# Free groups generated by two parabolic maps

Sagar B. Kalane and John R. Parker  Department of Mathematical Sciences,
Indian Institute of Science Education and Research (IISER) Pune, Dr Homi
Bhabha Rd, Ward No. 8, NCL Colony, Pashan, Pune, Maharashtra 411008
<EMAIL_ADDRESS><EMAIL_ADDRESS>Department of Mathematical
Sciences, Durham University, Upper Mountjoy, South Road, Durham DH1 3LE, UK
<EMAIL_ADDRESS>
###### Abstract.

In this paper we consider a group generated by two unipotent parabolic
elements of ${\rm SU}(2,1)$ 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.

###### Key words and phrases:

Complex hyperbolic geometry, Heisenberg translations, Discreteness.

###### 2020 Mathematics Subject Classification:

Primary 22E40; Secondary 51M10, 32M15, 20H10

First author is supported by NBHM post doctoral fellowship

##  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 ${\rm SU}(2,1)$ 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 [LU] and by Ignatov [Ign] giving conditions under which two
parabolic elements of ${\rm PSL}(2,{\mathbb{C}})$ 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 $\mathrm{PSL}(2,\mathbb{C})$ generated by two non-commuting
parabolic maps, see in [KS]. In [PW] 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 [PW] 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

(1)
$A=\left(\begin{matrix}1&-\sqrt{2}s_{1}e^{-i\theta_{1}}&-s_{1}^{2}+it_{1}\\\
0&1&\sqrt{2}s_{1}e^{i\theta_{1}}\\\ 0&0&1\end{matrix}\right),\quad
B=\left(\begin{matrix}1&0&0\\\ \sqrt{2}s_{2}e^{i\theta_{2}}&1&0\\\
-s_{2}^{2}+it_{2}&-\sqrt{2}s_{2}e^{-i\theta_{2}}&1\end{matrix}\right).$

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:

(2)
$(s_{1},t_{1},\theta_{1};s_{2},t_{2},\theta_{2})\longmapsto(s_{1}k,t_{1}k^{2},\theta_{1}+\psi;s_{2}/k,t_{2}/k^{2},\theta_{2}+\psi).$

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
$D_{A}\subset\partial{\bf H}^{2}_{\mathbb{C}}$ for $\langle A\rangle$ and
$D_{B}\subset\partial{\bf H}^{2}_{\mathbb{C}}$ for $\langle B\rangle$ satisfy
$D_{A}^{\circ}\cap D_{B}^{\circ}\neq\emptyset$ and
$\overline{D}_{A}\cup\overline{D}_{B}=\partial{\bf H}^{2}_{\mathbb{C}}$, 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 [GP1], Wyss-Gallifent [WG], Monaghan, Parker
and Pratoussevitch [MPP] or Jiang and Xie [XJ].

Our main theorem is:

######  Theorem 1.2.

Let $(s_{1}e^{i\theta_{1}},t_{1})$ and $(s_{2}e^{i\theta_{2}},t_{2})$ be non-
trivial elements of the Heisenberg group. Here, $\theta_{1}$ and $\theta_{2}$
are only defined when $s_{1}\neq 0$ and $s_{2}\neq 0$. 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
$-\pi/2\leq(\theta_{1}-\theta_{2})\leq\pi/2$. 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. (1)

$\displaystyle|s_{1}^{2}+it_{1}|^{1/2}|s_{2}^{2}+it_{2}|^{1/2}$
$\displaystyle\geq$ $\displaystyle
2^{1/2}\left(\left(1-\frac{t_{2}}{|s_{2}^{2}+it_{2}|}\right)^{1/3}+\left(1+\frac{t_{2}}{|s_{2}^{2}+it_{2}|}\right)^{1/3}\right)^{3/4}$
$\displaystyle\quad\times\left(\left(1-\frac{t_{1}}{|s_{1}^{2}+it_{1}|}\right)^{1/3}+\left(1+\frac{t_{1}}{|s_{1}^{2}+it_{1}|}\right)^{1/3}\right)^{3/4},$

  2. (2)

if $s_{1}\neq 0$ then $\displaystyle
s_{1}\,|s_{2}^{2}+it_{2}|^{1/2}\geq\begin{cases}\displaystyle\frac{2~{}{s_{2}}^{3}}{|s_{2}^{2}+it_{2}|^{3/2}}\,\cos(\theta_{1}-\theta_{2})+2&\hbox{if
}s_{2}\neq 0,\\\ 2&\hbox{if }s_{2}=0.\end{cases}$

  3. (3)

if $s_{2}\neq 0$ then
$\displaystyle|s_{1}^{2}+it_{1}|^{1/2}\,s_{2}\geq\begin{cases}\displaystyle\frac{2~{}{s_{1}}^{3}}{|s_{1}^{2}+it_{1}|^{3/2}}\,\cos(\theta_{1}-\theta_{2})+2&\hbox{if
}s_{1}\neq 0,\\\ 2&\hbox{if }s_{1}=0.\end{cases}$

  4. (4)

if both $s_{1},\,s_{2}\neq 0$ then $s_{1}s_{2}\geq
4\cos^{3}\bigl{(}(\theta_{1}-\theta_{2})/3\bigr{)}$.

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 Section 3 of
[Par2] and it is written down explicitly in Theorem 1.1 of Xie, Wang and Jiang
[XWJ]).

