The Joint Spectrum for a Commuting Pair of Isometries in Certain Cases

Abstract

We show that the joint spectrum of two commuting isometries can vary widely depending on various factors. It can range from being small (of measure zero or an analytic disc for example) to the full bidisc. En route, we discover a new model pair in the negative defect case and relate it to the modified bi-shift.

1 Introduction

An isometry V is called pure if $V^{*n}$ converges to 0 strongly as $n \rightarrow \infty$. This is equivalent to saying that V is the unilateral shift of multiplicity equal to the dimension of the range of the defect operator $I - VV^*$.

The famous Wold decomposition [17, 23] tells us that given an isometry V on a Hilbert space ${\mathcal {H}}$, the space ${\mathcal {H}}$ breaks uniquely into a direct sum ${\mathcal {H}} = {\mathcal {H}}_0 \oplus {\mathcal {H}}_0^\perp$ of reducing subspaces such that $V|_{{\mathcal {H}}_0}$ is a unitary and $V|_{{\mathcal {H}}_0^\perp }$ is a pure isometry. This immediately implies that for a non-unitary isometry V (i.e., when the defect operator is positive and not zero), the spectrum $\sigma (V)$ is the closed unit disc $\overline{{\mathbb {D}}}=\{z\in {\mathbb {C}}:|z|\le 1\}.$ The situation for a pair of commuting isometries is vastly different.

The topic of commuting isometries has been vigorously pursued in the last two decades, see [1, 2, 4,5,6, 11, 15, 16, 18, 20, 21] and the references therein. In [6] and [5], the novel idea of using graphs has led to a clear understanding of structures.

The defect operator $C(V_1,V_2)$ is introduced in [12] and [13], as

$$\begin{aligned} C(V_1,V_2)=I-V_1V_1^*-V_2V_2^*+V_1V_2V_2^*V_1^*. \end{aligned}$$

In [13] and [16], the authors provide the characterization of $(V_1,V_2),$ when the defect is positive, negative or zero. It is well known (see [11, 13]) that a pair has positive defect if and only if it is doubly commuting, and it has negative defect if and only if it is dual doubly commuting. In all the three cases, the defect is either a projection or negative of a projection. In general the defect is the difference of two projections; see [16].

In this paper we study the pairs of commuting isometries, whose defect is the difference of two mutually orthogonal projections. We characterize such pairs in Theorem 2.1 and we classify them in Table 1. We also provide the characterization for a few cases in Table 1; see Lemma 6.3 and Lemma 6.9. We rephrase the structure of $(V_1,V_2)$ in each case, which appears in Table 1, in a unified approach using the Berger-Coburn-Lebow (BCL) Theorem. The joint spectrum is studied in detail for all the cases except the last one appearing in Table 1.

There is the related concept of the fringe operators:

$$\begin{aligned} F_1: \ker V_1^*\rightarrow \ker V_1^* \quad \text { and }\quad F_2: \ker V_2^*\rightarrow \ker V_2^* \end{aligned}$$

defined by,

$$\begin{aligned} F_1(x)=P_{\ker V_1^*}V_2(x) \quad \text { and } \quad F_2(x)=P_{\ker V_2^*}V_1(x). \end{aligned}$$

(1.1)

In various characterizations of Table 1, we shall point out the criteria in terms of the fringe operators for possible use in examples.

1.1 The Joint Spectrum

If $(T_1,T_2)$ is a pair of commuting bounded operators on ${\mathcal {H}},$ then for defining (see [14, 22]) the Taylor joint spectrum $\sigma (T_1,T_2),$ one considers the Koszul complex $K(T_1,T_2)$:

$$\begin{aligned} 0\overset{\delta _0}{\rightarrow } {\mathcal {H}}\overset{\delta _1}{\rightarrow }{\mathcal {H}}\oplus {\mathcal {H}}\overset{\delta _2}{\rightarrow }{\mathcal {H}}\overset{\delta _3}{\rightarrow } 0 \end{aligned}$$

(1.2)

where $\delta _1(h)= (T_1h,T_2h)$ for $h\in {\mathcal {H}}$ and $\delta _2(h_1,h_2)= T_1h_2-T_2h_1$ for $h_1,h_2\in {\mathcal {H}}.$ From the way the complex is constructed, ${\text {ran}}\delta _{n-1} \subseteq \ker \delta _{n}.$ When ${\text {ran}}\delta _{n-1}= \ker \delta _{n}$ for all $n=1,2,3$ we say that the Koszul complex $K(T_1,T_2)$ is exact or the pair $(T_1,T_2)$ is non-singular. A pair $(\lambda _1,\lambda _2)\in {\mathbb {C}}^2$ is said to be in the joint spectrum $\sigma (T_1,T_2)$ if the pair $(T_1-\lambda _1I, T_2-\lambda _2I)$ is singular. In the case of a singular pair, we say that the non-singularity breaks at the stage n if ${\text {ran}}\delta _{n-1} \ne \ker \delta _{n}$.

Observe that the non-singularity breaks at stage 1 if and only if $(\lambda _1,\lambda _2)$ is a joint eigenvalue for $(T_1,T_2)$ and the non-singularity breaks at stage 3 if and only if the joint range of $(T_1-\lambda _1I, T_2-\lambda _2I)$ is not the whole space ${\mathcal {H}}.$ If $(\overline{\lambda _1},\overline{\lambda _2})$ is a joint eigenvalue of $(T_1^*,T_2^*)$, then by the fact that ${\text {ran}}T_1 + {\text {ran}}T_2 = {\mathcal {H}}$ implies $\ker T_1^* \cap \ker T_2^* = \{0\}$, the non-singularity of the Koszul complex $K(T_1-\lambda _1I,T_2-\lambda _2I)$ breaks at stage 3. There are a few elementary results which we record as a lemma so that we can refer to it later.

Lemma 1.1

Let ${\mathcal {H}}$ and ${\mathcal {K}}$ be two non-zero Hilbert spaces. Let $(T_1,T_2)$ be a pair of commuting bounded operators on ${\mathcal {H}}.$

• (1) $\sigma (T_1, T_2) \subseteq \sigma (T_1) \times \sigma (T_2)$.

• (2) If there is a non-trivial joint reducing subspace ${\mathcal {H}}_0$ for $(T_1,T_2),$ i.e., if ${\mathcal {H}}={\mathcal {H}}_0\oplus {\mathcal {H}}_0^\perp$ and

  

  then

  $$\begin{aligned} \sigma (T_1,T_2)=\sigma (T_{10},T_{20})\cup \sigma (T_{11},T_{21}). \end{aligned}$$

• (3) $(z_1, z_2) \in \sigma (T_1, T_2)$ if and only if $(\overline{z_1}, \overline{z_2}) \in \sigma (T_1^*, T_2^*)$.

• (4) $\sigma (I_{\mathcal {K}}\otimes T_1,I_{\mathcal {K}}\otimes T_2)=\sigma (T_1, T_2)=\sigma (T_1\otimes I_{\mathcal {K}},T_2\otimes I_{\mathcal {K}}).$

• (5) Let $(S_1,S_2)$ be a pair of commuting bounded operators on a Hilbert space ${\mathcal {K}}.$ If $(T_1,T_2)$ is jointly unitarily equivalent to $(S_1,S_2),$ then $\sigma (T_1,T_2)=\sigma (S_1,S_2).$

• (6) For any T in and S in , the joint spectrum $\sigma (T \otimes I_{\mathcal {K}}, I_{\mathcal {H}}\otimes S)$ is the Cartesian product $\sigma (T) \times \sigma (S)$.

Thus, for commuting isometries $V_1$ and $V_2$, we have $\sigma (V_1, V_2) \subseteq \overline{{{\mathbb {D}}}^2}$. The joint spectrum of a pair of commuting unitary operators is contained in the torus ${\mathbb {T}}^2,$ where ${\mathbb {T}}=\{z\in {\mathbb {C}}: |z|=1\}$ is the unit circle in the complex plane.

This note uses the fundamental pairs of isometries consisting of multiplication operators to describe the structure of $(V_1,V_2).$ These are fundamental in the sense that in each case the sign of the defect operator is dictated by the fundamental pair alone.

When the defect operator $C(V_1,V_2)$ of two commuting isometries is positive or negative, but not zero, then the whole space ${\mathcal {H}}$ breaks into a direct sum of reducing subspaces ${\mathcal {H}}= {\mathcal {H}}_0 \oplus {\mathcal {H}}_0^\perp$ in the style of Wold where the restriction of $(V_1,V_2)$ on the ${\mathcal {H}}_0$ part is the fundamental pair and the restriction of $(V_1,V_2)$ on the ${\mathcal {H}}_0^\perp$ part has defect zero. What the fundamental pair is depends on whether $C(V_1,V_2)$ is positive or negative. These are the contents of Theorem 4.11 and Theorem 5.10.

The fundamental pairs are such that in both cases (of $C(V_1,V_2)$ positive or negative), the joint spectrum of $(V_1,V_2)$ is the whole closed bidisc $\overline{{\mathbb {D}}^2}$. These are done in Theorem 4.11 and Theorem 5.11. If the defect operator $C(V_1, V_2)$ is zero, the joint spectrum of $(V_1,V_2)$ is contained in the topological boundary of the bidisc.

The structure theorem in the case ${\text {ran}}V_1={\text {ran}}V_2$, shows that $(V_1, V_2)$ is the direct sum of a prototypical pair (see Sect. 6.1.1) and a pair of commuting unitaries. The joint spectrum is computed. The joint spectrum of the prototypical pair of this case, is neither the closed bidisc nor contained inside the topological boundary of the bidisc.

In the case ${\text {ran}}V_2\subsetneq {\text {ran}}V_1,$ the joint spectrum $\sigma (V_1,V_2)\subseteq \{(z_1,z_1z_2): z_1,z_2\in \overline{{\mathbb {D}}}\}.$ The above inclusion is sharp; see Example 6.12, and it can be a strict inclusion; see Example 6.13. Note that $\{(z_1,z_1z_2): z_1,z_2\in \overline{{\mathbb {D}}}\}$ has measure non-zero and it is not equal to the closed bidisc.

In each case above except the case ${\text {ran}}V_2\subsetneq {\text {ran}}V_1,$ we point out the stage of the Koszul complex where non-singularity is broken.

1.2 The Berger-Coburn-Lebow Theorem

For a Hilbert space ${\mathcal {E}}$, the Hardy space of ${\mathcal {E}}$-valued functions on the unit disc in the complex plane is

$$\begin{aligned} H^2_{{\mathbb {D}}}({\mathcal {E}}) = \left\{ f : {\mathbb {D}} \rightarrow {\mathcal {E}} \mid f \text { is analytic and } f(z) = \sum _{n=0}^{\infty } a_n z^n \text { with } \sum _{n=0}^{\infty } \Vert a_n\Vert _{\mathcal {E}}^2 < \infty \right\} . \end{aligned}$$

Here the $a_n$ are from ${\mathcal {E}}$. This is a Hilbert space with the inner product

$$\begin{aligned} \left\langle \sum _{n=0}^{\infty } a_n z^n , \sum _{n=0}^{\infty } b_n z^n \right\rangle = \sum _{n=0}^{\infty } \langle a_n , b_n \rangle _{\mathcal {E}}\end{aligned}$$

and is identifiable with $H^2_{{\mathbb {D}}} \otimes {\mathcal {E}}$ where $H^2_{{{\mathbb {D}}}}$ stands for the Hardy space of scalar-valued functions on ${\mathbb {D}}$. We shall use this identification throughout the paper, often without any further mention, and $M_z$ denotes the multiplication operator by the coordinate function z on $H^2_{{\mathbb {D}}}.$ For $\lambda \in {{\mathbb {D}}}$, let $k_\lambda$ be the function in $H^2_{{{\mathbb {D}}}}$ given by

$$\begin{aligned} k_\lambda (z) = \sum _{n=0}^\infty z^n {\overline{\lambda }}^n = \frac{1}{1 - z{\overline{\lambda }}}. \end{aligned}$$

It is well-known that the span of $\{ k_\lambda : \lambda \in {{\mathbb {D}}}\}$ is dense in $H^2_{{{\mathbb {D}}}}$.

The space of -valued bounded analytic functions on ${{\mathbb {D}}}$ will be denoted by . Naturally, if , then it induces a multiplication operator $M_\varphi$ on $H^2_{{\mathbb {D}}}({\mathcal {E}})$. One of the main tools for us is the Berger-Coburn-Lebow (BCL) theorem [3].

Theorem 1.2

Let $(V_1,V_2)$ be a commuting pair of isometries acting on ${\mathcal {H}}.$ Then, up to unitary equivalence, the Hilbert space ${\mathcal {H}}$ breaks into a direct sum of reducing subspaces ${\mathcal {H}}={\mathcal {H}}_p\oplus {\mathcal {H}}_u$ such that

• (1) There is a unique (up to unitary equivalence) triple $({\mathcal {E}}, P, U)$ where ${\mathcal {E}}$ is a Hilbert space, P is a projection on ${\mathcal {E}}$ and U is a unitary on ${\mathcal {E}}$ such that ${\mathcal {H}}_p=H^2_{{{\mathbb {D}}}}({\mathcal {E}}),$ the functions $\varphi _1$ and $\varphi _2$ defined on ${{\mathbb {D}}}$ by

  $$\begin{aligned} \varphi _1(z)=U^*(P^\perp +zP) \quad \text { and }\quad \varphi _2(z)=(P+zP^\perp )U, \end{aligned}$$

  (1.3)

  are commuting multipliers in and $(V_1|_{{\mathcal {H}}_p},V_2|_{{\mathcal {H}}_p})$ is equal to $(M_{\varphi _1},M_{\varphi _2}).$

• (2) $V_1|_{{\mathcal {H}}_u}$ and $V_2|_{{\mathcal {H}}_u}$ are commuting unitary operators.

The result of the theorem above will be called the BCL representation of $(V_1, V_2)$. Using Theorem 1.2 we can compute the defect operator (see [13]), because $M_{\varphi _1}=I_{H^2_{{{\mathbb {D}}}}}\otimes U^*P^{\perp }+ M_z \otimes U^*P$ and $M_{\varphi _2}=I_{H^2_{{{\mathbb {D}}}}}\otimes PU+ M_z \otimes P^\perp U.$ Hence

$$\begin{aligned} C(M_{\varphi _1}, M_{\varphi _2})= (I- M_zM_z^*)\otimes (U^*PU-P)= E_0 \otimes (U^*PU-P) \end{aligned}$$

where $E_0$ is the one dimensional projection onto the space of constant functions in $H^2_{{\mathbb {D}}}$. Together with the fact that the defect operator of a pair of commuting unitary operators is zero, this means, in the decomposition ${\mathcal {H}}={\mathcal {H}}_p\oplus {\mathcal {H}}_u$ with ${\mathcal {H}}_p=H^2_{{{\mathbb {D}}}}({\mathcal {E}}),$

$$\begin{aligned} C(V_1, V_2) = (E_0\otimes (U^*PU-P))\oplus 0. \end{aligned}$$

(1.4)

Definition 1.3

• (1) A BCL triple $({\mathcal {E}},P,U)$ is a Hilbert space ${\mathcal {E}},$ along with a projection P and a unitary U. It is said to be the BCL triple for the pair of commuting isometries $(V_1,V_2)$ if ${\mathcal {E}},P$ and U are as in Theorem 1.2, part (1).

• (2) Given a BCL triple $({\mathcal {E}},P,U),$ the functions will always be as defined in (1.3).

• (3) A pair of commuting isometries $(V_1,V_2)$ is called pure if ${\mathcal {H}}_u = \{0\}$ in its BCL representation.

In fact, ${\mathcal {H}}_u$ is the unitary part of the product $V=V_1V_2$ in its Wold decomposition and ${\mathcal {E}}=\ker V^*$. Thus ${\mathcal {H}}_u=\{0\}$ if and only if V is pure.

The Berger-Coburn-Lebow theorem has an interesting consequence in the case when the part ${\mathcal {H}}_p$ is non-zero. We have $\varphi _1(z)\varphi _2(z) = z$ for every $z \in {\mathbb {D}}$ and hence $M_{\varphi _1} M_{\varphi _2} = M_z\otimes I_{\mathcal {E}}$. Hence, by the spectral mapping theorem for joint spectra (see for example [8]),

$$\begin{aligned} \{ z_1z_2 : (z_1, z_2) \in \sigma (M_{\varphi _1}, M_{\varphi _2}) \} = \sigma (M_z\otimes I_{\mathcal {E}}) = \overline{{\mathbb {D}}}. \end{aligned}$$

Thus, if ${\mathcal {H}}_p \ne \{0\}$, then $\sigma (M_{\varphi _1}, M_{\varphi _2})$ cannot be contained in the torus ${\mathbb {T}}^2$. Hence by Lemma 1.1 part (2), $\sigma (V_1,V_2)$ cannot be contained in the torus ${\mathbb {T}}^2.$

Let $({\mathcal {E}},P,U)$ be a BCL triple. It is easy to see that if the Koszul complex $K(\varphi _1(z)-\lambda _1I,\varphi _2(z)-\lambda _2I)$ breaks at stage 3 for some $z\in {{\mathbb {D}}},$ then the Koszul complex $K(M_{\varphi _1}-\lambda _1I,M_{\varphi _2}-\lambda _2I)$ breaks at stage 3. More generally, we show that, in all the cases under consideration in this note except the case ${\text {ran}}V_2\subsetneq {\text {ran}}V_1,$ we have

$$\begin{aligned} \overline{\cup _{z\in {{\mathbb {D}}}} \sigma (\varphi _1(z),\varphi _2(z))}=\sigma (M_{\varphi _1},M_{\varphi _2}). \end{aligned}$$

(1.5)

See Theorem 3.8, Theorem 4.15, Theorem 5.13 and Theorem 6.8.

In the following, when we consider the pair $(V_1,V_2)$ of commuting isometries, V denotes the product $V_1V_2,$ and $\{e_n:n\in {\mathbb {Z}}\}$ denotes the standard orthonormal basis of $l^2({\mathbb {Z}}).$

We end this section with the comment that one of the strong points of the BCL theorem is that it models the $V_i$ in terms of functions of one variable whereas $V_i$ could, a priori, be dependent on two variables (multipliers on the Hardy space of the bidisc, for example). This strength will be greatly exploited in this note.

