††2020 Mathematics Subject Classification: Primary 42C15, 47A15, 94A20.
Keywords: Frame-tuples, similarity, invariant subspaces, reducing subspaces,
shift operators, synthesis operator. This research was supported by grants:
UBACyT 20020170100430BA, PICT 2018-3399 (ANPCyT), PICT 2019-03968 (ANPCyT) and
CONICET PIP 11220110101018. Diana Carbajal was supported by the European
Union’s programme Horizon Europe, HORIZON-MSCA-2021-PF-01, Grant agreement
101064206.

# Frames by orbits of two operators that commute

A. Aguilera  Departamento de Matemática, Universidad de Buenos Aires,
Instituto de Matemática ”Luis Santaló” (IMAS-CONICET-UBA), Buenos Aires,
Argentina<EMAIL_ADDRESS>,  C. Cabrelli  Departamento de Matemática,
Universidad de Buenos Aires, Instituto de Matemática ”Luis Santaló” (IMAS-
CONICET-UBA), Buenos Aires, Argentina<EMAIL_ADDRESS>,  D.
Carbajal  Faculty of Mathematics, University of Vienna, Vienna, Austria
<EMAIL_ADDRESS>and  V. Paternostro  Departamento de
Matemática, Universidad de Buenos Aires, Instituto de Matemática ”Luis
Santaló” (IMAS-CONICET-UBA), Buenos Aires, Argentina<EMAIL_ADDRESS>
###### Abstract.

Frames formed by orbits of vectors through the iteration of a bounded operator
have recently attracted considerable attention, in particular due to its
applications to dynamical sampling. In this article, we consider two commuting
bounded operators acting on some separable Hilbert space $\mathcal{H}$. We
completely characterize operators $T$ and $L$ with $TL=LT$ and sets
$\Phi\subset\mathcal{H}$ such that the collection
$\\{T^{k}L^{j}\phi:k\in\mathbb{Z},j\in J,\phi\in\Phi\\}$ forms a frame of
$\mathcal{H}$. This is done in terms of model subspaces of the space of square
integrable functions defined on the torus and having values in some Hardy
space with multiplicity. The operators acting on these models are the
bilateral shift and the compression of the unilateral shift (acting
pointwisely). This context includes the case when the Hilbert space
$\mathcal{H}$ is a subspace of $L^{2}(\mathbb{R})$, invariant under
translations along the integers, where the operator $T$ is the translation by
one and $L$ is a shift-preserving operator.

##  1\. Introduction

Let $\mathcal{H}$ be a separable Hilbert space, and $\\{g_{i}\\}_{i\in I}$ a
countable set in $\mathcal{H}$. In sampling theory, every function $f$
belonging to $\mathcal{H}$ needs to be recovered from its samples $\\{\langle
f,g_{i}\rangle\\}_{i\in I}$. If we impose a stability condition on the
reconstruction, this problem is equivalent to the fact that the set
$\\{g_{i}\\}_{i\in I}$ is a frame of $\mathcal{H}$.

In dynamical sampling, it is assumed that the signal evolves in time under an
evolution operator $L$ defined on $\mathcal{H}$ and that we are able to sample
the functions $L^{t}f$ for $t=0,1,2,...,$ where $L^{t}f$ represents the
_evolved signal_ at time $t$. Thus, if the samples $\\{\langle
f,g_{i}\rangle\\}_{i\in I}$ are insufficient to reconstruct every function $f$
in the space, we hope to compensate the sparse data by sampling the evolved
signal as well, that is, considering space-time samples.

In other words, the problem consists in trying to recover any function $f$
from the samples $\\{\langle L^{t}f,g_{i}\rangle\\}_{i\in I,t=0,1,2,...}$ in a
stable way, which is equivalent to the fact that
$\\{(L^{*})^{t}g_{i}\\}_{i,t}$ is a frame of $\mathcal{H}$. This motivates the
study of the structure of orbits of operators in Hilbert spaces, i.e. to give
conditions on a Hilbert space $\mathcal{H}$, a bounded operator $L$ defined on
$\mathcal{H}$ and a set of functions $\Phi\subset\mathcal{H}$ in order that
the system $\\{L^{j}\varphi:j\in J,\varphi\in\Phi\\}$ forms a frame of
$\mathcal{H},$ for some index set $J.$

The dynamical sampling problem was first presented in [6, 7], taking
inspiration from the research of Lu, Vetterli, and their colleagues in [25,
27, 32]. In [27], the authors analyzed a linear diffusion (heat) equation
where the initial state evolves over time. They proposed a trade-off between
spatial and temporal sampling, with the goal of reconstructing the initial
state using only a few spatial sensors (samples) and compensating with
temporal samples. This can be considered as an inverse problem in differential
equations and has potential applications in various fields, such as weather
forecasting, ecology, and medical imaging between others.

Recently, there has been a lot of interest in this problem. See [4, 5, 6, 7,
10, 11, 15, 16, 19, 30] for foundations and theoretical developments, and [8,
9, 12, 18, 26, 28, 33] for applications.

In [3], dynamical sampling was considered for shift-invariant spaces: Let
$V\subset L^{2}(\mathbb{R})$ be a closed subspace, $T$ the translation
operator by one on $V$, i.e. $Tf(x)=f(x-1)$, and $L:V\rightarrow V$ a bounded
operator. We assume that $TV=V$ (that is, $V$ is shift invariant) and that
$TL=LT$ (that is, $L$ is shift preserving). The main question here is to
characterize $L$ and $\Phi\subset V$ such that $\\{L^{j}\varphi:j\in
J,\varphi\in\Phi\\}$ is a set of frame generators by integer translations of
$V$, which means that $\\{T^{k}(L^{j}\varphi):k\in\mathbb{Z},j\in
J,\varphi\in\Phi\\}$ forms a frame of $V.$

For the case where $L$ is normal and $V$ is finitely generated, necessary and
sufficient conditions were given in [3]. The conditions were obtained there
defining a special diagonalization for $L$, carefully developed in [1], and
then applying a result for the finite dimensional dynamical sampling problem.
The approach works for $L^{2}(\mathbb{R}^{d})$ and even when replacing
$\mathbb{R}^{d}$ with an LCA group.

In this paper we study a generalization of this case. We consider the set of
orbits $\mathcal{F}=\\{T^{k}L^{j}w_{i}:k\in\mathbb{Z},j\in J,i\in I\\}$ where
$\mathcal{H}$ is an abstract separable Hilbert space, $T$ and $L$ are
commuting bounded operators acting on $\mathcal{H}$, and $\\{w_{i}\\}_{i\in
I}$ is an at most countable set of vectors in $\mathcal{H}.$ We do not assume
that $T$ is unitary or that $L$ is normal. We do assume, however, that $T$ is
invertible. Then, we ask whether the set $\mathcal{F}$ forms a frame of
$\mathcal{H}$.

With this purpose, we define an equivalence relation between tuples of the
form $(\mathcal{H},T,L,\\{w_{i}\\}_{i\in I})$. We say that two tuples are
similar if the operators from one tuple are similar to the operators in the
other via a bounded isomorphism between the Hilbert spaces involved. The
isomorphism also maps the set of vectors of one into the set of vector of the
other. See Section 3.1 for a precise definition.

Our main result asserts that every tuple which generates a frame is similar to
a tuple of a particular class, the class of basic tuples. See Definition 3.4.
These basic tuples are simpler to study because of their structure and they
provide a rich platform to prove properties of these systems. In this way,
properties preserved by similarity can be obtained on a basic tuple and then
the results can be translated to the abstract setting of the general Hilbert
space.

The Hilbert space component of a basic tuple is a closed subspace of
$L^{2}({\mathbb{T}},H^{2}_{\mathcal{K}})$ where ${\mathbb{T}}$ denote the unit
circle, $\mathcal{K}$ a separable Hilbert space, and $H^{2}_{\mathcal{K}}$ the
Hardy space of $\mathcal{K}$-valued functions defined on the unit circle. See
Section 2.1.

The main idea comes from the observation that the kernel of the synthesis
operator of the system $\mathcal{F}$ can be seen as a closed subspace
$\mathcal{M}$ of $L^{2}({\mathbb{T}},H^{2}_{\mathcal{K}})$ that turns out to
be reducing for $U$ (the bilateral shift) and also invariant for
$\widehat{S}$, an operator that acts pointwisely as the unilateral shift in
$H^{2}_{\mathcal{K}}$. We then provide a functional representation for all the
systems $\mathcal{F}$ that form a frame, in terms of the model space
$\mathcal{N}_{\mathcal{M}}=L^{2}({\mathbb{T}},H^{2}_{\mathcal{K}})\ominus\mathcal{M}$,
with the bilateral shift and the compression of $\widehat{S}$ on
$\mathcal{N}_{\mathcal{M}}$ as the commuting operators.

We want to remark that our approach was used previously in [19] for the case
of just one operator $L$ acting on $\mathcal{H}$ on one function, and in [17]
for several functions.

Next we consider the following question: assume that $\mathcal{H}$, $T$ and
$L$ are given and satisfy the hypothesis of boundedness and commutativity.
Which are all vectors $w\in\mathcal{H}$ such that the iterations form a frame?
We were able to answer this question completely and the characterization is
similar to the one obtained before in [19] for the iteration of only one
operator. However, in our case, the proof is much more involved and strongly
uses the Helson’s characterization of reducing subspaces for the bilateral
shift. We finally show that this characterization does not work when we
consider iterations of multiple generators.

All the results are proved for the two cases: first when we consider forward
iterations of the operator $L$, and second, when we allow negative powers of
$L$, in which case, $L$ is invertible. The difference between these cases is
that the corresponding basic tuples live in different vector-valued
$L^{2}$-spaces.

The paper is organized as follows. We first introduce in Section 2 the
notation and the setting that we need in the paper. In Section 3 we study the
properties of basic tuples and prove the characterization of tuples that give
frames, splitting our analysis into two cases: unilateral tuples, that is,
when we consider forward iterations of the operator $L$, and bilateral tuples
when we allow integer iterations of $L$. Finally, in Section 4 we consider the
problem of characterization of all vectors that give frames by iterations.