######  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 $|t_{1}|^{1/2}\,|t_{2}|^{1/2}\geq 2$
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 $(s_{1}e^{i\theta_{1}},t_{1})$ and $(s_{2}e^{i\theta_{2}},t_{2})$. By
eliminating $s_{j}$ and $t_{j}$, we can get weakening of Theorem 1.2 as
follows:

######  Theorem 1.4.

Let $(s_{1}e^{i\theta_{1}},t_{1})$ and $(s_{2}e^{i\theta_{2}},t_{2})$ 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. (1’)

$\quad\bigl{|}s_{1}^{2}+it_{1}\bigr{|}^{1/2}\,\bigl{|}s_{2}^{2}+it_{2}\bigr{|}^{1/2}\geq
4$;

  2. (2’)

if $s_{1}\neq 0$ then $\quad s_{1}\,\bigl{|}s_{2}^{2}+it_{2}\bigr{|}^{1/2}\geq
4\cos^{2}\bigl{(}(\theta_{1}-\theta_{2})/2\bigr{)}$;

  3. (3’)

if $s_{2}\neq 0$ then $\quad\bigl{|}s_{1}^{2}+it_{1}\bigr{|}^{1/2}\,s_{2}\geq
4\cos^{2}\bigl{(}(\theta_{1}-\theta_{2})/2\bigr{)}$.

###### Proof.

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

$\bigl{(}(1-x)^{1/3}+(1+x)^{1/3}\bigr{)}^{3/4}\leq 2^{3/4}$

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

$\frac{2s_{2}^{3}}{|s_{2}^{2}+it_{2}|^{3/2}}\,\cos(\theta_{1}-\theta_{2})+2\leq
2\cos(\theta_{1}-\theta_{2})+2=4\cos^{2}\bigl{(}(\theta_{1}-\theta_{2})/2\bigr{)}.$

Thus (2’) follows from (2) and similarly (3’) follows from (3). ∎

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 $-\pi/2\leq\theta_{1}-\theta_{2}\leq\pi/2$ then

$4\cos^{3}\bigl{(}(\theta_{1}-\theta_{2})/3\bigr{)}\geq
4\cos^{2}\bigl{(}(\theta_{1}-\theta_{2})/2\bigr{)}$

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 $\phi$.

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

Therefore

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

The quadratic term is positive when $\cos(\phi)\geq\sqrt{3}/2$ 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

$\displaystyle s_{1}\,|s_{2}^{2}+it_{2}|^{1/2}$ $\displaystyle\geq$
$\displaystyle s_{1}s_{2}$ $\displaystyle\geq$ $\displaystyle
4\cos^{3}\bigl{(}(\theta_{1}-\theta_{2})/3\bigr{)}$ $\displaystyle\geq$
$\displaystyle 4\cos^{2}\bigl{(}(\theta_{1}-\theta_{2})/2\bigr{)}$
$\displaystyle=$ $\displaystyle 2\cos(\theta_{1}-\theta_{2})+2$
$\displaystyle\geq$
$\displaystyle\frac{2s_{2}^{3}}{|s_{2}^{2}+it_{2}|^{3/2}}\cos(\theta_{1}+\theta_{2})+2.$

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

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 ${\bf H}^{2}_{\mathbb{C}}$ and is the fixed point set of an anti-
holomorphic involution $\iota_{R}$ in the isometry group of ${\bf
H}^{2}_{\mathbb{C}}$. A ${\mathbb{C}}$-circle $C$ is the boundary of a totally
geodesic complex line of ${\bf H}^{2}_{\mathbb{C}}$. Thinking of $\partial{\bf
H}^{2}_{\mathbb{C}}$ 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 [Gol]. 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
$d\leq 2r$. 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

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

Observe that $d_{\alpha}(R)\leq 2r$, 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, $d_{\alpha}(R)\geq\sqrt{2}r$, 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 [Gol] 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

$\langle{\bf z},{\bf w}\rangle={\bf w}^{\ast}H{\bf
z}=\bar{w}_{3}z_{1}+\bar{w}_{2}z_{2}+\bar{w}_{1}z_{3},$

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

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

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

$\displaystyle V_{+}$ $\displaystyle=$ $\displaystyle\\{{\bf
z}\in\mathbb{C}^{2,1}:\langle{\bf z},{\bf z}\rangle>0\\},$ $\displaystyle
V_{-}$ $\displaystyle=$ $\displaystyle\\{{\bf
z}\in\mathbb{C}^{2,1}:\langle{\bf z},{\bf z}\rangle<0\\},$ $\displaystyle
V_{0}$ $\displaystyle=$ $\displaystyle\\{{\bf
z}\in\mathbb{C}^{2,1}\setminus\\{{\bf 0}\\}:\langle{\bf z},{\bf
z}\rangle=0\\}.$

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

${\bf H}_{\mathbb{C}}^{2}=\\{(z_{1},z_{2})\in\mathbb{C}^{2}:\
2\Re(z_{1})+|z_{2}|^{2}<0\\}.$

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

$\partial{\bf H}_{\mathbb{C}}^{2}-\\{\infty\\}=\mathbb{P}(V_{0}-\\{{\bf
q}_{\infty}\\})=\\{(z_{1},z_{2})\in\mathbb{C}^{2}:2\Re(z_{1})+|z_{2}|^{2}=0\\}.$

The origin is the point $o\in\partial{\bf
H}^{2}_{\mathbb{C}}={\mathbb{P}}(V_{0})$ with $(z_{1},z_{2})=(0,0)$.

