On Certain Commuting Isometries, Joint Invariant Subspaces and C ^∗-Algebras

Abstract

In this paper, motivated by the Berger, Coburn and Lebow and Bercovici, Douglas and Foias theory for tuples of commuting isometries, we study analytic representations and joint invariant subspaces of a class of n tuples of commuting isometries and prove that the C ^∗-algebra generated by the n-tuple of multiplication operators by the coordinate functions restricted to an invariant subspace of finite codimension in $H^2(\mathbb {D}^n)$ is unitarily equivalent to the C ^∗-algebra generated by the n-tuple of multiplication operators by the coordinate functions on $H^2(\mathbb {D}^n)$.

Dedicated to the memory of Ronald G. Douglas, our teacher, mentor and friend

FormalPara Notations

$\mathcal {H}$, $\mathcal {E}$, $\mathcal {E}_*$	Hilbert spaces
$\mathcal {B}(\mathcal {E}, \mathcal {E}_*)$	The space of all bounded linear operators from $\mathcal {E}$ to $\mathcal {E}_*$
$\mathcal {B}(\mathcal {E})$	The space of all bounded linear operators on $\mathcal {E}$
$\mathbb {D}^n$	Open unit polydisc in $\mathbb {C}^n$
$H^2(\mathbb {D}^n)$	Hardy space on $\mathbb {D}^n$
$H^2_{\mathcal {E}}(\mathbb {D}^n)$	$\mathcal {E}$-valued Hardy space on $\mathbb {D}^n$
$H^{\infty }_{\mathcal {B}(\mathcal {E},\mathcal {E}_*)}(\mathbb {D}^n)$	Set of all $\mathcal {B}(\mathcal {E},\mathcal {E}_*)$-valued bounded analytic functions on $\mathbb {D}^n$.
$(M_{z_1}, \ldots , M_{z_n})$	n-tuple of multiplication operator by the coordinate
	functions on $H^2(\mathbb {D}^n)$

1. (1) All Hilbert spaces are assumed to be over the complex numbers.
2. (2) For a closed subspace $\mathcal {S}$ of a Hilbert space $\mathcal {H}$, we denote by $P_{\mathcal {S}}$ the orthogonal projection of $\mathcal {H}$ onto $\mathcal {S}$.
3. (3) For nested closed subspaces $\mathcal {M}_1\subseteq \mathcal {M}_2\subseteq \mathcal {H}$, the orthogonal projection of $\mathcal {M}_2$ onto $\mathcal {M}_1$ is denoted by $P^{\mathcal {M}_2}_{\mathcal {M}_1}$.

1 Introduction

Tuples of commuting isometries on Hilbert spaces are cental objects of study in (multivariable) operator theory. This paper is concerned with the study of analytic representations, joint invariant subspaces and C ^∗-algebras of a certain class of tuples of commuting isometries.

To be precise, let $\mathcal {H}$ be a Hilbert space, and let (V ₁, …, V _n) be an n-tuple of commuting isometries on $\mathcal {H}$. In what follows, we always assume that n ≥ 2. Set

$$\displaystyle \begin{aligned} V = \operatorname*{\Pi}_{i=1}^n V_i. \end{aligned}$$

We say that (V ₁, …, V _n) is a pure n-isometry if V  is a unilateral shift. A closed subspace $\mathcal {S} \subseteq H^2(\mathbb {D}^n)$ is said to be an invariant subspace of $H^2(\mathbb {D}^n)$ if $M_{z_i}\mathcal {S}\subseteq \mathcal {S}$ for all i = 1, …, n where $M_{z_i}$ is the multiplication operator by the coordinate function z _i on $H^2(\mathbb {D}^n)$. Simpler (but complex enough) examples of pure n-isometry can be obtained by taking restrictions of the n-tuple of multiplication operators by coordinate functions $(M_{z_1}, \ldots , M_{z_n})$ on $H^2(\mathbb {D}^n)$ to invariant subspaces of $H^2(\mathbb {D}^n)$ as follows. Given an invariant subspace $\mathcal {S}$ of $H^2(\mathbb {D}^n)$, we let

$$\displaystyle \begin{aligned} R_{z_i} = M_{z_i}|{}_{\mathcal{S}} \in \mathcal{B}(\mathcal{S}) \quad \quad (i=1, \ldots, n). \end{aligned}$$

Then it is easy to see that $(R_{z_1}, \ldots , R_{z_n})$ is a pure n-isometry. We denote by $\mathcal {T}(\mathcal {S})$ the C ^∗-algebra generated by the commuting isometries $\{R_{z_1}, \ldots , R_{z_n}\}$. We simply say that $\mathcal {T}(\mathcal {S})$ is the C ^∗ -algebra corresponding to the invariant subspace $\mathcal {S}$.

In this paper we aim to address three basic issues of pure n-isometries: (i) analytic and canonical models for pure n-isometries, (ii) an abstract classification of joint invariant subspaces for pure n-isometries, and (iii) the nature of C ^∗-algebra $\mathcal {T}(\mathcal {S})$ where $\mathcal {S}$ is a finite codimensional invariant subspace in $H^2(\mathbb {D}^n)$. To that aim, for (i) and (ii), we consider the initial approach by Berger et al. [6] from a more modern point of view (due to Bercovici et al. [5]) along with the technique of [20]. For (iii), we will examine Seto’s approach [26] more closely from “subspace” approximation point of view.

We now briefly outline the setting and the main contributions of this paper. Let $\mathcal {E}$ be a Hilbert space, and let $\varphi \in H^{\infty }_{\mathcal {B}(\mathcal {E})}(\mathbb {D})$. We say that φ is an inner function if $\varphi (e^{it})^* \varphi (e^{it}) = I_{\mathcal {E}}$ for almost every t (cf. page 196, [21]). Recall that two n-tuples of commuting operators (A ₁, …, A _n) on $\mathcal {H}$ and (B ₁, …, B _n) on $\mathcal {K}$ are said to be unitarily equivalent if there exists a unitary operator $U : \mathcal {H} \rightarrow \mathcal {K}$ such that UA _i = B _i U for all i = 1, …, n. In [5], motivated by Berger et al. [6], Bercovici, Douglas and Foias proved the following result: A pure n-isometry is unitarily equivalent to a model pure n-isometry. The model pure n-isometries are defined as follows [5]: Consider a Hilbert space $\mathcal {E}$, unitary operators {U ₁, …, U _n} on $\mathcal {E}$ and orthogonal projections {P ₁, …, P _n} on $\mathcal {E}$. Let $\{\Phi _1, \ldots , \Phi _n\} \subseteq H^\infty _{\mathcal {B}(\mathcal {E})}(\mathbb {D})$ be bounded $\mathcal {B}(\mathcal {E})$-valued holomorphic functions (polynomials) on $\mathbb {D}$, where

$$\displaystyle \begin{aligned} \Phi_i (z) = U_i (P_i^\perp + z P_i) \quad \quad (z \in \mathbb{D}), \end{aligned}$$

and i = 1, …, n. Then the n-tuple of multiplication operators $(M_{\Phi _1}, \ldots , M_{\Phi _n})$ on $H^2_{\mathcal {E}}(\mathbb {D})$ is called a model pure n-isometry if the following conditions are satisfied:

• (a) U _i U _j = U _j U _i for all i, j = 1, …n;

• (b) $U_1 \cdots U_n = I_{\mathcal {E}}$;

• (c) $P_i + U_i^* P_j U_i = P_j + U_j^* P_i U_j \leq I_{\mathcal {E}}$ for all i ≠ j; and

• (d) $P_1 + U_1^* P_2 U_1 + U_1^* U_2^* P_3 U_2 U_1 + \cdots + U_1^* U_2^* \cdots U_{n-1}^* P_n U_{n-1}\cdots U_2 U_1 = I_{\mathcal {E}}$.

It is easy to see that a model pure n-isometry is also a pure n-isometry (see page 643 in [5]).

We refer to Bercovici et al. [3,4,5] and also [8,9,10, 12, 14, 15, 17, 19, 22, 26] and [27, 28] for more on pure n-isometries, n ≥ 2, and related topics.

Our first main result, Theorem 2.1, states that a pure n-isometry is unitarily equivalent to an explicit (and canonical) model pure n-isometry. In other words, given a pure n-isometry (V ₁, …, V _n) on $\mathcal {H}$, we explicitly solve the above conditions (a)–(d) for some Hilbert space $\mathcal {E}$, unitary operators {U ₁, …, U _n} on $\mathcal {E}$ and orthogonal projections {P ₁, …, P _n} on $\mathcal {E}$ so that the corresponding model pure n-isometry $(M_{\Phi _1}, \ldots , M_{\Phi _n})$ is unitarily equivalent to (V ₁, …, V _n). This also gives a new proof of Bercovici, Douglas and Foias theorem. On the one hand, our model pure n-isometry is explicit and canonical. On the other hand, our proof is perhaps more computational than the one in [5]. Another advantage of our approach is the proof of a list of useful equalities related to commuting isometries, which can be useful in other contexts.

Our second main result concerns a characterization of joint invariant subspaces of model pure n-isometries. To be precise, let $\mathcal {W}$ be a Hilbert space, and let $(M_{\Phi _1}, \ldots , M_{\Phi _n})$ be a model pure n-isometry on $H^2_{\mathcal {W}}(\mathbb {D})$. Let $\mathcal {S}$ be a closed subspace of $H^2_{\mathcal {W}}(\mathbb {D})$. In Theorem 3.1, we prove that $\mathcal {S}$ is invariant for $(M_{\Phi _1}, \ldots , M_{\Phi _n})$ on $H^2_{\mathcal {W}}(\mathbb {D})$ if and only if there exist a Hilbert space $\mathcal {W}_*$, an inner function $\Theta \in H^\infty _{\mathcal {B}(\mathcal {W}_*, \mathcal {W})}(\mathbb {D})$ and a model pure n-isometry $(M_{\Psi _1}, \ldots , M_{\Psi _n})$ on $H^2_{\mathcal {W}_*}(\mathbb {D})$ such that

$$\displaystyle \begin{aligned} \mathcal{S} = \Theta H^2_{\mathcal{W}_*}(\mathbb{D}), \end{aligned}$$

and

$$\displaystyle \begin{aligned} \Phi_i \Theta = \Theta \Psi_i, \end{aligned}$$

for all i = 1, …, n. Moreover, the above representation is unique in an appropriate sense (see the remark following Theorem 3.1).

The third and final result concerns C ^∗-algebras corresponding to finite codimensional invariant subspaces in $H^2(\mathbb {D}^n)$. To be more specific, recall that if n = 1 and $\mathcal {S}$ and $\mathcal {S}'$ are invariant subspaces of $H^2(\mathbb {D})$, then $U \mathcal {T}(\mathcal {S}) U^* = \mathcal {T}(\mathcal {S}')$ for some unitary $U : \mathcal {S} \rightarrow \mathcal {S}'$. Indeed, since $\mathcal {S} = \theta H^2(\mathbb {D})$ for some inner function $\theta \in H^\infty (\mathbb {D})$, it follows, by Beurling theorem, that $U: = M_{\theta } : H^2(\mathbb {D}) \rightarrow \mathcal {S}$ is a unitary and hence $U^* \mathcal {T}(\mathcal {S}) U = \mathcal {T}(H^2(\mathbb {D}))$. Clearly, the general case follows from this special case. For invariant subspaces $\mathcal {S}$ and $\mathcal {S}'$ of $H^2(\mathbb {D}^n)$, we say that $\mathcal {T}(\mathcal {S})$ and $\mathcal {T}(\mathcal {S}')$ are isomorphic as C ^∗-algebras if $U \mathcal {T}(\mathcal {S}) U^* = \mathcal {T}(\mathcal {S}')$ holds for some unitary $U : \mathcal {S} \rightarrow \mathcal {S}'$. It is then natural to ask: If n > 1 and $\mathcal {S}$ and $\mathcal {S}'$ are invariant subspaces of $H^2(\mathbb {D}^n)$, are $\mathcal {T}(\mathcal {S})$ and $\mathcal {T}(\mathcal {S}')$ isomorphic as C ^∗-algebras?

In the same paper [6], Berger, Coburn and Lebow asked whether $\mathcal {T}(\mathcal {S})$ is isomorphic to $\mathcal {T}(H^2(\mathbb {D}^2))$ for every finite codimensional invariant subspaces $\mathcal {S}$ in $H^2(\mathbb {D}^2)$. This question was recently answered positively by Seto in [26]. Here we extend Seto’s answer from $H^2(\mathbb {D}^2)$ to the general case $H^2(\mathbb {D}^n)$, n ≥ 2.

The rest of this paper is organized as follows. In Sect. 2 we study and review the analytic construction of pure n-isometries. We also examine a (canonical) model pure n-isometry. A characterization of invariant subspaces is given in Sect. 3. Finally, in Sect. 4, we prove that $\mathcal {T}(\mathcal {S})$ is isomorphic to $\mathcal {T}(H^2(\mathbb {D}^n))$ where $\mathcal {S}$ is a finite codimensional invariant subspaces in $H^2(\mathbb {D}^n)$.

