The configuration space of equidistant triples in the Heisenberg group

TL;DR

This paper characterizes the space of all equidistant triples in the Heisenberg group with the Korányi metric, showing it forms a specific hypersurface in three-dimensional real space.

Contribution

It establishes an explicit isomorphism between the configuration space of equidistant triples and a hypersurface in -dimensional real space, providing a geometric description.

Findings

Configuration space is isomorphic to a hypersurface in D

Provides a geometric description of equidistant triples

Advances understanding of the Heisenberg group's metric geometry

Abstract

We prove that the configuration space of equidistant triples on the Heisenberg group equipped with the Kor\'anyi metric, is isomorphic to a hypersurface of $\mathbb{R}^3$.

Equations

$$d(p_1,p_2)=d(p_2,p_3)=d(p_3,p_1).$$

$$G=\mathrm{Sim}(\mathfrak{H},d)=\mathfrak{H}\times \mathbb{R}\times {\mathbb{R}}_{+}^{\ast },$$

$$\left(g,(p_1,p_2,p_3)\right)\mapsto \left(g(p_1),g(p_2),g(p_3)\right).$$

$$\mathcal{E}=\left\{(a,b,c)\in [-2\pi /3,2\pi /3{]}^3|\cos a+\cos b+\cos c=\frac{3}{2}\right\}.$$

$$⟨\mathbf{z},\mathbf{w}⟩=z_1{\overline{w}}_3+z_2{\overline{w}}_2+z_3{\overline{w}}_1.$$

$$V_{-}=\Bigl{\{}{\bf z}\in{\mathbb{C}}^{2,1}\ :\ \langle{\bf z},\,{\bf z}\rangle<0\Bigr{\}},\quad V_{0}=\Bigl{\{}{\bf z}\in{\mathbb{C}}^{2,1}\setminus\{{\bf 0}\}\ :\ \langle{\bf z},\,{\bf z}\rangle=0\Bigr{\}}.$$

$${\mathbf{H}}_{\mathbb{C}}^2=\left\{(z_1,z_2)\in {\mathbb{C}}^2:2\Re (z_1)+|z_2{|}^2<0\right\},$$

$$\mathbf{o}=\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right],\infty =\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right].$$

$$\partial {\mathbf{H}}_{\mathbb{C}}^2\setminus \{\infty \}=\left\{(z_1,z_2)\in {\mathbb{C}}^2:2\Re (z_1)+|z_2{|}^2=0\right\},$$

$$\mathbf{z}=\left[\begin{array}{c}{z}_1\\ z_2\\ 1\end{array}\right].$$

$$\mathrm{PU}(2,1)=\mathrm{SU}(2,1)/\{I,\omega I,{\omega }^{2}I\},$$

$$\mathbf{z}=\left[\begin{array}{c}-|z{|}^2+it\\ \sqrt{2}z\\ 1\end{array}\right].$$

$$(z,t)\star (w,s)=(z+w,t+s+2\Im (z\overline{w})).$$

$${\left|(z,t)\right|}_{\mathfrak{H}}={\left|\mathcal{A}(z,t)\right|}^{1/2},\text{where}\mathcal{A}(z,t)=|z{|}^2-it.$$

$$d\left((z,t),(w,s)\right)={\left|(z,t{)}^{-1}\star (w,s)\right|}_{\mathfrak{H}}.$$

$$(z,t)\mapsto \left(\frac{z}{-|z{|}^2+it},-\frac{t}{|-|z{|}^2+it{|}^2|}\right),$$

$$d_{\mathfrak{H}}(R(p),R(p^{\prime }))=\frac{d_{\mathfrak{H}}(p,p^{\prime })}{d_{\mathfrak{H}}(p,o)d_{\mathfrak{H}}(p^{\prime },o)}.$$

$$\mathbb{A}(p_1,p_2,p_3)=\arg (-⟨{\mathbf{p}}_1,{\mathbf{p}}_2⟩⟨{\mathbf{p}}_2,{\mathbf{p}}_3⟩⟨{\mathbf{p}}_3,{\mathbf{p}}_1⟩),$$

