# Conjugacy classes of Automorphisms of the unit ball in a complex Hilbert
space

Rachna Aggarwal, Krishnendu Gongopadhyay and Mukund Madhav Mishra  Department
of Mathematical Sciences, Indian Institute of Science Education and Research
(IISER) Mohali, Knowledge City, S.A.S. Nagar, Sector 81, P.O. Manauli 140306,
India<EMAIL_ADDRESS>Department of Mathematical Sciences, Indian
Institute of Science Education and Research (IISER) Mohali, Knowledge City,
S.A.S. Nagar, Sector 81, P.O. Manauli 140306, India
<EMAIL_ADDRESS>Department of Mathematics, Hans Raj College,
University of Delhi, Delhi, India<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

In this article, we consider the ball model of an infinite dimensional complex
hyperbolic space, i.e. the open unit ball of a complex Hilbert space centered
at the origin equipped with the Carathéodory metric. We consider the group of
holomorphic automorphisms of the ball and classify the conjugacy classes of
automorphisms.We also compute the centralizers for elements in the group of
automorphisms.

###### Key words and phrases:

Hyperbolic space; Carathéodory metric; isometry group; dynamical types;
conjugacy classes; centralizer

###### 2020 Mathematics Subject Classification:

51M10; 51F25

##  1\. Introduction

The rank one symmetric spaces of non-compact type are given by real, complex
and the quaternionic hyperbolic spaces, and the Cayley plane. These spaces are
of primary interest to mathematicians for their broad horizons over different
branches of mathematical sciences. For a uniform and gentle introduction to
the first three of these spaces we refer to the article [5]. From geometric
perspective, the real hyperbolic spaces, especially in low dimensions, have
seen much attention due to their importance in the theory of Kleinian groups,
e.g. [1], [9]. The complex hyperbolic spaces also play an important role in
mathematics due to their connection with lattices in Lie groups and
deformation theory of discrete groups, cf. [11], [25], [26]. Recently
quaternionic hyperbolic spaces have also seen many investigations, e.g. [21],
[20], [8].

In spite of many works on the finite dimensional hyperbolic spaces, their
infinite dimensional counterparts have not seen much attention until Gromov
suggested investigations of these infinite dimensional hyperbolic spaces from
algebraic and geometrical point of views, cf. [16, pp. 121]. In the last two
decades, the literature has seen many contributions concerning the geometry of
the infinite dimensional real hyperbolic space, e.g. [2], [7], [24]. However,
the geometry of the infinite dimensional complex hyperbolic space is yet to
see comparable attentions.

We attempt investigation of the infinite dimensional complex hyperbolic space
by studying conjugacy classes of the isometries and their centralizers in this
article. We recall here that in the finite dimensional set up, dynamical
properties of the isometries are intimately related to the fixed point
classification of the isometries, and this is related to the conjugacy
classes. In the finite dimensional set up, the conjugacy classification can be
described completely using conjugacy invariants like the coefficients of the
characteristic polynomials, cf. [12], [13], [14], [15]. However, the situation
in infinite dimension is not simple not only due to lack of conjugacy
invariants, but also because of complexity in spectra. The starting point of
our work would be to use the ‘ball model’ of the complex hyperbolic space that
was introduced by Franzoni and Vesentini in [10, Chapter VI]. They have also
given a linear representation of the isometries of the ball model that we will
use extensively. Using similar ideas, we also introduce a ‘Siegel domain
model’ of the infinite dimensional complex hyperbolic space. We use these
different models to classify the conjugacy classes of the isometries, and
obtain discriptions for the centralizers.

In [22], Kulkarni proposed to use the centralizers as an internal ingredient
of the group of automorphisms in a geometry to classify the automorphisms into
disjoint classes which roughly reflect the ‘dynamical types’. This indicates
the importance of the conjugacy classes of centralizers that might be implicit
in geometry. The centralizers in finite dimensional set up have been
investigated in [13], [12], [4], [3]. In this paper, we describe the
centralizers for the elliptic and the hyperbolic elements. The conjugacy
classes of centralizers for such elements may be derived from this
description. We also provide descriptions of the centralizers for
‘translations’. As corollaries, we observe when certain isometries commute to
each other.

Before stating the main results we fix some notations. Let $K$ be a Hilbert
space and $\left<\,,\,\right>$ denote the inner product on $K$. For $x\in K$,
$\left<x\right>$ will denote the linear space generated by $x$. For a subspace
$M\subseteq K$, ${M}^{\perp}$ denotes the orthogonal complement of $M$ with
respect to the inner product on $K$. The space of all bounded linear operators
on $K$ will be denoted by $B(K)$ and for $T\in B(K)$, $\sigma(T)$ denotes the
spectrum of $T$. For a set $X\subseteq K$, $\overline{X}$ denotes the closure
of $X$ with respect to the norm on $K$ and $\partial{X}$, its boundary. The
unit circle in $\mathbb{C}$ is denoted by $S^{1}$. For any two operators $T$
and $S$ in $B(K)$, $T\longleftrightarrow S$ would mean $T$ and $S$ commute
with each other, i.e. $TS=ST$. An operator $T\in B(K)$ is called unitary if
$T^{*}=T^{-1}$ where $T^{*}$ is the Hilbert-adjoint of $T$ and $T$ is called
normal if $TT^{*}=T^{*}T$. It is called non-unitary normal if it is a normal
operator which is not unitary. The group of unitary operators in $B(K)$ will
be denoted by $\mathcal{U}(K)$.

Let $H$ be a complex Hilbert space. Suppose $B$ is the open unit ball in $H$
centered at the origin equipped with the Caratheódory metric. The group of bi-
holomorphic automorphisms of $B$ is denoted by $Aut(B)$. The elements of
$Aut(B)$ are isometries for the Carathéodory metric. An element in $Aut(B)$ is
called elliptic if it has a fixed point inside $B$, hyperbolic (resp.
parabolic) if it is not elliptic and has exactly two (resp. one) fixed points
in $\overline{B}$ lying on $\partial{B}$. In this article we mainly study the
conjugacy classes and centralizers of elements in the group $Aut(B)$.
Let $G$ be the group of all bijective bounded linear transformations on
$H\oplus\,\mathbb{C}$ which are isometries for the following sesquilinear form
defined on $H\oplus\,\mathbb{C}$.

(1.1)
$\displaystyle\mathcal{A}\left((x,z),\,(y,w)\right)=\left<x,y\right>-z\overline{w},\,\,\,\,\,\,\,\,(x,z),\,(y,w)\in
H\oplus\,\mathbb{C}.$

Let $\mathcal{Q}$ be the quadratic form determined by $\mathcal{A}$.

######  Proposition 1.1.

[23, Proposition 1] A general element of $G$ is of the form
$e^{i\theta}\left[{\begin{array}[]{cc}UA&U(\xi)\\\
\left<\boldsymbol{\cdot},\xi\right>&a\\\
\end{array}}\right]\,\text{where}\,\theta\in\mathbb{R}$, $\xi\in H$,
$a=\sqrt{1+\|\xi\|^{2}}$, $U\in\mathcal{U}(H)$ and A is a positive operator on
$H$ defined by $A=I$ on $\left<\xi\right>^{\perp}$ and $A(\xi)=a\xi$.

For a closed subspace $M\subseteq H\oplus\mathbb{C}$, let $G_{M}$ denote the
group of all bijective bounded linear mappings on $M$ leaving the form
$\mathcal{A}\restriction_{M}$ invariant. For a subspace $M\subseteq
H\oplus\,\mathbb{C}$, orthogonal complement of $M$ for the Hermitian form
$\mathcal{A}$ defined in (1.1) will be denoted by $M^{\dagger}$. A vector
$\mathbb{x}\in H\oplus\,\mathbb{C}$ is called time-like if
$\mathcal{Q}(\mathbb{x})<0$ and light-like if $\mathcal{Q}(\mathbb{x})=0$.
Isometries in $Aut(B)$ are projective transformations induced from elements in
$G$. We would use the word isometry to denote the elements of $Aut(B)$ and $G$
simultaneously.

An isometry in $G$ is called elliptic, hyperbolic or parabolic if a
corresponding element in $Aut(B)$ is of the respective type. An eigenvalue of
an isometry $T\in G$ is called time-like if its corresponding eigenspace has a
time-like vector. An elliptic isometry in $G$ is called boundary elliptic if
the geometric multiplicity of its time-like eigenvalue is atleast two,
otherwise it is called regular elliptic. The words ‘normal isometry’ and ‘non-
unitary normal isometry’ would mean isometries in $G$ which are normal and
non-unitary normal as Hilbert space linear operators. The center of the group
$G$ will be denoted by $Z_{G}$. The centralizer of an isometry $T\in G$ will
be denoted by $Z(G)$.

We will study parabolic isometries on a Siegel domain model which is a
counterpart of the half plane model in the classical case. Let
$e\in\partial{B}$. We write a general element $x\in H$ as
$x=\left<x,e\right>e+x^{\prime}$ where $x^{\prime}\in\left<e\right>^{\perp}$.
We consider the following sesquilinear form.

(1.2)
$\displaystyle\widehat{\mathcal{A}}\left((x,z),(y,w)\right)=-z\overline{\left<y,e\right>}-\left<x,e\right>\overline{w}+\left<x^{\prime},y^{\prime}\right>.$

The projective space of its negative vectors gives rise to the Siegel domain.
It is defined as

$\Sigma=\left\\{x\in
H\,\,:\,\,\text{Re}\,\left<x,e\right>>\frac{1}{2}\|x^{\prime}\|^{2}\right\\}.$

Hermitian forms (1.1) and (1.2) are equivalent via a unitary operator
$D=\left[\begin{array}[]{ccc}{}\dfrac{1}{\sqrt{2}}&{}0&{}-\dfrac{1}{\sqrt{2}}\\\
{}0&{}I&{}0\\\
{}\dfrac{1}{\sqrt{2}}&{}0&{}\dfrac{1}{\sqrt{2}}\end{array}\right].$ The
projective map induced by $D^{-1}$ is a Cayley map which is a biholomorphism
between the unit ball $B$ and the Siegel domain $\Sigma$. Let the image of $e$
under Cayley map be denoted by $\infty$ which is the point at infinity in the
completion of $\overline{\Sigma}\setminus\infty$. Let $\widehat{G}$ denote the
group of all bijective bounded linear mappings on $H\oplus\mathbb{C}$ leaving
$\widehat{\mathcal{A}}$ invariant. Similar to the $Aut(B)$ case, the
isometries of $Aut(\Sigma)$ are the projective maps induced by elements of
$\widehat{G}$.
For a subspace $M$ of $H\oplus\mathbb{C}$, we will use the same notation
$M^{\dagger}$ to denote the orthogonal complement of $M$ with respect to
$\widehat{\mathcal{A}}$. Let $\widehat{G}_{\infty}$ denote the stabilizer
group of infinity in $\widehat{G}$. A general element of
$\widehat{G}_{\infty}$ is of the form
$\left[\begin{array}[]{ccc}\lambda&\left<\boldsymbol{\cdot},U^{-1}(a^{\prime})\right>&\mu
s\\\ 0&U&\mu a^{\prime}\\\ 0&0&\mu\end{array}\right]$ where
$\lambda\in\mathbb{C}$ is such that $\lambda\,\overline{\mu}=1$,
$U\in\mathcal{U}(\left<e\right>^{\perp})$, $s\in\mathbb{C}$ and
$a^{\prime}\in\left<e\right>^{\perp}$ satisfy
$\text{Re}\,s=\dfrac{1}{2}\|a^{\prime}\|^{2}$. The form of a parabolic
isometry in $\widehat{G}_{\infty}$ having singleton spectrum is given by
$\lambda\left[\begin{array}[]{ccc}1&\left<\boldsymbol{\cdot},a^{\prime}\right>&s\\\
0&I&a^{\prime}\\\ 0&0&1\end{array}\right]$ which is denoted by
$(\lambda,a^{\prime},s)$ such that $s\neq 0$. Projective maps induced by such
isometries are Heisenberg translations with $a^{\prime}$ as component in the
horizontal direction. Thus a Heisenberg translation is called vertical if
$a^{\prime}=0$ otherwise it is called non-vertical.

###  1.1. Main Results

Following are the main results of this article. Theorem 1.2 describes an
elliptic isometry and its conjugacy class, Theorem 1.3 investigates a
hyperbolic isometry along with its conjugacy class and Theorem 1.4 explores
the conjugacy of parabolic isometries having singleton spectra. Centralizers
of the isometries are given in Theorems 1.5, 1.6 and 1.7 respectively.

######  Theorem 1.2 (Elliptic isometry).

Let $T\in G$ be an elliptic isometry. Then

  1. (1)

$T$ is unitary upto conjugacy.

  2. (2)

$\sigma(T)\subseteq S^{1}$.

  3. (3)

$T$ has a time-like eigenvector and vice versa.

  4. (4)

If $(x,1)$ is a time-like eigenvector of $T$ then $T=T_{1}\oplus\,T_{2}$ where
$T_{1}=T\restriction_{\left<(x,1)\right>}$ and
$T_{2}=T\restriction_{\left<{(x,1)}\right>^{\dagger}}$. Moreover
$T_{2}\in\mathcal{U}({\left<{(x,1)}\right>^{\dagger}})$.

######  Theorem 1.3 (Hyperbolic isometry).

Let $T\in G$ be a hyperbolic isometry. Then

  1. (1)

$T$ is non-unitary normal upto conjugacy.

  2. (2)

Spectrum of T has two eigenvalues of the form $re^{i\theta}$ and
$r^{-1}e^{i\theta}$ where $r>0$, $r\neq 1$. Eigenspaces with respect to these
eigenvalues are one dimensional, each generated by a light-like eigenvector.
Rest of the spectral values lie on $S^{1}$.

  3. (3)

$T=T_{1}\oplus T_{2}$ where $T_{1}=T\restriction_{M}$,
$M=\text{span}\,\left\\{(y_{1},1),(y_{2},1)\right\\}$, $(y_{1},1)$ and
$(y_{2},1)$ are light-like eigenvectors of $T$ and
$T_{2}=T\restriction_{{M}^{\dagger}}$. Moreover
$T_{2}\in\mathcal{U}(M^{\dagger})$.

######  Theorem 1.4 (Parabolic isometry).

Let $\widehat{S}\in\widehat{G}$ be a parabolic isometry such that
$\sigma(\widehat{S})=\\{\lambda\\}$. Then

  1. (1)

Upto conjugacy, $\widehat{S}$ is a Heisenberg translation.

  2. (2)

$\widehat{S}=\widehat{S}_{1}\oplus\widehat{S}_{2}$ where
$\widehat{S}_{1}=\widehat{S}\restriction_{K}$ and
$\widehat{S}_{2}=\widehat{S}\restriction_{K^{\dagger}}$. Here $K$ is a two or
three dimensional non-degenerate subspace of $H\oplus\mathbb{C}$ containing a
light-like eigenvector and $\widehat{S}_{2}=\lambda\,I$. The
$ker\,(\widehat{S_{1}}-\lambda\,I)$ is generated by the light-like eigenvector
and minimal polynomial of $\widehat{S}_{1}$ is $(x-\lambda)^{2}$ or
$(x-\lambda)^{3}$.

  3. (3)

All the parabolic isometries in $\widehat{G}$ having same singleton spectra
and degree of restricted minimal polynomial as $2$ (resp. $3$) get dispersed
into two conjugacy classes (resp. fall in a single conjugacy class).

######  Theorem 1.5 (Centralizer of an elliptic isometry).

Let $T$ be an elliptic isometry having $M$ as its time-like eigenspace. Then

$Z(T)=G_{M}\times Z(T\restriction_{M^{\dagger}})$

where $Z(T\restriction_{M^{\dagger}})\subseteq\mathcal{U}(M^{\dagger})$.

