Proof. Let \((a, x_1, x_2, x_3) \in \overline{\mathbb{H}}\). Then \((x_1, x_2, x_3) \in \overline{\mathbb{E}}\) and so, \(\overline{\mathbb{H}} \subset \overline{\mathbb{D}} \times \overline{\mathbb{E}}\). Let \(g \in \mathbb{C}[z_1,z_2, z_3]\) and define \(f(z_0, z_1, z_2, z_3)=g(z_1, z_2, z_3)\). Then \[\begin{align} \|g(X_1, X_2, X_3)\|=\|f(A, X_1, X_2, X_3)\| & \leq \sup\{|f(z_0, z_1, z_2, z_3)| : (z_0, z_1, z_2, z_3) \in \overline{\mathbb{H}} \}\\ & \leq \sup\{|f(z_0, z_1, z_2, z_3)| \;: \;z_0 \in \overline{\mathbb{D}}, \;(z_1, z_2, z_3) \in \overline{\mathbb{E}}\}\\ &=\sup\{|g(z_1, z_2, z_3)| : (z_1, z_2, z_3) \in \overline{\mathbb{E}}\}\\ \end{align}\] and so, \((X_1, X_2, X_3)\) is an \(\mathbb{E}\)-contraction. Now assume that \((X_1, X_2, X_3)\) is an \(\mathbb{E}\)-contraction. For any \(f \in \mathbb{C}[z_0, z_1, z_2, z_3]\), we define \(g(z_1, z_2, z_3)=f(0, z_1, z_2, z_3)\). Then \[\|f(0, X_1, X_2, X_3)\|=\|g(X_1, X_2, X_3)\| \leq \|g\|_{\infty, \overline{\mathbb{E}}}= \sup\{|f(0, z_1, z_2, z_3)| : (z_1, z_2, z_3) \in \overline{\mathbb{E}}\} \leq \|f\|_{\infty, \overline{\mathbb{H}}},\] where in the last inequality we have used the fact that \(\{0\} \times \overline{\mathbb{E}}\subseteq \overline{\mathbb{H}}\). Therefore, \((0, X_1, X_2, X_3)\) is an \(\mathbb{H}\)-contraction, which completes the proof. ◻

Proof. Let \(\alpha \in \overline{\mathbb{D}}\). The maps \(f_\alpha, g_\alpha : \overline{\mathbb{H}}\to \overline{\mathbb{H}}\) given by \(f_\alpha(a, x_1, x_2, x_3)=(\alpha a, \alpha x_1, \alpha x_2, \alpha^2 x_3)\) and \(g_\alpha(a, x_1, x_2, x_3)=(\alpha a, x_1, x_2, x_3)\) are both analytic. For any \(p \in \mathbb{C}[z_0, z_1, z_2, z_3]\), we have that \[\begin{align} \|p(\alpha A, \alpha X_1, \alpha X_2, \alpha^2 X_3)\| =\|p\circ f_{\alpha}(A, X_1, X_2, X_3)\| \leq \|p\circ f_\alpha\|_{\infty, \overline{\mathbb{H}}} \leq \|p\|_{\infty, \overline{\mathbb{H}}} \qquad \text{and} \end{align}\] \[\begin{align} \|p(\alpha A, X_1, X_2, X_3)\| =\|p\circ g_{\alpha}(A, X_1, X_2, X_3)\| \leq \|p\circ g_\alpha\|_{\infty, \overline{\mathbb{H}}} \leq \|p\|_{\infty, \overline{\mathbb{H}}}, \end{align}\] which completes the proof. ◻

Proof. It is evident that \(\overline{\mathbb{E}}\subseteq \overline{\mathbb{D}}^3\) and so, \(\overline{\mathbb{H}}\subseteq \overline{\mathbb{D}}\times \overline{\mathbb{E}}\subseteq \overline{\mathbb{D}}^4\). Let \(f(z_0, z_1, z_2, z_3)=z_0\). Suppose \((A, 0, 0, B)\) is an \(\mathbb{H}\)-contraction. By Proposition 5, \((0, 0, B)\) is an \(\mathbb{E}\)-contraction. Since \(\overline{\mathbb{E}}\subseteq \overline{\mathbb{D}}^3\), we have that the last component of an \(\mathbb{E}\)-contraction is a contraction (e.g. see [12]) and so, \(\|B\| \leq 1\). Moreover, we have that \(\|A\|= \|f(A, 0, 0, B)\| \leq \|f\|_{\infty, \overline{\mathbb{H}}} \leq \|f\|_{\infty, \overline{\mathbb{D}}\times \overline{\mathbb{E}}} \leq 1\). Conversely, let \((A, B)\) be a commuting pair of contractions. We have by Ando’s inequality (see Chapter I in [22]) that \[\label{Ando} \|p(A, B)\| \leq \|p\|_{\infty, \overline{\mathbb{D}}^2},\tag{7}\] for every \(p \in \mathbb{C}[z_1, z_2]\). It follows from Theorem 3 that \(\overline{\mathbb{D}}\times \{0\} \times \{0\} \times \overline{\mathbb{D}}\subseteq \overline{\mathbb{H}}\). Let \(f\) be a holomorphic polynomial in four variables and let \(g(z, w)=f(z, 0, 0, w)\). Using 7 , we have that \[\begin{align} \|f(A, 0, 0, B)\| =\|g(A, B)\| \leq \|g\|_{\infty, \overline{\mathbb{D}}^2} =\sup \{|f(\underline{z})| : \underline{z} \in \overline{\mathbb{D}}\times \{0\} \times \{0\} \times \overline{\mathbb{D}}\} \leq \|f\|_{\infty, \overline{\mathbb{H}}} \end{align}\] and so, \((A, 0, 0, B)\) is an \(\mathbb{H}\)-contraction. ◻

Proof. We have by Lemma 8 that the maps \(f_1, f_2: \overline{\mathbb{H}}\to \overline{\mathbb{B}}_2\) given by \(f_1(a, x_1, x_2, x_3)=(a, x_1)\) and \(f_2(a, x_1, x_2, x_3)=(a, x_2)\) are analytic. Let \(g\in \mathbb{C}[z_1, z_2]\) and let \(i \in \{1, 2\}\). Then \[\begin{align} \|g(A, X_i)\| =\|g\circ f_i(A, X_1, X_2, X_3)\| \leq \|g\circ f_i\|_{\infty, \overline{\mathbb{H}}} \leq \sup\left\{|g(a, x_i)| : (a, x_i) \in \overline{\mathbb{B}}_2\right\}= \|g\|_{\infty, \overline{\mathbb{B}}_2} \;. \end{align}\] Therefore, \((A, X_1), (A, X_2)\) are \(\mathbb{B}_2\)-contractions. We also have by Lemma 8 that the map \(f:\overline{\mathbb{H}}\to \overline{\mathbb{P}}\) given by \(f(a, x_1, x_2, x_3)=(a, x_1+x_2, x_3)\) is holomorphic. Let \(g \in \mathbb{C}[z_1, z_2, z_3]\). Then \[\begin{align} \|g(A, X_1+X_2, X_3)\|=\|g\circ f(A, X_1, X_2, X_3)\| & \leq \|g\circ f\|_{\infty, \overline{\mathbb{H}}} \\ &\leq \sup \{|g(a, x_1+x_2, x_3)|: (a, x_1, x_2, x_3) \in \overline{\mathbb{H}}\}\\ & \leq \|g\|_{\infty, \overline{\mathbb{P}}} \end{align}\] and so, \((A, X_1+X_2, X_3)\) is a \(\mathbb{P}\)-contraction. The proof is now complete. ◻

Proof. We have by Lemma 3.17 in [17] that \((a, s\slash 2) \in \overline{\mathbb{B}}_2\) if and only if there exists \(p \in \mathbb{T}\) such that \((a, s, p) \in \overline{\mathbb{P}}\). Let \((a, x_1)\in \overline{\mathbb{B}}_2\). Then there exists \(x_3 \in \mathbb{T}\) such that \((a, 2x_1, p) \in \overline{\mathbb{P}}\) and so, we have by Theorem 3 that \((a, x_1, x_1, x_3) \in \overline{\mathbb{H}}\). ◻

Proof. Since \(|x_1|, |x_2| \leq 1\), it follows from Theorem 11 that \((rx_1, rx_2, r^2x_1x_2) \in \mathbb{E}\) for \(0<r<1\) and so, \((x_1, x_2, x_1x_2) \in \overline{\mathbb{E}}\). If \(a=0\), then we have by Theorem 3 that \((a, x_1, x_2, x_1x_2) \in \overline{\mathbb{H}}\). If \(a \ne 0\), then \(|x_1|, |x_2| \leq \sqrt{1-|a|^2}\) and so, \(x_1, x_2 \in \mathbb{D}\). We discuss two cases from here onwards.

. Let \(|x_1|=|x_2|\). We recall from Theorem 7.1 in [4] that \((x_1, x_2, x_3) \in b\mathbb{E}\) if and only if \(x_1=\overline{x}_2x_3, |x_3|=1\) and \(|x_2| \leq 1\). In this case, one can find \(x_3 \in \mathbb{T}\) such that \(x_1=\overline{x}_2x_3\) and so, \((x_1, x_2, x_3) \in b\mathbb{E}\). It follows from Theorem 12 that \[\sup_{z_1, z_2 \in \mathbb{D}}|\Psi_{z_1, z_2}(a, x_1, x_2, x_3)|=\frac{|a|}{\sqrt{1-|x_1|^2}} \leq 1\] and so, by 6 , \((a, x_1, x_2, x_3) \in \overline{\mathbb{H}}\).