Suppose $z_{1},\,z_{2},\,z_{3}$ are distinct points in $\partial{\bf
H}^{2}_{\mathbb{C}}$ with standard lifts ${\bf z}_{1}$, ${\bf z}_{2}$, ${\bf
z}_{3}$ respectively, we define their Cartan angular invariant
${\mathbb{A}}(z_{1},z_{2},z_{3})$ to be

${\mathbb{A}}(z_{1},z_{2},z_{3})=\arg\bigl{(}-\langle{\bf z}_{1},{\bf
z}_{2}\rangle\langle{\bf z}_{2},{\bf z}_{3}\rangle\langle{\bf z}_{3},{\bf
z}_{1}\rangle\bigr{)}.$

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 $\partial{\bf H}_{\mathbb{C}}^{2}-\\{\infty\\}$ 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 ${\bf H}_{\mathbb{R}}^{3}$
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

$(\zeta_{1},v_{1})\cdot(\zeta_{2},v_{2})=\bigl{(}\zeta_{1}+\zeta_{2},v_{1}+v_{2}+2\Im(\zeta_{1}\bar{\zeta}_{2})\bigr{)}.$

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

${\bf z}=\begin{pmatrix}-|\zeta|^{2}+iv\\\ \sqrt{2}\zeta\\\ 1\end{pmatrix}\in
V_{0}.$

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

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

The map $(\tau,t)$ is a group homomorphism from ${\mathfrak{N}}$ to ${\rm
U}(2,1)$. 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

$\rho_{0}\big{(}(\zeta_{1},v_{1}),(\zeta_{2},v_{2})\big{)}=\big{|}|\zeta_{1}-\zeta_{2}|^{2}-iv_{1}+iv_{2}-2i\Im(\zeta_{1}\bar{\zeta}_{2})\big{|}^{\frac{1}{2}}.$

There is an easy way to compute the Cygan distance. If $z_{1},\,z_{2}$ are two
points in ${\mathfrak{N}}$ with standard lifts ${\bf z}_{1},\,{\bf z}_{2}\in
V_{0}$ respectively then

$\rho_{0}(z_{1},z_{2})=\bigl{|}\langle{\bf z}_{1},{\bf
z}_{2}\rangle\bigr{|}^{1/2}.$

We can define Cygan sphere $S_{(r,z_{0})}$ of radius $r$ and centre
$z_{0}\in\partial{\bf H}_{\mathbb{C}}^{2}$ by

$S_{(r,z_{0})}=\Bigl{\\{}z\in\partial{\bf
H}_{\mathbb{C}}^{2}:\rho_{0}(z,z_{0})=r\Bigr{\\}}.$

###  2.3. Special subsets of complex hyperbolic space and its boundary

There are two types of totally geodesic subspaces of ${\bf
H}_{\mathbb{C}}^{2}$ with real dimension 2. The first is the intersection with
${\bf H}_{\mathbb{C}}^{2}$ 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 ${\bf
H}_{\mathbb{C}}^{2}$. Their intersection with $\partial{\bf
H}_{\mathbb{C}}^{2}$ are called chains or $\mathbb{C}$-circles. The second
type of totally geodesic subspace is the intersection of ${\bf
H}_{\mathbb{C}}^{2}$ 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 ${\bf
H}_{\mathbb{C}}^{2}$. Their intersection with $\partial{\bf
H}_{\mathbb{C}}^{2}$ 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 ${\bf H}_{\mathbb{C}}^{2}$
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 $\partial{\bf
H}_{\mathbb{C}}^{2}={\mathfrak{N}}\cup\\{\infty\\}$.

###  2.4. Isometric spheres

In [Gol], 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 $\mathrm{PU}(2,1)$ which does not fix $\infty$, then
the isometric sphere corresponding to $P$ is given by

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

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

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

Such a map does not fix $\infty$ if and only if $g\neq 0$. 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

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

We will need the following lemma.

######  Lemma 2.1.

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

$\rho_{0}\big{(}P(z),P(\infty)\big{)}=\frac{r_{P}^{2}}{\rho_{0}\big{(}z,P^{-1}(\infty)\big{)}}.$

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

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

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

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

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

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

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

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

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

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 [GP2]. 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

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

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

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

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

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

###  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 $\partial{\bf
H}^{2}_{\mathbb{C}}={\mathfrak{N}}\cup\\{\infty\\}$ 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. (1)

the interiors of $S_{A}^{+},\,S_{A}^{-},\,S_{B}^{+},\,S_{B}^{-}$ are disjoint;

  2. (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. (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 [Par1] or the notion
of a group being compressing, due to Wyss-Gallifent in [WG] and used by
Monaghan, Parker and Pratoussevitch in [MPP].

We can consider how our construction relates to that of Parker and Will [PW].
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

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

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

$\iota_{R}:\left(\begin{matrix}z_{1}\\\ z_{2}\\\
z_{3}\end{matrix}\right)\longmapsto\left(\begin{matrix}\bar{z}_{3}r^{2}\\\
-i\bar{z}_{2}\\\ \bar{z}_{1}/r^{2}\end{matrix}\right).$

It is easy to check that $\langle\iota_{R}{\bf z},\iota_{R}{\bf
w}\rangle=\overline{\langle{\bf z},{\bf w}\rangle}$ for any ${\bf z}$, ${\bf
w}$ 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

${\bf p}_{\alpha.\epsilon}=\left(\begin{matrix}r^{2}ie^{2i\alpha}\\\ \epsilon
r\sqrt{2\sin(2\alpha)}e^{i\alpha}\\\ 1\end{matrix}\right)\in V_{0}.$

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

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

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

$\displaystyle\rho_{0}({p_{\alpha,\epsilon},p_{\theta,\eta}})$
$\displaystyle=$
$\displaystyle\Bigl{|}ir^{2}e^{2i\theta}-ir^{2}e^{-2i\alpha}+2\eta\epsilon
r^{2}\sqrt{\sin(2\alpha)\sin(2\theta)}e^{i\theta-i\alpha}\Bigr{|}^{1/2}$
$\displaystyle=$
$\displaystyle\Bigl{|}-2r^{2}\sin(\theta+\alpha)e^{i\theta-i\alpha}+2\eta\epsilon
r^{2}\sqrt{\sin(2\alpha)\sin(2\theta)}e^{i\theta-i\alpha}\Bigr{|}^{1/2}$
$\displaystyle=$ $\displaystyle
2^{1/2}r\Bigl{(}\sin(\alpha+\theta)-\eta\epsilon\sqrt{\sin(2\alpha)\sin(2\theta)}\Bigr{)}^{1/2}$
$\displaystyle=$ $\displaystyle
2^{1/2}r\Bigl{(}\sin(\alpha)\cos(\theta)+\cos(\alpha)\sin(\theta)-\eta\epsilon
2\sqrt{\sin(\alpha)\cos(\alpha)\sin(\theta)\cos(\theta)}\Bigr{)}^{1/2}$
$\displaystyle=$ $\displaystyle
2^{1/2}r\bigl{|}\sin^{1/2}(\alpha)\cos^{1/2}(\theta)-\eta\epsilon\cos^{1/2}(\alpha)\sin^{1/2}(\theta)\bigr{|}.$

