# Symmetries and reflections from composition operators in the disk

E. Andruchow, G. Corach, L. Recht

###### Abstract

We study the composition operators $C_{a}$ acting on the Hardy space $H^{2}$
of the unit disk, given by $C_{a}f=f\circ\varphi_{a}$, where

$\varphi_{a}(z)=\frac{a-z}{1-\bar{a}z},$

for $|a|<1$. These operators are reflections: $C_{a}^{2}=1$. We study their
eigenspaces $N(C_{a}\pm 1)$, their relative position (i.e., the intersections
between these spaces and their orthogonal complementes for $a\neq b$ in the
unit disk) and the symmetries induced by $C_{a}$ and these eigenspaces.

2020 MSC: 47A05, 47B15, 47B33 .

Keywords: Symmetries, Reflections, Projections, Subspaces in generic position
.

##  1 Introduction

Let $\mathbb{D}=\\{z\in\mathbb{C}:|z|\leq 1\\}$ be the unit disk and
$\mathbb{T}=\\{z\in\mathbb{C}:|z|=1\\}$ the unit circle. Consider the analytic
automorphisms $\varphi_{a}$ which map $\mathbb{D}$ onto $\mathbb{D}$ of the
form

$\varphi_{a}(z)=\frac{a-z}{1-\bar{a}z},$

for $a\in\mathbb{D}$. Save for a constant of module one, all automorphisms of
the disk are of this form. Note the fact that $\varphi_{a}(\varphi_{a}(z))=z$.
This implies that the composition operators induced by these automorphisms are
reflections (i.e., operators $C$ such that $C^{2}=1$). Namely, let
$H^{2}=H^{2}(\mathbb{D})$ be the Hardy space of the disk, i.e.

$H^{2}=\\{f:\mathbb{D}\to\mathbb{C}:f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}\hbox{
with }\sum_{n=0}^{\infty}|a_{n}|^{2}<\infty\\}.$

Then, an analytic map $\varphi:\mathbb{D}\to\mathbb{D}$ induces the (bounded
linear, see [6]) operator $C_{\varphi}:H^{2}\to H^{2}$,

$C_{\varphi}f=f\circ\varphi.$

In particular, for $a\in\mathbb{D}$, the operator $C_{a}:=C_{\varphi_{a}}$
satisfies $(C_{a})^{2}=1$, the identity operator. The eigenspaces of $C_{a}$
are

$N(C_{a}-1)=\\{f\in H^{2}:f\circ\varphi_{a}=f\\}\ \hbox{ and }\
N(C_{a}+1)=\\{g\in H^{2}:g\circ\varphi_{a}=-g\\},$

which verify that $N(C_{a}-1)\dot{+}N(C_{a}+1)=H^{2}$. Here $\dot{+}$ means
direct (non necessarily orthogonal) sum, we reserve the symbol $\oplus$ for
orthogonal sums.

Reflections which additionally are selfadoint are called symmetries: $S$ is a
symmetry if $S=S^{*}=S^{-1}$. Associated to a reflection $C$, there are three
natural symmetries: ${\bf r}(C)$, ${\bf n}(C)$ and $\rho(C)$. The first two
correspond to the decompositions $H^{2}=N(C-1)\oplus N(C-1)^{\perp}$ and
$H^{2}=N(C+1)\oplus N(C+1)^{\perp}$ respectively. The third is of differential
geometric nature, and is described below. The aim of this paper is the study
of the operators $C_{a}$ for $a\in\mathbb{D}$, the description of their
eigenspaces, their relative position, and the induced symmetries. In this
task, it will be important the role of the unique fixed point $\omega_{a}$ of
$\varphi_{a}$ inside the disk. Namely,

$\omega_{a}:=\frac{1}{\bar{a}}\\{1-\sqrt{1-|a|^{2}}\\}\ \hbox{ if }a\neq 0,\
\hbox{ and }\omega_{0}=0.$

The contents of the paper are the following. In Section 2 we recall basic
facts on the manifolds of reflections and symmetries, in particular the
condition for existence of geodesics between points in the latter space. In
Section 3 we state basic formulas concerning the operators $C_{a}$. In Section
4 we characterize the symmetries $\rho_{a}$, obtained as the unitary part of
the polar decomposition of $C_{a}$. For this task, we use Rosenblum’s
computation for the spectral measure of a selfadjoint Toeplitz operator [11].
Using a result by E. Berkson [3], we show that locally, the map
$a\mapsto\rho_{a}$ ($a\in\mathbb{D}$) is injective (it remains unanswered
wether it is globally injective in the disk $\mathbb{D}$). We also obtain
formulas for the range an nullspace symmetries of $C_{a}$, and a power series
expansion for $\rho_{a}$. The rest of the paper is devoted to the study of the
eigenspaces of $C_{a}$, and their relative position. If $a=0$, then the fixed
point of $\varphi_{0}$ is $\omega_{0}=0$ and $C_{0}$ is the reflection (and
symmetry) $f(z)\mapsto f(-z)$. Thus the eigenspaces of $C_{0}$ are the
subspaces ${\cal E}$ and ${\cal O}$ of even and odd functions of $H^{2}$. It
is elemenatry to see that for arbitrary $a\in\mathbb{D}$, the eigenspaces of
$C_{a}$ are

$N(C_{a}-1)=C_{\omega_{a}}({\cal E})\ \ \hbox{ and }\ \
N(C_{a}+1)=C_{\omega_{a}}({\cal O}).$

We then analyze the position of these eigenspaces for $a\neq b$. For instance
(Theorem 5.4),

$N(C_{a}-1)\cap N(C_{b}-1)=\mathbb{C}1\hbox{ and }N(C_{a}+1)\cap
N(C_{b}+1)=\\{0\\}.$

The computations of the intersections involving the orthogonal of these spaces
is more cumbersome, and we are only able to do it in the special case when
$b=0$ (Theorem 5.7). These facts, which are stated in Section 5, are used in
Section 6 to show which of these eigenspaces are conjugate with the
exponential of the Grassmann manifold of $H^{2}$.

This work was supported by the grant PICT2019 0460, from ANPCyT, Argentina.

##  2 Preliminaries, on reflections and symmetries

Denote the set of reflections by

${\cal Q}=\\{T\in{\cal B}(H^{2}):T^{2}=1\\}.$

The set ${\cal Q}$ has rich geometric structure (see for instance [4]): is it
an homogeneous C∞ submanifold of ${\cal B}(H^{2})$, carrying the action of the
group $Gl(H^{2})$ of invertible operators in $H^{2}$:

$G\cdot T=GTG^{-1},\ T\in{\cal Q},G\in Gl(H^{2}).$

The set ${\cal P}$ of selfadjoint reflections, or symmetries, is

${\cal P}=\\{V\in{\cal Q}:V^{*}=V\\}.$

Note that a symmetry $V$ is a selfadjoint unitary operator. Reflections and
symmetries can be viewed alternatively as oblique and orthogonal projections,
respectively. A reflection $T$ gives rise to an idempotent (or oblique
projection) with range equal to the eigenspace $\\{f\in H^{2}:Tf=f\\}$:
$Q_{+}=\frac{1}{2}(1+T)$ (and another with range equal to the other eigenspace
$\\{g\in H^{2}:Tg=-g\\}$ of $T$: $Q_{-}=\frac{1}{2}(1-T)$). If $S$ is a
symmetry, the corresponding idempotents $P_{+}$ and $P_{-}$ are orthogonal
projections.

The set ${\cal P}$, in turn, can be regarded as the Grassmann manifold of
$H^{2}$: to each reflection $V$ corresponds a unique projection
$P_{+}=\frac{1}{2}(1+V)$ and a unique subspace ${\cal S}$ such that
$R(P_{+})={\cal S}$. The geometry of the Grassmann manifold in this operator
theoretic context was developed in [4], [10]: ${\cal P}$ is presented as a
homogeneous space of the unitary group (as in the classical finite dimensional
setting), with a linear reductive connection and a Finsler metric. In [2] the
necessary and sufficient condition for the existence of a geodesic of this
connection between two subspaces ${\cal S}$ and ${\cal T}$ was stated: namely,
that

$\dim({\cal S}\cap{\cal T}^{\perp})=\dim({\cal S}^{\perp}\cap{\cal T}).$ (1)

Moreover, the geodesic is of the form $\delta(t)=e^{itX}{\cal S}$, for
$X^{*}=X$ co-diagonal with respect to both ${\cal S}$ and ${\cal T}$:

$X({\cal S})\subset{\cal S}^{\perp}\ \ \hbox{ and }\ \ X({\cal T})\subset{\cal
T}^{\perp}.$

The geodesic can be chosen of minimal length for the Finsler metric (see [10],
[4], [2]). This latter condition amounts to finding $X$ such that
$\|X\|\leq\pi/2$.

The condition for the existence of a unique minimal geodesic (up to
reparametrization) was given:

${\cal S}\cap{\cal T}^{\perp}=\\{0\\}={\cal S}^{\perp}\cap{\cal T}.$ (2)

In this case, the exponent $X=X_{{\cal S},{\cal T}}$ is unique with the above
mentioned conditions ($X_{{\cal S},{\cal T}}$ selfadjoint, codiagonal with
respect to ${\cal S}$ and ${\cal T}$, with norm less or equal then $\pi/2$,
satisfying $e^{iX_{{\cal S},{\cal T}}}{\cal S}={\cal T}$.).

In this paper we shall examine existence and uniqueness of geodesics of the
Grassmann manifold of $H^{2}$, for the eigenspaces of $C_{a}$.

One of the remarkable features of the space ${\cal Q}$ is the several natural
projection maps that it has onto ${\cal P}$. The natural projection maps
${\cal Q}\to{\cal P}$ are the range, nullspace and unitary part in the polar
decomposition:

  1. 1.

The range map ${\bf r}$, which maps $T\in{\cal Q}$ to the symmetry ${\bf
r}(T)=2P_{R(Q_{+})}-1$, i.e. the symmetry which is the identity on
$R(Q_{+})=\\{f\in H^{2}:Tf=f\\}$. We recall the formula for the orthogonal
projection $P_{R(Q)}$ onto the range $R(Q)$ of an idempotent $Q$ (see for
instance [1]):

$P_{R(Q)}=Q(Q+Q^{*}-1)^{-1}.$

Then

$P_{R(Q_{+})}=\frac{1}{2}(1+T)\\{\frac{1}{2}(1+T)+\frac{1}{2}(1+T^{*})-1\\}^{-1}=(1+T)\\{T+T^{*}\\}^{-1},$

and therefore

${\bf r}(T)=2(1+T)\\{T+T^{*}\\}^{-1}-1=(2+T-T^{*})\\{T+T^{*}\\}^{-1}.$ (3)

  2. 2.