2 Pure n-Isometries and Model Pure n-Isometries

In this section, we first derive an explicit analytic representation of a pure n-isometry. Then we propose a canonical model for pure n-isometries.

For motivation, let us recall that if X on $\mathcal {H}$ is a bounded linear operator, then X is a unilateral shift operator if and only if X and M _z on $H^2_{\mathcal {W}(X)}(\mathbb {D})$ are unitarily equivalent. Here

$$\displaystyle \begin{aligned} \mathcal{W}(X) = \ker X^* = \mathcal{H} \ominus X \mathcal{H}, \end{aligned}$$

is the wandering subspace for X (see Halmos [16]) and M _z denotes the multiplication operator by the coordinate function z on $H^2_{\mathcal {W}(X)}(\mathbb {D})$, that is, (M _z f)(w) = wf(w) for all $f \in H^2_{\mathcal {W}(X)}(\mathbb {D})$ and $w \in \mathbb {D}$. Explicitly, if X is a unilateral shift on $\mathcal {H}$, then

$$\displaystyle \begin{aligned} \mathcal{H} = \operatorname*{\oplus}_{m=0}^\infty X^m \mathcal{W}(X). \end{aligned}$$

Hence the natural map $\Pi _X : \mathcal {H} \rightarrow H^2_{\mathcal {W}(X)}(\mathbb {D})$ defined by

$$\displaystyle \begin{aligned} \Pi_X (X^m \eta) = z^m \eta, \end{aligned}$$

for all m ≥ 0 and $\eta \in \mathcal {W}(X)$, is a unitary operator and

$$\displaystyle \begin{aligned} \Pi_X X = M_z \Pi_X. \end{aligned}$$

We call Π_X the Wold-von Neumann decomposition of the shift X.

Now let $\mathcal {H}$ be a Hilbert space, and let (V ₁, …, V _n) be a pure n-isometry on $\mathcal {H}$. Throughout this paper, we shall use the following notation:

$$\displaystyle \begin{aligned} \tilde{V}_i = \operatorname*{\Pi}_{j \neq i} V_j, \end{aligned}$$

for all i = 1, …, n. For simplicity, we also use the notation

$$\displaystyle \begin{aligned} \mathcal{W} = \mathcal{W}(V), \end{aligned}$$

and

$$\displaystyle \begin{aligned} \mathcal{W}_i=\mathcal{W}(V_i)\ \text{ and } \tilde{\mathcal{W}}_i = \mathcal{W}(\tilde{V}_i), \end{aligned}$$

for all i = 1, …, n. Since $V=\Pi _{i=1}^nV_i$ and $\tilde {V}_i=V_i^*V$ for all i = 1, …, n, it is easy to see that

$$\displaystyle \begin{aligned} \mathcal{W}_i, \tilde{\mathcal{W}}_i \subseteq \mathcal{W}, \end{aligned}$$

for all i = 1, …, n. We denote by $P_{\mathcal {W}_i}$ and $P_{\tilde {\mathcal {W}}_i}$ the orthogonal projections of $\mathcal {W}$ onto the subspaces $\mathcal {W}_i$ and $\tilde {\mathcal {W}}_i$, respectively.

Theorem 2.1

Let (V ₁, …, V _n) be a pure n-isometry on a Hilbert space $\mathcal {H}$, $V = {\Pi }_{i=1}^n V_i$ , and let $\mathcal {W} = \mathcal {W}(V)$ . Let $\Pi _V : \mathcal {H} \rightarrow H^2_{\mathcal {W}}(\mathbb {D})$ be the Wold-von Neumann decomposition of V . If $\tilde {V}_i=V_i^*V$ and $\tilde {\mathcal {W}}_i = \mathcal {W}(\tilde {V}_i)$ , then

$$\displaystyle \begin{aligned} \Pi_V V_i = M_{\Phi_i} \Pi_V, \end{aligned}$$

where

$$\displaystyle \begin{aligned} \Phi_i(z) = U_i (P_{\tilde{\mathcal{W}}_i} + z P_{\tilde{\mathcal{W}}_i}^{\perp}), \end{aligned}$$

for all $z \in \mathbb {D}$ , and

$$\displaystyle \begin{aligned} U_i = (P_{\mathcal{W}} V_i + {\tilde{V_i}}^*)|{}_{\mathcal{W}}, \end{aligned}$$

is a unitary operator on $\mathcal {W}$ and i = 1, …, n. In particular, (V ₁, …, V _n) on $\mathcal {H}$ and $(M_{\Phi _1}, \ldots , M_{\Phi _n})$ on $H^2_{\mathcal {W}}(\mathbb {D})$ are unitarily equivalent.

Proof

Let $\Pi _V : \mathcal {H} \rightarrow H^2_{\mathcal {W}}(\mathbb {D})$ be the Wold-von Neumann decomposition of V . Then

$$\displaystyle \begin{aligned} \Pi_V V_i \Pi_V^* \in \{M_z\}', \end{aligned}$$

and hence there exists $\Phi _i \in H^{\infty }_{\mathcal {B}(\mathcal {W})}(\mathbb {D})$ [16, 21] such that $\Pi _V V_i \Pi _V^* = M_{\Phi _i}$ or, equivalently,

$$\displaystyle \begin{aligned} \Pi_V V_i = M_{\Phi_i} \Pi_V, \end{aligned}$$

for all i = 1, …, n. Note that $M_{\Phi _i}$ on $H^2_{\mathcal {W}}(\mathbb {D})$ is defined by

$$\displaystyle \begin{aligned} (M_{\Phi_i} f)(z) = {\Phi_i}(z) f(z), \end{aligned} $$

(2.1)

for all $f \in H^2_{\mathcal {W}}(\mathbb {D})$, $z \in \mathbb {D}$ and i = 1, …, n. We now proceed to compute the bounded analytic functions $\{\Phi _i\}_{i=1}^n$. Our method follows the construction in [20]. In fact, a close variant of Theorem 2.1 below follows from Theorems 3.4 and 3.5 of [20]. We will only sketch the construction, highlighting the essential ingredients for our present purpose. Let i ∈{1, …, n}, $z \in \mathbb {D}$ and $\eta \in \mathcal {W}$. By an abuse of notation, we will also denote the constant function η in $H^2_{\mathcal {W}}(\mathbb {D})$ corresponding to the vector $\eta \in \mathcal {W}$ by η itself. Then from (2.1), we have that

$$\displaystyle \begin{aligned} \Phi_i(z) \eta = (M_{\Phi_i} \eta)(z) = (\Pi_V V_i \Pi_V^* \eta)(z). \end{aligned}$$

Now it follows from the definition of Π_V that $\Pi _V^* \eta = \eta$, and hence Φ_i(z)η = ( Π_V V _i η)(z). But $I_{\mathcal {W}} = P_{\tilde {\mathcal {W}}_i} + \tilde {V}_i \tilde {V}_i^*|{ }_{\mathcal {W}}$ yields that $V_i \eta = V_i P_{\tilde {\mathcal {W}}_i} \eta + V \tilde {V}_i^* \eta$ and thus

$$\displaystyle \begin{aligned} \Pi_V V_i \eta & = \Pi_V(V_i P_{\tilde{\mathcal{W}}_i} \eta + V \tilde{V}_i^* \eta) \\ & = \Pi_V(V_i P_{\tilde{\mathcal{W}}_i} \eta) + \Pi_V(V \tilde{V}_i^* \eta) \\ & = \Pi_V(V_i P_{\tilde{\mathcal{W}}_i} \eta) + M_z \Pi_V (\tilde{V}_i^* \eta), \end{aligned}$$

as Π_V V = M _z Π_V. Now, since $V^* (V_i (I - \tilde {V}_i \tilde {V}_i^*)V_i^*) = 0$ and $V^* (\tilde {V}_i^* \eta ) = 0$, it follows that $V_i P_{\tilde {\mathcal {W}}_i} \eta \in \mathcal {W}$ and $\tilde {V}_i^* \eta \in \mathcal {W}$. This implies that

$$\displaystyle \begin{aligned} \Pi_V V_i \eta = V_i P_{\tilde{\mathcal{W}}_i} \eta + M_z \tilde{V}_i^* \eta, \end{aligned}$$

and so $\Phi _i(z) \eta = V_i P_{\tilde {\mathcal {W}}_i} \eta + z \tilde {V}_i^* \eta$. It follows that $\Phi _i(z) = V_i|{ }_{\tilde {\mathcal {W}}_i} + z \tilde {V}_i^*|{ }_{\tilde {V}_i \mathcal {W}_i}$ as $\mathcal {W} = \tilde {V}_i \mathcal {W}_i \oplus \tilde {\mathcal {W}}_i$. Finally, $\mathcal {W} = \mathcal {W}_i \oplus V_i \tilde {\mathcal {W}}_i$ implies that

$$\displaystyle \begin{aligned} U_i = \begin{bmatrix} \tilde{V}_i^*|{}_{\tilde{V}_i \mathcal{W}_i} & 0\\ 0 & V_i|{}_{\tilde{\mathcal{W}}_i} \end{bmatrix} : \begin{array}{c} \tilde{V}_i \mathcal{W}_i\\ \oplus\\ \tilde{\mathcal{W}}_i \end{array} \rightarrow \begin{array}{c} \mathcal{W}_i \\ \oplus\\ V_i \tilde{\mathcal{W}}_i \end{array}, \end{aligned}$$

is a unitary operator on $\mathcal {W}$. Therefore

$$\displaystyle \begin{aligned} \Phi_i(z) = U_i (P_{\tilde{\mathcal{W}}_i} + z P_{\tilde{\mathcal{W}}_i}^{\perp}), \end{aligned}$$

for all $z \in \mathbb {D}$. By definition of U _i, it follows that $U_i = (V_i P_{\tilde {{\mathcal {W}}_i}} + {\tilde {V_i}}^*)|{ }_{\mathcal {W}}$. This and

$$\displaystyle \begin{aligned} V_i P_{\tilde{\mathcal{W}}_i} = P_{\mathcal{W}} V_i, \end{aligned} $$

(2.2)

yields $U_i = (P_{\mathcal {W}} V_i + {\tilde {V_i}}^*)|{ }_{\mathcal {W}}$. ■

We now study the coefficients of the one-variable polynomials in Theorem 2.1 more closely and prove that the corresponding pure n-isometry $(M_{\Phi _1}, \ldots , M_{\Phi _n})$ on $H^2_{\mathcal {W}}(\mathbb {D})$ is a model pure n-isometry (see Sect. 1 for the definition of model pure n-isometries).

Let (V ₁, …, V _n) be a pure n-isometry on a Hilbert space $\mathcal {H}$. Consider the analytic representation $(M_{\Phi _1}, \ldots , M_{\Phi _n})$ on $H^2_{\mathcal {W}}(\mathbb {D})$ of (V ₁, …, V _n) as in Theorem 2.1. First we prove that $\{U_j\}_{j=1}^n$ is a commutative family. Let p, q ∈{1, …, n} and p ≠ q. As $\mathcal {W} = \ker V^*$, it follows that

$$\displaystyle \begin{aligned} \tilde{V}_p^* \tilde{V}_q^*|{}_{\mathcal{W}} = 0. \end{aligned}$$

Then using (2.2) we obtain

$$\displaystyle \begin{aligned} U_p U_q & = (P_{\mathcal{W}} V_p + \tilde{V}_p^*)(P_{\mathcal{W}} V_q + \tilde{V}_q^*)|{}_{\mathcal{W}} \\ & = (P_{\mathcal{W}} V_p P_{\mathcal{W}} V_q + \tilde{V}_p^* P_{\mathcal{W}} V_q + P_{\mathcal{W}} V_p \tilde{V}_q^*)|{}_{\mathcal{W}} \\ & = (P_{\mathcal{W}} V_p V_q + \operatorname*{\Pi}_{i \neq p,q} V_i^* P_{\tilde{\mathcal{W}}_q} + V_p P_{\tilde{\mathcal{W}}_p} \tilde{V}_q^*)|{}_{\mathcal{W}} \\ & = (P_{\mathcal{W}} V_p V_q + (\operatorname*{\Pi}_{i \neq p,q} V_i^*) (P_{\tilde{\mathcal{W}}_q} + \tilde{V}_q P_{\tilde{\mathcal{W}}_p} \tilde{V}_q^*))|{}_{\mathcal{W}} \\ & = (P_{\mathcal{W}} V_p V_q + (\operatorname*{\Pi}_{i \neq p,q} V_i^*))|{}_{\mathcal{W}}, \end{aligned}$$

as $(P_{\tilde {\mathcal {W}}_q} + \tilde {V}_q P_{\tilde {\mathcal {W}}_p} \tilde {V}^*_q)|{ }_{\mathcal {W}} = I_{\mathcal {W}}$, and hence

$$\displaystyle \begin{aligned} U_p U_q = U_q U_p,\end{aligned} $$

follows by symmetry. Now if I ⊆{1, …, n}, then the same line of arguments as above yields

$$\displaystyle \begin{aligned} \operatorname*{\Pi}_{i \in I} U_i = (P_{\mathcal{W}} (\operatorname*{\Pi}_{i \in I} V_i) + (\operatorname*{\Pi}_{i \in I^c} V_i^*))|{}_{\mathcal{W}}. \end{aligned} $$

(2.3)

In particular, since $P_{\mathcal {W}} V|{ }_{\mathcal {W}} = 0$, we have that

$$\displaystyle \begin{aligned} \operatorname*{\Pi}_{i=1}^n U_i = I_{\mathcal{W}}.\end{aligned} $$

The following lemma will be crucial in what follow.

