aligned}$$

• (4) if both $s_1,\,s_2\ne 0$ then $s_1s_2 \ge 4\cos ^3\bigl ((\theta _1-\theta _2)/3\bigr )$.

Observe that the expressions above are all invariant under the action of maps that fix both fixed points. Specifically, using the action of $(k,\psi )\in {\mathbb R}_+\times [0,2\pi )$ from (2) the left hand side in each case is a product of two terms, one scaling by k and the other by 1/k. Similarly, each term on the right hand side involving $(s_j,t_j)$ does not change when we scale by k and the only place $\theta _1$ and $\theta _2$ arise is via $(\theta _1-\theta _2)$, which does not change when we add $\psi$ to both angles.

Note that, each item in the above theorem involves different techniques of the proof. The first item of the theorem follows by considering fundamental domains bounded by Cygan spheres (special cases of bisectors), whereas last item follows by considering fundamental domains bounded by two fans. The middle two items have a mix of Cygan spheres and fans.

Substituting $s_1=s_2=0$ in Theorem 1.2 (1) we obtain the following corollary, which is well known (for example, it is implicit in Sect. 3 of [10] and it is written down explicitly in Theorem 1.1 of Xie, Wang and Jiang [14]).

Corollary 1.3

Let $(0,t_1)$ and $(0,t_2)$ be elements of the Heisenberg group. Let A and B given by (1) be the associated (vertical) Heisenberg translations fixing $\infty$ and o respectively. If $|t_1|^{1/2}\,|t_2|^{1/2} \ge 2$ then $\langle A,\,B\rangle$ is discrete and freely generated by A and B.

Note that the right hand sides of parts (1), (2) and (3) of Theorem 1.2 involve $(s_1e^{i\theta _1},t_1)$ and $(s_2e^{i\theta _2},t_2)$. By eliminating $s_j$ and $t_j$, we can get weakening of Theorem 1.2 as follows:

Theorem 1.4

Let $(s_1e^{i\theta _1},t_1)$ and $(s_2e^{i\theta _2},t_2)$ be as in Theorem 1.2 and let A and B given by (1). If one of the following three conditions is satisfied then $\langle A,\,B\rangle$ is discrete and freely generated by A and B. The conditions are

• (1’) $\quad \bigl |s_1^2+it_1\bigr |^{1/2}\,\bigl |s_2^2+it_2\bigr |^{1/2}\ge 4$;

• (2’) if $s_1\ne 0$ then $\quad s_1\,\bigl |s_2^2+it_2\bigr |^{1/2}\ge 4\cos ^2\bigl ((\theta _1-\theta _2)/2\bigr )$;

• (3’) if $s_2\ne 0$ then $\quad \bigl |s_1^2+it_1\bigr |^{1/2}\, s_2 \ge 4\cos ^2\bigl ((\theta _1-\theta _2)/2\bigr )$.

Proof

First note that for $-1\le x\le 1$ we have

$$\begin{aligned} \bigl ((1-x)^{1/3}+(1+x)^{1/3}\bigr )^{3/4}\le 2^{3/4} \end{aligned}$$

with equality if and only if $x=0$. Therefore, (1’) follows from (1). Secondly,

$$\begin{aligned} \frac{2s_2^3}{|s_2^2+it_2|^{3/2}}\,\cos (\theta _1-\theta _2)+2 \le 2\cos (\theta _1-\theta _2)+2 = 4\cos ^2\bigl ((\theta _1-\theta _2)/2\bigr ). \end{aligned}$$

Thus (2’) follows from (2) and similarly (3’) follows from (3). $\square$

The following lemma shows that part (4) of Theorem 1.2 follows from the other parts. Nevertheless, we will still include a direct geometrical proof of this in Section 5.

Lemma 1.5

If the condition of Theorem 1.2(4) holds then the conditions of Theorem 1.2(2) and (3) hold.

Proof

First we claim that if $-\pi /2\le \theta _1-\theta _2 \le \pi /2$ then

$$\begin{aligned} 4\cos ^3\bigl ((\theta _1-\theta _2)/3\bigr ) \ge 4\cos ^2\bigl ((\theta _1-\theta _2)/2\bigr ) \end{aligned}$$

with equality if and only if $\theta _1=\theta _2$. To see this, we define $\phi =(\theta _1-\theta _2)/3\in [-\pi /6,\pi /6]$, and write the right hand side in terms of $\phi$.

$$\begin{aligned} 4\cos ^2(3\phi /2) = 2\cos (3\phi )+2 = 8\cos ^3(\phi )-6\cos (\phi )+2. \end{aligned}$$

Therefore

$$\begin{aligned} 4\cos ^3\bigl (\phi \bigr ) - 4\cos ^2\bigl (3\phi /2\bigr )= & {} -4\cos ^3(\phi )+6\cos (\phi )-2 \\= & {} 2\bigl (1-\cos (\phi )\bigr )\bigl (2\cos ^2(\phi )+2\cos (\phi )-1\bigr ). \end{aligned}$$

The quadratic term is positive when $\cos (\phi )\ge \sqrt{3}/2$ and so for such values of $\phi$, this expression is non-negative with equality if and only if $\phi =0$, which proves the claim.

Therefore, if Theorem 1.2(4) holds, we have

$$\begin{aligned} s_1\,|s_2^2+it_2|^{1/2}\ge & {} s_1s_2 \\\ge & {} 4\cos ^3\bigl ((\theta _1-\theta _2)/3\bigr ) \\\ge & {} 4\cos ^2\bigl ((\theta _1-\theta _2)/2\bigr ) \\= & {} 2\cos (\theta _1-\theta _2)+2 \\\ge & {} \frac{2s_2^3}{|s_2^2+it_2|^{3/2}}\cos (\theta _1+\theta _2)+2. \end{aligned}$$

Hence Theorem 1.2(2) holds. A similar argument shows that if Theorem 1.2(4) then Theorem 1.2(3) holds too. $\square$

In order to prove Theorem 1.2 (1) we will use Lemma 1.6 below, which gives the maximum Cygan distance between a point on a finite ${\mathbb R}$-circle and any other point on the same ${\mathbb R}$-circle. We believe this will be of independent interest.

An ${\mathbb R}$-circle R is the boundary of a totally geodesic Lagrangian subspace of $\textbf{H}^2_{\mathbb C}$ and is the fixed point set of an anti-holomorphic involution $\iota _R$ in the isometry group of $\textbf{H}^2_{\mathbb C}$. A ${\mathbb C}$-circle C is the boundary of a totally geodesic complex line of $\textbf{H}^2_{\mathbb C}$. Thinking of $\partial \textbf{H}^2_{\mathbb C}$ as the one point compactification of the Heisenberg group, an ${\mathbb R}$-circle R is called finite if it does not contain the point $\infty$. Finite ${\mathbb R}$-circles are non-planar space curves with interesting geometric properties; see Goldman [1]. In particular, each finite ${\mathbb R}$-circle R is a meridian of a Cygan sphere. This Cygan sphere is preserved as a set by $\iota _R$ and its centre is $\iota _R(\infty )$. In particular, every point on R is the same Cygan distance from $\iota _R(\infty )$ and we call this distance r its radius. Given a point p of R we want to find a point q in R that maximises the Cygan distance from p among all points of R. We call this distance the diameter d of R at the point p. It is clear from the triangle inequality $d\le 2r$. However, the Cygan metric is not a geodesic metric, and so for most points p the diameter d is strictly less than 2r. In order to write points on R in an invariant way, we use the Cartan angular invariant ${\mathbb A}$. The lemma below gives a precise formula for the diameter of R at p.

Lemma 1.6

Let R be a finite ${\mathbb R}$-circle fixed by the anti-holomorphic involution $\iota _R$. Let r be the Cygan distance from $\iota _R(\infty )$ to any point of R. For $\alpha \in [0,\pi /2]$, let $p_\alpha$ be a point on R with ${\mathbb A}(p_\alpha ,\iota _R(\infty ),\infty )=2\alpha -\pi /2$. Then the maximum Cygan distance from $p_\alpha$ to any other point of R is given by

$$\begin{aligned} d_\alpha (R)=2^{1/2}r\bigl (\cos ^{2/3}(\alpha )+\sin ^{2/3}(\alpha )\bigr )^{3/4}. \end{aligned}$$

Observe that $d_\alpha (R)\le 2r$, which is attained if and only if $\alpha =\pi /4$, that is whenever $(p_\alpha , \iota _R(\infty ), \infty \big )$ lie on a common $\mathbb R$-circle. Also, $d_\alpha (R)\ge \sqrt{2}r$, which is attained if and only if $\alpha =0$ or $\alpha =\pi /2$, that is whenever $(p_\alpha , \iota _R(\infty ), \infty \big )$ lie on a common $\mathbb {C}$-circle.

2 Background

All material in this section is standard; see [1] for example unless otherwise indicated.

2.1 Complex Hyperbolic space

Let $\mathbb {C}^{2,1}$ be the 3-dimensional vector space over $\mathbb {C}$ equipped with the Hermitian form of signature (2, 1) given by

$$\begin{aligned} \langle \textbf{z},\textbf{w}\rangle =\textbf{w}^{*}H\textbf{z}={\bar{w}}_3 z_1+{\bar{w}}_2 z_2+{\bar{w}}_1 z_3, \end{aligned}$$

where $\textbf{z},\,\textbf{w}$ are column vectors in $\mathbb {C}^3$ and the matrix of the Hermitian form is given by

$$\begin{aligned} H =\left[ \begin{array}{cccc} 0 &{}\quad 0 &{}\quad 1\\ 0 &{} 1 &{} 0 \\ 1 &{} 0 &{} 0\\ \end{array}\right] . \end{aligned}$$

If $\textbf{z}\in \mathbb {C}^{2,1}$ then $\langle \textbf{z},\textbf{z}\rangle$ is real. Thus we may consider the following subsets of $\mathbb {C}^{2,1}\setminus \{{\textbf{0}}\}:$

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

A vector $\textbf{z}$ in $\mathbb {C}^{2,1}$ is called positive, negative or null depending on whether $\textbf{z}$ belongs to $V_+$, $V_-$ or $V_0$ respectively. Let $\mathbb P:\mathbb {C}^{2,1}\setminus \{\textbf{0}\}\longrightarrow \mathbb {C}{\mathbb P}^{2}$ be the projection map onto the complex projective space. The complex hyperbolic space is defined to be $\textbf{H}_{\mathbb {C}}^2=\mathbb P(V_{-})$. The ideal boundary of complex hyperbolic space is $\partial \textbf{H}_{\mathbb {C}}^2=\mathbb P(V_{0})$. Let $\textrm{U}(2,1)$ be the unitary group of above Hermitian form. The biholomorphic isometry group of $\textbf{H}_{\mathbb {C}}^2$ is the projective unitary group $\textrm{PU}(2,1)$. In addition, the map $\textbf{z}\longmapsto {\bar{\textbf{z}}}$ that sends each entry of $\textbf{z}$ to its complex conjugate ${\bar{\textbf{z}}}$ is an anti-holomorphic isometry of $\textbf{H}_{\mathbb {C}}^2$. Any other anti-holomorphic isometry may be written as the projectivisation of the composition of this map and an element of $\textrm{U}(2,1)$. All complex hyperbolic isometries are either holomorphic or anti-holomorphic.