The null-space map ${\bf n}$, which maps $T\in{\cal Q}$ to the symmetry which
is the identity on $R(Q_{-})=\\{g\in H^{2}:Tg=-g\\}$, which by similar
computations is given by

${\bf n}(T)=2(T-1)\\{T+T^{*}\\}^{-1}-1=(T-T^{*}-2)\\{T+T^{*}\\}^{-1}.$ (4)

.

  3. 3.

The unitary part $\rho$ in the polar decomposition, which maps $T$ to

$\rho(T)=T(T^{*}T)^{-1/2},$ (5)

the unitary part in the polar decomposition $T=\rho(T)(T^{*}T)^{1/2}$. We
refer the reader to [4] for the properties of this element $\rho(T)$. Among
them, the most remarkable, that $\rho(T)$ is a symmetry. We shall recall the
other properties of $\rho(T)$ in due course. Note, for instance, that
$(T^{*}T)^{-1}=TT^{*}$, so that

$(T^{*}T)^{-1/2}=(TT^{*})^{1/2}.$

Notice the following formulas:

######  Proposition 2.1.

Let $T\in{\cal Q}$ then

${\bf r}(T)=2(1+T)(T^{*}T+1)^{-1}\ \hbox{ and }\ \ {\bf
n}(T)=2(1-T)(T^{*}T+1)^{-1}.$

###### Proof.

Let $T=\rho(T)|T|$ be the polar decomposition. It is a straightforward
computation (or see [4]) that $|T|\rho(T)=\rho(T)|T^{*}|$. Also it is eassy to
see that since $T^{2}=1$, $|T^{*}|=|T|^{-1}$. Then

$T+T^{*}=\rho(T)|T|+|T|\rho(T)=\rho(T)(|T|+|T^{*}|)=\rho(T)(|T|+|T|^{-1}).$

Using again that $|T|\rho(T)=\rho(T)|T|^{-1}$ (and therefore also that
$\rho(T)|T|=|T|^{-1}\rho(T)$), we have that $\rho(T)$ commutes with
$|T|+|T|^{-1}$. Then

$(T+T^{*})^{-1}=\rho(T)(|T|+|T|^{-1})^{-1}=(|T|+|T|^{-1})^{-1}\rho(T).$

By an elementary functional calculus argument, we have that
$(|T|+|T|^{-1})^{-1}=|T|(|T|^{2}+1)^{-1}$. Then

$(T+T^{*})^{-1}=\rho(T)|T|(|T|^{2}+1)^{-1}=T(|T|^{2}+1)^{-1}.$

Thus,

${\bf r}(T)=2(T+1)T(|T|^{2}+1)^{-1}=2(1+T)(|T|^{2}+1)^{-1},$

and similarly

${\bf n}(T)=2(T-1)T(|T|^{2}+1)^{-1}=2(1-T)(|T|^{2}+1)^{-1}.$

∎

We shall return to these formulas for the case $T=C_{a}$ later, after we
further characterize $|C_{a}|$.

##  3 The operators $C_{a}$

It is not a trivial task to compute the adjoint of a composition operator,
however, for the special case of automorphisms of the disk, it was shown by
Cowen [5] (see also [6]) that

$C_{a}^{*}=(C_{\varphi_{a}})^{*}=M_{\frac{1}{1-\bar{a}z}}C_{a}(M_{1-\bar{a}z})^{*},$

where, for a bounded analytic function $g$ in $\mathbb{D}$, $M_{g}$ denotes
the multiplication operator. Equivalently,

$C_{a}^{*}=M_{\frac{1}{1-\bar{a}z}}C_{a}-aM_{\frac{1}{1-\bar{a}z}}C_{a}(M_{z})^{*},$
(6)

where $(M_{z})^{*}$ (or co-shift) is the adjoint of the shift operator
$S=M_{z}$.

In order to characterize the polar decomposition of $C_{a}$, it will be useful
to compute $C_{a}C_{a}^{*}$. Note that, for $f\in H^{2}$, after
straightforward computations,

$C_{a}C_{a}^{*}f(z)=\frac{1-\bar{a}z}{1-|a|^{2}}\\{f(z)-a\frac{f(z)-f(0)}{z}\\}.$
(7)

Also note how $C_{a}$ relates to the shift operator

$S=M_{z}:H^{2}\to H^{2},\ Sf(z)=zf(z),\ \hbox{ with adjoint
}S^{*}f(z)=\frac{f(z)-f(0)}{z}:$
$C_{a}C_{a}^{*}=\frac{1}{1-|a|^{2}}(1-\bar{a}S)(1-aS^{*}).$ (8)

For $a\in\mathbb{D}$, denote by $k_{a}$ the Szego kernel: for $f\in H^{2}$,
$\langle f,k_{a}\rangle=f(a)$, i.e.,

$k_{a}(z)=\displaystyle{\frac{1}{{1-\bar{a}z}}}.$ (9)

######  Remark 3.1.

Note the fact that

$C_{a}C_{a}^{*}(k_{a})=1.$

Indeed, this follows after a straightforward computation. Therefore, we have
also that

$C_{a}^{*}C_{a}(1)=k_{a}.$

For $a\in\mathbb{D}$, denote by

$\rho_{a}=\rho(C_{a}).$ (10)

Note that if $a=0$, $\varphi_{0}(z)=-z$ and $C_{0}f(z)=f(-z)$ is a symmetry,
thus $C_{0}^{*}=C_{0}$, $|C_{0}|=1$ and $\rho_{0}=C_{0}$.

Returning to the characterization of the modulus of $C_{a}$, we have that

######  Lemma 3.2.

With the current notations,

$|C_{a}^{*}|=\frac{1}{\sqrt{1-|a|^{2}}}|1-aS^{*}|.$

and

$\rho_{a}=\frac{1}{\sqrt{1-|a|^{2}}}C_{a}|1-aS^{*}|.$

######  Remark 3.3.

There is another symmetry related to $C_{a}$. In the book [6] (Exercise
2.1.9:), it is stated that for $a\in\mathbb{D}$, if we put

$\psi_{a}(z)=\frac{\sqrt{1-|a|^{2}}}{1-\bar{a}z}=\sqrt{1-|a|^{2}}\
k_{a}=\frac{k_{a}}{\|k_{a}\|_{2}},$

then the operator $W_{a}\in{\cal B}(H^{2})$, $W_{a}=M_{\psi_{a}}C_{a}$. i.e.,

$W_{a}f(z)=\psi_{a}(z)f(\varphi_{a}(z))$

is a unitary operator. In fact, it is straightforward to verify that
$W_{a}^{2}=1$, i.e., $W_{a}$ is a symmetry.

Note the relationship between $\rho_{a}$ and $W_{a}$:

$C_{a}=\frac{1}{\sqrt{1-|a|^{2}}}M_{1-\bar{a}z}W_{a}=\frac{1}{\sqrt{1-|a|^{2}}}(1-\bar{a}S)W_{a}.$
(11)

It follows that the symmetry $W_{a}$ intertwines $C_{a}C_{a}^{*}$ and
$C_{a}^{*}C_{a}$:

$C_{a}^{*}C_{a}=\frac{1}{1-|a|^{2}}W_{a}(1-aS^{*})(1-\bar{a}S)W_{a}=W_{a}(C_{a}^{*}C_{a})W_{a},$

thus

$|C_{a}|=\frac{1}{\sqrt{1-|a|^{2}}}W_{a}|1-\bar{a}S|W_{a}=W_{a}|C_{a}^{*}|W_{a},$
(12)

and $|C_{a}|^{-1}=\sqrt{1-|a|^{2}}W_{a}|1-\bar{a}S|^{-1}W_{a}$.

######  Remark 3.4.

Note that $C_{a}=\rho_{a}|C_{a}|$ implies that

$C_{a}C_{a}^{*}=\rho_{a}|C_{a}|^{2}\rho_{a}=\rho_{a}C_{a}^{*}C_{a}\rho_{a}.$

Then

$W_{a}\rho_{a}C_{a}^{*}C_{a}(W_{a}\rho_{a})^{*}=W_{a}\rho_{a}C_{a}^{*}C_{a}\rho_{a}W_{a}=W_{a}C_{a}C_{a}^{*}W_{a}=C_{a}^{*}C_{a},$

i.e., $W_{a}\rho_{a}$ commutes with $C_{a}^{*}C_{a}$ (and therefore also with
its inverse $C_{a}C_{a}^{*}$).

##  4 The symmetry $\rho_{a}$

If $\psi\in L^{\infty}(\mathbb{T})$, as is usual notation, let
$T_{\psi}\in{\cal B}(H^{2})$ be the Toeplitz operator with symbol $\psi$:
$T_{\psi}f=P_{H^{2}}(\psi f)$.

The following remark is certainly well known:

######  Lemma 4.1.

For $a\in\mathbb{D}$,

$W_{a}SW_{a}=T_{\varphi_{a}}=M_{\varphi_{a}}.$

###### Proof.

Straightforward computation:

$W_{a}SW_{a}f(z)=\sqrt{1-|a|^{2}}\
W_{a}(\frac{z}{1-\bar{a}z}f(\frac{a-z}{1-\bar{a}z}))=\frac{a-z}{1-\bar{a}z}f(z).$

∎

Therefore:

######  Theorem 4.2.

$|Ca|=\sqrt{1-|a|^{2}}\Big{(}T_{|1-\bar{a}z|^{-2}}\Big{)}^{1/2}=|T_{\psi_{a}}|.$

###### Proof.

As remarked above,

$|C_{a}|^{2}=C_{a}^{*}C_{a}=\frac{1}{1-|a|^{2}}W_{a}(1-aS^{*})(1-\bar{a}S)W_{a}=\frac{1}{1-|a|^{2}}W_{a}(1-aS^{*})W_{a}W_{a}(1-\bar{a}S)W_{a}$

which by Lemma 4.1 equals

$\frac{1}{1-|a|^{2}}(1-a(W_{a}SW_{a})^{*})(1-\bar{a}W_{a}SW_{a})=\frac{1}{1-|a|^{2}}(1-aT_{\varphi_{a}}^{*})(1-\bar{a}T_{\varphi_{a}})=\frac{1}{1-|a|^{2}}T_{1-\bar{a}\varphi_{a}}^{*}T_{1-\bar{a}\varphi_{a}}.$

Now we use the fact that $T_{\psi}^{*}=T_{\bar{\psi}}$ and that if
$\psi,\bar{\theta}\in H^{\infty}$, then $T_{\theta}T_{\psi}=T_{\theta\psi}$
(see chapter 7 of Douglas’ book [8], specifically Prop. 7.5 for the second
assertion). Then

$C_{a}^{*}C_{a}=\frac{1}{1-|a|^{2}}T_{(1-a\bar{\varphi}_{a})(1-\bar{a}\varphi_{a})}.$

Since $1-\bar{a}\varphi_{a}(z)=\frac{1-|a|^{2}}{1-\bar{a}z}$, it follows that
this expression above equals

$(1-|a|^{2})\displaystyle{T_{\frac{1}{|1-\bar{a}z|^{2}}}},$

and the proof follows. ∎

As a consequence, we may use the remarkable description of the spectral
decomposition of selfadjoint Toeplitz operators obtained by M. Rosenblum un
[11]. Let us quote in the next remark this description:

######  Remark 4.3.

Suppose that $\omega:\mathbb{T}\to\mathbb{R}$ is a measurable function that
satifies the following conditions:

  1. 1.