$$\mathbb{X}(p_1,p_2,p_3,p_4)=\frac{⟨{\mathbf{p}}_3,{\mathbf{p}}_1⟩⟨{\mathbf{p}}_4,{\mathbf{p}}_2⟩}{⟨{\mathbf{p}}_4,{\mathbf{p}}_1⟩⟨{\mathbf{p}}_3,{\mathbf{p}}_2⟩},$$

$$|\mathbb{X}(p_1,p_2,p_3,p_4){|}^{1/2}=\frac{d_{\mathfrak{H}}(p_4,p_2)\cdot d_{\mathfrak{H}}(p_3,p_1)}{d_{\mathfrak{H}}(p_4,p_1)\cdot d_{\mathfrak{H}}(p_3,p_2)}.$$

$${\mathbb{X}}_1(\mathfrak{p})=\mathbb{X}(p_1,p_2,p_3,p_4),{\mathbb{X}}_2(\mathfrak{p})=\mathbb{X}(p_1,p_3,p_2,p_4),{\mathbb{X}}_3(\mathfrak{p})=\mathbb{X}(p_2,p_3,p_1,p_4).$$

$$|{\mathbb{X}}_3{|}^2=|{\mathbb{X}}_2{|}^2/|{\mathbb{X}}_1{|}^2,$$

$$2|{\mathbb{X}}_1{|}^2\Re ({\mathbb{X}}_3)=|{\mathbb{X}}_1{|}^2+|{\mathbb{X}}_2{|}^2-2\Re ({\mathbb{X}}_1)-2\Re ({\mathbb{X}}_2)+1.$$

$${\mathfrak{X}}_{\mathbb{R}}=\{({\mathbb{X}}_1,{\mathbb{X}}_2)\in {\mathbb{R}}_{\ast }^2|{\mathbb{X}}_1+{\mathbb{X}}_2=1\},$$

$$a={\mathbb{A}}_1-{\mathbb{A}}_2,b=-{\mathbb{A}}_2-{\mathbb{A}}_4,c={\mathbb{A}}_4-{\mathbb{A}}_1.$$

$${\mathbb{A}}_1=\mathbb{A}(p_2,p_3,p_4),{\mathbb{A}}_2=\mathbb{A}(p_1,p_3,p_4),{\mathbb{A}}_4=\mathbb{A}(p_1,p_2,p_3).$$

$${\mathbb{A}}_3+{\mathbb{A}}_1={\mathbb{A}}_2+{\mathbb{A}}_4.$$

$$a=\arg ({\mathbb{X}}_1(\mathfrak{p})),b=\arg ({\mathbb{X}}_2(\mathfrak{p})),c=\arg ({\mathbb{X}}_3(\mathfrak{p})).$$

$$p_1=(0,0),p_2=(0,t),p_3=(0,s),t,s\in \mathbb{R}.$$

$$|t{|}^{1/2}=|s{|}^{1/2}=|t-s{|}^{1/2},$$

$$|{\mathbb{X}}_1(\mathfrak{p})|=|{\mathbb{X}}_2(\mathfrak{p})|=1.$$

Full text

1. Introduction

Let ${{\mathfrak{H}}}$ be the first Heisenberg group equipped with the Korányi distance $d$. An equidistant triple is a triple of points $P=(p_{1},p_{2},p_{3})\in{{\mathfrak{H}}}$ such that

[TABLE]

Denote by ${{\mathcal{E}}{\mathcal{T}}}$ the space of equidistant triples in ${{\mathfrak{H}}}$. Then the similarity group

[TABLE]

acts diagonally on ${{\mathcal{E}}{\mathcal{T}}}$:

[TABLE]

We denote by ${{\mathfrak{E}}{\mathfrak{T}}}$ the quotient of this action; this is the configuration space of $G$-equivalent equidistant triples in ${{\mathfrak{H}}}$. In this paper we are dealing with the problem of parametrising ${{\mathfrak{E}}{\mathfrak{T}}}$. The problem is addressed and solved in a different manner in [1] (see Proposition 4.6 there). We prove here the following theorem:

