Nonstandard Quantum Complex Projective Line

Nonstandard Quantum Complex Projective Line††This paper is a contribution to
the Special Issue on Noncommutative Manifolds and their Symmetries in honour
of Giovanni Landi. The full collection is available at
https://www.emis.de/journals/SIGMA/Landi.html

Nicola CICCOLI † and Albert Jeu-Liang SHEU ‡

N. Ciccoli and A.J.-L. Sheu

† Dipartimento di Matematica e Informatica, University of Perugia, Italy
<EMAIL_ADDRESS>
‡ Department of Mathematics, University of Kansas, Lawrence, KS 66045, USA
<EMAIL_ADDRESS>
Received March 06, 2020, in final form July 24, 2020; Published online August
03, 2020

In our attempt to explore how the quantum nonstandard complex projective
spaces $\mathbb{C}P_{q,c}^{n}$ studied by Korogodsky, Vaksman, Dijkhuizen, and
Noumi are related to those arising from the geometrically constructed
Bohr–Sommerfeld groupoids by Bonechi, Ciccoli, Qiu, Staffolani, and Tarlini,
we were led to establish the known identification of
$C\big{(}\mathbb{C}P_{q,c}^{1}\big{)}$ with the pull-back of two copies of the
Toeplitz $C^{*}$-algebra along the symbol map in a more direct way via an
operator theoretic analysis, which also provides some interesting non-obvious
details, such as a prominent generator of
$C\big{(}\mathbb{C}P_{q,c}^{1}\big{)}$ being a concrete weighted double shift.

quantum homogeneous space; Toeplitz algebra; weighted shift

58B32; 46L85

##  1 Introduction

In [9], the $C^{*}$-algebra $C\big{(}\mathbb{C}P_{q,c}^{n}\big{)}$ of
nonstandard quantum complex projective spaces studied by Korogodsky and
Vaksman [5] and Dijkhuizen and Noumi [3] is embedded in a concrete groupoid
$C^{*}$-algebra, and then shown to have $C\big{(}\mathbb{S}_{q}^{2n-1}\big{)}$
as a quotient algebra, which reflects the geometric observation [10] that the
nonstandard ${\rm SU}(n+1)$-covariant Poisson complex projective space
$\mathbb{C}P^{n}$ contains a copy of the standard Poisson sphere
$\mathbb{S}^{2n-1}$.

Although the work in [9] involves realizing
$C\big{(}\mathbb{C}P_{q,c}^{n}\big{)}$ as part of a concrete groupoid
$C^{*}$-algebra in order to analyze the algebra structure and extract useful
information, it is not clear whether one can actually realize
$C\big{(}\mathbb{C}P_{q,c}^{n}\big{)}$ as a groupoid $C^{*}$-algebra itself.
However from a purely differential geometric consideration, an elegant program
of constructing some quantum homogeneous spaces as the groupoid
$C^{*}$-algebras of geometrically constructed Bohr–Sommerfeld groupoids is
later successfully developed by Bonechi, Ciccoli, Qiu, Staffolani, Tarlini [1,
2]. Naturally, it is of great interest to decide whether the quantum complex
projective space arising from this new program is indeed the same as the known
version of $C\big{(}\mathbb{C}P_{q,c}^{n}\big{)}$. Indeed they are the same
for the case of $n=1$ because the underlying groupoids are shown to be
isomorphic in Proposition 7.2 of [1]. But the higher dimensional cases are far
from being settled.

It is hoped that by analyzing more carefully the embedding of
$C\big{(}\mathbb{C}P_{q,c}^{n}\big{)}$ in a concrete groupoid $C^{*}$-algebra
found in [9] via a representation theoretic approach, one can see some direct
connection with the geometrically constructed Bohr–Sommerfeld groupoid and
then possibly find a way to identify these two different versions of quantum
complex projective spaces. While attempting this approach, we come to
recognize the need of a more direct understanding of the algebra structure of
$C\big{(}\mathbb{C}P_{q,c}^{n}\big{)}$ based on some known representations of
the ambient algebra $C({\rm SU}_{q}(n+1))$.

In particular, for $n=1$, we want to directly derive the algebra structure of
$C\big{(}\mathbb{C}P_{q,c}^{1}\big{)}$ from the basic representations of
$C({\rm SU}_{q}(2))$, instead of via identifying
$C\big{(}\mathbb{C}P_{q,c}^{1}\big{)}$ with the algebra
$C\big{(}\mathbb{S}_{\mu c}^{2}\big{)}$ of the Podleś quantum 2-sphere [7] as
indicated in [3, 5]. In this note, we show how to accomplish it. Along the
way, our detailed analysis reveals some nontrivial hidden structures, for
example, a distinguished generator $x_{1}^{\ast}x_{2}$ of
$C\big{(}\mathbb{C}P_{q,c}^{1}\big{)}$ is a weighted double shift (on a core
Hilbert space that determines the $C^{*}$-algebra structure of
$C\big{(}\mathbb{C}P_{q,c}^{1}\big{)}$) with respect to an orthonormal basis,
and its weights are determined by a concrete formula.

##  2 Nonstandard quantum $\boldsymbol{\mathbb{C}P_{q,c}^{1}}$

We recall the description of $C\big{(}\mathbb{C}P_{q,c}^{n}\big{)}$ with
$c\in(0,\infty)$ and $q\in(1,\infty)$ obtained by Dijkhuizen and Noumi [3] as

$C\big{(}\mathbb{C}P_{q,c}^{n}\big{)}\cong
C^{\ast}(\\{x_{i}^{\ast}x_{j}\,|\,1\leq i,j\leq n+1\\})\subset C({\rm
SU}_{q}(n+1)),$

where

$x_{i}=\sqrt{c}u_{1,i}+u_{n+1,i}$

for the standard generators $\\{u_{i,j}\\}_{i,j=1}^{n+1}$ of $C({\rm
SU}_{q}(n+1))$.

In this paper, we focus on the case of $n=1$, with the goal to directly show
that $C\big{(}\mathbb{C}P_{q,c}^{1}\big{)}$ is the pullback
$\mathcal{T}\oplus_{C(\mathbb{T})}\mathcal{T}$ of two copies of the standard
symbol map

$\sigma\colon\ \mathcal{T}\rightarrow C(\mathbb{T})$

for the Toeplitz algebra $\mathcal{T}$ that is the $C^{*}$-algebra generated
by the (forward) unilateral shift $\mathcal{S}$ on
$\ell^{2}(\mathbb{Z}_{\geq})$, with
$\ker(\sigma)=\mathcal{K}\big{(}\ell^{2}(\mathbb{Z}_{\geq})\big{)}$ the ideal
of all compact operators.

For $C({\rm SU}_{q}(2))$, consider the known faithful representation $\pi$ of
$C({\rm SU}_{q}(2))$ determined by

$\pi(u)\equiv\left(\begin{matrix}\pi(u_{11})&\pi(u_{12})\\\
\pi(u_{21})&\pi(u_{22})\end{matrix}\right):=\left\\{\left(\begin{matrix}t_{1}\alpha&-q^{-1}t_{1}\gamma\\\
t_{2}\gamma&t_{2}\alpha^{\ast}\end{matrix}\right)\right\\}_{t_{2}=\overline{t_{1}}\in\mathbb{T}}$

as a $\mathbb{T}$-family of representations of $C({\rm SU}_{q}(2))$ on
$\ell^{2}(\mathbb{Z}_{\geq})$ with parameter
$t_{1}\equiv\overline{t_{2}}\in\mathbb{T}$, where

$\alpha=\left(\begin{matrix}0&\sqrt{1-q^{-2}}&0&&\\\ 0&0&\sqrt{1-q^{-4}}&0&\\\
0&0&0&\sqrt{1-q^{-6}}&\ddots\\\ &0&0&0&\ddots\\\
&&\ddots&\ddots&\ddots\end{matrix}\right)\in\mathcal{B}\big{(}\ell^{2}(\mathbb{Z}_{\geq})\big{)}$

and

$\gamma=\left(\begin{matrix}1&0&0&&\\\ 0&q^{-1}&0&0&\\\ 0&0&q^{-2}&0&\ddots\\\
&0&0&q^{-3}&\ddots\\\
&&\ddots&\ddots&\ddots\end{matrix}\right)\in\mathcal{B}\big{(}\ell^{2}(\mathbb{Z}_{\geq})\big{)}\
\text{\ \ self-adjoint}$

satisfying

$\alpha^{\ast}\alpha+\gamma\gamma^{\ast}\equiv\alpha^{\ast}\alpha+\gamma^{2}=I=\alpha\alpha^{\ast}+q^{-2}\gamma^{2}\equiv\alpha\alpha^{\ast}+q^{-2}\gamma^{\ast}\gamma$

and

$\gamma\alpha^{\ast}-q^{-1}\alpha^{\ast}\gamma=\alpha\gamma^{\ast}-q^{-1}\gamma^{\ast}\alpha=0\equiv\alpha\gamma-q^{-1}\gamma\alpha=\gamma^{\ast}\alpha^{\ast}-q^{-1}\alpha^{\ast}\gamma^{\ast}$

which ensure the required condition
$\pi(u)\pi(u)^{\ast}=I=\pi(u)^{\ast}\pi(u)$.

In this paper, we identify every element of $C({\rm SU}_{q}(2))\supset
C(\mathbb{C}P_{q,c}^{1})$ with a $\mathbb{T}$-family of operators on
$\ell^{2}(\mathbb{Z}_{\geq})$ via this faithful representation $\pi$, and we
analyze such a $\mathbb{T}$-family of operators pointwise at each fixed
$t_{1}\in\mathbb{T}$.

More explicitly, the generators
$x_{1}^{\ast}x_{2},x_{1}^{\ast}x_{1},x_{2}^{\ast}x_{2}$ of
$C\big{(}\mathbb{C}P_{q,c}^{1}\big{)}$ are $\mathbb{T}$-families of operators
with

$x_{1}:=\sqrt{c}t_{1}\alpha+t_{2}\gamma=\left(\begin{matrix}t_{2}&\sqrt{c}t_{1}\sqrt{1-q^{-2}}&0&&\\\
0&t_{2}q^{-1}&\sqrt{c}t_{1}\sqrt{1-q^{-4}}&0&\\\
0&0&t_{2}q^{-2}&\sqrt{c}t_{1}\sqrt{1-q^{-6}}&\ddots\\\
&0&0&t_{2}q^{-3}&\ddots\\\ &&\ddots&\ddots&\ddots\end{matrix}\right)$

and

$x_{2}:=-q^{-1}\sqrt{c}t_{1}\gamma+t_{2}\alpha^{\ast}=\left(\begin{matrix}-q^{-1}\sqrt{c}t_{1}&0&0&&\\\
t_{2}\sqrt{1-q^{-2}}&-q^{-2}\sqrt{c}t_{1}&0&0&\\\
0&t_{2}\sqrt{1-q^{-4}}&-q^{-3}\sqrt{c}t_{1}&0&\ddots\\\
&0&t_{2}\sqrt{1-q^{-6}}&-q^{-4}\sqrt{c}t_{1}&\ddots\\\
&&\ddots&\ddots&\ddots\end{matrix}\right).$

At any fixed $t_{1}\in\mathbb{T}$, it is easy to see that $\ker(x_{2})=0$
since

$\displaystyle
0=x_{2}\left(\sum_{n=0}^{\infty}z_{n}e_{n}\right)=-q^{-1}\sqrt{c}z_{0}e_{0}+\big{(}\sqrt{1-q^{-2}}z_{0}-q^{-2}\sqrt{c}z_{1}\big{)}e_{1}$
$\displaystyle\hphantom{0=x_{2}\left(\sum_{n=0}^{\infty}z_{n}e_{n}\right)=}{}+\big{(}\sqrt{1-q^{-4}}z_{1}-q^{-3}\sqrt{c}z_{2}\big{)}e_{2}+\cdots$

implies $z_{0}=z_{1}=z_{2}=\cdots=0$, and
$\dim(\operatorname{coker}(x_{2}))=1$ since $x_{2}\equiv
t_{2}\alpha^{\ast}\equiv t_{2}\mathcal{S}$ modulo $\mathcal{K}$ is a Fredholm
operator of index $-1$. On the other hand, $\dim(\ker(x_{1}))=1$ by solving
$0=x_{1}\Big{(}\sum\limits_{n=0}^{\infty}z_{n}e_{n}\Big{)}$ to get that if
$z_{0}=1$, then