We define Siegel domain model of complex hyperbolic space by taking the section defined by $z_3=1$ for the given Hermitian form. In other words, if $(z_1,z_2)$ is in $\mathbb {C}^2$, we define its standard lift to be $\textbf{z}=(z_1,z_2,1)^t$ in $\mathbb {C}^{2,1}$. The Siegel domain is the subset of $\mathbb {C}^2$ consisting of points whose standard lift lies in $V_-$. Specifically, the Siegel domain is:

$$\begin{aligned} \textbf{H}_{\mathbb {C}}^2 =\{(z_1,z_2)\in \mathbb {C}^2 : \ 2\Re (z_1)+|z_2|^2<0\}. \end{aligned}$$

Now consider $\textbf{q}_\infty =(1,0,0)^t$ be the column vector in $\mathbb {C}^{2,1}$. It is easy to see that $\textbf{q}_\infty \in V_0$. We define $\mathbb P(\textbf{q}_\infty )=\infty$, which lies in $\partial \textbf{H}_{\mathbb {C}}^2$. Any point in $\partial \textbf{H}_{\mathbb {C}}^2-\{\infty \}$ has a standard lift in $V_0$. That is,

$$\begin{aligned} \partial \textbf{H}_{\mathbb {C}}^2-\{\infty \}=\mathbb P(V_{0}-\{\textbf{q}_\infty \})=\{(z_1,z_2)\in \mathbb {C}^2 :2\Re (z_1)+|z_2|^2=0\}. \end{aligned}$$

The origin is the point $o\in \partial \textbf{H}^2_{\mathbb C}={\mathbb P}(V_0)$ with $(z_1,z_2)=(0,0)$.

Suppose $z_1,\,z_2,\,z_3$ are distinct points in $\partial \textbf{H}^2_{\mathbb C}$ with standard lifts $\textbf{z}_1$, $\textbf{z}_2$, $\textbf{z}_3$ respectively, we define their Cartan angular invariant ${\mathbb A}(z_1,z_2,z_3)$ to be

$$\begin{aligned} {\mathbb A}(z_1,z_2,z_3) =\arg \bigl (-\langle \textbf{z}_1,\textbf{z}_2\rangle \langle \textbf{z}_2,\textbf{z}_3\rangle \langle \textbf{z}_3,\textbf{z}_1\rangle \bigr ). \end{aligned}$$

We know that ${\mathbb A}(z_1,z_2,z_3)=\pm \frac{\pi }{2}$ (resp. ${\mathbb A}(z_1,z_2,z_3)=0$) if and only if $z_1,z_2,z_3$ lie on the same $\mathbb {C}$-circle (resp. $\mathbb R$-circle).

2.2 The Heisenberg group

The set $\partial \textbf{H}_{\mathbb {C}}^2-\{\infty \}$ naturally carries the structure of the Heisenberg group ${{\mathfrak {N}}}$. Thus, the boundary of complex hyperbolic space is the one-point compactification of the Heisenberg group, which should be thought of as a generalisation of the well known fact that the boundary of the upper half space model of $\textbf{H}_{\mathbb R}^3$ is the one point compactification of $\mathbb {C}$.

We recall that the Heisenberg group ${{\mathfrak {N}}}$ is $\mathbb {C}\times \mathbb R$ with the group law

$$\begin{aligned} (\zeta _1,v_1)\cdot (\zeta _2,v_2)=\bigl (\zeta _1+\zeta _2,v_1+v_2+2\Im (\zeta _1{\bar{\zeta }}_2)\bigr ). \end{aligned}$$

The identity element in the Heisenberg group is (0, 0), which we denote by o. The map $\Pi _V:{{\mathfrak {N}}}\longrightarrow {\mathbb C}$ given by $\Pi _V:(\zeta ,v)\longmapsto \zeta$ is called vertical projection. It is a homomorphism.

Given $z=(\zeta ,v)$ in the Heisenberg group, we define its standard lift to be

$$\begin{aligned} \textbf{z} = \begin{pmatrix} -|\zeta |^2 + iv \\ \sqrt{2}\zeta \\ 1 \end{pmatrix}\in V_0. \end{aligned}$$

Given $(\tau ,t)$ in the Heisenberg group, there is a unique, upper triangular unipotent element of $\textrm{U}(2,1)$ taking the standard lift of $o=(0,0)$ to the standard lift of $(\tau ,t)$, which is given by

$$\begin{aligned} T_{(\tau ,t)}=\begin{pmatrix} 1 &{}\quad -\sqrt{2} \bar{\tau } &{}\quad -|\tau |^2+it \\ 0 &{}\quad 1 &{} \quad \sqrt{2}\tau \\ 0 &{}\quad 0 &{}\quad 1 \end{pmatrix}. \end{aligned}$$

The map $(\tau ,t)$ is a group homomorphism from ${{\mathfrak {N}}}$ to $\textrm{U}(2,1)$. Thus, applying $T_{(\tau ,t)}$ to the standard lift of points in ${{\mathfrak {N}}}$ is equivalent to ${{\mathfrak {N}}}$ acting on itself by left translation; that is, the map $(\zeta ,v)\longmapsto (\tau ,t)\cdot (\zeta ,v)=\bigl (\tau +\zeta ,t+v+2\Im (\tau {\bar{\zeta }})\bigr )$.

We define Cygan metric on the Heisenberg group by

$$\begin{aligned} \rho _0\big ((\zeta _1,v_1), (\zeta _2,v_2)\big )= \big ||\zeta _1-\zeta _2|^2-iv_1+iv_2-2i\Im (\zeta _1{\bar{\zeta }}_2) \big |^{\frac{1}{2}}. \end{aligned}$$

There is an easy way to compute the Cygan distance. If $z_1,\,z_2$ are two points in ${{\mathfrak {N}}}$ with standard lifts $\textbf{z}_1,\,\textbf{z}_2\in V_0$ respectively then

$$\begin{aligned} \rho _0(z_1,z_2)=\bigl |\langle \textbf{z}_1,\textbf{z}_2\rangle \bigr |^{1/2}. \end{aligned}$$

We can define Cygan sphere $S_{(r,z_0)}$ of radius r and centre $z_0 \in \partial \textbf{H}_{\mathbb {C}}^2$ by

$$\begin{aligned} S_{(r,z_0)}= \Bigl \{z \in \partial \textbf{H}_{\mathbb {C}}^2: \rho _0(z,z_0)=r\Bigr \}. \end{aligned}$$

2.3 Special subsets of complex hyperbolic space and its boundary

There are two types of totally geodesic subspaces of $\textbf{H}_{\mathbb {C}}^2$ with real dimension 2. The first is the intersection with $\textbf{H}_{\mathbb {C}}^2$ of a complex line (copy of ${\mathbb C}{\mathbb P}^1$ inside ${\mathbb C}{\mathbb P}^2$). These are fixed by involutions that are holomorphic isometries of $\textbf{H}_{\mathbb {C}}^2$. Their intersection with $\partial \textbf{H}_{\mathbb {C}}^2$ are called chains or $\mathbb {C}$-circles. The second type of totally geodesic subspace is the intersection of $\textbf{H}_{\mathbb {C}}^2$ with a Lagrangian planes (copies of ${\mathbb R}{\mathbb P}^2$ inside ${\mathbb C}{\mathbb P}^2$). These are fixed by involutions that are anti-holomorphic isometries of $\textbf{H}_{\mathbb {C}}^2$. Their intersection with $\partial \textbf{H}_{\mathbb {C}}^2$ are called ${\mathbb R}$-circles. A ${\mathbb C}$-circle or an ${\mathbb R}$-circle is called infinite if it passes through $\infty$ and is called finite otherwise.

There are no totally geodesic real hypersurfaces in $\textbf{H}_{\mathbb {C}}^2$ and so it is necessary to make a choice for the hypersurfaces containing the sides of a fundamental domain. We make two choices in this paper. The first are Cygan spheres, which are a particular cases of the boundary of bisectors, and the second are fans. Both are foliated by complex lines and Lagrangian planes, and both are mapped to themselves by the involution fixing each complex line in this foliation and by the involution fixing each Lagrangian plane in the foliation. In what follows, we will discuss the boundary of these hypersurfaces in $\partial \textbf{H}_{\mathbb {C}}^2={{\mathfrak {N}}}\cup \{\infty \}$.

2.4 Isometric spheres

In [1], Goldman extended the definition of isometric spheres in real hyperbolic geometry to geometry of complex hyperbolic space. These are spheres in the Cygan metric and are a particular type of bisector.

Let P be any element of $\textrm{PU}(2,1)$ which does not fix $\infty$, then the isometric sphere corresponding to P is given by

$$\begin{aligned} I_P= \Bigl \{z \in \partial \textbf{H}_{\mathbb {C}}^2: \bigl |\langle \textbf{z}, q_\infty \rangle \bigr |= \bigl |\langle \textbf{z}, P^{-1}(q_\infty )\rangle \bigr |\Bigr \}. \end{aligned}$$

Let P be an element of $\textrm{PU}(2,1)$ then $P^{-1}$ has the following form:

$$\begin{aligned} P=\left( \begin{matrix} a &{}\quad b &{}\quad c \\ d &{}\quad e &{}\quad f \\ g &{}\quad h &{}\quad j \end{matrix}\right) , \qquad P^{-1}=\left( \begin{matrix} {\bar{j}} &{} {\bar{f}} &{}\quad {\bar{c}} \\ {\bar{h}} &{}\quad {\bar{e}} &{}\quad {\bar{b}} \\ {\bar{g}} &{}\quad {\bar{d}} &{}\quad {\bar{a}} \end{matrix}\right) , \end{aligned}$$

(3)

Such a map does not fix $\infty$ if and only if $g\ne 0$. The isometric sphere of P; denoted by $I_P$ is the sphere in Cygan metric with centre $P^{-1}(\infty )$ and radius $r_P=1/\sqrt{|g|}$. Similarly, the isometric sphere $I_{P^{-1}}$ is the Cygan sphere with centre $P(\infty )$ and radius $r_{P^{-1}}=r_P=1/\sqrt{|g|}$. In Heisenberg coordinates, the centres of these spheres are

$$\begin{aligned} P^{-1}(\infty )= \left( \frac{{\bar{h}}}{\sqrt{2}{\bar{g}}},-\Im (\frac{j}{g}) \right) , \qquad P(\infty )= \left( \frac{d}{\sqrt{2}{g}},\Im (\frac{a}{g}) \right) . \end{aligned}$$

We will need the following lemma.