##  2\. preliminaries

Let us begin by introducing some notation and definitions. All the Hilbert
spaces considered in this paper are separable. We will use the letters
$\mathcal{K}$ and $\mathcal{H}$ to denote complex Hilbert spaces. As usual,
$\mathcal{B}(\mathcal{H},\mathcal{K})$ will denote the set of linear bounded
operators from $\mathcal{H}$ into $\mathcal{K}$ and
$\mathcal{B}(\mathcal{H}):=\mathcal{B}(\mathcal{H},\mathcal{H})$. We will
write $\mathbb{N}_{0}:=\mathbb{N}\cup\\{0\\}$ and ${\mathbb{T}}$ for the
complex unit circle. For a closed subspace $\mathcal{N}$ of $\mathcal{H}$, we
will write $\mathcal{N}^{\perp}$ to denote the orthogonal complement of
$\mathcal{N}$ in $\mathcal{H}$ and $P_{\mathcal{N}}$ to denote the orthogonal
projection of $\mathcal{H}$ onto $\mathcal{N}$.

Let $\mathcal{M}$ be a closed subspace of a Hilbert space $\mathcal{H}$ and
$A\in\mathcal{B}(\mathcal{H})$. The subspace $\mathcal{M}$ is said to be
invariant for $A$ (or $A$-invariant) if $A(\mathcal{M})\subseteq\mathcal{M}$.
Furthermore, $\mathcal{M}$ is called reducing for $A$ (or $A$-reducing) if
$\mathcal{M}$ and $\mathcal{M}^{\perp}$ are $A$-invariant, which is equivalent
to say that $\mathcal{M}$ is invariant for $A$ and $A^{*}$. It is easy to
check that $\mathcal{M}$ is $A$-invariant if and only if
$P_{\mathcal{M}}AP_{\mathcal{M}}=AP_{\mathcal{M}}$ and $\mathcal{M}$ is
$A$-reducing if and only if $P_{\mathcal{M}}A=AP_{\mathcal{M}}$.

Given $T\in\mathcal{B}(\mathcal{H})$ and a closed subspace $\mathcal{M}$ of
$\mathcal{H}$, the operator defined by $P_{\mathcal{M}}T|_{\mathcal{M}}$ is
called the compression of $T$ to the subspace $\mathcal{M}$.

Particular attention has been put over the years in the study of invariant and
reducing subspaces under the action of sequential bilateral and unilateral
shift operators, that is

(1) $\displaystyle U:\\{\dots,a_{-1},(a_{0}),a_{1},\dots\\}$
$\displaystyle\mapsto\\{\dots,a_{-2},(a_{-1}),a_{0},\dots\\},\quad$
$\displaystyle\\{a_{i}\\}_{i\in\mathbb{Z}}\subseteq\ell^{2}(\mathbb{Z}),$ (2)
$\displaystyle S:\\{a_{0},a_{1},a_{2},\dots\\}$
$\displaystyle\mapsto\\{0,a_{0},a_{1},a_{2}\dots\\},$
$\displaystyle\\{a_{i}\\}_{i\in\mathbb{N}}\subseteq\ell^{2}(\mathbb{N}).$

Via the Fourier transform, these can be naturally represented as operators
acting over the functional spaces $L^{2}({\mathbb{T}})$ and the Hardy space
$H^{2}$, i.e., the subspace of $L^{2}({\mathbb{T}})$ consisting of all $f\in
L^{2}({\mathbb{T}})$ whose Fourier coefficients vanish for $n<0$, i.e.,

(3) $H^{2}:=\left\\{f\in
L^{2}({\mathbb{T}}):\int_{{\mathbb{T}}}f(z)z^{-n}\,dz=0\text{ for
}n<0\right\\}.$

Similarly, shift operators can be defined with multiplicity higher than $1$.
This means that instead of shifting a sequence of complex numbers, the
operators shift a sequence with values in a Hilbert space $\mathcal{K}$, where
$\dim(\mathcal{K})$ is the multiplicity. The representation of these operators
over functional spaces requires the introduction of vector-valued functions.

###  2.1. Vector-valued functions and Hardy spaces with multiplicity

Let $(\Omega,\mu)$ be a $\sigma$-finite measure space such that
$L^{2}(\Omega):=L^{2}(\Omega,\mu)$ is a separable Hilbert space. For our
purposes along this paper, we will be interested in the cases when
$\Omega={\mathbb{T}}$ or ${\mathbb{T}}^{2}$ and $\mu$ is the normalized
Lebesgue measure ($\mu({\mathbb{T}})=1$ or $\mu({\mathbb{T}}^{2})=1$,
respectively).

Let $\mathcal{K}$ be a separable Hilbert space. A vector-valued (or
$\mathcal{K}$-valued) function $f:\Omega\rightarrow\mathcal{K}$ is said to be
measurable if for each $x\in\mathcal{K}$, the complex-valued function
$\omega\mapsto\langle f(\omega),x\rangle_{\mathcal{K}}$ is measurable on
$\Omega$. We denote by $L^{2}(\Omega,\mathcal{K})$ the space of all measurable
$\mathcal{K}$-valued functions $f$ such that
$\int_{\Omega}\|f(\omega)\|_{\mathcal{K}}^{2}\,d\omega<\infty$, which is a
Hilbert space equipped with the inner product

(4) $\langle f,g\rangle=\int_{\Omega}\langle
f(\omega),g(\omega)\rangle_{\mathcal{K}}\,d\omega,\qquad f,g\in
L^{2}(\Omega,\mathcal{K}).$

Notice that the space $\mathcal{K}$ is naturally embedded in
$L^{2}(\Omega,\mathcal{K})$ when identified with the subset of constant
vector-valued functions of $L^{2}(\Omega,\mathcal{K})$. That is, given
$x\in\mathcal{K}$ define $\tilde{x}:\Omega\to\mathcal{K}$ as the function in
$L^{2}(\Omega,\mathcal{K})$ which is constantly $x$. For the sake of
simplicity, we will write $x$ instead of $\tilde{x}$, when it is clear from
the context that we mean the constant function. Given an orthonormal basis
$\mathcal{B}=\\{\varepsilon_{i}\\}_{i\in I}$ of $\mathcal{K}$ and a vector-
valued function $f\in L^{2}({\mathbb{T}},\mathcal{K})$, we can write

(5) $f(\lambda)=\sum_{i\in I}\langle
f(\lambda),\varepsilon_{i}\rangle_{\mathcal{K}}\,\varepsilon_{i},\quad\text{a.e.
}\lambda\in{\mathbb{T}}.$

We call $f_{i}:=\langle f(\cdot),\varepsilon_{i}\rangle_{\mathcal{K}}$ the ith
coordinate function of $f$ with respect to $\mathcal{B}$. It is easily seen
that $f_{i}$ belongs to $L^{2}({\mathbb{T}})$ for every $i\in I$.

The Hardy space with multiplicity $\alpha=\dim(\mathcal{K})$ is defined as the
closed subspace of $L^{2}({\mathbb{T}},\mathcal{K})$ consisting of all
functions $f\in L^{2}({\mathbb{T}},\mathcal{K})$ whose coordinate functions
$f_{i}$ belong to the Hardy space $H^{2}$. Note that this is equivalent to say
that $H^{2}_{\mathcal{K}}$ is the set of all functions $f\in
L^{2}({\mathbb{T}},\mathcal{K})$ such that $\langle
f(\cdot),u\rangle_{\mathcal{K}}\in H^{2}$ for every $u\in\mathcal{K}$. We will
denote it by $H^{2}_{\mathcal{K}}:=H^{2}({\mathbb{T}},\mathcal{K})$.

We are ready to formally define the shift operators with multiplicity acting
over these spaces.

######  Definition 2.1.

The operator $U:L^{2}({\mathbb{T}},\mathcal{K})\to
L^{2}({\mathbb{T}},\mathcal{K})$ defined by

(6) $(Uf)(\lambda)=\lambda f(\lambda),\quad\text{a.e.
}\lambda\in{\mathbb{T}},\,f\in L^{2}({\mathbb{T}},\mathcal{K}),$

is called the bilateral shift on $L^{2}({\mathbb{T}},\mathcal{K})$ with
multiplicity $\alpha=\dim(\mathcal{K})$.

######  Definition 2.2.

The operator $S:H^{2}_{\mathcal{K}}\rightarrow H^{2}_{\mathcal{K}}$ given by
the restriction of $U$ to $H^{2}_{\mathcal{K}}$ is called the unilateral shift
on $H^{2}_{\mathcal{K}}$ with multiplicity $\alpha=\dim(\mathcal{K})$.

The bilateral shift $U$ is unitary and its adjoint operator is given by
$(U^{*}f)(\lambda)=\overline{\lambda}f(\lambda)$, for a.e.
$\lambda\in{\mathbb{T}}$ and $f\in L^{2}({\mathbb{T}},\mathcal{K})$. Moreover,
since $H^{2}_{\mathcal{K}}$ is invariant under $U$, then the operator $S$ is
an isometry.

There is a simple way to construct an orthonormal basis of
$L^{2}({\mathbb{T}},\mathcal{K})$ or $H^{2}_{\mathcal{K}}$ through iterations
of the shift operators over an orthonormal basis of $\mathcal{K}$ that we will
sketch out now. Let $\mathcal{B}=\\{\varepsilon_{i}\\}_{i\in I}$ be an
orthonormal basis of $\mathcal{K}$. It is easy to see that
$\\{U^{k}{\varepsilon}_{i}\,:\,k\in\mathbb{Z},\,i\in I\\}$ is an orthonormal
basis of $L^{2}({\mathbb{T}},\mathcal{K})$. Indeed, the system is orthogonal
since for every $k,k^{\prime}\in\mathbb{Z}$ and $i,i^{\prime}\in I$

(7) $\langle
U^{k}{\varepsilon}_{i},U^{k^{\prime}}{\varepsilon}_{i^{\prime}}\rangle=\int_{\mathbb{T}}\langle\lambda^{k}\varepsilon_{i},\lambda^{k^{\prime}}\varepsilon_{i^{\prime}}\rangle_{\mathcal{K}}\,d\lambda=\langle\varepsilon_{i},\varepsilon_{i^{\prime}}\rangle_{\mathcal{K}}\int_{{\mathbb{T}}}\lambda^{k-k^{\prime}}\,d\lambda=\delta_{i,i^{\prime}}\delta_{k,k^{\prime}}.$

For the completeness, note that if $f\in L^{2}({\mathbb{T}},\mathcal{K})$ and
$f$ is orthogonal to each element in the basis $\mathcal{B}$, we have

(8) $0=\langle f,U^{k}\varepsilon_{i}\rangle=\int_{\mathbb{T}}\langle
f(\lambda),\varepsilon_{i}\rangle_{\mathcal{K}}\lambda^{-k}\,d\lambda,\quad\forall
k\in\mathbb{Z},\,\forall i\in I.$

Thus, the coordinate functions of $f$ respect to the basis $\mathcal{B}$,
$f_{i}(\lambda)=\langle f(\lambda),\varepsilon_{i}\rangle_{\mathcal{K}}$, are
zero for a.e. $\lambda\in{\mathbb{T}}$, implying that $f=0$.