$z_{n}=(-1)^{n}t_{2}^{n}q^{\frac{-n(n-1)}{2}}\sqrt{c}^{-n}t_{1}^{-n}\sqrt{1-q^{-2}}^{-1}\cdots\sqrt{1-q^{-2n}}^{-1}$

for all $n\in\mathbb{N}$, and $x_{1}$ is surjective since
$x_{1}\equiv\sqrt{c}t_{1}\alpha\equiv\sqrt{c}t_{1}\mathcal{S}^{\ast}$ modulo
$\mathcal{K}$ is a Fredholm operator of index $1$. So $x_{1}^{\ast}x_{2}$ is a
Fredholm operator of index $-2$. Actually, $\ker(x_{1}^{\ast}x_{2})=0$ (hence
$(x_{1}^{\ast}x_{2})^{\ast}(x_{1}^{\ast}x_{2})$ is invertible) and
$\dim(\operatorname{coker}(x_{1}^{\ast}x_{2}))=2$, and hence the partial
isometry $(x_{1}^{\ast}x_{2})|x_{1}^{\ast}x_{2}|^{-1}$ in the polar
decomposition of $x_{1}^{\ast}x_{2}$ is $\mathcal{S}\oplus\mathcal{S}$ (up to
a unitary direct summand) after a suitable choice of orthonormal basis. This
observation is consistent with our goal to show that
$C\big{(}\mathbb{C}P_{q,c}^{1}\big{)}$ is isomorphic to the pullback
$C^{*}$-algebra $\mathcal{T}\oplus_{C(\mathbb{T})}\mathcal{T}$, but is far
from sufficient to make such a conclusion. We need to do a much more detailed
analysis which starts with the following computation.

First we compute

$\displaystyle
x_{1}^{\ast}x_{1}=c\alpha^{\ast}\alpha+\sqrt{c}\overline{t_{1}}^{2}\alpha^{\ast}\gamma+\sqrt{c}t_{1}^{2}\gamma\alpha+\gamma^{2}$
$\displaystyle\hphantom{x_{1}^{\ast}x_{1}}{}=c+(1-c)\gamma^{2}+\sqrt{c}\overline{t_{1}}^{2}\alpha^{\ast}\gamma+\sqrt{c}t_{1}^{2}\gamma\alpha\equiv
c\quad\text{mod\ }\mathcal{K},$ $\displaystyle
x_{2}^{\ast}x_{2}=\alpha\alpha^{\ast}-q^{-1}\sqrt{c}t_{1}^{2}\alpha\gamma-q^{-1}\sqrt{c}\overline{t_{1}}^{2}\gamma\alpha^{\ast}+q^{-2}c\gamma^{2}$
$\displaystyle\hphantom{x_{2}^{\ast}x_{2}}{}=1+q^{-2}(c-1)\gamma^{2}-q^{-1}\sqrt{c}t_{1}^{2}\alpha\gamma-q^{-1}\sqrt{c}\overline{t_{1}}^{2}\gamma\alpha^{\ast}$
$\displaystyle\hphantom{x_{2}^{\ast}x_{2}}{}=1+q^{-2}(c-1)\gamma^{2}-q^{-2}\sqrt{c}t_{1}^{2}\gamma\alpha-q^{-2}\sqrt{c}\overline{t_{1}}^{2}\alpha^{\ast}\gamma\equiv
1\quad\text{mod\ }\mathcal{K},$ $\displaystyle
x_{1}^{\ast}x_{2}=\sqrt{c}\overline{t_{1}}^{2}(\alpha^{\ast})^{2}-cq^{-1}\alpha^{\ast}\gamma+\gamma\alpha^{\ast}-q^{-1}\sqrt{c}t_{1}^{2}\gamma^{2}\equiv\sqrt{c}\overline{t_{1}}^{2}\mathcal{S}^{2}\quad\text{mod\
}\mathcal{K},$ $\displaystyle
x_{2}^{\ast}x_{1}=\sqrt{c}t_{1}^{2}\alpha^{2}-cq^{-1}\gamma\alpha+\alpha\gamma-q^{-1}\sqrt{c}\overline{t_{1}}^{2}\gamma^{2},$

which imply

$x_{1}^{\ast}x_{1}+q^{2}x_{2}^{\ast}x_{2}=q^{2}+c.$

So the $C^{*}$-algebra $C\big{(}\mathbb{C}P_{q,c}^{1}\big{)}$ is generated by
$x_{1}^{\ast}x_{2}$ and $x_{1}^{\ast}x_{1}$, i.e.,

$C\big{(}\mathbb{C}P_{q,c}^{1}\big{)}=C^{\ast}(\\{x_{1}^{\ast}x_{2},x_{1}^{\ast}x_{1}\\}),$

since $x_{2}^{\ast}x_{1}=(x_{1}^{\ast}x_{2})^{\ast}$ and
$x_{2}^{\ast}x_{2}=1+cq^{-2}-q^{-2}x_{1}^{\ast}x_{1}$ are generated by
$x_{1}^{\ast}x_{2}$ (with $(x_{1}^{\ast}x_{2})^{\ast}(x_{1}^{\ast}x_{2})$
invertible) and $x_{1}^{\ast}x_{1}$. As a remark, we note that
$x_{1}^{\ast}x_{1}$ and $x_{2}^{\ast}x_{2}$ commute.

We also note that

$x_{1}x_{1}^{\ast}+x_{2}x_{2}^{\ast}=1+c$

and hence $x_{1}x_{1}^{\ast}$ and $x_{2}x_{2}^{\ast}$ commute. Indeed

$\displaystyle x_{1}x_{1}^{\ast}$
$\displaystyle=\big{(}\sqrt{c}t_{1}\alpha+t_{2}\gamma\big{)}\big{(}\sqrt{c}t_{2}\alpha^{\ast}+t_{1}\gamma\big{)}=c\alpha\alpha^{\ast}+\sqrt{c}t_{1}^{2}\alpha\gamma+\sqrt{c}t_{2}^{2}\gamma\alpha^{\ast}+\gamma^{2}$
$\displaystyle=c-cq^{-2}\gamma^{2}+\sqrt{c}t_{1}^{2}q^{-1}\gamma\alpha+\sqrt{c}t_{2}^{2}q^{-1}\alpha^{\ast}\gamma+\gamma^{2},$

while

$\displaystyle x_{2}x_{2}^{\ast}$
$\displaystyle=\big{(}t_{2}\alpha^{\ast}-q^{-1}\sqrt{c}t_{1}\gamma\big{)}\big{(}t_{1}\alpha-q^{-1}\sqrt{c}t_{2}\gamma\big{)}=\alpha^{\ast}\alpha-q^{-1}\sqrt{c}t_{2}^{2}\alpha^{\ast}\gamma-q^{-1}\sqrt{c}t_{1}^{2}\gamma\alpha+q^{-2}c\gamma^{2}$
$\displaystyle=1-\gamma^{2}-q^{-1}\sqrt{c}t_{2}^{2}\alpha^{\ast}\gamma-q^{-1}\sqrt{c}t_{1}^{2}\gamma\alpha+q^{-2}c\gamma^{2}.$

Before proceeding further, we recall some operator-theoretic properties often
used implicitly in the following analysis, including that
$\overline{\operatorname{range}(T)}=\ker(T^{\ast})^{\perp}$ and
$\ker(T^{\ast})=\ker(T^{\ast}T)$ for general bounded linear operators $T$ on a
Hilbert space $\mathcal{H}$ easily derived from $\langle
T^{\ast}(v),w\rangle=\langle v,T(w)\rangle$ and
$\langle(T^{\ast}T)(v),v\rangle=\langle T(v),T(v)\rangle$ for all
$v,w\in\mathcal{H}$ respectively. In $C^{*}$-algebra theory, a projection
refers to a self-adjoint idempotent. For an operator $T$ in the
$C^{*}$-algebra $\mathcal{B}(\mathcal{H})$ of all bounded linear operators on
$\mathcal{H}$, we recall that $T$ is a projection if and only if $T$ is
geometrically the orthogonal projection from $\mathcal{H}$ onto a closed
subspace of $\mathcal{H}$.

We will need some basic knowledge of Fredholm operators, i.e., operators
$T\in\mathcal{B}(\mathcal{H})$ with its quotient class $[T]$ an invertible
element of the Calkin algebra
$\mathcal{B}(\mathcal{H})/\mathcal{K}(\mathcal{H})$, which can be found in [4,
6]. Any such operator has closed finite-codimensional range and finite-
dimensional kernel, and the intersection of $\mathbb{R}\backslash\\{0\\}$ and
the spectrum $\operatorname{Sp}(T)$ of any positive Fredholm operator $T$ is a
compact subset of $(0,\infty)$. Also we note that the set of all Fredholm
operators is closed under taking adjoint and composition of operators. For any
positive Fredholm operator $T$ we will denote by $\underline{T^{-1/2}}$ the
positive operator $f(T)$ defined by functional calculus, where $f$ is the
nonnegative continuous function on $\\{0\\}\sqcup K$ such that
$f(\bullet)=\bullet^{-1/2}$ on $K:=\operatorname{Sp}(T)\backslash\\{0\\}$ and
$f(0)=0$.

Below we recall a folklore result with a proof.

######  Lemma 2.1.

For any Fredholm operator $T$ on a Hilbert space $\mathcal{H}$,

$\tilde{T}:=T\underline{(T^{\ast}T)^{-1/2}}$

is a partial isometry sending the closed subspace
$(\ker(T))^{\perp}\equiv\operatorname{range}(T^{\ast})$ isometrically onto the
closed subspace $\operatorname{range}(T)$ while annihilating $\ker(T)$.

######  Proof 2.2.

Note that since $T^{\ast}T$ is a positive Fredholm operator, the set $K$ is a
compact subset of $(0,\infty)$.

By the spectral theory of self-adjoint operators,

$\tilde{T}^{\ast}\tilde{T}\equiv\underline{(T^{\ast}T)^{-1/2}}T^{\ast}T\underline{(T^{\ast}T)^{-1/2}}\equiv
f(T^{\ast}T)(T^{\ast}T)f(T^{\ast}T)=\chi_{K}(T^{\ast}T)$

for the characteristic function $\chi_{K}$ on $\operatorname{Sp}(T^{\ast}T)$,
and hence $\tilde{T}^{\ast}\tilde{T}$ is the orthogonal projection from
$\mathcal{H}$ onto $\operatorname{range}(T^{\ast}T)$. Thus $\tilde{T}$
annihilates

$\ker\big{(}\tilde{T}\big{)}\equiv\ker\big{(}\tilde{T}^{\ast}\tilde{T}\big{)}\equiv\big{(}\operatorname{range}\big{(}\tilde{T}^{\ast}\tilde{T}\big{)}\big{)}^{\perp}=(\operatorname{range}(T^{\ast}T))^{\perp}\equiv\ker(T^{\ast}T)\equiv\ker(T)$

and is metric preserving on

$\operatorname{range}(T^{\ast}T)\equiv(\ker(T^{\ast}T))^{\perp}\equiv(\ker(T))^{\perp}\equiv\operatorname{range}(T^{\ast})$

due to

$\big{\langle}\tilde{T}(v),\tilde{T}(w)\big{\rangle}=\big{\langle}\big{(}\tilde{T}^{\ast}\tilde{T}\big{)}(v),w\big{\rangle}=\langle
v,w\rangle\qquad\text{for all }v,w\in\operatorname{range}(T^{\ast}T),$

i.e., $\tilde{T}$ is a partial isometry sending
$\operatorname{range}(T^{\ast}T)$ isometrically onto
$\operatorname{range}(\tilde{T})$ while annihilating $\ker(T)$.