2 The Defect Operator

Recall that ([12] and [13]) the defect operator of a pair of commuting isometries $(V_1,V_2)$ is defined as

$$\begin{aligned} C(V_1,V_2)=I-V_1V_1^*-V_2V_2^*+V_1V_2V_2^*V_1^*. \end{aligned}$$

(2.1)

It is easy to see that (see [16]) the defect

$$\begin{aligned} C(V_1,V_2) =P_{\ker V_1^*}-P_{V_2(\ker V_1^*)}=P_{\ker V_2^*}-P_{V_1(\ker V_2^*)}, \end{aligned}$$

(2.2)

and

$$\begin{aligned} \ker V_1^*\oplus V_1(\ker V_2^*)=\ker V_2^*\oplus V_2(\ker V_1^*)=\ker V_1^*V_2^*. \end{aligned}$$

(2.3)

Note that the defect operator lives on $\ker V^*$ in the sense that the defect operator is zero on the orthogonal component of $\ker V^*.$ Equation (2.2) shows that the defect is always a difference of two projections. Let

$$\begin{aligned} P_1=P_{\ker V_1^*}\quad \text { and }\quad P_2=P_{V_2(\ker V_1^*)}. \end{aligned}$$

(2.4)

Define

$$\begin{aligned} {\mathcal {H}}_1&:={\text {ran}}P_1\cap \ker P_2=\ker V_1^*\cap \ker V_2^*, \\ {\mathcal {H}}_2&:={\text {ran}}P_2\cap \ker P_1=V_1(\ker V_2^*)\cap V_2(\ker V_1^*),\\ {\mathcal {H}}_3&:={\text {ran}}P_1\cap {\text {ran}}P_2=\ker V_1^*\cap V_2(\ker V_1^*), \\ {\mathcal {H}}_4&:=\ker P_1\cap \ker P_2=V_1(\ker V_2^*)\cap \ker V_2^*. \end{aligned}$$

Notice that ${\mathcal {H}}_i\perp {\mathcal {H}}_j$ if $i\ne j,$ and the ${\mathcal {H}}_i$ are reducing for $P_1$ and $P_2$ and

$$\begin{aligned} {\mathcal {H}}_1\oplus {\mathcal {H}}_2\oplus {\mathcal {H}}_3\oplus {\mathcal {H}}_4\subseteq \ker V^*. \end{aligned}$$

(2.5)

Theorem 2.1

The following are equivalent:

• (a) The defect $C(V_1,V_2)$ is the difference of two mutually orthogonal projections.

• (b) $\ker V^*={\mathcal {H}}_1\oplus {\mathcal {H}}_2\oplus {\mathcal {H}}_3\oplus {\mathcal {H}}_4.$

• (c) $\ker V_1^*={\mathcal {H}}_1\oplus {\mathcal {H}}_3.$

• (d) $V_1(\ker V_2^*)={\mathcal {H}}_2\oplus {\mathcal {H}}_4.$

• (e) If $({\mathcal {E}},P,U)$ is the BCL triple for $(V_1,V_2),$

  $$\begin{aligned} U^*({\text {ran}}P)=(U^*({\text {ran}}P)\cap {\text {ran}}P) \oplus (U^*({\text {ran}}P)\cap {\text {ran}}P^\perp ). \end{aligned}$$

  (2.6)

Proof

Suppose

$$\begin{aligned} C(V_1,V_2)=Q_1-Q_2 \quad \text { with } \quad {\text {ran}}Q_1\perp {\text {ran}}Q_2 \end{aligned}$$

(2.7)

for some projections $Q_1,Q_2$ in $\ker V^*.$ Notice that the pair $(Q_1,Q_2)$ satisfying (2.7) is unique if it exists. Then for such a pair $(V_1,V_2)$ we have

$$\begin{aligned} C(V_1,V_2)=\begin{pmatrix} I_{{\text {ran}}Q_1}&{}0&{}0\\ 0&{}-I_{{\text {ran}}Q_2}&{}0\\ 0&{}0&{}0 \end{pmatrix}=P_1-P_2. \end{aligned}$$

(2.8)

For any two projections P and Q in ${\mathcal {H}}$ and $x\in {\mathcal {H}},$ $Px-Qx=x$ implies that $Px=x$ and $Qx=0.$ Using this fact and (2.8) we see that:

$$\begin{aligned} {\mathcal {H}}_1={\text {ran}}Q_1 \quad \text { and }\quad {\mathcal {H}}_2={\text {ran}}Q_2. \end{aligned}$$

Note that $P_1=P_2=\begin{pmatrix} I_{\mathcal {K}}&{}0\\ 0&{}0_{\mathcal {L}}\end{pmatrix}$ on ${\mathcal {K}}\oplus {\mathcal {L}}$ for some ${\mathcal {K}},{\mathcal {L}}$ that satisfy $\ker V^*= {\mathcal {H}}_1\oplus {\mathcal {H}}_2\oplus {\mathcal {K}}\oplus {\mathcal {L}}.$ Hence, by (2.8) ${\text {ran}}P_1={\mathcal {H}}_1\oplus {\mathcal {K}}$ and ${\text {ran}}P_2={\mathcal {H}}_2\oplus {\mathcal {K}}.$ Thus ${\mathcal {K}}={\text {ran}}P_1\cap {\text {ran}}P_2={\mathcal {H}}_3.$ Similarly ${\mathcal {L}}=\ker P_1\cap \ker P_2={\mathcal {H}}_4.$ Therefore

$$\begin{aligned} \ker V^*={\mathcal {H}}_1\oplus {\mathcal {H}}_2\oplus {\mathcal {H}}_3\oplus {\mathcal {H}}_4, \end{aligned}$$

and in this decomposition

$$\begin{aligned} P_1=\begin{pmatrix} I&{}0&{}0&{}0\\ 0&{}0&{}0&{}0\\ 0&{}0&{}I&{}0\\ 0&{}0&{}0&{}0 \end{pmatrix}, \quad P_2=\begin{pmatrix} 0&{}0&{}0&{}0\\ 0&{}I&{}0&{}0\\ 0&{}0&{}I&{}0\\ 0&{}0&{}0&{}0 \end{pmatrix}. \end{aligned}$$

(2.9)

This proves $(a)\implies (b),$ and $(b)\implies (a)$ is trivial.

Note the fact that, if ${\mathcal {E}}_1,{\mathcal {E}}_2,{\mathcal {E}}_3$ and ${\mathcal {E}}_4$ are any subspaces of ${\mathcal {E}}$ satisfying ${\mathcal {E}}_1\oplus {\mathcal {E}}_2={\mathcal {E}}={\mathcal {E}}_3\oplus {\mathcal {E}}_4,$ then ${\mathcal {E}}_1=({\mathcal {E}}_1\cap {\mathcal {E}}_3)\oplus ({\mathcal {E}}_1\cap {\mathcal {E}}_4)$ if and only if ${\mathcal {E}}_2=({\mathcal {E}}_2\cap {\mathcal {E}}_3)\oplus ({\mathcal {E}}_2\cap {\mathcal {E}}_4).$ Using the above fact the equivalence $(c)\iff (d)$ follows from (2.3).

The implication $(b)\implies (c)$ is clear. The implication $(c)\implies (b)$ follows from the equivalence of (c) and (d) and (2.3).

Note that $\ker M_{\varphi _1}^*={\text {ran}}(I-M_{\varphi _1}M_{\varphi _1}^*)=1\otimes {\text {ran}}(U^*PU)=1\otimes U^*({\text {ran}}P),$ $\ker M_{\varphi _2}^*=1\otimes {\text {ran}}P^\perp$ and $M_{\varphi _2}(\ker M_{\varphi _1}^*)= 1\otimes {\text {ran}}P.$ Now the proof of $(a)\iff (e)$ follows from the equivalence $(a)\iff (c).$ $\square$

Using the (e) part of the above theorem we easily get an example of a pair $(V_1,V_2)$ whose defect is not a difference of two mutually orthogonal projections.

Example 2.2

Let ${\mathcal {E}}={\mathbb {C}}^2,U=\frac{1}{\sqrt{2}}\begin{pmatrix} 1&{}1\\ -1&{}1 \end{pmatrix}$ and $P=\begin{pmatrix} 1&{}0\\ 0&{}0 \end{pmatrix}.$ The BCL triple $({\mathcal {E}},P,U)$ does not satisfy (2.6), hence the defect of the pair $(V_1,V_2)$ corresponding to this BCL triple, is not the difference of two mutually orthogonal projections.

The Table 1 gives a neat classification. We leave the proofs to the reader for the first two columns. The contents of the third column will unfold as we progress. One can notice that certain cases are not mentioned in the table. That is because those cases cannot occur.

Table 1 Classification

3 The Zero Defect Case

3.1 Structure

This subsection is mainly a rephrasing of known results. If one of the $V_i$’s is a unitary, then it is trivial to check that the defect $C(V_1, V_2)$ is zero. The following example is the prototypical example of a pure pair of commuting isometries with defect zero; this example serves as a building block in the general structure, see Theorem 3.4.

Example 3.1

Let ${\mathcal {L}}$ be a non-zero Hilbert space and W be a unitary on ${\mathcal {L}}.$ Consider the commuting pair of isometries $(M_{z}\otimes I_{\mathcal {L}}, I_{H^2_{{\mathbb {D}}}}\otimes W)$ on $H^2_{{{\mathbb {D}}}} \otimes {\mathcal {L}}$. As $I\otimes W$ is a unitary, the defect $C(M_{z}\otimes I_{\mathcal {L}}, I_{H^2_{{\mathbb {D}}}}\otimes W)=0.$ Also, $\sigma (M_{z}\otimes I_{\mathcal {L}}, I_{H^2_{{\mathbb {D}}}}\otimes W)=\overline{{\mathbb {D}}}\times \sigma (W),$ by Lemma 1.1 part (6). Also see [7].

Now consider the unitary $\Lambda :H^2_{{{\mathbb {D}}}}\otimes {\mathcal {L}}\rightarrow H^2_{{{\mathbb {D}}}}\otimes {\mathcal {L}}$ given by

$$\begin{aligned} \Lambda \left( \sum _{m= 0}^\infty a_mz^m\right) =\sum _{m= 0}^\infty W^m(a_m)z^m. \end{aligned}$$

(3.1)

Then, $\Lambda (M_z\otimes W^*)\Lambda ^*=M_z\otimes I$ and $\Lambda (I\otimes W)\Lambda ^*=I\otimes W.$ This says that $(M_z\otimes W^*,I\otimes W)$ and $(M_z\otimes I,I\otimes W)$ are jointly unitarily equivalent. In particular, we have $C(M_z\otimes W^*,I\otimes W)=0$ and

$$\begin{aligned} \sigma (M_z\otimes W^*,I\otimes W)=\overline{{\mathbb {D}}}\times \sigma (W) \end{aligned}$$

(3.2)

for any unitary W.

We proceed towards the structure of an arbitrary pair $(V_1, V_2)$ with $C(V_1, V_2) = 0$. A pair of commuting isometries $(V_1, V_2)$ is called doubly commuting if $V_1$ commutes with $V_2^*$. The following lemma is proved in [13] and [16], it is relating positivity of the defect operator $C(V_1,V_2)$ with double commutativity of the pair $(V_1,V_2).$ We give a short proof here.

Lemma 3.2

Let $(V_1,V_2)$ be a pair of commuting isometries on a Hilbert space ${\mathcal {H}}$. Then the following are equivalent:

• (a) $C(V_1,V_2)\ge 0.$

• (b) $(V_1,V_2)$ is doubly commuting.

• (c) If $({\mathcal {E}},P,U)$ is the BCL triple for $(V_1,V_2),$ then $U({\text {ran}}P)\subseteq {\text {ran}}P.$

Proof

Since commuting unitaries are always doubly commuting, it is enough to prove the equivalences when $(V_1,V_2) = (M_{\varphi _1},M_{\varphi _2})$. By virtue of (1.4), we have $C(V_1,V_2)\ge 0$ if and only if $U^*PU \ge P$ which happens if and only if ${\text {ran}}P$ is invariant under U. Now,

$$\begin{aligned} M_{\varphi _1}M_{\varphi _2}^*=M_{\varphi _2}^*M_{\varphi _1}&\text { if and only if } (I-M_zM_z^*)\otimes (U^*PU^*P^\perp )=0 \\&\text { if and only if } P^\perp UP=0\\&\text { if and only if } {\text {ran}}P \text { is invariant under } U. \end{aligned}$$

This completes the proof. $\square$

We recall some characterization results from [13] and add a few new ones.

Lemma 3.3

Let $(V_1,V_2)$ be a pair of commuting isometries on a Hilbert space ${\mathcal {H}}.$ Then the following are equivalent:

• (a) $C(V_1,V_2)=0.$

• (b) $\ker V_2^*$ is a reducing subspace for $V_1$ and $V_1|_{\ker V_2^*}$ is a unitary.

• (c) $\ker V_1^*$ is a reducing subspace for $V_2$ and $V_2|_{\ker V_1^*}$ is a unitary.

• (d) The fringe operators $F_{1}$ and $F_{2}$ are unitaries.

• (e) $\ker V_1^*$ and $\ker V_2^*$ are orthogonal and their direct sum is $\ker V^*.$

• (f) $({\text {ran}}V_1\ominus {\text {ran}}V)\oplus ({\text {ran}}V_2\ominus {\text {ran}}V)\oplus {\text {ran}}V={\mathcal {H}}.$

• (g) If $({\mathcal {E}},P,U)$ is the BCL triple for $(V_1,V_2),$ then ${\text {ran}}P$ reduces U.

Proof

The equivalences of (a),  (b),  (c),  (d) and (e) follows easily from (2.2) and (2.3).

$(e)\Rightarrow (f)$: Suppose (e) is true. We shall show that $\ker V_1^*=({\text {ran}}V_2\ominus {\text {ran}}V).$ Suppose $x\in \ker V_1^*$, which implies $x\in {\text {ran}}V_2$ and $x\in {({\text {ran}}V)}^{\perp }.$ So $x\in ({\text {ran}}V_2\ominus {\text {ran}}V).$ If $x\in ({\text {ran}}V_2\ominus {\text {ran}}V),$ then $x\in {\text {ran}}V_2$ and $x\in \ker V^*.$ So $x\in \ker V_1^*$. Similarly, $\ker V_2^*= {\text {ran}}V_1\ominus {\text {ran}}V.$ Hence (f) is true.

$(f) \Rightarrow (e)$: Suppose (f) is true. Since ${\text {ran}}V\subseteq {\text {ran}}V_1,$ we have ${\text {ran}}V_1= ({\text {ran}}V_1\ominus {\text {ran}}V)\oplus {\text {ran}}V.$ Therefore $\ker V_1^*=({\text {ran}}V_1)^{\perp }=({\text {ran}}V_2\ominus {\text {ran}}V).$ Similarly, $\ker V_2^*=({\text {ran}}V_1\ominus {\text {ran}}V).$ Hence $\ker V_1^*\oplus \ker V_2^*=\ker V^*.$

$(a) \Leftrightarrow (g)$: We use the formula $C(V_1,V_2)= (E_0\otimes (U^*PU-P))\oplus 0$ from (1.4). This gives

$$\begin{aligned} C(V_1,V_2)=0&\text { if and only if } U^*PU - P = 0\\&\text { if and only if } {\text {ran}}P \text { reduces } U. \end{aligned}$$

Thus, completes the proof. $\square$

We now write the structure theorem given in Popovici [18, Sec. 4], which highlights the importance of Example 3.1. We give a proof for the completeness.

Theorem 3.4

Let $(V_{1},V_{2})$ be a pair of commuting isometries on ${\mathcal {H}}$ with defect zero. Let $({\mathcal {E}},P,U)$ be the BCL triple for $(V_{1},V_{2}).$ Let ${\mathcal {E}}_1= {\text {ran}}P$ and ${\mathcal {E}}_2 = {\text {ran}}P^\perp$. Then ${\mathcal {E}}_1,{\mathcal {E}}_2$ are reducing subspaces for U, i.e.,

for some unitaries $U_1$ and $U_2$ on ${\mathcal {E}}_1$ and ${\mathcal {E}}_2$ respectively.

Also, ${\mathcal {H}}=(H^2_{{{\mathbb {D}}}}\otimes {\mathcal {E}}_1) \oplus (H^2_{{{\mathbb {D}}}}\otimes {\mathcal {E}}_2) \oplus {\mathcal {K}}$ where ${\mathcal {K}}=\underset{{n\ge 0}}{\bigcap } {\text {ran}}(V_1V_2)^n$ and in this decomposition,

$$\begin{aligned} V_1= \begin{pmatrix} M_z\otimes I_{{\mathcal {E}}_1}&{} 0&{}0\\ 0&{} I_{H^2_{{{\mathbb {D}}}}}\otimes U_2^* &{} 0\\ 0&{}0&{}W_1 \end{pmatrix}, \quad V_2= \begin{pmatrix} I_{H^2_{{{\mathbb {D}}}}}\otimes U_1 &{} 0&{}0\\ 0&{} M_z\otimes I_{{\mathcal {E}}_2} &{} 0\\ 0&{}0&{}W_2 \end{pmatrix}, \end{aligned}$$

up to unitarily equivalence, for some unitary $U_i$ on ${\mathcal {E}}_i, i=1,2$ and commuting unitaries $W_1, W_2$ on ${\mathcal {K}}.$

Proof

Let $({\mathcal {E}},P,U)$ be the BCL triple for $(V_1,V_2).$ By Lemma 3.3, we have ${\mathcal {E}}_1$ reduces U and in the decomposition ${\mathcal {E}}={\mathcal {E}}_1\oplus {\mathcal {E}}_2,$ we have

$$\begin{aligned} U=\begin{pmatrix} U_1 &{} 0 \\ 0 &{} U_2 \end{pmatrix} \quad \text { and } \quad P=\begin{pmatrix} I_{{\mathcal {E}}_1} &{} 0 \\ 0 &{} 0 \end{pmatrix}. \end{aligned}$$

In this case, $\varphi _1(z)=\begin{pmatrix} zU_1^{*} &{} 0 \\ 0 &{} U_2^{*} \end{pmatrix}$ and $\varphi _2(z)=\begin{pmatrix} U_1 &{} 0 \\ 0 &{} zU_2 \end{pmatrix},$ for $z\in {{\mathbb {D}}}$. Therefore,

$$\begin{aligned} M_{\varphi _1}=\begin{pmatrix}M_z\otimes U_1^*&{} 0\\ 0&{} I_{H^2_{{{\mathbb {D}}}}}\otimes U_2^*\end{pmatrix},\ M_{\varphi _2}=\begin{pmatrix}I_{H^2_{{{\mathbb {D}}}}}\otimes U_1&{} 0\\ 0&{} M_z\otimes U_2\end{pmatrix}. \end{aligned}$$

Note that $(M_z\otimes U_1^*,I_{H^2_{{{\mathbb {D}}}}}\otimes U_1)$ and $(M_z\otimes I_{{\mathcal {E}}_1},I_{H^2_{{{\mathbb {D}}}}}\otimes U_1)$ are jointly unitarily equivalent and $(I_{H^2_{{{\mathbb {D}}}}}\otimes U_2^*,M_z\otimes U_2)$ and $(I_{H^2_{{{\mathbb {D}}}}}\otimes U_2^*,M_z\otimes I_{{\mathcal {E}}_2})$ are jointly unitarily equivalent; see Example 3.1. This completes the proof, by Theorem 1.2. $\square$

Remark 3.5

In the structure theorem (Theorem 3.4) one can write the $U_1, U_2$ and ${\mathcal {E}}_1,{\mathcal {E}}_2$ explicitly in terms of $V_1$ and $V_2$ as follows: ${\mathcal {E}}_1=\ker V_1^*, {\mathcal {E}}_2=\ker V_2^*$ and $U_1=V_2|_{\ker V_1^*},U_2=V_1^*|_{\ker V_2^*}.$ This is because of Lemma 3.3.

A comment is in order. Over several decades Marek Słociński, first by himself and then with his collaborators, has developed a complete structure theorem on commuting pairs of isometries; see [5] and the references therein. One of his early results is the following.

Theorem 3.6

(M. Słociński [21]) Let $(V_1, V_2)$ be a pair of doubly commuting isometries on a Hilbert space ${\mathcal {H}}.$ Then there exists a unique decomposition

$$\begin{aligned} {\mathcal {H}}= {\mathcal {H}}_{ss}\oplus {\mathcal {H}}_{su}\oplus {\mathcal {H}}_{us}\oplus {\mathcal {H}}_{uu}, \end{aligned}$$

where the subspace ${\mathcal {H}}_{ij}$ reduces both $V_1$ and $V_2$ for all $i, j \in \{s, u\}$. Moreover, $V_1$ on ${\mathcal {H}}_{ij}$ is a shift if $i = s$ and unitary if $i = u$ and $V_2$ is a shift if $j = s$ and unitary if $j = u$.