Lemma 2.1

(Proposition 2.4 of [5]) Let P be any element of $\textrm{PU}(2,1)$ such that $P(\infty ) \ne \infty$, then there exist $r_P$ such that for all $z \in \partial {\textbf{H}^2_{\mathbb C}}\setminus \left\{ \infty , P^{-1}(\infty )\right\}$ we have

$$\begin{aligned} \rho _0\big (P(z),P(\infty )\big )=\frac{r_P^2}{\rho _0\big (z,P^{-1}(\infty )\big )}. \end{aligned}$$

Note that P maps $I_P$ to $I_{P^{-1}}$ and maps the component of $\overline{\textbf{H}^2_{\mathbb C}} \setminus I_P$ containing $\infty$ to the component of $\overline{\textbf{H}^2_{\mathbb C}} \setminus I_{P^{-1}}$ not containing $\infty$.

We use an involution $\iota$ swapping o and $\infty$. It is defined as

$$\begin{aligned} \iota =\left( \begin{matrix} 0 &{}\quad 0 &{}\quad 1 \\ 0 &{}\quad 1 &{}\quad 0 \\ 1 &{}\quad 0 &{}\quad 0 \end{matrix}\right) . \end{aligned}$$

(4)

For $(\zeta ,v)\ne (0,0)$ the involution $\iota$ is given in Heisenberg coordinates as

$$\begin{aligned} \iota (\zeta ,v)=\left( \frac{\zeta }{-|\zeta |^2+iv},\,\frac{-v}{\bigl ||\zeta |^2+iv\bigr |^2}\right) . \end{aligned}$$

As a consequence of Lemma 2.1, the involution $\iota$ maps the Cygan sphere with centre o and radius r to Cygan sphere with centre o and radius 1/r. Also, conjugating A by $\iota$ results in a matrix of the same form as B but where the indices of $s_j$, $t_j$, $\theta _j$ are all 1. Likewise, conjugating B by $\iota$ results in a matrix of the same form as A but where the indices of $s_j$, $t_j$, $\theta _j$ are all 2.

We can give geographical coordinates on isometric spheres. For convenience when proving Lemma 1.6 we modify the more usual coordinates by letting $\alpha$ vary in $[0,\pi /2]$. Specifically, we parametrise points on the Cygan sphere with centre o and radius $r>0$ by $s_{\alpha ,\beta }$ where $(\alpha ,\beta )\in [0,\pi /2]\times ({\mathbb R}/2\pi {\mathbb Z})$, given by

$$\begin{aligned} s_{\alpha ,\beta }=\Bigl (r\sqrt{\sin (2\alpha )}e^{i\alpha +i\beta },r^2\cos (2\alpha )\Bigr ). \end{aligned}$$

The point $s_{\alpha ,\beta }$ has standard lift

$$\begin{aligned} \textbf{s}_{\alpha .\beta }=\left( \begin{matrix} r^2ie^{2i\alpha } \\ r\sqrt{2\sin (2\alpha )}e^{i\alpha +i\beta } \\ 1 \end{matrix}\right) . \end{aligned}$$

Fixing $\alpha =\alpha _0$ gives a ${\mathbb C}$-circle. The union of the arcs where $\beta =\beta _0$ and $\beta =\beta _0+\pi$ gives an ${\mathbb R}$-circle. The involutions fixing these ${\mathbb C}$-circles and ${\mathbb R}$-circles all map the sphere to itself. For each $\beta _0$ this involution preserves the Cygan distance between points of the sphere, but the only value of $\alpha _0$ where this is true is $\alpha _0=\pi /4$, corresponding to the equator.

2.5 Fans

Fans are another class of surfaces in the Heisenberg group. They were introduced by Goldman and Parker in [3]. An infinite fan is a Euclidean plane in ${{\mathfrak {N}}}$ whose image under vertical projection $\Pi _V$ is an affine line in ${\mathbb C}$. Infinite fans are foliated by infinite ${\mathbb C}$-circles and infinite ${\mathbb R}$-circles. Given $ke^{i\phi }\in {\mathbb C}$, let $F^{(\infty )}_{ke^{i\phi }}$ be the fan whose image under vertical projection is the line given by the equation $x\cos (\phi )+y\sin (\phi )=k$, where $z=x+iy$. We can write points of

$$\begin{aligned} F^{(\infty )}_{ke^{i\phi }}=\Bigl \{ f_{a,b}=\bigl ((k+ia)e^{i\phi },\,b-2ka\bigr )\ :\ (a,b)\in {\mathbb R}^2\Bigr \}. \end{aligned}$$

The standard lift of $f_{a,b}$ is

$$\begin{aligned} \textbf{f}_{a,b}\left( \begin{matrix} -a^2-k^2+ib-2ika \\ \sqrt{2}(k+ia)e^{i\phi } \\ 1 \end{matrix}\right) . \end{aligned}$$

We can give this fan coordinates that resemble geographical coordinates. Fixing $a=a_0$ gives an infinite ${\mathbb C}$-circle and fixing $b=b_0$ gives an infinite ${\mathbb R}$-circle. The involutions that fixing both of these ${\mathbb C}$-circles and the ${\mathbb R}$-circles are all Cygan isometries.

We are also interested in fans that are the image of this one under the involution $\iota$. That is

$$\begin{aligned} F^{(o)}_{ke^{i\phi }}=\left\{ \left( \frac{-(k+ia)(a^2+k^2+ib-2ika)e^{i\phi }}{(a^2+k^2)^2+(b-2ka)^2},\, \frac{-b+2ka}{(a^2+k^2)^2+(b-2ka)^2}\right) \ :\ (a,b)\in {\mathbb R}^2\right\} . \end{aligned}$$

2.6 A discreteness criterion

In order to show the group $\langle A,B\rangle$ is discrete and free we will consider its action on $\partial \textbf{H}^2_{\mathbb C}={{\mathfrak {N}}}\cup \{\infty \}$ and we will use the Klein Combination theorem, Proposition 1.1.

The construction is the following. We will consider four topological spheres in ${{\mathfrak {N}}}\cup \{\infty \}$ called $S_A^+,\,S_A^-,\,S_B^+,\,S_B^-$. The complement of each of these spheres has two (open) components, which we call the interior and the exterior. We assume that:

• (1) the interiors of $S_A^+,\,S_A^-,\,S_B^+,\,S_B^-$ are disjoint;

• (2) A sends the exterior of $S_A^-$ onto the interior of $S_A^+$, and hence $A^{-1}$ sends the exterior of $S_A^+$ onto the interior of $S_A^-$;

• (3) B sends the exterior of $S_B^-$ onto the interior of $S_B^+$, and hence $B^{-1}$ sends the exterior of $S_B^+$ onto the interior of $S_B^-$.

Then the intersection of the exteriors, which we call D, is then a fundamental domain for $\langle A,B\rangle$. It is easy to see that if W is a reduced word in $A^{\pm 1}$ and $B^{\pm 1}$ (that is all consecutive occurrences of $A^{\pm 1}A^{\mp 1}$ and $B^{\pm 1}B^{\mp 1}$ have been cancelled) then W sends D into the interiors of one of $S_A^+,\,S_A^-,\,S_B^+,\,S_B^-$ corresponding to the last generator to be applied. In our constructions, the spheres will either be the Cygan spheres or they will be fans.

Many different versions of this result have been used for complex hyperbolic isometries; see, for example, Proposition 6.3 of Parker [9] or the notion of a group being compressing, due to Wyss-Gallifent in [12] and used by Monaghan, Parker and Pratoussevitch in [8].

We can consider how our construction relates to that of Parker and Will [11]. Suppose that the interiors of $S_A^+,\,S_A^-,\,S_B^+,\,S_B^-$ are disjoint, but that there are points $q_+=S_A^+\cap S_B^-$ and $q_-=S_A^-\cap S_B^+$. The existence of such points implies that we have equality in the relevant expression of Theorem 1.2. If furthermore $B(q_+)=q_-$ and $A(q_-)=q_+$ then $q_+$ is a fixed point of AB and $q_-$ is a fixed point of BA. The existence of points with these properties in necessary of AB is parabolic.

3 The Cygan diameter of a finite ${\mathbb R}$-circle

In this section we prove Lemma 1.6. Let R be any finite ${\mathbb R}$-circle and let $\iota _R$ be the anti-holomorphic involution fixing R. Let r be the radius of R, that is r is the Cygan distance from $\iota _R(\infty )$ to any point of R. Applying a Cygan isometry (Heisenberg translation and rotation) if necessary, we may assume that R has the following form

$$\begin{aligned} R=\Bigl \{p_{\alpha ,\epsilon }=\bigl (\epsilon r\sqrt{\sin (2\alpha )}e^{i\alpha },r^2\cos (2\alpha )\bigr )\ :\ \alpha \in [0,\pi /2],\ \epsilon =\pm 1\Bigr \}. \end{aligned}$$

Perhaps the easiest way to see that this collection of points comprise an ${\mathbb R}$-circle is to consider the following map $\iota _R$:

$$\begin{aligned} {\iota _R:\left( \begin{array}{l} z_1 \\ z_2 \\ z_3 \end{array}\right) \longmapsto \left( \begin{array}{l} \bar{z}_3r^2 \\ -i\bar{z}_2 \\ \bar{z}_1/r^2 \end{array}\right) .} \end{aligned}$$

It is easy to check that $\langle \iota _R\textbf{z},\iota _R\textbf{w}\rangle =\overline{\langle \textbf{z},\textbf{w}\rangle }$ for any $\textbf{z}$, $\textbf{w}$ in $\mathbb {C}^{2,1}$, and so $\iota _R$ is a complex hyperbolic isometry. Moreover, it is easy to check that $\iota _R^2$ is the identity. Hence, by construction, the subset of $V_0$ projectively fixed by $\iota _R$ is an ${\mathbb R}$-circle. For $\alpha \in [0,\pi /2]$ and $\epsilon =\pm 1$ consider

$$\begin{aligned} \textbf{p}_{\alpha .\epsilon }=\left( \begin{matrix} r^2ie^{2i\alpha } \\ \epsilon r\sqrt{2\sin (2\alpha )}e^{i\alpha } \\ 1 \end{matrix}\right) \in V_0. \end{aligned}$$

Observe that $\iota _R:\textbf{p}_{\alpha ,\epsilon }\longmapsto (-ie^{-2i\alpha })\textbf{p}_{\alpha ,\epsilon }$ and so $\textbf{p}_{\alpha ,\epsilon }$ is projectively fixed by $\iota _R$. We can express $p_{\alpha ,\epsilon }={\mathbb P}{} \textbf{p}_{\alpha ,\epsilon }$ in Heisenberg coordinates as:

$$\begin{aligned} p_{\alpha ,\epsilon }=\Bigl (\epsilon r\sqrt{\sin (2\alpha )}e^{i\alpha },r^2\cos (2\alpha )\Bigr ). \end{aligned}$$

This gives the result.