$\omega$ is bounded from below: $\omega(\theta)>-\infty$.

  2. 2.

For each $\lambda\in\mathbb{R}$, the set

$\Gamma_{\lambda}:=\\{e^{i\theta}\in\mathbb{T}:\omega(\theta)\geq\lambda\\}$

is a.e. an arc.

Then Rosenblum’s Theorem 3 in [11] states that: if $T_{\omega}$ is the
Toeplitz operator with symbol $\omega$, $\Lambda\subset\mathbb{R}$ is a Borel
subset and $E(\Lambda)$ is the spectral measure (of $T_{\omega}$) associated
to $\Lambda$, $u,v\in\mathbb{D}$, one has that

$\langle
E(\Lambda)k_{u},k_{v}\rangle=\int_{\Lambda}\Phi(\bar{u};\lambda)\overline{\Phi(\bar{v};\lambda)}dm(\lambda),$
(13)

where

$\Phi(u;\lambda)=\Psi(u;\lambda)\left(1-ue^{i\alpha(\lambda)}\right)^{-1/2}\left(1-ue^{i\beta(\lambda)}\right)^{-1/2},$
$\Psi(u;\lambda)=\exp\Big{(}-\int_{-\pi}^{\pi}\log|\omega(\theta)-\lambda|P(u,\theta)d\theta\Big{)},$
$P(u,\theta)=\frac{1}{4\pi}\frac{1+ue^{i\theta}}{1-ue^{i\theta}},$

$\alpha(\lambda)\leq\beta(\lambda)\in[-\pi,\pi]$ are such that

$\Gamma_{\lambda}=\\{e^{i\theta}:\alpha(\lambda)\leq\theta\leq\beta(\lambda)\\},$

and

$dm(\lambda)=\frac{1}{\pi}\sin(\frac{1}{2}(\beta(\lambda)-\alpha(\lambda)))d\lambda.$

In particular, note that the spectral measure of $T_{\omega}$ is absolutely
continuous with respect to the Lebesgue measure.

In our case, we must analize
$\omega(\theta)=\frac{1}{|1-\bar{a}e^{i\theta}|^{2}}=|k_{a}(e^{i\theta})|^{2}$.
We consider the case $a\neq 0$ (for $a=0$ recall that $\rho_{0}=C_{0}$). The
function $\omega$ is continuous, so condition 1. is fulfilled. With respect to
condition 2., note that, for $\lambda\leq 0$, $\Gamma_{\lambda}$ is empty, and
for $\lambda>0$

$\Gamma_{\lambda}=\\{e^{i\theta}:|\frac{a}{|a|^{2}}-e^{i\theta}|\leq\frac{1}{|a|\sqrt{\lambda}}\\}.$

Consider the following figure

Figure 1

Then clearly the spectral measure is zero if

  * •

$\displaystyle{\lambda>\frac{1}{(1-|a|)^{2}}}$ (here
$\alpha(\lambda)=\beta(\lambda)$ and $\Gamma_{\lambda}$ has measure zero), or
if

  * •

$\displaystyle{\lambda<\frac{1}{(1+|a|)^{2}}}$ (here $\alpha(\lambda)=-\pi$,
$\beta(\lambda)=\pi$ and $\Gamma_{\lambda}=\mathbb{T}$).

For
$\lambda\in\left[\displaystyle{\frac{1}{(1+|a|)^{2}},\frac{1}{(1-|a|)^{2}}}\right]$
we have the following figure:

Figure 2

Therefore, after elementary computations, on has that
$\beta(\lambda)=\arcsin(\gamma)$, $\alpha(\lambda)=-\arcsin(\gamma)$ and

$\sin\left(\frac{1}{2}(\beta(\lambda)-\alpha(\lambda))\right)=\sin(\gamma)=\sqrt{1-\frac{1}{4}\left(1+\frac{1}{|a|}(1-\frac{1}{\lambda})\right)^{2}}.$

Thus, we may characterize the function $\rho_{a}1$ (the symmetry $\rho_{a}$ at
the element $1\in H^{2}$). To this effect, recall that the set
$\\{k_{u}:u\in\mathbb{D}\\}$ is total in $H^{2}$.

######  Proposition 4.4.

With the current notations, for $v\in\mathbb{D}$, we have

$\langle\rho_{a}1,k_{v}\rangle=\frac{\sqrt{1-|a|^{1}}}{\pi}\displaystyle{\int_{\frac{1}{(1+|a|)^{2}}}^{\frac{1}{(1-|a|)^{2}}}}\lambda^{1/2}\Phi(0;\lambda)\overline{\Phi(\bar{v};\lambda)}\sqrt{1-\frac{1}{4}\left(1+\frac{1}{|a|}(1-\frac{1}{\lambda})\right)^{2}}d\lambda.$

###### Proof.

Recall that

$\rho_{a}=C_{a}(C_{a}^{*}C_{a})^{-1/2}=(C_{a}C_{a}^{*})^{-1/2}C_{a}=(C_{a}^{*}C_{a})^{1/2}C_{a},$

so that (since $1=k_{0}$)

$\rho_{a}1=|C_{a}|^{1/2}C_{a}1=|C_{a}|^{1/2}1=|C_{a}|^{1/2}k_{0},$

and then

$\langle\rho_{a}1,k_{v}\rangle=\sqrt{1-|a|^{2}}\langle
T_{|1-\bar{a}e^{i\theta}|^{-2}}^{1/2}k_{0},k_{v}\rangle,$

and the formula folllows applying Rosenblum’s result and the above elementary
computations. ∎

######  Remark 4.5.

In particular, we have that

$\rho_{a}1(0)=\langle\rho_{a}1,1\rangle=\frac{\sqrt{1-|a|^{2}}}{\pi}\int_{\frac{1}{(1+|a|)^{2}}}^{\frac{1}{(1-|a|)^{2}}}\lambda^{1/2}|\Phi(0,\lambda)|^{2}\sqrt{1-\frac{1}{4}\left(1+\frac{1}{|a|}(1-\frac{1}{\lambda})\right)^{2}}d\lambda,$

with

$|\Phi(0,\lambda)|^{2}=\exp\left(-\frac{1}{2\pi}\int_{-\pi}^{\pi}\log||1-\bar{a}e^{i\theta}|^{-2}-\lambda|d\theta\right).$

Clearly, if $A\subset\mathbb{D}$ is a finite set, then
$\\{k_{a}:a\in\mathbb{D}\setminus A\\}$ is also total in $H^{2}$. Therefore we
may characterize $\rho_{a}$ as follows:

######  Theorem 4.6.

With the current notations, for $a,u,v\in\mathbb{D}$, with $u\neq a$, we have

$\langle\rho_{a}k_{u},k_{v}\rangle=\frac{\bar{u}(|a|^{2}-1)^{3/2}}{\pi(\bar{u}-\bar{a})}\int_{\frac{1}{(1+|a|)^{2}}}^{\frac{1}{(1-|a|)^{2}}}\lambda^{1/2}\Phi(\varphi_{a}(u);\lambda)\overline{\Phi(\bar{v};\lambda)}\sqrt{1-\frac{1}{4}\left(1+\frac{1}{|a|}(1-\frac{1}{\lambda})\right)^{2}}d\lambda+$
$+\frac{\bar{a}}{\bar{a}-\bar{u}}\frac{\sqrt{1-|a|^{2}}}{\pi}\int_{\frac{1}{(1+|a|)^{2}}}^{\frac{1}{(1-|a|)^{2}}}\lambda^{1/2}\Phi(0;\lambda)\overline{\Phi(\bar{v};\lambda)}\sqrt{1-\frac{1}{4}\left(1+\frac{1}{|a|}(1-\frac{1}{\lambda})\right)^{2}}d\lambda.$

These inner products characterize $\rho_{a}$, because
$\\{k_{u}:u\in\mathbb{D},u\neq a\\}$ is a total set in $H^{2}$.

###### Proof.

The last assertion is clear.

Recall that

$\rho_{a}=C_{a}(C_{a}^{*}C_{a})^{-1/2}=(C_{a}C_{a}^{*})^{-1/2}C_{a}=(C_{a}^{*}C_{a})^{1/2}C_{a}.$

Note that

$C_{a}k_{u}(z)=\frac{1-\bar{a}z}{1-\bar{u}a-z(\bar{a}-\bar{u})}=\frac{1}{1-\bar{u}a}\frac{1-\bar{a}z}{1-\overline{\varphi_{a}(u)}z},$

which after routine computations (using that $a\neq u$, and $1=k_{0}$) yields

$C_{a}k_{u}=\frac{\bar{u}(1-|a|^{2})}{\bar{u}-\bar{a}}k_{\overline{\varphi_{a}(u)}}+\frac{\bar{a}}{\bar{a}-\bar{u}}k_{0}.$

Therefore,

$\rho_{a}k_{u}=(C_{a}^{*}C_{a})^{1/2}C_{a}k_{u}=(C_{a}^{*}C_{a})^{1/2}(\frac{\bar{u}(1-|a|^{2})}{\bar{u}-\bar{a}}k_{\overline{\varphi_{a}(u)}}+\frac{\bar{a}}{\bar{a}-\bar{u}}k_{0}),$

and thus

$\langle\rho_{a}k_{u},k_{v}\rangle=\sqrt{1-|a|^{2}}\langle
T_{|1-\bar{a}e^{i\theta}|^{-2}}^{1/2}(\frac{\bar{u}(1-|a|^{2})}{\bar{u}-\bar{a}}k_{\overline{\varphi_{a}(u)}}+\frac{\bar{a}}{\bar{a}-\bar{u}}k_{0}),k_{v}\rangle$
$=\sqrt{1-|a|^{2}}\\{\frac{\bar{u}(1-|a|^{2})}{\bar{u}-\bar{a}}\langle
T_{|1-\bar{a}e^{i\theta}|^{-2}}^{1/2}k_{\overline{\varphi_{a}(u)}},k_{v}\rangle+\frac{\bar{a}}{\bar{a}-\bar{u}}\langle
T_{|1-\bar{a}e^{i\theta}|^{-2}}^{1/2}k_{0},k_{v}\rangle\\}.$

The formula folllows applying Rosenblum’s result and the above elementary
computations. ∎

###  4.1 A result by E. Berkson

We are indebted to Daniel Suárez for pointing us the result below. In [3], E.
Berkson proved the following Theorem:

######  Theorem 4.7.

[3] Let $\varphi:\mathbb{D}\to\mathbb{D}$ be a bounded analytic map,
$\tilde{\varphi}$ its boundary function, and
$A=\tilde{\varphi}^{-1}(\mathbb{T})$. Suppose that $|A|>0$ ($=$ normalized
Lebesgue measure in $\mathbb{T}$). Let $\psi:\mathbb{D}\to\mathbb{D}$ be
another analytic map, and $C_{\varphi}$ and $C_{\psi}$ denote the composition
operators on $H^{p}(\mathbb{D})$, $1\leq p<\infty$. If

$\|C_{\psi}-C_{\varphi}\|<\left(\frac{|A|}{2}\right)^{1/p},$

then $\psi=\varphi$.