. Let \(|x_1|<|x_2|\) and define \(x_3=x_1\slash \overline{x}_2\). Note that \((x_1, x_2, x_3) \in \mathbb{D}^3\) and \(x_1=\overline{x}_2x_3\). We have by Theorem 11 that \((x_1, x_2, x_3) \in \mathbb{E}\). We have by Proposition 3.1 in [6] that \(\underset{z_1, z_2 \in \mathbb{D}}{\sup}|\Psi_{z_1, z_2}(a, x_1, x_2, x_3)|\) is attained at \((z_1, z_2)=(0, \overline{x}_2)\). Consequently, we have \[\sup_{z_1, z_2 \in \mathbb{D}}|\Psi_{z_1, z_2}(a, x_1, x_2, x_3)|= \bigg|\frac{a\sqrt{1-|x_2|^2}}{1-x_2\overline{x}_2}\bigg|=\frac{|a|}{\sqrt{1-|x_2|^2}}\leq 1.\] By 6 , \((a, x_1, x_2, x_3) \in \overline{\mathbb{H}}\). For the remaining case when \(|x_2|<|x_1|\), the above discussion shows that \((a, x_2, x_1, x_3) \in \overline{\mathbb{H}}\). It now follows from 6 that \((a, x_1, x_2, x_3) \in \overline{\mathbb{H}}\). ◻

Proof. We prove the parts (1), (3) and (4). Also, parts (2) and (5) follow from Proposition 5 and Proposition 7 respectively.

Proof of (1). Let \((A, X_1)\) be a \(\mathbb{B}_2\)-contraction. Take \(f \in \mathbb{C}[z_0, z_1,z_2,z_3]\) and define \(g(z_0, z_1)=f(z_0, z_1, 0, 0)\). Then \[\begin{align} \|f(A, X_1, 0, 0)\| =\|g(A, X_1)\| \leq \|g\|_{\infty, \overline{\mathbb{B}}_2} =\sup \{|f(z_0, z_1, 0, 0)|: (z_0, z_1) \in \overline{\mathbb{B}}_2\} \leq \|f\|_{\infty, \overline{\mathbb{H}}} , \end{align}\] where the last inequality follows from Theorem 3. Thus, \((A, X, 0, 0)\) is an \(\mathbb{H}\)-contraction. The converse to part (1) follows from Proposition 9.

Proof of (3). Let \((A, S, P)\) be a \(\mathbb{P}\)-contraction. Take \(f \in \mathbb{C}[z_0, z_1,z_2,z_3]\) and define \(g(z_0, z_1, z_3)=f(z_0, z_1\slash 2, z_1\slash 2, z_3)\). Then \[\begin{align} \|f(A, S\slash 2, S\slash 2, P)\| =\|g(A, S, P)\| \leq \|g\|_{\infty, \overline{\mathbb{P}}} &=\sup \{|f(z_0, z_1\slash 2, z_1\slash 2, z_3)|: (z_0, z_1, z_3) \in \overline{\mathbb{P}}\}\\ & \leq \|f\|_{\infty, \overline{\mathbb{H}}} , \end{align}\] where the last inequality follows from Theorem 3. Therefore, \((A, S\slash 2, S\slash 2, P)\) is an \(\mathbb{H}\)-contraction. The converse follows directly from part (3) of Theorem 10.

Proof of (4). We have by Proposition 3.11 in [17] that \((S, P)\) is a \(\Gamma\)-contraction if and only if \((0, S, P)\) is a \(\mathbb{P}\)-contraction. The latter is possible if and only if \((0, S\slash 2, S\slash 2, P)\) is an \(\mathbb{H}\)-contraction which follows from part (3). The proof is now complete. ◻

Proof. We prove this result here for the sake of completeness. Let \(AD_T^n=D_T^nB\) for some \(n \in \mathbb{N}\). Then \(Ap(D_T^n)=p(D_T^n)B\) for all \(p \in \mathbb{C}[z]\). Let \(p_k(z)\) be a sequence of polynomials converging uniformly to \(z^{1\slash n}\) on \([0, 1]\). Then the sequence \(p_k(D_T^n)\) converges to \(D_T\) in the operator norm topology. Consequently, we have that \(AD_T= \underset{k\to \infty}{\lim}Ap_k(D_T^n)=\underset{k\to \infty}{\lim} p_k(D_T^n)B=D_TB.\) ◻

Proof. The condition \((3) \implies (2)\) is trivial. Suppose condition \((2)\) holds. Then \((N_0, N_1)\) is a commuting pair of operators such that \(N_0^*\) is subnormal and \(N_1\) is normal. By Fuglede’s theorem [25], \(N_0\) doubly commutes with \(N_1\). Thus, \(N_0\) is quasinormal, i.e., \(N_0\) commutes with \(N_0^*N_0\), because \(N_0^*N_0=I-N_1^*N_1\). It is well-known that every quasinormal operator is subnormal and so, \(N_0\) is subnormal. Since both \(N_0, N_0^*\) are subnormal, it follows that \(N_0\) is normal. Hence, condition \((3)\) follows, thereby giving the equivalence of conditions \((2)\) and \((3)\). The equivalence of \((3)\) and \((4)\) follows from Theorem 19. We now prove the remaining implications.

\((1) \implies (2).\) Let \(\underline{N}\) be an \(\mathbb{H}\)-unitary. Then \(N_0, N_1, N_2, N_3\) are commuting normal operators such that \(\sigma_T(N_0, N_1, N_2, N_3) \subseteq b\mathbb{H}\). Let \((x_1, x_2, x_3) \in \sigma_T(N_1, N_2, N_3)\). We have by spectral mapping theorem that there exists some \(a \in \mathbb{C}\) such that \((a, x_1, x_2, x_3) \in \sigma_T(\underline{N}) \subseteq b\mathbb{H}\). It follows from Theorem 17 that \((x_1, x_2, x_3) \in b\mathbb{E}\). Thus, \(\sigma_T(N_1, N_2, N_3) \subseteq b\mathbb{E}\) and so, \((N_1, N_2, N_3)\) is an \(\mathbb{E}\)-unitary. The commutative \(C^*\)-algebra generated by \(N_0, N_1, N_2, N_3\) is isometrically isomorphic to the \(C(\sigma_T(\underline{N}))\) via the map that takes the coordinate function \(z_i\) to \(N_i\) for \(i=0, 1,2,3\). We have by Theorem 17 that \(|z_0|^2+|z_1|^2=1\) on \(b\mathbb{H}\) and so, \(N_0^*N_0=I-N_1^*N_1\).

\((2) \implies (1).\) It follows from the above discussion as in the proof of the implication \((2) \implies (3)\) that \(N_0\) is normal. Let \((a, x_1, x_2, x_3) \in \sigma_T(\underline{N})\). By the projection property of joint spectrum, we have that \((x_1, x_2, x_3) \in \sigma_T(N_1, N_2, N_3)\). Since \((N_1, N_2, N_3)\) is an \(\mathbb{E}\)-unitary, \((x_1, x_2, x_3)\in b\mathbb{E}\). Moreover, the map \(f(z_1, z_2, z_3, z_4)=|z_1|^2+|z_2|^2-1\) is continuous on \(\sigma_T(\underline{N})\). By continuous functional calculus, we must have \[f(N_0, N_1, N_2, N_3)=N_0^*N_0+N_1^*N_1-I=0 \quad \text{and so,} \quad \{0\}=\sigma_T(f(\underline{N})) =f(\sigma_T(\underline{N})),\] where the last equality follows from the spectral mapping principle. Therefore, \(|a|^2+|x_1|^2=1\) and so, \(\sigma_T(\underline{N}) \subseteq b\mathbb{H}\). Hence, \(\underline{N}\) is an \(\mathbb{H}\)-unitary.

\((2) \implies (5).\) Define \(A=[A_{ij}]_{i, j=1}^2=\begin{bmatrix} N_1 & -N_0^*N_3\\ N_0 & N_2 \end{bmatrix}.\) Since \(N_0, N_1, N_2, N_3\) are commuting normal operators, we have that \(A_{ij}\) are commuting normal operators. We are also given that \((N_1, N_2, N_3)\) is an \(\mathbb{E}\)-unitary. By Theorem 18, \(N_3\) is a unitary, \(N_1=N_2^*N_3\) and so, \(N_2^*N_2=N_1^*N_1\). Then \[\begin{align} AA^*=\begin{bmatrix} N_1N_1^*+N_0^*N_3N_3^*N_0 & N_1N_0^*-N_0^*N_3N_2^* \\ \\ N_0N_1^*-N_2N_3^*N_0 & N_0N_0^*+N_2N_2^* \end{bmatrix}=\begin{bmatrix} N_1N_1^*+N_0^*N_0 & N_0^*(N_1-N_2^*N_3) \\ \\ N_0(N_1^*-N_2N_3^*) & N_0^*N_0+N_1^*N_1 \end{bmatrix}=\begin{bmatrix} I & 0 \\ 0 & I\\ \end{bmatrix} \end{align}\] and \[\begin{align} A^*A=\begin{bmatrix} N_1^*N_1+N_0^*N_0 & -N_1^*N_0^*N_3+N_0^*N_2 \\ \\ -N_3^*N_0N_1+N_2^*N_0 & N_3^*N_0N_0^*N_3+N_2^*N_2 \end{bmatrix}=\begin{bmatrix} N_1N_1^*+N_0^*N_0 & N_0^*(N_2-N_1^*N_3) \\ \\ N_0(N_2^*-N_1N_3^*) & N_0^*N_0+N_1^*N_1 \end{bmatrix}=\begin{bmatrix} I & 0 \\ 0 & I\\ \end{bmatrix}. \end{align}\] Evidently, \((A_{21}, A_{11}, A_{22})=(N_0, N_1, N_2)\) and we have that \[A_{11}A_{22}-A_{12}A_{21}=N_1N_2+N_0^*N_0N_3=N_2^*N_2N_3+N_0^*N_0N_3=\left(N_2^*N_2+N_0^*N_0\right)N_3=N_3.\] Thus, \(\underline{N}=(A_{21}, A_{11}, A_{22}, A_{11}A_{22}-A_{12}A_{21})\) and so, \((5)\) holds.