Theorem 1.1**.**

The configuration space ${{\mathfrak{E}}{\mathfrak{T}}}$ of $G$-equivalent equidistant triples is in bijection with the hypersurface ${{\mathcal{E}}}$ of ${{\mathbb{R}}}^{3}$ which is defined by

[TABLE]

The proof of Theorem 1.1 relies upon the use of the cross-ratio variety ${{\mathfrak{X}}}$; this is a 4-dimensional variety parametrising the ${\rm PU}(2,1)$-configuration space of pairwise distinct quadruples on the boundary $\partial{{\bf H}}^{2}_{{\mathbb{C}}}$ of complex hyperbolic plane. In fact, ${{\mathfrak{E}}{\mathfrak{T}}}$ may be viewed as a 2-dimensional subvariety of ${{\mathfrak{X}}}$.

This paper is organised as follows: In Section 2 we review standard facts about complex hyperbolic plane and the Heisenberg group, as well as about cross-ratio variety. In Section 3 we prove Theorem 1.1 and discuss the particular case of equidistant triples lying in a ${{\mathbb{C}}}$-circle in Section 3.4.

Acknowledgements. I wish to thank Vassilis Chousionis for suggesting the problem to me, and also Viktor Schroeder for fruitful discussions.

2. Preliminaries

The material of this section is standard; a general reference is Goldman’s book, [5]. In Section 2.1 we review complex hyperbolic plane, its boundary and the Heisenberg group. Cartan’s angular invariant and complex cross-ratios are in Section 2.2. Finally, a brief overview of the cross-ratio variety and the ${\rm PU}(2,1)$-configuration of four pairwise distinct points in the boundary of complex hyperbolic plane is found in Section 2.3.

2.1. Complex hyperbolic plane and Heisenberg group

We consider the vector space $\mathbb{C}^{2,1}$, that is, $\mathbb{C}^{3}$ with the Hermitian form of signature $(2,1)$ given by

[TABLE]

We next consider the following subspaces of ${\mathbb{C}}^{2,1}$:

[TABLE]

Denote by ${\mathbb{P}}:{\mathbb{C}}^{2,1}\setminus\{0\}\longrightarrow{\mathbb{C}}P^{2}$ the canonical projection onto complex projective space. Then the complex hyperbolic plane ${\bf H}_{\mathbb{C}}^{2}$ is defined to be ${\mathbb{P}}V_{-}$ and its boundary $\partial{\bf H}^{2}_{\mathbb{C}}$ is ${\mathbb{P}}V_{0}$. Hence we have

[TABLE]

and in this manner, ${\bf H}^{2}_{\mathbb{C}}$ is the Siegel domain in ${\mathbb{C}}^{2}$.

There are two distinguished points in $V_{0}$ which we denote by ${\bf o}$ and ${{\bf\infty}}$:

[TABLE]

Let ${\mathbb{P}}{\bf o}=o$ and ${\mathbb{P}}{{\bf\infty}}=\infty$. Then

[TABLE]

and in particular, $o=(0,0)\in{{\mathbb{C}}}^{2}$.

Conversely, if we are given a point $z=(z_{1},z_{2})$ of ${\mathbb{C}}^{2}$, then the point

[TABLE]

is called the standard lift of $z$. Therefore the standard lifts of points of the complex hyperbolic plane and its boundary (except the point at infinity) are vectors of $V_{-}$ and $V_{0}$ respectively with the third inhomogeneous coordinate equal to 1.