By Theorem 3.4, and the fact that if one of the $V_i$’s is a unitary then the defect is zero, we have $C(V_1,V_2)=0$ if and only if $(V_1,V_2)$ is doubly commuting and ${\mathcal {H}}_{ss}$ in Theorem 3.6 is $\{0\}.$

3.2 Joint Spectrum

If $(V_1,V_2)$ is pure and defect $C(V_1,V_2)=0,$ then by Theorem 3.4 and Remark 3.5,

$$\begin{aligned} \sigma (V_1,V_2)={\left\{ \begin{array}{ll} \overline{{\mathbb {D}}}\times \sigma (U_1) &{}\text {if } V_2 \text { is unitary,}\\ \sigma (U_2^*)\times \overline{{\mathbb {D}}}&{}\text {if } V_1 \text { is unitary,}\\ \overline{{\mathbb {D}}}\times \sigma (U_1)\cup \sigma (U_2^*)\times \overline{{\mathbb {D}}}&{} \text {if neither }V_1 \text { nor } V_2 \text { is a unitary,} \end{array}\right. } \end{aligned}$$

(3.3)

where $U_1=V_2|_{\ker V_1^*},U_2=V_1^*|_{\ker V_2^*}.$

Lemma 3.7

Let $(V_1,V_2)$ be a pair of commuting isometries with defect zero and $\ker V^*\ne \{0\}.$ Let $({\mathcal {E}},P,U)$ be the BCL triple for $(V_1,V_2).$ Let $U_1=U|_{{\text {ran}}P}$ and $U_2=U|_{{\text {ran}}P^\perp }.$ Then

$$\begin{aligned} \sigma (\varphi _1(z),\varphi _2(z))= {\left\{ \begin{array}{ll} \{(z{\overline{\lambda }},\lambda ): \lambda \in \sigma (U_1)\}&{} \text { if } V_2 \text { is a unitary},\\ \{({\overline{\mu }},z\mu ):\mu \in \sigma (U_2)\}&{} \text { if } V_1 \text { is a unitary,}\\ \{(z{\overline{\lambda }},\lambda ), ({\overline{\mu }},z\mu ): \lambda \in \sigma (U_1),\ \mu \in \sigma (U_2)\}&{} \text { otherwise.} \end{array}\right. }\nonumber \\ \end{aligned}$$

(3.4)

Here, for every point in the joint spectrum, the non-singularity breaks at stage 3.

Proof

The proof in the case when neither $V_1$ nor $V_2$ is a unitary is done below in detail. The other cases follow similarly.

Letting ${\mathcal {E}}_1={\text {ran}}P$ and ${\mathcal {E}}_2={\text {ran}}P^\perp$, it is an easy check that both ${\mathcal {E}}_1$ and ${\mathcal {E}}_2$ are non-trivial. By Lemma 3.3, ${\mathcal {E}}_1$ reduces U. Hence in the decomposition ${\mathcal {E}}={\mathcal {E}}_1\oplus {\mathcal {E}}_2,$ we have

$$\begin{aligned} \varphi _1(z)=\begin{pmatrix} zU_1^{*} &{} 0 \\ 0 &{} U_2^{*} \end{pmatrix} \quad \text { and }\quad \varphi _2(z)=\begin{pmatrix} U_1 &{} 0 \\ 0 &{} zU_2 \end{pmatrix}, \end{aligned}$$

for $z\in {{\mathbb {D}}}.$ Then,

$$\begin{aligned} \sigma (\varphi _1(z))=\sigma (zU_1^{*}) \cup \sigma (U_2^{*}) \quad \text { and }\quad \sigma (\varphi _2(z))=\sigma (U_1) \cup \sigma (z U_2) \end{aligned}$$

for all $z\in {{\mathbb {D}}}$. For $z\in {{\mathbb {D}}},$ we have

$$\begin{aligned} \varphi _1(z)\varphi _2(z)=zI_{{\mathcal {E}}_1\oplus {\mathcal {E}}_2}=\varphi _2(z)\varphi _1(z). \end{aligned}$$

Therefore, for all $z\ne 0,$ by Lemma 1.1 part (1) and polynomial spectral mapping theorem, we get

$$\begin{aligned} \sigma (\varphi _1(z),\varphi _2(z))\subseteq \{(z{\overline{\lambda }},\lambda ), ({\overline{\mu }},z\mu ): \lambda \in \sigma (U_1),\ \mu \in \sigma (U_2)\}. \end{aligned}$$

For $z=0,$ it is easy to see that

$$\begin{aligned} \sigma (\varphi _1(0),\varphi _2(0))\subseteq \{(0,\lambda ), ({\overline{\mu }},0): \lambda \in \sigma (U_1),\ \mu \in \sigma (U_2)\}. \end{aligned}$$

To prove the other containments, let $\lambda \in \sigma (U_1).$ Then $U_1-\lambda I$ is not onto, because $U_1$ is normal. Let $h=\begin{pmatrix} h_1\\ h_2 \end{pmatrix}, k=\begin{pmatrix} k_1\\ k_2 \end{pmatrix}\in {\mathcal {E}}_1\oplus {\mathcal {E}}_2.$ Then,

$$\begin{aligned} (\varphi _1(z)-z{\overline{\lambda }}I)k+(\varphi _2(z)-{\lambda }I)h=\begin{pmatrix} z(U_1^{*}-{\overline{\lambda }}I) k_1+ (U_1-\lambda I)h_1 \\ (U_2^{*}-z {\overline{\lambda }}I)k_2+(zU_2-\lambda I)h_2 \end{pmatrix}. \end{aligned}$$

Since $U_1$ is a normal operator, ${\text {ran}}(U_1-\lambda I)={\text {ran}}(U_1^{*}-{\overline{\lambda }} I)$ and hence the first component of the above spans only ${\text {ran}}(U_1-\lambda I)$. Hence we have

$$\begin{aligned} {\text {ran}}(\varphi _1(z)-z{\overline{\lambda }}I)+\ {\text {ran}}(\varphi _2(z)-{\lambda }I)\ne {\mathcal {E}}_1\oplus {\mathcal {E}}_2. \end{aligned}$$

Therefore $(z{\overline{\lambda }},\lambda )\in \sigma (\varphi _1(z),\varphi _2(z))$.

Similarly, if $\mu \in \sigma (U_2),$ we have

$$\begin{aligned} {\text {ran}}(\varphi _1(z)-{\overline{\mu }}I)+\ {\text {ran}}(\varphi _2(z)-z{\mu }I)\ne {\mathcal {E}}_1\oplus {\mathcal {E}}_2. \end{aligned}$$

Therefore $({\overline{\mu }},z\mu )\in \sigma (\varphi _1(z),\varphi _2(z)).$ So,

$$\begin{aligned} \sigma (\varphi _1(z),\varphi _2(z))=\{(z{\overline{\lambda }},\lambda ), ({\overline{\mu }},z\mu ): \lambda \in \sigma (U_1),\ \mu \in \sigma (U_2)\} \text { for } z\in {{\mathbb {D}}}. \end{aligned}$$

$\square$

Theorem 3.8

With the hypothesis of Lemma 3.7, we also have

$$\begin{aligned} \sigma (M_{\varphi _1},M_{\varphi _2})=\overline{\cup _{z\in {{\mathbb {D}}}}\sigma (\varphi _1(z),\varphi _2(z))}= {\left\{ \begin{array}{ll} \overline{{\mathbb {D}}}\times \sigma (U_1) &{} \text { if } V_2 \text { is a unitary,}\\ \sigma (U_2^{*})\times \overline{{\mathbb {D}}}&{} \text { if } V_1 \text { is a unitary,}\\ \overline{{\mathbb {D}}}\times \sigma (U_1)\cup \sigma (U_2^{*})\times \overline{{\mathbb {D}}}&{} \text { otherwise.} \end{array}\right. } \end{aligned}$$

and, for every point $(z_1,z_2)\in {\left\{ \begin{array}{ll} {{\mathbb {D}}}\times \sigma (U_1) &{} \text { if } V_2 \text { is a unitary,}\\ \sigma (U_2^{*})\times {{\mathbb {D}}}&{} \text { if } V_1 \text{ is } \text{ a } \text{ unitary, }\\ {{\mathbb {D}}}\times \sigma (U_1)\cup \sigma (U_2^{*})\times {{\mathbb {D}}}&{} \text { otherwise,} \end{array}\right. }$ the non-singularity of $K(M_{\varphi _1}-z_1I,M_{\varphi _2}-z_2I)$ breaks at stage 3.

Proof

As in Lemma 3.7, we prove only the case when neither $V_1$ nor $V_2$ is a unitary, other cases follow similarly.

We saw in the proof of Lemma 3.7 that for any point $z\in {{\mathbb {D}}}$, a pair of points $(z_1,z_2)\in \sigma (\varphi _1(z),\varphi _2(z))$ if and only if

$$\begin{aligned} {\text {ran}}(\varphi _1(z)-z_1I)+\ {\text {ran}}(\varphi _2(z)-z_2I)\ne {\mathcal {E}}_1\oplus {\mathcal {E}}_2. \end{aligned}$$

Hence, if $(z_1,z_2)\in \sigma (\varphi _1(z),\varphi _2(z)),$

$$\begin{aligned} {\text {ran}}(M_{\varphi _1}-z_1I)+\ {\text {ran}}(M_{\varphi _2}-z_2I)\ne H^{2}({\mathcal {E}}_1\oplus {\mathcal {E}}_2). \end{aligned}$$

(3.5)

Therefore, $(z_1,z_2)\in \sigma (M_{\varphi _1},M_{\varphi _2}),$ which implies that

$$\begin{aligned} {\cup _{z\in {{\mathbb {D}}}}\sigma (\varphi _1(z),\varphi _2(z))}\subseteq \sigma (M_{\varphi _1},M_{\varphi _2}). \end{aligned}$$

Note that

$$\begin{aligned} {\cup _{z\in {{\mathbb {D}}}}\sigma (\varphi _1(z),\varphi _2(z))}&=\cup _{z\in {{\mathbb {D}}}}\{(z{\overline{\lambda }},\lambda ), ({\overline{\mu }},z\mu ): \lambda \in \sigma (U_1),\ \mu \in \sigma (U_2)\}\nonumber \\&={{{\mathbb {D}}}}\times \sigma (U_1)\cup \sigma (U_2^{*})\times {{{\mathbb {D}}}}.\end{aligned}$$

(3.6)

Since

$$\begin{aligned} M_{\varphi _1}=\begin{pmatrix} M_{z}\otimes U_1^{*} &{} 0 \\ 0 &{} I_{H^2_{{\mathbb {D}}}}\otimes U_2^{*} \end{pmatrix} \quad \text { and }\quad M_{\varphi _2}=\begin{pmatrix} I_{H^2_{{\mathbb {D}}}}\otimes U_1 &{} 0 \\ 0 &{} M_{z}\otimes U_2 \end{pmatrix}, \end{aligned}$$

we have

$$\begin{aligned} \sigma (M_{\varphi _1},M_{\varphi _2})=\sigma (M_{z}\otimes U_1^{*}, I_{H^2_{{\mathbb {D}}}}\otimes U_1) \cup \sigma (I_{H^2_{{\mathbb {D}}}}\otimes U_2^{*}, M_{z}\otimes U_2). \end{aligned}$$

By (3.2), we have

$$\begin{aligned} \sigma (M_{\varphi _1},M_{\varphi _2})= \overline{{\mathbb {D}}}\times \sigma (U_1)\cup \sigma (U_2^{*})\times \overline{{\mathbb {D}}}. \end{aligned}$$

(3.7)

Therefore from (3.6) and (3.7) we have

$$\begin{aligned} \sigma (M_{\varphi _1},M_{\varphi _2})=\overline{{\mathbb {D}}}\times \sigma (U_1)\cup \sigma (U_2^{*})\times \overline{{\mathbb {D}}}=\overline{\cup _{z\in {{\mathbb {D}}}}\sigma (\varphi _1(z),\varphi _2(z))}. \end{aligned}$$

The final thing to note is that for every point $(z_1,z_2)$ in the set ${{\mathbb {D}}}\times \sigma (U_1)\cup \sigma (U_2^{*})\times {{{\mathbb {D}}}},$ the non-singularity breaks at stage 3 and this is a direct consequence of (3.5). $\square$

To conclude the section, we note that by Theorem 1.2, $\sigma (V_1,V_2)=\sigma (M_{\varphi _1},M_{\varphi _2})\cup \sigma (V_1|_{{\mathcal {H}}_u},V_2|_{{\mathcal {H}}_u}).$ Hence Theorem 3.8 tells that

$$\begin{aligned} \overline{\cup _{z\in {{\mathbb {D}}}}\sigma (\varphi _1(z),\varphi _2(z))}\subseteq \sigma (V_1,V_2). \end{aligned}$$

(3.8)

The equality in (3.8) holds if and only if $\sigma (V_1|_{{\mathcal {H}}_u},V_2|_{{\mathcal {H}}_u})\subseteq \sigma (M_{\varphi _1},M_{\varphi _2}).$

4 The Negative Defect Case

4.1 The Prototypical Example

When the defect operator is negative, a fundamental example plays an important role in much the same way the unilateral shift plays its role in the Wold decomposition of a single isometry. We shall first describe this example and then show how it is a part of every pair of commuting isometries with negative defect.

The Hardy space of ${\mathcal {E}}$-valued functions on the bidisc ${\mathbb {D}}^2$ is

$$\begin{aligned} H^2_{{\mathbb {D}}^2}({{\mathcal {E}}})= & {} \left\{ f : {\mathbb {D}}^2 \rightarrow {\mathcal {E}} \mid f \text { is analytic and } f(z_1,z_2) = \sum _{m,n=0}^{\infty } a_{m,n} z_1^mz_2^n \right. \\&\left. \text { with } \sum _{m,n=0}^{\infty } \Vert a_{m,n}\Vert _{\mathcal {E}}^2 < \infty \right\} . \end{aligned}$$

This is a Hilbert space with the inner product

$$\begin{aligned} \left\langle \sum _{m,n=0}^{\infty } a_{m,n} z_1^mz_2^n , \sum _{m,n=0}^{\infty } b_{m,n} z_1^mz_2^n \right\rangle = \sum _{m,n=0}^{\infty } \langle a_{m,n} , b_{m,n} \rangle _{\mathcal {E}}\end{aligned}$$

and is identifiable with $H^2_{{\mathbb {D}}^2} \otimes {\mathcal {E}}$ where $H^2_{{\mathbb {D}}^2}$ stands for the Hardy space of scalar-valued functions on ${\mathbb {D}}^2$.

Let $U:H^2_{{{\mathbb {D}}}^2}\rightarrow H^2_{{{\mathbb {D}}}^2}$ be the unitary defined by

$$\begin{aligned} U(z_1^{m_1}z_2^{m_2})= {\left\{ \begin{array}{ll} z_1^{m_1+2}z_2^{m_2} &{} \text {if } m_1\ge m_2,\\ z_1^{m_1+1}z_2^{m_2-1} &{} \text {if } m_1+1=m_2,\\ z_1^{m_1}z_2^{m_2-2} &{} \text {if } m_1+2\le m_2. \end{array}\right. } \end{aligned}$$

(4.1)

on the orthonormal basis $\{z_1^{m_1}z_2^{m_2}\}_{m_1,m_2\ge 0}.$

The pair of multipliers by the coordinate functions $(M_{z_1}, M_{z_2})$ forms a pair of doubly commuting isometries on $H^2_{{\mathbb {D}}^2}$. There is a natural isomorphism between the Hilbert spaces $H^2_{{\mathbb {D}}^2}$ and $H^2_{{\mathbb {D}}} \otimes H^2_{{\mathbb {D}}}$ wherein $z_1^{m_1} z_2^{m_2}$ is identified with $z_1^{m_1} \otimes z_2^{m_2}$. In this identification, the pair of coordinate multipliers $(M_{z_1}, M_{z_2})$ is identified with $(M_z \otimes I, I \otimes M_z)$.

Definition 4.1

The pair of bounded operators $\tau _1:=U^*M_{z_1}$ and $\tau _2:=M_{z_2}U$ on the Hardy space of the bidisc $H^2_{{\mathbb {D}}^2}$ will be called the fundamental isometric pair with negative defect.

The following lemma justifies the name except the word fundamental which will be clear from Theorem 4.11.

Lemma 4.2

The pair $(\tau _1,\tau _2)$ is a pair of commuting isometries with defect negative and non-zero.

Proof

It is simple to check that the unitary U defined in (4.1) commutes with $M_{z_1z_2}$, the operator of multiplication by the function $z_1z_2$. That proves commutativity of $\tau _1$ and $\tau _2$. They are isometries because each is a product of an isometry and a unitary. Now, let

$$\begin{aligned} {\mathcal {W}}_1=\ker (\tau _1^*)=\overline{{\text {span}}}\{z_2^2,z_2^3,z_2^4,\dots \} \quad \text { and }\quad {\mathcal {W}}_2=\ker (\tau _2^*)=\overline{{\text {span}}}\{1,z_1,z_1^2,\dots \}. \end{aligned}$$

Then, $\tau _2({\mathcal {W}}_1)=\overline{{\text {span}}}\{z_2,z_2^2,z_2^3,\dots \}.$ Thus, we have

$$\begin{aligned} C(\tau _1,\tau _2)=P_{{\mathcal {W}}_1}-P_{\tau _2({\mathcal {W}}_1)}=-P_{{\text {span}}\{z_2\}}\le 0 .\end{aligned}$$

That completes the proof. $\square$

We shall prove the following lemma to compute the joint spectrum of $(\tau _1,\tau _2).$

Lemma 4.3

For any $\lambda \in {{\mathbb {D}}}$, we have

$$\begin{aligned} ({\text {ran}}(\tau _2-\lambda I))^\perp =\left\{ (I - {\overline{\lambda }} \tau _2)^{-1} x : x \in \ker M_{z_2}^*\right\} . \end{aligned}$$

(4.2)

Proof

Using the Neumann series $(I-{\bar{\lambda }} A)^{-1}=\sum _{n\ge 0}{\bar{\lambda }}^nA^n$ for $\lambda \in {{\mathbb {D}}}$ and any contraction A,  it is straightforward that the equality

$$\begin{aligned} ({\text {ran}}(A-\lambda I))^\perp =\{(I-{\bar{\lambda }}A)^{-1}x:x\in \ker A^*\} \end{aligned}$$

is satisfied when A is an isometry. The proof is complete by noting that $\ker \tau _2^*=\ker U^*M_{z_2}^*=\ker M_{z_2}^*.$ $\square$

Recall the Koszul complex for a pair of commuting bounded operators $(T_1, T_2)$ from (1.2):

$$\begin{aligned} 0\overset{\delta _0}{\rightarrow } {\mathcal {H}}\overset{\delta _1}{\rightarrow }{\mathcal {H}}\oplus {\mathcal {H}}\overset{\delta _2}{\rightarrow }{\mathcal {H}}\overset{\delta _3}{\rightarrow } 0. \end{aligned}$$

(4.3)

It is well-known that the most difficult stage to treat for the purpose of showing lack of exactness is the stage 2.

Proposition 4.4

The fundamental isometric pair with negative defect has the full closed bidisc $\overline{{{\mathbb {D}}}^2}$ as its joint spectrum. Moreover, for every point in the open bidisc ${{\mathbb {D}}}^2,$ the non-singularity breaks at stage 2.