######  Theorem 1.6 (Centralizer of a hyperbolic isometry).

Let $T$ be a hyperbolic isometry and $M$ be the two-dimensional subspace of
$H\oplus\mathbb{C}$ generated by two light-like eigenvectors of $T$. Then

$Z(T)=Z(T\restriction_{M})\times Z(T\restriction_{M^{\dagger}})$

where $Z(T\restriction_{M})$ gets identified with $S^{1}\times\mathbb{R}$ and
$Z(T\restriction_{M^{\dagger}})\subseteq\mathcal{U}(M^{\dagger})$.

The following two results also follow from the proofs of the above theorems.

######  Corollary 1.6.1.

Hyperbolic isometries either commute with boundary elliptic isometries or
hyperbolic isometries.

######  Corollary 1.6.2.

Two commuting hyperbolic isometries have the same fixed points.

######  Theorem 1.7 (Centralizer of Heisenberg translation).

Let $\widehat{T}=(\lambda,a^{\prime},s)$ be a Heisenberg translation. Then

  1. (1)

For $a^{\prime}=0$, centralizer of a vertical translation $\widehat{T}$
consists of all non-hyperbolic isometries in $\widehat{G}_{\infty}.$

  2. (2)

For a non-vertical translation $\widehat{T}$,

$Z(\widehat{T})=\left\\{\left[\begin{array}[]{ccc}\lambda^{\prime}&\left<\boldsymbol{\cdot},U^{-1}(b^{\prime})\right>&\lambda^{\prime}t\\\
\\\ 0&U&\lambda^{\prime}b^{\prime}\\\ \\\ 0&0&\lambda^{\prime}\\\
\end{array}\right]\in\widehat{G}_{\infty}\,\,:\,\,U(a^{\prime})=\lambda^{\prime}\,a^{\prime},\,\,\left<b^{\prime},a^{\prime}\right>\in\mathbb{R}\right\\}.$

Structure of the article. In section 2, we describe the group $Aut(B)$ along
with its linear representation. In section 3, we study some significant sub-
classes of $G$, provide a class of examples of each of elliptic, hyperbolic
and parabolic isometries and give infinite dimensional versions of some
general results in finite dimension. Section 4 investigates conjugacy classes
of elliptic and hyperbolic isometries. Section 5 studies parabolic isometries
along with the discussion of conjugacy within a subclass consisting of
singleton spectra. Section 6 investigates centralizers of elliptic,
hyperbolic, and parabolic isometries having singleton spectra.

### Acknowledgements

The authors would like to thank John R. Parker for many comments and
suggestions that has improved this work. The authors also acknowledge partial
support from the TARE project grant TAR/2019/000379 and the SERB core research
grant CRG/2022/003680.

##  2\. Holomorphic automorphisms of $B$

We now present the following set up as described in [10] which forms the basis
of our study.

######  Lemma 2.1.

[10, Proposition III.2.2.] An isometry in $Aut(B)$ fixing the origin is the
restriction of a linear bijection on $H$ to $B$.

######  Lemma 2.2.

[10, Proposition VI.1.2.] A linear map $U\in B(H)$ belongs to $Aut(B)$ if and
only if $U$ is unitary.

######  Theorem 2.3.

[10, Theorem VI.1.3.] A general element of $Aut(B)$ is of the form $F=U\circ
f_{b}$, $b\in B$, where $U\in\mathcal{U}(H)$ and $x\mapsto
T_{b}\left(\dfrac{x-{b}}{1-\left<x,b\right>}\right)$ which defines $f_{b}$ is
a holomorphic automorphism of $B$. The map $T_{b}:H\longrightarrow H$ is a
linear map expressed as

$T_{b}(x)=\dfrac{\left<x,b\right>}{1+\sqrt{1-\|b\|^{2}}}b+\sqrt{1-\|b\|^{2}}\,\,x.$

Here $f_{b}(b)=0$ and ${f_{b}}^{-1}=f_{-b}$.

$Aut(B)$ acts transitively on B [10, Proposition VI.1.5.].

We will be simultaneously using the same notation for inner products on
Hilbert spaces $H$ and $H\oplus\,\mathbb{C}$.

A linear representation of the isometries of $Aut(B)$ is described as follows
[10, Ch. VI, sec. 3]. Let $\mathcal{A}$ denote the Hermitian form on
$H\oplus\,\mathbb{C}$ defined by

(2.1) $\displaystyle\hskip
28.45274pt\mathcal{A}\left((x,z),\,(y,w)\right)=\left<x,y\right>-z\overline{w},\,\,\,\,\,\,\,\,(x,z),\,(y,w)\in
H\oplus\,\mathbb{C}$

and $\mathcal{Q}$ be the quadratic form determined by $\mathcal{A}$. Let $G$
be the group of all bijective bounded linear transformations on
$H\oplus\,\mathbb{C}$ leaving $\mathcal{A}$ invariant.
General form of an element of $G$ is $T=\left[{\begin{array}[]{cc}A&\xi\\\
\left<\boldsymbol{\cdot},\dfrac{A^{*}(\xi)}{a}\right>&a\\\
\end{array}}\right]$, cf. [10, Lemma VI.3.1], where $A\in B(H)$ is bijective,
$\xi\in H$, $a\in\mathbb{C}$, $|a|^{2}=1+\|\xi\|^{2}$, and

$\displaystyle
A^{*}A=I+\dfrac{1}{|a|^{2}}\left<\boldsymbol{\cdot},{A^{*}(\xi)}\right>A^{*}(\xi).$

The center $Z_{G}$ of $G$ is
$\\{e^{i\theta}{I},\,\,\,\,\theta\in\mathbb{R}\\}$, see [10, Lemma VI.3.4].
The following theorem tells that the group $G$ acts on $B$ surjectively, see
[23] for the proof.

######  Theorem 2.4.

[10, Theorem VI.3.5] The map $\phi:G\longrightarrow\text{Aut(B)}$ defined by
$T\mapsto\widetilde{T}$ is an onto homomorphism where
$T=\left[{\begin{array}[]{cc}A&\xi\\\
\left<\boldsymbol{\cdot},\dfrac{A^{*}(\xi)}{a}\right>&a\\\
\end{array}}\right]$ and
$\widetilde{T}(x)=\dfrac{A(x)+\xi}{\left<x,\dfrac{A^{*}(\xi)}{a}\right>+a}$,
$x\in B$.

Since ${ker}(\phi)=Z_{G}$, the map $\widetilde{\phi}:G/{Z_{G}}\longrightarrow
Aut(B)$ is an isomorphism.
Often we shall call $T$ as linearization of $\widetilde{T}$.

##  3\. Some special sub-classes of $G$

Proposition 1.1 gives a simplified form of elements of $G$. Notice that in the
proposition

(3.1) $\displaystyle a=\sqrt{1+\|\xi\|^{2}},$

thus $a\geq 1$. The Hilbert-adjoint operator for
$T=e^{i\theta}\left[{\begin{array}[]{cc}UA&U(\xi)\\\
\left<\boldsymbol{\cdot},\xi\right>&a\\\ \end{array}}\right]\in G$ is
$T^{*}=e^{-i\theta}\left[{\begin{array}[]{cc}(UA)^{*}&\xi\\\
\left<\boldsymbol{\cdot},U(\xi)\right>&a\\\ \end{array}}\right]$. Observe that
$\mathcal{A}\left((x,z),(y,w)\right)=\left<A^{\prime}(x,z),(y,w)\right>$ where

(3.4) $\displaystyle A^{\prime}=\left[{\begin{array}[]{cc}I&0\\\ 0&-1\\\
\end{array}}\right].$

For any two elements $\mathbb{x},\,\mathbb{y}\in H\oplus\,\mathbb{C}$ and
$T\in G$,
$\left<A^{\prime}\left(T(\mathbb{x})\right),T(\mathbb{y})\right>=\left<A^{\prime}(\mathbb{x}),\mathbb{y}\right>$,
i.e. $T^{*}A^{\prime}T=A^{\prime}$, i.e. $A^{\prime}T^{*}A^{\prime}={T}^{-1}$.
This gives $T^{-1}=\left[{\begin{array}[]{cc}(UA)^{*}&-\xi\\\
-\left<\boldsymbol{\cdot},U(\xi)\right>&a\\\ \end{array}}\right]$.

######  Remark 1.

Every isometry of $G$ in Proposition 1.1 is expressible as the product of a
unitary and a self-adjoint element of $G$, i.e.
$\left[{\begin{array}[]{cc}UA&U(\xi)\\\
\left<\boldsymbol{\cdot},\xi\right>&a\\\
\end{array}}\right]=\left[{\begin{array}[]{cc}U&0\\\ 0&1\\\
\end{array}}\right]\left[{\begin{array}[]{cc}A&\xi\\\
\left<\boldsymbol{\cdot},\xi\right>&a\\\ \end{array}}\right].$

Every isometry in $Aut(B)$ is the restriction of a holomorphic map of some
open ball containing $B$ into $H$ (see [10, Proposition VI.1.4]). It can be
observed from computations in [10, eq. (VI.1.6)-(VI.1.8)] that isometries of
$Aut(B)$ leave $\partial{B}$ invariant and are bijective on it. The following
theorem is due to Hayden and Suffridge which gives a fixed point
classification for elements in $Aut(B)$.

######  Theorem 3.1.

[19, Hayden and Suffridge] Every element $g\in Aut(B)$ has a fixed point in
$\overline{B}$. If $g\in\text{Aut(B)}$ has no fixed point in B, then the fixed
point set in $\overline{\text{B}}$ consists of atmost two points.

We call an isometry in $Aut(B)$ elliptic if has a fixed point in $B$,
hyperbolic (resp. parabolic) if it is not elliptic and has exactly two (resp.
one) fixed points on $\partial{B}$.

We say an isometry of $G$ is elliptic (resp. hyperbolic or parabolic) if it is
in the pre-image of an elliptic (resp. hyperbolic or parabolic) isometry of
$Aut(B)$ under the homomorphism $\phi$ in Theorem 2.4.

We call a vector $(x,z)\in H\oplus\,\mathbb{C}$ time-like if
$\mathcal{Q}(x,z)<0$,  light-like if $\mathcal{Q}(x,z)=0$ and  space-like if
$\mathcal{Q}(x,z)>0$. A linear subspace $W\subseteq H\oplus\,\mathbb{C}$ will
be called time-like space if it contains a time-like vector.

######  Remark 2.

For $x\in\overline{B}$, $x$ is a fixed point for an isometry
$\dfrac{UA(\boldsymbol{\cdot})+U(\xi)}{\left<\boldsymbol{\cdot},\xi\right>+a}\linebreak\in\text{Aut(B)}$
if and only if $(x,1)$ is an eigenvector for a corresponding isometry in $G$.

######  Lemma 3.2 ([23], Proposition 3).

Let $T=\left[{\begin{array}[]{cc}UA&{}r\xi\\\
\left<\boldsymbol{\cdot},\xi\right>&{}a\\\ \end{array}}\right]\in G$, $\xi\neq
0$, $|r|=1$. Then $T=T_{1}\oplus\,T_{2}$ where
$T_{1}=T\restriction_{\left<\xi\right>\oplus\,\mathbb{C}}$ and
$T_{2}=T\restriction_{(\left<\xi\right>\oplus\mathbb{C})^{\perp}}\equiv
U\restriction_{\left<\xi\right>^{\perp}}$. Also

  1. (1)

${\sigma(T)}=\\{\lambda_{1},\,\lambda_{2}\\}\cup{\sigma(U\restriction_{\left<\xi\right>^{\perp}})}$
where
$\lambda_{1},\,\lambda_{2}\,\,\text{are}\,\,\dfrac{a(r+1)\pm\sqrt{a^{2}(r+1)^{2}-4r}}{2}$.
The eigenspaces corresponding to the eigenvalues $\lambda_{1}$ and
$\lambda_{2}$ are generated by the eigenvectors $\left(k_{1}\xi,1\right)$ and
$(k_{2}\xi,1)$ respectively where $k_{1},\,k_{2}$ are
$\dfrac{a(r-1)\pm\sqrt{a^{2}(r+1)^{2}-4r}}{2\|\xi\|^{2}}$.

  2. (2)

$|\lambda_{1}|=\dfrac{1}{|\lambda_{2}|}$ and
$\|k_{1}\xi\|=\dfrac{1}{\|k_{2}\xi\|}$.

The isometries considered in the above lemma constitute of a subclass of $G$
whose elements decompose $H\oplus\,\mathbb{C}$ orthogonally into a two
dimensional subspace containing $\mathbb{C}$ and its orthogonal complement
(see [23, Proposition 2]).
The next result provides a class of examples of each of elliptic, hyperbolic
and parabolic isometries.

######  Proposition 3.3.

Let $T=\left[{\begin{array}[]{cc}UA&r\xi\\\
\left<\boldsymbol{\cdot},\xi\right>&a\\\ \end{array}}\right]\in G$, $\xi\neq
0$, $|r|=1$ be such that $T=T_{1}\oplus T_{2}$ (cf. Lemma 3.2). Then $T$ is
elliptic for $r=-1$, parabolic for $r=\left(\dfrac{2}{a^{2}}-1\right)\pm
i\left(\dfrac{2\|\xi\|}{a^{2}}\right)$ and hyperbolic for rest of the values
of $r$ on $S^{1}$.

###### Proof.

In view of Lemma 3.2, we perform the following analysis. Let $r=-1$. This
gives that $k_{1}=\dfrac{-2a+2}{2\|\xi\|^{2}}=\dfrac{1-a}{\|\xi\|^{2}}$. So
${k_{1}}^{2}=\dfrac{(1-a)^{2}}{\|\xi\|^{4}}$ and
$\|k_{1}\xi\|^{2}={k_{1}}^{2}\|\xi\|^{2}=\dfrac{(1-a)^{2}}{\|\xi\|^{2}}$.
Observe that by (3.1), $(1-a)^{2}-\|\xi\|^{2}=1+a^{2}-2a+1-a^{2}=2-2a<0$ as
$\xi\neq 0$. This implies that $(1-a)^{2}<\|\xi\|^{2}$, i.e.
$\dfrac{(1-a)^{2}}{\|\xi\|^{2}}<1$. Hence $T$ is elliptic by Remark 2.

We will see that $r=\left(\dfrac{2}{a^{2}}-1\right)\pm
i\left(\dfrac{2\|\xi\|}{a^{2}}\right)$ if and only if
$\lambda_{1}=\lambda_{2}$. From the value of $\lambda_{i}$, we have
$\lambda_{1}=\lambda_{2}$ if and only if $a^{2}(r+1)^{2}=4r$. This implies
$\text{Re r}=\dfrac{2}{a^{2}}-1$.