Complex hyperbolic plane ${\bf H}_{\mathbb{C}}^{2}$ is a Kähler manifold; its Kähler structure is given by the Bergman metric. The holomorphic sectional curvature equals to $-1$ and its real sectional curvature is pinched between $-1$ and $-1/4$. The full group of holomorphic isometries is the projective unitary group

[TABLE]

where $\omega$ is a non-real cube root of unity (that is ${\rm SU}(2,1)$ is a 3-fold covering of ${\rm PU}(2,1)$). There are two ways (up to ${\rm PU}(2,1)$ conjugacy) to embed real hyperbolic plane into complex hyperbolic plane; that is, as ${{\bf H}}^{1}_{{\mathbb{C}}}$ as well as ${{\bf H}}^{2}_{{\mathbb{R}}}$. These embeddings give rise to complex lines, i.e., isometric images of the embedding of ${{\bf H}}^{1}_{{\mathbb{C}}}$ into ${{\bf H}}^{2}_{{\mathbb{C}}}$ and Lagrangian planes, i.e., isometric images of ${{\bf H}}^{2}_{{\mathbb{R}}}$ into ${{\bf H}}^{2}_{{\mathbb{C}}}$, respectively.

There is an identification of the boundary of the Siegel domain with the one point compactification of ${{\mathbb{C}}}\times{{\mathbb{R}}}$: A finite point $z$ in the boundary of the Siegel domain has a standard lift of the form

[TABLE]

The unipotent stabiliser at infinity acts simply transitively and gives the set of these points the structure of a 2–step nilpotent Lie group, namely the Heisenberg group ${{\mathfrak{H}}}$. This is ${{\mathbb{C}}}\times{{\mathbb{R}}}$ with group law:

[TABLE]

The Heisenberg norm (Korányi gauge) is given by

[TABLE]

From this norm arises a metric, the Korányi-Cygan (K-C) metric, on ${{\mathfrak{H}}}$ by the relation

[TABLE]

The K-C metric is invariant under

1. a)

left-actions $L_{(w,s)}$ of ${{\mathfrak{H}}}$,   $(z,t)\to(w,s)\star(z,t)$,   $(w,s)\in{{\mathfrak{H}}}$; 2. b)

rotations $R_{\phi}$,   $(z,t)\mapsto(ze^{i\phi},t)$,   $\phi\in{{\mathbb{R}}}$;

1. c)

involution $j$, $j(z,t)=(\overline{z},-t)$.

These form the group ${\rm Isom}({{\mathfrak{H}}},d)$ of Heisenberg isometries. Note that all the above are orientation-preserving.

The K-C metric is also scaled up to multiplicative constants by the action of

1. d)

Heisenberg dilations $D_{r}$, $(z,t)\mapsto$ $(rz,r^{2}t)$, $r\in{{\mathbb{R}}}_{+}^{*}.$

and there is also an inversion, defined for each $p=(z,t)\in{{\mathfrak{H}}}$, $p\neq o$, by

[TABLE]

which satisfies

[TABLE]

The similarity group $G={\rm Sim}({{\mathfrak{H}}},d)$ comprises compositions of maps of the form a), b), d). Clearly, $G={{\mathfrak{H}}}\times{{\mathbb{R}}}\times{{\mathbb{R}}}^{*}_{+}$.

2.1.1. ${{\mathbb{R}}}$-circles and ${{\mathbb{C}}}$-circles

${{\mathbb{R}}}$-circles are boundaries of Lagrangian planes and ${{\mathbb{C}}}$-circles are boundaries of complex lines. They come in two flavours, infinite ones (i.e., containing the point at infinity) and finite ones. We refer to [5] for more more details about these curves.

2.2. Cartan’s Angular Invariant

Given a triple $(p_{1},p_{2},p_{3})$ of points at the boundary $\partial{{\bf H}}^{2}_{{\mathbb{C}}}$ the Cartan’s angular invariant ${{\mathbb{A}}}(p_{1},p_{2},p_{3})$ is defined by

[TABLE]

where ${{\bf p}}_{i}$ are lifts of $p_{i}$, $i=1,2,3$. The Cartan’s angular invariant lies in $[-\pi/2,\pi/2]$, is independent of the choice of the lifts and remains invariant under the diagonal action of ${\rm PU}(2,1)$. Any other permutation of points produces angular invariants which differ from the above possibly up to sign. The following propositions are in [5] to which we also refer the reader for further details:

Proposition 2.1**.**

Let $(p_{1},p_{2},p_{3})$ be a triple of points lying in $\partial{{\bf H}}^{2}_{{\mathbb{C}}}$ and let also ${{\mathbb{A}}}={{\mathbb{A}}}(p_{1},p_{2},$ $p_{3})$ be their Cartan’s angular invariant. Then:

1. (1)

All points lie in an ${{\mathbb{R}}}$-circle if and only if ${{\mathbb{A}}}=0$. 2. (2)

All points lie in a ${{\mathbb{C}}}$-circle if and only if ${{\mathbb{A}}}=\pm\pi/2$.