\((5) \implies (2).\) Assume that \(A\) is a unitary block matrix \([A_{ij}]_{i, j=1}^2\), where \(A_{ij}\) are commuting normal operators such that \(\underline{N}=(A_{21}, A_{11}, A_{22}, A_{11}A_{22}-A_{12}A_{21})\). Then \(\|N_j\| \leq \|A_{jj}\| \leq 1\) for \(j=1, 2\). The condition \(A^*A=I\) gives the following set of equations. \[\label{eq1} A_{11}^*A_{11}+A_{21}^*A_{21}=I, \quad A_{12}^*A_{12}+A_{22}^*A_{22}=I,\tag{8}\] \[\label{eq2} A_{12}^*A_{11}+A_{22}^*A_{21}=0, \quad A_{11}^*A_{12}+A_{21}^*A_{22}=0.\tag{9}\] Again, \(AA^*=I\) provides the following equations. \[\label{eq3} A_{11}A_{11}^*+A_{12}A_{12}^*=I, \quad A_{21}A_{21}^*+A_{22}A_{22}^*=I,\tag{10}\] \[\label{eq4} A_{21}A_{11}^*+A_{22}A_{12}^*=0, \quad A_{11}A_{21}^*+A_{12}A_{22}^*=0 .\tag{11}\] Using the above equations, we have the following. \[\begin{align} N_1^*N_3 =A_{11}^*(A_{11}A_{22}-A_{12}A_{21}) &=(A_{11}^*A_{11})A_{22}-A_{11}^*A_{12}A_{21} \\ & =(I-A_{21}^*A_{21})A_{22}-A_{11}^*A_{12}A_{21} \quad [\text{ by } {(\ref{eq1})}] \;\\ &=A_{22}-A_{22}A_{21}^*A_{21} -(A_{11}^*A_{12})A_{21}\\ &=A_{22}-A_{22}A_{21}^*A_{21} +(A_{22}A_{21}^*)A_{21} \;\quad [\text{ by } {(\ref{eq2})}]\\ &=N_2.\\ \end{align}\] To show that \(N_3\) is unitary, we just need to check that \(N_3^*N_3= I\), because \(N_3\) is a normal operator. \[\begin{align} N_3^*N_3 &=(A_{11}^*A_{22}^*-A_{12}^*A_{21}^*)(A_{11}A_{22}-A_{12}A_{21})\\ &=A_{11}^*A_{11}A_{22}^*A_{22}-(A_{11}^*A_{12})A_{22}^*A_{21}-(A_{12}^*A_{11})A_{21}^*A_{22}+A_{12}^*A_{21}^*A_{12}A_{21}\\ & =A_{11}^*A_{11}A_{22}^*A_{22}+(A_{21}^*A_{22})A_{22}^*A_{21}+(A_{22}^*A_{21})A_{21}^*A_{22}+A_{12}^*A_{21}^*A_{12}A_{21} \quad [\text{ by } (\ref{eq2})]\\ &=(A_{11}^*A_{11}+A_{21}^*A_{21})A_{22}^*A_{22}+(A_{22}^*A_{22}+A_{12}^*A_{12})A_{21}^*A_{21}\\ & =A_{22}^*A_{22}+A_{21}^*A_{21} \quad [\text{ by } (\ref{eq1})]\\ & =I. \quad [\text{ by } (\ref{eq3})] \end{align}\] Hence, \(N_1, N_2, N_3\) are commuting normal operators satisfying \(\|N_2\| \leq 1, N_1^*N_3=N_2\) and \(N_3^*N_3=N_3^*N_3=I\). It follows from Theorem 18 that \((N_1, N_2,N_3)\) is an \(\mathbb{E}\)-unitary. Moreover, we have by 8 that \(N_0^*N_0+N_1^*N_1=A_{21}^*A_{21}+A_{11}^*A_{11}=I.\) The proof is now complete. ◻

Proof. The equivalence of \((1)\) and \((2)\) follows directly from Theorem 21 and Theorem 18. Also, \((3) \implies (1)\) follows from Theorem 21. We prove \((1) \implies (3)\). Let \((A, X_1)\) be a \(\mathbb{B}_2\)-unitary. Then \(A\) and \(X_1\) are commuting normal operators and so, \((A, X_1, |X_1|, U)\) is a commuting quadruple of normal contractions. By Theorem 18, \((X_1, |X_1|, U)\) is an \(\mathbb{E}\)-unitary and so, by Theorem 21, \((A, X_1, |X_1|, U)\) is an \(\mathbb{H}\)-unitary. The proof is now complete. ◻

Proof. Let \((N_1, N_2, N_3)\) be an \(\mathbb{E}\)-unitary. By Theorem 18, \(\|N_1\|, \|N_2\| \leq 1\) and \(N_1^*N_1=N_2^*N_2\). Since \(N_1, N_2, N_3\) are commuting normal operators, we have by Fuglede’s theorem [25] that \(N_1, N_2, N_3\) doubly commute with each other. Thus, \(D_{N_1}\) commutes with \(N_1, N_2\) and \(N_3\). Consequently, the quadruple \((D_{N_1}, N_1, N_2, N_3)\) consists of commuting normal operators such that \((N_1, N_2, N_3)\) is an \(\mathbb{E}\)-unitary and \(D_{N_1}^*D_{N_1}+N_1^*N_1=D_{N_1}^2+N_1^*N_1=I.\) We have by Theorem 21 that \((D_{N_1}, N_1, N_2, N_3)\) is an \(\mathbb{H}\)-unitary. The converse follows from Theorem 21. ◻

Proof. Assume that \((N_1, N_2, N_3)\) is an \(\mathbb{E}\)-unitary and \(U\) is a unitary on \(\mathcal{H}\) such that \(UN_j=N_jU\) for \(j=1, 2, 3\). By Fuglede’s theorem, \(UN_j^*=N_j^*U\) for \(j=1, 2, 3\) and so, \(UD_{N_1}^2=D_{N_1}^2U\). By Lemma 20, we have that \(UD_{N_1}=D_{N_1}U\). Take \(N_0=UD_{N_1}\). It is easy to see that \(N_0, N_1, N_2, N_3\) are commuting normal operators and \(N_0^*N_0+N_1^*N_1=I\). We have by Theorem 21 that \((N_0, N_1, N_2, N_3)\) is an \(\mathbb{H}\)-unitary. Conversely, suppose that \((N_0, N_1, N_2, N_3)\) is an \(\mathbb{H}\)-unitary. It follows from Theorem 21 that \((N_1, N_2, N_3)\) is an \(\mathbb{E}\)-unitary and \(N_0^*N_0=I-N_1^*N_1=D_{N_1}^2\). So, \((N_0^*N_0)^{1\slash 2}=D_{N_1}\). The polar decomposition theorem (see the discussion after Corollary 25) ensures the existence of a unitary \(U\) on \(\mathcal{H}\) such that \(UN_j=N_jU\) for \(j=1, 2, 3\) and \(N_0=U(N_0^*N_0)^{1\slash 2}=(N_0^*N_0)^{1\slash 2}U.\) Therefore, \(N_0=UD_{N_1}=D_{N_1}U\) which completes the proof. ◻

Proof. We have by Theorem 2.2 in [9] that a commuting pair \((S, P)\) of normal operators is a \(\Gamma\)-unitary if and only if \(S=S^*P\), \(P\) is a unitary and \(\|S\| \leq 2\). It follows from Theorem 18 that \((S\slash 2, S\slash 2, P)\) is an \(\mathbb{E}\)-unitary if and only if \((S, P)\) is a \(\Gamma\)-unitary. The desired conclusion now follows from Proposition 27. ◻

Proof. \((1) \implies (2).\) Let \((V_0, V_1,V_2,V_3)\) be an \(\mathbb{H}\)-isometry on a Hilbert space \(\mathcal{H}\). By definition, there exists an \(\mathbb{H}\)-unitary \((U_0, U_1, U_2, U_3)\) acting on a larger Hilbert space \(\mathcal{K}\) containing \(\mathcal{H}\) such that \(\mathcal{H}\) is an invariant subspace for each \(U_j\) and \((V_0, V_1, V_2, V_3)=(U_0|_\mathcal{H}, U_1|_{\mathcal{H}}, U_2|_{\mathcal{H}} ,U_3|_{\mathcal{H}})\). We have by Theorem 21 that \((U_0, U_1)\) is a \(\mathbb{B}_2\)-unitary and \((U_1, U_2, U_3)\) is an \(\mathbb{E}\)-unitary. Therefore, \((V_1, V_2, V_3)\) is an \(\mathbb{E}\)-isometry. Since each \(V_j=U_j|_\mathcal{H}\), we have that \(V_j^*V_j=P_\mathcal{H}U_j^*U_j|_\mathcal{H}\). Again by Theorem 21, we have \[I-V_0^*V_0-V_1^*V_1=P_{\mathcal{H}}\bigg(I-U_0^*U_0-U_1^*U_1\bigg)\bigg|_\mathcal{H}=0.\] It follows from Theorem 19 that \((V_0, V_1)\) is a \(\mathbb{B}_2\)-isometry.

\((2) \implies (3).\) We only need to show that \((V_0, V_2)\) is a \(\mathbb{B}_2\)-isometry. Since \((V_1, V_2, V_3)\) is an \(\mathbb{E}\)-isometry, we have by Theorem 32 that \(V_1=V_2^*V_3\) and \(V_3\) is an isometry. Then \[V_1^*V_1=V_3^*V_2V_1=V_3^*V_1V_2=V_3^*V_2^*V_3V_2 =V_2^*V_3^*V_3V_2=V_2^*V_2.\] Since \((V_0, V_1)\) is a \(\mathbb{B}_2\)-isometry and \(V_1^*V_1=V_2^*V_2\), we have that \(V_0^*V_0+V_2^*V_2=V_0^*V_0+V_1^*V_1=I\) and by Theorem 19, \((V_0, V_2)\) is a \(\mathbb{B}_2\)-isometry.