It is easy to check that ${\mathbb A}(p_{\alpha ,\epsilon },o,\infty )=2\alpha -\pi /2$. Thus, there are two points $p_\alpha$ satisfying the condition ${\mathbb A}(p_{\alpha },\iota _R(\infty ), \infty )=2\alpha -\pi /2$, namely $p_{\alpha ,+1}$ and $p_{\alpha ,-1}$. Given $\theta \in [0,\pi /2]$ and $\eta =\pm 1$ the Cygan distance from $p_{\alpha ,\epsilon }$ to $p_{\theta ,\eta }$ is given by

$$\begin{aligned} \rho _0({p_{\alpha ,\epsilon }, p_{\theta ,\eta }})= & {} \Bigl | ir^2e^{2i\theta }-ir^2e^{-2i\alpha } +2\eta \epsilon r^2\sqrt{\sin (2\alpha )\sin (2\theta )}e^{i\theta -i\alpha }\Bigr |^{1/2} \\= & {} \Bigl | -2r^2\sin (\theta +\alpha )e^{i\theta -i\alpha } +2\eta \epsilon r^2\sqrt{\sin (2\alpha )\sin (2\theta )}e^{i\theta -i\alpha }\Bigr |^{1/2} \\= & {} 2^{1/2}r\Bigl ( \sin (\alpha +\theta )-\eta \epsilon \sqrt{\sin (2\alpha )\sin (2\theta )}\Bigr )^{1/2} \\= & {} 2^{1/2}r\Bigl ( \sin (\alpha )\cos (\theta )+\cos (\alpha )\sin (\theta )-\eta \epsilon 2\sqrt{\sin (\alpha )\cos (\alpha )\sin (\theta )\cos (\theta )}\Bigr )^{1/2} \\= & {} 2^{1/2}r\bigl |\sin ^{1/2}(\alpha )\cos ^{1/2}(\theta )-\eta \epsilon \cos ^{1/2}(\alpha )\sin ^{1/2}(\theta )\bigr |. \end{aligned}$$

We need to maximize this quantity. Taking positive square roots of all the trigonometric functions, we see that this maximum arises when $\eta =-\epsilon$. We now use calculus to find the resulting maximum. Putting all this together, gives a proof of Lemma 1.6.

Lemma 3.1

Given $\alpha \in [0,\pi /2]$ define $f_\alpha :[0,\pi /2]\longrightarrow {\mathbb R}$ by

$$\begin{aligned} f_\alpha (\theta )=\sin ^{1/2}(\alpha )\cos ^{1/2}(\theta )+\cos ^{1/2}(\alpha )\sin ^{1/2}(\theta ). \end{aligned}$$

Then

$$\begin{aligned} f_\alpha (\theta )\le \bigl (\sin ^{2/3}(\alpha )+\cos ^{2/3}(\alpha )\bigr )^{3/4}. \end{aligned}$$

Proof

First observe

$$\begin{aligned} f_\alpha (0)= & {} \sin ^{1/2}(\alpha )\le \bigl (\sin ^{2/3}(\alpha )+\cos ^{2/3}(\alpha )\bigr )^{3/4}, \\ f_\alpha (\pi /2)= & {} \cos ^{1/2}(\alpha )\le \bigl (\sin ^{2/3}(\alpha )+\cos ^{2/3}(\alpha )\bigr )^{3/4}. \end{aligned}$$

In the first case, there is equality if and only if $\alpha =\pi /2$ and in the second case if and only if $\alpha =0$.

Now suppose $\theta \in (0,\pi /2)$. Differentiating with respect to $\theta$ we have

$$\begin{aligned} f'_\alpha (\theta )=\frac{-\sin (\theta )\sin ^{1/2}(\alpha )}{2\cos ^{1/2}(\theta )} +\frac{\cos (\theta )\cos ^{1/2}(\alpha )}{2\sin ^{1/2}(\theta )}. \end{aligned}$$

Therefore, if $\theta _0$ is a value of $\theta$ for which $f_\alpha '(\theta _0)=0$ we have

$$\begin{aligned} \sin ^{3/2}(\theta _0)\sin ^{1/2}(\alpha )=\cos ^{3/2}(\theta _0)\cos ^{1/2}(\alpha ). \end{aligned}$$

In other words, $\cos ^{1/2}(\theta _0)=k\sin ^{1/6}(\alpha )$ and $\sin ^{1/2}(\theta _0)=k\cos ^{1/6}(\alpha )$ for some constant k, depending on $\alpha$. Using $1=\cos ^2(\theta _0)+\sin ^2(\theta _0)$ we have:

$$\begin{aligned} 1=k^4\bigl (\sin ^{2/3}(\alpha )+\cos ^{2/3}(\alpha )\bigr ). \end{aligned}$$

Hence:

$$\begin{aligned} \cos ^{1/2}(\theta _0)= & {} \frac{\sin ^{1/6}(\alpha )}{\bigl (\sin ^{2/3}(\alpha )+\cos ^{2/3}(\alpha )\bigr )^{1/4}}, \\ \sin ^{1/2}(\theta _0)= & {} \frac{\cos ^{1/6}(\alpha )}{\bigl (\sin ^{2/3}(\alpha )+\cos ^{2/3}(\alpha )\bigr )^{1/4}}. \end{aligned}$$

Therefore

$$\begin{aligned} f_\alpha (\theta _0)= & {} \sin ^{1/2}(\alpha )\cos ^{1/2}(\theta _0)+\cos ^{1/2}(\alpha )\sin ^{1/2}(\theta ) \\= & {} \frac{\sin ^{2/3}(\alpha )}{\bigl (\sin ^{2/3}(\alpha )+\cos ^{2/3}(\alpha )\bigr )^{1/4}} +\frac{\cos ^{2/3}(\alpha )}{\bigl (\sin ^{2/3}(\alpha )+\cos ^{2/3}(\alpha )\bigr )^{1/4}} \\= & {} \bigl (\sin ^{2/3}(\alpha )+\cos ^{2/3}(\alpha )\bigr )^{3/4}. \end{aligned}$$

$\square$

4 Fundamental domain bounded by Cygan spheres

First consider B as given in equation (1). The isometric sphere of B is a Cygan sphere of radius $r_B=1/|s_2^2+it_2|^{1/2}$ with centre $B^{-1}(\infty )$. Likewise, the isometric sphere of $B^{-1}$ is a Cygan sphere of the same radius with centre $B(\infty )$. By construction o, the fixed point of B, lies on both these spheres and they are both tangent at this point. In this section, we use the result on the diameters of an finite ${\mathbb R}$-circle to find the smallest Cygan sphere centred at o containing both these isometric spheres.

Now consider $\iota A\iota$ and do a similar thing. The isometric sphere of $\iota A\iota$ is a Cygan sphere of radius $r_A=1/|s_1^2+it_1|^{1/2}$ with centre $\iota A^{-1}\iota (\infty )=\iota A^{-1}(o)$ and the isometric sphere of $\iota A^{-1}\iota$ is a Cygan sphere of the same radius with centre $\iota A(o)$. Again, we want to find the smallest Cygan sphere centred at o containing both these isometric spheres. Now apply $\iota$ to find the largest Cygan sphere with the images of these isometric spheres in its exterior. If this sphere has large enough radius, these two spheres will be disjoint from the isometric spheres of B and $B^{-1}$. Thus the interiors of these four spheres will be disjoint, and the result will follow as in Sect. 2.6.

4.1 A Cygan ball containing the isometric spheres of B and $B^{-1}$

Proposition 4.1

The isometric spheres of B and $B^{-1}$ are contained in the Cygan ball ${{\mathcal {B}}}$ with centre o and radius $d_B$ where

$$\begin{aligned} d_B=\frac{2^{1/4}}{|s_2^2+it_2|^{1/2}} \left( \left( 1-\frac{t_2}{|s_2^2+it_2|}\right) ^{1/3}+\left( 1+\frac{t_2}{|s_2^2+it_2|}\right) ^{1/3}\right) ^{3/4}. \end{aligned}$$

Furthermore, let $p_+$ and $p_-$ be the points of these two isometric spheres on the boundary of ${{\mathcal {B}}}$. Then B sends $p_+$ to $p_-$.

Proof

We want to find the largest Cygan distance from o to another point on the isometric sphere of B, respectively $B^{-1}$. Since Cygan spheres are strictly convex, each of these points $p_+$, respectively $p_-$, must be unique. Let $R_+$, respectively $R_-$, be the meridian of the isometric sphere of B, respectively $B^{-1}$, passing through o. We claim that $p_+$ lies on $R_+$. Observe that there is an anti-holomorphic involution $\iota _{R_+}$ whose fixed point set is $R_+$ which maps the isometric sphere of B to itself isometrically. Thus if $p_+$ does not lie on $R_+$ then $\iota _{R_+}(p_+)$ is a point of the isometric sphere of B different from $p_+$ and the same distance from o. This contradicts uniqueness of $p_+$. Thus $p_+$ lies on $R_+$ and similarly $p_-$ lies on $R_-$.