Proof

Let $\lambda _1,\lambda _2\in {{\mathbb {D}}}.$ We shall find a non-zero function $h_2\in ({\text {ran}}(\tau _2-\lambda _2I))^\perp$ such that

$$\begin{aligned} (\tau _1-\lambda _1I)h_2\in {\text {ran}}(\tau _2-\lambda _2I).\end{aligned}$$

(4.4)

This would imply that there exists $h_1\in {\mathcal {H}}=H^2_{{{\mathbb {D}}}^2}$ such that

$$\begin{aligned} (\tau _1-\lambda _1I)h_2=(\tau _2-\lambda _2I)h_1\end{aligned}$$

producing a pair $(h_1,h_2)$ in $\ker \delta _2$ which would not be in ${\text {ran}}\delta _1$.

To that end, we shall use the description of $({\text {ran}}(\tau _2-\lambda I))^\perp$ obtained in Lemma 4.3. Since any element from $\ker \tau _2^*=\ker M_{z_2}^*$ is of the form $\sum _{m=0}^\infty a_mz_1^m$ for a square summable sequence $\{a_m\}_{m\ge 0}$, it follows from Lemma 4.3 that

$$\begin{aligned} ({\text {ran}}(\tau _2-\lambda _2I))^\perp&=\left\{ (I - \overline{\lambda _2} \tau _2)^{-1} \sum _{m= 0}^\infty a_mz_1^m : \sum _{m= 0}^\infty |a_m|^2< \infty \right\} \nonumber \\&=\left\{ \sum _{n= 0}^\infty (\overline{\lambda _2}M_{z_2} U)^n \left( \sum _{m= 0}^\infty a_mz_1^m\right) :\sum _{m= 0}^\infty |a_m|^2< \infty \right\} \nonumber \\&=\left\{ \sum _{m,n=0}^{\infty }\overline{\lambda _2}^na_mz_1^{m+2n}z_2^n:\sum _{m= 0}^\infty |a_m|^2 < \infty \right\} . \end{aligned}$$

(4.5)

Our candidate for $h_2$ to satisfy (4.4) is

$$\begin{aligned} h_2= \frac{1}{(1 - \overline{\lambda _2}z_1^2z_2)(1 - \lambda _1z_1)} = \sum _{m,n=0}^{\infty }\overline{\lambda _2}^n \lambda _1^m z_1^{m+2n}z_2^n. \end{aligned}$$

By (4.5), this function is in $({\text {ran}}(\tau _2-\lambda _2I))^\perp$. We shall verify below that $(\tau _1-\lambda _1I)h_2$ is in the closure of ${\text {ran}}(\tau _2-\lambda _2I)$. Since $|\lambda _2|<1$ and $\tau _2$ is an isometry, $(\tau _2-\lambda _2 I)$ is bounded below and hence its range is closed. That will complete the proof.

First note that

$$\begin{aligned} U \sum _{m,n=0}^{\infty } {\overline{\lambda _2}}^n a_m z_1^{m+2n} z_2^n = \sum _{m,n=0}^{\infty } {\overline{\lambda _2}}^n a_m z_1^{m+2n+2} z_2^n = M_{z_1}^2 \sum _{m,n=0}^{\infty } {\overline{\lambda _2}}^n a_m z_1^{m+2n} z_2^n\nonumber \\ \end{aligned}$$

(4.6)

and

$$\begin{aligned} (M_{z_1}^*-\lambda _1 I)h_2&= M_{z_1}^*(h_2)-\lambda _1 h_2 \nonumber \\&= \underset{(m,n)\ne (0,0)}{\sum _{m,n=0}^{\infty }}\overline{\lambda _2}^n \lambda _1^m z_1^{m+2n-1}z_2^n-\sum _{m,n=0}^{\infty } \overline{\lambda _2}^n \lambda _1^{m+1} z_1^{m+2n}z_2^n \nonumber \\&= \sum _{m=1,n=0}^{\infty }\overline{\lambda _2}^n \lambda _1^m z_1^{m+2n-1}z_2^n+ \sum _{n=1}^{\infty }\overline{\lambda _2}^n z_1^{2n-1}z_2^n-\sum _{m,n=0}^{\infty } \overline{\lambda _2}^n \lambda _1^{m+1} z_1^{m+2n}z_2^n \nonumber \\&= \sum _{m,n=0}^{\infty } \overline{\lambda _2}^n \lambda _1^{m+1} z_1^{m+2n}z_2^n + \sum _{n=1}^{\infty }\overline{\lambda _2}^n z_1^{2n-1}z_2^n-\sum _{m,n=0}^{\infty } \overline{\lambda _2}^n \lambda _1^{m+1} z_1^{m+2n}z_2^n \nonumber \\&= \sum _{n=1}^{\infty }\overline{\lambda _2}^n z_1^{2n-1}z_2^n . \end{aligned}$$

(4.7)

We now compute the inner product between a typical element of $({\text {ran}}(\tau _2-\lambda _2I))^\perp$ and $(\tau _1-\lambda _1I)h_2$ by using the two equations above.

$$\begin{aligned}&\left\langle (\tau _1-\lambda _1I)h_2, \sum _{m,n=0}^{\infty } \overline{\lambda _2}^{n}a_{m}z_1^{m+2n}z_2^{n}\right\rangle \\&\quad =\left\langle (U^*M_{z_1}-\lambda _1I)h_2, \sum _{m,n=0}^{\infty } \overline{\lambda _2}^{n}a_{m}z_1^{m+2n}z_2^{n} \right\rangle \\&\quad =\left\langle M_{z_1}h_2, U\sum _{m,n=0}^{\infty } \overline{\lambda _2}^{n}a_{m}z_1^{m+2n}z_2^{n} \right\rangle - \lambda _1\left\langle h_2,\sum _{m,n=0}^{\infty } \overline{\lambda _2}^{n}a_{m}z_1^{m+2n}z_2^{n} \right\rangle \\&\quad = \left\langle M_{z_1}h_2, M_{z_1}^2\sum _{m,n=0}^{\infty } \overline{\lambda _2}^{n}a_{m}z_1^{m+2n}z_2^{n} \right\rangle - \lambda _1\left\langle h_2,\sum _{m,n=0}^{\infty } \overline{\lambda _2}^{n}a_{m}z_1^{m+2n}z_2^{n} \right\rangle \\&\quad = \left\langle (M_{z_1}^*-\lambda _1 I)h_2, \sum _{m,n=0}^{\infty } \overline{\lambda _2}^{n}a_{m}z_1^{m+2n}z_2^{n} \right\rangle \\&\quad = \left\langle \sum _{n=1}^{\infty }\overline{\lambda _2}^n z_1^{2n-1}z_2^n, \sum _{m,n=0}^{\infty } \overline{\lambda _2}^{n}a_{m}z_1^{m+2n}z_2^{n} \right\rangle = 0. \end{aligned}$$

This shows that $(\tau _1-\lambda _1I)h_2\in {\overline{{\text {ran}}}}(\tau _2-\lambda _2I) = {\text {ran}}(\tau _2-\lambda _2I)$ and hence completes the proof.

$\square$

Note 4.5

In Remark 5.4, we shall see that there is a joint invariant subspace ${\mathcal {M}}$ for $(\tau _1, \tau _2)$ such that the defect operator of $(\tau _1|_{{\mathcal {M}}}, \tau _2|_{{\mathcal {M}}})$ is positive (and not zero).

4.2 General Theory for the Negative Defect Case

Here we shall show that the fundamental example above is a typical example. This helps us to compute the joint spectrum of any commuting pair of isometries with negative defect.

In [11], Gaspar and Gaspar introduced the dual doubly commuting pairs. If $(\bar{V_1},\bar{V_2})$ is the minimal unitary extension to ${\bar{{\mathcal {H}}}}$ of $(V_1,V_2)$ acting on ${\mathcal {H}}$, then the pair of commuting isometries $(\bar{V_1}^*|_{{{\bar{{\mathcal {H}}}}} \ominus {\mathcal {H}}}, \bar{V_2}^*|_{{{\bar{{\mathcal {H}}}}} \ominus {\mathcal {H}}})$ is called the dual of $(V_1,V_2)$. If the dual is doubly commuting, then $(V_1,V_2)$ is called a dual doubly commuting pair.

A pair $(V_1,V_2)$ of commuting isometries is called a bi-shift (see [19]) if there is a wandering subspace ${\mathcal {R}}$ (i.e., $V_1^{p_1}V_2^{p_2}({\mathcal {R}})\perp V_1^{q_1}V_2^{q_2}({\mathcal {R}})$ if $(p_1, p_2), (q_1, q_2)\in {\mathbb {Z}}_+^2$ and $(p_1, p_2) \ne (q_1, q_2)$) such that

$$\begin{aligned} {\mathcal {H}}=\bigoplus _{(n_1, n_2) \in {\mathbb {Z}}_+^2 } V_1^{n_1}V_2^{n_2}({\mathcal {R}}). \end{aligned}$$

$(V_1,V_2)$ is called a modified bi-shift if it is pure and its dual is a bi-shift.

Popovici used the concepts above greatly in his papers [19] and [20]. We are thankful to him for sending us his papers. First we shall give a characterizing lemma for this case; see also [13].

Lemma 4.6

Let $(V_1,V_2)$ be a pair of commuting isometries on a Hilbert space ${\mathcal {H}}$. Then the following are equivalent:

• (a) $C(V_1,V_2)\le 0$ and $C(V_1,V_2)\ne 0.$

• (b) $V_2(\ker V_1^*)\supsetneq \ker V_1^*.$

• (c) $V_1(\ker V_2^*)\supsetneq \ker V_2^*.$

• (d) The adjoint of the fringe operators are isometries and not unitaries.

• (e) $(V_1,V_2)$ is dual doubly commuting and $C(V_1,V_2)\ne 0.$

• (f) $\ker V_1^*$ is orthogonal to $\ker V_2^*$ and $\ker V_1^*\oplus \ker V_2^*\ne \ker V^*.$

• (g) $C(V_1,V_2)$ is the negative of a non-zero projection.

• (h) If $({\mathcal {E}},P,U)$ is the BCL triple for $(V_1,V_2),$ then $U({\text {ran}}P^\perp )\subsetneq {\text {ran}}P^\perp .$

Proof

The equivalence $(e) \Leftrightarrow (h)$ is proved in [11]. All other proof are along the same lines as the proofs of various parts of Lemma 3.3. $\square$

The geometrical structure and a model for dual doubly commuting isometries is known due to [11, 18]. Here we observe that the multiplication operators $M_{\varphi _i},i=1,2$ associated to $(V_1,V_2)$ as in Theorem 1.2, have some special forms in this case. This also helps us getting a model in the Hardy space of the bidisc. Some steps of the proof are used to obtain Theorem 4.15. The wandering space arguments used in the proof of the following theorem, appears in [18, Thm 4.3].

Let be the multipliers associated with the BCL triple $(l^2({\mathbb {Z}}),p_-,\omega )$ where $p_-$ is the projection onto $\overline{{\text {span}}}\{e_n:n<0\}$ and $\omega$ is the bilateral shift on $l^2({\mathbb {Z}})$.

Theorem 4.7

Let $(V_1,V_2)$ be a pair of commuting isometries such that $C(V_1,V_2)$ is a non-zero negative operator. Let $({\mathcal {E}},P,U)$ be the BCL triple for $(V_1,V_2).$ Then, up to unitary equivalence

$$\begin{aligned} {\mathcal {E}}= (l^2({\mathbb {Z}})\otimes {\mathcal {L}}) \oplus {\mathcal {E}}_2 \end{aligned}$$

for some non-trivial closed subspace ${\mathcal {L}}$ and a closed subspace ${\mathcal {E}}_2$ of ${\mathcal {E}}.$ Moreover,

(4.8)

with $C({M_{\varphi _1}}|_{H^2_{{{\mathbb {D}}}}({\mathcal {E}}_2)},{M_{\varphi _2}}|_{H^2_{{{\mathbb {D}}}}({\mathcal {E}}_2)})=0.$ In particular,

$$\begin{aligned} \sigma (M_{\psi _1},M_{\psi _2})= \sigma ( M_{\psi _1}\otimes I_{\mathcal {L}},M_{\psi _2}\otimes I_{\mathcal {L}})\subseteq \sigma (M_{\varphi _1},M_{\varphi _2})\subseteq \sigma (V_1,V_2). \end{aligned}$$

(4.9)

Proof

Since $C(V_1,V_2)\le 0 \text { and }C(V_1,V_2)\ne 0$, by Lemma 4.6, $U({\text {ran}}P^\perp ) \subsetneq {\text {ran}}P^\perp .$ Consider ${\mathcal {L}}:=({\text {ran}}P^\perp \ominus U({\text {ran}}P^\perp ))\ne \{0\}.$ Let $m,n\in {\mathbb {Z}}$ and $m> n$. Then for $x,y\in {\mathcal {L}}$ we have

$$\begin{aligned} \langle U^m x, U^ny\rangle = \langle U^{m-n}x, y\rangle =0,\end{aligned}$$

because $U^{m-n}x\in U({\text {ran}}P^\perp ).$ Therefore $U^m({\mathcal {L}})\perp U^n({\mathcal {L}})$ if $m, n\in {\mathbb {Z}}$ and $m\ne n.$

Set ${\mathcal {E}}_1:=\oplus _{n\in {\mathbb {Z}}} U^n({\mathcal {L}}).$ Clearly U reduces ${\mathcal {E}}_1.$ Now $U({\text {ran}}P^\perp )\subsetneq {\text {ran}}P^\perp$ implies:

$$\begin{aligned} \oplus _{n\ge 0}U^n({\mathcal {L}})\subseteq {\text {ran}}P^\perp . \end{aligned}$$

(4.10)

For all $x\in {\mathcal {L}}, y\in {\text {ran}}P^\perp$ and $n< 0,$ $\langle U^nx,y\rangle = \langle x, U^{-n}y\rangle =0$ implies that

$$\begin{aligned} \oplus _{n< 0}U^n({\mathcal {L}})\subseteq {\text {ran}}P. \end{aligned}$$

(4.11)

It is clear from inclusions (4.10) and (4.11) that P also reduces ${\mathcal {E}}_1.$ Now $U|_{{\mathcal {E}}_1}=W_{\mathcal {L}},$ the bilateral shift on ${\mathcal {E}}_1$ with the wandering subspace ${\mathcal {L}}$ and $P|_{{\mathcal {E}}_1}=P_{{\mathcal {E}}_1^-},$ the projection on ${\mathcal {E}}_1^-:=\oplus _{n< 0}U^n({\mathcal {L}})$ in ${\mathcal {E}}_1.$

Let ${\mathcal {E}}=\ker V^*= {\mathcal {E}}_1\oplus {\mathcal {E}}_2.$ In this decomposition we have,

$$\begin{aligned} U=\begin{pmatrix} W_{\mathcal {L}}&{}0 \\ 0 &{} U|_{{\mathcal {E}}_2}\end{pmatrix} \quad \text { and } \quad P=\begin{pmatrix}P_{{\mathcal {E}}_1^-}&{}0 \\ 0 &{} P|_{{\mathcal {E}}_2} \end{pmatrix}. \end{aligned}$$

Let $\Theta :{\mathcal {E}}_1\rightarrow l^2({\mathbb {Z}})\otimes {\mathcal {L}}$ be the unitary given by

$$\begin{aligned} \Theta (U^nx)=e_n\otimes x \text { for } x\in {\mathcal {L}}, n\in {\mathbb {Z}}. \end{aligned}$$

Then, with this identification (U, P) is jointly unitarily equivalent to

$$\begin{aligned} \left( \begin{pmatrix} \omega \otimes I_{\mathcal {L}}&{}0 \\ 0 &{} U|_{{\mathcal {E}}_2}\end{pmatrix} ,\begin{pmatrix} p_-\otimes I_{\mathcal {L}}&{}0 \\ 0 &{} P|_{{\mathcal {E}}_2} \end{pmatrix}\right) \end{aligned}$$

where $\omega$ is the bilateral shift on $l^2({\mathbb {Z}})$ and $p_-$ is the projection in $l^2({\mathbb {Z}})$ onto $\overline{{\text {span}}}\{e_n:n<0\}.$ Therefore,

$$\begin{aligned} \varphi _1(z)&=\begin{pmatrix} \omega ^*\otimes I_{\mathcal {L}}&{}0 \\ 0 &{} U^*|_{{\mathcal {E}}_2}\end{pmatrix} \left[ \begin{pmatrix} p_-^\perp \otimes I_{\mathcal {L}}&{}0 \\ 0 &{} P^\perp |_{{\mathcal {E}}_2}\end{pmatrix}+z\begin{pmatrix} p_- \otimes I_{\mathcal {L}}&{}0 \\ 0 &{} P|_{{\mathcal {E}}_2}\end{pmatrix}\right] \nonumber \\&=\begin{pmatrix} \omega ^*(p_-^\perp +zp_-)\otimes I_{\mathcal {L}}&{}0 \\ 0 &{} U^*(P^\perp +zP)|_{{\mathcal {E}}_2} \end{pmatrix}=\begin{pmatrix} \psi _1(z)\otimes I_{\mathcal {L}}&{}0 \\ 0 &{} \varphi _1(z)|_{{\mathcal {E}}_2} \end{pmatrix}, \end{aligned}$$

(4.12)

and

$$\begin{aligned} \varphi _2(z)&=\left[ \begin{pmatrix} p_- \otimes I_{\mathcal {L}}&{}0 \\ 0 &{}P|_{{\mathcal {E}}_2}\end{pmatrix}+z\begin{pmatrix} p_-^\perp \otimes I_{\mathcal {L}}&{}0 \\ 0 &{} P^\perp |_{{\mathcal {E}}_2} \end{pmatrix}\right] \begin{pmatrix} \omega \otimes I_{\mathcal {L}}&{}0 \\ 0 &{} U|_{{\mathcal {E}}_2} \end{pmatrix}\nonumber \\&=\begin{pmatrix} (p_-+zp_-^\perp )\omega \otimes I_{\mathcal {L}}&{}0 \\ 0 &{} (P+zP^\perp )U|_{{\mathcal {E}}_2} \end{pmatrix}=\begin{pmatrix} \psi _2(z)\otimes I_{\mathcal {L}}&{}0 \\ 0 &{} \varphi _2(z)|_{{\mathcal {E}}_2} \end{pmatrix}, \end{aligned}$$

(4.13)

where $p_-^\perp$ denotes the projection $I_{l^2({\mathbb {Z}})}-p_-$ in $l^2({\mathbb {Z}}).$ This in particular says that ${\mathcal {E}}_1$ reduces $\varphi _i(z)$ for $z\in {{\mathbb {D}}},i=1,2.$ Therefore $H^2_{{\mathbb {D}}}({\mathcal {E}}_1)$ reduces $M_{\varphi _i},$ for $i=1,2,$ and

$$\begin{aligned} M_{\varphi _i}=\begin{pmatrix} M_{\psi _i\otimes I_{\mathcal {L}}}&{}0\\ 0&{} M_{{\varphi _i|_{{\mathcal {E}}_2}}} \end{pmatrix} =\begin{pmatrix} M_{\psi _i}\otimes I_{\mathcal {L}}&{}0\\ 0&{} M_{\varphi _i}|_{H^2_{{\mathbb {D}}}({\mathcal {E}}_2)} \end{pmatrix}, \ i=1,2, \end{aligned}$$

(4.14)

where

$$\begin{aligned} \psi _1(z)=\omega ^*(p_-^\perp +zp_-),\ \psi _2(z)= (p_-+zp_-^\perp )\omega \end{aligned}$$

and for $z\in {{\mathbb {D}}}$ and $i=1,2.$

It remains to prove that $C({M_{\varphi _1}}|_{H^2_{{{\mathbb {D}}}}({\mathcal {E}}_2)},{M_{\varphi _2}}|_{H^2_{{{\mathbb {D}}}}({\mathcal {E}}_2)})=0$. To that end, note that by Lemma 3.3, it is enough to show that $U({\text {ran}}P|_{{\mathcal {E}}_2})={\text {ran}}P|_{{\mathcal {E}}_2}.$ Notice from (4.10) and (4.11) that ${\text {ran}}P|_{{\mathcal {E}}_2}={\text {ran}}P \ominus (\oplus _{n< 0}U^n({\mathcal {L}})).$ Hence

$$\begin{aligned} U({\text {ran}}P|_{{\mathcal {E}}_2})&=U({\text {ran}}P \ominus (\oplus _{n< 0}U^n({\mathcal {L}})))\\&=U({\text {ran}}P)\ominus (\oplus _{n< 0}U^{n+1}({\mathcal {L}}))\\&=(U({\text {ran}}P^\perp ))^\perp \ominus ({\mathcal {L}}\oplus _{n< 0}U^n({\mathcal {L}}))\\&=({\text {ran}}P \oplus {\mathcal {L}})\ominus ({\mathcal {L}}\oplus _{n< 0}U^n({\mathcal {L}}))\quad (\because {\text {ran}}P^\perp ={\mathcal {L}}\oplus U({\text {ran}}P^\perp ))\\&={\text {ran}}P \ominus (\oplus _{n< 0}U^n({\mathcal {L}}))={\text {ran}}P|_{{\mathcal {E}}_2}. \end{aligned}$$

It remains only to prove (4.9), and it follows directly from (4.8) by Lemma 1.1. $\square$

Lemma 4.8

The pair $(M_{\psi _1},M_{\psi _2})$ is jointly unitarily equivalent to $(\tau _1,\tau _2).$ In particular,

$$\begin{aligned} \sigma (M_{\psi _1},M_{\psi _2})=\overline{{{\mathbb {D}}}^2} \end{aligned}$$

and for every point in ${{\mathbb {D}}}^2,$ the non-singularity breaks at stage 2.