Proposition 2.2**.**

Suppose that $p_{i}$ and $p^{\prime}_{i}$, $i=1,2,3$, are points in $\partial{{{\bf H}}_{{\mathbb{C}}}^{2}}$. If there exists a holomorphic isometry $g$ of ${{\bf H}}^{2}_{{\mathbb{C}}}$ such that $g(p_{i})=p^{\prime}_{i}$, $i=1,2,3$, then ${{\mathbb{A}}}(p_{1},p_{2},p_{3})={{\mathbb{A}}}(p^{\prime}_{1},p^{\prime}_{2},p^{\prime}_{3})$. Conversely, if ${{\mathbb{A}}}(p_{1},p_{2},p_{3})={{\mathbb{A}}}(p^{\prime}_{1},p^{\prime}_{2},p^{\prime}_{3})$, then there exists a holomorphic isometry $g$ of ${{\bf H}}^{2}_{{\mathbb{C}}}$ such that $g(p_{i})=p^{\prime}_{i}$, $i=1,2,3$. This isometry is unique unless $p_{i}$, $i=1,2,3$, lie in a ${{\mathbb{C}}}$-circle.

2.3. Cross-ratio variety and the configuration space

Given a quadruple of pairwise distinct points ${{\mathfrak{p}}}=(p_{1},p_{2},p_{3},p_{4})$ in $\partial{{\bf H}}_{{\mathbb{C}}}^{2}$, we define their complex cross-ratio as follows:

[TABLE]

where ${{\bf p}}_{i}$ are lifts of $p_{i}$, $i=1,2,3,4$, see also [6], [7], [8]. The cross-ratio is independent of the choice of lifts and remains invariant under the diagonal action of ${\rm PU}(2,1)$. We stress here that for points in the Heisenberg group, the square root of its absolute value is

[TABLE]

Given a quadruple ${{\mathfrak{p}}}=(p_{1},p_{2},$ $p_{3},p_{4})$ of pairwise distinct points in the boundary $\partial{{\bf H}}^{2}_{{\mathbb{C}}}$, all possible permutations of points gives us 24 complex cross-ratios corresponding to ${{\mathfrak{p}}}$. Due to symmetries, see [3], Falbel showed that all cross-ratios corresponding to a quadruple of points depend on three cross-ratios which satisfy two real equations. Indeed, the following proposition holds; for its proof, see for instance [7].

Proposition 2.3**.**

Let ${{\mathfrak{p}}}=(p_{1},p_{2},p_{3},p_{4})$ be any quadruple of pairwise distinct points in $\partial{{\bf H}}^{2}_{{\mathbb{C}}}$. Let

[TABLE]

Then

[TABLE]

Equations (2.1) and (2.2) define a 4-dimensional real subvariety of ${{\mathbb{C}}}^{3}$ which we call the cross-ratio variety ${{\mathfrak{X}}}$. This variety is isomorphic to the subset $\mathfrak{F^{\prime}}$ of the ${\rm PU}(2,1)$ configuration space $\mathfrak{F}$ of pairwise disjoint quadruples of points in $\partial{{\bf H}}^{2}_{{\mathbb{C}}}$, comprising quadruples whose points do not all lie in the same ${{\mathbb{C}}}$-circle. In the latter case, we have a 2–1 map between the subset $\mathfrak{F}_{{\mathbb{R}}}$ comprising of quadruples whose points all lie in a ${{\mathbb{C}}}$-circle and the subvariety ${{\mathfrak{X}}}_{{\mathbb{R}}}$ of ${{\mathfrak{X}}}$ defined by