This means we can use Lemma 1.6 to find the distances betwen o and $p_\pm$. The Cygan isometric sphere of B has radius $r_B=1/|s_2^2+it_2|^{1/2}$ and centre

$$\begin{aligned} B^{-1}(\infty )=\left( \frac{s_2 e^{i\theta _2}}{s_2^2+it_2},\ \frac{t_2}{|s_2^2+it_2|^2}\right) . \end{aligned}$$

Simillarly, The Cygan isometric sphere of $B^{-1}$ has radius $r_{B^{-1}}=r_B =1/|s_2^2+it_2|^{1/2}$ and centre

$$\begin{aligned} B(\infty )=\left( \frac{s_2 e^{i\theta _2}}{-s_2^2+it_2},\ \frac{-t_2}{|s_2^2+it_2|^2}\right) . \end{aligned}$$

By construction, o lies on both these spheres. It is easy to see that

$$\begin{aligned} {\mathbb A}(o,B^{-1}(\infty ),\infty )=\arg (s_2^2+it_2),\quad {\mathbb A}(o,B(\infty ),\infty )=\arg (s_2^2-it_2). \end{aligned}$$

Define $\alpha$ by $2\alpha -\pi /2={\mathbb A}(o,B^{-1}(\infty ),\infty )$. Then ${\mathbb A}(o,B(\infty ),\infty )=-2\alpha +\pi /2$. This means that

$$\begin{aligned} \cos (2\alpha )=\frac{t_2}{|s_2^2+it_2|}. \end{aligned}$$

Hence

$$\begin{aligned} \sin ^2(\alpha )= & {} \frac{1-\cos (2\alpha )}{2} = \frac{1}{2}\left( 1-\frac{t_2}{|s_2^2+it_2|}\right) , \\ \cos ^2(\alpha )= & {} \frac{1+\cos (2\alpha )}{2} = \frac{1}{2}\left( 1+\frac{t_2}{|s_2^2+it_2|}\right) . \end{aligned}$$

Therefore the isometric sphere of B is contained in the ball with centre o and radius

$$\begin{aligned} d_B= & {} 2^{1/2}r_B\,\bigl (\cos ^{2/3}(\alpha )+\sin ^{2/3}(\alpha )\bigr )^{3/4} \\= & {} \frac{2^{1/4}}{|s_2^2+it_2|^{1/2}} \left( \left( 1-\frac{t_2}{|s_2^2+it_2|}\right) ^{1/3}+\left( 1+\frac{t_2}{|s_2^2+it_2|}\right) ^{1/3}\right) ^{3/4}. \end{aligned}$$

The same is true for the isometric sphere of $B^{-1}$. This completes the proof of the first part.

To prove the second part, we observe that there is an infinite ${\mathbb R}$-circle $R_0$ passing through o so that the inversion $\iota _{R_0}$ fixing $R_0$ interchanges $B^{-1}(\infty )$ and $B(\infty )$. Hence this inversion also interchanges the isometric spheres of B and $B^{-1}$. Since it fixes the origin, it must also interchange $p_+$ and $p_-$. Assume that $s_2\ne 0$. Then $R_+$ and $R_-$ are the unique meridians of these two isometric spheres passing through o. Hence $B=\iota _{R_0}\iota _{R_+}=\iota _{R_-}\iota _{R_0}$. Thus by construction

$$\begin{aligned} B(p_+)=\iota _{R_0}\iota _{R_+}(p_+)=\iota _{R_0}(p_+)=p_-. \end{aligned}$$

Finally, when $s_2=0$ then assume without loss of generality that $t_2>0$ (otherwise replace B with $B^{-1}$). In this case, $p_+$, o are the north and south poles of the isometric sphere of B and $p_-$, o are the south and north poles of the isometric sphere of $B^{-1}$. It is clear that B sends $p_+$ to $p_-$. $\square$

4.2 Proof of Theorem 1.2(1)

To obtain condition (1) of Theorem 1.2, we do a similar thing for A using a Cygan metric where o is the infinite point. We use the involution $\iota$ given in (4). Recall that $\iota$ maps the Cygan sphere centred at o of radius d to the Cygan sphere centred at o with radius 1/d.

Proof

By a similar argument to Proposition 4.1 shows that the isometric spheres of $\iota A\iota$ and $\iota A^{-1}\iota$ are contained in the Cygan ball with centre o and radius $d_A$ where

$$\begin{aligned} d_A=\frac{2^{1/4}}{|s_1^2+it_1|^{1/2}} \left( \left( 1-\frac{t_1}{|s_1^2+it_1|}\right) ^{1/3}+\left( 1+\frac{t_1}{|s_1^2+it_1|}\right) ^{1/3}\right) ^{3/4}. \end{aligned}$$

Therefore, the image under $\iota$ of these spheres is contained in the exterior of a Cygan sphere with centre o and radius $1/d_A$. Hence, if $d_B\le 1/d_A$ then we can use the Klein combination theorem, Proposition 1.1, to conclude that A and B freely generate $\langle A,\,B\rangle$. The condition $d_B\le 1/d_A$ is equivalent to

$$\begin{aligned} 1\ge d_Ad_B= & {} \frac{2^{1/2}}{|s_2^2+it_2|^{1/2}|s_1^2+it_1|^{1/2}} \left( \left( 1-\frac{t_2}{|s_2^2+it_2|}\right) ^{1/3}+\left( 1+\frac{t_2}{|s_2^2+it_2|}\right) ^{1/3}\right) ^{3/4}\\{} & {} \times \left( \left( 1-\frac{t_1}{|s_1^2+it_1|}\right) ^{1/3}+\left( 1+\frac{t_1}{|s_1^2+it_1|}\right) ^{1/3}\right) ^{3/4}. \end{aligned}$$

Multiplying through by $|s_1^2+it_1|^{1/2}|s_2^2+it_2|^{1/2}$ we obtain condition (1) of Theorem 1.2. $\square$

Note, that if we simply wanted to obtain condition (1’) of Theorem 1.4, we could use the triangle inequality to say that any two points on a Cygan sphere of radius r are a distance at most 2r apart. Therefore all points on the isometric spheres of B and $B^{-1}$ lie within a Cygan distance $2/|s_2^2+it_2|^{1/2}$. Similarly, all points on the isometric spheres of $\iota A\iota$ and $\iota A^{-1}\iota$ lie within a a Cygan distance $2/|s_1^2+it_1|^{1/2}$. Arguing as above, but with these weaker bounds, we obtain the condition.

$$\begin{aligned} 1 \ge \frac{2}{|s_1^2+it_1|^{1/2}}\,\frac{2}{|s_2^2+it_2|^{1/2}}. \end{aligned}$$

Multiplying through by $|s_1^2+it_1|^{1/2}|s_2^2+it_2|^{1/2}$ we obtain condition (1’) of Theorem 1.4.

4.3 Criteria for equality in Theorem 1.2(1)

We now briefly discuss what happens when we have equality in the criterion of Theorem 1.2(1). By construction, there are points $p_+$ and $p_-=B(p_+)$ on the isometric spheres of B and $B^{-1}$ so that both $p_+$ and $p_-$ are a distance $d_B$ from o. Similarly, there are points $q_+$ and $q_-$ on the $\iota$ images of the isometric spheres of $\iota A\iota$ and $\iota A^{-1}\iota$ so that A sends $q_+$ to $q_-=A(q_+)$ and $q_+$ and $q_-$ are a distance $1/d_A=d_B$ from o. If we have $q_-=p_+$ and $q_+=p_-$ then $p_+$ is a fixed point of AB and $p_-$ is a fixed point of BA. We claim this only happens when either (a) $s_1=s_2=0$ and $t_1t_2=4$ or (b) $t_1=t_2=0$, $\theta _1=\theta _2$ and $s_1s_2=4$. In the first case AB is screw parabolic with angle $\pi$ and in the second case AB is unipotent.

Write $-s_2^2+it_2=ir_2^2e^{2i\alpha _2}$. Define $\phi _2$ by

$$\begin{aligned} \cos (\phi _2)=\frac{\sin ^{1/3}(\alpha _2)}{\bigl (\sin ^{2/3}(\alpha _2)+\cos ^{2/3}(\alpha _2)\bigr )^{1/2}},\quad \sin (\phi _2)=\frac{\cos ^{1/3}(\alpha _2)}{\bigl (\sin ^{2/3}(\alpha _2)+\cos ^{2/3}(\alpha _2)\bigr )^{1/2}} \end{aligned}$$

and $d_B$ by

$$\begin{aligned} d_B=r_2^{-1}\bigl (\sqrt{2\cos (\phi _2)\sin (\alpha _2)}+\sqrt{2\sin (\phi _2)\cos (\alpha _2)}\bigr ). \end{aligned}$$

Let $T_+$, respectively $T_-$, be the Heisenberg translation taking o to $B^{-1}(\infty )$, respectively $B(\infty )$. Then

$$\begin{aligned} T_+^{-1}(o)=\left( \begin{matrix} -ir_2^{-2}e^{-2i\alpha _2} \\ i\sqrt{2\sin (2\alpha _2)}r_2^{-1}e^{2i\alpha _2+i\theta _2} \\ 1 \end{matrix}\right) , \quad T_-^{-1}(o)=\left( \begin{matrix} ir_2^{-2}e^{2i\alpha _2} \\ i\sqrt{2\sin (2\alpha _2)}r_2^{-1}e^{-2i\alpha _2+i\theta _2} \\ 1 \end{matrix}\right) . \end{aligned}$$

The points on the ${\mathbb R}$-circle furthest from these two points are

$$\begin{aligned} \left( \begin{matrix} -ir_2^{-2}e^{-2i\phi _2} \\ -i\sqrt{2\sin (2\phi _2)}r_2^{-1}e^{-i\phi _2+3i\alpha _2+i\theta _2} \\ 1 \end{matrix}\right) , \quad \left( \begin{matrix} ir_2^{-2}e^{2i\phi _2} \\ i\sqrt{2\sin (2\phi _2)}r_2^{-1}e^{i\phi _2-3i\alpha _2+i\theta _2} \\ 1 \end{matrix}\right) . \end{aligned}$$