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

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

Then

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

###### Proof.

First observe

$\displaystyle f_{\alpha}(0)$ $\displaystyle=$
$\displaystyle\sin^{1/2}(\alpha)\leq\bigl{(}\sin^{2/3}(\alpha)+\cos^{2/3}(\alpha)\bigr{)}^{3/4},$
$\displaystyle f_{\alpha}(\pi/2)$ $\displaystyle=$
$\displaystyle\cos^{1/2}(\alpha)\leq\bigl{(}\sin^{2/3}(\alpha)+\cos^{2/3}(\alpha)\bigr{)}^{3/4}.$

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

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

Therefore, if $\theta_{0}$ is a value of $\theta$ for which
$f_{\alpha}^{\prime}(\theta_{0})=0$ we have

$\sin^{3/2}(\theta_{0})\sin^{1/2}(\alpha)=\cos^{3/2}(\theta_{0})\cos^{1/2}(\alpha).$

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:

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

Hence:

$\displaystyle\cos^{1/2}(\theta_{0})$ $\displaystyle=$
$\displaystyle\frac{\sin^{1/6}(\alpha)}{\bigl{(}\sin^{2/3}(\alpha)+\cos^{2/3}(\alpha)\bigr{)}^{1/4}},$
$\displaystyle\sin^{1/2}(\theta_{0})$ $\displaystyle=$
$\displaystyle\frac{\cos^{1/6}(\alpha)}{\bigl{(}\sin^{2/3}(\alpha)+\cos^{2/3}(\alpha)\bigr{)}^{1/4}}.$

Therefore

$\displaystyle f_{\alpha}(\theta_{0})$ $\displaystyle=$
$\displaystyle\sin^{1/2}(\alpha)\cos^{1/2}(\theta_{0})+\cos^{1/2}(\alpha)\sin^{1/2}(\theta)$
$\displaystyle=$
$\displaystyle\frac{\sin^{2/3}(\alpha)}{\bigl{(}\sin^{2/3}(\alpha)+\cos^{2/3}(\alpha)\bigr{)}^{1/4}}+\frac{\cos^{2/3}(\alpha)}{\bigl{(}\sin^{2/3}(\alpha)+\cos^{2/3}(\alpha)\bigr{)}^{1/4}}$
$\displaystyle=$
$\displaystyle\bigl{(}\sin^{2/3}(\alpha)+\cos^{2/3}(\alpha)\bigr{)}^{3/4}.$

∎

##  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 Section 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

$d_{B}=\frac{2^{1/4}}{|s_{2}^{2}+it_{2}|^{1/2}}\left(\left(1-\frac{t_{2}}{|s_{2}^{2}+it_{2}|}\right)^{1/3}+\left(1+\frac{t_{2}}{|s_{2}^{2}+it_{2}|}\right)^{1/3}\right)^{3/4}.$

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

$B^{-1}(\infty)=\left(\frac{s_{2}e^{i\theta_{2}}}{s_{2}^{2}+it_{2}},\
\frac{t_{2}}{|s_{2}^{2}+it_{2}|^{2}}\right).$

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

$B(\infty)=\left(\frac{s_{2}e^{i\theta_{2}}}{-s_{2}^{2}+it_{2}},\
\frac{-t_{2}}{|s_{2}^{2}+it_{2}|^{2}}\right).$

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

${\mathbb{A}}(o,B^{-1}(\infty),\infty)=\arg(s_{2}^{2}+it_{2}),\quad{\mathbb{A}}(o,B(\infty),\infty)=\arg(s_{2}^{2}-it_{2}).$

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

$\cos(2\alpha)=\frac{t_{2}}{|s_{2}^{2}+it_{2}|}.$

Hence

$\displaystyle\sin^{2}(\alpha)$ $\displaystyle=$
$\displaystyle\frac{1-\cos(2\alpha)}{2}=\frac{1}{2}\left(1-\frac{t_{2}}{|s_{2}^{2}+it_{2}|}\right),$
$\displaystyle\cos^{2}(\alpha)$ $\displaystyle=$
$\displaystyle\frac{1+\cos(2\alpha)}{2}=\frac{1}{2}\left(1+\frac{t_{2}}{|s_{2}^{2}+it_{2}|}\right).$

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

$\displaystyle d_{B}$ $\displaystyle=$ $\displaystyle
2^{1/2}r_{B}\,\bigl{(}\cos^{2/3}(\alpha)+\sin^{2/3}(\alpha)\bigr{)}^{3/4}$
$\displaystyle=$
$\displaystyle\frac{2^{1/4}}{|s_{2}^{2}+it_{2}|^{1/2}}\left(\left(1-\frac{t_{2}}{|s_{2}^{2}+it_{2}|}\right)^{1/3}+\left(1+\frac{t_{2}}{|s_{2}^{2}+it_{2}|}\right)^{1/3}\right)^{3/4}.$

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 $s_{2}\neq 0$. 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