As a consequence, for $a\neq b\in\mathbb{D}$ we have that ($p=2$):

$\|C_{a}-C_{b}\|\geq\frac{1}{\sqrt{2}}$ (14)

On the other hand, it is a consequence of Theorem 4.2 that

$C_{a}^{*}C_{a}-C_{b}^{*}C_{b}=\displaystyle{T_{\frac{1-|a|^{2}}{1-\bar{a}z}}-T_{\frac{1-|b|^{2}}{1-\bar{b}z}}=T_{\delta_{a,b}}},$

where
$\delta_{a,b}(z)=\displaystyle{\frac{1-|a|^{2}}{1-\bar{a}z}-\frac{1-|b|^{2}}{1-\bar{b}z}}$.
Thus

$\|C^{*}_{a}C_{a}-C_{b}^{*}C_{b}\|=\|\delta_{a,b}\|_{\infty}=\sup\\{|\delta_{a,b}(z)|:z\in\mathbb{T}\\}.$

######  Remark 4.8.

Note that, after an elementary computation, $\delta_{a,b}$ also equals

$\delta_{a,b}(z)=\bar{a}\varphi_{a}(z)-\bar{b}\varphi_{b}(z).$

So that we have

$||a|-|b||\leq||\delta_{a,b}\|_{\infty}\leq|a|+|b|.$

Moreover,

$|\delta_{a,b}(z)|=\Big{|}\frac{1-|a|^{2}}{1-\bar{a}z}-\frac{1-|b|^{2}}{1-\bar{b}z}\Big{|}\leq\Big{|}\frac{1-|a|^{2}}{1-\bar{a}z}-\frac{1-|b|^{2}}{1-\bar{a}z}\Big{|}+\Big{|}\frac{1-|b|^{2}}{1-\bar{a}z}-\frac{1-|b|^{2}}{1-\bar{b}z}\Big{|}=$
$=\frac{1}{|1-\bar{a}z|}||a|^{2}-|b|^{2}|+(1-|b|^{2})\frac{|z||a-b|}{|1-\bar{a}z||1-\bar{b}z|}\leq\frac{1}{1-|a|}(|a|+|b|)|a-b|+(1+|b|)\frac{|a-b|}{1-|a|}\leq$
$\leq\frac{4}{1-|a|}|a-b|.$

In particular, contrary to what happens to $C_{b}$ and $C_{a}$, if $b\to a$,
then both $C_{b}^{*}C_{b}\to C_{a}^{*}C_{a}$ and $|C_{b}|\to|C_{a}|$.
Therefore, we have the following:

######  Proposition 4.9.

Fix $a\in\mathbb{D}$ and $r<\frac{1}{\sqrt{2}}$, consider the open
neighbourhood ${\cal B}_{r}(a)$ of $a$ in $\mathbb{D}$ given by

${\cal B}_{r}(a):=\\{b\in\mathbb{D}:\||C_{b}|-|C_{a}|\|<r\\}.$

Then, if $b\in{\cal B}_{r}(a)$, $b\neq a$, we have that

$\|\rho_{b}-\rho_{a}\|\geq\left(\frac{1}{\sqrt{2}}-r\right)\frac{1+|a|}{\sqrt{1-|a|^{2}}}.$

###### Proof.

By Berkson’s Theorem, if $a\neq b$

$\frac{1}{\sqrt{2}}\leq\|C_{a}-C_{b}\|=\|\rho_{a}|C_{a}|-\rho_{b}|C_{b}|\|\leq\|\rho_{a}|C_{a}|-\rho_{b}|C_{a}|\|+\|\rho_{b}|C_{a}|-\rho_{b}|C_{b}|\|$
$\leq\||C_{a}|\|\|\rho_{a}-\rho_{b}\|+\||C_{a}|-|C_{b}|\|,$

because $\rho_{b}$ is a unitary operator. If $b\in{\cal B}_{r}(a)$,

$\frac{1}{\sqrt{2}}\leq\|C_{a}\|\|\rho_{a}-\rho_{b}\|+r.$

The proof follows recalling that
$\||C_{a}|\|=\|C_{a}\|=\frac{\sqrt{1-|a|^{2}}}{1+|a|}$. ∎

###  4.2 Formulas for ${\bf r}(C_{a})$ and ${\bf n}(C_{a})$.

Using Theorem 4.2 we can refine the formulas for ${\bf r}(T)$ and ${\bf n}(T)$
obtained in Proposition 2.1, the range and nullspace symmetries induced by a
reflection $T$, to the case when $T=C_{a}$:

######  Corollary 4.10.

We have

${\bf r}(C_{a})=2(1+C_{a})T_{{\bf g}_{a}}^{-1}\ \hbox{ and }\ \ {\bf
r}(C_{a})=2(1-C_{a})T_{{\bf g}_{a}}^{-1},$

where $T_{{\bf g}_{a}}$ is the Toeplitz operator with symbol

${\bf g}_{a}(z)=1+\frac{1-|a|^{2}}{|1-\bar{a}z|^{2}}.$

###### Proof.

Note that for $T=C_{a}$ we have ${\bf
n}(C_{a})=2(1+C_{a})(|C_{a}|^{2}+1)^{-1}$, and from Theorem 4.2 we know that

$|C_{a}|^{2}=(1-|a|^{2})\displaystyle{T_{\frac{1}{|1-\bar{a}z|^{2}}}}.$

Then

$|C_{a}|^{2}+1=(1-|a|^{2})\displaystyle{T_{\frac{1}{|1-\bar{a}z|^{2}}}}+1=\displaystyle{T_{1+\frac{1-|a|^{2}}{|1-\bar{a}z|^{2}}}}=T_{{\bf
g}_{a}}.$

The computation of ${\bf r}(C_{a})$ is similar.

∎

###  4.3 A power series expansion for $\rho_{a}$

Let us further consider $|1-\bar{a}S|^{-1}$. Note that

$|1-\bar{a}S|^{2}=(1-aS^{*})(1-\bar{a}S)=1+|a|^{2}-2{\rm Re}(\bar{a}S),$

where ${\rm Re}T=\frac{1}{2}(T+T^{*})$, for $T\in{\cal B}(H^{2})$, as is usual
notation. Then

$|1-\bar{a}S|^{2}=(1+|a|^{2})\left(1-\frac{2}{1+|a|^{2}}{\rm
Re}(\bar{a}S)\right)=(1+|a|^{2})\left(1-\frac{2|a|}{1+|a|^{2}}T\right),$

where $a=|a|\omega$ and $T={\rm Re}(\bar{\omega}S)$ is a contraction. Using
the power series expansion
$(1-kt)^{-1/2}=1+\sum_{n=1}^{\infty}(2n-1)(2n-3)\dots
1(\frac{k}{2})^{n}t^{n}$, we get

######  Lemma 4.11.

With the current notations, i.e. $T={\rm Re}(\bar{\omega}S)$, $a=|a|\omega$,
we have that

  1. 1.

$|1-\bar{a}S|^{-1}=\frac{1}{\sqrt{1+|a|^{2}}}\left(1+\sum_{n=1}^{\infty}(2n-1)(2n-3)\dots
1(\frac{|a|}{1+|a|^{2}})^{n}T^{n}\right),$

where $T=\frac{1}{2}(\bar{\omega}S+\omega S^{*})$ and $a=|a|\omega$.

  2. 2.

$|C_{a}|^{-1}=\sqrt{1-|a|^{2}}W_{a}\\{1+\sum_{n=1}^{\infty}(2n-1)(2n-3)\dots
1(\frac{|a|}{1+|a|^{2}})^{n}T^{n}\\}W_{a}$
$=\sqrt{1-|a|^{2}}\left(1+\sum_{n=1}^{\infty}(2n-1)(2n-3)\dots
1(\frac{|a|}{1+|a|^{2}})^{n}(W_{a}TW_{a})^{n}\right).$

  3. 3.

$\rho_{a}=(1-\bar{a}S)\\{1+\sum_{n=1}^{\infty}(2n-1)(2n-3)\dots
1(\frac{|a|}{1+|a|^{2}})^{n}T^{n}\\}W_{a}=\Big{(}\mu(1-\bar{a}S)\Big{)}W_{a},$

where $\mu(A)=$ unitary part in the polar decomposition of $A$: $A=\mu(A)|A|$.

###### Proof.

Straightforward computations. ∎

Next we see that the map $\mathbb{D}\ni a\mapsto|C_{a}|$ is one to one:

######  Proposition 4.12.

Let $a,b\in\mathbb{D}$. Then $|C_{a}|=|C_{b}|$ if and only if
$|C_{a}^{*}|=|C_{b}^{*}|$ if and only if $a=b$

###### Proof.

Recall that $(C_{a}^{*}C_{a})^{-1}=C_{a}C_{a}^{*}$, and thus
$|C_{a}|^{-1}=|C_{a}^{*}|$. By uniqueness of the positive square root of
operators, clearly $|C_{a}^{*}|=|C_{b}^{*}|$ if and only if
$C_{a}C_{a}^{*}=C_{b}C_{b}^{*}$. Next note that at the constant function $1\in
H^{2}$, we have (since $S^{*}1=0$)

$C_{a}C_{a}^{*}(1)=\frac{1}{1-|a|^{2}}(1-\bar{a}S)(1-aS^{*})(1)=\frac{1}{1-|a|^{2}}(1-\bar{a}S)(1)=\frac{1-\bar{a}z}{1-|a|^{2}}.$

Evaluating at $z=0$, we get that $C_{a}C_{a}^{*}=C_{b}C_{b}^{*}$ implies that
$|a|=|b|$, and thus $1-\bar{a}z=1-\bar{b}z$ for all $z\in\mathbb{D}$, i.e.,
$a=b$. ∎

######  Question 4.13.

Proposition 4.9 states that given $a\in\mathbb{D}$, there is an open
neighbourhood of $a$ such that for $b$ in this neighbourhood,
$\rho_{a}=\rho_{b}$ implies $a=b$. We do not now though if globally the map
$\mathbb{D}\ni a\mapsto\rho_{a}\in{\cal B}(H^{2})$ is injective.