The images of these two points under $T_+$ and $T_-$ are $p_+$ and $p_-$. As vectors in standard form these are:

$$\begin{aligned}{} & {} \textbf{p}_+=\left( \begin{matrix} -d_B^2 e^{-i\phi _2+i\alpha _2} \\ d_B\sqrt{2\cos (\phi _2-\alpha _2)}e^{-2i\phi _2+2i\alpha _2+i\theta _2} \\ 1 \end{matrix}\right) , \\{} & {} \textbf{p}_-=\left( \begin{matrix} -d_B^2 e^{i\phi _2-i\alpha _2} \\ -d_B\sqrt{2\cos (\phi _2-\alpha _2)}e^{2i\phi _2-2i\alpha _2+i\theta _2} \\ 1 \end{matrix}\right) . \end{aligned}$$

A similar caculation shows that

$$\begin{aligned} B:\textbf{p}_+ \longmapsto e^{-2i\phi _2+2i\alpha _2}{} \textbf{p}_-. \end{aligned}$$

Thus, B sends the vector $\textbf{p}_+$ to the vector $\textbf{p}_-$ with a non-trivial multiplier. Applying the involution $\iota$ and changing all the indices from 2 to 1 gives

$$\begin{aligned}{} & {} \iota \textbf{q}_+=\left( \begin{matrix} -d_A^2 e^{-i\phi _1+i\alpha _1} \\ d_A\sqrt{2\cos (\phi _1-\alpha _1)}e^{-2i\phi _1+2i\alpha _1+i\theta _1} \\ 1 \end{matrix}\right) , \\{} & {} \iota \textbf{q}_-=\left( \begin{matrix} -d_A^2 e^{i\phi _1-i\alpha _1} \\ -d_A\sqrt{2\cos (\phi _1-\alpha _1)}e^{2i\phi _1-2i\alpha _1+i\theta _1} \\ 1 \end{matrix}\right) , \end{aligned}$$

where

$$\begin{aligned} \cos (\phi _1)= & {} \frac{\sin ^{1/3}(\alpha _1)}{\bigl (\sin ^{2/3}(\alpha _1)+\cos ^{2/3}(\alpha _1)\bigr )^{1/2}},\quad \sin (\phi _1) \ =\ \frac{\cos ^{1/3}(\alpha _1)}{\bigl (\sin ^{2/3}(\alpha _1)+\cos ^{2/3}(\alpha _1)\bigr )^{1/2}}, \\ d_A= & {} r_1^{-1}\bigl (\sqrt{2\cos (\phi _1)\sin (\alpha _1)}+\sqrt{2\sin (\phi _1)\cos (\alpha _1)}\bigr ). \end{aligned}$$

Applying $\iota$ we see that

$$\begin{aligned} \textbf{q}_+= & {} \left( \begin{matrix} 1 \\ d_A\sqrt{2\cos (\phi _1-\alpha _1)}e^{-2i\phi _1+2i\alpha _1+i\theta _1} \\ -d_A^2 e^{-i\phi _1+i\alpha _1} \end{matrix}\right) \sim \left( \begin{matrix} -d_A^{-2} e^{i\phi _1-i\alpha _1} \\ -d_A^{-1} \sqrt{2\cos (\phi _1-\alpha _1)}e^{-i\phi _1+i\alpha _1+i\theta _1} \\ 1 \end{matrix}\right) , \\ \\ \textbf{q}_-= & {} \left( \begin{matrix} 1 \\ -d_A\sqrt{2\cos (\phi _1-\alpha _1)}e^{2i\phi _1-2i\alpha _1+i\theta _1} \\ -d_A^2 e^{i\phi _1-i\alpha _1} \end{matrix}\right) \sim \left( \begin{matrix} -d_A^{-2} e^{-i\phi _1+i\alpha _1} \\ d_A^{-1} \sqrt{2\cos (\phi _1-\alpha _1)}e^{i\phi _1-i\alpha _1+i\theta _1} \\ 1 \end{matrix}\right) . \end{aligned}$$

Thus we have

$$\begin{aligned} A:\textbf{q}_+ \longmapsto \textbf{q}_-. \end{aligned}$$

Hence, A sends the vector $\textbf{q}_+$ to the vector $\textbf{q}_-$ with a multiplier 1. Now, $\textbf{p}_+ = \textbf{q}_-$ and $\textbf{p}_- = \textbf{q}_+$ if and only if either (a) $s_1=s_2=0$ and $t_1t_2=4$ or (b) $t_1=t_2=0$, $\theta _1=\theta _2$ and $s_1s_2=4$. In the first case AB is a screw parabolic with angle $\pi$ and having fixed point ${ p}_+$. This is because the non-trivial multiplier is an eigenvalue of AB associated to its fixed point. Also, in the second case AB is an unipotent element.

5 Domain bounded by fans

Consider the infinite fan $F^{(\infty )}_{ke^{i\theta _1}}$ where $\theta _1$ is the angle associated to the Heisenberg translation A given by (1) and k is any real number. We claim that A sends $F_{ke^{i\theta _1}}^{(\infty )}$ to $F_{(k+s_1)e^{i\theta _1}}^{(\infty )}$. This is most easily seen by considering the standard lifts of points on the fans (Fig. 1).

$$\begin{aligned}{} & {} \left( \begin{matrix} 1 &{}\quad -\sqrt{2}s_1e^{-i\theta _1} &{}\quad -s_1^2+it_1 \\ 0 &{} 1 &{} \sqrt{2}s_1e^{i\theta _1} \\ 0 &{} 0 &{} 1 \end{matrix}\right) \left( \begin{matrix} -k^2-a^2+ib-2ika \\ \sqrt{2}(k+ia)e^{i\theta _1} \\ 1 \end{matrix}\right) \\{} & {} \quad = \left( \begin{matrix} -(k+s_1)^2-a^2+ib-2i(k+s_1)a+it_1 \\ \sqrt{2}(k+s_1+ia)e^{i\theta _1} \\ 1 \end{matrix}\right) . \end{aligned}$$

The images of $F^{(\infty )}_{ke^{i\theta _1}}$ and $F^{(\infty )}_{(k+s_1)e^{i\theta _1}}$ under vertical projection $\Pi _V$ are the lines

$$\begin{aligned} x\cos (\theta _1)+y\sin (\theta _1)=k,\quad x\cos (\theta _1)+y\sin (\theta _1)=k+s_1. \end{aligned}$$

The slab bounded by the fans $F^{(\infty )}_{ke^{i\theta _1}}$ and $F^{(\infty )}_{(k+s_1)e^{i\theta _1}}$ is a fundamental region for $\langle A\rangle$. The image of the slab under vertical projection $\Pi _V$ is the strip

$$\begin{aligned} S_A(k)=\left\{ (x,y)\ :\ k\le x\cos (\theta _1)+y\sin (\theta _1) \le k+s_1 \right\} . \end{aligned}$$

Fig. 1

The vertical projection of the isometric spheres of B and $B^{-1}$ and strip which is the vertical projection of a fundamental domain for $\langle A\rangle$ bounded by two infinite fans. We need the projections of both isometric spheres to lie in the strip

Therefore, if we can find a fundamental domain $D_B$ for $\langle B\rangle$ containing the complement of this slab then we can apply the Klein combination theorem to conclude that $\langle A,B\rangle$ is freely generated by A and B. In particular, this is true if the vertical projection of $\partial D_B$ is contained in $S_A$ and the vertical projection of $D_B$ contains the complement of $S_A$. To see this, consider a point in the complement of $S_A$. By construction, its pre-image under vertical projection contains at least one point of $D_B$ and no points in its boundary. Since it is connected and path connected (the fibres of vertical projection are copies of the real line) all points in this preimage are contained in $D_B$ as required.

In what follows, we give two different fundamental domains for $\langle B\rangle$.

5.1 Proof of Theorem 1.2 parts (2) and (3)

Proposition 5.1

Let $(s_1e^{i\theta _1},t_1)$ and $(s_2e^{i\theta _2},t_2)$ be elements of the Heisenberg group with $s_1$ and $s_2$ both non-zero. Replacing one of them by its inverse if necessary, we suppose $-\pi /2\le (\theta _1-\theta _2)\le \pi /2$. Let A and B given by (1) be the associated Heisenberg translations fixing $\infty$ and o respectively. Suppose that

$$\begin{aligned} s_1\,|s_2^2+it_2|^{1/2} \ge \frac{2~s_2}{|s_2^2+it_2|^{1/2}}\,\cos (\theta _1-\theta _2)+2. \end{aligned}$$

Then the vertical projection of the isometric spheres of B and $B^{-1}$ are contained in the strip $S_A$.

Proof

(Proposition 5.1) The vertical projection of the isometric sphere of B is a circle with radius $r_2=1/|s_2^2+it_2|^{1/2}$ and centre,

$$\begin{aligned} \frac{s_2e^{i\theta _2}}{s_2^2+it_2}=\frac{s_2e^{i\theta _2}(s_2^2-it_2)}{|s_2^2+it_2|^2}. \end{aligned}$$

Similarly, The vertical projection of the isometric sphere of $B^{-1}$ is a circle with radius $r_2=1/|s_2^2-it_2|^{1/2}$ and centre,

$$\begin{aligned} \frac{s_2e^{i\theta _2}}{-s_2^2+it_2}=\frac{-s_2e^{i\theta _2}(s_2^2+it_2)}{|s_2^2+it_2|^2}. \end{aligned}$$

Now, for a point (x, y) on the vertical projection of the isometric sphere of B we have

$$\begin{aligned} x= & {} \frac{s_2(s_2^2\cos (\theta _2)+t_2\sin (\theta _2))}{|s_2^2+it_2|^2}+\frac{\cos (\phi )}{|s_2^2+it_2|^{1/2}},\\ y= & {} \frac{s_2(s_2^2\sin (\theta _2)-t_2\cos (\theta _2))}{|s_2^2+it_2|^2}+\frac{\sin (\phi )}{|s_2^2+it_2|^{1/2}}. \end{aligned}$$

Therefore

$$\begin{aligned} x\cos (\theta _1)+y\sin (\theta _1) = \frac{s_2^3\cos (\theta _1-\theta _2)-s_2t_2\sin (\theta _1-\theta _2)}{|s_2^2+it_2|^2}+\frac{\cos (\phi -\theta _1)}{|s_2^2+it_2|^{1/2}}. \end{aligned}$$

Similarly, for a point on the vertical projection of the isometric sphere of $B^{-1}$ we have

$$\begin{aligned} x\cos (\theta _1)+y\sin (\theta _1) = \frac{-s_2^3\cos (\theta _1-\theta _2)-s_2t_2\sin (\theta _1-\theta _2)}{|s_2^2+it_2|^2}+\frac{\cos (\phi -\theta _1)}{|s_2^2+it_2|^{1/2}}. \end{aligned}$$