\((3) \implies (4).\) One can easily employ similar techniques as above in \((2) \implies (3)\) to show that \((V_0, V_1)\) is a \(\mathbb{B}_2\)-isometry.

\((4) \implies (1).\) We first show that \((V_0, V_1, V_2, V_3)\) has a simultaneous normal extension. Since \((V_0, V_1)\) and \((V_0, V_2)\) are \(\mathbb{B}_2\)-isometries and \((V_1, V_2, V_3)\) is an \(\mathbb{E}\)-isometry, we have by Theorems 19 & 32 that \(V_0^*V_0+V_1^*V_1=I, V_0^*V_0+V_2^*V_2=I\) and \(V_3^*V_3=I\). Therefore, \[2V_0^*V_0+V_1^*V_1+V_2^*V_2+V_3^*V_3=3I \quad \text{and so,} \quad \frac{2}{3}V_0^*V_0+\frac{1}{3}V_1^*V_1+\frac{1}{3}V_2^*V_2+\frac{1}{3}V_3^*V_3=I.\] We have by Lemma 31 that \((V_0, V_1, V_2, V_3)\) admits a simultaneous normal extension. Thus, there exist a Hilbert space \(\mathcal{K}\) containing \(\mathcal{H}\) and a commuting quadruple \((U_0, U_1, U_2, U_3)\) of normal operators acting on \(\mathcal{K}\) such that \(\mathcal{H}\) is a joint invariant subspace for \(U_0, U_1, U_2, U_3\) and \(V_j=U_j|_\mathcal{H}\) for \(0 \leq j \leq 3\). Without loss of generality, we assume that \((U_0, U_1, U_2, U_3)\) on \(\mathcal{K}\) is the minimal normal extension of the triple \((V_0, V_1, V_2, V_3)\). Then \[\mathcal{K}= \overline{span}\{U_0^{*j_0}U_1^{*j_1}U_2^{*j_2} U_3^{*j_3}h \;| \;j_0, j_1, j_2, j_3 \geq 0, \;h \in \mathcal{H}\}.\] We prove that \((U_0, U_1, U_2, U_3)\) acting on \(\mathcal{K}\) is an \(\mathbb{H}\)-unitary. We have by Lemma 30 that \(U_j\) is unitarily equivalent to the minimal normal extension of \(V_j\) for \(j=0, 1, 2, 3\). Consequently, \(U_3\) is a unitary by being the minimal normal extension of the isometry \(V_3\). Choose a holomorphic polynomial \(h\) in three variables and define \(f(z_0, z_1, z_2, z_3)=h(z_1, z_2, z_3)\). Evidently, \(S'=f(V_0, V_1, V_2, V_3)\) is a subnormal operator and let \(N'\) be its minimal normal extension. By Lemma 30, \(f(U_0, U_1, U_2, U_3)\) and \(N'\) are unitarily equivalent. Bram proved in [30] that the spectral inclusion relation holds for subnormal operators, i.e., \(\sigma(N') \subseteq \sigma(S')\). Using the normality of \(N'\), we have that \[\begin{align} \|N'\|=\sup\{|z| : z \in \sigma(N')\} \leq \sup\{|z| : z \in \sigma(S')\} & = \sup\{|z| : z \in \sigma(f(V_0, V_1, V_2, V_3))\}\\ & = \sup\{|z| : z \in \sigma(h(V_1, V_2, V_3))\}\\ & = \sup\{|z| : z \in h(\sigma_T(V_1, V_2, V_3))\}\\ & \leq \sup\{|z| : z \in h(\overline{\mathbb{E}})\}\\ &=\|h\|_{\infty, \overline{\mathbb{E}}}. \end{align}\] Since \(h(U_1, U_2, U_3)=f(U_0, U_1, U_2, U_3)\) and \(f(U_0, U_1, U_2, U_3)\) is unitarily equivalent to \(N'\), it follows that \(\|h(U_1, U_2, U_3)\| =\|N'\| \leq \|h\|_{\infty, \overline{\mathbb{E}}}.\) Thus, \((U_1, U_2, U_3)\) is an \(\mathbb{E}\)-contraction and by Theorem 18, \((U_1, U_2, U_3)\) is an \(\mathbb{E}\)-unitary. Since \((V_0, V_1)\) is a \(\mathbb{B}_2\)-isometry, we have by Theorem 19 that \[\label{eqn95601} \|V_0y\|^2+\|V_1y\|^2-\|y\|^2=\langle (V_0^*V_0-I+V_1^*V_1)y, y \rangle=0\tag{12}\] for every \(y \in \mathcal{H}\). Let \(x \in \mathcal{H}\). Then

where the last equality follows from 12 . Thus, \((I-U_0^*U_0-U_1^*U_1)x=0\) for all \(x \in \mathcal{H}\). It follows from the definition of \(\mathcal{K}\) that \(I-U_0^*U_0-U_1^*U_1=0\) on \(\mathcal{K}\). By Theorem 21, \((U_0, U_1, U_2, U_3)\) is an \(\mathbb{H}\)-unitary and so, \((V_0, V_1, V_2, V_3)\) is an \(\mathbb{H}\)-isometry. The proof is now complete. ◻

Proof. We have by Theorem 33 that \((V_1, V_2, V_3)\) is an \(\mathbb{E}\)-isometry and so, by Theorem 32, \(V_3\) is an isometry. Let \(\mathcal{H}_1 \oplus \mathcal{H}_2\) be the Wold type decomposition of \(V_3\) such that \(V_3|_{\mathcal{H}_1}\) is a unitary and \(V_3|_{\mathcal{H}_2}\) is a pure isometry. Indeed, the space \(\mathcal{H}_1\) is given by \[\mathcal{H}_1=\{x \in \mathcal{H}: V_3^{*n}V_3^n x=V_3^nV_3^{*n}x=x \;\;\text{for} \;\;n=0, 1, 2, \dotsc\}.\] By Theorem 38, \(\mathcal{H}_1, \mathcal{H}_2\) are joint reducing subspaces for \(V_1, V_2, V_3\) such that \((V_1|_{\mathcal{H}_1}, V_2|_{\mathcal{H}_1}, V_3|_{\mathcal{H}_1})\) is an \(\mathbb{E}\)-unitary and \((V_1|_{\mathcal{H}_2}, V_2|_{\mathcal{H}_2}, V_3|_{\mathcal{H}_2})\) is a pure \(\mathbb{E}\)-isometry. Let \[V_0=\begin{bmatrix} A_{0} & C\\ C' & B_{0}\\ \end{bmatrix}, \quad V_1= \begin{bmatrix} A_1 & 0\\ 0 & B_1\\ \end{bmatrix}, \quad V_2= \begin{bmatrix} A_2 & 0\\ 0 & B_2\\ \end{bmatrix} \quad \text{and} \quad V_3=\begin{bmatrix} A_3 & 0\\ 0 & B_3\\ \end{bmatrix}.\] with respect to \(\mathcal{H}=\mathcal{H}_1 \oplus \mathcal{H}_2\). Since \(V_0V_3=V_3V_0\), we have that \(CB_3=A_3C\) and \(C'A_3=B_3C'\). Such an intertwining relation is possible only when \(C=C'=0\) since \(A_3\) is a unitary and \(B_3^{*n} \to 0\) strongly as \(n\) tends to infinity. Thus, \(\mathcal{H}_1, \mathcal{H}_2\) are reducing subspaces for \(V_0\) and so, \(V_0=A_0 \oplus B_0\) with respect to \(\mathcal{H}=\mathcal{H}_1 \oplus \mathcal{H}_2\). Evidently, \(A_0\) is subnormal operator acting on \(\mathcal{H}_1\) and so, a further decomposition of \(A_0\) into normal part and completely non-normal part is possible. Indeed, Lemma 3.1 in [31] gives that the space \[\mathcal{H}^{(u)}=\overset{\infty}{\underset{n=0}{\bigcap}}\text{Ker}\left(A_0^{*n}A_0^n-A_0^nA_0^{*n}\right)=\left\{x \in \mathcal{H}_1 \;: \;A_0^{*n}A_0^nx=A_0^nA_0^{*n}x \;\;\text{for} \;\;n=0, 1, 2, \dotsc\right\}\] is the maximal reducing subspace of \(A_0\) such that \(A_0|_{\mathcal{H}^{(u)}}\) is normal. Note that \[\mathcal{H}^{(u)}=\bigg\{x \in \mathcal{H}\;: \;V_3^{*n}V_3^n x=V_3^nV_3^{*n}x=x \;\;\& \;\;V_0^{*n}V_0^nx=V_0^nV_0^{*n}x \;\;\text{for} \;\;n=0, 1, 2, \dotsc \bigg\} \subseteq \mathcal{H}_1.\] Evidently, \((A_1, A_2, A_3)\) is a commuting triple of normal operators acting on \(\mathcal{H}_1\) and \(A_jA_0=A_0A_j\) for \(1 \leq j \leq 3\). We have by Fuglede’s theorem [25] that \(A_j^*A_0=A_0A_j^*\) for \(1 \leq j \leq 3\) and so, \((A_0, A_1, A_2, A_3)\) is a doubly commuting quadruple of operators on \(\mathcal{H}_1\). Then \[A_j\left(A_0^{*n}A_0^n-A_0^nA_0^{*n}\right)=\left(A_0^{*n}A_0^n-A_0^nA_0^{*n}\right)A_j \quad \text{and} \quad A_j^*\left(A_0^{*n}A_0^n-A_0^nA_0^{*n}\right)=\left(A_0^{*n}A_0^n-A_0^nA_0^{*n}\right)A_j^*\] for \(1 \leq j \leq 3\). Therefore, \(\mathcal{H}^{(u)}\) is a joint reducing subspace of \(A_0, A_1, A_2\) and \(A_3\). Let us define \[\underline{U}=(U_0, U_1, U_2, U_3 )= (A_0|_{\mathcal{H}^{(u)}}, A_1|_{\mathcal{H}^{(u)}}, A_2|_{\mathcal{H}^{(u)}}, A_3|_{\mathcal{H}^{(u)}}).\] Then \(\underline{U}\) is a commuting quadruple of normal operators and \((A_1|_{\mathcal{H}^{(u)}}, A_2|_{\mathcal{H}^{(u)}}, A_3|_{\mathcal{H}^{(u)}})\) is an \(\mathbb{E}\)-unitary. For \(x \in \mathcal{H}^{(u)}\), we have that \(U_0^*U_0x+U_1^*U_1x=A_0^*A_0x+A_1^*A_1x=x\). It now follows from Theorem 21 that \(\underline{U}\) is an \(\mathbb{H}\)-unitary. Let \(\mathcal{H}' \subseteq \mathcal{H}\) be a joint reducing subspace for \(V_0, V_1, V_2, V_3\) such that \(\underline{U}'=(V_0|_{\mathcal{H}'}, V_1|_{\mathcal{H}'}, V_2|_{\mathcal{H}'}, V_3|_{\mathcal{H}'})\) is an \(\mathbb{H}\)-unitary. Let \(U_j'=V_j|_{\mathcal{H}'}\) for \(0 \leq j \leq 3\). By Theorem 21, \((U_1', U_2', U_3')\) is an \(\mathbb{E}\)-unitary and by Theorem 18, \(U_3'\) is a unitary. Since \(\mathcal{H}_1\) is the maximal closed subspace of \(\mathcal{H}\) that reduces \(V_3\) to a unitary, we have that \(\mathcal{H}' \subseteq \mathcal{H}_1\). Also, \(A_0|_{\mathcal{H}'}=V_0|_{\mathcal{H}'}=U_0'\) is a normal operator. Since \(\mathcal{H}^{(u)}\) is the maximal closed subspace of \(\mathcal{H}_1\) reducing \(A_0\) to a normal operator, we have that \(\mathcal{H}' \subseteq \mathcal{H}^{(u)}\). Hence, \(\mathcal{H}^{(u)}\) is the maximal joint reducing subspace for \(V_0, V_1, V_2, V_3\) restricted to which \((V_0, V_1, V_2, V_3)\) acts as an \(\mathbb{H}\)-unitary. Let \(\mathcal{H}^{(c)}=\mathcal{H}\ominus \mathcal{H}^{(u)}\). Then \(\mathcal{H}^{(c)}=(\mathcal{H}_1\ominus \mathcal{H}_1^{(u)}) \oplus \mathcal{H}_2\) and \(V_3|_{\mathcal{H}_2}\) is a pure isometry. The maximality of \(\mathcal{H}^{(u)}\) implies that \((V_0|_{\mathcal{H}^{(c)}}, V_1|_{\mathcal{H}^{(c)}}, V_2|_{\mathcal{H}^{(c)}}, V_3|_{\mathcal{H}^{(c)}})\) is a pure \(\mathbb{H}\)-isometry. The uniqueness part is also an immediate consequence of the maximality of \(\mathcal{H}^{(u)}\). The proof is now complete. ◻