##  5 The eigenspaces of $C_{a}$

Denote by ${\cal E}$ and ${\cal O}$ the closed subspaces of even and odd
functions in $H^{2}$. Note that they are, respectively, ${\cal E}=N(C_{0}-1)$
and ${\cal O}=N(C_{0}+1)$. For general $a\in\mathbb{D}$, the eigenspaces of
$C_{a}$ are

$N(C_{a}-1)=\\{f\in H^{2}:f\circ\varphi_{a}=f\\}\ \ \hbox{ and }\ \
N(C_{a}+1)=\\{g\in H^{2}:g\circ\varphi_{a}=g\\}.$

For $a\in\mathbb{D}$, denote by $\omega_{a}$ the fixed point of $\varphi_{a}$
inside $\mathbb{D}$. Explicitely:

$\omega_{a}=\frac{1}{\bar{a}}\\{1-\sqrt{1-|a|^{2}}\\}.$ (15)

Elementary computations shows that

$\varphi_{\omega_{a}}\circ\varphi_{a}=-\varphi_{\omega_{a}}$ (16)

which at $z=0$ gives

$\varphi_{\omega_{a}}(a)=-\omega_{a}.$ (17)

######  Theorem 5.1.

For $a\in\mathbb{D}$, the eigenspaces of $C_{a}$ are

$N(C_{a}-1)=\\{f=\sum_{n=0}^{\infty}\alpha_{n}(\varphi_{\omega_{a}})^{2n}:(\alpha_{n})\in\ell^{2}\\}=C_{\omega_{a}}({\cal
E}),$ (18)

and

$N(C_{a}+1)=\\{g=\sum_{n=0}^{\infty}\alpha_{n}(\varphi_{\omega_{a}})^{2n+1}:(\alpha_{n})\in\ell^{2}\\}=C_{\omega_{a}}({\cal
O}).$ (19)

###### Proof.

It follows from (16) that the even powers of $\varphi_{\omega_{a}}$ belong to
$N(C_{a}-1)$:

$(\varphi_{\omega_{a}})^{2n}\circ\varphi_{a}=(\varphi_{\omega_{a}})^{2n},$

and the odd powers belong to $N(C_{a}+1)$:

$(\varphi_{\omega_{a}})^{2n+1}\circ\varphi_{a}=-(\varphi_{\omega_{a}})^{2n+1}.$

Therefore, any sequence of coefficients $(\alpha_{n})\in\ell^{2}$ gives an
element

$f=\sum_{n=0}^{\infty}\alpha_{n}(\varphi_{\omega_{a}})^{2n}\in N(C_{a}-1),$

and an element

$g=\sum_{n=0}^{\infty}\alpha_{n}(\varphi_{\omega_{a}})^{2n+1}\in N(C_{a}+1).$

Conversely, suppose that $f\in N(C_{a}-1)$. Using (17)

$f\circ\varphi_{\omega_{a}}=f\circ\varphi_{a}\circ\varphi_{\omega_{a}},$

and since
$\varphi_{a}\circ\varphi_{\omega_{a}}=\frac{a\bar{\omega}_{a}-1}{1-\bar{a}\omega_{a}}\varphi_{\varphi_{\omega_{a}}(a)}=-\varphi_{-\omega_{a}}$,
we get

$f\circ\varphi_{\omega_{a}}(z)=f\circ\varphi_{\omega_{a}}(-z),$

i.e., $f\circ\varphi_{\omega_{a}}\in{\cal E}$. The fact for odd functions is
similar. ∎

Note that if we denote $h(z)=\sum_{n=0}^{\infty}\alpha_{n}z^{2n}$, which is an
arbitrary even function in $H^{2}$, we have that
$f=h\circ\varphi_{\omega_{a}}=C_{\omega_{a}}h$. And similarly if
$k(z)=\sum_{n=0}^{\infty}\alpha_{n}z^{2n+1}$ is an arbitrary odd function in
$H^{2}$, $g=C_{\omega_{a}}k$. Then

$C_{\omega_{a}}\big{|}_{\cal E}:{\cal E}\to N(C_{a}-1)\ \hbox{ and }\
C_{\omega_{a}}\big{|}_{\cal O}:{\cal O}\to N(C_{a}+1).$

######  Theorem 5.2.

The restrictions $C_{\omega_{a}}\big{|}_{\cal E}$ and
$C_{\omega_{a}}\big{|}_{\cal O}$ are bounded linear isomorphisms. Their
inverses are, respectively, $C_{\omega_{a}}\big{|}_{N(C_{a}-1)}$ and
$C_{\omega_{a}}\big{|}_{N(C_{a}+1)}$.

###### Proof.

Note that

$H^{2}=C_{\omega_{a}}({\cal E}\oplus{\cal O})=C_{\omega_{a}}({\cal
E})\dot{+}C_{\omega_{a}}({\cal O})\subset N(C_{a}-1)\dot{+}N(C_{a}+1),$

were $\dot{+}$ denotes direct (non necessarily orthogonal) sum. It follows
that $C_{\omega_{a}}({\cal E})=N(C_{a}-1)$ and $C_{\omega_{a}}({\cal
O})=N(C_{a}+1)$. This completes the proof, since $C_{a}$ is its own inverse. ∎

######  Remark 5.3.

Clearly, if $p,g\in H^{2}$ are, respectively, inner and outer functions, then
$C_{a}p=p\circ\varphi_{a}$ and $C_{a}g=g\circ\varphi_{a}$ are also,
respectively, inner and outer. Therefore, if $f\in N(C_{a}-1)$, and $f=pg$ is
the inner/outer factorization of $f$, then $f=C_{a}p\cdot C_{a}g$ is another
inner/outer factorization. By uniqueness, it must be $C_{a}p=\omega p$ for
some $\omega\in\mathbb{T}$. But then $p$ is an eigenfunction of $C_{a}$, and
so it must be $\omega=\pm 1$. Therefore, if $f\in N(C_{a}-1)$, then either a)
$p,g\in N(C_{a}-1)$ or b) $p,g\in N(C_{a}+1)$. The latter case cannot happen:
the outer function $g$ verifies that $C_{\omega_{a}}g$ is odd, and therefore
it vanishes at $z=0$,

$0=C_{\omega_{a}}g(0)=g(\omega_{a}).$

A similar consideration can be done for $N(C_{a}+1)$. If $f=pg$ is the
inner/outer factorization of $f\in N(C_{a}+1)$, then again $C_{a}p=\pm p$. If
$C_{a}p=p$, then

$-f=-pg=f\circ\varphi_{a}=(p\circ\varphi_{a})(g\circ\varphi_{a})$

implies $p\circ\varphi_{a}=\pm p$. If $p\circ\varphi_{a}=p$, then
$g\circ\varphi_{a}=-g$, and therefore the outer function $g$ vanishes, a
contradiction. Thus $p\in N(C_{a}+1)$ and $g\in N(C_{a}-1)$.

Let us examine the position of the subspaces $N(C_{a}\pm 1)$ and their
orthogonal complements.

######  Theorem 5.4.

Let $a\neq b$ in $\mathbb{D}$. Then

  1. 1.

$N(C_{a}-1)\cap N(C_{b}-1)=\mathbb{C}1,$

where $1\in H^{2}$ is the constant function.

  2. 2.

$N(C_{a}+1)\cap N(C_{b}+1)=\\{0\\}.$

###### Proof.

Let us first prove 1. As seen above, the reflection $C_{\omega_{a}}$ carries
$N(C_{a}-1)$ onto the space ${\cal E}$ of even functions. Another way of
putting this, is that

$C_{\omega_{a}}C_{a}C_{\omega_{a}}=C_{0}.$

Note that since $C_{b}^{-1}=C_{b}$, this product is in fact a conjugation. A
straigtforward computation shows that in general, for $b,d\in\mathbb{D}$

$\varphi_{d}\circ\varphi_{b}\circ\varphi_{d}=\varphi_{d\bullet b},\hbox{ where
}\ d\bullet b:=\frac{2d-b-\bar{b}d^{2}}{1+|d|^{2}-\bar{b}d-b\bar{d}}.$ (20)

Note that

$C_{\omega_{a}}\left(N(C_{b}-1)\right)=N(C_{\omega_{a}}C_{b}C_{\omega_{a}}-1)=N(C_{\omega\bullet
b}-1).$

Then $N(C_{a}-1)\cap N(C_{b}-1)=\mathbb{C}1$ if and only if

$\mathbb{C}1=C_{\omega_{a}}(N(C_{a}-1)\cap N(C_{b}-1))={\cal E}\cap
N(C_{\omega_{a}\bullet b}-1),$

i.e., we have reduced to the case when one of the to points is the origin. Let
us prove that $N(C_{a}-1)\cap{\cal E}=\mathbb{C}1$. Let $f\in
N(C_{a}-1)\cap{\cal E}$, for $a\in\mathbb{D}$, $a\neq 0$. Then

$f=f\circ\varphi_{0}=f\circ\varphi_{a}.$

In particular, this implies that

$f=f\circ\varphi_{a}\circ\varphi_{0}\circ\dots\circ\varphi_{a}=f\circ(\varphi_{a}\circ\varphi_{0})^{(n)}\circ\varphi_{a},$

for all $n\geq 1$ (here $(\varphi_{a}\circ\varphi_{0})^{(n)}$ denotes the
composition of $\varphi_{a}\circ\varphi_{0}$ with itself $n$ times). We shall
need the following computation:

######  Lemma 5.5.

$(\varphi_{a}\circ\varphi_{0})^{(n)}\varphi_{a}=\varphi_{a_{n}},$

where

$a_{n}=\displaystyle{\frac{a}{|a|}\
\frac{1-\left(\frac{1-|a|}{1+|a|}\right)^{n+1}}{1+\left(\frac{1-|a|}{1+|a|}\right)^{n+1}}}.$

###### Proof.

Our claim is equivalent to

$a_{n}=\frac{a}{|a|}\
\frac{(1+|a|)^{n+1}-(1-|a|)^{n+1}}{(1+|a|)^{n+1}+(1-|a|)^{n+1}}.$

The proof is by induction in $n$. It is an elementary computation. For $n=1$,
we have that

$\varphi_{a}\circ\varphi_{0}\circ\varphi_{a}(z)=\varphi_{a}(-\frac{a-z}{1-\bar{a}z})=\frac{a+\frac{a-z}{1-\bar{a}z}}{1+\bar{a}\frac{a-z}{1-\bar{a}z}}=\frac{2a-(1+|a|^{2})z}{1+|a|^{2}-2az}=\frac{\frac{2a}{1+|a|^{2}}-z}{1-\frac{2\bar{a}}{1+|a|^{2}}z}=\varphi_{\frac{2a}{1+|a|^{1}}}(z).$