$B(p_{+})=\iota_{R_{0}}\iota_{R_{+}}(p_{+})=\iota_{R_{0}}(p_{+})=p_{-}.$

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 $p_{-}$. ∎

###  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

$d_{A}=\frac{2^{1/4}}{|s_{1}^{2}+it_{1}|^{1/2}}\left(\left(1-\frac{t_{1}}{|s_{1}^{2}+it_{1}|}\right)^{1/3}+\left(1+\frac{t_{1}}{|s_{1}^{2}+it_{1}|}\right)^{1/3}\right)^{3/4}.$

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
$d_{B}\leq 1/d_{A}$ 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 $d_{B}\leq 1/d_{A}$ is equivalent to

$1\geq
d_{A}d_{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}\\\
$

$\times\left(\left(1-\frac{t_{1}}{|s_{1}^{2}+it_{1}|}\right)^{1/3}+\left(1+\frac{t_{1}}{|s_{1}^{2}+it_{1}|}\right)^{1/3}\right)^{3/4}.$

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. ∎

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.

$1\geq\frac{2}{|s_{1}^{2}+it_{1}|^{1/2}}\,\frac{2}{|s_{2}^{2}+it_{2}|^{1/2}}.$

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 $t_{1}t_{2}=4$ or (b) $t_{1}=t_{2}=0$,
$\theta_{1}=\theta_{2}$ and $s_{1}s_{2}=4$. In the first case $AB$ is screw
parabolic with angle $\pi$ and in the second case $AB$ is unipotent.

Write $-s_{2}^{2}+it_{2}=ir_{2}^{2}e^{2i\alpha_{2}}$. Define $\phi_{2}$ by

$\cos(\phi_{2})=\frac{\sin^{1/3}(\alpha_{2})}{\bigl{(}\sin^{2/3}(\alpha_{2})+\cos^{2/3}(\alpha_{2})\bigr{)}^{1/2}},\quad\sin(\phi_{2})=\frac{\cos^{1/3}(\alpha_{2})}{\bigl{(}\sin^{2/3}(\alpha_{2})+\cos^{2/3}(\alpha_{2})\bigr{)}^{1/2}}$

and $d_{B}$ by

$d_{B}=r_{2}^{-1}\bigl{(}\sqrt{2\cos(\phi_{2})\sin(\alpha_{2})}+\sqrt{2\sin(\phi_{2})\cos(\alpha_{2})}\bigr{)}.$

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

$T_{+}^{-1}(o)=\left(\begin{matrix}-ir_{2}^{-2}e^{-2i\alpha_{2}}\\\
i\sqrt{2\sin(2\alpha_{2})}r_{2}^{-1}e^{2i\alpha_{2}+i\theta_{2}}\\\
1\end{matrix}\right),\quad
T_{-}^{-1}(o)=\left(\begin{matrix}ir_{2}^{-2}e^{2i\alpha_{2}}\\\
i\sqrt{2\sin(2\alpha_{2})}r_{2}^{-1}e^{-2i\alpha_{2}+i\theta_{2}}\\\
1\end{matrix}\right).$

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

$\left(\begin{matrix}-ir_{2}^{-2}e^{-2i\phi_{2}}\\\
-i\sqrt{2\sin(2\phi_{2})}r_{2}^{-1}e^{-i\phi_{2}+3i\alpha_{2}+i\theta_{2}}\\\
1\end{matrix}\right),\quad\left(\begin{matrix}ir_{2}^{-2}e^{2i\phi_{2}}\\\
i\sqrt{2\sin(2\phi_{2})}r_{2}^{-1}e^{i\phi_{2}-3i\alpha_{2}+i\theta_{2}}\\\
1\end{matrix}\right).$

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

${\bf p}_{+}=\left(\begin{matrix}-d_{B}^{2}e^{-i\phi_{2}+i\alpha_{2}}\\\
d_{B}\sqrt{2\cos(\phi_{2}-\alpha_{2})}e^{-2i\phi_{2}+2i\alpha_{2}+i\theta_{2}}\\\
1\end{matrix}\right),\quad{\bf
p}_{-}=\left(\begin{matrix}-d_{B}^{2}e^{i\phi_{2}-i\alpha_{2}}\\\
-d_{B}\sqrt{2\cos(\phi_{2}-\alpha_{2})}e^{2i\phi_{2}-2i\alpha_{2}+i\theta_{2}}\\\
1\end{matrix}\right).$

A similar caculation shows that

$B:{\bf p}_{+}\longmapsto e^{-2i\phi_{2}+2i\alpha_{2}}{\bf p}_{-}.$

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

$\iota{\bf q}_{+}=\left(\begin{matrix}-d_{A}^{2}e^{-i\phi_{1}+i\alpha_{1}}\\\
d_{A}\sqrt{2\cos(\phi_{1}-\alpha_{1})}e^{-2i\phi_{1}+2i\alpha_{1}+i\theta_{1}}\\\
1\end{matrix}\right),\quad\iota{\bf
q}_{-}=\left(\begin{matrix}-d_{A}^{2}e^{i\phi_{1}-i\alpha_{1}}\\\
-d_{A}\sqrt{2\cos(\phi_{1}-\alpha_{1})}e^{2i\phi_{1}-2i\alpha_{1}+i\theta_{1}}\\\
1\end{matrix}\right),$

where

$\displaystyle\cos(\phi_{1})$ $\displaystyle=$
$\displaystyle\frac{\sin^{1/3}(\alpha_{1})}{\bigl{(}\sin^{2/3}(\alpha_{1})+\cos^{2/3}(\alpha_{1})\bigr{)}^{1/2}},\quad\sin(\phi_{1})\
=\
\frac{\cos^{1/3}(\alpha_{1})}{\bigl{(}\sin^{2/3}(\alpha_{1})+\cos^{2/3}(\alpha_{1})\bigr{)}^{1/2}},$
$\displaystyle d_{A}$ $\displaystyle=$ $\displaystyle
r_{1}^{-1}\bigl{(}\sqrt{2\cos(\phi_{1})\sin(\alpha_{1})}+\sqrt{2\sin(\phi_{1})\cos(\alpha_{1})}\bigr{)}.$