Analogously, we can see that the system
$\\{S^{j}{\varepsilon}_{i}\,:\,j\in\mathbb{N}_{0},\,i\in I\\}$ is an
orthonormal basis of $H^{2}_{\mathcal{K}}$.

##  3\. Systems of iterations

Let $\mathcal{H}$ be a separable Hilbert space,
$T,L\in\mathcal{B}(\mathcal{H})$ such that $T$ is invertible and $TL=LT$, and
$\\{w_{i}\\}_{i\in I}\subset\mathcal{H}$ an at most countable collection of
vectors. In this section, we provide a characterization of all systems of
iterations $\left\\{T^{k}L^{j}w_{i}:k\in\mathbb{Z},j\in J,i\in I\right\\}$
that form a frame, Parseval frame or Riesz basis of $\mathcal{H}$. We will see
that these systems are associated to a particular class of systems of
iterations called basic, which have much more structure and consist of two
shift operators acting on a vector-valued $L^{2}$-subspace.

For this purpose, let us consider the family of all tuples
$(\mathcal{H},T,L,\\{w_{i}\\}_{i\in I})$ with $\mathcal{H},T,L$ and
$\\{w_{i}\\}_{i\in I}$ as above. We will say that two tuples
$(\mathcal{H}_{1},T_{1},L_{1},\\{v_{i}\\}_{i\in I})$ and
$(\mathcal{H}_{2},T_{2},L_{2},\\{w_{i}\\}_{i\in I})$ are similar if there
exists a bounded isomorphism $C:\mathcal{H}_{1}\rightarrow\mathcal{H}_{2}$
that satisfies $C(v_{i})=w_{i}$ for every $i\in I$, and has the following
intertwining properties:

$T_{2}C=CT_{1}\quad\textrm{and}\quad L_{2}C=CL_{1}.$

It can be easily seen that the similarity is an equivalence relation.
Moreover, two similar tuples are said to be unitarily equivalent if the
associate isomorphism $C$ is unitary.

Furthermore, we will say that a tuple $(\mathcal{H},T,L,\\{w_{i}\\}_{i\in I})$
is a frame-tuple (Parseval-tuple or Riesz-tuple) if the collection

(9) $\left\\{T^{k}L^{j}w_{i}:k\in\mathbb{Z},j\in J,i\in I\right\\}$

forms a frame (a Parseval frame or a Riesz basis, respectively) of
$\mathcal{H}.$ In this case, the vectors $\\{w_{i}\\}_{i\in I}$ will be called
generators. The index set $J$ will be $\mathbb{N}_{0}$ or $\mathbb{Z}$.

When the iterations of $L$ are taken over $\mathbb{N}_{0}$, we call the tuple
$(\mathcal{H},T,L,\\{w_{i}\\}_{i\in I})$ a unilateral tuple. Observe that when
the iterations are taken over $\mathbb{Z}$, the operator $L$ must be
invertible. In that case we call $(\mathcal{H},T,L,\\{w_{i}\\}_{i\in I})$ a
bilateral tuple.

The following two lemmas hold regardless of the type of tuple (unilateral or
bilateral).

######  Lemma 3.1.

The similarity relation between tuples preserves the frame property.

###### Proof.

Assume that the tuples $(\mathcal{H}_{1},T_{1},L_{1},\\{v_{i}\\}_{i\in
I})\text{ and }(\mathcal{H}_{2},T_{2},L_{2},\\{w_{i}\\}_{i\in I})$ are similar
via the isomorphism $C\in\mathcal{B}(\mathcal{H}_{1},\mathcal{H}_{2}).$ Let
$g\in\mathcal{H}_{1}$. It is sufficient to observe that

$\displaystyle\sum_{k,j,i}|\langle g,T_{2}^{k}L_{2}^{j}w_{i}\rangle|^{2}$
$\displaystyle=\sum_{k,j,i}|\langle
g,(CT_{1}^{k}C^{-1})(CL_{1}^{j}C^{-1})Cv_{i}\rangle|^{2}$
$\displaystyle=\sum_{k,j,i}|\langle
C^{*}g,T_{1}^{k}L_{1}^{j}v_{i}\rangle|^{2}$

where the sum is indexed over $k\in\mathbb{Z},\,j\in J$, where
$J=\mathbb{N}_{0}$ or $\mathbb{Z}$ depending on the case (unilateral or
bilateral), and $i\in I$. Since $C$ is a bounded isomorphism,
$(\mathcal{H}_{1},T_{1},L_{1},\\{v_{i}\\}_{i\in I})$ is a frame tuple if and
only if so is $(\mathcal{H}_{2},T_{2},L_{2},\\{w_{i}\\}_{i\in I})$. ∎

######  Lemma 3.2.

If two Parseval-tuples are similar, then they are unitarily equivalent.

###### Proof.

Let $(\mathcal{H}_{1},T_{1},L_{1},\\{v_{i}\\}_{i\in I})$ and
$(\mathcal{H}_{2},T_{2},L_{2},\\{w_{i}\\}_{i\in I})$ be two similar Parseval-
tuples via the isomorphism $C\in\mathcal{B}(\mathcal{H}_{1},\mathcal{H}_{2})$.
Then, by the intertwining properties of $C$, we have that
$C(T_{1}^{k}L_{1}^{j}v_{i})=T_{2}^{k}L_{2}^{j}w_{i}$, for every
$k\in\mathbb{Z}$, $j\in J$, where $J=\mathbb{N}_{0}$ or $\mathbb{Z}$, and
$i\in I$. In particular, $C$ sends a Parseval frame into another Parseval
frame, thus, it must be unitary. To see this, assume that
$C\in\mathcal{B}(\mathcal{H}_{1},\mathcal{H}_{2})$ is an isomorphism and
$C(f_{k})=g_{k}$, where $\\{f_{k}\\}_{k}$ and $\\{g_{k}\\}_{k}$ are Parseval
frames of the Hilbert spaces $\mathcal{H}_{1}$ and $\mathcal{H}_{2},$
respectively. Set $\widetilde{C}:=C^{-1}$. Then we have:

$\|f\|^{2}=\sum_{k}|\langle f,f_{k}\rangle|^{2}=\sum_{k}|\langle
f,\widetilde{C}Cf_{k}\rangle|^{2}=\sum_{k}|\langle\widetilde{C}^{*}f,g_{k}\rangle|^{2}=\|\widetilde{C}^{*}f\|^{2}.$

Thus, $\widetilde{C}^{*}$ is unitary, and therefore so is $C$. This completes
the proof. ∎

Our goal is to identify all frame-tuples, Parseval-tuples and Riesz-tuples, up
to similarity. Let us begin by presenting a particular class of frame-tuples
which will be crucial for our purpose.

###  3.1.  Basic tuples

In this subsection we define two types of tuples, the unilateral basic tuple
and the bilateral basic tuple.

For the unilateral case, we will consider subspaces of
$L^{2}({\mathbb{T}},\mathcal{K})$ with $\mathcal{K}=H^{2}_{\ell^{2}(I)}$,
where $I$ is an at most countable index set. So, for $f\in
L^{2}({\mathbb{T}},H^{2}_{\ell^{2}(I)})$, we have $f(\lambda)\in
H^{2}_{\ell^{2}(I)}$ and $f(\lambda)(z)\in\ell^{2}(I)$ for a.e.
$\lambda,z\in{\mathbb{T}}$. Let us define a new operator acting on
$L^{2}({\mathbb{T}},H^{2}_{\ell^{2}(I)})$ which will play the role of the
pointwise unilateral shift operator.

######  Definition 3.3.

Define $\widehat{S}:L^{2}({\mathbb{T}},H^{2}_{\ell^{2}(I)})\to
L^{2}({\mathbb{T}},H^{2}_{\ell^{2}(I)})$ as

(10) $\widehat{S}f(\lambda)=S(f(\lambda)),\quad\text{a.e.
}\lambda\in{\mathbb{T}},\,f\in L^{2}({\mathbb{T}},H^{2}_{\ell^{2}(I)}).$

More precisely, for $f\in L^{2}({\mathbb{T}},H^{2}_{\ell^{2}(I)})$ and for
a.e. $\lambda,z\in{\mathbb{T}}$

$\widehat{S}f(\lambda)(z)=S(f(\lambda))(z)=zf(\lambda)(z).$

From the definition of the bilateral shift
$U\in\mathcal{B}(L^{2}({\mathbb{T}},H^{2}_{\ell^{2}(I)}))$ and $\widehat{S}$,
it is easy to see that they commute with each other. Moreover, $\widehat{S}$
is an isometry, since so is $S$. Observe that with this operator in hand, we
can construct an orthonormal basis of
$L^{2}({\mathbb{T}},H^{2}_{\ell^{2}(I)})$ from an orthonormal basis
$\\{\varepsilon_{i}\\}_{i\in I}$ of $\ell^{2}(I)$. Indeed, as discussed in the
previous section, $\\{S^{j}{\varepsilon}_{i}\,:\,j\in\mathbb{N}_{0},\,i\in
I\\}$ is an orthonormal basis of $H^{2}_{\ell^{2}(I)}$ and it is clear that
the same system can be seen inside of
$L^{2}({\mathbb{T}},H^{2}_{\ell^{2}(I)})$ as
$\left\\{\widehat{S}^{j}{\varepsilon}_{i}\,:\,j\in\mathbb{N}_{0},\,i\in
I\right\\}$. Hence, the system

(11)
$\left\\{U^{k}\widehat{S}^{j}{\varepsilon}_{i}\,:\,k\in\mathbb{Z},\,j\in\mathbb{N}_{0},\,i\in
I\right\\}$

is an orthonormal basis of $L^{2}({\mathbb{T}},H^{2}_{\ell^{2}(I)})$. Observe
that (11) is a Fourier basis,
$(U^{k}\widehat{S}^{j}\varepsilon_{i})(\lambda)(z)=\lambda^{k}z^{j}\varepsilon_{i}$
for a.e. $\lambda,z\in{\mathbb{T}}$ and $i\in I$.