Lemma 2.2

Fix 1 ≤ j ≤ n. Let I ⊆{1, …, n}, and let j∉I. Then

$$\displaystyle \begin{aligned} (\operatorname*{\Pi}_{i \in I} U_i^*) P^\perp_{\tilde{\mathcal{W}}_j} (\operatorname*{\Pi}_{i \in I} U_i) = (\operatorname*{\Pi}_{i \in I^c \setminus \{j\}} V_i) (\operatorname*{\Pi}_{i \in I^c \setminus \{j\}} V_i^*)|{}_{\mathcal{W}}- (\operatorname*{\Pi}_{i \in I^c} V_i) (\operatorname*{\Pi}_{i \in I^c} V_i^*)|{}_{\mathcal{W}}. \end{aligned}$$

Proof

Since $P_{\tilde {\mathcal {W}}_j}=I_{\mathcal {W}}-P_{\mathcal {W}}\tilde {V}_j\tilde {V}_j^*|{ }_{\mathcal {W}}$, we have $P^\perp _{\tilde {\mathcal {W}}_j} = P_{\mathcal {W}}\tilde {V}_j \tilde {V}_j^*|{ }_{\mathcal {W}}=\tilde {V}_j \tilde {V}_j^*|{ }_{\mathcal {W}}$. By once again using the fact that $V^*|{ }_{\mathcal {W}} = P_{\mathcal {W}} V|{ }_{\mathcal {W}} = 0$, and by (2.3), one sees that

$$\displaystyle \begin{aligned} (\operatorname*{\Pi}_{i \in I} U_i^*) P^\perp_{\tilde{\mathcal{W}}_j} (\operatorname*{\Pi}_{i \in I} U_i) & = [(\operatorname*{\Pi}_{i \in I} V_i^*) + P_{\mathcal{W}}(\operatorname*{\Pi}_{i \in I^c} V_i)] \tilde{V}_j \tilde{V}_j^* [P_{\mathcal{W}} (\operatorname*{\Pi}_{i \in I} V_i) + (\operatorname*{\Pi}_{i \in I^c} V_i^*)]|{}_{\mathcal{W}} \\ & = (\operatorname*{\Pi}_{i \in I^c\setminus\{j\}} V_i) \tilde{V}_j^* P_{\mathcal{W}} (\operatorname*{\Pi}_{i \in I} V_i)|{}_{\mathcal{W}} \\ &= (\operatorname*{\Pi}_{i \in I^c\setminus\{j\}} V_i)\tilde{V}_j^*(I-VV^*) (\operatorname*{\Pi}_{i \in I} V_i)|{}_{\mathcal{W}} \\ & = (\operatorname*{\Pi}_{i \in I^c\setminus\{j\}} V_i) (\operatorname*{\Pi}_{i \in I^c\setminus \{j\}} V_i^*)|{}_{\mathcal{W}}- (\operatorname*{\Pi}_{i \in I^c} V_i) (\operatorname*{\Pi}_{i \in I^c} V_i^*)|{}_{\mathcal{W}} \end{aligned}$$

This completes the proof of the lemma. ■

Theorem 2.3

If (V ₁, …, V _n) be an n-isometry on a Hilbert space $\mathcal {H}$ , and let U ₁, …, U _n be unitary operators as in Theorem 2.1 . Then

• (a) U _p U _q = U _q U _p for p, q = 1, …n,

• (b) $\prod _{p=1}^{n} U_p =I_{\mathcal {W}}$,

• (c) $(P_{\tilde {\mathcal {W}}_i}^\perp + U_i^* P_{\tilde {\mathcal {W}}_j}^\perp U_i) = (P_{\tilde {\mathcal {W}}_j}^\perp + U_j^* P_{\tilde {\mathcal {W}}_i}^\perp U_j) \leq I_{\mathcal {W}}$ (1 ≤ i < j ≤ n),

• (d) $P^\perp _{\tilde {\mathcal {W}}_1} + U_1^* P^\perp _{\tilde {\mathcal {W}}_2} U_1 + U_1^* U_2^* P^\perp _{\tilde {\mathcal {W}}_2} U_2 U_1 + \cdots + ( \operatorname *{\Pi }_{i=1}^{n-1} U^*_i) P^\perp _{\tilde {\mathcal {W}}_n} ( \operatorname *{\Pi }_{i=1}^{n-1} U_i) = I_{\mathcal {W}}$.

Proof

By Lemma 2.2 applied to I = {p} and j = q, where p, q ∈{1, …, n} and p ≠ q, we have

$$\displaystyle \begin{aligned} U_p^* P^\perp_{\tilde{\mathcal{W}}_q} U_p = (\operatorname*{\Pi}_{i \neq p,q} V_i)(\operatorname*{\Pi}_{i \neq p,q} V_i^*)|{}_{\mathcal{W}}- \tilde{V_p}\tilde{V_p}^*|{}_{\mathcal{W}}, \end{aligned}$$

hence

$$\displaystyle \begin{aligned} (P_{\tilde{\mathcal{W}}_p}^\perp + U_p^* P_{\tilde{\mathcal{W}}_q}^\perp U_p) & =P_{\mathcal{W}}\tilde{V_p}\tilde{V_p}^*|{}_{\mathcal{W}} + (\operatorname*{\Pi}_{i \neq p,q} V_i)(\operatorname*{\Pi}_{i \neq p,q} V_i^*)|{}_{\mathcal{W}}- P_{\mathcal{W}}\tilde{V_p}\tilde{V_p}^*|{}_{\mathcal{W}} \\ &= (\operatorname*{\Pi}_{i \neq p,q} V_i) (\operatorname*{\Pi}_{i \neq p,q} V_i^*)|{}_{\mathcal{W}} \\ &\le I_{\mathcal{W}}. \end{aligned}$$

Therefore by symmetry, we have

$$\displaystyle \begin{aligned} (P_{\tilde{\mathcal{W}}_p}^\perp + U_p^* P_{\tilde{\mathcal{W}}_q}^\perp U_p) = (P_{\tilde{\mathcal{W}}_q}^\perp + U_q^* P_{\tilde{\mathcal{W}}_p}^\perp U_q) \leq I_{\mathcal{W}}. \end{aligned}$$

Finally, we let I _j = {1, …, j − 1} for all 1 < j ≤ n and I _n+1 = {1, …, n}. Then Lemma 2.2 implies that for 1 < j ≤ n,

$$\displaystyle \begin{aligned} (\operatorname*{\Pi}_{i \in I_j} U_i) P^\perp_{\tilde{\mathcal{W}}_j} (\operatorname*{\Pi}_{i \in I_j} U_i^*) = [(\operatorname*{\Pi}_{i \in I_{j+1}^c} V_i) (\operatorname*{\Pi}_{i \in I_{j+1}^c} V_i^*) - (\operatorname*{\Pi}_{i \in I_{j}^c} V_i) (\operatorname*{\Pi}_{i \in I_{j}^c} V_i^*)]|{}_{\mathcal{W}}. \end{aligned}$$

This and $P^\perp _{\tilde {\mathcal {W}}_1} = \tilde {V}_{1} \tilde {V}^*_{1}|{ }_{\mathcal {W}}$ imply that

$$\displaystyle \begin{aligned} P^\perp_{\tilde{\mathcal{W}}_1} + U_1^* P^\perp_{\tilde{\mathcal{W}}_2} U_1 + U_1^* U_2^* P^\perp_{\tilde{\mathcal{W}}_3} U_2 U_1 + \cdots + (\operatorname*{\Pi}_{i=1}^{n-1} U^*_i) P^\perp_{\tilde{\mathcal{W}}_n} (\operatorname*{\Pi}_{i=1}^{n-1} U_i) = I_{\mathcal{W}}. \end{aligned}$$

This completes the proof of the theorem. ■

As a corollary, we have:

Corollary 2.4

Let $\mathcal {H}$ be a Hilbert space and (V ₁, …, V _n) be a pure n-isometry on $\mathcal {H}$ . Let $(M_{\Phi _1}, \ldots , M_{\Phi _n})$ be the pure n-isometry as constructed in Theorem 2.1 , and let $(M_{\Psi _1}, \ldots , M_{\Psi _n})$ on $H^2_{\tilde {\mathcal {W}}}(\mathbb {D})$ , for some Hilbert space $\tilde {\mathcal {W}}$ , unitary operators $\{\tilde {U}_i\}_{i=1}^n$ and orthogonal projections $\{P_i\}_{i=1}^n$ on $\tilde {\mathcal {W}}$ , be a model pure n-isometry. Then:

• (a) $(M_{\Phi _1}, \ldots , M_{\Phi _n})$ is a model pure n-isometry.

• (b) (V ₁, …, V _n) and $(M_{\Phi _1}, \ldots , M_{\Phi _n})$ are unitarily equivalent.

• (c) (V ₁, …, V _n) and $(M_{\Psi _1}, \ldots , M_{\Psi _n})$ are unitarily equivalent if and only if there exists a unitary operator $W : \mathcal {W} \rightarrow \tilde {\mathcal {W}}$ such that $W U_i = \tilde {U}_i W$ and $W P_i = \tilde {P}_i W$ for all i = 1, …, n.

Proof

Parts (a) and (b) follows directly from the previous theorem. The third part is easy and readily follows from Theorem 4.1 in [20] or Theorem 2.9 in [5]. ■

Combining Corollary 2.4 with Theorem 2.3, we have the following characterization of commutative isometric factors of shift operators.

Corollary 2.5

Let $\mathcal {E}$ be a Hilbert space, and let $\{\Phi _i\}_{i=1}^n \subseteq H^\infty _{\mathcal {B}(\mathcal {E})}(\mathbb {D})$ be a commutative family of isometric multipliers. Then

$$\displaystyle \begin{aligned} M_z = \operatorname*{\Pi}_{i = 1}^n M_{\Phi_j},\end{aligned} $$

or, equivalently

$$\displaystyle \begin{aligned} \operatorname*{\Pi}_{i = 1}^n {\Phi_j}(z) = z I_{\mathcal{E}},\quad (z\in\mathbb{D})\end{aligned} $$

if and only if, up to unitary equivalence, $(M_{\Phi _1}, \ldots , M_{\Phi _n})$ is a model pure n-isometry.

In other words, $z I_{\mathcal {E}}$ factors as n commuting isometric multipliers $\{\Phi _i\}_{i=1}^n$ in $H^\infty _{\mathcal {B}(\mathcal {E})}(\mathbb {D})$ if and only if there exist unitary operators $\{U_i\}_{i=1}^n$ on $\mathcal {E}$ and orthogonal projections $\{P_i\}_{i=1}^n$ on $\mathcal {E}$ satisfying the properties (a)–(d) in Theorem 2.3 such that $\Phi _i(z) = U_i(P_i^\perp + z P_i)$ for all i = 1, …, n.