As $|r|^{2}=(\text{Re}\,r)^{2}+(\text{Im}\,r)^{2}=1$,
$(\text{Im}\,r)^{2}=1-\left(\dfrac{2}{a^{2}}-1\right)^{2}=\dfrac{4}{a^{2}}-\dfrac{4}{a^{4}}=\dfrac{4(a^{2}-1)}{a^{4}}=\dfrac{4\|\xi\|^{2}}{a^{4}}$
(by (3.1)) which gives that $r=\left(\dfrac{2}{a^{2}}-1\right)\pm
i\left(\dfrac{2\|\xi\|}{a^{2}}\right)$. Thus $\lambda_{1}=\lambda_{2}$ if and
only if $r=\left(\dfrac{2}{a^{2}}-1\right)\pm
i\left(\dfrac{2\|\xi\|}{a^{2}}\right)$. Also
$k_{i}=\dfrac{a(r-1)}{2\|\xi\|^{2}}$. This implies

$\displaystyle\|k_{i}\xi\|^{2}$ $\displaystyle=|k_{i}|^{2}\|\xi\|^{2}$
$\displaystyle=\dfrac{a^{2}|r-1|^{2}}{4\|\xi\|^{2}}$
$\displaystyle=\dfrac{a^{2}(r-1)(\overline{r}-1)}{4\|\xi\|^{2}}$
$\displaystyle=\dfrac{a^{2}(1-r-\overline{r}+1)}{4\|\xi\|^{2}}$
$\displaystyle=\dfrac{a^{2}(2-2\,\text{Re}\,r)}{4\|\xi\|^{2}}$
$\displaystyle=\dfrac{a^{2}(1-\,\text{Re}\,r)}{2\|\xi\|^{2}}$
$\displaystyle=1$

as
$1-\,\text{Re}\,r=1-\dfrac{2}{a^{2}}+1=2-\dfrac{2}{a^{2}}=\dfrac{2\|\xi\|^{2}}{a^{2}}$.
So for the case of identical eigenvalues of $T_{1}$, it has a fixed point on
$\partial{B}$ in view of Remark 2. We will see that $T$ cannot have a fixed
point inside the ball and has exactly one fixed point on $\partial{B}$. There
will be two possibilities. Either $\sigma(T_{1})\cap\sigma(T_{2})=\emptyset$
or $\sigma(T_{1})\cap\sigma(T_{2})\neq\emptyset$. In the former case, the
eigenvectors of $T$ will come strictly either from
$\left<\xi\right>\oplus\mathbb{C}$ or from $\left<\xi\right>^{\perp}$. There
is exactly one eigenvector of $T_{1}$ of the form $(x,1)$ for which
$x=k_{i}\xi\in\partial{B}$ and eigenvectors of $T_{2}$ are of the form
$(y,0)$, if any, for $y\in\left<\xi\right>^{\perp}$. Thus in this case, $T$
does not have a fixed point inside the ball and cannot have more than one
fixed point on the boundary in view of Remark 2. In the latter case, let
$(y,0)$ be an eigenvector of $T_{2}$ corresponding to the common eigenvalue.
Thus $T$ has an eigenvector of the type $(k_{i}\xi,1)+(y,0)=(k_{i}\xi+y,1)$
but $\|k_{i}\xi+y\|^{2}=\|k_{i}\xi\|^{2}+\|y\|^{2}>1$ as $\|k_{i}\xi\|=1$ and
$y\neq 0$. So for this case as well, $T$ cannot have a fixed point inside the
ball and more than one fixed point on boundary of the ball in view of Remark
2. Hence $T$ is parabolic.
Let $\lambda_{1}\neq\lambda_{2}$ and $r\neq-1$. This gives that

$\displaystyle\|k_{1}\xi\|^{2}$ $\displaystyle=$
$\displaystyle\dfrac{\left|a(r-1)+\sqrt{a^{2}(r+1)^{2}-4r}\right|^{2}}{4\|\xi\|^{2}}$
$\displaystyle=$
$\displaystyle\dfrac{\left(a(r-1)+\sqrt{a^{2}(r+1)^{2}-4r}\right)\left(a\left(\frac{1}{r}-1\right)+\sqrt{a^{2}\left(\frac{1}{r}+1\right)^{2}-\frac{4}{r}}\right)}{4\|\xi\|^{2}}$
$\displaystyle=$
$\displaystyle\dfrac{\left(a(r-1)+\sqrt{a^{2}(r+1)^{2}-4r}\right)\left(a(1-r)+\sqrt{a^{2}(1+r)^{2}-4r}\right)}{4r\|\xi\|^{2}}$
$\displaystyle=$
$\displaystyle\dfrac{-a^{2}(r-1)^{2}+a^{2}(r+1)^{2}-4r}{4r\|\xi\|^{2}}$
$\displaystyle=$ $\displaystyle\dfrac{4ra^{2}-4r}{4r\|\xi\|^{2}}$
$\displaystyle=$ $\displaystyle 1\,\,\text{by}\,\,(\ref{eq:7})$
$\displaystyle=$
$\displaystyle\|k_{2}\xi\|^{2}\,\,\,\,\,\,\text{by}\,\,\text{Lemma}\,\,\ref{n}\,(2).$

In view of Remark 2, above computation shows that $T$ has two fixed points on
$\partial{B}$. Also, by the same logic as given for the parabolic case, it
cannot have a fixed point inside the ball, hence $T$ is hyperbolic.

∎

Next result explores normal isometries in $G$.

######  Theorem 3.4.

[23, Theorem 3] Let $T=e^{i\theta}\left[{\begin{array}[]{cc}UA&U(\xi)\\\
\left<\boldsymbol{\cdot},\xi\right>&a\\\ \end{array}}\right]\in G$,
$\theta\in\mathbb{R}$. Then

  1. (1)

$T$ is normal if and only if $T=e^{i\theta}\left[{\begin{array}[]{cc}UA&\xi\\\
\left<\boldsymbol{\cdot},\xi\right>&a\\\ \end{array}}\right]$, i.e.
$U(\xi)=\xi$.

  2. (2)

$T$ is unitary if and only if $T=e^{i\theta}\left[{\begin{array}[]{cc}U&0\\\
0&1\\\ \end{array}}\right]$, i.e. $\xi=0$.

  3. (3)

If $T$ is normal and $\xi\neq 0$, then
$\sigma(T)=e^{i\theta}\\{a\pm\|\xi\|\\}\cup\sigma\left(e^{i\theta}U\restriction_{\left<\xi\right>^{\perp}}\right)$,
where $a\pm\|\xi\|$ are both positive, different from $1$ and inverses of each
other. Eigenspaces corresponding to the eigenvalues $e^{i\theta}(a\pm\|\xi\|)$
are spanned by the eigenvectors $\left(\pm\dfrac{\xi}{\|\xi\|},1\right)$
respectively.

  4. (4)

In $G$, a unitary element is elliptic and a non-unitary normal isometry is
hyperbolic.

The following lemma is motivated from [4, Lemma 2.13].

######  Lemma 3.5.

For every $T\in G$, if $M$ is a subspace of $H\oplus\mathbb{C}$ such that
$T:M\longrightarrow M$ is a bijection, then $T$ maps $M^{\dagger}$ to
$M^{\dagger}$ bijectively.

###### Proof.

For $u\in M^{\dagger}$ and $v\in M$,

$\displaystyle\mathcal{A}\left(T(u),v\right)$
$\displaystyle=\mathcal{A}\left(T^{-1}\left(T(u)\right),T^{-1}(v)\right)$
$\displaystyle=\mathcal{A}\left(u,T^{-1}(v)\right)$ $\displaystyle=0$

as the isometry $T:M\longrightarrow M$ is a bijection. Using the similar
logic, we can prove that $M^{\dagger}$ is invariant under $T^{-1}$. This
proves the result. ∎

######  Proposition 3.6.

[17, pp. 20, eq. (21)] If $M$ is a non-degenerate, finite dimensional subspace
of $H\oplus\mathbb{C}$, then $H\oplus\mathbb{C}=M\oplus M^{\dagger}$.

The next lemma asserts the validity of [27, Lemma 3.10] in infinite
dimensional set up. It talks about orthogonality of time-like and light-like
vectors.

######  Lemma 3.7.

Let $(x,z),\,(y,w)\in(H\oplus\mathbb{C})\setminus\\{0\\}$ such that
$\mathcal{Q}(x,z)\leq 0$ and $\mathcal{Q}(y,w)\leq 0$. Then either
$(x,z)=k(y,w)$ for some $k\in\mathbb{C}$ or
$\mathcal{A}\left((x,z),(y,w)\right)\neq 0$.

###### Proof.

We have $\|x\|^{2}\,\,\leq|z|^{2}$ and $\|y\|^{2}\,\,\leq|w|^{2}$, $z,\,w\neq
0$. Suppose that $\mathcal{A}\left((x,z),(y,w)\right)=0$, i.e.
$\left<x,y\right>=z\overline{w}$. We will show that $(x,z)$ and $(y,w)$ are
linearly dependent. For arbitrary $k\in\mathbb{C}$,

$\displaystyle\left|z-kw\right|^{2}$ $\displaystyle=$
$\displaystyle(z-kw)(\overline{z}-\overline{kw})$ $\displaystyle=$
$\displaystyle|z|^{2}-z\overline{kw}-kw\overline{z}+\lvert kw\rvert^{2}$
$\displaystyle=$
$\displaystyle|z|^{2}-\overline{k}\left<x,y\right>-k\left<y,x\right>+\lvert
kw\rvert^{2}$ $\displaystyle\geq$
$\displaystyle\|x\|^{2}-\overline{k}\left<x,y\right>-k\left<y,x\right>+\|ky\|^{2}$
$\displaystyle=$ $\displaystyle\left<x-ky,x-ky\right>$ $\displaystyle=$
$\displaystyle||x-ky||^{2}.$

Choosing $k=\dfrac{z}{w}$ makes left hand side of the above inequality zero
and hence right hand side is also $0$ which gives that $x=ky$, i.e.
$(x,z)=(ky,z)=\left((z/w)y,z\right)=(z/w)(y,w)=k(y,w)$. ∎

######  Theorem 3.8.

[17, Corollary 3 - Witt’s Theorem, Section 2, Chapter XV] Let $(E,\psi)$ be a
complex vector space with a Hermitian form $\psi$ and $F$ and $G$ be two
finite dimensional subspaces of $E$. If $D:F\longrightarrow G$ is an onto
linear isometry, then $D$ can be extended to a surjective linear isometry of
$E$ (see the paragraph below Theorem 2 on page 380 in [17]).

##  4\. Conjugacy classes for elliptic and hyperbolic isometries

Recall that for a subspace $M$ of $H\oplus\mathbb{C}$, $M^{\dagger}$ denotes
the orthogonal complement of $M$ with respect to the Hermitian form
$\mathcal{A}$.

Proof of Theorem 1.2. Let $h\in Aut(B)$ be the isometry corresponding to $T$
(cf. Theorem 2.4) such that $h$ fixes $b\in B$.
(1) Using transitivity of elements of $Aut(B)$, there exists an isometry $m\in
Aut(B)$ such that $m(0)=b$. Then $m^{-1}h\,m(0)=0$ thereby making $m^{-1}h\,m$
unitary using Lemma 2.1 and Lemma 2.2. Hence $T$ is unitary upto conjugacy.
(2) Let $S$ be the linearization of $m$ and $S^{-1}TS=\widetilde{U}$ where
$\widetilde{U}$ is a unitary operator. As conjugacy preserves spectrum,
$\sigma(T)=\sigma(\widetilde{U})\subseteq S^{1}$.
(3) Since $b\in B$, $(b,1)$ is a time-like eigenvector of $T$ (cf. Remark 2).
Conversely, if $(x,z)$ is a time-like eigenvector of an isometry
$T^{\prime}\in Aut(B)$, then $z\neq 0$ as $\|x\|<|z|$. Thus $(z^{-1}x,1)$ is a
time-like eigenvector of $T^{\prime}$ where $\lvert z^{-1}x\rvert<1$ thereby
making $T^{\prime}$ an elliptic isometry using Remark 2.
(4) Without loss of generality, a time-like vector is of the form $(x,1)$
where $x\in B$. Using Proposition 3.6,
$H\oplus\,\mathbb{C}=\left<(x,1)\right>\oplus\,\left<(x,1)\right>^{\dagger}$.
By Lemma 3.5, $T$ maps ${\left<(x,1)\right>^{\dagger}}$ to
${\left<(x,1)\right>^{\dagger}}$ bijectively. Thus $T=T_{1}\oplus T_{2}$ where
$T_{1}=T\restriction_{\left<(x,1)\right>}$ and
$T_{2}=T\restriction_{\left<(x,1)\right>^{\dagger}}$. Also
$\mathcal{Q}\restriction_{\left<(x,1)\right>^{\dagger}}$ is positive definite
by Lemma 3.7 making $\left<(x,1)\right>^{\dagger}$ an inner product space. We
can construct a linear isometry taking $\left<(x,1)\right>$ to $\mathbb{C}$.
By Witt’s theorem (Theorem 3.8), there exists a linear isometry from
$\left<(x,1)\right>^{\dagger}$ onto $\mathbb{C}^{\dagger}=H$. This makes
$\left<(x,1)\right>^{\dagger}$ complete as $H$ is so. Hence $T_{2}$ is a
unitary operator. ∎

###  4.1. Proof of Theorem 1.3

The following result characterizes non-unitary normal isometries on the basis
of fixed points.

######  Lemma 4.1.

A non-unitary isometry of $G$ is normal if and only if it has two eigenvectors
of the type $(\pm x_{0},1)$. Moreover the projections $\pm x_{0}$ lie on
$\partial{B}$ (cf. Remark 2).

###### Proof.

Forward part follows from Theorem 3.4 (3). For the other way, let
$T=\left[{\begin{array}[]{cc}UA&U(\xi)\\\
\left<\boldsymbol{\cdot},\xi\right>&a\end{array}}\right]$, $\xi\neq 0$, be
such that $T(\pm x_{0},1)=\lambda_{i}(\pm x_{0},1)$, $i=1,\,2$ for some
$\lambda_{1},\,\lambda_{2}\in\mathbb{C}$. This gives that $\pm
UA(x_{0})+U(\xi)=\pm\lambda_{i}x_{0}$ and
$\pm\left<x_{0},\xi\right>+a=\lambda_{i}$. Substituting the value of
$\lambda_{i}$ from one equation into the other, we get $\pm
UA(x_{0})+U(\xi)=\pm\left(\pm\left<x_{0},\xi\right>+a\right)x_{0}$, i.e.

(4.1) $\displaystyle UA(x_{0})+U(\xi)$
$\displaystyle=\left<x_{0},\xi\right>x_{0}+ax_{0}\,\,\,\,\text{and}$ (4.2)
$\displaystyle-UA(x_{0})+U(\xi)$
$\displaystyle=\left<x_{0},\xi\right>x_{0}-ax_{0}.$

Observe that the fact that $T$ is non-unitary forces $x_{0}$ to be non-zero.
On solving (4.1) and (4.2) we get $U(\xi)=\left<x_{0},\xi\right>x_{0}$ and
$UA(x_{0})=ax_{0}$. Since $\left<x_{0},\xi\right>\neq 0$, we further have
$U^{-1}(x_{0})=\dfrac{\xi}{\left<x_{0},\xi\right>}=\dfrac{A(x_{0})}{a}$. As
$A(\xi)=a\xi$ (cf. Proposition 1.1),
$\xi=aA^{-1}(\xi)=\left<x_{0},\xi\right>x_{0}=U(\xi)$ thereby making $T$ a
normal isometry (cf. Theorem 3.4 (1)). Now as
$\left<x_{0},\xi\right>x_{0}=\xi$,
$\|\xi\|^{2}=\left<\xi,\xi\right>=\left<\left<x_{0},\xi\right>x_{0},\xi\right>={\left(\left<x_{0},\xi\right>\right)}^{2}$.
Hence $\left<x_{0},\xi\right>=\pm\|\xi\|$ and
$x_{0}=\pm\dfrac{\xi}{\|\xi\|}\in\partial{B}$. ∎

######  Lemma 4.2.

$Aut(B)$ acts bi-transitively on $\partial B$.