Proof

Define the unitary $\Lambda :H^2_{{{\mathbb {D}}}^2}\rightarrow H^2_{{\mathbb {D}}}(l^2({\mathbb {Z}}))$ by

$$\begin{aligned} \Lambda \left( \sum _{m,n=0}^{\infty }a_{m,n}z_1^mz_2^n\right) =\sum _{k=0}^\infty \left( \sum _{m= 0}^\infty a_{m+k,k}e_m+\sum _{m= 1}^\infty a_{k,m+k}e_{-m}\right) z^k. \end{aligned}$$

(4.15)

Then, $\Lambda \tau _i\Lambda ^*=M_{\psi _i}$ for $i=1,2.$ That is, $(M_{\psi _1},M_{\psi _2})$ is jointly unitarily equivalent to $(\tau _1,\tau _2).$ In particular,

$$\begin{aligned} \sigma (M_{\psi _1},M_{\psi _2})=\sigma (\tau _1,\tau _2)=\overline{{{\mathbb {D}}}^2} \end{aligned}$$

(4.16)

and for every point in ${{\mathbb {D}}}^2,$ the non-singularity breaks at stage 2 by Proposition 4.4. $\square$

We shall use the following lemma proved in [11] and [9].

Lemma 4.9

Let $({\mathcal {E}},P,U)$ be a BCL triple. Then the pair $(M_{\varphi _1},M_{\varphi _2})$ has a non-trivial joint reducing subspace if and only if (P, U) has a non-trivial joint reducing subspace.

Lemma 4.10

The pair $(\tau _1,\tau _2)$ does not have any non-trivial joint reducing subspace.

Proof

By Lemma 4.9, $(\tau _1,\tau _2)$ has a non-trivial joint reducing subspace if and only if $(p_-,\omega )$ has a non-trivial joint reducing subspace in $l^2({\mathbb {Z}}).$ Let ${\mathcal {E}}_{0}\ne \{0\}$ be a joint reducing subspace for $(p_-,\omega ).$ Let $f=\sum _{n\in {\mathbb {Z}}} a_ne_n\in {\mathcal {E}}_{0},$ $a_{n_0}\ne 0$ for some $n_0\in {\mathbb {Z}}.$ Now

$$\begin{aligned} {\omega ^*}^{n_0}(f)=\sum _{n\in {\mathbb {Z}}} a_ne_{n-n_0},\ p_-{\omega ^*}^{n_0}(f)=\sum _{n<n_0} a_ne_{n-n_0}\in {\mathcal {E}}_{0}. \end{aligned}$$

Hence $\sum _{n\ge n_0} a_ne_{n-n_0}\in {\mathcal {E}}_{0}$ and

$$\begin{aligned} p_-\omega ^*\left( \sum _{n\ge n_0} a_ne_{n-n_0}\right) =a_{n_0}e_{-1}\in {\mathcal {E}}_{0}. \end{aligned}$$

Since ${\mathcal {E}}_{0}$ is reducing for $\omega$ and $e_{-1}\in {\mathcal {E}}_{0}$ we have ${\mathcal {E}}_{0}=l^2({\mathbb {Z}}).$ $\square$

The following theorem on the structure and the joint spectrum of commuting pair of isometries with negative defect follows directly from Theorem 1.2, Theorem 4.7, Lemma 1.1 and Lemma 4.8 and shows that when the defect of $(V_1,V_2)$ is negative, then apart from a reduced part which has defect zero, $(V_1,V_2)$ is the fundamental isometric pair with negative defect, albeit with a higher multiplicity.

Theorem 4.11

Let $(V_1,V_2)$ be a pair of commuting isometries such that $C(V_1,V_2)\le 0$ and $C(V_1,V_2)\ne 0.$ Then, there is a non-trivial subspace ${\mathcal {L}}\subsetneq \ker V^*$ such that, up to unitary equivalence, ${\mathcal {H}}={\mathcal {H}}_0\oplus {\mathcal {H}}_0^\perp ,$ where ${\mathcal {H}}_0= H^2_{{{\mathbb {D}}}^2}\otimes {\mathcal {L}}$ and in this decomposition

$$\begin{aligned} V_i = \begin{pmatrix} \tau _i\otimes I_{\mathcal {L}}&{}0\\ 0&{} V_i|_{{\mathcal {H}}_0^\perp } \end{pmatrix}, i=1,2 \quad \text { and }\quad C(V_1|_{{\mathcal {H}}_0^\perp },V_2|_{{\mathcal {H}}_0^\perp })=0, \end{aligned}$$

where the dimension of ${\mathcal {L}}$ is same as the dimension of the range of $C(V_1,V_2)$ and $(\tau _1, \tau _2)$ is the fundamental isometric pair with negative defect. Moreover,

$$\begin{aligned} \sigma (V_1,V_2)=\overline{{{\mathbb {D}}}^2} \end{aligned}$$

and for every point in ${{\mathbb {D}}}^2,$ the non-singularity breaks at stage 2.

The pair $(\tau _1, \tau _2)$ serves as a model in many ways.

Theorem 4.12

A pair of commuting isometries $(V_1, V_2)$ is a modified bi-shift if and only if $(V_1, V_2)$ is jointly unitarily equivalent to $(\tau _1 \otimes I_{\mathcal {L}}, \tau _2 \otimes I_{\mathcal {L}})$ for some Hilbert space ${\mathcal {L}}$.

Proof

Let $(V_1,V_2)=(\tau _1 \otimes I_{\mathcal {L}}, \tau _2 \otimes I_{\mathcal {L}}).$ Then

$$\begin{aligned} \ker V_1^*=\overline{{\text {span}}}\{z_2^2,z_2^3,\dots \}\otimes {\mathcal {L}},\ \ker V_2^*=\overline{{\text {span}}}\{1,z_1,z_1^2,\dots \}\otimes {\mathcal {L}}. \end{aligned}$$

Clearly $V_1^*|_{\ker V_2^*}, V_2^*|_{\ker V_1^*}$ and $V_1V_2$ are shifts. Therefore, by [20, Prop. 3.8] $(V_1,V_2)$ is a modified bi-shift.

Suppose $(V_1,V_2)$ is a modified bi-shift. If $({\mathcal {E}},P,U)$ is the BCL triple for $(V_1,V_2),$ then by [11, Thm. 2.4], $U({\text {ran}}P^\perp )\subsetneq {\text {ran}}P^\perp .$ Hence by Lemma 4.6, $C(V_1,V_2)\le 0$ and $C(V_1,V_2)\ne 0.$ Therefore, by Theorem 4.11, up to unitary equivalence, $V_1$ and $V_2$ are

$$\begin{aligned} V_1=\begin{pmatrix} \tau _1\otimes I_{\mathcal {L}}&{}0&{} 0\\ 0&{}M_z\otimes I_{{\mathcal {E}}_1}&{} 0\\ 0&{}0&{} I_{H^2_{{\mathbb {D}}}}\otimes U_2 \end{pmatrix}, V_2=\begin{pmatrix} \tau _2\otimes I_{\mathcal {L}}&{}0&{} 0\\ 0&{}I_{H^2_{{\mathbb {D}}}}\otimes U_1&{} 0\\ 0&{}0&{} M_z\otimes I_{{\mathcal {E}}_2} \end{pmatrix}. \end{aligned}$$

By [20, Prop. 3.8] we have ${\mathcal {E}}_1={\mathcal {E}}_2=\{0\}.$ Hence $(V_1,V_2)$ is jointly unitarily equivalent to $(\tau _1\otimes I_{\mathcal {L}},\tau _2\otimes I_{\mathcal {L}}).$ $\square$

Remark 4.13

$(l^2({\mathbb {Z}})\otimes {\mathcal {L}}, p_-\otimes I_{\mathcal {L}},\omega \otimes I_{\mathcal {L}})$ is the BCL triple for a modified bi-shift $(V_1,V_2),$ where $\dim {\mathcal {L}}=\dim ({\text {ran}}C(V_1,V_2)).$ (See also [11, Thm. 2.4]).

Some other descriptions of modified bi-shift are given in [11], by Gaspar and Gaspar. From the results in [11], it is clear that a dual doubly commuting pair is a direct sum of a zero defect part and a modified bi-shift. (caution: the W in [11] is $U^*$ for us).

In the rest of this section we shall see the relation between the joint spectrum of a pair of commuting isometries $(V_1,V_2)$ with defect negative and non-zero, and the joint spectrum of $(\varphi _1(z),\varphi _2(z)),z\in {{\mathbb {D}}},$ where $\varphi _i$’s are the multipliers given in the BCL representation of $(V_1,V_2).$ Indeed, we prove that

$$\begin{aligned} \sigma (V_1,V_2)=\sigma (M_{\varphi _1},M_{\varphi _2})=\overline{\cup _{z\in {{\mathbb {D}}}}\sigma (\varphi _1(z),\varphi _2(z))}=\overline{{{\mathbb {D}}}^2}. \end{aligned}$$

(4.17)

Lemma 4.14

Let the operator valued functions $\psi _1$ and $\psi _2$ be as in Theorem 4.7. Then, For $z\in {{\mathbb {D}}},$ $(\lambda _1,\lambda _2)\in {{\mathbb {D}}}^2$ is a joint eigenvalue for $(\psi _1(z),\psi _2(z))$ if $\lambda _1\lambda _2=z.$ Also

$$\begin{aligned} \sigma (\psi _1(z),\psi _2(z))=\{(\lambda _1,\lambda _2)\in \overline{{{\mathbb {D}}}^2}:\lambda _1\lambda _2=z\}, \end{aligned}$$

(4.18)

and

$$\begin{aligned} \overline{\cup _{z\in {{\mathbb {D}}}}\sigma (\psi _1(z),\psi _2(z))} =\overline{{{\mathbb {D}}}^2}=\sigma (M_{\psi _1},M_{\psi _2}). \end{aligned}$$

(4.19)

Proof

Let $z\in {{\mathbb {D}}}\setminus \{0\}$ and $\lambda _1,\lambda _2\in {{\mathbb {D}}}$ be such that $\lambda _1\lambda _2=z.$ Consider

$$\begin{aligned} x_{\lambda _1,\lambda _2}=\sum _{n= 0}^\infty \lambda _1^ne_n+\frac{1}{\lambda _1}\sum _{n= 1}^\infty \lambda _2^{n-1}e_{-n}. \end{aligned}$$

Note that

$$\begin{aligned} \psi _i(z)x_{\lambda _1,\lambda _2}=\lambda _ix_{\lambda _1,\lambda _2}, i=1,2. \end{aligned}$$

That is, $(\lambda _1,\lambda _2)$ is a joint eigenvalue for $(\psi _1(z), \psi _2(z))$ with eigenvector $x_{\lambda _1,\lambda _2},$ where $z\ne 0$ and $\lambda _1,\lambda _2\in {{\mathbb {D}}}$ are such that $\lambda _1\lambda _2=z.$ Therefore,

$$\begin{aligned} \sigma (\psi _1(z),\psi _2(z))\supseteq \{(\lambda _1,\lambda _2)\in \overline{{{\mathbb {D}}}^2}:\lambda _1\lambda _2=z\} \text { for } z\ne 0. \end{aligned}$$

As $\psi _1(z)$ and $\psi _2(z)$ are commuting contractions, $\sigma (\psi _1(z),\psi _2(z))\subseteq \overline{{{\mathbb {D}}}^2}.$ Since $\psi _1(z)\psi _2(z)=zI,$ by spectral mapping theorem, we have the other inclusion. Hence

$$\begin{aligned} \sigma (\psi _1(z),\psi _2(z))=\{(\lambda _1,\lambda _2)\in \overline{{{\mathbb {D}}}^2}:\lambda _1\lambda _2=z\} \text { for } z\ne 0. \end{aligned}$$

(4.20)

For the case $z=0,$ consider for $\lambda _1,\lambda _2\in {{\mathbb {D}}}\setminus \{0\},$

$$\begin{aligned} x_{\lambda _1,0}=\sum _{n= 0}^\infty \lambda _1^ne_n+\frac{1}{\lambda _1}e_{-1},\ x_{0,\lambda _2}=\sum _{n= 1}^\infty \lambda _2^{n-1}e_{-n}\quad \text { and }\quad x_{0,0}=e_{-1} . \end{aligned}$$

Then $(\lambda _1,0),(0,\lambda _2)$ and (0, 0) are joint eigenvalues for $(\psi _1(0),\psi _2(0))$ with the joint eigenvectors $x_{\lambda _1,0},x_{0,\lambda _2}$ and $x_{0,0}$ respectively. Therefore, by the similar reasoning as in the case of $z\ne 0,$ we get

$$\begin{aligned} \sigma (\psi _1(0),\psi _2(0))=\overline{{\mathbb {D}}}\times \{0\}\cup \{0\}\times \overline{{\mathbb {D}}}=\{(\lambda _1,\lambda _2)\in \overline{{{\mathbb {D}}}^2}:\lambda _1\lambda _2=0\}. \end{aligned}$$

(4.21)

Hence, by (4.20), (4.21) and Lemma 4.8, we get (4.19).

Indeed, one can show that the joint eigen spaces are one dimensional, for all the joint eigenvalues of $(\psi _1(z),\psi _2(z)), z\in {{\mathbb {D}}}.$ $\square$

Theorem 4.15

Let $(V_1,V_2)$ be a pair of commuting isometries such that $C(V_1,V_2)\le 0$ and $C(V_1,V_2)\ne 0.$ Let $({\mathcal {E}},P,U)$ be the BCL triple for $(V_1,V_2).$ Then for $z\in {{\mathbb {D}}}$, $(\lambda _1,\lambda _2)\in {{\mathbb {D}}}^2$ is a joint eigenvalue for $(\varphi _1(z),\varphi _2(z))$ if $\lambda _1\lambda _2=z,$

$$\begin{aligned} \sigma (\varphi _1(z),\varphi _2(z))=\{(\lambda _1,\lambda _2)\in \overline{{{\mathbb {D}}}^2}:\lambda _1\lambda _2=z\}, \end{aligned}$$

and

$$\begin{aligned} \sigma (V_1,V_2)=\sigma (M_{\varphi _1},M_{\varphi _2})=\overline{\cup _{z\in {{\mathbb {D}}}}\sigma (\varphi _1(z),\varphi _2(z))}=\overline{{{\mathbb {D}}}^2}. \end{aligned}$$

Moreover, the non-singularity breaks at stage 2.

Proof

As $\varphi _1(z)$ and $\varphi _2(z)$ are commuting contractions, $\sigma (\varphi _1(z),\varphi _2(z))\subseteq \overline{{{\mathbb {D}}}^2}.$ Since $\varphi _1(z)\varphi _2(z)=zI,$ by spectral mapping theorem,

$$\begin{aligned} \sigma (\varphi _1(z),\varphi _2(z))\subseteq \{(\lambda _1,\lambda _2)\in \overline{{{\mathbb {D}}}^2}:\lambda _1\lambda _2=z\}. \end{aligned}$$

From (4.12) and (4.13) note that:

Now the proof follows from Lemma 4.14, Lemma 1.1 and Lemma 4.8. $\square$

5 The Positive Defect Case

By Lemma 3.2 (see also [13, 16]), $C(V_1,V_2)\ge 0$ if and only if $(V_1,V_2)$ is doubly commuting. The structure of doubly commuting pair of isometries are well understood in the literature; see [11, 18, 21]. We rephrase and shine some of the existing results using the defect operator. We also study the joint spectrum in detail.

5.1 The Prototypical Example

Definition 5.1

The fundamental isometric pair of positive defect is the bi-shift $(M_{z_1}, M_{z_2}),$ the pair of multiplication by the coordinate functions on $H^2_{{{\mathbb {D}}}^2}$.

It is folklore that the $M_{z_i}$ are isometries and the defect is positive because of the following lemma. It will be clear from Theorem 5.10 why we call it fundamental.

Lemma 5.2

The defect operator $C(M_{z_1}, M_{z_2})$ is the projection onto the one dimensional space of constant functions.

Proof

The proof is a straightforward computation. $\square$

If $(V_1,V_2)$ is a bi-shift on ${\mathcal {H}}$ with the wandering subspace ${\mathcal {R}},$ then $\Lambda :{\mathcal {H}}\rightarrow H^2_{{{\mathbb {D}}}^2}\otimes {\mathcal {R}}$ given by $\Lambda (V_1^{n_1}V_2^{n_2}x)= z_1^{n_1}z_2^{n_2}\otimes x, x\in {\mathcal {R}}$ is a unitary and $\Lambda V_i \Lambda ^*= M_{z_i}\otimes I_{{\mathcal {R}}}$ for $i=1,2.$ Also, by the above lemma $\dim ({\text {ran}}C(V_1,V_2))=\dim {\mathcal {R}}.$ The following lemma is well known.

Lemma 5.3

The joint spectrum $\sigma (M_{z_1}, M_{z_2})$ is the whole bidisc $\overline{{\mathbb {D}}^2}$. Indeed, every point in the open bidisc is a joint eigenvalue for $(M_{z_1}^*,M_{z_2}^*).$ In particular, for every $(w_1, w_2) \in {{\mathbb {D}}}^2$, the non-singularity of $K(M_{z_1}-w_1I,M_{z_2}-w_2I)$ is broken at the third stage.