It remains to show that
$\operatorname{range}(\tilde{T})=\operatorname{range}(T)$. In fact, since
$\frac{1}{f|_{K}}$ is a well-defined continuous function on $K$, the
restriction of $(T^{\ast}T)^{-1/2}\equiv f(T^{\ast}T)$ to
$\operatorname{range}(T^{\ast}T)$ is an invertible linear operator on
$\operatorname{range}(T^{\ast}T)$ and hence

$\operatorname{range}\big{(}(T^{\ast}T)^{-1/2}\big{)}=\operatorname{range}(T^{\ast}T)\equiv(\ker(T))^{\perp}.$

Thus we get

$\operatorname{range}(\tilde{T})\equiv\operatorname{range}\big{(}T(T^{\ast}T)^{-1/2}\big{)}=T((\ker(T))^{\perp})=\operatorname{range}(T).$

At each fixed $t_{1}\in\mathbb{T}$, by applying Lemma 2.1 to the Fredholm
operator values of the norm continuous $\mathbb{T}$-families $x_{1}$ and
$x_{2}$, we get two partial isometries

$\tilde{x}_{1}:=x_{1}\underline{(x_{1}^{\ast}x_{1})^{-1/2}}$

and

$\tilde{x}_{2}:=x_{2}\underline{(x_{2}^{\ast}x_{2})^{-1/2}}=x_{2}(x_{2}^{\ast}x_{2})^{-1/2}$

where

$\underline{(x_{2}^{\ast}x_{2})^{-1/2}}=(x_{2}^{\ast}x_{2})^{-1/2}\equiv\big{(}(x_{2}^{\ast}x_{2})^{-1}\big{)}^{1/2}\equiv\sqrt{(x_{2}^{\ast}x_{2})^{-1}}$

is a well-defined invertible operator on $\ell^{2}(\mathbb{Z}_{\geq})$ since
$\ker(x_{2})=0$ and hence the spectrum of the positive Fredholm operator
$x_{2}^{\ast}x_{2}$ is a compact subset of $(0,\infty)$, implying the
invertibility of $x_{2}^{\ast}x_{2}$ and making the functional calculus
$(x_{2}^{\ast}x_{2})^{-1/2}\equiv\big{(}(x_{2}^{\ast}x_{2})^{-1}\big{)}^{1/2}$
meaningful.

######  Theorem 2.3.

The $C^{*}$-algebra $C\big{(}\mathbb{C}P_{q,c}^{1}\big{)}$ coincides with
$C^{\ast}(\\{\tilde{x}_{1}^{\ast}\tilde{x}_{2},x_{1}^{\ast}x_{1}\\})$, i.e.
the $C^{*}$-algebra generated by two $\mathbb{T}$-families
$x_{1}^{\ast}x_{1}\geq 0$ and $\tilde{x}_{1}^{\ast}\tilde{x}_{2}$ of operators
on $\ell^{2}(\mathbb{Z}_{\geq})$ such that at each fixed
$t_{1}\mathbb{\in}\mathbb{T}$, $\tilde{x}_{1}^{\ast}\tilde{x}_{2}$ is an
isometry of index $-2$ $($with a zero kernel and a range of codimension $2)$,
where $\tilde{x}_{1}^{\ast}$ and
$\tilde{x}_{2}=x_{2}(x_{2}^{\ast}x_{2})^{-1/2}$ are isometries of index $-1$
while $\tilde{x}_{1}=x_{1}\underline{(x_{1}^{\ast}x_{1})^{-1/2}}$ and
$\tilde{x}_{2}^{\ast}$ are surjective partial isometries of index $1$.

######  Proof 2.4.

By Lemma 2.1, the surjective Fredholm operator $x_{1}$ with kernel of
dimension $1$ yields a surjective partial isometry
$\tilde{x}_{1}=x_{1}\underline{(x_{1}^{\ast}x_{1})^{-1/2}}$ of index $1$ and
the injective Fredholm $x_{2}$ with cokernel of dimension $1$ yields an
isometry $\tilde{x}_{2}=x_{2}(x_{2}^{\ast}x_{2})^{-1/2}$ of index $-1$.

Now both $\tilde{x}_{2}$ and the adjoint $\tilde{x}_{1}^{\ast}$ of the
surjective partial isometry $\tilde{x}_{1}$ are isometries of index $-1$, and
hence $\tilde{x}_{1}^{\ast}\tilde{x}_{2}$ is an isometry of index $-2$ with
$(\tilde{x}_{1}^{\ast}\tilde{x}_{2})^{\ast}(\tilde{x}_{1}^{\ast}\tilde{x}_{2})=1$.

From the definition of $\tilde{x}_{i}$, we get

$\tilde{x}_{1}^{\ast}\tilde{x}_{2}=\underline{(x_{1}^{\ast}x_{1})^{-1/2}}x_{1}^{\ast}x_{2}(x_{2}^{\ast}x_{2})^{-1/2}\in
C^{\ast}(\\{x_{1}^{\ast}x_{2},x_{1}^{\ast}x_{1},x_{2}^{\ast}x_{2}\\})=C\big{(}\mathbb{C}P_{q,c}^{1}\big{)}.$

Since $(x_{1}^{\ast}x_{1})^{1/2}\underline{(x_{1}^{\ast}x_{1})^{-1/2}}$ is the
orthogonal projection onto
$\operatorname{range}(x_{1}^{\ast}x_{1})\equiv\operatorname{range}(x_{1}^{\ast})$,
by spectral theory:

$x_{1}^{\ast}x_{2}=(x_{1}^{\ast}x_{1})^{1/2}\underline{(x_{1}^{\ast}x_{1})^{-1/2}}x_{1}^{\ast}x_{2}=(x_{1}^{\ast}x_{1})^{1/2}\tilde{x}_{1}^{\ast}\tilde{x}_{2}(x_{2}^{\ast}x_{2})^{1/2}\in
C^{\ast}(\\{\tilde{x}_{1}^{\ast}\tilde{x}_{2},x_{1}^{\ast}x_{1},x_{2}^{\ast}x_{2}\\}).$

So we get

$C\big{(}\mathbb{C}P_{q,c}^{1}\big{)}\equiv
C^{\ast}(\\{x_{1}^{\ast}x_{2},x_{1}^{\ast}x_{1},x_{2}^{\ast}x_{2}\\})=C^{\ast}(\\{\tilde{x}_{1}^{\ast}\tilde{x}_{2},x_{1}^{\ast}x_{1},x_{2}^{\ast}x_{2}\\}).$

Furthermore the generator $x_{2}^{\ast}x_{2}$ is redundant since

$x_{2}^{\ast}x_{2}=1+cq^{-2}-q^{-2}x_{1}^{\ast}x_{1}=\big{(}1+cq^{-2}\big{)}(\tilde{x}_{1}^{\ast}\tilde{x}_{2})^{\ast}(\tilde{x}_{1}^{\ast}\tilde{x}_{2})-q^{-2}x_{1}^{\ast}x_{1}$

can be generated by $\tilde{x}_{1}^{\ast}\tilde{x}_{2}$ and
$x_{1}^{\ast}x_{1}$. Thus

$C\big{(}\mathbb{C}P_{q,c}^{1}\big{)}\equiv
C^{\ast}(\\{\tilde{x}_{1}^{\ast}\tilde{x}_{2},x_{1}^{\ast}x_{1},x_{2}^{\ast}x_{2}\\})=C^{\ast}(\\{\tilde{x}_{1}^{\ast}\tilde{x}_{2},x_{1}^{\ast}x_{1}\\}).$

We remark that for each $i\in\\{1,2\\}$,

$\tilde{x}_{i}^{\ast}\tilde{x}_{i}=\underline{(x_{i}^{\ast}x_{i})^{-1/2}}x_{i}^{\ast}x_{i}\underline{(x_{i}^{\ast}x_{i})^{-1/2}}=\chi_{\operatorname{Sp}(x_{i}^{\ast}x_{i})\backslash\\{0\\}}(x_{i}^{\ast}x_{i})\in
C^{\ast}(\\{x_{i}^{\ast}x_{i}\\})$

and hence belongs to $C\big{(}\mathbb{C}P_{q,c}^{1}\big{)}\equiv
C^{\ast}(\\{\tilde{x}_{1}^{\ast}\tilde{x}_{2},x_{1}^{\ast}x_{1},x_{2}^{\ast}x_{2}\\})$,
since any $C^{*}$-algebra is closed under functional calculus by continuous
functions vanishing at $0$.

##  3 Invariant subspace decomposition

In this section, fixing an arbitrary value of the parameter
$t_{1}\equiv\overline{t_{2}}\in\mathbb{T}$, we study and treat any
$\mathbb{T}$-family of operators on $\ell^{2}(\mathbb{Z}_{\geq})$, including
any element of $C\big{(}\mathbb{C}P_{q,c}^{1}\big{)}$, as an operator in
$\mathcal{B}\big{(}\ell^{2}(\mathbb{Z}_{\geq})\big{)}$.

As such the index-$1$ surjective partial isometry $\tilde{x}_{1}$ has
$\ker(\tilde{x}_{1})=\ker(x_{1})=\mathbb{C}v_{1}$ for some unit vector
$v_{1}$, and

$p_{1}:=1-\tilde{x}_{1}^{\ast}\tilde{x}_{1}\in
C^{\ast}(\\{\tilde{x}_{1}^{\ast}\tilde{x}_{2},\tilde{x}_{1}^{\ast}\tilde{x}_{1}\\})\subset
C\big{(}\mathbb{C}P_{q,c}^{1}\big{)}$

is the rank-$1$ orthogonal projection onto $\mathbb{C}v_{1}$. Note that
$(\tilde{x}_{1}^{\ast}\tilde{x}_{1})(v_{1})=0$ is equivalent to
$\tilde{x}_{1}(v_{1})=0$ (or equivalently
$v_{1}\perp\operatorname{range}(\tilde{x}_{1}^{\ast})=\operatorname{range}(\tilde{x}_{1}^{\ast}\tilde{x}_{1})$).

Note that

$p_{2}:=\tilde{x}_{1}^{\ast}\tilde{x}_{1}-(\tilde{x}_{2}^{\ast}\tilde{x}_{1})^{\ast}(\tilde{x}_{2}^{\ast}\tilde{x}_{1})=\tilde{x}_{1}^{\ast}\tilde{x}_{1}-(\tilde{x}_{1}^{\ast}\tilde{x}_{2})(\tilde{x}_{2}^{\ast}\tilde{x}_{1})=\tilde{x}_{1}^{\ast}(1-\tilde{x}_{2}\tilde{x}_{2}^{\ast})\tilde{x}_{1}$

is also a rank-1 projection onto $\mathbb{C}v_{2}$ for some unit vector
$v_{2}$. In fact $p_{2}$ clearly annihilates $\ker(\tilde{x}_{1})$ and can be
viewed as the conjugation of the rank-$1$ projection
$1-\tilde{x}_{2}\tilde{x}_{2}^{\ast}$ (onto the kernel of
$\tilde{x}_{2}^{\ast}$) by the unitary operator
$\tilde{x}_{1}|_{(\ker(\tilde{x}_{1}))^{\perp}}$, from
$(\ker(\tilde{x}_{1}))^{\perp}$ onto $\ell^{2}(\mathbb{Z}_{\geq})$. In an
explicit description, $v_{2}$ can be taken as the inverse image under
$\tilde{x}_{1}|_{(\ker(\tilde{x}_{1}))^{\perp}}$, of any unit vector in the
$1$-dimensional range of $1-\tilde{x}_{2}\tilde{x}_{2}^{\ast}$, and, in
particular,
$v_{2}\in(\ker(\tilde{x}_{1}))^{\perp}\equiv\operatorname{range}(\tilde{x}_{1}^{\ast}\tilde{x}_{1})$.
The inequalities

$0\leq
p_{2}=\tilde{x}_{1}^{\ast}\tilde{x}_{1}-(\tilde{x}_{1}^{\ast}\tilde{x}_{2})(\tilde{x}_{2}^{\ast}\tilde{x}_{1})\leq\tilde{x}_{1}^{\ast}\tilde{x}_{1}$

relate the three projections $p_{2}$,
$(\tilde{x}_{1}^{\ast}\tilde{x}_{2})(\tilde{x}_{2}^{\ast}\tilde{x}_{1})$, and
$\tilde{x}_{1}^{\ast}\tilde{x}_{1}$ in
$C^{\ast}(\\{\tilde{x}_{1}^{\ast}\tilde{x}_{2},\tilde{x}_{1}^{\ast}\tilde{x}_{1}\\})$,
and clarify their geometric relation: $p_{2}$ and
$(\tilde{x}_{1}^{\ast}\tilde{x}_{2})(\tilde{x}_{2}^{\ast}\tilde{x}_{1})$ are
projections onto two mutually orthogonal subspaces which add up to the range
of the projection $\tilde{x}_{1}^{\ast}\tilde{x}_{1}$, i.e.,

$\operatorname{range}(p_{2})\oplus^{\perp}\operatorname{range}((\tilde{x}_{1}^{\ast}\tilde{x}_{2})(\tilde{x}_{2}^{\ast}\tilde{x}_{1}))=\operatorname{range}(\tilde{x}_{1}^{\ast}\tilde{x}_{1}),$

indicating, in particular,
$v_{2}\in\operatorname{range}(p_{2})\subset\operatorname{range}(\tilde{x}_{1}^{\ast}\tilde{x}_{1})$.