###### Proof.

The broad approach of the proof is based on [5, Proposition 2.1.3]. Let
$x_{1},\,y_{1},\,x_{2},y_{2}\in\partial B$ be such that $x_{1}\neq x_{2}$ and
$y_{1}\neq y_{2}$. We have to show the existence of an element $m\in Aut(B)$
satisfying $m(x_{1})=y_{1}$ and $m(x_{2})=y_{2}$. Clearly the elements
$(x_{i},1),\,(y_{i},1),\,\,i=1,2$ are light-like vectors. Let
$W=span\,\\{(x_{1},1),(x_{2},1)\\}$. Define a linear map $T:W\longrightarrow
H\oplus\,\mathbb{C}$ as

$T(x_{1},1)=\mu(y_{1},1)$ $T(x_{2},1)=(y_{2},1)$

where
$\mu=\dfrac{\mathcal{A}\Big{(}(x_{1},1),(x_{2},1)\Big{)}}{\mathcal{A}\Big{(}(y_{1},1),(y_{2},1)\Big{)}}$
which makes sense by Lemma 3.7. We will see that with this choice of $\mu$,
$T$ becomes an isometry. As $H\oplus\mathbb{C}$ is a complex Hilbert space, it
is enough to show that $\mathcal{Q}(T(\mathbb{x}))=\mathcal{Q}(\mathbb{x})$
for every $\mathbb{x}\in W$. Further as $(x_{1},1)$ and $(x_{2},1)$ are light-
like vectors, it suffices to show that
$\mathcal{A}\Big{(}T(x_{1},1),T(x_{2},1)\Big{)}=\mathcal{A}((x_{1},1),(x_{2},1))$.
Consider

$\displaystyle\mathcal{A}\Big{(}T(x_{1},1),T(x_{2},1)\Big{)}$
$\displaystyle=\mathcal{A}(\mu(y_{1},1),(y_{2},1))$
$\displaystyle=\mu\mathcal{A}((y_{1},1),(y_{2},1))$
$\displaystyle=\mathcal{A}((x_{1},1),(x_{2},1)).$

Using Witt’s theorem (Theorem 3.8), $T$ can be extended to an isometry
$\widetilde{T}$ of $H\oplus\,\mathbb{C}$. So, we have an element
$\widetilde{T}\in G$. Thus it is of the form some
$e^{i\theta}\left[{\begin{array}[]{cc}UA&U(\xi)\\\
\left<\boldsymbol{\cdot},\xi\right>&a\\\ \end{array}}\right]$ such that
$e^{i\theta}\left[{\begin{array}[]{cc}UA&U(\xi)\\\
\left<\boldsymbol{\cdot},\xi\right>&a\\\
\end{array}}\right]\left[{\begin{array}[]{c}x_{i}\\\ 1\\\
\end{array}}\right]=\mu_{i}\left[{\begin{array}[]{c}y_{i}\\\ 1\\\
\end{array}}\right]$ where $\mu_{1}=\mu$ and $\mu_{2}=1$. This gives that
$e^{i\theta}\Big{(}UA(x_{i})+U(\xi)\Big{)}=\mu_{i}y_{i}$ and
$e^{i\theta}\Big{(}\left<x_{i},\xi\right>+a\Big{)}=\mu_{i}$ which implies that
$UA(x_{i})+U(\xi)=\Big{(}\left<x_{i},\xi\right>+a\Big{)}y_{i}$. We have
$x\mapsto\dfrac{UA(x)+U(\xi)}{\left<x,\xi\right>+a}=m(x)$ say, is the
corresponding isometry in $Aut(B)$ satisfying $m(x_{i})=y_{i},\,\,i=1,2.$ ∎

Proof of Theorem 1.3. Let $h\in Aut(B)$ be the isometry corresponding to $T$
which fixes $y_{1},\,y_{2}\in\partial{B}$.
(1) Lemma 4.2 says that there exists $m\in Aut(B)$ such that for $\pm
x\in\partial{B}$, $m(y_{1})=x$ and $m(y_{2})=-x$. This implies the isometry
$m\,h\,m^{-1}$ fixes $\pm x$. It is non-unitary being conjugate to a
hyperbolic isometry and hence its linearization is a normal isometry by Remark
2 and Lemma 4.1.
(2) Let $R$ be a linearization of $m\,h\,m^{-1}$. So $R$ is a non-unitary
normal isometry. As conjugacy preserves spectrum, $\sigma(T)=\sigma(R)$. Rest
follows by part (3) of Theorem 3.4.
(3) By Remark 2, $(y_{1},1)$ and $(y_{2},1)$ are eigenvectors of $T$.
Let $M=span\,\\{(y_{1},1),(y_{2},1)\\}$. Then $M$ is non-degenerate. This
gives that $H\oplus\,\mathbb{C}=M\oplus\,M^{\dagger}$ using Proposition 3.6.
Also Lemma 3.5 implies that $T$ maps $M^{\dagger}$ to $M^{\dagger}$
bijectively, hence $T=T_{1}\oplus\,T_{2}$ where $T_{1}=T\restriction_{M}$ and
$T_{2}=T\restriction_{M^{\dagger}}$. By Lemma 3.7,
$\mathcal{Q}\restriction_{M^{\dagger}}$ is positive definite making
$M^{\dagger}$ an inner product space. For $x\in\partial{B}$, we can construct
a linear isometry mapping $\left<(y_{1},1)\right>$ to $\left<(x,1)\right>$ and
$\left<(y_{2},1)\right>$ to $\left<(-x,1)\right>$. Using Witt’s theorem
(Theorem 3.8), there exists a linear isometry from $M^{\dagger}$ onto
$(\left<x\right>\oplus\mathbb{C})^{\dagger}$. As
$(\left<x\right>\oplus\mathbb{C})^{\dagger}=(\left<x\right>\oplus\mathbb{C})^{\perp}$
and $(\left<x\right>\oplus\mathbb{C})^{\perp}$ is complete, this makes
$M^{\dagger}$ complete and hence $T_{2}$ is a unitary operator.∎

##  5\. Conjugacy classes for parabolic isometries

###  5.1. Stabilizer of infinity

We will be following [5] in our approach and study parabolic isometries on
Siegel domain model.
Let $e\in\partial{B}$. We write
$H\oplus\,\mathbb{C}=\left<e\right>\oplus\,\left<e\right>^{\perp}\oplus\,\mathbb{C}$
where $\left<e\right>^{\perp}$ has $\\{e_{j},\,\,j\in I\\}$ as orthonormal
basis where $I$ is an indexing set. A general element $(x,z)\in
H\oplus\,\mathbb{C}$ has the representation
$(x,z)=\left<x,e\right>(e,0)+(x^{\prime},0)+z(0,1)$.
In the following, we shall describe the Siegel domain model along with the
isometries of it and then examine those isometries of the model which fix
$\infty$.
Consider the sesquilinear form

(5.1)
$\displaystyle\widehat{\mathcal{A}}\left((x,z),(y,w)\right)=-z\overline{\left<y,e\right>}-\left<x,e\right>\overline{w}+\left<x^{\prime},y^{\prime}\right>.$

Observe that
$\widehat{\mathcal{A}}\left((x,z),(y,w)\right)=\left<\widehat{A^{\prime}}(x,z),(y,w)\right>$
where

$\displaystyle\widehat{A^{\prime}}(x,z)=(-ze+x^{\prime},-\left<x,e\right>).$

Hence

(5.5) $\displaystyle\widehat{A^{\prime}}=\left[\begin{array}[]{ccc}0&0&-1\\\
0&I&0\\\ -1&0&0\end{array}\right]$

Let $\widehat{\mathcal{Q}}$ be the quadratic form determined by
$\widehat{\mathcal{A}}$. So

(5.6)
$\displaystyle\widehat{\mathcal{Q}}(x,z)=-2\text{Re}\,(\overline{z}\left<x,e\right>)+\|x^{\prime}\|^{2}.$

If $\widehat{\mathcal{Q}}(x,z)<0$, i.e.
$\text{Re}\,(\overline{z}\left<x,e\right>)>\dfrac{1}{2}\|x^{\prime}\|^{2}$
then $z\neq 0$. Hence the projective space induced from the negative vectors
is given by

$\Sigma=\left\\{x\in
H\,\,:\,\,\text{Re}\,\left<x,e\right>>\dfrac{1}{2}\|x^{\prime}\|^{2}\right\\}.$

We call it the Siegel domain. Consider a linear transformation $D$ on
$H\oplus\mathbb{C}$ which takes $(e,0)$ to $\dfrac{(e,1)}{\sqrt{2}}$, $(0,1)$
to $\dfrac{(-e,1)}{\sqrt{2}}$ and is identity on the rest. Notice that such
$D$ is unitary as it takes an orthonormal basis to an orthonormal basis. The
operator $D$ has the following representation.

$\displaystyle
D=\left[\begin{array}[]{ccc}{}\dfrac{1}{\sqrt{2}}&{}0&{}-\dfrac{1}{\sqrt{2}}\\\
{}0&{}I&{}0\\\
{}\dfrac{1}{\sqrt{2}}&{}0&{}\dfrac{1}{\sqrt{2}}\end{array}\right].$

As $D$ is unitary,
$D^{-1}=D^{*}=\left[\begin{array}[]{ccc}{}\dfrac{1}{\sqrt{2}}&{}0&{}\dfrac{1}{\sqrt{2}}\\\
{}0&{}I&{}0\\\
{}-\dfrac{1}{\sqrt{2}}&{}0&{}\dfrac{1}{\sqrt{2}}\end{array}\right]$.

######  Proposition 5.1.

The Hermitian forms $\mathcal{A}$ and $\widehat{\mathcal{A}}$ are equivalent,
i.e. $D^{-1}A^{\prime}D=\widehat{A^{\prime}}$ (see (3.4) and (5.5)).

###### Proof.

$\displaystyle D^{-1}A^{\prime}D$ $\displaystyle=$
$\displaystyle\left[\begin{array}[]{ccc}{}\dfrac{1}{\sqrt{2}}&{}0&{}\dfrac{1}{\sqrt{2}}\\\
{}0&{}I&{}0\\\
{}-\dfrac{1}{\sqrt{2}}&{}0&{}\dfrac{1}{\sqrt{2}}\end{array}\right]\left[\begin{array}[]{ccc}1&0&0\\\
0&I&0\\\
0&0&-1\end{array}\right]\left[\begin{array}[]{ccc}{}\dfrac{1}{\sqrt{2}}&{}0&{}-\dfrac{1}{\sqrt{2}}\\\
{}0&{}I&{}0\\\
{}\dfrac{1}{\sqrt{2}}&{}0&{}\dfrac{1}{\sqrt{2}}\end{array}\right]$
$\displaystyle=$
$\displaystyle\left[\begin{array}[]{ccc}{}\dfrac{1}{\sqrt{2}}&{}0&{}-\dfrac{1}{\sqrt{2}}\\\
{}0&{}I&{}0\\\
-{}\dfrac{1}{\sqrt{2}}&{}0&-{}\dfrac{1}{\sqrt{2}}\end{array}\right]\left[\begin{array}[]{ccc}{}\dfrac{1}{\sqrt{2}}&{}0&{}-\dfrac{1}{\sqrt{2}}\\\
{}0&{}I&{}0\\\
{}\dfrac{1}{\sqrt{2}}&{}0&{}\dfrac{1}{\sqrt{2}}\end{array}\right]$
$\displaystyle=$ $\displaystyle\left[\begin{array}[]{ccc}0&0&-1\\\ 0&I&0\\\
-1&0&0\end{array}\right].$

∎

Next we will see that the projectivization of $D^{-1}$ is a Cayley map which
is a biholomorphism between the unit ball $B$ and the Siegel domain $\Sigma$.
For a time-like vector $(x,1)$,

$\displaystyle D^{-1}(x,1)$ $\displaystyle=$
$\displaystyle\left(\dfrac{\left<x,e\right>e}{\sqrt{2}}+x^{\prime}+\dfrac{e}{\sqrt{2}},-\dfrac{\left<x,e\right>}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}\right)$
$\displaystyle=$
$\displaystyle\dfrac{\left<x,e\right>+1}{\sqrt{2}}(e,0)+(x^{\prime},0)+\dfrac{1-\left<x,e\right>}{\sqrt{2}}(0,1)$
$\displaystyle=$
$\displaystyle\dfrac{1-\left<x,e\right>}{\sqrt{2}}\left(\dfrac{1+\left<x,e\right>}{1-\left<x,e\right>}(e,0)+\dfrac{\sqrt{2}}{1-\left<x,e\right>}(x^{\prime},0)+(0,1)\right)$

which is well defined as Cauchy-Schwarz inequality gives
$|\left<x,e\right>|\leq\|x\|\|e\|=\|x\|<1$, hence $1-\left<x,e\right>\neq 0$.
$D^{-1}$ induces the following projective transformation

$\mathcal{D}(x)=\dfrac{1+\left<x,e\right>}{1-\left<x,e\right>}e+\dfrac{\sqrt{2}}{1-\left<x,e\right>}x^{\prime}.$

######  Proposition 5.2.

The projective map $\mathcal{D}:B\longrightarrow\Sigma$ defined by
$x\mapsto\dfrac{1+\left<x,e\right>}{1-\left<x,e\right>}e+\dfrac{\sqrt{2}}{1-\left<x,e\right>}x^{\prime}$
is a bi-holomorphism.

###### Proof.

First we will show that $\mathcal{D}(B)\subseteq\Sigma$. Consider for $x\in B$

$\displaystyle\text{Re}\,\left(\dfrac{1+\left<x,e\right>}{1-\left<x,e\right>}\right)$
$\displaystyle=$
$\displaystyle\text{Re}\,\left(\left(\dfrac{1+\left<x,e\right>}{1-\left<x,e\right>}\right)\left(\dfrac{1-\left<e,x\right>}{1-\left<e,x\right>}\right)\right)$
$\displaystyle=$
$\displaystyle\text{Re}\,\left(\dfrac{1-|\left<x,e\right>|^{2}+2i\,\text{Im}\,\left<x,e\right>}{\lvert
1-\left<x,e\right>\rvert^{2}}\right)$ $\displaystyle=$
$\displaystyle\dfrac{1-|\left<x,e\right>|^{2}}{\lvert
1-\left<x,e\right>\rvert^{2}}.$

As $x\in B$,
$\|x\|^{2}=\|\left<x,e\right>e+x^{\prime}\|^{2}=|\left<x,e\right>|^{2}+\|x^{\prime}\|^{2}<1$,
i.e. $1-|\left<x,e\right>|^{2}>\|x^{\prime}\|^{2}$. Hence

$\text{Re}\,\left(\dfrac{1+\left<x,e\right>}{1-\left<x,e\right>}\right)>\dfrac{\|x^{\prime}\|^{2}}{\lvert
1-\left<x,e\right>\rvert^{2}}=\dfrac{1}{2}\left\|\dfrac{\sqrt{2}}{1-\left<x,e\right>}x^{\prime}\right\|^{2}.$

As $D^{-1}(x,1)=\alpha(x)(\mathcal{D}(x),1)$ for some
$\alpha(x)\in\mathbb{C}$, bijectivity of $D^{-1}$ implies that of
$\mathcal{D}$. The holomorphicity of the map is easy to establish.

∎

######  Remark 3.

Let $\mathcal{D}(e)$ be denoted by $\infty$ and
$\overline{\Sigma}=\Big{\\{}x\in
H\,\,:\,\,\text{Re}\,\left<x,e\right>\geq\frac{1}{2}\|x^{\prime}\|^{2}\Big{\\}}\cup\\{\infty\\}$.
So, $\overline{\Sigma}$ is the completion of
$\overline{\Sigma}\setminus\infty$. Observe that if for $x\in\partial{B}$,
$\left<x,e\right>=1$, then $1=|\left<x,e\right>|\leq\|x\|=1$ and hence $x=ce$,
for some $c\in\mathbb{C}$, by Cauchy-Schwarz inequality. This gives that $x=e$
and we have $1-\left<x,e\right>\neq 0$ for $x\in\partial{B}\setminus\\{e\\}$.
Proof of the above Proposition tells that $\mathcal{D}$ maps $\partial{B}$ to
$\partial{\Sigma}$ bijectively. Also
$\mathcal{D}:\overline{B}\setminus\\{e\\}\longrightarrow\overline{\Sigma}\setminus\\{\infty\\}$
is a homeomorphism. We define neighbourhoods of $\infty$ to be images of
neighbourhoods of $e$. This makes $\overline{\Sigma}$ a topological space and
$\mathcal{D}:\overline{B}\longrightarrow\overline{\Sigma}$, a homeomorphism.
The projective transformation $\mathcal{D}$ is called the Cayley map. Metric
on $\Sigma$ can be defined as the pull back of the Carathéodory metric on $B$
via Cayley map.