3 Joint Invariant Subspaces

Let $\mathcal {W}$ be a Hilbert space. Let $(M_{\Phi _1}, \ldots , M_{\Phi _n})$ be a model pure n-isometry on $H^2_{\mathcal {W}}(\mathbb {D})$, and let $\mathcal {S}$ be a closed invariant subspace for $(M_{\Phi _1}, \ldots , M_{\Phi _n})$ on $H^2_{\mathcal {W}}(\mathbb {D})$, that is

$$\displaystyle \begin{aligned} M_{\Phi_i} \mathcal{S} \subseteq \mathcal{S}, \end{aligned}$$

for all i = 1, …, n. Then $(M_{\Phi _1}|{ }_{\mathcal {S}}, \ldots , M_{\Phi _n}|{ }_{\mathcal {S}})$ is an n-tuple of commuting isometries on $\mathcal {S}$. Clearly

$$\displaystyle \begin{aligned} \operatorname*{\Pi}_{i=1}^n (M_{\Phi_i}|{}_{\mathcal{S}}) = (\operatorname*{\Pi}_{i=1}^n M_{\Phi_i})|{}_{\mathcal{S}}, \end{aligned}$$

and since

$$\displaystyle \begin{aligned} \operatorname*{\Pi}_{j=1}^n M_{\Phi_j} = M_z, \end{aligned}$$

it follows that

$$\displaystyle \begin{aligned} (\operatorname*{\Pi}_{i=1}^n M_{\Phi_i})|{}_{\mathcal{S}} = M_z|{}_{\mathcal{S}}, \end{aligned} $$

(3.1)

that is, $\mathcal {S}$ is a invariant subspace for M _z on $H^2_{\mathcal {W}}(\mathbb {D})$. Moreover, since $M_z|{ }_{\mathcal {S}}$ is a unilateral shift on $\mathcal {S}$, the tuple $(M_{\Phi _1}|{ }_{\mathcal {S}}, \ldots , M_{\Phi _n}|{ }_{\mathcal {S}})$ is a pure n-isometry on $\mathcal {S}$. Then by Corollary 2.4 there is a model pure n-isometry $(M_{\Psi _1}, \ldots , M_{\Psi _n})$ on $H^2_{\tilde {\mathcal {W}}}(\mathbb {D})$, for some Hilbert space $\tilde {\mathcal {W}}$, such that $(M_{\Phi _1}|{ }_{\mathcal {S}}, \ldots , M_{\Phi _n}|{ }_{\mathcal {S}})$ and $(M_{\Psi _1}, \ldots , M_{\Psi _n})$ are unitarily equivalent. The main purpose of this section is to describe the invariant subspaces $\mathcal {S}$ in terms of the model pure n-isometry $(M_{\Psi _1}, \ldots , M_{\Psi _n})$.

As a motivational example, consider the classical n = 1 case. Here the model pure 1-isometry is the multiplication operator M _z on $H^2_{\mathcal {W}}(\mathbb {D})$ for some Hilbert space $\mathcal {W}$. Let $\mathcal {S}$ be a closed subspace of $H^2_{\mathcal {W}}(\mathbb {D})$. Then by the Beurling [7], Lax [18] and Halmos [16] theorem (or see page 239, Theorem 2.1 in [13]), $\mathcal {S}$ is invariant for M _z if and only if there exist a Hilbert space $\mathcal {W}_*$ and an inner function $\Theta \in H^\infty _{\mathcal {B}(\mathcal {W}_*, \mathcal {W})}(\mathbb {D})$ such that

$$\displaystyle \begin{aligned} \mathcal{S} = \Theta H^2_{\mathcal{W}_*}(\mathbb{D}). \end{aligned}$$

Moreover, in this case, if we set

$$\displaystyle \begin{aligned} V = M_z|{}_{\mathcal{S}}, \end{aligned}$$

then $\mathcal {W}_* = \mathcal {S} \ominus z \mathcal {S}$ and V  on $\mathcal {S}$ and M _z on $H^2_{\mathcal {W}_*}(\mathbb {D})$ are unitarily equivalent. This follows directly from the above representation of $\mathcal {S}$. Indeed, it follows that $X = M_{\Theta } : H^2_{\mathcal {W}_*}(\mathbb {D}) \rightarrow \mbox{ran} M_{\Theta } = \mathcal {S}$ is a unitary operator and

$$\displaystyle \begin{aligned} X M_z = V X. \end{aligned}$$

Now, we proceed with the general case.

Theorem 3.1

Let n > 1. Let $\mathcal {W}$ be a Hilbert space, $(M_{\Phi _1}, \ldots , M_{\Phi _n})$ be a model pure n-isometry on $H^2_{\mathcal {W}}(\mathbb {D})$ , and let $\mathcal {S}$ be a closed subspace of $H^2_{\mathcal {W}}(\mathbb {D})$ . Then $\mathcal {S}$ is invariant for $(M_{\Phi _1}, \ldots , M_{\Phi _n})$ on $H^2_{\mathcal {W}}(\mathbb {D})$ if and only if there exist a Hilbert space $\mathcal {W}_*$ , an inner function $\Theta \in H^\infty _{\mathcal {B}(\mathcal {W}_*, \mathcal {W})}(\mathbb {D})$ and a model pure n-isometry $(M_{\Psi _1}, \ldots , M_{\Psi _n})$ on $H^2_{\mathcal {W}_*}(\mathbb {D})$ such that

$$\displaystyle \begin{aligned} \mathcal{S} = \Theta H^2_{\mathcal{W}_*}(\mathbb{D}), \end{aligned}$$

and

$$\displaystyle \begin{aligned} \Phi_j \Theta = \Theta \Psi_j, \end{aligned}$$

for all j = 1, …, n.

Proof

Let $(M_{\Phi _1}, \ldots , M_{\Phi _n})$ be a model pure n-isometry on $H^2_{\mathcal {W}}(\mathbb {D})$, and let $\mathcal {S}$ be a closed invariant subspace for $(M_{\Phi _1}, \ldots , M_{\Phi _n})$ on $H^2_{\mathcal {W}}(\mathbb {D})$. Let

$$\displaystyle \begin{aligned} \mathcal{W}_* = \mathcal{S} \ominus z \mathcal{S}. \end{aligned}$$

Since $\mathcal {S}$ is an invariant subspace for M _z on $H^2_{\mathcal {W}}(\mathbb {D})$ (see Eq. (3.1)), by Beurling, Lax and Halmos theorem, there exists an inner function $\Theta \in H^\infty _{\mathcal {B}(\mathcal {W}_*, \mathcal {W})}(\mathbb {D})$ such that $\mathcal {S}$ can be represented as

$$\displaystyle \begin{aligned} \mathcal{S} = \Theta H^2_{\mathcal{W}_*}(\mathbb{D}), \end{aligned}$$

If 1 ≤ j ≤ n, then

$$\displaystyle \begin{aligned} \Phi_j \mathcal{S} \subseteq \mathcal{S}, \end{aligned}$$

implies that $\mbox{ran } (M_{\Phi _j} M_{\Theta }) \subseteq \mbox{ran } M_{\Theta }$, and so by Douglas’s range and inclusion theorem [11]

$$\displaystyle \begin{aligned} M_{\Phi_j} M_{\Theta} = M_{\Theta} M_{\Psi_j}, \end{aligned}$$

for some $\Psi _j \in H^{\infty }_{\mathcal {B}(\mathcal {W}_*)}(\mathbb {D})$. Note that $M_{\Phi _j} M_{\Theta }$ is an isometry and ∥ Θ Ψ_j f∥ = ∥ Ψ_j f∥ for each $f \in H^2_{\mathcal {W}_*}(\mathbb {D})$. But then $\|M_{\Psi _j} f\| = \|f\|$ implies that $M_{\Psi _j}$ is an isometry, that is, Ψ_j is an inner function, and hence

$$\displaystyle \begin{aligned} M_{\Psi_j} = M_{\Theta}^* M_{\Phi_j} M_{\Theta}, \end{aligned}$$

for all j = 1, …, n. So

$$\displaystyle \begin{aligned} \operatorname*{\Pi}_{i=1}^n M_{\Psi_i} = (M_{\Theta}^* M_{\Phi_1} M_{\Theta}) \cdots (M_{\Theta}^* M_{\Phi_n} M_{\Theta}). \end{aligned}$$

Now $P_{\text{ran } M_{\Theta }} = M_{\Theta } M_{\Theta }^*$ and ${\Phi _j} \Theta H^2_{\mathcal {W}_*}(\mathbb {D}) \subseteq \Theta H^2_{\mathcal {W}_*}(\mathbb {D})$ implies that

$$\displaystyle \begin{aligned} M_{\Theta} M_{\Theta}^* M_{\Phi_j} M_{\Theta} = M_{\Phi_j} M_{\Theta}, \end{aligned}$$

for all j = 1, …, n. Consequently

$$\displaystyle \begin{aligned} \operatorname*{\Pi}_{j=1}^n M_{\Psi_j} = M_{\Theta}^* (\operatorname*{\Pi}_{j=1}^n M_{\Phi_j}) M_{\Theta}^* = M_{\Theta}^* M_z M_{\Theta} = M_{\Theta}^* M_{\Theta} M_z = M_z, \end{aligned}$$

that is, $(M_{\Psi _1}, \ldots , M_{\Psi _n})$ is a pure n-isometry on $H^2_{\mathcal {W}_*}(\mathbb {D})$. In view of Corollary 2.5, this also implies that the tuple $(M_{\Psi _1}, \ldots , M_{\Psi _n})$ is a model pure n-isometry. This completes the proof of the theorem. ■

The representation of $\mathcal {S}$ is unique in the following sense: if there exist a Hilbert space $\hat {\mathcal {W}}$, an inner multiplier $\hat {\Theta } \in H^\infty _{\mathcal {B}(\hat {\mathcal {W}}, \mathcal {W})}(\mathbb {D})$ and a model pure n-isometry $( M_{\hat {\Psi }_1}, \ldots , M_{ \hat {\Psi }_n})$ on $H^2_{\hat {\mathcal {W}}}(\mathbb {D})$ such that $\mathcal {S} = \hat {\Theta } H^2_{\hat {\mathcal {W}}}(\mathbb {D})$ and $\Phi _i \hat {\Theta } = \hat {\Theta } \hat {\Psi }_i$ for all i = 1, …, n, then there exists a unitary $\tau : \mathcal {W}_* \rightarrow \hat {\mathcal {W}}$ such that

$$\displaystyle \begin{aligned} \Theta = \hat{\Theta} \tau, \end{aligned}$$

and

$$\displaystyle \begin{aligned} \hat{\Psi}_j\tau = \tau \Psi_j \quad \quad (j = 1, \ldots, n). \end{aligned}$$

In other words, the model pure n-isometries $( M_{\hat {\Psi }_1}, \ldots , M_{ \hat {\Psi }_n})$ on $H^2_{\hat {\mathcal {W}}}(\mathbb {D})$ and $(M_{\Psi _1}, \ldots , M_{\Psi _n})$ on $H^2_{\mathcal {W}_*}(\mathbb {D})$ are unitary equivalent (under the same unitary τ). Indeed, the existence of the unitary τ along with the first equality follows from the uniqueness of the Beurling, Lax and Halmos theorem (cf. page 239, Theorem 2.1 in [13]). For the second equality, observe that (see the uniqueness part in [19])

$$\displaystyle \begin{aligned} \hat{\Theta} \tau \Psi_i = \Theta \Psi_i = \Phi_i \Theta = \Phi_i \hat{\Theta} \tau, \end{aligned}$$

that is $\hat {\Theta } \tau \Psi _i = \hat {\Theta } \hat {\Psi }_i\tau$, and so

$$\displaystyle \begin{aligned} \tau \Psi_i =\hat{\Psi}_i \tau, \end{aligned}$$

for all i = 1, …, n.

It is curious to note that the content of Theorem 3.1 is related to the question [1] and its answer [24] on the classifications of invariant subspaces of Γ-isometries. A similar result also holds for invariant subspaces for the multiplication operator tuple on the Hardy space over the unit polydisc in $\mathbb {C}^n$ (see [19]).

Our approach to pure n-isometries has other applications to n-tuples, n ≥ 2, of commuting contractions (cf. see [9]) that we will explore in a future paper.

4 C ^∗-Algebras Generated by Commuting Isometries

In this section, we extend Seto’s result [26] on isomorphic C ^∗-algebras of invariant subspaces of finite codimension in $H^2(\mathbb {D}^2)$ to that in $H^2(\mathbb {D}^n)$, n ≥ 2. Given a Hilbert space $\mathcal {H}$, the set of all compact operators from $\mathcal {H}$ to itself is denoted by $K(\mathcal {H})$. Recall that, for a closed subspace $\mathcal {S} \subseteq H^2(\mathbb {D}^n)$, we say that $\mathcal {S}$ is an invariant subspace of $H^2(\mathbb {D}^n)$ if $M_{z_i} \mathcal {S} \subseteq \mathcal {S}$ for all i = 1, …, n. Also recall that in the case of an invariant subspace $\mathcal {S}$ of $H^2(\mathbb {D}^n)$, $(R_{z_1}, \ldots , R_{z_n})$ is an n-isometry on $\mathcal {S}$ where

$$\displaystyle \begin{aligned} R_{z_i} = M_{z_i}|{}_{\mathcal{S}} \in \mathcal{B}(\mathcal{S}) \quad \quad (i=1, \ldots, n). \end{aligned}$$

Lemma 4.1

If $\mathcal {S}$ is an invariant subspace of finite codimension in $H^2(\mathbb {D}^n)$ , then $K(\mathcal {S}) \subseteq \mathcal T(\mathcal {S})$.