######  Definition 3.4.

A tuple $(\mathcal{N},U|_{\mathcal{N}},A,\\{\varphi_{i}\\}_{i\in I})$ is
called a unilateral basic tuple if $\mathcal{N}$ is a closed subspace of
$L^{2}({\mathbb{T}},H^{2}_{\ell^{2}(I)})$ that is $U$-reducing and
$\widehat{S}^{*}$-invariant, $A:\mathcal{N}\rightarrow\mathcal{N}$ is the
compression of $\widehat{S}$ to $\mathcal{N}$, i.e.,
$A:=P_{\mathcal{N}}\widehat{S}|_{\mathcal{N}},$ and
$\varphi_{i}:=P_{\mathcal{N}}\varepsilon_{i},$ where
$\\{\varepsilon_{i}\\}_{i\in I}$ is the canonical orthonormal basis of
$\ell^{2}(I).$

Observe that in Definition 3.4 we are assuming that $U|_{\mathcal{N}}$
commutes with $A$. Indeed, for every $f\in\mathcal{N}$

$U|_{\mathcal{N}}Af=U|_{\mathcal{N}}P_{\mathcal{N}}\widehat{S}f=P_{\mathcal{N}}U|_{\mathcal{N}}\widehat{S}f=P_{\mathcal{N}}\widehat{S}U|_{\mathcal{N}}f=AU|_{\mathcal{N}}f.$

Here we used that $U$ and $P_{\mathcal{N}}$ commute because $\mathcal{N}$ is
$U$-reducing, and that $U$ and $\widehat{S}$ also commute.

We now turn to the bilateral case. These tuples will be constructed from
subspaces belonging to $L^{2}({\mathbb{T}}^{2},\ell^{2}(I))$ that are reducing
for the bilateral shift in each variable, defined as follows. For $i=1,2$, let
$U_{i}:L^{2}({\mathbb{T}}^{2},\ell^{2}(I))\to
L^{2}({\mathbb{T}}^{2},\ell^{2}(I))$ be the operator given by

(12) $(U_{i}f)(z_{1},z_{2})=z_{i}f(z_{1},z_{2}),\quad\textrm{a.e.
}(z_{1},z_{2})\in{\mathbb{T}}^{2}.$

Observe that $U_{1}$ and $U_{2}$ commute and
$\\{U_{1}^{k}U_{2}^{j}\varepsilon_{i}:k,j\in\mathbb{Z},i\in I\\}$ is an
orthonormal basis of $L^{2}({\mathbb{T}}^{2},\ell^{2}(I))$.

######  Definition 3.5.

A tuple
$(\mathcal{N},U_{1}|_{\mathcal{N}},U_{2}|_{\mathcal{N}},\\{\varphi_{i}\\}_{i\in
I})$ is called a bilateral basic tuple if $\mathcal{N}$ is a closed subspace
of $L^{2}({\mathbb{T}}^{2},\ell^{2}(I))$ that is reducing for $U_{1}$ and
$U_{2}$ and $\varphi_{i}:=P_{\mathcal{N}}\varepsilon_{i},$ where
$\\{\varepsilon_{i}\\}_{i\in I}$ is the canonical orthonormal basis of
$\ell^{2}(I)$.

Basic tuples are always frame-tuples. In fact, they are Parseval-tuples as we
show next.

######  Proposition 3.6.

Every basic tuple is a Parseval-tuple.

###### Proof.

Unilateral case. We first note that the restriction of $U$ to $\mathcal{N}$,
$U|_{\mathcal{N}}:\mathcal{N}\to\mathcal{N}$, is unitary. To prove the
proposition we have to see that the system

(13)
$\\{U|_{\mathcal{N}}^{k}A^{j}\varphi_{i}:k\in\mathbb{Z},j\in\mathbb{N}_{0},i\in
I\\}$

is a Parseval frame of $\mathcal{N}$. We observe that (13) is the orthogonal
projection onto $\mathcal{N}$ of the orthonormal basis
$\\{U^{k}\widehat{S}^{j}\varepsilon_{i}:k\in\mathbb{Z},j\in\mathbb{N}_{0},i\in
I\\}$. Indeed, since $\mathcal{N}$ is $\widehat{S}^{*}$-invariant, we have
$\widehat{S}^{*}P_{\mathcal{N}}=P_{\mathcal{N}}\widehat{S}^{*}P_{\mathcal{N}}$
and then
$P_{\mathcal{N}}\widehat{S}=P_{\mathcal{N}}\widehat{S}P_{\mathcal{N}}$. As
$\mathcal{N}$ is also $U$-reducing it follows that

(14)
$P_{\mathcal{N}}U^{k}\widehat{S}^{j}\varepsilon_{i}=U^{k}P_{\mathcal{N}}\widehat{S}^{j}\varepsilon_{i}=U^{k}P_{\mathcal{N}}\widehat{S}^{j}P_{\mathcal{N}}\varepsilon_{i}=U|_{\mathcal{N}}^{k}A^{j}\varphi_{i}.$

Thus, the proposition for the unilateral case is proved.

Bilateral case. This is very similar to the unilateral case. We just need to
see that
$\\{U_{1}|_{\mathcal{N}}^{k}U_{2}^{j}|_{\mathcal{N}}\varphi_{i}:k\in\mathbb{Z},j\in\mathbb{Z},i\in
I\\}$ is a Parseval frame of $\mathcal{N}$. Since
$\\{U_{1}^{k}U_{2}^{j}\varepsilon_{i}:k,j\in\mathbb{Z},i\in I\\}$ is an
orthonormal basis of $L^{2}({\mathbb{T}}^{2},\ell^{2}(I))$, then
$\\{P_{\mathcal{N}}U_{1}^{k}U_{2}^{j}\varepsilon_{i}:k,j\in\mathbb{Z},i\in
I\\}$ is a Parseval frame for $\mathcal{N}$. But
$P_{\mathcal{N}}U_{1}^{k}U_{2}^{j}\varepsilon_{i}=U_{1}|_{\mathcal{N}}^{k}U_{2}|_{\mathcal{N}}^{j}P_{\mathcal{N}}\varepsilon_{i}$
for every $k,j\in\mathbb{Z},\,i\in I$. Therefore,
$(\mathcal{N},U_{1}|_{\mathcal{N}},U_{2}|_{\mathcal{N}},\\{\varphi_{i}\\}_{i\in
I})$ is a Parseval-tuple. ∎

Since basic tuples are Parseval-tuples, then two similar basic tuples must be
unitarily equivalent by Lemma 3.2. Even more, they must be equal, as show
next.

######  Theorem 3.7.

If two basic tuples are similar, then they are equal.

###### Proof.

We will just prove the unilateral case, as the bilateral case will be
analogous. Let
$(\mathcal{N}_{1},U|_{\mathcal{N}_{1}},A_{1},\\{P_{\mathcal{N}_{1}}\varepsilon_{i}\\}_{i\in
I})$ and
$(\mathcal{N}_{2},U|_{\mathcal{N}_{2}},A_{2},\\{P_{\mathcal{N}_{2}}\varepsilon_{i}\\}_{i\in
I})$ be two similar basic tuples. Then, there exists a unitary isomorphism
$C:\mathcal{N}_{1}\to\mathcal{N}_{2}$ such that $CA_{1}=A_{2}C$,
$CU|_{\mathcal{N}_{1}}=U|_{\mathcal{N}_{2}}C$ and
$CP_{\mathcal{N}_{1}}\varepsilon_{i}=P_{\mathcal{N}_{2}}\varepsilon_{i}$ for
every $i\in I$.

Moreover, by the intertwining properties of $C$ and (14) we have that

(15) $\displaystyle
C(P_{\mathcal{N}_{1}}(U^{k}\widehat{S}^{j}\varepsilon_{i}))$
$\displaystyle=C(U^{k}A_{1}^{j}P_{\mathcal{N}_{1}}\varepsilon_{i})=U^{k}A_{2}^{j}C(P_{\mathcal{N}_{1}}\varepsilon_{i})=U^{k}A_{2}^{j}P_{\mathcal{N}_{2}}\varepsilon_{i}$
(16) $\displaystyle=P_{\mathcal{N}_{2}}(U^{k}\widehat{S}^{j}\varepsilon_{i}),$

for every $k\in\mathbb{Z}$, $j\in\mathbb{N}_{0}$ and $i\in I$. Since
$\\{U^{k}\widehat{S}^{j}\varepsilon_{i}\\}_{k,j,i}$ is an orthonormal basis of
$L^{2}({\mathbb{T}},H^{2}_{\ell^{2}(I)})$, then [23, Proposition 2.6] implies
that $P_{\mathcal{N}_{1}}=P_{\mathcal{N}_{2}}$, and so
$\mathcal{N}_{1}=\mathcal{N}_{2}$ and the tuples must be equal. ∎

In what follows we will show that, in fact, any frame-tuple
$(\mathcal{H},T,L,\\{w_{i}\\}_{i\in I})$ must be similar to a unique basic
tuple.

Since the $U$-reducing and $\widehat{S}^{*}$-invariant subspaces of
$L^{2}({\mathbb{T}},H^{2}_{\ell^{2}(I)})$ serve as models for frames of
iterations of two commuting operators -one of them acting unilateraly- the
study of their structure, that is very rich, is of independent interest. This
is done in great detail in [2].

On the other hand, for the bilateral case, the model subspaces are in
$L^{2}({\mathbb{T}}^{2},\ell^{2}(I))$ and they are reducing for $U_{1}$ and
$U_{2}$. This implies that they are multiplication-invariant subspaces, and
their structure has been very well studied, see for instance [13, 14].

###  3.2. Unilateral frame-tuples

Let $\mathcal{H}$ be a separable Hilbert space. The main purpose of this
subsection is to characterize the unilateral frame-tuples in $\mathcal{H}$.
That is, those tuples $(\mathcal{H},T,L,\\{w_{i}\\}_{i\in I})$ where
$T,L\in\mathcal{B}(\mathcal{H})$, $T$ is invertible, $TL=LT$,
$\\{w_{i}\\}_{i\in I}\subset\mathcal{H}$ is an at most countable set, and the
system

(17) $\left\\{T^{k}L^{j}w_{i}:k\in\mathbb{Z},j\in\mathbb{N}_{0},i\in
I\right\\}$

forms a frame for $\mathcal{H}$.