Let $\widehat{G}$ be the group of all bijective bounded linear operators on
$H\oplus\mathbb{C}$ leaving $\widehat{\mathcal{A}}$ invariant. Proposition 5.1
tells that

$\widehat{G}=\\{\widehat{T}\in
B(H\oplus\mathbb{C})\,\,:\,\,\widehat{T}\,\,\text{is bijective
and}\,\,\widehat{T}=D^{-1}TD,\,\,T\in G\\}.$

Also elements of the group $Aut(\Sigma)$ are the projective maps induced from
elements of $\widehat{G}$ where the group $Aut(\Sigma)$ is defined as follows.

$Aut(\Sigma)=\\{\mathcal{D}h\mathcal{D}^{-1},\,\,\,\,h\in Aut(B)\\}.$

######  Remark 4.

Cayley map gives fixed point classification for isometries in $Aut(\Sigma)$.
We call a vector $(x,z)$ time-like, light-like or space-like with respect to
the Hermitian form $\widehat{\mathcal{A}}$ defined in (5.1) if
$\widehat{Q}(x,z)<,=\text{or}>0$ respectively. We shall refer to these vectors
also as time-like, light-like or space-like. Meaning would be clear from the
context.

Let us denote the stabilizer group of infinity in $\boldsymbol{\widehat{G}}$
by $\boldsymbol{\widehat{G}_{\infty}}$. $\mathcal{D}(e)=\infty$. $(e,1)$ is
the lift of $e$ and $D^{-1}(e,1)={\sqrt{2}}(e,0)$. Thus $(e,0)$ will be
considered as standard lift of $\infty$.

######  Proposition 5.3 (Stabilizer group of infinity).

Let $\widehat{T}\in\widehat{G}$ satisfy $\widehat{T}(e,0)=\lambda(e,0)$, then

(5.14)
$\displaystyle\widehat{T}=\left[\begin{array}[]{ccc}\lambda&\left<\boldsymbol{\cdot},U^{-1}(a^{\prime})\right>&\mu\,s\\\
0&U&\mu\,a^{\prime}\\\ 0&0&\mu\end{array}\right]$

where $a^{\prime}\in\left<e\right>^{\perp}$ in $H$,
$U\in\mathcal{U}(\left<e\right>^{\perp})$, $\lambda,\,s\in\mathbb{C}$ obey
$\lambda\overline{\mu}=1$ and $Re\,s=\frac{1}{2}\|a^{\prime}\|^{2}$.

###### Proof.

The fact that $\widehat{T}$ leaves $\widehat{\mathcal{A}}$ invariant will be
used to establish the result. Let
$\widehat{T}(e_{j},0)=\left(b_{j}\,e+{w_{j}}^{\prime},z_{j}\right)$ and
$\widehat{T}(0,1)=\left(te+c^{\prime},\mu\right)$, where
$b_{j},\,z_{j},\,t,\,\mu\in\mathbb{C}$ and
${w_{j}}^{\prime},\,c^{\prime}\in\left<e\right>^{\perp},\,j\in I$. Using (5.1)
and (5.6), we get $\widehat{{Q}}(e,0)=\widehat{{Q}}(0,1)=0$,
$\widehat{{Q}}(e_{j},0)=1$ and
$\widehat{\mathcal{A}}\Big{(}(e,0),(e_{j},0)\Big{)}=\widehat{\mathcal{A}}\Big{(}(0,1),(e_{j},0)\Big{)}=\widehat{\mathcal{A}}\Big{(}(e_{i},0),(e_{j},0)\Big{)}\\\
=0$, $i\neq j$ and $\widehat{\mathcal{A}}\Big{(}(e,0),(0,1)\Big{)}=-1$.

Also
$\widehat{{Q}}\Big{(}\widehat{T}(e,0)\Big{)}=\widehat{{Q}}\Big{(}\lambda(e,0)\Big{)}=0$.
Since $\widehat{T}$ leaves $\widehat{\mathcal{A}}$ invariant, we have
$\widehat{{Q}}\Big{(}\widehat{T}(0,1)\Big{)}=\widehat{{Q}}(0,1)$, i.e.
$\widehat{{Q}}\left(te+c^{\prime},\mu\right)=-2\text{Re}\,(\overline{\mu}t)+\|c^{\prime}\|^{2}=0$
which gives that

$\text{Re}\,(\overline{\mu}t)=\dfrac{1}{2}\|c^{\prime}\|^{2}.$

Similarly
$\widehat{\mathcal{A}}\Big{(}\widehat{T}(e,0),\widehat{T}(0,1)\Big{)}=\widehat{\mathcal{A}}\Big{(}(e,0),(0,1)\Big{)}$,
i.e.
$\widehat{\mathcal{A}}\Big{(}\lambda(e,0),\left(te+c^{\prime},\mu\right)\Big{)}=-\lambda\overline{\mu}=-1$
implying

$\lambda\overline{\mu}=1.$

$\widehat{\mathcal{A}}\Big{(}\widehat{T}(e,0),\widehat{T}(e_{j},0)\Big{)}=\widehat{\mathcal{A}}\Big{(}(e,0),(e_{j},0)\Big{)}$,
i.e.
$\Big{(}\lambda(e,0),\left(b_{j}\,e+{w_{j}}^{\prime},z_{j}\right)\Big{)}=-\lambda\overline{z_{j}}=0$
giving that

$z_{j}=0.$

$\widehat{\mathcal{A}}\Big{(}\widehat{T}(0,1),\widehat{T}(e_{j},0)\Big{)}=\widehat{\mathcal{A}}\Big{(}(0,1),(e_{j},0)\Big{)}$
, i.e.
$\widehat{\mathcal{A}}\Big{(}\left(te+c^{\prime},\mu\right),\left(b_{j}\,e+{w_{j}}^{\prime},0\right)\Big{)}=-\mu\overline{b_{j}}+\left<c^{\prime},{w_{j}}^{\prime}\right>=0$
which yields that
$\overline{\mu}b_{j}=\left<{w_{j}}^{\prime},c^{\prime}\right>$, i.e.

(5.15) $\displaystyle b_{j}=\lambda\left<{w_{j}}^{\prime},c^{\prime}\right>.$

$\widehat{{Q}}\Big{(}\widehat{T}(e_{j},0)\Big{)}=\widehat{{Q}}(e_{j},0)$, i.e.
$\widehat{{Q}}\left(b_{j}\,e+{w_{j}}^{\prime},0\right)=\|{w_{j}}^{\prime}\|^{2}=1$.

Lastly for $i\neq j$,
$\widehat{\mathcal{A}}\Big{(}\widehat{T}(e_{i},0),\widehat{T}(e_{j},0)\Big{)}=\widehat{\mathcal{A}}\Big{(}(e_{i},0),(e_{j},0)\Big{)}$,
i.e.
$\widehat{\mathcal{A}}\Big{(}\left(b_{i}\,e+{w_{i}}^{\prime},0\right),\left(b_{j}\,e+{w_{j}}^{\prime},0\right)\Big{)}=\left<{w_{i}}^{\prime},{w_{j}}^{\prime}\right>=0$.

The above computation asserts that the collection
$\\{{w_{j}}^{\prime},\,\,j\in I\\}$ is orthonormal in
$\left<e\right>^{\perp}$.
Define a linear operator
$U:\left<e\right>^{\perp}\longrightarrow\left<e\right>^{\perp}$ as
$U(e_{j})={w_{j}}^{\prime}$. Then $U$ is bounded and norm preserving. Let
$x\in H$ where
$x^{\prime}=\sum\limits_{j=1_{x}}^{n_{x}}\left<x^{\prime},e_{j}\right>e_{j}$.
Observe that

$\displaystyle\sum\limits_{j=1_{x}}^{n_{x}}\left<x^{\prime},e_{j}\right>b_{j}$
$\displaystyle=\sum\limits_{j=1_{x}}^{n_{x}}\left<x^{\prime},e_{j}\right>\lambda\left<U(e_{j}),c^{\prime}\right>\,\,by\,\,(\ref{bj})$
$\displaystyle=\lambda\left<U\left(\sum\limits_{j=1_{x}}^{n_{x}}\left<x^{\prime},e_{j}\right>e_{j}\right),c^{\prime}\right>$
$\displaystyle=\lambda\left<U(x^{\prime}),c^{\prime}\right>$
$\displaystyle=\lambda\left<x^{\prime},U^{-1}(c^{\prime})\right>.$

Consider

$\displaystyle\widehat{T}(x,z)$
$\displaystyle=\widehat{T}\Big{(}\left<x,e\right>(e,0)+(x^{\prime},0)+z(0,1)\Big{)}$
$\displaystyle=\left(\lambda\,\left<x,e\right>(e,0)+\sum\limits_{j=1_{x}}^{n_{x}}\left<x^{\prime},e_{j}\right>\Big{(}b_{j}\,e+U(e_{j}),0\Big{)}+z\Big{(}te+c^{\prime},\mu\Big{)}\right)$
$\displaystyle=\left(\Big{(}\lambda\left<x,e\right>+\sum\limits_{j=1_{x}}^{n_{x}}\left<x^{\prime},e_{j}\right>b_{j}+z\,t\Big{)}e+\sum\limits_{j=1_{x}}^{n_{x}}\left<x^{\prime},e_{j}\right>U(e_{j})+zc^{\prime},z\mu\right)$
$\displaystyle=\bigg{(}\Big{(}\lambda\left<x,e\right>+\lambda\left<x^{\prime},U^{-1}(c^{\prime})\right>+z\,t\Big{)}e+U(x^{\prime})+zc^{\prime},z\mu\bigg{)}$
$\displaystyle=\left[\begin{array}[]{ccc}\lambda&\lambda\left<\boldsymbol{\cdot},U^{-1}(c^{\prime})\right>&t\\\
0&U&c^{\prime}\\\
0&0&\mu\end{array}\right]\left[\begin{array}[]{c}\left<x,e\right>\\\
x^{\prime}\\\ z\end{array}\right].$

We will see that $U$ is surjective which will make it a unitary operator. As
$\widehat{T}$ is surjective, for $y^{\prime}\in\left<e\right>^{\perp}$, there
exists $(x,z)\in H\oplus\mathbb{C}$ such that $\widehat{T}(x,z)=y^{\prime}$.
This implies that $U(x^{\prime})+zc^{\prime}=y^{\prime}$ and $z\mu=0$. This
further yields that $z=0$ and hence $U(x^{\prime})=y^{\prime}$ thereby making
$U$ a unitary operator.
$\widehat{T}$ can also be written as
$\left[\begin{array}[]{ccc}\lambda&\lambda\left<\boldsymbol{\cdot},U^{-1}\left(\dfrac{\mu
c^{\prime}}{\mu}\right)\right>&\dfrac{\mu t}{\mu}\\\ 0&U&\dfrac{\mu
c^{\prime}}{\mu}\\\ 0&0&\mu\end{array}\right]$. By taking
$\dfrac{c^{\prime}}{\mu}=a^{\prime}$ and $\dfrac{t}{\mu}=s$, we get

$\widehat{T}=\left[\begin{array}[]{ccc}\lambda&\left<\boldsymbol{\cdot},U^{-1}(a^{\prime})\right>&\mu\,s\\\
0&U&\mu\,a^{\prime}\\\ 0&0&\mu\end{array}\right]$

where ${\lambda}\overline{\mu}=1$ and $\text{Re}(\overline{\mu}\mu
s)=\dfrac{1}{2}\|\mu a^{\prime}\|^{2}$, i.e.
$\text{Re}\,s=\dfrac{1}{2}\|a^{\prime}\|^{2}$.

∎

######  Corollary 5.3.1.

If $\widehat{T}\in\widehat{G}_{\infty}$ is such that $s=0$, cf. (5.14) then
$\widehat{T}$ is non-parabolic.

###### Proof.

Observe from Proposition 5.3 that if $s=0$, $a^{\prime}=0$, so
$\widehat{T}=\left[\begin{array}[]{ccc}\lambda&0&0\\\ 0&U&0\\\
0&0&\mu\end{array}\right]$ which has $(0,1)$ and $(e,0)$ as light-like
eigenvectors. Hence $\widehat{T}$ is non-parabolic. ∎

######  Remark 5.

In Proposition 5.3, $\lambda\overline{\mu}=1$, i.e
${\mu}=\dfrac{1}{\overline{\lambda}}$. This gives that $|\lambda|=1$ if and
only if $\lambda=\mu$.

######  Corollary 5.3.2 (Iwasawa decomposition).

Every element in $\widehat{G}_{\infty}$ is the product of a translation
$(1,a^{\prime},s)=\left[\begin{array}[]{ccc}1&\left<\boldsymbol{\cdot},a^{\prime}\right>&s\\\
0&I&a^{\prime}\\\ 0&0&1\\\ \end{array}\right]$
($\text{Re}\,s=\dfrac{1}{2}\|a^{\prime}\|^{2}$), a rotation
$R_{U}=\left[\begin{array}[]{ccc}1&0&0\\\ 0&U&0\\\ 0&0&1\end{array}\right]$
where $U$ is a unitary operator and a dilation
$D_{\lambda}=\left[\begin{array}[]{ccc}\lambda&0&0\\\ 0&I&0\\\
0&0&\mu\end{array}\right]$ where $\lambda\overline{\mu}=1$.

###### Proof.

We have
$(1,a^{\prime},s)R_{U}D_{\lambda}=\left[\begin{array}[]{ccc}1&\left<\boldsymbol{\cdot},a^{\prime}\right>&s\\\
0&I&a^{\prime}\\\ 0&0&1\end{array}\right]\left[\begin{array}[]{ccc}1&0&0\\\
0&U&0\\\ 0&0&1\end{array}\right]\left[\begin{array}[]{ccc}\lambda&0&0\\\
0&I&0\\\ 0&0&\mu\end{array}\right]$
$=\left[\begin{array}[]{ccc}1&\left<\boldsymbol{\cdot},U^{-1}(a^{\prime})\right>&s\\\
0&U&a^{\prime}\\\
0&0&1\end{array}\right]\left[\begin{array}[]{ccc}\lambda&0&0\\\ 0&I&0\\\
0&0&\mu\end{array}\right]=\left[\begin{array}[]{ccc}\lambda&\left<\boldsymbol{\cdot},U^{-1}(a^{\prime})\right>&\mu\,s\\\
0&U&\mu\,a^{\prime}\\\ 0&0&\mu\end{array}\right]$. ∎

######  Proposition 5.4.

The projective transformation induced by $\widehat{T}\in\widehat{G}_{\infty}$
satisfying (5.14) is an affine map
$\mathcal{\widehat{T}}:\Sigma\longrightarrow\Sigma$ given by

$\mathcal{\widehat{T}}(x)=\overline{\lambda}\left[\begin{array}[]{cc}\lambda&\left<\boldsymbol{\cdot},U^{-1}(a^{\prime})\right>\\\
0&U\\\ \end{array}\right]\left[\begin{array}[]{c}\left<x,e\right>\\\
x^{\prime}\\\ \end{array}\right]+\left[\begin{array}[]{c}s\\\ a^{\prime}\\\
\end{array}\right].$

###### Proof.

Since the $\mathbb{C}$-component of a time-like vector with respect to
$\widehat{Q}$ is non-vanishing, so without loss of generality, let $(x,1)$ be
a time-like vector with respect to $\widehat{Q}$, i.e. $x\in\Sigma$. Then
$\widehat{T}(x,1)=\bigg{(}\Big{(}\lambda\left<x,e\right>+\left<U(x^{\prime}),a^{\prime}\right>+\mu
s\Big{)}e+U(x^{\prime})+\mu
a^{\prime},\mu\bigg{)}=\mu\Bigg{(}\bigg{(}|\lambda|^{2}\left<x,e\right>+\overline{\lambda}\left<U(x^{\prime}),a^{\prime}\right>+s\bigg{)}e+\overline{\lambda}U(x^{\prime})+a^{\prime},1\Bigg{)}$.
Hence the projective map say
$\mathcal{\widehat{T}}:\Sigma\longrightarrow\Sigma$, induced by $\widehat{T}$
is given by

$\displaystyle\mathcal{\widehat{T}}(x)$
$\displaystyle=\bigg{(}|\lambda|^{2}\left<x,e\right>+\overline{\lambda}\left<U(x^{\prime}),a^{\prime}\right>+s\bigg{)}e+\overline{\lambda}U(x^{\prime})+a^{\prime}$
$\displaystyle=\overline{\lambda}\left[\begin{array}[]{cc}\lambda&\left<\boldsymbol{\cdot},U^{-1}(a^{\prime})\right>\\\
0&U\\\ \end{array}\right]\left[\begin{array}[]{c}\left<x,e\right>\\\
x^{\prime}\\\ \end{array}\right]+\left[\begin{array}[]{c}s\\\ a^{\prime}\\\
\end{array}\right].$

∎

Now we compute the spectrum of the isometry obtained in Proposition 5.3.