[TABLE]

see [3], [2]. For further reference, we shall need the following:

Remark 2.4**.**

Let ${{\mathfrak{p}}}$ a quadruple of pairwise distinct points in $\partial{{\bf H}}^{2}_{{\mathbb{C}}}$ which do not all lie in the same ${{\mathbb{C}}}$-circle and let also ${{\mathbb{X}}}_{i}({{\mathfrak{p}}})$, $i=1,2,3$ be as above. Let also $a=\arg({{\mathbb{X}}}_{1})$, $b=\arg({{\mathbb{X}}}_{2})$, $c=\arg({{\mathbb{X}}}_{3})$. We have

[TABLE]

Here,

[TABLE]

As for ${{\mathbb{A}}}_{3}={{\mathbb{A}}}(p_{1},p_{2},p_{4})$ we have

[TABLE]

3. The Configuration Space of Equidistant Triples

In this section we are going to prove Theorem 1.1. The proof will follow after a series of lemmas which follow below.

3.1. The lemmas

Throughout this section we will have the following notation: We shall denote by $P$ a triple $(p_{1},p_{2},p_{3})$ of pairwise distinct points in the Heisenberg group ${{\mathfrak{H}}}$. We will consider also the quadruple ${{\mathfrak{p}}}=(p_{1},\infty,p_{2},p_{3})$; let ${{\mathbb{X}}}_{i}({{\mathfrak{p}}})$, $i=1,2,3$ be the corresponding point on the cross-ratio variety ${{\mathfrak{X}}}$ and let

[TABLE]

There is an important note here: points of ${{\mathfrak{p}}}$ cannot all lie in the same (infinite) ${{\mathbb{C}}}$-circle. To see this, normalise so that

[TABLE]

Then conditions $d(p_{1},p_{2})=d(p_{1},p_{3})=d(p_{2},p_{3})$ deduce

[TABLE]

which cannot happen.

Lemma 3.1**.**

The triple $P=(p_{1},p_{2},p_{3})$ is equidistant if and only if

[TABLE]

Proof.

Since $P=(p_{1},p_{2},p_{3})$ is an equidistant triple, we have

[TABLE]

where $d$ is the Korányi distance. The result follows from the formulae

[TABLE]

Note that the above imply as well $|{{\mathbb{X}}}_{3}({{\mathfrak{p}}})|=1$. ∎

Lemma 3.2**.**

If $P=(p_{1},p_{2},p_{3})$ is an equidistant triple then $a,b,c$ satisfy

[TABLE]

Proof.

Consider the cross-ratio variety equations (2.1) and (2.2) as in the previous section. We may rewrite (2.2) equivalently as

[TABLE]

If $P$ is an equidistant triple then $|{{\mathbb{X}}}_{i}({{\mathfrak{p}}})|=1$, $i=1,2,3$ and thus (3.2) reduces to (3.1). ∎

Lemma 3.3**.**

If $P=(p_{1},p_{2},p_{3})$ is an equidistant triple, ${{\mathfrak{p}}}=(p_{1},\infty,p_{2},p_{3})$ and $(a,b,c)$ as above. If $g\in G={\rm Sim}({{\mathfrak{H}}})$, we set

[TABLE]

Then

[TABLE]

Proof.

The proof is immediate by invariance of cross-ratios. ∎

Lemma 3.4**.**

Let $(a,b,c)\in[-2\pi/3,2\pi/3]^{3}$ such that it satisfies

[TABLE]

Then there exists an equidistant triple $P=(p_{1},p_{2},p_{3})$ such that if ${{\mathfrak{p}}}=(p_{1},\infty,p_{2},p_{3})$ then

[TABLE]

Proof.