Assume the the system (17) is a Bessel sequence. We then consider its
associated synthesis operator defined on
$L^{2}({\mathbb{T}},H^{2}_{\ell^{2}(I)})$, that is

(18)
$C:L^{2}({\mathbb{T}},H^{2}_{\ell^{2}(I)})\rightarrow\mathcal{H},\;\;\;\;\;\;Cf=\sum_{k,j,i}f_{kj}^{i}T^{k}L^{j}w_{i},$

where

$f^{i}_{kj}:=\langle
f,U^{k}\widehat{S}^{j}\varepsilon_{i}\rangle=\int_{{\mathbb{T}}}\int_{{\mathbb{T}}}\langle
f(\lambda)(z),\varepsilon_{i}\rangle\,\lambda^{-k}z^{-j}\,dz\,d\lambda,$

for $k\in\mathbb{Z},j\in\mathbb{N}_{0},i\in I$ (the Fourier coefficients in
the basis (11)). Note that, since the system (17) is a Bessel sequence, the
series converges unconditionally.

######  Remark 3.8.

Usually, the synthesis operator of a system like in (17) would be defined on
the sequence space $\ell^{2}(\mathbb{Z}\times\mathbb{N}_{0}\times I).$ Here,
we precomposed the usual synthesis operator with the isometric isomorphism
between $L^{2}(\mathbb{T},H^{2}_{\ell^{2}(I)})$ and
$\ell^{2}(\mathbb{Z}\times\mathbb{N}_{0}\times I)$ defined by
$f\mapsto\\{\langle f,U^{k}\widehat{S}^{j}\varepsilon_{i}\rangle\\}_{k,j,i}$.

Now we are ready to prove the characterization of unilateral frame-tuples.

######  Theorem 3.9.

A tuple $(\mathcal{H},T,L,\\{w_{i}\\}_{i\in I})$ is a unilateral frame-tuple
if and only if it is similar to a unilateral basic tuple. Moreover, this basic
tuple is unique.

###### Proof.

Assuming that $(\mathcal{H},T,L,\\{w_{i}\\}_{i\in I})$ is a unilateral frame-
tuple we will find a similar unilateral basic tuple
$(\mathcal{N},U,A,\\{\varphi_{i}\\}_{i\in I})$.

Let $C$ be the synthesis operator defined as in (18). If $f\in
L^{2}({\mathbb{T}},H^{2}_{\ell^{2}(I)})$ we have:

(19) $\langle
Uf,U^{k}\widehat{S}^{j}\varepsilon_{i}\rangle=f^{i}_{k-1,j}\quad\text{and}\quad\langle
U^{*}f,U^{k}S^{j}\varepsilon_{i}\rangle=f^{i}_{k+1,j},$

for every $(k,j,i)\in\mathbb{Z}\times\mathbb{N}_{0}\times I$. Analogously, we
also have
$\langle\widehat{S}f,U^{k}\widehat{S}^{j}\varepsilon_{i}\rangle=f^{i}_{k,j-1}$.
Thus, it is easy to see that the following intertwining relations hold:

(20) $TC=CU,\quad T^{-1}C=CU^{*}\quad\text{and}\quad LC=C\widehat{S}.$

From here we deduce that $\ker(C)\subseteq
L^{2}({\mathbb{T}},H^{2}_{\ell(I)})$ is reducing for $U$ and invariant for
$\widehat{S}$ or equivalently, $\mathcal{N}:=\ker(C)^{\perp}$ is reducing for
$U$ and invariant for $\widehat{S}^{*}$.

Given that the system (17) is a frame of $\mathcal{H}$, $C$ is a bounded
surjective operator. Then, the restriction of $C$ to $\mathcal{N}$

(21) $C|_{\mathcal{N}}:\mathcal{N}\to\mathcal{H},$

is a bounded isomorphism. On the other hand, let
$\varphi_{i}:=P_{\mathcal{N}}\varepsilon_{i}$ for every $i\in I$ and observe
that $C(P_{\mathcal{N}^{\perp}}\varepsilon_{i^{\prime}})=0$ for every
$i^{\prime}\in I$, and that $\varepsilon_{i^{\prime}}$ is orthogonal to
$U^{k}\widehat{S}^{j}\varepsilon_{i}$ for every
$(k,j,i)\in\mathbb{Z}\times\mathbb{N}_{0}\times I$ except for $k=j=0$ and
$i=i^{\prime}$ in which case
$\langle\varepsilon_{i^{\prime}},\varepsilon_{i^{\prime}}\rangle=1$. Then, we
obtain

(22)
$C|_{\mathcal{N}}(\varphi_{i^{\prime}})=C\varepsilon_{i^{\prime}}=\sum_{k,j,i}\langle\varepsilon_{i^{\prime}},U^{k}\widehat{S}^{j}\varepsilon_{i}\rangle
T^{k}L^{j}w_{i}=w_{i^{\prime}},\quad\text{ for every }i^{\prime}\in I.$

Since relations in (20) still hold when restricting $C$ to $\mathcal{N}$, we
conclude that $(\mathcal{H},T,L,\\{w_{i}\\}_{i\in I})$ is similar to
$(\mathcal{N},U,A,\\{\varphi_{i}\\}_{i\in I})$.

The converse is a consequence of Lemma 3.1 and Proposition 3.6. The unicity
follows from Theorem 3.7. That is, if there exists another basic tuple similar
to $(\mathcal{H},T,L,\\{w_{i}\\}_{i\in I})$, then by transitivity it must be
similar to $(\mathcal{N},U,A,\\{\varphi_{i}\\}_{i\in I})$, and therefore
equal. ∎

If $\\#I=1$ in Theorem 3.9, then $\ell^{2}(I)=\mathbb{C}$ and
$H^{2}_{\ell^{2}(I)}=H^{2}$. Thus, we have the following:

######  Corollary 3.10.

A tuple $(\mathcal{H},T,L,w)$ is a unilateral frame-tuple if and only if there
exists a $U$-reducing and $\widehat{S}^{*}$-invariant subspace
$\mathcal{N}\subseteq L^{2}({\mathbb{T}},H^{2})$ such that
$(\mathcal{H},T,L,w)$ is similar to $(\mathcal{N},U,A,\varphi)$, where $A$ is
the compression of $\widehat{S}$ to $\mathcal{N}$ and
$\varphi=P_{\mathcal{N}}1$.

Another implication of Theorem 3.9 is the characterization of Parseval-tuples.

######  Corollary 3.11.

A tuple $(\mathcal{H},T,L,\\{w_{i}\\}_{i\in I})$ is a unilateral Parseval-
tuple if and only if it is unitarily equivalent to a unilateral basic tuple.

###### Proof.

Let us assume that $(\mathcal{H},T,L,\\{w_{i}\\}_{i\in I})$ is a unilateral
Parseval-tuple. Then, from Theorem 3.9, it is similar to a basic tuple, that
by Proposition 3.6 is a Parseval-tuple. Thus, by Lemma 3.2, they must be
unitarily equivalent. The other implication is immediate since Parseval frames
are preserved under unitary operators. ∎

In the next result, we show that any unilateral Riesz-tuple must be similar to
a basic tuple where the Hilbert space is the whole
$L^{2}({\mathbb{T}},H^{2}_{\ell^{2}(I)})$.

######  Proposition 3.12.

A tuple $(\mathcal{H},T,L,\\{w_{i}\\}_{i\in I})$ is a unilateral Riesz-tuple
if and only if it is similar to
$(L^{2}({\mathbb{T}},H^{2}_{\ell^{2}(I)}),U,\widehat{S},\\{\varepsilon_{i}\\}_{i\in
I})$, where $\\{\varepsilon_{i}\\}_{i\in I}$ is the canonical basis of
$\ell^{2}(I)$.

###### Proof.

If $(\mathcal{H},T,L,\\{w_{i}\\}_{i\in I})$ is a unilateral Riesz-tuple, the
kernel of the synthesis operator is the trivial subspace $\\{0\\}.$ Thus, its
orthogonal complement is $L^{2}({\mathbb{T}},H^{2}_{\ell^{2}(I)})$ and by the
proof of Theorem 3.9 we obtain the claim in the proposition. The converse is
straightforward. ∎

###  3.3. Bilateral frame-tuples

When the operator $L$ is invertible, it is possible to consider integer
iterations of it. Given a tuple $(\mathcal{H},T,L,\\{w_{i}\\}_{i\in I})$ with
$L$ being invertible, we seek for the frame condition on the set

$\left\\{T^{k}L^{j}w_{i}:k,j\in\mathbb{Z},i\in I\right\\}.$

As well as in the previous subsection, we obtain that the set of iterations of
$T$ and $L$ generated by $\\{w_{i}\\}_{i\in I}$ is a frame for $\mathcal{H}$
if and only if the tuple $(\mathcal{H},T,L,\\{w_{i}\\}_{i\in I})$ is similar
to a unique bilateral basic tuple. This is proved by following the same
arguments as in the proof of Theorem 3.9 replacing the synthesis operator by
the corresponding one for this case, that is,
$C:L^{2}({\mathbb{T}}^{2},\ell^{2}(I))\rightarrow\mathcal{H}$

$Cf=\sum_{k,j,i}f_{kj}^{i}T^{k}L^{j}w_{i}$

with $f_{kj}^{i}=\langle f,U_{1}^{k}U_{2}^{j}\varepsilon_{i}\rangle,$ for
$k,j\in\mathbb{Z},i\in I$. Hence, we just state the following theorem without
proof.

######  Theorem 3.13.

A tuple $(\mathcal{H},T,L,\\{w_{i}\\}_{i\in I})$ is a bilateral frame-tuple if
and only if it is similar to a bilateral basic tuple. Moreover, this basic
tuple is unique.

As in the case of unilateral iterations, we have the following particular case
of Theorem 3.13 when $\\#I=1$.