On the other hand,

$a_{1}=\frac{a}{|a|}\
\frac{(1+|a|)^{2}-(1+|a|)^{2}}{(1+|a|)^{2}+(1-|a|)^{2}}=\frac{2a}{1+|a|^{2}}.$

Suppose the formula valid for $n$. Then

$(\varphi\circ\varphi_{0})^{n+1}\circ\varphi_{a}(z)=(\varphi\circ\varphi_{0})^{n}\circ\varphi_{a}\circ(\varphi_{a}\circ\varphi_{0})(z)=\varphi_{a_{n}}(-\frac{a-z}{1-\bar{a}z})$
$=\displaystyle{\frac{a_{n}+\frac{a-z}{1-\bar{a}z}}{1+\bar{a}_{n}\frac{a-z}{1-\bar{a}z}}}=\displaystyle{\frac{\frac{a}{|a|}{\bf
f}_{n}+\frac{a-z}{1-\bar{a}z}}{1-\frac{a}{|a|}{\bf
f}_{n}\frac{a-z}{1-\bar{a}z}}}=\frac{a(\frac{{\bf f}_{n}}{|a|}+1)-(|a|{\bf
f}_{n}+1)z}{|a|{\bf f}_{n}+1-\bar{a}(\frac{{\bf
f}_{n}}{|a|}+1)z}=\frac{\beta_{n}-z}{1-\bar{\beta}_{n}z}=\varphi_{\beta_{n}}(z),$

where

$\beta_{n}=a\ \frac{(\frac{{\bf f}_{n}}{|a|}+1)}{|a|{\bf f}_{n}+1}\ \hbox{ and
}\ {\bf
f}_{n}=\frac{(1+|a|)^{n+1}-(1-|a|)^{n+1}}{(1+|a|)^{n+1}+(1-|a|)^{n+1}}.$

Thus, we have to show that $\beta_{n}=a_{n}$. Note that

$\beta_{n}=\frac{a}{|a|}\ \frac{{\bf f}_{n}+|a|}{|a|{\bf f}_{n}+1}$

and that

$\frac{{\bf f}_{n}+|a|}{|a|{\bf
f}_{n}+1}=\frac{(1+|a|)^{n+1}-(1-|a|)^{n+1}+|a|(1-|a|)^{n+1}+|a|(1+|a|)^{n+1}}{-|a|(1-|a|)^{n+1}+(1+|a|)^{n+1}+(1-|a|)^{n+1}+|a|(1+|a|)^{n+1}}$
$=\frac{(1+|a|)^{n+2}-(1-|a|)^{n+2}}{(1+|a|)^{n+2}+(1-|a|)^{n+2}},$

which completes the proof of the lemma. ∎

Returning to the proof of the theorem, it is clear that constant functions
belong to ${\cal E}\cap N(C_{a}-1)$. Suppose that there is a non constant
$f\in{\cal E}\cap N(C_{a}-1)$. Then $f_{0}=f-f(0)\in{\cal E}\cap N(C_{a}-1)$
as well. As remarked above, $f_{0}=f_{0}\circ\varphi_{a_{n}}$ for all $n\geq
0$ (for $n=0$, $a_{0}=a$). It follows that $0$ and $a_{n}$, $n\geq 0$ are
zeros of $f_{0}$. Since $f_{0}$ is also even, also $-a_{n}$, $\geq 0$ occur as
zeros of $f_{0}$. Consider $f_{0}=BSg$ the factorization of $f_{0}$ with $B$ a
Blaschke product, $S$ singular inner and $g$ outer. Then the pairs of factors

$\varphi_{a_{n}}\cdot\varphi_{-a_{n}}$

appear in the expression of $B$. Since $f_{0}=f_{0}\circ\varphi_{a}$, and
$S\circ\varphi_{a}$ and $g\circ\varphi_{a}$ are non vanishing in $\mathbb{D}$,
it follows that

$(\varphi_{a_{n}}\circ\varphi_{a})\cdot(\varphi_{-a_{n}}\circ\varphi_{a})$

must also appear in the expression of $B$. Note that

$\varphi_{a_{n}}\circ\varphi_{a}=(\varphi_{a}\circ\varphi_{0})^{(n)}\circ\varphi_{a}\circ\varphi_{a}=((\varphi_{a}\circ\varphi_{0})^{(n)}=(\varphi_{a}\circ\varphi_{0})^{(n-1)}\circ\varphi_{a}\circ\varphi_{0}=\varphi_{a_{n-1}}\circ\varphi_{0}.$

Also

$\varphi_{-a_{n}}(z)=-\frac{a_{n}+z}{1+\bar{a}_{n}z}=-\varphi_{a_{n}}(-z)=\varphi_{0}\circ\varphi_{a_{n}}\circ\varphi_{0}=\varphi_{0}\circ(\varphi_{a}\circ\varphi_{0})^{(n)}\circ\varphi_{a}\circ\varphi_{0}=\varphi_{0}\circ(\varphi_{a}\circ\varphi_{0})^{(n+1)}.$

Then

$\varphi_{-a_{n}}\circ\varphi_{a}=\varphi_{0}\circ(\varphi_{a}\circ\varphi_{0})^{(n+1)}\circ\varphi_{a}=\varphi_{0}\circ\varphi_{a_{n+1}}.$

Note the effect of $C_{a}$ on the following pairs of factors of $B$:

$z\cdot z=z^{2}=\varphi_{0}^{2}\stackrel{{\scriptstyle
C_{a}}}{{\longrightarrow}}(\varphi_{0}\circ\varphi_{a})^{2}=\varphi_{a}^{2},$
$\varphi_{a}\cdot\varphi_{-a}\stackrel{{\scriptstyle
C_{a}}}{{\longrightarrow}}(\varphi_{a}\circ\varphi_{a})\cdot(\varphi_{-a}\circ\varphi_{a})=z\cdot(-\varphi_{a_{1}})=-z\varphi_{a_{1}},$

and

$\varphi_{a_{1}}\cdot\varphi_{a_{-1}}\stackrel{{\scriptstyle
C_{a}}}{{\longrightarrow}}(\varphi_{a_{1}}\circ\varphi_{a})\cdot(\varphi_{a_{-1}}\circ\varphi_{a})=(\varphi_{a}\circ\varphi_{0})\cdot\varphi_{a_{2}}=-\varphi_{-a}\cdot\varphi_{a_{2}}.$

Other pairs of factors in the expression of $B$, after applying $C_{a}$, do
not involve $\varphi_{a}$ or $\varphi_{0}$, due to the spreading of the
indices. Summarizing, after applying $C_{a}$, we get the products

$(\varphi_{a})^{2},\ -z\varphi_{a_{1}}\hbox{ and
}-\varphi_{-a}\cdot\varphi_{a_{2}},$

i.e., we do not recover the original factors $z^{2}$ and
$\varphi_{a}\cdot\varphi_{-a}$. It follows that $f$ is contant.

To prove 2., a similar trick as above allows us to reduce to the case of
$a\neq 0$ and $b=0$, i.e., we must prove that there are no nontrivial odd
functions in $N(C_{a}+1)$. Let $f\in H^{2}$ be odd such that
$f\circ\varphi_{a}=-f$. Then $f^{2}=f\cdot f$ is even and
$(f(\varphi_{a}(z)))^{2}=(-f(z))^{2}=(f(z))^{2}$, i.e., $f^{2}\in N(C_{a}-1)$.
Therefore, by the previous case, $f^{2}$ is constant, and therefore $f\equiv
0$. ∎

######  Corollary 5.6.

The maps $\mathbb{D}\to{\cal P}$ given by

$a\mapsto{\bf r}(C_{a})\ \hbox{ and }\ a\mapsto{\bf n}(C_{a})$

re one to one.

Let us further proceed with the study of the position of the subspaces
$N(C_{a}\pm 1)$ and $N(C_{b}\pm 1)$ for $a\neq b$, considering now their
orthogonal complements. We shall restrict to the case $b=0$. The conditions
look similar, but as far as we could figure it out, some of the proofs may be
quite different.

######  Theorem 5.7.

Let $a\in\mathbb{D}$, $a\neq 0$.

  1. 1.

$N(C_{0}-1)^{\perp}\cap N(C_{a}-1)=\\{0\\}=N(C_{0}-1)\cap N(C_{a}-1)^{\perp},$

  2. 2.

$N(C_{0}+1)^{\perp}\cap N(C_{a}+1)=\\{0\\}=N(C_{0}+1)\cap N(C_{a}+1)^{\perp},$

  3. 3.

$N(C_{0}-1)^{\perp}\cap N(C_{a}-1)^{\perp}=\\{0\\}=N(C_{0}+1)^{\perp}\cap
N(C_{a}+1)^{\perp}.$

###### Proof.

Assertion 1.: for the left hand equality, let $f\in N(C_{0}-1)^{\perp}={\cal
O}$ such that $f\circ\varphi_{a}=f$. Then, by the above results,
$f^{2}\in{\cal E}\cap N(C_{a}-1)$, and therefore $f^{2}$ is constant. Then
$f$, being constant and odd, is zero.