Set $2\eta=\arg(1-e^{ia}-e^{ib})$. We have

[TABLE]

This is strictly positive unless

[TABLE]

Assume first that $(a,b,c)\notin B$ and set

[TABLE]

with $A_{1},A_{4}\in(-\pi/2,\pi/2)$. Notice that we have

[TABLE]

Consider the triple $P=(p_{1},p_{2},p_{3})$ of points in ${{\mathfrak{H}}}$ where

[TABLE]

and the quadruple ${{\mathfrak{p}}}=(p_{1},\infty,p_{2},p_{3})$, with lifts:

[TABLE]

Now,

[TABLE]

and

[TABLE]

This gives

[TABLE]

which proves our claim.

Finally, we consider $(a,b,c)\in B$; we shall only treat the case where $a=\pi/3$, $b=-\pi/3$ and $c=\pi/3$, in other words when ${{\mathbb{A}}}_{4}=\pi/2$ and ${{\mathbb{A}}}_{1}=\pi/6$. Then we set

[TABLE]

All other cases can be treated in a similar manner. ∎

Lemma 3.5**.**

Let $(a,b,c)\in{{\mathcal{E}}}$ and consider from Lemma 3.4 the equidistant triple $P=(p_{1},p_{2},p_{3})$ of points in ${{\mathfrak{H}}}$ which is such that

[TABLE]

Then any other equidistant triple $P^{\prime}=(p_{1}^{\prime},p_{2}^{\prime},p_{3}^{\prime})$ such that

[TABLE]

is of the form $P^{\prime}=g(P)$, where $g\in G={\rm Sim}({{\mathfrak{H}}})$.

Proof.

We consider $P$, $P^{\prime}$ and ${{\mathfrak{p}}}$, ${{\mathfrak{p}}}^{\prime}$, respectively. Since ${{\mathbb{X}}}_{i}({{\mathfrak{p}}})={{\mathbb{X}}}_{i}({{\mathfrak{p}}}^{\prime})$, $i=1,2,3$, we have from Proposition 5.10 in [7] and Lemma 5.5 in [3] that since not all points in ${{\mathfrak{p}}}$ lie in the same ${{\mathbb{C}}}$-circle, there exists a $g\in{\rm PU}(2,1)$ such that $g(p_{i})=p_{i}^{\prime}$, $i=1,2,3$ and $g(\infty)=\infty$. That is, $g\in G$ and the proof is complete. ∎

3.2. The Equidistant surface

Equation (3.1) is the equation of a hypersurface in ${{\mathbb{R}}}^{3}$ which we shall call equidistant hypersurface and denote it by ${{\mathcal{E}}}$. This hypersurface comprises of infinitely many connected components. Notice that we have

[TABLE]

On the other hand, since for instance

[TABLE]

we have that $\cos a$, and in the same manner $\cos b$ and $\cos c$, are $\geq-1/2$. We deduce that the connected components of ${{\mathcal{E}}}$ may be taken by transporting the central component where

[TABLE]

by multiples of $2\pi$ in all possible directions, see the figure where the central component of ${{\mathcal{E}}}$ is clearly shown.

3.3. Proof of Theorem 1.1

Given an equidistant triple $P=(p_{1},p_{2},p_{3})$ we consider the quadruple ${{\mathfrak{p}}}=(p_{1},\infty,p_{2},p_{3})$ in the ${\rm PU}(2,1)$-configuration space of pairwise distinct points on the boundary $\partial{{\bf H}}^{2}_{{\mathbb{C}}}$ of complex hyperbolic plane. Let ${{\mathbb{X}}}_{i}({{\mathfrak{p}}})$, $i=1,2,3$ be the cross-ratios associated to ${{\mathfrak{p}}}$; the triple $({{\mathbb{X}}}_{1}({{\mathfrak{p}}}),{{\mathbb{X}}}_{2}({{\mathfrak{p}}}),{{\mathbb{X}}}_{3}({{\mathfrak{p}}}))$ defines a point in the cross=ratio variety ${{\mathfrak{X}}}$. In particular, in this case we have by Lemma 3.1 that $|{{\mathbb{X}}}_{i}({{\mathfrak{p}}})|=1$, $i=1,2,3$ and moreover, if

[TABLE]

then from Lemma 3.2

[TABLE]

By Lemma 3.3 this equation is invariant by the diagonal action of $G$ on the space of equidistant triples ${{\mathcal{E}}{\mathcal{T}}}$: This proves that the map

[TABLE]

is well-defined.