######  Corollary 3.14.

A tuple $(\mathcal{H},T,L,w)$ is a bilateral frame-tuple if and only if there
exists a subspace $\mathcal{N}\subseteq L^{2}({\mathbb{T}}^{2})$ which is
reducing for $U_{1}$ and $U_{2}$ such that $(\mathcal{H},T,L,w)$ is similar to
$(\mathcal{N},U_{1}|_{\mathcal{N}},U_{2}|_{\mathcal{N}},P_{\mathcal{N}}1)$,
where $1\in L^{2}({\mathbb{T}}^{2})$ is the function which is constantly one.

Also, we have the next results regarding Parseval-tuples and Riesz-tuples.

######  Corollary 3.15.

A tuple $(\mathcal{H},T,L,\\{w_{i}\\}_{i\in I})$ is a bilateral Parseval-tuple
if and only if it is unitarily equivalent to a bilateral basic tuple.

######  Proposition 3.16.

A tuple $(\mathcal{H},T,L,\\{w_{i}\\}_{i\in I})$ is a bilateral Riesz-tuple if
and only if it is similar to
$(L^{2}({\mathbb{T}}^{2},\ell^{2}(I)),U_{1},U_{2},\\{\varepsilon_{i}\\}_{i\in
I})$, where $\\{\varepsilon_{i}\\}_{i\in I}$ is the canonical basis of
$\ell^{2}(I)$.

##  4\. Frames of orbits of a single vector

Let $\mathcal{H}$ be a separable Hilbert space,
$T,L\in\mathcal{B}(\mathcal{H})$ such that $T$ is invertible and $TL=LT$.
Assume that $\mathcal{H},T,L$ and the index set $I=\\{1,\dots,n\\}$ are fixed
and consider the set

(23) $\mathcal{V}_{n}:=\big{\\{}\\{v_{i}\\}_{i\in
I}\subset\mathcal{H}:(\mathcal{H},T,L,\\{v_{i}\\}_{i\in I})\text{ is a
unilateral (bilateral) frame-tuple}\big{\\}}.$

Now, define an equivalent relation in $\mathcal{V}_{n}$ given by
$\\{v_{i}\\}_{i\in I}\sim\\{w_{i}\\}_{i\in I}$ if and only if
$(\mathcal{H},T,L,\\{v_{i}\\}_{i\in I})$ is similar to
$(\mathcal{H},T,L,\\{w_{i}\\}_{i\in I})$.

In this section we will prove that when $\\#I=1$ there is only one equivalence
class (whenever $\mathcal{V}_{1}$ is not empty). We will later show that this
is not true for multiple generators (see Remark 4.9). Specifically, for the
unilateral case the claim read as follows.

######  Theorem 4.1.

Let $\mathcal{H}$ be a Hilbert space, $T,L\in\mathcal{B}(\mathcal{H})$ such
that $T$ is invertible and $LT=TL$ and consider the set

$\mathcal{V}:=\left\\{v\in\mathcal{H}:(\mathcal{H},T,L,v)\text{ is a
unilateral frame-tuple}\right\\}.$

Assume that $w\in\mathcal{V}.$ Then, $v\in\mathcal{V}$ if and only if
$(\mathcal{H},T,L,v)$ is similar to $(\mathcal{H},T,L,w)$, i.e.

$\mathcal{V}=\left\\{Bw:B\in\mathcal{B}(\mathcal{H}),\,B\text{ is invertible
and commutes with }T\text{ and }L\right\\}.$

In order to prove Theorem 4.1, we will show that for $\\#I=1$ a weaker
condition than similarity is sufficient for two unilateral basic-tuples to be
equal.

######  Theorem 4.2.

For $i=1,2$, let $\mathcal{N}_{i}\subseteq L^{2}({\mathbb{T}},H^{2})$ be a
reducing subspace for $U$ and invariant for $\widehat{S}^{*}$, and let
$A_{i}=P_{\mathcal{N}_{i}}\widehat{S}$. If there exists an isomorphism
$\Psi:\mathcal{N}_{1}\rightarrow\mathcal{N}_{2}$ such that

(24) $\Psi A_{1}=A_{2}\Psi\quad\text{and}\quad\Psi
U|_{\mathcal{N}_{1}}=U|_{\mathcal{N}_{2}}\Psi,$

then $\mathcal{N}_{1}=\mathcal{N}_{2}$ and $A_{1}=A_{2}$.

Before we prove Theorem 4.2 we need to recall some notions on the structure of
reducing subspaces of $L^{2}({\mathbb{T}},\mathcal{K})$, where $\mathcal{K}$
is a separable Hilbert space. For more details on this topic, we refer the
reader to [14] where the $U$-reducing subspaces of
$L^{2}({\mathbb{T}},\mathcal{K})$ are the multiplicative-invariant subspaces
of $L^{2}({\mathbb{T}},\mathcal{K})$ with respect to the determining set
$\\{\gamma^{j}\\}_{j\in\mathbb{Z}}$, with $\gamma(\lambda)=\lambda$ a.e
$\lambda\in{\mathbb{T}}$, see [14, Definition 2.2 and 2.3].

Helson proved in [24] that $U$-reducing subspaces of
$L^{2}({\mathbb{T}},\mathcal{K})$ can be characterized in terms of range
functions. Later, Bownik and Ross gave in [14, Theorem 2.4] an extended
version of this result.

A range function in $\mathcal{K}$ is a mapping
$J:{\mathbb{T}}\rightarrow\\{\text{closed subspaces of }\mathcal{K}\\}$. It is
measurable if for each $x,y\in\mathcal{K}$, the complex-valued function
$\lambda\mapsto\langle P_{J(\lambda)}x,y\rangle$ is measurable.

######  Theorem 4.3.

[14, Theorem 2.4] Let $\mathcal{M}$ be a closed subspace of
$L^{2}({\mathbb{T}},\mathcal{K})$. The following are equivalent:

  1. i)

$\mathcal{M}$ is $U$-reducing,

  2. ii)

there exists a measurable range function such that

$\mathcal{M}=\\{f\in L^{2}({\mathbb{T}},\mathcal{K}):f(\lambda)\in
J(\lambda)\textrm{ for a.e. }\lambda\in{\mathbb{T}}\\}.$

The correspondence between $U$-reducing subspaces and measurable range
functions is one-to-one and onto, assuming that range functions which are
equal almost everywhere are identified.

Moreover, when $\mathcal{M}$ is $U$-reducing, since
$L^{2}({\mathbb{T}},\mathcal{K})$ is separable, there is a countable set
$\mathcal{A}\subset L^{2}({\mathbb{T}},\mathcal{K})$ such that
$\mathcal{M}=\overline{{\rm
span}}\\{U^{k}f:f\in\mathcal{A},k\in\mathbb{Z}\\}$. Then, the measurable range
function $J$ associated to $\mathcal{M}$ is given by

(25) $J(\lambda)=\overline{{\rm span}}\\{f(\lambda):f\in\mathcal{A}\\}.$

In [24], Helson also proved the following property in terms of projections: if
$\mathcal{M}$ is $U$-reducing and $J$ is its range function, then

(26) $P_{\mathcal{M}}f(\lambda)=P_{J(\lambda)}(f(\lambda)),\quad\text{a.e.
}\lambda\in{\mathbb{T}},\,f\in L^{2}({\mathbb{T}},\mathcal{K}).$

For proving Theorem 4.2 we need a characterization of subspaces of
$L^{2}({\mathbb{T}},H^{2})$ that are reducing for $U$ and invariant for
$\widehat{S}^{*}$, that we include below. Its proof can be found in [2,
Corollary 3.10]. Recall that a function $h\in H^{2}$ is inner if $|h(z)|=1$
a.e. $z\in{\mathbb{T}}$.

######  Theorem 4.4.

Let $\mathcal{N}\subseteq L^{2}({\mathbb{T}},H^{2})$ be a closed subspace. The
following statements are equivalent:

  1. i)

$\mathcal{N}$ is $U$-reducing and $\widehat{S}^{*}$-invariant,

  2. ii)

There exists $\phi\in L^{2}({\mathbb{T}},H^{2})$ such that $\phi(\lambda)$ is
an inner function for a.e. $\lambda\in\sigma(\mathcal{N}^{\perp})$ and
$\mathcal{N}^{\perp}=\phi L^{2}({\mathbb{T}},H^{2}):=\left\\{\phi f:f\in
L^{2}({\mathbb{T}},H^{2})\right\\}$, where
$\sigma(\mathcal{N}^{\perp})=\\{\lambda\in{\mathbb{T}}:J_{\mathcal{N}^{\perp}}(\lambda)\neq\\{0\\}\\}$.

Morever, it can be proved that the measurable range function associated to a
$U$-reducing subspace of the form $\mathcal{M}=\phi L^{2}({\mathbb{T}},H^{2})$
for $\phi\in L^{2}({\mathbb{T}},H^{2})$ is given by
$J_{\mathcal{M}}(\lambda)=\phi(\lambda)H^{2}$ for a.e.
$\lambda\in{\mathbb{T}}$ (see [2, Proposition 3.11]).

######  Remark 4.5.

A note about the terminology. If $H^{2}_{\mathcal{K}}$ is the Hardy space of
$\mathcal{K}$-valued functions, and $\mathcal{Q}\subset H^{2}_{\mathcal{K}}$
is a non-trivial closed subspace, invariant by $S^{*}$, the adjoint of the
unilateral shift $S$, then $\mathcal{Q}$ is called a model space. These spaces
had proven to be very important in operator theory since the compression of
the shift on a model space serves as a model  for a certain class of
contractions. (See [20] for a very comprehensive and detailed introduction to
the theory of model spaces).

The Hilbert space $\mathcal{N}$ of a basic tuple is a subspace of
$L^{2}({\mathbb{T}},H^{2}_{\mathcal{K}})$ that is $U$-reducing and
$\widehat{S}^{*}$-invariant. After Theorem 4.3, we know that $\mathcal{N}$ has
a range function. The invariance under $\widehat{S}^{*}$ tells us that the
values of the range function are subspaces of $H^{2}_{\mathcal{K}}$ invariant
under $S^{*}$, that is, they are model spaces.

Thus, $\mathcal{N}$ has a range function whose values are model spaces or, in
the terminology of direct integrals, it is the direct integral over
${\mathbb{T}}$ of model spaces. This connection justifies calling the space
$\mathcal{N}$, in the $L^{2}$-context, a model space.

Another ingredient that we need for the proof of Theorem 4.2 is the concept of
operator-valued function. We give now the definition and some properties.