Proof. We shall frequently use the natural isomorphism between the Hardy space \(H^2(\mathcal{L})\) of \(\mathcal{L}\)-valued functions on the unit disc \(\mathbb{D}\) and \(\ell^2(\mathcal{L})\). We use this identification without any further mention. We can write the operators as

with respect to the decomposition \(\ell^2(\mathcal{L})=\mathcal{L} \oplus \mathcal{L} \oplus \dotsc\). It is easy to see that \(V_3\) commutes with \(V_0, V_1\) and \(V_2\). A few steps of simple calculations give that the operators of the form \[\begin{bmatrix} P & 0 & 0 & \dotsc \\ Q & P & 0 & \dotsc \\ 0 & Q & P & \dotsc\\ \dotsc & \dotsc & \dotsc & \dotsc \\ \end{bmatrix} \quad \text{and} \quad \begin{bmatrix} R & 0 & 0 & \dotsc \\ S & R & 0 & \dotsc \\ 0 & S & R & \dotsc\\ \dotsc & \dotsc & \dotsc & \dotsc \\ \end{bmatrix} \;\] commute if and only if \([P, R]=[Q, S]=0,\) and \([Q, R]=[S, P]\). Consequently, \(V_0V_1=V_1V_0\) if and only if conditions \((6)-(8)\) in the statement of the theorem hold. Similarly, \(V_0V_2=V_2V_0\) if and only if conditions \((3)-(5)\) of the statement hold. The last commutativity condition is \(V_1V_2=V_2V_1\) which holds if and only if conditions \((1)\) and \((2)\) of the statement hold. Again, some simple computations give that \[V_0^*V_0=\begin{bmatrix} G_0^*G_0+G_1^*G_1 & G_1^*G_0 & 0 & \dotsc \\ G_0^*G_1 & G_0^*G_0+G_1^*G_1 & G_1^*G_0 & \dotsc \\ 0 & G_0^*G_1 & G_0^*G_0+G_1^*G_1 & \dotsc \\ \dotsc & \dotsc & \dotsc & \dotsc \\ \end{bmatrix}\] and \[V_1^*V_1=\begin{bmatrix} F_1F_1^*+F_2^*F_2 & F_2^*F_1^* & 0 & \dotsc \\ F_1F_2 & F_1F_1^*+F_2^*F_2 & F_2^*F_1^* & \dotsc \\ 0 & F_1F_2 & F_1F_1^*+F_2^*F_2 & \dotsc \\ \dotsc & \dotsc & \dotsc & \dotsc \\ \end{bmatrix}.\] Hence, \(V_0^*V_0+V_1^*V_1=I\) if and only if conditions \((9)\) and \((10)\) in the statement of the theorem holds. We use these equivalent conditions to obtain the desired conclusion.

Assume that \(\underline{V}\) is a pure \(\mathbb{H}\)-isometry. The commutativity hypothesis of the operators in \(\underline{V}\) gives conditions \((1)-(8)\) of the statement. By Theorem 33, \(V_0^*V_0+V_1^*V_1=I\) which gives conditions \((9)-(10)\). It follows from Theorem 33 and Remark 37 that \((V_1, V_2, V_3)\) is a pure \(\mathbb{E}\)-isometry. We have by Theorem 42 that \(\|F_1^*+F_2z\|_{\infty, \overline{\mathbb{D}}} \leq 1\). Conversely, conditions \((1), (2)\) and \(\|F_1^*+F_2z\|_{\infty, \overline{\mathbb{D}}} \leq 1\) implies that \((V_1, V_2, V_3)\) is a pure \(\mathbb{E}\)-isometry by virtue of Theorem 42. Also, conditions \((1)-(8)\) give that \(\underline{V}\) is a commuting quadruple and conditions \((9)-(10)\) yield that \(V_0^*V_0+V_1^*V_1=I\). By Theorem 33 and Proposition 40, \(\underline{V}\) is a pure \(\mathbb{H}\)-isometry. The proof is now complete. ◻

Proof. It suffices to find \(G_0, G_1 \in \mathcal{B}(\mathcal{L})\) satisfying all the equations in the statement of Theorem 43. By spectral theorem, there exists a unique spectral measure \(E\) on \(\sigma_T(F_1, F_2)\) such that \[F_1=\int_{\sigma_T(F_1, F_2)} \boldsymbol{z}_1 dE \quad \text{and} \quad F_2=\int_{\sigma_T(F_1, F_2)} \boldsymbol{z}_2dE,\] where \(\boldsymbol{z}_1, \boldsymbol{z}_2\) are the natural coordinate maps on \(\mathbb{C}^2\). Let \((\alpha_1, \alpha_2) \in \sigma_T(F_1, F_2)\). We show that \(|\alpha_1|+|\alpha_2| \leq 1\). Let \(z \in \overline{\mathbb{D}}\). We have by spectral mapping principle that \(\overline{\alpha}_1+z\alpha_2 \in \sigma(F_1^*+F_2z)\). Since \(\|F_1^*+F_2z\| \leq 1\), we have that \(|\overline{\alpha}_1+z\alpha_2| \leq 1\) for all \(z \in \overline{\mathbb{D}}\). Thus, \(|\alpha_1| +|\alpha_2| \leq 1\) if at least one of \(\alpha_1\) or \(\alpha_2\) is zero. Assume that both \(\alpha_1\) and \(\alpha_2\) are non-zero. Define \(\displaystyle z=\frac{\overline{\alpha}_2|\alpha_1|}{|\alpha_2|\alpha_1}\). Then \(|\alpha_1|+|\alpha_2|=|\overline{\alpha}_1+z\alpha_2| \leq 1\) and so, \(\sigma_T(F_1, F_2) \subseteq \{(\alpha_1, \alpha_2) \in \mathbb{C}^2 : |\alpha_1|+|\alpha_2| \leq 1\}\). Consequently, the real-valued maps \(\alpha, \beta\) on \(\sigma_T(F_1, F_2)\) given by \(\displaystyle \alpha(z_1, z_2)=\sqrt{1-(|z_1|-|z_2|)^2}\) and \(\displaystyle \beta(z_1, z_2)=\sqrt{1-(|z_1|+|z_2|)^2}\) are continuous. Consider the functions on \(\sigma_T(F_1, F_2)\) defined as \[f(z_1, z_2)=\frac{1}{2}\left[\alpha(z_1, z_2)+\beta(z_1, z_2)\right] \quad \text{and} \quad g(z_1, z_2)=\left\{ \begin{array}{ll} \displaystyle \frac{-z_1z_2}{2|z_1z_2|}\left[\alpha(z_1, z_2)-\beta(z_1, z_2)\right], & z_1z_2 \neq 0 \\ \medskip 0, & z_1z_2=0 \,.\\ \end{array} \right.\] Clearly, \(f\) is continuous on \(\sigma_T(F_1, F_2)\). Also, \(g\) is Borel measurable on \(\sigma_T(F_1, F_2)\) since the map \[h: \sigma_T(F_1, F_2) \to \mathbb{C}, \quad h(z_1, z_2)=\left\{ \begin{array}{ll} \displaystyle \frac{-z_1z_2}{|z_1z_2|}, & z_1z_2 \neq 0 \\ \medskip 0, & z_1z_2=0 \,.\\ \end{array} \right.\] is Borel measurable. A few tedious but routine computations give that \[\label{eqn95norm} |f(z_1, z_2)|^2+|g(z_1, z_2)|^2=1-|z_1|^2-|z_2|^2 \quad \text{and} \quad \overline{f(z_1, z_2)}g(z_1, z_2)+z_1z_2=0\tag{13}\] for all \((z_1, z_2) \in \sigma_T(F_1, F_2)\). Let \(G_0=f(F_1, F_2)\) and \(G_1=g(F_1, F_2)\). Consequently, \(G_0, G_1, F_1, F_2\) are commuting normal operators and so, conditions \((1)-(8)\) in Theorem 43 hold. An application of the spectral theorem for commuting normal operators and 13 gives that \(G_0^*G_0+G_1^*G_1=I-F_1^*F_1-F_2^*F_2\) and \(G_0^*G_1+F_1F_2=0\), which are conditions \((9)-(10)\) of Theorem 43. ◻