The right hand equality: suppose $f\in N(C_{0}-1)\cap N(C_{a}-1)^{\perp}$ is
$\neq 0$, i.e., $f$ is even and $\langle f,C_{\omega_{a}}(z^{2k})\rangle=0$
for $k\geq 0$ (in particular, when $n=0$ we get $f(0)=0$). Thus
$C^{*}_{\omega_{a}}(f)$ is odd. Recall that

$C^{*}_{\omega_{a}}(f)=\displaystyle{\frac{1}{1-\bar{\omega}_{a}z}f(\frac{\omega_{a}-z}{1-\bar{\omega}_{a}z})-\frac{\omega_{a}}{1-\bar{\omega}_{a}z}\frac{f(\frac{\omega_{a}-z}{1-\bar{\omega}_{a}z})-f(0)}{\frac{\omega_{a}-z}{1-\bar{\omega}_{a}z}}}$
$=\frac{1}{1-\bar{\omega}_{a}z}\left(f(\frac{\omega_{a}-z}{1-\bar{\omega}_{a}z})-\omega_{a}\frac{f(\frac{\omega_{a}-z}{1-\bar{\omega}_{a}z})}{\frac{\omega_{a}-z}{1-\bar{\omega}_{a}z}}\right).$

Since $f$ is even with $f(0)=0$, put
$f(z)=\sum_{n=1}^{\infty}\alpha_{n}z^{2n}$. Then, after routine computations
we get

$C^{*}_{\omega_{a}}(f)=z\frac{|\omega_{a}|^{2}-1}{(1-\bar{\omega}_{a}z)^{2}}\sum_{n=1}^{\infty}\alpha_{n}(\frac{\omega_{a}-z}{1-\bar{\omega}_{a}z})^{2n-1}.$

The fact that $C^{*}_{\omega_{a}}(f)$ is odd, implies that

$A(z)=\frac{1}{(1-\bar{\omega}_{a}z)^{2}}\sum_{n=1}^{\infty}\alpha_{n}(\frac{\omega_{a}-z}{1-\bar{\omega}_{a}z})^{2n-1}$

is even. Note that therefore

$C_{\omega_{a}}(A)=\frac{(1-\bar{\omega}_{a}z)^{2}}{(1-|\omega_{a}|^{2})^{2}}\sum_{n=1}^{\infty}\alpha_{n}z^{2n-1}\in
N(C_{a}-1).$

Let us abreviate $\alpha(z)=\sum_{n=1}^{\infty}\alpha_{n}z^{2n-1}$, which is
an odd function. Thus

$(1-\bar{\omega}_{a}z)^{2}\alpha\in N(C_{a}-1).$

Note that $(1-\bar{\omega}_{a}z)^{2}$ is outer. Therefore, if $\alpha=pg$ is
the inner/outer factorization of $\alpha$, then

$(1-\bar{\omega}_{a}z)^{2}\alpha=p\left((1-\bar{\omega}_{a}z)^{2}g\right)$

is also an inner/outer factorization. Then, by Remark 5.3, we have $p\in
N(C_{a}-1)$. By a similar argument, since $\alpha$ is odd it follows that $p$
is either odd or even. Note that $p$ even would imply $g$ odd, and thus
vanishing at $z=0$, which cannot happen. Thus $p\in N(C_{a}-1)\cap{\cal
O}=\\{0\\}$, which is the first assertion of this theorem. Clearly this
implies that $f=0$.

Assertion 2.: the proof of the second assertion is similar. Let us sketch it
underlining the differences. The left hand equality: suppose that $f$ is odd
and $f\perp N(C_{a}+1)$. Then
$f=C_{\omega_{a}}\iota=\iota(\varphi_{\omega_{a}})$ for some odd function
$\iota$. Then $f^{2}=\iota^{2}(\varphi_{\omega_{a}})\in N(C_{a}-1)$. Then, by
the first part of Theorem 5.4, we have that $f^{2}$ is constant, then $f$ is
constant, and the fact that $f=\iota(\varphi_{\omega_{a}})$ with $\iota$ odd
implies that $f=0$.

The right hand equality of the second assertion, if $f\in N(C_{0}+1)\cap
N(C_{a}+1)^{\perp}$, then $f$ is odd, $f(z)=\sum_{k\geq 0}\beta_{k}z^{2k+1}$
and $C_{\omega_{a}}^{*}f$ is even. Similarly as above,

$C_{\omega_{a}}^{*}f(z)=z\frac{|\omega_{a}|^{2}-1}{(1-\bar{\omega}_{a}z)^{2}}\sum_{k\geq
0}\beta_{k}\left(\frac{\omega_{a}-z}{1-\bar{\omega}_{a}z}\right)^{2k},$

and thus $B(z)=\frac{1}{(1-\bar{\omega}_{a}z)^{2}}\sum_{k\geq
0}\beta_{k}\left(\frac{\omega_{a}-z}{1-\bar{\omega}_{a}z}\right)^{2k}$ is odd.
Therefore, if $\beta(z):=\sum_{k\geq 0}\beta_{k}z^{2k}$, we have

$C_{\omega_{a}}(B)=\frac{(1-\bar{\omega}_{a}z)^{2}}{(1-\bar{\omega}_{a}z)^{2}}\beta\in
N(C_{a}+1),\ i.e.,\ (1-\bar{\omega}_{a}z)^{2}\beta\in N(C_{a}+1).$

If $\beta=qh$ is the inner/outer factorization, then $q$ and $h$ are even, and

$(1-\bar{\omega}_{a}z)^{2}\beta=q\left((1-\bar{\omega}_{a}z)^{2}h\right)$

is the inner/outer factorization of an element in $N(C_{a}+1)$. Then, again by
Remark 5.3, $q\in N(C_{a}+1)$. Then $q^{2}$ is even and lies in $N(C_{a}-1)$,
and therefore is constant, by the first part of Theorem 5.4. Thus $q$ is
constant in $N(C_{a}+1)$, which implies that $q=0$, and then $f=0$.

Assertion 3.: For the left hand equality: $f\in N(C_{0}-1)^{\perp}\cap
N(C_{a}-1)^{\perp}$ is odd, $f(z)=\sum_{n\geq 0}\beta_{n}z^{2n+1}$, and
similarly as above,

$C^{*}_{\omega_{a}}f(z)=\frac{z(\bar{\omega}_{a}-1)}{(1-\bar{\omega}_{a}z)^{2}}\sum_{n\geq
0}\beta_{n}\left(\frac{\omega_{a}-z}{1-\bar{\omega}_{a}z}\right)^{2n}\ \hbox{
is odd},$

so that $D(z)=\frac{1}{(1-\bar{\omega}_{a}z)^{2}}\sum_{n\geq
0}\beta_{n}\left(\frac{\omega_{a}-z}{1-\bar{\omega}_{a}z}\right)^{2n}$ is
even, and

$C_{\omega_{a}}D=\frac{(1-\bar{\omega}_{a}z)^{2}}{(1-|\omega_{|}^{2})^{2}}\sum_{n\geq
0}\beta_{n}z^{2n}\in N(C_{a}-1).$

Denote $\delta(z)=\sum_{n\geq 0}\beta_{n}z^{2n}$, so that
$(1-\bar{\omega}_{a}z)^{2}\delta\in N(C_{a}-1)$.

Note that $f(z)=z\delta(z)$ Then we have

$(1-\bar{\omega}_{a}z)^{2}\delta=(1+(\bar{\omega}_{a}z)^{2})\delta-2\bar{\omega}_{a}f$

is an orthogonal sum: the left hand term is even and the right hand term is
odd. One the other hand, rewriting this equality, we have

$(1+(\bar{\omega}_{a}z)^{2})\delta=(1-\bar{\omega}_{a}z)^{2}\delta+2\bar{\omega}_{a}f$

is also and orthogonal sum: the left hand term belongs to $N(C_{a}-1)$ and the
right hand term is orthogonal to $N(C_{a}-1)$. Then we have

$\|(1-\bar{\omega}_{a}z)^{2}\delta\|^{2}=\|(1+\bar{\omega}_{a}z)^{2})\delta\|^{2}+\|2\bar{\omega}_{a}f\|^{2}\
\hbox{ and }\
\|(1+(\bar{\omega}_{a}z)^{2})\delta\|^{2}=\|(1-\bar{\omega}_{a}z)^{2}\delta\|^{2}+\|2\bar{\omega}_{a}f\|^{2}.$

These imply that $f=0$.

The right hand equality: let $f\in N(C_{a}-1)^{\perp}$ be even, and suppose
first that $f(0)=0$. Then $f(z)=\sum_{n\geq 1}\alpha_{n}z^{2n}$. We proceed
similarly as in the third assertion, we sketch the proof. We know that

$C_{\omega_{a}}^{*}(f)(z)=\frac{z(\bar{\omega}_{a}^{2}-1)}{(1-\bar{\omega}_{a}z)^{2}}\sum_{n\geq
1}\alpha_{n}(\frac{\omega_{a}-z}{1-\bar{\omega}_{a}z})^{2n-1}$

is even, so that $E(z)=\frac{1}{(1-\bar{\omega}_{a}z)^{2}}\sum_{n\geq
1}\alpha_{n}(\frac{\omega_{a}-z}{1-\bar{\omega}_{a}z})^{2n-1}$ is odd. Then

$h(z):=C_{\omega_{a}}(E)(z)=\frac{(1-\bar{\omega}_{a}z)^{2}}{1-|\omega_{a}|^{2}}\sum_{n\geq
1}\alpha_{n}z^{2n-1}\in N(C_{a}+1).$

Note that $\sum_{n\geq 1}\alpha_{n}z^{2n-1}=\frac{f(z)}{z}$. Then we have on
one hand that

$(1-|\omega_{a}|^{2})h(z)=(1+(\bar{\omega}_{a}z)^{2})\sum_{n\geq
1}\alpha_{n}z^{2n-1}+2\bar{\omega}_{a}f(z)$

is an orthogonal sum, the left hand summand is odd and the right hand summand
is even. Thus

$\|(1-|\omega_{a}|^{2})h\|^{2}=\|(1+(\bar{\omega}_{a}z)^{2})\sum_{n\geq
1}\alpha_{n}z^{2n-1}\|^{2}+\|2\bar{\omega}_{a}f\|^{2}.$

On the other hand. the above also means that

$(1+(\bar{\omega}_{a}z)^{2})\sum_{n\geq
1}\alpha_{n}z^{2n-1}=(1-|\omega_{a}|^{2})h(z)+2\bar{\omega}_{a}f(z)$

is also an orthogonal sum, the left hanf summand belongs to $N(C_{a}+1)$ and
the right hand summand belongs to $N(C_{a}+1)^{\perp}$. Then

$\|(1+(\bar{\omega}_{a}z)^{2})\sum_{n\geq
1}\alpha_{n}z^{2n-1}\|^{2}=\|(1-|\omega_{a}|^{2})h\|^{2}+\|2\bar{\omega}_{a}f\|^{2}.$

These two norm identities imply that $f=0$. Suppose now that $f(0)\neq 0$, by
considering a multiple of $f$, we may assume $f(0)=1$, i.e.,
$f(z)=1+\sum_{n\geq 1}\alpha_{n}z^{2b}$. Then

$g(z):=C^{*}_{\omega_{a}}f(z)=\frac{1}{1-\bar{\omega}_{a}z}+(\bar{\omega}_{a}-1)\frac{z}{1-\bar{\omega}_{a}z}\sum_{n\geq
1}\alpha_{n}\left(\frac{\omega_{a}-z}{1-\bar{\omega}_{a}z}\right)^{2n-1},$

which is also even. Then $g^{\prime}(z)$ is odd and $g^{\prime}(0)=0$. Note
that

$g^{\prime}(z)=\frac{\bar{\omega}_{a}}{(1-\bar{\omega}_{a}z)^{2}}+\frac{\bar{\omega}_{a}-1}{(1-\bar{\omega}_{a}z)^{2}}\sum_{n\geq
1}\alpha_{n}\left(\frac{\omega_{a}-z}{1-\bar{\omega}_{a}z}\right)^{2n-1}+$
$+(\bar{\omega}_{a}-1)(|\omega_{a}|^{2}-1)\frac{z}{(1-\bar{\omega}_{a}z)^{3}}\sum_{n\geq
1}\alpha_{n}\left(\frac{\omega_{a}-z}{1-\bar{\omega}_{a}z}\right)^{2n-2},$

so that

$0=g^{\prime}(0)=\bar{\omega}_{a}+(\bar{\omega}_{a}-1)\sum_{n\geq
1}\alpha_{n}\omega_{a}^{2n-1}.$

Note that $f(\omega_{a})=1+\sum_{n\geq
1}\alpha_{n}\omega_{a}^{2n}=1+\omega_{a}\sum_{n\geq
1}\alpha_{n}\omega_{a}^{2n-1}$, i.e.,

$0=\bar{\omega}_{a}+(\bar{\omega}_{a}-1)\left(\frac{f(\omega_{a})-1}{\omega_{a}}\right),$

or
$f(\omega_{a})=\displaystyle{\frac{|\omega_{a}|^{2}}{1-\bar{\omega}_{a}}+1}$.
Since $f$ is even, $f(\omega_{a})=f(-\omega_{a})$, i.e.,
$\frac{1}{1-\bar{\omega}_{a}}=\frac{1}{1+\bar{\omega}_{a}}$, or $\omega_{a}=0$
(which cannot happen because $a\neq 0$). It follows that $f\equiv 0$. ∎

######  Question 5.8.

A natural question is wether these properties above hold for arbitrary $a\neq
b\in\mathbb{D}$.