######  Lemma 5.5.

Let $\widehat{T}\in\widehat{G}_{\infty}$ be of the form (5.14) such that
$s\neq 0$. Then
$\sigma(\widehat{T})\setminus\\{\lambda,\mu\\}=\sigma(U)\setminus\\{\lambda,\mu\\}.$

###### Proof.

We will show that for any $r\in\mathbb{C}\setminus\\{\lambda,\mu\\}$,
$\widehat{T}-rI$ is bijective if and only if $U-rI$ is bijective. Let
$r\in\mathbb{C}\setminus\\{\lambda,\mu\\}$. If $U-rI$ is one-one,
$\widehat{T}-rI$ is one-one, for
$(\widehat{T}-rI)(x,z)=\left[\begin{array}[]{ccc}\lambda-r&\left<\boldsymbol{\cdot},U^{-1}(a^{\prime})\right>&\mu
s\\\ 0&U-rI&\mu a^{\prime}\\\
0&0&\mu-r\end{array}\right]\left[\begin{array}[]{c}\left<x,e\right>\\\
x^{\prime}\\\ z\end{array}\right]\equiv 0$ implies $z=0$, $x^{\prime}=0$ as
$U-rI$ is injective and hence $\left<x,e\right>=0$. Conversely, suppose
$\widehat{T}-rI$ is one-one and $(U-rI)(x^{\prime})=0$ for some
$x^{\prime}\in\left<e\right>^{\perp}$. We have
$(\widehat{T}-rI)\left({\dfrac{\left<x^{\prime},U^{-1}(a^{\prime})\right>}{r-\lambda}}e+x^{\prime},0\right)=\begin{bmatrix}\lambda-r&\left<\boldsymbol{\cdot},U^{-1}(a^{\prime})\right>&\mu
s\\\ 0&U-rI&\mu a^{\prime}\\\
0&0&\mu-r\end{bmatrix}\begin{bmatrix}\dfrac{\left<x^{\prime},U^{-1}(a^{\prime})\right>}{r-\lambda}\\\
x^{\prime}\\\ 0\end{bmatrix}\equiv 0$ which implies $x^{\prime}=0$. Thus
$U-rI$ is one-one. Now, suppose $U-rI$ is onto. We will see that
$\widehat{T}-rI$ is onto. Let $\Big{(}\left<y,e\right>e+y^{\prime},w\Big{)}\in
H\oplus\mathbb{C}$. Since $U-rI$ is onto, for $y^{\prime}-\dfrac{\mu
wa^{\prime}}{\mu-r}\in\left<e\right>^{\perp}$, there exists
$x^{\prime}\in\left<e\right>^{\perp}$ such that
$(U-rI)(x^{\prime})=y^{\prime}-\dfrac{\mu wa^{\prime}}{\mu-r}$. Let
$\left<x,e\right>=\dfrac{\left<y,e\right>-\dfrac{\mu
sw}{\mu-r}-\left<x^{\prime},U^{-1}(a^{\prime})\right>}{\lambda-r}$. With this
choice, consider
$(\widehat{T}-rI)\left(x,\dfrac{w}{\mu-r}\right)=\left[\begin{array}[]{ccc}\lambda-r&\left<\boldsymbol{\cdot},U^{-1}(a^{\prime})\right>&\mu
s\\\ 0&U-rI&\mu a^{\prime}\\\
0&0&\mu-r\end{array}\right]\left[\begin{array}[]{c}\left<x,e\right>\\\
x^{\prime}\\\ \dfrac{w}{\mu-r}\\\ \end{array}\right]=\\\
\Bigg{(}\bigg{(}(\lambda-r)\left<x,e\right>+\left<x^{\prime},U^{-1}(a^{\prime})\right>+\dfrac{\mu
sw}{\mu-r}\bigg{)}e+(U-rI)(x^{\prime})+\dfrac{\mu
wa^{\prime}}{\mu-r},w\Bigg{)}=\Big{(}\left<y,e\right>e+y^{\prime},w\Big{)}$.
Conversely, let $\widehat{T}-rI$ be surjective and
$y^{\prime}\in\left<e\right>^{\perp}$. Then $(y^{\prime},0)\in
H\oplus\mathbb{C}$ and by surjectivity of $\widehat{T}-rI$ there exists
$(x,z)=\Big{(}\left<x,e\right>e+x^{\prime},z\Big{)}\in H\oplus\mathbb{C}$ such
that

$\displaystyle(\widehat{T}-rI)(x,z)$
$\displaystyle=\left[\begin{array}[]{ccc}\lambda-r&\left<\boldsymbol{\cdot},U^{-1}(a^{\prime})\right>&\mu
s\\\ 0&U-rI&\mu a^{\prime}\\\
0&0&\mu-r\end{array}\right]\left[\begin{array}[]{c}\left<x,e\right>\\\
x^{\prime}\\\ z\end{array}\right]$
$\displaystyle=\left[\begin{array}[]{c}(\lambda-r)\left<x,e\right>+\left<x^{\prime},U^{-1}(a^{\prime})\right>+\mu
sz\\\ (U-rI)x^{\prime}+\mu a^{\prime}z\\\
(\mu-r)z\end{array}\right]=\left[\begin{array}[]{c}0\\\ y^{\prime}\\\
0\end{array}\right].$

This implies $z=0$ ($\mu\neq r$) and hence $(U-rI)(x^{\prime})=y^{\prime}$
making $U-rI$ surjective. This proves the lemma.

∎

######  Proposition 5.6.

If $\widehat{T}\in\widehat{G}_{\infty}$ is of the form (5.14) such that $s\neq
0$, then $\sigma(\widehat{T})=\\{\lambda,\mu\\}\cup\sigma(U)$.

###### Proof.

Clearly, $\lambda\in\sigma(\widehat{T})$. For $\mu\neq\lambda$, we will see
that $\mu$ is also an eigenvalue of $\widehat{T}$, i.e.
${\text{ker}}(\widehat{T}-\mu I)\neq\\{0\\}$.

$(\widehat{T}-\mu
I)(x,z)=\left[\begin{array}[]{ccc}\lambda-\mu&\left<\boldsymbol{\cdot},U^{-1}(a^{\prime})\right>&\mu
s\\\ 0&U-\mu I&\mu a^{\prime}\\\ 0&0&0\\\
\end{array}\right]\left[\begin{array}[]{c}\left<x,e\right>\\\ x^{\prime}\\\
z\\\ \end{array}\right]=0$

implies

(5.16)
$\displaystyle(\lambda-\mu)\left<x,e\right>+\left<x^{\prime},U^{-1}(a^{\prime})\right>+\mu
sz=0$

and

(5.17) $\displaystyle(U-\mu I)(x^{\prime})+\mu a^{\prime}z=0.$

Remark 5 tells that $\mu\notin S^{1}$ and hence $U-\mu I$ is invertible as
$\sigma(U)\subseteq S^{1}$. So by (5.17),

$x^{\prime}=-\mu z(U-\mu I)^{-1}(a^{\prime})$

and then (5.16) yields that $\left<x,e\right>=\dfrac{\mu
sz+\left<x^{\prime},U^{-1}(a^{\prime})\right>}{\mu-\lambda}.$ If we choose
$z\neq 0$ then we always get a non-zero vector
$\Big{(}\left<x,e\right>e+x^{\prime},z\Big{)}\in{\text{ker}}(\widehat{T}-\mu
I)$ as $s\neq 0$. This yields
$\sigma(\widehat{T})=\bigg{(}\sigma(\widehat{T})\setminus\\{\lambda,\mu\\}\bigg{)}\cup\\{\lambda,\mu\\}=\bigg{(}\sigma(U)\setminus\\{\lambda,\mu\\}\bigg{)}\cup\\{\lambda,\mu\\}=\sigma{(U)}\cup\\{\lambda,\mu\\}$
using Lemma 5.5.

∎

###  5.2. Parabolic isometries

Observe that if $\widehat{T}\in\widehat{G}$ is a parabolic isometry of the
form (5.14), then $s\neq 0$ in view of Corollary 5.3.1.

######  Lemma 5.7.

If $\widehat{T}$ satisfying (5.14) is a parabolic isometry then $\lambda=\mu$.

###### Proof.

We will prove the result by contradiction. Suppose $\lambda\neq\mu$. As proof
of Proposition 5.6 yields that $\mu$ is an eigenvalue ($s\neq 0$, Corollary
5.3.1), let the corresponding eigenvector be $w$. Clearly, $w$ cannot be time-
like or light-like eigenvector as it would contradict the parabolic nature of
$\widehat{T}$. So, $w$ is a space-like vector. Also,
$\mathcal{\widehat{A}}(w,w)=\mathcal{\widehat{A}}\Big{(}\widehat{T}(w),\widehat{T}(w)\Big{)}=|\mu|^{2}\mathcal{\widehat{A}}(w,w)$.
Eigenvector $w$ being space-like gives $|\mu|=1$ which is a contradiction in
view of Remark 5. ∎

We shall record the preceding analysis in the following proposition.

######  Proposition 5.8.

Spectrum of a parabolic isometry is contained in the unit circle.

###### Proof.

For a parabolic isometry $\widehat{T}\in\widehat{G}$, Proposition 5.6 and
Lemma 5.7 yield $\sigma({\widehat{T}})=\\{\lambda\\}\cup\sigma(U)$. Hence
$\sigma(\widehat{T})\subseteq S^{1}$ by Remark 5. ∎

Next we have a spectral characterization of hyperbolic isometries.

######  Proposition 5.9.

An isometry is hyperbolic if and only if its spectrum is not contained in the
unit circle.

###### Proof.

Theorem 1.2 (2), Theorem 1.3 (2) and Proposition 5.8 yield the required
result. ∎

From now onwards, we will investigate parabolic isometries having singleton
spectra.

######  Lemma 5.10.

Let $\widehat{T}$ be a parabolic isometry in $\widehat{G}_{\infty}$ having
singleton spectrum (see (5.14)). Then
$\widehat{T}=\lambda\left[\begin{array}[]{ccc}1&\left<\boldsymbol{\cdot},a^{\prime}\right>&s\\\
0&I&a^{\prime}\\\ 0&0&1\end{array}\right]$ where $s\neq 0$, $|\lambda|=1$ and
$\text{Re}\,s=\dfrac{1}{2}\|a^{\prime}\|^{2}$.

###### Proof.

Let $\widehat{T}$ satisfying (5.14) have singleton spectrum. Then
$\sigma(\widehat{T})=\\{\lambda\\}$. By Proposition 5.6,
$\sigma(U)=\\{\lambda\\}$ and hence $U=\lambda\,I$. Corollary 5.3.1 tells that
$s\neq 0$. Hence the result follows. ∎

We will denote an isometry described in the above lemma by
$(\lambda,a^{\prime},s)$.

######  Proposition 5.11.

The projective maps induced by parabolic isometries in $\widehat{G}_{\infty}$
having singleton spectra are Heisenberg translations defined as
$\mathcal{\widehat{T}}_{u}:\Sigma\longrightarrow\Sigma$ such that

$\displaystyle\widehat{\mathcal{T}}_{u}(x)=\Big{(}\left<x,e\right>+\left<u,e\right>+\left<x^{\prime},u^{\prime}\right>\Big{)}e+(x^{\prime}+u^{\prime}),$

$u=\left<u,e\right>e+u^{\prime}\in\partial{\Sigma}$.

###### Proof.

Let $\widehat{T}=(\lambda,a^{\prime},s)$ be a parabolic isometry in
$\widehat{G}_{\infty}$ having singleton spectrum. Then by Proposition 5.4, its
projective map is given by
$\mathcal{\widehat{T}}(x)=\left[\begin{array}[]{cc}1&\left<\boldsymbol{\cdot},a^{\prime}\right>\\\
0&I\end{array}\right]\left[\begin{array}[]{c}\left<x,e\right>\\\
x^{\prime}\end{array}\right]+\left[\begin{array}[]{c}s\\\
a^{\prime}\end{array}\right]=\Big{(}\left<x,e\right>+s+\left<x^{\prime},a^{\prime}\right>\Big{)}e+x^{\prime}+a^{\prime}=\bigg{(}\left<x,e\right>+\Big{<}(se+a^{\prime}),e\Big{>}+\left<x^{\prime},a^{\prime}\right>\bigg{)}e+x^{\prime}+a^{\prime}$.
Let $\mathcal{\widehat{T}}=\mathcal{\widehat{T}}_{u}$ where
$u=se+a^{\prime}\in\partial{\Sigma}$. Then

$\widehat{\mathcal{T}}_{u}(x)=\Big{(}\left<x,e\right>+\left<u,e\right>+\left<x^{\prime},u^{\prime}\right>\Big{)}e+(x^{\prime}+u^{\prime}).$

∎

From now on, we shall refer to an isometry of the form
$(\lambda,a^{\prime},s)$ as Heisenberg translation. The above proposition
tells that $u^{\prime}=a^{\prime}$ is the component in the horizontal
direction. Hence

######  Definition 1.

A Heisenberg translation of the form $(\lambda,a^{\prime},s)$ is called
vertical translation if $a^{\prime}=0$ and non-vertical translation if
$a^{\prime}\neq 0$.

Recall that for a subspace $M$ of $H\oplus\mathbb{C}$, we use the same
notation $M^{\dagger}$ to denote the orthogonal complement of $M$ for the
Hermitian form $\widehat{\mathcal{A}}$ as well.

######  Proposition 5.12.

Let $\widehat{T}=(\lambda,a^{\prime},s)$ be a Heisenberg translation. Then
$\widehat{T}$ decomposes $H\oplus\mathbb{C}$ orthogonally into a two or three
dimensional subspace say $K$ and $K^{\dagger}$ such that
$\widehat{T}\restriction_{{K}^{\dagger}}=\lambda\,I$. The eigenspace
corresponding to $\lambda$ for the operator $\widehat{T}\restriction_{K}$ is
generated by a light-like eigenvector and its minimal polynomial is of maximal
degree.