Proof

Since $\mathcal T (\mathcal {S})$ is an irreducible C ^∗-algebra (cf. [26, Proposition 2.2]), it is enough to prove that $\mathcal T (\mathcal {S})$ contains a non-zero compact operator. As

$$\displaystyle \begin{aligned} \operatorname*{\Pi}_{i=1}^n (I_{H^2(\mathbb{D}^n)} - M_{z_i}M_{z_i}^*) = P_{\mathbb C}\in \mathcal{T}(H^2(\mathbb{D}^n)), \end{aligned}$$

we are done when $\mathcal {S} = H^2(\mathbb {D}^n)$. Let us now suppose that $\mathcal {S}$ is a proper subspace of $H^2(\mathbb {D}^n)$. For arbitrary 1 ≤ i < j ≤ n, we have

$$\displaystyle \begin{aligned}{}[R_{z_i}^*, R_{z_j}] = P_{\mathcal{S}}M_{z_i}^*M_{z_j}|{}_{\mathcal{S}}- P_{\mathcal{S}}M_{z_j}P_{\mathcal{S}}M_{z_i}^*|{}_{\mathcal{S}}= P_{\mathcal{S}} M_{z_j} P_{\mathcal{S}^{\perp}} M_{z_i}^*|{}_{\mathcal{S}} \in K(\mathcal{S}), \end{aligned}$$

as $\mathcal {S}^{\perp }$ is finite dimensional. It remains for us to prove that $[R_{z_i}^*, R_{z_j}] \neq 0$ for some 1 ≤ i < j ≤ n. If not, then $\mathcal {S}$ is a proper doubly commuting invariant subspace with finite codimension. As a result, we would have $\mathcal {S} = \varphi H^2(\mathbb {D}^n)$ for some inner function $\varphi \in H^\infty (\mathbb {D}^n)$ ([25]) and hence $\mathcal {S}$ has infinite codimension (see the corollary in page 969, [2]), a contradiction. ■

In what follows, a finite rank operator on a Hilbert space will be denoted by F (without referring to the ambient Hilbert space). Also, if $\mathcal {M}$ is an invariant subspaces of $H^2(\mathbb {D}^n)$, then we set

$$\displaystyle \begin{aligned} R_{z_i}^{\mathcal{M}} = M_{z_i}|{}_{\mathcal{M}} \in \mathcal{B}(\mathcal{M}),\end{aligned} $$

and simply write $R_{z_i}$, i = 1, …, n, when $\mathcal {M}$ is clear from the context.

Lemma 4.2

Suppose $\mathcal {M}_1$ and $\mathcal {M}_2$ are invariant subspaces of $H^2(\mathbb {D}^n)$, $\mathcal {M}_1 \subseteq \mathcal {M}_2$ and $\mathit{\mbox{dim}}(\mathcal {M}_2 \ominus \mathcal {M}_1) < \infty$ . Then $\mathcal {T}(\mathcal {M}_1) = \{P_{\mathcal {M}_1} T|{ }_{\mathcal {M}_1}: T \in \mathcal {T}(\mathcal {M}_2)\}$ . Moreover, if $\mathcal {L}$ is a closed subspace of $\mathcal {M}_1$ and $P^{\mathcal {M}_2}_{\mathcal {L}} \in \mathcal {T}(\mathcal {M}_2)$ , then $P^{\mathcal {M}_1}_{\mathcal {L}} \in \mathcal {T}(\mathcal {M}_1)$.

Proof

Note that $R_{z_i}^{\mathcal {M}_2}|{ }_{\mathcal {M}_1}= R_{z_i}^{\mathcal {M}_1}$ and so, by taking adjoint, we have

$$\displaystyle \begin{aligned} P_{\mathcal{M}_1}(R_{z_i}^{\mathcal{M}_2})^*|{}_{\mathcal{M}_1}= (R_{z_i}^{\mathcal{M}_1})^*, \end{aligned}$$

for all i = 1, …, n. Then $R_{z_i}^{\mathcal {M}_1}(R_{z_j}^{\mathcal {M}_1})^* = P_{\mathcal {M}_1}R_{z_i}^{\mathcal {M}_2}P_{\mathcal {M}_1}^{\mathcal {M}_2}(R_{z_j}^{\mathcal {M}_2})^*|{ }_{\mathcal {M}_1}$, i = 1, …, n. This yields

$$\displaystyle \begin{aligned} R_{z_i}^{\mathcal{M}_1}(R_{z_j}^{\mathcal{M}_1})^* & =P_{\mathcal{M}_1}R_{z_i}^{\mathcal{M}_2} I_{\mathcal{M}_2} (R_{z_j}^{\mathcal{M}_2})^*|{}_{\mathcal{M}_1} - P_{\mathcal{M}_1} R_{z_i}^{\mathcal{M}_2} P_{\mathcal{M}_2 \ominus \mathcal{M}_1}^{\mathcal{M}_2} (R_{z_i}^{\mathcal{M}_2})^*|{}_{\mathcal{M}_1} \\ &=P_{\mathcal{M}_1}R_{z_i}^{\mathcal{M}_2}(R_{z_j}^{\mathcal{M}_2})^*|{}_{\mathcal{M}_1} + F, \end{aligned} $$

for all i, j = 1, …, n, as $\mbox{dim} (\mathcal {M}_2 \ominus \mathcal {M}_1) < \infty$. Similarly $(R_{z_j}^{\mathcal {M}_1})^* R_{z_i}^{\mathcal {M}_1} = P_{\mathcal {M}_1} (R_{z_j}^{\mathcal {M}_2})^* R_{z_i}^{\mathcal {M}_2}|{ }_{\mathcal {M}_1} + F$ for all i, j = 1, …, n. Now let $T_1 \in \mathcal {T}(\mathcal {M}_1)$ be a finite word formed from the symbols

$$\displaystyle \begin{aligned} \{R_{z_i}^{\mathcal{M}_1}, (R_{z_i}^{\mathcal{M}_1})^*: i =1, \ldots, n\}, \end{aligned}$$

and let $T_2 \in \mathcal {T}(\mathcal {M}_2)$ be the same word but formed from the corresponding symbols in

$$\displaystyle \begin{aligned} \{R_{z_i}^{\mathcal{M}_2}, (R_{z_i}^{\mathcal{M}_2})^*: i =1, \ldots, n\}. \end{aligned}$$

Then $T_1 = P_{\mathcal {M}_1} T_2|{ }_{\mathcal {M}_1} + F$. Since both $\mathcal {T}(\mathcal {M}_1)$ and $\{P_{\mathcal {M}_1} T|{ }_{\mathcal {M}_1}: T \in \mathcal {T}(\mathcal {M}_2)\}$ are closed subspaces of $\mathcal {B}(\mathcal {M}_1)$ and both contain all the compact operators in $\mathcal {B}(\mathcal {M}_1)$, it follows that $\mathcal {T}(\mathcal {M}_1) = \{P_{\mathcal {M}_1} T|{ }_{\mathcal {M}_1}: T \in \mathcal {T}(\mathcal {M}_2)\}$. The second assertion now clearly follows from the first one. ■

A thorough understanding of co-doubly commuting invariant subspaces of finite codimension is important to analyze C ^∗-algebras of invariant subspaces of finite codimension in $H^2(\mathbb {D}^n)$. If $\mathcal {S}$ is a closed invariant subspace of $H^2(\mathbb {D})$, then we know that $\mathcal {S} = \theta H^2(\mathbb {D})$ for some inner function $\theta \in H^\infty (\mathbb {D})$. To simplify notations, for a given inner function $\theta \in H^\infty (\mathbb {D})$, we denote

$$\displaystyle \begin{aligned} \mathcal{S}_{\theta} = \theta H^2(\mathbb{D}), \quad \mbox{and} \quad \mathcal{Q}_{\theta} = H^2(\mathbb{D}) \ominus \theta H^2(\mathbb{D}). \end{aligned}$$

Also, given an inner function $\theta _i \in H^\infty (\mathbb {D})$, 1 ≤ i ≤ n, denote by $M_{\theta _i}$ the multiplication operator

$$\displaystyle \begin{aligned} (M_{\theta_i} f)(z_1, \ldots, z_n) = \theta_i(z_i) f(z_1, \ldots, z_n) \end{aligned}$$

for all $f \in H^2(\mathbb {D}^n)$ and $(z_1, \ldots , z_n) \in \mathbb {D}^n$. Recall now that an invariant subspace $\mathcal {S}$ of $H^2(\mathbb {D}^n)$ is said to be co-doubly commuting [23] if $\mathcal {S} = \mathcal {S}_{\Phi }$ where

$$\displaystyle \begin{aligned} \mathcal{S}_{\Phi} =(\mathcal{Q}_{\varphi_1}\otimes\cdots\otimes \mathcal{Q}_{\varphi_n})^{\perp}, \end{aligned} $$

(4.1)

and φ _i, i = 1, …, n, is either inner or the zero function. We warn the reader that the suffix Φ in $\mathcal {S}_{\Phi }$ refers to the finite Blaschke products $\{\Phi _i\}_{i=1}^n$. Here, in view of (4.1) (or see [23]), we have

$$\displaystyle \begin{aligned} (M_{\varphi_p} M_{\varphi_p}^*) (M_{\varphi_q} M_{\varphi_q}^*) = (M_{\varphi_q} M_{\varphi_q}^*) (M_{\varphi_p} M_{\varphi_p}^*), \end{aligned}$$

for all p, q = 1, …, n, and

$$\displaystyle \begin{aligned} P_{\mathcal{S}_{\Phi}} = I_{H^2(\mathbb{D}^n)} - \operatorname*{\Pi}_{i=1}^n (I_{H^2(\mathbb{D}^n)} - M_{\varphi_i} M_{\varphi_i}^*). \end{aligned} $$

(4.2)

It also follows that

$$\displaystyle \begin{aligned} \mathcal{S}_{\Phi} = M_{\varphi_1} H^2(\mathbb{D}^n) + \cdots + M_{\varphi_n} H^2(\mathbb{D}^n). \end{aligned}$$

Therefore, $\mathcal {S}_{\Phi }$ has finite codimension if and only if φ _i is a finite Blashcke product for all i = 1, …, n. Moreover, it can be proved following the same line of argument as Lemma 3.1 in [26] that if $\mathcal {S}$ is an invariant subspace of $H^2(\mathbb {D}^n)$ then $\mathcal {S}$ is of finite codimension if and only if there exist finite Blaschke products φ ₁, …, φ _n such that

$$\displaystyle \begin{aligned} \mathcal{S}_{\Phi} \subseteq \mathcal{S}. \end{aligned}$$

Given $\mathcal {S}_{\Phi }$ as in (4.1) and 1 ≤ i < j ≤ n, we define $\mathcal {Q}_{\Phi }{[i,j]}$ by

$$\displaystyle \begin{aligned} \mathcal{Q}_{\Phi}{[i,j]} = \mathcal{Q}_{\varphi_i}\otimes\mathcal{Q}_{\varphi_{i+1}}\otimes\cdots\otimes\mathcal{Q}_{\varphi_j} \subseteq H^2(\mathbb{D}^{j-i+1}). \end{aligned}$$

Lemma 4.3

Let $\{\Phi _i\}_{i=1}^n$ be finite Blaschke products. If

$$\displaystyle \begin{aligned} \mathcal{L}_1 = {\mathcal{Q}_{\Phi}{[1, n-1]}}^\perp \otimes H^2(\mathbb{D}), \ \mathcal{L}_2 = {\mathcal{Q}_{\Phi}{[1, n-1]}} \otimes \mathcal{S}_{\varphi_n},\end{aligned}$$

$$\displaystyle \begin{aligned} \mathcal{L}_3 = {\mathcal{Q}_{\Phi}{[1, n-1]}} \otimes H^2(\mathbb{D}), \mathcal{L}_2^{\prime} = {\mathcal{Q}_{\Phi}[1, n-1]} \otimes \varphi_n \mathcal{S}_{\varphi_n}\end{aligned}$$

and

$$\displaystyle \begin{aligned} \mathcal{L}_2^{\prime\prime} = {\mathcal{Q}_{\Phi}[1, n-1]} \otimes \varphi_n \mathcal{Q}_{\varphi_n}, \end{aligned}$$

then $P_{\mathcal {L}_1}, P_{\mathcal {L}_2}, P_{\mathcal {L}_2^{\prime }}$ and $P_{\mathcal {L}_2^{\prime \prime }}$ are in $\mathcal T (H^2(\mathbb {D}^n))$ and $P_{\mathcal {L}_1}^{\mathcal {S}_{\Phi }}, P_{\mathcal {L}_2}^{\mathcal {S}_{\Phi }}, P_{\mathcal {L}_2^{\prime }}^{\mathcal {S}_{\Phi }}$ and $P_{\mathcal {L}_2^{\prime \prime }}^{\mathcal {S}_{\Phi }}$ are in $\mathcal T (\mathcal {S}_{\Phi })$.