Proof. Since the last component of an \(\mathbb{E}\)-contraction is a contraction, we have that \(\|X_3\| \leq 1\). Clearly, \(D_{X_3}=D_{X_3^*}\) and so, \(\mathcal{D}_{X_3}=\mathcal{D}_{X_3^*}\). Consider the characteristic function of \(X_3\) given by \[\Theta_{X_3}(z)=[-X_3+zD_{X_3^*}(I-zX_3^*)^{-1}D_{X_3}]|_{\mathcal{D}_{X_3}} \quad \text{ for z \in \mathbb{D}}.\] Let \(G_1, G_2\) be the fundamental operators of \((X_1^*, X_2^*, X_3^*)\). It follows from Theorem 3 in [33] that \[\begin{align} \label{eqn95Lata} (G_1^*+G_2z)\Theta_{X_3}(z)=\Theta_{X_3}(z)(F_1+F_2^*z) \quad \text{and} \quad (G_2^*+G_1z)\Theta_{X_3}(z)=\Theta_{X_3}(z)(F_2+F_1^*z) \end{align}\tag{14}\] for all \(z \in \mathbb{D}\). Since \(X_1, X_2, X_3\) are commuting normal operators, we have that \[D_{X_3}G_1D_{X_3}=(X_1-X_2^*X_3)^*=D_{X_3}F_1^*D_{X_3} \quad \text{and} \quad D_{X_3}G_2D_{X_3}=(X_2-X_1^*X_3)^*=D_{X_3}F_2^*D_{X_3}.\] By uniqueness of fundamental operators, \((G_1, G_2)=(F_1^*, F_2^*)\). Substituting \(z=0\) in 14 , we have \[\begin{align} \label{eqn95504} F_1X_3=X_3F_1 \quad \text{and} \quad F_2X_3=X_3F_2 \quad \text{on \mathcal{D}_{X_3}}. \end{align}\tag{15}\] Since \(X_1, X_2, X_3\) are commuting normal operators, \(D_{X_3}X_j=X_jD_{X_3}\) for \(1 \leq j \leq 3\) and so, \(\mathcal{D}_{X_3}\) is a joint reducing subspace for \(X_1, X_2, X_3\). By 15 and Fuglede’s theorem [25], we have that \[\begin{align} \label{eqn95505} F_1^*X_3=X_3F_1^* \quad \text{and} \quad F_2^*X_3=X_3F_2^* \quad \text{on \mathcal{D}_{X_3}}. \end{align}\tag{16}\] It follows from 15 and 16 that \(D_{X_3}^2\) doubly commutes with \(F_1\) and \(F_2\). By Lemma 20, we have \[\begin{align} \label{eqn95506} D_{X_3}F_j=F_jD_{X_3} \quad \text{and} \quad D_{X_3}F_j^*=F_j^*D_{X_3} \quad \text{on \mathcal{D}_{X_3}} \end{align}\tag{17}\] for\(j=1, 2\). For the normal operators \(N=X_1-X_2^*X_3\) and \(M=X_2-X_1^*X_3\), we have that \[\begin{align} \begin{aligned} N^*N=(X_1^*-X_2X_3^*)(X_1-X_2^*X_3) =D_{X_3}F_1^*D_{X_3}^2F_1D_{X_3} =F_1^*F_1D_{X_3}^4 \quad [\text{by} \;\eqref{eqn95506}], \end{aligned} \end{align}\] and \[\begin{align} \begin{aligned} M^*M=(X_2^*-X_1X_3^*)(X_2-X_1^*X_3) =D_{X_3}F_2^*D_{X_3}^2F_2D_{X_3} =F_2^*F_2D_{X_3}^4 \quad [\text{by} \;\eqref{eqn95506}]. \end{aligned} \end{align}\] Similarly, one can prove that \(NN^*=F_1F_1^*D_{X_3}^4\) and \(MM^*=F_2F_2^*D_{X_3}^4\). Thus, \(F_i^*F_iD_{X_3}^4=F_iF_i^*D_{X_3}^4\) and by Lemma 20, \(F_i^*F_iD_{X_3}=F_iF_i^*D_{X_3}\) for \(i=1,2\). Since \(F_1, F_2\) are operators on \(\mathcal{D}_{X_3}\), it follows that \(F_1\) and \(F_2\) are normal operators. It is clear that \(NM=MN\) since \(X_1, X_2, X_3\) are commuting normal operators. Also, we have \[\begin{align} \begin{aligned} NM=(X_1-X_2^*X_3)(X_2-X_1^*X_3) =D_{X_3}F_1D_{X_3}^2F_2D_{X_3} =F_1F_2D_{X_3}^4 \quad [\text{by} \;\eqref{eqn95506}] \end{aligned} \end{align}\] and \[\begin{align} \begin{aligned} MN=(X_2-X_1^*X_3)(X_1-X_2^*X_3) =D_{X_3}F_2D_{X_3}^2F_1D_{X_3} =F_2F_1D_{X_3}^4 \quad [\text{by} \;\eqref{eqn95506}] \end{aligned} \end{align}\] which gives that \(F_2F_1D_{X_3}^4=F_1F_2D_{X_3}^4\). Again by Lemma 20, \(F_2F_1D_{X_3}=F_1F_2D_{X_3}\) and so, \(F_1F_2=F_2F_1\). It is remaining to show that \(\|F_1^*+F_2z\|_{\infty, \overline{\mathbb{D}}} \leq 1\). Since \(F_1, F_2\) are fundamental operators of \((X_1, X_2, X_3)\), the operator \(F_1+F_2z\) has numerical radius at most \(1\) for all \(z \in \overline{\mathbb{D}}\). It follows from Lemma 2.6 in [14] that \(\omega(F_1^*+F_2z) \leq 1\) for all \(z \in \mathbb{T}\). Since \(F_1^*+F_2z\) is a normal operator for any \(z \in \overline{\mathbb{D}}\), its norm is same as its numerical radius and so, \(\|F_1^*+F_2z\| \leq 1\) for all \(z \in \mathbb{T}\). The desired conclusion now follows from the maximum modulus principle. ◻

Proof. The converse is trivial. We prove that the forward part. Suppose \((A, X_1, X_2, X_3)\) is an \(\mathbb{H}\)-contraction on a Hilbert space \(\mathcal{H}\). Let \((V, V_1, V_2, V_3)\) acting on a Hilbert space \(\mathcal{K} \supseteq \mathcal{H}\) be an \(\mathbb{H}\)-isometric dilation of \((A, X_1, X_2, X_3)\). Consider the space given by \[\mathcal{K}_0=\overline{\text{span}}\left\{V^{i}V_1^{j}V_2^{k}V_3^{\ell}h : \;h \in \mathcal{H} \;\text{and} \;i, j, k, \ell \in \mathbb{N} \cup \{0\} \right\}.\] Clearly, \(\mathcal{K}_0\) is a joint invariant subspace for \(V, V_1, V_2, V_3\) and \(\mathcal{H}\subseteq \mathcal{K}_0 \subseteq \mathcal{K}\). Let us define \((W, W_1, W_2, W_3)=(V|_{\mathcal{K}_0}, V_1|_{\mathcal{K}_0}, V_2|_{\mathcal{K}_0}, V_3|_{\mathcal{K}_0})\). Then \[A^{i}X_1^{j}X_2^{k}X_3^{\ell}=P_\mathcal{H}W^{i}W_1^{j}W_2^{k}W_3^{\ell}|_\mathcal{H}\] for all \(i, j, k, \ell \in \mathbb{N}\cup \{0\}\). Since \((V, V_1, V_2, V_3)\) on \(\mathcal{K}\) is an \(\mathbb{H}\)-isometry, there is an \(\mathbb{H}\)-unitary \((U, U_1, U_2, U_3)\) acting on a Hilbert space \(\widetilde{\mathcal{K}}\) containing \(\mathcal{K}\) such that \(\mathcal{K}\) is a joint invariant subspace for \((U, U_1, U_2, U_3)\) and \((V, V_1, V_2, V_3)=(U|_{\mathcal{K}}, U_1|_{\mathcal{K}}, U_2|_{\mathcal{K}}, U_3|_{\mathcal{K}})\). Thus, we have that \[(W, W_1, W_2, W_3)=(V|_{\mathcal{K}_0}, V_2|_{\mathcal{K}_0}, V_3|_{\mathcal{K}_0}, V_4|_{\mathcal{K}_0})=(U|_{\mathcal{K}_0}, U_1|_{\mathcal{K}_0}, U_2|_{\mathcal{K}_0}, U_3|_{\mathcal{K}_0})\] and so, \((W, W_1, W_2, W_3)\) on \(\mathcal{K}_0\) is a minimal \(\mathbb{H}\)-isometric dilation of \((A, X_1, X_2, X_3)\). ◻