Remark 5.4

Recall the pair $(\tau _1, \tau _2)$ defined in Sect. 4.1. One can observe that $\{\tau _1^m\tau _2^n(1):m,n\ge 0\}$ is an orthonormal subset of $H^2_{{{\mathbb {D}}}^2}.$ Let ${\mathcal {M}}=\overline{{\text {span}}}\{\tau _1^m\tau _2^n(1):m,n\ge 0\}\subset H^2_{{{\mathbb {D}}}^2}.$ Clearly it is a joint invariant subspace (but not reducing) for $(\tau _1,\tau _2).$ Identify ${\mathcal {M}}$ and $H^2_{{{\mathbb {D}}}^2}$ via $\tau _1^m\tau _2^n(1)\mapsto z_1^mz_2^n.$ Clearly, the pair of isometries $(\tau _1|_{\mathcal {M}},\tau _2|_{\mathcal {M}})$ gets identified with $(M_{z_1},M_{z_2})$ on $H^2_{{{\mathbb {D}}}^2}$. Since $C(M_{z_1},M_{z_2})\ge 0$ and $C(M_{z_1},M_{z_2})\ne 0$, the same is true for the pair $(\tau _1|_{\mathcal {M}},\tau _2|_{\mathcal {M}})$. Thus, a pair of commuting isometries with negative defect, when restricted to a joint invariant subspace, can have positive defect.

5.2 The General Case of the Positive Defect Operator

We start with a characterization available in [13] and [16].

Lemma 5.5

Let $(V_1,V_2)$ be a pair of commuting isometries on a Hilbert space ${\mathcal {H}}$. Then the following are equivalent:

• (a) $C(V_1,V_2)\ge 0$ and $C(V_1,V_2)\ne 0.$

• (b) $V_2(\ker V_1^*)\subsetneq \ker V_1^*.$

• (c) $V_1(\ker V_2^*)\subsetneq \ker V_2^*.$

• (d) The fringe operators are isometries and not unitaries.

• (e) $(V_1,V_2)$ is doubly commuting and $C(V_1,V_2)\ne 0.$

• (f) $C(V_1,V_2)$ is a non-zero projection.

• (g) If $({\mathcal {E}},P,U)$ is the BCL triple for $(V_1,V_2),$ then $U({\text {ran}}P)\subsetneq {\text {ran}}P.$

Let be the multipliers associated with the BCL triple $(l^2({\mathbb {Z}}),p_{0+},\omega )$ where $p_{0+}$ is the projection onto $\overline{{\text {span}}}\{e_n:n\ge 0\}$ and $\omega$ is the bilateral shift on $l^2({\mathbb {Z}})$. Now, we obtain a structure theorem which has its own independent interest and is applied later to obtain Theorem 5.13.

Theorem 5.6

Let $(V_1,V_2)$ be a pair of commuting isometries such that $C(V_1,V_2)$ is a non-zero positive operator. Let $({\mathcal {E}},P,U)$ be the BCL triple for $(V_1,V_2).$ Then, up to unitary equivalence

$$\begin{aligned} {\mathcal {E}}= (l^2({\mathbb {Z}})\otimes {\mathcal {L}}) \oplus {\mathcal {E}}_2 \end{aligned}$$

for some non-trivial closed subspace ${\mathcal {L}}$ and a closed subspace ${\mathcal {E}}_2$ of ${\mathcal {E}}.$ Moreover

(5.1)

with $C({M_{\varphi _1}}|_{H^2_{{{\mathbb {D}}}}({\mathcal {E}}_2)},{M_{\varphi _2}}|_{H^2_{{{\mathbb {D}}}}({\mathcal {E}}_2)})=0.$ In particular,

$$\begin{aligned} \sigma (M_{\eta _1},M_{\eta _2})= \sigma (M_{\eta _1}\otimes I_{\mathcal {L}}, M_{\eta _2}\otimes I_{\mathcal {L}})\subseteq \sigma (M_{\varphi _1},M_{\varphi _2})\subseteq \sigma (V_1,V_2). \end{aligned}$$

(5.2)

Proof

We shall not give details of this proof because it follows the same line as the proof of Theorem 4.7. $\square$

Lemma 5.7

The pair $(M_{\eta _1},M_{\eta _2})$ is jointly unitarily equivalent to $(M_{z_1},M_{z_2}).$ In particular, $\sigma (M_{\eta _1},M_{\eta _2})=\overline{{{\mathbb {D}}}^2}.$

Proof

Define the unitary $\Lambda :H^2_{{{\mathbb {D}}}^2}\rightarrow H^2_{{\mathbb {D}}}(l^2({\mathbb {Z}}))$ by

$$\begin{aligned} \Lambda \left( \sum _{m,n=0}^{\infty }a_{m,n}z_1^mz_2^n\right) =\sum _{k=0}^\infty \left( \sum _{m= 0}^\infty a_{m+k,k}e_{-(m+1)}+\sum _{m= 1}^\infty a_{k,m+k}e_{m-1}\right) z^k. \end{aligned}$$

(5.3)

Then, $\Lambda M_{z_i}\Lambda ^*=M_{\eta _i}$ for $i=1,2.$ That is, $(M_{\eta _1},M_{\eta _2})$ is jointly unitarily equivalent to $(M_{z_1},M_{z_2}).$

In particular, $\sigma (M_{\eta _1},M_{\eta _2})=\sigma (M_{z_1},M_{z_2})=\overline{{{\mathbb {D}}}^2},$ by Lemma 5.3. $\square$

Remark 5.8

$(l^2({\mathbb {Z}})\otimes {\mathcal {L}}, p_{0+}\otimes I_{\mathcal {L}},\omega \otimes I_{\mathcal {L}})$ is the BCL triple for the bi-shift with wandering subspace ${\mathcal {L}}.$ (Compare this with the Remark 4.13).

Lemma 5.9

The pair $(M_{z_1},M_{z_2})$ does not have any non-trivial joint reducing subspace.

The proof of the above lemma is similar to the proof of Lemma 4.10. Now the following structure theorem for $C(V_1,V_2)\ge 0$ follows from Theorem 1.2, Theorem 5.6, Lemma 5.7 and Lemma 5.2. It shows the role played by the bi-shift of a possibly higher multiplicity and is a rephrasing of some results in [11, 21] using the defect operator.

Theorem 5.10

Let $(V_1, V_2)$ be a pair of commuting isometries on ${\mathcal {H}}$ with $C(V_1, V_2)\ge 0$ and $C(V_1, V_2)\ne 0$. Then, there is a non-trivial Hilbert space ${\mathcal {L}}\subsetneq \ker V^*$ such that, up to unitary equivalence, ${\mathcal {H}}={\mathcal {H}}_0\oplus {\mathcal {H}}_0^\perp ,$ where ${\mathcal {H}}_0= H^2_{{{\mathbb {D}}}^2}\otimes {\mathcal {L}}.$ In this decomposition

$$\begin{aligned} V_i = \left( \begin{array}{cc} M_{z_i}\otimes I_{\mathcal {L}}&{} 0 \\ 0 &{} V_{i0}\end{array} \right) \end{aligned}$$

and the defect operator $C(V_{10}, V_{20})$ is zero. Moreover, the dimension of ${\mathcal {L}}$ is the same as the dimension of the range of $C(V_1, V_2)$.

Theorem 5.11

Let $(V_1,V_2)$ be a pair of commuting isometries on a Hilbert space ${\mathcal {H}}$ with $C(V_1,V_2)\ge 0$ and $C(V_1,V_2)\ne 0.$ Then, $\sigma (V_1,V_2)=\overline{{{\mathbb {D}}}^2}$ and every point $(w_1,w_2)\in {{\mathbb {D}}}^2$ is a joint eigenvalue of $(V_1^*,V_2^*)$. In particular, for every $(w_1, w_2) \in {{\mathbb {D}}}^2$, the non-singularity of $K(V_1-w_1I,V_2-w_2I)$ is broken at the third stage.

Proof

The proof follows from Theorem 5.11 and Lemma 5.3. $\square$

The following lemma is useful to see the relation between the joint spectrum of a pair of commuting isometries $(V_1,V_2)$ with defect positive and non-zero, and the joint spectra $\sigma (\varphi _1(z),\varphi _2(z)),$ $z\in {{\mathbb {D}}},$ where $\varphi _i$’s are the multipliers given in the BCL theorem.

Lemma 5.12

Let the operator valued functions $\eta _1$ and $\eta _2$ be as in Theorem 5.6. Then, for $z\in {{\mathbb {D}}},$ $(\overline{\lambda _1},\overline{\lambda _2})\in {{\mathbb {D}}}^2$ is a joint eigenvalue for $(\eta _1(z)^*,\eta _2(z)^*)$ if $\lambda _1\lambda _2=z.$ Also,

$$\begin{aligned} \sigma (\eta _1(z),\eta _2(z))=\{(\lambda _1,\lambda _2)\in \overline{{{\mathbb {D}}}^2}:\lambda _1\lambda _2=z\}, \end{aligned}$$

(5.4)

and

$$\begin{aligned} \overline{\cup _{z\in {{\mathbb {D}}}}\sigma (\eta _1(z),\eta _2(z))} =\overline{{{\mathbb {D}}}^2}=\sigma (M_{\eta _1},M_{\eta _2}). \end{aligned}$$

(5.5)

Proof

Let $z\in {{\mathbb {D}}}\setminus \{0\}$ and $\lambda _1,\lambda _2\in {{\mathbb {D}}}$ be such that $\lambda _1\lambda _2=z.$ Consider

$$\begin{aligned} x_{\lambda _1,\lambda _2}=\sum _{n= 0}^\infty \overline{\lambda _2}^{n}e_n+\frac{1}{\overline{\lambda _2}}\sum _{n= 1}^\infty \overline{\lambda _1}^{n-1}e_{-n}. \end{aligned}$$

Note that

$$\begin{aligned} \eta _i(z)^*x_{\lambda _1,\lambda _2}={\overline{\lambda }}_ix_{\lambda _1,\lambda _2}, i=1,2. \end{aligned}$$

That is, $(\overline{\lambda _1},\overline{\lambda _2})$ is a joint eigenvalue for $(\eta _1(z)^*, \eta _2(z)^*)$ with eigenvector $x_{\lambda _1,\lambda _2},$ where $z\ne 0$ and $\lambda _1,\lambda _2\in {{\mathbb {D}}}$ are such that $\lambda _1\lambda _2=z.$ Therefore,

$$\begin{aligned} \sigma (\eta _1(z),\eta _2(z))\supseteq \{(\lambda _1,\lambda _2)\in \overline{{{\mathbb {D}}}^2}:\lambda _1\lambda _2=z\} \text { for } z\ne 0. \end{aligned}$$

As $\eta _1(z)$ and $\eta _2(z)$ are commuting contractions, we have $\sigma (\eta _1(z),\eta _2(z))\subseteq \overline{{{\mathbb {D}}}^2}.$ Since $\eta _1(z)\eta _2(z)=zI,$ by spectral mapping theorem, we have the other inclusion. Hence

$$\begin{aligned} \sigma (\eta _1(z),\eta _2(z))=\{(\lambda _1,\lambda _2)\in \overline{{{\mathbb {D}}}^2}:\lambda _1\lambda _2=z\} \text { for } z\ne 0. \end{aligned}$$

(5.6)

For the case $z=0,$ consider for $\lambda _1,\lambda _2\in {{\mathbb {D}}}\setminus \{0\},$

$$\begin{aligned} x_{\lambda _1,0}=\sum _{n=1}^\infty {\overline{\lambda _1}}^{n-1}e_{-n},\ x_{0,\lambda _2}=\sum _{n=0}^\infty \overline{\lambda _2}^ne_{n}+\frac{1}{\overline{\lambda _2}}e_{-1}\text { and } x_{0,0}=e_{-1} . \end{aligned}$$

Then $(\overline{\lambda _1},0),(0,\overline{\lambda _2})$ and (0, 0) are joint eigenvalues for $(\eta _1(0)^*,\eta _2(0)^*)$ with the joint eigenvectors $x_{\lambda _1,0},x_{0,\lambda _2}$ and $x_{0,0}$ respectively. Therefore, by the similar reasoning as in the case of $z\ne 0,$ we get

$$\begin{aligned} \sigma (\eta _1(0),\eta _2(0))=\overline{{\mathbb {D}}}\times \{0\}\cup \{0\}\times \overline{{\mathbb {D}}}=\{(\lambda _1,\lambda _2)\in \overline{{{\mathbb {D}}}^2}:\lambda _1\lambda _2=0\}. \end{aligned}$$

(5.7)

Hence, by (5.6), (5.7) and Lemma 5.7, we get (5.5).

Indeed, one can show that the joint eigenspaces are one dimensional, for all the joint eigenvalues of $(\eta _1(z)^*,\eta _2(z)^*), z\in {{\mathbb {D}}}.$ $\square$

Theorem 5.13

Let $(V_1,V_2)$ be a pair of commuting isometries such that $C(V_1,V_2)\ge 0$ and $C(V_1,V_2)\ne 0.$ Let $({\mathcal {E}},P,U)$ be the BCL triple for $(V_1,V_2).$ Then, for $z\in {{\mathbb {D}}},$ $(\overline{\lambda _1},\overline{\lambda _2})\in {{\mathbb {D}}}^2$ is a joint eigenvalue for $(\varphi _1(z)^*,\varphi _2(z)^*)$ if $\lambda _1\lambda _2=z,$ also

$$\begin{aligned} \sigma (\varphi _1(z),\varphi _2(z))=\{(\lambda _1,\lambda _2)\in \overline{{{\mathbb {D}}}^2}:\lambda _1\lambda _2=z\}. \end{aligned}$$

(5.8)

Moreover, every point in the open bidisc is a joint eigenvalue for $(M_{\varphi _1}^*,M_{\varphi _2}^*),$ and

$$\begin{aligned} \sigma (V_1,V_2)=\sigma (M_{\varphi _1},M_{\varphi _2})=\overline{\cup _{z\in {{\mathbb {D}}}}\sigma (\varphi _1(z),\varphi _2(z))}=\overline{{{\mathbb {D}}}^2}. \end{aligned}$$

(5.9)

Proof

Since $\varphi _1(z)$ and $\varphi _2(z)$ are commuting contractions, $\sigma (\varphi _1(z),\varphi _2(z))\subseteq \overline{{{\mathbb {D}}}^2}.$ As $\varphi _1(z)\varphi _2(z)=zI,$ by spectral mapping theorem,

$$\begin{aligned} \sigma (\varphi _1(z),\varphi _2(z))\subseteq \{(\lambda _1,\lambda _2)\in \overline{{{\mathbb {D}}}^2}:\lambda _1\lambda _2=z\}. \end{aligned}$$

In the same manner as in the proof of Theorem 4.7, we get

where ${\mathcal {L}}={\text {ran}}P\ominus U({\text {ran}}P)$ and ${\mathcal {E}}_2=\ker V^*\ominus ( l^2({\mathbb {Z}})\otimes {\mathcal {L}}).$ Now the proof follows from Lemma 5.12, Lemma 5.7 and Lemma 5.3. $\square$

6 Towards the General Defect Operator

This section deals with the cases of ${\text {ran}}V_1={\text {ran}}V_2$ and ${\text {ran}}V_2\subsetneq {\text {ran}}V_1.$ We provide the characterization and study the joint spectrum for both the cases. At the end of this section we give an example for the unknown case of ${\mathcal {H}}_i\ne 0$ for all $i=1,2,3,4.$

6.1 Range of $V_1$ Equal to the Range of $V_2$

In this case the defect operator is the difference of two mutually orthogonal projections whose ranges together span the kernel of $V^*$. The structure in this case, known from [4], is briefly recalled below because it is needed for deciphering the joint spectrum.

6.1.1 The Prototypical Family of Examples

Let ${\mathcal {L}}$ be a Hilbert space and W be a unitary on ${\mathcal {L}}$. Consider the pair of commuting isometries $(M_z\otimes I, M_z\otimes W)$ on $H^2_{{{\mathbb {D}}}}\otimes {\mathcal {L}}$. Clearly ${\text {ran}}(M_z\otimes I)={\text {ran}}(M_z\otimes W).$

Lemma 6.1

Let ${\mathcal {L}}$ and W be as above. If $(V_1, V_2) = (M_z\otimes I, M_z\otimes W)$, then there are two projections P and Q such that

• (1) they are mutually orthogonal to each other,

• (2) the dimensions of ranges of P and Q are same,

• (3) the span of the ranges of P and Q is the kernel of $V^*$ and

• (4) the defect operator $C(V_1, V_2)$ on $H^2_{{{\mathbb {D}}}}\otimes {\mathcal {L}}$ is $P - Q$.

Proof

In keeping with the notation $E_0$ for the projection onto the one dimensional subspace of constants in $H^2_{{{\mathbb {D}}}}$, let us denote by $E_1$ the projection onto the one dimensional space spanned by z. The kernel of $V^*$ is

$$\begin{aligned} {\text {ran}}(I \otimes I_{\mathcal {L}}- M_z^2(M_z^2)^* \otimes I_{\mathcal {L}}) = {\text {ran}}((E_0 + E_1) \otimes I_{\mathcal {L}})=\overline{{\text {span}}}\{1,z\}\otimes {\mathcal {L}}\end{aligned}$$

and the defect operator is

$$\begin{aligned} C(V_1, V_2)&= I \otimes I_{\mathcal {L}}- M_zM_z^* \otimes I_{\mathcal {L}}- M_zM_z^* \otimes I_{\mathcal {L}}+ M_z^2(M_z^2)^* \otimes I_{\mathcal {L}}\\&= E_0 \otimes I_{\mathcal {L}}- M_zE_0M_z^* \otimes I_{\mathcal {L}}\\&= E_0 \otimes I_{\mathcal {L}}- E_1 \otimes I_{\mathcal {L}}. \end{aligned}$$

Thus, setting $P = E_0 \otimes I_{\mathcal {L}}$ and $Q = E_1 \otimes I_{\mathcal {L}}$, we are done. $\square$

Lemma 6.2

If ${\mathcal {L}}\ne \{0\},$ then the joint spectrum of $(M_{z}\otimes I_{{\mathcal {L}}}, M_{z}\otimes W)$ is

$$\begin{aligned} \sigma (M_{z}\otimes I_{{\mathcal {L}}}, M_{z}\otimes W)=\{z(1,\alpha ): z\in \overline{{\mathbb {D}}}, \alpha \in \sigma (W)\}. \end{aligned}$$

Proof

This is a straightforward application of the polynomial spectral mapping theorem. Consider the polynomial $f(x_1,x_2)=(x_1, x_1x_2).$ Then,

$$\begin{aligned} \sigma (M_{z}\otimes I_{{\mathcal {L}}}, M_{z}\otimes W)&=f(\sigma (M_{z}\otimes I_{{\mathcal {L}}}, I_{H^2_{{\mathbb {D}}}}\otimes W))\nonumber \\&=f(\sigma (M_{z})\times \sigma (W)) \nonumber \\&=\{z(1,\alpha ): z\in \overline{{\mathbb {D}}}, \alpha \in \sigma (W)\}. \end{aligned}$$

(6.1)

$\square$

6.1.2 General Theory

Theorem 6.3

Let $(V_1,V_2)$ be a pair of commuting isometries on a Hilbert space ${\mathcal {H}}$. Then the following are equivalent:

• (a) ${\text {ran}}V_1={\text {ran}}V_2.$

• (b) $V_1(\ker V_2^*)=V_2(\ker V_1^*).$

• (c) The defect operator $C(V_1,V_2)$ is a difference of two mutually orthogonal projections $Q_1, Q_2$ with ${\text {ran}}Q_1\oplus {\text {ran}}Q_2=\ker V^*.$

• (d) The fringe operators $F_1$ and $F_2$ are zero.