Proof

Clearly $\mathcal {S}_{\Phi } = \mathcal {L}_1 \oplus \mathcal {L}_2$, $H^2(\mathbb {D}^n) = \mathcal {L}_1 \oplus \mathcal {L}_3$ and $\mathcal {L}_2 = \mathcal {L}_2^{\prime } \oplus \mathcal {L}_2^{\prime \prime }$. By virtue of Lemma 4.2, we only prove the lemma for $H^2(\mathbb {D}^n)$. Since $\mathcal {L}_2^{\prime \prime }$ is finite-dimensional, it follows, by Lemma 4.1, that $P_{\mathcal {L}_2^{\prime \prime }} \in \mathcal {T}(H^2(\mathbb {D}^n))$. Since $\varphi _i \in H^\infty (\mathbb {D})$ is a finite Blaschke product, it follows that φ _i is holomorphic in an open set containing the closure of the disc, and hence $M_{\varphi _i} = \varphi _i(M_{z_i})\in \mathcal {T}(H^2(\mathbb {D}^n))$ for all i = 1, …, n. Then, by (4.2), $P_{\mathcal {S}_{\Phi }} \in \mathcal {T}(H^2(\mathbb {D}^n))$. In view of $\mathcal {S}_{\Phi } = \mathcal {L}_1 \oplus \mathcal {L}_2$, it is then enough to prove only that $P_{\mathcal {L}_2} \in \mathcal {T}(H^2(\mathbb {D}^n))$. This readily follows from the equality

$$\displaystyle \begin{aligned} P_{\mathcal{L}_2} = \Big(\operatorname*{\Pi}_{i=1}^{n-1} (I_{H^2(\mathbb{D}^n)} - M_{\varphi_i} M_{\varphi_i}^*)\Big) M_{\varphi_n} M_{\varphi_n}^*. \end{aligned}$$

This completes the proof of the lemma. ■

In particular, $\mathcal {T}(\mathcal {S}_{\Phi })$ contains a wealth of orthogonal projections. This leads to some further observations concerning the C ^∗-algebra $\mathcal {T}(\mathcal {S}_{\Phi })$. First, given $\mathcal {S}_{\Phi }$ as in (4.1), we consider the unitary operator $U: H^2(\mathbb {D}^n) \rightarrow \mathcal {S}_{\Phi }$ defined by

Then $U = P_{\mathcal {L}_1} + M_{\varphi _n} P_{\mathcal {L}_3}$ and $U^* = P^{\mathcal {S}_{\Phi }}_{\mathcal {L}_1} + M_{\varphi _n}^* P^{\mathcal {S}_{\Phi }}_{\mathcal {L}_2}$. We have the following result:

Theorem 4.4

If $\{\Phi _i\}_{i=1}^n$ are finite Blaschke products, then

$$\displaystyle \begin{aligned} U^*\mathcal{T}(\mathcal{S}_{\Phi}) U = \mathcal T(H^2(\mathbb{D}^n)). \end{aligned}$$

In particular, $\mathcal {T}(\mathcal {S}_{\Phi })$ and $\mathcal T(H^2(\mathbb {D}^n))$ are unitarily equivalent.

Proof

A simple computation first confirms that

$$\displaystyle \begin{aligned} U^* R_{z_n} U= M_{z_n}\in \mathcal T(H^2(\mathbb{D}^n)), \end{aligned}$$

that is

$$\displaystyle \begin{aligned} M_{z_n} \in U^*\mathcal{T}(\mathcal{S}_{\Phi})U \quad \mbox{and} \quad R_{z_n} \in U\mathcal{T}(H^2(\mathbb{D}^n))U^*. \end{aligned}$$

Next, let i = 1, …, n − 1. Then

$$\displaystyle \begin{aligned} R_{z_i} U = M_{z_i} P_{\mathcal{L}_1} + R_{z_i} M_{\varphi_n} P_{\mathcal{L}_3}=M_{z_i} P_{\mathcal{L}_1} + M_{z_i} M_{\varphi_n} P_{\mathcal{L}_3}, \end{aligned}$$

as $M_{\varphi _n} \mathcal {L}_3 = \mathcal {L}_2 \subseteq \mathcal {S}_{\Phi }$, and so

$$\displaystyle \begin{aligned} U^*R_{z_i} U & = (P^{\mathcal{S}_{\Phi}}_{\mathcal{L}_1} + M_{\varphi_n}^* P^{\mathcal{S}_{\Phi}}_{\mathcal{L}_2}) (M_{z_i} P_{\mathcal{L}_1} + M_{z_i} M_{\varphi_n} P_{\mathcal{L}_3}) \\ & = M_{z_i}P_{\mathcal{L}_1} + P_{\mathcal{L}_1} M_{z_i} M_{\varphi_n} P_{\mathcal{L}_3} + M_{\varphi_n}^* P_{\mathcal{L}_2}M_{z_i} M_{\varphi_n} P_{\mathcal{L}_3}, \end{aligned} $$

as $M_{z_i}\mathcal {L}_1\subseteq \mathcal {L}_1$ and $M_{z_i}M_{\varphi _n} \mathcal {L}_3 = M_{z_i}\mathcal {L}_2 \subseteq \mathcal {S}_{\Phi }$. Then $U^*R_{z_i} U\in \mathcal T(H^2(\mathbb {D}^n))$ for al i = 1, …, n, by Lemma 4.3. In particular

$$\displaystyle \begin{aligned} U^*\mathcal T (\mathcal{S}_{\Phi}) U\subseteq \mathcal T (H^2(\mathbb{D}^n)). \end{aligned}$$

On the other hand, since $\mathcal {L}_2 = \mathcal {L}_2^{\prime } \oplus \mathcal {L}_2^{\prime \prime }$ and $\mathcal {L}_2^{\prime \prime }$ is finite dimensional, it follows that $P_{\mathcal {L}_2} = P_{\mathcal {L}^{\prime }_2} + F$, and thus $U^* = U^*|{ }_{\mathcal {L}_1} + U^*|{ }_{\mathcal {L}_2^{\prime }} + F$. Now $U M_{z_i} U^*|{ }_{\mathcal {L}_1} = U M_{z_i}|{ }_{\mathcal {L}_1} = M_{z_i}|{ }_{\mathcal {L}_1}$ as $z_i \mathcal {L}_1 \subseteq \mathcal {L}_1$ and hence

$$\displaystyle \begin{aligned} U M_{z_i} U^*|{}_{\mathcal{L}_1} = R_{z_i}|{}_{\mathcal{L}_1}, \end{aligned}$$

and on the other hand

$$\displaystyle \begin{aligned} U M_{z_i} U^*|{}_{\mathcal{L}^{\prime}_2} = U(M_{z_i} M_{\varphi_n}^*|{}_{\mathcal{L}^{\prime}_2}) = U(M_{z_i} P_{\mathcal{S}_{\Phi}} M_{\varphi_n}^*)|{}_{\mathcal{L}^{\prime}_2} = U (R_{z_i} R_{\varphi_n}^*)|{}_{\mathcal{L}^{\prime}_2}, \end{aligned}$$

where $R_{\varphi _n} = M_{\varphi _n}|{ }_{\mathcal {S}_{\Phi }}$. Moreover, since $\mathcal {L}_3 = \mathcal {L}_2 \oplus \mathcal {S}_{\Phi }^\perp$ and $\mathcal {S}_{\Phi }^\perp$ is finite dimensional, it follows that $P_{\mathcal {L}_3} = P_{\mathcal {L}_2} + F$, and thus

$$\displaystyle \begin{aligned} U M_{z_i} U^*|{}_{\mathcal{L}^{\prime}_2} & = P_{\mathcal{L}_1} R_{z_i} R_{\varphi_n}^*|{}_{\mathcal{L}^{\prime}_2} + M_{\varphi_n} P_{\mathcal{L}_3} R_{z_i} R_{\varphi_n}^*|{}_{\mathcal{L}^{\prime}_2} \\ & = P_{\mathcal{L}_1} R_{z_i} R_{\varphi_n}^*|{}_{\mathcal{L}^{\prime}_2} + M_{\varphi_n} P_{\mathcal{L}_2} R_{z_i} R_{\varphi_n}^*|{}_{\mathcal{L}^{\prime}_2} + F \\ & = P^{\mathcal{S}_{\Phi}}_{\mathcal{L}_1} R_{z_i} R_{\varphi_n}^*|{}_{\mathcal{L}^{\prime}_2} + R_{\varphi_n} P^{\mathcal{S}_{\Phi}}_{\mathcal{L}_2} R_{z_i} R_{\varphi_n}^*|{}_{\mathcal{L}^{\prime}_2} + F, \end{aligned}$$

and hence

$$\displaystyle \begin{aligned} U M_{z_i}U^* = R_{z_i}P^{\mathcal{S}_{\Phi}}_{\mathcal{L}_1} + P^{\mathcal{S}_{\Phi}}_{\mathcal{L}_1} R_{z_i} R_{\varphi_n}^*P^{\mathcal{S}_{\Phi}}_{\mathcal{L}^{\prime}_2}+ R_{\varphi_n} P^{\mathcal{S}_{\Phi}}_{\mathcal{L}_2} R_{z_i} R_{\varphi_n}^*P^{\mathcal{S}_{\Phi}}_{\mathcal{L}^{\prime}_2} + F. \end{aligned}$$

By Lemma 4.3, it follows then that $U M_{z_i}U^* \in \mathcal {T}(\mathcal {S}_{\Phi })$ and so

$$\displaystyle \begin{aligned}U\mathcal T(H^2(\mathbb{D}^n))U^* \subseteq \mathcal T(\mathcal{S}_{\Phi}).\end{aligned}$$

Therefore, the conclusion follows from the fact that $U^* R_{z_n} U= M_{z_n}\in \mathcal T(H^2(\mathbb {D}^n))$. ■

Now let $\mathcal {S}$ be an invariant subspace of finite codimension, and let $\mathcal {S}_{\Phi } \subseteq \mathcal {S}$, as in (4.1), for some finite Blashcke products $\{\Phi _i\}_{i=1}^n$. We proceed to prove that $\mathcal {T}(\mathcal {S})$ is unitarily equivalent to $\mathcal {T}(\mathcal {S}_{\Phi })$. Let

$$\displaystyle \begin{aligned} m :=\dim (\mathcal{S}\ominus \mathcal{S}_{\Phi}). \end{aligned}$$

Observe that

$$\displaystyle \begin{aligned} P_{\mathcal{S}_{\Phi}} = M_{\varphi_1} M_{\varphi_1}^* + (I_{H^2(\mathbb{D}^n)} - M_{\varphi_1} M_{\varphi_1}^*) \Big(I_{H^2(\mathbb{D}^n)} - \operatorname*{\Pi}_{i=2}^n (I_{H^2(\mathbb{D}^n)} - M_{\varphi_i} M_{\varphi_i}^*) \Big), \end{aligned}$$

and so

$$\displaystyle \begin{aligned} \mathcal{S}_{\Phi} = \Big(\mathcal{S}_{\varphi_1}\otimes H^2(\mathbb{D}^{n-1}) \Big) \oplus \Big(\mathcal{Q}_{\varphi_1}\otimes {\mathcal{Q}_{\Phi}[2,n]}^{\perp}\Big). \end{aligned}$$

Lemma 4.5

$P_{\mathcal {S}_{\varphi _1}\otimes H^2(\mathbb {D}^{n-1})}^{\mathcal {S}}, P_{\mathcal {Q}_{\varphi _1}\otimes {\mathcal {Q}_{\Phi }[2,n]}^{\perp }}^{\mathcal {S}} \in \mathcal T(\mathcal {S})$ and

$$\displaystyle \begin{aligned}P_{\mathcal{S}_{\varphi_1}\otimes H^2(\mathbb{D}^{n-1})}^{\mathcal{S}_{\Phi}}, P_{\mathcal{Q}_{\varphi_1}\otimes {\mathcal{Q}_{\Phi}[2,n]}^{\perp}}^{\mathcal{S}_{\Phi}} \in \mathcal T(\mathcal{S}_{\Phi}).\end{aligned}$$

Proof

First one observes that, by virtue of Lemma 4.2, it is enough to prove the result for $\mathcal {S}$. Note that $M_{\varphi _1} \mathcal {S} \subseteq \mathcal {S}$. Define $R_{\varphi _1} \in \mathcal {B}(\mathcal {S})$ by $R_{\varphi _1} = M_{\varphi _1}|{ }_{\mathcal {S}}$. Then $R_{\varphi _1} = \varphi _1(M_{z_1})|{ }_{\mathcal {S}}\in \mathcal T(\mathcal {S})$ and

$$\displaystyle \begin{aligned} P_{M_{\varphi_1} \mathcal{S}} = R_{\varphi_1} R_{\varphi_1}^* \in \mathcal T(\mathcal{S}). \end{aligned}$$