Now $v_{2}\perp v_{1}$ since $v_{2}$ is in the range of the self-adjoint
operator $\tilde{x}_{1}^{\ast}\tilde{x}_{1}$ and hence is perpendicular to
$\ker(\tilde{x}_{1}^{\ast}\tilde{x}_{1})=\mathbb{C}v_{1}$. Furthermore,

$v_{i+2n}:=(\tilde{x}_{1}^{\ast}\tilde{x}_{2})^{n}(v_{i})$

with $n\geq 0$ and $i\in\\{1,2\\}$ are orthonormal vectors, since

$v_{1}\perp\operatorname{range}(\tilde{x}_{1}^{\ast})\supset\operatorname{range}(\tilde{x}_{1}^{\ast}\tilde{x}_{2})\ni(\tilde{x}_{1}^{\ast}\tilde{x}_{2})(v_{1})$

and

$v_{2}\perp\operatorname{range}((\tilde{x}_{1}^{\ast}\tilde{x}_{2})(\tilde{x}_{2}^{\ast}\tilde{x}_{1}))=\operatorname{range}(\tilde{x}_{1}^{\ast}\tilde{x}_{2})\qquad\text{with\
\ }v_{1}\perp v_{2}\,.$

Thus
$\mathcal{V}:=\operatorname{Span}\\{v_{1},v_{2}\\}\perp\operatorname{range}(\tilde{x}_{1}^{\ast}\tilde{x}_{2})$
or more precisely, by combining with the fact that the index-$(-2)$ isometry
$\tilde{x}_{1}^{\ast}\tilde{x}_{2}$ has
$\operatorname{range}(\tilde{x}_{1}^{\ast}\tilde{x}_{2})$ of codimension $2$,

$\mathcal{H}=\mathcal{V~{}}\oplus^{\perp}\operatorname{range}(\tilde{x}_{1}^{\ast}\tilde{x}_{2})\qquad\text{as
Hilbert space orthogonal direct sum.}$

Hence, since $\tilde{x}_{1}^{\ast}\tilde{x}_{2}$ is an isometry, we
inductively get:

$\operatorname{range}(\tilde{x}_{1}^{\ast}\tilde{x}_{2})^{k-1}=(\tilde{x}_{1}^{\ast}\tilde{x}_{2})^{k-1}(\mathcal{V})\oplus^{\perp}\operatorname{range}(\tilde{x}_{1}^{\ast}\tilde{x}_{2})^{k}$

for all $k\geq 1$.

Clearly, for each $i\in\\{1,2\\}$, the operator
$\tilde{x}_{1}^{\ast}\tilde{x}_{2}$ restricted to the closed linear span
$\mathcal{H}_{i}\subset\ell^{2}(\mathbb{Z}_{\geq})$ of $\\{v_{i+2n}\colon
n\geq 0\\}$ is a unilateral shift $\mathcal{S}$, while the orthogonal
projection onto $\mathbb{C}v_{i}$ is $p_{i}\in
C^{\ast}(\\{\tilde{x}_{1}^{\ast}\tilde{x}_{2},\tilde{x}_{1}^{\ast}\tilde{x}_{1}\\})$.

Since $\tilde{x}_{1}^{\ast}\tilde{x}_{2}$ is a unilateral shift simultaneously
on both $\mathcal{H}_{1}$ and $\mathcal{H}_{2}$, it generates a
$C^{*}$-algebra
$C^{\ast}(\\{\tilde{x}_{1}^{\ast}\tilde{x}_{2}\\}|_{\mathcal{H}_{1}\oplus\mathcal{H}_{2}})$
containing two “synchronized” copies of the ideal of compact operators, i.e.,

$C^{\ast}(\\{\tilde{x}_{1}^{\ast}\tilde{x}_{2}\\}|_{\mathcal{H}_{1}\oplus\mathcal{H}_{2}})\supset\big{\\{}T\oplus
T\colon
T\in\mathcal{K}\big{(}\ell^{2}(\mathbb{Z}_{\geq})\big{)}\big{\\}}\cong\mathcal{K}\big{(}\ell^{2}(\mathbb{Z}_{\geq})\big{)},$

where $\mathcal{H}_{1}$ and $\mathcal{H}_{2}$ are identified with the same
Hilbert space $\ell^{2}(\mathbb{Z}_{\geq})$ in a canonical way, i.e.,
identifying $v_{i+2n}$ for $i\in\\{1,2\\}$ with the canonical orthonormal
basis vector $e_{n}\in\ell^{2}(\mathbb{Z}_{\geq})$.

However our goal is to show that $C\big{(}\mathbb{C}P_{q,c}^{1}\big{)}$ or for
now
$C^{\ast}(\\{\tilde{x}_{1}^{\ast}\tilde{x}_{2},\tilde{x}_{1}^{\ast}\tilde{x}_{1}\\}|_{\mathcal{H}_{1}\oplus\mathcal{H}_{2}})$
contains the direct sum
$\mathcal{K}(\mathcal{H}_{1})\oplus\mathcal{K}(\mathcal{H}_{2})$ of all “non-
synchronized” pairs of compact operators. This can be achieved by noticing
that for any $k,m\in\mathbb{Z}_{\geq}$,

$\varepsilon_{k,m}^{(1)}:=(\tilde{x}_{1}^{\ast}\tilde{x}_{2})^{k}p_{1}((\tilde{x}_{1}^{\ast}\tilde{x}_{2})^{\ast})^{m}|_{\mathcal{H}_{1}\oplus\mathcal{H}_{2}}\in
C^{\ast}(\\{\tilde{x}_{1}^{\ast}\tilde{x}_{2},\tilde{x}_{1}^{\ast}\tilde{x}_{1}\\}|_{\mathcal{H}_{1}\oplus\mathcal{H}_{2}})$

is a typical matrix unit in $\mathcal{K}(\mathcal{H}_{1})\oplus 0$ sending
$v_{1+2m}$ to $v_{1+2k}$ while eliminating all other $v_{i+2n}$ with $i+2n\neq
1+2m$, and we get $\mathcal{K}(\mathcal{H}_{1})\oplus 0$ as the closure of the
linear span of $\varepsilon_{k,m}^{(1)}$ with $k,m\in\mathbb{Z}_{\geq}$.
Similarly the elements

$\varepsilon_{k,m}^{(2)}:=(\tilde{x}_{1}^{\ast}\tilde{x}_{2})^{k}p_{2}((\tilde{x}_{1}^{\ast}\tilde{x}_{2})^{\ast})^{m}|_{\mathcal{H}_{1}\oplus\mathcal{H}_{2}}\in
C^{\ast}(\\{\tilde{x}_{1}^{\ast}\tilde{x}_{2},\tilde{x}_{1}^{\ast}\tilde{x}_{1}\\}|_{\mathcal{H}_{1}\oplus\mathcal{H}_{2}})$

with $k,m\in\mathbb{Z}_{\geq}$ linearly span a dense subspace of
$0\oplus\mathcal{K}(\mathcal{H}_{2})$. Thus

$\mathcal{K}(\mathcal{H}_{1})\oplus\mathcal{K}(\mathcal{H}_{2})=(\mathcal{K}(\mathcal{H}_{1})\oplus
0)+(0\oplus\mathcal{K}(\mathcal{H}_{2}))\subset
C^{\ast}(\\{\tilde{x}_{1}^{\ast}\tilde{x}_{2},\tilde{x}_{1}^{\ast}\tilde{x}_{1}\\}|_{\mathcal{H}_{1}\oplus\mathcal{H}_{2}}).$

Next we want to show that each $v_{k}$ with $k\geq 1$ is an eigenvector of
$x_{1}^{\ast}x_{1}$ or equivalently of
$x_{2}^{\ast}x_{2}=1+cq^{-2}-q^{-2}x_{1}^{\ast}x_{1}$, and hence each
$\mathcal{H}_{i}$ is invariant under $x_{1}^{\ast}x_{1}$ and
$x_{2}^{\ast}x_{2}$.

######  Proposition 3.1.

The isometry $\tilde{x}_{1}^{\ast}\tilde{x}_{2}$ intertwines the positive
operators $x_{1}^{\ast}x_{1}$ and $(1+c)-x_{2}^{\ast}x_{2}$, i.e.,

$(x_{1}^{\ast}x_{1})(\tilde{x}_{1}^{\ast}\tilde{x}_{2})=(\tilde{x}_{1}^{\ast}\tilde{x}_{2})(1+c-x_{2}^{\ast}x_{2}).$

######  Proof 3.2.

A direct computation shows

$\displaystyle(x_{1}^{\ast}x_{1})(\tilde{x}_{1}^{\ast}\tilde{x}_{2})=x_{1}^{\ast}x_{1}\underline{(x_{1}^{\ast}x_{1})^{-1/2}}x_{1}^{\ast}\tilde{x}_{2}=\underline{(x_{1}^{\ast}x_{1})^{-1/2}}x_{1}^{\ast}x_{1}x_{1}^{\ast}\tilde{x}_{2}$
$\displaystyle\hphantom{(x_{1}^{\ast}x_{1})(\tilde{x}_{1}^{\ast}\tilde{x}_{2})}{}=\underline{(x_{1}^{\ast}x_{1})^{-1/2}}x_{1}^{\ast}(1+c-x_{2}x_{2}^{\ast})\tilde{x}_{2}=\tilde{x}_{1}^{\ast}(1+c-x_{2}x_{2}^{\ast})\tilde{x}_{2}$
$\displaystyle\hphantom{(x_{1}^{\ast}x_{1})(\tilde{x}_{1}^{\ast}\tilde{x}_{2})}{}=(1+c)\tilde{x}_{1}^{\ast}\tilde{x}_{2}-\tilde{x}_{1}^{\ast}x_{2}x_{2}^{\ast}\tilde{x}_{2}=(1+c)\tilde{x}_{1}^{\ast}\tilde{x}_{2}-\tilde{x}_{1}^{\ast}x_{2}x_{2}^{\ast}x_{2}(x_{2}^{\ast}x_{2})^{-1/2}$
$\displaystyle\hphantom{(x_{1}^{\ast}x_{1})(\tilde{x}_{1}^{\ast}\tilde{x}_{2})}{}=(1+c)\tilde{x}_{1}^{\ast}\tilde{x}_{2}-\tilde{x}_{1}^{\ast}x_{2}(x_{2}^{\ast}x_{2})^{-1/2}x_{2}^{\ast}x_{2}=(1+c)\tilde{x}_{1}^{\ast}\tilde{x}_{2}-(\tilde{x}_{1}^{\ast}\tilde{x}_{2})(x_{2}^{\ast}x_{2})$
$\displaystyle\hphantom{(x_{1}^{\ast}x_{1})(\tilde{x}_{1}^{\ast}\tilde{x}_{2})}{}=(\tilde{x}_{1}^{\ast}\tilde{x}_{2})(1+c-x_{2}^{\ast}x_{2}).$
∎

######  Proposition 3.3.

The isometry $\tilde{x}_{1}^{\ast}\tilde{x}_{2}$ intertwines the $($possibly
degenerate$)$ eigenspaces $E_{\lambda}(x_{2}^{\ast}x_{2})$ and
$E_{1+c-\lambda}(x_{1}^{\ast}x_{1})$, where $E_{\lambda}(T):=\ker(\lambda-T)$
for linear operators $T$ and $\lambda\in\mathbb{C}$. More precisely,

$(\tilde{x}_{1}^{\ast}\tilde{x}_{2})(E_{\lambda}(x_{2}^{\ast}x_{2}))\subset
E_{1+c-\lambda}(x_{1}^{\ast}x_{1}),$

and

$(\tilde{x}_{1}^{\ast}\tilde{x}_{2})^{-1}(E_{1+c-\lambda}(x_{1}^{\ast}x_{1}))\subset(E_{\lambda}(x_{2}^{\ast}x_{2})),$

where
$(\tilde{x}_{1}^{\ast}\tilde{x}_{2})^{-1}(E_{1+c-\lambda}(x_{1}^{\ast}x_{1}))$
is the inverse image of $E_{1+c-\lambda}(x_{1}^{\ast}x_{1})$ under $($the non-
surjective$)$ $\tilde{x}_{1}^{\ast}\tilde{x}_{2}$.