Applying $\iota$ we see that

$\displaystyle{\bf q}_{+}$ $\displaystyle=$
$\displaystyle\left(\begin{matrix}1\\\
d_{A}\sqrt{2\cos(\phi_{1}-\alpha_{1})}e^{-2i\phi_{1}+2i\alpha_{1}+i\theta_{1}}\\\
-d_{A}^{2}e^{-i\phi_{1}+i\alpha_{1}}\end{matrix}\right)\sim\left(\begin{matrix}-d_{A}^{-2}e^{i\phi_{1}-i\alpha_{1}}\\\
-d_{A}^{-1}\sqrt{2\cos(\phi_{1}-\alpha_{1})}e^{-i\phi_{1}+i\alpha_{1}+i\theta_{1}}\\\
1\end{matrix}\right),$ $\displaystyle{\bf q}_{-}$ $\displaystyle=$
$\displaystyle\left(\begin{matrix}1\\\
-d_{A}\sqrt{2\cos(\phi_{1}-\alpha_{1})}e^{2i\phi_{1}-2i\alpha_{1}+i\theta_{1}}\\\
-d_{A}^{2}e^{i\phi_{1}-i\alpha_{1}}\end{matrix}\right)\sim\left(\begin{matrix}-d_{A}^{-2}e^{-i\phi_{1}+i\alpha_{1}}\\\
d_{A}^{-1}\sqrt{2\cos(\phi_{1}-\alpha_{1})}e^{i\phi_{1}-i\alpha_{1}+i\theta_{1}}\\\
1\end{matrix}\right).$

Thus we have

$A:{\bf q}_{+}\longmapsto{\bf q}_{-}.$

Hence, $A$ sends the vector ${\bf q}_{+}$ to the vector ${\bf q}_{-}$ with a
multiplier $1$. Now, ${\bf p}_{+}={\bf q}_{-}$ and ${\bf p}_{-}={\bf q}_{+}$
if and only if either (a) $s_{1}=s_{2}=0$ and $t_{1}t_{2}=4$ or (b)
$t_{1}=t_{2}=0$, $\theta_{1}=\theta_{2}$ and $s_{1}s_{2}=4$. 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.

$\displaystyle\left(\begin{matrix}1&-\sqrt{2}s_{1}e^{-i\theta_{1}}&-s_{1}^{2}+it_{1}\\\
0&1&\sqrt{2}s_{1}e^{i\theta_{1}}\\\
0&0&1\end{matrix}\right)\left(\begin{matrix}-k^{2}-a^{2}+ib-2ika\\\
\sqrt{2}(k+ia)e^{i\theta_{1}}\\\ 1\end{matrix}\right)$ $\displaystyle=$
$\displaystyle\left(\begin{matrix}-(k+s_{1})^{2}-a^{2}+ib-2i(k+s_{1})a+it_{1}\\\
\sqrt{2}(k+s_{1}+ia)e^{i\theta_{1}}\\\ 1\end{matrix}\right).$

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

$x\cos(\theta_{1})+y\sin(\theta_{1})=k,\quad
x\cos(\theta_{1})+y\sin(\theta_{1})=k+s_{1}.$

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

$S_{A}(k)=\left\\{(x,y)\ :\ k\leq x\cos(\theta_{1})+y\sin(\theta_{1})\leq
k+s_{1}\right\\}.$

Figure 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 $(s_{1}e^{i\theta_{1}},t_{1})$ and $(s_{2}e^{i\theta_{2}},t_{2})$ 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
$-\pi/2\leq(\theta_{1}-\theta_{2})\leq\pi/2$. Let $A$ and $B$ given by (1) be
the associated Heisenberg translations fixing $\infty$ and $o$ respectively.
Suppose that

$s_{1}\,|s_{2}^{2}+it_{2}|^{1/2}\geq\frac{2~{}s_{2}}{|s_{2}^{2}+it_{2}|^{1/2}}\,\cos(\theta_{1}-\theta_{2})+2.$

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,

$\frac{s_{2}e^{i\theta_{2}}}{s_{2}^{2}+it_{2}}=\frac{s_{2}e^{i\theta_{2}}(s_{2}^{2}-it_{2})}{|s_{2}^{2}+it_{2}|^{2}}.$

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,

$\frac{s_{2}e^{i\theta_{2}}}{-s_{2}^{2}+it_{2}}=\frac{-s_{2}e^{i\theta_{2}}(s_{2}^{2}+it_{2})}{|s_{2}^{2}+it_{2}|^{2}}.$

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

$\displaystyle x$ $\displaystyle=$
$\displaystyle\frac{s_{2}(s_{2}^{2}\cos(\theta_{2})+t_{2}\sin(\theta_{2}))}{|s_{2}^{2}+it_{2}|^{2}}+\frac{\cos(\phi)}{|s_{2}^{2}+it_{2}|^{1/2}},$
$\displaystyle y$ $\displaystyle=$
$\displaystyle\frac{s_{2}(s_{2}^{2}\sin(\theta_{2})-t_{2}\cos(\theta_{2}))}{|s_{2}^{2}+it_{2}|^{2}}+\frac{\sin(\phi)}{|s_{2}^{2}+it_{2}|^{1/2}}.$

Therefore

$x\cos(\theta_{1})+y\sin(\theta_{1})=\frac{s_{2}^{3}\cos(\theta_{1}-\theta_{2})-s_{2}t_{2}\sin(\theta_{1}-\theta_{2})}{|s_{2}^{2}+it_{2}|^{2}}+\frac{\cos(\phi-\theta_{1})}{|s_{2}^{2}+it_{2}|^{1/2}}.$