• (e) If $({\mathcal {E}},P,U)$ is the BCL triple for $(V_1,V_2),$ then $U({\text {ran}}P)={\text {ran}}P^\perp$ (or equivalently $U({\text {ran}}P^\perp )={\text {ran}}P$).

Proof

$(a)\Rightarrow (b)$ This is immediate from the equality

$$\begin{aligned} \ker V_1^*\oplus V_1(\ker V_2^*)= \ker V^*=\ker V_2^*\oplus V_2(\ker V_1^*) \end{aligned}$$

after we note that ${\text {ran}}V_1={\text {ran}}V_2$ implies that $\ker V_1^* = \ker V_2^*.$

$(b)\Rightarrow (c)$ Set $P_1 = P_{\ker V_1^*}$ and $P_2 = P_{V_2(\ker V_1^*)}.$ By (2.2), $C(V_1,V_2)=P_1- P_2$. Moreover,

$$\begin{aligned} {\text {ran}}P_1\oplus {\text {ran}}P_2&=\ker V_1^*\oplus V_2(\ker V_1^*)\\&=\ker V_1^*\oplus V_1(\ker V_2^*) \quad \text {(by (b))}\\&=\ker V^*. \end{aligned}$$

$(c)\Rightarrow (d)$ It is immediate by noticing that

$$\begin{aligned} Q_1=P_{\ker V_1^*}=P_{\ker V_2^*}\quad \text { and }\quad Q_2=P_{V_2(\ker V_1^*)}=P_{V_1(\ker V_2^*)} \end{aligned}$$

from the discussions in Sect. 2.

$(c) \Rightarrow (a)$ Immediate from the last line above.

$(d)\Rightarrow (c)$ $F_1=0$ and $F_2=0$ implies that $\ker V_1^*\perp V_2(\ker V_1^*)$ and $\ker V_2^*\perp V_1(\ker V_2^*).$ Now from (2.3), we see that $V_1(\ker V_2^*)=V_2(\ker V_1^*).$ Set $P_1=P_{\ker V_1^*}$ and $P_2=P_{V_2(\ker V_1^*)},$ to obtain (c).

Thus, we have shown the equivalences of $(a),\ (b),\ (c)$ and (d).

$(a)\iff (e)$ If $({\mathcal {E}},P,U)$ is the BCL triple for $(V_1,V_2),$ then

$$\begin{aligned} {\text {ran}}V_1 = {\text {ran}}V_2 \end{aligned}$$

if and only if

$$\begin{aligned} \ker V_1^* = \ker V_2^* \end{aligned}$$

if and only if

$$\begin{aligned} \ker M_{\varphi _1}^* = \ker M_{\varphi _2}^* \end{aligned}$$

if and only if

$$\begin{aligned} {\text {ran}}E_0\otimes U^*PU={\text {ran}}E_0\otimes P^\perp \end{aligned}$$

if and only if

$$\begin{aligned} U^*PU = P^\perp . \end{aligned}$$

$\square$

The following result is a special case of Remark 4.2 in [4], which shows that the pair $(M_z\otimes I,M_z\otimes W),$ studied in Sect. 6.1.1, is the prototypical pair in this case. We give a proof which is different from the one in [4].

Theorem 6.4

Let $(V_1,V_2)$ be a pair of commuting isometries on ${\mathcal {H}}$ satisfying any of the equivalent conditions in Theorem 6.3. Then, there exist Hilbert spaces ${\mathcal {L}}$ and ${\mathcal {K}}$ such that up to unitarily equivalence ${\mathcal {H}}=(H^2_{{{\mathbb {D}}}}\otimes {\mathcal {L}}) \oplus {\mathcal {K}}$ and in this decomposition,

$$\begin{aligned} V_1=\begin{pmatrix} M_z\otimes I_{{\mathcal {L}}}&{} 0\\ 0&{}W_1 \end{pmatrix}, \quad V_2= \begin{pmatrix} M_z\otimes W&{} 0\\ 0&{}W_2 \end{pmatrix}, \end{aligned}$$

for some unitary W on ${\mathcal {L}}$ and commuting unitaries $W_1, W_2$ on ${\mathcal {K}}.$

Proof

Let $({\mathcal {E}},P,U)$ be the BCL triple for $(V_1,V_2).$ Let ${\mathcal {L}}={\text {ran}}P^\perp .$ By Theorem 6.3 (e), we have $U({\mathcal {L}})={{\mathcal {L}}}^{\perp }$ and $U({\mathcal {L}}^{\perp })={{\mathcal {L}}}.$ Let $U_1=U|_{\mathcal {L}}:{\mathcal {L}}\rightarrow {\mathcal {L}}^\perp$ and $U_2=U|_{{\mathcal {L}}^\perp }:{\mathcal {L}}^\perp \rightarrow {\mathcal {L}}.$ In the decomposition ${\mathcal {E}}={{\mathcal {L}}}^{\perp }\oplus {\mathcal {L}},$ we have

$$\begin{aligned} U=\begin{pmatrix} 0 &{} U_1 \\ U_2 &{} 0 \end{pmatrix} \quad \text { and }\quad P=\begin{pmatrix} I_{{\mathcal {L}}^{\perp }} &{} 0 \\ 0 &{} 0 \end{pmatrix}. \end{aligned}$$

Hence $\varphi _1(z)=\begin{pmatrix} 0 &{} U_2^{*} \\ z U_1^{*} &{} 0 \end{pmatrix}$ and $\varphi _2(z)=\begin{pmatrix} 0 &{} U_1 \\ z U_2 &{} 0 \end{pmatrix}$ for $z\in {{\mathbb {D}}},$ and

$$\begin{aligned} M_{\varphi _1}=\begin{pmatrix} 0 &{} I\otimes U_2^* \\ M_{z}\otimes U_1^* &{} 0 \end{pmatrix} \quad \text { and } \quad M_{\varphi _2}=\begin{pmatrix} 0 &{} I\otimes U_1 \\ M_z\otimes U_2 &{} 0 \end{pmatrix}. \end{aligned}$$

Let us define the unitary

$$\begin{aligned} \Lambda : (H^2_{{\mathbb {D}}}\otimes {{\mathcal {L}}^\perp }) \oplus (H^2_{{\mathbb {D}}}\otimes {\mathcal {L}})\rightarrow H^2_{{\mathbb {D}}}\otimes {\mathcal {L}}\end{aligned}$$

given by,

$$\begin{aligned} \Lambda \left( \begin{pmatrix} \sum _{n=0}^{\infty } a_{n}z^{n} \\ \sum _{n=0}^{\infty } b_{n}z^{n} \end{pmatrix}\right) :=\sum _{n \text { is even }} {(U_2U_1)}^{\frac{n}{2}}(b_{\frac{n}{2}})z^{n}+\sum _{n \text { is odd }} {(U_2U_1)}^{\frac{n-1}{2}}U_2(a_{\frac{n-1}{2}})z^{n}, \end{aligned}$$

$a_n\in {\mathcal {L}}^\perp ,b_n\in {\mathcal {L}}.$ One can see that $\Lambda M_{\varphi _1}{\Lambda }^{*}= M_{z}\otimes I_{{\mathcal {L}}}$ and $\Lambda M_{\varphi _2}{\Lambda }^{*}= M_{z}\otimes W,$ where $W=U_2U_1.$ This completes the proof by Theorem 1.2. $\square$

Corollary 6.5

Let $(V_1,V_2)$ be a pair of commuting isometries satisfying any of the equivalent conditions in Theorem 6.3. If $V_1V_2$ is pure, then both $V_1$ and $V_2$ are pure.

Since the BCL triple is unique (up to unitary equivalence), we shall use the following convenient choice of BCL triple due to A. Maji et al in [16] for a pair of commuting isometries.

Theorem 6.6

Let $(V_1,V_2)$ be a pair of commuting isometries on a Hilbert space ${\mathcal {H}}$. Then, $(\ker V^*,P,U_0)$ is the BCL triple for $(V_1,V_2),$ where is the orthogonal projection onto $V_2(\ker V_1^*)$ and

$$\begin{aligned} U_0=\begin{pmatrix} V_2|_{\ker V_1^*} &{} 0 \\ 0 &{} V_1^*|_{V_1(\ker V_2^*)} \end{pmatrix}: \begin{array}{ccc} \ker V_1^*&{} &{}V_2(\ker V_1^*) \\ \oplus &{}\rightarrow &{}\oplus \\ V_1(\ker V_2^*)&{} &{}\ker V_2^* \end{array} \end{aligned}$$

is a unitary operator on $\ker V^*.$

Remark 6.7

In Theorem 6.4, we can write the W and ${\mathcal {L}}$ explicitly in terms of $V_1$ and $V_2$ as follows:

By Theorem 6.6 and Theorem 6.3, ${\mathcal {L}}={\text {ran}}P^\perp =\ker V_1^*,$ $U_1:\ker V_1^*\rightarrow V_1(\ker V_1^*)$ given by $U_1=V_2|_{\ker V_1^*}$ and $U_2:V_1(\ker V_1^*)\rightarrow \ker V_1^*$ given by $U_2=V_1^*|_{V_1(\ker V_1^*)}.$ Therefore $W=V_1^*|_{V_1(\ker V_1^*)}V_2|_{\ker V_1^*}.$

6.1.3 Joint Spectrum

If $(V_1,V_2)$ is pure and satisfying any of the equivalent conditions in Theorem 6.3, then by Theorem 6.4, Remark 6.7 and by Sect. 6.1, we have

$$\begin{aligned} \sigma (V_1,V_2) =\{z(1,e^{i\theta }): z\in {\overline{{{\mathbb {D}}}}}, \ e^{i\theta }\in \sigma (U_2U_1)\}, \end{aligned}$$

(6.2)

where $U_1=V_2|_{\ker V_1^*},U_2=V_1^*|_{V_1(\ker V_1^*)}.$

The final theorem of this section tells us the nature of elements in the joint spectrum and the relation between the joint spectrum of the commuting isometries satisfying any of the equivalent conditions in Theorem 6.3 and the joint spectra of the associated multipliers at every point of ${{\mathbb {D}}}.$

Theorem 6.8

Let $(V_1,V_2)$ be a pair of commuting isometries on ${\mathcal {H}}$ satisfying any of the equivalent conditions in Theorem 6.3 and $\ker V^*\ne \{0\}.$ Let $({\mathcal {E}},P,U)$ be the BCL triple for $(V_1,V_2).$ Let $U_1:{\text {ran}}P^\perp \rightarrow {\text {ran}}P \text { be the unitary given by } U_1(x)=U(x), x\in {\text {ran}}P^\perp$ and $U_2:{\text {ran}}P \rightarrow {\text {ran}}P^\perp \text { be the unitary given by } U_2(y)=U(y), y\in {\text {ran}}P.$ Then

• $\sigma (\varphi _1(z),\varphi _2(z))=\{\pm \sqrt{z} (e^{-i \frac{\theta }{2}} , e^{i \frac{\theta }{2}}) : e^{i \theta } \in \sigma (U_1U_2) \}.$ Here for every point in the joint spectrum, the non-singularity breaks at stage 3.

• $\sigma (M_{\varphi _1},M_{\varphi _2})=\overline{\cup _{z\in {{\mathbb {D}}}}\sigma (\varphi _1(z),\varphi _2(z))}= \{z(1,e^{i\theta }): z\in {\overline{{{\mathbb {D}}}}}, \ e^{i\theta }\in \sigma (U_1U_2)\}.$ Here, for every point $(z_1,z_2)$ in the set $\{z(1,e^{i\theta }): z\in {{\mathbb {D}}}, \ e^{i\theta }\in \sigma (U_1U_2)\}$, the non-singularity in the Koszul complex $K(M_{\varphi _1}-z_1I,M_{\varphi _2}-z_2I)$ breaks at stage 3.

Proof

Let ${\mathcal {E}}_1={\text {ran}}P$ and ${\mathcal {E}}_2={\text {ran}}P^\perp .$ We have by Theorem 6.3 that $U({\mathcal {E}}_1)={\mathcal {E}}_2$ and $U({\mathcal {E}}_2)={\mathcal {E}}_1.$ Hence in the decomposition ${\mathcal {E}}={\mathcal {E}}_1\oplus {\mathcal {E}}_2,$ we have $\varphi _1(z)=\begin{pmatrix} 0 &{} {U_2^{*}} \\ z {U_1^{*}} &{} 0 \end{pmatrix}$ and $\varphi _2(z)=\begin{pmatrix} 0 &{} U_1 \\ z U_2 &{} 0 \end{pmatrix},$ for $z\in {{\mathbb {D}}},$ where $U_1$ and $U_2$ are as in the statement.

Now, we shall show that: for $z\in {{\mathbb {D}}},$

$$\begin{aligned} \sigma (\varphi _1(z))&=\{w\in {{\mathbb {D}}}: w^{2}={\overline{\alpha }}z\text { for some } \alpha \in \sigma (U_1U_2)\}, \end{aligned}$$

(6.3)

$$\begin{aligned} \sigma (\varphi _2(z))&=\{w\in {{\mathbb {D}}}: w^{2}={\alpha }z\text { for some } \alpha \in \sigma (U_1U_2)\}. \end{aligned}$$

(6.4)

Clearly (6.3) and (6.4) hold for $z=0$. So assume that $z\ne 0$.

Assertion 1

If $\lambda$ is an eigenvalue of $\varphi _1(z)$ then $\frac{\overline{\lambda ^{2}}}{{\overline{z}}}$ is an eigenvalue of $U_1U_2$. In particular, $\lambda \in {{\mathbb {D}}}$.

Proof of Assertion 1

If $\lambda$ is an eigenvalue of $\varphi _1(z),$ there exists a non-zero vector $h_1\oplus h_2 \in {\mathcal {E}}_1\oplus {\mathcal {E}}_2$ such that $\varphi _1(z)\begin{pmatrix} h_1 \\ h_2 \end{pmatrix}=\lambda \begin{pmatrix} h_1 \\ h_2 \end{pmatrix}$ which implies ${U_2}^{*}(h_2)=\lambda h_1$ and $\frac{z}{\lambda } {U_1^{*}}(h_1)= h_2.$ Hence ${(U_1U_2)^{*}}h_1=\frac{{\lambda ^{2}}}{z}h_1$. Notice that $h_1\ne 0$, otherwise ${U_2}^{*}(h_2)=\lambda h_1$ implies $h_2=0$. $\square$

Assertion 2

For $\lambda \in {\mathbb {C}}$, $(\varphi _1(z)-\lambda I_{{\mathcal {E}}})$ is not onto if and only if $({(U_1U_2)^{*}}-\frac{{\lambda ^{2}}}{z} I_{{\mathcal {E}}_1})$ is not onto. In particular, if $(\varphi _1(z)-\lambda I_{{\mathcal {E}}})$ is not onto then $\lambda \in {{\mathbb {D}}}$. Also

$$\begin{aligned} {\text {ran}}(\varphi _1(z)-\lambda I)=\left\{ \begin{pmatrix} h_1 \\ h_2 \end{pmatrix}\in {\mathcal {E}}_1\oplus {\mathcal {E}}_2: \lambda h_1+ U_2^* h_2\in {\text {ran}}((U_1U_2)^*-\frac{\lambda ^2}{z} I) \right\} .\nonumber \\ \end{aligned}$$

(6.5)

Proof of Assertion 2

Let $\begin{pmatrix} h_1 \\ h_2 \end{pmatrix}\in {\mathcal {E}}_1\oplus {\mathcal {E}}_2.$ We have $(\varphi _1(z)-\lambda I)\begin{pmatrix} h_1 \\ h_2 \end{pmatrix}=\begin{pmatrix} -\lambda h_1+{U_2}^{*} h_2 \\ z {U_1}^{*} h_1- \lambda h_2 \end{pmatrix}.$ Let $k=\begin{pmatrix} k_1 \\ k_2 \end{pmatrix}\in {\mathcal {E}}_1\oplus {\mathcal {E}}_2.$ Now $(\varphi _1(z)-\lambda I)\begin{pmatrix} h_1 \\ h_2 \end{pmatrix}=\begin{pmatrix} k_1 \\ k_2 \end{pmatrix}$ if and only if $-\lambda h_1 +{U_2}^{*} (h_2)=k_1$ and

$$\begin{aligned} ({(U_1U_2)^{*}}-\frac{{\lambda ^{2}}}{z} I_{{\mathcal {E}}_1})(h_1)=\frac{\lambda k_1+{U_2}^{*}(k_2)}{z}. \end{aligned}$$

This proves Assertion 2.