An operator-valued function in $\mathcal{K}$ is a function
$F:{\mathbb{T}}\to\mathcal{B}(\mathcal{K})$. It is said to be measurable if
for every $x\in\mathcal{K}$, $\lambda\mapsto F(\lambda)(x)$ is measurable. The
norm of an operator-valued function in $\mathcal{K}$ is defined as
$\|F\|_{\infty}={\rm ess\,sup}_{\lambda\in{\mathbb{T}}}\|F(\lambda)\|_{op}$.

If $F:{\mathbb{T}}\to\mathcal{B}(\mathcal{K})$ is a measurable operator-valued
function in $\mathcal{K}$ such that $\|F\|_{\infty}<\infty$, we denote by
$\widehat{F}:L^{2}({\mathbb{T}},\mathcal{K})\to
L^{2}({\mathbb{T}},\mathcal{K})$ the operator defined by

$\widehat{F}f(\lambda)=F(\lambda)f(\lambda),\quad\text{ for a.e.
}\lambda\in{\mathbb{T}},\,\,f\in L^{2}({\mathbb{T}},\mathcal{K})$

and by $\mathcal{F}$ the class of all measurable functions
$F:{\mathbb{T}}\to\mathcal{B}(\mathcal{K})$ such that $\|F\|_{\infty}<\infty$,
and $\widehat{\mathcal{F}}=\\{\widehat{F}:F\in\mathcal{F}\\}$. In [31, Theorem
3.17], it is shown that the correspondence $F\mapsto\widehat{F}$ is an
adjoint-preserving algebra isomorphism. Moreover, $\widehat{F}$ is normal
(self-adjoint, unitary or a projection), if and only if $F(\lambda)$ is normal
(self-adjoint, unitary or a projection) for a.e. $\lambda\in{\mathbb{T}}$.
Also, in [31, Corollary 3.19] the authors showed that the commutant of $U$
acting on $L^{2}({\mathbb{T}},\mathcal{K})$ is $\widehat{\mathcal{F}}$.

We will make use of the following properties of operator-valued functions and
reducing subspaces for $U$ acting on $L^{2}({\mathbb{T}},\mathcal{K})$ whose
proofs follows from [13, Theorem 4.1 and Lemma 4.2].

######  Lemma 4.6.

Let $\mathcal{M},\mathcal{N}\subseteq L^{2}({\mathbb{T}},\mathcal{K})$ be
reducing subspaces for $U$ with range functions
$J_{\mathcal{M}},J_{\mathcal{N}}$ respectively, and let
$\widehat{F}\in\widehat{\mathcal{F}}$. Then, we have:

  1. i)

If $\widehat{F}|_{\mathcal{M}}=0$, then,
$F(\lambda)|_{J_{\mathcal{M}}(\lambda)}=0$ for a.e. $\lambda\in{\mathbb{T}}$.

  2. ii)

If $\widehat{F}|_{\mathcal{M}}:\mathcal{M}\to\mathcal{N}$ is an isomorphism,
then so is $F(\lambda)|_{J_{\mathcal{M}}(\lambda)}:J_{\mathcal{M}}(\lambda)\to
J_{\mathcal{N}}(\lambda)$, for a.e. $\lambda\in{\mathbb{T}}$.

Now we are ready to prove Theorem 4.2.

###### Proof of Theorem 4.2.

Since $\mathcal{N}_{i}\subseteq L^{2}({\mathbb{T}},H^{2})$ for $i=1,2$ is
$U$-reducing and $\widehat{S}^{*}$-invariant, then Theorem 4.4 says that there
exists $\phi_{i}\in L^{2}({\mathbb{T}},H^{2})$ such that $\phi_{i}(\lambda)$
is an inner function for a.e. $\lambda\in\sigma(\mathcal{N}_{i}^{\perp})$ and
$\mathcal{N}_{i}=(\phi_{i}L^{2}({\mathbb{T}},H^{2}))^{\perp}$. Thus, its range
function $J_{i}$ is given by

$J_{i}:\lambda\mapsto J_{i}(\lambda)=(\phi_{i}(\lambda)H^{2})^{\perp}.$

To conclude that $\mathcal{N}_{1}=\mathcal{N}_{2}$, and hence $A_{1}=A_{2}$,
it is sufficient to show that $\phi_{1}=\phi_{2}$.

Observe first that if we extend $\Psi$ as zero on $\mathcal{N}_{1}^{\perp}$,
from the second equation in (24) we get that this extension commutes with $U$
and therefore, by [31, Corollary 3.19], there is
$\widehat{F}\in\widehat{\mathcal{F}}$ such that $\widehat{F}=\Psi$ on
$\mathcal{N}_{1}$ and $\widehat{F}=0$ on $\mathcal{N}_{1}^{\perp}$. Moreover,
since
$\widehat{F}|_{\mathcal{N}_{1}}=\Psi:\mathcal{N}_{1}\rightarrow\mathcal{N}_{2}$
is an isomorphism, ii) of Lemma 4.6 implies that
$F(\lambda)|_{J_{1}(\lambda)}:J_{1}(\lambda)\rightarrow J_{2}(\lambda)$ is an
isomorphism for a.e. $\lambda\in{\mathbb{T}}$.

On the other hand, the first equation in (24) can be rewritten in terms of
$\widehat{F}|_{\mathcal{N}_{1}}$ as

(27)
$\widehat{F}|_{\mathcal{N}_{1}}A_{1}=A_{2}\widehat{F}|_{\mathcal{N}_{1}}.$

We want to see now that $A_{1}$ and $A_{2}$ belong to $\widehat{\mathcal{F}}$.
To do this we define for a.e. $\lambda\in{\mathbb{T}}$ and for $i=1,2$ the
function $G_{i}:{\mathbb{T}}\rightarrow\mathcal{B}(H^{2})$ which for every
$\lambda\in{\mathbb{T}}$ is defined as the operator
$G_{i}(\lambda):=P_{J_{i}(\lambda)}S$. Then, $\lambda\mapsto G_{i}(\lambda)$
is measurable since so is $\lambda\mapsto P_{J_{i}(\lambda)}$ and
$\|G_{i}\|_{\infty}={\mathrm{ess}\sup}_{\lambda\in{\mathbb{T}}}\|G_{i}(\lambda)\|_{op}={\mathrm{ess}\sup}_{\lambda\in{\mathbb{T}}}\|P_{J_{i}(\lambda)}S\|_{op}\leq
1<\infty$. This proves that $G_{i}\in\mathcal{F}$.

Besides, for $i=1,2$ we have that $\widehat{G}_{i}$ and $A_{i}$ coincide. This
is a consequence of (26) as the following computation shows:

(28)
$\widehat{G}_{i}f(\lambda)=G_{i}(\lambda)(f(\lambda))=P_{J_{i}(\lambda)}(Sf(\lambda))=P_{J_{i}(\lambda)}((\widehat{S}f)(\lambda))=P_{\mathcal{N}_{i}}(\widehat{S}f)(\lambda)=A_{i}f(\lambda)$

for every $f\in L^{2}({\mathbb{T}},H^{2})$ and for a.e
$\lambda\in{\mathbb{T}}$.

Hence, for a.e. $\lambda\in{\mathbb{T}}$ it is obtained from (27) that

(29)
$\begin{split}F(\lambda)|_{J_{1}(\lambda)}G_{1}(\lambda)=G_{2}(\lambda)F(\lambda)|_{J_{1}(\lambda)}.\end{split}$

The equation above says that $F(\lambda)|_{J_{1}(\lambda)}$ intertwines
$G_{1}(\lambda)$ and $G_{2}(\lambda)$ which are two compressions of the
unilateral shift operator acting in the subspaces
$J_{1}(\lambda)=(\phi_{1}(\lambda)H^{2})^{\perp}$ and
$J_{2}(\lambda)=(\phi_{2}(\lambda)H^{2})^{\perp}$ respectively.

At this point we use some results about quasi-similar $C_{0}$ contractions and
minimal functions which are developed in [29, Chapter III].

First, we observe that by [29, Proposition 4.3, Chapter III], for $i=1,2$ and
for a.e. $\lambda\in{\mathbb{T}}$, the operator $G_{i}(\lambda)$ is a $C_{0}$
contraction whose minimal function is $\phi_{i}(\lambda)$. Second, since
$G_{1}(\lambda)$ and $G_{2}(\lambda)$ are quasi-similar, i.e, there exists a
bounded isomorphism that intertwines $G_{1}(\lambda)$ and $G_{2}(\lambda)$ for
a.e. $\lambda\in{\mathbb{T}}$, then by [29, Proposition 4.6, Chapter III] we
get that the inner functions $\phi_{1}(\lambda)$ and $\phi_{2}(\lambda)$
coincide a.e. $\lambda\in{\mathbb{T}}$ and therefore $\phi_{1}=\phi_{2}$.
Consequently, from the definitions of $\mathcal{N}_{i}$ and $A_{i}$ for
$i=1,2$, the result follows. ∎

Finally, we prove the main results of this section.

###### Proof of Theorem 4.1.

Observe that if $v\in\mathcal{H}$ is such that $(\mathcal{H},T,L,v)$ is
similar to $(\mathcal{H},T,L,w)$, then by Lemma 3.1, $(\mathcal{H},T,L,v)$ is
a frame-tuple. Hence $v\in\mathcal{V}$.

It remains to show that if $v\in\mathcal{V}$, then it can be defined a linear
bounded operator $B:\mathcal{H}\to\mathcal{H}$ with $BT=TB$ and $BL=LB$ such
that $v=Bw$.