Now on the one hand

$$\displaystyle \begin{aligned} \mathcal{S}_{\varphi_1}\otimes H^2(\mathbb{D}^{n-1}) = M_{\varphi_1} H^2(\mathbb{D}^n) = M_{\varphi_1} \mathcal{S} \oplus \Big(M_{\varphi_1} H^2(\mathbb{D}^n) \ominus M_{\varphi_1} \mathcal{S} \Big), \end{aligned}$$

also, $M_{\varphi _1} H^2(\mathbb {D}^n) \ominus M_{\varphi _1} \mathcal {S} = M_{\varphi _1} (H^2(\mathbb {D}^n) \ominus \mathcal {S})$ is finite dimensional, and hence we conclude $P_{\mathcal {S}_{\varphi _1}\otimes H^2(\mathbb {D}^{n-1})} \in \mathcal T(\mathcal {S})$. This along with $\mbox{dim }(\mathcal {S}\ominus \mathcal {S}_{\Phi }) < \infty$ and the decomposition

$$\displaystyle \begin{aligned} \mathcal{S}=(\mathcal{S}_{\varphi_1}\otimes H^2(\mathbb{D}^{n-1})) \oplus (\mathcal{Q}_{\varphi_1}\otimes {\mathcal{Q}_{\Phi}[2,n]}^{\perp}) \oplus (\mathcal{S}\ominus \mathcal{S}_{\Phi}), \end{aligned}$$

implies that $P_{\mathcal {Q}_{\varphi _1}\otimes {\mathcal {Q}_{\Phi }[2,n]}^{\perp }} \in \mathcal T(\mathcal {S})$. This completes the proof of the lemma. ■

For simplicity, let us introduce some more notation. Given $q \in \mathbb {N}$, let us denote

$$\displaystyle \begin{aligned} \mathbb{C}^{\otimes q} = \mathbb{C} \otimes \cdots \otimes \mathbb{C} \subseteq H^2(\mathbb{D}^q). \end{aligned}$$

Note that $\mathbb {C}^{\otimes q}$ is the one-dimensional subspace consisting of the constant functions in $H^2(\mathbb {D}^q)$. Recalling $\dim (\mathcal {S}\ominus \mathcal {S}_{\Phi }) = m (< \infty )$, we consider the orthogonal decomposition of $\mathcal {S}_{\varphi _1}\otimes H^2(\mathbb {D}^{n-1})$ as:

$$\displaystyle \begin{aligned} \mathcal{S}_{\varphi_1}\otimes H^2(\mathbb{D}^{n-1})= \mathcal{S}_1 \oplus \mathcal{S}_2 \oplus \mathcal{S}_3, \end{aligned}$$

where

$$\displaystyle \begin{aligned} \begin{cases} & \mathcal{S}_1 = (\varphi_1\mathcal{Q}_{z^m})\otimes \mathbb{C}^{\otimes (n-2)} \otimes H^2(\mathbb{D}) \\ & \mathcal{S}_2 = \mathcal{S}_{z^m\varphi_1}\otimes \mathbb{C}^{\otimes (n-2)} \otimes H^2(\mathbb{D}) \\ & \mathcal{S}_3 = \mathcal{S}_{\varphi_1}\otimes (\mathbb{C}^{\otimes (n-2)} )^{\perp}\otimes H^2(\mathbb{D}). \end{cases} \end{aligned}$$

Finally, we define

$$\displaystyle \begin{aligned} \mathcal{L} = \mathcal{S}_2 \oplus \mathcal{S}_3 \oplus \Big(\mathcal{Q}_{\varphi_1}\otimes {\mathcal{Q}_{\Phi}[2,n]}^{\perp}\Big). \end{aligned}$$

With this notation we have

$$\displaystyle \begin{aligned} \mathcal{S}_{\Phi} = \mathcal{S}_1 \oplus \mathcal{L}, \end{aligned}$$

and

$$\displaystyle \begin{aligned} \mathcal{S} =(\mathcal{S} \ominus \mathcal{S}_{\Phi}) \oplus \mathcal{S}_1 \oplus \mathcal{L}. \end{aligned}$$

Lemma 4.6

$P_{\mathcal {S}_i}^{\mathcal {S}} \in \mathcal T(\mathcal {S})$ and $P_{\mathcal {S}_i}^{\mathcal {S}_{\Phi }} \in \mathcal T (\mathcal {S}_{\Phi })$ for all i = 1, 2, 3.

Proof

In view of Lemma 4.2, it is enough to prove that $P_{\mathcal {S}_i}^{\mathcal {S}} \in \mathcal T(\mathcal {S})$, i = 1, 2, 3. Note that $P_{\mathcal {S}_{\varphi _1} \otimes \mathbb {C}^{\otimes (n-2)} \otimes H^2(\mathbb {D})} \in \mathcal {T}(\mathcal {S})$ as

$$\displaystyle \begin{aligned} P_{\mathcal{S}_{\varphi_1} \otimes \mathbb{C}^{\otimes (n-2)} \otimes H^2(\mathbb{D})} = P_{\mathcal{S}_{\varphi_1} \otimes H^2(\mathbb{D}^{n-1})} (I_{\mathcal{S}} - X) P_{\mathcal{S}_{\varphi_1} \otimes H^2(\mathbb{D}^{n-1})}, \end{aligned}$$

where

$$\displaystyle \begin{aligned} X = \sum \limits_{2 \leq i_1 < \cdots < i_k \leq n-1} (-1)^{k+1} R_{z_{i_1}} \cdots R_{z_{i_k}} R_{z_{i_1}}^* \cdots R_{z_{i_k}}^*. \end{aligned}$$

Therefore

$$\displaystyle \begin{aligned} P_{\mathcal{S}_3} = P_{\mathcal{S}_{\varphi_1}\otimes H^2(\mathbb{D}^{n-1})} - P_{\mathcal{S}_{\varphi_1}\otimes \mathbb{C}^{\otimes (n-2)} \otimes H^2(\mathbb{D})} \in \mathcal{T}(\mathcal{S}). \end{aligned}$$

Finally, since $P_{\mathcal {S}_2}= R_{z_1}^m P_{\mathcal {S}_{\varphi _1} \otimes \mathbb {C}^{\otimes (n-2)} \otimes H^2(\mathbb {D})}R_{z_1}^{*m}$ and $\mathcal {S}_1 \oplus \mathcal {S}_2 = \mathcal {S}_{\varphi _1} \otimes \mathbb {C}^{\otimes (n-2)} \otimes H^2(\mathbb {D})$, it follows that $P_{\mathcal {S}_1}$ and $P_{\mathcal {S}_2}$ are in $\mathcal {T}(\mathcal {S})$. ■

Before we proceed to the unitary equivalence of the C ^∗-algebras $\mathcal T (\mathcal {S})$ and $\mathcal T (\mathcal {S}_{\Phi })$ we note that

$$\displaystyle \begin{aligned} \varphi_1\mathcal{Q}_{z^m} = \mbox{span } \{\varphi_1, \varphi_1 z, \ldots, \varphi_1 z^{m-1}\}. \end{aligned}$$

Theorem 4.7

If $\mathcal {S}$ is a finite co-dimensional invariant subspace of $H^2(\mathbb {D}^n)$ and $\mathcal {S}_{\Phi }\subseteq \mathcal {S}$ for some finite Blaschke products $\{\Phi _i\}_{i=1}^n$ , then $\mathcal T (\mathcal {S})$ and $\mathcal T (\mathcal {S}_{\Phi })$ are unitarily equivalent.

Proof

By noting that $H^2(\mathbb {D}) =\mathbb {C} \oplus \mathcal {S}_z$, we decompose $\mathcal {S}_1$ as $\mathcal {S}_1 = \mathcal {F}_1 \oplus \mathcal {M}_1$ where

$$\displaystyle \begin{aligned} \mathcal{F}_1 =(\varphi_1\mathcal{Q}_{z^m}) \otimes \mathbb{C}^{\otimes (n-1)}, \quad \mbox{and} \quad \mathcal{M}_1 = (\varphi_1\mathcal{Q}_{z^m}) \otimes \mathbb{C}^{\otimes (n-2)} \otimes \mathcal{S}_{z}. \end{aligned}$$

Taking into consideration $\mbox{dim} \mathcal {F}_1 = \mbox{dim } (\mathcal {S}\ominus \mathcal {S}_{\Phi })$, we have a unitary $V: \mathcal {F}_1 \to \mathcal {S}\ominus \mathcal {S}_{\Phi }$, and then, using the decompositions

$$\displaystyle \begin{aligned} \mathcal{S}_{\Phi} = \mathcal{F}_1 \oplus \mathcal{M}_1 \oplus \mathcal{L}. \end{aligned}$$

and

$$\displaystyle \begin{aligned} \mathcal{S} =(\mathcal{S} \ominus \mathcal{S}_{\Phi}) \oplus \mathcal{S}_1 \oplus \mathcal{L}, \end{aligned}$$

we see that

$$\displaystyle \begin{aligned} U = \begin{bmatrix} V & 0 & 0 \\ 0 & M_{z_n}^* & 0 \\ 0 & 0 & I_{\mathcal{L}} \end{bmatrix} : \mathcal{F}_1 \oplus \mathcal{M}_1 \oplus \mathcal{L} \rightarrow (\mathcal{S} \ominus \mathcal{S}_{\Phi}) \oplus \mathcal{S}_1 \oplus \mathcal{L}, \end{aligned}$$

defines a unitary from $\mathcal {S}_{\Phi }$ to $\mathcal {S}$. We claim that $U^*\mathcal T (\mathcal {S})U=\mathcal T (\mathcal {S}_{\Phi })$. First we prove that $U^*\mathcal T (\mathcal {S})U \subseteq \mathcal T (\mathcal {S}_{\Phi })$. Since $\mbox{dim} \mathcal {F}_1 < \infty$, it suffices to prove that $U^*R_{z_i}^{\mathcal {S}}U|{ }_{\mathcal {M}_1 \oplus \mathcal {L}} \in \mathcal {T}(\mathcal {S}_{\Phi })$ for all i = 1, ⋯ , n. Observe first that $U \mathcal {M}_1 = M_{z_n}^* \mathcal {M}_1 = \mathcal {S}_1 \subseteq \mathcal {S}_{\Phi }$, $M_{z_n} \mathcal {S}_1 \subseteq \mathcal {S}_1$ and $M_{z_n} \mathcal {L} \subseteq \mathcal {L}$. Since

$$\displaystyle \begin{aligned} U^*R_{z_n}^{\mathcal{S}}U|{}_{\mathcal{M}_1 \oplus \mathcal{L}} = U^* M_{z_n} M_{z_n}^*|{}_{\mathcal{M}_1} + M_{z_n}|{}_{\mathcal{L}}, \end{aligned}$$

and $U^* M_{z_n} M_{z_n}^*|{ }_{\mathcal {M}_1} = M_{z_n}^2 M_{z_n}^*|{ }_{\mathcal {M}_1} = M_{z_n}^2 P_{\mathcal {S}_{\Phi }} M_{z_n}^*|{ }_{\mathcal {M}_1}$, it follows that

$$\displaystyle \begin{aligned} U^*R_{z_n}^{\mathcal{S}}U|{}_{\mathcal{M}_1 \oplus \mathcal{L}} = (R_{z_n}^{\mathcal{S}_{\Phi}})^2 (R_{z_n}^{\mathcal{S}_{\Phi}})^* P^{\mathcal{S}_{\Phi}}_{\mathcal{M}_1} + R_{z_n}^{\mathcal{S}_{\Phi}} P^{\mathcal{S}_{\Phi}}_{\mathcal{L}} \in \mathcal{T}(\mathcal{S}_{\Phi}). \end{aligned}$$

Now for 1 < i < n, we have

$$\displaystyle \begin{aligned} U^*R_{z_i}^{\mathcal{S}}U|{}_{\mathcal{M}_1 \oplus \mathcal{L}} = U^* M_{z_i} M_{z_n}^*|{}_{\mathcal{M}_1} + U^* M_{z_i}|{}_{\mathcal{L}}, \end{aligned}$$

where $U^* M_{z_i} M_{z_n}^*|{ }_{\mathcal {M}_1} = M_{z_i} M_{z_n}^*|{ }_{\mathcal {M}_1}$ as $z_i \mathcal {S}_1 \subseteq \mathcal {S}_3 \subseteq \mathcal {L}$. On the other hand, since $z_i \mathcal {S}_2 \subseteq \mathcal {S}_3$ we have $z_i \mathcal {L} \subseteq \mathcal {L}$ and hence $U^* M_{z_i}|{ }_{\mathcal {L}} = M_{z_i}|{ }_{\mathcal {L}}$, whence

$$\displaystyle \begin{aligned} U^*R_{z_i}^{\mathcal{S}}U|{}_{\mathcal{M}_1 \oplus \mathcal{L}} = R_{z_i}^{\mathcal{S}_{\Phi}} (R_{z_n}^{\mathcal{S}_{\Phi}})^* P^{\mathcal{S}_{\Phi}}_{\mathcal{M}_1} + R_{z_i}^{\mathcal{S}_{\Phi}} P^{\mathcal{S}_{\Phi}}_{\mathcal{L}} \in \mathcal{T}(\mathcal{S}_{\Phi}). \end{aligned}$$