As $\cos (\theta _1-\theta _2)\ge 0$, we can see that for all points (x, y) in the vertical projections of the isometric spheres of B and $B^{-1}$ we have

$$\begin{aligned}{} & {} \frac{-s_2^3\cos (\theta _1-\theta _2)}{|s_2^2+it_2|^2}-\frac{1}{|s_2^2+it_2|^{1/2}} -\frac{s_2t_2\sin (\theta _1-\theta _2)}{|s_2^2+it_2|^2} \\{} & {} \quad \le x\cos (\theta _1)+y\sin (\theta _1) \\{} & {} \quad \le \frac{s_2^3\cos (\theta _1-\theta _2)}{|s_2^2+it_2|^2} +\frac{1}{|s_2^2+it_2|^{1/2}} -\frac{s_2t_2\sin (\theta _1-\theta _2)}{|s_2^2+it_2|^2}. \end{aligned}$$

Hence these vertical projections lie in a strip of the form $S_A(k)$ provided

$$\begin{aligned} \frac{2s_2^3\cos (\theta _1-\theta _2)}{|s_2^2+it_2|^2} +\frac{2}{|s_2^2+it_2|^{1/2}}\le s_1. \end{aligned}$$

Multiplying through by $|s_2^2+it_2|^{1/2}$, we get following condition

$$\begin{aligned} s_1\,|s_2^2+it_2|^{1/2} \ge \frac{2~s_2}{|s_2^2+it_2|^{1/2}}\,\cos (\theta _1-\theta _2)+2. \end{aligned}$$

$\square$

The case of Theorem 1.2 arising from condition (2) follows immediately by the above argument. Swapping the roles of A and B also gives the case arising from condition (3) of Theorem 1.2.

Proposition 5.2

Let $(s_1e^{i\theta _1},t_1)$ and $(0,t_2)$ be elements of the Heisenberg group. Let A and B given by (1) be the associated Heisenberg translations fixing $\infty$ and o respectively with $s_2=0$. Suppose that

$$\begin{aligned} s_1\,|t_2|^{1/2} \ge 2. \end{aligned}$$

Then the vertical projection of the isometric spheres of B and $B^{-1}$ are contained in the strip $S_A$.

Proof

This follows from the proof of Proposition 5.1 but is much simpler. In this case, the vertical projections of the isometric spheres of both B and $B^{-1}$ are centred at the origin.

If (x, y) is a point on the vertical projection of either of these two isometric spheres are given respectively by

$$\begin{aligned} x=\frac{\cos (\phi )}{|t_2|^{1/2}},\quad y=\frac{\sin (\phi )}{|t_2|^{1/2}}. \end{aligned}$$

Hence

$$\begin{aligned} \frac{-1}{|t_2|^{1/2}}\le x\cos (\theta _1)+y\sin (\theta _1)=\frac{\cos (\phi -\theta _1)}{|t_2|^{1/2}}\le \frac{1}{|t_2|^{1/2}}. \end{aligned}$$

For these two Cygan spheres to lie in a strip of the form $S_A(k)$ we must have $2/|t_2|^{1/2}\le s_1$. This gives the result. $\square$

5.2 Criteria for equality in Theorem 1.2(2)

When we have equality in Proposition 5.1, the bisectors for B and the fans for A are tangent at the two points

$$\begin{aligned} \textbf{q}_+= & {} \left( \begin{matrix} -1/|s_2^2+it_2|-2s_2(s_2^2+it_2)e^{i\theta _1-i\theta _2}/|s_2^2+it_2|^{5/2}-(s_2^2-it_2)/|s_2^2+it_2|^2 \\ \sqrt{2}s_2(s_2^2-it_2)e^{i\theta _2}/|s_2^2+it_2|^2+\sqrt{2}e^{i\theta _1}/|s_2^2+it_2|^{1/2} \\ 1 \end{matrix}\right) , \\ \textbf{q}_-= & {} \left( \begin{matrix} -1/|s_2^2+it_2|-2s_2(s_2^2-it_2)e^{i\theta _1-i\theta _2}/|s_2^2+it_2|^{5/2}-(s_2^2+it_2)/|s_2^2+it_2|^2 \\ -\sqrt{2}s_2(s_2^2+it_2)e^{i\theta _2}/|s_2^2+it_2|^2-\sqrt{2}e^{i\theta _1}/|s_2^2+it_2|^{1/2} \\ 1 \end{matrix}\right) . \end{aligned}$$

We can easily verify that $\textbf{q}_+$ (resp. $\textbf{q}_-$) lies on the isometric sphere of B (resp. $B^{-1}$). A calculation shows that

$$\begin{aligned} B\textbf{q}_+= & {} \left( \begin{matrix} -1/|s_2^2+it_2|-2s_2(s_2^2+it_2)e^{i\theta _1-i\theta _2}/|s_2^2+it_2|^{5/2}-(s_2^2-it_2)/|s_2^2+it_2|^2 \\ \sqrt{2}s_2e^{i\theta _2}/|s_2^2+it_2|+\sqrt{2}(s_2^2+it_2)^2e^{i\theta _1}/|s_2^2+it_2|^{5/2} \\ (s_2^2-it_2)/|s_2^2+it_2| \end{matrix}\right) \\= & {} \frac{s_2^2-it_2}{|s_2^2+it_2|} \left( \begin{matrix} -1/|s_2^2+it_2|-2s_2(s_2^2+it_2)^2e^{i\theta _1-i\theta _2}/|s_2^2+it_2|^{7/2}-(s_2^2+it_2)/|s_2^2+it_2|^2 \\ -\sqrt{2}s_2(s_2^2+it_2)e^{i\theta _2}/|s_2^2+it_2|^2-\sqrt{2}(s_2^2+it_2)^3e^{i\theta _1}/|s_2^2+it_2|^{7/2} \\ 1 \end{matrix}\right) . \end{aligned}$$

This shows that, B projectively maps $\textbf{q}_+$ to $\textbf{q}_-$ if and only if $(s_2^2+it_2)^3=|s_2^2+it_2|^3$ which is equivalent to $t_2=0$ as $s_2^2\ge 0$. Similarly, $A\textbf{q}_-=\textbf{q}_+$ if and only if both

$$\begin{aligned} (s_1^2-it_1)|s_2^2+it_2|^{1/2}= & {} 2s_1+\frac{2s_1s_2(s_2^2+it_2)e^{i\theta _2-i\theta _1}}{|s_2^2+it_2|^{3/2}} +\frac{4s_2it_2e^{i\theta _1-\theta _2}}{|s_2^2+it_2|^2}-\frac{2it_2}{|s_2^2+it_2|^{3/2}}, \\ s_1|s_2^2+it_2|^{1/2}= & {} 2+\frac{2s_2^3e^{i\theta _2-i\theta _1}}{|s_2^2+it_2|^{3/2}}. \end{aligned}$$

Thus, $\textbf{q}_-$ is mapped to $\textbf{q}_+$ by A only when $t_1=t_2=0$, $\theta _1=\theta _1$ and $s_1s_2=4$.

5.3 Proof of Theorem 1.2(4)

We now construct a fundamental domain for $\langle B\rangle$ bounded by two fans. To do this, we apply the involution $\iota$ given by (4).

Fix $k\in {\mathbb R}-\{0\}$. Consider the fan with with vertex the origin o given by

$$\begin{aligned} F_k^{(o)}=\left\{ \left( \begin{matrix} 1 \\ \sqrt{2}(k+ia) \\ -k^2-a^2+ib-2ika \end{matrix}\right) \ :\ (a,b)\in {\mathbb R}^2\right\} . \end{aligned}$$

Putting this into Heisenberg coordinates $(x+iy,v)$ we find

$$\begin{aligned}{} & {} x=\frac{-k(k^2+a^2)+a(b-2ka)}{(k^2+a^2)^2+(b-2ka)^2},\quad y=\frac{-a(k^2+a^2)-k(b-2ka)}{(k^2+a^2)^2+(b-2ka)^2},\\{} & {} \quad v=\frac{-b+2ka}{(k^2+a^2)^2+(b-2ka)^2}. \end{aligned}$$

Lemma 5.3

Suppose that $k>0$. The images of $F_k^{(o)}$ and $F_{-k}^{(o)}$ under vertical projection are, respectively, the sets given in polar coordinates $(r,\theta )$ by

$$\begin{aligned} r\le \frac{1-\cos (\theta )}{2k}\quad \hbox { and }\quad r\le \frac{1+\cos (\theta )}{2k}. \end{aligned}$$

In Cartesian coordinates, these are

$$\begin{aligned} y^2-4kx(x^2+y^2)-4k^2(x^2+y^2)^2\ge 0\quad \hbox { and }\quad y^2+4kx(x^2+y^2)-4k^2(x^2+y^2)^2\ge 0. \end{aligned}$$

These are the interiors of two cardioids.

Proof

We want to find out the region in the (x, y) plane which is the image of $F_k^{(o)}$ under vertical projection. To do this, we consider local coordinates (a, b) on $F_k^{(o)}$ and we seek points where the map $(a,b)\longmapsto (x,y)$ is not a local diffeomorphism. In other words, we need to find where the Jacobian J of this map vanishes. A calculation shows this Jacobian is

$$\begin{aligned} J=\frac{-a(k^2+a^2)+k(b-2ka)}{((k^2+a^2)^2+(b-2ka)^2)^2}. \end{aligned}$$

Therefore, the Jacobian vanishes precisely when

$$\begin{aligned} b=2ka+\frac{a}{k}(k^2+a^2). \end{aligned}$$

We can parameterise this curve using a as:

$$\begin{aligned} x=\frac{-k(k^2-a^2)}{(k^2+a^2)^2},\quad y=\frac{-2k^2a}{(k^2+a^2)^2},\quad v=\frac{-ka}{(k^2+a^2)^2}. \end{aligned}$$

Writing it in polar coordinates $x+iy=re^{i\theta }$ and eliminating a gives

$$\begin{aligned} r^2=x^2+y^2=\frac{k^2}{(k^2+a^2)^2},\quad x=r\cos (\theta )=\frac{k}{(k^2+a^2)}\left( 1-\frac{2k^2}{(k^2+a^2)}\right) . \end{aligned}$$

Using $k>0$ we see that $r=k/(k^2+a^2)$ and $2kr=1-\cos (\theta )$.