###### Proof.

Let
$\widehat{T}=\lambda\left[\begin{array}[]{ccc}1&\left<\boldsymbol{\cdot},a^{\prime}\right>&s\\\
0&I&a^{\prime}\\\ 0&0&1\end{array}\right]$ be a Heisenberg translation.
Let $K=span\,\Big{\\{}(e,0),\,(a^{\prime},0),\,(0,1)\Big{\\}}$. It is easy to
see that $\widehat{T}:K\longrightarrow K$ is a bijection as
$\widehat{T}(e,0)=\lambda(e,0)$, $\widehat{T}(0,1)=\lambda(se+a^{\prime},1)$
and $\widehat{T}(a^{\prime},0)=\lambda(\|a^{\prime}\|^{2}e+a^{\prime},0)$. In
view of Lemma 3.7, $K$ is a non-degenerate subspace as it contains two light-
like vectors. We have $H\oplus\mathbb{C}=K\oplus K^{\dagger}$ by Lemma 3.6.
Observe that
$K^{\dagger}=\\{(x^{\prime},0)\,\,:\,\,\left<x^{\prime},a^{\prime}\right>=0\\}$
and ${\widehat{T}\restriction_{K^{\dagger}}}=\lambda\,I$. Let
$\widehat{T}=\widehat{T}_{1}\oplus\widehat{T}_{2}$ where
$\widehat{T}_{1}=\widehat{T}\restriction_{K}$ and
$\widehat{T}_{2}=\widehat{T}\restriction_{K^{\dagger}}$. Next we see that
$ker(\widehat{T}_{1}-\lambda\,I)=\left<(e,0)\right>$. Observe that
$(\widehat{T}-\lambda\,I)(x,z)=\lambda\left[\begin{array}[]{ccc}0&\left<\boldsymbol{\cdot},a^{\prime}\right>&s\\\
0&0&a^{\prime}\\\
0&0&0\end{array}\right]\left[\begin{array}[]{c}\left<x,e\right>\\\
x^{\prime}\\\ z\end{array}\right]=\left[\begin{array}[]{c}0\\\ 0\\\
0\end{array}\right]$
implies $\left<x^{\prime},a^{\prime}\right>+sz=0$ and $a^{\prime}z=0$ giving
$z=0$ and $\left<x^{\prime},a^{\prime}\right>=0$. Hence
$ker\,(\widehat{T}-\lambda\,I)=\left<(e,0)\right>\oplus K^{\dagger}$ and thus
$ker\,{(\widehat{T}_{1}-\lambda\,I)}=\left<(e,0)\right>$. Consider
${(\widehat{T}_{1}-\lambda\,I)}^{2}(0,1)=\lambda^{2}\left[\begin{array}[]{ccc}0&\left<\boldsymbol{\cdot},a^{\prime}\right>&s\\\
0&0&a^{\prime}\\\
0&0&0\end{array}\right]\left[\begin{array}[]{ccc}0&\left<\boldsymbol{\cdot},a^{\prime}\right>&s\\\
0&0&a^{\prime}\\\ 0&0&0\end{array}\right]\left[\begin{array}[]{c}0\\\ 0\\\
1\end{array}\right]=\\\
\lambda^{2}\left[\begin{array}[]{ccc}0&\left<\boldsymbol{\cdot},a^{\prime}\right>&s\\\
0&0&a^{\prime}\\\ 0&0&0\end{array}\right]\left[\begin{array}[]{c}s\\\
a^{\prime}\\\
0\end{array}\right]=\lambda^{2}\left[\begin{array}[]{c}\|a^{\prime}\|^{2}\\\
0\\\ 0\end{array}\right]\not\equiv 0$ for $a^{\prime}\neq 0$. Hence for a non-
vertical translation, minimal polynomial of $\widehat{T}_{1}$ is
$(x-\lambda)^{3}$ as $K$ is a three dimensional subspace. Observe that if
$\widehat{T}$ is a vertical translation, i.e. $a^{\prime}=0$ then
$K=\left<(e,0)\right>\oplus\mathbb{C}$ is a two dimensional subspace and hence
minimal polynomial is $(x-\lambda)^{2}$. ∎

######  Proposition 5.13.

(1) All non-vertical Heisenberg translations having same spectra fall in a
single conjugacy class.
(2) All vertical Heisenberg translations having same spectra get dispersed
into two conjugacy classes.

###### Proof.

Observe that any isometry in $\widehat{G}$ conjugating one Heisenberg
translation into the other has to fix $\infty$.
(1) We will see that any two non-vertical Heisenberg translations of the form
$(\lambda,a^{\prime},s)$ and $(\lambda,b^{\prime},t)$ are conjugate to each
other, i.e. to show the existence of an isometry
$\widehat{R}=\left[\begin{array}[]{ccc}\lambda^{\prime}&\left<\boldsymbol{\cdot},U^{-1}(c^{\prime})\right>&\mu^{\prime}s_{1}\\\
0&U&\mu^{\prime}c^{\prime}\\\
0&0&\mu^{\prime}\end{array}\right]\in\widehat{G}_{\infty}$ such that
$\widehat{R}(\lambda,a^{\prime},s)\widehat{R}^{-1}=(\lambda,b^{\prime},t)$.
This is same thing as showing
$\widehat{R}(1,a^{\prime},s)\widehat{R}^{-1}=(1,b^{\prime},t)$. We have
$\widehat{R}(1,a^{\prime},s)=\left[\begin{array}[]{ccc}\lambda^{\prime}&\left<\boldsymbol{\cdot},U^{-1}(c^{\prime})\right>&\mu^{\prime}s_{1}\\\
0&U&\mu^{\prime}c^{\prime}\\\
0&0&\mu^{\prime}\end{array}\right]\left[\begin{array}[]{ccc}1&\left<\boldsymbol{\cdot},a^{\prime}\right>&s\\\
0&I&a^{\prime}\\\ 0&0&1\end{array}\right]$
=$\left[\begin{array}[]{ccc}\lambda^{\prime}&\lambda^{\prime}\left<\boldsymbol{\cdot},a^{\prime}\right>+\left<\boldsymbol{\cdot},U^{-1}(c^{\prime})\right>&\lambda^{\prime}\,s+\left<U(a^{\prime}),c^{\prime}\right>+\mu^{\prime}\,s_{1}\\\
0&U&U(a^{\prime})+\mu^{\prime}\,c^{\prime}\\\
0&0&\mu^{\prime}\end{array}\right]$
and
$(1,b^{\prime},t)\widehat{R}=\left[\begin{array}[]{ccc}1&\left<\boldsymbol{\cdot},b^{\prime}\right>&t\\\
0&I&b^{\prime}\\\
0&0&1\end{array}\right]\left[\begin{array}[]{ccc}\lambda^{\prime}&\left<\boldsymbol{\cdot},U^{-1}(c^{\prime})\right>&\mu^{\prime}\,s_{1}\\\
0&U&\mu^{\prime}c^{\prime}\\\ 0&0&\mu^{\prime}\end{array}\right]$
=$\left[\begin{array}[]{ccc}\lambda^{\prime}&\left<\boldsymbol{\cdot},U^{-1}(c^{\prime})\right>+\left<\boldsymbol{\cdot},U^{-1}(b^{\prime})\right>&\mu^{\prime}\,s_{1}+\left<\mu^{\prime}\,c^{\prime},b^{\prime}\right>+\mu^{\prime}\,t\\\
0&U&\mu^{\prime}\,c^{\prime}+\mu^{\prime}\,b^{\prime}\\\
0&0&\mu^{\prime}\end{array}\right]$.
From the above computations we see that
$\widehat{R}(1,a^{\prime},s)=(1,b^{\prime},t)\widehat{R}$ if and only if

(5.18) $\displaystyle U(a^{\prime})$
$\displaystyle=\mu^{\prime}\,b^{\prime}\,\,\,\,\text{and}$ (5.19)
$\displaystyle\lambda^{\prime}\,s+\left<U(a^{\prime}),c^{\prime}\right>$
$\displaystyle=\left<\mu^{\prime}\,c^{\prime},b^{\prime}\right>+\mu^{\prime}\,t.$

Choose $\mu^{\prime}=\dfrac{\|a^{\prime}\|}{\|b^{\prime}\|}$ and a unitary
operator $U$ which takes an orthonormal basis in
$\left<a^{\prime}\right>^{\perp}$ to an orthonormal basis in
$\left<b^{\prime}\right>^{\perp}$. We need to show the existence of
$c^{\prime}$ satisfying
$\lambda^{\prime}\,s+\mu^{\prime}\left<b^{\prime},c^{\prime}\right>=\mu^{\prime}\,\left<c^{\prime},b^{\prime}\right>+\mu^{\prime}\,t$,
i.e.
$|\lambda^{\prime}|^{2}\,s-t=\left<c^{\prime},b^{\prime}\right>-\left<b^{\prime},c^{\prime}\right>=2i\,\text{Im}\,(\left<c^{\prime},b^{\prime}\right>)$.
By putting $\lambda^{\prime}=\dfrac{\|b^{\prime}\|}{\|a^{\prime}\|}$ and using
$\text{Re}\,s=\dfrac{1}{2}\|a^{\prime}\|^{2}$ and
$\text{Re}\,t=\dfrac{1}{2}\|b^{\prime}\|^{2}$, we are able to find a suitable
$c^{\prime}$.

(2) Equations (5.18) and (5.19) tell that two vertical Heisenberg translations
$(\lambda,0,s)$ and $(\lambda,0,t)$ are conjugate to each other if and only if
there exists $\lambda^{\prime}$ satisfying $|\lambda^{\prime}|^{2}\,s=t$, i.e.
$|\lambda^{\prime}|^{2}=\dfrac{t}{s}=\dfrac{\text{Im}\,t}{\text{Im}\,s}$ as
$\text{Re}\,s=\text{Re}\,t=0$. Hence all vertical translations of the form
$(\lambda,0,s)$ with $\text{Im}\,s>0$ fall in one conjugacy class and those
with $\text{Im}\,s<0$ fall in the other conjugacy class.
It is easy to see that $(\lambda,a^{\prime},s)$ cannot be conjugated to
$(\lambda,0,t)$ as (5.18) tells $b^{\prime}=0$ if and only if $a^{\prime}=0$.
∎

Proof of Theorem 1.4. (1) If $\widehat{S}$ is a parabolic isometry having
singleton spectrum then by Lemma 4.2, $\widehat{S}$ can be conjugated to a
parabolic isometry in $\widehat{G}_{\infty}$ having singleton spectrum, i.e. a
Heisenberg translation.
(2) Follows from Proposition 5.12 in view of (1).
(3) By (2) and (1), a parabolic isometry having singleton spectrum and degree
of restricted minimal polynomial as three (resp. two) is conjugate to a non-
vertical (resp. vertical) Heisenberg translation. Now Proposition 5.13 asserts
the claim.∎

##  6\. Centralizers of isometries

Centralizer of an element $T\in G$ will be denoted by $Z(T)$. Recall that for
any two bounded linear operators $T$ and $S$ on a Hilbert space,
$T\longleftrightarrow S$ means $T$ and $S$ commute with each other. Some
results of this section are motivated from [3] and [13]. Using the terminology
of [3] we have the following.

######  Definition 2.

For an isometry $T\in G$, an eigenvalue of $T$ is called time-like eigenvalue
if its corresponding eigenspace is time-like. We call an elliptic isometry
boundary elliptic if the geometric multiplicity of its time-like eigenvalue is
atleast two, otherwise it is called regular elliptic.

######  Lemma 6.1.

Let $T,\,S\in G$. If $T\longleftrightarrow S$ and $E$ is an eigenspace of $T$
then $S(E)=E$.

###### Proof.

The fact that $S^{-1}$ also commutes with $T$ proves the claim. ∎

######  Corollary 6.1.1.

Regular elliptic isometries commute only with elliptic isometries.

######  Lemma 6.2.

Let $M$ be a non-degenerate subspace of $H\oplus\mathbb{C}$ such that
$H\oplus\mathbb{C}=M\oplus M^{\dagger}$. Then for any $R\in G$,
$H\oplus\mathbb{C}=R(M)\oplus{R(M)}^{\dagger}$.

###### Proof.

Clearly, $R(M)\oplus R(M)^{\dagger}\subseteq H\oplus\mathbb{C}$. We will show
that $H\oplus\mathbb{C}\subseteq R(M)\oplus R(M)^{\dagger}$. As $M$ is non-
degenerate, $R(M)\cap{R(M)}^{\dagger}=\\{0\\}$. Let $y\in H\oplus\mathbb{C}$
and $R(m)\in R(M)$ for some $m\in M$. We will see that $y-R(m)\in
R(M)^{\dagger}$. For $x\in M$, consider
$\mathcal{A}(y-R(m),R(x))=\mathcal{A}(y,R(x))-\mathcal{A}(R(m),R(x))=\mathcal{A}(R^{-1}(y),x)-\mathcal{A}(m,x)=\mathcal{A}(R^{-1}(y)-m,x)=0$
as $H\oplus\mathbb{C}=M\oplus M^{\dagger}$, $x\in M$, $R^{-1}(y)\in
H\oplus\mathbb{C}$ and thus $R^{-1}(y)-m\in M^{\dagger}$. ∎

Recall that for a closed subspace $M\subseteq H\oplus\mathbb{C}$, $G_{M}$
denotes the group of all bijective bounded linear mappings on $M$ leaving the
form $\mathcal{A}\restriction_{M}$ invariant.

Proof of Theorem 1.5. First we will show that $T$ decomposes
$H\oplus\mathbb{C}$ into $M$ and $M^{\dagger}$. Let $T=R\widetilde{U}R^{-1}$
where $\widetilde{U}\in G$ is of the form
$e^{i\theta}\left[\begin{array}[]{cc}U&0\\\ 0&1\\\ \end{array}\right]$.
Observe that $e^{i\theta}$ is the only time-like eigenvalue of $T$ and
$\widetilde{U}$ both. As $M$ is a time-like eigenspace of $T$ (corresponding
to eigenvalue $e^{i\theta}$), $R^{-1}(M)$ is an eigenspace of $\widetilde{U}$
corresponding to eigenvalue $e^{i\theta}$ and hence this eigenspace contains
$\mathbb{C}$. So $R^{-1}(M)$ is a closed subspace of $H\oplus\mathbb{C}$ . As
$R^{-1}(M)^{\dagger}=R^{-1}(M)^{\perp}$ ($R^{-1}(M)^{\perp}\subseteq H$),
$H\oplus\mathbb{C}=R^{-1}(M)\oplus(R^{-1}(M))^{\dagger}$. Lemma 6.2 tells
$H\oplus\mathbb{C}=M\oplus M^{\dagger}$. Also $T(M)=M$ gives that
$T(M^{\dagger})=M^{\dagger}$ (Lemma 3.5). Hence $T$ decomposes
$H\oplus\mathbb{C}$ into $M$ and $M^{\dagger}$. Let $S\in Z(T)$. Lemma 6.1
tells that $S(M)=M$ and hence $S(M^{\dagger})=M^{\dagger}$ by Lemma 3.5. Now
we show that $M^{\dagger}$ is a closed subspace of $H\oplus\mathbb{C}$. As
$\mathcal{A}\restriction_{{M}^{\dagger}}$ is positive definite by Lemma 3.7,
it is an inner product space. Since
$R^{-1}(M^{\dagger})=R^{-1}(M)^{\dagger}=R^{-1}(M)^{\perp}$,
$R^{-1}:M^{\dagger}\longrightarrow(R^{-1}(M))^{\perp}$ is a surjective linear
isometry. This gives that $M^{\dagger}$ is a Hilbert space as
$(R^{-1}(M))^{\perp}$ is complete. The above analysis yields
$Z(T)=Z(T\restriction_{M})\times Z(T\restriction_{{M}^{\dagger}})$ where
$Z(T\restriction_{{M}^{\dagger}})\subseteq\mathcal{U}(M^{\dagger})$. As
$T\restriction_{M}$ is a scalar multiple of identity,
$Z(T\restriction_{M})=G_{M}$. ∎