Now we decompose $\mathcal {M}_1$ as $\mathcal {M}_1 = \mathcal {K}_1 \oplus \tilde {\mathcal {K}}_1$ where

$$\displaystyle \begin{aligned} \mathcal{K}_1 = (\varphi_1\mathcal{Q}_{z^{m-1}})\otimes \mathbb{C}^{\otimes (n-2)} \otimes \mathcal{S}_{z} \quad \mbox{and} \quad \tilde{\mathcal{K}}_1 = (\varphi_1 {z^{m-1}} \mathbb{C})\otimes\mathbb{C}^{\otimes (n-2)} \otimes \mathcal{S}_{z}. \end{aligned}$$

Then

$$\displaystyle \begin{aligned} U^*R_{z_1}^{\mathcal{S}}U|{}_{\mathcal{M}_1} = U^*M_{z_1} M_{z_n}^*|{}_{\mathcal{K}_1} + U^*M_{z_1} M_{z_n}^*|{}_{\tilde{\mathcal{K}}_1} = M_{z_n} M_{z_1} M_{z_n}^*|{}_{\mathcal{K}_1} + M_{z_1} M_{z_n}^*|{}_{\tilde{\mathcal{K}}_1}, \end{aligned}$$

as $M_{z_1} M_{z_n}^* \mathcal {K}_1 \subseteq \mathcal {S}_1$ and $M_{z_1} M_{z_n}^* \tilde {\mathcal {K}}_1 \subseteq \mathcal {S}_2$. On the other hand, $U^*R_{z_1}^{\mathcal {S}}U|{ }_{\mathcal {S}_2 \oplus \mathcal {S}_3} = M_{z_1}|{ }_{\mathcal {S}_2 \oplus \mathcal {S}_3}$ as $M_{z_1} (\mathcal {S}_2 \oplus \mathcal {S}_3) \subseteq \mathcal {S}_2 \oplus \mathcal {S}_3 \subseteq \mathcal {L}$, and finally, by denoting $\mathcal {N} = \mathcal {Q}_{\varphi _1}\otimes \mathcal {Q}_{\Phi }{[2,n]}^{\perp }$, we have

$$\displaystyle \begin{aligned} U^*R_{z_1}^{\mathcal{S}}U|{}_{\mathcal{N}} = U^* M_{z_1}|{}_{\mathcal{N}} = U^*(I_{\mathcal{S}} - P^{\mathcal{S}}_{\mathcal{S}_1}) M_{z_1}|{}_{\mathcal{N}} + U^* P^{\mathcal{S}}_{\mathcal{S}_1} M_{z_1}|{}_{\mathcal{N}}. \end{aligned}$$

Then $\mathcal {S} \ominus \mathcal {S}_1 = (\mathcal {S} \ominus \mathcal {S}_{\Phi }) \oplus \mathcal {L}$ and $M_{z_1}\mathcal {N}\subseteq \mathcal {S}_{\Phi }$ implies that

$$\displaystyle \begin{aligned} U^*R_{z_1}^{\mathcal{S}}U|{}_{\mathcal{N}} = P^{\mathcal{S}_{\Phi}}_{\mathcal{L}} M_{z_1}|{}_{\mathcal{N}} + M_{z_n} P^{\mathcal{S}_{\Phi}}_{\mathcal{S}_1} M_{z_1}|{}_{\mathcal{N}}, \end{aligned}$$

and so

$$\displaystyle \begin{aligned} U^*R_{z_1}^{\mathcal{S}}U|{}_{\mathcal{M}_1 \oplus \mathcal{L}} & = R_{z_n}^{\mathcal{S}_{\Phi}} R_{z_1}^{\mathcal{S}_{\Phi}} (R_{z_n}^{\mathcal{S}_{\Phi}})^* P^{\mathcal{S}_{\Phi}}_{\mathcal{K}_1} + R_{z_1}^{\mathcal{S}_{\Phi}} (R_{z_n}^{\mathcal{S}_{\Phi}})^* P^{\mathcal{S}_{\Phi}}_{\tilde{\mathcal{K}}_1} +R_{z_1}^{\mathcal{S}_{\Phi}} P^{\mathcal{S}_{\Phi}}_{\mathcal{S}_2 \oplus \mathcal{S}_3} \\ & \quad + P^{\mathcal{S}_{\Phi}}_{\mathcal{L}} R_{z_1}^{\mathcal{S}_{\Phi}} P^{\mathcal{S}_{\Phi}}_{\mathcal{N}} + R_{z_n}^{\mathcal{S}_{\Phi}} P^{\mathcal{S}_{\Phi}}_{\mathcal{S}_1} R_{z_1}^{\mathcal{S}_{\Phi}} P^{\mathcal{S}_{\Phi}}_{\mathcal{N}} + F. \end{aligned}$$

This implies that $U^*R_{z_1}^{\mathcal {S}}U \in \mathcal {T}(\mathcal {S}_{\Phi })$, and therefore $U^*\mathcal T (\mathcal {S})U \subseteq \mathcal T (\mathcal {S}_{\Phi })$. We now proceed to prove the reverse inclusion $U\mathcal {T}(\mathcal {S}_{\Phi })U^*\in \mathcal {T}(\mathcal {S})$. Since $\mbox{dim} (\mathcal {S}\ominus \mathcal {S}_{\Phi }) < \infty$, it is enough to prove that $UR_{z_i}^{\mathcal {S}_{\Phi }}U^*|{ }_{\mathcal {S}_1 \oplus \mathcal {L}} \in \mathcal {T}(\mathcal {S})$ for all i = 1, …, n. Once again, note that $U^*\mathcal {S}_1=\mathcal {M}_1\subseteq \mathcal {S}_{\Phi }$, $z_n\mathcal {M}_1\subseteq \mathcal {M}_1$, $z_n \mathcal {S}_1 \subseteq \mathcal {S}_1$ and $z_n\mathcal {L}\subseteq \mathcal {L}$. Hence

$$\displaystyle \begin{aligned} UR^{\mathcal{S}_{\Phi}}_{z_n}U^*|{}_{\mathcal{S}_1\oplus\mathcal{L}} = UM_{z_n}^2|{}_{\mathcal{S}_1}+ U M_{z_n}|{}_{\mathcal{L}} = M_{z_n}|{}_{\mathcal{S}_1}+ M_{z_n}|{}_{\mathcal{L}}, \end{aligned}$$

that is

$$\displaystyle \begin{aligned} UR^{\mathcal{S}_{\Phi}}_{z_n}U^*|{}_{\mathcal{S}_1\oplus\mathcal{L}} = R_{z_n}^{\mathcal{S}}P^{\mathcal{S}}_{\mathcal{S}_1\oplus \mathcal{L}}\in \mathcal{T}(\mathcal{S}). \end{aligned}$$

Now, for fixed 1 < i < n, we have $z_i\mathcal {M}_1\subseteq \mathcal {S}_3$ and $z_i\mathcal {L}\subseteq \mathcal {L}$. Then

$$\displaystyle \begin{aligned} U R_{z_i}^{\mathcal{S}_{\Phi}}U^*|{}_{\mathcal{S}_1\oplus\mathcal{L}} & = U M_{z_i}M_{z_n}|{}_{\mathcal{S}_1} + UM_{z_i}|{}_{\mathcal{L}}\\ &=M_{z_i}M_{z_n}|{}_{\mathcal{S}_1} + M_{z_i}|{}_{\mathcal{L}}\\ &= R^{\mathcal{S}}_{z_i}R_{z_n}^{\mathcal{S}}P^{\mathcal{S}}_{\mathcal{S}_1}+ R_{z_i}^{\mathcal{S}}P_{\mathcal{L}} \in \mathcal{T}(\mathcal{S}). \end{aligned} $$

Finally, we consider the decomposition $\mathcal {S}_1= \mathcal {S}_1^{\prime }\oplus \mathcal {S}_1^{\prime \prime }$ where

$$\displaystyle \begin{aligned} \mathcal{S}_1^{\prime} = (\varphi_1\mathcal{Q}_{z^{m-1}})\otimes \mathbb{C}^{\otimes (n-2)} \otimes H^2(\mathbb{D})\ \mbox{and} \ \mathcal{S}_1^{\prime\prime} = (\varphi_1 {z^{m-1}} \mathbb{C})\otimes \mathbb{C}^{\otimes (n-2)} \otimes H^2(\mathbb{D}). \end{aligned}$$

Then

$$\displaystyle \begin{aligned} UR^{\mathcal{S}_{\Phi}}_{z_1}U^*|{}_{\mathcal{S}_1}& = U M_{z_1}M_{z_n}|{}_{\mathcal{S}_1^{\prime}}+ U M_{z_1}M_{z_n}|{}_{\mathcal{S}_1^{\prime\prime}} \\ & =M_{z_n}^*M_{z_1}M_{z_n}|{}_{\mathcal{S}_1^{\prime}} +M_{z_1}M_{z_n}|{}_{\mathcal{S}_1^{\prime\prime}} \\ & =M_{z_1}|{}_{\mathcal{S}_1^{\prime}}+M_{z_1}M_{z_n}|{}_{\mathcal{S}_1^{\prime\prime}}, \end{aligned}$$

as $z_1z_n\mathcal {S}_1^{\prime }\subseteq \mathcal {M}_1$ and $z_1z_n\mathcal {S}_1^{\prime \prime }\subseteq \mathcal {S}_2$. Moreover

$$\displaystyle \begin{aligned} UR^{\mathcal{S}_{\Phi}}_{z_1}U^*|{}_{\mathcal{S}_2\oplus\mathcal{S}_3}= UM_{z_1}|{}_{\mathcal{S}_2\oplus\mathcal{S}_3}=M_{z_1}|{}_{\mathcal{S}_2\oplus\mathcal{S}_3}, \end{aligned}$$

as $z_1(\mathcal {S}_2\oplus \mathcal {S}_3)\subseteq \mathcal {S}_2\oplus \mathcal {S}_3$. From the definition of $\mathcal {N}$, it follows that

$$\displaystyle \begin{aligned} UR^{\mathcal{S}_{\Phi}}_{z_1}U^*|{}_{\mathcal{N}}= UP_{\mathcal{M}_1}^{\mathcal{S}_{\Phi}}M_{z_1}|{}_{\mathcal{N}}+ U(I_{\mathcal{S}_{\Phi}}-P_{\mathcal{M}_1}^{\mathcal{S}_{\Phi}})M_{z_1}|{}_{\mathcal{N}}, \end{aligned}$$

this in turn implies that

$$\displaystyle \begin{aligned}UR^{\mathcal{S}_{\Phi}}_{z_1}U^*|{}_{\mathcal{N}}=M_{z_n}^*P_{\mathcal{M}_1}^{\mathcal{S}}M_{z_1}|{}_{\mathcal{N}} + P_{\mathcal{L}}^{\mathcal{S}}M_{z_1}|{}_{\mathcal{N}}+ F, \end{aligned}$$

as $\mathcal {S}_{\Phi }\ominus \mathcal {M}_1=\mathcal {F}_1\oplus \mathcal {L}$ and $\mathcal {F}_1$ is finite dimensional. Therefore

$$\displaystyle \begin{aligned} UR^{\mathcal{S}_{\Phi}}_{z_1}U^*|{}_{\mathcal{S}_1\oplus \mathcal{L}} & = R_{z_1}^{\mathcal{S}}P^{\mathcal{S}}_{\mathcal{S}_1^{\prime}}+ R^{\mathcal{S}}_{z_1}R^{\mathcal{S}}_{z_n} P^{\mathcal{S}}_{\mathcal{S}_1^{\prime\prime}} + R_{z_1}^{\mathcal{S}}P^{\mathcal{S}}_{\mathcal{S}_2\oplus\mathcal{S}_3} \\ & \quad +(R^{\mathcal{S}}_{z_n})^*P_{\mathcal{M}_1}^{\mathcal{S}}M_{z_1}P^{\mathcal{S}}_{\mathcal{N}} + P_{\mathcal{L}}^{\mathcal{S}}R^{\mathcal{S}}_{z_1}P^{\mathcal{S}}_{\mathcal{N}}+ F \in\mathcal{T}(\mathcal{S}). \end{aligned}$$

This completes the proof of the theorem. ■

On combining Theorems 4.4 and 4.7, we have the following:

Theorem 4.8

If $\mathcal {S}$ is a finite co-dimensional invariant subspace of $H^2(\mathbb {D}^n)$ , then $\mathcal T (\mathcal {S})$ and $\mathcal T (H^2(\mathbb {D}^n))$ are unitarily equivalent.

In the case n = 2, the proof of the above result is considerably simpler and direct than the one by Seto [26] (for instance, if n = 2, then 1 < i < n case does not appear in the proof of Theorem 4.7).