######  Proof 3.4.

The commutation relation

$(x_{1}^{\ast}x_{1})(\tilde{x}_{1}^{\ast}\tilde{x}_{2})=(\tilde{x}_{1}^{\ast}\tilde{x}_{2})((1+c)-x_{2}^{\ast}x_{2})$

implies that if $v\in E_{\lambda}(x_{2}^{\ast}x_{2})$ then
$(\tilde{x}_{1}^{\ast}\tilde{x}_{2})(v)\in
E_{1+c-\lambda}(x_{1}^{\ast}x_{1})$, because

$\displaystyle(x_{1}^{\ast}x_{1})((\tilde{x}_{1}^{\ast}\tilde{x}_{2})(v))=(\tilde{x}_{1}^{\ast}\tilde{x}_{2})((1+c)-x_{2}^{\ast}x_{2})(v)$
$\displaystyle\hphantom{(x_{1}^{\ast}x_{1})((\tilde{x}_{1}^{\ast}\tilde{x}_{2})(v))}{}=(\tilde{x}_{1}^{\ast}\tilde{x}_{2})((1+c)-\lambda)v=((1+c)-\lambda)((\tilde{x}_{1}^{\ast}\tilde{x}_{2})(v)).$

On the other hand, if $(\tilde{x}_{1}^{\ast}\tilde{x}_{2})(v)\in
E_{1+c-\lambda}(x_{1}^{\ast}x_{1})$, then

$\displaystyle((1+c)-\lambda)((\tilde{x}_{1}^{\ast}\tilde{x}_{2})(v))=(x_{1}^{\ast}x_{1})((\tilde{x}_{1}^{\ast}\tilde{x}_{2})(v))=(\tilde{x}_{1}^{\ast}\tilde{x}_{2})((1+c)-x_{2}^{\ast}x_{2})(v)$
$\displaystyle\hphantom{((1+c)-\lambda)((\tilde{x}_{1}^{\ast}\tilde{x}_{2})(v))}{}=(1+c)(\tilde{x}_{1}^{\ast}\tilde{x}_{2})(v)-(\tilde{x}_{1}^{\ast}\tilde{x}_{2})((x_{2}^{\ast}x_{2})(v)),$

and hence $(\tilde{x}_{1}^{\ast}\tilde{x}_{2})(\lambda
v)=(\tilde{x}_{1}^{\ast}\tilde{x}_{2})((x_{2}^{\ast}x_{2})(v))$. Since
$\tilde{x}_{1}^{\ast}\tilde{x}_{2}$ is injective, we get $\lambda
v=(x_{2}^{\ast}x_{2})(v)$, i.e., $v\in E_{\lambda}(x_{2}^{\ast}x_{2})$.

######  Corollary 3.5.

If $\lambda$ is an eigenvalue of $x_{2}^{\ast}x_{2}$, then $1+c-\lambda$ is an
eigenvalue of $x_{1}^{\ast}x_{1}$.

######  Proof 3.6.

If $E_{\lambda}(x_{2}^{\ast}x_{2})\neq 0$ then
$E_{1+c-\lambda}(x_{1}^{\ast}x_{1})\supset(\tilde{x}_{1}^{\ast}\tilde{x}_{2})(E_{\lambda}(x_{2}^{\ast}x_{2}))\neq
0$ since $\tilde{x}_{1}^{\ast}\tilde{x}_{2}$ is injective.

The equality $x_{1}^{\ast}x_{1}+q^{2}x_{2}^{\ast}x_{2}=q^{2}+c$ implies

$E_{\lambda}(x_{2}^{\ast}x_{2})=E_{q^{2}+c-q^{2}\lambda}(x_{1}^{\ast}x_{1})$

for any $\lambda\in\mathbb{R}$, or equivalently

$E_{\lambda}(x_{1}^{\ast}x_{1})=E_{q^{-2}(q^{2}+c-\lambda)}(x_{2}^{\ast}x_{2}).$

######  Proposition 3.7.

The orthonormal vectors $v_{k}$, $k\in\mathbb{N}$, are eigenvectors of
$x_{1}^{\ast}x_{1}$ $($and of $x_{2}^{\ast}x_{2}\equiv
1+q^{-2}c-q^{-2}x_{1}^{\ast}x_{1})$, and hence each of $\mathcal{H}_{1}$ and
$\mathcal{H}_{2}$ is invariant under all of the generators
$\tilde{x}_{1}^{\ast}\tilde{x}_{2}$, $x_{1}^{\ast}x_{1}$, and
$x_{2}^{\ast}x_{2}$ of $C\big{(}\mathbb{C}P_{q,c}^{1}\big{)}$. More
explicitly, $(x_{1}^{\ast}x_{1})(v_{k})=c_{k}v_{k}$ for all $k\in\mathbb{N}$,
where $c_{k}$ is defined recursively by

$c_{k+2}=c-q^{-2}c+q^{-2}c_{k}\qquad\text{for }k\in\mathbb{N},\text{\ with\
}c_{2}=1+c\text{\ and }c_{1}=0,$

which can be rewritten as

$c_{2n}=q^{-2(n-1)}+c\qquad\text{and}\qquad c_{2n+1}=\big{(}1-q^{-2n}\big{)}c$

for all $n\in\mathbb{Z}_{\geq}$.

######  Proof 3.8.

We prove $(x_{1}^{\ast}x_{1})(v_{k})=c_{k}v_{k}$ and the formula
$c_{k+2}=c-q^{-2}c+q^{-2}c_{k}$ inductively on $k$.

First
$v_{1}\in(\operatorname{range}(\tilde{x}_{1}^{\ast}))^{\perp}=\ker(\tilde{x}_{1})$,
so

$(x_{1}^{\ast}x_{1})(v_{1})=\big{(}(x_{1}^{\ast}x_{1})^{1/2}x_{1}^{\ast}x_{1}\underline{(x_{1}^{\ast}x_{1})^{-1/2}}\big{)}(v_{1})=\big{(}(x_{1}^{\ast}x_{1})^{1/2}x_{1}^{\ast}\tilde{x}_{1}\big{)}(v_{1})=0.$

Next since
$v_{2}\in\operatorname{range}(\tilde{x}_{1}^{\ast}(1-\tilde{x}_{2}\tilde{x}_{2}^{\ast})\tilde{x}_{1})$,
so $v_{2}=\tilde{x}_{1}^{\ast}(w)$ for some unit vector

$w\in\operatorname{range}(1-\tilde{x}_{2}\tilde{x}_{2}^{\ast})=\ker(\tilde{x}_{2}\tilde{x}_{2}^{\ast})=\ker(\tilde{x}_{2}^{\ast})=\ker(x_{2}^{\ast})$

and hence

$\displaystyle(x_{1}^{\ast}x_{1})(v_{2})=(x_{1}^{\ast}x_{1})(\tilde{x}_{1}^{\ast}(w))=(x_{1}^{\ast}x_{1})\underline{(x_{1}^{\ast}x_{1})^{-1/2}}x_{1}^{\ast}(w)$
$\displaystyle\hphantom{(x_{1}^{\ast}x_{1})(v_{2})}{}=\underline{(x_{1}^{\ast}x_{1})^{-1/2}}(x_{1}^{\ast}x_{1})x_{1}^{\ast}(w)=\underline{(x_{1}^{\ast}x_{1})^{-1/2}}x_{1}^{\ast}(1+c-x_{2}x_{2}^{\ast})(w)$
$\displaystyle\hphantom{(x_{1}^{\ast}x_{1})(v_{2})}{}=\tilde{x}_{1}^{\ast}((1+c)w-0)=(1+c)\tilde{x}_{1}^{\ast}(w)=(1+c)v_{2}.$

Now assume that $(x_{1}^{\ast}x_{1})(v_{k})=c_{k}v_{k}$, i.e., $v_{k}\in
E_{c_{k}}(x_{1}^{\ast}x_{1})$, for $k\in\mathbb{N}$. Then

$\displaystyle
v_{k+2}\equiv(\tilde{x}_{1}^{\ast}\tilde{x}_{2})(v_{k})\in(\tilde{x}_{1}^{\ast}\tilde{x}_{2})(E_{c_{k}}(x_{1}^{\ast}x_{1}))\equiv(\tilde{x}_{1}^{\ast}\tilde{x}_{2})(E_{q^{-2}(q^{2}+c-c_{k})}(x_{2}^{\ast}x_{2}))$
$\displaystyle\hphantom{v_{k+2}}{}\subset
E_{1+c-q^{-2}(q^{2}+c-c_{k})}(x_{1}^{\ast}x_{1})=E_{c-q^{-2}c+q^{-2}c_{k}}(x_{1}^{\ast}x_{1}),$

and hence $(x_{1}^{\ast}x_{1})(v_{k+2})=c_{k+2}v_{k+2}$ for
$c_{k+2}:=c-q^{-2}c+q^{-2}c_{k}$.

The recursive formula $c_{k+2}=c-q^{-2}c+q^{-2}c_{k}$ rewritten as
$c_{k+2}-c=q^{-2}(c_{k}-c)$ immediately leads to $c_{i+2n}-c=q^{-2n}(c_{i}-c)$
and hence

$c_{i+2n}=q^{-2n}(c_{i}-c)+c$

for any $i\in\\{1,2\\}$ and $n\in\mathbb{N}$. More explicitly, we have
$c_{2n}=q^{-2(n-1)}+c$ and $c_{2n+1}=\big{(}1-q^{-2n}\big{)}c$ for all
$n\in\mathbb{N}$.

######  Corollary 3.9.

The element
$x_{1}^{\ast}x_{2}=(x_{1}^{\ast}x_{1})^{1/2}\tilde{x}_{1}^{\ast}\tilde{x}_{2}(x_{2}^{\ast}x_{2})^{1/2}$
is a weighted shift on $\mathcal{H}_{1}$ and $\mathcal{H}_{2}$ with respect to
the orthonormal bases $\\{v_{2n-1}\\}_{n\geq 1}$ and $\\{v_{2n}\\}_{n\geq 1}$
respectively. More precisely,

$(x_{1}^{\ast}x_{2})(v_{k})=\sqrt{c_{k+2}}\sqrt{1+q^{-2}c-q^{-2}c_{k}}v_{k+2}$

for the constants $c_{k}$ specified in the above proposition.