######  Remark 5.9.

A straightforward computation shows that if $a\in\mathbb{D}$, the unique
$b\in\mathbb{D}$ such that the fixed point $\omega_{b}$ of $\varphi_{b}$ (in
$\mathbb{D}$) is $a$ is given by $b=\frac{2a}{1+|a|^{2}}$. Let us denote this
element by $\Omega_{a}$. One may iterate this computation: denote by
$\Omega^{2}_{a}:=\Omega_{\Omega_{a}}$, and in general
$\Omega_{a}^{n+1}:=\Omega_{\Omega_{a}^{n}}$. Then it is easy to see that

$\Omega_{a}^{n}=a_{2^{n}-1},$

where $a_{k}\in\mathbb{D}$ are the numbers obtained in Lemma 5.5. Note that
all these iterations $\Omega_{a}^{n}$ are multiples of $a$, with increasing
moduli, and $\Omega_{a}^{n}\to\frac{a}{|a|}$ as $n\to\infty$.

Moreover, it is easy to see that the sequence $a_{n}$ is an interpolating
sequence: it consists of multiples of $\frac{1-r^{n+1}}{1+r^{n+1}}$ by the
number $\frac{a}{|a|}$ of modulus one, with $r<1$. Therefore $\Omega_{a}^{n}$
is an interpolating sequence.

##  6 Geodesics between Eigenspaces of $C_{a}$

Recall from the introduction the condition for the existence of a geodesic of
the Grassmann manifold of $H^{2}$ that joins two given subspaces ${\cal S}$
and ${\cal T}$, namely, that

$\dim({\cal S}\cap{\cal T}^{\perp})=\dim({\cal T}\cap{\cal S}^{\perp}).$

This condition clearly holds for ${\cal E}=N(C_{0}-1)$ and ${\cal
O}=N(C_{0}+1)={\cal E}^{\perp}$: both intersections are, respectively, ${\cal
E}\cap{\cal O}^{\perp}={\cal E}$ and ${\cal O}\cap{\cal E}^{\perp}={\cal O}$,
and have the same (infinite) dimension. Our first observation is that this no
longer holds for $N(C_{a}-1)$ and $N(C_{a}+1)$ when $a\neq 0$:

######  Proposition 6.1.

If $0\neq a\in\mathbb{D}$, then there does not exist a geodesic of the
Grassmann manifold of $H^{2}$ joining $N(C_{a}-1)$ and $N(C_{a}+1)$.

###### Proof.

The proof follows by direct computation. First, we claim that

$N(C_{a}+1)\cap N(C_{a}-1)^{\perp}=\\{0\\}.$ (21)

Note that $f\in N(C_{a}-1)^{\perp}$ if and only if $\langle f,g\rangle=0$ for
all $g\in N(C_{a}-1)=C_{\omega_{a}}({\cal E})$, i.e.,

$0=\langle C_{\omega_{a}}^{*}f,g\rangle,$

for all $g\in{\cal E}$. This is equivalent to $C_{\omega_{a}}^{*}f\in{\cal
O}$, or also that $f\in C_{\omega_{a}}^{*}({\cal O})$.

Using the operator $C_{\omega_{a}}$, our claim (21) is equivalent to

$\\{0\\}=C_{\omega_{a}}(N(C_{a}+1))\cap C_{\omega_{a}}C_{\omega_{a}}^{*}({\cal
O})={\cal O}\cap C_{\omega_{a}}C_{\omega_{a}}^{*}({\cal O}),$

where the last equality follows from the fact
$C_{\omega_{a}}(N(C_{a}+1))={\cal O}$ observed before. Let $f\in{\cal O}$.
Then (since $f(0)=0$)

$g(z)=C_{\omega_{a}}C_{\omega_{a}}^{*}f(z)=\frac{1-\bar{\omega}_{a}z}{1-|\omega_{a}|^{2}}\left(f(z)-\omega_{a}\frac{f(z)}{z}\right)$
$=\frac{1}{1-|\omega_{a}|^{2}}\left(f(z)(1+|\omega_{a}|^{2})-\left(\omega_{a}\frac{f(z)}{z}+\bar{\omega}_{a}zf(z)\right)\right).$

Then, since $g$ and the first summand are odd, and the second summand is even,
the second summand is zero, which implies that $f\equiv 0$.

On the other hand, a similar computation shows that

$\dim\left(N(C_{a}-1)\cap N(C_{a}+1)^{\perp}\right)=1,$

which would conclude the proof. Indeed, by a similar argument as above, it
suffices to show that

$\dim({\cal E}\cap C_{\omega_{a}}C_{\omega_{a}}^{*}({\cal E}))=1.$

Let $g,f$ be even functions such that

$g(z)=C_{\omega_{a}}C_{\omega_{a}}^{*}f(z)=\frac{1-\bar{\omega}_{a}z}{1-|\omega_{a}|^{2}}\left(f(z)-\omega_{a}\frac{f(z)-f(0)}{z}\right)$
$=\frac{1}{1-|\omega_{a}|^{2}}\left(\left(f(z)+|\omega_{a}|^{2}(f(z)-f(0))\right)-\left(\bar{\omega}_{a}f(z)z+\omega_{a}\frac{f(z)-f(0)}{z}\right)\right).$

It follows that

$\bar{\omega}_{a}f(z)z+\omega_{a}\frac{f(z)-f(0)}{z}\equiv 0,$

i.e., $f(z)=\displaystyle{\frac{c}{\omega_{a}+\bar{\omega}_{a}z^{2}}}$. This
implies that

${\cal E}\cap C_{\omega_{a}}C_{\omega_{a}}^{*}({\cal
E})=\langle\frac{1}{\omega_{a}+\bar{\omega}_{a}z^{2}}\rangle.$

∎

Note though that the orthogonal projections onto $N(C_{a}-1)$ and $N(C_{a}+1)$
are unitarilly equivalent: both subspaces are infinite dimensional and
infinite co-dimensional.

Also on the negative side, the subspaces ${\cal O}$ and $N(C_{a}-1)$, for
$a\neq 0$, cannot be joined by a geodesic:

######  Corollary 6.2.

There exist no geodesics of the Grassmann manifold of $H^{2}$ joining
$N(C_{0}+1)$ and $N(C_{a}+1)$, for $a\neq 0$.

###### Proof.

Note that, by Theorem 5.4, part 1, for $b=0$:

$N(C_{0}+1)^{\perp}\cap N(C_{a}-1)=N(C_{0}-1)\cap N(C_{a}-1)=\mathbb{C}1;$

whereas by Theorem 5.7, Assertion 3, left hand identity, we have that

$N(C_{0}+1)\cap N(C_{a}-1)^{\perp}=N(C_{0}-1)^{\perp}\cap
N(C_{a}-1)^{\perp}=\\{0\\}.$

∎

On the affirmative side, a direct consequence of the results in the previous
section is the existence of unique normalized geodesics of the Grassmann
manifold joining ${\cal E}=N(C_{0}-1)$ with $N(C_{a}-1)$, ${\cal
O}=N(C_{0}+1)$ with $N(C_{a}+1)$, and ${\cal E}$ with $N(C_{a}+1)$:

######  Corollary 6.3.

Let $a\in\mathbb{D}$, $a\neq 0$.

  1. 1.

There exists a unique (geodesic) curve
$\delta^{-}_{0,a}(t)=e^{tZ^{-}_{0,a}}{\cal E}$ of the Grassmann manifold of
$H^{2}$, with $(Z^{-}_{0,a})^{*}=-Z^{-}_{0,a}$, $Z^{-}_{0,a}{\cal
E}\subset{\cal O}$ and $\|Z^{-}_{0,a}\|\leq\pi/2$, such that

$e^{Z^{-}_{0,a}}{\cal E}=N(C_{a}-1).$

  2. 2.

There exists a unique (geodesic) curve
$\delta^{+}_{0,a}(t)=e^{tZ^{+}_{0,a}}{\cal E}$ of the Grassmann manifold of
$H^{2}$, with $(Z^{+}_{0,a})^{*}=-Z^{+}_{0,a}$, $Z^{+}_{0,a}{\cal
E}\subset{\cal O}$ and $\|Z^{+}_{0,a}\|\leq\pi/2$, such that

$e^{Z^{+}_{0,a}}{\cal O}=N(C_{a}+1).$

  3. 3.

There exists a unique (geodesic) curve
$\delta^{+,-}_{0,a}(t)=e^{tZ^{+,-}_{0,a}}{\cal E}$ of the Grassmann manifold
of $H^{2}$, with $(Z^{+,-}_{0,a})^{*}=-Z^{+,-}_{0,a}$, $Z^{+,-}_{0,a}{\cal
O}\subset{\cal E}$ and $\|Z^{+,-}_{0,a}\|\leq\pi/2$, such that

$e^{Z^{+,-}_{0,a}}{\cal O}=N(C_{a}-1).$

###### Proof.

  1. 1.

Follows from assertion 1 in Theorem 5.7.

  2. 2.

Follows from assertion 2 in Theorem 5.7.

  3. 3.

$N(C_{0}-1)\cap N(C_{a}+1)^{\perp}=\\{0\\},$

is the right hand side of assertion 2 in Theorem 5.7.

$N(C_{0}-1)^{\perp}\cap N(C_{a}+1)=N(C_{0}+1)\cap N(C_{a}+1)=\\{0\\},$

is part 2. of Theorem 5.4 for $b=0$.

∎

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