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

$x\cos(\theta_{1})+y\sin(\theta_{1})=\frac{-s_{2}^{3}\cos(\theta_{1}-\theta_{2})-s_{2}t_{2}\sin(\theta_{1}-\theta_{2})}{|s_{2}^{2}+it_{2}|^{2}}+\frac{\cos(\phi-\theta_{1})}{|s_{2}^{2}+it_{2}|^{1/2}}.$

As $\cos(\theta_{1}-\theta_{2})\geq 0$, we can see that for all points $(x,y)$
in the vertical projections of the isometric spheres of $B$ and $B^{-1}$ we
have

$\displaystyle\frac{-s_{2}^{3}\cos(\theta_{1}-\theta_{2})}{|s_{2}^{2}+it_{2}|^{2}}-\frac{1}{|s_{2}^{2}+it_{2}|^{1/2}}-\frac{s_{2}t_{2}\sin(\theta_{1}-\theta_{2})}{|s_{2}^{2}+it_{2}|^{2}}$
$\displaystyle\leq$ $\displaystyle x\cos(\theta_{1})+y\sin(\theta_{1})$
$\displaystyle\leq$
$\displaystyle\frac{s_{2}^{3}\cos(\theta_{1}-\theta_{2})}{|s_{2}^{2}+it_{2}|^{2}}+\frac{1}{|s_{2}^{2}+it_{2}|^{1/2}}-\frac{s_{2}t_{2}\sin(\theta_{1}-\theta_{2})}{|s_{2}^{2}+it_{2}|^{2}}.$

Hence these vertical projections lie in a strip of the form $S_{A}(k)$
provided

$\frac{2s_{2}^{3}\cos(\theta_{1}-\theta_{2})}{|s_{2}^{2}+it_{2}|^{2}}+\frac{2}{|s_{2}^{2}+it_{2}|^{1/2}}\leq
s_{1}.$

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

$s_{1}\,|s_{2}^{2}+it_{2}|^{1/2}\geq\frac{2~{}s_{2}}{|s_{2}^{2}+it_{2}|^{1/2}}\,\cos(\theta_{1}-\theta_{2})+2.$

∎

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 $(s_{1}e^{i\theta_{1}},t_{1})$ 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

$s_{1}\,|t_{2}|^{1/2}\geq 2.$

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

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

Hence

$\frac{-1}{|t_{2}|^{1/2}}\leq
x\cos(\theta_{1})+y\sin(\theta_{1})=\frac{\cos(\phi-\theta_{1})}{|t_{2}|^{1/2}}\leq\frac{1}{|t_{2}|^{1/2}}.$

For these two Cygan spheres to lie in a strip of the form $S_{A}(k)$ we must
have $2/|t_{2}|^{1/2}\leq s_{1}$. This gives the result. ∎

###  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

$\displaystyle{\bf q}_{+}$ $\displaystyle=$
$\displaystyle\left(\begin{matrix}-1/|s_{2}^{2}+it_{2}|-2s_{2}(s_{2}^{2}+it_{2})e^{i\theta_{1}-i\theta_{2}}/|s_{2}^{2}+it_{2}|^{5/2}-(s_{2}^{2}-it_{2})/|s_{2}^{2}+it_{2}|^{2}\\\
\sqrt{2}s_{2}(s_{2}^{2}-it_{2})e^{i\theta_{2}}/|s_{2}^{2}+it_{2}|^{2}+\sqrt{2}e^{i\theta_{1}}/|s_{2}^{2}+it_{2}|^{1/2}\\\
1\end{matrix}\right),$ $\displaystyle{\bf q}_{-}$ $\displaystyle=$
$\displaystyle\left(\begin{matrix}-1/|s_{2}^{2}+it_{2}|-2s_{2}(s_{2}^{2}-it_{2})e^{i\theta_{1}-i\theta_{2}}/|s_{2}^{2}+it_{2}|^{5/2}-(s_{2}^{2}+it_{2})/|s_{2}^{2}+it_{2}|^{2}\\\
-\sqrt{2}s_{2}(s_{2}^{2}+it_{2})e^{i\theta_{2}}/|s_{2}^{2}+it_{2}|^{2}-\sqrt{2}e^{i\theta_{1}}/|s_{2}^{2}+it_{2}|^{1/2}\\\
1\end{matrix}\right).$

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

$\displaystyle B{\bf q}_{+}$ $\displaystyle=$
$\displaystyle\left(\begin{matrix}-1/|s_{2}^{2}+it_{2}|-2s_{2}(s_{2}^{2}+it_{2})e^{i\theta_{1}-i\theta_{2}}/|s_{2}^{2}+it_{2}|^{5/2}-(s_{2}^{2}-it_{2})/|s_{2}^{2}+it_{2}|^{2}\\\
\sqrt{2}s_{2}e^{i\theta_{2}}/|s_{2}^{2}+it_{2}|+\sqrt{2}(s_{2}^{2}+it_{2})^{2}e^{i\theta_{1}}/|s_{2}^{2}+it_{2}|^{5/2}\\\
(s_{2}^{2}-it_{2})/|s_{2}^{2}+it_{2}|\end{matrix}\right)$ $\displaystyle=$
$\displaystyle\frac{s_{2}^{2}-it_{2}}{|s_{2}^{2}+it_{2}|}\left(\begin{matrix}-1/|s_{2}^{2}+it_{2}|-2s_{2}(s_{2}^{2}+it_{2})^{2}e^{i\theta_{1}-i\theta_{2}}/|s_{2}^{2}+it_{2}|^{7/2}-(s_{2}^{2}+it_{2})/|s_{2}^{2}+it_{2}|^{2}\\\
-\sqrt{2}s_{2}(s_{2}^{2}+it_{2})e^{i\theta_{2}}/|s_{2}^{2}+it_{2}|^{2}-\sqrt{2}(s_{2}^{2}+it_{2})^{3}e^{i\theta_{1}}/|s_{2}^{2}+it_{2}|^{7/2}\\\
1\end{matrix}\right).$