Conversely, if $(a,b,c)\in{{\mathcal{E}}}$ where $(a,b,c)\in[-2\pi/3,2\pi/3]$, by Lemma 3.4 there exists a $P=(p_{1},p_{2},p_{3})\in{{\mathcal{E}}{\mathcal{T}}}$ such that if ${{\mathfrak{p}}}=(p_{1},\infty,p_{2},p_{3})$ then $a=\arg({{\mathbb{X}}}_{1}({{\mathfrak{p}}}))$, $b=\arg({{\mathbb{X}}}_{2}({{\mathfrak{p}}}))$ and $c=\arg({{\mathbb{X}}}_{3}({{\mathfrak{p}}}))$; therefore ${{\mathfrak{E}}{\mathfrak{T}}}\to{{\mathcal{E}}}$ is onto. Finally, by Lemma 3.5 ${{\mathfrak{E}}{\mathfrak{T}}}\to{{\mathcal{E}}}$ is 1–1 when the points in ${{\mathfrak{p}}}$ do not all lie in the same ${{\mathbb{C}}}$-circle and 2–1 when all points in ${{\mathfrak{p}}}$ lie in the same ${{\mathbb{C}}}$-circle. This concludes the proof of Theorem 1.1.∎

3.4. The ${{\mathbb{C}}}$-circle case

The subset ${{\mathfrak{E}}{\mathfrak{T}}}_{{\mathbb{C}}}$ of ${{\mathfrak{E}}{\mathfrak{T}}}$ comprising equivalent equidistant triples of points lying on a ${{\mathbb{C}}}$-circle is of special interest. We can show in an elementary way that ${{\mathfrak{E}}{\mathfrak{T}}}_{{\mathbb{C}}}$ is just two points on the equidistant hypersurface ${{\mathcal{E}}}$. We start with a lemma:

Lemma 3.6**.**

With the assumptions of Section 3.1 let $P=(p_{1},p_{2},p_{3})$ and ${{\mathfrak{p}}}=(p_{1},\infty,p_{2},p_{3})$. Then

[TABLE]

Proof.

We have

[TABLE]

∎

We now prove

Proposition 3.7**.**

The subset ${{\mathfrak{E}}{\mathfrak{T}}}_{{\mathbb{C}}}$ of ${{\mathfrak{E}}{\mathfrak{T}}}$ comprising equivalence classes of equidistant triples of points lying on the same ${{\mathbb{C}}}$-circle is on bijection with the points $(\pi/3,\pi/3,\pi/3)$ or $(-\pi/3,-\pi/3,-\pi/3)$ of the equidistant surface ${{\mathcal{E}}}$.

Proof.

We will show that three equidistant points lie on a ${{\mathbb{C}}}$-circle if and only if the equidistant hypersurface reduces to the point $(\pi/3,\pi/3,\pi/3)$ or $(-\pi/3,-\pi/3,-\pi/3)$. We start by assuming that the three points lie on a ${{\mathbb{C}}}$-circle, that is, ${{\mathbb{A}}}=\pm\pi/2$. Here, ${{\mathbb{A}}}={{\mathbb{A}}}(p_{1},p_{2},p_{3})$. Since from Lemma 3.6 we have

[TABLE]

Equation (3.1) becomes

[TABLE]

This is written equivalently as

[TABLE]

or,

[TABLE]

that is,

[TABLE]

Therefore we obtain,

[TABLE]

This gives

[TABLE]

Conversely, suppose that three points lie on a ${{\mathbb{C}}}$-circle and $\arg({{\mathbb{X}}}_{i})=\pm\frac{\pi}{3}$. Then $p_{i}$ are equidistant. Indeed, from equation (3.2) we have

[TABLE]

Factoring out we may write equivalently

[TABLE]

This gives $|{{\mathbb{X}}}_{1}|=|{{\mathbb{X}}}_{2}|=1$ and therefore the points are equidistant. ∎