###  6.1. Centralizers of hyperbolic isometries

In this section we provide proof of Theorem 1.6.
Proof of Theorem 1.6. As $M$ is a non-degenerate subspace by Lemma 3.7, $T$
decomposes $H\oplus\mathbb{C}$ into $M$ and $M^{\dagger}$ (Lemma 3.6 and Lemma
3.5). Let $S\in Z(T)$. Lemma 6.1 tells that $S$ leaves each one-dimensional
eigenspace invariant. Hence $S(M)=M$ and $S(M^{\dagger})=M^{\dagger}$ (Lemma
3.5). The same argument as in the proof of Theorem 1.3 (3) shows that
$M^{\dagger}$ is a Hilbert space. So we have $Z(T)=Z(T\restriction_{M})\times
Z(T\restriction_{M^{\dagger}})$ where $Z(T\restriction_{M})\subseteq G_{M}$
and $Z(T\restriction_{M^{\dagger}})\subseteq\mathcal{U}(M^{\dagger})$. We will
show that $Z(T\restriction_{M})$ gets identified with $S^{1}\times\mathbb{R}$.
Let $M=\text{span}\\{(y,1),(z,1)\\}$ where $(y,1)$ and $(z,1)$ are two light-
like eigenvectors of $T$. Let $S^{\prime}=S\restriction_{M}\in
Z(T\restriction_{M})$. Then $S^{\prime}(y,1)=\alpha(y,1)$ and
$S^{\prime}(z,1)=\beta(z,1)$ for some $\alpha,\,\beta\in\mathbb{C}$. This
implies that $S^{\prime}$ is completely determined by its eigenvalues. As
$S^{\prime}\in G_{M}$ too,
$\mathcal{A}((y,1),(z,1))=\mathcal{A}(S^{\prime}(y,1),S^{\prime}(z,1))$, i.e.
$\left<y,z\right>-1=\alpha\overline{\beta}(\left<y,z\right>-1)$ if and only if
$\alpha\overline{\beta}=1$ by Lemma 3.7. Now there are two possibilities
either $\alpha=\beta$ or $\alpha\neq\beta$. $\alpha=\beta$ if and only if
$\alpha\in S^{1}$ by Remark 5. $\alpha\neq\beta$ if and only if $S^{\prime}$
is a hyperbolic isometry and hence $\alpha$ and $\beta$ are of the form
$re^{i\theta}$ and $r^{-1}e^{i\theta}$ by Theorem 1.3 (2). This tells that for
$re^{i\theta}\in S^{1}\times\mathbb{R}$, there corresponds a unique
$S^{\prime}\in Z(T\restriction_{M})$ such that
$S^{\prime}(y,1)=re^{i\theta}(y,1)$ and
$S^{\prime}(z,1)=r^{-1}e^{i\theta}(z,1)$. Hence every element of
$S^{1}\times\mathbb{R}$ uniquely determines an element of
$Z(T\restriction_{M})$. ∎

Following is the description of centralizers in the group of unitary operators
defined on a separable Hilbert space $H$. We note it here without proof.
Using the multiplicity theory for normal operators on a separable Hilbert
space $H$ [6, Chapter IX, section 10], we will compute centralizer of a
unitary operator in $\mathcal{U}(H)$.

The integral representation of normal operators with respect to spectral
measure yields that if a normal operator $N$ is star-cyclic [6, Chapter IX,
Definition 3.1], then it is unitarily equivalent to the multiplication
operator ( multiplication by identity) defined on $L^{2}(\sigma(N),\,\mu)$
where $\mu$ is a scalar valued spectral measure. Now, decomposing $H$ into
countable orthogonal pieces such that the normal operator restricted to each
piece is star-cyclic gives that $N$ is unitarily equivalent to the direct sum
of multiplication operators. In precise form, we have the following.

######  Theorem 6.3.

[6, Chaper IX, Theorem 10.20] If $N$ is a normal operator on $H$, then there
are mutually singular measures $\mu_{\infty}$, $\mu_{1}$,
$\mu_{2},\,\cdot\cdot\cdot$ and an isomorphism $U:H\longrightarrow
L^{2}(\mu_{\infty};\,H_{\infty})\oplus L^{2}(\mu_{1})\oplus
L^{2}(\mu_{2};\,H_{2})\oplus\,\cdot\cdot\cdot$ such that
$UNU^{-1}=N_{\infty}\oplus N_{1}\oplus N_{2}\oplus\,\cdot\cdot\cdot$ where
$H_{n}$ is an n-dimensional Hilbert space, $L^{2}(\mu_{n};\,H_{n})$ is the
space of square integrable $H_{n}$ valued functions and $N_{n}$ is
multiplication by $z$ on $L^{2}(\mu_{n};\,H_{n})$.

Also, if $N$ is multiplication by $z$ on $L^{2}(\mu;\,H_{n})$ and $M_{\phi}$
is multiplication by $\phi$, then

$\\{N\\}^{\prime}=\left\\{M_{\phi}\,\,:\,\,\phi\in
L^{\infty}(\mu;\,B(H_{n}))\right\\},$

see [6, Chapter IX, Corollary 6.9]. This gives

$\\{N_{\infty}\oplus N_{1}\oplus
N_{2}\oplus\,\cdot\cdot\cdot\\}^{\prime}=L^{\infty}(\mu_{\infty};\,B(H_{\infty}))\oplus
L^{\infty}(\mu_{1})\oplus
L^{\infty}(\mu_{2};\,B(H_{2}))\oplus\,\cdot\cdot\cdot.$

As a corollary to the above analysis, we have the following result.

######  Proposition 6.4.

If $V$ is a unitary operator on a separable Hilbert space $H$, then

$Z(V)=U^{-1}Z(V_{\infty}\oplus V_{1}\oplus V_{2}\oplus...)U$

where $U$ is as in the preceding theorem and $Z(V_{\infty}\oplus V_{1}\oplus
V_{2}\oplus...)=U(H_{\infty})\oplus U(H_{1})\oplus U(H_{2})\oplus...$,
$U(H_{n})$ is the group of unitary elements in $L^{\infty}(\mu;\,B(H_{n})).$

For the non-separable case, one may refer to [18].

###  6.2. Centralizers of Heisenberg translations

######  Definition 3.

Two Heisenberg translations $(\lambda,a^{\prime},s)$ and
$(\lambda^{\prime},b^{\prime},t)$ are said to be isotropic if
$\left<b^{\prime},a^{\prime}\right>\in\mathbb{R}$, i.e their translations in
the horizontal direction have real product.

$\widehat{T}=\lambda\left[\begin{array}[]{ccc}1&\left<\boldsymbol{\cdot},a^{\prime}\right>&s\\\
0&I&a^{\prime}\\\ 0&0&1\end{array}\right]=(\lambda,a^{\prime},s)$ will denote
a Heisenberg translation, where $s\neq 0$, $|\lambda|=1$ and
$\text{Re}\,s=\dfrac{1}{2}{\|a^{\prime}\|}^{2},$ throughout the subsection.

######  Lemma 6.5.

If $\widehat{S}\longleftrightarrow\widehat{T}$ then $\widehat{S}$ is either a
boundary elliptic or parabolic isometry in $\widehat{G}_{\infty}$.

###### Proof.

Corollary 6.1.1 and Corollary 1.6.1 tell that parabolic isometries commute
either with boundary elliptics or parabolics. Also $\widehat{T}$ fixes only
$\infty$ and $\widehat{S}\longleftrightarrow\widehat{T}$. This implies
$\widehat{S}$ fixes $\infty$ and hence $\widehat{S}$ lies in
$\widehat{G}_{\infty}$. ∎

######  Proposition 6.6.

Two Heisenberg translations commute with each other if and only if they are
isotropic.

###### Proof.

Let $(\lambda,a^{\prime},s)$ and $(\lambda^{\prime},b^{\prime},t)$ be two
Heisenberg translations. We have

$\displaystyle(\lambda,a^{\prime},s)(\lambda^{\prime},b^{\prime},t)$
$\displaystyle=\lambda\lambda^{\prime}\left[\begin{array}[]{ccc}1&\left<\boldsymbol{\cdot},a^{\prime}\right>&s\\\
0&I&a^{\prime}\\\
0&0&1\end{array}\right]\left[\begin{array}[]{ccc}1&\left<\boldsymbol{\cdot},b^{\prime}\right>&t\\\
0&I&b^{\prime}\\\ 0&0&1\end{array}\right]$
$\displaystyle=\lambda\lambda^{\prime}\left[\begin{array}[]{ccc}1&\left<\boldsymbol{\cdot},b^{\prime}\right>+\left<\boldsymbol{\cdot},a^{\prime}\right>&t+\left<b^{\prime},a^{\prime}\right>+s\\\
0&I&b^{\prime}+a^{\prime}\\\ 0&0&1\end{array}\right]\,\,\text{and}$
$\displaystyle(\lambda,b^{\prime},t)(\lambda,a^{\prime},s)$
$\displaystyle=\lambda^{\prime}\lambda\left[\begin{array}[]{ccc}1&\left<\boldsymbol{\cdot},b^{\prime}\right>&t\\\
0&I&b^{\prime}\\\
0&0&1\end{array}\right]\left[\begin{array}[]{ccc}1&\left<\boldsymbol{\cdot},a^{\prime}\right>&s\\\
0&I&a^{\prime}\\\ 0&0&1\end{array}\right]$
$\displaystyle=\lambda^{\prime}\lambda\left[\begin{array}[]{ccc}1&\left<\boldsymbol{\cdot},a^{\prime}\right>+\left<\boldsymbol{\cdot},b^{\prime}\right>&s+\left<a^{\prime},b^{\prime}\right>+t\\\
0&I&a^{\prime}+b^{\prime}\\\ 0&0&1\end{array}\right].$

This gives that
$(\lambda,a^{\prime},s)\longleftrightarrow(\lambda,b^{\prime},t)$ if and only
if they are isotropic. ∎

Proof of Theorem 1.7. Let $\widehat{S}\in\widehat{G}$ be such that
$\widehat{S}\longleftrightarrow\widehat{T}$. Lemma 6.5 tells that
$\widehat{S}\in\widehat{G}_{\infty}$ and it is either boundary elliptic or
parabolic. This implies $\sigma(\widehat{S})\subseteq S^{1}$. As
$\widehat{G}_{\infty}$ contains only boundary elliptic, hyperbolic and
parabolic isometries fixing infinity, its general isometry
$\left[\begin{array}[]{ccc}\lambda^{\prime}&\left<\boldsymbol{\cdot},U^{-1}(b^{\prime})\right>&\mu^{\prime}\,t\\\
0&U&\mu^{\prime}\,b^{\prime}\\\ 0&0&\mu^{\prime}\end{array}\right]$ is
hyperbolic if and only if $\lambda^{\prime}\neq\mu^{\prime}$ in view of
Proposition 5.9, Proposition 5.6 and Remark 5. Hence a boundary elliptic or
parabolic isometry in $\widehat{G}_{\infty}$ is of the form
$\left[\begin{array}[]{ccc}\lambda^{\prime}&\left<\boldsymbol{\cdot},U^{-1}(b^{\prime})\right>&\lambda^{\prime}\,t\\\
0&U&\lambda^{\prime}\,b^{\prime}\\\
0&0&\lambda^{\prime}\end{array}\right]=\widehat{S}$ say.

  1. (1)

For $\widehat{T}=\lambda\left[\begin{array}[]{ccc}1&0&s\\\ 0&I&0\\\
0&0&1\end{array}\right]$,

$\displaystyle\widehat{S}\widehat{T}$
$\displaystyle=\lambda\left[\begin{array}[]{ccc}\lambda^{\prime}&\left<\boldsymbol{\cdot},U^{-1}(b^{\prime})\right>&\lambda^{\prime}\,t\\\
0&U&\lambda^{\prime}\,b^{\prime}\\\
0&0&\lambda^{\prime}\end{array}\right]\left[\begin{array}[]{ccc}1&0&s\\\
0&I&0\\\ 0&0&1\end{array}\right]$
$\displaystyle=\lambda\left[\begin{array}[]{ccc}\lambda^{\prime}&\left<\boldsymbol{\cdot},U^{-1}(b^{\prime})\right>&\lambda^{\prime}\,s+\lambda^{\prime}\,t\\\
0&U&\lambda^{\prime}\,b^{\prime}\\\ 0&0&\lambda^{\prime}\end{array}\right]$
$\displaystyle=\widehat{T}\widehat{S}$
$\displaystyle=\lambda\left[\begin{array}[]{ccc}1&0&s\\\ 0&I&0\\\
0&0&1\end{array}\right]\left[\begin{array}[]{ccc}\lambda^{\prime}&\left<\boldsymbol{\cdot},U^{-1}(b^{\prime})\right>&\lambda^{\prime}\,t\\\
0&U&\lambda^{\prime}\,b^{\prime}\\\ 0&0&\lambda^{\prime}\end{array}\right].$

Hence the result.

  2. (2)

For
$\widehat{T}=\lambda\left[\begin{array}[]{ccc}1&\left<\boldsymbol{\cdot},a^{\prime}\right>&s\\\
0&I&a^{\prime}\\\ 0&0&1\end{array}\right]$, $a^{\prime}\neq 0$,

$\displaystyle\widehat{S}\widehat{T}$
$\displaystyle=\lambda\left[\begin{array}[]{ccc}\lambda^{\prime}&\left<\boldsymbol{\cdot},U^{-1}(b^{\prime})\right>&\lambda^{\prime}\,t\\\
0&U&\lambda^{\prime}\,b^{\prime}\\\
0&0&\lambda^{\prime}\end{array}\right]\left[\begin{array}[]{ccc}1&\left<\boldsymbol{\cdot},a^{\prime}\right>&s\\\
0&I&a^{\prime}\\\ 0&0&1\end{array}\right]$
$\displaystyle=\lambda\left[\begin{array}[]{ccc}\lambda^{\prime}&\lambda^{\prime}\left<\boldsymbol{\cdot},a^{\prime}\right>+\left<\boldsymbol{\cdot},U^{-1}(b^{\prime})\right>&\lambda^{\prime}\,s+\left<U(a^{\prime}),b^{\prime}\right>+\lambda^{\prime}\,t\\\
0&U&U(a^{\prime})+\lambda^{\prime}\,b^{\prime}\\\
0&0&\lambda^{\prime}\end{array}\right]$

and

$\displaystyle\widehat{T}\widehat{S}$
$\displaystyle=\lambda\left[\begin{array}[]{ccc}1&\left<\boldsymbol{\cdot},a^{\prime}\right>&s\\\
0&I&a^{\prime}\\\
0&0&1\end{array}\right]\left[\begin{array}[]{ccc}\lambda^{\prime}&\left<\boldsymbol{\cdot},U^{-1}(b^{\prime})\right>&\lambda^{\prime}\,t\\\
0&U&\lambda^{\prime}\,b^{\prime}\\\ 0&0&\lambda^{\prime}\end{array}\right]$
$\displaystyle=\lambda\left[\begin{array}[]{ccc}\lambda^{\prime}&\left<\boldsymbol{\cdot},U^{-1}(b^{\prime})\right>+\left<\boldsymbol{\cdot},U^{-1}(a^{\prime})\right>&\lambda^{\prime}\,t+\left<\lambda^{\prime}\,b^{\prime},a^{\prime}\right>+\lambda^{\prime}\,s\\\
0&U&\lambda^{\prime}\,b^{\prime}+\lambda^{\prime}\,a^{\prime}\\\
0&0&\lambda^{\prime}\end{array}\right].$

This tells that $\widehat{S}\longleftrightarrow\widehat{T}$ if and only if
$U(a^{\prime})=\lambda^{\prime}\,a^{\prime}$ and
$\left<a^{\prime},b^{\prime}\right>\in\mathbb{R}$. ∎

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