######  Proof 3.10.

This is a simple consequence of $(x_{1}^{\ast}x_{1})(v_{k})=c_{k}v_{k}$ and

$(x_{2}^{\ast}x_{2})(v_{k})\equiv\big{(}1+q^{-2}c-q^{-2}x_{1}^{\ast}x_{1}\big{)}(v_{k})=\big{(}1+q^{-2}c-q^{-2}c_{k}\big{)}v_{k}.$

With each of $\mathcal{H}_{1}$ and $\mathcal{H}_{2}$ invariant under the self-
adjoint operators $x_{i}^{\ast}x_{i}$, it is clear that the orthogonal
complement $\mathcal{H}_{0}:=(\mathcal{H}_{1}\oplus\mathcal{H}_{2})^{\perp}$
in $\ell^{2}(\mathbb{Z}_{\geq})$ is also invariant under each
$x_{i}^{\ast}x_{i}$. On the other hand, since we know the orthonormal vectors
$v_{1},v_{2}\in(\operatorname{range}(\tilde{x}_{1}^{\ast}\tilde{x}_{2}))^{\perp}$
for the index-$(-2)$ isometry $\tilde{x}_{1}^{\ast}\tilde{x}_{2}$, we get a
Wold-von Neumann decomposition (Theorem 3.5.17 of [6]) for the isometry
$\tilde{x}_{1}^{\ast}\tilde{x}_{2}$ as

$\tilde{x}_{1}^{\ast}\tilde{x}_{2}=\tilde{x}_{1}^{\ast}\tilde{x}_{2}|_{\mathcal{H}_{0}}\oplus\tilde{x}_{1}^{\ast}\tilde{x}_{2}|_{\mathcal{H}_{1}}\oplus\tilde{x}_{1}^{\ast}\tilde{x}_{2}|_{\mathcal{H}_{2}}$

with

$\mathcal{H}_{0}\equiv\big{(}\operatorname{Span}\big{(}\big{\\{}(\tilde{x}_{1}^{\ast}\tilde{x}_{2})^{k}(v_{i})\colon
i\in\\{1,2\\}\text{ and
}k\in\mathbb{Z}_{\geq}\big{\\}}\big{)}\big{)}^{\perp}.$

Here $(\tilde{x}_{1}^{\ast}\tilde{x}_{2})|_{\mathcal{H}_{0}}$ is a unitary
operator on $\mathcal{H}_{0}$ (if $\mathcal{H}_{0}\neq 0$) since
$\tilde{x}_{1}^{\ast}\tilde{x}_{2}|_{\mathcal{H}_{0}}$ is an index-$0$
isometry in view of $\tilde{x}_{1}^{\ast}\tilde{x}_{2}|_{\mathcal{H}_{i}}$
being an index-$(-1)$ isometry for each $i\in\\{1,2\\}$.

It is not clear whether $\mathcal{H}_{0}$ is actually trivial or not, so we
remark that any discussion involving $\mathcal{H}_{0}$ below is only needed
and valid when $\mathcal{H}_{0}\neq 0$.

We already know that $(\tilde{x}_{1}^{\ast}\tilde{x}_{2})|_{\mathcal{H}_{i}}$
is a unilateral shift for each $i\in\\{1,2\\}$. So with respect to the
decomposition

$\ell^{2}(\mathbb{Z}_{\geq})=\mathcal{H}_{0}\oplus\mathcal{H}_{1}\oplus\mathcal{H}_{2}$

into orthogonal subspaces, the generators $\tilde{x}_{1}^{\ast}\tilde{x}_{2}$,
$x_{1}^{\ast}x_{1}$, $x_{2}^{\ast}x_{2}$ and hence all elements of
$C\big{(}\mathbb{C}P_{q,c}^{1}\big{)}$ can be viewed as block diagonal
operators. Then it is easy to see that Propositions 3.1 and 3.3 hold for the
restrictions of $\tilde{x}_{1}^{\ast}\tilde{x}_{2}$, $x_{1}^{\ast}x_{1}$,
$x_{2}^{\ast}x_{2}$ to each $\mathcal{H}_{i}$.

######  Lemma 3.11.

The spectrum $\operatorname{Sp}(x_{1}^{\ast}x_{1}|_{\mathcal{H}_{0}})$ of
$x_{1}^{\ast}x_{1}|_{\mathcal{H}_{0}}$ is invariant under the function

$f_{1}\colon\ s\mapsto c-q^{-2}c+q^{-2}s\equiv c+q^{-2}(s-c)$

and its inverse function. Similarly, the spectrum
$\operatorname{Sp}(x_{2}^{\ast}x_{2}|_{\mathcal{H}_{0}})$ of
$x_{2}^{\ast}x_{2}|_{\mathcal{H}_{0}}$ is invariant under the function

$f_{2}\colon\ s\mapsto 1-q^{-2}+q^{-2}s\equiv 1+q^{-2}(s-1)$

and its inverse function.

######  Proof 3.12.

By Proposition 3.1,

$(x_{1}^{\ast}x_{1}|_{\mathcal{H}_{0}})(\tilde{x}_{1}^{\ast}\tilde{x}_{2}|_{\mathcal{H}_{0}})=(\tilde{x}_{1}^{\ast}\tilde{x}_{2}|_{\mathcal{H}_{0}})(1+c-x_{2}^{\ast}x_{2}|_{\mathcal{H}_{0}})$

with $\tilde{x}_{1}^{\ast}\tilde{x}_{2}|_{\mathcal{H}_{0}}$ unitary, we get
$x_{1}^{\ast}x_{1}|_{\mathcal{H}_{0}}$ and
$1+c-x_{2}^{\ast}x_{2}|_{\mathcal{H}_{0}}$ unitarily equivalent and hence

$\operatorname{Sp}(x_{1}^{\ast}x_{1}|_{\mathcal{H}_{0}})=\operatorname{Sp}(1+c-x_{2}^{\ast}x_{2}|_{\mathcal{H}_{0}})=1+c-\operatorname{Sp}(x_{2}^{\ast}x_{2}|_{\mathcal{H}_{0}}).$

On the other hand, from $x_{1}^{\ast}x_{1}+q^{2}x_{2}^{\ast}x_{2}=q^{2}+c$, we
have

$\operatorname{Sp}(x_{2}^{\ast}x_{2}|_{\mathcal{H}_{0}})=q^{-2}\big{(}q^{2}+c-\operatorname{Sp}(x_{1}^{\ast}x_{1}|_{\mathcal{H}_{0}})\big{)}=1+q^{-2}c-q^{-2}\operatorname{Sp}(x_{1}^{\ast}x_{1}|_{\mathcal{H}_{0}}).$

Hence

$\operatorname{Sp}(x_{1}^{\ast}x_{1}|_{\mathcal{H}_{0}})=1+c-\big{(}1+q^{-2}c-q^{-2}\operatorname{Sp}(x_{1}^{\ast}x_{1}|_{\mathcal{H}_{0}})\big{)}=c-q^{-2}c+q^{-2}\operatorname{Sp}(x_{1}^{\ast}x_{1}|_{\mathcal{H}_{0}}),$

which shows that under the invertible function $f_{1}\colon
s\in\mathbb{R}\mapsto c-q^{-2}c+q^{-2}s\in\mathbb{R}$, the set
$\operatorname{Sp}(x_{1}^{\ast}x_{1}|_{\mathcal{H}_{0}})\subset\mathbb{R}$
equals itself and hence the inverse function $(f_{1})^{-1}$ maps
$\operatorname{Sp}(x_{1}^{\ast}x_{1}|_{\mathcal{H}_{0}})$ onto itself too.

Since the invertible function $g\colon s\in\mathbb{R}\mapsto
1+c-s\in\mathbb{R}$ maps
$\operatorname{Sp}(x_{2}^{\ast}x_{2}|_{\mathcal{H}_{0}})$ onto
$\operatorname{Sp}(x_{1}^{\ast}x_{1}|_{\mathcal{H}_{0}})$, the conjugate
$g^{-1}\circ f_{1}\circ g$ and its inverse function map
$\operatorname{Sp}(x_{2}^{\ast}x_{2}|_{\mathcal{H}_{0}})$ onto itself, where

$\displaystyle\big{(}g^{-1}\circ f_{1}\circ g\big{)}(s)=1+c-f_{1}(1+c-s)$
$\displaystyle\hphantom{\big{(}g^{-1}\circ f_{1}\circ
g\big{)}(s)}{}=1+c-\big{(}c-q^{-2}c+q^{-2}(1+c-s)\big{)}=1-q^{-2}+q^{-2}s.$ ∎

Note that $f_{1}(s)-c=q^{-2}(s-c)$ and $f_{2}(s)-1=q^{-2}(s-1)$ for all
$s\in\mathbb{R}$ with $q>1$. So the only bounded backward $f_{1}$-orbit is the
constant $f_{1}$-orbit $\\{c\\}$, and similarly the only bounded backward
$f_{2}$-orbit is the constant $f_{2}$-orbit $\\{1\\}$, where by a backward
$f_{i}$-orbit, we mean the set $\\{(f_{i})^{-n}(s)\colon n\in\mathbb{N}\\}$
for a point $s\in\mathbb{R}$. On the other hand, any forward $f_{i}$-orbit
converges to the constant $f_{i}$-orbit, i.e.,
$\lim\limits_{n\rightarrow\infty}(f_{1})^{n}(s)=c$ and
$\lim\limits_{n\rightarrow\infty}(f_{2})^{n}(s)=1$ for any $s$. Since each
spectrum $\operatorname{Sp}(x_{i}^{\ast}x_{i}|_{\mathcal{H}_{0}})$ is a
compact and hence bounded set that is invariant under backward iterations of
$f_{i}$, it can contain only the constant $f_{i}$-orbit. We have, therefore

######  Corollary 3.13.

$\operatorname{Sp}(x_{1}^{\ast}x_{1}|_{\mathcal{H}_{0}})=\\{c\\}$ and
$\operatorname{Sp}(x_{2}^{\ast}x_{2}|_{\mathcal{H}_{0}})=\\{1\\}$, i.e.,
$x_{1}^{\ast}x_{1}|_{\mathcal{H}_{0}}=c\,\mathrm{id}$ and
$x_{2}^{\ast}x_{2}|_{\mathcal{H}_{0}}=\mathrm{id}$.

######  Proposition 3.14.

There is a unital $C^{*}$-algebra isomorphism from
$C^{\ast}(\\{\tilde{x}_{1}^{\ast}\tilde{x}_{2},x_{1}^{\ast}x_{1}\\})|_{\mathcal{H}_{1}\oplus\mathcal{H}_{2}}$
to the pullback $\mathcal{T}\oplus_{C(\mathbb{T})}\mathcal{T}$ of two copies
of $\mathcal{T}\overset{\sigma}{\rightarrow}C(\mathbb{T})$, sending
$\tilde{x}_{1}^{\ast}\tilde{x}_{2}|_{\mathcal{H}_{1}\oplus\mathcal{H}_{2}}$ to
$\mathcal{S}\oplus\mathcal{S}$. This isomorphism provides an exact sequence

$0\rightarrow\mathcal{K}(\mathcal{H}_{1})\oplus\mathcal{K}(\mathcal{H}_{2})\rightarrow
C^{\ast}(\\{\tilde{x}_{1}^{\ast}\tilde{x}_{2},x_{1}^{\ast}x_{1}\\})|_{\mathcal{H}_{1}\oplus\mathcal{H}_{2}}\cong\mathcal{T}\oplus_{C(\mathbb{T})}\mathcal{T}\overset{\sigma}{\rightarrow}C(\mathbb{T})\rightarrow
0$

of $C^{*}$-algebras with
$\sigma(\tilde{x}_{1}^{\ast}\tilde{x}_{2}|_{\mathcal{H}_{1}\oplus\mathcal{H}_{2}})=\mathrm{id}_{\mathbb{T}}$,
$\sigma(x_{1}^{\ast}x_{1}|_{\mathcal{H}_{1}\oplus\mathcal{H}_{2}})=c$, and
$\sigma(x_{2}^{\ast}x_{2}|_{\mathcal{H}_{1}\oplus\mathcal{H}_{2}})=1$.