Similarly, in terms of x and y the region is given by

$$\begin{aligned} y^2-4kx(x^2+y^2)-4k^2(x^2+y^2)^2 =\left( \frac{a(k^2+a^2)-k(b-2ka)}{(k^2+a^2)^2+(b-2ka)^2}\right) ^2\ge 0. \end{aligned}$$

A similar argument holds for $F_{-k}^{(o)}$ by replacing k with $-k$ throughout. Only here we have

$$\begin{aligned} x=r\cos (\theta )=\frac{k}{k^2+a^2}\left( \frac{2k^2}{k^2+a^2}-1\right) \end{aligned}$$

and so $2kr=1+\cos (\theta )$. $\square$

Fig. 2

Cardiods which are the vertical projection of finite fans bounding a fundamental domain for $\langle B\rangle$ and strip which is the vertical projection of a fundamental domain for $\langle A\rangle$ bounded by two infinite fans. We need the cardioids to both lie in the strip

For $k>0$ let $C_k$ and $C_{-k}$ be the cardioids given in polar coordinates $(r,\theta )$ and Cartesian coordinates (x, y) by

$$\begin{aligned} C_k= & {} \left\{ (r,\theta )\ : \ r=\frac{1-\cos (\theta )}{2k}\right\} =\left\{ (x,y)\ :\ y^2-4kx(x^2+y^2)-4k^2(x^2+y^2)^2=0\right\} , \\ C_{-k}= & {} \left\{ (r,\theta )\ : \ r=\frac{1+\cos (\theta )}{2k}\right\} =\left\{ (x,y)\ :\ y^2+4kx(x^2+y^2)-4k^2(x^2+y^2)^2=0\right\} . \end{aligned}$$

Lemma 5.4

For each $k>0$ and each $\phi$ with $-\pi /2\le \phi \le \pi /2$, the cardioids $C_k$ and $C_{-k}$ are both contained in the strip $S_\phi$ given by

$$\begin{aligned} S_{k,\phi }=\left\{ (x,y)\ :\ \Bigl |x\cos (\phi )+y\sin (\phi )\Bigr | \le \frac{\cos ^3(\phi /3)}{k}\right\} . \end{aligned}$$

Proof

(See Fig. 2.) The cardioid $C_k$ is singular at the origin $r=0$. Using the polar parametrisation of $C_k$ we can parametrise its non-singular points by $\theta \in (0,2\pi )$. A point on $C_k$ are given in parametric representation by

$$\begin{aligned} \bigl (x(\theta ),\,y(\theta )\bigr ) = \left( \frac{(1-\cos (\theta ))\cos (\theta )}{2k},\,\frac{(1-\cos (\theta ))\sin (\theta )}{2k}\right) . \end{aligned}$$

Consider

$$\begin{aligned} f_\phi (\theta )=k\bigl (x(\theta )\cos (\phi )+y(\theta )\sin (\phi )\bigr )=\frac{(1-\cos (\theta ))\cos (\theta -\phi )}{2}. \end{aligned}$$

Differentiating with respect to $\theta$ we have

$$\begin{aligned} f'_\phi (\theta )= & {} \frac{-(1-\cos (\theta ))\sin (\theta -\phi )+\sin (\theta )\cos (\theta -\phi )}{2} \\= & {} \sin (\theta /2)\cos (3\theta /2-\phi ). \end{aligned}$$

Since $0<\theta /2<\pi$ we see $\sin (\theta /2)\ne 0$ and so the maximum and minimum values of $f_\phi (\theta )$ occur when $3\theta /2-\phi =\pi /2+n\pi$ where n is an integer. That is, $\theta =2\phi /3+\pi /3+2n\pi /3$. We have

$$\begin{aligned} f_\phi (2\phi /3+\pi /3+2n\pi /3)=\cos ^2(\phi /3+(n-1)\pi /3)\cos (\phi /3-(2n+1)\pi /3). \end{aligned}$$

Since $-\pi /2\le \phi \le \pi /2$ we see that the maximum value of $|f_\phi (\theta )|$ occurs when $n=1$, that is when $\theta =2\phi /3+\pi$. The maximum value is

$$\begin{aligned} |f_{\phi }(2\phi /3+\pi )|=\cos ^3(\phi /3). \end{aligned}$$

Similarly for $C_{-2k}$. Here we need to maximise the absolute value of

$$\begin{aligned} g_\phi (\theta )=\frac{(1+\cos (\theta ))\cos (\theta -\phi )}{2}. \end{aligned}$$

This occurs when $\theta =2\phi /3$. The maximum value is

$$\begin{aligned} |g_{\phi }(2\phi /3)|=\cos ^3(\phi /3). \end{aligned}$$

$\square$

If we have equality in Lemma 5.4, we see that the points on the fans $F_k^{(o)}$ and $F_{-k}^{(o)}$ that project to points on the boundary of the strip are, respectively:

$$\begin{aligned} \left( \begin{matrix} -\cos ^3(\phi /3)e^{i\phi /3}/k^2 \\ -\sqrt{2}\cos ^2(\phi /3)e^{2i\phi /3}/k \\ 1 \end{matrix}\right) , \quad \left( \begin{matrix} -\cos ^3(\phi /3)e^{i\phi /3}/k^2 \\ \sqrt{2}\cos ^2(\phi /3)e^{2i\phi /3}/k \\ 1 \end{matrix}\right) . \end{aligned}$$

(5)

Proof

(Theorem 1.2(4)) We now prove the condition (4) of Theorem 1.2. Conjugating A and B by a Heisenberg rotation fixing o and $\infty$, if necessary, we assume that $\theta _2=0$.

It is clear that (independent of $t_1$) the map A sends $F_{{-\frac{s_1}{2}}e^{i\theta _1}}^{(\infty )}$ to $F_{{\frac{s_1}{2}}e^{i\theta _1}}^{(\infty )}$:

$$\begin{aligned}{} & {} \left( \begin{matrix} 1 &{}\quad -\sqrt{2}s_1e^{-i\theta _1} &{}\quad -s_1^2+it_1 \\ 0 &{}\quad 1 &{}\quad \sqrt{2}s_1e^{i\theta _1} \\ 0 &{}\quad 0 &{}\quad 1 \end{matrix}\right) \left( \begin{matrix} -\frac{s_1^2}{4}-a^2+ib-is_1a \\ \sqrt{2}(-\frac{s_1}{2}+ia)e^{i\theta _1} \\ 1 \end{matrix}\right) \\{} & {} \quad =\left( \begin{matrix} -\frac{s_1^2}{4}-a^2+ib-ias_1+it_1 \\ \sqrt{2}(\frac{s_1}{2}+ia)e^{i\theta _1} \\ 1 \end{matrix}\right) \end{aligned}$$

Therefore the slab bounded by these two fans is a fundamental region for $\langle A\rangle$. This slab is

$$\begin{aligned} D_A=\left\{ (x+iy,v)\in {{\mathfrak {N}}}\ : \ \bigl | x\cos (\theta _1)+y\cos (\theta _1)\bigr |\le \frac{s_1}{2} \right\} \cup \{\infty \}. \end{aligned}$$

The image of $D_A$ under vertical projection is the strip

$$\begin{aligned} S_A=\left\{ (x,y)\ :\ \bigl | x\cos (\theta _1)+y\cos (\theta _1)\bigr |\le \frac{s_1}{2} \right\} . \end{aligned}$$

Similarly, (independent of $t_2$) the map B sends the fan $F_{-\frac{s_2}{2}}^{(o)}$ to the fan $F_{\frac{s_2}{2}}^{(o)}$:

$$\begin{aligned} \left( \begin{matrix} 1 &{}\quad 0 &{}\quad 0 \\ \sqrt{2}s_2 &{}\quad 1 &{}\quad 0 \\ -s_2^2+it_2 &{}\quad -\sqrt{2}s_2 &{} \quad 1 \end{matrix}\right) \left( \begin{matrix} 1 \\ \sqrt{2}(-\frac{s_2}{2}+ia) \\ -\frac{s_2^2}{4}-a^2+ib-is_1a \end{matrix}\right) =\left( \begin{matrix} 1 \\ \sqrt{2}(\frac{s_2}{2}+ia) \\ -\frac{s_2^2}{4}-a^2+ib -ias_2+it_2\end{matrix}\right) . \end{aligned}$$

That is, $\iota B\iota$ sends the fan $F_{-\frac{s_2}{2}}^{(\infty )}$ to the fan $F_{\frac{s_2}{2}}^{(\infty )}$. So the slab $D_{\iota B\iota }$ bounded by these two fans is a fundamental domain for $\langle \iota B\iota \rangle$, where

$$\begin{aligned} D_{\iota B\iota }=\left\{ (x+iy,v)\in {{\mathfrak {N}}}\ :\ \bigl | x\bigr |\le \frac{s_2}{2} \right\} \cup \{\infty \}. \end{aligned}$$

Thus the image of $D_{\iota B\iota }$ under $\iota$ is a fundamental domain for $\langle B\rangle$. This consists of all points in the exterior of the fans $F_{-\frac{s_2}{2}}^{(o)}$ and $F_{\frac{s_2}{2}}^{(o)}$.

In order to show that $\langle A,B\rangle$ is free it is sufficient to show that the fans $F_{\pm \frac{s_2}{2}}^{(o)}$ are contained in the slab $D_A$ between $F_{{-\frac{s_1}{2}}e^{i\theta _1}}^{(\infty )}$ to $F_{{\frac{s_1}{2}}e^{i\theta _1}}^{(\infty )}$. It suffices to show that the vertical projections of $F_{\pm \frac{s_2}{2}}^{(o)}$ are contained in the strip $S_A$. The vertical projections of these fans are given in polar coordinates $(r,\theta )$ by

$$\begin{aligned} r\le \frac{1-\cos (\theta )}{s_2},\quad r\le \frac{1-\cos (\theta )}{-s_2}. \end{aligned}$$

Using Lemma 5.4 with $\phi =\theta _1$ and $k=s_2/2$ (see Figure 2) we see that this is true provided

$$\begin{aligned} \frac{\cos ^3(\frac{\theta _1}{3})}{s_2/2}\le \frac{s_1}{2}. \end{aligned}$$

The result follows. $\square$

Plugging in $k=s_2/2$ and $\phi =\theta _1$ into the formulae for the two extreme points (5), we see that B maps one to the other if and only if $\theta _1=0$ and $t_2=0$. Similarly, A maps one to the other if and only if $s_1s_2=4$, $\theta _1=0$ and $t_1=0$.