Proof. Assuming the minimality of \((V, V_1, V_2, V_3)\), we have that \[\mathcal{K}=\overline{\text{span}}\left\{V^{i}V_1^{j}V_2^{k}V_3^{\ell}h : \;h \in \mathcal{H} \;\text{and} \;i, j, k, \ell \in \mathbb{N} \cup \{0\} \right\}.\] Let \(h \in \mathcal{H}\). Then \[AP_\mathcal{H}(V^{i}V_1^{j}V_2^{k}V_3^{\ell}h)=A(A^{i}X_1^{j}X_2^{k}X_3^{\ell}h)=A^{i+1}X_1^{j}X_2^{k}X_3^{\ell}h=P_\mathcal{H}(V^{i+1}V_1^{j}V_2^{k}V_3^{\ell}h)=P_\mathcal{H}V(V^{i}V_1^{j}V_2^{k}V_3^{\ell}h)\] and so, \(AP_\mathcal{H}=P_\mathcal{H}V\). Also, one can show that \(X_nP_\mathcal{H}=P_\mathcal{H}V_n\) for \(n=1,2,3\). Moreover, for \(h \in \mathcal{H}\) and \(k \in \mathcal{K}\), we have that \(\langle A^*h, k \rangle =\langle A^*h, P_\mathcal{H}k \rangle =\langle h, AP_\mathcal{H}k \rangle =\langle h, P_\mathcal{H}Vk \rangle =\langle V^*h, k \rangle.\) Therefore, \(A^*=V^*|_\mathcal{H}\) and similarly \(X_n^*=V_n^*|_\mathcal{H}\) for \(n=1, 2, 3\). The proof is complete. ◻

Proof. The minimality follows trivially if we prove \((V, V_1, V_2, V_3)\) is an \(\mathbb{H}\)-isometric dilation of \((A, X_1, X_2, X_3)\). This is because \(V_3\) on \(H^2(\mathbb{D}) \otimes \mathcal{D}_{X_3^*}\) is the minimal isometric dilation of \(X_3\). Since \(X_3\) is a pure contraction, it follows from the proof of Theorem 3.2 in [14] that the map given by \[W: \mathcal{H}\to H^2(\mathbb{D}) \otimes \mathcal{D}_{X_3^*}, \quad Wh=\overset{\infty}{\underset{n=0}{\sum}}z^n \otimes D_{X_3^*}X_3^{*n}h.\] is an isometry. Also, for a basis vector \(z^n \otimes y\) of \(H^2(\mathbb{D}) \otimes \mathcal{D}_{X_3^*}\), we have that \[\label{eqn95W42} W^*(z^n \otimes y)=X_3^nD_{X_3^*}y, \quad \text{for} \;n=0, 1, 2, \dotsc.\tag{18}\] Indeed, it was proved in Theorem 3.2 of [14] that \(V_j^*|_{W(\mathcal{H})}=WX_j^*W^*|_{W(\mathcal{H})}\) for \(j=1, 2, 3\) and so, \((V_1, V_2, V_3)\) is an \(\mathbb{E}\)-isometric dilation of \((X_1, X_2, X_3)\). Consequently, \(V_2\) is a contraction and so, \(\|G_1^*+G_2z\|_{\infty, \overline{\mathbb{D}}} \leq 1\). It is easy to see that \((V, V_1, V_2, V_3)\) is unitarily equivalent to the quadruple \((T_{A_0+A_1z}, T_{G_1^*+G_2z}, T_{G_2^*+G_1z}, T_z)\) on \(H^2(\mathcal{D}_{X_3^*})\) via the natural identification map. It is evident that conditions \((1)\)-\((10)\) in the statement of this theorem and that of Theorem 43 are same. Thus, by Theorem 43, \((V, V_1, V_2, V_3)\) is a pure \(\mathbb{H}\)-isometry. For a basis vector \(z^n \otimes y\) of \(H^2(\mathbb{D}) \otimes \mathcal{D}_{X_3^*}\), we have that \[\begin{align} \begin{aligned} W^*V(z^n \otimes y) =W^*(z^n \otimes A_0y)+W^*(z^{n+1}\otimes A_1y) &=X_3^nD_{X_3^*}A_0y+X_3^{n+1}D_{X_3^*}A_1y \qquad \quad \;[\text{by} \;\eqref{eqn95W42}]\\ &=X_3^n(D_{X_3^*}A_0+X_3D_{X_3^*}A_1)y\\ &=X_3^nAD_{X_3^*}y \qquad \qquad \qquad \;\; [\text{by condition (11)}] \\ &=AX_3^nD_{X_3^*}y\\ &=AW^*(z^n \otimes y). \qquad \qquad \qquad \qquad \quad [\text{by} \;\eqref{eqn95W42}]\\ \end{aligned} \end{align}\] Therefore, \(W^*V=AW^*\) and so, \(V^*|_{W(\mathcal{H})}=WA^*W^*|_{W(\mathcal{H})}\). The proof is now complete. ◻

Proof. It follows from Lemma 4 that the commuting quadruple \((B, Y_1, Y_2, Y_3)=(A^*, X_1^*, X_2^*, X_3^*)\) acting on \(\mathcal{H}\) is an \(\mathbb{H}\)-contraction. By hypothesis, \(Y_3\) is a pure contraction and \(G_1, G_2\) are the fundamental operators of \((Y_1^*, Y_2^*, Y_3^*)\). Again by hypothesis, there exist \(A_0, A_1 \in \mathcal{B}(\mathcal{D}_{Y_3^*})\) such that the operator equations associated with \((B, Y_1, Y_2, Y_3)\) in Theorem 50 are satisfied. We have by Theorem 50 that \(W\) is an isometry and the operator quadruple on \(H^2(\mathbb{D}) \otimes \mathcal{D}_{X_3}\) \[(V, V_1, V_2, V_3)=\left(I\otimes A_0+T_z \otimes A_1, \;I\otimes G_1^*+T_z \otimes G_2, \;I\otimes G_2^*+T_z \otimes G_1, \;T_z \otimes I \right)\] is a minimal \(\mathbb{H}\)-isometric dilation of \((A^*, X_1^*, X_2^*, X_3^*)\). By Proposition 49, \((V^*, V_1^*, V_2^*, V_3^*)\) is an \(\mathbb{H}\)-co-isometric extension of \((A, X_1, X_2, X_3)\), which completes the proof. ◻

Proof. Assume that \((A, 0 , 0, X_3)\) is an \(\mathbb{H}\)-contraction. By Theorem 16, \((A, X_3)\) is a commuting pair of contractions. A well-known result due to Ando (see Chapter I of [22]) gives that \((A, X_3)\) dilates to a pair of commuting isometries \((V, V_3)\). Finally, we have by Corollary 36 that \((V, 0, 0, V_3)\) is an \(\mathbb{H}\)-isometric dilation of \((A, 0, 0, X_3)\). ◻

Proof. Let \((0, S\slash 2, S\slash 2, P)\) be an \(\mathbb{H}\)-contraction on a Hilbert space \(\mathcal{H}\). Take a matricial polynomial \(f=[f_{ij}]\) and define \(g=[g_{ij}]\) with \(g_{ij}(z_1, z_2)=f_{ij}(0, z_1\slash 2, z_1\slash 2, z_2)\) for \(1 \leq i, j \leq n\). We have by Theorem 16 that \((S, P)\) is a \(\Gamma\)-contraction. Furthermore, Theorem 4.3 in [10] ensures the existence of a \(\Gamma\)-unitary dilation \((T, U)\) of \((S, P)\). It follows from Arveson’s dilation theorem (see Theorem 1.2.2 in [35]) that \(\Gamma\) is a complete spectral set for \((T, U)\) and so, the matricial von Neumann’s inequality 5 holds for \((T, U)\). Also, by Theorem 16, \((0, T\slash 2, T\slash 2, U)\) is a normal \(\mathbb{H}\)-contraction that dilates \((0, S \slash 2, S\slash 2, P)\). Then \[\begin{align} \begin{aligned} \left\|[f_{ij}(0, S\slash 2, S\slash 2, P)]\right\| \leq \left\|[f_{ij}(0, T\slash 2, T\slash 2, U)]\right\| &=\left\|[g_{ij}(T, U)]\right\|\\ & \leq \sup \{\|[g_{ij}(z_1, z_2)]\| : (z_1, z_2) \in \Gamma \}\\ & =\sup \{\|[f_{ij}(0, z_1\slash 2, z_1\slash 2, z_2)]\| : (z_1, z_2) \in \Gamma \}\\ & \leq \sup\{\|[f_{ij}(\underline{z})] \| : \underline{z} \in \overline{\mathbb{H}}\}, \end{aligned} \end{align}\] where the last inequality follows from Theorem 3. Hence, \(\overline{\mathbb{H}}\) is a complete spectral set for the \(\mathbb{H}\)-contraction \((0, S\slash 2, S\slash 2, P)\). The desired conclusion follows from Arveson’s dilation theorem. ◻