Using Assertions  1 and  2 and the fact that for any bounded normal operator T if $\alpha \in \sigma (T)$ then $T- \alpha I$ is not onto, we get (6.3) for $z\ne 0$. In a similar way, one can show (6.4) and for $z\ne 0$

$$\begin{aligned} {\text {ran}}(\varphi _2(z)-\lambda I)=\left\{ \begin{pmatrix} k_1 \\ k_2 \end{pmatrix}\in {\mathcal {E}}_1\oplus {\mathcal {E}}_2: \lambda k_1+ U_1 k_2\in \ {\text {ran}}((U_1U_2)-\frac{{\lambda ^{2}}}{z} I) \right\} .\nonumber \\ \end{aligned}$$

(6.6)

For $z\in {{\mathbb {D}}},$ since

$$\begin{aligned} \sigma (\varphi _1(z),\varphi _2(z))\subseteq \sigma (\varphi _1(z))\times \sigma (\varphi _2(z)) \end{aligned}$$

and

$$\begin{aligned} \varphi _1(z)\varphi _2(z)=zI_{{\mathcal {E}}_1\oplus {\mathcal {E}}_2}=\varphi _2(z)\varphi _1(z), \end{aligned}$$

by spectral mapping theorem we get,

$$\begin{aligned}&\sigma (\varphi _1(z),\varphi _2(z))\\&\subseteq \{(z_1,z_2)\in {{\mathbb {D}}}^2: {z_1}^{2}={\overline{\alpha }}z, {z_2}^{2}=\beta z, z_1z_2=z, \text { for some } \alpha ,\beta \in \sigma (U_1U_2)\}.\\&= \{(z_1,z_2)\in {{\mathbb {D}}}^2: {z_1}^{2}={\overline{\alpha }}z, {z_2}^{2}=\alpha z, z_1z_2=z, \text { for some } \alpha \in \sigma (U_1U_2)\}. \end{aligned}$$

Let $z\ne 0$ and $z_1,z_2\in {{\mathbb {D}}}$ be such that ${z_1}^{2}={\overline{\alpha }}z,\ {z_2}^{2}=\alpha z$ and $z_1z_2=z$ for some $\alpha \in \sigma (U_1U_2)$. We shall show that ${\text {ran}}(\varphi _1(z)- z_1 I) +{\text {ran}}(\varphi _2(z)- z_2 I) \ne {\mathcal {E}}_1\oplus {\mathcal {E}}_2$. For this, let $y\notin {\text {ran}}(U_1U_2-\alpha I)$. Suppose $\begin{pmatrix} y \\ 0 \end{pmatrix}\in {\text {ran}}(\varphi _1(z)-z_1 I)+ {\text {ran}}(\varphi _2(z)-z_2 I).$ Then there exists $\begin{pmatrix} h_1 \\ h_2 \end{pmatrix}\in {\text {ran}}(\varphi _1(z)- z_1 I)$ and $\begin{pmatrix} k_1 \\ k_2 \end{pmatrix}\in {\text {ran}}(\varphi _2(z)- z_2 I)$ such that $\begin{pmatrix} h_1 \\ h_2 \end{pmatrix} +\begin{pmatrix} k_1 \\ k_2 \end{pmatrix}=\begin{pmatrix} y \\ 0 \end{pmatrix}.$ Since ${\text {ran}}T={\text {ran}}T^*$ for any bounded normal operator T,  note that from (6.5) and (6.6), we have $z_1h_1+ {U_2}^{*} h_2, z_2k_1+U_1 k_2 \in {\text {ran}}(U_1U_2-\alpha I).$ Since $y\notin {\text {ran}}(U_1U_2-\alpha I),$ we have $(z_1 U_1- z_2 {U_2}^{*})(k_2)\notin {\text {ran}}(U_1U_2-\alpha I).$ So ${(U_1U_2-\alpha I)}^{*}(U_1(k_2))\notin {\text {ran}}(U_1U_2-\alpha I).$ Which is a contradiction, as ${\text {ran}}(U_1U_2-\alpha I)={\text {ran}}(U_1U_2-\alpha I)^*.$ So ${\text {ran}}(\varphi _1(z)- z_1 I) +{\text {ran}}(\varphi _2(z)- z_2 I) \ne {\mathcal {E}}_1\oplus {\mathcal {E}}_2$ for all $z\ne 0,$ $z_1,z_2\in {{\mathbb {D}}}$ such that ${z_1}^{2}={\overline{\alpha }}z,\ {z_2}^{2}=\alpha z$ and $z_1z_2=z$ for some $\alpha \in \sigma (U_1U_2)$. Also, note that ${\text {ran}}(\varphi _1(0)) +{\text {ran}}(\varphi _2(0)) = {\mathcal {E}}_1\oplus 0.$ Hence, we have

$$\begin{aligned}&\sigma (\varphi _1(z),\varphi _2(z))\\ {}&=\{(z_1,z_2)\in {{\mathbb {D}}}^2: {z_1}^{2}={\overline{\alpha }}z, {z_2}^{2}=\alpha z, z_1z_2=z, \text { for some } \alpha \in \sigma (U_1U_2)\}\\&=\{\pm \sqrt{z} (e^{-i \frac{\theta }{2}} , e^{i \frac{\theta }{2}}) : e^{i \theta } \in \sigma (U_1U_2) \} \end{aligned}$$

for $z\in {{\mathbb {D}}}.$

As we saw, for any point $(z_1,z_2)\in \sigma (\varphi _1(z),\varphi _2(z))$, $z\in {{\mathbb {D}}}$,

$$\begin{aligned} {\text {ran}}(\varphi _1(z)-z_1I)+\ {\text {ran}}(\varphi _2(z)-z_2I)\ne {\mathcal {E}}_1\oplus {\mathcal {E}}_2. \end{aligned}$$

Hence

$$\begin{aligned} {\text {ran}}(M_{\varphi _1}-z_1I)+\ {\text {ran}}(M_{\varphi _2}-z_2I)\ne H^{2}({\mathcal {E}}_1\oplus {\mathcal {E}}_2). \end{aligned}$$

Therefore $(z_1,z_2)\in \sigma (M_{\varphi _1},M_{\varphi _2}),$ which implies ${\cup _{z\in {{\mathbb {D}}}}\sigma (\varphi _1(z),\varphi _2(z))}\subseteq \sigma (M_{\varphi _1},M_{\varphi _2}).$ Notice that for every point $(z_1,z_2)$ in ${\cup _{z\in {{\mathbb {D}}}}\sigma (\varphi _1(z),\varphi _2(z))},$ the non-singularity of $K(M_{\varphi _1}-z_1I,M_{\varphi _2}-z_2I)$ breaks at stage 3.

Take $z\in {{\mathbb {D}}}$ and $e^{i\theta }\in \sigma (U_1U_2).$ Let $w=ze^{\frac{i \theta }{2}}, z_1=we^{-\frac{i \theta }{2}}$ and $z_2=we^{\frac{i \theta }{2}}.$ So $(z_1, z_2)\in \sigma (\varphi _1(w^{2}),\varphi _2(w^{2}))$ and $(z_1,z_2)=z(1,e^{i\theta })$. Hence we have the equality:

$$\begin{aligned}&{\cup _{z\in {{\mathbb {D}}}}\sigma (\varphi _1(z),\varphi _2(z))}\nonumber \\&=\cup _{z\in {{\mathbb {D}}}}\{(z_1,z_2)\in {{\mathbb {D}}}^2: {z_1}^{2}={\overline{\alpha }}z, {z_2}^{2}=\alpha z, z_1z_2=z, \text { for some } \alpha \in \sigma (U_1U_2)\}\nonumber \\&=\{z(1,\alpha ): z\in {{\mathbb {D}}}, \alpha \in \sigma (U_1 U_2)\}. \end{aligned}$$

(6.7)

Now as in the proof of Theorem 6.4, we have $(M_{\varphi _1}, M_{\varphi _2})$ is jointly unitarily equivalent to $(M_{z}\otimes I_{{\mathcal {E}}_2}, M_{z}\otimes U_2U_1),$ hence from (6.1) and (6.7), we have

$$\begin{aligned} \sigma (M_{\varphi _1}, M_{\varphi _2})=\sigma (M_{z}\otimes I_{{\mathcal {E}}_2}, M_{z}\otimes U_2U_1)&=\{z(1,\alpha ):z\in \overline{{\mathbb {D}}}, \alpha \in \sigma (U_1U_2)\}\\&=\overline{\cup _{z\in {{\mathbb {D}}}}\sigma (\varphi _1(z),\varphi _2(z))}. \end{aligned}$$

$\square$

Note that $\sigma (V_1,V_2)=\sigma (M_{\varphi _1},M_{\varphi _2})\cup \sigma (V_1|_{{\mathcal {H}}_u},V_2|_{{\mathcal {H}}_u}),$ by Theorem 1.2. Hence by Theorem 6.8,

$$\begin{aligned} \sigma (M_{\varphi _1},M_{\varphi _2})=\overline{\cup _{z\in {{\mathbb {D}}}}\sigma (\varphi _1(z),\varphi _2(z))}\subseteq \sigma (V_1,V_2). \end{aligned}$$

(6.8)

The above inclusion is an equality if and only if $\sigma (V_1|_{{\mathcal {H}}_u},V_2|_{{\mathcal {H}}_u})\subseteq \sigma (M_{\varphi _1},M_{\varphi _2}).$

6.2 Range of one Isometry is Strictly Contained in the Range of Other

If $(V_1,V_3)$ is any pair of commuting isometries with $V_3$ is not unitary, set $V_2=V_1V_3.$ Then $(V_1,V_2)$ satisfy ${\text {ran}}V_2\subsetneq {\text {ran}}V_1.$ Since the study of the case ${\text {ran}}V_1\subsetneq {\text {ran}}V_2$ is equivalent to the case of ${\text {ran}}V_2\subsetneq {\text {ran}}V_1,$ we consider only the case ${\text {ran}}V_2\subsetneq {\text {ran}}V_1.$

Lemma 6.9

Let $(V_1,V_2)$ be a pair of commuting isometries on a Hilbert space ${\mathcal {H}}$. Then the following are equivalent:

• (a) ${\text {ran}}V_2\subsetneq {\text {ran}}V_1.$

• (b) $V_2(\ker V_1^*)\subsetneq V_1(\ker V_2^*).$

• (c) The fringe operator $F_1=0$ and $F_2\ne 0.$

• (d) If $({\mathcal {E}},P,U)$ is the BCL triple for $(V_1,V_2),$ then $U({\text {ran}}P)\subsetneq {\text {ran}}P^\perp$ (or equivalently $U^*({\text {ran}}P)\subsetneq {\text {ran}}P^\perp$).

• (e) If ${\mathcal {F}}$ is the defect space of $V_2$ and if $(M_z\otimes I_{\mathcal {F}})\oplus W_2$ is the Wold decomposition of $V_2,$ then $V_1=M_\varphi \oplus W_1,$ with $\varphi (z)=W^*(Q^\perp +zQ)$ for some unitary W and projection $Q\ne I$ in , and a unitary $W_1$ on ${\mathcal {H}}\ominus H^2_{{\mathbb {D}}}({\mathcal {F}}).$

Proof

$(a)\iff (b)$ and $(a)\iff (c)$ follows from (2.3).

$(a)\iff (d)$: Let $({\mathcal {E}},P,U)$ be the BCL triple for $(V_1,V_2).$ Note that $\ker V_1^*\subsetneq \ker V_2^*$ if and only if $\ker M_{\phi _1}^{*}\subsetneq \ker M_{\phi _2}^{*}$ if and only if $U^{*}PU\le P^{\perp }$ and $U^{*}PU\ne P^{\perp }$ if and only if $U({\text {ran}}P)\subsetneq {\text {ran}}P^\perp ,$ because $\ker M_{\phi _1}^{*}=1\otimes {\text {ran}}(U^{*}PU)$ and $\ker M_{\phi _2}^{*}=1\otimes {\text {ran}}P^\perp .$

$(a)\iff (e)$: By Douglas Lemma ( [10]), we have $V_2=V_1V_3$ for some isometry (non-unitary) $V_3.$ Notice that $V_3$ commutes with $V_1.$ Let $({\mathcal {F}}, Q, W)$ be the BCL triple for $(V_1,V_3).$ Then, (see Theorem 1.2)

where $\phi (z)=W^*(Q^\perp +zQ)$ and $\psi (z)=(Q+zQ^\perp )W$ for all $z\in {{\mathbb {D}}}.$ Now since $V_2=V_1V_3$ and $\varphi (z)\psi (z)=z$ for all $z\in {{\mathbb {D}}}$ we have

$$\begin{aligned} V_2=M_z\otimes I_{\mathcal {F}}\oplus W_2, \end{aligned}$$

where $W_2=W_1W_3.$ This completes the proof of $(a)\implies (e),$ because $V_3$ is not a unitary implies that $Q\ne I.$ $(e)\implies (a)$ is trivial. $\square$

6.2.1 Joint Spectrum

Consider the pair

$$\begin{aligned} (V_1,V_2)=(U^kV^n,U^lV^m) \end{aligned}$$

(6.9)

with V an isometry (not unitary) and U a unitary which commutes with V,  for some non-negative integers n, m, l, k and $n<m,$ as in [4, Sec. 4]. Since U is a unitary commuting with V,  we have

$$\begin{aligned} {\text {ran}}V_2={\text {ran}}U^lV^m={\text {ran}}V^m\subsetneq {\text {ran}}V^n ={\text {ran}}U^kV^n={\text {ran}}V_1. \end{aligned}$$

In [4], Burdak has given the model for such pairs, viz.,

$$\begin{aligned} V_1=(M_z^n\otimes W^k)\oplus W_1, V_2=(M_z^m\otimes W^l)\oplus W_2, \end{aligned}$$

(6.10)

for some unitary $W,W_1,W_2.$ We shall now compute the joint spectrum $\sigma (M_z^n\otimes W^k,M_z^m\otimes W^l)$ of the pure part, as an easy application of the polynomial spectral mapping theorem, viz., consider the polynomial $f:{{\mathbb {C}}}^{2}\rightarrow {{\mathbb {C}}}^{2}$ given by,

$$\begin{aligned} f(z_1,z_2)=(z_1^nz_2^k, z_1^mz_2^l). \end{aligned}$$

By the polynomial spectral mapping theorem,

$$\begin{aligned} \sigma (M_z^n\otimes W^k,M_z^m\otimes W^l)&=\sigma ( f(M_{z}\otimes I, I\otimes W))\\&=f(\sigma (M_{z}\otimes I, I\otimes W) \\&=f(\overline{{\mathbb {D}}}\times \sigma (W))\\&=\{(z^n\alpha ^k,z^m\alpha ^l): z\in \overline{{\mathbb {D}}}, \alpha \in \sigma (W)\}. \end{aligned}$$

Following example shows that the class considered in [4], namely, the class of pairs of the type given in (6.9), is only a subclass of this case.

Example 6.10

Let ${\mathcal {H}}=H^2_{{{\mathbb {D}}}^2}$ and $(V_1,V_2)=(M_{z_1},M_{z_1z_2}).$ Then there is no unitary U commuting with both $V_1$ and $V_2$ such that $V_1^m=UV_2^n$ for any m, n. In particular, $(V_1,V_2)$ is not of the form given in (6.9). But clearly ${\text {ran}}V_2\subsetneq {\text {ran}}V_1.$

Lemma 6.11

Let $(V_1,V_2)$ be a pair of commuting isometries, with ${\text {ran}}V_2\subsetneq {\text {ran}}V_1.$ Then,

$$\begin{aligned} \sigma (V_1,V_2)\subseteq \{(z_1,z_2): |z_2|\le |z_1|, z_1\in \overline{{\mathbb {D}}}\}. \end{aligned}$$

(6.11)

Proof

By Douglas Lemma ( [10]), we have $V_2=V_1V_3$ for some isometry $V_3,$ which commutes with $V_1.$ Consider the polynomial $p:{\mathbb {C}}^2\rightarrow {\mathbb {C}}^2$ given by $p(z_1,z_2)=(z_1,z_1z_2).$ By the spectral mapping theorem,

$$\begin{aligned} \sigma (V_1,V_2)=\sigma (p(V_1,V_3))&=p(\sigma (V_1,V_3))\\&\subseteq p(\overline{{\mathbb {D}}}\times \overline{{\mathbb {D}}})=\{(z_1,z_1z_2): z_1,z_2\in \overline{{\mathbb {D}}}\}\\ {}&=\{(z_1,z_2): |z_2|\le |z_1|, z_1\in \overline{{\mathbb {D}}}\}. \end{aligned}$$

$\square$

The inclusion in (6.11) is sharp; see Example 6.12, and it can be a strict inclusion; see Example 6.13.

Example 6.12

Consider the pair $(M_{z_1},M_{z_1z_2})$ of commuting isometries in $H^2({{\mathbb {D}}}^2).$ Notice that

$$\begin{aligned} {\text {ran}}(M_{z_1z_2})&=\left\{ \sum _{m,n\ge 1}^\infty a_{m,n}z_1^mz_2^n:\sum _{m,n\ge 1 }^\infty |a_{m,n}|^2<\infty \right\} \\ {}&\subsetneq \left\{ \sum _{m\ge 1,n\ge 0}^\infty a_{m,n}z_1^mz_2^n:\sum _{m\ge 1,n\ge 0 }^\infty |a_{m,n}|^2<\infty \right\} ={\text {ran}}M_{z_1}. \end{aligned}$$

Consider the polynomial $p:{\mathbb {C}}^2\rightarrow {\mathbb {C}}^2$ given by $p(z_1,z_2)=(z_1,z_1z_2).$ By the spectral mapping theorem, we have

$$\begin{aligned} \sigma (M_{z_1},M_{z_1z_2}) =\sigma (p(M_{z_1},M_{z_2}))&=p(\sigma (M_{z_1},M_{z_2}))= p(\overline{{\mathbb {D}}}\times \overline{{\mathbb {D}}})\\&=\{(z_1,z_2): |z_2|\le |z_1|, z_1\in \overline{{\mathbb {D}}}\}. \end{aligned}$$

The measure of the above spectrum is non-zero, indeed it is $\frac{\pi ^2}{2}.$

Example 6.13

Let $0\le m<n.$ Consider $(M_z^m,M_z^n).$ Then,

$$\begin{aligned} {\text {ran}}M_z^n=\left\{ \sum _{k=n}^\infty a_kz^k:\sum _{k=n}^\infty |a_k|^2<\infty \right\} \subsetneq \left\{ \sum _{k=m}^\infty a_kz^k:\sum _{k=m}^\infty |a_k|^2<\infty \right\} ={\text {ran}}M_z^m. \end{aligned}$$

By the spectral mapping theorem,

$$\begin{aligned} \sigma (M_z^m,M_z^n)=\{(z^m,z^n): z\in \overline{{\mathbb {D}}}\}\subsetneq \{(z_1,z_1z_2): z_1,z_2\in \overline{{\mathbb {D}}}\}. \end{aligned}$$

Example 6.14

Let ${\mathcal {H}}=H^2_{{{\mathbb {D}}}^2}\oplus H^2_{{{\mathbb {D}}}^2},$ and $V_i=M_{z_i}\oplus \tau _i$ for $i=1,2.$ Then $(V_1,V_2)$ is a pair of commuting isometries with defect $E_0\oplus -P_{{\text {span}}\{z_2\}}.$ Clearly the defect is a difference of two mutually orthogonal projections and $(V_1,V_2)$ lies in the unknown case given in Table 1.

The following example gives an irreducible pair of commuting isometries lying in the unknown case given in Table 1.

Example 6.15

Consider the pure pair $(V_1,V_2)$ with the BCL triple $(l^2({\mathbb {Z}}), p_{01},\omega )$ where $\omega$ is the bilateral shift and $p_{01}$ is the projection in $l^2({\mathbb {Z}})$ onto ${\text {span}}\{e_0,e_1\}.$ Then

$$\begin{aligned} C(V_1,V_2)=E_0\otimes (\omega ^*p_{01}\omega -p_{01})=E_0\otimes p_{{\text {span}}\{e_{-1}\}}-E_0\otimes p_{{\text {span}}\{e_1\}},\qquad \end{aligned}$$

(6.12)

where $E_0$ is the one dimensional projection onto the space of constant functions in $H^2_{{\mathbb {D}}}$. The pair $(V_1,V_2)$ is irreducible. To show this, we shall show that the pair $(p_{01},\omega )$ is irreducible; see [9, 11]. Suppose ${\mathcal {E}}_0\ne \{0\}$ is a reducing subspace for $(p_{01},\omega ).$ Let $\sum _{n\in {\mathbb {Z}}}a_ne_n\in {\mathcal {E}}_0$ and $a_{n_0}\ne 0$ for some $n_0\in {\mathbb {Z}}.$ Consider

$$\begin{aligned} {\omega ^*}^{n_0}\sum _{n\in {\mathbb {Z}}}a_ne_n=\sum _{n\in {\mathbb {Z}}}a_ne_{n-n_0}\in {\mathcal {E}}_0. \end{aligned}$$

This implies that $p_{01}\sum _{n\in {\mathbb {Z}}}a_ne_{n-n_0}=a_{n_0}e_0+a_{n_0+1}e_1\in {\mathcal {E}}_0.$ Hence $\omega ^*(a_{n_0}e_0+a_{n_0+1}e_1)=a_{n_0}e_{-1}+a_{n_0+1}e_0\in {\mathcal {E}}_0.$ So $a_{n_0}e_{-1}\in {\mathcal {E}}_0.$ Thus $e_{-1}\in {\mathcal {E}}_0,$ shows that ${\mathcal {E}}_0=l^2({\mathbb {Z}}).$

The pair $(V_1,V_2)$ lies in the unknown case of ${\mathcal {H}}_i\ne 0$ for all $i=1,2,3,4$ mentioned in Table 1. To see this first note that (6.12) shows that $C(V_1,V_2)$ is a difference of two mutually orthogonal projections and one can see that, it is not in any other case of the Table 1, using the characterization in terms of BCL given for those cases.