Since $w$, $v\in\mathcal{V}$, $(\mathcal{H},T,L,w)$ and $(\mathcal{H},T,L,v)$
are unilateral frame-tuples. Therefore, by Corollary 3.10, there exist two
basic tuples $(\mathcal{N}_{1},U,A_{1},\varphi_{1})$ and
$(\mathcal{N}_{2},U,A_{2},\varphi_{2})$ that are similar to
$(\mathcal{H},T,L,w)$ and $(\mathcal{H},T,L,v)$ respectively. Recall that the
isomorphism $C_{i}\in\mathcal{B}(\mathcal{N}_{i},\mathcal{H})$ for $i=1,2$
given by the similarity relation satisfies that

(30) $C_{1}A_{1}=LC_{1},\quad
C_{1}U|_{\mathcal{N}_{1}}=TC_{1}\quad\text{and}\quad C_{1}\varphi_{1}=w$

and

(31) $C_{2}A_{2}=LC_{2},\quad
C_{2}U|_{\mathcal{N}_{2}}=TC_{2}\quad\text{and}\quad C_{2}\varphi_{2}=v.$

Now, from equations (30) and (31) we have that

(32) $C_{1}A_{1}C_{1}^{-1}=C_{2}A_{2}C_{2}^{-1},\quad
C_{1}U|_{\mathcal{N}_{1}}C_{1}^{-1}=C_{2}U|_{\mathcal{N}_{2}}C_{2}^{-1}.$

which is equivalent to

$\Psi A_{1}=A_{2}\Psi,\quad\Psi U|_{\mathcal{N}_{1}}=U|_{\mathcal{N}_{2}}\Psi$

where $\Psi=C_{2}^{-1}C_{1}:\mathcal{N}_{1}\to\mathcal{N}_{2}$. Since $\Psi$
is an isomorphism, Theorem 4.2 gives that $\mathcal{N}_{1}=\mathcal{N}_{2}$,
$A_{1}=A_{2}$ and consequently $\varphi_{1}=\varphi_{2}$.

Now, let $B:\mathcal{H}\rightarrow\mathcal{H}$ be defined as
$B:=C_{2}C_{1}^{-1}$. Observe that $B$ is bounded, invertible and

$v=C_{2}\varphi_{2}=C_{2}C_{1}^{-1}C_{1}\varphi_{1}=Bw.$

Moreover,

$BL=C_{2}C_{1}^{-1}L=C_{2}A_{1}C_{1}^{-1}=C_{2}A_{2}C_{1}^{-1}=LC_{2}C_{1}^{-1}=LB,$

and then, $B$ and $L$ commute. Analogously, it can be seen that $BT=TB$. This
completes the proof. ∎

The same characterization given in Theorem 4.1 can be obtained when we
consider bilateral frame-tuples. More precisely, we have:

######  Theorem 4.7.

Let $\mathcal{H}$ be a Hilbert space, $T,L\in\mathcal{B}(\mathcal{H})$ such
that $LT=TL$ and consider the set

$\mathcal{V}:=\left\\{v\in\mathcal{H}:(\mathcal{H},T,L,v)\text{ is a bilateral
frame-tuple}\right\\}.$

Assume that $w\in\mathcal{V}.$ Then, $v\in\mathcal{V}$ if and only if
$(\mathcal{H},T,L,v)$ is similar to $(\mathcal{H},T,L,w)$, i.e.

$\mathcal{V}=\left\\{Bw:B\in\mathcal{B}(\mathcal{H}),\,B\text{ is invertible
and commutes with }T\text{ and }L\right\\}.$

The proof of Theorem 4.7 is based in Corollary 3.14 and the following
proposition that characterize reducing subspaces of $L^{2}({\mathbb{T}}^{2})$,
and whose proof can be found in [21].

######  Proposition 4.8.

A subspace $\mathcal{M}\subset L^{2}({\mathbb{T}}^{2})$ is reducing for
$U_{i}$, $i=1,2$ if and only if there exists a Borel set
$E\subseteq{\mathbb{T}}^{2}$ such that
$\mathcal{M}=\mathcal{X}_{E}L^{2}({\mathbb{T}}^{2})$.

###### Proof of Theorem 4.7.

Let $w\in\mathcal{V}$. Since $(\mathcal{H},T,L,w)$ and $(\mathcal{H},T,L,v)$
are bilateral frame-tuples, then by Corollary 3.14 they are similar to basic
frame-tuples
$(\mathcal{M},U_{1}|_{\mathcal{M}},U_{2}|_{\mathcal{M}},P_{\mathcal{M}}1)$ and
$(\mathcal{N},U_{1}|_{\mathcal{N}},U_{2}|_{\mathcal{N}},P_{\mathcal{N}}1)$
respectively.

Analogously as in the proof of the case of unilateral iterations, the key is
to show that the subspaces $\mathcal{M}$ and $\mathcal{N}$ are equal.

From the isomorphisms given by the similarity relations, we can define an
isomorphism $V:\mathcal{M}\rightarrow\mathcal{N}$ satisfying

(33) $V\,U_{1}|_{\mathcal{M}}=U_{1}|_{\mathcal{N}}V\quad\text{and}\quad
V\,U_{2}|_{\mathcal{M}}=U_{2}|_{\mathcal{N}}V.$

Since $\mathcal{M}$ and $\mathcal{N}$ are reducing for $U_{1}$ and $U_{2}$, by
Proposition 4.8 we have that there exist Borel sets
$E_{1},E_{2}\subseteq{\mathbb{T}}^{2}$ such that
$\mathcal{M}=\mathcal{X}_{E_{1}}L^{2}({\mathbb{T}}^{2})$ and
$\mathcal{N}=\mathcal{X}_{E_{2}}L^{2}({\mathbb{T}}^{2})$.

In order to show that $\mathcal{M}=\mathcal{N}$, we will see that the
characteristic functions $\mathcal{X}_{E_{1}}$ and $\mathcal{X}_{E_{2}}$
coincide a.e. on ${\mathbb{T}}^{2}$. To do this, take
$g=\mathcal{X}_{E_{1}\setminus E_{2}}\in\mathcal{M}\subset
L^{2}({\mathbb{T}}^{2})$. Since
$\\{U_{1}^{k}U_{2}^{j}1\,:\,k,j\in\mathbb{Z}\\}$ is an orthonormal basis of
$L^{2}({\mathbb{T}}^{2})$ we have the expansion

$g=\sum_{k\in\mathbb{Z}}\sum_{j\in\mathbb{Z}}\beta_{kj}U_{1}^{k}U_{2}^{j}1,$

whose coefficients satisfy $\\{\beta_{kj}\\}\in\ell^{2}(\mathbb{Z}^{2})$. By
applying the isomorphism $V$ and using the relations in (33) we get that

$\displaystyle Vg=VP_{\mathcal{M}}g$
$\displaystyle=V\left(\sum_{k\in\mathbb{Z}}\sum_{j\in\mathbb{Z}}\beta_{kj}\,U_{1}^{k}\,U_{2}^{j}P_{\mathcal{M}}1\right)=\sum_{k\in\mathbb{Z}}\sum_{j\in\mathbb{Z}}\beta_{kj}\,U_{1}^{k}|_{\mathcal{N}}\,U_{2}^{j}|_{\mathcal{N}}VP_{\mathcal{N}}1.$

Let $h=VP_{\mathcal{N}}1$ and observe that ${\rm supp\,}(g)\subseteq
E_{1}\setminus E_{2}$ and ${\rm supp\,}(h)\subseteq E_{2}$. Then, it is
obtained

$Vg=gh=0.$

Since $V$ is an isomorphism, we conclude that $g=0$. A similar argument can be
used to show that $\mathcal{X}_{E_{2}\setminus E_{1}}=0$. Thus,
$|E_{2}\setminus E_{1}|=0=|E_{1}\setminus E_{2}|$, and therefore
$\mathcal{X}_{E_{1}}$ coincide with $\mathcal{X}_{E_{2}}$ a.e.
$(z_{1},z_{2})\in{\mathbb{T}}^{2}$. As a consequence,
$\mathcal{M}=\mathcal{N}$.

The proof concludes in the same manner of Theorem 4.1.

∎

######  Remark 4.9.

  1. (a)

It is not true that in the case of more than one generator there exists only
one equivalent class of vectors $\\{v_{i}\\}\in\mathcal{V}_{n}$ as we show
next. Suppose that $\mathcal{V}_{n}$ is not empty for some $n>1$. Let
$\\{v_{i}\\}_{i\in I}\in\mathcal{V}_{n}$, that is,
$(\mathcal{H},T,L,\\{v_{i}\\}_{i\in I})$ is a unilateral frame-tuple, and
assume that $\\{v_{1},v_{2}\\}$ is linearly independent. Then, it is easy to
check that $\\{v_{i}\\}_{i\in I}\cup\\{v_{1}+v_{2}\\}$ and $\\{v_{i}\\}_{i\in
I}\cup\\{v_{1}-v_{2}\\}$ belong to $\mathcal{V}_{n+1}$. Suppose that the
corresponding tuples $(\mathcal{H},T,L,\\{v_{i}\\}_{i\in
I}\cup\\{v_{1}+v_{2}\\})$ and $(\mathcal{H},T,L,\\{v_{i}\\}_{i\in
I}\cup\\{v_{1}-v_{2}\\})$ are similar via an isomorphism
$C:\mathcal{H}\to\mathcal{H}$. Then, in particular, it must hold that

$C(v_{1})=v_{1},\;C(v_{2})=v_{2},\;\dots,\,C(v_{1}+v_{2})=v_{1}-v_{2}$

which is not possible.

  2. (b)

As we already mentioned, if we fix the Hilbert space $\mathcal{H}$ and the
operator $T$ and $L$ in $\mathcal{B}(\mathcal{H})$, Theorems 4.1 and 4.7 say
that all the frame-tuples with a single generator are similar (when there is
at least one). Before, we saw that this is no longer true for more generators.
At this point one could ask if there is some relation between $\\#I$ and the
number of equivalent classes of the relation defined at the beginning of this
section. What we know is that if $\\#I=1$ and $\mathcal{V}_{1}\neq\emptyset$,
there is only one equivalence class. But a slight modification in part a shows
that for $\\#I=n>1$ there are infinitely many equivalence classes. Indeed,
suppose that $n>1$ and that $\mathcal{V}_{n}\neq\emptyset$. Consider
$\\{v_{i}\\}_{i\in I}\in\mathcal{V}_{n}$ and assume that $\\{v_{1},v_{2}\\}$
is linearly independent. Then, for $a,b\in\mathbb{C}\setminus\\{0\\}$, we have
that $\\{v_{i}\\}_{i\in I}\cup\\{av_{1}+bv_{2}\\}\in\mathcal{V}_{n+1}$. As in
a, it can be seen that each generator set $\\{v_{i}\\}_{i\in
I}\cup\\{av_{1}+bv_{2}\\}$ belongs to a different equivalence class. Then,
there are infinitely many when we have 3 generators or more. The same idea
works for 2 generators, since if $v\in\mathcal{V}_{1}$,
$\\{v,av\\}\in\mathcal{V}_{2}$ for every $a\in\mathbb{C}\setminus\\{0\\}$ and
they all belong to different equivalence classes.

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