This shows that, $B$ projectively maps ${\bf q}_{+}$ to ${\bf q}_{-}$ if and
only if $(s_{2}^{2}+it_{2})^{3}=|s_{2}^{2}+it_{2}|^{3}$ which is equivalent to
$t_{2}=0$ as $s_{2}^{2}\geq 0$. Similarly, $A{\bf q}_{-}={\bf q}_{+}$ if and
only if both

$\displaystyle(s_{1}^{2}-it_{1})|s_{2}^{2}+it_{2}|^{1/2}$ $\displaystyle=$
$\displaystyle
2s_{1}+\frac{2s_{1}s_{2}(s_{2}^{2}+it_{2})e^{i\theta_{2}-i\theta_{1}}}{|s_{2}^{2}+it_{2}|^{3/2}}+\frac{4s_{2}it_{2}e^{i\theta_{1}-\theta_{2}}}{|s_{2}^{2}+it_{2}|^{2}}-\frac{2it_{2}}{|s_{2}^{2}+it_{2}|^{3/2}},$
$\displaystyle s_{1}|s_{2}^{2}+it_{2}|^{1/2}$ $\displaystyle=$ $\displaystyle
2+\frac{2s_{2}^{3}e^{i\theta_{2}-i\theta_{1}}}{|s_{2}^{2}+it_{2}|^{3/2}}.$

Thus, ${\bf q}_{-}$ is mapped to ${\bf q}_{+}$ by $A$ only when
$t_{1}=t_{2}=0$, $\theta_{1}=\theta_{1}$ and $s_{1}s_{2}=4$.

###  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

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

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

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

######  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

$r\leq\frac{1-\cos(\theta)}{2k}\quad\hbox{ and }\quad
r\leq\frac{1+\cos(\theta)}{2k}.$

In Cartesian coordinates, these are

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

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

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

Therefore, the Jacobian vanishes precisely when

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

We can parameterise this curve using $a$ as:

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

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

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

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

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

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

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

and so $2kr=1+\cos(\theta)$. ∎

Figure 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

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

######  Lemma 5.4.

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

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

###### Proof.

(See Figure 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

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

Consider

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

Differentiating with respect to $\theta$ we have

$\displaystyle f^{\prime}_{\phi}(\theta)$ $\displaystyle=$
$\displaystyle\frac{-(1-\cos(\theta))\sin(\theta-\phi)+\sin(\theta)\cos(\theta-\phi)}{2}$
$\displaystyle=$ $\displaystyle\sin(\theta/2)\cos(3\theta/2-\phi).$

Since $0<\theta/2<\pi$ we see $\sin(\theta/2)\neq 0$ 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

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

Since $-\pi/2\leq\phi\leq\pi/2$ 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

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

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

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

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

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

∎

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:

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

###### 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
$F_{{\frac{s_{1}}{2}}e^{i\theta_{1}}}^{(\infty)}$:

$\left(\begin{matrix}1&-\sqrt{2}s_{1}e^{-i\theta_{1}}&-s_{1}^{2}+it_{1}\\\
0&1&\sqrt{2}s_{1}e^{i\theta_{1}}\\\
0&0&1\end{matrix}\right)\left(\begin{matrix}-\frac{s_{1}^{2}}{4}-a^{2}+ib-
is_{1}a\\\ \sqrt{2}(-\frac{s_{1}}{2}+ia)e^{i\theta_{1}}\\\
1\end{matrix}\right)=\left(\begin{matrix}-\frac{s_{1}^{2}}{4}-a^{2}+ib-
ias_{1}+it_{1}\\\ \sqrt{2}(\frac{s_{1}}{2}+ia)e^{i\theta_{1}}\\\
1\end{matrix}\right)$

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

$D_{A}=\left\\{(x+iy,v)\in{\mathfrak{N}}\ :\
\bigl{|}x\cos(\theta_{1})+y\cos(\theta_{1})\bigr{|}\leq\frac{s_{1}}{2}\right\\}\cup\\{\infty\\}.$

The image of $D_{A}$ under vertical projection is the strip

$S_{A}=\left\\{(x,y)\ :\
\bigl{|}x\cos(\theta_{1})+y\cos(\theta_{1})\bigr{|}\leq\frac{s_{1}}{2}\right\\}.$

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

$\left(\begin{matrix}1&0&0\\\ \sqrt{2}s_{2}&1&0\\\
-s_{2}^{2}+it_{2}&-\sqrt{2}s_{2}&1\end{matrix}\right)\left(\begin{matrix}1\\\
\sqrt{2}(-\frac{s_{2}}{2}+ia)\\\ -\frac{s_{2}^{2}}{4}-a^{2}+ib-
is_{1}a\end{matrix}\right)=\left(\begin{matrix}1\\\
\sqrt{2}(\frac{s_{2}}{2}+ia)\\\ -\frac{s_{2}^{2}}{4}-a^{2}+ib-
ias_{2}+it_{2}\end{matrix}\right).$

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

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

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

$r\leq\frac{1-\cos(\theta)}{s_{2}},\quad r\leq\frac{1-\cos(\theta)}{-s_{2}}.$

Using Lemma 5.4 with $\phi=\theta_{1}$ and $k=s_{2}/2$ (see Figure 2) we see
that this is true provided

$\frac{\cos^{3}(\frac{\theta_{1}}{3})}{s_{2}/2}\leq\frac{s_{1}}{2}.$

The result follows. ∎

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 $s_{1}s_{2}=4$, $\theta_{1}=0$ and $t_{1}=0$.

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