######  Proof 3.15.

We note that the eigenvalues $c_{k}$ of
$x_{1}^{\ast}x_{1}|_{\mathcal{H}_{1}\oplus\mathcal{H}_{2}}$ satisfying
$c_{k+2}=c-q^{-2}c+q^{-2}c_{k}=f_{1}(c_{k})$ form two forward $f_{1}$-orbits
and hence $\lim\limits_{k\rightarrow\infty}c_{k}=c$. This limit is also clear
from the explicit formulae of $c_{2n}$ and $c_{2n+1}$ given in Proposition
3.7. Similarly, one can verify that the eigenvalues $c_{k}^{\prime}$ of
$x_{2}^{\ast}x_{2}|_{\mathcal{H}_{1}\oplus\mathcal{H}_{2}}$ form two forward
$f_{2}$-orbit and hence $\lim\limits_{k\rightarrow\infty}c_{k}^{\prime}=1$.

So $x_{1}^{\ast}x_{1}|_{\mathcal{H}_{1}\oplus\mathcal{H}_{2}}\equiv c\oplus c$
mod $\mathcal{K}(\mathcal{H}_{1})\oplus\mathcal{K}(\mathcal{H}_{2})$ and
$x_{2}^{\ast}x_{2}|_{\mathcal{H}_{1}\oplus\mathcal{H}_{2}}\equiv 1\oplus 1$
mod $\mathcal{K}(\mathcal{H}_{1})\oplus\mathcal{K}(\mathcal{H}_{2})$, while
$\tilde{x}_{1}^{\ast}\tilde{x}_{2}|_{\mathcal{H}_{1}\oplus\mathcal{H}_{2}}=\mathcal{S}_{\mathcal{H}_{1}}\oplus\mathcal{S}_{\mathcal{H}_{2}}$
for copies $\mathcal{S}_{\mathcal{H}_{i}}$ of the unilateral shift.

It has been shown earlier that
$\mathcal{K}(\mathcal{H}_{1})\oplus\mathcal{K}(\mathcal{H}_{2})\subset
C^{\ast}(\\{\tilde{x}_{1}^{\ast}\tilde{x}_{2},x_{1}^{\ast}x_{1}\\})|_{\mathcal{H}_{1}\oplus\mathcal{H}_{2}}$,
so it is not hard to see that
$C^{\ast}(\\{\tilde{x}_{1}^{\ast}\tilde{x}_{2},x_{1}^{\ast}x_{1}\\})|_{\mathcal{H}_{1}\oplus\mathcal{H}_{2}}$
is the pullback of two copies of
$\mathcal{T}\overset{\sigma}{\rightarrow}C(\mathbb{T})$. In fact,
$\mathcal{S}_{\mathcal{H}_{1}}\oplus\mathcal{S}_{\mathcal{H}_{2}}\equiv\tilde{x}_{1}^{\ast}\tilde{x}_{2}|_{\mathcal{H}_{1}\oplus\mathcal{H}_{2}}$
generates $\\{T\oplus T\colon T\in\mathcal{T}\\}$ as a $C^{*}$-subalgebra of
$C^{\ast}(\\{\tilde{x}_{1}^{\ast}\tilde{x}_{2},x_{1}^{\ast}x_{1}\\})|_{\mathcal{H}_{1}\oplus\mathcal{H}_{2}}$
and hence

$\tilde{x}_{1}^{\ast}\tilde{x}_{2}|_{\mathcal{H}_{1}\oplus\mathcal{H}_{2}}\in\\{T\oplus
T\colon
T\in\mathcal{T}\\}+(\mathcal{K}(\mathcal{H}_{1})\oplus\mathcal{K}(\mathcal{H}_{2}))\subset
C^{\ast}(\\{\tilde{x}_{1}^{\ast}\tilde{x}_{2},x_{1}^{\ast}x_{1}\\})|_{\mathcal{H}_{1}\oplus\mathcal{H}_{2}}.$

On the other hand,

$x_{1}^{\ast}x_{1}|_{\mathcal{H}_{1}\oplus\mathcal{H}_{2}}\in(c\oplus
c)+(\mathcal{K}(\mathcal{H}_{1})\oplus\mathcal{K}(\mathcal{H}_{2}))\subset\\{T\oplus
T\colon
T\in\mathcal{T}\\}+(\mathcal{K}(\mathcal{H}_{1})\oplus\mathcal{K}(\mathcal{H}_{2}))$

and hence

$C^{\ast}(\\{\tilde{x}_{1}^{\ast}\tilde{x}_{2},x_{1}^{\ast}x_{1}\\})|_{\mathcal{H}_{1}\oplus\mathcal{H}_{2}}\subset\\{T\oplus
T\colon
T\in\mathcal{T}\\}+(\mathcal{K}(\mathcal{H}_{1})\oplus\mathcal{K}(\mathcal{H}_{2})).$

So we get

$C^{\ast}(\\{\tilde{x}_{1}^{\ast}\tilde{x}_{2},x_{1}^{\ast}x_{1}\\})|_{\mathcal{H}_{1}\oplus\mathcal{H}_{2}}=\\{T\oplus
T\colon
T\in\mathcal{T}\\}+(\mathcal{K}(\mathcal{H}_{1})\oplus\mathcal{K}(\mathcal{H}_{2}))=\mathcal{T}\oplus_{C(\mathbb{T})}\mathcal{T},$

where the second equality is due to that any $S\oplus
T\in\mathcal{T}\oplus\mathcal{T}$ with $\sigma(S)=\sigma(T)$ can be written as

$S\oplus T=(T\oplus T)+((S-T)\oplus 0)\in T\oplus
T+(\mathcal{K}(\mathcal{H}_{1})\oplus\mathcal{K}(\mathcal{H}_{2})).$

Replacing $\mathcal{T}\oplus_{C(\mathbb{T})}\mathcal{T}$ in the canonical
exact sequence

$0\rightarrow\mathcal{K}(\mathcal{H}_{1})\oplus\mathcal{K}(\mathcal{H}_{2})\rightarrow\mathcal{T}\oplus_{C(\mathbb{T})}\mathcal{T}\overset{\sigma}{\rightarrow}C(\mathbb{T})\rightarrow
0$

by the isomorphic $C^{*}$-algebra
$C^{\ast}(\\{\tilde{x}_{1}^{\ast}\tilde{x}_{2},x_{1}^{\ast}x_{1}\\})|_{\mathcal{H}_{1}\oplus\mathcal{H}_{2}}$,
we get the stated exact sequence with
$\sigma(\tilde{x}_{1}^{\ast}\tilde{x}_{2}|_{\mathcal{H}_{1}\oplus\mathcal{H}_{2}})=\mathrm{id}_{\mathbb{T}}$,
$\sigma(x_{1}^{\ast}x_{1}|_{\mathcal{H}_{1}\oplus\mathcal{H}_{2}})=c$, and
$\sigma(x_{2}^{\ast}x_{2}|_{\mathcal{H}_{1}\oplus\mathcal{H}_{2}})=1$.

######  Theorem 3.16.

The restriction map

$T\in
C^{\ast}(\\{\tilde{x}_{1}^{\ast}\tilde{x}_{2},x_{1}^{\ast}x_{1}\\})\mapsto
T|_{\mathcal{H}_{1}\oplus\mathcal{H}_{2}}\in
C^{\ast}(\\{\tilde{x}_{1}^{\ast}\tilde{x}_{2},x_{1}^{\ast}x_{1}\\})|_{\mathcal{H}_{1}\oplus\mathcal{H}_{2}}$

is a $C^{*}$-algebra isomorphism, and hence
$C^{\ast}(\\{\tilde{x}_{1}^{\ast}\tilde{x}_{2},x_{1}^{\ast}x_{1}\\})$ is
isomorphic to the pullback $\mathcal{T}\oplus_{C(\mathbb{T})}\mathcal{T}$ of
two copies of $\mathcal{T}\overset{\sigma}{\rightarrow}C(\mathbb{T})$ with
$\tilde{x}_{1}^{\ast}\tilde{x}_{2}$ corresponding to
$\mathcal{S}\oplus\mathcal{S}$.

######  Proof 3.17.

Clearly we only need to consider the case with $\mathcal{H}_{0}\neq 0$.

Since $\tilde{x}_{1}^{\ast}\tilde{x}_{2}|_{\mathcal{H}_{0}}$ is unitary, as
shown in the above discussion of Wold–von Neumann decomposition, and
$C(\mathbb{T})$ is the universal $C^{*}$-algebra generated by a single unitary
generator, there is a unique $C^{*}$-algebra homomorphism

$h\colon\ C(\mathbb{T})\rightarrow
C^{\ast}(\\{\tilde{x}_{1}^{\ast}\tilde{x}_{2}|_{\mathcal{H}_{0}}\\})$

sending $\mathrm{id}_{\mathbb{T}}$ to
$\tilde{x}_{1}^{\ast}\tilde{x}_{2}|_{\mathcal{H}_{0}}$ while fixing all
scalars in $\mathbb{C}\subset C(\mathbb{T})$.

Clearly with $x_{1}^{\ast}x_{1}|_{\mathcal{H}_{0}}=1$ and
$x_{2}^{\ast}x_{2}|_{\mathcal{H}_{0}}=c$,

$h\circ\sigma\colon\
C^{\ast}(\\{\tilde{x}_{1}^{\ast}\tilde{x}_{2},x_{1}^{\ast}x_{1}\\})|_{\mathcal{H}_{1}\oplus\mathcal{H}_{2}}\rightarrow
C^{\ast}(\\{\tilde{x}_{1}^{\ast}\tilde{x}_{2}|_{\mathcal{H}_{0}}\\})=C^{\ast}(\\{\tilde{x}_{1}^{\ast}\tilde{x}_{2},x_{1}^{\ast}x_{1}\\})|_{\mathcal{H}_{0}}$

is a well-defined $C^{*}$-algebra homomorphism sending
$\tilde{x}_{1}^{\ast}\tilde{x}_{2}|_{\mathcal{H}_{1}\oplus\mathcal{H}_{2}}$ to
$\tilde{x}_{1}^{\ast}\tilde{x}_{2}|_{\mathcal{H}_{0}}$ and
$x_{i}^{\ast}x_{i}|_{\mathcal{H}_{1}\oplus\mathcal{H}_{2}}$ to
$x_{i}^{\ast}x_{i}|_{\mathcal{H}_{0}}$ for $i\in\\{1,2\\}$. Hence the
restriction map

$T\in
C^{\ast}(\\{\tilde{x}_{1}^{\ast}\tilde{x}_{2},x_{1}^{\ast}x_{1}\\})\mapsto
T|_{\mathcal{H}_{1}\oplus\mathcal{H}_{2}}\in
C^{\ast}(\\{\tilde{x}_{1}^{\ast}\tilde{x}_{2},x_{1}^{\ast}x_{1}\\})|_{\mathcal{H}_{1}\oplus\mathcal{H}_{2}}$

gives a well-defined isomorphism.

In Theorem 3.16, we treat elements of
$C^{\ast}(\\{\tilde{x}_{1}^{\ast}\tilde{x}_{2},x_{1}^{\ast}x_{1}\\})$ as
operators instead of families of operators by fixing implicitly the value of
$\mathbb{T}$-parameter at any $t_{1}\in\mathbb{T}$, i.e., the statement of
Theorem 3.16 is a pointwise result at any $t_{1}\in\mathbb{T}$. It is clear
that collectively the restriction map

$C\big{(}\mathbb{C}P_{q,c}^{1}\big{)}\rightarrow
C\big{(}\mathbb{C}P_{q,c}^{1}\big{)}\big{|}_{\widetilde{\mathcal{H}_{1}}\oplus\widetilde{\mathcal{H}_{2}}}$

is still a $C^{*}$-algebra isomorphism where elements of
$C\big{(}\mathbb{C}P_{q,c}^{1}\big{)}$ are $\mathbb{T}$-families of operators
on $\ell^{2}(\mathbb{Z}_{\geq})$ and
$\widetilde{\mathcal{H}_{1}}\oplus\widetilde{\mathcal{H}_{2}}$ represents a
$\mathbb{T}$-family of Hilbert subspaces
$\mathcal{H}_{1}\oplus\mathcal{H}_{2}$ of $\ell^{2}(\mathbb{Z}_{\geq})$
constructed pointwise for each $t_{1}\in\mathbb{T}$ as described above.