Proof. Let \((X_1, X_2, X_3)\) be an \(\mathbb{E}\)-contraction and let \((N_1, N_2, N_3)\) be its \(\mathbb{E}\)-isometric dilation. Take \(f=[f_{ij}]\) be a matricial polynomial and consider \(g=[g_{ij}]\) with \(g_{ij}(z_1, z_2, z_3)=f_{ij}(0, z_1, z_2, z_3)\). By Theorem 16, both \((0, X_1, X_2, X_3)\) and \((0, N_1, N_2, N_3)\) are \(\mathbb{H}\)-contractions. It follows from Arveson’s dilation theorem that \(\overline{\mathbb{E}}\) is a complete spectral set for \((X_1, X_2, X_3)\). Then \[\begin{align} \|[f_{ij}(0, X_1, X_2, X_3)]\|=\|[g_{ij}(X_1, X_2, X_3)]\| & \leq \sup\{\|[g_{ij}(z_1, z_2, z_3)]\|: (z_1, z_2, z_3) \in \overline{\mathbb{E}}\}\\ & \leq \sup\{\|[f_{ij}(0, z_1, z_2, z_3)]\|: (z_1, z_2, z_3) \in \overline{\mathbb{E}}\}\\ & \leq \sup\{\|[f_{ij}(\underline{z})]\| : \underline{z} \in \overline{\mathbb{H}}\}, \end{align}\] where we have used Theorem 3 in the last inequality. Consequently, \(\overline{\mathbb{H}}\) is a complete spectral set for \((0, X_1, X_2, X_3)\) and by Arveson’s dilation theorem, \((0, X_1, X_2, X_3)\) admits an \(\mathbb{H}\)-isometric dilation. For the converse, one can directly apply Theorems 16 & 33. ◻

Proof. Let \((A, X_1)\) be a \(\mathbb{B}_2\)-contraction and let \((V, V_1)\) be its \(\mathbb{B}_2\)-isometric dilation. Take \(f=[f_{ij}]\) be a matricial polynomial and consider \(g=[g_{ij}]\) with \(g_{ij}(z_1, z_2)=f_{ij}(z_1, z_2, 0, 0)\). By Theorem 16, \((A, X_1, 0, 0)\) and \((V, V_1, 0, 0)\) are \(\mathbb{H}\)-contractions. Since \((A, X_1)\) admits a rational dilation to \((V, V_1)\), we have by Arveson’s dilation theorem that \(\overline{\mathbb{B}}_2\) is a complete spectral set for \((A, X_1)\). Then \[\begin{align} \begin{aligned} \left\|[f_{ij}(A, X_1, 0, 0)]\right\| =\left\|[g_{ij}(A, X_1)]\right\| & \leq \sup \{\|[g_{ij}(z_1, z_2)]\| : (z_1, z_2) \in \overline{\mathbb{B}}_2\}\\ & =\sup \{\|[f_{ij}(z_1, z_2,0, 0)]\| : (z_1, z_2) \in \overline{\mathbb{B}}_2\}\\ & \leq \sup\{\|[f_{ij}(\underline{z})] \| : \underline{z} \in \overline{\mathbb{H}}\}, \end{aligned} \end{align}\] where the last inequality follows from Theorem 3. Hence, \(\overline{\mathbb{H}}\) is a complete spectral set for \((A, X_1, 0, 0)\) and so, by Arveson’s dilation theorem, it admits an \(\mathbb{H}\)-isometric dilation. The converse follows directly from Theorems 16 & 33. ◻

Proof. We have by Proposition 5 that \((N_1, N_2, N_3)\) is an \(\mathbb{E}\)-contraction acting on \(\mathcal{H}\). Assume that \(\mathcal{H}=\mathcal{H}_1\oplus \mathcal{H}_2\) is the canonical decomposition of the contraction \(N_3\). A simple application of Theorem 58 gives that \(\mathcal{H}_1, \mathcal{H}_2\) are joint reducing subspaces for \(N_0, N_1, N_2, N_3\). Suppose that \[N_0=\begin{bmatrix} P_0 & 0 \\ 0 & Q_0 \end{bmatrix}, \quad N_1=\begin{bmatrix} P_1 & 0 \\ 0 & Q_1 \end{bmatrix}, \quad N_2=\begin{bmatrix} P_2 & 0 \\ 0 & Q_2 \end{bmatrix} \quad \text{and} \quad N_3=\begin{bmatrix} P_3 & 0 \\ 0 & Q_3 \end{bmatrix}\] with respect to \(\mathcal{H}=\mathcal{H}_1\oplus \mathcal{H}_2\). Note that \(P_3\) is a unitary and \(Q_3\) is a completely non-unitary contraction. We have by Theorem 59 that \((P_1, P_2, P_3)\) is an \(\mathbb{E}\)-unitary on \(\mathcal{H}_1\). Let us define \[\mathcal{H}^{(u)}= Ker(I-P_0^{*}P_0-P_1^*P_1)=\{x \in \mathcal{H}_1 : P_0^*P_0x+P_1^*P_1x=x \}.\] We have by Fuglede’s theorem that \((P_0, P_1, P_2, P_3)\) is a doubly commuting quadruple of operators. Consequently, \(\mathcal{H}^{(u)}\) is a reducing subspace for \(P_j\) and so, for \(N_j\) with \(j=0, 1, 2, 3\). Let us define \(\underline{U}=(U_0, U_1, U_2, U_3 )= (P_0|_{\mathcal{H}^{(u)}}, P_1|_{\mathcal{H}^{(u)}}, P_2|_{\mathcal{H}^{(u)}}, P_3|_{\mathcal{H}^{(u)}})\). Then \((P_1|_{\mathcal{H}^{(u)}}, P_2|_{\mathcal{H}^{(u)}}, P_3|_{\mathcal{H}^{(u)}})\) is an \(\mathbb{E}\)-unitary and \(U_0^*U_0x+U_1^*U_1x=P_0^*P_0x+P_1^*P_1x=x\) for all \(x \in \mathcal{H}^{(u)}\). By Theorem 21, \(\underline{U}\) is an \(\mathbb{H}\)-unitary. Let \(\mathcal{H}' \subseteq \mathcal{H}\) be a joint reducing subspace for \(N_0, N_1, N_2, N_3\) such that \(N'=(N_0|_{\mathcal{H}'}, N_1|_{\mathcal{H}'}, N_2|_{\mathcal{H}'}, N_3|_{\mathcal{H}'})\) is an \(\mathbb{H}\)-unitary. Let \(N_j'=N_j|_{\mathcal{H}'}\) for \(0 \leq j \leq 3\). By Theorem 21, \((N_1, N_2, N_3)\) is an \(\mathbb{E}\)-unitary and so, \(N_3\) is a unitary. Since \(\mathcal{H}_1\) is the maximal closed subspace of \(\mathcal{H}\) that reduces \(N_3\) to unitary, we have that \(\mathcal{H}' \subseteq \mathcal{H}_1\). Consequently, \(N_j|_{\mathcal{H}'}=P_j|_{\mathcal{H}'}\) for \(j=0, 1\). Since \((N_0, N_1, N_2, N_3)\) on \(\mathcal{H}'\) is an \(\mathbb{H}\)-unitary, we have by Theorem 21 that \(P_0^*P_0x+P_1^*P_1x=N_0^*N_0x+N_1^*N_1x=x\) for all \(x \in \mathcal{H}'\). Hence, \(\mathcal{H}' \subseteq \mathcal{H}^{(u)}\) and so, \(\mathcal{H}^{(u)}\) is the maximal closed joint reducing subspace for \(N_0, N_1, N_2, N_3\) restricted to which \((N_0, N_1, N_2, N_3)\) is an \(\mathbb{H}\)-unitary. Let \(\mathcal{H}^{(c)}=\mathcal{H}\ominus \mathcal{H}^{(u)}\). The desired conclusion now follows from the maximality of \(\mathcal{H}^{(u)}\). ◻

Proof. Let \(\underline{T}=(A, X_1, X_2, X_3)\) be an \(\mathbb{H}\)-contraction on a Hilbert space \(\mathcal{H}\). Define \[\mathcal{H}_0=\bigcap_{s \in \mathbb{N}^4}\bigcap_{t \in \mathbb{N}^4}Ker\left(\underline{T}^s\underline{T}^{*t}-\underline{T}^{*t}\underline{T}^{s}\right).\] We have by Corollary 4.2 in [37] that \(\mathcal{H}_0\) is the largest joint reducing subspace of \(A, X_1, X_2, X_3\) restricted to which \((A, X_1, X_2, X_3)\) is a commuting quadruple of normal operators. Let us define \(\underline{N}=(N_0, N_1, N_2, N_3)=(A|_{\mathcal{H}_0}, X_1|_{\mathcal{H}_0}, X_2|_{\mathcal{H}_0}, X_3A|_{\mathcal{H}_0})\) on \(\mathcal{H}_0\). Then \(\underline{N}\) is a normal \(\mathbb{H}\)-contraction. By Proposition 60, there is a maximal closed joint reducing subspace \(\mathcal{H}^{(u)}\) for \(N_0, N_1, N_2, N_3\) contained in \(\mathcal{H}_0\) such that \((N_0|_{\mathcal{H}^{(u)}}, N_1|_{\mathcal{H}^{(u)}}, N_2|_{\mathcal{H}^{(u)}}, N_3|_{\mathcal{H}^{(u)}})\) is an \(\mathbb{H}\)-unitary. One can employ similar method as in Theorem 60 and prove that \(\mathcal{H}^{(u)}\) is the maximal closed joint reducing subspace of \(A, X_1, X_3, X_4\) restricted to which \(\underline{T}\) is an \(\mathbb{H}\)-unitary. Let \(\mathcal{H}^{(c)}=\mathcal{H}\ominus \mathcal{H}^{(u)}\). The remaining part of the theorem follows from the maximality of \(\mathcal{H}^{(u)}\). ◻