##  4 Superfluous circle parameter

In this section, we show that the $\mathbb{T}$-parameter is superfluous for
the $C^{*}$-algebra
$C\big{(}\mathbb{C}P_{q,c}^{1}\big{)}\big{|}_{\widetilde{\mathcal{H}_{1}}\oplus\widetilde{\mathcal{H}_{2}}}$
consisting of $\mathbb{T}$-families of operators on
$\mathcal{H}_{1}\oplus\mathcal{H}_{2}$, and hence
$C\big{(}\mathbb{C}P_{q,c}^{1}\big{)}|_{\widetilde{\mathcal{H}_{1}}\oplus\widetilde{\mathcal{H}_{2}}}\cong
C\big{(}\mathbb{C}P_{q,c}^{1}\big{)}$ is isomorphic to
$C^{\ast}(\\{\tilde{x}_{1}^{\ast}\tilde{x}_{2},x_{1}^{\ast}x_{1}\\})|_{\mathcal{H}_{1}\oplus\mathcal{H}_{2}}\cong\mathcal{T}\oplus_{C(\mathbb{T})}\mathcal{T}$
(for any $t_{1}\in\mathbb{T}$ fixed) as obtained in Proposition 3.14.

Recall that by a simple change of orthonormal basis $e_{k}\rightsquigarrow
t^{k}e_{k}$ of $\ell^{2}(\mathbb{Z}_{\geq})$ for any fixed $t\in\mathbb{T}$,
the weighted shift operator $\alpha$ becomes $\tilde{\alpha}=t\alpha$ with
respect to the new orthonormal basis, while the self-adjoint operator $\gamma$
remains the same operator $\tilde{\gamma}=\gamma$.

Note that the earlier concrete description of $x_{1}^{\ast}x_{2}$,
$x_{1}^{\ast}x_{1}$, and $x_{2}^{\ast}x_{2}$ as families of operators
parametrized by $t_{1}\in\mathbb{T}$ (with $t_{2}=\overline{t_{1}}$) viewed as
a representation of $C\big{(}\mathbb{C}P_{q,c}^{1}\big{)}\equiv
C^{\ast}(\\{x_{1}^{\ast}x_{2},x_{1}^{\ast}x_{1},\allowbreak
x_{2}^{\ast}x_{2}\\})$ can be first “consolidated” by a change of orthonormal
basis converting $\alpha$ to $\tilde{\alpha}:=t_{1}^{2}\alpha$ and $\gamma$ to
$\tilde{\gamma}=\gamma$, so that we can rewrite the description as

$\displaystyle
x_{1}^{\ast}x_{1}=c+(1-c)\gamma^{2}+\sqrt{c}\overline{t_{1}}^{2}\alpha^{\ast}\gamma+\sqrt{c}t_{1}^{2}\gamma\alpha$
$\displaystyle\hphantom{x_{1}^{\ast}x_{1}}{}=c+(1-c)\tilde{\gamma}^{2}+\sqrt{c}\tilde{\alpha}^{\ast}\tilde{\gamma}+\sqrt{c}\tilde{\gamma}\tilde{\alpha},$
$\displaystyle
x_{2}^{\ast}x_{2}=1+q^{-2}(c-1)\gamma^{2}-q^{-2}\sqrt{c}t_{1}^{2}\gamma\alpha-q^{-2}\sqrt{c}\overline{t_{1}}^{2}\alpha^{\ast}\gamma$
$\displaystyle\hphantom{x_{2}^{\ast}x_{2}}{}=1+q^{-2}(c-1)\tilde{\gamma}^{2}-q^{-2}\sqrt{c}\tilde{\gamma}\tilde{\alpha}-q^{-2}\sqrt{c}\tilde{\alpha}^{\ast}\tilde{\gamma},$
$\displaystyle
x_{1}^{\ast}x_{2}=\sqrt{c}\overline{t_{1}}^{2}(\alpha^{\ast})^{2}-cq^{-1}\alpha^{\ast}\gamma+\gamma\alpha^{\ast}-q^{-1}\sqrt{c}t_{1}^{2}\gamma^{2}$
$\displaystyle\hphantom{x_{1}^{\ast}x_{2}}{}=t_{1}^{2}\big{(}\sqrt{c}\overline{t_{1}}^{4}(\alpha^{\ast})^{2}-cq^{-1}\overline{t_{1}}^{2}\alpha^{\ast}\gamma+\overline{t_{1}}^{2}\gamma\alpha^{\ast}-q^{-1}\sqrt{c}\gamma^{2}\big{)}$
$\displaystyle\hphantom{x_{1}^{\ast}x_{2}}{}=t_{1}^{2}\big{(}\sqrt{c}(\tilde{\alpha}^{\ast})^{2}-cq^{-1}\tilde{\alpha}^{\ast}\tilde{\gamma}+\tilde{\gamma}\tilde{\alpha}^{\ast}-q^{-1}\sqrt{c}\tilde{\gamma}^{2}\big{)},$

where it is understood that $\tilde{\alpha},\tilde{\gamma}$ with respect to
suitable orthonormal basis of $\ell^{2}(\mathbb{Z}_{\geq})$ are the same
familiar matrix operators $\alpha,\gamma$, and hence we can simply replace
$\tilde{\alpha},\tilde{\gamma}$ by $\alpha,\gamma$ in the above formulas for
$x_{1}^{\ast}x_{2}$, $x_{1}^{\ast}x_{1}$, and $x_{2}^{\ast}x_{2}$.

So we have

$\displaystyle
x_{1}^{\ast}x_{1}=c+(1-c)\gamma^{2}+\sqrt{c}\alpha^{\ast}\gamma+\sqrt{c}\gamma\alpha,$
$\displaystyle
x_{2}^{\ast}x_{2}=1+q^{-2}(c-1)\gamma^{2}-q^{-2}\sqrt{c}\gamma\alpha-q^{-2}\sqrt{c}\alpha^{\ast}\gamma,$
$\displaystyle
x_{1}^{\ast}x_{2}=t_{1}^{2}\big{(}\sqrt{c}(\alpha^{\ast})^{2}-cq^{-1}\alpha^{\ast}\gamma+\gamma\alpha^{\ast}-q^{-1}\sqrt{c}\gamma^{2}\big{)},$

where only $x_{1}^{\ast}x_{2}$ still involves $t_{1}=\overline{t_{2}}$ as a
factor. From Proposition 3.7 and Corollary 3.9, there is an orthonormal basis
$\\{v_{2k},v_{2k-1}\colon k\in\mathbb{N}\\}$ of
$\mathcal{H}_{1}\oplus\mathcal{H}_{2}$ consisting of eigenvectors of
$x_{1}^{\ast}x_{1}$ and $x_{2}^{\ast}x_{2}$ and with respect to which
$x_{1}^{\ast}x_{2}$ is a double weighted shift. So after the change of
orthonormal basis $v_{2k}\rightsquigarrow\big{(}t_{1}^{2}\big{)}^{k}v_{2k}$
and $v_{2k-1}\rightsquigarrow\big{(}t_{1}^{2}\big{)}^{k}v_{2k-1}$, the factor
$t_{1}^{2}$ in the formula of $x_{1}^{\ast}x_{2}$ can be dropped while the
formulas of $x_{1}^{\ast}x_{1}$ and $x_{2}^{\ast}x_{2}$ remain the same, i.e.,
we have
$C^{\ast}(\\{x_{1}^{\ast}x_{2},x_{1}^{\ast}x_{1}\\})|_{\mathcal{H}_{1}\oplus\mathcal{H}_{2}}$
for any fixed $t_{1}\in\mathbb{T}$ unitarily equivalent to
$C^{\ast}(\\{x_{1}^{\ast}x_{2},x_{1}^{\ast}x_{1}\\})|_{\mathcal{H}_{1}\oplus\mathcal{H}_{2}}$
for $t_{1}:=1$. So we conclude that the parameter $t_{1}\in\mathbb{T}$ is
“edundant” in the sense that representations of the generators
$x_{1}^{\ast}x_{2}$, $x_{1}^{\ast}x_{1}$, and $x_{2}^{\ast}x_{2}$ of
$C\big{(}\mathbb{C}P_{q,c}^{1}\big{)}$ as operators on
$\ell^{2}(\mathbb{Z}_{\geq})$ by the above formulas for different $t_{1}$’s in
$\mathbb{T}$ are unitarily equivalent representations.

So we can now say that $C\big{(}\mathbb{C}P_{q,c}^{1}\big{)}\cong
C^{\ast}(\\{x_{1}^{\ast}x_{2},x_{1}^{\ast}x_{1}\\})$ where
$C^{\ast}(\\{x_{1}^{\ast}x_{2},x_{1}^{\ast}x_{1}\\})$ is considered as in the
previous section for the operators $x_{1}^{\ast}x_{2},x_{1}^{\ast}x_{1}$
without specifying any value of the $t_{1}$-parameter. Thus we conclude that
$C\big{(}\mathbb{C}P_{q,c}^{1}\big{)}$ is isomorphic to the pullback
$\mathcal{T}\oplus_{C(\mathbb{T})}\mathcal{T}$ of two copies of
$\mathcal{T}\overset{\sigma}{\rightarrow}C(\mathbb{T})$ by Proposition 3.14,
and hence is isomorphic to the algebra $C\big{(}\mathbb{S}_{\mu c}^{2}\big{)}$
of Podleś quantum 2-sphere by the result of [8].

We now summarize our conclusion in the following theorem, where the operators
$X_{1}:=\sqrt{c}\alpha+\gamma$ and
$X_{2}:=-q^{-1}\sqrt{c}\gamma+\alpha^{\ast}$ on $\ell^{2}(\mathbb{Z}_{\geq})$
are respectively the values of the $\mathbb{T}$-families $x_{1}$ and $x_{2}$
at $t_{1}=1=t_{2}$.

######  Theorem 4.1.

The $C^{*}$-algebra $C\big{(}\mathbb{C}P_{q,c}^{1}\big{)}\cong
C^{\ast}(\\{X_{1}^{\ast}X_{2},X_{1}^{\ast}X_{1}\\})$ for the linear operators
$X_{1}:=\sqrt{c}\alpha+\gamma$ and
$X_{2}:=-q^{-1}\sqrt{c}\gamma+\alpha^{\ast}$ on $\ell^{2}(\mathbb{Z}_{\geq})$,
and is isomorphic to the pullback

$\mathcal{T}\oplus_{C(\mathbb{T})}\mathcal{T}\equiv\big{\\{}(T,S)\in\mathcal{T}\oplus\mathcal{T}\colon\sigma(T)=\sigma(S)\big{\\}}$

of two copies of the standard Toeplitz $C^{*}$-algebra $\mathcal{T}$ along the
symbol map $\mathcal{T}\overset{\sigma}{\rightarrow}C(\mathbb{T})$.

We remark that the above change of orthonormal basis
$v_{2k}\rightsquigarrow\big{(}t_{1}^{2}\big{)}^{k}v_{2k}$ and
$v_{2k-1}\rightsquigarrow\big{(}t_{1}^{2}\big{)}^{k}v_{2k-1}$ is “compatible”
and hence works well with the elements $x_{1}^{\ast}x_{2}$,
$x_{1}^{\ast}x_{1}$, and $x_{2}^{\ast}x_{2}$ of
$C\big{(}\mathbb{C}P_{q,c}^{1}\big{)}$, but is not suitable for manipulating
more fundamental elements like $x_{1}$ and $x_{2}$ in
$C\big{(}\mathbb{S}_{q}^{3}\big{)}\equiv C({\rm SU}_{q}(2))$.

### Acknowledgements

N. Ciccoli was partially supported by INDAM-GNSAGA and Fondo Ricerca di Base
2017 “Geometria della quantizzazione”. A.J.-L. Sheu was partially supported by
University of Perugia – Visiting Researcher Program, the grant H2020-MSCA-
RISE-2015-691246-QUANTUM DYNAMICS, and the Polish government grant
3542/H2020/2016/2.

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